.%., ' vl IftA-FKAHKim GIFT or A» J.Rathbone \ ^* NEW AND COMPLETE SYSTEM OF ARITHMETICK. COMPOSED FOR THE USE OF THE CITIZENS bF THE UNITED STATES. BY NICOLAS PIKE, A. M. A. A. S. atJID MlfNtTS REIPUBLIC^E MAJUS MEI-IUSVE AFFERRE POSSITMUS, aiTAM SI JUVENTUTEM DOCEMUS, ET BENE EUXTDIMUS ? E VARUS SUMENDUM EST OPTIMUM CICERO. FOURTH EDITION; REVISED, CORRECTED, AND IMPROVED, BY CHESTER DEWEY, A.A.S. PROFESSOR OF MATHEMATICKS, AND NATURAL PHILOSOPHY IW WILLIAMS COLLEGE. TROY, N. y. PRINTED AND PUBLISHED BY WM. S. PAP.KER, S07D AT THE TI?OV BOOKSTORE, AND BY THE PRINCIPAL BOOKSELLET? 5 !W THE UNITED STATES. EDUCATION DEPT. .VORTHB^RM' DISTRICT OF XfEW TORS', TO WIT: BE IT REMEMBERED, That on the twenty first day of September, in the forty seventh year of the Independence of the United States of America, A. D, 1822, William S. Parker of the said District, hrts deposited in this Office the title of a Book, the right whereof he claims as proprietor, in the words follow- ing', to wit : " A new and comple system of Arithmetick, composed for the use of the citi • zens of the United States. By Nicholas Pike, a. m. a. a. s. Quid munus reipublicffi majus meliusve afferre possumus, quam si juventutem docemus, et bene erudimus ? E variis sumendum est optimum. — Cicero. Fourth Edition ; revised, corrected, and improved, by Chester Dev»^et, a. a. s. Professor of Mathematicks, and Natural Philosophy in Williams College." In conformity to the act of the Congi-ess of the United States, entitled " An act for the encouragement of learning, by securing the copies of Maps, Charts, and Books, to the authors and proprietors of such copies, during the times there- in menfioned ;" and also, to the act entitled " An act supplementary to an act /mtitled ' An act for the encourag-ement of learning, by securing the copies of Maps, Charts, and Books, to the authors and proprietors of such copies during the times thereiii mentioned,' and extending the , bQuefits thereof to the arts ot Designing, Engraving anlCt^tltjng hisXotJcal.aqd;ot|ier prints." ' v^ .RICHMD'.R. LANSING, CUx^i qf J^LG Northern District of Js'ew YorJi- PREFACE TO THE FIRST EDITION. It maji, perhaps, by some, be thought needles?, when Au- thors are so muhiphed, to attempt pubhshing any thina^ further on Arithmetick, as it may be imagined (here can be nothing more than the repetition of a subject already exhdusted. It is however the opinion of not a few, who are conspicuous for their knowledge in the Mathematicks, that the books, now in use among us, are gen- erally deficient in the illustration and application of the rules ; of the truth of which, the general complaint among Schoolmasters is a strong confirmation. And not only so, but as the United States are now an independent nation, it was judged that a System might b,e calculated more suitable to our meridian, than those heretofore published. Although I had sufficient reason to distrust my abilities for so ar- duous a task, yet not knowing any one who would take upon him- self the trouble, and appreht-nding I could not render the publick more essential service, than by an attempt to remove the difficul- ties complained of, with diffidence 1 devoted myself to the work. . I have availed my?elf of the best authors which could be obtained but have followed none particularly, except Bonnycastle's Method of Demonstration. Although I have arranged the work in such order as appeared to me the most regular and natural, the student is not obliged to pay a strict adherence to it ; but may pass from one Rule to another, as his inclination or opportunity for study, may reqiiire. The Fpderal Coin, being purely decimal, most naturally falls in after Decimal Fractions. t I have given several methods of extracting the Cube Root, and am indebted to a learned friend, who declines having his name made publick, for the investigation of two very concise Algebraick Theorems for the extraction of all Hoots, and of a particular The- orem for the Sursolid. Amnng the Miscellaneous Questions, I have given some of a phi- losophical nature, as vv.ll with a vi(;vv to inspire the pupil with 9k- relish for philosophical studies, as to the usefulness of thew in the common business of life. Being sensible the followirjg Treatise will stand or fall, accord* ing to its real merit or demerit, I submit it to the judgment of the candid- With pleasure I embrace this opportunity, to express my grati- tude to those learned Gentlemen, who have honoured tliis Treatise with their approbation, as well as to such Gentlemen, as ha.e en- couraged it by their subscriptions ; and to request the reader to excuse any errours he may meet with ; for although great pains have been taken in correcting, yet it is difficult to prevent errours from creeping into the press, and some may have escaped my own ob servation ; in cither ca*;e, a hint from ihe candid will much oblige their Most obedient, And humble Servant, THE AUTHOR ir>^r%itr%^ ¥%nr% PREFACE TO THE FOfFRTH EDITION. Pike's Anrr^HMETicK is universally acknowledged to be the most complete system ever published in the United States. It early obtained a very high reputation, and has continued to receive the approbation of the publick, wherever it has been used. It is de- signed for the instruction of our youth in academies and higher schools, as well as for the use of the man of business and the gen- tleman. All those rules, which are so frequently employed in the various departments of business, are introduced into it. It is the source too, from which the later Arithmelicks have chiefly been compiled. By them, however, it has not been superseded, so much more full and extensive are its rules and their application. In the demonstration and illustration of the rules, it stands pre- eminent. The continued demand for the work has induced the publisher and proprietor of the copy right, to present to the publick a new and improved edition. In the revision of the work much labour has been bestowed, and in the language of a Mathematician well acquainted with the work, " to excellent purpose. It is still Pike's Arithmetick, but altogether more perfect than it was before. As a complete system, it may be pronounced superior to any ever published." The imperfections of the previous editions, which have been noticed by the most distinguished teachers of Arithme- tick, are to a great degree remedied in the present edition. The alterations and improvements consist in the following par- ticulars. Several rules have been added, as well as a variety of Tables, of much practical importance. Some Tables have been corrected and others have been enlarged. Several simple and obvious rules were redundant and have been omitted. The Rule of Three and Interest have been much improved* .Demonstra- tions of a large proportion of the rules were riot givein by Mr. Pike : where the subject would readily admit, they have been supplied. The illustrations of the Kules are more copious, and in many cases simplilied. Most of the Algebraiok demonstrations, which are useless to the mere student in Arithmetick, have been exchanged for arithmetical illustrations. I^ogarithms, TrigonomC" try, Algebra, and Conic Sectiofis, are omitted. These subjects were so briefly treated by Mr. Pike, as to possess little value. As they require a large volume of themselves, and are very fully treated of in Day's Course of Alathematicks, and in the system of Mathematicks now publishing at the University in Massachusetts, the publisher has been uniformly advised to omit them entirely. A concise System of Book Keeping by single and double Entry, has been added to the work, which, we hesitate not to say, will ginatly enhance its value. U is confidently believed that this edition will merit the appro- bation of the publick, and receive that patronage vyhich has been so liberally bestowed on the previous editions. THE PUBLISHER. Troy, October 31, 1821:. RECOMMENDATIONS. Dartmouth University, 1786. At the request of Nicolas Pike, Esq. we have inspected his System of Arith- metiek, which we cheerfully recommend to the publick, as easy, accurate, and complete. And we apprehend there is no treatise of the kind extant, from which so great utility may arise to Schools. B. WOODWARD, Math, and Phil. Prof. JOHN SMITH, Prof, of the Learned Languagea. I do most sincerely concur in the preceding recommendation. J. WHEEJ^OCK, President of the University. Providence, Rhode Island, 1785. Whoever may have the perusal of this treatise on Ai'ithmetick may natur- ally conclude I might have spared myself the trouble of giving it this recom- mendation, as the work will speak more for itself than the most elaborate rec- ommendation from my pen can speak for it : But as I have always been much delighted with the coutempljftion of mathematical subjects, and at the same time fully sensible of the utility of a work of this nature, 1 was willing to render every assistance in my power to bring it to the publick view : And should the student read it with the same pleasure with which I perused the sheet? before they went to the press, I am persuaded he will not fail of reaping that benefit from it which he may expect, or wish for, to satisfy his curiosity in a subject of this nature. The author, in treating on numbers, has done it with so much perspicu- ity and singular address, that I am convinced the study thereof will become more a pleasure than a task. The arrangement of the work, and the method by which he leads the fi/ro into the first principles of xiumbcrs, are novelties I have not met with in any book i have seen. Wingate, Hutton, Ward, Hill, and many other authors whose names might be adduced, if necessary, have claimed a considerable siiare of mer- it ; but when brought into a compariktive point of view with this treatise, they are inadequate and defective. This volume contains, besides what is useful and necessary in tlie common affairs of life, a great fund for amusement and en- tertainment. The Mechanick will find in it mvich more than he may have oc- casion for ; the Lawyer, Merchant and Mathematician, will find an ample field for the exercise of their genius ; and I am well assured it may be read to great advantage by students of every class, from the lowest school to the University. More than this need not be said by me, and to have said less, would be keeping- back a tribute justly due to the merit lof this work. BENJAMIN WEST. University in Cambridge, 1786. Having, by the desire of Nicolas Pike, Esq. inspected the following volume in manuscript, we beg leave to acquaint the publick, that in our opinion it is a work well executed, and contains a complete system of Arithmetick. The rules are plain, and the demonstrations perspicuous and satisfactory ; and we es- teem it the best calculated, of any single piece we have met with, to lead youth, by natural aad easy gradations, into a methodical and thorough acquaintance with the science of figures. Persons of all descriptions may find in it every thing, respecting numbers, necessary to their business; and not only so, but if they have a speculative turn, and mathematical taste, may meet v/itb much for their entertciinmeat at a leisure hour. vi RECOMMENDATIOiNS. We are happy to see so useful an American production, which, if it should meet with the encouragement it deserves, among the inhabitants of the United States, will save much money in the country, which would otherwise be sent to Europe, for publications of this kind. We heartily recommend it to schools, and to the communily at large, and wish that the industry and skill of the Author may be rewarded, for so benefi- cial a work, by meeting with the general approbation and encouragement of the publick. JOSEPH WILLARD, D. D. President of the University. E. WIGGLESWORTII, S. T. P. HoUis. S. VVILLTAMS, L. L. D. Math, et Phil. Nat. Prof. HoUis. Yale College, 1786. Upon examining Mr. Pike's System of Arithmetick and Geometry, in manu- script, I find it to be a work of such mathematical ingenuity, that I esteem myself honoured in joining with the Rev. President Willard, and other learned gentle- men, in recommending it to the publick us a production of genius, interspersed with originality in this part of learning, and as a book, suitable to be- taught ia schools : of utility to the merchant, and well adapted even for the University in- struction, I consider it of such merit, as that it will probably gain a very gene- ral reception and use throughout the republick of letters. EZRA STILES, President. BosTOBT, 1786. From the known character of the Gentlemen who have recommended Mr. Pike's System of Arithmetick, there can be no room to doubt, that it is a valua- ble performance ; and will be, if published, a very useful one. I therefore wish him success in its publication. JAMES BOWDOIN. Union College, Oct. 10, 1822. Pike's Arithmetick is too well known and too highly appreciated to re? quire any recommendation ; and by furnishing an edition of that work, in which oommon language is :;ubstituled for algebraic signs, Professor Dewey has confer- red a favcar on those who may wish to acquire or teach Aritlimetick witho i:f jUgebra ; by whom it is presumed this edition will be patronised. E. NOTT, President. Schenectady, Oct. 16, 1822. Mr. Wm. S. Parker, I have for many years been fully acquainted with P^7rc'5 Si/stem ofArithme- tick, and am persuaded of its excellence ; I do not know of any treatise of more practical utility ; the arrangements of its parts is natural, its rules are plain and easily understood and applied, and it contains all that is ofany importance to the Mercantile or Scientific Arithmetician. • To those who have not the elementary knowledge of Algebra, the translation of the Algebraic expression into plain Arithmetical language must be very acceptable and profitable. This improve- ment, together with the notes and emendations of Professor Dewey, cannot fail to ensure the public confidence and patronage, A hand so able as his, cannot touch without improving an elementary treatise, and wherever he is known, his «'ime must be a sufficient credential. Wishing you all success, and abundant remuneration for your labours, I am, Sir, your friend and servant. T. M'AULEY, S. T. D. Late Professor of Mathematicks. Union College. RECOMMENDATIONS. vii Amherst, Mass, Feb. 9, 1822. I HAVE long been acquainted with Pike's Arithmetic, and think it the best of any extant, for those who wish to acquire a thorough knowledge of Arith- metic as a science and an art. The plan of improvement adopted and pursued by Professor Dewey, in the present edition, is, in my opinion, such as to render the work more perfect and more useful. By supplying defects, omitting redun- dancies, and illustrating what was obscure, he has given to the present edition u superior value. I cheerfully recommend the work to the patronage of the pub- He, and especially to the patronage of the Instructors of youth in Academies and Siohools, as combining more excellences than any other Arithmetic now hi use. ZEPH. SWIFT MOORE, President of the Collegiate Institution, at Amherst, Mass. Lewox, Ms. April 20, 1822. I HAVE seen Pike's Arithmetic revised by Mr. Professor Dewey of Williams College. I entirely approve of all the alterations, fidditions, and jllustrations. I cannot but believe, that the work thus presented to the public, will be duperi- or to any thing of the kind extant. While it initiates the scholar into the theory of this science, it is distinguished for a happy conciseness, lucid method, and graceful -simplicity, which cannot fail to make it a valuable companion for rtbe Merchant, Mechanic, or Fanner. LEVI GLEZEN, Preceptor of Lenox Academy. Extract of a Letter from Mr. Benedict, Tutor of JVilliams College, to the Publisher, dated Williams College, Janitary 2, 1822. Mr. Parker, " From the experience which I have had in instructing youth, I have had occa- sion to acquaint myself with many, if not most of the Systems of Arithmetick in use in this country. I can therefore speak with some more coniidence than I otherwise should, iVom having proved their excellences and defects by actual trial of them. It is most certain that as a complete System on this important part of education, the work under consideration stands preeminent. It is impos- sible that Arithmetick should be so treated ol", as not to leave much to be done by the instructor. Still, as I think. Pike's System will enable the teacher to ben- efit his scholars, to give them sound theoretical and practical knowledge in this branch, to induce them to think and reason closely, and increase their power of arithmetical invention, far more than any one within the compass of my knowl- edge. Excellent as it was when it came from its author, it had its defects. By the revfsion of it by Lord, little else was done than to change the sterling to fed- eral notation. Much remained to be done. In some parts, Mr. Pike had been needlessly minute, and loaded the work with a multiplicity of rules on one sub- ject, which the accountant could not but make for himself, as occasion demanded, with perfect ease. Though his illustrations and demonstrations are usually very good, in some cases tliey were obscure ; and in some parts, as for instance that of interest, there was a great deficiency. I have examined the work with Mr. Dewey's corrections, with considerable care. He has bestowed great labour upon it, and I think to excellent purpose. It is still Pike's Arithmetick ; but altogether more perfect than it was before. I do beheve that as a complete Sys- tem, it may be pronounced superior to any one ever published. I most earnestly wish you success in its publication ; and I feel a confidence that good judges will not hesitate on perusing it, tog;ive it an unqualified recommendation." Yours respectfully, GEORGE W. RKXEDTCT. CONTENTS. Page. NUMERATION .... ^^ Simple Addition u . . - 20 , . Subtraction - . , . ^4 Multiplication . - . . ^5 Division - - . . 22 Tables in Compound Addition _ * - ^^ Compound Addition - - - "50 — Subtraction - - - " 54 Problems resulting from the preceding Rules - - ^^ Reduction - - - - "60 Vulgar Fractions - - - - 7^ Decimal Fractions .... ^ Decimal Tables - - - - \q\ Compound Multiplication ^ . . -jq^ ■ Division - - . -' joS Reduction of Coins - - - - 114 Duodecimals, or Cross Multiplication - - - 126 Single Rule of Three - - - - 129 The method of making Taxes, in a Note - - 136 Single Rule of Three Direct in Vulgar Fractions - - 137 To find the value of Gold in New England currency - - 141 Single Rule of Three Direct and Inverse - - 144 Rule of Three Inverse - - -145 Abbreviation of Vulgar Fractions - - - 147 Double Rule of Three - - - 148 Conjoined Proportion - - - - 152 Arbitration of Exchanges - - - . 154 Single Fellowship .... 155 Double Fellowship - - - - 139 Practice ----- 161 Form of a Bill of Parcels - - - - 184 Tare and Tret - - - - 184 Involution - - - - - 186 Evolution - - - - - 188 Table of Powers - - - - 189 Extraction of the Square Root - - - 190 Application and Use of the Square Root - - 194 Extraction of the Cube Root _ _ - 197 Application and use of the Cube Root - - - 203 Extraction of the Biquadrate Root _ - - 205 of the Sui-solid bv Approximation - - 206 Of the Roots of all Powers _ - - 507 Surds - - - - - 210 Proportion in General - - - - 217 Arithmetical Proportion - - - - 219 Progression _ _ - - 220 Geometrical Proportion _ _. _ - 233 Progression - - - - 235 Simple Interest _ ^ _ - 253 Commission or Factorage .. _ _ - 255 Brokerage - _ _ - 256 Buying and Selling Storks - _ _ - 256 Simple Interest by Decimals _ _ _ 261 in Federal Money _ _ _ 257 CONTENTS. ix Rule for Interest on Bonds, &c. - *■ - 268 in Massachusetts ^ - - - 269 in Connecticut - - " * - 270 in New York ----- 270 Discount . , . - - 275 by Decilnals . - - - 279 Duties . » . - _ 281 Barter _ . ^ - - - 282 Loss and Gain - - - - - 284 Equation of Payments . - - _ 2B9 by Decimals - - - 291 Exchange - - - - - 293 Policies of Insurance _ - - -' 297 Compound Interest . . . - 3Q4 by Decimals - - - 3Q7 Discount by Compound Interest - - - 312 Annuities - - - - - 313 Annuities, or Pensions in Arrears, at Compound Interest - 31s Present Worth of Annuities at Compound Interest - 320 Annuities in Reversion at Compound Interest - - 324 Purchasing Annuities forever, or Freehold Estates - 330 Purchasing Freehold Estates in Reversion - - 332 Table, shewing the amount of jGl or $1 from 1 year to 50 - 334 shewing the present worth of £l or $1 from I year to 40 336 shewing the amount of £1 or $\ Annuity, &c. - 337 shewing the present worth of £1 or$l Annuity, &c. - 338 shewing the Annuity which £1 or $1 will purchase, &c. 339 value of an Annuity of £1 or $\ at different rates and payable at different periods 340 value of an Annuity of £1 or $1 for a single life - 341 Expectation of Life at several ages . _ _ 342 Circulating Decimals . . _ - 342 Alligation Medial - - - . -^ 348 Alligation Alternate - - . - 3^49 Single Position - - - . _ 354 Double Position - - - . . 356 Permutations und Combinations - . _ 359 A short method of reducing a Vulgar Fraction to a Decimal - 364 of finding the Duplicate, Triplicate, &c. Ratio of two Num- bers, whose difference is small - - oge To estimate the Distance of Objects - - ogiy To estimate the Height of Objects - . . ^g Miscellaneous Questions , . ^^o Of Gravity - . . I^^'' Ofthe Fall of Bodies . - , «77 Of Pendulums - _ . _ '^ Of the Lever or Steelyard - _ _ Ofthe Wheel and Axle - . . %o Ofthe Screw .... Of the Specifick Gravities of Bodies Table of Specifick Gravities - - _ 30^ Use ofthe Barometer in Measuring Heights - ^ 29"^ Legal value of Foreign Coins - - - 394 Tableoftiic Weight of Money - . . 395 of Exchange - - _ . 39^5 ofthe Value of Sundry Pieces in the several States - 390 of Commission or Brokerage - . - 399 . of the net PrOcoe^ls, after the Commispiolis at ^h and 5 Pf T cent ?.rc cleductM 381 383 ?84 386 B JO'^ X CONTENTS. Page. Table shewing the number of Days from any day in one month to any day in any other month 401 Table of the Measure of Length of the principal places in Europe compared with the American Yard 402 Table directing how to buy and sell by the Hundred Weight 403 Comparison of the American Foot with the Feet of other Countries 404 Table to cast up Wages or expenses for a yr. at so much per day, week, or mo. 405 Table to find Wages, or Expenses for a mo. week or day, at so much per yr. 405 A Perpetual Almanack _ _ _ . 4Qg Tables reducing Troy Weight to Avoirdupois, and the contrary 407 of the Money of Commercial Countries - - 408 Jews, Greeks, and Romans - 418 of Measures of Length, Capacity, &c. of Various countries 419 An Account of the Gregorian Calendar, or New Style - 425 Chronological Problems 426 Problem 1. To find in which Century the last year is to be Leap year, and the contrary 426 Prob. 2. To find, with regard to any other Years, whether any Year be Leap Year, or not 426 3. To find the Dominical Letter according to the Julian Method 426 4. To find the same according to the Gregorian Method 427 5. To find the Prime, or Golden Number - 428 6. To find the Juhan Epact - - 428 7. To find the Gregorian Epact - - 429 To find the same forever _ _ _ 430 8. To calculate the Moon's Age on any given Day - 430 9. To find the Times ofthe New and Full Moon, and first and last Qrs. 431 10. Having the time of the Moon's Southing given, to find her Age 432 11. To find the time ofthe Moon's Southing 432 12. To find on what day of the week any given day in any rfio. will fall 433 13. To find the Cycle ofthe Sun 434 Table ofthe Dominical Letters according to the Cycle of the Sun 434 Prob. 14. To find the year of the Dionysiau Period - 434 15. To find the year of Indiction - - 435 16. To find the Julian period - _ _ 435 17. To find the Cycle of the Sun, Golden Number and Indiction, &c. 435 18. To find the time of High Water - - 436 19. To find on Avhat day Easter will happen - - 436 Table to find Easter from the year 1753 to 4199 - - 438 Plane Geometry _ , _. _ 439 Mensuration of Superficies and Solids _ - - 443 Section I. Of Superficies _ _ _ 443 Article 1 . To measure a Square _ _ _ 443 2. To measure a Parallelogmm, or Long Square 443 3- When the Breadth of a Superficies is given to find how much in Length will make a Square Foot, Yard, &c. 444 4. To measure a Rhombus - - - . 444 5. To measure a Rhomboides - - 445 6. To measure a Triangle - - 445 7. Another method of finding the Area of the Triangle 447 8. To measure a' Trapezium - _ _ 447 9. To measure any Irregular Figure - - 448 10- To measure a Trapezoid - ^ 449 11. To measure any Regular Polygon - - 449 12. Having the Diameter of a Circle, to find the Circumference 451 13. Having the Circumference, to find the Diameter 452 14. To find the Area of a Circle 45S 15. Having the Diam. to find the Aroa, without the Circumference 453 1 G. Having the Circum. to find the Area without the Diameter 454 CONTENTS, s\ Page. Article 17. Having the Dimensions of any of the parts of a Circle, to find the Side of a Square, equal to the Circle - 454, 18. Having the Area of a Circle, to find the Diameter - 455 19. Having the Area, to find the Circumference -• 455 20. Having the Side of a Square, to find the Diameter of a Circle, which shall be equal to the Square whose side is given 45a 21. Having the Side of a Square, to find the Circumference of a Cir- cle equal to the given Square ' 45fi 22. Having the Diam, of a Circle, to find the Area of a Semicircle 456 23. Having the segment of a circle, to find the length of the Arch Line 45G 24. Having the Chord and Versed Sine of a Segment, to find the Di- ameter of a Circle 457 25. To measure a Sector - - 457 26. To measure the Segment of a Circle - - 458 27. To measure an Ellipsis - - 460 Directions for applying Superficies to Surveying - 460 Section n. Of Solids - - - 461 Art. 28. To measure a Cube _ _ - 4G1 29. To measure a Parallelopipedon - - 463 Having the Side of a Square Solid, to find what Length will make a Solid Foot 465 30. To measure a Cylinder - - - 465 Having the Diameter of a Cylinder given, to find what length will make a Solid Foot 466 f To find how much a round tree, which is equally thick, from end to end, will hew to when made square - - 46t» 31. To measure a Prism _ _ - 467 .32. To measure a Pyramid or Cone - - 467 33. To measure the Frustum of a Pyramid or Cone - 469 34. To measure a Sphere or Globe _ - - 472 35. To measure a Frustum or Segment of a Globe - 473 36. To measure the middle Zone of a Globe — 473 37. To measure a Spheroid — - 474 38. To measure the middle Frustum of a Spheroid - 474 39. To measure a Segment, or Frustum of a Spheroid - 474 40. To measure a Parabolick Conoid - - 474 41. To measure the lower Frustum of a Parabolick Conoid 475 42. To measure a Parabolick Spindle 475 43. To measure the middle Zone, or Frustum of a Parabolick Spindle 4T5 -44. To measure aCylindroidor Prismoid - — 475 45. To measure a solid Ring _ _ _ 476 46. To measure the Solidity of any ijiregular Body whose Dimensions cannot be taken _ _ _ 47c Of the five Regular Bodies - - 477 47. To measure a Tetracdron - _ _ 47C 48. To measure an Octaedron _ _ „ 47C 49. To measure a Dodecaedron - - 478 ■tO. To measure an Eicosiedron - - 478 51. To gauge a Cask _ _ _. 479 52. To gauge a Mash Tub - - - 480 53. Having the Difference of Diameters, Height and Content of a Mash Tub, to find the Diameteis at Top and Bottom 480 54. To ullage a Cask, lying on one side, by the Gauging Rod 481 55. To find a Ship's Tonnage _ _ ^ 481 The Proportions and Tonnage of Noah's Ark - - 481' Questions in Mensuration _ _, _ 48.. Book Keeping by Single Entry _ _ - 488 %- ~ — ^ Double Entrv - -^ «-> 505 EXPLANATION OF THii CHARACTERS tiADE USE 01" IN THIS TREATISB. :-- The sign of equality : as 12 pence = 1 shilling, signifies that 1.2 pence n.T0 ft'qual to one shilling ; and, in general, that whatever precedes it is equal to what follows. + The sign of Addition : as 5-f-5— 10, that is, 5 added to 5 is equal to 10. Read 5 plus 5, or 5 more 5 equal to 10. — Tl)^. sign of Subtraction : as, 12 — 4=8, that is, 12 lessened by 4 is equal to 8, or 4 from 12 and 8 remains. Read 12 minus 4, or 12 less 4 equal to 8. X The sign of Multiplication : as 6 X 5=30, that is, 6 multiplied by 5 is equal to 30. Read 6 into 5 equal to 30. -Tt or 5)30( The sign of Division : as 30-f-5=6, that is, 30 divided by 5 is equal to 6. Read 30 by 5 equal to 6. 17 "" — ~ Numbers placed fractionwise, do likewise denote division, the numerator or upper number being the dividend, and the denominator or lower number, the divisor ; thus, — is the same as 875-^25=35. 2b : :: : The sign of proportion, thus, 2 : 4 :: 8 : 16, that is, as 2 is to 4 so is S to 1^. ~ Signifies Geometrical Progression. 9 — 24-6=il3 Shews that the difference between 2 and 9 added to 6 is equat to 13. Read 9 minus 2 plus 6 equal to 13. And that the line above (called a Ftnculimi) connects all the numbers over which it is drawn. 12 — 34-4=:5 Signifies that the sum of 3 and 4 taken from 12 leaves or is equal tV 5. I " Signifies the second power, or Square, I ^ Signifies the third power, or Cube. — !"^ . . — 'o I Signifies any power in general, as 6l2=rsquare of 6 ; and 50{ =:cUDe *jf 50, (fee. thus m signifies either the square or cube, or any other power. v'l or i^ Prefixed to any number or quantity, signifies that the square root oi' that number is required. It likewise (as also the character for any other root) stands for the expression of the root -of that number or quantity to which it is prefixed. As ^36=6, and ^108-f 36=12, and 36 1 2=6, &c. 3 i_ y/f or Is Prefixed to any number, s'gnifies that the cube root of that number is required, or expressed. As y/216=6, and >/oiaH-216=9, &c. or 216 j^ = 6, &c. IV " I n — i — i / t)i' l_- Signifies any root in general. As 301 2=2:squarQ root,2l6i s^^eube v |m ^ ° ' * n , , root, (fcc. Thus, — signifies either the square root, cube root, or any oth root wh;#ever. ah cd When streral letters are set together, they are supposed to be multi- plied into each other; as those in the m,ai-gin are the same as aXbXcXd, and pc]:>resent the continual product of quantities or nttmbers. 1 ft . . b — Is the ro'ipr'ocal of rr, and — is the reciprocal of — . // h a If a be (he root, tlsrn aXcc=aa or «2 is the square of a, and aXaXct=aaa or (7 3 is the cube of r , iJcc. .Yote. The figure above is called the index of the power. It is usual to write shillings at the left hand o^ a stroke, and pt nee at the right ; thus, 13/4 is thirteen shillings and four pence. .!\'ofe. The use of these charactci-s must be perfectly inders1:(»d ly tlie pu- T'jI. as he may have occasion foi: \hcm'.. NEW AND COMPLETE SYSTEM OF ARITHMETICK. AHITIIMETICK is tlie Art or Science of computing by num- bers, and consists both in I'heory and Practice. The Theo- ry considers the nature and quality of number?, and demonstrates the rea'ion of praciical operations. The l^ractice is that, which •^hews the method of working: by numbers, so as to be most useful and 'expeditious lor business, and is comprised under five principal or fundamental Kules, viz. Notation or Numeration, Addition, Subtraction, Multiplication, and Division ; the knowledge of which is so necessary, that scarcely any thing in life, and nothing in trade can be done without it. NUMERATION 1. TEACH CS the different value of figures by their different places, and to read or write any sum or number by these tenchar^ acters, 0, 1, 2, 3, 4, 5, b\ 7, 0, 9. — is called a cypher, and all the rest are called figures or digits.* The names and significations of these characters, and the origin or generation of the numbers they stand for, are as follow ; nothing ; 1 one, or a single thing called an unit; 1 + 1=2, two ; 2-}- 1=S, three ; 3-1-1=4, lour ; 4+1=^5, iive ; o-|-l=6, six; 6-F 1=7, seven ; 7-}- 1=8, eight ; 84-1=9, nine; 9-|-l = 10 ten ; which has no single character ; and thus, by the continual addition of one, all numbers are generated. 2. The value of figures when alone, is called their sm/)/g value, and is invariable. Besides the simple value, they have a local val- ue, that is, a value which varies according to the place they stand * VVitie figures or digits were ol>taIned from the Arabians, and were introduc- ed into Europe in the ninth century. The Arabs probably derived the deci- j»al notation from hiJia. Tlie fcxagefimal divifion had previoufly been in gen- <:ral ufe in Europe. This mode of divifion is yet retained in a few cafes, as in the divifion of time, whcrt; ftxty minutes make an hoMT,ftxty feconds a minute, &c. The figures are doubtlefs called digits from digitus, a fin^^er, hecaufc cotmf- ing ufcd to he performed on the fingers. V 18 iNUMEllATION. in when counecteil together. In a combination of figures, reckon^ ing from the right to the left, the figirre in the first place represents its simple value ; that in the second place, ten times its simple val~ ue, and so on ; each succeeding figure being ten limes the value of it in the place immediately preceding. There is no reason in the nature of numbers that their locaF value should vary according to this law. They might have been made to increase in 3, 4, 5, &c. fold, or in any other ratio. The tenfold increa'se is assumed because it is most convenient. 3. The values of the places are estimated according to their or- der : The first is denominated the place of units ; the second, tens ; the third, hundreds ; ^an(l yo,opj as in the table. 'J'hus in the num- ber — 5293467 ; 7, in the first place signifies only seven ; 6, in the second' placeV s'!<>'Difi€g 6' tens; of sixty ; 4, in th« third place, four hundred; 3, in' the' fourth place, three thousand; 9, in the fifth place, ninety thoiisand ; 2, in the sixth piace, two hundred thou- sand ; 5, in the seventh place, is five millions ; and the whole, tak- en together, is read thus ; five millions, two hundred and ninetv three thousand, four hundred and sixty seven. The process of Numeration may be more clearly seen by th^ following TABLE, tn tfi o .2 f.i £ . '-A ^ , • 'TO cC '^3 -g w. ^ 3 S2 ^ = .2 - if' fn *- 'O t» .J3 ti? ~ -t3 m c >« ~ -o 'i^ •X3 en ~ "w cfl O "3 tf) S "w t« .2 "^ fi " "^ (A "'' M2r-~XE^WEHE-"2:r-^S: — H^I:•^:_• 2 4 5, 9 3 8. 6 7 5, 2 6 7. {5 9 1,3 4 d: Six places of figures, beginning on the right, are called a period, and each successive six places another period. Each period is con- sidered as divided into two half periods of three figures each. These are distinguished by the comma, antl the point for a period. There is an obvious reason for this division into periods, for at the begin- ning of each period, there is anew denomination of units, of which the tens, hundreds, thousands, fiic. are numerated as in the fir^ period. 4. A cypher, though it is of no signification itself, yet, it pos- sesses a place, and, when set on the right hand of figures, in whole NUMERATION. 19 numbers, increases their value in the same tenfold proportion ; thus, 9 signifies only nine ; but if a cypher is placed on its right hand, thus, 90, it then becomes ninety ; and, if two cyphers be placed on its right, thus, 900, it is nine hundred, &c. 6. To icnuijierate any parcel of figures, observe the following Knle. First, :Committhe words at the head of the table, viz. unit?, tens, hundreds, Lc U) memory, then, to Ihe simple value oi each figure, join the name of its place, beginning at the I'^it hand, and reading towards the right.— More particularly — I. Place a dot under the right hand tigure of the 2tl, 4th, Gth, 8th, &c. half period?, and ihe figure over such dot will, universally, have the name of thousands. — 2. Place the fiijures, 1,2, 3, 4, &c. as indices over the 2d, 3d. i'th, &,c. period. These indices will then shew the number of times the millions are incrt^ased. — Ihe figure under 1, bearing the name of millions, that under 2, the naipe of billions (or millions of mill- ions) that under 3, trillions. EXAMPLE. Scxtlllions. Quintilli. Quatrill. Trillions. Billions. Millions. Units. rvA-*n Ow^--^ ro*-^ r^A-^ n^/^o rvA-o r^^.^-n th. un. th. un. tli. un. th. un. th. un. tU. uu. c.x.t.c.x.\j. r*/-nfV«*^ r».««of^^^> r^^^^rx-^^^ r»-»«<» f^^-/^^r>^'^\ t'K\^fs^.^^ 6 5 4 3 2 1 :}13,'208,000,341, 620,057,219,356,809,379,120,406,1 29,763 H H H H H H H s- =r D- sr cr tr zr o o o o o o o 1= c c c a a c c/i :n ro » en (u rn w V V to at » (» « 3 3 D D a o a C« CLi Crf Ci !^ CU CU CA en CA (fj ci< tn en NoTis 1. Billions is substituted for millions of millions : Trillions, for millions of millions of million* ; Q,uatrillions, for millions of mill- ions of millions of millions, and so on. These names of periods of figures, derived from the Latin nume- ral*!, may he continued without end. They are as follows, for twenty periods, viz. Units, Millions, Uillions, Trillions, Quatrill- ion«, Quiutillions, Sextillions, Septillions, Octitlions, Nonillions, Do- minions, L^idecillions, Duodecillions, 1 redecillions, Q,uatuordecili • ions, Quindecillions, Sexdecillions, Septendecjllions, Octodecillion-', Noremdecillions, Vigintil lions. The Application. Wrhe (iQwn, in proper Jigures, ihefoUazmitig numbers. Fifteen. -- ^i Two hundred and feyenty nine. ------- 279 Three thoufand four hundred and three. - - - - - - 'i'hirty fevcnthoufand, five hundred and fixty fcvcn, - - - 375^7 Four hundred, one thoufand and twenty eight. _ - - Nine millions, ftvcnty two ti)Ouf^nd snd two hundred. - - 907x200 rifty five millions, three hundred, nine thoufand and nine. - Eight hundred millions, forty four thoufand, and fifty five. Two thoufand, five hundred and forty three millions, four? a?414^l"OJ hundred and thirty one thoufand, feven hundred and two. 5 . < J / * 20 SIMPLE ADDITION. Write down in words at length the following numbers. 8 437 709040 3476194 7584397647 17 3010 879066 84094007 49163189186 129 76506 4091875 690748591 500098400700 Notation by Roman Letters. I. One. XV. Fifteen. CC. Two hundrea. H. Two. XVI. Sixteen. CCC. 'Jl^ree hundre^L Ilf. Three. XVII. Seventeen. CCCC. Four hundred. IV. Four. XVIll. Eighteen. D. or Ij- Five hundred. V. Five. XIX. Nineteen. DC. Six hundred. VI. Six. XX. Twenty. l)CC. Seven hundred. VII. Seven. XXX. Thirty. DCCC. Eight hundred. VIII. Eight. XL. Forty. DCCCC Nine hundred. IX. Nine. L. Fifty. M. or €13. One Thousand. X. Fen. LX. Sixty. ^DO- I'^ive Fhousand. XI. Eleven. LXX. Seventy. ^OOO- Fifty thousand. XII. Twelve. LXXX. Eighty. I OOO'O )3 Five hund. thou XIII. Thirteen. XC. Ninety. MDCCCV'llI. One Thousand, XIV. Fourteen. C. Hundred. eight hundred and eight. A less literal number placed after a greater, always augments the value of the greater ; if put before, it diminishes it. 1 hus, VI. is 6 ; IV. is 4 ; XI. is 11 ; IX. is 9, &c. The practice of counting on the lingers doubtless originated the method of Notation by Roman Letters. The letter I was taken ibr one finger, or one ; and hence 11, for two ; 111, tor three ; Illl, ibr four; and V, as representing the opening between the thumb and fore finger, and bfing also an easier combination of the marks for the fingers, was taken for five. As IV is a sim[)ler expression for four than the above, it was doubtless adopted lor this reason, and on the general principle too that a less literal number placeil before a greater shotild diminish the greater so much, and, plaroj after a greatershould augment it so much. Hence as IV ,\>fovr ; VI is six ; Vlll is eighty and so. on. Ten was expressed by X, lie- cause it \s two V^s united, and tvyice five is ten. i'^ifu was express- ed by L, because it is half of C or E, as it was anciently written, and C is the initial of the Latin centum., one hundred. Five hundred is expressed by D, because it is half of the Goth; >C CD or M., the initial oi^ m.ilU, one thousand. ADDITION iS il;e puttina: together of two or more numbers, or sum?, to make them one total, or whole sum. SIMPLE ADDITKIY Is the adding of several integers or whole numbers together, which are all of one kind, or sort ; as 7 pounds, 12 pounds, and '10 poun(j^;.Jieing added '.oget.herj their aggregate, or sum total, i^ ."V pounds.* ; SIMPLE ADDITION. 2i Rule. llavinf^ placed units under units, tens under lens, &;c. draw aline underneath, and begin with the units; after adding up every iig- ure in that column, consider how many tens are contained in their sum, and placing the excess under the units, carry so many as you have tens, to the next column, of tens : Proceed in the same man- ner through every column, or row, and set down the whole amount of the last row.* * This Rule as well as the method of proof, is founded on the known axiom, " the whole is equal to the fum of all its parts." The method of placing the numbers, and carrying for the tens, is evident from the nature of notation ; for, any other difpofition of the numbers would alter their value ; and carrying one, for every ten, from an inferiour to a fuperiour column, is, evidently, right; be- canfc one unit in the latter cafe is equal to the value often units in the former. Kefides the method of proof here given, there is another, by carting out the nines ; thus : X. Add the figures in the upper row together, and find how many nines are contained in their fum. 2. Reject the nines, and fet, down the remainder, directly even with the fig- ures in the row. 3. Do the fame with each of the given numbers, and fct all the exccfles of nines in a column, and find their fum ; then, if the excefs of nines in this fum, found, as before, is equal to the excefs of nines in the fum total ; the queftion is fuppofed to be right. Example. 573 u 5"J This method depends upon a property of the number ^ I 9, which, except 3, belongs to no other digit whatever; r viz, that any number divided by 9, will leave the fame ^ 1 remainder, as the fum of its figures, or digits, divided by 9456 8471 53*4 ^ ^6891 »< 6j 9: which may be thus demonftrated. Demonflration. Let there be any number, as 5432; this,fcparated into itsfeveral parts, becomes 5000 f4COf 30 i a; but 5000 -5 Xi 000 --.5X999-1-1 — 5X999 -j-5. In like manner 400 4 99 f 4, and 30— 3x9 ' 3. Therefore, 543* — 5x999 5» : 4 99-f4 »+3 9 *_3j:*=J- 999-1-4X99 1-3 X9r5-f4H-3H- 3- 543* 5x999 1-4x99 f 3 X9-!-5H-4+3-1 a And u^ ■ ; but 5 X999-f4X99-h3 X9 is .9 9 divilible by 9 ; therefore, 5432, divided by 9, will leave the fame remainder, as 5-i 4 '-34-*> divided by 9; and the fame will hold good of any other number whatever. The fame property belongs to the number 3 : However, this inconveniency attends this method, that, although the work will always prove right, when it is l"o ; it will not, always, be right, when it proves fo ; I have, therefore, given this demonflration more for the fake of the curious, than for any real advantage. In cafling out the nines, proceed thus. Begin with the uppermoft row of the r,xample at the left hand ; 5 and 7 are la, from which take out nine, and 3 re- mains : 3 added to 3 make 6, which muft be added to the 8, becaufe 6 is lefs than 9, and the fum is 14 : caft out nir\e and 5 remains, which is to bt placed at the right againfl the row, as in the example. In the next row, 9 the firft fig- ure, may be omitted becaufe it is 9 ; then 1 and 5 make 6, which added to the 6, make 1 a, from which take out 9. and 3 remains to be placed on the right of the row as liefore. Proceed thus with all the rows and with the fum at bottom. Then add the remainders ajainfV thefevrral ro^ys, cafting out 9 as often as i: 22 SIMPLE ADDITION. Proof. Begin at the top of the sum and reckon the figures down; wards, in the same manner as ihey were added upwards, and, if it be right, this aggregate will be equal to the first. Or, cut off the upper line of figures, and find the amount of the rest; then, if the amount and upper line, when added, be equal to the sum total, the work is supposed to be right. Addition and Subtraction Table. j 1 1 2 1 - 4 1 ^ R ' 7 8 1 • 10 12 |2 3 ~4 ~~b ~6 '"9 To 4 i ^ ^ 1 '7 8 9 10 1 >> 1 '^ i 13 14 ~i5 5 1 ^ ' 1 « 9 10 11 1 1 2 1 .1-3 8 1 '^ 8 1 9 10 1 " 12 13 14 ' 1 -. 16 1 7 1 ^ 9 10 " 12 13 1 1 -' 1 ■;i i 14 15 1 16 1 17 1 18 11:* 20 1 ": ! 1 2 When you would add two numbers, look one of them in the left hand column and the other at top, and in the common angle of meeting, or, at the right hand of the first, apd under the second, you will find the sum — as, 6 and 8 is 13. When you would subtract : Find the number to be subtracted in t!ie left hnnf all the fimilar parts, he equal to the difference of the whole. 1, When any figure in the greater number is lefs than its correfpondcnt fi;;- lire in the lefs, the ten which is added by the Rule, is the value of an unit in t'le next higher plice, l)y the nature of notation; and the one which is adilcd to the next place of the lefs number, is to dinilnifli the correfpondcnt place oi ». he greater, accordingly ; which is only taking from oiic place and adding at^ irmch to another, wlicrcby the total is never changed : And, by this means, th:: greater is refolved into fiich parts, a* are, eacli, greater than, or equal to, the 1'.inil;-ir pjrt of the Itfs ; ;ind rhc difference of the correfpondcnt figufC8, taken together, will, evidently, make up the difference of the whole. The truth of the rr.ethod of proof is evident ; lor tlie difTcrercc of two nuir- Iv-r^ -.iddcd ti. tlu- I(..'V, !« nntiifoft.'v. equ.il to the greater. SIMPLE MULTIPLICATION. 25 The operation on the first three Examples is Fufficiently plain. In the 4th Ex. I begin on the right hand, and lake 2 from 4, and set down the difference 2, under the column. As 8 is greater than .3, I add 10 to 3, which makes 13, and from it take the 8, and 6 is the diflerence to be set down. As I add 10 to the 3, I now add 1 to the 1 in the next higher place, because 10 in one place is equal only to 1 in the next higher place, and take the 2 from the 9, and the difference is 7. The rest of the work is obvious. The same proofs must be followed in every similar case. 7. 8. 9. 100200300400500600700800900 10000 1000000 980760.54032011023043067089 9999 1 10. What is the difference of 40875 and 38968? Ans. 1907. 11. What number must be added to 6892, so that the sum shall be 8265? Ans. 1373. 12. America was discovered in 1492 ; how many years have elapsed since ? 13. If you lend your friend 3646 dollars, and afterwards are paid 2998 dollars ; how much is still due ? Ans. 648 dollars. 14. If a man was seventy live years old in the year 1821, in what year was he born ? Ans. 1746. 15. The Independence of the United States was declared July 4th, 1776 ; how many years have passed since ? Ans. 16. Sir Isaac Newton died in the year 1727, aged eighty five ; in what year was he born? Ans. 1642. MULTIPLICATION TEACHES to find the amount of one number increased as many times as there are units in another, and thus performs the work of many additions in the most compendious manner ; brings numbers of great denominations into small, as pounds into shillings, pence or farthings, &c. and, by knowing the value of one thing, we find the value of many. The number given to be multiplied, is called the Multiplicand. The number given to multiply by, is called the Multiplier. The multiplicand and multiplier are otten called /oc^or^. The result of the operation, or the number found by multiplying, is called the Product. Multiplication is distinguished into Simple and Compound. SIMPLE MULTIPLICATlok Is the multiplying of any two numbers together, withput having regard to their signification ; as 7 tiines 8 i^ 56, &r. D 26 iSlMPLE ML'LTIPLICATION. Multiplication and Division Table. 1 2 3 4 5 6 7| M 9| 10 1 111 12 2 4 6| 8| 10 1 12 1 14 J 16 1 18 I 20.1 22 1 24 3 6 9 12 15 18 21 1 24 1 27 1 30 1 33 1 36 4 8 12 16 1 20 24 28 1 32 1 36 1 40 1 44 1 48 5 10 15 20 25 30 35 40 45 1 50 1 55 1 60 6 12 18 1 24 1 30 36 42 1 48 1 54 1 60 1 66 1 72 7| 14 1 21 1 28 i 35 1 42 i 49 1 56 1 63 1 70 1 77 1 84 8| 16 1 24 1 32 1 40 1 48 1 56 1 64 1 72 1 80 1 88 1 96 9 18 27 36 45 1 54 1 63 1 72 1 81 1 90 1 99 1 108 10 20 30 i 40 1 50 j 60 70 1 80 1 90 1 100 110 1 120 n 1 22 33 44 55 1 66 |77| 88 ! 99 1 110 1 121 1 132 12 1 24 36 48 60 72 84 1 96 1 108 1 120 1 132 1 144 To learn this Table for Multiplication : Find your multiplier hi the left hand column, and your multiplicand at toj), and in the com- mon angle of meeting, or against your multiplier, along at the right hand, and under your multiplicand, you will find the product, or answer. To learn it for Division : Find the divisor in the left hand col- umn, and run your eye along the row to the right hand until you find the dividend ; then, directly over the dividend, at top, you will find the quotient, shewing how often the divisor is contained in the dividend. Rule. Having placed the multiplier under the multiplicand so that unit* stand under units, tens under tens, &c. and drawn a line undei them, then, 1. IVhcn the multiplier does not exceed 12; begin at the right hand of the multiplicand and multiply each figure by the multiplier, set- ting down the unit figure under units, and so on, and carrying for the tens to the next place, as in addition, and the work is done.* 2. When the multiplier exceeds 12 ; multiply each figure of the multiplicand by every figure in the multiplier as before, placing the first figure of each product exactly under its multiplier : then • Dem. When the multiplier is a finglc digit, it is plain that we find the pro- duiSl ; for, bymuhiplying every figure, that is, every part of the multiplicand, we multiply the whole ; and, the writing down of the products, which arc lei>. than ten, ortheextefs of tens, in the places of the figures multiplied, and car- rying the number of tens to the product of the next place, is only gathering tO' gether the limilar parts of the refpective products, and is, therefore, the fame, in effect, as though we wrote down the multiplicand as often as the multiplier txprefies, and added them together; for the fum of every column is the pro- duct of the figures in the place of that column ; and the products, collected to- gether, arc evidently equal to the whole required product. SIMPLE MULTIPI^CATION. ad<^ together these several |)roducts as they stand, an4 their sum will be the total product.* * If the multiplier be a number, made up of more than one figure; after we have found the product of the multiplicand by the firft figure of the multiplier, 38 above, we fuppofe the multiplier divided into parts, and, after the fame manner, find the product of the multiplicand by the fecond figure of the mul- tiplier; but as the figure, by which we are multiplying, ftands in the place of tens, the product muft be ten times its fimple value ; and; therefore, the firft fig- ure in this product muft be noted in the place of tens, or, which is the fame, directly under the figure we are multiplying by. And, proceeding in the fame manner with all the figures of the multiplier, feparately, it is evident we fliall multiply all the parts of the multiplicand by all the parts of the multiplier; therefore, thcfe fcveral products, being added together, will be equal to the whole required product. The reafon of the method of proof, depends upon this propofition, that if two numbers arc to be multiplied together, cither of them may be made the multiplier or multiplicand, and the product will be the fame. A fmall attention to the nature of numbers will make this truth evident ; for 5X9 = 45 =9xr,; and, in general, 2X)X4X5X6, &c. == 3X2X6 XoX-*,&c. without any regard to the order of the terms ; and this is true of any number of fadtors whatever. The following examples arc fubjolncd, to make the reafon of the rule appear as clearly as pofllblc. 23795(> 3728 1903648 = 8 times the multiplicand. 475912 r= 20 times ditto. 1665692 = 700 times ditto. 713868 = 3000 times ditto. 64753 5 15= 3X5 25 = 50X5 35 = 700X5 20 = 4000X5 30 =60000X5 323 ;65=t]4753 X 5 887099068=3728 tiroes ditto. Another method of proving the rule is as follows. Let the factors be 6475I; .md ■>. Now ;)4753=OOU004-4000-1-7004-50-|-:.'. The fum of the products of thcfe quantities feverally multiplied by 5, is the true product. Then 60000-1- 4U00+700^-504-;3 is one faAOr. 5 the multiplier the other faaor. 300000^-20000^-3500-f 250+ 15=32'i765=64753 X 5. Or let the faaors be 45 and 24. Then 45= 40-f 5, and 24=20+4, and 40+5 multiplicand. l.ct the favSlors be 24 and 2^1. Then, 20+4 multiplier. 20+4 800+100=45 X 20 '^^+-^ 160+20=45X4 400+80 =24x20 800+260+20= 1080= 15 X 2 1. . .^^+16 = 24 X 4. 400+160+16 = 576=24X21. Multiplication may alfo be proved, by cafting out the nines ; but is liable to the inconvenience before mentioned. It may likewifc be, very naturally, proved by divifion ; for the produtSt, be- ing divided by eitlier of the fa(Slors,wIII, evidently, give the other ; and it might not be amifs for the pupil to be taught divifion,at the fame time with multipli- cation ; as it not only fcrves f'^r proof; but alfo gives him a readier knowledge of them both. But it would have been contrary to good method to have giv- en this rule in the text, bccaufc the pupil is fuppofed, as yet, to be unacquain'- ed with divifion. 28 SIMPLE MULTIPLICATION. Proof. Multiply the multiplier by the multiplicand. Multiply 3851 by 3. By addition. 3851 Multiplicand. 3851 3 Multiplier. 3851 3851 11553 Product. 11553 Sum. Having placed the numbers according to the rule,— then say, 3 times 1 is 3, and place 3 directly under units ; then 3 times 5 is 15, set down 5 and carry the one to the next product. Then, 3 times 8 is 21, to which the 1 is to b^ added, making 25 ; set down 5 and carry 2. Then 3 times 3 is 9, and the 2 to be carried, make 11, which set down, and the work is done. The result is thcsame as is obtained by addition. Multiply 6053 by 11. 11 Prod. 66583 Proceeding as before, multiply 3 by II, and of the product, 33, set down 3 under units, and carry 3 ; then 5.by 11, and to the pro- iluct, 55, add the 3 to be carried, set down 8, and carry 5 ; then O by 1,1, and as the product is 0, set down the 5, which was to be carried ; then 6 by U, and, as there is none to carry, set down the product, 66, and the operation is finished. Multiply 67013 by 29. 67013 MuItipHcand. 29 Multiplier. 603117 Product by 9, the units of the multiplier. 134026 Product by 2, the tens of the multiplier. 1943377 Product or answer. in this example, the multiplicand is first multiplied by 9, the units of the multiplier, and the product set down, as in the preceding ex- amples. The multiplicand is then multiplied by 2, the tens of the muUiplier, as before, the first figure of the product is placed under the 2, in the place of tens. The t\¥o products are then added, and their sum is the whole product or answer. Examples. f. 2. 3. . 4. 37934 769308 4980076 763896 2 3 4 5 Prod. 75868. SIMPLE MULTIPLICATION. 29 5. 67589 6 6. 503'; '64 : 7 7. 8. 3818295 9164785 8 9 Prod. 405534 9. 4879567 10 10. 5864794 11 11. €583478646 12 Prod. 64512734 12. 6357534 47 13. 8324629 59 14. 46293845 106 44502738 25430136 277763070 46293845 Prod. 298804098 4907147570 15. 647906 4873 16. 760483 9152 17. 91867584 6875 ^157245938 18. Multiply 103 by sixty seven. Ans. 6901. 19. Said Jack to Harry, you have only 77 chesnuts, but I have seven times as raany ; how many have I ? Ans. 539. 20. If four bushels of wheat make a barrel of 6our, and the price of wheat be one dollar a bushel, what will 225 barrels of Hour cost ? Ans. 900 dolls. 21. Eighty nine men shared equally in a prize, and received 17 dolls, each ; how much was the prize ? Ans. 1513 dolls. 22. Multiply 308879 by twenty thousand five hundred and three. Ans. 6332946137. In some cases the operations of multiplication are shortened by particular rules. Several Cases follow. Note. A composite number is the product of two or more num- bers, as 27, which 3x9, and, as 315, which =? 5X7X9. CASE I. When the multiplier is a composite number, multiply the multi- plicand by one of those figures, first, and that product by the other, kc. and the last product will be the total required.* * The rcafon of this method is obvious : For any number, multiplied by the component parts of another number, muft give the fame prodiitft, as though it were multiplied by that number at once: Thus, in example firft, 5 times the product of 7, multiplied into the given number, makea 33 times that given number, as plainly as 5 times 7 makes 35- m SIMPLE MULTIPLICATION. 1. Mult. 69375 by 35. 7 7x5 = 35 415625 5 2078125 Examples, 2. 39187 by 48. 3. 91632 by 56- 4. 3065 by 63. 6. 14567 by 144. 6. 6061 by 12K CASE n. When there are cyphers on the right hand of either the multipfteand, or multiplier, or both : Neglect those cyphers ; then place the sig- Dificant figures under one another, and multiply by them only ; add them together, as before directed, and place to the right hand as many cyphers as there are in both the factors. Examples. 1. 2. 3. 67910 956700 930137000 5600 320 9500 Prod. 380296000 306144000 8836301500000 CASE in. To multiply by 10, 100, 1000, «^c. : Set down the multiplicand uq- derneath, and join the cyphers in your multiplier to the right hand of them.* Examples. 1. 2. 3. 4. 57935 84935 613975 8473965 10 100 1000 10000 Prod. 679350 CASE l\^. 7b multiply by 9^9, 999, «^'c, in one line ; Place as many dots at the right hand of the multiplicand, as there are figures of 9 in your multiplier, which dots suppose to be cyphers ; then, beginning with the right hand dot, subtract the given multiplicand from the new ane, and the remainder will be the product. t * This is evident from the nature of numbers, fince every cypher annexed to :iic right of a number incrcafes the value of that number in a tenfold proportion. f Here It may eafily ht fcen that, if you multiply any fum by 9, the produift will be hut 9 tenths of the producSk of the fame fum, multiplied by 10 : and as the annexing of a clot or cypher, to the right hand of the multiplicand, fuppof- es it to be increafed tenfold : therefore, fubtradling the given multiplicand from the tenfold multiplicand, it is evident that the remainder will be ninefold jhe faid given multiplicand, equal to the produ^b of the fame by 9; and the *am.c will hold true of any number of nines. SiMPLE MULTIPLICATION. 31 Examples. 1. 2. 3. 6473.. 867389 . . . 5384976... ^ 99 999 9999 640827 53844376024 That these examples may appear as clear as possible, I will il« lustrate them by giving another. Mult. 371967.., by 999 371595033 C According to the rule, ) ( it will stand thus. ^ 371967... Minuend. 371967 Subtrah. 371696033 Rem. or total Pro. CASE V. To multiply by 13, 14, 15^ 4-c. to 19; also from 101 to 109, from iOOl to 1009, ^c. : Multiply with the unit figure only, of the mul- tiplier, removing the product one place to the right hand of the multiplicand, and so many places further as there may be cyphers between the significant figures ; then add all together, and their sum will be the product.* Examples. 1. 2. 3. 75964X13 7598x104 G735X100. 227892 30392 33675 Prod. 987532 CASE VI. To multiply by 21, 31, 41, 4'c. to 91, also by the same figures with any number of cyphers betzeeen them : Multiply by the left hand fig- ure, only, of the multiplier, and set the unit figure of the product one place to the left, and as many places further as there are cy phers between the significant figures ; and add the numbers togeth er for the product. Examples. 1. 2. 3, 73918 X21 66934 X301 45936 X400i 147836 170802 Prod. 1552278 17137134 4. 3167X500001. * The reafon of this Rule, and of the following one alfo, will be cvideut or infpedking an example under each rule, 32 DIVISION, CASE. VII, To multiply any number^ by any number^ giving only the Product ! Put down the product figure of the first figure of the multiplicand by thej^r^^ of the multiplier. To the product of the second of the multiplicand by the Jirst of the multiplier, add the number to be carried, and the product of the^;*si of the multiplicand by the second of the multiplier v then, carrying for the tens in the sam, put down the rest. To the product of the third of the multiplicand by the first of the multiplier, add the number to be carried, and the pro- duct of the second of the multiplicand by the second o( the multipli- er, also the product of the first of the multiplicand by the third of tlie multiplier, carry the tens, and put down the rest, and so pro- ceed till you have multiplied the highest of the multiplicand by the lorn-est of the multiplier. Then multiply the highest of the multi- plicand by the second of the multiplier : Add the number to be carried, and the product of the last but one of the multiplicand by the third of the multiplier, and the product of the last but two of the multiplicand by the fourth of the multiplier, &c. Then to the product of the last but one of the multiplicand by the fourth of the multiplier ; and so proceed till you have multiplied the last of the multiplicand by the last of the multiplier, which finishes the work. Example. Explanation, , , Mult. 632141S — By 2354 5x4—20 Prod. 12526610910 1X44-24-5X5=31 4x4-f3-r 1X54-5X3=39 1x44-3+4x5+1x34-5x2=40 2x4+4+1x5+4x3+1x2=31 3x4+3+2x5+1x3+4x2=36 5x4+3+3x5+2x3+ 1x2=46 6X 5+4+3X 3+2X 2=42 5x 3+4+3X 2=25 '. 5x2+2=12 DIVISION TEACHES to separate any number or quantity given, into any number of parts assigned ; or to find how often one number is con- tained in another ; and is a concise way of performing several Sub- tractions. SIMPLE DIVISION. 33 There are four parts to be noticed in Division, \u. The Dividend, is the number given to be divided. The Divisor^ ii the nuoiber given to divide by. The Quotient, or answer to the question, shows how many times the divisor is contained in the dividend. If there be any thing left after the operation is performed, it is called the Remainder ; sometimes there is a remainder and some- limes there is not. The remainder, when there is any, is of the same denomination as the dividend. Division is both Simple and Compound. Proof. Multiply the divisor and quotient tog^elher, and add the remain- der, if there be any, to the product ; if the work be right, that sum will be equal to the dividend. SIMPLE DIVISION Is the dividinj^ of one number by another, without regard to (heir values: As 56, divided by 8, produces 7 in the quotient: That is, 8 is contained 7 times in 56.* Rule. Having drawn a curve line on each side of the dividend and placed the divisor on the left hand of it, Seek how many times the divisor is contained in the least num- ber of the figures of the dividend on the left hand that do contain ♦ According to the rule, we refolv^c the dividend Into parts, and find, by tri- al, the number of times the divlfor Is contained In each of thofc parts ; and the only thing which remains to be proved, Is, that thefeveral figures of the quo- tient, taken as one number, accordlno to the order, in which they arc placed, are the true quotient of the whole dividend by the dlvifor ; which may be thus dcmonftrated. Dent. The complete value of the firft part of the dividend, is, by the nature of notation, 10, 101', 1000, &c. times the fimple value of what is taken in the' operation; accordingly, as there are 1, 2, or o, &c. figures ftandlng before it ; and, confequently, the true value of the quotient figure, belonging to that part of the dividend, is alfo iO, 100, 1000, &c. times its fimple value ; but the true Value of the quotient figure, belonging to that part of the dividend, found by the rule, is alfo U», iOO, 10(X), &c. times Its fimple value; for there are as many fig- ures fD. .... 87513671 4 2 2 3 9 5* 112517577 112517577 2 7 1 — . . .... 2 3 7* '. '. . . .' .* . 3 4* 9 8 7 6 5 4 3 2 1 Proof by Addition. We need only to refer to the example, except for the proof by addition ; where it nviy be remarked, that the Afterifms fliew the numbers to be added, and the dotted lines their order. f The reafon of this contradlion it is cafy to conceive; for cutting off the fame figures from each, is the fame as dividing each of them by 10, 100, 1000, &c. and it is evident, that as often as the whole divifor is contained in the whole dividend, fo often mud any part of the divifor l>e contained in the like part of the dividend; this method is only to avoid a needlefs repetition of cyphers-, which would happen in the common way, as may be fecn by working one of the examples of this cafe in the common way without cutting off the cyphers. SIMPLE DIVISION. 37. Examples, 1. 2. 65|00)3794326|75(58374 5193|000)8937643|893( 325 544 3. 520 917|0)47658|3( 243 195 4. 87^j00P)91764789430|00Q( 482 '455 276 260 1675 Rem. 5. 6. 7. Q.uot. Rem. Q,uot. Rem. Q,uot. Rem. 1|0)9584|(5 ' I|00)76495l80 li0OO)93751339|462 Note. In dividing by 10, 100, 1000, &c. when you cut off as many figures from the dividend, as there are cyphers in the divi- sor, your work is done ; those figures, cut off at the right hand,, aj^e the remainder, and those on the leA, the quotient, as above. CASE II. Short Division may be used when the divisor does not e;(ceed 12. it is performed by the following Rule, First, seek how often the divisor can be had in the first figure, or figures of the dividend ; which, when found, place in the quo- tient; then, mentally, multiply your divisor by the figure placed in the quotient, and subtract the product from the like number of the left hand figures of your dividend, and the remaining units, if any, must be accounted so many tens, which you must suppose to stand at the left hand of the next figure in the dividend, and to be reck- oned with it; then, seek how often you can have your divisor is those two figures ; but, if nothinq^ remain, you must then seek how often your divisor is contained in the next figure, or figures, and thus proceed till you have done. Examples. Divisor. Dividend. 2. 3. 4. 6. 2)71935 3)51903 5)633795 6)8471937 7)193847 Quot. 35967—1 rem. ^ iDlMPLE DIVISION. 6. 7. 8. d. 3)5437846 9)45963784 11)91843756 12)1196437847536 CASE III. When the divisor is such a number thai any t7a;o, or more, figures in zhe Table, being multiplied together, wiU produce it: Divide tire giv- en dividend by one of those figures; the quotient, thence arising, by (be other, and so on ; and the last quotient will be the answer.'^ ♦ This follows from the contradtion in cafe 3d, of Simple Multiplication, of which it Is only the rcverfe ; for the fourth part of the half of any thing is evi- dently the fame as the eighth part of the whole ; and fo of any other number. As the learner at prcfcnt is fuppofed to be unacquainted with the nature of fradlions, and as the quotient is incomplete without the remainder ; I fliall here give a rule for finding the true remainder, without having recourfc to fratftions. Rule I. JifuUiply the quotient by the divifor : Subtradl the product from the divi- dend and therefult will be the true remainder. The Rule, which is moft commonly made ufc of, when tbc divifor is a com* jjoiite number, is Rule II. Multiply thelaft remainder by the preceding divifor, or laft but one, and to the produ<5t add the preceding remainder ; multiply this fum by the next pre- ceding divifor, and to the product add the next preceding remainder ; and fc on, till you have gone through all the diviforsand rcniainders,to the firft. Example. 6)35397 divided by 15,Q 1 the laft remainder. — multiply by ♦,') the laft divifor but one, 5)14'2r>2— 5 — .')'2^46— -2 add 2 the fecond remainder, 560—1 7 multiply by 6 the firft divifor. Ans. 509-i-^;V 42 add 5 the firft remainder. 47 the true remainder, To explain this rule from the example, we may obfervc, that every unit in the L'rft quotient may be looked upon as containing 6 of the units In the given divi- ilcnd; conferuently, every unit which remains, will contain the fame; there- i^ore, tbii remainder muft be multiplied by 6, to find the units it contains of the j^lvf n dividend. Again, each unit in the next quotient will contain 5 of the pre- i'cding ones, or 30 of the firfl, that is, 6 times 5 ; therefore, what remains muft b^ iTtultiplied by 30, or, which is tlie fame thing, by G and 5 continually : Now, this is the fame as the Rule ; for inftead of finding the remainders, fcparately , they are reduced from the bottom, upwards, ftcp by ftep, to one another, and che remaining units, of the fame clals, taken as they occur. iJiMPLE DIVISION. 1st. method, 9)196473 8)21830 ^uoU 2728— ■57 Examples, 2c?. method, 8)196473 9)24559 Qnot. 2728-^57 3f/. methods 72)196473(2728 Qiwt 144 524 604 207 1^4 633 676 57 RtimaintJer. I have wrought the above question three way?, that the learner may understand the method of finding the true re.naindcr, accord- ing to this case. In the ^trsr, in dividing by 9, 3 remains, and by 8, 6 remains; which being the last remainder, 1 mtiUiply it by the first Jivisor 9, and add in the first remainder 3, and ihey make 57, the true remainder. In the second method, dividing by 8, 1 re- mains, and by 9, 7 remains ; I therefore, multiply 7, the last re- mainder, by 8, adding in the 1, and they make 57 as before. The third method is self evident, and shews that the other remainders are true. 2. 3. 4. 5. 36)79638 25)197835 87)93975 54)93738764 6. 7. 8. 121)75323939 132)38473692 , 144)891376429732 CASE. IV. IVIien (he divisor is a whx)le nvtnbcr zvith some part of unity, as S-^, 5}, 6|, &c. proceed by either of the tbllowing methods. 1. Multiply the whole ntimber in the divisor by the number of parts into which unity is divided in the fraction, and to the product add the number of parts of unity taken in the fraction, and the di- visor will be reduced to the parts indicated by the fraction, for a new divisor. Multiply the dividend by the parts into which unity is divided in the fraction, for a ne^v dividend. Divide the new div- idend by the new divisor, and the quotient will be the answer. Ex. 1. Divide 1820 by 21. Here, I multiply 2 by 3, and add 1 to the product, and have 7, for the new divisor. Then I multiply 1820 also by 3, and have 5460 for the new dividend. Then 5460-7-7=780, the answer, or 21 is contained in 1820, exactly 780 times. jVote. It is obvious that the dividend and divisor are propor- tionally increased by the multiplication, so that the quotient wil? be the same as if they had not been thus increased. Ex. 2. Divide G375 by 5^ . 40 SIMPLE DIVISION. Proceediog as before, the new divisor k J I, and the new divi- dend is 12760. Then 12760—11=1159 and 1 remains, or 1169^^- ihe number of times 5i is contained in 6375. 3. Divide 10142 by 3-|. Ans. 2766. 4. Divide 170 by 2f . 6. Divide 168765 by 15|. II. Proceed according to the general rule for division, being careful to add the value of the fractions to the several remainders, at every step of the process. Ex. 1. Divide 1820 by 2^. 2^)1820(78(5 Here 7x2i=16|, which being subtracted from 18, leaves 1|. [Now f belongs to the place of hundredths, and is f of 10 for the place of tenths. But I of 10=6|, to which add Ihe 2, and we have 8|to be annexed to the 1 remainder, and we have 18| for the true remainder. Now 2} is contain- ed in 18| exactly 8 times. The rest of the pro- cess is evident. Ex. 2. Divide 6375 by 6^. 51)6375(11 69 Jy — Here once 6|- taken Irom 6, leaves a. But |- l in thousandths place is 5 in the place of hun- — dredlhs, and 5 added to the 3 hundredths is 8. 8 Taking from 8, once 5^, 2i remain. As 5 J before, the i becomes 5 in the place of tenths, — whicJh added to 7 tenths, make 12, which 2| added to the remainder 2 in the preceding ' place, give 32. From this 6 times 5} are to 32 be taken, and 4i remain. But 4^ in the place -7]^ of tenths, is 45, which added to 5 make 50. — - From 50 take 9 times 5i, and i remains. Now 4i in 5i are 11 halves, so that ^ is ^\, which an nexed to the quotient, gives the whole quo- 60 tient. 49i Ex. 3. Divide 347 by 2| Ans 144-i%. 4. Divide 13567 by 13f. Supplement to Contractions in Multiplication. 1. The shortest method of multiplication, when the multiplier is any even part of 100, 1000, &c. is by division : For if the mnliipli- candbe increased by a number of cyphers equal to the number of places in the multiplier, and a part of that product taken for the same proportion, which the multiplier bears to 1, an 2. Divide 2625 dolls, among 35 soldiers. Ans. 73 dolls, each. 3. If 360 cents are to be divided equally among eighteen poor persons, how much vviil each have ? Ans. 20 cents. 4. If a field of 19 acres produce 513 bushels of wheat, liow much is that for one third of an acre ? Ans 9 bushels. 5. A man receives 1095 pounds a year, what is it for a day ? Ans. 3 pounds. 6. There are 5^ yds. in a rod ; how many yards are there in 40 rods ? Ans. 220 yds. 7. What number must you multiply by 47, to produce 298804098 ? Ans. 6357534. 8. What number must be multiplied by 379 to produce 33789678 ? Ans. 9. In a rod are 16i feet ; how many feet in 320 rods ? 10. If 3650 pounds of bread are to be divided equally among 365 soldiers for 10 days, how much will each receive a day ? Ans. 1 pound, n. Multiply 34678 by 250. Ans. 866950- 12. Multiply 125 by 77777. Ans, TABLES IN COMPOUND ADDITION. 43 TABLES IN COMPOUND ADDITION. 1. Federal Money.* marked. mills. 10 Mills ^ g rCent tn.c. ^ 10= 1 cent. 10 Cents f ® iDime d. f 100= 10= 1 dime. 10 Dimes (^ jDollar g. ( 1000= 100:= 10= 1 dollar. 10 Dollars) g (Eagle E.) 10000=1000=100=10=1 eagle, 2. English Money. marked 4 Farthings ^ C Penny grs. d. 12 Pence > make one < Shilling s. 20 Shillings ) ' ( Pound. £. * The following account is abftractcd from the " Act cflablifliing a Mint, and regulating the Coins of the United States," paflcd April and, 1792. The money of account of the United States (hall he cxprcflcd in dollars or units, difmcs or tenths of a dollar, cents or hundredths of a dollar, and mills or thoufandths of a dollar. The coins of gold, silver, and copper of the U. S. fliall he of the following de- nominations, viz. . r 1. Eagle, of the value of ten, dollars. '3 ■< 2. Half Eagle,;... .Jive dolls. O C -'. QvARTZk EAGLZy..........tzvo and a ha// do\h. 4. Dollar, of the value of the Spanjb w///(ri dollar. >. Half Dollar,..m... ». ...Aa^the dollar. > i f). Quarter DoLLARy«.........~....<»n of the doll. B. Half Dime,.- ~~.~ one ttvent'utb of the doll. CUj^ C .9. Cent, .......ik.......r,...one hundredth ol X\\t Ao^.. y c. c lO. Half Cent, -...-.«. one half the cent. The Jlandard for all gold coins of the U. S fliall be eleven parts of pure gold and one part of alloy in tivelve pirts of the coin. The alloy is to be filvcr and copper, but the filvcr is not to exceed one half in the alloy. The Eagle fliall contain two hundred ahd forty feven and a half grains of pure gold, or two hundred and feventy grains of ftandard gold; and the other gold coins in the fame proportion. The Jiandard for all filvcr coins of the U. S. fliall be one thoufand four hun- dred and eighty five parts of pure filvcr and one hundred and fcvcnty nine parts alloy; and tne alloy fliall be pure copper. The Dollar fliall contain three hundred and feventy one and one fourth grains of pure filver, or of four hundred and fixteen grains of ftandard filver; and the other filver coins in the fame proportion. The Copper coins are to be pure copper. The Cent is to contain eleven p cn- ny weights of copper; and the Half Cent in proportion. The proportional value of gold to filver in all coins current by law In the U. S. fliall be fifteen to one, or fifteen pounds weight of pure filver fliall be equal . to one pound weight of pure gold. All coins of gold and filver, ilfued from the Mint of the U. S. fliall be a law- ful tender in all payments at the preceding values when of full weight, and If Rot of full weight, of proportional values. 44 TABLES IN COMPOUND ADDITION. Farthings. 4 = 1 Penny. 48 = 12 = 1 Shilling. 960 = 240 = 20 = 1 Pound. A groat is 4c?. FfiNCE Tables. cl. S. d. d. s. d. s. d. s. d. 20 = 1 8 120 = 10 1 = 12 11 = 132 30 = 2 6 130 = 10 10 ^ 2 = 24 12^ = 144 40 = 3 4 140 = 11 8 3= 36 13 = 166 50 = 4 2 150 = 12 6 4 = 48 14 = 168 60 = ^ 160 = 13 4 6 = 60 15 = 180 70 = 5 10 170 = 14 2 6 == 72 16 = 192 80 = 6 8 180 = 15 7 = 84 17 = 204 90 = 7 6 190 = 15,10 8 = 96 18 = 216 100 = 8 4 200 = 16 8 9 = 108 19 = 228 no ^ 9 o 240 = 20 10 = 120 20 = 240 3. Troy Weight.* 24 Grains make one Pennyweight, marked grs. put 20 Pcnnyweig hts - - Ounce, oz. 12 Ounces - =. - Poi] nd, - ^ m \G Drams 16 Ounces 28 Pounds - 4 Quarters ~ 20 Hundred wt Grains. 24 = 1 Pennyvveighto 480 = 20 = 1 Ounce. 5760 = 240 = 12 = 1 Pound. 4. Avoirdupois Weight.! make 1 Ounce, marked dr. oz. Pound, ----- iti - Quarter of a hundred wt, - qr. Hundred wt. or 112 pounds, - Cwt. - - Ton, T. * By this weight are weighed Gold, Silver, Jewels, Eledluaries, and all liquors. All ounce of gold is divided into 24. parts, called carats^ and an ounce of fil- vcr, into 20 parts, called pennyweights; therefore, to diftinguiili finenefs of metals, inch gold as will abide the fire without lofs, is accounted 24 carats fine -. If it lofe % carats in triiil, it is tailed az carats fine, &c. A pound of filver which, lofes nothing in trial, is iz ounces fine; but, if it Jofe 3 pennyweights, it is ii oz. 17 pwts. fine, &c. Alloy is fome bafe metal with which gold or filver is mixed to abate its fine- nefs ; 22 car-its of gold, and 2 carats of copper, are cftcemed the true ftandard for gold coin in England, the alloy being one eleventh part of the fine gold : and n oz 2 pwts. of fine filver, melted vt^ith 18 pwts. of copper, make the true flandard for filver coin. Note. 175 Troy ounces, are precifely equal to 192 Avoirdupois ounces, and 175 Troy pounds are equal to 144 Avoirdupois, x lb. Troy =: 5760 grain?, and I lb. Avoirdupois = 70CO grains. •{• By Avoirdupois arc weighed all coarfe and drofly goods, grocery and chand- lery wares ; bread, and all metals, except gold and filver. A barrel of pork weighs 200 lb. A barrel of beef, 2Qo lb. A f^nintal of fifli. TABLES IN COMPOU^fD ADDITIOX. 45 Drams. 16 = 1 Ounce. 256 = 16 = 1 Pound. 7168 = 448 = 28 = 1 Quarter. 28672= 1792= 112= 4= 1 Hund. wt. 573440 = 35840 = 2240 = 80 = 20 = 1 Ton. 5. Apothecaries' Weight.* 20 Grains 3 Scruples 8 Drams 12 Ounces make one Grains. 20 = 60 = 480 = 5760 = Scruple, Dram,. Ounce, Pound, marked gr. 9 - 3 - lb. 1 Scruple. 3 = 1 Dram. 24 = 8 = 1 Ounce. ;88 = 96 = 12 = 1 Pound. 6. Cloth Measure.! 2 Inches, and one foupth 4 Nails, or 9 Inches 4 Quarters of a yard, or 36 Inches 3 Quarters of a yard, or 27 Inches 5 Quarters of a yard, or 45 Inches 6 Quarters of a yard, or 54 Inches 4 Quarters, 1 Inch & one 5th, or 37 Inches and one fifth 3 Quarters and two thirds make 1 Nail, marked in. na. Quarter of a yard, qr. Yard, Ell Flemish, Ell English, Ell French, Ell Scotch, Spanish Var. yd. E. Fl. E. E. E. Fr, p. Sc. 1 Cwt. Avoirdupois, ja particular things make one dozen ; la dozen i grofs, and 144 dozen i great grofs. ao particular things make 1 fcore, lb. A Stone of Iron, fliot,7 lb. 56 orhorfeman's weight,^ I4 94 Butcher's Meat, 8 30 A gallon of Train Oil 7^ 2.06 A Tod is - - - - 28 112 A Weigh - - - - 182 1120 A Sack - - - - 364 A Firkin of Forelgt A Barrel of — ■ ^ Punch, of- A Fother of- |Butter %oap Anchovies Soap Raiiins Prunes Lead 19^ Cwt. A Sack Ahft 4368 * All the weights now ufed by Apothecaries, above grains, are Avoirdupois. The Apothecaries' pound and ounce, and the pound and ounce Troy are the fame, only differently divided and fubdivided. t All Scotch and Irifli linens are bought by the Englifli or American yard, which is the fame, and all Dutch linens by the Ell Fiemifli ; but are all fold in America by the American yard; though the Dutch liucns are fold in EugUnc^ >y the Ell Englifli, and the Scotch and Irifli linens, as in America. The Scotch allow one Englifli yard in every fcore yards. 4Q TABLES IN COMPOUND ADDITION Nails, 4 = 1 Quarter. 16 = 4 = 1 Yard. 12 = 3 = 1 Flemish Ell. 20 = 5 = 1 English £11. 24 = 6 = 1 French Ell. 7. Long Measure.* 3 Barley corns -r make 1 Inch, 12 Inches - - . ^ Foot, 3 Feet, - . - - - Yard, - 51 Yards, or 16i feet - - Koc^ Perch, or Pole, 40 Poles Furlong, 8 Furlongs - - - . Mile, C91 Statute miles, nearly \ C^^gree of a ^ f great Circle, A great Circle of the Earth. Or^ in Measuring Distances. make 1 Link. marked bar. in. ft. ■ yd. pol. fur. mile. deg. 360 Degrees 7yW Inches 25 Links - 100 Links 10 Chains - 8 Furlongs Bar- corn?, 3 — 1 Inch 36 = 12 = 108 = 36 == 694 = 198 = 23760 = 7920 = 660 = 190080 = 63360 = 6280 Inches, 1^^-^ = 1 Link, Pole. Chain. Furlong,. Mile. 1 Foot. 3 = 161- 1 Yard. 61 = 1 Pole. 220 = 40 = I Furlong. 1760 = 320 = 8 = 1 M. 198 = 25 = 1 Pole or Perch. 792 = 100 = 4=1 Chain. 7920 = 1000 = 40 = 10 = 1 Furlong. 63360 r=: 8000 = 320 = 80 = 8 = 1 Mile. 8. TiME.t make 1 Minute, marked s, Houri; . Day, - Week, - Month, '. t!0 Seconds - - - - make J Minute, marked s. m. GO Minutes HourJ - - h. 24 Hours - - - - - " Day, - - d. 7 Days Week, - - w. 4 Weeks .-.-.- Month, - - mo. 23 Months, 1 day and 6 hours - - Julian year, yr. • The ufe of l p , 2721 Feet J " 40 Poles . - - . . Rood. 4 Roods, or 160 Rods, > ^ ^^^.^ or 4840 yards J 640 Acres - - - Mile. Inches, 144= 1 Foot. 1296= 9= 1 Yard. 39204= 2721= 30i= 1 Pole. 1568160= 10890= 1210= 40= 1 Rood. 6272640= 43560= 4840= 160= 4= 1 Acre. ' 4014489600=27878400=3097600=102400=2560=640=1 Mile. • The civil folar year of 365 days, being (hort of the true by 5h. 48m. 48fec. occafioned the beginning of the year to run forward through the fcafons nearly i day in 4 years. On this acrouct, Julius Ctfar ordained that one day fhould be added to February, every fourth year, by caufing the a4th day to be reckoned twice; and hccaufe this 24lh day was the fixth, (fextilis) before the kalenda of March, there were in this year, two of thefe fcxtiles, which gave the name of Biflextile to this year, which being thus corretStcd, was from thence called the Julian year. t A juft and equal meafure of the year is called the periodical year, as being the time of the earth's period about the fun ; in departing from any fixed point in the heavens, and returning to the fame again. I The fcveral points of the ecliptick having a retrograde, or backward mo- tion, the equinox will, as it were, meet the fun ; by which mean the fun will ar- rive at the Equinox, or firft point of Aries, before his revolution is completed, and this fpace of time is called the tropical year. § The Zodiack is a great circle of the fphere, containing the 12 figns, througK Tvfilch the fun pafle?. Ton or Load. 48 TABLES IN COMPOUND ADDITION. n. Solid Measure.* I72« Indies - - make 1 Foot. 27 Feet - - . - Yard. 40 Feet of round Timber, or } 60 feet of hewn Timber, ) 128 Solid Feet, i. e. 8 in length, 4} i-. , r «t < in breadth and 4 in height, I ^""'^ <>^ ^ood. 12. Wine Measure.! 2 Pints - make 1 Quart, marked pts. qts. 4 Quarts - - Gallon, - gal. 10 Gallons - - Anchor of Brandy, anc. 18 Gallons - - Runlet, - - run. 311 Gallons - Half an Hogshead, ihhd. 42 Gallons - ' Tierce, - tier. 63 Gallons - Hogshead, - hhd. 2 Hogsheads - Pipe or butt, P. or B. 2 Pipes - - Tun, - - Tun Cubick Inches. 28f = 1 Pint. 57| = 2 = 1 Quart. 231 = 8 = 4 = 1 Gallon. 9702 = 336 = 168 = 42 = 1 Tierce. 14553 == 604 = 262 = 63 = li= 1 Hogshead. t9404 == 672 = 336 = 84 = 2 = 11-= 1 Puncheon. 29106 = 1008 = 604 = 126 = 3 = 2 = li= 1 Pipe. 38212 = 2016 = 1008 =262 = 6=z4=3=2 = l Tun. 15. Ale or Beer Measure.| 2 Pints - - make 1 Quart, marked ptS. qts. 4 Quarts - - Gallon, - - gal. 8 Gallons - - - Firkin of Ale in Lond. A. fir. S^Gallons * ■^ Firkin of Ale or Beer. 9 Gallons - - - Firkin of Beer in Lond. B. fir. 2 Firkins - - Kilderkin, - - kil. 2 Kilderkins - - Barrel, - - bar. H Barrel, or 64 Gallons Hogshead of Beer, hhd. 2 Barrels - - Puncheon, - - puii. 3 Barrels or 2 Hogsheads Butt, - - butt. * By Solid Mcafurc are meafurcd all things that have length, breadth and depth. f All Brandies, Spirits, Perry, Cider, Mead, Vinegar,Honey and Oil, arc meaf- urcd by Wine Meafurc : Honey is commonly fold by the pound Avoirdupois. \ Milk is fold by the Beer quart. A barrel of Mickarel, and other barrelled fifli, by hw in Maflachufctts, is to contain not lefs than 30 gallons ; in Connedlicut and New York the Shad and Salmon Barrel muft contain aoo lb. In England, a barrel of Salmon or Eels is 42 gallons, and a barrel of Herrings 3a gallons. The gallon> appointed to be ufed for meafuring all kinds of Liquors, n Ireland^'is two hundred «ind feventcen cubick indies, and fix tenths. TABLES IN COMPOUND ADDITION. 49 Beer. Cubick Inches. I 331 = 1 Pint. 701 = 2 = 1 Quart. 282 = 8 = 4 = 1 Gallon. 2538 = 72 = 36 = 9=1 Firkin. 5076 ^144= 72= 18= 2 = 1 Kilderkin. 10152 = 288 = 144 = 36 = 4 = 2 = 1 Barrel. 15228 = 432 = 216 = 54 = 6=3 = li= 1 Hogshead. 20304 = 576 = 288 = 72 = 8 = 4 = 2= 1^= 1 Puncheon, 30456 = 864 = 432 = 108 = 12 = 6 = 3 =»= 2 = li= 1 Butt. Ale. Cubick Inches. 351 = 1 Pint. 70X = 2 = 1 Quart. 282 = 8 = 4=1 Gallon. 2256 = 64 = 32 = 8 = 1 Firkin. 4512 = 128 = 64 = 16 = 2 = 1 Kilderkin. 9024 = 256 = 128 = 32 = 4 = 2 = 1 Barrel. 13536 = 384 = 192 = 48 2^ 6 = 3 = li= 1 Hogshead. 16. Dry Measure.* 2 Pints - - mpke 1 Quart, marked pfs. qts. 2 Quarts - - - Pottle, - - pot. 2 Pottles - - Gallon - - gal. 2 Gallons - - - Peck, - - pk. 4 Pecks - - Bqshel, - - bu. 2 Bushels - - - Strike, - - str. 2 Strikes - - Coom, - - co. 2 Cooms - - - Quarter, - - qr. 4 Quarters - - Chaldron, - ch. 4i Quarters - - Chaldron in London. 5 Quarters - - Wey, - - wey. 2 Weys > - - Last, - - last. Cubick Inches. 268| = 1 Gallon. 537f = 2=1 Peck. 2150| = 8 = 4=1 Bushel. 4300| = 16 = 8=2=1 Strike. 8601f =32 =16 =4=2=1 Coom. 172031 = 64= 32= 8=4= 2= 1 Quarter. 86016 = 320 = 160 = 40 = 20 = 10 = 6 = 1 Wey. 172032 = 640 = 320 = 80 = 40 = 20 = 10 = 2 = 1 Last. * This measure is applied to all dry goods, as Corn, Seed, Fruit, Roots, Salt, Sind, Oysters and Coals. A Winchester bushel, is iS^ inches diameter, and 8 inches dcrp, a m COMPOUND ADDITION COMPOUND ADDITION IS the adding of several numbers together, having different de- nominations as Pounds, Shillings, Pence, &c. Tons, Hundreds, Quarters, &c. Rule.* I. Place the numbers so that those of the same denomination may stand directly under each other. il. Add the first cohimn or denominalion together as in whole numbers ; then divide the sum by as many of the same denomina- tion as make one of the next greater, setting down the remainder under the column added, and carry the quotient to the next supe- riour denomination, continuing the same to the last, which add as in simple addition. Examples. J, Federal Moke v. 1. 2, 3. E. D. d. c. m. D. c. m. D. c. m. 7 3 8 9 6 49 18 7 375 2 12 6 25 32 1 29 18 9006 93 76 '7 12 5 3 6 2 5 13 25 199 18 7 7 14 8 97 2 30 01 24 1 10 3 !. English Money • 1. 2. 3, 4. £ s. d. £ s. d. qr. £ J. d. qr. £ s. d. qr. 9 16 10 47 17 6 2 847 11 11 2 915 10 10 2 7 10 9 - 3 9 10 3 491 19 6 1 64 8 9 1 18 6 75 13 9 1 59 6 10 5 16 11 3 5 11 11 4 11 11 747 16 1 2 419 2 10 2 6 8 16 8 2 849 12 11 3 491 19 11 3 6 9 10 17 6 2 1 741 17 8 2 762 17 6 1 35 8 As the denominations of Federal Money increase iike whoh. numbers, in a ten fold ratio, the operation is the same as in whole * The reason of this rule is evident from what has been said in Simple Ad- dition : For, in addition of money, as i, in the pence is equal to 4 in the far- things; I, ill the shiUings, to la in the pence; and i, in the pounds, to ap in the shillings; therefore, carrying as directed, is the arranging the money, aris- ing from each column, properly, in the fcale of denominations ; and this rea'» soning will hold good in the addition of compound numbers, of any denomi- nation whatever. COMPOUND ADDITION. ^1 numbers. But in denominations which do not increase in the same manner, the operations are somewhat 'tifferent. ihus, in Ex. 1. of English Money, I find the sum of the pence to be 64. Now 51 pence are 4 shillings and 6 pence ; therefore, I set down G under the pence, and carry 4 to the shillings, which I then find to be 68. But 68 shillings are 3 pounds and 8 shillings. I set doxvn the 8 under the shillings, and carry 3 to the pounds, and the sura of the pounds is 35, which I set down. The su.n of the whole is then 35 pounds, 8 shillings and 6 pence. The process is similar in each Example. In all sums oC different denominations, the student should be careful to find the numbers by which the denominations in the Table increase, for by Ihem he is to carry from one denom- ination to another. 3. Troy Weight. I. 2. 3. li. oz. fiut. gr. ii. oz. fivt. gr. th. oz. pv>U gr. 767 10 17 22 649 11 19 20 859 9 15 20 39 6 9 17 32 9 6 5 437 10 17 22 417 11 16 18 841 10 11 19 640 11 6 935 9 17 19 473 9 17 23 738 9 12 18 478 10 17 22 764 11 8 9 49 16 17 387 9 16 15 165 6 10 19 584 10 9 3027 11 16 17 la the 1st Ex. I find the 8um of the grains to be 113. Now 113 grs. are 4 pwts. and 17 grs. because 24 is contained in 113, four times, and 17 is the remainder. Then 1 set down 17 under the grs. and carry 4 to the pwts. and their sum is 96. Now 96 pwts. are 4 oz. and 16 pwts. for 20 pwts. make 1 oz. ; therefore I set 16 under the pwts. and carry 4 to the ounces, which makes their sum 69. But 59 oz. are 4 lbs. and 11 oz. because 12 oz. make a lb. ; therefore I set down 11 oz. and carry 4 to the lbs. which makes their sum 3027. The answer, then is 3027 lbs. 11 oz, 16 pwts. and 17 grs. 4. Avoirdupois Weight. 1. 2. 3. 4. IL oz, Jr. Cwt. qrs. lb. T. Civt. qrs. lb. r. Cwt. grs. lb. oz. dr. 19 13 12 17 3 19 59 13 2 17 91 17 2 25 13 15 21 9 6 18 1 27 6 17 1 21 19 9 17 10 12 4 15 15 9 2 9 45 11 3 25 14 13 2 9 11 22 10 5 14 3 16 57 16 2 19 47 11 3 19 14 18 13 12 12 6 75 17 3 17 69 1 12 6 11 10 15 2 6 19 26 77 19 3 27 15 11 94 10 12 5iJ COMPOUND ADDITION. 6. Apothecaries' Weight. 1. 2. 3. 4. 3 Bgr, 5 3 9 ^'•• ft 3 3 9 gr. ft 3 3 B^r. 9 1 17 10 7 2 19 12 11 6 1 15 5 9 3 2 13 3 2 19 ■ 6 3 12 4 9 10 12 4 8 6 19 6 1 17 7 6 1 17 91 10 7 2 16 9 10 5 2 12 4 6 9 5 2 12 4 8 12 19 6 6 6 1 17 5 2 12 6 1 16 6 1 10 8 9 4 8 1 10 9 3 2 18 4 9 2 1 6 7 1 1 17 33 2 1 6. Cloth Measure . 1. 2. 3. 4. 5. Yd. qr. n. /?.jE. jr. «. jE.jF/. yr. n. E,Fr. qr. n. Yds. qr. n. 76 2 3 91 3 2 75 2 1 49 3 3 914 2 3 3 3 1 49 4 3 7 1 3 19 5 2 49 2 1 42 3 3 6 2 3 84 2 24 2 1 561 3 67 2 2 84 4 1 76 2 3 67 4 3 84 2 16 3 3 7 48 2 2 48 2 2 649^3 1 49 2 2 61 2 1 7. ; 9 2 3 6 3 3 617 1 3 Long Measure. 1. 2. 3. 4. 5. It, in. bar. Yd .//. »«. Po/. ft. /«. Milfur.pol. Deg, mi. fur. fol. ft. in. he. 9 11 2 7 2 11 12 11 10 9 7 36 759 bQ 6 29 15 10 2 6 9 1 4 1 6 ! 3 10 9 7 3 19 317 39 1 36 11 6 1 7 2 6 10 1 8 12 11 4 1 24 497 63 7 24 9 8 1 8 10 7 2 9 7 15 6 6 5 12 562 17 11 13 11 9 6 2 8 1 10 4 14 9 4 6 9 64 48 5 17 9 4 2 7 10 2 9 2 11 5 11 11 5 1 10 764 52 4 19 15 11 1 8. Time. 1. 2. 3. IV.d. h. m. s. Mo.d. /j. m. r. >^. d. 3^. mo. IV. d. h. tn. s. 3 6 22 57 42 5 24 19 43 19 10 17 67 1 1 3 6. 23 29 55 1 5 19 31 28 4 27 21 35 7 9 27 4 8 1 1 19 45 38 2 3 17 9 15 9 18 12 4 8 16 29 9 2 3 17 18 19 3 9 17 58 4 19 23 19 1 11 14 46 10 2 5 11 50 13 1 1 16 19 10 8 11 12 13 17 6 9 W 9 2 1 16 18 17 2 2 20 53 48 9 19 8 29 12 5 20 46 9 3 6 18 17 59 D. Motion. 1. 2. 3. 17*» 55' 48" 25° 49' 51" 9s 2&° 35' 53" 1 37 51 6 21 36 10 18 31 29 19 45 19 47 18 4 17 13 42 19 19 37 25 25 39 6 19 60 COMPOUND ADDITION. 63 10. Land or Square Measure. 1. 2. 3. FoLfeet. in. Yds. ft. in. Acres, rood. poU feet, in. 36 179 137 28 7 119 756 3 37 245 120 19 248 119 9 3 75 29 1 28 93 25 12 96 75 29 6 120 516 3 31 128 119 18 110 122 4 8 12 37 1 19 218 20 9 269 24 9 1 119 61 92 103 25 221 143 8 3 43 Measure. 191 1 25 129 136 11. Solid M 1 ' . 2. 3. Ton. feet. in. rjif. /^/r/. in. Cord. feet. in. 29 36 1229 75 22 1412 37 119 1016 12 19 64 9 26 195 9 110 159 18 11 917 3 19 1091 48 127 1017 19 8 1001 28 15 1110 8 111 956 6 523 49 24 218 21 9 27 17 39 1119 18 17 1225 9 28 1091 12. Wine Measure. 1. 2. 3. Tier 37 . gal. qts. pts, 36 3 1 nbd. gal. qts. 51 58 1 t>ts. 1 Ton. 37 bbd, gal. qts. 2 37 2 9 17 2 1 27 39 3 19 1 59 1 35 28 9 9 18 1 28 2 32 19 1 1 9 2 1 19 47 1 9 3 1 16 24 1 1 37 1 17 3 12 40 1 1 1 15. 5 3 URE. 14 2 48 2 Ale and Beer Meas 1. 2. 3. A.B. f. gal, 49 3 7 £. B. fr. gal. 29 1 8 Hhd. 379 gal. 53 qtu 3 26 2 3 19 3 5 19 1 9 4 16 3 121 37 2 17 3 9 1 8 467 19 1 27 1 6 14 2 561 16 19 3 7 17 1 5 75 2 16. Dry Measure. 2. ^rs. bu. P- qts. Bus. /• qts. pts. Cb. bu. />• qts. 64 7 3 7 37 2 5 1 37 27 3 7 9 4 1 5 19 3 7 1 6 29 1 5 19 6 2 1 \% 2 15 30 4 2 5 1 6 1 4 11 3 9 17 3 6 9 3 5 1 9 5 3 4 19 3 1 2 2 1 54 COMFOUND SUBTRACTION. COMPOUND SUBTRACTION TEACHES to find the difference, inequality, or excess, between any two sums of divers denominations. Place those numbers under each other, which are of the same denomination, the less being below the greater ; begin with the least denomination, and, if it exceed the figure over it, add as ma- ny units as make one of tlie next greater ; subtract it therefrom ? and to the difference add the upper figure, remembering, always, to add one to the next superior denomination, for that which you edded before. Examples. Federal Money. 1. f $ c. m. From 39 15 5 Take 23 17 2 Diff. 10 98 3 c. 18 E. 21 10 $ 8 7 c. m. $ 1 2 100 5 48 c. m. 87 5 Borrowed 100 Paid 29 m. Lent 200 Received 145 c. 60 He mains to pay Due tome $ Borrowed 3000 c. $ c. m. Lent 7159 12 8 Paid C 195 at \lllo 49 several j 247 37 5 Received C 245 37 5 at >3112 15 several i 2000 limes. 995 12 5 times. 1092 92 Paid in all 2552 99 Received in all Remains to pay 447 01 Remains due English Money. 1. JC s. d. qr. Borrowed 349 15 tj 1 Paid 195 11 8 1 2. £ s. d. qr. Lent 791 9 8 1 Received 197 16 4 2 iiem. to pay 154 3 10 Proof Due to me "* The reafon of tbis Rule will readily appear, from what was faid in Simple .^abtraaion; for the adding depends upon the fime principle, and is only dif- •ferent.a? the numbers to b'^ fubtra'^ed are of different denominations. COMPOUND SUBTRACTION. 5^ In the 1st Ex. of English Money, I take 1 qr. from 1 qr. and set down 0, the remainder. Because I cannot take eight from 6 pence. I add to 6, 12 pence which make a shilling, and from 18 take 8, and set down 10, the difference. As I added 12 pence = 1 shilling to the upper pence, I now carry 1 shilling to the lower shillings, and take 12 from 15, and set down 3, the remainder. The rest of the process is evident. It is obvious that a similar course must be pur- sued in the Examples under the several weights and measures. 3. Tnoy Weight. I. 2. 3. lb. oz. pwt. dr. lb. oz. pwt. gr. lb. oz. pwt. gr. Bought 749 5 13 16 189 8 12 10 543 3 9 13 Sold 96 9 19 13 148 4 16 19 179 1 16 10 Rem. 652 ' 7 14 [ 03 1. lb. oz. Bought 7 9 Sold 3 12 dr. 12 9 4. Avoirdupois Weight 2. 3. C. rr. lb. T.cwt.qr. lb. 8 2 13 6 13 1 12 4 1 15 1 12 2 17 4. T.cwt.qr. lb. oz. dr. 9 11 3 17 5 12 3 12 1 19 10 9 Rem. 3 13 3 13 16 1. Ife 3 3 9 71 9.3 1 37 8 4 1 5. Apothecaries' Weight. 2. 3. ife 3 5 9 gr. ft 3 3 9 gr. 65 10 6 2 10 84 1 1 1 1 31 8 4 2 9 65 9 3 1 17 34 6 2 17 I Pol. 21 9 1. Yd8.qr. n. 35 1 2 19 1 3 6. Cloth Measure. 2. 3. :. E. qr. 11. E. Fl. qr. 167 3 1 765 1 >91 3 2 149 2 4. n. E.Fr. qr. n, 3 549 4 2 1 197 4 '^ 15 3 3 1. Yds. ft. in. 28 2 10 17 2 11 7. Long Measure. 2. 3. ft. in. Mil.fur.poJ. Beg. 11 9 76 3 11 38 13 8 27 3 21 19 4. m. fun p. yds,ft.in.bar. 41 3 29 2 1 7 2 35 5 31 3 1 9 1 10 2 11 1. Mo. d. h. m. G 17 13 27 1 21 16 41 s. 19 '35 44 8. Time. 2. 3. Mo.w.d. h. Y.mo. d. 9 2 5 15 7 3 13 4 3 5 15 4 2 19 4. Y. mo.w.d. h. m. s, 48 9 2 5 19 27 31 19 9 3 4 20 19 49 4 25 20 45 5G ' COMPOUND SUBTRACTION. 1. 79° 21' sr 41 41 52 9. Motion. 2. 6s 11° 12' 48" 3 8 39 29 Mea A. 56 29 J. 4s 19° 41' 22'^ 1 22 19 46 10. 1. A. R. Pol. 29 1 10 24 1 25 Land or Square 2. A. R. Pol. 29 2 17 17 1 56 SURE. 3. R. Pol. ft. i"n. 3 19 27 110 21 210 129 I. Tons. ft. in. 49 19 1100 38 36 1296 11. Solid Measure. 2. Ydu. ft. in. 79 11 917 17 25 1095 3. Cords. ft. in. 349 97 1250 192 127 1349 12. Wine Measure. 1. 2. 3. 4. Hhd. gaK qts. pts. Tier. gal. qts. Hhd. gal. qts. Tun. hhd. gal. 79 21 2 1 19 17 1 375 41 2 532 1 19 38 61 3 1 12 29 2 197 36 3 197 1 47 13, Ale and Beer Measure. 1. 2. 3. A.B. fir. gal. qts. B.B. fir. gal. qts. pts. Hhds. gal. qts. 39 1 2 1 21 3 5 2 769 17 1 24 3 6 2 19 1 7 2 1 391 42 3 14. Dry Measure. 1. 2. 3. Qa. bu. pk. qts. Bu. pk. qts. pts. Ciial. bo. pk. qts. 56 2 2 1 91 1 3 2 39 12 2 1 39 3 1 2 29 2 1 1 24 25 3 2 PROBLEMS. # «T PROBLEMS nESULTING FROSl A COMPARISON OF THE PRECEDING RULES. Prob. 1. Having the sum of two numbers, and one of them giv- eti, to find the other. Rule. Subtract the given number from the gi vert sum, and the re- mainder will be the number required. Let-SSS be the sum of two num- From 288 the Sum, hers ; one of which is 1 15, the oth- Take 1 15 the given number. <'r is required ? Rem. 1 73 the other. pROB. 2. Having the greater of two numbers, and the differ- ence between that iind the less given, to find the less. Rule. Subtract the one from the other. Let the greater number be 325, and From 325 the greater, the difference between that and the Take 198 the difference. Other, 198: What is the other? Rem. 127 the less. Prob. 3. Having the least of two numbers given, and the dif- ference between that and a greater, to find the greater. Rule. Add them together. p. ^ 127 the less number. uiven ^ 193 ^i^^ difference. Sum 325 the greater number required. Prob. 4. Having the sum and difference of two numbers given, to find those numbers. Rule. To half the sum add half the difference, and the sum U the greater, and from half the sum take half the difference, and the remainder is th« less. Or, from the sum take the difference, and half the remainder is the least : to the least add the given differ- ence, and the sum is the greatest. What are those two numbers, whose sum is 48, and difference 14 ? 2)48 2)14 24-f 7=31 the jrreater,and 24— 7=17 the less. i.s„m=24 }m.^ Ov46'—l4-^2=\l, & 17+14=3J. This rule is obvious on considering any exanaple in the following manner. Thus, let the two numbers be S2 an^2=4G-f 324^46— 32--r-e=46"-f-4G--f-2=J6, the gfreater. And 4G-|-32-~2— 46-{-32~-2 =4t)-f32~46-f 32-^-2 = 32-r32-4-2= 32, the less. For, in the first case, the 32 to be subtracted from 32, leaves nothing ; and, in the latter, .the 46 is balanced by the other 46, and you have only 32-j-32-r-2=32. The preceding and following probjoms are evident from the rules of Additiuri and Subtraction, Multiplication and Division. Prob. 5. Having the product of two numbers, and one of theffs given, to find the other. Rule. Divide the product by the given numb^^.r, and the quotient wfll be the number required. If 5S P MiOliLEMS. Let the product of two numbers be 288 8)28b' and one of tliem 8 ; I demand the other ? Answer^ 36 PiipB. G. Having the dividend and quotient, to find the divisor Rule. Divide the dividend by the quotient. Cor. Hence we s^et another method of proving' Division. C'von 5^^^ ^^^® Dividend. 30)288(8 Divisor. } 36 the Quotient. 288 Required the Divisor. pROB. 7. Having the Divisor and Quotient given, to find the Dividend. Rule. Multiply them together. ^. ^8 the Divisor. JG uiven < oo .1 /^ .• , o ^ 3G the Quotient. 8 Required the Dividend. -»--^ — 288 the dividend. By a due conrsideralion and apphcation of these Problems only many questions (of which kind are some of the following) may be resolved in a short and elegant manner, although some of them are generally supposed to belong to higher rules. APPLICATION OF THE PRECEDING RULES. 1. The least of two numbers is 19418, and the difierence be= tween them is 2384 : What is the greater, and sum of both ? 19418-f2384=21802 greater, and 1941.8-f 21802=41220 sum. 2 Suppose a man born in the year 1743 ; when will he be 77 years of age ? 1743-f77=1820 Ansmer. 3. Wbat number is that, which, being added to 19418, will make 21802? 2384 c^n5. 4. Gen. Washington was born in 1732 ; what was his age in 1799 ? 67 Ms, 6. America was discovered by Columbus in 1492 and its inde- pendence declared in 1776 : How many years elapsed betweeti those two eras ? 284 A7is. 6. 1 he Massacre at Boston, by the British troops, happened March 5th, 1770, and the Battle at Lexington, April 19lh, 1775; How long between ? April 19th, 1775~March 5th, 177C=5 y. 1 m. 14 d. Jlns. 7. Gon. Burgoyne and his army were captured October 17th, 1777, and Karl Cornwallis and his army, October 19th, 1781 : What space of time between ? 4 years and 2 days. Ans. 8. The war between America and Kngland commenced April I9th, 1775, and a general peace took place .Tannary 20th, 1783: How lonix did the war continue ? "7 y. 9 m. 1 d. Ann. 9. A, h, C and D ptircliased a quantity of goods in partncrsbip; A paid £ 12 10.9. a doilHr* and a crnvvnt piece ; P),35s. C, 29s. lOi/. and D, 79*;. ; What did the good:, co.sl ? .i/n.v. i.' \0 hi 1. l^KOBLEMS. 59 10. A man borrowed, at different limes, these several sums, viz. £29 5s. £18 17s. Gd. £45 I2s. £98, 3 dollars, one crown piece and an half; Pray how much was he in debt? Ans. £ 193 2 G 11. There are four numbers ; the first 317, the second 912, the third 1229, and the fourth as much as the other three, abating 97 : What is the sum of them all ? Ans. 4819. 12. Bought a quantity of goods for -£125 10s. paid for truckage 45s. for freight 79s. 6d. for duties 35s. lOd. and mv expences were j3s. 9d. : What did the goods sfan ^ ^ ' £16 An^. # 60 REDUCTION- 26. If the Federal Debt, which is 42 million dollars, be equally divided between the 13 States : What will be the share of each ? Ans. 3230769^3 dollars. 27. If 9000 men march in a column of 750 deep : How many march abreast? ]2 Jlns. 28. What number, deducted from the 32d part of 3072, will leave the 96th part of the same ? 64 Jlns. 29. What number is that, which> multiplied by 3589, will pro- duce 92050672 ? 25648 Ans. 30. Suppose the quotient arising from the division of two num- bers to be 5379, the divisor 37625: What is the dividend, if the remainder came out 9357 ? 202394232 Ans. 31. There is a certain number, which being divided by 7, the quotient resulting multiplied by 3, that product divided by 5, irom ihe quotient 20 being subtracted, and 30 added to the remainder, the half sum shall make 35: Can you tell me the number? 700 Ans. 32. A sheepfold was robbed three nights successively ; the fir«t night, half tha sheep were stolen, and half a sheep m.ore ; the second half the remainder were lost, and half a sheep more ; the last night they took half what were left and half a sheep more ; by which time they were reduced to 30: How many were there at first ? Begiti with 30, and, reckoning back fiom the last night (o Ih^ first, you will find that 31 were stolen the 3d night, 62 the 2d, atid 124 the first. Ans. 247. 33. Two boys, A and B, had 850 chesnuts between them; but A had 150 more than B : How njany had each. 85C-r-2=425 half sum, and 150-t-2^75 half di(f. ; thtn 425-^75 =500 A's, and 425—75==350 B's. S4. What number added to the 27th part of 6615, will make 570* 325^ Ans. REDUCTION T]'..\('\l{]S t>^ bring nntnlirrs of one (!onomuia'.i(in to ulhors ol dillVrerit dtnominatiotif?, retaijiing" the same value, h is of /ci'o feorts, \\a Descending and Ascending. RKDUCTiO:: DIISCENDIKG Toiclir-^ to c!iarge nnmiiors from i hia^her to a lower dcnorai- nation. It is performed by mnltiplic.a'.irtj. REDUCTION. 61 Rule* Multiply the highest denomination given, by so many of the !iext less as make one of that greater, and thus continue until yod have brought it down as low as your question requires. Proof. Change the order of the question, and divide your last product by the last multiplier, and so on. Note. From this rule and Case Vf. of Simple Multiplication, it appears, that Federal Money is reduced from higher to lower de-^ nominations by annexing as many cyphers as there are places from the denomination given, to that required ; or, if the given sum be of different denominations, by annexing the several figures of all the denominations in their order, and continuing with cyphers, (if necessary,) to the denomination required ; or, what amounts to the same thing, by reading the whole number from the left to the re- quired denomination, as one number in the required denomination. Examples. 1. In 3 eagles 2 dollars, how many mills ? Ans. 32000 m. 2. In 91 dollars 75 cents, how many cents ? Am. 9175 c. .3. In 50 eagles, how many dollars ? Ans, 600 D 4. In 44 dollars, 1 cent, 4 mills, how many mills ? 5. In 9 dollars, 31 cents, 7 mills, how many mills? 6. How many cents in 39 dollars 5 cents ? 7. In 28 dollars 17 cents, 5 mills, how many mills? 8. In £27 15s. ^)(l. 2qrs. how many farthings? £ s. d. qr. 27 11 9 2 multiplied by 20=shillings in a pound. 55b=shininss 'by I2=pence in a shilling. G6e9=pence. ~^y 4=forthings in a penny. Ans. =2GG78 farthinsrs jYote. In multiplying by 20, I added In t!ie 15s. by It, the $<3. And by 4, the 2ijrs. which must always be done in like cases. To prove the ab )ve question, change the order of it, and it wi^ stand thus : In 20678 farthings how many pounds ? * Tlie reason of this Rule is exceedingly obvious ; for pounds are br'ouj'ht into shillings by multiplying tlicm by 20 ; shillings into pence by multiplying them by 12; and pence into farthings by multiplying them by 4; and the contrary by division ; and this will be trnr in the rfluction of numbers con- ■iisting of any dcnomii.ation whatever Die rule for Reduction ascending is simply the reverse of this, and equally evident. 6^1 K£DUCT|ON. 4)26678 12)6669 2qrs. 2|0)55|5 9 d. Answer, £27 15 9 2 C. In £36 12s. lOtl. Iqr. how many fartljingg ? Ans. 35177. 10. In £95 Us. 5(1. 3qrs. how m^ny farthings ? Ans. 91751. 1 1. Jn £ 719 9s. 1 Id. how many halfpence ? Ans. 345358. 12. In 29 guineas, at 28s. how many pence ? Ans. 9744. 13. In 37 pistoles, at 22s. how many shillings, pence, and farthings ? Ans. 814s. 9768d. 39072qrs. 14. In 49 half Johannes, at 48s. how many sixpences ? Ans. 4704-. 15. In 473 French crowns, at 6s. 8d. how many threej>ences ? An^. 126131 \Q. In 53 moidores, at 36s. how many shillings, pence and farthings? Ans. 1908s. 22896d. 91584qrs. 17. In ^29 how many groats, threepences, pence, and farthings ? Ans. 1740 groats, 2320 threepences, 6960d. 27840qr3. 18 Reduce 47 guineas and one fourth of a guinea into shillings^ sixpences, groats, threepences, twopences, pence and farthings. Ans. 1323 shillings, 2046 sixpences, 3969 groats, 5292 three- pences, 7938 twopences, 15876 pence, and 63504 qrs. REDUCTION ASCENDLYG Teaches to change numbers from a lower to a higher denorni nation. It is performed by division. Rule. Divide the lowest denomination given, by so many of that name, as make one of the next higher, and thus continue till you have brought it into that deno.nination which your question requires. Note. From this rule and the note under Case II. of Simple Di- vision, it appears, that Federal Money is reduced from lower to iiighor deiiominations by cutting off as many places as the given denomination stands to the right of that required ; the figures cut off belonging to their rc^pective denominations. Examples, 1. How many eagles in 42000 mills ? Ans. 4 E. jj 2 2. In 3175 cents, how many dollars ? Ans. $ 31 75 c. .1. In 500 dollars how many Eagles ? Ans. SQ 4. In 44^4 mills, how many dimes ? i». In 9317 mills, how many dollars ? -;. How many dollars in 28175 mills? T. In 547325 (\irthingSj how many pence, 'shillings, and potjiuls' REDUCTION. 68 Farthings in a penny = 4)547325 Pence in a shilling = 12)136831 1 qr. Shillings in a pound = 2|0)1140|2 7d. £570 2s. 7d. 1 qr. Ans. 136831d. 11402s. and £570. Note. The remainder is always of the sarae name as the dividend. 8. Bring 35177 farthings into pounds. 9. Bring 91751 farthings into pence, &c. 10. Bring 345358 halfpence into pence, shilling«<, and pounds. 11. Reduce 9744 pence to guineas, at 28s. per guinea. 12. In 39072 farthings, how many pistoles, at 22s. 13. In 4704 sixpences, how many half Johannes ? 34. In 12613i threepences, how many French crowns, at 6s. Sd.-? 15. In 91584 tarlhings, how many moidores, at 36s. ? 16. In 27840 farthings, how many pence, threepences, groats, shillings and pounds ? 17. In 63504 farthings, how many pence, twopences, threepenc- es, groats, sixpences, shillings and guineas ? Note. The preceding questions may serre as proofs to those in Reduction descending. REDUCTION DESCENDING AND ASCENDING. 1. Money. 1. In £97 how many pence and English or French crowns, at. Gs. 8d. ? Ans. 23280d. and 291 crowns. 2. In 947 English crowns, at 6s. 8il. how many shillings and Eng- lish guineas? Ans. 6313s. 4d. and 225 guineas 13s. 4(i. 3. In 519 English half crowns, how many pence and pounds? Ans. 20760d. and £ 86 10s. 4. In 1259 groats, how many farthings, pence, shillings, and guin- eas ? Ans. 20144qrs. 5036d. 419s. 8d. and 14 guin, 27s. 8d. 5. In 75 pistoles, how many pounds ? Ans. £ 82 10s. 6. In 735 French crowns, how many shillings and French guin- eas, at 26s. 8d. ? Ans. 4900s. and 183 guin. 24s. 7. In 5793 pence, how many farUiings, pounds, and pistoles? Ans. 23172qrs. £24 2s. 9(1, and 21 pistoles, 20s. 9d. 8. In £ 99, how many shillings, and half Johannes, at 48s. ? Ans. 1980s, and 41 half joes. 12s. 9. In £ 179, how many guineas ? Ans. 127 guin. 24s. 10. In £ 345 how many moidores ? Ans. 191 moid. 24s. 11. In 59 half joes, 37 moidores, 45 guineas, 63 pistoles, 24 Eng- lish crowns, and 19 dollars ; how many pounds, half joes, moidores, guineas, pistoles, English crowns, dollars, shillings, pence, and far things ? Ans. £354 4s. 147 half Joes, 28s. 196 moidores, 28s. 253 guin- eas, 322 pistoles, 1062 English crowns. 4-, 1 180 dollar?, 4^ ':p>p--% shillings, 85008d. and 340032qrs. 64 REDUCTION. When it is required lo know how many sorts of coin, of differ- ent values, and of equal number, are contained in any number of another kind ; reduce the several sorts of coin into the lowest de- nomination mentioned, and add them together for a divisor ; then reduce the money given, into the same denomination, for a divi- dend, and the quotient, arising from the division, will be the num- ber required. Note. Observe the same direction in weights and measures. 1. In 275 half Johannes, how many moidores, guineas, pistoles, dollars, shillings and sixpences, of each the like number ? A moidore is 36s. } „r> • 275 half joes. that is I 72 sixpences. 48 sl.il. in ajohau. A guinea is 28s. that is A pistole is 22s. that is A dollar is 6s. that is One shilling has > 56 ditto. 44 ditto. 2200 1100 13200 shillings. sixp. in a shill. dividend=26400 sixpences. ^ 12 ditto 2 do. 187)26400(141 of each and 33 ?ixp. or 1 do. " Divisor=187 sixpences. 2. A Gentleman distributed £37 10s. between 4 poor persons, JQ the following manner, viz. that as often as the first had 209. the second should have 15s. the third, 10s. and the fourth, 5s. What did each person receive? Ans. The first man J£15, second £11 5s. third £7 10s. fourth £3 15s. 2. Troy Weight. 1. How many grs. in a silver bowl, that weighs 31b. 10 oz. 12 pwt. ft oz. pwt. 3 10 12 12 ounces in a pound. 46 ounces. 20 pennyweights in an ounce. 932 pennyweights. 24 grains in one pwt. 3728 1864 i'roof. 24)22368 grains, answer. 2|0)93|2 12)46—12 pwt. Jfe 3—10 oz. REDUCTION. 65 t. In 4870ZS. how many pwts. and grs. ? Ans. 9740pwt. and 233760gr. 3. In 13 ingots of gold, each weighing 9oz. Spwt how many grains ? Ans. 67720gr. 4. In 97397grs. how many pounds ? Ans. 16ife lOoz. I8pwt 5gr. 5. How many rings, each weighing 5pwt. 7gr. may be made of lib. 5oz. 16pwt. 2gr. of gold. Ans. 158. 3. Avoirdupois Weight. Cwt qrs. ib oz. 1. In 91 3 17 14 how many ounces? 4 367 quarters. Proof. 28 16)164702 2943 28)10293 14oz. 735 - 4)367 171b. 10293 pounds. 16 Cwt. 91 3qrs. 61762 10294 164702 ounces!. 2. In 12 tons, 15cvvt. Iqr. 19!b. 6oz. 12dr. how many drams ? Ans. 7323500dr. 3. In 241b. 1 loz. 9dr. how many drams ? Ans. 6329dr. 4. In 44800 pounds, how many drams and tons ? Ans. 1 1468800dr. and 20 tons. 5. In 281b. Avoirdupois how many pounds Troy ? 28 7000 grains in 1 lb. Avoirdupois. f ;'^- '"J =576!0)19600!0(34}fe Ub.lr.^ i ^j,728 6. In 47lb. 9oz. 13pwl. 17gr. Troy, how many pounds Avoirdupois ? 2.320 '47 9 13 1 2304 12 160 673 12 20 ,76|0)192|0(Ooz. 11473 20 24 .^76|0)3840|0)6pwt. 45899 3456 22947 3840 carried over. 275369 carried ovfi-n-. I 0(5 REDU CTION. Brought over. 3840 24 7l000)275l369(39fe Br'ghl OTCr. 21 1536 66 768 63 576|0)9216|0(16gr. 576 2369 16 3456 14214 3456 2369 7 [000)37 1 904 (5oz. 35 2904 16 17424 2904 7|000)46|464(64ff|(lr„ 42 4464 4. Apothecaries' Weight. 1. How many grains are there in 37ife 63 ? ft 5 Proof. 37 6 2|0)21600|G 12 3)10800 450 ounces. , — 8 8)3600 3600 drams. 12)450 3 37 ife 65 10800 scruples* 20 Ans. 216000 grains. 2. In 9ft 83 13 29 19gr. how many grains ? Ans. 55799gi'. 3. In 55799 grains, how many pounds, &c. ? Ans. 91fe 83 13 29 19gr. 5. Cloth Measure. 1. In 127 yards, how many quarters and nails ? 4 Proof. Ans. 508 qrs. 4)203 2 4 4)508 Ans. 2032 naili. 127 yards.. REDUCTION- er 5. In 9173 nails, how many yards ? Ans. 573yds. Iqr. In, 3. In 75 ells English, how many quarters and nails ? Ans. 2^75qrS. 1500n, 4. In 56 ells Flemish, how many quarters and nails ? Ans. 168qrs. 672n. 6. In 39 ells French, how many quarters and nails ? Ans. 234qrs. 936n. 6. In 7248 nails, how many yards, ells Flemish, ells English, and ells French ? Ans. 453yds. 604 ells Flem. 362 ells Eng. 2qrs. 302 ells French. 7. In 19 pieces of cloth, each 15 yards, 2 quarters, how many yards, quarters and nails ? Ans. 294yds. 2qrs. 1 178qrs. and 4712n. 6. Long Measure. 1. How many barley corns will reach fromNewburyportto Bos- ton, it being 43 miles ? 43 miles. 8 3)8173440 proof. Here I divide by 11, and multiply the quotient by 2 because twice 5i is 11 ; or I might first have multipli- ed by 2, and, then, have divided the product by 11. 344 furlongs, 40 12)2724480 3)227040 U)75680 13760 rods. 6880Q 6880 6880 2 75680 yards. 3 4 10) 137610 8)344 43 227040 feet. 12 2724480 inches. 3 8173440 Answer. 2. How many barley corns will reach round the globe, it being 360 degrees ? Ans. 4756801600. 3. How many inches from Newburyport to London, it being 2700 miles ? Ans. 171072000. 4. How often will a wheel, of 16 feet and 6 inches circumfer- ence, turn round in the distance from Newburyport to Cambridge, U being 42 miles ? Ans. 13440 times. 5. In 190080 inches, how many yards and leaguef^ ? Ans. 5280yds. and 1 league. 68 REDUCTION. 7. Time. 1. In 20 years how many seconds ? (1. h. 365 6 in a ye^r. 24 1466 730 8766 hours in 1 year. 20 176320 hours in 20 years. 60 Proof. 610)631 ]6200|O 6|0)1051920|O 2[0)17632|0 4X6)8766 4)1461 365ci. 6h: 10519200 minutes in ditto. 60- 631152000 seconds in ditto. 2. Suppose your age to be 15y. 19d. llh. 37m. 45s. how many seconds are there in it, allowing 365 days and 6 hours to the year ? Ans. 475047465. 3. In 31536000 seconds how many years? Ans. 1 year. 4. How many minutes from the first day of January to the 14th day of August, inclusively ? Ans. 325440. 5. How many days since the commencement of the ChristianiEra ? 6. How many minutes since the commencement of the American war, which happened on the 19th day of April, 1775? 7. How many seconds between the commencement of the war», April 19th, 1775, and the independence of the United States of A- merica, which took place the 4th day of July, 1776* ? Ans. 38188800. 8. Motion. 1. In 9 si^ns, 13° 25*, how many seconds ? 9s 13<» 25' * 6|0)102030|0 Proof. 30 — ^ — : 283 degrees. 60 6|0)1700|5 3|0)2{ 17005 minutes. 60 9s 13° 25' 1020300 seconds. 1776 was a leap year. REDUCTION; 69 9. Land or Sgiuare Measure. 1. In 29 acres, 3 roods, 19 poles, how many roods and perches? Acres. R. Poles. Proof. 29 3 19 4|0)477|9 — - 4)119~19p. 119 roods. • 40 29ac. 3 roods. Answer 4779 perches. 2. In 1997 poles how many acres ? Am. 12a. Ir. 37p. 3. In 89763 square yards how many acres, &c. ? Ans. 18a. 2r. 7p. lOlft. 36in. 4. How many square feet, square yards, and square poles, in a square mile ? Ans. 27878400 feet, 3097600 yards, and 102100 poles. 10. Solid Measure. 1. In 15 tons of hewn timber how many solid inches ? 15 tons. Proof. 50 5|0 — 1728)1296000(75|0 750 feet. 1209a 1728 15 tons, 8640 6000 , 8640 1500 5250 750 Ans. 1296000 inches. 2. In 9 tons of round timber how many inches? Ans. 622080. 3. In 25 cords of wood how many inches ? Ans. 5529600. Grindstones are usually sold by the solid foot, and the contents are found by the following Rule ; — Multiply the sum of the whole diameter and of the half of the diameter, by the half diameter, and this product by the thickness, and you have the contents in cubic inches. 4. What is the content of a grindstone, whose diameter is 32 inches and its thickness 3 inches ? 32 diameter. 1728)2304(1 foot. 16 half diameter. 1728 48 576 16 3 768 )1728( 1 third. 3 thickness, 1723 2304 solid inches. Ans. 1 foot and ^ foot. # 70 REDUCTION. 5. How many solid feet in a grindstone, whose diameter is 40 inches and thickness 4 inches ? Ans. 21 feet. Note. This rule is not designed to give the solid contents with perfect accuracy. For the true rule, see Mensuration, Art. 30. 11. Wine Measure, 1. In 9hhds. ISgalls. 3qts. of wine how many quarts? hhds. gal. qts. Proof. 9 15 3 4)2331 63 65)582— 3qt3. 32 65 9hhds — 15gals. 582 gallons 4 Ans. 2331 quarts. 2. In 12 pipes of wine how many pints ? Ans. 1209(J. 3. In 9758 pints of braudv how many pipes ? Ans. 9p. Ihhd. 22gal. 3qt^ 4. In 1008 quarts of cyder how many tons? Ans. 1 ton, ^ 12. Ale or Beer Measure. 1. In 29hhds. beer how many pints ? hhds. Proof. 29 2)12528 54 4)6264 116 14i? 54)1560 15G6 galion:> 29 hhd*. 4 6264 quarts. , 2 I Ans. 12528 pints. >. In 47bar. ISgal. of ale how many pints? Ans. !3680i 3. In 36 puncheons of beer how many butts? Ans. 24. 13. Dry Measure. I, In 42 chaldrons of coals how many pecks ? Chaldrons. Proof. 42 4)5376 32 32)1344(42 "B4 1^> 126 ^^4 'l344 bushels. I'l 4 Ans. 5376 peck?. "^-iw-- VULGAR FRACTIONS. 71 2. In 75 bushels of corn how many pints ? Ana* 4800. 3. In 9376 quarts how many bushels ? Ans. 293. FRACTIONS. Parts of a thing are expressed by figures, as well as whole things. When a whole is expressed by figures, the number is called an in- teger. But when a part, or some parts of a thing, are denoted by figures, as one fourth, tu^o thirds^ four sevenths, three tenths, &c. of a thing, the expressions of these parts by figures are called Fractions, The term, fraction, is derived from a Latin word, which signifies to break, as an integer or unity is supposed to be broken or divided into a certain number of equal parts, one or more of which parts are denoted by the fraction. Thus one fourth denotes one of the four equal parts, and three tenths denotes three of the ten equal parts, into which a thing is broken or an integer divided. Fractions arise naturally from the operations of Division, when the divisor is not contained a certain number of times exactly in the dividend. For the remainder after the division is performed, is a part of the dividend which has not been divided ; the divisor being the number of parts into which the integer is divided, and the remainder- showing the number of those parts expressed by the fraction. Thus 4 is contained in 9, two and one fourth iimeSy and, hence the quotient cannot be folly expressed in such cases, except by a whole number and a fraction. Fractions are divided into two kinds, Vulgar, and Decimal. VULGAR FRACTIONS. Vnlgar Fractions are expressions for any assignable parts of « unit, or whole number ; and are represented by two numbers plac ed one above another, with a line drawn between them, thus : f , |, &c. signifying five eighths, four thirds. The figure above the line is called the numerator, and that beto^^ it the denominator. The denominator shews how many parts the integer is divided into ; and the numerator shews how many of those parts are mean: by the fraction. Fractions are either proper, improper, single, compound, or mixed. 1. A single or simple fraction is a fraction expressed in a simple form ; as a. f , ^^, &c. 2. A compound fraction is a fraction expressed in a compound form, being a fraction of a fraction ; as ^of-^, f of-\ of if, which are read thus, one half of three fourths, twosevenllis of five elev enths of nineteen twentielhs, &c. 72 VULGAR FRACTIONS. 3. A proper fraction is a fraction whose numerator is less than its denominator ; as |, |, &c. 4. An improper traction is a fraction, whose numerator exceeds its denominator ; as |, |, &c. 5. A mixed number is composed ofa whole number and a fraction, as 7|, 35y''j, &;c. that is, seven and three fifths, &c. 6. A fraction is said to be in its least, or lowest terms, when it is expressed by the least numbers possible. 7. The common measure of two, or more numbers, is that num- ber which will divide each of them without a remainder : Thus, 5 is (he common measure of 10, 20 and 30 ; and the greatest number, which will do this, is called the greatest common measure. 8. A number, which can be measured by two, or more numbersj is called their common multiple : And, if it be the least number, which can be so measured, it is called the least common multiple; thus, 40, 60, 80, 100, are multiples of 4 and 5 : but their least com- mon multiple is 20. Note. The product of two or more numbers is a common mul- tiple of those numbers. Thus, 3x4x5=60, and 60, or 3x4x5, is evidently divisible, without remainder, by each of those numbers. a\nd the same must be true in every similar case. 9. A prime number is one, which can be measured only by itself or a unit, as, 3, 7, 23, &;c. 10. A perfect number is equal to the Sum of all its aliquot parts.* An aliquot part ofa number is contained a certain number of times exactly in the number. / Problem I.j To find the greatest common measure of two^ or 2nore, numhets. Rule. I. If there be two numbers only, divide the greater by the less, and this divisor by the remainder, and so on, always dividing the * The foUowljig perfect numbers are all which are, at present, known. 6 850 72)108(1 216)324'(1 72 216 — Last greatest com. meas.=36)72(2 Common meas.=108)21G(2 72 216 — Therf fore, 36 is the answer required. 2. What is the greatest common measure of 1224 and 1080 ? Ans. 72. o'. What is the greatest common measAire of 1440, 672 and 3472 ? Ans. 16. • Problem II.* To find the l^aat ^mmon multiple of tu^o or more numbers. Rule. I. Divide hy any number that ^ill divide two, or more, of the ijiven numbers withnut a remainder, and set the quotients, togethr er with the undivi«led numbers, in a line beneath. II. Divide the second line, a© before, and so on, till there are no two luimbers that can t>e divided; Uien, the continued product of the divisors and quotient* will give the multiple required. * The reason af this rule nwyalso l>e shewn froth the first example: Thu^, it is evident that 6xioxi6x'JO = 19100 triiy be divided by 6, lO, 16 ana 20, without a remainder ; hut 20 is a multiple of 5 ; therefore, 6x10X16x4, or 3840, is also divisible by 6, 10, i6 and 20. Also, 16 is a multiple of 4; therefore 6x10x4X4-960, is ajso divisible by 6, 10, 16 and to. Also, to is a multiple of 2; therefore, 6x5x4X4 - 480, is also divisible by 6, 10, i6and so. Also, 6 is a multiple of 2; therefore, 3 X5 ^ 4'<'4 -'*40, is also divisible bv 6j 10, i6, ■luU z&y and isWidcntly the least iiuvober fhat can t^c 59 divided* K ^2)0 2 16 4 ^2)3 1 8 2 *3 1 *4 1 VULGAR FRACTIONS. Examples. t. What is the least common multiple of 6, 10, 16 and 20? I survey my given numbers, and find *5)6 10 16 20 that five will divide two of them, viz. 10 and 20, which I divide by 5, bringing into a line with the quotients the num- bers which 5 will not measure : Again, 1 view the numbers in the second line, and find 2 will measure them all, and get 3, 1, 8, 2 in the third line, and find that two will measure 8 and 2, and in the fourth line get 3, 1, 4, 1, all prime ; if- * * * * I ^[jgn multiply the prime numbers and 5x2x2x3X4=240 Ans. the divisors continually into each other, for the number sought, and find it to be 240. 2. What is the least common multiple of 6 and 8 ? Ans. 24. 3. What is the least number that 3, 6, 8 and 10 will measure ? Ans. 120. 4. What is the least number which can be divided by the 9 di- gits, separately without a remainder ? Ans. 2520. REDUCTION OF VULGAR FRACTIONS Is the bringing of them out of one form into another, in order to prepare them for the operations of Addition, Subtraction, &c. CASE I.* To abbreviate, cr reduce fractions to their lowest terms. Rule. Divide the terms of the given fraction by any number, whicli will divide them without a remainder, and the quotients, again, in • That dividing both the numerator and denominator of the fraction by the same number, will give anottier fraction of equal value, is evident, because both patts arc diminished proportionally, and if both parts of the equal frat- lion be multiplied by the divisor, the original fraction will be formed again. a88 8 36 36 8 288 Thus olH-T^T ^^^^'Aq^'q^^'^'' ■^"'^ if *hc divisions be performed as of ten as can be done, or the common divisor be the greatest possible, the terms of the resulting fraction must be the least possible. Note I. Any number, ending with an even number or cypher, is divisible by 2. a. Any number, ending with 5 or o, is divisible by 5. 3. If the right hand place of any number be o, the whole is divisible by ic. 4. If the two right hand figures of any number be divisible by 4, the whole is divisible by 4. 5. If the three right hand figures of any number be divisible by 8, the whole is divisible by 8. 6. If the sum of the digits, constituting any uumber, be divisible by 3 or 9, the whole is divisible by 3 or 9. VULGAR FRACTIONS, 75 the same manner ; and so on, till it appears that there is no nura- ber greater than 1, which will divide them, and the fraction will be in its lowest terms. Or, Divide both the terms of the fraction by their greatest common measure, and the quotients vvill be the terms of the fraction le- quired. Examples. 1. Reduce f|f to its lowest terms. ■ (4) (3) « lm=U=T%==i the answer.* Or thus : 288)480(1 Therefore 96 is the greatest commoii 288 measure. and 96 J f If =? the same as before. 192)288(1 192 Com. meas. 96)192(2 192 2. Reduce ^^\ to its lowest terms. Ans. y\. 3. Reduce y^sV ^o its lowest terms. Ans. ^, 4. Reduce j^j\ to its lowest terms. Ans. J . 7. Tf a number cannot be divided by some number less tl^an (he square root thereof, that number is a prime. 8- A\\ prime numbers, except a and 5, have 1, 3, 7, or 9 in the place of units : and all other numbers are compossite. 9. When numbers, with the sign of Addition or Subtraction between them, are to be divided by any numbers, each of the numbers must be .divided ; Thus 64-9-t-i a--a-f-3-(-4r-9, or 6-f 9-f ia=a7^9. 3 3 T 10. But if the numbers have the sign of Mul..ipHcation between them ; then 4x6x10 ax6xio ax6x* only one of tliem must be divided : Thus, — — t- — -= — 7 ~ — ■= — _ . , :=^ axj ^ ^^ S AX* * Hence if both parts of a fraction be multiplied by the same number, it» , ^ 3 3 3 9 9 4 36 8 288 value IS no altered. For -=7X7=,;= X"— ,^X-?=7o-:, and so on. If 5 b i T5 15 4 60 8 480 fractions be multiplied together, in which equal terms occur ia the numerator and denominator, these equal terms may be cxpunnged or cancelled, for their quotient would be i, which as a factor would not alter the value of the fraction, ^^^ 4 J 4X5 41 I3a4i3*4i ^.. Thus-Xg'-^-^-g=-.3.--. and -x-x-X7:.-X^X jX-=7=i. Anthmet- ic?I operations are often much shortened by observing what quantities may be expunged, and by omitting them in the operations. For the same object, ex- pressions may be changed to equivalent ones, and quantities expunged, Thn? ^ 7>^ \ ^Xi5 I 8 a88 8 9x31 4 < 8 _8_ % '^ 45 "^^ ^3x15 ~ 3'*"^ 9 ^ 480^9 ^ 8x6o=4Xi5"T^" 7.0 VULGAR FRACTIONS. 5. Reduce y^i ^^ *ls lowest terms. Ans. -]v 6. Reduce ^||| to its lowest terms. Ans. v. »Yo^e. If the numerator of a fraction be multiplied, or it? denom- inator divided, by a whole Dumber, the value of the fraction will 1X3 Im; so many times increased. Thus, i multiplied by 3.— ( ' = 311 -=-= _ Hence, to multiply a fraction by an integer, is to y fj tj' I • 'O . multiply the numerator, or divide the denominator of the fraction by the integer. 1. Multiply /j- by 7. Ans. |f 2. Increase the value of yij, nineteen times. Ans. -^. 3. Increase the value of ^, seven fold. Ans. 3. If the numerator of a fraction he divided, or its denominator multiplied, by a whole number, the value of the fraction will be so 3-4-3 1 3 many times diminished. Thus, f divided by 3,—-^- — =-=———-=; 2T~9' Henc<», to divide a fraction by an integer, is to divide the numerator, or multiply the denominator of the fraction, by the whole number. 1. Divide f by 7. » Ana. ^%. 2. Diminish the value of-}, seven times. Ans. ^V- 3. Diminish the value of J, four times. Ans. j. Note. The reason of man"y operations will be evident from an attention to the M\ovi'\n^ self -evident truths. 1. If equals b« added to equals, their sums will be eq\ial. Thus, 34-44-9=8x2. Let 7 be added to each, and 3+4+9+7=8x2! +7=23. 2. If equals be subtracted from equals, the remainders will be equal. Thus, 3+11=7x2. Let 3 be taken from each, and 3+ 11—3=7x2—3=11. 3. If equals be multiplied by the same quantity, the products will be equal. Thus, let 5 + 7=6x2, be multiplied by 6, and .54:7x6=6x2x6=72. 4. If equivalent quantities be divided by the same quantity, the quotients will be equal, 'i'hus, let 43+17=129<5 be divided by 6, 43+17 12X5 and 43+17-~5=12x5r^5,= 12, or =—7— =12. CASE II. To reduce a mixed number to its equivalent improper fracti(jrh Rule.* Multiply the whole number by the denominator of the fraction, * All fi'a(?llon9 rcprcfent a divifion of a numerator by the denominator, and are taken altojirtlier as proper and adequate cxprcflions of the quotient. Thus tlie quotient of 3, divided by4,is-|; from whence the rule is manifcft ; forif any number is multiplied and divided by the fame number, i>is cvidtnt ihf ,tjfuotitut nuift be ihc fame as the quantity firft given. VULGAR FRACTIONS. 77 and add the numerator of the fraction to the product ; under which euhjoin the denominator, and it will form the fraction require*!. Examples. h Reduce 36f to its equivalent improper fraction. 36 I multiply 36 by 8, and adding the nu- X8-{-5 merator 5 to the product, as I mul- tiply, the sum 293 is the numerator^ Ans. 293 of the fraction sought, and 8 the de* nominator: So that 2 13 i^ the im- 8 proper fraction, equal to 36f. 36x8+5=293 Answer as before. Or, g ~8~ 2. Reduce 127 j\ to its equivalent improper fraction. Ans. ^{^',. 3. Reduce 653,^^^ to its equivalent improper fraction. ^ Ans. *2^«». CASE Ill.t To reduce a tzhole number to an equivalent fraction having a given denominator. Rule. Multiply the whole number by the given denominators Place the product over the said denominator, and it will form the fraction required. Examples. 1. Reduce 6 to a fraction, whose denominator shall be 8. 6X8=48, and V the Ans — Proof V =48-^-8=6. 2. Reduce 15 to a fraction, whose denominator Shall be 12. Ans. V^^ 3. Reduce 100 to a fraction, whose denominator shall be 70. Ans. '^^^''=»|o=100. A whole number is made a fraction by drawing a line under it, 9x1 and putting unity or 1, for a denominator, as r=— j — by therule, ^nd 12 i.s V^*c> CASE IV.J To reduce an improper fraction to its equivalent whole, or mixed number. Rule. Divide the numerator by the denominator: the quotient will b« the whole number, and the remainder, if any, will be the numera- tor to the given denominator. f Multiplication and Divifion arc here equally ufed, and confequently the re- sult is the fame as the quantity firft propofcd. i This cafe is, evidently, the revcrfc of qafc ^d, and his its reafon m the na- ture •f €«ram«Bi divifioB. 78 VULGAR FRACTIONS. 1. Reduce 2|3 to its equivalent whole, or mixed number. 8)293(36|- Ans. 24 53 48 — Or, 2|3— 293-~8=36f as before. 5 2. Reduce ^ijs to its equivalent whole, or mixed number. Ans. 127^t, 3» Reduce *^^9^° to its equivalent whole, or mixed number. Ans. 653/^. 4. Reduce Y to its equivalent whole number. Ans. 9. CASE v.* To reduce a compound fraction to an equivalent simple one. Rule. Multiply all the numerators continually together for a new na- me rator, and all the denominators, for a new denominator, and they will form (he simple fraction required. If part of the compound fraction be a whole or mixed number, it must be reduced to an improper fraction, by case 2d, or 3d. if the denominator of any member of a compound fraction be equal to the numerator of another member thereof, these equal numerators and denominators may be expunged, and the other members continually multi{)iied, as by the rule, will produce the iractioDS required in lower terms. E XAxMPLES. 1. Reduce -I of | of f off to a simple fraction. 1X2x3x4 24 1 *7. — z — : — r=T7i7r=7 the Answer. 2x3x4X5 120 5 Or, by expunxiiig the equal numerators and denominators, it will give ] as before. 2. Reduce | of 5 off of |^ to a simple fraction. 3X4X5X11 660 n • 4y^/Jy>d2'^T44i)""24 '^"'' ^'*' ^'^ expungmg the equal nu- 3X1 1 ^jeratr>rs and denominators, it will be =5b=^i 2is before. 6X12 ^^ ^* * TWil a conipouud fradlJonmay be reprefcnted by a fimple one is very cvi- u'iit ; fincc a part of a part muft be equal to fome part of the whole. The ruth of the rule for this redudlion may he fliown as follows. l,et the compound fra(5lion to be reduced, be 1 of _«_. Then | of _6_=_6_ t. i_=: Q, and confequentlv -5 of r, =A.vS=-li? the lame as by tie rule. ^ -_ . ^ ' i IT) 4 -^ 4 ^ . i{ the compound fratflion confifts of more numbers than two, the firft two raay be reduced to one, and that one and the tliird will be the fame as a frac- ■'x>n of two Bunibers, and fo oi». VULGAR FRACTIOxXS. 7S 3. Reduce f of ^ of |f to a a simple fraction. Ans. ||f. 4. Reduce j\ of H of ,^ of 20 to a simple fraction. Ans 3 2±=z9_2_ ^"^' 30« 51- 5. Reduce i of ^ of | of 12^ to a simple fraction. Ans. U^Hh CASE. VI. To reduce fractions of different denominators to equivalent fractions having a common denominator. Rule I.* Multiply each numerator into all the denominators except its own, for a new numerator, and all the denominators into each oth- ^r, continually, for a common denominator. Examples. 1. Reduce a, f , and f to equivalent fractions having a common denominator. 1X5 x 8=40 the new numerator for |. ~X4x8= 64 the new numerator for |. 5x4x5=100 ditto for |. 4x5x8=160 the common denominator. Therefore the new equivalent fractions are -j^o* tVo ^^^ lih the answer. 2. Reduce |, |, f, |^ and {- to fractions having a common denoDj- inafnr " Ans -fJ-^-, -'L6_8_ jlbjl JUUL l«i>8 indlOr. /ins. 1^X5 2 > 1152' IIS2» 1T52» Tl52* 3. Reduce ^, | of |, 7|, and j\, to a common denominator. Ana (>3C 1040 14503 432 ^^^- T8T2» T»?2' T8T"2 > TsTi 4. Reduce |], f of 2i, y'j, and |-, to a common denominator. Ans -"AO- 216 0. JL72JL _T3 •""''• ild20» 11J20J 11S20> IIoIt>' Rule H. To reduce any given fractions to others^ which shall have the least common denominator. 1. By Problems, Page 73, find the least common multiple of all the denominators of the given fractions, and it will be th€ common denominator required. 2. Divide the common denominator by the denominator of eacL fraction, and multiply the quotient by the numerator, and the pro- duct will be the numerator of the fraction required. * By placing the numbers multiplied, properly under one another, it will be seen that the numerator and denominator of every fraction are multiplied by the very same number, and consequently their values are not altered. Thus, m the first example. 1|X5X8 4|x5xa In the second X4X! X4XH X4X5 X4X5 rule, tlie common denominator is a multiple of all the denom- inators, and consctjucntly will divide by any of them ; Therefore, proper part« may be taken for aiK the numerators as required. 80 VULGAR FRACTIbNS. Examples. 1. Reduce ^, ^i and |- to fractions having the least common de nominator possible. 4)3 4 8 4X3X2=24= least common de- nominator. 3 1 2 24-r-3 xl=8 the first numerator; 24-~-4X3=l^ the second uu« nierator; 24-^8x7=21 the third numerator. Whence, the required fractions are gl^, If, §}. 2. Reduce J, |, f, and 4 to fractions having the least common denominator. " " Ans. ^a, aj, ia, and |^. CASE VII. To reduce a fraction of one denomination to an equivalent fraction of a higher denomination. Rule.*- Blultiply the given denominator by the parts in the several i\c^ nominations between it and that denomination to which it is to be reduced, for a new denominator, which is to be placed under the given numerator : Or, compare the given fraction with the several denominations between it and that denomination to which it is to be reduced, and then, by case 6th, reduce the compound fraction thus formed, loa single one, and the equivalent fraction of the required denomination will be obtained. Let this fraction be reduced to its lowest terms. * The reafon of the rule may be foea in the following manner. As there arc la pence in a fhilling, foiw-fifths of one penny can be only a t-welfth part as much of la pence or a fliilling,as it is of one penny. Hence, to reduce four fifths of a penny to the fraAion of a fliilling, the given fradlion muft be di- miniflied la times, or one twelfth of it will be the equivalent fradlion of a fhilling A fradlion is diminiflied in value, according to the note to Cafe 1. by multiplying the denominator by the whole number. Thus four fifths of a pen- 4 4 14 14 ny =r - — — -of afliil!ing=:— y— =— of — =: — of a fliilling. For the fame rea-- fon,fow fixticths of a fliilling can be only one twentieth a^ much of a pound, pound. Put thefe two operations together, and you have four-fifths of a penny. The fam« operation might have been performed thus. In a pound there are 4 4 1 210 pence. Then, four-fifths of a penny = ,— of a pound, == - of — :~k ' 5X'^40 a 240 -- - as before. And in general the fradkion of one denomination muft be as much diminifliedto be an equivalent fra G 6 4 penny, =^d.= 67; d. and, therefore [r-s. =8s.C-;;;d. But -d.=-- of 7 of * ^ / < 7 V til 21 .. 3 4 3 a farthing, = -- qr. = 3yqr. Tliercfore, 7^£ — 8-9. = 8s. G-d.— .88.6d.:i7qr. The same process is obviously applicable to every similar case. Or, the process 12 4 2880 :> may be conducted thus; ~X=^-':7 of —- of — of —=;—;:;- qr. ^= 411— qrs. ==: 102d. oTqrs. -- Cs. Od. ::;rqrs. VULGAR FRAeTIQNS. m vemain, multiply it by the next inferior denomination, and divide by the denominator as before, and so on, as far as necessary ; and the quotients placed after one another, in their order, wUl be the answer required ; or, reduce the numerator, as if it were a whole number, to the lowest denomination, and divide the result by the denominator ; the quotient will be the number of the lowest denom- ination, (which must be .brought into higher denominations as far as it will go,) and the remainder will be a numerator to be placed overthe given denominator for a fraction of the lowest denomination. Note. From this rule, in connexion with what has been said o[ Reduction of Federal Money, it appears, that, annexing to the giv- en numerator as many cyphers, as wjH fill all the places to the low- est denomination, and dividing the number so formed by the de- nominator, the quotient will be the answer in the several denomi- nations, and the remainder a numerator to be placed overthe given denominator, forming a fraction of the lowest denomination. Examples. 1. What is the value off of a dollar ? By the general rule. 5 10 By the note. $ d. c. m. 8) 5 8)50( 2 $e 10 Ans. 6d. 2c. 5m. 8)20( 4 or 62c. 5m. c, 2 10 6 2 5 8)40 m. 5 Or thus. j^5=5000m. and « o_pom.=:625m.=62c. 5m. 2. What is the value of ^ of a dollar? g d. c. m. 64)17 Ans, as before. 128 420 384 360 320 (2d. 6c. 5|m. or 26c. 5f m. Ans. Or, ^17=17000ra. And »V7"m.=265fm.^ 40 64 $6c. 5|m. Ans. as before. J, What is the value of/,- of an eagle ? Ans. $1 87c. 5ra. Sd VULGAR FKACTiONS. 4. What is Uie value of y"^ of a dollar ? Atis. 4, 5. What is the value ot 4 ^f a pound ? Ans. 14s. 3d. Ifqr. (5. What is the value of ^^^ ^^^ shilling? Ans. 4id. 7. What is the value o\\\ of a £ ? Ans. ^. Gt]. 8. What is the value of f| of a pistole ? Ans. 13s. 6d. IK What is the value of J.^ of a Cwt. ? Ans. 2 qrs. 9fe 10 oz. 7|idi% 10. What is the value of| of a ife Avoirdupois? Ans. 12oz. 12|dr. n. W^hat is the value off of a ife Troy ? Ans. 7oz. 4pwt. .12. What is the value of ^3 of a ton ? Ans. 4cvvt. Sqrs. 12Ife. 14oz. 12y\dr. 13. What is the value of f of a yard ? Ans. 2qrs. 2|n. M. What is the value of i of an ell English ? Ans. 4qrs. l|n. 15. W'hat is the value off of a mile ? Ans. 6fur. 26[>. Hi]. 16. What is the value oi\% of a day ? Ans. 16h. 36m. 65yy. 17. The value of -}| of a Julian year is required ? Ans. 257d. i9h. -I5m. 52ffs. 18. 'i'he value of y\ of a guinea h demanded h-j Ans- 18.-;. 19. What is the value of jS-ofa dollar. ' Ans. 6s. 7|d. 520. What is the value of f of a moidore ? Ans. 21s. ^^. 21. What is the value of 5- of an acre ? Ans. 3r. l/'/p. CASE X. To reduce any givc7i quantily io the fraction, of any greater iJenomi^ nation of the same kind. Reduce the given quantity to the lowest term mentioned, for a nuuieralpr; then reduce the integral part to the same term for <♦ detiominator ; which wiil be the IVaclion required- Note. It appears fiom this rule and what has been said boforc, that,. in Federal Money, where the given quantily contains no frac- tion of its lowest denomination, the annexing of as many cyphers {q i of ihe required denojnination, as will extend to the lowest denoni.. ination in the given quantily, will forma denominator, which plac- ed under the given quantity used as one number for a numerator, %vi!l make the answer, which may be reduced to its; lowest term-j. Or, if there be a fraction of (lie lowest denomination, multiply the given whole numbers by its denominator, adding its numerator, for a numerator ; and let the denominator itself at ihe left of as many cyphers a^ were mentioned above be a denominator ; the fraction so f)rmed will be the answer; winch may be J^educed to its lowest terms. * This cafe in tlic rcvcrfc of llic furnKT, and tlie proof evident from that. Note. If there he a fraction jiivcn wlt)i the fjiid quaDflty,it mufl he fartiic. TcJiiud to the dcnon.iuative partb thereof, adding iLtreto the numcratqr. VULGAR FRACTIONS. 85 Examples. 1. Reduce 6d. 2c. 6m. to the fraction of a dollar. Bv the general rule. Cd. " lOd. int. pt. XlO+2 10 (32 100 By the note. Xl0-f5 10 $ d. c. m. 6 2 6 625 1000 =1$. 10 And, TVoV=t$ Ans. Ans. as before. 'i,'. Reduce 26c. 5|«i. to the fraction of a dollar. By the general rule. By the note. 26c. iOOc. Int. pt. g d. c. m. XlO+5 10 265x84-5=^ 1 2 5 265 1000 And §1x8=8 X84-5 8 =m- Ans. as before. 2125 8000 And,fie=HSAns. 3. Reduce $1 87c. 5m. to the fraction of an eagle. Ans. ^j"^-. 4. Reduce 43c. l^m. to the fraction of a dollar. Ans. y'^^. 5. Reduce 143. Sid.^ to the fraction of a pound. Ans.m^=f£. 6. Reduce 4|d. to the fraction of a shilling. Ans. |%. 7. Reduce 3s. Gd. to the fraction of a pound. Ans. -^q£, 8. Reduce I3s. 6d. to the fraction of a pistole. Ans. ||pistole. 9. Reduce 2qrs. 9fe. lOoz. 7|4dr. to the fraction of a cwt. Ans. ||cwt. iO. ReJuce 12oz. 12|dr. to the fraction of a ife Avoirdupois. Ans. |-ife. 11. Reduce 7oz. 4pwt. to the fraction of a Ife Troy. Ans |lb. ^2. Reduce 4cwt. 2qr3. 12ife 14oz. 12y«^dr. to the fraction ofaton^ Ans. vs^on. 13. Reduce 2qrs. 2?jn. to the fraction of a yard. Ans. fyd. 14. Reduce 4qra. llu. to the fraction of an ell English. Ans. |E. E. 15. Reduce 6fur. 26po. lift, to the fraction of a mile. Ans. fm. 16. lieduce 16b. 36m. 55/3S.to the fraction of a day. Ans. j\day. 17. Reduce 257d. 19h. 45in 52}-«s. to the fraction of a Julian year, Ans. If J. y, 1^. Reduce I83. to the fraction of a guinea. Ans. Yi^^- 19. Keduce 5s. T^d to the fraction ol' a dollar. Ans. y|dol. 20. Reduce 21s. Tid. to the fraction of amoidore. Answer fmoidorc. 21. Reduce 3r. 17:r>. to the fraction of an acre. An:-, facre^ S^ ^ VULGAR FRACTIONS. ADDITION OF VULGAR FRACTIONS. Rule.* v Reduce compound fractions to single ones ; mixed numbers to improper fractions ; fractions of difterent integers to those of the same ; and all of them to a common denominator ; then the sum of the numerators written over the common denominator will be the sum of the fractions required. Examples. 1. Add 7|, -^- off, and 7 together. Flr-st 74 = s_9 5 of ^=iA and 7=^ Then the fractions are ^j II» ^^^ t > therefore, 39x56x 1=2184 15X 5X 1= 75 7X 6X56=1960 4219 Or thus, 2184-1-75+1960 280 — — — 1 ^ 1 ^ 5x56x1=280 2. Add -f, 91, and | of i together. Ans. ^^\. 3. What is the sum of |, f of | of i, and 8/3 ? Ans. 9^2^^. 4. What is the sum of /^ of 4f , f of i, and 91 ? Ans. 12ff. 5. Add together fE. |g and lie. Ans. p 53c. 24m. f>. Add together ij fc. ^^^ c. and ^ra. Ans. 20c. 9m. 7. Add £1 -fs. and fd. together. Ans. 2s. 8yV_d. Z. What is the sum off of £17, £9f and I of 1 of £A? Ans. £16 12s. 3-fd. 9. Adr I of a yard, 1 of a foot, and f of a mile together. Ans. 1100yds. 2ft. 7inchei.u ^ 3J1..___2_0_=::_6J_=:4 rpmainrlpr 4X28 = 112 com. den. J 112 112 112 7 2. From ff take f. .^W5. if^, 3. From 371 take 19f .^ns. n^f. 4. From 13i take | of 15. Ans, 2yV 5. From g^ take |c. Ans. 49c. l^m. G. Take 3ic. irom } of2i$. Ans, 43ic. 7. From f of f of g5, take -f of 96c. added to ^ of l^g. .^«s. 96c. 9*-ra. 8. From intake y\s. j3nj. 49. lid-. 9. From -foz. take |pvvt. *5ns. 13pwt. 12fa^r. 10. From i of a league take | of a mile. .'Z/is. Imi. Ifur 11. From 5 weeks take 19j days. ^'^w^. 15da. 4ho. 48min. MULTIPLICATION OF VULGAR FRACTIONS. Rule.* Reduce compound fractions to simple ones, and mixed numberf to improper fractions ; then the prodnct of the numerators will be the numerator, and the product of the denominators, the denomi- nator of the product required. Note. Where several fractions are to be multiplied, if the nu- merator of one fraction be equal to the denominator of another, their equal numerators and deoominators may be omitted. Examples. 1. What is the continued product of 4', 4, 1 off, and 6. 1X7" 4^=¥> } of 1=. =^, and 6=f 4X8 -whole number and the fraction thus formed will be the remainder : but, when the numerator in the fubtrahend is greater than that in the minuend, fubtract the numerator in the fubtrahend from the common denominator, adding the numerator in the minuend, and carrying i to the integer of the ful)trahend. Hence, a fracftion is fubtradted from .1 wiiole nurnher, by taking the numera-* tor of the fraAion from its denominator, and placing the remainder over the denominator, then taking one from the whole number. Examples. From 12| 12f 12 Take 7i 7? ?. Rem. 5' 4 A n * Multiplication of a fraction implies the taking of some part or parts -c^ii" the multiplicand, and therefore may truly be expressed by a compound fraction. Thus 1 multiplied by :i is the same as 4 of Ji ; aud as the directions of the rule agfee with the method already given, to reduce these fractions to simple ones, it is shown to be ri?.ht. 88 VULGAR FKACTiONS. 13xlX ''X6 Then V^XiX-^'^Xf— =III = 1U ^^^e Answer. 3X5X32X1 - ' 2. Multiply -^ by 2^. Ans. //,, ,1. Multiply 51 by }. Ans. |. 4. Multiply i of 6 by I off. Ans. /^. 5. Multiply -f of f by | of i of 1 If Ans. -j^^.. 6. Multiply 9|, i of f, and 12^ continually together. Ans. 24i|. 7. What is the continual product of f of |, 5|, 7 and | of f ? Ans. 4^V- n. What is the continual product of 7, J, f of |, and 3i ? Ans. i;^ Snother method for the Muliiplicaiion of mixed Cluantilies. CASE I. To multiply a whole number by a fraction^ or a fraciion by a -whole number. Rule. Multiply the whole number by the numerator of the fraction and tJivide the product by the denominator : But if the numerator be 1, difide by the denominator only. 3. 4. 5. G. 7. 28 36 48 325 25-1 1 2 :l 5 t 'S a 4: » 12 1. 2. Mult. 8 15 Byi i Prod. 2 7} 9} 3)72 4)144 8)1625 12)1813 Prod. 24 36 203} 151 ^v^ CASE 11. To multiply a zvkole number by a luixed 07ie. Rule. Multiply by the fraction as in Case 1st ; l!?en multijjly by the whole number, and add the two products, as h the examples — or, to multiply a mixed number by a whole one, clvange the place ot^ the factors, and proceed as the rule directs. — St.e example C 1. 2. 3. 4. . 5. 6. 'Mult. 15 35 68 42 12:) J ,t By 3X. 5^ 7ji 9-} ' G,^ 2.'i 7A. ]1| 748 126 645 Mull. 24 ^^ ]2^' ^^'tV 18 l;af r>y i^. ?r5d. 52A 186| 476 378 1032 15)l68~ Prod. 538-j.\ 396 11121 11/^ £4 IVod. 35.^—1. VULGAR FRACTIONS, 89 CASE in. To multiply a mixed number by a mixed member. Rule. Multi(jly the integral part of the mnltipircand by the denomina- tor of its fractional part, and add thereto its numerator: Then mnUiply by the mixed multiplier, by Case 2d, and divide the pro- ddct by the denominator of the fractional part of the multiplicand, as in the following example. Mtdt. 422 j 1st. 42f=213^ By Qdr ^ which mult, by 8| After this manner may feet and 3)426 inches be multiplied, calling 1 inch j^ of a foot, 2 inches ^, 3 inches {, 142 }► 4 inches i, 5 inches jV, G inches -}, 1704 7 inches -jV, 8 inches |, 9 inches ^ 10 inches |, 1 1 inches {^ of a foot. 1846 I'roduct =3694 j DIVISION OF VULGAR FRACTIONS. Rule.* Prepare the fiactions as before: then, invert the divisor and proceed exactly as in Multiplication: The products will be the quotient required. Examples. 1. Divide ^ of 17 by | off 1X17 A of 17 = i of V = = y and | of f = i| = i ; there- 3x 1 17x2 ¥ ^ ^H ^^'^ quotient required. Ans. 1^^. 3. Divide 12} bv i of 7. ' Ans. 5>j, 4. Divide b\- by'7f Ans.-*,}, 5. Divide f by 9. Ans. -jV- 6. Divide i of } of f by 1 of f. Ans. *. 7. Divide 7 by f Ao^. 18|. 8. Divide 4204} by J of 112. Ans. 42|»f. * The reason of the rule may be shewn thus. Svippose it were required to 4 2 4 J 4 4 2 ; divide - by 7. Nnw - — 2 is manifestly "of -- 'I- '^ ^ Z ZZ -Zl Q^ C CU ~ .C • — 2 1 -2 3 4 5 o 7 8 * It will be x'cry appnrcnt to tlie learner from tlie mture of decimals, and what has been said of Federal il^ doll, and 50 dollars, 5 cents, thus 5005 — 50 .5 _ doll, wherefore, it is, in all respfctK, addtd, subtracted, multiplied and divided, the same as decimals ; and, of all coins, it is the most simple. • It may also be obst rvcd that the sum exhibits the particular number of each different piece of money contained in it, viz. 455997 mills -45599 " cents =• E. D. d. cm. 4559.^^" dimes — 455^9_s t dolhrs — 45^fi_p.i)_^^^^ J 5 9 9 7- AKo,the names of tl;c coins, less llian a dollar, are significantpf tluir value? For the mUl, winch strands in tl.e 3d place at the rio;|.t hand of the separatxir DECIMAL FRACnONS. 91 From this table it ia evident, that in decimal*, as vvell as in whole nunibers, pach figure, takes its value by its distance trom unit's '][>lace : If it be in the first place alter units "l^or the separating point) it signifies tenths ; if in the second, hundredths, kc. decreas- ing in each place in a teiifuld proportion. Every single figure expressing a decimal, has for its denomina- tor a unit or 1, with so many cyphers as its place is distant irom unit's place : Thus 2 in the decimal part of the table = /^ ; 3 = T'o I ^"^ ^^ Tuoo-' ^^' ^^^ '^ '^ decimal be expressed by several rigures, the denominator is 1, with so many cyphers as the lowest iigure is distant from unit's place. So -357 dignities jV/u? and 0053 Cyphers, placed at the right hand ol a decimal faction, do not alter its value, since every signiticant Iigure continues to possess the same place: So -5, 50, and 500, are all of the same vajue, anu eacu — lo^— loo — Tooo — f But cyphers, placed at the left hand of a decimal, do alter its value, every cypher depressing it to ^^ of the value it had before, by removing every signiticant figure one place further from the place of uniis. So 5, -05, 005, all express dilfercnt decimals, viz. •5, t'o- ; '^5, -jI^ ; -005, j^%^. IJence may be observed the contrary eftects of cyphers being annexed to whole numbers, and decimals. It is likewise evident from the table, that since the places of de- cimals decrease in a tenfold proportion from units downwards, so they consequently increase in a tenfold proportioo from the right hand toward the left, as the places of whole numbers do : For, tea hundredth parts make one tenth, ten tenths make 1 ; ten units, ten; ten tens, one hundred, &c. viz. yVd^^rV' "io^^^» ^^*^ 1X10=^ 10, which proves that decimals are subject to the same law of No- tation, and consequently of operation, as whole numbers are. Decimal fractions of unequal denominators are reduced to one common denominator, when there are annexed to the right hand of those, which have fewer places, so many cypi)ers,a8 make them equal in phces with that which has the most. So these decima!«^, •5, 06, '455, may he reduced to the decimals, "500, '060, and '465, which have, all, 1000 for their denominator. Of Decimals, that is the greatest, whose highest figure i^ great- est, wliether they consist of an equal or unequal number of places : Thus, -5 is greater than, -459, for if it be reduced to the SJkipe do- nominator with -459, it will be 500. or place of thousandths, is contracted f com milie the Latin (or thousand : Ce/fi, which occupico tlie second place, or pUce of hundredths, is an abbreviiticn of cenlam, the Latin for hundred : And i//>w, which is in the first place or place uf tenths, is derived {i\ny\ d'ume, ilie French for tinth. Such being the nafure of Federal Money, its operations can in no ot'icr way Ire ;.o well understood as in obtaining a good knowledge of decimals, and applyiii^j rhcir several rules to the various cases of monev matters. In sums of Federal Money, it is customary to read it in dollars, cents and mills. Thus the ubove example is read /J55 dolls. 99 cents and 7 mills. 9^ DECIMAL FRACTIONS. A mixed number, viz. a whole number, with a decimal annexed, is equal to an improper fraction, whose numerator is all the figuress of the mixed number, taken as one whole number, and the denom- inator, that of the decimal part. So 45*309 is equal to VoW, as is evident from the method given to reduce a mixed number to an improper fracliop : Thus, 4oxlOOO-'r-309=''/(?/o^ ^^ above. ADDITION OF DECIMALS, 1. Place the numbers, whetlier mixed, or pure decimals, under each other, according to the value of their places. 2. Find their sum as in whole numbers, and point oft" so many places for decimals, as are equal to the greatest number of deci- mal places in any of the given numbers. Examples. 1. Find the sum of 19 073-l-2-3597+223+-0197681-|-347eiT- 12-338. 19073 2-3597 223- •0197581 3478 1 12-358 3734.9104581 the sum. 2. Required the sum of 429-4-21 •37-f-355003^}-l 07+1 7 ?' Ans. 808•148 3. Required the sum of 5-3-|- 1 1 •973-f49H-94- 1 •7314H-343 f Ans. 103-2044^ . 4. Required the sum of 973-|-19-|-l-75-f 93-7164-f-9501 ? Ans. 1088.'U65, SUBTRACTION OF DECIMALS. Rule. riacp the numbers according to their value ; then subtract as in whole numbers, and point oil the decimals as in Addition. Examples. I. Find the difference of 1793-13 and 817C5G93^ From 1793-13 Take 817 05C93 P.emainder 976*07307 i. From 17 J -195 take 125-917C. Ans. 45 2 774. 75. From 219-1384 take 195-01. Ans. 23-2284. a. From 480 take 21 3-0075. Ans. 234-9f>2,5. DECIMAL FRACTIONS, 93 MULTimCATION OF DECIMALS. I CASE I, Rule. 1. Whether they b^ mixed numbers, or pure decimals, place the factors L'.nd multiply tliem as in whole numbers. '2. Point off so many figures from the product as there are deci- mal places in both thi* factors ; and if there be not go many places ill the product, supply the defect by prefixing cyphers. The reason for pointing off the figures for decimals, is evident from substituting the equivalent vulgar fractions, as in the ibllow- ing example- Or, 'Hie reason of the rule for pointing off the fig- ures for decimals, is evident from the notation of decimals. Thus .5X-5=-25 ; for -Sxl^S, or once five tenths. But, as the multi- plier is less than unity, or tenths, multiplying is only taking tenths o^ tenths, and so ma^iy tenths of tenths are evidently so niany hun- dredths. So also, tenths ofhundredths would be thousandths ; hun- dredths of hundredths, would be ten thousandths, and so on. Examples. 1. Multiply 02345 23^5 by -00163 -02345= 100000 7035 163 14070 .00163= 2345 lOOOOe 382235 •0000382235 the product,: 10000000000 S. Multiply 25-238 by 12-17. Ans. 307-14646. '6. Multiply -3759 by -945. Ans. -3552255, 4. Multiply -84179 by -0385. Ans. -032408915. To multiply by 10, 100, 1000, &c. remove the separating point .350 many places to the right hand, as the multiplier has cyphers- (10 ) (3 45 So -345 Multiplied by \ 100 \ makes \ 345 (1000) (345 ' For -345x10 is 3-45P, &c. CASE II. To contract the operalion^ so as to retain so many decimal places \u the Product «« moy be thought necessary. Rule. 1. Write the unit's place of the multiplier under that figure of the multiplicand, whose place you would reserve in the product ; and dispone of the rer^t of the figures in a contrary order to what they are usually placed in. 2- In multiplying, reject all the figures which are on the right hand of the multiplying digit^ and ict down the products, po tha^ m DECIMAL FKACTiONiS. their right hand figures may fall in a straighlt line below each ether ; observing? to increase the first figure of evesrj line with what would arise b}^ carrying 1 from 5 to 15, 2 from I,':" to 26, 3 from 25 to 35, kc. from the preceding figures, when you begin to multiply, and the sum will be the product required. Examples. 1. It is required to multiply 5G-7534916 % 6-376928, and to re- ia'm only live place© of decimals in the product. Contracted way. Common way. 66-7534916 66-75340 16 129673 6 376928 28376746 . 1702605 . 397274 . .34052 . 5108 . 113 . 45 . 305- 15943 45 40279328 113 5069832 6107|314244 34052 09496 397274 4412 1702604 748 8376745 80 305-15943 80818048 By the operation in the common umy, it is evident that all the ngures which are cut off at the right hand, by the perpendicular iine, are, wholly omitted in the coniracted way, and the last product here is the first there ; consequently the rea^^on of placing the ?nultip!ier in a reverse order, must appear very plain. difisiOj^ of decimals. Rule.* , I, The places of decimal parts in the divisor and quotient cu-d'..i- ed togeth.er, must always be equal to those in the dividend ; there- fore, divide as in whole numbers, and, from the right hand of the qtiotient, point ofi' so many {daces for decimals, as the decimal pla- ces in the dividend exceed those in the divisor. 2. If the places of the quotient be not so many as the rule re- quires, stippiy the defect liy pretixing cyjjhers to the left hand. 3. If at any time there be a remainder, or the, tlecimal places in tiie 9'567)G96271 Here, the units place of the divisor 597402 in the first step falls under 7 in the place of hundredths in the dividend, (here- ^t9'56) 98869 . , fore, I put 4, the first quotient figure, 89604 in the place of hundredths, by prefixing a cypher. :^9'5) 9265 1 have set down every divisor, to ex* 8935 plain the work ; but you need only put a dash over every figure rejected, a^ 0.9) 310 you pfoceed, to show it is omitted. 297 9) 13 '' 9 Remainder 4 When decimals or whole num- bers are to be divided by 10, 100, Exaiui 1000, &c. (viz. unity with cyphers) it is performed by removing the 9eparatrijf,inthe dividend, so ma- ny places toward the left hand as there are cyphers in the divisor. REDUCTION OF DECIMALS. CASE I. To reduce a Uulgar ^(action to its equiva'^'nt Decimal. Rule.* DiviJe the numerator by the denominator, as In division of do »imal*, and the quotient will be the decimal required : — Or, su many cyphers as you annex to the given numerator, so many pla- * By annexing one, two, three, &c. cyphers to the numerator,, the value of the traaion is increafcd ten, a hundred, &c. times. After dividing, the quotient will of courfe, be ten.a hundred, &o. times too much -. the quotient muft therc- *bre be divided by ten, a hundred, &c. to give the true quotient or fraaioo. In the f:rft example, ^, is mfldc i «LP « arr » 3.* , which divided bv lOOU, is _.yj,T. %''i25; and is t1»c ntlc. 0-. DECIMAL FRACTIONS, 97 ces must be pointed off in the quotient, and if there be not so many places of figures in the q'lotient, the deficiency must be supplied by prefixing so many ciphers before the quotient figures. Examples. 1. Reduce i to a decimal. 8)1.000 •125 Ans. 2. Reduce f , f and | to decimals. Answers, STS^ 625, -6664- « 3. Reduce ^, i, |, i, |, f and | to decimals. Answers, -25, -6, 75, -SSS-f , -8, -833+, -875. I. Reduce -,%, f|, 4W, and /g- to decimals. Answers, -263+, •692-f , -025, -25. 5. Reduce ^^j, j-f^^, ahd ^/tt to decimals. Answers, 0186+, -00797+, -00266+. CASE H. To reduce numbers of different denominations^ as of Money^ Weight and Measure^ to their equivalent decimal values. Rule.* I. Write the given numbers perpendicularly nnder each other, for dividends ; proceeding orderly from the least to the greatest. II. Opposite to each dividend, on the left hand, place such a number, for a divisor, as will bring it to the next superiour denom- ination, and draw a line perpendicularly between them. III. Begin with the highest, and write the quotient of each di- vision as decimal parts on the right hand of the dividend next be- low it, and so on, until they are all used, and the last quotient will be the decimal sought. Examples. 1. Reduce 17s. 8fd. to the decimal of a pound. 4 12 20 3- 8-75 17-729166, &c, •886458, &c. the decimal required* Here, in dividing 3 by 4, I suppose 2 cyphers to be annexed to the 3, which make it 300, and -75 is the quotient, which I write against 8 in the next line ; this quotient, viz. 875 feing pence and decimal parts of a penny, I divide them by 12, which brings them to snillings and decimal parts, 1 therefore divide by 20, and, there being no whole number, the quotient is decimal parts of a pound. * The reason of the rule may be explained from the first clample ; Thus, three farthings are ^ of a penny, which, reduced to a decimal, is, -75; conse- quently, 8fd. may be expressed, 8. 75d. but 8.75 is 81| of a penny = y3_7_5_ of a shilling, which reduced to a decimal, is, •7291663.+ In like manner, 17749166$.+ are i i/J?^J^^6,. ^ ii^|a ^^^j^ == .88/>4^8:+-a»b7 the role. w 98 DECIMAL FRACTIONS. 2. Reduce 1, 2, 3, 4, and so on to 19 shillings, to decimals. Shillings. 123450789 10 Answers. -05, -1, -16, -2, -25, -3, -36, -4, -45, '5, Shillings. 11 12 13 14 15 16 17 18 19 Answers. -55, -6, -65, -7, -75, -8, -85, -9, -95. Here, when the shillings are even, half the number, with a point prefixed, is their decimal expression ; but if the number be odd, annex a cypher to the shillings, and then halving them, you will have their decimal expression. 3. *Keduce 1, 2, 3, and so on to 11 pence, to the decimals of a shilling. Pence. 12 3 4 5 6 Answers. -083-}-, -166, -25^ -SSS-f, -416-1-, -5, Pence. 7 8 9 10 11 Answers. •583-f, -Oee-f, -75, -8334-, -916+. 4. Reduce 1, 2, 3, &c. to 11 pence, to the decimals of a pound. Pence. 1 2 3 4 5 Answers. -00416+, -00834-, '0125, •Oi666-f, -0208+, Pence. 6 7 8 9 10 11 Answers. -025, -02916+, -0333+, -0375, 0416+, -04583+. 5. Reduce 1, 2 and 3 farthings to the decimals of a penny. lqr. = -25d. 2qr. = -5d. and 3qr =-75d. Answers. 6. Reduce 1, 2 and 3 farthings to the decimals of a shilling. Answers, lqr.^-02083+s. 2qrs. = -04166+s. 3qrs.=-0625s. 7. Reduce 1, 2 and 3 farthmgs to the decimals of a pound. Ans. lqr. = -0010416+£. 2qra. =-002083+ £. 3qrs.=-003125£. 8. Reduce 13s. 5Jd. to the decimal of a pound Ans. -6729+. 9. Reduce 7Cwt. 3qrs. 17ib. lOoz. 12dr. to the decimal of a ton. Am. -39538+. 10. Reduce lOoz. 13pwt. 9gr. to the decimal of a pound Trow Ans. -8890625." 11. Reduce 3qrs. 3n. to the decimal of a yard. Ans. '9375. 12. Reduce 5fur. 12po. to the decinial of a mile. Ans. '6625^ 33. Reduce 55m. 37sec. to the decimal of a day. Ans. -03862+ . CASE III. To find the decimal rf any number of shillings, pence and farthings, by inspection. RuLE.t 1. Write half the greatest even number of shillings for the first decimal figure. * The answers to this question arc the same as the decimal parts of a foot. •j-The invention of the rule is as follows : As shillings are 30 many loths of a pound, half of them must be so many tenth's, and consequently take the place of tenths in the decimals; but when they arc odd, their half will always consist of two figures, the first of which will be half the even number, next less, and the second a 5 ; Again, farthings are so many 96oths of a pound, and had it happened that icoo, instead ©f 960, had made a pound, it is plain any . DECIMAL FRACTIOKS. <)9 II, Let the farthings in the given pence and farthings possess the secpnd and third places : observing to increase the second place or place of hundredths, by 5 if the shillings be odd, and the third place by 1, when the farthings exceed 12, and by 2 when they es- ceed 36. Examples. 1. Find the decinaal of 13.s. 9|d. by inspection. •6.. =1 of 12s. 6 for the odd shilling. 39 =the farthings in 93d. Add 2 for the excess of 3C. •691 = decimal required. 2. Find, by inspection, the decimal expressions of 18s. 3id. and 17s. 8id. Ans. £-914 and £-885. 4. Value the following sums, by inspection, and find their total, viz. 15s. 3d,4-8s. lUd.-f 10s. G^d.-f-ls. S-^d.+^d.+^^d. Ans. "je 1-834 the total. CASE. IV. To find the value of any given decimal in the terms of the integer. Rule. L Multiply the decimal by the number of parts in the next less denomination, and cut off so many places fof a remainder, to the right hand, as there are places in the given decimal. H. Multiply the remainder by the nei^t inferiour denomination^ and cut off a remainder as before. III. Proceed in this manner through all the parts of the integer, and the several denominations, standing on the left hand, make the answer. This case is the reverse of Case U. and the reason of the rule is hence obvious. Examples. 1. Find the value of '73968 of a pound. , 20 14-79360 12 9*52320 4 2»0928O Ans. 14s. 9id. number of farthings would have made so many thousandths, and might have taken their place in the decimal accordingly. But 960 increased by J- part, of itself, is :=:iooo, consequently any number of farthings, increased by their _i_ part, will be an exact decimal expression for them : Whence, if the number of farthings be more than la, oj P*""' '^ greater than jqr. and, therefore i must be added ; and when the number of farthings is more than z^t -^^ P*rt U greater than i^qf. for which 2 must be added. 100 DECIMAL FRACTIONS. 2. What is the value of -679 of a shilling? Ans. 8-148d. 3. What is the value of -DgGDje? Ans 19s. llfd/i^ or£l. 4. Wiiat is the value of -617 of a Cvvt. ? Ans. 2qrs. 131b. loz. lO^^dr. 5. What is the value of -8593 of a lb. Troy ? Ans. lOoz. 6pwt. 5gr. 6. What IS the value of -397 of a yard? Ans. Iqr. 2-352n. 7. What is the value of '8469 of a degree ? Ans. 5Sm. 6fur. 35po. Oft. llin. 8. What is the value of '569 of a year ? Ans. 207da. 16ho. 26m. 24sec. 9. What is the value of -713 of a day ? Ans. 17h. 6no. 43sec. CASE V. To find the value of any decimal of a pound by inspection. Rule.* Double the first figure, ov place of tenths, for shillings, and if the second figure he 5, or more than 5, reckon another shilling ; then, after the 3 is deducted, call the figures in the second and third places so many farthings, abating 1 when they are above 12, and 2 when above 36, and the result will be the answer. JVote. When the Decimal has but 2 figures, if any thing remain after the shillings are taken out, a cypher must be annexed to the right hand, or supposed to be so. Examples. 1. Find the value of •876£ by inspection. 16s. = double of 8. Is. for the 6 in the second place, which is to be taken And 6id.=26 farthings remain to be added. [out of 7. Deduct ^d. for th^ excess of 12. 17s. 6id. the Ans. 2. Find, by inspection, the value of •49J6. 8s. - - = double of 4. Is. - - for the 5 in the place of hundredths. lOd. = 40 farthings, a being annexed to the remaining 4 Ded. id. for the excess of 36. 9s. 9id. the Answer. 3. Find^the value of •097je by Inspection. Ans. U n]d. 4. Value the following decimals bv Inspection, and find their ?nm, viz. •785£ -f •537i: -f '916£ + •74i: + '5£ + 'SojC -f- •09JS 4- -OOSje. Ans. £3 16s. 6d. • As this rule is the reverse of the rule, Case HI. the reason is apparent from the demonstration of that rule. DECIMAL FRACTIONS. 201 i DECIMAL TABLES OF COIN. WEIGHT. AND MEASURE . TABLE I. Coin. £ 1 the Integer. dec. •95 Shil 19 i8 17 i6 ^5 '75 Ul 7 '6s •6 ■55 •5 Shil 9 8 7 6 5 4 3 2 X dec. •45 •4 •35 •3 *»5 •a •15 •I •05 TABLE III. Troy Weight. lib. the Integer. Ounces the fame as Table II Pence. It 10 9 S 7 6 5 4 3 a 1 Pwts. Decimals. 10 '041666 9 -0375 8 -035333 7 '029166 6 -025 5 -020833 4 -016666 3 •01*5 a -008333 Decimals. 1 '00 4166 •045833 ioTalns. DecTniaJsr •C41666 •0375 •033383 •019166 •025 '020833 •016666 •0125 •008333 •004166 Earth's 3 2 I Decimals. '003125 '0020833 •0010416 12 II 10 9 8 7 6 S 4 3 2 I 'O02083 •00191 •001736 •001562 '001389 '001215 •001042 •000868 •000694 •000521 •000347 •000173 Pounds. 27 26 a5 24 ^3 22 21 20 19 18 17 16 Jt5 14 13 12 XX 10 9 8 7 6 5 4 3 2 t Decimals •241071 •232143 •223214 •214286 •ao5357 •196428 •1875 '178571 •16^643 •160^14 •151786 •142857 •1339*8 •125 •110671 •107143 •098x14 '089286 •080357 '071428 '0625 •0535^1 '044643 •035714 '026786 •017857 •008928 Ounces. I Decimals. 12 .-75 XI 10 9 8 7 6 5 4 3 2 X •6875 '625 •5625 •5 •4375 •375 '3125 •25 •1875 •125 '0625 j Drams. I Decimals. 15 -058593 TABLE II. Coin &LongMeas. I Shil and i Fool the Integer. P^°«^ Decimals. Inches. IX 7 6 5 4 3 2 I •916666 •833333 •75 •666666 •583333 •5 '4166661 •333333 •15 •166666 I Oz. the Inttgcr. Pennyweights the fame as Shillings in the firft Table. Decimals. •025 •022916 •020833 •01875 •016666 •014583 •0125 •010416 •008333 •00625 •004166 •002083 Grains. 12 XI 10 9 8 7 6 5 4 3 2 I Ounces. jDecimals. 15 -008370 14 1 •007812 13 , -007254 X2 I '006696 11 I -006138 14 13 12 II 10 9 8 7 6 5 4 3 2 I •054687 •050781 046875 •642968 •039062 •035T56 •03125 027343 •023437 •019531 •015625 •011718 •007812 '003906 10 9 8 7 6 J 4 3 2 I 1 '00558 •005022 •004464 •003906 '003348 '00279 •002232 '001674 •001116 •000558 TABLE VI ClothMeasuure I Yard the Integer Decimals. •75 •5 •*5 Farth's 3 2 X TABLE IV. Avoirdupois Wt. •0 83333 1 iialb. the Integer. qr.ofoz8.|Dccimals 3 ' '000418 2 '000279 T I •000139 Decimals, •0625 ■041666 •020383 Qrs. 3 2 I Decimals. '75 •5 •25 Qrs. 3 2 X Nails. Decimals. 3 •1875 2 •125 I '0625 TABLE VII Long Measure. I Mile the Integer, Decimals. ^•568182 •511364 Yard! 1000 900 TABLE V. AVOIRUUFOIS Wt. lib. the Integer. Decimals. Ounces 15 14 13 •9375 •875 -8125 800 700 600 500 4C0 300 200 100 •454545 •397717 •34 •284091 •227272 •^70454 •113636 •056818 Carried over. 102 DECJIVIAL FRACTIONS. Yards. becimals. TABLE Vlir. Days. 30 20 Decimals. 1 90 80 •(J 5 1136 •045454 LiquiD Measvre. •082192 054794 Hours. 11 Decimals. 458333 70 •039773 I Gal. the Integer. 10 •027397 10 •416666 60 034091 Quarts the fame as 9 •024657 9 •375 50 40 028409 •022727 qrs. in Table VI. ■ Pint. -1'^''^ 8 7 021918 019178 8 7 •333333 •291666 30 20 •017045 011364 o Gills. 2 •09375 •0625 6 5 •016438 013698 I •25 •208333 10 •005682 1 nQirtr 4 •010959 4 •166666 9 .0051 14 1 yj^ic-y 3 •008219 •005479 3 ■125 8 •004545 TABLE iX. 2 2 •083333 7 •003977 1 •0f)9739 1 •041666 6 3 2 •003409 .002841 .002273 .001704 .001136 Time. 1 Year the Integer. Months the fame as Pence in Table II. 1 I iDay Hours. 23 the Integer. Decimals. •958333 minutes 30 20 . Decimals. •020833 •013888 1 000568 Days. Decimals. 22 •916666 10 •006944 Feet. 2 1 Decimals. ■0G03787 0001892 365 300 200 100 90 roooooo •821928 •547945 •273973 •246575 21 20 19 18 17 •875 •833333 •791^66 •75 •708333 9 8 7 6 5 •00625 •005555 •004861 •004166 003472 laches. Decimals. 6 0000947 80 •219178 16 •666666 4 •002777 5 •000079 70 •191781 15 •625 3 •002083 4 •0000632 . 60 •164383 14 •583333 2 .001388 3 •0000474 50 •136986 13 •541666 1 •000694 2 0000316 40 109589 12 •5 1 •0000158 1 General Rule. To find the value of goods in Federal Money. — Multiply the price and quantity together; poi.jt off in the product, for denomi- nations lower than dollars, as nrjany places as there are in the giv- en price ; or, if there be decimal places in the quantity also, ac- cording to the rule for multipllicatioi\ of decimals. This is reaUy mulliplication of decimals, the dollar being considered the unit. Examples. 1. Find the cost of 823 yards, at .^1.29c. a yard. 823xg 1.29c. =^106 1.67c. Ans. 2. Find the value of 56yds. 2qrs. at ,^3.11r. per vard. 56yds. 2qr8.=56-5; and 66-5X3-1 l=^175.7"^Ic. 5m. Ans. •;. What must I pay for 6iyda. at ^5.50c. per yard ' Ans. ,$33.68c. 7ro. -5. 4. Bought 7}}yd3. at 34 cents per yard: what did I pay for the vffcole ? Ans. g2.62J |e. COMPOUND MULTIPLICATION. lOr* COMPOUND MULTIPLICATION* IS extremely useful in finding the value of Goods, &c. And as in Compound Addition, we carry from the lowest denomination to the next higher, so we begin and carry in Compound Multiplication : One general rule being to multiply the price by the quantity. The reason of the following rules is obvious. CASE I. fVhen the quantity does not exceed 12 yards, -pounds yS,'c : Set down the price of 1, and place the quantity underneath the least denomi- nation, for the multiplier, and, in multiplying by it, observe the same rules for carrying from one denomination to another as ia' Compound Addition. Introductory Examples. Multiply by 15 31 I. S. 17 14 d. 1 2 2 £ 25 2. 8. 12 d. 8 3 D. d. 8 5 3. c. 1 m. 7 4 4. £ s. d. 67 18 6-i. 5 Prod. 34 6 8 6. D. c. 4 75 6 3 13. 6. s. 12 d. 11 7 s i £ 31 7. s. d. 16 8J- 8 m. 5 12 E, 2 8. D. d. c. m. 7 8 9 1 9. £ s. d, 4 13 4; 10 £ 8 10. s. d. 15 11^ 11 11. D. c. 35 87 12. £ s. d. 14 17 8A 9 133 19 2-1. In the last example, I say, 9 times 1 is 9 farthing8=2id. I set down \ and carry 2, saying, 9 times 8 is 72, and 2 1 carry makes 74 pence, =6s. 2d. 1 set down 2 in the pence and carry 6 ; then, 9 times 7 (the unit figure in the shillings) is 63, and 6 I carry is 69, * The produ(Sl of a number, confifting of fcvcral parts or denominations, by any liraple number whatever, will be exprcflcd by taking the proJudt of that fimple number, and each part by itfclf, as fo many diftintSt quefHons : Thus ^eSri 15s. *jd. multiplied by 5, will be \k)'o iFrs. 45d.— (by taking the Oiillings from the pence, and the pounds from the fliillings, and placing them in the fhillings and pounds rcfped:ively,) £168 18s. Cd. and this will be true when the multiplicand is any compound number whatever. 204 COMPOUND MULTIPLICATION. I set down 9 under the unit figure of the shillings, and carry 6, .saying, 9 times 1 is 9, and 6 i carry is 15, then I halve 16, which is 7 and 1 over, which I set in the ten's place in the shillings, and that, with the 9, makes 19 shillings ; then I carry the 7 as pounds : Lastly, 9 times 4 is 36, and 7 I carry, are 43 pounds : I set down 3 and carry 4, saying, 9 times 1 is 9, and 4 1 carry makes 13, which I set down, and the product is X133 19s. 2\d. Practical Questions. Kote, The facility of reckoning in the Federal Money compare ^(\ wilh pounds, shillings^ &c. may be seen from the examples in this and the following cases ; where the same questions are given in both the currencies, as near as can be, avoiding small fractions. In the following examples in this and the succeeding cases, the price in pounds, or shillings, &c. is in the currency of New Jersey, Pennsylvania, Delaware, and Maryland, where the dollar is 7s. 6d. in the last example in each case ; in the last example but one, the price is in the currency of New York and North Carolina, where the dollar is 8s. ; and in the other examples, in the currency of New England, where the dollar is 6s. Thus in the 3d example, the price, 9s. lOd. in the currency of New York, &,c. is equal to 122c. 9m. ; and in example 4th, the price 13s. 7-^d. = 181c. 7m. 1. What will 9 yards of cloth at < j.^^ * q * ? per yard, come to ? £0 63. 4d. price of one yard, '^^c, 9m. Multiplied by 9 yards. 9 66c. 8m. Ans. £2 8 price of 9 yds. §8 -00 1 ^ ^ . ,S 15s. 4d. I I _ S ^2 6s. 2. 3 yards at ^ ^^ ^^^^ ^^ ^ per yard - J ^^ .. ^ 59s. lOd. } _ J £2 19s. I •'* \p 22c. 9m. i " ■ " " ~ ^ ^7 37c. 4m. J \ 13s. 7id. I _ ^ J £6 2s. 7id. } ^- ^ " " " l$\ 81c. 7m. 5 'I pQ 35c. 3m. S ' CASE IL When the multiplier, that is, the quantity, is abose 12 : You must multiply by two such numbers, as, when multiplied together, will produce the given quantity. Examples. 1. What will 144 yards of cloth cost at \ ^^'^ ^^^^ I per yard ? £ s. d. c. m. 3 61 price of 1 yard. -5764 Or, -5764 Multiply by 12 144 12 Produces 2 1 6 price of 12 yards. 23056 69168 Multiplied by 12 23066 12' 6764 - Answer £24 18 price of 144 yards. 83'OOIG Ans. 1^83-00 u; COIWPOUND MULTrPLICATION. 105 Questions. Answers, o o^ . C63. 3|d. } . ^ £7 lis. 6d. } 2. 24 yards at J ^^ , ^^^ ^m. ^ P" ^"^^ = } $25 24c. 8m. I .^ -,« 5 9s, lOd. / \ £13 53 6d. } J ,?1 22c. 9m. S ^ ' ' ' ' }p3 18c. 3m. S . .. \ 13s. 7id. i ] £29 19s.. 6(\ I ^ iJl 81c;7m. ^ * - - - - ^ ^79 94c. 8m. S CASE III. When (he quantity is such a number, as that no two nitmbers in the table will prodnce it exactly : 'i'hen multiply by two such numbers as come the nearest to it; and for the number wanting, multiply the given price of one yard by the said number of yards wanting-, and add the products together for the answer ; but if the product of the two numbers exceed the given quantity, then find the value of the overplus, which subtract from the last product, aud the re- mainder will be the answer. Examples. rds < £ s. d. 1. What will 47 yards of cloth, at I 173. 9d. I ^^ ^,^, co*ne to ? I g2 95c. 8m. | ^^^ ^^^^' 17 9 price of 1 yard. Multiplied by 6 Produces 4 8 9 price of 6 yards. Multiplied by 9 Ans. g2-958 47 20706 11832 Produces 39 18 9 price of 45 vards. Add 1 15 6 price of 2 yards. gl39-02C Ans. £41 14 3 price of 47 yards. Xote. This may be performed by first finding the value of 48 yards, from which if you subtract the price of 1, the remainder will be the answer as above. Questions. Answers. 2 75 yards at J ^'' ^^'^' \ oer yard = \ ^^^ ^'- ^^'^- 1 ^. 75 yard., at ^ ^3^ ^^^ ^ per yard - ^ ^^^ 3^^^ gim. \ ., ^^, \ 16s. 4d. I _ \ £55 2s. 6d. \ ^' ^^^ ' ' ' ' }f,2 4c. ^ - ' " ' -^gi37 81c. i . Q ^ \0i. Od. i i £29 10s. Od. I " " ■ " < gl 23cf S ' ' " ■ "^ < S^S 66c. 6m. S CASE. IV. When the quantity is any number above the Mulfiplicoiinn TabU : Multiply the price of 1 yard by 10, vviiich will produce ihe price of 10 yards: This product, multiplied by 10, wiji give the price of 100 yards ; then, you must multiply the price of ^-r.ft hundred by the number of hundreds in your question ; (he price often, by liie number of tens ; aod the price of unity, or 1, by the number of O lOe COMPOUND MULTIPLICATION. units : Lastly, add these several products together, and the sum will be the answer. Examples. 1. What will 369 yards af cloth, at 5 4s. 7|d. } , mount to ? I lie. Im. \ P^^ ^^^^' *' -£ s. d. c. m. 4 7i price of 1 yard. -771 — 359^ 10 ■ — 6939 2 6 3 price of fO y^rds. 3855 10 2313 23 2 6 price 100yds. Ans. g276-78^ % 69 7 6 price of 300 yards. 5 time* the price of 10 yds. =i 11 11 3 price of 50 yards. S times the price of 1 yd.= 2 1 71 price of 9 yards. Answer JGSS 4| price of 359 yards. t 297 yards at \ ^^'' ^"^^' ? per yard = \ f ^'^^ ^^'' '^^- 1 z. z\ii yaras at ^ ^^ gg^ ^^^ ^ per yara ^ ^^^ ^^^ ^^ ^ J 9s. Hid. > _ . . .^^ £235 Os.Sid. } ? §1 65c 61m. i ^ I ^783 40c. 6im. \ i 5s. lOd. > _^ £149 6s. 8d. \ ^- ^1^ - - ■ \ $0 72iic. 5 I $313 331c. < ^ _^ ^ 18s. 9d. > _ < £717 3s. 9d. > ^' ^^^ ' " " ^ ^2 50c. I . - - - ~ ^ gjgj2 50c. I CASE V. To find the value of one hundred th eight : As 112 is the gross hundred, so 112 farthings are —2s. 4d. and 112 pence — 9«. 4d. ; therefore, if the price be farthings, or not more than 3d. multiply 2s. 4d. by the farthings in the price of 1 lb. or, if the price he pence, multiply 9s. 4d. by the pence in the price of I lb. aud in either case, the product will be the answer. Examples. 1. What will 1 Cwt. of chalk come to at J'g^^l'^ > per pound? 112 farthings = 024 price of 1 Cwt. at } per lb. -021 J^d. = 6 farthings in the price. 112 AiisvTcr £0 14 price of 1 Cwt. at li per lb. 42 .... ■' 21 21 "^-* « Ads, 2-352 COMFOUND MULTIPLICATION. 1^7 8. d. 2. 1 Cwt. of tin at SJd. per lb. ? 2 4 price of 1 Cwt. at id. per lb. •03125 9 Jarthings in the price of 1 ib. 112 625.0 3125 3125 Ans. ^l 1 price ©f 1 Cwt. at 2i per lb. f 8. d. Ans. what cosi 1 Ife ? 8. If Sj'cwt cost g3.64, what cost 1 ft? Ans. Ic. Note. This Case proves the 6th in Compound MuUipHcation. CASE IV. Having the price of any number of yards, ^c. to find the price of 1 yard: Divide the price by the quanUt}'^ beginning at the highest denomination, and, if any thing remain, reduce it into the next, and every inferiour denomination, and. at each reduction, divide as before, remembering each time, to add the odd shillings, pence, &c. if there be any, and you will have the value of unity rt quired. Note. If there be j, | or | of a yard, pound, &c. multiply both the price and quantity by 4, and then proceed as above directed j or, in federal money, work by decimals. Examples. n If db^m of figs cost J ^^^ 11^ l^^l j what are they per fe ? Ife £' 's. d. (I^uantity =: 95J- - Price = 16 13 6f Mult, by 4 4 Produces 382 for a divisor. Product£66 14 3 for a dividend. 382)66 14 3 (0 3 6f |f| per fe. 20 $ cm. dec, c m. 95i=:95-5)55-69376(-58 2-f Ans. 382)1334(3 " 4775 , 1146 7843 188 7640 12 ■ 2037 :^82)2259(5 1910 1910 349 ,:^82) 1396(3 1146 250 1275 COMPOUND DIVISION. 113 2. If 147 bushels of rye cost £47 12s. 6d. what is it per bush- el ? Ans. 6s. 5fd. 3. If 331 yards of baize cost £23 13s. 9id. what is it per yard ? Ans. 15s, 5id. y^%. 4. If 147lbs. cost gl58 76c. what is the price of 1 pound ? Ans. $1 8c. 5. Bought 331yds. of cloth for g85 63c. 2m. ; what did I pay a yard ? Ans. $2 57c. 5m. Note. This proves the 3d and 4th Cases in Multiplication. Practical Questions in Money. 1. Divide < ^g. . r^' ^ > among 5 men and 4 womeD, and give the men twice as much as the women. Men. Women. £ ss. d. £ s d. 5 and 4 Divide by 14)273 9 4(19 10 8=1 woman's share. ult. by2 14 4 women. 10 shares. Add 4 women's shares 14 the number of shares in the =Divi8or. 133 78 1 1 <^R . ._ 2 8=women's share. £19 equal 7 10 8 2 — £39 14)149(10 14 — — £195 9 78 12 1 4=1 man's share. 5 men. 6 8= men's share. 2 8=women's share — £273 9 4 Proof. 14)112(8 112 D 14)911-555(65.111+ =1 woman's share. 84 4 women. 71 260444= women's share. 70 65-1114- 15 . 2 14 130 222-1- =1 man's share, 15 5 14 — 6511114- =men's share. 15 260-4444- =vvomen's share. 14 911-5554- Proof P Hi REDUCTION OF COINS. 2. Divide J£120 17;?. 4d. among 7 men and 7 women, and give the women 3 times so much as the men. Aus. £4 6s 4d.==a man's share. £12 19s. =a woman's share. 3. Divide £39 12s. 5d. among 4 men, G women, and 9 boys : Give each man double to a woman, and each woman double to a boy. Ans. £1 Is. 5d.=a boy's share. £2 2s. 10d.=:a woman's share. £4 6s. 8d.=a man's ditto. 4. Divide 5 guineas among 8 men : — Give A 8d. more than F, and B 8d. more than C, &c. Ans. H's share=153. 2d. REDUCTION OF COINS. RULES for reducing the Federal Coin, and (he currencies of the several United States ; also English, Irish, Canada, Nova-Scotia, Livres Tournois and Spanish milled Dollars, each to the par of all the others.* * "^ I. To reduce New Hampshire^ 1. Reduce £349 19«. Id. to Massachusetts t Rhode Island^ dollars. Connecticut^ and Virginia cur- '9 =^ the shillings. rency : '05 = odd shillings. 1. To Federal Money. '004 =: qrs. in Id. Rule. — Reduce the shillings, pence and farthings, to decimals, '954 = decimal, by Inspection (Case 3d, Dec. 3)349 954 D. c. m. Frac.) divide ^the whole by 3, 1166 513=1166 51 3 Ans. putting the comma one figure 2. Reduce 19s. Ifd, to dollars- further to the right hand in the ^3 19c. Ans. quotient, than in the pounds of 3. Reduce Is. to cents, the dividend, and the quotient Is. = -05 then will be the answer in dollars, 3)0 5 c. m. cents and mills, 0- 166| = 16 6|. • The Rules for the reduction of money depends upon the relative value of the currency of different States, &c. This value is given in the several rules. Thus 4 pounds of the currency of New York and North Carolina, are equal to 3 pounds of New England and Virginia; 4 of New England. and Virginia, are equal to 3 of England ; 4 of New England &c. are equal to 5 of Pennsyl- vania, New Jersey, &c.; and i pound of New England, &c. is equal to seven ninths of a pound of South Carolina and Georgia, and thus of the others. The rules are therefore obvious. In some cases, the process is contracted, and the contraction is given for the rule, because the operation is simpl.ified. Thus the first rule is equivalent to multiplying the pounds and .decimals by 20 and dividing the product by 6, the shillings in a dollar. But as 2^1=; li , the sum is multiplied by 10 in removing the separatrix one figure to the right, before the division by 3 is made. The relative value of a i.', depends on the number of shillings reckoned to the dollar, — the greater the uumbrr of shillings in a dollar, the less the £, and the reverse. REDUCTION OF COINS. 15 4. Reduce Id. Ans. Ic. 3Am. 5, Reduce 1 qr. Iqr. = -OOlOll and 3)'0 01 041 000 347 = SyW »«»*^*- 2. To New York and North Carolina currency. Rule. — Add one third to the New Hampshire, &c. sum, and the sum total will be the New York, &c. currency. Reduce £100 NewHampshire, &c. to New York, kc. £ 3)100 -f 33 6 8 4^100 — 25 £75 Answer. 6. To Irish Money. Rule. Multiply the New Hampshire, &c. sum by 13, and divide the product by 16. Reduce £100 NewHampshire, &c. to Irish Money. 100 4x3-l-the given sum. 400 £133 6 8 Ans. 3. To Pennsylvania, New Jer- sey, Delaware and Maryland cur- rency. Rule. — Add one fourth to the New Hampshire, &c. sum. Reduce £100 New Hampshire, ^c. to Pennsylvania, £(c. 4)100 H- 25 £125 Ans.. 4. To South Carolina and Geot' gia currency. Rule. — -^Multiply the New Hampshire, &c. sum by 7, and divide the product by 9, and the quotient is the answer. Reduce £lOONew Hampshire, &c. to South Carolina, ^c. 100 7 ■ 9)700 £77 15 6| Ans. 5. To English Money. Rule. — Deduct one 4th from the New Hampshire, &c. sum Reduce £100 New Hampshire, ^c. to English Money. 1200 -1-100 16=4X4)1300 4)325 £81 5 Ans.. 7. To Canada and Nova Scotia currency. Rule. Multiply the New Hampshire, &lc. sum by 6, and divide the product by 6. Reduce £100 New Hampshire, &c. to Canada, &(c. 100 5 6)500 £83 6 8 Answer. 8. To Uvres Tournois. Note. 12 deniers, or pence, make 1 sol, or shilling, 20 sols, or sous, 1 livre, or pound. Rule.- — Multiply the New Hampshire, &c. pounds, by 17i, and the product will be livres : Or, multiply the sum in shillings by 7 : Divide the product by 8, and the quotient will be Uvres, sous, &€, ^ 116 REDUCTION OF COINS. Reduce jGlOONew Hampshire, &c. to Livres J ournois. 100 Or, 100 174- 20 2000 7 8)14000 Ans. 1760 Viv. ld.==lsou. 5ideniers Ans. 1750 livres. 17^ sous. 17i livres. £l II. To reduce Federal Moneys to JVcra England and Virginia currency. Rule. — Multiply the Federal money by 3, and if it consist of dollars only, Cut oflf 1 figure, if of cents also, cut off 3, and if of mills, 4 figures at the right hand ; then reduce the figures so cut off to farthings each time cutting off as ai first and the left hand figures are pounds, shil- lings, &c. Or, reduce them by inspection. 1. Reduce gl 166 61c. 3m. to New England currency, gem. 1166-61 3 3 £349 963 9 20 s.19'0780 12 III. To reduce New Jersey^ Penn- sylvania^ Delaware and Mary- land currency. 1. To JVew Hampshire, Massa- chusetts, Rhode Island, Connecti- cut, and Virginia currency. Rule. Deduct one fifth from the New Jersey, &c. sum, and the remainder will be New- Hampshire, &c. currency. Heduce £lOONew-Jersey,&c. to New-Hampshire, &c. 5)100 — 20 £80 Answer. 2. To New York and North Ca- rolina currency. Rule. Add one fifteenth to the New Jersey, &c. sum. Reduce £100 New Jersey, &c. to New York, &c. 15=3x6)100 3)20 -1- 6 13 4-f giv. sum. £106 13 4 the Answer. 3. To South Carolina andGeor' gia currency. hule. Multiply the New Jer- sey, &c. sum by 28, and divide the product by 45, and the quo- tient is South Carolina &c. Reduce £100 New Jersey, &C: to South Carolina, &:c. 100 •936 =ld. nearly. Or, 18s. = double of 9. Is. =5inthe2d place. Id. =3-9or4qrs.that 45= 4X7=28 400 7 19s. Id. Ans. 2. Reduce 45 dollars. £13 10s. Ans. C. Reduce gl2 7c. to N. E money. £3 12s. 504. =5X9)2800 5)311 2 2| £62 4 61 Ad8. REDUCTION OF COINS. 11' 4. To English Money. Rule. Multiply the New Jer- sey, 4^0. by 3, and divide the pro- duct by 5. Reduce £100 New Jersey, &c. to English money. 100 3 6)300 £60 Answer. 5. To Irish Money. Rule. Multiply the New Jer- sey, &c. by 13, and divide the product by 20. Reduce £100 New Jersey, &c. to Irish. 100 4x3-i-lhe giv. sum. by 10, and the quotient will be li- vres, sous, &c. Reduce JElOONewJerseyj^'C, to Livres Tournois, lOO 14 400 100 Or ld.=:I?f0US. bus. »I007 id.=iffoi ao> is"i4fot 3 i-i=i4Hv. aooo 7 Ans. 1 400UV. 10)14000 X400 8. To Spanish milled dollars Rule. Multiply the New Jer- sey, &c. pounds by 2| and the product will be dollars— Or, mul- tiply them by 8 : Divide the pro- duct by 3, and the quotient will be dollars. — If there be shillings in the given sum, for every 7s. 6d. add 1 dollar to the quotient. Reduce £100 10s. New Jer- 400 gey, &c. to dollars. 3 100 Or 100 8 2 1200 + 100 3)800 200 100X|= 66| 20=4X5)1300 266| 10s.= H 10s. = lJ- 4)260 268 as be- Ans. 268 dol. [fore. £65 Answer. 6. To Canada and JVova Scotia IV. To reduce New York and K. currency. Carolina currency. R«i-le. Deduct one third from 1. To JVew Hampshire, Massa- the New Jersey, &c. chusetts, Rhode Island, Connecti' Reduce £100New Jersey, &c. cut, and Firginia currency. to Canada, &c. Rule. Deduct one fourth from 3)100 the New York, &c. ^ -^33 C 8 Reduce £100 New York, &c. to New Hampshire, &;c. £66 13 4 Ans. 4)100 7. To Livres Tournois. —25 Rule. Multiply the New Jer- sey, &c. pounds by 14, and the product will be Livres Tournois, — or multiply the sum in shil- lifigs by 7 ; divide the product £75 Answer. 2. To New Jersey, Pennsylva- nia, Delaware^and Maryland cur- rency. 118 REDUCTION OF COINS, Rule. Deduct one sixteenth from the New York, &c. sum. Reduce £100 New York, &c. to New Jersey, &,c. 16=4x4)100 4)25 — £6 6 £93 15 Answer. 3. To South Carolina and €leor- gia currency. Rule. Multiply the New York, &c. sum by 7, and divide the pro- duct by 12: The quotient is South Carolina, &c. Reduce £100 New York,&c. to South Carolina, kc, 100 7 12)700 £58 6 8 Answer. 4. To English Money. Rule. Multiply the New York, kc. sura by 9 : Divide the pro- duct by 16, and the quotient is English. Reduce £100 New York, kc. to English Money. 100 9 100 6X6-f thrice the giv. sum. 600 6 3600 4-300=100x3 64=8X8)3900 8)487 10 £60 18 9 Ans. 6. To Canada and J^ova Scotie. currency.. Rule. Multiply the New York, &c. sum by 5, and divide the pro- duct by 8. Reduce £100 New York, &c. to Canada, &,c. 100 8)500 £62 10 Ans. 7. 7\) Livres Tournois. Rule. Multiply the,NewYork» &c. sum in shillings by 21 : Di- vide the product by 32, and the quotient will be livres, sous, &c. Reduce £lOO New York, kc. to Livres Tournois. 100 Note. 20 ld.=l^%sou. 16=4X4)900 ls.=13isou. 2000 ll.=13}liv. 4)225 21 £o6 5 Answer. 2000 # 5. To Irish Money. 4000 Rule. Multiply the New York, kc sum by 39 : Divide the pro- 32=4X8)42000 duct by 64, and the quotient is l- rish. 4)5250 Reduce £100 New York, &.c. to Irish money. Ans, 1312^ livres. REDUCTION OF COINS. 119 8. To Spanish milled Dollars Rule. If the New York sum be pounds only, annex a cypher to ihem, then divide by 4, and the quotient is dollars : But if it be pounds and shillings, annex half the shillings to the pounds, and divide as before, and the quo- tient is dollars. Reduce £100 New York, &c. to Dollars. 4)1000 250 Doll. Ans. Reduce £100 8s. to Dollars. 4) 1004 251 Dolls. Ans. V. To reduce South Carolina and Georgia currency. 1. To JVesy Hampshire^ Massa- chusetts, Rhode Island, Connecticut and Firginia currency. Rule. Multiply the South Ca- rolina, &c. sum by 9, and divide the product by 7. Reduce £100 South Carolina, &c. to New Hampsliire, kc, 100 9 7)900 1 1 5} Ans. £128 2. To New Jersey Pennsylva- nia, Delaware and Maryland cur- rency. Rule. Multiply the South Car- olina, &.C. sum by 45, and divide the product by 28. Reduce £100 South Carolina lie. to New Jersey, &c. 100 9X5=45 900 5 28—4x7)4500 4)642 17 H 3. To New YorJz and North Carolina currency. Rule. Multiply the South Car- olina, &c. sum by 12, and divide the product by 7. Reduce £100 South Corolioa, &c. to New York, &c. 100 12 7)1200 £171 8 Gf Ans. 4. To English Money. Rule. From the South Caroli- na, &:c. sum, deduct one twenty- eighth. Reduce £100 South Carolina, &.C. to English Money. 28=4X7)100 4)14 5 8f 3 11 54 from 100. £96 8 6f Ans. 5. To Irish Money. Rule. Multiply the South Ca- rolina, 4*c. sum by 117, and di- vide the product by 112. Reduce £100 South Carolina, 4*c. to Irish. 100 12x94-9 times [the giv.sum. 1200 9 10800 4-100x9=900 112= =4X4X7)11700 4)1671 8 H 4)417 17 1* £160 14 3? Ans. £104 9 3 1 Ans. C. To Canada and Nova Scotia, currency. 120 REDUCTION OF COINS. Rule. Multiply the South Ca- rolina, 4*0. sum by 15, and divide the product by 14. Reduce JGIOO South Carolina, 4'C. to Canada, ^c. 100 14^=2x7)1500 2)214 5 84 £107 2 lOf Answer. 7. To Livres Tournois. Rule. Multiply the South Ca- rolina, 4'C. pounds by 22i, and the product will be livres. Reduce 1001. South Carolina, ^•c. to Livres. 100 Note. ld.=lf sous. 22i ls.=Ulivre. — "" 11. :22i livres. 200 200 50 Ans. 2250 livres. 8. To Spanish milled Dollars, Rule. Multiply the South Car- olina, ^c. pounds by 30, and di- vide the product by 7, and if there be shillings, turn them into dollars, and add them. Reduce £100 South Carolina, t^c. to Dollars. 100 10x3=30 7)3000 Dollars 428^. Note. 4—83. VI, To reduce English Money, 1. To New Hampshire, Massa- chusetts, Rhode Island, Connecti- cut, and Virginia currency. Rule. To the English sum, add one third. Reduce £100 English to New Hampshire, &c. 3)100 + 33 6 8 £133 6 8 Ans. 2. To New Jersey, Pennsyha^ nia, Delaware and Maryland cur- rency Rule. Multiply the English money by 5, and divide the pro- duct by 3. Reduce £100 English, to New Jersey, 4^0. 100 5 3)500 £166 13 4 3. To New York and North Carolina currency. Rule. Multiply the English money by 16, and divide the pro- duct by 9. Reduce £100 English, to New York, ^c. 100 4X4 400 4 9)1600 # £177 15 6| Answer. 4. To South Carolina and Geor- gia currency. Rule. To the English money add one twenty seventh. Reduce £100 English.to South Carolina, 4-c. REDUCTION OF COINS. 121 27^3X9)100 3)11 2 2 50 40 -f 3 14 Of £103 14 Of Ans. 6. To Irish Money. Rule. To the English sum adu one twelfth. Reduce JGlOOEnglish money to Irish Money. 12)100 + 868 £108 6 8 Ans. 0. To Canada and Nova Scotia currency. Rule. To the English sum add one ninth. Reduce £100 English, to Can- ada, &c. 9)100 4- 11 2 2| £111 2 2| Answer. 7. To Livres Tournois. Rule. Multiply the English pounds by 23i, and the product will be livres. Reduce £100 English to Li- vres Tournois. 100 Note. ld. = I|f'50us. 23| Is =lJ|ivre. £l=23ilivres. 300 200 331 Liv. sou. den. Ans. 23331 Liv.=2333 6 8. 8. To Federal Money. Rule. Multiply the pounds, or pounds and decimals of a pound by 40 and divide the product by 9, and the quotient will be the dolls or dollars and cents. 1. Reduce £50sterling to Fed- eral money. 9)2000 g222.22|cts. Ans. ^. Reduce £36 lOs. 9d. sterl- ling to dollars and cents. £36 10s. 9d.= £36.525, an 36.523X40 -,^^„„ :l_-==;4l62.33icts. Ans. 9 •* 3. Reduce £l sterling to Fed- eral money. Ans. ^4.44fc. 4. Reduce £1003.5 sterling to Federal money. Ans. 4460 VII. To reduce Irish Money. 1. To A''ezio Hampshire y Massa- chusetts^ Rhode Island^ Connecticut and Virginia currency. Rule. Multiply the Irish sum bv 16, and divide the product by lb. Reduce £100 Irish, to New Hampshire, &c. 100 4X4 400 4 13)1600 £123 1 Gj\ Ans. 2. To New Jersey^ Fennsylva' nia, Delaware and Maryland cur- rency. Kule. Multiply the Irish sum by 20, and divide the product by 13. Reduce £100 Irish to New Jersey, &c. 100 4x5=20 400 13)2000(163 16 11^3 Answer. ^ 122 REDUCTION Oi' C0iN:5. 3. 3o A'cw York and North Carolina currency, Kule. Multiply the Irish sum by 64, and divide the product by 39. Reduce -6100 Irish to New York, &c. 100 8x8=^64 800 8 s. 39)6400(164 2/^ Answer. 4. To Souih Carolina and Geor- gia currency. Rule. Multiply the Irish sum by 1 12, and divide the product by 117. Reduce £100 Irish to South Carolina, &;c. 100 7X4X4=112 700 4 2800 4 _, £ S. d. 117)11200(95 14 6-pVt Ans. 5. To English Money. Rule. From the Irish sum de- duct one thirteenth. Reduce £100 Irish to English money. £ s. d. 13)100(7 13 10-/3 100 —7 13 10/^ 92 6 Iji-Ans. 6. To Canada and JVova Scotia currency. Rule. To the Irish sum add one thirty ninth. Reduce £100 Irish to Canada, kc. ■ *. £ s. d. 39)100(2 11 3^% 100 . +2 11 3-/^ 102 11 3/^Ans. 7. To Livres Tournois. Rule. Multiply the Irish sum, in pence, by 70 ;"divide that pro- duct by 39, and the quotient wilt be sous, which, divided by 20. will be livres. Reduce £100 Iri&b to Livre.« Tournois. 1 00X20X1 2=24000d. 70 2-0- 39)1680000(4307|6 sou. Ans. Livres. 2153 16^1 ld.==lfisous. ls.=21J7_sous. £1=21 liv. lOifsous. VI [ I. To reduce Canada and No- va Scotia currency. 1. To JVew Hampshire, J^fassa- chusetis, Rhode Island, Connecti- cut, and Virginia currency. Rule. To the Canada, &c. sum add one fifth. Reduce £100 Canada, &c. to New Hampshire, &c. 5)100 + 20 £120 Answer. 2. To Nerv York and North Carolina currency. Rule. Multiply the" Canada, kc. sum by 8, and divide the pro- duct by 5. Reduce £100 Canada, kc. to New York, &c. 100 8 5)800 £1G0 Answer. REDUCTION OF COINS. 3. To KexR} Jersey^ Pennsylva- nia^ Delaware jUnd Maryland cur- rency. Rule. TotheCanada, &c. sum add one half. Reduce £100 Canada, ^c. to New Jersey', &c. 2)100 4- 50 ^150 Answer. 4. To South Carolina and Geor- gia currency. Rule. From the Canada, &c. sum deduct one fifteenth. Reduce £100 Canada, kc. to South Carolina, &c. 15=3X5)100 3)20 — 6 13 4 £93 6 8 Answer. ,5. To English Money. Rule. From the Canada, &c. deduct one tenth. Reduce £100 Canada, kc. to Knglish money. 10)100 — 10 £90 Answer. C. To Irish Money. Rule. From the Canada, &i,c, deduct one fortieth. Reduce £100 Canada, kc. to Irish money. 40)100 — 2 10 £97 10 Answer. 7. To Livress Tournois. Rule. Multiply the Canada, &c. pounds by 21, and the product will be livres Reduce £100 Canada, &c. to !'vres Tournois'. 100 7X3=21 700 ld. = l|sous. 3 ls.=21sous. £l=2Hivres. Ans. 2100 8. To Spanish milled Dollars. Rule. Reduce the Canada, &c. sum to shillings : Divide them by 5, and the quotient is dollars. Or, Multiply the pounds by 4, and the product is dollars : And if there be shillings turn them in- to dollars, and add them to the product. Reduce £100 Canada, kc. to dollars. 100 155 15 20 4 620 4- 3=153. Ans. $623 Ans. IX. To reduce Livres Tournois. 1. To JVew Hampshire^ Massa- chusetts., Rhode Island, Connecti- cut, and Virginia currency. Rule. Multiply the livres by 2: Divide the product by 35, and ihe quotient will be pounds. Or, Multiply the livres by 8 : Divide the product by 7, and the quo- tient will be shillings. Reduce 1760 livres to New- Hampshire, &c. currency. 1750 Or, 1750 2 8 35)3500(100Ans.7)14000 35 2|0)200|0 00 £100 as bef. 2. ToMrc York and J^orth Ca- rolina currency. 124 REDUCTION OF COINS. Rule. Multiply the livres by 32: Divide the product by 21, and the quotient will be shillings. Reduce 13121 livres to New York, &LC. currency. 1312-5 32 26250 39375 •20 21)42000(200|0 £100 Answer. 3. To JVew Jersey^ Pennsylva' nia, Delaware and Maryland cur- rency. Rule. Divide the livres by 14, and the quotient will be pounds. Or, multiply the livres by 10: Divide the product by 7, and the quotient will be shillings. Reduce 1400 livres to New Jersey, &c. currency. 1400 10 _- — , Or, 7)14000 14)1400(1001. 14 20)200j0 00 £100 Ans. 4. To South Carolina and Geor' gia currency. Rule. Multiply the livres by 2, divide the product by 45, and the quotient will be pounds. Or, deduct one ninth, and the remain- der will be shillings. Reduce 2250 livres to South* Carolina, &c. currency. 2250 Or, 2 9)2260 £ — 250 45)4600(100 Ans. 45 2|0)2000lO 00 JClOOasbef. 5. To English Money. * Rule. Multiply the livres by G: Divide the product by 7, and the quotient is shillings : Or, de- duct one seventh from the livres, and the remainder will be shil- lings. Reduce 2333ilivres to English money. Or, 7)2333^ — 333i 2l0)200|0 £lOOasbef. 2333-1- 7; 14000 2|0)200|0 Ans. £100 6. To Irish Money. Rule. Reduce the livres to sous, then multiply them by 39 : divide this product by 70, and the quotient will be pence. Reduce 2153 Hv. 16i| so. tc Irish money. 20 43076|| 39 387720 129228 710)168000|0 12)24000 2j0)200|0 £100 Answer. 7. To Spanish milled Dollars, or to Federal Dollars. Rule. Multiply the livres by 4 : Divide the product by 21, and the quotient will be Spanish or Federal Dollars. Reduce 1000 livres to dollars. 1000 Or, 1000 4 4 21)4000(190 \ ^p;;;2i)4ooo(i9p \ j^'j; 21 190=19011 $189 10 6 — s. d. q. 21)60(2 10 1 21 190 1G9 10 10 d. c. m. 2 REDUCTION OF COINS. 125 X. To reduce Spanish milled Dol- lars. 1. To JVew Hampshire, Massa- chusetts^ Rhode Island, Connecti- cut, and Virginia currency. Rule. Multiply the Dollars by 3, and double the right hand fig- ure of the product, for shillings ; the left hand figures are poands. Reduce 629 dollars to New Hampshire, &c. 529 3 £158 14 Answer. 2. To New York and North Carolina currency. Rule. Multiply the number of dollars by 4 : Double the right hand figure of the product for shillings, and the left hand fig- ures are pounds. Reduce 529 dollars' to New- York, &c. 4 £211 12 Answer. 3. To New Jersey Pennsylva- nia, Delaware and Maryland cur- rency. Rule. Multiply the number of dollars by 3, and divide by 8. Reduce 629 dollars to New Jersey, &c. 529 3 £ s.d. 8)1687(198 76 Answer. Or, 8)1587 Reduce 529 dollars io South Carolina, &c. 529 7 £198f Ans. 4. To South Carolina and Geor- gia currency. Rule. Multiply the number of dollars by 7, and divide by 30. 3|0)370|3 £l23if Answer. *5. To English Money, at 4s. 6d. per dollar. Rule. Multiply the dollars by 9, and divide by 40. Reduce 529 dollars to English money. 529 9 4;0)476|1 £119^V Answer. 6. To Canada and Nova Scotia currency. Rule. Divide the dollars by 4, Reduce 529 dollars to Canada, &c. 4)529 . £1321 Answer. 7. To Livres Tournois. Rule. Multiply the dollars by 5}, and the product will be li- vres. Or, multiply them by 2 J ; divide by 4, and the quotient wiH be livres. Reduce 100 Spanish dollars to livres. 100 500 100X1=25 Or, 100 21 4)2100 Ans. 525 livres. 625 as bef. * Note, that in England dollars are Bullion, that is, they are bought and* sold by weight, and their value varies as other articles of merchandize. Note. { 1 Cent = l^V Sous. ) M Dime =101 Sous. > (1 Dollar = 51 Livres. i l^fei DUODECIMALS. DUODECIMALS, OR CROSS MULTIPLICATION, IS a Rule, maile «pe of by workmen and artificers in casting up the contents of their works. Dimensions are generally taken in feet, inches and parts. Inches and parts are sometimes called primes, seconds, thirds, &c. and are marked thus; inches or primes ('), seconds ('), thirds ('"), fourths (""), &c. This method of multiplying is not confined to twelves ; but may be greatly extended : For any number, whether its inferiour de- nominations decrease from the integer in the same ratio, or not, may be multiplied crosswise • and, for the better understanding of it, the learner must observe, that if he multiplies any denomina- tion by an integer, the value of an unit in the product will be equal to the value of an unit in the multiplicand ; but if he multiplies by any number of an inferiour denomination, the value of an unit in the product will be so much inferiour to the value of an unit in the multiplicand as an unit of the multiplier is less than an integer. Thus, pounds, multiplied by pounds, are pounds ; pounds, multi- plied by shillings, are shillings, &c. shillings, multiplied by shillings are twentieths of a shilling ; shillings, multiplied by pence, are twentieths of a penny ; pence, multiplied by pence, are 240ths of 3 penny, &c. Rule.* }, Under the multiplicand write the corresponding denomina- tiwis of the multiplier. *■ The reason of this rule is evident by considering^ the denominations beluw the integer, as fractional parts of the integer, and multiplying as in Vulgar Frac- rions. Thus inches or primes aro 12ths of a foot, seconds are 12ths of an inch, cirl44tlis of a foot, and so on. Then feet multiplied by inches would give inchr 3 6 3 6 Xj2-^ ^'\ for 2 feet X —=—=6 inches ; inches by inches give seconds^ ^^^Tc) 18 n n-{-6 12 6 1 6 ^.o thirds ^Ives fourths Jor -^p< u.^=-^^7^Y:^.f=^j;7^+J^^^ r=zl* and G"', and so on. A similar process will show the correctness of the Rule, when the denomina- tions do not decrease uniformly by 12 or any one number, as in pounds, shillings >.,rA peace, where 1 shilling would be _.l. of a pound, and 1 penny _J , of a ^.ound, and so on. Note. It is evident that when the denominations decrease by any one num- ber, as 12, the denomination of the product is the sum of the denominations of Ihe factors. Thus primes into primes give seconds, 2 being the sum of l-f- 1, the ■ifraominations of the factors , seconds into thirds give f:i'tl'.s. 2-\-'j^-^i ; second": 'jIo fourths jive sixths, and io mi. DUODECIMALS. iST 2. Multiply each term in the multiplicand, beginning at the low- est, by the highest denomination in the multiplier, and write the result of each under its respective term, observing, in duodeci- mals, to carry an unit for every 12, from each lower denomination to its next superiour, and for other numbers accordingly. 3. In the same manner multiply all the multiplicand by the primes or second denomination in the multiplier, and set the re- sult of each term one place removed to the right hand of those in the multiplicand. 4. Do the same with the seconds in the multiplier, setting the result of each term two places to the right hand of those in the multiplicand. 5. Proceed in like manner with all the rest of the denomini^- tions, and their sum will be the answer required. Examples. I. Mnlti ply H feet by 2| feet. Or til F. ' 2r> 2 6 Or thus. 2o 2 6 21 21- 2 125 5 60 1 3 11 Ans. 6.25 Ans. 6 3 — Ans. 6i square feet = 6ft. 36in. iSo that the 3 is not 3 inches, hxf. 36 inches, or | of a square foot, 2, Multiply 9f. 8' 6" by 7f. 9' 3'' F. ' " 9 8 6 7 9 3 67 11 6 ~ Product by the feet in the multiplier. 7 3 4 6"' = ditto by the primes. 2 6 1 6"" = ditto by the seconds. 75 5 3 7 6 Answer. 3. How many square feet in a board 17 feet 7 inches long, and 1 foot 5 inches wide ? Ans. 24ft. 10' 11' 4. How many cubick feet in a stick of timber 12 feet 10 inches long, 1 foot 7 inches ^vide, and 1 foot 9 inches thick ? Ans. 35ft. 6' 8'' 6"' 5. How many cubick feet of wood in a load 6 feet 7 inche? long, 3 feet 5 inches high, and 3 feet 8 inches wide ? Ans. 82ft. 6' 8" 4" 6. There is a house with 4 tiers of windows, and 4 windows irr a tier; the height of the first tier is 6ft. 8'; of the second, 5ft. 9' ; of the third, 4ft. 6'; and of the fourth, 3ft. 10'; and the breadth of each is 3it. 5' ; how many square feel do Ihey contain in th*' '^hole ? An3, ^ooft. 7 J 28 DUODECIMALS. The two following questions are Sexcesimals. 7. If 2 places diflfer in longitude 2** 12' ; what is their diflerence of lime ? Mult. £° 12' 00" 00'' by 3' 69' 20'" the time in which the sun passe? through 1' 8' 46" 32'" Answer. 8. Two places differ in longitude 3P 27' 30"; What is the dif ference, in time, of the sun's coming to the meridian of those pla- ces, the sun passing through 16** m an hour? 31° 37' 30" 4' 00" In 4' of a solar day, or day of 24 hours, the sun passes 1* £«» 6' 30" 00'" Answer. 9. Bought a load of wood, which was 3 feet wide, 2 feet 8 inch- es high, and 8 feet long ; what part of a cord of wood did it con- tain ? Ans. Half a cord. 10. A load of wood was 4 feet 6 inches wide, 3 feet 10 inches high, and 7 feet 8 inches long ; how many feet more than a cord did it contain ? Ans. 4i feet. 11. A stick of timber is 1 foot 8 inches in depth, and 2 feet 3 inches in width, and 42 feet 8 inches long ; how many solid feet of timber does it contain? Ans. 160. 12. Multiply £3 6 8 by £2 5 7. £ s. d. 3 6 8 2 5 7. ■ 13. A, Band C bought a drove of i^3X-£2=£6 =600 sheep in company ; ApaidJ£l46s. t>s.XJe2=12s. = 12 B, £13 10s. and C, £11 6s. They 8d.x£2=16d. =014 agreed to dispose of them at the £3x6s. =15s. = 15 market; that each man should take l3S.x6s. =|^s. = 16 18s. as pay for his time, &c. and that $d.x6s. =|^d. =002 the remainder should be divided in £3x7d. =21d. = 19 proportion to their several stocks : 6s.X7d. =f|d. = 00 2j\ At the close of the sale, they found 8d.x7d. =^Yo<''= 0^'V themselves possessed of £46 5s. __. what was each man's gain, exclu- Ans. 7 11 111 sive of the pay for his lime, &;c. £14 6-|-£l3 10-|-£11 6=£39, and £46 5— £.39=£7 5, and £7 5~18s.x3=£4 lis. whole gain, and £4 ll"~39=2s. 4d. gain m the pound. £14 5 £13 10 £115 X24 X2 4 X24 £s. d. il 13 3 MUG 8 6 17 12 6 Proof. .... 4<>' 46 3 9 (163 A.£l 13 3 B.£l 11 6 C.£l 6 3 £4 11 SINGLE RULE OF THREE. 12? THE SINGLE RULE OF THREE, IS SO calleiK because three numbers are given to find a fourth, which shall have the same ratio to one of the given numbers, as there is betwieen the other two. It is usually distinguished into Direct and Inverse. The reason of this distinction, and the par- ticular rules, will be given hereafter. It will be more easy how- ever, for the student to proceed according to the following General Hule for stating and working questions in the Rule of Three. General Rule* 1. Place that number, which is of the same name or quality as the answer sought, for the second term. 2. Consider whether the answer should be greater or less than the second term. If it must be greater, place the greater of the two remaining numbers in the question on the right for the third * This Rule, on account of its j^rent and extensive usefulness, is sometimes called the Golden Rule •/ Proportion : For, on a proper application of it and the preceding rules, the whole business of Arithnietick, as well as every mathe- matical enquiry depends. The rule itself is founded on this obvious principle, that the magnitude or quantity of any effect varies constantly ih proportion to ♦he varying part of the cause : Thus, the quantity of goods bought, is in pro- portion to the money laid out ; the space gone over by an uniform motion, is iu proportion to the time, &;c. * As the idna, annexed to the term, proportion^ is easily conceived, the truth of the rule, as applied to ordinary inquiries, may be made evident by attending t<"> principles, already explained. It has been shewn, in IVlidtiplication of Money, that the price of one, multi- plied by the quantity, is the price of the whole ; and in Division, that the price of the whole, divide s. d. As 8 : 537 :: 3 : 201 7 6 Ans. 26. If j\- of a ship cost gll63 what is the whole worth? §2058 28c. 5ai. An5J. 27. Bought a cask of wine at 7Gc. 5fn. per g^allon, for §125: llow much did it contain ? 163o;al. Iqt. 1 ,\pt. Ans. 28. What comes the insnrance of JC537 155. to, at£4i per cent- nm? £ £ £ s. £ s. d. As 100 : 4} :: 537 15 : 24 3 111 ^\ Ans. 29. What come the commissions of £785 to at 3^ guineas per cent? £38 9s. 3^, /-d. Ans. 30. A merchant honght 9 pachages of clolh, at 3 gtiineas for 7 yards: each package contained 8 parcelts, each parcel 12 pieces, and each piece 20 yards ; how many dollars came the whole to, and how many per yard ? Vds. guin. pack. § As 7 : 3 :: 9 : 34560 Ans. for the whole cost. Yds. guin. yd. § As 7 : 3 :: 1:2 Ans. per yard. 31. A merchant honght 49 tuns of wine for §910 ; freight co«t |90 ; duties §40 ; cellar §31 G7c. ; other charges §50 and he would gain §185 by the bargain ; what must I give him tor 23 tuns " 'I'un^- $ $ $ ^$ ^- $ $ ^'"n«. $ As 49 : 9 10+90-f 40-4-31 67-f 50-f 185 :: 23 : 613 33c An^. 32. If §100 gain §6 in a year, what will §475 gain in that time " Ans. §28 50c. 33. The earth being 360 degrees in circumference, turns round on its axis in 24 hours ; how far does it turn in one minute, in the 43(1 parallel of latitude ; the tlej^ree of longitude, in this latitude, being about 51 statute miles? II. D. M. M. M. As 24 : 360 X 51 :: 1 : 12^3 Ans. 34. Shipt for the West Indies 225 quintals offish, at 15sv 6d. per quintal ; 37000 feet of boards, at 8i dolls, per 1000; 12000 shin- gles, at 1 guin. per 1000 ; 19000 hoops at §1^ per 1000, and 53 half joes ; and in return, I have had 3000 galls, of rum, at Is. 3d. per gallon ; 2700 gallons of molasses, at 5^4. per gallon : ISOOife oi coffee, at 8^4. perih; and 19cwt. of su^ar, at 12s. 34. percwt. and my charges on the voyage were £37 12s. pray, did I gain or lose, and how much by the voyage ? Ans. lost £134 9s. 9d. 35. If a statr, 4 feet long, cast a shade (on level ground) 7 feet ; what is the height of ihat steeple, whose shade, at the same time, measures 198 feet ?^ F.sh. F.hei. F.sh. F.hei. As 7:4:: 190 : 113} Ans. * As the ra*ys of lij;ht from the sun may be considered parallel, the lengths o: the shadows must be proportioned to the hrigh/t cf tho ob'f>r'<. H'-ii?^ th" r.-a son of the; statement of this question. U^Q SINGLE RULE OF THREE. *36. Suppose a tax of ^755 be laid on a town, and the intentdry of all the estates in the town amounts to ^9345, what must A pay ^vhose estate is gl49? $ $ $ $? . c. m. As 9345 : 755 :: 149 : 12 12 7 Ans, * It may not be amiss to show the general method of assessing town or parish tiaxes. Y'irst, then, an inventory of the v^lue of all the estates, both real and per- sonal, arid the number of polls for which each person is rateable, must* be taken in separate columns : The most concise way is then to make the total value of the inventory the first term, the tax to be assessed the second, and $1 the third, and the quotient will show the value on the dollar : 2dly, make a table, by mul- tipliplying the value on the dollar by 1, % 3, 4, &;c.^-3dly. From the inventory take the real and personal estates of each man, and find them separately in the tUble, which wul shew you each man's proportional share of tho tax for real and personal estates. JVote. If any part of the tax is averaged on the polls, or otherwise, before stat- ing to find the value on the dollar, you must deduct the sum of the average tax from the whole sum to be assessed : for which average you must have a sepa- ! ate column, as well as for the real and personal estates. EXAMPLK. Suppose the General Court should grant a tax of o!: ,11 25o. 03 ch fre 3 7i 75 -f$10 2.2=:r$22 68c. 7im. or, $22 6Cc. An« SINGLE RULE OF THREE. 137 31. If 50 gallons of water, in one hour, fall into a cistern, con- taininsf 230 gallons, and by a pipe in the cistern 35 gallons run out in an hour ; in what time will it be filled? Ans. lo^h. 38» A butcher went with £41G, to buy cattle : Oxen, at £22 each, cows atJ£4, steers at £3 lOs. and calves at £2 10s. and of each a like number ; how many of each could he purchase with that sum ? Ans- 13 each. 39. Said Harry to Dick, my purse and money are worth 3| guin- eas, but the money is worth eleven times as much as the purse ; pray, how much money is there in it? Ans. £4 3s. 5d. 40.* If I of a yard cost | of a j£, what will -pj of a yard cost ? A^ f : i :: tV • fX^-^f ^fH^ Answer. 41. There is a cistern, having four cocks ; the first will empty it in ten minutes ; the second in 20 minutes ; the third in 40, and the fourth in 80 minutes j in what time will all four, running to- gether empty it ? i}^) Cist. xMin. C^ ) Cist. Min. Cist. Min. '^■^ \li)} : 1 - CO : /"rj^V As 111 : 60 ;; i ; 5} Ans. \^^J \ i) that is-r- : 60 :: 1 : -pr-=5f 4 4o * 11} Cist. 42. A and C depart from the same place, and travel the same road ; but A goes 5 days before B, at the rate of 20 miles per day ; B follows at the rate of 25 miles per day : In what time and dis- tance will he overtake A? M. M. I). M. D. D. D. M. D. M. As 25—20 : 1 :: 20x5 : 20. And, As 1 : 25 :: 20 : 500 43. If the earth revolves 36(> times in 305 days, in what timfj does it perform one revolution ? Ans. 23h. 56' 3" 56"'-f = 1 Sidereal day.t 44. If the earth makes one complete revolution in 23h. 56'3"-{-, in what time does it pass through one degree ? Ans. 3' 55" 20'" 45. If the earth performs its ditirnal revolution in a solar day,;}; or 24 hours ; in what time does it move one degree ? Ans. 4' 46. Sold a c;Mgo of fiax j^eed in Ireland, for jC 1 795 10s. Iri«h money ; what does that amount to, in Massachusetts currency, £81 6s. Irish being equiil to £100 Massachusetts. Ans. £2209 I63. lid. * If the first term of \he statement be a Vulg-ar Fraction, whether the other terms are or not, after the first and third terms are reduced to the same dc nomi- nation, invert the fjrst term as in division of Vulg'ar Fractions, and the product of the three terms will of course be t'ne answer. The student should work the questions in V'ulg'ar, or Decimal Fractions, ac . cording as the rules for fractirtus require. t A sidereal day is tlic space of time which happens between the departure of a star from, and its return to the same meridian again. I The solar day is that space of time wliich intervenes between Boston for 3750 Num. £ s. As 8 : 12 :: 5 : 7 10 Ans. If the question were wrought with the fractions, it is evident that the denominator would belong both to the dividend and divisor, and thus destroy each other. Then in the example under Rule 1. the statement would be, 9xH As Jf : 9 :: -'*- : the answer=f|x9xiJ-=— -. i o And under Rule H. the statement would be, 8x12x5 12x5 12 :: # : answer=- 8 8 8 Whence the reason of the rules. ()5. If '625 of a yard cost £ 25, what will 4'75yds. come to ? Yds. £ Yds. 4-75X-25 ^ -^ ^^ '^"5 : -SQ :: ^75 : > ,^^^ -c^l-9=:l 18 Ans. 140 SINGLE RULE OF THREE. 66. If 9-75y(ls. cost ^11 25c. what will -Syds. cost? Ans. 57c. eifm. 67. There is a cistern, which has 3 cocks ; the lirst will empty it in -25 hour, the second in -75 of an hour, and the third in 1*5 hour : in what time will it be emptied if all three run together? H. Cist. H. Cist. ^ (-25 : 1 :: 1 : 4 As ? -75 : 1 :: 1 : 1-3334- ( 1-5 : 1 :: 1 : 0-6G7-- 6 Cist. N Cist. il. Cist. M. H. m. As G : 1 :: 1 : |=:0 1G7— = 10 Ans. G8. A conduit has a cock, which will fill a cistern in '2 of an hour : this cistern has 3 cocks ; the first will empty it in 1-25 hour, the second in '025 of an hour, and the third in '5 hour. In what time will the cistern be filled, if all four run together? Ans. Ih. 40m. GO. If 19yds. cost $25 75c. what will 435-5yds. come to. Ans. ;^590 21c. 7^,-m. 70. If 345yd3. of tape cost g5 17c. 5m. what will one yard cost? Ans. Oc. Im. 'd. 71. If I give gl2 82c. 5m. for 675 tops, how many taps will 19 mills buy ? Ans. 1 top. 72. If when wheat is <^1 per bushel, the two penny loaf weigh 9Goz. what ought it to weigh when wheat is gl 25c. per bushel? Ans. 7oz. ISpwt. 14grs. 73. How much in length, that is 8^ inches broad, will make a foot square ? Ans. l6p^ inches. 74. What number of men must he employed to finish in 9 days» what 15 men would perform in 30 days? Ans. 60 men. 75. If 9 men can btiild a wall in 5 months by working 14 hours a day, in what time will the same men do it, when they work only 10 hours a day ? Ans. 7 months. 76. How many yards of carpet, 2| feet wide, will cover a floor, which is 18 feet long and IG feet wide ? Ans. 34ifyds. 77. If 745 soldiers are to be clothed, and each suit is to contain 31yds. of cloth If yd. wide, and to be lined with shalloon |yd. wide : houMuany yards of shalloon will be necessary ? Ans. 4097|yds, 78. If a man count 100 cents in a minute for 10 hours in a day : in how many days will hp count a million of cents? Ans. 1G| days. 79. Proceeding to coimt at the same rate as in the last question ; how many men must be employed for 100 years of 3G5 days each, to count niie trillion ?. " Ans. 45GG21 004^^1 m^n, 80 The number of inhabitants on the earth is computed to be 750000000 ; suppose they had each counted one lor every second fiom tlie creation to this time or GOOO years of 305 days each : how many would they have counted ? Ans. 14I9J12000 billions. SINGLE RULE OF THREE. 141 SI. In a certain school, J^th of the pupils study Greek, ^J^ study Latin, f study Arilhmelick, \ read and write, and 20 attend to oth- er things ; what is the number of pupils ? oV+rV+l+i^A, then 20=/^ and j\ : 20 :: if : 100. Ans. To find the value of Gold in Massachusetts currency- Prob. 1. Given the weight of any quantity of gold, to find its value. Oz. £ Oz. je pwt. s. gr. d. 2| Theorem 1 . As 1 : 5i : : 12 : 64 :: 1 : 6-» :: 1 :: 2|(Case 1.)=— (Case 2.)=^ (Case 3.)=f, Therefore, Rule 1. — If the given quantity be in grains ; say, As the denom- inator is to the number of grains ; so is the numerator to their va- lue in pence. 1. What is the value of 18 grains of gold? By Case 1. By Case ?. By Case 3. Gr. Gr. Gr. As 1 18 :: ^ As 2 ; : 18 :: 51 As 3 : 18 :: 8 02 ^ 8 36 90 3)144 12 6 --_ — 48d.=4 12)48(4s. Ana. 2)96(48d.=4s. Rule 2. — If the given quantity consist of ounces, pennyweights, and grains, halve the grains, and then proceed as in multiplication of pounds, shillings and pence, making the numerator in Case 2d, the multiplier. 1. What is the value of 7oz. 8pwt. 16gr. of gold? Gr. gr. oz. pwt. gr. 16-r-2 = 8, then, 7 8 8 ^i 37 2 3 9 4 ^ £39 12 lOf Ans. Rule 3. — If the given quantity consist of poumls only, multiply by 64, and the product will be the answer ; but, if it consist of pound?, ounces, &-c. it will be most convenient to reduce the pounds to ounces, and proceed by Rule 2. 1. What is the value of 36!b. of gold, at £64 per lb. ? 64 144 216 £2304 Ans. 4;^ SINGLE RULE OF THREE. 2. What is the value of 16lb. 903. 12pvvt. 18gr. of gold ? 12 —- pwt. gr. gr. oz. 189 12 9 == 18-~e 948 3 9 63 4 3 £1011 , 8 Aus. pROB. 2. To ascertain the value of any given quantity of gold ic Spanish milled dollars, or federal money. Theorem 2. Ipwt. of gold — 5Js. 1 dollar «= 6s. And, "^=f^=|. Therefore, Rule. Reduce the given quantity of gold to pennyweights; then, as the denominator is to the given quantity ; so is the nume- rator to the answer in dollars. Or, Divide by the denominator, and multiply the quotient by the nu- merator. Or, Divide by the denominator and subtract the quotient from the dividend. In either case, you will have the answer. 1. What is the value of 6oz. 6pwt. of gold, in Spanish dollars ? 20 pwt. As 9 : 126 :: 8 |26 pwt. Or, 8 ^^)]26 Or, —14 9)1008 9)126 112 Ans. Ans. 112 Dolls. 14x8 = 112Ans. 2. In 7oz. 13pwt. 17gr. how many dollars ? oz, pwt. gr. 7 13 20 24 619 307 3689 ^16)29512 216 791 648 3432 1296 SINGLE RULE O^ THREE. 143 To find the value of this remainder. 1. In shillings, &c. 2. In Federal Money. 136 Annex cyphers, as in division 6 of decimals ; the two quotient, places next to dollars, will be 216)816(3s. cents; the third, mills ; the oth- 648 ers, decimals of a mill ; or the . remainder with the divisor, will 168 form a fraction of a mill. 12 216)1360(62c. 9Um 1296 216)20]6(9d, 944 640 432 72 4 2080 1944 216)288(liqr. 216 113. — 1 " 2 10 2'i' 2 16 ¥ Prob. 3. To ascertain the weight of gold equivalent to any given sum, currency. Rule 1. If the given sum be in pence, reverse Rule 1. Theo- rem 1. that is ; As the numerator 8 is to the given sum in pence -. 90 is the denominators to the weight required, in grain«. What weight of gold is equal to 4s. ? d. 12 As 8 : 48 :: r> — 3 4r. 8)144 Ans. 18 grains. Rule 2. If the given sum be in pounds, shillings and pence. As 1 is equal to y ; therefore, divide the given sum by 8, and that quotient by 2 ; add the two quotients together, double the last denomination, and you will have the answer. What quantity of gold is equivalent to £45 13s. 4d. oz. pwt. gr. Mark the pounds, shillings and ^ 8)45 13 4 pence, as oz. pwt. and gr. ^ 2)5 14 2> . ,, 2 17 1 ^ ^^'^'^' 8 11 3+: Oz. 8 11 6 Ans. Pros. 4. To find the value i>f gold equivalent to ar^y^given sujr in Federal money, 144 RULE OF THREE DIRECT. Rule. As the numerator 8 is to the number of dollars ; so is the denominator 9 to the answer in pennyweights : Or, divide the dol- lars by the numerator 8, and add the quotient to the dividend. Or, divide as before, and multiply the quotient by the denomina- tor 9. In either case you will have the answer. 1. Required the weight of gold equal to 16 dollars. As 8 : 76 :: 9 Or thus, 8)76 Or, 9-ix9=85Apwt. 9 9i 8)684 Ans. 85ipwt. oz. pwt. gr. Ans. 85ipwt.=4 5 12 2. Reqiiired the weight of gold equal gl59 7uc. As 8 : 159 75 :: 9 : 179pwt. 17igr. Ans. 9 8)1437-75 179-71875 24 287500 143750 17-25 grains. Or, 159-75>^8 -f- 159-75 = 179pwt. 17]gr. An?. Or, 159-75-r-8 X 9 = 179pwt. 17igr. as before. RULE OF THREE DIRECT AND INVERSE. Though Direct and Inverse Proporlion, are properly only parts of the same rule, yet for the use of those who may desire i(, the common distinctions will be made and the common rules given. The Rule of Three Direct teaches, by having three numbers given, to find 2i fourth, which shall have the same ratio to the se- cond, as the third has to the Jirst. The Rule of Three Inverse teaches, by having three numbers given, to find a fourth, which shall have I he same ratio to the se- cond, as the first has to the third. It is also called reciprocal or indirect proportion. If more require more, or less require less, the question belongs to the Rule of Three Direct. But if more require hss, or less require more, the question belongs to the Rule of riupe Inverse. The principal ditficully, which will embarrass the learner, will be to distinguish when the proportion is direct, and when inverse. This must be done by an attentive considf ration of the qut.'tion proposed. For more requires more, wlien the third term is greater than the tirst, and the question requires the fourth term to be greater than the second ; and less requires less, when the third term i-- less ihau the tirst, and the fourth is required to be lets than the second. RULE OF THREE INVERSE. 145 More 13 said to require less^ when the third term is greater than the first, and the question requires the fourth to be less than the second ; and less requires mure, when the third term is less than the first, and the fourth is required to be greater than the second. RULE OF THREE DIRECT. Rule. 1. State the question by making that number, which asks* the question, the third term ; that which is of the same name or quality as the demand, the first term ; and that, which is of the same name or quality with the answer required, the second terra. 2. Divide the product of the second and third terms by the first term and the quotient will be the answer. JVote. The directions under the General Rule, as well as the demonstration, apply to this rule. Examples. X. If 61bs. of sugar cost lOs. what will 331bs. cost at the same rate ? lbs. s. lbs. As 6 : 10 :: 33 : the answer. 10 6)330 65s. = £2 15s. Ans. In this example 331bs. asks the question, and is made the third term ; 6lbs. being of the same quality, is made the first term ; and 10s. being of the quality of the answer required, is placed for the second term. To invert the question, say, s. lbs. s. lbs. As 10 : 6 :: 55 : 33 the Ans. 2. If 100yds. of cloth cost $66 what will 1 yard cost? Ans. 66c. 3. If my income be Jl75t) a year, and I spend 19s. 7d. a day, how much shall I have saved at the end of the year? Ans. £167 12s. Id. RULE OF THREE IJVFERSE, OR RECIPROCAL PROPORTIOA', RuLE.t State and reduce the terms as in the Rule of Three Direct ; then, multiply the first and second terms together, and divide the product * The term which asks or moves the question, has generally some -words like these before it, viz. What will? what cost? How many? how long? how much ? &;c. t The reason of this rule may be explained from the principles of Compound Maltiplicatioa and C^mpouiitl Dinaion, in th* same manner as the direct rule.— 146 RULE OF THREE INVERSE. by the thinl ; the quotient will be the answer in the same denonvina tion as the middle term was reduced into. If there be fractions in your question, they must be stated as be- fore directed, and if they be vulgar, invert the third term : Then multiply the three terms continually together, and th6 product will be the answer. Examples. 1. How mucfi shalloon, that is f yard wide, will line 6| yards ef cloth which is 1} yard wide ? vd. yds. qrs. qrs. qrs. qrs. As *|i : 6f ;: 3 As 5 : 27 :: 3 4 4 6 5 27 3)136 4)45 Hi yards, Answi The same by Vulgar Fractions. First. li~f, 6i=\\ and 3qrs.=f. Then, 5x27x4 As I : V :: I And |xVX|=-^f^^5^-=W=V = IUj'^s- Ans. The same by Decimal Fractions. 1^=:1.25, 6f==6-75 and 3qrs. = '75. Then, As 1-25 : 6-75 :: '75 1-25 3375 1350 675 ■' 2. What length of board 7 ^75)8-4375(1 l-26yds. Ans. inches wide, will make a square 7 5 foot? In.br. in.len. in.br. in.len. As 12 : 12 :: 7| : J9i Ans. For example^ if 4 men can do a piece of work in 12 days, in >vhat time will 8 ineu "** " ' As 4 men : 12 days :: 8 men : ^=6 days, the Answer. And here the product of the first and second terms, that is, 4 times 12, or 48, is evidently the time in which one man would perform the work. Therefore, ii men will do it in one eighth part of the time, or 6 days. RULE OF THREE INVERSE. U7 3. Suppose I lend a friend £350 for 6 months, he promising the like kindness ; but, when requested, can spare but £125, how long- may I keep it to balance the favour ? £ Mo. £ Mo. As 350 : 5 :: 125 : 14 Ans. 4. Suppose 450 men afe in a garrison, and their provisions are calculated to last but 5 months ; how many must leave the garrison, that the same provisions may be sufficient for those who remain 9 months ? Mo. M. Mq. M. M. As 5 '450 :: 9 : 250, and 450—250=200 men, Ans. 5. If a nian perform a journey in 15 days, when the day is 12 hours long, in hovy many days will he do it, when the day is but 10 hours ? Ans. 18 days. 6. If a piece of land, 40 rods in length, and 4 in breadth, make an acre, how wide must it be, when it is but 19 rods long to make an acre ? Ans. Crods 6ft. HyV"- 7. If a piece of board be 30 inches in length, what breadth will make 1| square foot? Ans. 7-2 inches. 8. A \yal1, which was to be built 24 feet high, was raised 8 feet by 6 men, in 12 days : llow many men must be employed to finish the wall in four tlays ? n. ft. m. And, ft. m. ft. m. As 8 : q :: 24—8 : d. m. As 12 : 12 ; 12 to finish it in 12 days. d. m. :: 4 : 3G to finish in 4 days. 9. There is a cistern having a pipe, which will empty it in 6 hours : How many pipes of th^ same capacity will empty it in 20 minutes ? h. pi. mi. pi. As G : 1 :: 20 : 18 Ans. 10. If a field will feed G cows 91 days, hovy long will it feed 21 cows ? Ans. 26 days. 1 1. How much in length, that is 13| poles in breadth, will make a square acre ? Ans. lly^J*- poles. 12. If a suit of clothes can be made of4i yards of cloth, 1| yard wide ; how many yards of coating-^ of a yard wide, will it require for the same person ? Ans. 6yds. Iqr. 340. Abbreviations. To kno-^ rvhether a fraction, when abbreviated^ be equivalent in all respects to the 07'iginal fraction. Rule. As the numerator qi' the fraction, in its lowest terms, is to its de- nominator ; so will the numerator of the original fraction be to its own denominator. Or, as one numerator is to the other ; so will one denominator be to the other, &c. A owes B £75 13s. 6d. ; now £100 of A's money is equa!to£l40 of B's ; what must A pay to satisfy the said debt? I 148 DOUBLE RULE OF THREE, £ s. d. ]-U=h therefore, 75 13 6 5 7)378 7 G £54 1 Of Ans. Now, to prove whether ^ be equal |||. Num. Den. Num. Den. Num. Num. Den. Den. As 5 : 7 :: 100 : 140 Or, as 5 : 100 :: 7 : 140. All questions in the Rule of Three Direct or Inverse, may be wrought by the following Rule. State the questions as directed in the Rule of Three Direct ; then multiply the second term by the first or third term according- ly as the answer ought to be greater or less ; divide the product by the other term, and the quotient will be the answer. COMPOUND PROPORTION, OR DOUBLE RULE OF THREE, TEACHES to resolve such questions as require two or more statements by Single Proportion, and hence its name. There is always an odd number of terms given, as five, seven, &c. All questions in Compound Proportion may be stated and wrought by the following General Rule.* 1. Place that term, which is of the same kind or quality with the answer sought, for the second term. 2. Then, of the two terms in the question of the same kind, place the greater or less on the right for the third term, and the other on the left for the first term, according to the directions un- der the General Rule for Simple Proportion. Arrange the two remaining terms under the first and third, on the same principle. 3. Find the fourth term from the first statement, and place it for the second term in the second part of the statement, and find thp fourth term from this statement, and it will be the answer required- Note. If there be more than five terms in the question, the sam^i mode of statement must be continued, and a third proportion form- ed, and so on, and the fourth term found from the last statement, will be the answer as before. * This nile is evident from the General Rule of Three, for each statement is a particular statement under that Rule. If, then, all the separate dividend?* be collected into one dividend, and all the divisors into one divisor, their quo- tient must be the answer. Thus, in Ex. 1 . D. Int. D. Int. M. Int. M. 400X6 400X6 400x6x9 As 100 : :: 400 : -j^, and as 12 :-^y - :: 9 : l^]o^,-=$lB Int. Aus. DOUBLE RULE OF THREE. 149 Examples. 1. If a principal of glOO, gain $6 interest in one year, what will ^400 gain in 9 months. Statement and Operation. D. Int. D. M. Int. M. As 100 : 6 :: 400 Or, 12 : 6 :: 9 M. M. D. 12 : :: 9 100 : :: 400. 400X6 400x6 Then 100 : 6 :: 400 : ^^^ =24. And 12 : -J^ : 9 : 18 Int. Ans. Or 12 : 24 : 9 : 18 Ans. 6X9 6X9 6X9X400 Then 12 :6::9 : — =4^ and 100 : -j^ ^ 400 : -j^^- = 1 8 Ans. Or 100 : 4^ : 400 : 18 Int. Ans. In this question, the answer sought is interest, and therefore $(> must be the second term. As ^400 will gain more interest in the same lime than glOO, ^^400 must be placed on the right for the third term, and glOO on the left for the first term. And as the sam>2 sum will gain more interest in 12 months than in 9 months, the 9 must be placed under the third term, and the 12 under the first term. The operation is obvious on inspecting it. Note. Instead of working two proportions, the whole may be reduced to one, by multiplying the Jirst terms together, and also the third terms, and using their products for the first and third terms. This is merely changing the order of the operation, as *-wilI be seen in the preceding example. D. Int. D. 100 : 6 :: 400? ^,.. , . , ,, 12 • •• 9 ( becomes, evidently, 400X0X6 100X12 : 6 :: 400x9 : ^^^^^ -=18, as before. The work may also frequently be contracted by dividing thes f.rst and third terms by a common divisor, or the ^;^r5/ and second terms, and using their quotients, for the divisor will diminish thfe terms in the same ratio, and the proportion be still preserved. 'J hus, in the preceding example, 100 : 6 :: 400 becomes 1 : 6 :: 4, by dividing by 100. 12 : :: 9 4 : :: 3, 3. And . I ■' „ [ becomes 1 : 6 : 3 : 18, Ans, as before. Ex. 2. If 950 soldiers consume 350 quarters of wheat in 7 months, how many soldiers will consume 1464 quarters in 1 month? Ans. 27816 soldiers. Ex. 3. If 1464 quarters of wheat be used by 27816 soldiers in a month, in what time will 950 soldiers consume 350 quarters ? Ans. 7 months. Ex. 4. If 144 men, in 6 days of 12 hours each, dig a trench 200 feet long, 3 wide and 2 deep, how many hours long is the day, iov I>OUBLE KULE OF THREE. when 30 men dig a trench 360 feet long, 6 wide and 3 deep, in 259-2 days ? Ans. 7 hours. The following Rule for the Double Rule of Three, involves the consideration of Direct and Inverse Proportion. Though the General Rule will enable the student to solve all questions with ease, this Rule is retained for the satisfoclion of those who might desire to use it. RULK. Always place the three conditional terms in this order : That number, which is the principal cause of gain, loss or action, pos- sesses the first place ; that, which denotes the space of lime, dis- tance of place, rate, medium or mean of action, the second ; and that, which is the gain, loss or action, the third : This being done, place the other two terms which move the question, under those of the same name, and if the blank place, or term sought, fall tjnder the third place, then the question is in direct proportion : therefore, Rule I * Bltiltiply the three last terms together, for a dividend, and the two first for a divisor : — But, if the blank fall under the first or second place ; then, the proportion is inverse ; therefore, Rule II. Multiply the first, second and last terms together for a dividend, and the other two for a divisor, and the quotient will be the an- swer. Examples. 1. If g 100 gam $6 in a year ; vvhat will §400 gain in 9 months ? D. P. Mo. D. Int. 100 : 12 :: 6 Terms in the supposition, or conditional terms. 400 : 9 Terms which move the question. Mere, the blank falling under the third place, the question is in direct proportion, and the answer must be found by the first Rule ; therefore, 400X 9x6—21600 For the dividend, and, i 00X1 2 =1200 For the divisor. ■• 1. When the blank falls under the tl^ini term by this mode of stalciner.i. >l in obvious on injpcctinjr the statement that the proportion is direct^ and tlio same terms are taken to form the dividend and divisor as in the preceding rule, or, by two statements in the Sing^le Rule of Three direct. 2. But in Example 2nd, and when the blank falls under the first or second term, the proportion is inverse. In thip Example, 7nore principal and interest require /^S5 "time, and, every statement according to the rule will make more req\are less. The operation by the rule is the same as from two statements by the Single Rule of Three inverse. These statements on this Example would he iliua 100X12 100 ; 12 :: 400 : "-~ 100X12 100XJ2X"10 ^•fl : — -rr ■• 6 : — rr,;rr77 — "-?- {*b3 illustrate? the rule. 4'X) i-COxO m DOUBLE RULE OF THREE. I5i D.Pr. 100: 400 : 9 See Mo. 12 ; 9 the work at larsre. D. Int :: ^> 100 12 3600 6 12j00)216|00( 12 18D. Ans. 9G 96 2. If glOO will gain $0 in a 3rear ; in what time will J5400 gam ^18? D. Mo. D. 100 : 12 :: 6 Terms in the supposition. 400 : :: 18 Terms which move the question. Here, the blank falling under the 2d place, the question is in reciprocal or inverse Proportion, and the answer must be sought by the second rule ; therefore, 100x12x18=21600 For the dividend, 400X 6 = 2400 For the divisor. D.Pr. Mo. D.Int. 100: 12 :: 6 400 : :: 18 6 12 2400 21G 100 24|00)216|00(9 months, Ans. 216 ^. What principal, at 6 per 4. If g400 gain $\S in 9 cent, per ann. will gain ^18 in months ; what is the rate per 9 months ? cent, per annum ? Pr. Mo. Int. Pr. Mo. Int. 100 : 12 :: 6 400 : 9 :: 18 9 :: 18 100 : 12 :: $6 Ans. 12 9 216 6 100 D. 64)21600(400 Ans. 216 00 \n^ COMPARISON OF WEIGHTS, Lc. Here, the blank falling under the first place, the proportion i* inverse, and the answer found by the second rule, as in the last example. 5. If 8 men spend £32 in 13 weeks ; what will 24 men spend in 52 weeks? Ans, £384. 6. If the freight of 9hhds. of sugar, each weighing 12cwt. 20 leagues, cost ^50; what must be paid for the freight of 50 tierces ditto, each weighing 2|cwt. 100 leagues ? Ans. ^289 35c. Iffm, 7. There was a certain edifice completed in a year by 20 work- men ; but the same being demolished, it is necessary that just such an one should be built in 5 months. I demand the number of men to be employed about it? Ans. 48 men. 8. If 6 men build a wall 20 feet long, 6 feet high and 4 feet thick, in 16 days, in what time will 24 men build one 200 feet long, n feet high, and 6 feet thick ? Ans. 80 days. COMPARISOJV OF WEIGHTS AJVD MEASURES. Examples. 1. If 78 pence Massachusetts be worth 1 French crown, how niiany Massachusetts pence are worth 320 French crowns ? F. cr. d. F cr. As 1 : 78 :: 320 78 2560 2240 24960 Ans. 2. If 24 yards at Boston make 16 ells at Paris, how many ells at Paris will make 128 yards at Boston ? Bost. Par. Bost. Par. As 24yds. : 16ells :: 128yds. : 85iells, Ans. 3. If 60lfe at Boston make 561fe at Amsterdam, how many pounds at Boston will be equal to 350 at Amsterdam ? Ans. 375ife Boston. 4. If 95ife Flemish make 1001b American, how many American pounds are equal to 550fe Flemish ? Ans. 578 J fife American. CONJOINED PROPORTION, IS when the coins, weights or measures of several countries are compared in the same question ; or, in other words, it is joining many proportions together, and by the relation, which several an- tecedents have to their consequents, the proportion between the first antecedent and the last consequent is discovered, as well as the proportion between the others ia their several respects. I CONJOINED PPxOPORTfON. 153 This rule may generally be so abridged by cancelling equal quantities on both sides, and abbreviating commensurables^ that the whole operation may be performed with very little trouble, and it may be proved by as many statings in the Single Rule ot" Three, as the nature of the question may require. CASE I. When it is required to find how many of the first sort of coin, weight, or measure, mentioned in the question, are equal to a given quantity of the last. Rule. Place the numbers alternately, that is, the antecedents al the left hand, and the consequents at the right, and let the last num- ber stand on the left hand ; then multiply the left hand column continually for a dividend, and the right hand for a divisor, and the quotient will be the answer. Examples. 1. Suppose 100 yards of America=100 yards of England, and 100 yards of England=50 canes of Thoulouse, and 100 canes of Thoulouse = 160 ells of Geneva, and 100 eJls of Geneva=200 ells of Hamburgh : How many yards of America are equal to 379 ells of Hamburgh ? Antecedents. Consequents. Abriged. 100 of America = 100 of England. Ant. Con. 100 of England = 50 of Thoulouse. 5 8 100 of Thoulouse = 160 of Geneva. 379 J 00 of Geneva = 200 of Hamburgh. .379 of Hamburgh ? 370X5 Therefore, » =23G|yds. of America=379 ells of Hamburgh, Illustration. The two 100s of both sides cancel each other. Let the last cy- phers of the next three antecedents and consequents be cancelled, which is dividing by 10. Then divide the second antecedent and consequent by 5, and the quotients will be 2 on the side of the antecedents, and 1 on the side of the consequents ; then 2 will measure the third antecedent and consequent, and the quotients will be 5 and 8. 10 will measure the 4th antecedent and conse- quent, and the quotients will be 1 and 2. Now, there being 2 left on each side, they cancel each other, and as there is no farther room for abridging by reason of the odd number 379, the opera- tion is finished, nnd the answer found, as before. 2. If 20tfe at Boston make 23ife at Antwerp, and 155 at Antwerp make 180 at Leghorn : How many at Boston are equal to 144 art Leghorn? Ans. 107iflfe. 3. If 121fe at Boston make lOib at Amsterdam, lOlb at Amsterdam 12Jfe at Paris : How many pounds at Boston are equal to 80ife at Paris ? Ans. 80ft3,. I lo4 ARlirTKATlON OF EXCHANGES, 4. If 140 braces at Venice be equal to 150 braces at Legborn, and 7 braces at Leghorn be equal lo 4 American yards : How many Venetian braces are eqtial to 32 American yards ? Ans. 52~^*j. 5. If 40ife at Newburyport make 36 at Amsterdam, and 90fe at Amsterdam make 116 at Dat>rzick : How many pounds at Newbn ryport are equal to 260ife at Dantzick ? Ans. 2242^- CASE II. When it is required to tind bow many of the last Sort of coin, weight or measure, mentioned in the question, are equal lo a given quantity of the first. KULK. Place the numbers alternately, begihning at the left band, and let the last number stand on the right hand ; then multiply the first row for a divisor, and the second for a dividend. Examples. 1. Suppose 100 yard^ of America--=100 yards of England, and 100 yards of England-=50 canes of Thoulouse, and 100 canes of Thoulouse=^160 ells of Geneva, and 100 ells of Geneva=200 e\U of Hamburgh.: How many eli.i of Hamburgh are equal lo 236" yards of America ? Ant. Con. Abridged, 100 Amer. == 100 Eng. Ant. Con. 100 Eng. = 50 Thoul. 6 100 Thoul. = 160 Gen. 236| 100 Gen. == 200 Hamb. 236 I-X 8 = 379 Ham. 2361 Amer. 5 Ans. This needs no further illustration. The learner will readily see, that this case being the reverse of the former, they are proofs to each other. 2. If 20fe at Boston make 23115 at Antwerp, and 155 at Antwerp make 180 at Leghorn : How many at Leghorn are equal to 144 at Boston ? An-s. 144^. 3. If 12ife at Boston m?ike lOife at Amsterdam, and lOOife at Am- sterdam 120f6 at Paris: How «»any at Paris are equal to oOlfe at Boston ? Ans. 80lfe. 4. If 140 braces at Venice be equal to 150 braces at Leghorn, and 7 braces at Leghorn he equal to 4 American yards : How many American yards are equalto 52,-^^ Venetian braces? Ans. 32 yards. , 5. If 40'lfe at Newburyport make 36 at Amsterdam, and OOfe at Amsterdam make 116 at Drmtzick : How many pounds at Oant- zick are equal to 244 at Newburyport ? Ans 283^|lfe. JIRBITRATION OF EXCHANGES. By this term is understood how to choose, or determine the best way of remHting money from abroad with advantage; which i? performed by conjoined proportion : 'i'hus, FELLOWSHIP. Voo 1. Suppose a merchant lias effects at Amsterdam lo tlie amount org3530, which he can remit hj way of Lisbon at 840 rees per tlollar, and thence to Boston, at 8s. Id. per milree (or 1000 rees ;) Or, by way ofNantz, at 5| livres per dollar, and thence to Boston at 6s. 8d. per crown ; It is required to arbitrate these cxchangci^, that is, to choose that which is most advantageous ? 1 dollar at Amsterdam = 840 rees at Lisbon. 1000 rees at Lisbon == 97d. at Boston. 3530 dollars at Amsterdam. 840x97x3530 rrTTTTT-; =£1198 8s. 8yV'- by wav of Lisbon. 1000x1 1 dollar at Amsterdam = 5| livres at Nantz. 6 livres at Nantz = 80 pence at Boston. 3530 dollars at Amsterdam. 5|x 80x3530 ]v6 =ii^l059 by way of Nantz. Here it may be observed that tlie difference is J£139 8s. 8y*-„d. in favour of remitting by way of Lisbon rather than by Nanlz, which depends on the course of exchange, at that lime ; but the course may vary so, that^ in a short lime by way of Nantz may be better; hence appears the necessity and advantage of an ex- tensive correspondence, to acquire a thorough knowledge in the courses of exchange, to make this kind of remittance. ? 2. A merchant in England can draw directly for 1000 piastres ih Leghorn at 50d. sterling per piastre ; but he chooses to remit the sum to Cadiz at 19 piastres for 7000 maravedies ; thence to Am- sterdam at 189d. Flemish for 680 maravedies; and thence to Liv- erpool at 9d. Flemish for 5d. sterling : what is gained by this cir- •;u!ar remittance, and vyhat is the value of a piastre to him ? Ans. Gain £28 |49. sterling nearly. Value of a piastre 5Gd. 3-55qr. sterling. 3. A merchant in New York orders JC500 sterling, doe him nt London at 54d. sterling per dollar, to be sent by the following cir- cuit; to Hamburgh at 15 niarks banco per pound sterling ; thence to Copenhagen at 100 marks banco for 33 rix dollars ; thence to Bourdeaux at one rix dollar for G francs ; thence to Lisbon at 125 francs for 18 milrees ; and thence to New York at ^1^ per milree : did he gain or lose by this circular remittance, and what was the arbitrated value of a dollar by this remittance ? Ans. He gained. Value of a dollar was 69d. sterling nearly. FELLOWSHIP. THE Rules of Fellowship are those by which the accompts of :^cveral merchants or other persons, trading in partnership, are so adjusted, ths^t each may have his share of the gain, or sustain his 156 SINGLE FELLOWSHIP. share of the loss, in proportion to his share of the joint stock, to- gether with the time of its continuance in trade. SIJVGLE FELLOWSHIP Is, when the stocks are employed for any certain equal time. As the whole stock is to the whole gain or loss, so is each man's particular stock to his particular share of the gain, or loss. Proof. Add all the particular shares of the gain or loss together, and, if it be right, the sum will be equal to the whole gain or loss. Examples. 1. Divide the number 360 jnto four parts, which shall be to each other, as 3, 4, 6 and 6. As 3-{-4-f-6-l-6 : 3G0 :: Answer. 360 Proof 2. A, B, C, and D companied ; A put in £145; B, £219; C, £378, and D,£417, with which they gained £569 : What was the share of each ? £ s. d. Whole stock Gain C ^^^ * '^^ 3 8^ |H-| A's share. A«14^ • 91Q LAftZAi7. VfiQ . )219 : 107 10 31 J^ B's ditto. Asl45-r219-ho78+417:569:: ^3^3 . ^^^ ^j g- ^ C's ditto. (417 : 204 14 51 Tj'j% D's ditto. £569 — Proof. 3. A, B, C, and D are concerned in a joint stock of ^1000; of which A's part is §150 ; B's $250 ; C's g276, and D's ^325. Up- on the adjustment of their accompts, they have lost ^337 50^. What is the loss of each ? Ans. A's loss ^50 62Ac. B's g84 37J-C. C's $92 811c. and D's gl09 685c. 4. A and B companied ; A out in £45, and took | of the gain ; "What did B put in ? 5—3—2. Thon, As 3 : 45 :: 2 : 30 Ans. 6. A, B and C freighted a ship with 68900 feet of boards ; A put in 16520 feet; B 28750; and C the rest; but iqa storm, the cap- tain threw overboard 26450 feet : How much must each sustain of the loss? ,\ns. A, 6341f feet. P, n036f and C, 90711 do. 6. A gentleman died, leavir)g three sons and a daughter, to whom he bequeathed his estate in the following manner : To the eldest son, he gave 312 moidores, to the second, 312 guineas, to the third, * That thoir gain oi- loss, in this rule, is in proportion to their stocks is .evi- dent : For, as the times, in which the stocks are in trade, aio equal, il" I put in I of the whole stock, I ou^ht to have -i of the g^ain : li my part of the stock be j, my share oi the gain or loss ought to he i also. And generally the same ra- tio that the whole stock has to the whole fi:ain or loss, must e^h person's pav- : f icular stock have to iiis respective gain or loss. SINGLE FELLOWSHIP. ' 157 312 pistoles, and to the daughter, 312 dollars ; but when his debts were paid, there were but 312 half joes left: What mu!*t each have in proportion to the legacies which had been bequeathed them? Ans. 1st son £293 Os. 3d.— 2d. son £227 17s. lOfd.— 3d. son X 179 Is. 2id. and the daughter £48 16s. 8id. 7. A ship, worth ^3000, being lost at sea, of v\hich } belonged to A, i to B, and the rest to C : What loss will each sustain, sup- posing ^450 to have been insured upon her? Ans A's loss $312 50c. B's 937 50 C*s 625 8. A and B venturing equal sums of money, cleared by joint trade ^140: By agreement, as A executed the business, he was to have 8 per cent, and B was to have 5 per cent. : What was A allowed for his trouble ? ^ (fi (3 (3 As 84-5 : 140 :: 8 : 86^2- And, as 8+5 : 140 :: 5 : 53if Ans. $32 30c. ^j^nu 9. A bankrupt is indebted to A £120, to B £230, to C £340. and to D£450, and his whole estate amounts only to £560 : How must it be divided among the creditors ? Ans. A, £58 18s. ll^d. B,£ll2 19s.7fd. C,£167 Os. 4d. and D. £221 Is. Oid. 10. A, B, and C put their money into a joint stock ; A put in $40 ; B and C together $170 : They gained $126, of which B took,g42 : What did A and G gain, and B and C put in respectively ? Ans. $24 A's gain, $70 B's stock, $100 C's stock. $60 C's gain 11. A, B, and C companiod ; A put in £40 ; B 60, and C a sum unknown : They gained £72 ; of which C took £32 for his share ; What did A and B g:ain, and C put in ? Ans. £16 A's gain, £24 B's gain, and £80 C's stock. 12. A, B, and C put in $720, and gained $540, of which, so oft- ten as A took up $3, B took 5, and C 7 : What did each put in and gain ? Instead of the above rule, you may find a common multiplier to multiply the proportions by, or multiplicand to be multiplied by the given proportions, thus, 15)720(48 multiplicand to find the stock?. And 15)540(36 multiplicand to find the gains. ^ ' $ 48x3=144 A's stock, i C 36x3=108 A's jrain. 48X5=240 B's ditto. > And ? 36x5=180 B's ditto. 48X7=336 C's ditto. ) ( 36x7=252 C's ditto, as before. 13. A, B, C, and D companied ; and gnin. d a sum of monry of which A, B and C took£i20, B. C and D,£l80, C, D and A, £160, ^nd D, A and B,£140 : What dij>tinrt gain had e xrh ? The sum of the^e 4 num'ifrs i.s£6G0. and as each man's money is named 3 ti.res. th'^refjue |^ viz. £200 is \hr whole gain Therefore £200— £ 120 A'fi B'-' and C's ^air. = ]e80 D's gain ;— And £200— .£ 180 B's, C*s and D'5 -air.= £20 A'x gsin.--£ '^OO— £160 C's. D's, and A's gain=£40 B's gain.— And £200— £ 140 D's, A's and B's gain=£60 C's gain. \b^ SINGLE FELLOWSHIP. 14. Two merchants companied ; A put in £40, and B 288 do- cats. They gained £135, of which A took £60. What was the value of a ducat ? As £60, A's gain : £40, his stock :: £135 the whole gain— £ 60, A's gain : £50, B's stock. Due. £ Due. s. d. And, as 288 : 60 :: 1 : 3 5| Ans. 15. Four men spent, a^a reckoning, 20 shillings, of which they agreed that A should pay |, B, ^, C, |, and D, |. What must each liay in that proportion ? s. d. 9 2-j G 1] ^, Answer. IG. A, B,andC companied ; A put in £40-25; B £80*5 ; and C £ 161 r they gained £120. What is each man's share ? ££££££ 4i;)25+80-5+161 : 120 :: 40-25 ; 17-142475 = A's 34-28495 =B's G8-5699 =C's Proof £119-997325 17. A, B, C, and D gain ^200 in trade, of which as often as A has ^6, B must have glO, C gl4, and D g20 : What is the share of t'ach ? Ans. A's share ^24, B^s jg40, C's $56, and D's $Q0. 18. An insolvent estate of ^633 60c. is indebted to A, ^312 75c. to B, ^297, to C, $50 25c.toD,g0 25c.toE,g200 toF,gl42 50c. and to G, ^^21 25c. ; what proportion will each creditor receive ' $ c. ■ Ans. A's share = 193 51 41 B's - - 183 76-^1 C's - 31 09-2:> D's - 15-41 E's - 123 75- F's • - 88 17-18 G*s - 13 14-1^7 Proof ^633 59-97 I'?. A sHtp vias driven on ?hore in a gale, and in Iighle?)ing and getting her afloat again and in reloading, an expense of <;§763 was incurred ; the ship was valued at ^10000, freight at Ji3200, molas- ses owned by A, at ^5200, sugar owned by B, at j^4700, and rum oivned by C, at J2500 : how much is this loss on every $100, and Ii?)w much must each parly pay of it *? DOUBLE FELLOWSHIP. 15^ ^ ^ $ $ $ 0000+3200-1-52004-4700+2500=23600. As 25600 ; 768 :: lOO:.^' g g g § Ans. on each $100. Then, As 100 : 3 :: 1000 : 300 to be paid by the ship, 320 : 96 - - -freight, 5200 : 156 - - A, 4700 : 141 . - - B, 2500 : 75 - - C, Ans. 768 Proof. 20. A vessel, valued at g 13000 was laden with hardware for E vahied at J3000, with cordage for F", at goOOO, with dry goods for G, at ^3200, with goods for H, at J7900, and for \, at $4400 ; the. captain was obhged to prevent sinking in a storm to throw over- board three fifths of the hardware, and two fitlhs of the cordage, with goods of IT valued at $2700; allowing the freight ti be $3500, what will be the average of the loss on 100 dolls, and what must be paid to E, F, and U, for their property thrown overboard ? Ans. $16 25cts. on $100, and E, F, and II must receive togeth- er $5443 75cts. Note. If the property of E, F, and 11, had "been insured, the re- mainder of their loss must be paid by the insurers. See Policiej' of Insurance. DOUBLE FELLOWSHIP;'' Or, Fellowship with Time, is occasioned by the shares of part- ners being continued unequal times. KULF.. Multiply each man's stock, or share by the time It was continu- ed in trade. Then, As the whole sum of the products, is to the whole gain or loss, %o is each man's particular product, to his particular share of the gain or loss. Examples. 1. A, B, and C hold a pasture in common^ for which they pay £40 per annum. A put in 9 oxen for 5 weeks ; B, 12 oxen for 7 weeks, and C 8 oxen for 16 weeks. What must each pay of the rent ? 0x5=45. 12x7=84, and 8x16 = 128, then 1284-844-45=257, As 257 : 40 :: 45 As 257 : 40 :: 84 As 257 : 40 :: 128 45 84 40 £ s. d. 200 160 257)5120:i9 18 5-/A 160 320 . £ s. (1. £ g. d. 257)1800(7 0|f»- 257)3360(13 1 5i|| * When times are equal, the shares of the gain or loss are evidently as the stocks, as in Single Fellowship ; and when the stocks are equal, the shares *jr<- as the times ; wliereforc, when rjeither aro eqnnl. the shav-s nv\-t hp a,' ^h"'' products. 60 DOUBLE FELLOWSHIP. £ s. (1. A's = 7 o|M B's = 13 1 HU C's = 19 18 h%\ Proof 40 2. Four merchants traded in company ; A put in ^400 tor i\vv, months, B, gGOO for 7 months, C, g960 for 8 months, and D, ^1200 for 9 months ; but by misfortunes at sea, they lost ^750. What must each man sustain of the loss. Answer \ ^' ^^^ ^^"- ^^>' ^' ^^^^ ^^^^ Vo^- I ^n:,wer, ^ g 142 40 6-^V ^ 284 81 Ol| ^ 3. A, with a capital of £100 began trade January 1st, 1787, and meeting with success in his business, he took in B as a partner, on the 1st day of March following, with a capital of £150. Three months after that, they admit C as a third partner, who brought into stock £180, and after trading together until the 1st of Janua- ry, 1788, they found there had been gained since A's* commencing business £177 13s. How must this be divided among the partners '.' Ans. A, £53 IGs. 8d B,£67 5s. lOd. C,£56 10s. 6d. 4. Two merchants entered into partnership for 18 months ; A, at first, put into stock ^400, and at the end of 8 months he put in ^200 more ; B, at first, put in gllOO. and at 4 months' end took out §280. Now at the expiration of the time, they found they had gained gl052. What is each man's just share ? Ans. A, §385 90c. B, $66G 10c. 3. A and B companied ; A put in the 1st of January, £150 ; but B could not put in any until the 1st of May : What did he then put' in, to have an equal share with A at the year's end ? Ans. £225. 6. E, F, and G companied ; E put in, the first of March, £30, F, the first of May, put in 80 yards of broadcloth ; and on the 1st of June, G put in §120. On the 1st of January following, they reckoned their gains, of which E and F took £228. F and G£215 10s. and G and E £187 10s. What was the whole gain, and tit gain of each ? What did they value a yard of cloth at ? and, what was G's dollar worth ? 2281.-1-2151. 10s.-f-187l. 103=6311. and 6311 -f-2=315]. 10s. the whole gain; then, 3161. 10s.— 228=871. 10s. G's gain. 3151. 10s. —2151. 10s. = 1001. E's gain, and 3151. 10s —1871. 10s.= 1281. F's gain. To find the value of one yard of cloth, say, A>* 1001. E's gain : 301. his stock :: 1281. F's gain : 381. 8s. ; then, inversely. As 19 months : 381. 83. :: 8 months : 481. the value of the whole cloth. As 80yds. : 481, :: 1yd. : 12s. answer. Now, to find the value of a dollar. As 1001. E's gain : 301. his stock :: 871. 10s. G's gain : 261. 3s. ; then, inversely, As 10 months : 261. 6s. :: 7 months : 371. 10s. = 120 dollars. Lastly r As 120 dollars : 371. 10s. :: 1 dollar : 6s. 3d. Answer. PRACTICE. 161 7. E, F and G companied ; E put in ^400 for '75 of a year ; F 4id. per yard come to ? per yard come to ? 379 149 3 8)1137 1)0)44|7 2|0)14|2 1-t Ans. £44 14 Ans. £7 2 U- Questions. Answers. Yds. s. d. £ s. d. 3. 127 at 71 per yard. 3 19 4| 4. 159 — 8 6 6 5. 173 — 9 6 9 9 (]. 241 — 10 10 10 7. 249 — 7 6 93 7 6 8. 357 — 12 6 223 2 6 CASE VIII. When the price is any even number of shillings, if it be required to know what quantity of any thing rnay be bought for so much money : Annex a cypher to the money, and divide it by half the price, and the quotient will be the quantity to be purchased. PPAchCE. 17 Examples, 1. How many yards of cloth, at 18s. per yard, may I have for 345? Half the price =^9)3450==:money with a cypher annexed. i;345 3831 yards, Ans * Questions. 2. How many yds. at 2 per yd. for 427 ? 3. 4 312 4. 6 917 6. 8 195 Answers. Yds. 4270 1560 3050| 487i CASE IX. To find the value of goods sold by particular Quantities, viz. I. By the score. H. Round timber. III. By 5 score to the hundred. IV. By 112 to the hundred. V. By 6 score to the hundred. VI. By the great gross^ VII. By the thousand. I. To ^nd the value of goods sold by the score. The price of one is given, to find the price of one score. If the given price be shillings and pence, or only pence, divide the given price, in pence, by. 12. The quotient will be the an* swer in pounds, and the remainder will be so many times. Is. 8 J. for a score is 20, and 20d. is Is. 8d. Examples. 1. At 9d. each: What is that per score ? 12)9d.(-75=£0 15 Ans. Or by inverting the question. 1 score=20=ls. 8d. 9 2. At 4s. 9d. each: What is that per score ? £4 15 Ans. 15s. It may be remarked, that when the price is shillings and pence, the answer will be just so many pounds as there are shillings, and so many times Is. 8d. as there are pence. If farthings are given, for ^d. reckon 6d. for |d. lOd. and for Jd. Is. 3d. 2 is tV 5-i IQUOT PARTS. 20 THE INTEGER. 12 is f\ 14 -t\ 15- -J 16 is A 18 - /^ M * At £1 a yard, the quantity would be 345yds. But as the price is 18s. you want only l.8:=_9_^ of the qunntity. This explains the rule ; for the course is similar /or any other case. 172 PRACTICE. 3. What cost 7 ; at 2s. 9d. per score? s. d. 2 9 4. What cost 17; at 19s. lOd, per score ? 8i 16s. lOid. Ans. 3i 7=0 111 II. Round Timber. Forty feet make a load or ton of round timber. If the given price of a foot be shillings, Rule. Multiply the given price by 2, and the product will be the an- swer in pounds. 5. What cost a ton at 3s. per 6. What cost a ton at 9s. per foot? 3s.x2=61. Ans.* foot? 9s.x2=181. Ans. If the given price of 1 foot be pence only, or shillings and pence, divide the given price, in pence, by 6. The quotient will be the answer in pounds, and the remainder will be so many times 3s. 4d. 7th. What cost 40 feet, at 17d. per foot? €th. At Is. 9d. per foot: What 6)17 cost a ton? £3 10 Ans. £2 16 8 Ans. If the given price of a foot be farthings only, or pence and far- things, divide the given price in farthings, by 6 ; then divide ihat quotient by 4, and this last quotient will be the answer. 9th. At '|d. per foot : What cost a ton ? 6)3 10th. At 131 per foot: What cost a ton ? 4)0 10 £2 4 2 Ans. £0 2 6 Ans. ♦ If a ton of timber had boon 20 feet, then 3s. a foot would have mauo £.Z ; but as a ton has twice 20 feet, the answer in pounds must be twice the number of shiUings. In like manner for any number of shillings. If the price be pence 40X17 20X17 17 a foot, as in Ex. 7, the rule is — =: = — for pound?, witli a remainder 12 tj 6 ^ 209. of 5X -7- =5 times 2s. 4d. tl 40x3 When the price is farthings a foot, the rule is as in Ex. 9, for shillicg.?, 20-f-3 3 . , ^^ for pound?. 0X4 6X4 *^ PRACTICE. 173 Or, suppose every shilling in the price to be 21. every penny to be 3s. 4d. and every farthing to be lOd. 12lh. What cost 40 at 16id. per foot ? nth. What cost 40 feet at s» d. |d. per foot? lOx 2=£2 |d. X 10 £0 2 6 Ans. 3 4 x 3= 10 4- X 10= 1 8 £2 n III.* To find the value of goods sold hy 5 score to the hundred. 1st, If the given price be pounds and shillings, or shillings only. Rule. Multiply the given price in shillings, by 5, and the quotient will be the answer in pounds, for lOOs. are £5. 13th. At 19s. per yard, what cost 100 yards ? 14th. At 41. 135. per cwt. what 193, cost 100 cwt. or 5 tons ? 5 ' J£465 Ans. £95 Ans. 2d. If the given price of 1 be pence only, or shillings and pence. Rule. Multiply the given price in pence, by 5 ; then divide that pro- duct by 12. The quotient will be pounds ; and the remainder so many times Is. 8d. 16th. What cost 100 bushels, at 35s. 4d. per bushel ? 15th. Ifl yard cost 9d. what s d. Or, cost 100 vards ? 35 4 35s. 4d. '9 J^ J4d.|i| 5 — 424 175 12)45 5 1 13 4 £3 15 Ans. 12)2120 £176 13 4 £17G 13 4 Ans. Here 5 is di- vided by i. * In Federal Money. — Remove the decimal point two places to the right for the answer. Examples. 1. What cost 100 yards at $2 50c. per yard.? $2-50 X 100=$250- An?. 2. What cost 100 yards at 75c. per yard ? $-75 X 100=:$75- Ans. •3. What cost 100 yards at 5c. 64m. per yard .? $•05625 X 100=$5-625 An*. 4. What cost 100 yards at 37c. 5m. per yard ? Ans. $37 50c. 6 What cost 100 yard=* at 68c. 7ira. per yard ^ Ans. $68 75c. 174 PRACTICE. 3. If the given price of 1 be shillings and pence ; Multiply the price by 6, and the product under the place of shillings, will be the answer in pounds, and the product under the place of pence, will be so many times Is. 8d. 17th. At 2s. 6d. per bushel: what cost 100 bushels ? s. d. 2 5 18th. At 25s 3d. per ton ^ 5 what cost 100 tons ? £126 5 Ans, 12 1 £12 1 8 Ans. 4.* To find the price of owe at so much per hundred of 5 score. General Kule. Multiply the given price by 12 ; divide the product hy 5, ant! the quotient will be the answer in pence. But if the price be pounds only ;= Kule. Divide the given price by 5, and the quotient will be the answer in shillings. 19th. If 100yds. cost £6ii, what 21st. If 100 yards cost £11 '^. cost 1yd? 9d. what cost 1 yard. 5)65 £ s. d, — 11 7 9 13s. Ans. 12 20th, If lOOyds. cost £2 18s. 5)136 13 a 4d. what is that per yard ? £ s. d. 12)27 6 7 2 18 4 ' 12 2s. 3id. Ans. In dividing 27 by 12 (in the 5)35 21st question) the quotient is 2s- and the remainder 3d. the 6 is. 7d. Answer. 2^^- of a penny = one farthing, ■ — and the 7 is of no account. TARLE OF ALIQ;UOT PARTS. 100 THE INTEGEPw 5 is A!25 is 10 ,VPO 3 20 50 is I 60 A 75 is I- 80 -,\ 90 /o *' In Federal Money. — Renicre (.lie decimal poiiit two places to tlie left for the answer. Examples. I. Tf 100 yards cost |i>50, v.rhat cost 1 yard .? $250~100=::$2v>0 Ans. -2. in 00 yards cost $7b' what cost 1 yard .=> $75'~100==$-75 Ans. 3. If 100 yards cost $3 62c. 5m. what cost 1 yard ? $5-625-M00~|-05C25=r5c. 6i^. Ans. 4. M 100 yards cost .$37 50c. what cost 1 yard.^ Ans. 37c. 5m. 5. If 100 yards cost |ca 75e. what cost 1 yard ? Ans. 68o. 7^m. PRACTICE. 175 22d.* At £3 7s. 6d. per 100 : What will 23 cost ? 20 3 7 6 13 6 1 4 > Add 8 23= £0 15 6 Ans. 23d. At £2 Is. lOd. per 100 : 24lh. At £5 9s. 6d. ner 100 i What cost 18? je s. d. 20 * 2 1 10 I 8 4f 10 What cost 35 ? £ p. d. 5 9 6 3 Sub. 18= £0 7 6fADs. 10)16 8 6 6 1 12 10 5 6f Add. £1 18 3^ Ans. IV. To find the value of goods, sold by 1 12ife the Cwt. The price of life is given to find the value of 1 cwt. Rule. For a farthing, account 2s. 4d. per cwt. For a half a penny, 4s. 8d. For three farthings, 7s. And for every penny 9s. 4d. per cwt. 25th.. What cost Icwt. at 3id. 26th. At 8|d. per ft : What per ft ? cost 1 cwt ? Ans. £4 Is. 8d, At Id. per ft s. d. Icwt. costs 9 4 At 3d. £18 At id. 1 8 0^ 4 85 Add. £1 12 8 Ans. * To find the value of any number at a given price per 100, in federal money,— Multiply the price per 100 by the given quantity, and point oft" two right hand figures, in the product more than required by multiplication of decimals. Or, point off" the two right hand places in the given quantity, and multiply, and point, as in multiplication of decimals. EXAM-PLES. t. What cost 56 yards at $87 50c. per 100 yards ? $87-5 xsa ^~— 149, Ans, Or, $87*5 X "56=i549, as before. iOO ivtJ PRACTICE. V. To find the value of goods sold by 6 score, to ike hundred. The price of 1 is given to find the price of 1 hundred. Rule. Suppose every penny in the price to be so many pounds, and for the farthings, such a part of a pound, as they are of a penny j then, half of that sum will be the answer. 27th. At 4-td. per yard : What 28th. At 16s. Ojd. per yard : cost 120 yards ? " What cost 120 yards. £ s. Ans.jeiao 12s. 6d. 2)4 10 £2 6 Ans. To find the price of one, at so much per hundred of 6 score. Rule. Blulliply the price by 2, then call the pounds so many pencCj and the shillings, such a part of a penny, as they are of a pound, and you will have the answer. 29th. If 120 yards cost i:3 12s.: 30th. If 120 yards costce^/ What cost 1 yard ? 18s. 6d. : What cost 1 yard'' £ 8. Ans. ll^d.+l of farthing. 3 12 7 4 Ans. 7id. TABLE OF ALIQUOT PARTS. 120 THE INTEGER. Also, 6 is 2V 10 - -,v 12 - tV 20— A 24 is 1 36 is j\ 72 is 1 30 is 1 45 - 1 75— f 40 is i 48— 1 80-1 60 is 1 50 - j% 84 -^V 70-^^ 90— f 96 100 105 108 ,4 5 IS 4 :Ust. At £3 17s. 6d. per hundred, what cost 14 £ s. d. 12 j\ 3 17 e 1 31 P^^^- 14= £0 9 Oi Ads. i>. What cost 45 ^Ib. beef at $5i per 100-' 4f^.y v'45'5 -=$2-5025=$2 50c. 2hm. Ans. Or, $5-5 X •455lb.=$S'5025, as before- 100 3. What cost 375 yards at $375 per 100 yards? Ans. $1404 25c. 4. What cost 54 yards at $16 per 100 ? Ans. $8 64p. ■}. What cost 512 vards at $6 25c. per 100 yards ? Abb. p^ J PRACTICE. Ill 32. At£2 135. 6id. per hua- 33. At£l 19s. 3(1. per hun- dred, what cost 49 ? dr^d, what cost 75 ? £ s. d. Ans.JSl 4s. 6Ad. 40 2 13 61 17 10 0?- 3 6 Si 5 1" 49 =£1 1 10 1 Ans. VI. !ro y?rtrf . ,, |r^^' 9f CASE X. When the price of 1 is any number of dollars and parts of a dollar : Multiply the quantity by the number of dollars ; and, finding, by the general rule, the price at the parts of gl, the sum of the \YhoIe is the answer Examples. 1. What cost 395 yards at g3 24c. per yard ? c. k c. 20 of 20c. 395 3 1185 79 15 = price al. gl = ditto at = ditto at 80 = ditto at 20c. 4 Ans. $1279 80 = ditto at ^3 24c. 2. What cost 269 yards at $2 60c. per yard ? Ans. ^699 40c, 3. 694 12 10 8397 40 4. 318 4 121 1311 75 5. 175 4 44 . 777 CASE XL When the price of 1 contains the same aliquot part of a dollar antj number of times exactly ; or, in other -wordsy when the price of 1 has an aliquot part, which is also an aliquot part of a dollar : First, {iml the value of the given quantity at the aliquot part j then multiply KiJils. Or, point off the three right hand places in the givejo. quantity : and multiply and point as in multiplication of decimals. Examples. 1. Vv'hat cost 875 at $13 per lOCO ? 875X13=11375; and n375-?-1000=n-375=$n 37c. 5m. Ans. '2. What cost 39175 feet of boards, at $16 per 1000 ? Ans. $626 OOo. •;. What cost 325 nails at $1 50c. per 1000 .^ Aus. 48c. "l^m., or, 48{q, PRACTICE. 18! this by the number of times which the aliquot part is contained in the given sum, for the answer. Or, Since the price in this case is alwa;ys such a number, as, being divided by the aliquot part, will make the numerator of a fraction, of which the denominator is the denominator of that fraction, which the aliquot part is of a dollar ; Multiply the quantity by the nu- merator, and divide the product by the denominator, (or, when convenient, divide the quantity by the denominator, and multiply the quotient by the numerator,) for the answer.* Examples. J. 2 1. What cost 384 yards at 87| cents per yard ? 12ic. = ^ of -875 = I $i I 384- = price at $1 48- =F ditto at -12^ X7 X 7*^ Ans. $336- = ditto at -37^ Or thus, .875=31, and 384x|=='«V'^'(=='f*X7)— g336, Ans. as before 2. What cost 842 yards at 66|c. per yard ? Ans. §561 33^c. 3. What cost 912 yards at 55c. per yard ? Ans. g501 60c. MISCELLANEOUS QUESTIONS IN PRACTICE. I. What cost 300 yards at 27c. per yard ? Ans. g81 917 $1 12 5m. 1031 62c. 5in. 351 35 12 32 862^ ft. boards at gl2 per M. ? 10 34 6 32159 13 75c. 442 18 6^ CASE XII. L Since 2s. is ^V o^ *^^' the decimal of 2s. is •! : Wherefore any quantity being given at 2s. per lb. yard, &c. the price is found in pounds and decimal parts of a pound, by separating the unit figure of the given quantity from the rest, for a decimal. Let it be required to find the value of 356 yards at 2s. per yard ^ By pointing off the unit figure 6 for a decimal, I find the > £oc.f; _. lount to be £356, which is known to be equal to 351. 12s. ^ II. Consequently, if the price be a rnultiple of 2s. (viz. any even number of shillings) the amount at 2s. being first found in pounds and decimal parts, as above, and that amount multiplied by the number which shows how often 2s. is contained in the given price, the product will be the amount required in pounds and deci-mal parts of a pound. What cost 427 gallons of wine, at 8s. per gallon ? * Some of the prices wliich apply to this case, are to be found in the seconU table of parts of a dollar. amo 182 PRACTICE. £42 7 amount at 2s. per gallon. 4 Ans. £170-8 or 1701. 16s. The examples in Case 1st. may be worked in this manner. Likewise, if the price be pounds and even shillings. 754 yards at 11 8s. 76-4 amount at 2s. 14x2=28s. 3016 764 Or, 734 75-4x4=301-6 Add. £l05d-6 Ang. £1065-6=]0551. 12s. III. If the price be an aliquot part of 2s. : Find the amount at 2s. »nd divide it by the denominator of the part, and the quotient wiU be the answer. At 8d. per lb. : What cost 976 lb.? I 8d. U I 97-6 £32 533= £32 10 8 Ans. IV. If the price be not an aliquot part : Divide it into aliquot parts. 7235 yards, at 7d. 4d. 3d. 723'5 120 583 90-437 211-02 =£211 4| Ans. V. If the price be pounds and shillings, or pounds, shillings and, pence: Reduce the shillings, &c. to the decimal of a pound, and multiply the quantity thereby, or the price by the quantity. At 151. 12s. 6d. per Cwt : VVhat cost 75 Cwt. ? £15 12 6 = £1^-625 75 78125 109375 1171-875 £1171 17 6 Ans. ' VL Jf the quantity likewise be of divers denominations : Reduce the less denominations to the decimal of that, whereof the orice is PRACTICE. n$ :^b. lOoz. of silk, at £4 5 9 =£4-287 mb. lOoz. =9-625 21435 8574 25722 38583 41-262375 £415 3 Ans, teases 6th. and 7th. may be wrought io this manner. Or, you may take parts for the lower denominations. 8oz. 2oz. 4-287 9 38-583 2-1435 •535875 41-262375 £41 VII. When the price is any odd number of shillings : h it be re- tq^uired to know what quantity of any thing may be bought for any sum of money, in pounds : Annex two cyphers to the money, and divide it by half the price. Note. As half a shilling (or 6 pence) is '5, therefore, to halve any odd number of shillings, is only to annex '5 to half of the great- est even number in the price. 1st. How many yds. at 7s. per yd. may I have for 4351. ? Half=3-5)43500(1242||-yds.Ans. 35 85 70 150 140 2d. How many pounds of tea, at 5s. per lb. for £37? 1481b. Ans. 3d. How many yards at 9s. per yard may I have for 5401. ? 'An^, 1200 yards. 100 70 30 14 TARE AND TRET. BlLL OF PARCELS. Newburyport, January 1st, 1808. Mr. Timothy Huckster Bought o{ Samuel Merchant, ?5|^ife Bohea tea, at 3s. 6(1. per ft. 48ft Cheese, at 9d. per ft. 15 Pair worsted hose, at 6s. 8d. per pain 4^ Dozen vvomen's gloves, at 36s. 6d. per dozen. 19 Dozen knives and forks, at 6s. 9d. per dozen. 9 Grindstones at 15§. 9d. per stone. ^ Cwt. Brown sugar, at 51s. percwt. 31 ft Loaf Sugar, at Is* O^d. per ft. — . £34 3 . Received payment in full, Samuel Merchant, TARE AND TRET. TARE and Tret are practical rules for deducting certain allow- ances, which are made by merchants and tradesmen in selling their goods by weight. Tare is an allowance, made to the buyer, for the weight of the box, barrel or bag, &c. which contains the goods bought, and is either at so much per box, &c» at so much per cwt. or at so much in the gross weight. Tret is an allowance of 4ft in every 104ft for waste, dust, kc. Cloff is an allowance of 2ft upon every 3cwt. Gross Weight is the whole weight of any sort of goods, together with the box, barrel, or bag, &c. which contains them. Sutlle is, when part of the allowance is deducted from the gross. Neat weight is what remains after all allowances are made. CASE I.* When the tare is at so much per box.^ barrel or bagy ^'C. : Multiply the number of boxes, barrels, &c. by the tare, and subtract the product from the gross, and the remainder will be the neat weight required. ExAMlPLES. 1. In 6 hogsheads of sugar, each weighing 9cwt. 2qrs. lOlb. gross, tare 251b. per hogshead ; how much neat ? * This, as well as every other case in this rule, is bnly an applieation of tb^ rales of Proportion and Practice. TARE AND TRET. 185 Cwt. qr. ft Cwt. qr. ft 25x&=l 1 10 9 2 10 gross wt. of Ihhd. 6 57 2 4 gross. 1 1 10 tare. Ans. 56 22 aeat. 2. In 5 bags of cotton, marked with the gross weight as follows, tare 23ft per bag ; what neat weight? Cwt. qr. ft A=7 1 19 B=3 3 2l C=5 1 12 D=6 13 E=8 1 Cwt. qr. ft — Ans. 30 14 neat. 3. What is the neat weight of 15 hogsheads of tobacco, each 7cvvt. Iqr. 13ft, tare 100ft per hogshead? Ans. 97cwt. Oqr. lift. CASE II. When the tare is at so much per cwt. : Divide the gross weight by the aliquot parts of a cwt. subtract the quotient from the gross, and the remainder will be the neat weight. Examples. 1. In 129cvvt. 3qrs. 161b. gross, tare 14Ib. per ewt. what neat weight ? Cwt. qr. lb. 141b. I i I 129 3 16 gross. I 16 O 26i tare. Ans. 113 2 171^ neat. 2. In 97cwt. Iqr. 71b. gross, 3. What is the neat weight of tare 201b. per cwt. what neat 9 barrels of potash, each weigh- weight. ing 3051b, gross, tare 121b. per Ans. 79Cwt. 3qrs. 211b. neat. cwt. ? Ans. 2450ft 14oz. 44dr. 4. What is the value of the neat weight of 7hhd8. of tobacco, at 51. 7s. 6d. per cwt. each weighing 8cwt. 3qrs. 10ft gross, tare 21ft per cwt. ? Ans. £270 4' 4} reckoning the odd ounces. CASE HI. When tret is allowed with tare : Divide the suttle weight by 26, and the quotient will be the tret, which subtract from the suttle, and the remainder will be the neat.-^ * Tret is 41b. in 104, which is -_4_= i . And CIofF i? 51b. in 3€wt. or • Hh. which IS _|-=_J_. Z \m INVOLUTION. Examples. 1. In 247cu;t. 2qt:s. 15il5 gross, tare 281fe per cwt. aiid tret 4fe for every 104!fe what neat weight ? I 28 I I I 247C.2qr.l5ife gross. 61 3 17 12 tare, subtract. I 4 I 2V I 185 2 25 4 suttle. 7 16 tret, subtract. Ans. 178 2 9 4 neat. 2. What Is the neat vveight of 4hh(ls of tobacco, weighing as fal- low: The 1st. 5cvvt. Iqr. 121fe gioss, tare 65ife per hhd. ; the 2d. 3cvvt Oqr. 191fe gross, tare 75ife : the 3d 6cwt. 3qrs. gross, tare 49ife ; and the 4lh, 4cwt. 2qrs. 9fe gross, tare 35ife, and allowing tret to each as usual ? Ans. 17cwt. Oqr. 19ife4- CASE IV. When tare, tret, and cUrff^ are allowed : Deduct the tare and tret as before, and divide the sattle by 168, and the quotient will be the cloff, which subtract from the suttle, and the remainder will be the neat. Examples. 1. What is the neat weight of Ihhd. of tobacco, weighing 16cwt. 2qrs. 20lb gross, tare 141fe per cwt. tret 4ife per 104, and cloflf2it per 3cwt. ? 14feisi)16 2 20 gross. 2 9 8 tare, subtract. 43feis ^\)U 2 10 8 2 6 13 tret, subtract. 2ife is lis)^-* ^ 3^1 *"^^'^- 9 5 cloff, subtract. Ans. 13 3 22 6 neat. 2. If 9hhds. of tobacco/ contain 85cwt. Oqr. 2ife, tare 30ife per hhd. tret and cloff as usual, what will the neat weight come to at €^d. per fe after deducting for duties and other charges, 511. lis. 8d ? Ans.Jt:iiJ7 18s. 5d. INVOLUTION TEACHES the method ofiinding ihe powers of numbers. A power is the product arising from multiplying any number in- to itself contmnally a certain number of times. Thus, any number* is called the ^rs^ power, as 3 ; if it be multiplied by itself, the pro- duct is called the second power or square, as 3x3 ; if the second power be multiplied by the first power, the product is called the INVOLUTION. 18-7 i^rc/ power, or cube, as 3x3x3; if the third power be multiplied by the first power, the product is the fourtk power, or biquadrate, as 3x3x3x3, or 81 is the fourth power of 3, and so on. The power is often denoted by a figure placed at the right and a little above the number, which figure is called the index or ex- ponent of that power. The index or exponent is always one more than the number of multiplications to produce the power, or is equal to the number of times the given number is used as a factor in producing the power. Thus the square of 3, = 3 X 3 = 32 ; and the cube of 3, =3x3x3 = 33; and the 4th power of 3, =3x3x3x3 = 3*; and the 5th power of 3, = 3 X 3 X 3 x 3 X 3=3^ , and so on. In producing the square of 3, for instance, there is only one mul- tiplication, or two factors ; ia producing the cube, there are two multiplications or three factors, and so on. Hence, Involution is performed by the following Rule. Multiply the given number, or fi,rst power continually by itself, till the number of multiplications be 1 less than the index of the power to be found, and the last product will be the power required. Note. Whence, because fractions are multiplied by taking the products of their numerators, and of their denominators, they will be involved by raising each of their terhis (o the power required, and if a mixed number be proposed, either reduce it to an impro- per fraction, or reduce the vulgar fraction to a decimal, and pro- ceed by the rule. Examples. 1. What is the 5th power of 9 ? 9 9 8I=2d. power. 729=3d. power. 9 G561-— 4th. power. 9 59049=5th power, or ansvver=95. 2. What is the 5th power of f ? Ans. 3^^"^,. 3. What is the fourth power of -045 ? Ans. -000004100625. Here we see, that in rai>iing a fraction to a higher power, we decrease its value. 188 EVOLUTION. EVOLUTION, OR THE EXTRACTION OF ROOTS. THE Root is a number whose continual multiplication into itself produces the power, and is denominated the square, cube, biqiia- drate, or 2d. 3d. 4th. root, &c. accordingly as it is, when raised to the 2d. 3d. &c. power, equal to that power. Thus, 4 is the square root of 16, because 4x4=16, and 3 is the cube root of 27, because 3x3x3=^:27, and so on. Although there is no number of which we cannot find any power exactly, yet there are many numbers, of which precise roots can never be determined. But, by the help of decimals, we can ap- proximate towards the i^ot to any assigned degree of exactness. The roots, which approximate, are called surd roots, and those which are perfectly accurate, are called rational roots. Roots are sometimes denoted by writing the character v' before the power, with the index of the power over it ; thus the 3d root 3 of 36 is expressed >/ 36, and the 2d root of 36 is V 36, the in- dex 2 being omitted when the square root is designed. If the power be expressed by several numbers, with the sign -f- or — between them, a line is drawn from the top of the sign over all the parts of it. Thus th e 3d ro ot of 47-f 22 is -^47+22, and the 2d root of 69 — 17 is ^69—1 7, &c. Sometimes roots are designated like powers, with fractional in- dices. Thus, the squre root of 15, is ]5'2, the cube root of 21 is 2P» and 4th root of 37 — r- 20 is 37— 20* , &c. The denominator shows the root which is to be extracted, and the numerator shows the power to which that root is to be raised. Or the number may be raised to the power indicated b}^ the numerator, and the root, in- 2. 2 dicated by the denominator, then extracted. Thus 64"^=4 f=^lC. :? 3 3 T=^Q4 =V^4096=16. Hence the square of the cube root of any quantity is the same as the cube root of the square of the same qjiantity. The index or exponent of the root is one more than the number of muhiplications, required to produce the given number orpow«?r. EVOLUTION". 189 w i 1 >-< n r c i hO g .1 Q '(5 1 1 c 1 C/3 O^ CD c fa CJJ i 1 03 1 o 1 1 • ft , ! 1 1 o 1 1 o •s o O 1 c •-t o p o 1 o CO S o 1^ J5 o •-! CI o p" o c 1 o ^, _, i^ih^i^I— 1 -: ~| — 1 ^|,— 1 -Alw-. -. — CO 05 CO 05 i-3 ■ o Cn Si ^S 2 ^ o> j:^. 4 4i. CO i CO i en CO CO O C5 O-i CO cn CO -4 i 2 -. CO 03 1 CO Cti 05 -J o 1 CO o -J O'l >£^ CT) *-• g C5 4-. 1 CO o i i CO -a i CO CO CO O-i . 1 03 to Ci 8 l^ en 1 CO 1 -1 CO O CO 6S 8 i -J o CD -4 ^ o o ^ -3 -J C5 05 ^ 05 i CO ^^ CD CO CO CO § 1. 2 Oi o: )0 s 0-1 5 CO o CO i 03 -J CO 1 o j^ -1 CO 1 i 1 i o en CO i 9 CO § CO u o s CO 1 12 i 0> i 1 s C5 On CD li 1 11 2 ^3 s CO CO (-0 8 i CO 1 o s 1 3. 1 -J CO a> CO on 00 4i- 4i. CO CO CD 1 CO CO K O 190 EXTRACTION OF THE SQUARE ROOTo THE EXTRACTIOxX OF THE SQUARE ROOT. Rule. *1. Distinguish the given number into periods of two figures each, by putting a point over the place of units, another over the place of hundreds, and so on, which points shew the number of figures the root will consist of, 2. Find the greatest gquare number in the first, or left hand period, place the root of it at the right hand of the given number, (after the manner of a quotient in division) for the first figure of the root, and the square number, under the period, and subtract it therefrom, and to the remainder bring down the next period for a dividend. 3. Place the double of the root, already found, on the left hand of the dividend for a divisor. 4. Seek how often the divisor is contained in the dividend, (ex- cept the right hand figure) and place the answer in the root for the second figure of it, and likewise on the right hand of the divi- sor : Multiply the divisor with the figure last annexed by the figure last placed in the root, and subtract the product from the dividend ; To the remainder join the next period for a new dividend. 5. Double the figures already found in the root, for a new divi- sor, (or, bring down your last divisor for a new one, doubling the light hand figure of it) and from these, find the next figure in the root as last directed, and continue the operation, in the same man- ner, till you have brought down all the periods. Note 1. If when the given power is pointed off as the power requires, the left hand figure should be deficient, it must neverthe- less stand as the first period. Note 2. If there be decimals in the given number, it must b6 pointed both ways from the place ot units : If, when there are ^ In or;ler to slaew the reason of the rale, it will be proper to premise the following Lemma. The product of any two numbers can have, at most, but so niaiiy places of figures as ra-e in both the factors, and at least but one less. Dcvwnstraiion. Take two.r.umbers consisting of any number of places ; but let. them be the least possible of those placss, viz. Unity with cyphers, as 100 und 10 : Then their product will be 1 with so many cyphers annexed as are in. t)oth the aumbers, viz. 1000 ; but 1000 has one place less than 100 and ' 10 to- gether have : And siace 100 and 10 wei-e taken the least possible, the product of sny other two numbers, of the same number of places, will be greater than li)00 ; consequently, the product of any two numbers can have, at least, but one place less than both the factors. Again, take two numbers, of any number of places, which shall be the grcaJ- f :t possible of those places, as 03 and 9. ■■ Now, 99 X 9 is less than 99 X 10 ; but 'JO X 10 (:c:990) contains only so many places of figures as are in 99 and 9 ; therelore, 99 X 9, or the product of any other two numbers, consisting of the -rme number of places^ canaot liave Baore places of figures, than are in both its factors. .' - \, ('crollary 1; A square uuiTibcr cannot have more places of figures th%n,double ♦•♦e j-daoee of the' root, aiid at Ica^t but one lesfe. ; .'". . ^oroUci-ij 2. A cube number cannot have more pla.ccs of figftrcs Ihafi triple tlie rVv:rf of thsroo*. rr'' ;-^ > .* i ■ ^ ' , . -, ,. EXTRACTION OF THE SQUARE ROOT. 191 integers, the first period ia the decimals he deficient, it may be completed by annexing so many cyphers as the power requires : And the root must be made to consist of so many whole numbers and decimals as there are periods belonging to each ; and when the periods belonging to the given number are exhausted, the ope- ration may be continued at pleasure by annexing cyphers. Examples. 1st. Required the square root of 30138696025 ? 30138696026(173605 the root. 1 1st. Divisor=27)201 189 2d. Divisor=343)1238 1029 3d. Divisor=3466)20969 20796 4th. Divisor=347205) 1736025 1736025* I 2d. Required the square root of 5755 ? 675'50(23'98+, the root, 4 43)175 129 469)4650 4221 4788)42900 38304 4596 Remainder. ■^ The Rule for the extraction of the square root, may be illustrated by at- tending to the process by which any number is raised to the square. The seve- ral products of the multiplication are to be kept separate, as in the proof of the rule for Simple Multiplication. Let 37 be the number to be raised to the square, 37 X 37= 1369====37 X 37 37 37 49=72 49=73 21 .=3X7 > _2x3x7 210=30X7 ) ^o^^Q^f 21 .=3X7 S — ^'^-^^ ' 210=30X7 \ -^"^"^ 9 . .=32 <)00=302 ... (37 (30-4-7=37 fX3)42 .=2X3X7 " * 49^=72 rCarried over. 192 EXTRACTION OF THE SQUARE IlOOt. oti. What is the square root of 10342656 ? Ans. 3216. •nil. What is the square root of 964-5192360241 ? Ans. 31-05B7L ' 5th. What is the square root of 234 09 ? Ans. 15-3. 6th. What is the square root of -0000316969 ? Ans. -00563. 7th What is the square root of -046369 ? Ans. -213. Rules For the Square Root of Vulgar Fractions and Mixed J^umlers. x\fter reducing- the fraction to its lowest terms, for this and all other roots ; then, 1st. Extract the root of the numerator for a new numerator^ and the root of the denominator iov a new denominator, which is the best method, provided the denominator be a complete power. But if it be not, 2d. Multiply the numerator and denominator together ; and the root of this product being- made the numerator to the denominator of the given factor, or made the denominator to the numerator of it, will form the fractional part required.* Or, Now, it is evident that 9, in the place of hundredths, is the greatest square iis this product ; put its root, 3, in the quotient, and 900 is taken from the product. The next products are 21-f21=2x3x 7, for a dividend. Double the root al- ready found, and it is 2 X 3, for a divisor, which gives 7 for the quotient, whicli annexed to the divisor, and the whole then multiplied by it, gives 2x3x7 (=42) -f-7x7 (=49) which placed in their proper places, completely exliausts the- re- mainder of the square. The same may be shown in any other case, and the rule become? obvious. Perhaps the following may be considered more simple and plain. Let 37,=: 30-f-7, be multiplied, as in th^^monstration of simple multiplication, and the products kept sei^arate, 30+7 30+7 900+30x7 30X7+49 900+2x30x7+49=1369, the sum, and square. 900 (30+7 The root of 900 is 30, and leaves the two other terms, which are ex- 2X30+7x7)2x30x7+49 hausted by a divisor, formed and 2 X 30 X 7+49 multiplied as directed in the rule. /7x'^ 7 '"^ Let the fraction be >/7^ then by the rule, >/l= — ^~~"" — "- =^l'o7+. • -' ^ -- v/7X2 The reason of which is, that the value of a fraction is not altered by multiplyino^ l)oth its parts by the same quajitity. Thus y/l~ ^ ^ ^ ^' But v'7X\/2= " V7Xx/ 7_ 7 V7X2, and ^2Xv/2=2 evidently. Ajid thus also, ^^x^y~ n'-ycf /7 X2^ -and is the rule. Sco Sards. EXTRACTION OF THE SQUARE ROOT. 193 od. Reduce the vulgar fraction to a decimal, and extract its root. 4th. Mixed numbers may either be reduced to improper fractions, and extracted by the first or second rule, or the vulgar fraction may be reduced to a decimal, then joined to the integer, and the root of the whole extracted. Examples. 1st. What is the square root of y^flg ? By Rule 1. TiHo^TaaT ^^C"* *'°°^ of the numerator. 16 1681(41 root of the denominator. 16 ol)81 Therefore, -/y=the root of the given fraction. 81 By Rule 2. 16x1681=26896, and ^26896=164. Then, TVVT=TV4=TV=09756-f By Rule 3. IG81)16(-0095181439H-. And ^/.0095181439=-097364-. 2d. What is the square root of f|f| ? Ans. -f,^. 3d. What is the square root of 42{ ? Ans. 61-. Note. In extracting the square or cube root of any surd num- ber, there is always a remainder or fraction left, when the root is found. To find the value of which, the common method is, to an- nex pairs of cyphers to the resolvend, for the square, and ternaries of C)'[)hers to that of the cube, which makes it tedious to discover ihe value of the remainder, especially in the cube, whereas this trouble might be saved if the true denominator could be discovered. As in division the divisor is always the denominator to its own fraction, so likewise it is in the square and cube, each of their di- visors being ihe denominators to their own particular fractions or numerators. In the square the quotient is always doubled for a new divisor ; therefore, when the work is completed, the root doubled is the true divisor or den(iminator to its own fraction ; as, if the root be 12, the denominator will be 24, to be placed under the remainder, which vulgar fraction, or its equivalent decimal, must be annexed to the quotient or root, to complete it."'^ If to the remainder, either of the square or cube, cyphers be an- nexed, and divided by their respective denominators, the quotient will produce the decimals belon,^insr to the root. ^* Although these denominators give a small matter too much in the square root, and too little in the cube, yet they will be sufficient in common ii?e. i\n?iitioiT^ than the operation -with cvpher?. A a 191 APFLICATION AND USE APPLICATION AND USE OF THE SQUARE ROOT. Prob, I. To find a 7nean proportional between two numbers. Rule, Blultiply the given numbers together, and extract the square root of the product ; which root will be the mean propor- tional sought. Note. When the first is to the second as the second 13 to the third, the second is called a mean proportional between the other two. Thus, 4 is a mean proportional between 2 and 8, for 2 : 4 :; 4X4 4 :: ~^=B, or 4 is as much greater than 2, as 8 is greater than 4. Bj' Theorem I. of Geometrical Proportion 2x8=4x4=4 . To iind a mean proportional between 2 and 8, take the square root of their product. The same must be true in every case, and is the rule. Example;. What is the mean proportional between 24 and 96 ? V^96x24=48. Answer. Prob. II. To find the side of a square equal in area to arty given superficies whatever. Rule. Find the area, and the square root is the side of the Sijuare sought.* Examples. 1st. If the area of a circle be 184125, What is the side of a square equal in area thereto ? v/184 126=13-569-f Answer. 2d. If the area of a triangle be 160, what is the side of a square equal in area thereto? \/lG0=12*649-f- Answer. Prob. III. A certain general has an army of 6625 men : pray How many must he place in rank and file, to form them into a square ? V'5625=75 Answer.' Prob. IV. Let 10952 men be so formed, as that the number in rank may be double the file. 74 in file, and 148 in rank. Prob V. If it be required to place 2016 men so as that there may be 56 in rank and 36 in file, and to stand 4 feet distance in rank and as much in file, How much ground do they stand on? To answer this, or any of the kind, use the following propor- tion : As unity : the distance :: so is the number in rank less by one : to a fourth number ; next, do the same by the file, and mul- * A square i.s a figure of four equal sides, each pair meeting perpentliculaily. or, a fl,^are whose length and breadth are equal. As the area, or number oi cquare feet, inches, &c. in a square, is equal to the product of two sides whicli are equal, the second power is called the square. Hence tlie rule of Prob. II. is evident. t If you would have the number of men be double, triple, or quadruple, &c. as many in rank as in file, extract the square root of ^, i, ^, &;c. of the given num- ber of men, and that will be the number of men in file, whioli double, triplf. quadruple, &c. and the product will be the number in rank. I OF THE SQUARE ROOJ, 195 tiply the two numbers together, found by the above proportion, and the product will be the answer.* As 1 : 4 :: 56~-l : 220. And, as 1 : 4 :: 36—1 : 140. Then. 220X140=30800 square feet, the Answer. Prob. VI. Suppose I would set ont an orchard of 600 trees, so that the length shall be to the breadth as 3 to 2, and the distance of each tree, one from the other, 7 yards : Mow many trees must it be in length, and how many in breadth? and, How many square yards of ground do they stand on ? To resolve any question of this nature, say, as the ratio in length : is to the ratio in breadth :: so is the number of trees : to a fourth number, whose square root is the number in breadth. And as the ratio in breadth : is to the ratio in length :: so is the number of trees : to a fourth, whose root is the number in length. As 3 : 2 :: 600 : 400. And v^400=20=n umber in breadth. As 2 ; 3 :: 600 : 900. And v'900=30=nuraber in length. As 1 ; 7 :: 30—1 : 203. And as 1:7:: 20—1 : to 133. And. 203x133=26999 square yards, the Answer. Prob. VII. Admit a leaden pipe ^ inch diameter tvill fill a cistern in 3 hours; I demand the diameter of another pipe which will fill the same cistern in 1 hour. Rule. As the given time is to the square of the given diameter, so is the required time to the square of the required, diameter j 1= 75 : and •75x-75= 5625. Then, as 3h. : -5625 :: Ih. : 1*6875 inversely, and \/l'6875=l 3, inches nearly, Ans, Prob VHI. If a pipe whose diameter is 1 5 inches, fill a cistern iu 5 hours, in what time will a pipe whose diameter, is 3-5 inches fill the same ? 1-5x15=2 25; and 3-5 x3'5=l 2-25. Then, as 2-25 : 5 :: 12'25 : 918-f-hour, inversely =^55 min. 5 sec. Answer. Prob. IX. If a pipe 6 inches ])ore, will be 4 hours in, running off a certain quantity of water. In what time will 3 pipes, each four inches bore, be in discharging double the quantity ? 6x6=36. 4X4=16, and 16x3=48. Then, as 36 : 4h. :: 48 :^ 3h. inversely, and as Iw. : 3h. :: 2w. : 6h. Answer. Prob X. Given the diameter of a circle, to make another circle which shall be 2, 3, 4, &c, tiroes greater or less than the given circle. Rur.E. Square the given diameter, and if the required circle bo greater, multiply the square of the diameter by the given pro- portion, and the square root of the product will be the required di- ameter. But if the required circle be less, divide the square of the diameter by the given proportion, and the root of the quotient will be the diameter required. * The abovo rule will be found useful ia pUmting trees, having the distance of ground between each given. t For more water will run as the area of the pipe is grcalei-, nml the areas of circular pipes vaj:y as the square of tlxeir diameter-. 196 USE, &c. OF THE SQUARE ROOT. There is a circle whose diameter is 4 inches ; I demand the di" ameter of a circle 3 times as large ? 4X4=16 ; and 16x3=48; and ^48=6-928-1- inches, Answer. Prob. XI. To find the diameter of a circle equal in area, to an ellipsis, (or oval) whose transverse and conjugate diameters are given.* Rule. Multiply th<» two diameters of the ellipsis together, and the square root of that product will be the diameter of a circle equal to the ellipsis. Let the transverse diameter of an ellipsis be 48, and the conju- gate 36 : What is the diameter of an equal circle ? 48x36=1728, and v'l 728=41. 569-{-the Answer. Note. The square of the hypolhenusfe, or the longest side of a right angled triangle, is equal to the sum of the squares of the other two sides; and consequently the difterence of the squares of the hypothenuse and either of the other sides is the square of the remaining side. Prob. XII. A line 36 yards long will exactly reach from the top of a fort to the opposite bank of a river, known to be 24 yards broad. The height of the wall is required? 36x36 = 1296 ; and 24x24=576. Then, 1296—576=720, and y'720=2e'J3-l-yards, the Answer. Prob. XIII. The height of a tree growing in the centre of a cir- cular island 44 feet in diameter, is 75 feet, and a line stretched from the top of it over to the hither edge of the water, is 256 iept. What is the breadth of the stream, provided the land on each side of the water be level ? 256x256=65536: and 75:^75=5625 : Then, 65536— -5625= 59911 and V'59911=244-76-(- and 244'76— ^=222-76 feet, Ans. Prob. XIV. Suppose a ladder 60 feet long be so planted as to reach a window 37 feet from the ground, on one side of the street, and without moving it at the foot, will reach a window 23 feet high on the other side ; I demand the breadth of the street ? 102.64 feet the Answer. Prob. XV. Two ships sail from the same port ; one goes due north 45 leagues, and the other due west 76 leagues : How far are they asunder?! 88*32 league*?, Answer. Prob. XVI. Given the sum of two numbers, and the difierencr. of \heir squares, to find those numbers Rule. Divide the diflerenr.e of their squares by the sum of the numbers, and the quotient will be their diiierence. The two num- * The tmnsverse and conjugate are the longest and shortest diameters of an <5llip!?is ; they pass throngh the centre, and cross each other at right angles, and the diameter of the equal circle is the square root of the product of the diame- ters of the ellipsis. t The square root ma)' in the same manner he applied to navigation ; and, when deprived of other means of solving problems of that nature, the following proportion will serve to fuid the course. As the sura of the hypothenuse (or distance) and half the greater leg (wheth- er difference of latitude or departure) is to the less leg ; so is 86, to the, angle Ojjpositc the less leg. EXTRACTION OF THE CUBE ROOT. 197 bers may then be found, from their sum and difference, by Prob. 4, page 57. Ex. The sum of two numbers is 32, and the difference of their squares is 256, what are the numbers ? Ans. The greater is 20. The less 12. Prob. XVII. Given the difference of two numbers, and the dif- ference of their squares, to find the numbers. Rule. Divide the difference of the squares by the difference of the numbers, and the quotient will be their sum. Then pro- ceed by Prob. 4, p. 57. Ex. The difference of two numbers is 20, and the difference of their squares is 2000 ; what are the numbers ? Ans. GO the greater. 40 the less. Examples for the two preceding problems. 1. A and B played at marbles, having 14 apiece at first ; B hav- ing lost some, would play no longer, and the difference of the squares of the numbers which each then had, was 336 ; pray how many did B lose ? Ans. B lost 6. 2. Said Harry to Charles, my father gave me 12 apples more than he gave brother Jack, and the difference of the squares of our separate parcels was 288 ; Now, tell me how many he gave us, and you shall have half of mine. Ans. Harry's share 12. Jack's share 6. EXTRACTION OF THE CUBE ROOT. A cube is any number multiplied by its square. To extract the cube toot, is to find a number which, being multiplied into its square, shall produce the given number, FIRST METHOD. Rule. *1. Separate, the given number into periods of three figures each, by putting a point over the unit figure, and every third fi- gure beyond the place of units. 2. Find the greatest cube in the left hand period, and put its root in the quotient. 3. Subtract the ctibe, thus found, from the said period, and to the remainder bring down the next period, and call this the divi' dend. 4 Alaltiply the square of the quotient by 300, calling it the triple square, and the quotient by 30, calling it the triple quotient, and the sum of these call the divisor. 5. Seek how often the divisor may be had in the dividend, and place the result in the quotient. *= The reason of pointing the given number, as directed in the rule, is obvious from Coroll. 2, to the Lemma made use of in demonstrating the square root. The process for extracting the Cube Root may be iUustrated in the same mauaer as that for the Square Root. Jake the fame number 37, and multiply }m EXTRACTION OF THE CUBE ROOT. 6. Multiply the triple square by the last quotient figure, and write the product under the dividend ; multiply the square of the last quotient figure by the triple quotient, and place this product under the last ; under all, set the cube of the last quotient figure and call their sum the subtrahend. as before, collecting the twice 21 into one sum, as they belong to ami the operation will be simplified, 37 3=50653. the same place C 49=72 :< 42-=2> 49=72 X3x7 420=2X30X' 900=303 7 the multiplier. 37 ;73=^ f 343=73 343=7 3 294 -=2X3X72 2940=2x30x7 6:i ■ -=33 X 7 0300=302 X 7 1 47 -=3 X 7 2 1470=30 X 7 2 1 20 • -=2 X 32 X 7 1 2600=2 X 30 " X ' 27 •••=3^ 27000=303 27- •• (37 ' 27000 (30x7 ;X32M89* -=3x32x7 3 X 302 \1 8900=3 X 302 x' ,X3 J 441—3x3x72 3x30/ 4410=3X30X7 As 27 or 27000 is the greatest cube, its root is 3 or 30, and that part of tiie cube is exhausted by this extraction. Collect those terms which belong to the same places, and we have 32 x7=G3, and 2x32 X 7=126, and 63-1-126=3x32 X7=189; and2x3x72=294 and 3x72=147, and 294-f 147=441=3X3X72, for a clixddend, which divided by the divisor, formed according to the rule, the quotient is 7, lor the next figure in the root. And it is evident on inspecting the work, that that part of the cube not exhausted is composed of the several products which form the suhlrahend^ according to the rule. The same may be shov/n in any other case, and the universality of the rule hence inferred. The other method of illustration, employed on the square root, is equally applicable in Ihis'case. 37:--;- 0-1- 7, and 30- ., 2x30x7-}-72 30-| 7 the multiplier. 30"-f 2X302 x7-j- 30X72 302x7-f2x30x72-}-73 :.73:-50653=30;5-i-3x302x7-l-3x30x72 -1-7 3 (30-1-7=3'; 30^ i;.vi:> .^ 3.C0O- ..,,. ..J :::x302 x7-f-3X 30x72-1-7 3 dividend. ;:x302 X 74-3x50x73 -{-7 3 subtrahend. It is evident that 303 is the greatest cube. When its root is extracted, llie Bext three terms eonslltute, the dividend ; and, tbe several products formed by means of the quotient or second figure in the root, are precisely ^equal to the remainitig parts of the povier, whose root was to be found. The arithmetical demonstrations of the Rules for extracting either the square or cube root, are not only more consistent with the plan of an Arithmctick than demonstrations on the figure, called a square, and the solid, called a cube^ but they are much more readily undereluo'i 1'}' thoco unaccustomed lothe mathcmati- «ulconside": '^'i"> ' 3, it makes 432, to which add 36, the triple number of the root, and it produces 468 for a denominator.* SECOND METHOD. Rule. 1. Having pointed the given number into periods of three figures iftacb, find the greatest cube in the left hand period, subtracting it therefrom and placing'its root in the quotient ; to the remainder bring down the next period and call it the dividend. 2. Under this dividend write the triple square of the root, so that units in the latter may stand under the place of hundreds in the former ; and under the said triple square, write the triple root, removed one place to the right hand, and call the sum of these the divisor. 3. Seek how often the divisor may be had in the dividend, ex- clusive of the place of units, and write the result in the quotient. 4. Under the divisor write the product of the triple square of the root by the last quotient figure, setting down the unit's place of this line, under the place of tens in the divisor ; under this line, write the product of the triple root by the square of the last quotient figure, so as to be removed one place beyond the right hand figure of the former ; and, under this line, removed one place forward to the right hand, writedown the cube of the last quotient figure, and call their sum the subtrahend. 5. Subtract the subtrahend from the dividend, and to the re- mainder bring down the next period for a new dividend, with which proceed as before, and so on until the whole be finished. Example. Required the Cube Root of 16194277 ? * It may not be amiss to remark here, that t]ie denominators, botli of the square and cube, shew how many numbers they are denominators to, that is, what numbers are contained between any square or cube number and the next succcediir^ square or cube number, exclusive of botli numbers, for a complete number, of either, leaves no fraction, when the root is extracted, and conse- quently has no use for a denominator, but all the numbers contained between them have occasion for it : Suppose the square root to be 12, then its square is 144, and the denominator 2^^, which will be a denominator to all the succeeding numbers, mitil we come to the next square number, vis. 169, whose root is lij, with which it has nothinjj to do, for between the square numbers 144 and 169 are contained 24 numbers excluding;- both the square numbers. It is the same in the cube ; for, suppose the root to be 6, the cube number is 216, and its de- nominator 126 will be a denominator to all the succeeding numbers, until avc .-•ome to the next cube number, viz. 343, whose root is 7, with which it has noth- ing to do, as ceasing then to be a denominator; for between the cube 343 and 216 arc 126 numbers, excluding both cubes. And so it is with all other denom- inatoi's, cither in the square or cube. EXTRACTION O^ THE CUBE ROOT. 201 16194277(253=Root. 8 8194 = First dividend. 12 = Triple square of 2. 6 =a: Triple of 2. » 126 = First divisor. 60 == Triple square of 2, multiplied by 5. 150 = Triple of 2 multiplied by the square of 5, 125 = Cube of 5. 7625 == First Subtrahend. 569277 = Second dividend. 1875 = Triple square of 25. 75 ::= Triple of 25. 18825 = Second divisor. B625 = Triple square of 25 multiplied by 3. 675 = Triple of 25 multiplied by the square of 3. 27 == Cube of 3. 569277 =^ Second subtrahend. FIRST METHOD BY APPROXIMATION. Rule. 1. Find, by trial, a cube near to the given number, and call it the supposed cube. 2. Then as twice the supposed cube, added to the given num- ber, is to twice the given number, added to the supposed cube, so is the root of the supposed cube, to the true root, or an ap- proximation to it. 3. By taking the cube of the root, thus found, for the supposed cube, and repeating the operation, the root will be had to a great- er degree of exactness. Example. It is required to find the cube root of 54854153 ? Let 64000000=supposed cube, whose root is 400 ; Then, 64000000 64854153 2 2 128000000 109708306 5485415 3 64000000 As 182854153 : 173708306":: 400 400 182854153)69483322400(379— root nearly. B b 202 EXTRACTION OF THE CUBE ROOT. Again, let 54439939 = supposed cube, whose root is 370. Then, 54439939 54854153 2 2 108879878 109708306 54854153 54439939 As 163734031 : 164148246 :: 37J> 379 1477334205 1149037715 492444735 l63734031)62212184855(379-958793+=rootcorFected. SECO? 20 4 2 I EXTRACTION OF THE ROOTS. 207 A Oeneral Rule for Extracting the Roots of all Powers. 1.* Prepare the given number, for extraction, by pointing off from the unit's place, as the required root directs. 2. Find the first figure of the root by trial, or by inspection into the table Of powers, and subtract its power from the left hand pe- riod. 3. To the remainder bring down the first figure in the next peri- od, and call it the dividend. 4. Involve the root to the next inferiour power to that which is ■given, and multiply it by the number denoting tiie given power, for a divisor. 6. Find how many times the divisor may be had in the dividend, and the quotient will be another figure of the root. 6. Involve the whole root to the given power, and subtract it from the giv€7i number, as before. 7. Bring down the first figure of the next period to the remain der for a new dividend, to which find a new divisor, as before, aiul., in like manner, proceed till the whole be finished. Examples. 1st. What is the cube root of 20346417 ? 20346417(273 2X2x2=8 root of the 1st. period, or 1st. Subtrahend. 23 = 8 z=i 1st. Subtrah.2x2=4(=next inferiour power,) and, — 4x3=(the index of the given pow.)= 12 Ist. Divisor. 22x3==:12)123=Dividend 27x27x27= 19683=2d. Subtrahend. 27x27=729 (next inferior power) and, 073 = 19683=2d. Subt.729x3(=index of the given pow.) = 2187=2d. Divisor. 272x3=2I87)6634=2d.Di.273x273x273=27346417=3d. Subtra. 2733 = 20346417=3d. Subtrahend, * The extracting of roo^s of very high powers will, by this rule, be a tedious operatiori : The following method, when practicable, will be much more convoi" ient. When the index of the power, whose root is to be extracted, is a composite number, take any two or more indices, Whose product is equal to the given indez\, and extract out of the given number a root answering to another of the indices, and so on to the last. Thus, the fourth root=:square root of the square root ; the sixth root=:square root of the cube root; the eighth root=:;square root of the fourth root; the ninth root=the cube root of the cube root ; the truth root=;square root of the fifth root ; the twelfth root=the cube root of tlie fourth, &c. The general rule for extracting the roots of all powers, may be illustrated in the same way, as those for the square and cube roots. Any student may at once see the truth of the rule, in exhausting the several products of the case illustrat- :ng the rule for the cube root. And thn same will be evident by raisiaa: ihc iH'-li'^r to any hi2:her power. 20^ EXTRACTION OF ROOTS Sd. What is the biquadrate root of 34827998976 ? Ans.431-9'(' 3d. Extract the sursolid, or fifth root of 281950621 876? Ans. 195. 4th. Extract the square cubed, or sixth root of 1178420166016625? Ans. 326. A General* Rule for Extracting Roots by Approximation. 1. Subtract one from the exponent of the root required, and multiply half of the remainder by that exponent, and this product by that power of the assumed root, whose exponent is two less than that of the root required. * The general theorem for the extraction of all roots, by approximation, from whence the rule was taken, and the Theorems deducible from it, as high as tfee twelfth power. Let G=:resolvend whose root is to be extracted. p'-^e= root required ; r being assumed as near the true root, and w=:exponent of the pow- cr-^then the equation will stand thus. m — 2r^ m — 2 Hence, 1 G r f wi — 2r^ m — 2 1 ^ ^V — ... .. -^ r. m-1^ ^|m-.2 2 mm^ll"^ ^— i sm for the cube root r-\-i y/G rr r 3r 12 2 For the Biquadrate - - y/G rr 2r 6»T 18 3 For the Sursolid - - - y/G 3rr 2r 10r3 80 4 For the squared cube root y/G 2rr 3r 15r4 75 5 For the second sursolid - ^G brr 5r 21r5 252 6 . y/G 3rr 6r For the squared Biquadrate 1 28r6 196 7 For the cubed cube - - y/G 7rr 7r 36r^ 576 8 For the squared sursolid ^/G Arr 8r 45r8 405 9 For the third sursolid - v/G 9rr Or 55r9 1100 10 y/G 5rr 10 r For the squared square cube — 1 &c. 66rio 726 11 :j: By this Theorem the fraction is obtained in numbers to the lowest terms in all the odd powers ; and in the even powers o»ly by having the numerator and denominator found by this equation. BY APPROXIMATION. 209 2. Divide the given number by the last product ; and from the quotient subtract a traction, whose numerator is obtained by sub- tracting two from the exponent, and multiplying the remainder by the square of the assumed root ; and whose denominator is found by subtracting one from the exponent and multiplying the square of the remainder by the exponent. 3. After this subtraction is made, extract the square root of the remainder. 4. From the exponent subtract two, and place the remainder as a numerator ; then subtract one from the exponent, and place the remainder under the numerator for a denominator. 5. Multiply this fraction by the assumed root ; add the product to the square root, before found, and the sum will be the root re- quired, or an approximation to it. Example. What is the square cubed root of 1178420166015625 ? Let the assumed root = 300 p 1 Exponent of the required root is 6. Then, x6=15. 2 3004=8100000000 and this multiplied by 15=121500000000. 1178 42016 6015625-r-123 500000000=96989314, from this 6—2x3002 Subtract — —2400 6x6—12 And x/72y8y314=85-43 6—2 To which add 7^ — rX300= 240 o — 1 ^ And the sum is the approximated root= 325^43 For the 2d operation, let 325-43 = assumed root. ANOTHER METHOD BY APPROXIMATION." Rule. 1. Having assumed the root in the usual way, involve it to that power denoted by the. exponent less 1. * A rational formula for extracting the root of any pure power by approxi- mation. Let the resolvend be called G, and let r-\-e be the required root, r being as* sumed in the usual way. 1 G m~l Let G — be required ; then r-\-e = 1 —r the general Theorem. in m — 1 m mr G 2 Hence, For the cube root r-\-e = 1 r. G 3 For th^ bis/3. Aho, 1^32—1/16x^=1/ IS XV'2=2V'2. Reduce \/l2o to its most simple terms. * Ans. 5\/5. 3 • 3 Reduce \/3384 to its most simple terms. Ans. 8n/7. 4 Reduce V^^j to its simplest terms. Ans. |Vf- 3 7 Reduce \/481, v/351, and \/896 to their most simple term?. Reduce 5\/20 to its simplest terms. Ans. 10\/5. Hence, it is obvious, that if a factor be multiplied into a surd, the whole may be reduced tn the form of a surd, by raising the factor to the power denoted by the surd, multiplying the power in- to the surd, and placing the product under the radical sign. Thus 3v/3=^v/32xV3=:v79x3--=v'27 ; and 8V7= V'8 ^xV 7= .7512x7== >/3584. 3. Surds of the same radical sign may be added together, when the quantities under the radical sign are the same, by prefixing the sum of the rational parts to the surd quantity. Thus l\/24-l\/2, or N/2-f v/2=2x/2, or twice V2 ; and oV5+W5~lV5. If the surds are not already in their most simple terms, they may often be added after the reduction is made. Thus ^/20-|-v^80= 2N/5-f-4v'5==6v'5 ; and, v/162-fV1350=3V2-f 5^2=8^2. What is the sum of VBG and ^3584 ? Ans. 10v^7. 4. Surds of the same radical '•ign may be subtracted, if the surd part he the game, by placing the difference of the rational parts before tlie $?:rd. ff the quantities are not already in their simpierj-; SURDS. 213 terms, they should be reduced to this form. Thus n/4 — \/4=0 ; and 3x/3~-2v^3=lv/3 or x/3. Also 7^/5—4^5=3^5. 4 4 4 What is the difference between v/1350 and v/162 ? Ans. 2v/2. J\''ote 1. Surds, apparently incapable of addition or subtraction except by their signs, may sometimes be reduced to a commoa surd, by the following process, and their sum and difference readi- ly found. Thus let the surds be \/| and \/f As v/|=v/|xl=^ Vi=^2Vh and as V^=V^\=Wi^ then ^i+v^f =2Vi+fVi==: Wi=W^\='2XW^=TW^* their sum: And 2Vi-iVi^Wi=' jjVQy their difference. What is the sum and difference of \/| and f\/| ? Ans. Their sum is ^VQ. Their diff. is -^\VS. What is the sum of ^^^v/lS and \/^^^ ? Ans. V|. What is the difference of v^f and n//^ ? Ans. ^v^lo. JVote 2. If the same quantity is under different radical signs, or if the same radical sign has different quantities under it when the surds are in their simplest terms, the surds can be added or sub- tracted only by the signs of addition or subtraction. Thus it is 3 evident, that y/2-\-V2, is neither twice the square root of 2 nor 3 twice the cube root of 2 ; and that 3\/3— 2\/3, is neither the square root of 3 nor the cube root of 3. It is equally obvious, that 2y'3 -|-2\/2, is neither four times the square of 3 nor of 2 ; and that 4v/2 — 2v/3, is neither twice the square of 2 nor of 3. 5. Surds of the same radical sign are multiplied like other num- bers, but the product must be placed under the same radical sign, Thu| v/27Xn/64 = V'27x64=n/1728=12, for v'27=3, and ^04 =4, and ^27x^64=3x4=12. And V'2x V3=V"^xS = VG. Also 3v/3x4n/5=12V15 or v^27X\/80, and V27X\/80=x/27x80 = 12n/15=v/2160. • Sometimes the product of the surds becomes a complete power of that root, and the root should then be extracted, as in the first of the preceding examples. Also in this example ; v'2Xv'200= v/400=20. It is evident from the first example in this section, that, when quantities are under the same radical sign, the root of the product of qvantities is equal to the product of their roots. If a surd be raised to a power denoted by the index of the root, the power will be rational. Thus, \/3X\/3, or 32x3-'^=3. In this example 2 is the index of the root, and the surd is raised to 3 3 3 the second power or square. Also \/4X\/4xv'4=4. If fractional indices be used, in order to multiply surds of the same root, you l_ JL X _3 have only to add the indices. Thus 43x43x4^=43=4' or 4, uni- ty being the implied index of 4, or of the first power of any num- ber. In all cases when the sum of the numerators contains the com- ^'14 bUKDS. 7non denominator a certain number of times exactly ^ the product tt'u'/ be rational. Thus 3^x3^x3^=3 3 =3^=32=8, and 72x7'-- -:r7-T~=7 - =75. As 5"* may be expressed according to the notation ot' powers, 4 3 4. ii 4±'i 8 thus, 6*,and 53 by 5', hence 5^X5'=5 ]'=5'=o». Therefore, to multiply different powers of the same root, you have only to add the indices of the i^iven root, and place the sum for the index oi' tlie power which is produced. Thus 32x32=3'*, or the square of a number multiplied by itself produces the fourth power; the cube by the cube, the sixth power, and so on. Thus also 2*x2^ — 5_L. y 8 2 6. Surds of the same radical sign are divided like whole num- bers, but the quotient must be placed under the same radical sign. Thus -/1728--^^64=vi^--S34=V2'7==3; and V^-^V^— ^t)-^3=\/2. Sometimes the quotient becomes a complete power, as in the lir>t example, in which case the root should be extracted. So also in the following; ^400—^ 100= v'^i^^-^^^t)= V4=2. As Vl'J'28 — 12, and ^64— 4, then ^ 1720-^v'64=12~4=3=^ 3 ^27. Hence, the quotient of the roots of quantities is the same as the root of their 'quotient ^ if the quantities are under the same radi- cal sign. Divide ^ \0^ by >^6. Xow V 108-4-VG= V^08^^= V 18 = Now 9^^100=^8100, and 3^2 = V9x2=3-v^2. Divide 9^100 by 3^2. V 18, and v/8i00H-v^l8=v/810Q~-18=v/ 450=1 5v/2. Or Qv/lOO -^3v^^=9-T-3x V 400-^^2 = 9~-3x ^ 100-r-2=9H-3 X a/ bQ=^ 3v/50=Sx5v/2=15v/2. Divide -^V'yW ^v ^Vf Now-}-f-^=|; and \^ ii^-^ V 1'= \^ hS ^f v''f =^^^^2 ; and |xiv/2=fv/2. Divide ^"48 by v'-.V ; W60 by 3^15 ; and |>/-^ by iv'!- If the quantities under the radical sign be the same, the quotient will be found by dividing the rational parts only. Thus \/2~x/2 3 3 =v/li=l, or 5/2 is contaliied in v/2 once. Also iv/3+iv/3=2, and 2v/5+5\/5=3^. To divide one power by another of the same root, place the dif- ference of the indices for the index of the given root. 'J'his is mere- ly reversing a process given in t)je preceding section. The rea- son of the process may aljjO be seen in the following manner. Thu.^ 2** 2^X22 3 8 3 ^■".--C!^:^; — =r. — r=:2- =2'-'3^ A!^o. 22-!-2* =2''-;"2* , ]>v reduc SURDS. 215 4 ji-9 |_v/2^ ing the indices to a common denominator ; and 2* 4|2^ 4 r^^X^S 4 4 5. If the index of the divisor exceed that of the dividend, the index of the quotient will be their difference with the sign of subtraction before it. Thus, 5--r-5^=5--'^=5-^. Now, as 52-t-3s=— = 52 ^ 1 1 ——-=—, 5-5=—., Hence a power whose index has the sign of subtraction before it, is the same powerof the reciprocal of that quantity. Hence, there is an obvious method of transferring pow- ers from the numerator to the denominator of a fraction. Thus, 1 3-3 1 2 2^=2-2^ and -7-=. ^23' and ^37=2x3-. There is also an obvious method of finding the value of a quantity whose index has the sign of subtraction before it. Thus 2-3=-~=_ and 7[~''=T[~''= l^- 11 _i 1 1 v/2 orr25-=:^^=.^5^=16. And 2 ^=-7=;^-=— 7. To raise surds to any power, multiply the index of the surd by the index of the required power. Thus 2^ raised to the square is 22 + 2=23=^22^^=2 ; and the cube of 3* =3*"^ *'''*= 3"*''^'''' If there be rational parts with the surd, they must be raised to the given power, and prefixed to the required power of the surd. Thus, 3v/3, or 3X3^ raised to the square, is 3= X3'^^2=32x3^ =32X33^^=9n/32, or 9x/9. And the cube of 22=2^'^3+i=r 22X3=2^=>/23=v/8=:2V'2, when reduced to its simplest terms. Also, the fourth powerof Ax/2 is 2i(r^22=^^-. Required the fifth power of Ix/j;. 8. To extract any root of a surd quantity, divide the index of the quantity by the index of the required root. Thus, the square 1^ _l_::-0 1. 6 ?. 3-:-'s root of 23 is 23 * "==2« or \/2, and the cube root of 3* is' 3* ' ' 3 --=3^x4=34 Q. ^3. If there be rational parts with the surd, their root must be pre- fixed to the required root of the irrational part. Thus, the square root of 9x/9, or 9x93=92v/32=3v/3. The process must evident- ly be the reverse of that in the preceding section, and the reason of it is obvious. 216 S'URDb. What is the cube root of-^V^S ? Ans. |. What is the square root of 10* ? Ans. lO^ or 100^/10. Examples. -■ "f JL 2 1. MuUiplj G* by G% and the product is G-^^ or v/G^. 2. Divide 6"^ by G% and the quotient i§ 6^^ or -v/6. 3 3 3 :3. Add \/32 and \/108, and multiply the same by x/yf^. Ans IG^orgv/S, 3 3 3 4. Addv32 and \/100, and divide (he sum by \/,4j- Ans. 5^. 5, Find the shortest method of dividing 3 by v/2, to any given place of decimal?. 3 3Xx/2 3v/2 x/18 4-242640 &c ^"^'^ ~v/7=;7iW2="^"^-*^= i =2 121320 &c. 6. Find tlie sum of x/'j and x/aV' ^"^ ^^^^ their difference. 3 3 ;{ 3 Ans. Their surii is 1^/54, or f >/2. Their diff. is |v/2, or \/^. 3 3 7. What is the sum and difference of n/| and \//^. Ans. Their sum is /^v/lB. Their diff. ^\s/lQ. 8. There are four spheres each 4 inches in diameter, lying so as to touch each ether in the form of a square, and on the middle of this square is put a fifth ball of the same diameter ; what is the perpendicular distance between the two horizontal planes which pass through the centres of the balls? 4 4v/2 Ans. -^-=— ^=2v/2=:V'8=2'8284-f inches; JVote, It may be seen from this example that the diameter of the ball divided by v^2, will give the distance between the planes, whatever be the diameter of the ball, or, which is the same, half the diameter of the ball multiplied by the square root of 2. 9. I'here are two balls, each four inches in diameter, which touch each other, and another, of the same dinmetcr is so placed ])etween them that their centres are in the same vertical plane ; what is the distance between the horizontal planes which pass through their centres ? Ans. r— =|V3=2V'3 inches. jYoic. It is evident from this example, that in all similar cases, half the diameter of the ball multiplied by the square root of 3, gives the distance between the planes. 10. There is a quantity to whose square i is to be added ; of the sum the square root is to be taken and raised to the cube ; to this power -*- are to be added, and the sum will be yVlS ; what i« thai quantity ? Ang. VL PROPORTION. 217 OF PROPORTION IN GENERAL. NUMBERS are compared together to discover the relations they have to each other. There must be two numbers to form a comparison : the number, which is compared, being written first, is called the ajitecedent ; and that, to which it is compared, the consequent. Numbers are compared with each other two different ways : The one comparison considers the dijjference of the two numbers, and is called arithmetical relation, the difference being sometimes named the arithmetical ratio ; and the other considers their quotient^ which is termed geometrical relation, and the quotient, the geometrical ratio. Thus, of the numbers 12 and 4, the difference or arith- 12 metical ratio is 12 — 4=8 ; and the geonxetrical ratio is -7- =3, and of 2 to 3 is \. \( two, or more, couplets of numbers have equal ratio'?, or dif- ferences, the equality is termed proportion ; and their terms, simi- larly posited, that is, either all the greater, or all the less taken as antecedents, and the rest as consequents, are called proportion- als. So the two couplets 2. 4, and G, 8, taken thus, 2, 4, 6, 8, or thus, 4, 2, 8, G, are arithmetical proportionals; and the two coup- lets, 2, 4, and 8, 16, taken thus 2, 4, 8, 16, or thus, 4, 2, IG, 8, are geometrical proportionals.* * To denote numbers as being- geometrically proportional, the couplets are separated by a double colon, and a colon is written between the terms of each couplet; we may, also, denote arithmetical proportionals by separating the couplets by a double colon, and writing a colon turned horizontally between the terms of each couplet. So the above arithmeticals may be written thus, 2.-4 :: 6 .. 8, and 4 .. "2 :: « .- 6 ; where the first antecedent is less or greater than its consequent by just so much as the second antecedent is less or greater than it;* consequent : And the geometricals tlius, 2 : 4 :: 8 : 16, and 4 :• 2 :: 16:8; where the first antecedent is contained in, or contains its consequent, just So often, as the second is contained in, or contains its consequent. Four numbers are said to be reriprocalh/ or inversely proportional, when the fourth is less than the second, by as many times, as the third is greater than the first, or when the first is to the third, as the fourth to the second, and vice versa. Thus 2, 9, 6 and 3, are reciprocal proportionals. Note. It is common to read the geometricals 2 : 4 :: 8 : 16, thus, 2 is to 4 as 8 to 16, or. As 2 to 4 so is 8 to 16. Harmonical proportion is that, which is between those numbers which assign the lengths of musical intervals, or the lengths of strings sounding musical notes ; and of tkree numbers it is, when the Jirst is to the third, as the difference belweeu the first and second is to the difference between the second and third., as the num- bers 3, 4, 6. Thus, if the lengths of strings be as these numbers, they will sound an octave 3 to 6, a fiftli 2 to 3, and a fourth 3 to 4. Again, between 4 numbers, when the Jir.it is to the fourth, as the difference be- tween the first and second is to the difference between the third and fourth, as in the numbers 5, 6, 8, 10 ; for strings of sych lengths will sound an octave 5 to 10 ; u .sixth greater, 6 to 10 ; a third greater 8 to 10 ; a third less 5 to 6 ; a sixth less 5 to 8 ; and a fourth 6 to 8. Let 10, 12, and 15, be three numbers in harmonical proportion, then by the precpim°; defiftition. 10 •. 15 :: 12 — 10 : 15—12, and bvTlicorem I. of Geometri-n P d iiii> i^KOPORTlON. Proportion is tlistinguisheil into continued and discontinued, if. of several couplets of proportionals, written down in a series, the difference or ratio of each consequent, and the antecedent of the next following couplet, be the i^ame as the common difference or ratio of the couplets, the proportion is said to be continued, and the numbers themselves, a series of continued arithmetical or ge- cal Proportion, lOx lo— iii=15X l:ii— 10, or 10X15— 10x12— 15X12— 15X 10, whence if any two of the three terms be given, the other may be found ia tlie following manner. Case 1. Given the 1st and 2d terms to find the 3d. As lOX 15— IPX 12— 15X 12— 15X 10, then j O X 15- 15X 12-}-15X 10=:10X 12, or 2X10X15—12X15=10X12, or, "^iu— 12 X 15=10 X 1 2, and 15= 10X12 —. that is, 15, the ihird is equal to the product of the first and second 2X10—12 ttrms^ divided by the difference of twice the first term and the second term. 2. Given the 1st and 3d to find the second term. From the same equivalent expression, we get 2X lOX 15=15X 12-f lOX 12= j_ 2X10X15 15-|- 10X12, and — — — :--, =^12, that is, the 5gconrf term is equal to twice the lU— |- lo prod^ict of the first and third terms^ divided by the sum of the first and second terms. 3. Given the second and third to find the fii-st term. From the same expression, we get2XlOXl5 — 10X12=15X12, or 2X15—12 15X12 X 10=15X12, and 10 ^ — - _ ■■-— , that is^ the first term is equal to the product of the second and third terms^ divided by the difference of twice the third term and the second term. Ex. Find from tliird term, or monochord, 50, and the first term, or octave^ 25, the second term. 2 X 25 X 50 2500 By Case 2, — — '—=^—^ — =33'33, the second term, and is the length cif 25+50 .75 that chord, which is called a fifth. If there be four harmonical proportionals, as, 5, 6r B and 10 ; then, according to the definition, 5 : 10 :: 6—3 : 10—8, and as before, 5 X 10— 8=10X0— 5, or 5X 10 — 5X0=10X6^ — 10X5. From this expression, we may find any one of four harmonical proportionals from the other three. Thus, the first three being giv 5X8 en to find the fourth; 2x10X5— 10 XG=5X 8, and 10= --, that is, the 2Xi> — 6 fourth term is equal to the product of the first and third divided _ by the difference of twice the first teryn and the second term. In the same manner, it may be shown, that the third term of four harmonical proportionals is equal to the difference of tic ice the product of the first andfouvth terms and t/te product of the second and fnirih terms., divided by the first term. ''^X5Xl0 6x10 If the terms be 5, 6, 8, and 10, then 8=^^—-^ ■ . 5 Also, The second term is eqiial to the difference of twice the fourth and tJif third temiy multiplied *>y the quotient of the fir at divided by the fourth term, h the terms be as before, C=2x 10 — 8x— • Also, The .//«/ term is equal to the product of the second and fourth terms, divided by the diffcrancc of twice the fourth and the third frrvi. Thw^ 5— cxio f^xio— t;" ARITHMETICAL PROPORTION. 219 jinetiical proportionals. So 2, 4, 6, 8, form an arilhmelical pro- gression ; for 4 — 2=6 — 4=8 — 6=2 ; and 2, 4, 8, 16, a geometric cal progression ; for |=f=V^=^2. But, if the jlifference or ratio of the consequent of one couplet, and the antecedent of the next couplet be not the same as the com- mon difference or ratio of the couplets, the proportion is said to be discontinued. So 4, 2, 8, 6, are in discontinued arithmetical proportion ; for 4—2=8 — 6=2=rommon difference of the coup- lets, 8 — 2=:6=difference of the consequent of one couplet and the antecedent of the next; also, 4, 2, 16, 8, are in discontinued 4 16 ■ geometrical proportion ; for ~=-g-=^2=common ratio of the coup- 16 . ' lets, and -^=8=ratio of the consequent of one couplet and the antecedent of the next. ARITHMETICAL PROPORTION. Theorem I. IF any four quantities 2, 4, 6, 8 be in arithmetical proportion,* the sum of the two means is equal to the sum of the two ex- tremes.! And if any three quantities, 2, 4, 6, be in arithmetical propor- tion, the double of the mean is equal to the sum of the extremes. Theorem II. In any continued Arithmetical Proportion (1, 3, 5, 7, 9, 11) the sum of the two extremes, and that of every other two term.^, equally distant from them, are equal. Thus, 1 -{-1 l=3-{-9=5-|-7.J When the number of terms is odd, as in the proportion 3. 8. 13. 18. 23, then, the sum of the tW4> extremes being double to the mean or middle term, the sum of any other two terms, equally remote from the extremes, must likewise be double to the mean, * Althouj^h in the comparison of quantities according to their diiTerences, the term proportion is used : yet the word progression., is frequently .substituted in its room, and is indeed more proper ; the former form beinj]^, in the common ac- ceptation of it, synonymous with ratio, which is only used in the other kind of comparison. t For since 4 — 2=8—6, therefore 4-f 6=2+8. ij: Since, by the nature of jjrogressionals, the second term exceeds the first by just so much as its corresponding; term, the last but one, wants of the last, it is evident that when these corresponding terms are added, the excess of the one will make good the defect of the other, and so their sum be exactly the same w;th that of the two extremes, and in the same manner it will appear that the sum of any two other corresponding torm? must be equal to that of the two extreme". 220 ARITHMETICAL PROGRESSION. Theorem III. In any continued Arithmetical Proportion, as 4, 44.2, 44-4, 44-6, 4-|-8,&c. the last or greatest term is equal to the sum of the first or least term and the common difference of the terms, multi- plied by the number of the terms less one.* Theorem iV. The sum of any rank, or series of quantities in continued Arith- metical Proportion (1. 3. 5. 7. 9. 11.) is equal to the sum of the two extremes multiplied into half the number of terms.t ARITHxMETICAL PROGRESSION. ANY rank of numbers, more .than two, increasing by a common excess, or decreasing by a common difference, is said to be in Arithmetical Progression. If the succeeding terms of a progression exceed each other, it isc?ll"d an ascending series or progression ; if the contrary, a de- scending series. o ^ U. 2. 4. 6. 8. 10, &c. is an ascending arithmetical series. } 1. 2. 4. 8. 16. 32, &c. is an ascending geometrical series. . , i 10. 8. 6. 4. 2. 0, &c. is a descending arithmetical series. ( 32. 16. 8. 4. 2. 1, &c. is a descending geometrical series. * For since each term, after the first, exceeds that preceding it by the com- mon difference, it is plain that the last must exceed the first by so many times the common difference as there are terms after the first ; and therefore must be equal to the first, and the common difference repeated that number of times. t For, because (by the second Theorem) the sum of the two extremes, and that of every other two terms, equally remote from them are equal, the whole Feries, consisting of half so many such equal sums as there are terms, will there- fore be equal to tlie sum of the two extremes, repeated half as many times as there are terms. The same thing also holds, when the number of terms is odd, ns in the scric? A^ 8, 12, 16, 20 ; for then, the mean, or middle term, being equal to half the sura of any two terms, equally distant from it on contrary sides, it is obvious that the value of the whole series is the same as if every term thereof were equal to the mean, and therefore is equal to the mean (or half the sum of the two extremes) multiplied by the whole number of terms ; or to the sum of the extremes mul- tij^lied by half the number of terms. '['he sum of any number of terms of the aritlimetical series of odd numbers^ 1, 3, 5, 7, 9, fec.is equal to the square of that number. For, 0+1 or the sum of 1 term = 12 or 1 1+3 or the sum of 2 terms = 22 or 4 4+5 or the sum of 3 terms = 32 or 9 9+7 or the sum of 4 terms = 4? or IG 16+9 or the sura of f> terms =: .53 or 2.5, &c. By continuing the addition, tlie rule would be true for any number of term?. EXAMPI,K. ']']ie fa-st term, the ratio, and number of terms given, to find the sum of the series, A g:entleman travelled 29 days, the first day he went but 1 mile, and increased every day's travel 2 miios ; How far did he travel ? 29X29^=^41 miles, Ans. ARITHMETICAL PROGRESSION. 221 The numbers which form the series, are called the terms of the progression. Note, The first and last terms of a progression are called the extremes, and the other terms the means. Any three of the five following things being given, the other two may he easily found. 1. The first term. 2. The last term. 3. The number of terms. 4. The common difference. 5. The sum of all the terms. ^ Problem F. The first ierm^ the last term, and the number of terms beifig given, (o find the common difference. Rule.* Divide the difference of the extremes by the number of terms less 1, and the quotient will be the comtnon difference sought. Examples. 1st. The extremes are 3 and 39, and the number of terms is 19 : What is the common difference ? Extremes. 39 > -3^ Divide by the number of terms less 1 = 19-— 1 = 18)36(2 Ans, 36 39—3 — Or, =2. 19—1 2d. A man had 10 sons, whose several ages differed alike ; the youngest was 3 years old, and the eldest 48 : What was the com- mon difference of their ages? Ans. 5. 3d. A man is to travel from Boston to a certain place in 9 days, and to go but 5 miles the first day, increasing every day by an equal excess, so that the last day's journey may be 37 miles : Required the daily increase ? Ans. 4. Problem IF. The first iermy the last term, and the number of terms being given, to find the sum of all the terms. RuLE.f Multiply the sum of the extremes by the number of terms, and half the product will be the answer. * The difference of the first and last terms evidently shews the incrrase of the first terra by all the subsequent additions, till it becomes equal to the last ; and as the number of those additions was one less than the number of terms, and the increase, by every addition, equal, it is plain that the total increase, divided by the number of additions, must give the difference of every one separately j whence the rule is manifest. t Suppose another series of the same kind with the given one be placed under / in an inverse order ; then will the sum of any two corresponding terms be the 222J ARITHMETICAL PROGRESSION. Examples. 1st. The extremes of an arithmetical series are 3 and 39^ aitd the number of terms 19 : Required the sum of the series ? , g > Extremes. Sum=42 Number of terms = x 1 9 378 42 2)798 39-f3xl9 Or, =399. 399 Ans. 2 2J. It is required to find how many strokes the liammer*'of a dock would strike in a week, or 168 hours, provided it increased 1 at each hour? Ans. 14196. 3 and the number of terms 1760. 4tb. A man bought 25 yards of linen in Arithmetical Progress- ion ; for the 4(h yard he gave 12 cents, and for the last yard 75 cents : What did the whole amount to, and what did it average per yard ? 75—12 — =3 the common difference hy which the first term is found 22— 1 [to be 3. 75+3x25 Then — =J9 75b. and the average price is39cts. per yard 2 6th. Required the sum of the first 1000 numbers in ther natural order? Ans. 500500. dme as that of the first and last ; consequently, any one of those sums, multiplied .'y the number of terms, must give the whole sum of the two series. Let 1, 2, 3, 4, 5, 6, 7, 8, be the given series. And 8, 7, 6, 5, 4, 3, 2, 1, the same inverted. Then, 9+9+9+9-f 9+9+9+9=9 X 8=72, and 72 1+2+3-f4+5-fC+7-fO=— =36. ARITHMETICAL PROGRESSIO^J:. ' 223 Problem III. Given the extremes and the common difference^ to find the namh^r of terms. Rule.* Divide Ihe difference of the extremes by the com- mon difference, and the quotient increased by 1 will be the num- ber of terms required. Examples. 1st. The extremes are 3 and 39, and the common difference 2: What is the number of terms ? on i Extremes. 39? - 3^ Common difference =2)36 Q.uotient=18 Add 1 3g__3 Or,- h 1^19. 19 Ans. 2 2d.. A man going a journey, travelled the first day 7 miles, tUc last day 51 miles, and each day increased his journey by 4 miles : How many days did he travel, and bow far ? Ans. !2 days^ and 348 miles. Problem IV*. TTie extremes and common difference given^ to find the sum of all the series. Rule. Multiply the sum of the extremes by their difference in- creased by the common difference, and the product divided by twice the common difference will give the sum.j Examples. 1st. If the extremes are 3 and 39, and the common difference 2 : What is the sum of the series ? * By the first Problem, the difference of the extremes, divided by the number of terms less 1, gave the common difference ; consequently the same divided by the common diflerence, must give the number of terms less 1 ; hence, this quo- tient, augmented by 1, must be the answer to the questicm. t By the 3d Problem find the number of terms, and then, with the number of terms and the extremes, find, by Prob. 2, the sum of the series. This is the rule, 39 3 v/hich is contracted in the text. Thus in the 1st Example, by Problem 3, |-l=rthe number of terms, and by Prob. 2, 39-f-3X39 — 3-j-l=twice the sum of ■D ^ 30— T . , 39^3_i-2 ^ 3y+3Xri^'— ^H-!iI !;ie series. But —- 1-1 is also ~— Therefore. — — Tv"; '" 'i-^ ?iim of the eerie?, and i« the rule. 224 ARITHMETICAL PROGRESSION. 39-1-3=42 sum of the extremes. 39 — 3=='36=diiference of extremes. 364-2=;38=difference of extremes increased by the coo>moL difference. 42 X 38 336 126 Twice the common difference=4)1596 399 39+3x39— 3-{-2 Or, =399. 2X2 2d. A owes B a certain sum, to be discharged in a year, by pay- ing 6d. the first week, 18d. the second, and thus to increase every weekly payment by a shilling, till the last payment be 21. 1 Is. 6d. : What is the debt ? Ans. £67 12:.. Pkoblem V. The extremes and sum of the series given, to find the number of tenm Rule. Twice the sum of the series, divided by the Sum of the extremes, will give the number of terms.* Examples. 1st. Let the extremes be 3 and 39, and the sum of the series 399 : What is the number of terms ? Sum of the series=399 X 2 Sum of the extremes=39+3=42)798(19 Ans. 42 .378 399x2 378 Or, =19. 39+3 2d. A owes B 671. 12s. to be paid weekly in Arithmetical Pro- ofression, the first payment to be 6d. and the last to be 51s. 6d. : How many payments will there be, and how long will he be in dis- charging the debt ? Ans. 52 payments, and as many weeks. * This Problem is the reverse of Prob. II. and the reason of the rule is ob^ i ous from tlie demonstration of the Rule, Prob. II, ARITHMETICAL PROGRESSION. 22^, Problem VI* Che extremes and the sum of the series given, to Jind the common difference. Rule. Divide the product of the sum and difference of the extremes, by the difference of twice the sum of the series, and the sum of the extremes, and the quotient will be the common difference.* Examples. 1st. Let the extremes be 3 and 39, and the sum 399 : What is the common difference ? Sum of the extremes = 39 + 3 = 42 Diff. of the extremes = 39—3 = X 36 252 126 399X2—42=756)1512(2 Anav 1512 3y_j-3x 39—3 Or, _-- .-=2. 399x2—39-1-3 2d. A owes BJ£67 12s. to be discharged in a year, by weekly ipayments ; the first payment to be 6d. and the last, JC2 lis 6d. ; What is the common difference of the payments, and what will each payment he ? 61*5-1- 5x51*5 — '5 — . — . =18. and 6d.4'l3.= ls. 6d.=2d payment, 1352x2—51-54-5 Is. 6d.-{-ls.=2s, 6d.=3d payment, &c. Problem VII. The first term, the common difference, and the number of terms given, to jind the last term. Rule. The number of terms less 1, multiplied by the common differ- ence, and the first term added to the product, will give the last term.t Examples. 1st. If the first term be 3, the common difference 2, and the number of terms 19 : What is the last term ? * This rale is only a contraction of the following process. By Prob. V. find the number of terms, and, then, from the extremes and number of terms, find by JProb. I. the common difference. t By Prob. I. the diffcreixee of the last and first terms divided by the number of terms less 1, gives tjie common difierence, whence the common ditference multiplied by the number of terms less 1, and the product increased by the first, term, must give the last term. K e :?26 ARITHMETICAL PROGRESSION. Number of terms — 19 — 1 Number of terras less 1 = 18 Common difference = X 2 36 First term = + 3 39 the Ans. Or, 19— lXi-^+3=r39. 2d. A owes B a certain sum to be paid in Arithmetical Prop:rei3- slon ; the first payment is 6d. the number of payments 52, and the common difference of the payments is 12d. : What is the last pay- ment? Ans. £2 lis. 6d. Problem Vllf. Thejirsi term, common diff'erence^ and number of terms gmen^ to find the $uni of the series. Rule. To the first term add the product of the number of terms less I by half the common difference, and their sum, multipHed by the number of terms, will give the sum of the progression.* Examples. 1st. If the first term be 3, the common difference 2, and number of terms 19 : What is the sum of the series ? First term = 3 Add the product of the number of terms } _.iq iw| in less 1 by ^ common difference 5 Their sum 21 Multiply by the number of terms = li' 189 21 Or, 19x3+19—1x1=399 Ans. = 399 2d. Sixteen persons gave charity to a poor man ; the first gave^ 7c. and the second 12c. and so on in arithmetical progression ; I demand what sum the last person gave, and how much the poor man received in all ? Ans. 82c. the last gave, and ^7 12c. the whole sura. * Find by Prob. VII. the last term, and then by Prob. II. Uxe sum of the pro - §*ression. Tl*e rule is merely a contraction of this process. ARITHMETICAL PROGRESSION. 1^27 Problem IX. Given the first term^ number of terms, and the sum of the serieSy to find the last term. Rule. Divide twice the sum by the number of terras ; from the quo- tient take the first term, and the remainder will be the last.* Examples. 1st. If the first term be 3, the number of terms 19, and the sum 399 ; What is the last term ? Sum of the terms == 399 Multiply by 2 I)ivide by the number of terms = 19)798 Quotient = 42 ' Subtract the first term = 3 Or, -^^^^^ — 3=^39. 19 2d. A merchant being indebted to 12 creditors ^2460, ordered his clerk to pay the first ^40, and: the rest increasing in arithmeti- cal progression: i demand the difference of the payments, and the last payment ? Ans. g3i)=diff"; and ^370. last payment. Problem X. Given the last term^ the number of terms, ai(d the mm of the termSf to find the first term. Rule. Divide twice the sum by the number of terms ; from the quo- tient subtract the last term, and the remainder will be the first. t Examples. 1. If the last term be 39, the number of terms 19, and the sum (if the series 399 ; what is the first term ? Sum of the series = 399 Multiply by 2 Divide by the number of terms == 19)798 Quotient = 42 From the quotient take the last term = 39 Or, ..^^^^^ — 39=3. Remainder = 3 Ans. 19 "= By Prbb. II. the product of the.sum of the series and the number of term?, divided by 2, gives the sum of the series ; whence twice the sum of the series divided by the number of terms, and the quotient diminished by tlie first term, will ^We the last term. t By Prob. II. half the product of the sum of the extremes and the number of terms, gives the sum of the terms ; whence, twice the sum of the terms di- vided by the number of terms, and the quotient diraiaished by the last term, must give ^he first term. 228 ARITHMETICAL PROGRESSION. 2. A man had 10 sons, whose several ages differed ^like ; the eldest was 48 years old, and the sum of all their ages was 255 : What was the age of the youngest ? Ans. 3 years. Problem Xf. The common difference, number of terms , and the last term given, to find the first term. Rule. From the last term subtract the product of the terms less 1 bv the common difference, and the remainder will be the first term.* Examples. 1. If the common difierence be 2, the number of terms 19, and the last term 30 ; what is the fir?t ? Last term = 39 Subtract the number of-terms less 1 > __-j- o — q/> multiplied by the common difference S ~ ~ Remains 3 An?, Or, 39—19—1 x^ = 3. 2. A man travelled 6 days, each day going 4 miles farther than on the preceding day, till the last day's journey was 40 miles ; how far did he ride the first day ? ' Ans. 20 miles. Problem XII. The common difference^ tJie number of terms ^ and last term given, to find the sum of the series. Rule. From the last term take the number of terms minus 1, multipii^ ed by half the common difference, and the remainder, multiplied by the number of terms, will give the sum-t Examples. 1. If the common difference be 2, number of terms 19, and tho last term 39 ; what is the sum of the series ? Last term=39 Subtract the number of terms [ess 1 ? __,y , i = ip multiplied 'by | the common difference 5 — Remainder = 21 Multiply by the number of terms = 19 189 21 — Answer, 399 Or, 19x39—19—1 X 1=399 -' By Prob. I. the difference of the extreme? divided by the number of terms less 1, gives the common difference, whence the last term diminished by the product of the common diflerence and the number of terms less 1, must give the first term. < t By Prob. XI. find the f.rst term, and then by Prob. VIII. find the sum of the pi'ogression. The rule is only a contraction of this process, as may be seen in working an example, and keeping the several terms separate in the operatioQ. ARITHMETICAL PROGRESSION. ^29 2. A man performed a journey in 6 days, and, each day, travell- ed 4 miles farther than on the preceding day, till his last day's travel was 40 miles ; how far did he travel in the whole ? Ans. 180 miles. Problem XIII. TJie sum of the terms, the number of terms, and the common diff'er- ence given, to find the first term. Rule. Divii^ the sum by the number of terms ; from the quotient talie half the product of the number of terms, minus unity, by the com- mon difference, and the remainder will be the first term.* Examples. 1. If the sum of the series be 399, the number of terms 19, and the common difference 2 ; what is the first term ? Number of terms = 19)399==sum- Quotient = 21 Subtract | the product of the number of> __ ■ 9 — i<^ terms, less 1, by the common difference \ -! ~ ^ Ans. 3 399 2x 19—1 2. A man travelled 180 miles in G days ; he increased his jour- ney, each day by 4 miles : how far did he travel the first day } Ans. 20 mites. Problem XIV\ TJie sum of the termSy number of terms, and the cormnon difference given, to find the last term. Rule. Divide the sum of the series by the number of terms ; to the quotient add half the product of the number of terms minus unity by the common difference, and the sum will be the last term.j Examples. 1. If the sum of the series be 399, the number of terms 19, and the common difference 2 ; what is the last term ? Divide by the number of terms = 19)399 sum. Quotient = 21 Add 1 the product of the number of/ __19 — 1x2 terms, Jess 1, by ihp common difference J ^ =1C 399 2Xi>-»— T Ans. =39 * By Pro!-). VIII. the product of the number of terms less !, ftntl of half the common difference, added to the first term, and the sum multijjlied by the num- ber of terms, gives the sum of the progression, whence divide the sum of the se- ries by the number of terms, and diminish the quotient by the product of th« number of terms less ] and half the common difference, or by half tlie product . of the number of terms less 1 and the common diflerence, and you have the first term. t This rule is obtained from the rule of Prob. XII. in a similar manner as the preceeding rule from Prob, VIII. 2'SO ARITHMETICAL PROGRESSION. 2. A person bought a farm for £510 to be paid monthly in arithmetical progression, and to be completed in a year, each pay ment to exceed that preceding by £5: What were the first and last payments ? Ans. £15 the first payment, and £70 the last payment. T^e following Table contains a summary of the "isihole doctrine of Arithmetical Progression. Note. The table contains several cases, whose rules are not given in the text, because they are not very easily demonstrated without the aid of Algebra. Each of these cases however is illus- trated by examples, which follow the table, so that the expresjiion for the process in the table, may be more intelligible to the learner. It should be observed that where two letters or a figure with a letter or letters, occur in the rules in the table, without a sign be- tween them, the product of the quantities is intended. Thus, d(> means t/Xo, and ^ds signities 8Xc/X5. CASES OF ARITHMETICAL PROGRESSION, Case I Givfcfi I KftjViir-o J?(/IullOfi a, /, n, / — a n— 1 Prob. I. a-\-lxn Prob. II, 3. a,l,^''d a, /, s, < -- I — a + 1 Prob. III. l-{-aXl — a-\~d " 2d Prob. IV -,-; Prob. V. l-\-ax l~a ^-=:z Prob. VI. 2s~lJf.a 4. ct, d, s, < - n 2d I x/2a- ^'^ l^X^^^ ©usly equal, or =. — — — Hence by the definition of geometrical propor- ^ ^ 6X5 10X12 ^ *= tionals, 2X3:6X5:: 6x4 : lOx 12, and is the theorem. • .Hence, if four quantities are proportional, their squares, cubes, &c. will likewise beproportional. Thus, let the terms be 2 : 6 :: 4 : 12, then 2x2 : 6 X6 :: 4x4 : 12X12, or 2^ :6^ :: 4^ : 12", and hence also, 2^ :6^ :: 4^ : 12^ and 2* ; 6^ :: 4^ ::12'. :j: For since 2 : 4 :: 4 : 'fi. thence will 2X8=:4x4, by Theorem 1 ; and therefore 2 X 2 X C=2 X 4 X 4, by equal m ultiplication ; consequently, 2^2:4x4 :: 2 : 8, by Theorem 2. In like manner it may be proved that, of four quantities continually propor- tional the mtlfc of the iu\X k to llial of the second, as the first quantity to the foyjrth. GEOMETRICAL PROGRESSION. 2^ TliEpREM 7. In any continued Geometrical Proportion, | , 3, 9, 27, 81 , &c. the product ofthe two extremes, and that of every clher two term^ equally distant from them are equal.* Theorem 8. The sum of any number of quantities, in continued Geometrical Proportion, is equal to the difference ofthe product ofthe second^ and last terms, and the square of the tirst, divided by the differ- ence ofthe first and second terms.! GEOMETRICAL PROGRESSION. A GEOMETRICAL Progression is, when a rank or scries of numbers increases, or decreases, by the continual multiplication, or division, of some equal number, which i$ called the ratio. ^ For, the ratio of the .first term to the second being the same as that of f.he last but one to the last, these four terms are in proportion ; and there- fore by Theorem 1, the product ofthe extremes is equal to that of their two ad- jacent terms ; and after the same manner, it will appear that the product of the third and last but two is equal to that of their two adjacent teims, the sec- ond and last but ontj, and so of the rest ; whence the truth of the proposi- tion is manifest. t Take any series of continued geometrical proportionals, as 2, 6, 18, 51, 162, 486, and its sum is, 2+6-f lb-|-54H-162H-486,=728, by addition. Multi- ply the whole, by that number by which any term of the series is multiplied or 4ivided to form the succeeding term, which is in this example, 3, and you have, ..6+184-54+162+486+1458=2184. Subtract the first series, 2+ 6+18 +54+162+486 - - - ::r^728. As all the terms in the up- per series are cancelled by those in th§ lower, except the last in the former and the first in the latter, those two terms become — 2+1458, or, 1458— 2=r 2184 — 728. Now, the iipper series exceeds the lower three times, and, of course, from thrice the series there has been taken once the series, and the remain- der must be iicice^ or 3 — 1 times, the series and equal to twice or 3r-rl.tiiHe?, the sum of the series, that is, -~^ — r- = 728, the sum of the series. As ^^53__2 3X486 2 -^ is also — pr ^, multiply botli parts of the fraction by 2, which. •^r\ ,. .. 1 ^^\ 1 2X3X486—2X2 6x486—4 ... Will not alter its value, and you have : — ; — or , winch 2x3—1 ^~^ must also be equal to tlie sum (jf the series, and is the rule. For 6 is the second temx of the given series ; 486, the greatest term ; 4, the square of tlio second term ; £^nd the divisor, 6 — 2, is the difference of the first and scconcT terms. 3X486 2 In this deniQustration it is shown that '—-. j; — =rtl)esum of the series ; that js, that the greatest term multiplied by the number by which the series in- creases, and the product diminished by the least term, and this divided by e number one less than that by which the series increases, tlsa quotient is tJie sum of the series. Let the series be 1, 4, 16, 64, 256, 1024, whose multi? 4X10'? 4 1 " plicr i3 4. Then — — ^—1365, the rum of the series. 236 GEOMETRICAL PROGRESSION. Problem I. Given one of the extremes^ the ratio^ and the number of the terms of a geometrical series^ to find the ether extreme. Rule. Multiply or divide, (as the case may require) the given extreme by such power of the ratio, whose exponent* is equal to the num- ber of terms less 1, and the product or quotient, will be the other extreme.! * As the last term or uny term near the last, is very tedious to be found, by continual multiplication, it will often be necessary in order to ascertain them, to have a series of numbers in Arithmetical Proportion, called indices^ or exponents^ beo^innin^ with a cypher, or a unit whose common difference is one. When the first term of the series a-nd the ratio are equals the indices must begin with a unit, and, in this case, the product of any two terms is equal to that term signified by the sum of their indices. This is obvious on inspection of this example. rpi < 1. 2. 3. 4. 5. 6, &;c. indices, or arithmetical series. ' ( 2. 4. 8. 1 6. 32. 64, &;c. geometrical series (leading terms.) Now, 64-6=:12=the index of the twelfth term, and 64X64=4096— the twelfth term. But, when theirs/ term of the series and the ratio are different^ the indices must begin with a cypher, and the sum of the indices., made choice of, must be one less than the number of terr^u^ given in the question ; because 1 in the indices stands over the seeond ttrm^ and 2 in the indices., over the third term, &;c. And, in this case, the product of anV two terms divided by the ^r5/, is equal- to that term beyond the first, signified by the sum of their indices^ as is obvious from this example. rp, < 0. 1. 2. 3. 4. 5. 6, &c. indices. ' ^ 1. 3. 9. 27. 81. 243. 729, &c. geometrical series. Here, 6 + 5 = 11 the index of the 12th term. 729X243=177147 the 12th term, because the first term of the series and the ratio are different, by which means a cypher stands over the first term. Thus, by the help of these indices, and a few of the first terms in any geomet-. yical series, any term, whose distance from the first term is assigned, though it were ever so remote, may be obtained without producing all the terms. • t The rule is evident from the manner in which a geometrical progi-ession i^ formed. Let 2, 4, 8, 1 6, 32, Sic. be the series, whose ratio is 2. The second term 3ii formed by multiplying the first term by the ratio ; the third term, by multiply- ing the second by the ratio, and so on. The series may therefore be written thus, 2,^2X2, 2X2X2, 2X2X2X2, 2X2X2X2X2, &c. or thus 2, 2x2i, 2X:^^, 2X2^,2X24, and so on. Any term after the first is evidently that power of the ratio whose index is one less than the number of the term multiplied by the first lertn. Thus, the 3d term is 2X23 ; the 4th term is 2X2^, and the 8th term would be 2X2 "7, and so on. In an ascending series, iherefore^ multiply the first term by that power of the ratio whose index is one less than the number of the term souglit, and the product is the term bought. In a descending series, as 243, 81, 27, 9, 3, 1, whose ratio is 3, and which is also 1X35 1 X3», 1X31,1 X33, 1 X32, 1 X31, 1, the last term, 1, =^-— — . Hence the rule li evident, whatever be the fir?i term or the ratio. JSTote I. If the ratio of any geometrical series be 2, the difference of the great- est and least terms is equal to the sum of all the terms exfiept tlie greatest ; if the ratio be 3, the difference is double the sum of all the terms except the greatest; if the ratio be 4, tlie difference is triple the sum, itc. Let the series be 1,2, 4, 8, 10, 32, whose ratio is 2, and whose sum is 63, and the sum of whose fii"st five terms is 31. Now 32—1, the differeiice of the extremes, is e^'wa/ to the suim > ^ GEOMETRICAL PROGRESSION. 237 Examples. 1. If the first term be 4, the ratio 4, and the rrnmber of terms 9 ; Wliat is the last term ? 1. 2. 3. 44- 4= 8 4. 16. 64. 256.x256=65636=power of the ratio, whose exponent •3 less by 1, than the number of terms. 65536=8th power of the ratio Multiply by 4=firsl term. 262144=last term. Or, 4x48=262144=the Answer. 2. If the last term be 262144, the ratio 4, and the number of terms 9, what is the first term ? 4=lhe first term. the first five terms. For, by the principle proved under Theorem 8, of G^omet- i>Xl6=:l 32 I rical Proportion, the sum of the first five terms is , or r=32 — h the difference of the ^eatest and least terms. Let the series be, 3, 9, 27, 81, 243, whose ratio is 3, then 243--3=twice th« ^JX81 — 3 sum of all the terms except the greatest. Now by the same principle — ^ 243 3 243- 3 =sum of all the terms except the last, =.--- — —=: — , that is, the sum of all 3 — 1 2 , the terms except the greatest is half the difference of the greatest and least terms, or this difference is twice the sum of all the terms except the greatest. Let the series be 1,4, 16, 64, 256, whose ratio is 4, as before = the 05^ 1 256 1 sum of all the terms except the greatest, =^- 1=--— 1 or tbe difference of the greatest and least terms is equal to thrice the sura of the series except the last term. JVo/e 2. In any geometrical progression decreasing; to infinity, or beyond any Hssignable limit, the square of the Jirst terrn, divided by the difference between ih& first and second terms^will be the sum of the series. Let the series be i, 1, 1, -'-, &:c. to infinity. By Theorem 8 of Geometrical Proportion, the sum of any number of terms, an the first four terms, is ^— ^ ^ yy -i-i, or frohn the equare of the firat term the pi-oduct of the seciond and fourth terras is to be taken and the remainder divided by the difference of the first and second terms. But if the series be infinite, the last term is infinitely small and must be considered ; and then the product to be taken from the square of th<» first term is 0.. For when _L. i? supposed infinitely small or 0, then { X J- be- oomes ixO=::0; and the expression becomes l-^^i, -or ^Xi=l, the sumof the above series continued without end. Hence the rule is manifest. The above expression ^ — |XJ~^isals6^ — 'I'^tV"^-^' tbat \s, multiph^ the last term by the raiio^i and divide the difference between this product and the first term by the difference between 1 and the ratio^ and the quotient Avill be the sum of the series. Let the series be 1 , i^ j. jl^ whose sum is, by addition.; '^f Tiien by the rale, THi X J^-f-l —1= 1 _1 xy-^~|,=|f X|=:^|-, as before. If the series were continued infinitely, then as before the product to be 'subtracted from thefirst term would be (}•, and ]_f.2=;lx^=l" ^'^ald be the ^um. That is, if the series descend to infinity, divide the frsf fna by the difference between, unify and (he ratio, and the quotif ut will be tlio c .vu of th« 2Sii GEOMETRICAL PKOGRESSION. Jigain^ given the first terrn, and the ratiOf to find any other tena y assigned. Rule I * When the indices begin with a unit. 1. Write down a few of the leading terms of the series, and place their indices over them. 2. Add together such indices, whose suai shall make up the en- tire index to the term required. 3. Multiply the terms of the geometrical series, belonging to those indices, together, and the product will be the term sought. Examples. 1. If the first term be 2, apd the ratio 2, what is the 13th term ? 1. 2. 3. 4. 64-6x3= 13 2. 4. 8. \Q. 32x32x8=8192 Ans. Or^ 2x2^2=^8192. 2. A merchant wanting to purchase a cargo of horses for the West-Indies, a jockey told him he would take all the trouble and expense upon himself, of collecting and purchasing 30 horses for the voyage, if he would give him what the last horse would come to by doubling the whole number by a halfpenny, that is, two far- things for the first, four for the second, eight for the third, &c. to which, the merchant, thinking he had made a very good bargain, readily agreed : Pray what did the last horse come to, and what did the horses, one with another, cost the merchant? 1. 2. 3. 4. 5. 6-f- 6=12th. 12-f 12+ 6=last term. 2, 4. 8. 16. 32. 64x64=4096, and 4096x4096x64 = 1073741824qrs.= £11 18481 Is. 4d. and their average price war^ £37282 14s. O^d. a piece. RuleII.* When the indices begin -with a CTj-pher. 1. Write down a few of the leading terms of the series, as be fore and place their indices over them. iaihiite series. Thus, the sum of — --, , , — --- ,&c. to hiTiuIiv 1-05 ' 1-06:^' 1-U53' 1-06 i 1111 And the sum of four tcriiis of the series, , , , — — 1-05 1-053 1-053' 1-05' 1-05 1-055 • i W^ ^^^ ^ ^^'"1-05" 1-055 ■ 1-05 ^ ^^ ^1-05 1-055 1-05 , • u • . i 1 , . . ^ 1 1 1 X '-I which IS 1 X , which is 1 — ■ X — ,= 1-05—1 l-Uo4 '^ iU5— 1' ■ 1-054 ^- -05' -05 05 ^ Thcsf^ rules sire only particular cases of the preceding gjeneral rul?, and th288-8, which, however incredi- ble it may appear to some, is more than four hundred and fifteen millions of miU- ions, nine hundred and thirty thousand, eight hundred and ninety-nine millions, eight hundred and forty thousand, two hundred and eighty-eight times larger than the globe we inhabit.* For the solution of the four following questions, see last part of note under Problem I. 9. A frigate pursues a ship at 8 leagues distance, and sails twice as fast as the ship ; how far noust the frigate sail, before she comes up with her? First, 8. 4. 2. I. f i. &c. 8x8=64, and 64^8—- 4=16 leagues, Ans. 10. Suppose a ball to be put in motion by a force which impels it 10 rods the first minute, 8 the second, and so on decreasing by a ratio of 1-25 each minute to infinity; what space would it move through ? Ans. 60 rods. 11. Required the value of -999, to infinity, or -91? The first 9 or •9,=fV, the secon d, or - 9=^1^ ; therefore, •9x9~-9-— 09=1 Ans. 12. Required the sum of |, i, i, &c. to iefinity ? Ans. 1. 13. What is the sum of i, |, J^* ^^- ^o infinity ? Ans. -i. H. What is the sum of i, yV' A' ^^- ^^ infinity ? Ans. ^-. 16. What is the sum of -1, -01, -001, &c. to infinity ? Ans. i. 16. What is the sum of 1, ^, ^\, ^^^ ^^- ^^ infinity ? Ans. l-J. 17. What is the sum off, |-, ^, /y, for 7 terms ? Ans. 43.12. 1 " 1 1 18. What is the sum of ^— , p^.j,f^ ^c. to mfinity ? Ans. 16|. Problem III. The fust term, the last term (or the extremes) and the ratio given^ tb find the sum of the series.^ Rule I. Divide the difference of tlie extremes by the ratio less by 1 ; add the greater extreme to the quotient, and the result will be the ■sura of all the terms. Rule 11. Or, Multiply the greatest term by the ratio, from the pfoduct jjubtract the lea^^t term ; then divide the remainder by the ratio, less by 1, and the quotient will be the sum of all the terms. ** To fiiil the solid content of.it globe.- See Art. 34th.. of Mensuration of Solids. Note, that -523598 is two thirds of -785398 the area of a circle, whose diameter is 1. t It will be ^een, when we come to rirrulating decimals, tli^t -9 Ls the manner of expressing -999, fee, to infinity. % Rule 1. In Note 1, under Prob. I. it is shown that the difference of the reatest and least terms divided by tiie ratio less 1, gives the sum of the series ex- pt the last term. Tp tiiis quotient add the last term, and the sum will be the urn. of the series. llulft 2 and 3, are demonstrated umler Theorehr 8 of Geometrical Proportion. 242 GEOMETRICAL PROGRESSION. Rule III. Or, When all the terms are given, then, from the product of the second and last terms, subtract the square of the first term ; this remainder being divided by the second term less the ^rs^, will give the sum of the series. Examples. 1. If the series be 2. 6. 18. 54. 162. 486. 1458. 4374. what is its sum total ? First Method. From the greatest term = 4374 Subtract the least = 2 Pivid€ by the ratio, less 1=3— 1=2)4372 diC of extremes. Quotienl= 2186 Add the greater extreme= 4374 6560 4374—2 Or, h4374=6560 Ans. 3—1 Second Method. Greatest term=4374 Multiply by the ratio= 3 Product= 13122 Subtract the least term= 2 Divide by the ratio, less by 1=3—1=2)13120 6560 Ans. 4374 X 3—2 Or, =6560 3—1 Third Method. Greatest term=4374?^ Multiply by the second term= 6 Product=26244i Subtract the square of the first term=2x2= 4 Divide the remainder by the 2d. term less the lirst=6— 2=4)26240- Ans. 656^] 4374x6—4 Or, =6560., 6—2 GEOMETRICAL PROGRESSION. 243 2. A man travelled 6 days, the first day he went 4 miles, and each day doubling his day's travel, his last day's ride was 128 miles ; how far did he go in the whole ? Ans. 252 miles. 3. A gentleman, dying, left 5 fons, to whom he bequeathed his estate as follows, viz. to his youngest son £1000 ; to the eldest £5062 lOs and ordered ,that each son should exceed the next younger by the equal ratio of 1| ; vyhat did the several legacies amount to? Ans. £13187 lOg. PROBLEril IV. Given the extremes and ratio, to find the number of terms. Rule. - Divide the greatest term by the least ; find what power of the ratio is equal to the quotient, then, add one to the index oi that power, and the sum will be the number of terms.* Examples. 1. If the least term be % the greatest term 4374, and the raN'e 3 ; what is the number oi terms ? Divide by the least term=2)4374=grealest term. 3x3x3x3x3x3x3=quotient, 2187==7th pow. then 7H-1=8, Ans. 2. A gentleman travelled 252 miles; the first, day he rode 4 miles ; the last 128, and each day's journey was double to the pre- ceding one : How many days was he in performing the journey ? Ans. 6 days. Problem V. Given the least terniy the ratio, and the sum of the series, to find the last term. RuLK. Multiply the sum of the series by the ratio less l,to that f)roduct add the first term, and the result, divided by the ratio, will give the last term.j Examples. 1. If the first term be 2, the ratio 3, and the sum of the series 6560 : What is the last term ? *Lettheseriesbel,2,4,8, lG,or, 1, 1x2, 1X22, 1x23, 1x24. Divide the 1X24 -greatest terni by the least, and =:24. But the exponent is always 1 le^s than the number of terms, whence 4-j- 1, or 1 added to the index of the last term will give the number of term?. Though the index of the last is not common y given, yet as the quotient arising from dividing the last term by the first term, is. ^ always some power of the ratio, it is readily found by multiplication, as in the 1st Example. t This rule follows directly from Rule 2, Prob. III. For, as the product of the last term and ratio, diminished by the first term, and the remainder divided by' tlie ratio diminished by 1, gives the sum of the series ; multiply the sum of the series by the ratio diminished by 1, to the product add the fir;t term, and di- T'ide the sum by the ratio, and you will have the last term. 244 GEOMETRICAL PROGRESSION. Sum of the series=6560 Multiply by the ratio less 1= 2 Producl=13120 Atld the least tertn= 2 Divide their sum by the ratio=3)r3122 3-^1x6560-1-2 ^ 4374 Ans, Or, =4374 Ans. 3 2. A gentleman performed a journey of 262 miles ; the first day he rode 4 miles, and each day after the first, twice so far as the day before : How far did he ride the last day ? Ans, 128 miles. Problem VI. Given the least term, the ratio, and the sum of the series, to Jind the number of terms. Rule. To the product of the sum of the series, and the ratio minus 1, add the first term ; which sum, divided by the first term, will give that power of the ratio signified by the number of terms.* Example. If the first term be 2, the ratio 3, and the sum of the series 80; What is the number of terms? Sum=80 Multiply by the ratio less 1=3 — 1= 2 160 Add the first term= 2 . Divide by the first term=r2)162 81 which, found in the Ta- ble of Powers, is the fourth power of the ratio, therefore the num- ber of terms is 4. Problem VII. Given the extremes, and the su7n of the sei'ies, to Jind the ratio. Rule. From the sum of the series subtract the least term ; di- vide the remainder by the sum of the scries minus the greatesi term, and the quotient will be the ratio t * By Prob. II. the difference between 1 and that power of the ratio indicated by the number of terms, divided by the ratio less 1, and the quotient multiplied by the first term, gives the sum of the series. Whence, if the sum of the series be multiplied by the ratio less 1, and the product be added to the least term, you will have the product of the first term and that power of the ratio signified by the number of terms. Divide then the former product by tlie least term, and tbe quotient will be that power of the ratio signified by the number of terms. t This rule is deduced from Rule 2, Prob, HI. in the easiest manner. m: GEOMETRICAL PROGRESSION. 245 Examples. 1. If the least term be 2, the greatest term 4374, and the sum of the series 6560 : What is the ratio ? Sum of the series=6560 Subtract the least term= 2 Divide Ad& idethe re. by thesumofthe ? =6560-4374=2 186)6Si?(3 series, mmus greatest term y ^ g^^r 2. A debt of ^252 was paid in Geometrical Progression, the first payment was $4 and the last gl28 : In what ratio did the payments exceed each other ? Ans. 2, viz. a double ratio. Problem VIII. The first term, the number of terms^ and the last term given^ to find the ratio. Rule. Divide the greater extreme by the less, and extract such root of the quotient, whose index is equal to the number of terms, less 1. Or, find the quotient in the Table of Powers, the root of which is the answer.* Examples. 1. Given the extremes 2 and 4374, and the number of terms 8 : What is the ratio ? Divide by the least term=2)4374=greatest term. I -^2187=3 4374 Or, 2 1 =3, Ans. Problem IX. I'he extremes and number of terms given, to find the sum of the series. Rule. 1. Subtract the least term from the greatest, and make the dif- ference a dividend. 2. Divide the greatest term by the least, and extract such root of the quotient, whose index is equal to the number of terms less 1 ; take 1 from the said root, and make the remainder a divisor. (Or find the quotient in the table of powers, which will shew the root, from which subtract 1.) * Let the series be 1, 1x3, 1x32, 1 x33, 1 X34. Divide the last term by 1 X34 the least term, and -. =34 , and the quotient is that power of the ratio, whos» index is 1 less than the number of terms. Extract that root of the quotient, whos^ index is 1 lecF than the number of terms, and that root will be the ratio. -MG GEOMETRICAL PROGRESSION. 3. Divide the dividend by the divisor, and the greatest term, added to the quotient, will give the sum of the series.* Examples . Given the extremes 2 and 4374, and the number of terms 8 : What is the sum of the series ? From the greatest term=4374 Take the least= 2 Blakc this remainder a dividend 4372 Divide the greatest term by the least 2)4374 7 And extract the 7lh root of the quotient, ^^2187=3 ; Then» 3—1=2)4372 Q,uotient=2186 A^ l-j-4-f42-f 43-1-44 +45 — 4095. But, we have seen that 3XI+4-I-43-I-434-44 4095—3 is contained in just 1023 times, and 3X l+4-}-43 -{-43+44 is the series 4 ' except the last term 3 X 45 . Therefore 4095—1023=3072, the last term. * This rale is deduced from Rule 1, Prob. IJI. and will he evident on atti^ndi^^- to that rule . H h f50 GEOMETRICAL PROGRESSION. Divide the sum of the series by the difference between the sum and the last term ; the quotient will give the ratio, and the remain- der, after the operation, the first term.* Example. If the number of terms be 4, the last term 64, and the sum of the series 80 ; required the first term and the ratio ? From the sum=80 Take the last term=64 Divide by the difference=26)80(3 the ratio. 78 The first term= 2 T/ie following Table exhibits a summary vierss of the doctrine of Geo- metrical Progression. CASES OF GFOMETKICAL f^ROGKESSlON. Cas" 1 Given j Pir.qaiivd | Solutiot* 1. am I 7i— 1 ar Prob. I. s n xa Prob. II. r— 1 2. ad s I— a H Prob. lU. r—l n LI— La . . . I 1 Prnh IV L.r * By the second rule, Prob. XIII. it is found that if the sum of the scries be di- vided by the ratio, the quotient is the difference between the sum of the serie-j and the greatest term, and the remainder is the first term. Therefore, if the sum of the series be divided by the difference between the sum and the last term, th^ quotient must be the ratio, and the remainder the least term, hi tlie example 4095 there taken, the ^catest term divided by the ratio, or — — =:=1023 with a remam- der=3 the first term. But as 1023=4095—3072 or the difference between the 4095 sum of the series and the last term, therefore -~: — —---=4, the ratio, and leaves a remainder 3,r=thc first term. The same must hold true in every series, where the ratio is greater than the least term. GEOMETRICAL PROGRESSION, 251 Cas- 1 Giveu 1 i.equiied Solntiofi. 3. ars / r— lX.s+« r Prob. V. n L.r—iXs-\-a — L.a L.r Prob. VI. 4. ah r s — a Prob. VII. . n L L.l-'L.a,^ .s — a — L.6 — / 5. ans r rs 11 r a s — a a i I lXs—l\ =aXs—a\ 6. anl r 1 a /I— 1 Prob. VIII. s a /—a Prob. IX. 7. ml a I n—l Prob. X. r s H / I, n-,1 Prob. XI. r r-1 8. 1 1 rns a r— 1 Xs Prob. XII. n r —1 I 5 71 n^ — 1 • •ys Prob XIII INTEREST. Case I Given | Required | Solution. \ 9. rh a s—rxs—i Prob. XIV. n L.I- I L.s — r X s — I 1 1 L.r '^ 10. nls a n—l in— 1 1 =lxs—t\ aXs—a r n s n — 1 / J 1 ■ r — l—s I—s ^ a=tir8t or least term. 1 /=Iast or greatest term. TT !s=sum of all the terms. ^^^ }n=number of terms. 1 r=ratio. J L=lo^arithm. Examples in Geometrical Progression, To be solved by those formulas in the Table, of which the rules are not given in the Problems. 1. The least term is 2, the greatest term 4374, and the sum of the series 6560 ; required the number of terras. Ans. 8. This is solved by the 2nd rule of Case 4 in the table. 2. If the ratio be 4, the last term 3072, and the sum of the se- ries 4095 ; what is the first term and the number of terms ? Ans, The first term 3, the number of terms 6. Wrought by Case 9. 3. Given the number of terms 4, the last term 54, and the sum of the seVies 80, to find the first term and ratio. Ans. The first term 2, the ratio 3. Wrought by Case 9 in the table. INTEREST. INTEREST is a premium allowed by ihe Borrower to the Lender, for the n>e of his nionev. It is e.-timated at a certain number of pounds, dollars, &,c. for each huudrrd pounds, dollars, &:c. for a year; and hi ihh same proportion, for a less or greater «um, or for a longer or &hoder time. He.ice, interest is sa^id lo be ^0 much jtjer cent, ov per centum ^ per annum. SIMPLE INTEREST. 5253 The Principal is the sum lent. The Rate is the premium per cent, agreed on.* The Amount is the sum of the principal and interest. Interest is of two kinds, Simple and Compound. SIMPLE JJVTEREST. Simple Interest is that which is allowed on the Principal only. JVote. By this Rule, Commission, Brokerage, Insurance, pur- chasing Stocks, or any thing else, rated at so much per cent, ars calculated. General Rule. 1. Multiply the Principal by the Rate, and divide by 100 (or cut off the two right hand figures in the Pounds) and the quotient, or left hand figures will be the answer in Pounds, &c. the right hand figures being reduced and cut off as at first. If the principal b«* dollars, the right hand figures will be cents.f 2. For more years than one, muHiply the Interest of one year by the number of years. 3. For any number of months take the aliquot parts of a year ; and for days, the aliquot parts of 30. JVote. When ^ | multiply f ^ | of the given number of the rate percent. < . [» the prin- ^ ^ [.months, and you will have per annum is o j cipal by | J | the interest for the given Examples. 1. What is the interest of £573 ISs. 9id. at G per cent per innum ? Ans. £34 8s. 5d. * Laivful or legal interest is that whicli is permitted by the laws of the Slate. It is different under different governments. In England tlie Rate is^'t-e perjceut. in the New England States, it is six^ and, in the State of New York it is seven per cent. The courts of the United States allow interest according to the prac- tice of the State where the suit was commenced. The rules of the Courts in the States of Massachusetts, Connecticut, and New York, for computing lesal interest, will be given immediately before Discount. t This rale is a contraction of a process in the Rule of Three. Thus in the £, Rate. Principal. 14. Example. A', 100 :: 6 :: £573 13s. 9^d. : the interest for a year. And — : ;~:.T~ — —-=£34 8s. 5d. the interest. The remainder of the rule is 100 «)V>vious. The reason of the rule in the Note above is, that as 9 is J of J 2 months, if tl,c rate be 9 per cent, multiply the jjrincipal by f of the months ; if the rate '.e 8 per cent, by -| of tl^o niunber of' months ; and if 6, by half the months, 254 SIMPLE INTEREST. £573 13 H 6 £34-42 20 2 9 8-42 12 613 4 •52 2. What is the interest of £329 17s. 6^d. for 3 years, 7 months, and 12 days, at 5 per cent, per annum. Ans. £59 13s. £329 17 6id. 16-49 20 7 81 987 .12 10-52 4 6 mons. ^ 1 mon. 10 days 2 days Then, 16 9 101 interest of 1 year 3. 49 9 7|do. of 3 years. 8 4 111 do. of 6 months 1 7 5f do. of 1 month 9 If do. of 10 days 1 9f do. of 2 days Ans. £59 13 do. of 3y. 7m. 12d, 210 Or thus : £5 6 months 1 month 10 days 2 days £ s. d. j\ 329 17 6| i 16 9 3 49 9 71 i 8 4 Hi 1. 3 1 7 6f 1 9 If 1 91 £59 13 Ans. 3. What is the interest of 347 dollars 50 cen.ts, at G per cent. p^r annum for a year. p47-50 Or ^100 : 6 :: 347.50c. : §20 85c. 6 20-8500 Ans. J^20 85c. 4. What is the interest of §797 13c. at 6 per cent, per annum, far 8 months ? SIMPLE INTEREST. 255 §797-13 4 31-8852 Ans. pi S^c, 5j\m, 5. What is the interest of §649 17c. at 6 per cent, per annuiHj for 15 months ? Ans. g48 68c. 7fm. 6. Required the amount of £725 12s. 6d. at 5 per cent, per annum, for a year? 5=^j\\126 12 6 36 6 n Ans. £761 18 1^ 7. What is the amount of §560 50c. at 6 per cent, for 16 months.' §560 50 8 44-84 660-50 Ans. §605-34c. 8. What is the interest of §150 75c. for 1 month, at 6 per cent, per annum ? ijl50-75 -75375 Ans. 75cts. 3f mills. So that any number of dollars, considered as so many cents, is the interest for 2 months, at 6 per cent, and the half of it is the monthly interest. coM^usslo^\ or factorage. Is an allowance of so much per cent, to a Factor or Corres- pondent, for buying and selling goods. 9. Required the commission on £436 9s. 6d. at 3i per cent, £436 9 6 3X 1309 8 218 4 6 9 1^-27 13 20 ^ 5-63 12 6-39 4 1-56 Ans. £15 6 6}. 1^> SIMFLE INTEREST. to. Kequireu the commission on ^649 75c. at 1^ percent. 649-75 324*875 162-4375 11370625 Ans. pi 37o. Ofm. BROKERAGE h an alloTVance of so much per cent, to a person called a Dro- ker, for assisting merchants in purchasing or selling goods. 11. Required the Brokerage on £911 12s. at 5s. or J percent. 5s.=i|911 12 / . 2 27 IG 20 5-58 Ans. £2 5s. G^d. 12 6-96 4 3-84 12. Required the Brokerage on g876 21c. at 33} cents, or at percent. i|876-21 292-07 Ans. $2 92c. Of^m. BmiNG AND SELLING STOCKS. Stock is a general name for the capitals of trading companies, or, of a fund established hy government, the value of which is va- riable 13. Required the amount of £375 15s bank stock, at £75 per cent ? 375 15 187 17 6 93 18 9 Or thus |25|i|375 In Subtract 93 18 9 As befora^£281 16 3 Ans. £281 16 3 14. Required the amount of ^2195 50c. bank stock, at 125 per cent. |25;i|21 95-50 Add 548-875 Ans. ^2744-375 15. What is the value of $6950 at 105 per cent ? "Ans. jf;7297 50cts. 46. Value of £225 of stock at 95 per cent. ? Ajis. £213 15g. SIMPLE INTEREST. 257 TO CALCULATE INTEREST FOR DAYS, Rule I. Multiply the principal by the days, and that product by the rate, and divide the last product by 365x100,* 15. Required the interest of £360 10s. for 175 days, at 6 per f:ent. 360-5x175x6 100X365 =^10-^=^10 7s. 4|d. Rule for making a divisor for any Rate. Multiply 365 by 100, and divide by the rate. Thus, for 6 per 365X100 cent. z =6083 divisor. o 365X100 ^. . For 5 per cent. =7300 divisor. 365X100 For 7 per Cent. ;:: =5214 divisor, and so for any other rate. Therefore, * This rule is the result of a statement in the Double Rule of Three, as follows. £ Rate; PHn. As 100 : 6 :: 360-5 : the interest for 1 year. Days. Days. And as 365 : :: 175 : the interest for 175 days. Wrought according to the Rule we have ^ =£10 7s. 4 |d. the interest required. 365 X 1 00 The rule for finding a divisor for any Rate, is a contraction of this result. Kor -^^Q'^ X 17 J X Q _3QQ.5 ^^^ ^ -— ^— - . But dividing both parts of the 365X100 365X100 " ^ fi fi 1 last fraction by 6, the Rate, we have -— — — ■=— — jrr.-;- Therefore 360-5 ■^ 36^X100 36500 6083 a ^.^ .« 1 360-5x175 ^ , X 175 X -=360-5 X 175 X — -= . In tlie same way divisors 365X100 6083 60i'.3 ^ are formed for any other rate. Hence too, the 2d Rule is obvious, for ^' ^ 6083 r=the interest, and is the product of the principal and days divided by the divi- sor formed as above; When the time is given in months.^ the divisor is formed in a similar manner. Suppose in the last example the time had been 1 1 months. Then, As 100 : 6 :: 360*5 : the interest for a year, and as 12 : :: 11 months : the inter- est for 11 months. Then --;^. '^- = the interest =: 360*5 X 1 1 X — . 152X100 1200 ^ — ■— . If the Rate were 5, tlien -~ —=:-—. Hence the rule is evident. 200 ISOO 240 JVb/e. As 365 days : 5 per cent. :: 7300 days : 100 per cent. And as 12 months t 5 per cent. :: 240 months : 190 per cent. Hence it is evident that if the Rate be 5^ any principal will gain 100 per cent, that is, will double in 7300 days or 240 months. And at 6 per cent, any sum will double in 6083 days, or 200 months, ant at 7 pfn- cent, in 5214^ days, or 17 IC- months. 1 i 35B^ SIMPLE INTEREST. Rule II. Multiply the princrpal by the days ; divide by 6083 for 6 per cent, and 7300 for 5 per cent, (the days in which any sum will double at those rates) and the quotient is the interest. For months, maltiply the principal by them, and divide by 200 for € per cent, or 240 for 5 per cent, (the months in which any sum will double at those r;»tes) and the quotient is the answer. Hence, when interest is to be calculated on cash accounts, or accounts current, where partial payments are made, or partial debts contracted; multiply the several balances into the days they are at interesty which should be done at every transaction, and thf: sum of these products divided by 6083 and 7300^ will give the in tcrest at 6 and 5 per cent. For any other rate, make the proper addition or deduction, or find a divisor as before directed. When partial payments are made at short periods, subtract the several payments from the original suih in their order, placing their dates in the margin. 16. Suppose abillofg35i>wasdae January 1, 1807; that ^75 wa-* paid February 3d, ^50 March 5th, $80 April 9th, and June 7th, ^145 : What interest is due at 5, 6 and 7 per cent. ? Dates. 1 Bill. Days. Products. January 1. Feb. 3, paid p50 75 33 30 35 59 11550 Balance, March 5, paid 275 50 8250 Balance, April 9, paid Balance, ■June 7, paid 225 80 145 145 7875 8555 7300)36230(4'963' Ans. $4 96c. 3m. at 5 per cent. 6083)36230(5-955 Ans. ^5 95ic. at 6 per cent. 52 [4)36230(6 -948 Ans. $Q 94c. 8m. at 7 per cent. After the dates are placed in the margin, the number of days in each of those periods is to be computed and marked against it3 respective sum : lastly, divide the sum of the products by 6083, &c. Interest on accounts current is calculated nearly in the same manner. 17. Compute the interest at 6 per cent, on the following ac c<^unt to August lOlh. SIMPLE INTEREST. 258 Dr. Mr. A. Jones, bis account current, with B. Carr, Or. 1807. D. 1807. D. Jan. 1, To Cash, - - 560 March 10, By Cash, - - 120 Feb. 10, To do. - 300 April 25, By do. - .,- 130 May 15, To do. - 140 June 16, By do. - 450 July 25, To do. - 100 July SI, By do. - ^50 1807. IDs. Days. 1 Products 1 Dr. Cr. Jan. Ij Dr. 560 "~4(r 22400 ^560 120 Feb. 10, Dr. 300 300 130 140 450 Dr. 860 28 24080 100 150 March 10, Cr. 120 1100 850 Dr. 740 46 34040 850 April 25, Cr. 130 250 Balance. Dr Dr. 610 140 90 12200 May 15, /i\j January 30 Dr. 750 32 24000 February 23 March 31 June 16, Cr. 450 April 30 Dr. 300 35 10500 May 31 June 30 July 21, Cr. 150 July 31 Dr. 150 4 600 August 10 July 25, Dr. Dr. 100 250 16 4000 Days 221 Aug. 10, 0083)131820(21-67'2 .Ans g21 67c. 2m. 221 131820 Here the sums on either side are introduced according to the order of their dates ; those on the Dr. side are added to the for- mer balance, and those on the Cr. side subtracted. Before we i:alculate the days, we try if the last sum ^250 be equal to the bal- ance of the account, which proves the additions and subtractions. And before multiplying \ye try if the sum of the column of days be equal to the number of days from January 1 to August 10. When payments are made at considerably distant periods, it is usual to calculate the interest to the date of each payment, and add it to the principal, and then subtract the payment from the .amount. 18. A note was given for ^540 the 18th August, 1804, and there was paid the 19th of March, 1805, g.50, and the 19th of December, 1805, $25 ; and the 23d of September, 180G, |^25 ; and the 18th of August, 1807, gllO: Required the interest, and balance due on .he '. Iih of November, 1807, at ^ per cent. ? ^ blMPLE INTEREbl. A note given 18th August, 1804, for interest to 19th March, 1805, 218 days, gl9-35 Paid 19th March, 1805, Balance due 19th March, 1805, Interest to I9th Dec. 1805, 276 days, Balance due 19th Dec. 1805, Paid 19th Dec. 1805, Balance due 19th Dec. 1805, Interest to 23d. Sept. 1806, 278 days, Balance due 23d Sept. 1806, Paid 23d Sept. 1806, Balance due 23d Sept. 1806, Interest to 18th Aug. 1807, 329. days, Balance due 18th Aug. 1807, Paid 18th Aug. 1807, Balance due 18th Aug. 1807, Interest to 11th Nov. 1807, 85 daya, Balance due 11th Nov. 1807, Amount of interest paid, ^98-823 19. A owes B the following sum?, wi(h interest at 6 per cent, per annum : $60 for 7 months, gl50 for 9 months, J75-50 for S months, g365-25 for 8 months, and 510-20 for 5 months : Kef^'ui-cd the amount? g 60 X7=: 420 150 X9=1350 75-50x3= 226-^0 365-23x8=2922 510-20X5=2551 1160-95 200)7469-50(37«347 Interest. 1160-95 Principal. 19-352 ;Jo40 19-352 559-352 50 23-022 609-352 23-022 532-374 2500 23-197 507-374 23-197 530-571 25000 27343 505 571 27-343 532-914 110- 5-909 422914 5-909 428-823 Ans. §1198 297 Amount. 20. A note for glOOO is given January 1, 1803, with interest at 6 per cent per annum ; February 19, 1803, glOO are paid ; June 7, 1803, giaO; April 14, 1804, g37-50; July U, 1804, §76; Sept. 29, 1804, ^,250; Dec. 17, 1805, g39 ; March 4, 1806, ^175; Aug. 7, 1806, jil05; Oct. 30, 1806, g50 ; May 12, 1807, gS40, and Nov. 17, 1807, g72 : How much is due, January 1, 1808 't SIMPLE INTEREST BY DECIMALS. 26 J SIMFLE INTEREST BY DECIMALS. A TABLE OF RATIOS, FROM ONE POUND, &C. TO TEN POUNDS. j Kate per cent. ratios. rate per cent. ratios. rate per cent. ratios. 1 '01 4 •04 7 •07 H 0125 n •0425 ^ •0725 H •015 H •045 ^\ •075 n •0175 4 •0475 n •0775 2 •02 5' •05 8 •08 n •0225 H •0525 8i •0825 f 2i •025 ^ •055 8i •085 2£ •0275 5} •0575 H •0875 3 •03 6 •06 9 •09 3} •0325 6^ •0625 9i •0925 3^ •035 H •065 9i •095 3^ •0376 6i •0675 9^ •0975 10 I 1 Ratio .*-L.« .^ the Simple Interest of £l or ^1 for 1 year, at the rate per cent, agreed on, and is found by dividing the rate by 100, and reducing it to a decimal. Thus, 7^0 = ^^» *'^^» rf o="05, and so on. A TABLE for the ready finding of the decimal parts of a year, equal to any number of days, or quarters of a year. l^ays. I decimal parts. | days. | decimal parts. | days. | dec. parts, j •00274 •005479 •008219 •010959 •013699 •016438 •019178 •021918 •024657 10 20 30 40 50 60 70 80 90 •027397 •054794 •082192 •109589 •136986 •164383 •191781 •219178 •246575 100 200 300 365 •273973 •547945 •821918 1 -000000 I ol a yeai= 25 J- of a vear: ^ of a year= 7; CASE I.* Tlie principal, time, and ratio given, to find the interest and amount. Rule. Multiply the principal, time and ratio continually together, and the last product will be the interest, commission, brokerage, &c. to which add the principal, and the sum will be the amount. * This is a contraction of the Geneml Rule for Simple Interest. If the iuter- ffst on £30 or $30 was required for 2 years at 6 per cent, by the general rule, the interest is -^-rr^XS— 30x^:^^X2— 30 X^ 06X2, which is the product of 100 ivrincipal, ratio, and time 100 And the amount=30+30X'06X2=je33^6 or$. m^ SlMi^LE INTEREST BY PECIMALS. Examples. 1. Required tbe amount of £537 10s. at£6 per cent, per annum^ fi>f 5 j'cars ? Principal 537-5 Multiply by the ratio= -06 Product 32-250 Multiply by the time=^ 5 . In(erest=161-250 Add tbe piincipal=537-5 Amount=698-75 ' 20 _. „ 1500 Ans. £698 159, Or, 537-5X06x5-f537-5— £698 15s. 2. What is the simple interest of £917 16s. at £5 per cent, ppk- ^nnnom, for 7 years ? Ans.£321 4 7. S. What is the amount of£391 17s. at£4| per cent, per annum, for 3} years ? Ans. £449 3 If 4. What is the amount of £235 3s. 9d. at£5| per cent, per an- mim, from March 5th, 1784, to Nov. 23d, 1784 ? Ans. £244 81. 5. If my correspondent is to ha?e£2| per cent ; what will" his commission on £785 13s. amount to ? Ans. £19 12 10|. 6. What will be the interest and amount of £445 lOs. in 3 years and 129 days, at £8-} per cent, per annum ? Ans. Interest, £126 19 8i, and the amount=£572 9 8i. 7. If a broker disposes of a cargo for me, to the amount of.t!i)o7 lO.-i. on commission at£li per cent, and procures me anothercargo of the value £817 15s. on commission at£l| per cent. ; what wiii liis commission, on both cargoes, amount to ? Ans. £22 5 7. 8. What is the simple interest of $^6-666 for 1| years at 7 per ^ent. ? Ans. ^8 16e. 6m. 9. Find the amount of §1 for 9 years s^nd 200 days, computing interest at 7 per cent. ? Ans. $1 66c. 8m. 10. What is the interest of jj23G at 5 per cent, for one year and MO days ? 11. lloquired the interest on g648a at 6 percent, for two year§, six months and 20 days. CASE II. T/te amount, titnc, and ratio given, iojind the principal. Rule. MuhipTy the ratio by the time ; add unity to the product for a divisor, by which sum divide the amount, and the quotient will be the principal.* * In the demonstration of the R-iilr A.r Cji.«e I. it was proved thutlhc amonrl r^^ic principal added to tho product ofthe principal, ratio, and time, or. tsikii if SIMPLE INTEREST BY DECIMALS. ms Examples. 1. What principal will amount to £1045 14s. in 7 years, at £^ per cent, per annum ? Ratio=^-OD Multiply by the time=: 7 Product=-42 Add 1- Divisor==l-42)lO45'7(736.40844-— £736 8 2, 1045-7 Or, =£736 8 2 Ans. •06x74-1 2. What principal will amount to £3810, in 6 years, at £4i per cent, per annum ? Ans. £3000^ 3. What principal will amount to £^}66 Ps- 0^- in 3^ years^ at £ f)l per cent, per annum ? Ans. £563. . 4. What principal will amount to £335 7s. 3d. in 3 years and 9T days, at £9^ per cent, per annum ? Ans. £255 ID 0-\ CASE HI. The amount J principal, and time given, to find the ratio. Rule. Subtract the principal from the amount; divide the remainder hy the product of the time and principal, and the quotient will be the ratio.* Examples. I. At what rate per cent, will £543 amount to £705 I8«. in ^'■■ rears » l^'rora the amount=705-9 Take the principal=543' Divide by 543x5=^27 15)1 62. 90(0^ 162-90 Ike same example, the amount, 33'6=30-f 30X*0t>X2, or which is the same };hing,= 1 -f •()(> X !^ X 30. Divide both by the same quantity^ 1 + -Ob X 2, and the • ^^^ .-11 u 111 ^'^ rPoGxixSG expression will still be equal, and we have — ; — ; — — = ; thca cancel the equal terms in the last fraction, and — ; ■ — =30, that is, the a- mount divided by the product of the ratio and time increased by 1, gives a quo- tient, which is the principal. The same may be shown in any other example, and, hence the rule is general. * Under case I. it was shown that the amount, 33'ar=:30+3U X "UO X ii. T-dk« the principal, 30, from both sides, and 33-G— 30r=z30X '06x2, or 3-6=:30X2X'06. _ . , , ,3-6 30x2X.0e Divide both parts by the product of time and principal, and -— — r,=n77CTi — ^ — or — =-06, the ratio, and illustrates the rule. 30X2 364 SLUPLE INTEREST BY DECfMALS. 705-9— 543 . Or, =-OG~£GAns. 543x5 2. At what rate per cent, will £391 17s. amount to £449 os. l|d, 74qr. in 3} years ? Ans. £4^.. 3. At what rate per cent, will £4 13 12s. 6(1. amount to £546 4s. lt)id. in 4| years ? Ans. £6^. 4. At what rate percent. will£3000 amount to £3810 in 6 years? Ans. £4|. CASE IV. The amounty principal^ and rate per cent, giverty to Jind ike time' Rule. Subtract the principal from the amount ; divide the remainder by the product of the ratio and principal ; and the quotient will he. the time.* Examples. 1. In what time will £543 amount to £705 IGs. at £6 per cent- ner annum ? From the amount=705-9 Take the principal=543 Divide by 543X-06=:32-58) 162-9(5 years, Ans. 162 9 2. In what lime will £3000 amount to £3810, at ^ per cent, per annum ? Ans. 6 years. 3. In what lime will £391 17s. amount to £449 3s. l|d. at£4i per cent, per annum ? Ans. 3} years. 7b ^rind the Interest of amj Sum^ at 6 per cent, per annum, for any number of months. Rule. If the months be an even number, multiply the principal by half that number ; and if the months be uneven, halve the even months, to which annex /^ ; thus the half of 19 is 9 5 ; and multiply the principal as before, dividing by 100 or cutting oft' two figures more at the right hand, than there are decimals in both factors, which reduce to farthings, each time cutting oft' as at first. 4. What is the interest of £345 16s. 6d. for 9 years and 11 months, at 6 per cent, per annum ? .^ ^- 3-6 30X'06X2 ' 3-fi ' ^^ ;. _ .„ ^ * Also, =- — ' , and -=2, the time, and is an ilUustj.T- 30X-0« 30X-0U mx'06 tion of Ihc Rule for Case IV. SIMPLE INTEREST BY DECIMALS. 9.&i y. m. 9 11 12 345-825 2)119 months. 59-5 59'5=i No. of months. 1729125 3112425 1729125 je205'765875=: £205 15 3^ Ans. Principal='je345 16 6 Amount: £551 11 92 A Table of decimal parts for every day in the twelfth part of a year, which consists of 36 5^ days. 1 dai/s. \dec. pts.\ days. \d€c. pts.\ days. dec. pts.\ daifs. dec. pts.\ days. \dtc. pts.\ 1 •033 7 •230 13 A21 19 •624 25 •821 2 •066 8 •263 14 •460 20 •657 26 •854 3 •098 9 •296 15 •493 21 •690 27 •887 4 •131 10 •328 16 •526 22 •723 28 •920 5 •164 11 •361 17 •558 23 •756 29 •953 6 •197 12 •394 18 •591 24 •788 30 •986 To find the Interest of any Sum, either for Months, or Months and Days at 6 per cent, per annum. Rule. Multiply the principal by the number of months, (or months and parts, answering to the ^\vea number of days in the table) and cut off one tig'ure at the right hand of the product more than is re- quired by the rule in decimals, and the product will be the inter- est for the given time, in shillings and decimal parts of a shilling.* * In the .iVb/e, under the General Rule for Simple Interest, it is shown that ■v^'hen the Rate is 6 per cent, the product of the principal and ?ial/ the number of months divided by 100, ^ives the Interest* Whence, the product of the prin- cipal and the number of months divided by 100, must give twice the Interest. 30*5 X 1 7 Let then the principal £30*5 be put to interest for 17 months. Then — — £5*185=2X £2-5925=twice the interest, and the interest is 2*5925. Multi- ply by 20 and the interest will be reduced to shillings and the decimal parts ol' 30'5 X 1 7 X '^0 a shilling, and we have — -"— =2x£2-5925X20. Divide by 2, and 30-5X17X10 ^^ ' Tm =£2-5925X20, and dividing both parts of the fraction by 10, 30-5 X 17 10 — £2-5925 X 20, that is, multiply the principal by the number of months ■^ni\ divide the product by 10, or, cut off only one figure more than the rule for K k ii)60, SIMPLE INTEREST BY DEGIMALSv Examples. 1. What is the inlercst of 2. What is the interest of £ 100, for a year ? £ 250 10s. for 19 months and 7 Principal=100 dajs ? Blult.by the months=12 Principal= £250-5 Ans. s.l20|0=£6 Time= 19-23 7615 JVote. This Table may also be 5010 tised for the parts of a year^ in 22545 Compound Interest, after having 2505 Worked for whole years. The shillings, &c. in the principal must Ans. s. 481-7115 iirst be reduced to the decimal — £24 1 8^ of a pound. Another Method of calculating lutenst for Months, at £6 per cent, psr annum. RtJLE. If the principal consists of pounds onfy, cut off the unit figure, and, as it then stands, it will be the interest for one month in shil- lings and decimal parts : — If it consist of pounds, shillings, &;c. re- duce the shillings, &c. to decimals, which, with the unit figure of the pounds, will be decimal parts of a shillmg.* Examples. 1. What is the interest of 2. What is the interest of £ 175, for 5 months ? £255 I6s. for 7 months ? £175=17-5 shill.=interest for Shill. 1 month, 17-5 £255 16=25-58 int. for 1 mr. Multiply by the time= 5 7 2|0)87-5 2|0)179 06 An&.=£4 7 6 Ans. £€ ^19 0| decimals requires, and you have the inlerest in shillings and decimal parts of a shilling. If there be months and days, the days being made decimals from the table, the same rule would manifestly apply. * This rule is only a contraction of the following process by the Double Rule of Three, to find the interest of any sum, e. g, £36, for 1 month. As 100 : 6 :: 36 12 : :: 1 month : the interest. 6 X 36 6 X 36 X '^O Hence -'— — ^^=^£*18=:3'6s. the interest for 1 montli. But ^^ ~ 12X100 ... • 1-2X100 "" the interest in shillings, or shillings and decimals, and cancelling the equal pari? 6X36X20 1X36X2 1X36X1 36 ^ ^ , .„. ' , , , , ^^ *^"^' •iisaob-"=^-237ir =-13^0- ="1^=^^*^ '^^^'^-'' ^' ^^"^^ ^' ''': cimals of a pound, the rule would be equally correct,r SIMPLE INTEREST IN FEDERAL MPNEY. 2G7 SIMPl^E INTEREST IN FEDERAL MONEY. Pboblem I. When the principal is given in JWw England pounds^ shillings, ^-c. and the anni^iil interest is required in federal money, at 6 per cent. Rule. Reduce the shillings, &c. to their equivalent decimal, hy inspec- iion, divide the whole by 5, and the quotient is the annual inter- est : Or, multiply the principal by 2, and the product (having the tinlt figure of the pounds cut off) will be the interest as before.* Examples. I . Required the annual interest of £517 3s. 74d. at 6 i>€r cent. ? 3s. = -15 6)517181 7id.= 030 D. cm. Excess of 12=001 103-436=103 43 6 Ans. . Or, 517181 •181 2 D. c. m. 103-4362=103 43 6y\. ^. Required the annual interest of £1, in cents ? Ans. 20 cents. Problem 11. When the principal i^ given in New England currency, and. the in- terest and amount are required in federal money at 6 per cent. Rule. Reduce the New England money to federal, then divide the principal by 20 and that quotient by 5 ; add those quotients to- gether, and they are the interest; or add thetn to the principal, did their sum is the amount. t Examples. 1. Required the s^mount of £425 I63. 8?,d. for 1 year, at 6 per •ent. ? * This rule is a contraction of the following process. By the General Rulf; for Simple Interest, (in the first example) the annual interest^:" — — — 100 This, multiplied by 20, would be reduced to shillings and deciiAals of a shillin;^, •and divided by ^, would be reduced to dollars and decimals of a dollar. Then, .M728LX6x50^51!:l£L=li=il!l»i = iJ,03 43c. 6^.^™. lOOXti 5X1 5 * '« 100X6 10X1 10 * + Dividing the principal by 20, gives the interest at 5 per cent, because 5 is >ae twentieth of a hundred; then dividing this quotient by 5, evidently gives ■« he interest for 1 per cent. Then, as 5-{- 1=6, the sum of the two quotients will be the interest at 6 per cent. ■ ' - Interest at 6 per cent, may often be calculated most easily, by finding the in- terest at 5 per c^nt. and adding one fifth of this interest to itself for 6 per cent. And a'id two fifihs of it to itself, and vmi Viave tho interest at 7 per cent. 268 INTEREST ON BONDS AND OBLIGATIONS. •8 3)425-835 •034 20)1419-460 •001 5)70-9725 •835 H1945 D. c. m. 1504-6170=1504-61 7. Ans. 2. Required the amount of £112 4s. 6d. for one year ? Ans. g396'52c 7io. 9dec. Problem III. When the principal is JVew England currency, and the monthly r?j- terest is required in federal money. Rule. Reduce the shillings, &c. to decimals, by inspection, then sepa- rate the right han«l figure of the pounds with the decimals, divide by 6, and the quotient is the answer in dollars, cents, &c.* Example. Required the monthly interest of £425 16s. S^d. in federal mo- ney? • -8 •034 6)42-5835 •001 D. c. m. 709725==7'09 71 Ans. £•835 • . INTEREST ON BONDS AND OBLIGATIONS, HAVING PARTIAL PAYMENTS ENDORSED. As there is much diversity of opinion relative to the computation Q>{ laxsoful interest in these cases, several Rules will be given, con- nected with the practice of the Courts of several of the States. The difference of the rules depends on the principle assumed in respect to the time Zi'lien interest becomes due. Rule L The foUovving rule is generally thought to allow too little inter- est. It is, however, adopted in some of the Slates. Find the amount of the principal and interest for the whole time ; then find also the amount q( each endorsement for the time it has been paid. From the lirst amount deduct the sum of the amounts of thr; several endorsements, and the remainder is the balance due. This rule involves the following points; 1. Interest is not due until the obligation is paid. ^42.vrCiX20 • This rule is a contraciion of the foilowinjf process. * ^—^=: tliR 6 ..,.,,, ^, 425-r.n5y?0Xf) 4e'5-8n5XlXl 4^vte A'^-^P.SS r-nncipal in clOii?. I hen ■ ~ — == — — =r t=:-.-- — i ^' . ^ 6XKHJXl'i 1X10X6 lUXt. C th6 "mpnlhly interest In ilollars and decimals of a dcllar. INTEREST ON BONDS AND OBLIGATIONS. iLC9 * 2. Hence, interest must be allowed on the endorsements from *he time they were severally made. Note. A shorter method of computing interest according to this rule, may be seen in examples 16 and 17 of Simple Interest. The absurdity of this rule may be seen in the following manner. Sup- pose I borrow glOO at 6 per cent, for ten years, and pay six dol- lars at the end of each year, what will be due at the end of 10 years ? The amount of glOO is gl60. But the first endorsement of g6 has borne interest for 9 years ; the second, for 8 years ; the third, for 7 years, and so on ; so that six dollars have in fact been drawing interest for forty Jive years, and thus produced g 16-20 of interest. This, added to the nine endorsements of §6 each, gives g70-20. That is, while I have paid only the annual interest of §6. the 'principal has actually been reduced §16 20. By paying J6 annually for 25 years, and computing interest on the several en- dorsements by this rule, the whole principal would be paid, and the lender would be indebted to the borrower the sum of §2, while in fact the lender had received no part of the sura lent. Rule II. The following rule was drafted by the late Judge Sedgwick, and ordered by the Court of Com.Tion Pleas for the County of Berk- shire in 1791, and is the Rule now used by the Courts of Massa- 'chusetts. ** On all contracts carrying interest, on which partial payments may have been made, the principal sum with the interest thereof shall be formed, at the time of payment, into an aggregate sum, from which shall be deducted the sum paid, if one year's interest shall have become due, and if not, then the interest shall be cast to the end of the year, and the aggregate formed as aforesaid, and from the same the payment or payments and the interest thereof from the time or times of payment shall be deducted, and the bal- ance or balances thus formed shall bear interest, and so from time to time, provided, that in »J0 instance^ interest shall be cast on a greater sum than the principal sum nor on any interest accrued." [Records of the Court.] This rule involves the following principles. 1. That, when an obligation has borne interest for one or more years, interest is not due at intervals of time less than one year. 2. Interest is to be computed to that endorsement, which, to- gether with the preceding endorsement or endorsements and its or their interest since the time of payment, shall be equal to or <^'xceed the interest on the principal when this endorsement is made, provided one year's interest shall have accrued. The re- mainder is a new principal, on which interest is to be computed as before. 3. Interest is allowed on all endorsements from the time of their payment, until the year has elapsed, or until an endorsement is made beyond the year, which with the preceding endorsement or ^endorsements and its or their intercut, eqnah or exceeds the in- merest due at snch endorsement. 27d INTEREST ON BONDS AND OBLIGATIONS. 4. Interest is not allowed on interest, because the deduction, when made, is intended to pay the interest then due, and the ex- cess, if any, goes to reduce the principal. 6. The design of the rule is to give the interest as nearly annt.'- aVIy as the endorsements will admit. But if endorsements are not made on the oblip^ation, the rule implies that the interest is not due until the obligation is paid ; and, it is well known, that the Courts will not sustain an action for the payment of the interest from year to year, uniess the obligation contains the express con- dition that the interest shall be paid annually. Note. In (he " Scholar's Adthmetick," and in the ** Mercan- tile Arithmetick," the Rule, said io be-established by the Courts of JMassachusetts, is precisely the same as that established in the State ef New York, which will be found on a following page, It will b6 evident from a comparison of the preceding Rule and that of New York, that the computation of interest ougfi^ to be considerably dif- ferent in the two States. Rule III. The f the State of New York. " The Rule for ca-^ling Interest, when partial payments have Itcvn rnadp, is to apply the payment, in the fir.«t place, to the dis- charge of the interest then due. If the payment exceeds the in- terest, the surplus goes towards discharging the principal, and the INTEREST ON BONDS AND OBLIGATIONS. 271 subsequent interest is to be computed on the balance of principal remaining due. If the payment be less than the interest, the sur- plus of interest must not be taken to augment the principal; but interest continues on the former principal until the period wbea the payments, taken together, exceed the interest due, and then the surplus is to be applied towards discharging the principal ; and interest is to be computed on the balance of principal as aforesaid." [Johnson's Chancery Reports, Vol. I. page 17.] This Rule contains the following principles. 1. rhat Interest is due at any time when a payment is made, if the payment is equal to the interest to that time : if not, the pay- ment is to be added to the following payment or payments, until their sum is equal to or exceeds the interest then accrued, and the balance is a new principal. 2. Interest is not allowed on any entlorsement. 3. Interest is not taken on interest, because the interest is due when the endorsement or endorsements are made. Note. Example 18 in Simple Interest is calculated by this rule. Example. I. To be calculated according to the preceding Rules. For value received I promise to pay James Penny or order one thousand dollars on demand with interest. James Gold. January 1st, 1815. The endorsements were, March 1, 1816, received on the above Note, 75 dolls. July 1, 1816, 20 do. Sept. 1, 1817, . - - . - 20 erore it becomes due, and is the difference between that sum, due some iimo hence, and its present worth. The present worth of any sum or debt, due some time hence, is such a sum, as if put to interest, would in that time an\' if I keep the moniey in my own hands till the debt shall become due, it is plain i .nay make an advantage of it by putting it out to interest for that time ; but if 1 pay it before it is due, I give that benefit to another ; therefore we have only lu inquire what discount ought to be allov/ed. And here, many suppose that, since by not paying till it becomes d'ue tlicy may employ it at interest ; there- fore, by paying it before due-, the)^ sliall lose that interest, apd for that reason all such interest ought to be discounted ; but the supposition is false, for they caiv not be said to lose that interest till the time arrives, when the debt becomes due ; whereas we are to consider what woul4 be lost, at present, by paying the debt before it becomes due ; this can, in point of equity, be no otlier than such asuD% which being put out at iviterest till the debt ^Iiall become due, would amount to the interest of the debt for the sarr.e time. It is" besides plain, that the advan- tage arising from discharging a debt due some tim.e hence, by a present payment, according to the principles above mentioned, is exactly the same as employing the whole sum at interest till the time when the debt becomes due, arrives : for, if the discount allowed for present payment be put out at interest ibr that time, its amount will be the same as the interest of the vholc debt for the same time ; thus the discount of ij 106, due one year hence, reckohihg interest at £G per cent, will be £6, and £6 put out to interest at £6 per cent, for one year, vrill amount to £6 7s. 2id. Vv^iich is exactly equal to the inierest of jG106» tor one year at £6 per cent. The truth of the rule for working is eviden^t from the nature of Simple Inter- (^< t ; for since the debt may be considered as the amount of some principal (call- .' 1 here the present wf)rth) at a certain rate per cent, and for the given time, ■Jjat amount mii?t be iii the ?a:'^:: r-.- ;>->ruoa eitlior io it- pjincipal or interest, S76 • DISCOUiST. Subtract the present worth from the given sum, and. the remain- der will be the discount required. Or, As the amount of £100 for the given rate and time, is to the interest of £100 for that time : so is the given sum or debt to the discount required. Or, In federal money, divide the given sum by the amount of glOO for the given time and rate ; point off from the right of the quotient, two places less than in division of decimals for the pren- eat worth. i ^xru-t :, iU-. j:>. ,. ^e> '^-'ooo i i tj pf;r cent. lii Biiiik Discount, the interest is considered as the discoiuit ABBREVIATIONS IN DISCOUNT. 277 5. What ready money will discharge a debt of ^1595 due 5 ajonths and twenty days hence, at 6 per cent. ? Ans. gl541 32c. 6m. 6. Bought a quantity ofgoods for §250, ready money, and sold Ihem for gSOO payable 9 months hence : What was the gain, in ready money, supposing discount to be made at 6 per cent. ? Ans. $37 8c. l^m. 7. What is the present wojth of §960, payable as follows ; viz. J at 3 months, ^ at 6 months, and the rest at 9 months, supposing the discount to be made at 6 per cent. ? Ans. g936 70c. Rule H. As any sum of money, at 6 per cent, per annum, will double, at simple interest, in 200 months ; therefore, To 200 add the number of months for which the discount is required, by which sum divide the product of the money and time, ; in months,) and the quotient will be the discount. Examples. h What is the discount of §150 75c. for a year ? 200 150-75 -f 12 Xl2 § c. m. 212)1809'Q0(8'633=:;:8-53 3 discount, Ans. 1696 1130 1060 150-75 8-533 700 G36 Present worth 142-217 640 636 4 2. Wha.t is the present worth of §426 55c. at 6 per cent, to be paid 8 months hence ? Ans. §410 14c. 6m. 3. What is the discount of £361 15s. 6d. to be paid 1 year and 8 months hence ? Ans. J£32 17s, 9id. ABBREVUTIOm IN DISCOUNT. Any principal to be discounted for one year, at any of the fol- lowing rales, (or for any rate and time, whose product is equal to any of the following rates) being multiplied by the multiplier, (if ^ny,) and divided by the corresponding divisor, the quotient will be the discount. 2^7S. .ABBREVIATIONS IN DISCOUNT. Rates.* ^- 11 ~-81 (or by 9 and 9) 2 -r-51 21-^41 ' 4 -r-26 5 -~21 (or by 7 and 3) ..a u -^63 andxS r 74-h43andx3 -27 andx2 (orX2 and^9 and 3) -13 -n -28 andx3 (orX3, and~^7 and 4) -^ 9 Examples. 1. IJow much must I abate of £5394 jOs. due 3 yeai^ hence^ a^t 2| per cent, per annum ? £5394 10s. ■ ■ ■ 2 10 U2i 2-1 X3 27~-3=9j 10789 8, therefore, x2, and-f'27 3)1198 lb 6i £399 11 10 Ans. 2. What is the discount of $546 62c. 5m. for 8i years, at 1 per cent, per annum, (or for 1 year, at 8i per cent, per annum ?) $ c. m. 13)546-62 5 Ans. g42-04 8 3. What is the discount of ^125 at IJ- per cent, per annum, fo^ four years, (or, at 4 per cent, per annum, for 1| years ?) Ans. ^5 95c. 2m. * These contractions are obvious from any example, wrought accordir.^ tc. the General Piule. I'hus, l6t the sum to be discounted be 300 dollar?. 1. At 1^ per cent. Then, by the E.ule, lOU 405 'IT H :: $300 : discount, or, 5 405 : 5 :: 300 : = — , and is the Iluie 405 81 ol : J :: 300 : the discount req.uired. '2. AtSpercoit. Then, 102 : 2 :: 300 : discount, or, 300 51 : 1 :: 300 : discounts — . 51 3. At 2i percent. Then, t02'5 : 2-5 :: 300 : discount, or 41 :. 1 :: 300 : Ai:s\ver required. 4. At o pe^r cent.. Then, 108-: 8 :: 300 : discount, or, 27 ; 2 ;: 300 : Answer required. In the game way, may all the contractions be made, Tt.e eonlractions {uv' uvade on the two terms of tl^e proportion ^vhich are invariable, when the rcls is given. DISCOUNT Bt DECIMALS. ^79 DISCOUNT BY DECIMALS. * The sum io he discounted, the time and the ratio given, to find tJie present worth. Rule. Multiply (he ratio by the time, add unity to the product for a divisor ; by which sum divide the sum to be discounted, and the jquotient w^ill be tlie present worth.* Subtract the present worth from the principal, or sum to be dis- counted, and the remainder will be the discount. Or, as the amount of £l for the given lime, is to £\, so is llie "interest of the debt for the said time, to tlie discount required. Subtract the discount from the principal, and the remainder will be the present worth. Examples. 1. What is the present worth of 6001. due 3 years hence, at 01. per cent, per annum ? First iMethod. Ratio~06 Multiply by the time= 3 rroduct=-18 Add 1- Divisor=M8)6O0(508-4745 present worth. 600 br, =:X508 9s. 5^d. Ans. •06x3+1 Present worth=508*4745=JC50o 9s. S^^d. which, subtracted from the principal, will give the discount— £91 lOs. G^d. * As the sum to be discounted is, in fact, the amount of some principal at tiie S^iven rate and time ; to find the principal, which is now the present worthy yovi have only to employ the rule for Case 2, in Simple hiterest by decimals. This is the rule in the text. Thus in Ex. 1, by said Case 2, — j ; — ;-=the -principal, iii this case the present worth. The remninder of the rule is evident from %vhat has been said under Discount, Rule 1 . •. jYo/e 1 . In the method used in Simple Interest by Decimals, yoU may easily find rules for obtaining either of the four terms, ji^rcA'cnf worthy ratio, time, or sjitn. io be discounted, when the other three are given. J^ote 2. When the ratio is -06, or six per cent, per annum, and the given time yj,:i3 expressed in months, if the debt be divided by unity added to half as many c^ hundredths of an unit as there are months in the given time, the quotient will I;, ^e the present worth. Thus for 3 years or 3G month?, the divisor, wo have jmr. Been to be 1+-06X3,— 1-f •18,or 1 added to half as many hundredths as there ^re months ; if the time be 3^ years or 42 months, the divisor is H--0Gx3"5.,-- 1 + -21 ; if 10 months, then l-l--06XlA=l-f--05i as before ; if 2 months, thra l-h-C6 X-rj,=l-f -01 ; if 1 month, then l-J-'OGX i-*-,— 1-'-00>, and „^ c^ !gO DISCOUNT BY DECIMALS. Second Method. Ratio=-06 Multiply by 3 As 118 : 1 :: 108 1 Add I- M8)108'00(91*5254. Amount of 11. for the given time And 600X'06x3=108~interest of the debt for the given time. — Discount=9I-5264=j£91 10s. 6d. which taken from the principal will leave the prestent worth=£508 9s. 6d. 2. What is the present worth of §558 62c. 5m. due 2 years hence, at 4^ per cent, per annum? First Method, Ratio=-045 X Time= 2- •09 -hi' Divisor=l'09)558'625(512'5— present worth. 545 136 109 t) 1 o X. 1 o 558625 545 ^)r, — =g512*5Ans. 545 •045x2+1 — - And g558 625--g5 12-5=^46 12c. 5m.=discr)unt, Or, As ^1-09 (=amount ofgl for the given time) : $1 :: g50 27625 (= interest of the debt for the given time) : g4G125=;li.'=coijnt as above. And, g558-625—g46125=g512'5=present worth, as above. 3. Required the present worth and discount of §4125 50c. at 6^ per cent, per annum, due 18 months hence ? . C present worth §3746 19c. l^m. ^°^- I discount 379 30 21. 4. What ready money will discharge a debt of 13541. 8s. due "^ years, 3 months, and 12 days hence, at 6^1. per cent, per annum ? Ans.jeil35 7s. 9d. DUriES. 281 DUTIES. DUTIES are assessed upon articles imported into the country, at a certain rate per pound, hundred, ton, gallon, &c. without re- spect to the value of the articles ; or upon articles according to their actual cost. The latter are called ad valorem duties. The duties are computed in the former case, on the most obvious prin- ciples, as will be seen in the following Examples. 1 . Required the duty on 987Ife of chocolate at 3 cents per pound. 3 ^29-61 cents, Ans. 2. If the duty on molasses is 5 cents on a gallon when imported in an American vessel, and 10 per cent, more in a foreign vessel, what is the duty on 3960 gallons in both vessels ? Ans. $197-50, and g217-25. 3. Required the duty on lOCwt. 3qrs. 14Ife of cordage at g2-25 per Cwt. in an American vessel, and at 10 per cent, more in a for- eign vessel? Ans. g24-47 nearly, and §26-81. 4. What is the duty on Ghhds. of brown sugar, weighing 53Cwt. 2qrs. 201fe tare 12lfe per 100, at 2|-cts. per pound in an American vessel, and at 10 per cent, more in a foreign vessel? Ans. §132, and gl45-20. Duties ad valorem are estimated by adding 20 per cent, to the actual cost of the goods, &c. when imported from or beyond the Cape of Good Hope, and 10 per cent, when imported trom any other places. Insurance, commission, &c. do not belong to the ^actual cost. The duties are computed in the following obvious manner. When the cost is reduced to Federal Money, add the per cent, to the cost, and then find the duty per cent, ad valorem. Examples. 1. What will be the duty on an invoice of goods, which cost £786 15s. sterling, at 15 per cent, ad valorem when imported in an American vessel, or at 10 per cent, more when imported in a foreign vessel from England ? £ s. $ 786 15=:3493-17 Add 10 per ceot.= 349-317 3842-487 10 per cent.= 384 2487 6 do. =^ 192-12435 15 per ceut.=g576-37-305 Ans. in Am. vessel 16A per ct.=§G34-01 Ans. in Fgr. ship. M m 282 BARTER. 2. What is the duty on goods, which cost in India, g2780oO, imported in an American ship, at 12 ^ per cent, ad valorem? Ans. $417-075. BARTER IS the exchanging of one commodity for another, and teaclres traders to propdltion their quantities vv'ithout loss.* Case i. When the quantity of one commodity is given, with its value, or that of its integer, that is, of lib. lca?t. lyd. ^'C. as also the value of the integer of some other commodity , to be given for it, to find the quan-^ tity of ifiis ; or, having the quantity thereof given, to find the tat^ of selling it. Rule. Find the value of the given quantity by the concisest method, then find what quantity of the other, at the rate proposed, you may have for the same money : Or, if the quantity be given, find from thence the rate of selling it. Or, As the quantity of one article is to its price, so, inversely, is the quantity of the other to its price. Or, as the price of one article is to its quantity ; so inversely fi»ihe price of the other to its quantity. Examples. 1. How much tea at 9s. 6d. per fe must be given in barter for 156 gallons of wirte, at 12s. 3id. per gallon ? Galls. 3d. 156 12 9s. 6d. = 114d 1872 39 6 6 1917 G 12 23010 d. ft oz. As 114 : 1 :: 23010 :: 201 13//y Ans. price, quan. price. quan. s. d. gals. s. d. ft oz. Or, As 12 3-^- : 166 :, 9 G- : 201 IS/^-V Ans. as before. * The Rules in Barter are only applications of the Rule of Three, and are ea- sily un(J.erstood. BARTER. 28J 2. How much clotb, at 15s. 8d. per yard, must he gjven for 5cvyt. Sqrs. 191bs. of steel, at 5 guineas per cwt? Ans. 52yds. 3qrs. 2n. 3. Suppose A has 350 yards of linen, at 25c. per yard, which he would barter with B for sugar, at $5 per cwt. How much sugar will the linen come to ? Ans. ITcwt. 2 qrs. 4. A has broadcloths at g44 per piece, and B has mace, at gl 42c. per ife : How many pounds of piace must B give A for 35 pieces of cloth ? Ans. 1084|^ife. 5. A has 7icwt. of sugar at 12 cents per ife for which B gave Jiira :i2^cwt. of flour : What was the flour rated at per ife ? 4"^. 7c. 2m. CASE H, If the quantites of two commodities be given ^ and the rate of selling them, to find, in case of inequality, how much of some other com- modity mpst be given. Rule. Find the separate values of the two comiHodjties ; subtract the ,ess from the greater, and the difference will be the amount of ^he third commodity, whose quality and rate may be easily found. t Examples. 1. Two merchants barter ; A has SOcwt. of cheese, at 23s. 6d. per cwt. and B has 9 pieces of broadcloth, at 31. 15s. per piece : Which must receive money and how much? Ans. B must pay AjGl 10s. 2. A and B would barter ; A has 150 bushels of wheat, at jjl 25c. per bushel, for which B gives G5 bushels of barley, worth 62 ic. per bushel, and the balance in oats at 37Jc. per bushel: What quantity of oats must A receive from B ? Ans. 391| bushels. CASE III. Sometimes, in bartering, one commodity is rated above the ready money price ; then, to find the bartering price of the other, say, As the ready money price of the one, is to its bartering price ; so \i that of the other, to its bartering price : Next, find the quanti- ty required, according to either the bartering or ready money price. Examples. 1. A has ribbands at 2s. per yard ready money ; but in barter ha will have 2s. 3d. B has broadcloths at 323. 6d. per yard ready money ; at what rate mu»t B value his cloth per yard, to be equi- valent to A's bartering price, and how many yards of ribband, at 2s. 3d. per yard, must then be given by A for 488 yards of B's broad- doth ? Ans. B's broadcloth, at£l iGs. 6id. per yd. 7930 yds. ribband. 2. A and B barter ; A has 150 gallons of brandy, at ^1 37|c. per gallon ready money, but in barter he will have $1 50c. fB has 284 LOSS AND GAIN. linen at 44c. per yard ready money ; how mustB sell his liuen per yard in proportion to A's bartering price, and how many yards are equal to A's brandy ? Ans. barter price is 48c. and he must give A 468yds. 3qrs. 3. F and Q barter; F has Irish linen, at 60c. per yard, butiu^ barter he will have 64c. Q delivers him broadcloth at $6 per yard, worth only ^5 50c. per yard : Fray which has the advan- tage in barter, and how much linen does P give Q,for 148 yards of broadcloth ? c. c. $ c. $ c. As 60 : 64 :: 6 60 : 6 86|; therefore Q, by selling at $6 has the advantage. Then, $ yds. c. yds. qrs. As 6 : 148 :: 64 : 1387 2 linen, Ans. 4. A has 2Q0 yards of linen, at Is. 6d. ready money per yard, which he barters with B, at Is. 9d. per yard, taking buttons at 7id. per gross, which are worth but 6d. : How many gross of buttons will pay for the linen, who gets the best bargain, and by how much, both in the whole, and per cent. ? Yd. d. Yds. d. . d. Gross, d. Gross. Yd. d. Yds. £ As 1 : 21 :: 200 : 4200. As 71 : 1 :: 4200 : 660. As 1 : 18 :: 200: 15. gr. d. gr. £ [value of A's linen. As 1 : 6 :: 660 : 14 value of B's goods. So that B gains ll. of A. £ £ £ £s- d. As 14 : 1 :: 100 : 7 2 10 per cent. 5. A has linen clojh, at 30c. per yard, ready money, in barter 36c. B has 3610 yards of ribband, at 22c. per yard ready money, and would have of A ^200 in ready money, and the rest in linen cloth ; what rate does the ribband bear in barter per yard, and how much linen ntiust A give B ? Ans. The rate of ribband is 26c- 4m. per yard, and B naust re- ceive 1980| yards of linen, and ^200 in cash. LOSS AND GAIN IS an excellent rule, by which merchants and traders discover r their profit, or loss per cent, or by the gross : It also instructs them to raise or fall the price of their goods, so as to gain or lose go much percent, &c. The rules are only particular applications of the Rule of Three. CASE I. To know Zi)hat is gained or lost per cent. Rule. First see what the gain or lo^s is, by subtraction ; then, a§ the price it cost, is to the gain or loss : so is 1001. to the gain or loss T)nr cent. LOSS AND GAIN. 285 Or, in federal money, annex two cyphers to the gain or loss, and ^iivide by the cost for the gain or loss per cent. Examples. 1. If I buy serge at 90g. per yard, and sell it again at gl 2c. per I yard : What do I gain per cent, or in laying out glOO ? c. c. S J^ r Sold for gl-02 As 9Q : 12 :: 100 : 13^ per cent, gain, Ans. ^' Cost -90 •Gain -12 per yard. 12-00 Or, 1-02 — •90=12— gain per yard ; and =13i per cent, gain, '^ [as before. N. B. The first questions in the several cases, serve to elucidate each other. 2. If I buy serge at $1 2c. per yard, and sell it again at 90c. per yard : What do I lose per cent, or in laying out JlOO? ^ C. ^ C. C. ^ ^ C. DO. Cost 102 As 1 02 : 12 :: 100 : 11 76 5 per cent, loss, Ans. Sold for -90 — 12.00 Loss 12 Or, y— 3"— 11-765 per cent, loss, Ans. as before. 3. If I buy a cwt. of tobacco for £9 6s. 8d. and sell it again at 3s. lOd. per ffe do 1 gain or lose, and what per cent. ? m JS s. , d. : ' 112 Sold for 10 5 4 -— Cost 9. 6 8 £ • j2d.|YV|l I 4 value at 2s. per ife. 18 8 gained in the gross. 18 8 value at 2d. per Ife. 10 5 4 value at Is. lOd. per ife. £ s. d. s. d. £ £ As 9 6 8 : 18 8 :: 100 : 10 Ans. 10 per cent. gain. 4. A draper bought 60 yards of cloth at $4 50c. per yard, an« 38 yards of cloth at jJ2 50c. per yard, and sold them, one with another, at s. d. 9 per cwt. as 3 8 s. d. £ £ s. d. 16 11 :: 100 : 27 17 8^ [loss per cent. Ans. Lost .go 16 11 in the whole. 6. At 4s. 6d. io the pound profit : How much per cent. ? X ?. a. £ £ s. As 1 : 4 6 :: 100 : 22 10 Ans. 7. If I buy candles at Is. 6d. per fe and sell them again at 2s. per fe and allow 3 months for payment : What do 1 gain per cent. ? d. d. £ £ s. d. Mo. j^ Mo. £ s. As 18 : 24 :: 100 : 133 6 8 ; then by discount. As 12 : 6 :: 3 : 1 10 £ s. £ s. £ s. d. £ s. d. Then, as 101 10 : 1 10 :: 133 6 8 : 1 19 4f, which taken from X133 Gs. 8d. Ieaves£l31 7s. 3id. therefore, Ans. £31 7s. 3|d. 8. If I buy cloth at 13s. per yard, on 8 months credit, and sell it a^ain at 12s. ready money, do 1 gain, or lose, and what per cent. ? Ans. lost £4 per cent, or 6d. in the yard. 9. If I buy gloves at^l 25c. per pair: How long credit mu^t 1 have, to gain ^13 per cent, when I sell them at $1 36c. per pair? '^ c. ^ c. c. ^ ^ c Sold at VsG As 1-25 : -U : 100 : 8-80 gain per cent. rdy. mo. Prime cost 1-2^ $ ^ c. g c. Then, 13— 8-80=4'20 Now, Gained -11 per pair. $ Mo. g c. Mo. days. As 6 : 12 :: 4-20 : 8 12 Ans. - In casting up tlie a^i^'Jiit of goods bought, imported or export- ed : to the prime cost of such goods we must add all the charges upon them, in order to fix the price they stand ns in. 10. Suppose 1 import from France, 12 bales of cloth, containing 10 pieces each, which, with the charges there, amounted to J360 : I pay duty here 92c. per piece, for freight §12 and portage gl 25c. ; What does it stand me in per piece, and how must 1 sell it per piece to gain §10 per cent. ? Ans. §4 43 3 the price at which it must be sold per piece, CASE IL To knoio hojp a commodiiy must be soldy to gain or lose so much per' cent. KuLF.. As £100 is to the price ; so is £100 with the profit added, or loss Subtracted, lo the gaining or losing price. Or, In federal money, multiply 100 dollars added to the gain, or less by the loss per cent, by the cost ; and pointing off the two i ight hand figures of the product gives the answer. Examples. 1. If I buy a quantity of serge at 90c. per yard: How R)ust 1 It it per yard to gain 13} per cent. ? ** s' C. C. *J( c As 100 : !!3 33} :: 00 : 1 2 Ans. LOSS AND GAm. 287 : $ '*• G. $ .Or, 113 33ix90=102; and pointing oflf two right hadd places, ^1 02, Ans. as before. 2. If a barrel of powder cost J£4, how must it be sold to lose X 10 per cent. ? £ £ £ Or thus : As 100 : 4 :: 90 90 4 4 100)360(3 £3160 300 20 60 9.12|00 20 100)1200(12 Ans. £3 i2s. 3. Bought cloth, at $2 50c. per yard, which not proving so good as I expected, I am content to lose 17} per cent, by it : How must -I sell it per yard? Ans. $2 6c. 2Jm. ' 4. If 120ife of steel cost £7, how must I sell it per Ife to gain £13; per cent. ? Ans. Is. 4d. per Ife. 5. A gentleman bought 10 tons of iron for £200, the freight and duties came to £25, and his own charges to £8 6s. 8d. ; How must he sell it per Ife to gain £20 per cent, by it ? £ £ £ s. d. £ s. d. £ s. d. £ s. d. £ As 100 : 20 :: 233 6 8 : 46 13 4 Then, 233 6 8-f46 13 4=280, Tons. £ lb. d. As 10 : 280 :: 1 : 3 per ft Ans. 6. If a bag of cotton, weighing 8cwt. Oqrs. 201fe cost §45 55c. how must it be sold per cwt. to lose g8 per cent. ? /- Ans. $5 12c. 3m. 7. Bought fish in Newburyport, at 10s. per quintal, and sold it at Philadelphia, at 17s. 6d. per quintal ; now, allowing the charges at an average, or one with another, to be 2s. 6d. per quintal, and considering I must lose £20 per cent, by remitting my money home ; what do I gain per cent. ? Selling price 17 6 Philadelphia currenc}^ per quintal. Charges 2 6 ditto. 15 ditto. £ 5. £ ?. As 100 : 15 :: 80 : 12 New England currency. Sold at 12s. per quintal. Bought at 10s. per quintal. Gained 2s. per quintal, s. s. £ £ As 10 : 2 :: 100 : 20 per cent, gained, Ans. 0. Bought 50 gallons of brandy, at 75c. per gallon, but, by acci- dent, 10 gallons leaked out: At what rate must I sell the remain- der per gallon, to gain upon the whole prime cost, at the rate o; 10 per cent. ? Ans, $1 3c. l^m. 288 LOSS AND GAIN. CASE III. When there is gain or loss per cent, to know what the commodity cost- Rule. As .£100 with the gain per cent, added, or loss per cent, sub- 'racted, is to the price ; so is £100 to the prime cost. Or, In federal money, divide the price with two cyphers annexed by §100 added to the gain, or less by the loss, per cent, for the answer. Examples. 1. in yard of cloth be sold, at ^1 2c. and there is gained 13} per cent, what did the yard cost ? * 4 C. ^ C As T0O+T3I : 1 2 :: 100 : 90 prime cost, Ans. 10200 Or, ;j^^^_^='9, Ans. as before. 2. If 12 yards of cloth are sold at 15s. per yard, and there vi £1 10s. loss per cent in the sale : What is the prime cost of the whole. Yds. s. Yds. £ £ s. £ £ £ s. u. As 1 : 15 :: 12 : 9 As 92 10 : 9 :: 100 : 9 14 7 Ans. 3. If 401fe of chocolate be sold at 25c. per Ife and I gain 9 per cent, what did the whole cost me ? Ans. $9 17c. 4m. -p 4. If 19icwt. sugar be sold at gl4 50c. per cwt. and I gain gl5 per cent. : What did it cost per cwt. ? Ans. gl2 60c. 8m. CASE IV. If by 'wares sold at such a rate, there is so much gcined or lost per cent, to know what would be gained or lost per cent, if sold at another rate. R ULE. As the first price is to £100 with the profit per cent, added, or loss per cent, subtracted ; so is the other price to the gain or loss per cent, at the other rate. N. B. If your answer exceed 100, the excess is your gain per cent, but if it be less than 100, the deficiency is your loss per cent. Examples. 1. If cloth, sold at ^1 2c. per yard, be 13J- profit per cent^ what gain or loss per cent, shall I have, if I sell the same at ?0c. per yard ? $ c. $ c. $ As 1 2 : 113i :: 90 : ,100 And, 100 — 100=0, Ans. I neither gain, nor lose. 2. If cloth, sold at 4s. per yard, be £10 per cent, profit : What shall I gain or tose per cent, if sold at 3s. 6d. per yard? EQUATION OF PAYMENTS, 289 3, £ s. d. As 4: 110 :: 3 6 12 12 JS £ £ ,^ — Then, 100— 961=3^ 48 42 no \ 48)4620(961 Ans. I lostJCSf per cent, by (he last sale. 432 300 288 12 3. If I sell a gallon of wine for ^I 50c. and thereby lose 12 per cent. : What shall I gain or lose per cent, if I sell 4 gallons of the ^anie wine for ^6 75c. ? Ans. 1 per cent. loss. 4. 1 sold a watch for 501. and by so doing, lost 171. per cent- whereas in trading I ought to have cleared 201. per cent. How much was it sold under its real value ? Ans. 221. 5s. 9|d. EQUATION OF PAYMENTS IS the finding a time to pay, at once, several debts, due at differ- ent tiroes, so that no loss shall be sustained by either party. Rule 1.* Multiply each payment by the time at which it is due ; then di- vide the sum of the products by the sum of the payments, and the quotient will be the equaled time, or that required. * This rule is founded upon H supposition that the sum of the interests of the several debts, which are payable before the equated time, from their terms to that time, ought to be equal to the sum of the interests of Ihe debts payable after the e luated time, from that time to their terms. Some, who defend this princi- ple, have endeavoured to prove it to be ri2;ht by this argument ; that what is {gained by keeping some of the debts after they are due, is lost by paying others before they are due ; but this cannot be the case ; for though, by keeping a debt after it is due, tiiere is gained the interest of it for that time ; yet, by paying a. debt before it is duo, the payer does not lose the interest for that time, but the discount only, which is less tiian the interest, ond therefore the rule is not accu- rately true ; however, in most questions, which occur in business, the eiTour is /o trilling, that it will always be made use of as the most elligible method. From the principle assumed in tl-iis rule, the rule may be derived in the follow- ing manner. Thus in Example 1, where W months is found to be the equated time, let the interest be supposed at any rate, as 6 per cent. Then the first payment .' , . \. 100X0X8*-^ IS to be at mterest for 8—6 or 2 mouthy and by the rule for interest, -— - =it.s interest. The second sum i^ to bo on interest ^- — "■ of 1 nionth, end N n 290 EQUATIOiN" OF PAYMENTS. Examples. 1. A owes B g380 to be paid as follows, viz. glOO in 6 monlh?, ^120 iQ 7 months, and gl60 in IQ monihs : What is the equatet^ time for the payment of the whole debt ? HK)x C= 600 120X 7= 840 1(30X10=1600 100+1204-160—380)3040(8 Months, Aus. 3040 2. A owes B f04l. 15s. to be paid in 4 J months, 1611. to be paid in 3^ months, and 1621. 6s. to be paid in 5 months : What is the equated time for the paj^ment of the whole ? Ans. 4 months and 8 days. 3. There rs owing to a merchant 9981. to be paid, 1781. ready money, 200K at 3 months, and 3201. rn 8 months ; I demand the irr different time for tbe payment of the vvhoFe ? Ans. 4^ months 4. The sum of ^164 I6c. 6m. is to be paid, ^in 6 months, i in C months, and ■} in 12 months : what is the mean time for the pay- ment of the whole ? Ans. 7| months. Rule U. See, by rule 1st, at what time the first man, mentioned, ought to pay in his whole money; then, as his money is to his time, so is the other'^s money, to bis time, inversely, which, when found, must be added to, or subtracted from, the time at which the second ought to have paid in his money, as the case may require, and the sum? or remainder, will be the true time of the second's payment. Examples. 1 . P is indebted to Q, $ t50 to be paid, $50 at 4 monihs, and g 100 at 8 months : Q. owes F ^250 to be paid at 10 months : tt is agreed =;it3 interest to the equated time. The sum' of the interest of 100 ^ these two payineuts is, by the assumed principle, to be equal to the interest of . 160x6xiu — i:. the third payment, or j£lSO for 10 — 8 or 2 months, which is — IOOxGxh"-^ , l^OxGxiT-^ 160X6X10— H ^^ Then, ' = ^ -—' Now, as the rate per 100 100 100 ^ cent, and 100 will be factors common to every term in every case, they may be expunged from every term, and then we have, lOOXH— + 120 X;{-^=160X 10— B. From thio equivalent expression, it is easy to find the equated time; for, 100x8— 100x6-fl20Xt!— 120x7=100 XlO— 160xr„ or 100x8-1-^-^0X8+160x3— 100x6-|-120x7-fl60xl0, or, 8 Xl(X)+lt'0+ 160—100x6+120x7+160x10, and 8=100X6+120X7+160X10 ,.,.,,,, ^^ , which IS the rule. The same may be shown loo+ix-o+ioo ^ in every similar case, and tlio general rule inferred. This rule is manifestly incorrect. The iriie rule will bo given in Equation o'.' I^aymenta by Decimal'-. Kt^UATION OF PAYMENTS BY PECIMALS. 291 between them that P shall make preser*! pay of his whole defct, and that Q, shall pay his so much the sooner, as to balance that fa- vour ; I demand the time at which Qmust pay the $250 reckonin,^ simple interest. ■ 50X4=200 100X8=800 60-f-100=15lO)100|0(6| months, P*s equated time. 90 10 L). mo. D. mo. mo. mo. mo. As 150 : 6| :: 250 : 4. Then, 10—4=6 time ofQ's payment. 2. A merchant has 1201. due to him, to be paid at 7 months ; but the debtor agrees to pay ^ ready money, and ^at 4 months ; I de- inand the time he must have to pay in the rest, at simple interest^ §0 that neither party may have the advantage of the other ? Debt £120 1=60 must be paid down. ^=40 must be paid at 4 months. i=20 unpaid. Now, as he pays 601. 7 months, and 401. 3 months before they are respectively due, say, as the interest of 201. for 1 month, is to 1 month, so is the sum of the interest of 601. for 7 mqnths, and of 401. for 3 months, to a fourth number, which, added to the 7 months, will give the time for which the 201. ought to be retained. Ans. 2 years and IQ mouths. 3. A merchant has §1200 due to him, to be paid i at 2 months, } at 3 months, and the rest at 6 months ; but the debtor agrees to pay i down : How long may the debtor detain the other half, so that neither party may sustain loss ? ' Now as i was paid 4^ months before it was due, it is reasonable Jhat he should detain the other ^, 4^. months after it became due, which added, gives 8| months, the true time for the second pay- ment. Equated time=4^ months. EQUATIOJV OF PAYmNTS BY DECIMALS. Rule,* 1. To the sum of both payments add the continual product of the tirst payment, the ratio, and the time between the payments, and call this the first number. * Suppose a sum of money be due immediately, and another at the expiration of a certain given time for^^arJ, and it is proposed to find a time, so that" neither party ghall sustain loss. Now, it is plain that the equated time must fall between the two payments ; and that what is gotten by keeping the first debt after it is due, should be equal •J) what it l(^st by paying the seccud deU before it is dwe ; but the gain arising £92 EQUATION' OF PAYMENTS BY DECIMALS. 2. Multiply twice the first payment by the ratio, and call this the second number. 3. Divide the first number by the second, and call the quotient the third number. 4. Call the square of the third number the fourth number. 5. Divide the product of the second payment and time between the payments by the product of the first payment and the ratio, and call the quotient the fit^th number. 6. From the fourth number take the fifth, and call the square root of the difference the sixth number. 7. Then the difference of the third and sixth numbers is the equated time, after the first payment. Examples. There are glOO payable in 2 years, and gl06 at 6 years hence ; what is the equated time, allowing simple interest, at 6 per cent. per annum ? 1st payment=^100 1st payment 100 Ratio=-06 Multiply by 2 6-00 200 Time between the payments=4ys. MnU.bytheratio= 06 Add both payments= j Div. by the 2d num. = 12)230=lst number. 24 12-00?=2dnuai- 100 106 19-166-f-=3d number. 19-166+ 3d number squared— 367-333556=;lth number. 2^1 payment-- 106 Multiplied by the time=^ 4 Ui payment >nult. by .he raUo=6)42^. \ ^'i^J^^r;::^^ 70 666-f— 5th number. From the 4th number=367 333556 Take the 6th number= 70 666666 296 668890(17-221sqr.rool~6thniifiK Carried up. from the keeping ofu sum of money after it is due, is evidently equal to tlie in- terest of the debt for that time : And the loss, which is sustained by the payiui of a sum of money before it is due, is evidently equal to the discount of the del for that time : Therefore it is obvious that the debtor must retain the sum in mediately due, or the first payment, till its interest shall be equal to the discouv.t of the second sum for the time it is paid before due ; because in that case the gaiii and loss will be equal, and consequently neither party can be a loser. EXCHANGE. 293 From the 3d number=191 66 Brought up. Take the Gth number=17-224 1 ■942==equated time from the first pay- ment ; therefore 3 942 years =3y. 11m. 9d.= whole equat- ed time. 100+ 106+ 100 X -06X4 100+ 106+ 100 X '06X4 Or, 106X412 =1-942. 100 X -06 1 100X2X-06 100X2X-06 2. There are glOO payable one year hence, and gl06 to be paid six years hence ; what is the equated time, computing interest at 6 per cent. ? Ans. 3. A debt of ^1000 is to be paid, one half in three years and the other half in 6 years ;"what is the equated time for paying both, computing interest at 7 per cent ? Ans. EXCHANGE. THE object of Exchange is to ascertain what sum of money ought to be paid in one country for a sum of different denomina- tions or of different relative value received in another, according to the course of exchange. The par of exchange respects the intrinsic value of the money of different countries compared with each other. Thus a pound sterling is equal to 4 dolls, and 44 cents in the United States ; the mark banco of Hamburgh, to 331 cents ; 40 marks banco to £3 sterling. If the exchange be made at the intrinsic value of the money of different countries, it is said to be at par ; but if the mo- ney of one country be estimated at less or more than its intrinsic value, the exchange is said to be above par, or below par. '^ Owing to o2i ------- 2^ 75 - - - 2i U7i 2i 100 - - ~ 2^ 125 - - 3 150 -------- 34 175 2|. 200 - - 4 i liese multipliers are thus formed. Let the advance be 25 per cent, on £100 ; then, as 25:=i of a hundred, lOOX — = — =:tho sum with the advance. This 4 4 IS to be reduced to New England currency by increasing it by one third of itself. Thus —- X:^ — :="T<:r^^^^'3' pounds ; which is evidently the same as to mu^- 4 3X4 12 tiply 100 by 1 J. In the same way may the other multipliers be four EXCHANGE. ^95 5. A Bill of Exchange of £75 16s. is sold at Boston at 26s. New England currency per pound sterling ; ivhat is the value in Fed nral Money of a pound sterling at this rate of exchange ? Ans. 2, OF FRANCE. The money of account is livres, sols, and deniers. 12 deniers make 1 sol or shilling. 20 sols 1 livre or pound. The livre is estimated at 18^- cents in the U. S. The crown of exchange is 3 livres, or livres tbumois, aiid i^ equal to 55| cents. The present money of account is francs and centimes or hun- dredths. 80 francs~8i livres, or a franc=||^ livre. 1. To reduce francs to livres, or the contrary, multiply the francs by 81 and divide the product by 80 for livres ; or Hvultiply the li- vres by 80 and divide the product by 81 for francs. 2156x81 Thus 215G francs= ^7; =2183 \Wtes 19 sols. And 2341 2341x80 iivre9= ' ^. =2312 francs, 09|| centimes. 2. To reduce livres to dollars and cents : multiply the livres by the <;ents in a livre at the course of exchange. • Examples. 1. If the livre be 20 cents in exchange, what is the amount of 2150 livres in Federal money, and what is the per cent, above par at this exchange ? Ans. Amount is g430. And above par 83*^ per cent. 2. If the livre be 18 cents in exchange, required the amount of 3580 livres 16 sols, in dolls, and cents, and the rate per cent, be- low par. Ans. 644-54/^ cents, and 2?^ per cent. belo\v par. 3. If a crown be valued in exchange at 18d. sterling, required the livres in £100 sterling, and the amount also iu Federal money at par. d. liv. £ liv. As 31ivres=l crown, 18 : 3 :: 100 : 4000 and 4000x1 8-Ji=g740, 4. In 2583 francs, how many dollars ? " 2583x55A = 1433dol]s. 561 cents. 5. A bill of exchange on a merchant in New York of ^730 65cts= was bought at Paris at I^ per cent, advance ; what is the amount in francs, and what was the estimated value of a franc at this es.^ change ? Ans. 3. OF SPAIN. 4 Maravadies make 1 quarto. 81 quartos=34 raarav, 1 rial plate. 8 rials plate 1 piastre or current dollar^ 375 maravadies 1 ducat of exchange. Hard or plate dollars arc 88-/^ per cent, above current dollars qt money of vellon, o'™ ^96 ElTHANGE. 100 rials pla(e =188/^ rials vellon. 17 do. =::32 do. The rial plate is 10 cents, and the rial vellon 6 cents in the U. States. To reduce rials plate to rials vellon, or the contrary, multiply the rials plate by 32 and divide the product by 17, for rials vellon ; ov multiply the rials vellon by 17 and divide the product by 32, for rials plate. 1100x32 1. Thus 1100 rials plate= jp^ rials vellon=::2070i^ Ans. 100x17 And 100 rials vellon= — ^^ rials plate=53| rials plate. Ani-'. Note. The rules to reduce rials plate or vellon to Federal Mo ney are obvious and need no examples. 2. In the sale of a bill of exchange of 1563 rials plate, the rial plate was estimated at 9f cents ; how much per cent, was the rial below par and how much the loss ? Ans 4A per cent. $6'94| the loss. 3. If the piastre be valued in exchange at 81 cents, what is the. percent, above par on a bill of 1672 piastres 5 rials plate, and wha'. is the advance on the bill in Federal Money ? Ans. 4. OF HAMBURGH. 12 deniers=2 grotes make 1 shilling lubs, or stiver. lo shilling lubs=32 grotes 1 mark banco-* 3 )vioUs. The difference between 100 and the rate per cent, will be the first term, 100 the second, and the sum to be insured the third term of a proportion, and the rule is merely a particular application of the Rule of Three. In the first example, the proportion would stand thus, 100 — 8 ; 100 :: 759 : the policy= -—- — ^-—-£825. Now the premiun[i on £825, is, ——-—£66, and 66-f- lOO — 8 100 750=:£825, tlie policy. The rule for decimals is evidently a contraction of this rule. In Case III. the last three ternis ip the preceding proportion are given to find the rate. Those three terms evidently give the diflfererice between 100 and the rate, and the rule is obvious. In Case IV. the first two terms and the last term of the preceding proportion are given, to find the third term or sum covered, and the reason of the operation J5 \.l\in from tlio con«idenilion of that proportion. :?00 POLICIES OF INSURANCE 1230 : 500 :: 100 100 1250)60000(40 and £100— 40=:i:60, Ads. 50000 Or, =40, &c. as before. 1250 2. It'a policy be taken out for g781-25, to cover ^625 : Requir- ed the premium per cent. ? *|> c. k S § c. As 781-25 : 623 :: 100 : 87-30. And, 100— 87-5=12-5, or 121 62500 [per cent, premiunn, Ans. Or, =87-5, &c. as before. 781-25 CASE IV. When the policy for covering any sum and the premium per cent, are given, tojind the sum to be covered, RULK. Deduct the premium per cent, from 100, and say, As IQO is to the remainder, so is the policy to the sum required to be covered. Or, In decimals^ Multiply the policy by the remainder found a^ before, and point off two right hand places in the product for the an- swer. Examples. 1. If a policy be taken out for 12501. at 60 per cent. : What is the adventure or sum to be covered ? 100 60 100 : 40 :: 1250 Or, 1250x100—60=50000, and, < 40 pointing off two places, 500 00 ^ £ Ans. as before. 100)50000(500 Ans. 2. If a policy be taken out for $781 25c. at V2.\ pe«-cent. requir ed the sum covered ? 781-25X100—12^ As 100 : 100—121 :; 781-25 : ^=$625, Ans. 100 Or, 781 •25x100—12 5=62500 ; and 625-00, Ans. as before. CASE V. When a given sum. is adventured several voyages round from one place : to another^ either at the same, or di^/ferent risks, from place to place,' and it is required to take out a policy for such a sum as will cover \ the adventure all round, supposing the risk out and home to he equal ' and tantamount to thz several given risks. Rule. 1. Raise 1001. or ^100 to that power denoted by the number o{ risks, and multiply the said power by the sum adventured, (or to be covered) fur a dividend. POLICIES OF INSURANCE. 301 2. Subtract the several premiums, each, from 1001. and multi- ply the several remainders continually together for a divisor, and ^ the quotient, arising from this division, will give the policy to cov- I er the adventure the voyage round.* Example. A merchant adventured glSOO from Boston to Philadelphia, at 3 per cent, from thence to Guadaloupe, at 4, from thence to Nantz, at o, and from thence home at 6 per cent. ; For what sum niust he take out a policy to cover his adventure the vogage round, suppos ing the risk to be equal out and home, and tantamount to the sev eral given risks ? 100 X 100 X 100 X 100x1500 - ■ :. :== ..- =g 1803-835, Ans. 100— 3X100— 4X100— 5X100--6 "^ CASE VI. When a given sum is adventured several voyages round, as in the last case^ either at the same^ or different risks, from port to port, and the premium for the. voyage round is required, tantamount to the several given rates per cent. * It is evident that the policy to be takeij out for tlie first voyage becomes the sum for which a policy is to be taken out for the second voyage, and so on. Hence the examples of this case are to be solved by the rule for Case II. making the sum in the policy for the fir^t voyage, the sum for which a policy is to be tak- en out for the second voyage. Therefore the operation on tlie given example would be as follow*. 100 — 3 : 100 :: 1500 : policy for 1st voyage=: -—. Now as — ^-- 100 — 3 100 — 3 is the sum to be insured on the second voyage, we have, .nr. . inn 100x1500 ^ , ^. 100x100x1500 100--4 : 100 :: — — - ; 2nd policy^. — . ^ ^"^'^ 1 00— yj X J 00—4 » 1 .nn r inn 100X100X1500 ^^ ^. 100X100X100X1500 And 100 — 5 : 100 :: zrriizr ■ ■ ■ ■ - : 3d policyrir: — — — . 1 00— 3 X 100—4 100—3 X 100—4 X 100— S . , ^^ . ^^ 100X100X100X1500 ^^, And 100—6 : 100 :: ■. : 4th policy= 100—3 X 100—4 X 100—3 100X100 X100X100X1500 1004X1500 100—3 X 100— 4 X Too— 5 xToO— 6 100^3 X 100—4 X 100— ^X 100— 6 which is the Rule. The same may be shown by the Double Rule of Three, thas, 100—3 : 100 :: 1500 : V, ,. 100* X1500 100—4 : 100 :: : I ^^^"^ polxcy--= ._ , : _ ^ 100—5 • 100 •• • ( 100—3 X 100—4 X 100—5 X 100— 1> 100—6 : 100 :: : j ==$1803 83c. 5m. It is plain that hov/ever numerous the voyages, the power of 100 must be equal to their number, and that the divisor must always be the continued product of the differences between 100 and the several rates of insurance. If tlie rate of in- , ,,_ ., 1004X1500 snraace had been the same on each of the voyo^'cs, then the p-,\'-.- - .— _ — li the rate had been 6 per cc-nt. 30S POLICIES OF INSURANCE. Rule.* 1. Find the sum for which the policy must be taken, by the last case. 2. Multiply the sum adventured by J 00, and divide that product by the policy. 3. Take the quotient from 100, and the remainder will be the premium per cent, on the policy, tantamount to the several pre- miums given ID the question. Example. A merchant adventured ^1500 from Boston to Philadelphia, at 3 per cent.: from thence to Guadaloupe, at 4; thence to Nantz, at 5; and thence home, at 6 percent.: What will b^ the premium, tantamount to those given in the question, on a poli- cy for covering the first adventure, the whole voyage, supposing the risks out and home equal ? In Case V. we found the policy, which would cover the adventure 1500x100 the vo-yage round, to be 00 ^O0^^X10O-4X1O()-r.x ■ POLICIES OF INSURANCE. Si^j i. Divide the last product by the policy. 3. Extract that root of the quotient denoted by the number of risks. 4. Take this root from 100, and the remainder will be the equal premium from one port to the other. Example. A merchant adventured 36)35 18 6 6|02 20 12 7]18 o;28 12 4 ■ — 1 — . 2'22 1)14 • £ s. d. iTincipal for the 5th year 605 19 9 Interest for ditto 36 7 2 Amount for 5 years 642 6 11 v'^nbtract the first principal 480 Compound interest for 5 years 162 6 11 'n federal money, thus : The principal beinsT §1600 for five ycaps, Principal for the 1st year §1600- Rate of interest 6 Interest 1st year 96 00 Amount 1st and prin. 2d year 169d- 6 Interest 2d year 101-76 Amount 2d year, prin. 3d 1797*76 6 Interest 3d year 107-8656 Araouiit 3d, principal 4th 1905-6256 6 Interest 4th year 1 14-33753t Amount 4th, principal 5th ve^r 2019 963136 6 P p Carried over. 30G COMPOUND INTEREST. Brought over. Interest 5th year 121'197788}6 Amount for 5 years 214M6092416 Subtract Ist principal 1600- Compound Interest for 5 years= 541-16092416 Or thus : 1st principal J 1600* 6 Interest 96-00 2d principal 1696- 6 Interest 101-76 3d principal 1797-76 6 Interest 107-866 4th principal 1905626 6 Interest 114-338 6th principal 2019-964 6 Interest 121-198 Amount 214 M62 Jst principal 1600" Compound Interest 541-162 nearly, as before. 2. What is the compound interest of ^740 for 6 years, at 4 per cent, per annum? Ans. ^196 33c. 6m. 3. What will £400 amount to in 5 years, at £4 per cent, per annum? Ans. £48^ 13s. 2id. 4. What will £150 amount to in a year, at 2 *per cent, per^ month? Ans. £190 4s. 5d. 5 What is the compound interest of ^500 at 2 per cent, a month^ for one year? Ans. $134 12c. Im. 6. What is the amount of glOO at 6 per cent, compound inter- est for 3 years ? 7. What is the compound interest of g 100 at 7 per cent, for 3 years? COMPOUND INTEREST. 307 Methop II. When the rate is at 5 per cent, per annum. 1. Divide the principal by 20, and this quotient, added to the principal, will be the amount for the iirst year, and the principal Tor the second. 2. In like manner find the amount for every succeeding year. }Vheji the rate is at 6 per cent, per annum. 1. Divide the principal by 20, and that quotient by 5: these quotients, added to the principal, ivill be the amount for the first vear, and the principal for the second. • 2, In like manner obtain the amount for every succeeding year. Examples 1. What is the amount of £480 at 6 per cent, per annum, for 5 years ? 20)480 5) 24 4 10 20)508 16 amount of 1st year. 5) 25 8 91 5 1 9 J0)539 6 6i ditto of 2d. 5) 26 19 3f 7 lOJ- 20)571 13 8^ ditto of 3d. 5} 28 11 8-1 5 14 4 20)605 19 83 ditto of 4th. 5) 30 5 llf- 6 1 2i 2. Of the same sum at 6 per cent, per ^nnum, for 5 years. £ 20)480 24 20)504 amount of 1st year 25 4 20)529 4 ditto of 2d. 26 9 2|- 20)555 13 21 27 15 7| 20)683 8 JO ditto of 4th. 29 3 51- £612 12 31 do. of 6th. Ans, JVote. The same may be done in federal money, but the first method is generally more easy. .0612 6 10| do.of5tb. Ans. COMPOUND INTEREST BY DECIMALS. '\ Table of tlxo Amount of J^l or ^], at -^ per cent, per month, as practised at the Banks. Months. Dec. parts- — r- Months. Jt: or $ Dec. parts. Months. Ji or S Dec. parts. 1 I ! 1 005 1 01 1 015 I 02 5 6 7 8 1025 103 1035 104 9 10 11 12 1 045 105 1 055 106 308 COMPOUND INTEREST Bt DECIMALS. A Table of the Amount of £1 cr $\, from 1 Day to 31 Days, at 6 per cent per annum. ... Days. Dec. parts. Days. £ or $ Dec. parts. Days. £ or $ • Dec. parts. 1 1-00016 12 100197 22 1 00361 2 1-00032 13 ' 1-00213 23 1 00378 3 1-00049 14 1-0023 24 1-00394 4 1 00065 15 1 -00246 25 1-0041 5 1 -00082 16 1-00263 26 1 -00427 6 1-00098 17 1 00279 27 1- 00443 1 7 1-00113 18 1-00295 28 10046 8 1-00131 19 1 003 12 29 1-00476 9 1 00147 20 1-00328 30 1 00493 10 1-00164 21 1-00345 31 1 00509 11 1-00180 These tables are formed by adding the interest of £1 or ^1, to £ 1 or $1, for the given rate and tinne. Thus, hy rule for Simple Interest, the interest of £1 or $1 fur 1 day, is, -00016438-1-, and the amount is 100016438-f . CASE I.* When the principal^ the rate of interest, and time, are given, to Jin d either the amount or interest. Rule. 1. Find the aniount of £1 or gl for one year at the given rate per cent. 2. Involve the amount, thus found, to such power, as is denoted by the number of years ; or, in Table i. at the end of Annuities, * The reason of the rule may be seen by the following process. If the rate be -6 per cent, the amount of £l or $1 for 1 year, is, by the rule for Simple In- tere&t by Decimals, r06. This is the principal for the second year, and its amount is by the same rule, 1-06-f 1-06 X '00=^4^06 X l-0G=l-06 X VOe^Toy" . That is, the amount of £1 or $\ for tioo years is equal to the square, of the amount of £l or $i ior one year. This is tlie principal for the third year, and its amount is, VOtJ ""-f Tos" X •06=l-i-'06 X ^0(7"= 1-06 X K)6 =1-06 , thai is the amount for three years is the cube of the amount for 1 year. In the Kanie Avay it maybe shown, that the amount ior four years is the /, be required for tlie pailg of ft A'ca'r, it nxixy be deteraiincd as foliuAj's : COMPOUND INTEREST BY DECIMALS. 309 under the rate, and against the given number of years, you will lind the power.* 3. Multiply this power by the principal, or given sum, and the product will be the amount required, from which if you subtract the principal, the remainder will be the interest. Examples. 1. What is the compound interest of £600 for 4 years, at 6 per cent, per annum ? C amount of £l for 1 year, at 6 per -,-.,- ^ ^„ i cent, per annum. Multiply by 1*06 ^ ^ M236=2d power. Multiply by 1-1236 l-26247696=4lh power- Multiply by COO=principal. 757-48617600~amount. Subtract 600 157-486176=£157 9s. 8id.=interest required. By Table I. Tabular amnt. of £1 for 4 years, at 6 percent. per ann.=l '2624760 Multiply by the principal= 600 Amount=757-4861400 2. What is the amount of gl500 for 12 years, at 3i per cent, per annum ? gl 0.35— amount of J^l for 1 year at 3^- per cent, per annum. And, 1035»2xl506=^2266 60c. nearly, Ans. Another method of working compound interest for years^ months, Mid days, which is much more concise than the preceding method. I. When the time is an aliquot part of a year. Rule 1 . Find the amount of £1 for 1 year, us before, and that root of it, vvhich is denoted by \he aliquot part, will be the amount of j£l for the time sought. 2. Multiply the amount, thus found, by the principal, and it will be the amount of the given sum required. 11. When the time is not an aliquot part of a year. RuT.E 1. Reduce the time into days, and the 3t)5th root of the amount of iJl for 1 year is the amoiuit for 1 day. 2. Raise this amount to that power, whose index is equal to the number of days, and it will be the aniovmt of £l for the given time. 3. Multiply this amount by the principal, and it will be the amount of the given sum required. * Tire amounts of £l or $1 in this treble, are so many powers of the amount of £1 or $1 for 1 year ; whose indices are denoted by the number of years. Note. When the given time consists of years and months, or years, months, and days ; first seek the amount of £l or $1 in the table of years, then in the table of months, &c. multiply these several amounts and the principal continu- ally together, and the last product will be the amount required. Thus, if the amount of £480 in 5^ years, at 6 per cent, per annum, were re- quired ; the amount of £l for 5 vears=£ 1-33822, dittc for 6 months— £l "0:2900. Now, l-33822Xl-029G6x4U0=:£66i-2341 Answr-r. ^iO COMPOUND INTEREST BY DECIMALS. Rule. To the logarithm of the principal, found in any Table of loga- rithms, add the several logarithms, answering to the number of years, months and days found in the following tables, and their sum tviH be the logarithm of the amount for tlie given lime, which be- ing found in any table of logarithms, the natural number corres- ponding thereto will be the answer.* iOCARITHMICK TABLES, SIX I'ER CEJVT. PER ANNUM, FOR YEARS, MONTHS AND DAYS. Years. Dec. pis. Y. Dec. pts. Y.' Dec. pts. Y. Dec. pts. Months Dec. pts. 1 •025306 n •278366 21 •531426 :il -784586 1 •002166 1 2 •050612 12 •303672 22 •556732 32 •809792 2 •004321 3 •075918 13 •328978 23 •582038 33 -835098 3 -006466 4 ' -101224 14 ■354284 24 •607344 34 •860404 4 •0086 5 •12653 15 •37969 26 •63265 35 •88571 5 •010724 6 1 -151836 16 •404896 26 •657956 ,36 •911016 6 •012837 7 -177142 17 •430202 27 •683262 37 •936322 7 •01494 8 •202448 18 •455508 28 •708568 ■38 •961628 8 •017033 9 -227754 IG •480814 29 •733974 39 •986934 9 •019116 10 •25306 20 •50612 30 •75938 40 1^01224 10 11 •021189 •023252 ! Days. D. D. 14 D. 20 D. ! 1 •000071 •000571 •000999 •001426 26 •001852 2 •000143 9 •000642 15 •00107 21 •001497 27 •001923 3 •000215 |10 •000713 16 •001142^22 •001568 28 •001994 4 •000287 11 •000785 17 •001213 23 •001639 29 •002065 5 •000358 12 •000857 18 •001284 24 •00171 30 •002136 1 6 •000429 13 •000928 19 -CO 1355 ^5 •001781 31 •00.2207 1 7 i •0005 __. What is the amount of 1321. 10s. C^ years, 8 months, and 15 da^s ? at 6 per cent, per annum, To the log f^oir. for of £132-5 9 years Add ( ditto for 8 months (ditto for 15 days =212221G = •227764 = -017033 = -00107 2-3t]8073 per (cause 8 month? are pai?*, deduct! ) __. .rjrynn.ioo cerjt. upon the logarithm 'f 15 dajs 3 Remains 23G80302, the nearer! to which., to the table of logarilhais, U 2 368101, and the natural number answerios: thereto is 233-4= X233 8s. Ans. i * Aitt'ion^'li tlirro i* a small erroi.fr in the logarithm for days, yet they are ex- act enough for common use. And if after the first month you deduct ^ per cent. •(.^^ each mouth past (that is, ^ per cent, after 1 month, H per cfent. after 3 /u;nithsT fee.) frojn the logarithm of tlie number of days, it will give ^iie true an- fcW(.r. Wo'fe, That, after 1 mouth, ^ per cent, on the logaritlim of 1 day is -000000355, iyn 2 day?M, is -000090710 : After 2 months, 1 per cent, on the ]ogaritl»m of 1 dav, }.-< '-00000071, on Sday^^, -00000143 :" After 10 iuonths, 5 per cent. '>u the lo^rarithin ri;r 1 day, is -(XKiOOSiy, on 6 day?, is, -0000^^145, &-•. COMPOUND INTEREST BV DECIMALS. CASE II. Whtn the amonnt, rate and time^ are givcriy to find theprincipaL Rui-E. Divide the amount by the amount of£l or gl for the given tim-e, and the quotient will be the principal.* Or, If you multiply the present value of£l or §1 for the given number of years, at the given rate per cent, by the amount, the product will be the principal or present worth, j Examples. 1. Wtiat is the present worth of 7571. 9s. 8-}d. due 4 years hence, dsscounting at the rate of 61. per cent, per annum ? By Table I. Divide by the tabular ^ _, onnAnaa^^r'^ ioni4r,nr-PCTin 4«« . rH r 4 ^ =j*2G2476y)7a/-486i400(i/6OO Ar». amount ot II. tor 4 years, S By Table 11. Mult, by the present worth of 11. } "^""^^^"^ -920936 for 4 years, at C per cent per ann. ^ _' _^ Ans. 599-999923582704+=J1:60G; 2. What principal must be put to interest G years, at 51 per ct, per annum, to amount to $689-4214033809453125 ? Ans. J!500. CASE III. When the principal^ rate andameunt, are gizerij to find the time. Rule. Divide the amount by the principal: then divide thfs quotient by the amount of £1 or $1 for I year, this quotient by the same, till nothing remain, and the number of the divisions will show the time. I Or, Divide the amount by the principal, and the quotient will be the amount of £l or gl for the given time, which seek under the given rate in Table 1, and, in a line with it, you will see the time. * By Case I. the amount is equal to the prh>cipal multiplied by that power o{ the amount of jSl or $1 for 1 year at the given rate, which is indicated by the number of years : therefore, if the amount be divided by this power of the amount; cf j£l or i^l for 1 year, the quotient must be the principal. Thus, in the exam- ple in the proof of Case I. l* 05 XlOO:=:the amount; therefore, '——■ 1-05 •' 100, the principal. t See Table II. shewing the present value of jGl, discounting at the rates of 4^, 4j^, tSzic. per cent, the construction of which is thus : Amount. Pres. worth. Amount. Pres. worth. As 1-06 : 1 :: 1 : -9433962, and so on, for any other rate pe^ ■eut. and time. . I By the example in the proof of Case L VOo^ X 100— the amount ; divide Ihif by the principal, 100, and the quotient will be 1^^ . This quotient divided Iv the ratio, and this quotient by the ratio, and so on, v.^ill be exhamte.l by fix-p d ■ ^i'?ion«, which shows th^ number of years. 312 DISCOUNT BY COMPOUND INTEREST. Example. In what time will g500 amount to $689 42c. Im.-f-, at 6i per cent, per annum ? 500 '1-056 1-056 1065 1-056 1-056 1-056 689-421-f lvJ79-~ m a 1-307-- 1 -239— > -5 M74-I- mount of 11. or %\ for the given time ; then, extract such root as the time denotes, and that root will be the amount of 11. or gl for 1 year, from which subtract unity, and the remainder will be the ratio.* Or, Having found the amount of 11. or gl for the time as above directed, look for it in Table 1st, even with the given time, and di- rectly over the amount you will find the ratio. Example. At what rate per cent, per annum will ^600 amount to ^689-421 403-f- in 6 years ? *;89-4214034- ^ TKF^ = 1-378843—; and V« 1378843^ =1-055. Then i 055— l-=^055=ratio num, Answer. Hence the rate is b\ per cefnt. per an- DISCOUNT BY COMPOUND L\TEREST.] The sum, or debt to be discounted, the time and rate, given, to find the present worth. Rule. Divide the debt by that power of the amount of 11. or gl for 1 year, denoted by the time, and the quotient will be the pres- ent worth, which, subtracted from the debt, will leave the discount. * Proceeding as in the preceding demonstration, and extracting that root of the quotient, which is shown by the number of years, we have the amount of £1 or .^1 for 1 year. From this subtract 1, and the remainder is tho ratio. Thus ih s — — r- the preceding example, i/ 1*05 =1-05, and 1*05 — 1='05, the ratio. t As the present worth is such a principal, as at the given rate and time, would ainount to the debt, this rule must be the same as that of Case H. of Compound Interest, the principal being in this case \he present worthy and the amoiint the sum orer of yeaJs ; thus, if l=amount at 6 per cent, far 4 years, then, =:the principal which will produce the amount at the rate apid time. 10&4 r r r ^ Therefore, if l=:the sum to be discounted at that rate and time, then ■ ■ -• ; ^ - '= its l'0o4 present worth, and is the rule. Note, The present worth of 11. or $1 for any time and rate, is the recipracal ef the aiaount of 11. cr ^l for the same time. ^14 AiNNUlTlES. The amount is the sum of the annuities for the time it has been forborne, with the interest due on each. CASE I. 7^0 *Jincl the amount of an annuity at Simple Interest. Rule. Multiply the sum of the natural series of numbers, 1, 2, 3, 4, &c. to the number of years less 1, by the interest of the annuity for one year, and the product will be the interest which is due ou the annuity. Multiply the annuity by the time, and the sum of the two pro- ducts, will be the amount.* Examples. 1. What is the amount of an annuity of £100 for four years^ computing interest at 6 per cent. ? 1 -{-24-3=6, sum of the natural series to the number of years less 1. 61. interest of annuity for 1 year. 6x6=361. the whole interest. 100x4=4001. product of annuity and time. Ans, 4361. amount. 2. If a pension of g20 be continued unpaid for six years, what is its amount at 6 and 7 per cent. ? Ans. At 6 per cent. ^138. At 7 per cent. $141. 3. If an annuity of ^20 to be paid halfeach half year is forborne for six years ; what is its amount at 6 per cent. ? Ans. $169 60c. 4. If a pension of £33 is forborne for 12 years, at 7 per cent, what is the amount ? Ans. CASE II. To find the present worth of an annuity at Simple filtered. M Rule. fl Let the present worth of each year be found by itself, discount- ing from the time it is due ; then, the sum of all these will be the present worth.! * It is plain that upon the first year's annuity there will be due so many year's interest, as the given number of years less one, and gradually one year less upon each succeeding year, to that preceding the last, which has but one year's interest, and the last bears none. • There i^, therefore, due in the whole as many years' interest of the annuity as the sum of the series, 1, 2, 3, &;c. to the number of years diminished one. It is evident then, that the wliole inter- est due must equal this sum of the natural series multiplied by the interest for one year ; and that the amount will be all the annuities or tlie product of thq annuity and time added to the whole interest. This is the rule. t This rule depends on the principles of discount. The annuity may be con- sidered for each year, as a debt, due 1, % 3,&;c. years hence, wf which tlie pres- ANNUITIES OR PENSIONS IN ARREARS, &p, 315 Examples. 1. Find the present worth of an annuity of glOO continued five years at six per cent. i As 106 : 100 :: 100 : 943396, the present worth for I year. 112 : 100 :: 100 ; 89-2857, 2 years. 1 18 : 100 :: 100 : 84-7457, 3 years. 124 : 100 :: 100 : 806451, 4 years. 130 : 100 ::'lOO : 76-9230, 5 years. ^425-9301, present wortli required. 2. Find the present tvorth of an annuity of 751. continued for 4 years at 7 per cent. ? Ans. 3. What is the present worth of a pension of ^20 to be continu- ed for 6 years, at 6 per cent. ? Ans. ANNUITIES OR PENSIONS. IN ARREARS, AT COMPOUND INTEREST. CASE I. ^Fu€U the annuity i or pension^ the time it continues^ and the rate per cent, are given, to find the amount. Rule I.* 1. Make 1 the first term of a Geometrical Progression, and the amount of j£l or §1 for 1 year at the given rate per cent, the ratio. 2. Carry the series to so many terms as the number of years, and find its sum. mt worth is to be found. Hence the sum of the present worth for the several [rears, must be the present worth for the whole. This rule is very absurd in practice. It is obvious on inspecting the operation of 'Ex. 1 . that the difference between the present worth of the several years is I'.ontinually diminishing. Whence, after a certain number of years, the present worth of an annuity of $100 would produce more than $100 interest in one year, which is greater than the annuity to be purchased. * I. From the nature of an amiuity, as explained in the proof of the rule, Case I. of Annuities at Simple Interest, there is due one year's interest less than the number of years the annuity has been continued. Now, by Case I. of Com- pound Interest, the amount of JCI or $1 at the given rate, is equal to that pow- er of the amount for one year, which is indicated by the number of years. This amount is obtained for one less than the number of years, by forming the geo- metrical series as directed in the Rule, or beginning with unity. Thus in" Ex. 1, the series is, 1, 1-06, 1*062, 1-063, and the last term is the amount of J2l or SI for one less than four^ the number of years. The sum of this series is the amount at Compound Interest, of an annuity of £,\ or $1 for four years. The amount of any other annuity for the same time and rate, will be as much great- er or less, as the annuity is greater or less than £1 or $1, that is, the amount of the annuity of £l or $1 must be multiplied by the annuity to obtain its amount. (Hence, the rule is manifestly correct. In Ex, ], the above series amounts, by Prob. HI. of Geometrical Profession, to — —H-. and this multiplied by the 316 ANNUITIES OR PENSIONS IN ARREARS, 3. Multiply the sum thus found by the given annuity, and the product will be the amount sought. Rule II. Or, multiply the amount of £l or ^1 for 1 year into itself so many times as there are years less by 1 ; then multiply this pro- duct by the annuity ; and subtract the annuity therefrom. Lastly, divide the remainder by the ratio less 1, and the quatient will be the amount. Examples. 1. What will an annuity of 601. per annum, payable yearly, amount to in 4 years, at 61. per cent. ? First Method. l4-l'06-|-F66|'4-T'06p=4-374616=sum. Multiply by 60=annuily« 262-476960 20 9-53920 12 6-4704 4 1-8816 Ans.£262 9s. 6id. Or, l-fl-06-{-l-06r4-l-06'''x60==je262 9s. 6^-d, Second Method. l-06Xl*06x l-06xl-06 = l-26247 Multiply by 60 annuity. 75-74820 Subtract 60 Carried up. i-OCi — I annuity, GO, gives the amount required,^: X60=2b2-4769C. •06 ]-064 — 1 II. The second rule is derived from the expression, x60 ; for it i •06 , 1-064x60—1x60 , , ,. . , also, — =the above amount, and is tlie rule. *0o Because the amounts of annuities, at the same rate and for tlie same time, ar as the annuities, if the amount be divided by the amount of j£l or $1 for the same time and rate, the quotient will be the annuity. This is the 2d Rule under Case II. And the 2d Rule of Case III. is readily inferred from the same principle. AT COMPOUND INTEREST. 317 Brought up. Divide by 1'06— l=-06)15'7482(262'47=262l. 9s. 4|d. Ans, 12 37 36 14 12 28 24 42 42 1-06x1 -06x1 06X1 -06X60— 60 Or, jTSel^ =£262.47. Or, bv Table III.* Multiply tbe tabular number under the rate, and opposite to tht lime, by the annuity, and the product will be the amount. 2. What will an annuity of 601. per annum amount to in 20 years, allowing 61. percent, compound interest? Under 61. per cent, and opposite 20, in table 3d, you will find, Tabular number=36-78559 Multiply by 60=annuity. 2207-13540=22071. 2s. Bid. Ans. 3. What will a pension of §75 per annum, payable yearly, amount to in 9 years at 5 per cent, compound interest ? Ans. $826 99 233^m. 4. If a salary of 1001. per annum, to be paid yearly, be forborne 5 years, at 61. per cent. What is the amount ? Ans. 5631. 14s. 2d, 6. What will wages of $25 per month, amount to in a year, at|' per cent, per month ? Ans. $308 38c. 9m. CASE n. When the amount y rate per cent, and time are given, to find the annuity ^ pension^ ^c. Rule I. Multiply the whole amount by the amount of U. or $1 for a year, from which subtract the whole amount, divide the remainder by that power of the amount of 11. or $1 for a year, signified by the number of years, made less by unity, and the quotient will be the answer. * Table 3 is calculated thus : Take the first year's amount, which is 11. mul* tiply it by l-06+l=2-06=:second year's amount, which also multiply by 1-06-f l:=:3-1836=third year's amount, &:c. and iu this manner proceed in calculating tables at any other rates. 5iS ANNUITIES OR PENSIONS IN AHREAIIS, Rule II. Or, imd the amount of an annuity of 11. or $1 for the given time and rate (by Case 1 ;) divide the given sum by this amount j and the quotient will be the annuity required. Examples. 1. What annuity, being forborne 4 years, will amount to -€262*47696, at 61. per cent, compound interest? 262.47696=amount. Multiply by 106=amount of H. for 1 year. 157486176 106 262476960 106 278-2255776 636 Subtract 262-47696 1060 =26247696)15-7486176fje60 Ans. 1-1236 15-7486176' 106 -67416 112360 1,191016 262'47696xl -06— 262-47696 1-06 Or, =60. l-06xl-06xl*06xl06— 1 7146096 11910160 1-26247696 Subtract 1- Divisor=-26247696 Or, thus. Aioount of all annuity of 11. for 4 years at 6 per cent, per annum 26247696 n;:4.374616 (by Case 1 ;) and 4.37461^ ^''^^Q ^"^• Or, by Table HI. the amount of U. is found to be 4374616 ; and the answer is found, as before. 2. What annuity, being forborne 20 years, will amount to ^2207-1354, at 6 per cent, compound interest ? Amount of an annuity of ^1 for 20 years at 6 per cent, per an- T}um=^36'78559. And, 36-78559)2207-13540(g60, Ans. 2207 1354 CASE HI. iVken th€ annuity f amount and ratio arc givcTiy iofind the time. HULE I. Multiply the amount by the ratio, to this product add the annui- jY, and from the sum subtract the amount ; this remainder bein^ AT COMPOUND INTEREST. 319 uivitled by the annuity, the quotient will be that power of the ra- tio signitied by the time, which being divided by the amount of 11. for 1 year, and this quotient by the same, till nothing remain, the number of those divisions will be equal to the time. Or, look for this number under the given rate in table 1, and in a line with it, you will see the time. Or, Rule H. Divide the amount by the annuity ; from the quotient subtract 1 , from the remainder subtract the ratio ; from successive remainders subtract the square, cube, &c. of the ratio, till nothing remain ; and the whole number of the subtractions will be the answer. Or, find the quotient in Table III. under the rate, and in a line with it stands the answer. Examples. 1. In what time will 601. per annum, payable yearly, amount to £2C2-47G96, allowing 61. per cent, compound interest, for (he for- bearance of payment? 26247696=amount. Multiply by 106=ratio. 157486176 262476960 278-2255776 Add 60" =annnuity. Or thus ; Annuity— 60)262-47696=amt. 333 2255776 *" -- Subtract 262-47696 4-374616 1, Subtract 1* Divide by 60)76-7486176 337461G 3. 2-314616 Subtract M236 = =ratlo! =raUol 2 4. 1-191016 Subtract 1-191016= 3^ rm 4 vpars. Divide by 106 )1-96247696 2. Subtract 1-06 =ratio. Divide by 1 06)1-191016 Divide by 106)11236 Divide by 1-06)106 1 The number of divisions by ^ Or, fooking into Table III. un- 106, being 4, gives the number f ' /if;, '!^^'/' the quotient of years ~ 4, the answer. t^'^^^^' '^'"^^' '»"•"'' "^ ^^^''' Ans. as before. Or, in Table I. under the given rate, you will find 1'262476, and m a line under years, you will find 4. ^ 2. In what time will an annuity of g60 payable yearly, amoun> to §2207-1354, allowing 6 per cent, for the forbearance of par n^ent? Ans. 20 years/ 320 PKESENT WORTH OF ANNUITIES, ^c. PRESENT WORTH OF ANNUITIES, 4'Q. AT COMPOUND IN TEREST. CASE I. JVhen the annuityy 4'c. rate and time arc given to find the present worth. Rule I.* 1. Divide the annuity by the amount of jjl or <£! for 1 year, and the quotient will be the present worth of 1 year's annuity. * This rule depends on the rule for finding the present worth in Discount at Compound hiterest. For each year the present worth is to be found by that rule. Then, the sum of the present worth for the several years, must evidently be the present worth of the whole, and is the rule. Or, suppose the annuity to be 11. or $1 at 6 per cent, then — is the present l*Ot) ■^vorth for one year ; for two years ; for three years ; . fov ^ ' l-06a ^ 1-063 ^ 1-06.4 tour years, and so on. Then the sum, or 1 1 — -4 , will be i\ic whole present worth. Let any annuity be substituted for the numerator of ( hese several fractions, and you have the rule in the text. By Note 2, Prob. I. of Geometrical Progression, the sum of the series, ' ^ 1-U6 - J 1 . "^ - . 1 11 1 1 >x ...., L , is 1 — - X — \ or X . ^ ow \i W\F. 1-063^1-063' 1-064 1-064 -06 -06 '06 1-064 ramuity were to continue forever, or the number of years were infinite, then the index of the denominator of the last expression would be infinite, and the value of the fraction would be infinitely diminished or become notliing, and — - would -06 be the present worth of an annuity of 11. or $1 to continue forever ^t 6 per ccnL Hence, if an aimuity is is. perpetuity^ oris to continue forever, its present Avorth is found by dividing the annuity by the ratio, or the interest of 11. or ^1 for a year at the given rate. The present worth of an annuity of $1 to continue forever at 5 per cent, is — z==— r =$20, and an annuity of $100 at 5 per cent, to continue forever, would now be worth $2000, and at 7 per cent. $1 428A. Rule II. is derived from the expression 1 X-— , when the annuity is 1-064 'Ob ^1 or 11. and the rate 6 percent. That is wlien the annuity is $1 or 11. divide the annuity by that power of the ratio indicated by the number of years, and subtract the quotient from the annuity ; the remainder divided by the ratio of the series less 1, will be the present worth. But the present worth of annuities va- ries as the annuity. Hence the rule is manifest. Note. Another rule for obtaining the present worth may be derived from the preceding. Thus, the sum of the series, -J-, —-. ~, —Z is, by Note 2, of Prob. I, of Geometrical ProgrGssion, 1 X , which is alro ** 1.064 1-06—1 X •= — : =rtbe present worth of 11. or gl for four 1-4)64 l-ed--l 1-06*— 1064 ^ ' year^ at 6 per cent. That i?. divide the difference between unity and that power AT COMPOUND INTEREST. 321 2. Divide the annuity by the square of the ratio, anil the quo- tient will be the present worth for two years. 3, In like manner, find the present worth of each year by itself, and the sura of all these will be the present value of the annuity sought. Rule II. Or, divide the annuity, &lc. by that power of the ratio signified by the number of years, and subtract the quotient from the annui- ty ; this remainder being divided by the ratio less 1, the quotient will be the present worth. ElXAMPLES. 1.* What ready money will purchase an annuity of 601. to con- tinue 4 years, at 61. per cent, compound interest ? First Method. Ratio = 106)60-00000(56-603=present worth for 1 year. Kaibl^=r 11236)6000000(53-399= do. for 2 years. Katio l''= M91O16)i30OO000(50-377= do. for 3 years. Kaliol^ =l-26247696)60-00000(47-525=: do. for 4 years. 207-904=i:2O7 13s. 0|d. Ans. be^raUo^^^ =l-26247696)6O0000000(47-525 Second Method 4th the From 60 ^^ Subtract 47-625 Or, 4--^ =47-525 GO— 47-525=12-475 -Odl 12'475 Divis.l-OG— 1=-06)12-475 And =r:207-916- -06 2 07-916=je207 18s. 3^-d. Ans. of the ratio which is indica'ted by the number of years, by the difference between that power of the ratio which is one greater than the number of years and that power of the ratio which is equal to the number of years, and the quotient is tlie present worth of 11. or ^1. Then, as annuities are as their present worth, multi- ply this quotient by the given annuity, and the product is its present worth. The rules for the next Case are derived directly from this rule, and need no further illustration. * The amount of an annuity may also be found for years and parts of a year, thus : 1. Find the amount for the whole years, as before. 2. Find the interest of that amount for the given parts of a year. 3. Add this interest to the former account, and it will give the whole amount required. The present worth of an annuity for years and parts of a year may be found thus : 1. Find the present worth for the whole years, as before. 2. Find the present worth of this pi-esent worth, discounting for the given parts of a ystr, and it will be tlie whole present worth required. ■Questions in this case may also be answered by first finding the amount of ihG given annuity by Case I. of annuities in arrears, page 315, and then the present worth, or principal, by Case II. of Compound Interest, -page 311. R r 322 _ PRESENT WORTH OF ANNUITIES, Lc. By Table III. Under 61. per cent, and opposite 4, we find 4-3746 l=amount of 11. annuity for 4 year?. Multiply by 60=annuity. 262-47660=amount of 601. for 4 years. Then, opposite 4 years, and under 61. per cent, in Table 2(1. We have -792093 Multiply by 262-7466 4762558 4752568 3168372 5544651 1584186 4752558 1584186 208.1 197426338= £208 2s. 4id. Or, opposite 4 years, and under 61. per cent, in Table 1st, we have l-26247=the anjount of 11. for 4 years : Then, 262-7466-t-I -26247=208- 1209= £208 2s. 5d. Ans. By Table IV.* Multiply the tabular nuKiber, under the rate, and opposite the time, into the annuity, and the product will be the present worth. Thus, in Example 1st. What ready money will purchase £60 annuity, to continue 4 years, at £6 per cent, compound interest ? Under 61. per cent, and even with 4 years. We have 3-465 l=present worth £l for 4 year?. Multiply by 60=annuity. Ans.=207-9060= £207 18s. lid. 2. What is the present worth of an annuity of ^60 per annum, to continue 20 years, at 6 per cent, compound interest ? Ans. $688-65 (nearly.) CASE II. When the present tvorth, time^ and rate are given) to find the annuity', rent, 4*c. Rule. 1. From that power of the ratio, denoted by the number of years plus I, subtract that power of it denoted by the number pf years. * Table 4th is thus made : Divide £1 by l-06='94339 the present worth of the first year, which, divided by 1-06, is equal to -88999, which, added to the first year's present wortli, is = 1-83339, the second year's present worth, then -88999, divided by 1-06, and the quotient added to 1-83339, gives 2-6701 for the tliirU year's present worth, &c. i AT COMPOUND INTEREST. 323 2. Divide the remainder by that power of the ratio, signified by the time made less by unity. 3. Multiply the present worth into this quotient, and the product will be the annuity, pension, rent, &c. Or, 1. Multiply that power of the ratio, denoted by the number of years plus 1, by \he present worth. 4. Multiply that power of the ratio, denoted by the time, by the present worth, and subtract this product from the former. 5. Divide the remainder by that power of the ratio, denoted by the time made less by unity, and the quotient will be the annuity. Examples. 1. What annuity, to continue 4 years, will £207*904 purchase, compound interest, at £6 per cent. ? First Method. From l-06xl-06xl-06xl'06xl-06=l-3382255776 Subt. l-06xl-06xl-06xl-06 =1-26247696 Divide by 1-06|4— 1=-26247696)'0767486176('2885898 •2885898 Multiply by 207-9 present worth. 25973082 20201286 57717960 Ans. 59-99781942= £60. Second Method. From 1-06X1-06XI-06X1-06X t-06x207-9=278-21709757;3 Take 1-Oex 1*06 X 1-06 Xl'06x207-9 =262-468959984 Divide by 1-06|* — 1 =-26247696)15-748137589(59'998=601, By Table V.* Multiply the tabular number corresponding with the rate and time, by the purchase money, and the product will be the annuity. Under £6 per cent, and opposite 4 years, you will find •28859=annuity which £l will purchase in 4 years. Multiply by 207-9 259731 202013 577180 59-997861 = £60. 2. What salary, to continue 20 years, will g688 66c. purchase, at 6 per cent, compound interest ? Ans. ^60. * Table 5th is made in this manner : Divide £1 by the present worth of jgl for 1 year, and the quotient will be the annuity, which £l will purchase for 1 year : divide £l by the present worth of £l for 2 years, and the quotient will be the annuity, which Jgl will purchase for 2 years, &c. 224 ANNUITIES, kc. IN REVERSION CASE III. When the annuity ^ present worth and ratiOy are given^ to find the time. Rule.* Divide the annuity by the product of the present worth and ra- tio subtracted from the sum of the present worth and annuity, and the quotient will be that power of the ratio, denoted by the num- ber of yearg, which, being divided by the ratio, and this quotient by the same, till nothing remain, the number of divisions will show the time : Or, the above quotient being sought in Table 1st under the given rate, in a hue with it, you will see the time. Examples. 1. For bow long may an annuity of £60 per annum be purchas- ed for £207-906336762", at £6 per cent, compound interest ? Multiply 207-906336762 To 207-906336762==present worth, by 1-06 Add 60- =annuity. 1247438020572 From 267-906336762 2079063367620 Subt. 220-380716967 220-38071696772 47-525619793=divisov. 47 6256 1 9795)60-000000000( 1 '26247696 Divide by 1-06)1-26247696 1 06)1-191016 1-06)1 •1236-. 1-06) 1-06 J } The number of divisions P^ ' =time=4 years. Or ■ — = 1-26247696, '207-906336762-f60— 207-906336762x1 -06 which heing sought in Table 1, under the given rate, in a line with it, is 4=4 years. 2. How long may a lease of ^300 yearly rent, be had for «s2132-341 allowing 5 per cent, compound interest, to the purchas- er? Ans, 9 years. AKHUITIES, LEASES, 4-c. TAKEjY I/Y REVERSI0,X AT COM- POUND INTEREST. CASE I. When the annvilij, time and ratio, are gi;&en^ to find the present Torth of the annuity in reversion. RULF, I. 1. Divide the annuity by that power of the ratio denoted by the tir?te of its continuance. ■ Tius rale lo derived directly from Rule II. Case L I AT COMPOUND INTEREST. 325 2. Subtract this quotient from tlie annuity: divide by the ratio less 1, and the quotient will be the present worth, to commence immediately. 3. Divide this quotient by that power of the ratio denoted by the time of reversion, (or, time to come, before the annuity com- mences) and the quotient will be the present worth of the annuity in reversion. EULE II. Or, 1. Multiply the annuity by that power of the ratio denoted by the time of its continuance, minus unity, for a dividend. 2. Multiply that power of the ratio denoted by the time of its continuance, that power of it denoted by the time of reversion, and the ratio less 1, continually together for a divisor, and the quo- tient arising from the division of these two numbers will be the present worth of the annuity in reversion. Rule III. Find the present worth of the annuity for the number cf year- before it is to begin and is to be continued ; find also the present worth of the annuity for ihe number of years before the annuit v commences : and the difterence between the present worth for these two periods, is the present worth of the annuity in rever- sion.* Examples. 1. What is the present worth of 601. payable yearly, for 4 years ; Tiut not to commence till two years hence, at 61. per cent. ? First Method. Ratio=l-06 Or, in Table 4th, find the prescriL 1'06 value of 11. at the given rate, both for the time in being and the time in re- 636 version added togetlier. and subtract 1060 the present worth of the ti.ne in be- ing from the other, multiply the re- 2d. power=l'1236 mainder by the annuity, and the pro Carried over. 11236 duct will be the answer. * Rule III. is merely an expression of this truth, viz. the present worth of au annuity continued for the sum of the' years before the annuity is to commence, and is to continue after it begins, is evidently too much by the present worth of the same annuity for the time of reversion or the time before the annuity is to commence, Rvue I. The first two steps are Rule II. of Case I. to find the present worth of Annuities &lc. at Compound Interest, for iJio time of reversion and continu- ance of the 'annuity. ' But as this present worth" would evidently bs Ico much, it must be discounted at compouud interest for the time of reversion, which in the f rst example is 2 years. The third step in R,ul3 I. is, tlierefore, meroly tlie rule for Discount at Compound Interest. Hence the process is obvious. ''W^^i^ - 4HBp w o2G ANNUITIES, &c. IN REVERSION Brought OFer. 1-1236 Pres. worth of the time in being and reversion 5 '^^ ^^"^ 64416 Present worth of the i _^ ^^ 33708 time in being ) ""ilirri. 22472 308402 11236 60 11236 £18504120 Div, by 4th pow.=l-26247696)60'O00000000000(47'525619794281 Subtract the quotient=47-625619794281 Divide by 1'06— 1='06)12-474380206719 Divide by l-06xl-06=M236;)207-9063367619(185'035899=185l. Os. 8id.=the present worth of the annuity in reversion. 60 60—47-5256 ^^' ri62i^96=47-5256 -^^-^=207-906 207-906 And J. j^^g =185-035899 Second Method. •26247696=4th power— 1 Multiply by 60=annuity. 15-74861760=dividend, •08511115)15-74861760(185-03a l-26247696=4th power. [Ans. M236=2d power. 757486176 378743088 1-06|4— 1x60 .252495392 Or, "7:7^^,4^-7:7^13^.^^-^=125 -036 126247696 ^^^1 X^^^' Xl*06— 1 126247696 1-418519112256 •06=ratio— 1 •0851 1 1 14673536=divisor. Third Method. 11 60x1— ■f:56lX:5^=295038-^ the present worth for- 6 years. 60x1— -j^^X.-^= 11 0-002 -}- the present worth for 2 years. £l85-035-f =the present worth in reversion. Example 2. What is the present worth of a reversion of a lease of ^60 per annum, to continue 20 years, but not to commence till the end of 8 \earsj allowing 6 per cent to the purchaser? Ans. ^431-782 (nearly.; AT COMPOUND INTEREST. 327 3. An annuity of §1 in reversion is to commence at the end of 20 years, and is to continue 15 years ; what is its present worth at 4 percent.? ' Ans. §60743 nearly, 4. An annuity of gl in reversion is to commence after 5 years, and to continue forever ; what is its present worth at 6 per cent. ? Ans. gl2 45c. 43. w3rt annuity i several times in reversion^ and rate being given, to find the several present values. Find the present value of£l 6rf,l by Table 4, at the given rate, and for the several given times, which, being severally multiplied by the annuity, the products will be the several present values of that annuity, for the several times given ; subtract the several present values, the one from the other, and the several remainders will answer the question. 6. A has a term of 6 years in an estate at 601. per annum. B has a term of 14 years in the same estate, iu reversion, after the 6 years are expired ; and C has a further term of 16 years, after the expiration of 20 years. I demand the present values of the seve- ral terms at 6 per cent. ? £ s. d. Pres. value of £1 for 36y.==14-61722x60==877 7-2- Ditto of ditto for 20 years =11-46992x60=688 3 10^- Ditto of ditto for 6 years = 4-91732x60=295 9a=A's term. Therefore, 877 7f--688 3 10|=je 188 16 9 C's term, and 688 3 10^—295 9i=£393 3 U=B>8 term. 6. For a lease of certain profits for 7 years, A offers to pay ^300 gratuity, and §300 per annum, B offers §800 gratuity and §250 per annum, C bids §1300 gratuity and §200 per annum, and D bids §2500 for the whole purchase, without any yearly rent ; which is the best offer, computing at 6 per cent. ? § By Table 4, the present worth of §300 per annum } -^-. «, . for 7 years, at 6 per cent is S ^*"'**' ^ To which add 300* Value of A's offer=1974-714 Present worth of §250 per annum for" 7 years=1395-595 To which add 800- Value of B's offer=2195-595 Present worth of §200 per annum for 7 years=ll 16-476 To which add 1300- Value of C's offer=24 16-476 D's offer=2500- Hence it appears that D-s offer is the best. The above questions may be answered by the 4ih arid 2d Tables. 3i5 ANNUITIES, &c. IN REVERSION Take question Isi.for Example. 1. Blultiply the tabular number in Table 4,corricsponding to the rate and the time of continuance, into the annuity, and the pro- duct will t>e the present worth, to commence immediately. 2. Multiply this present worth by the tabular number in Table 2, corresponding to the rate and the time of reversion, and the pro- duct will be the present worth of the annuity ia reversion. In Table 4th we have 3-4651 Multiply by 60=annuity. 207-90G0 in Table 2d we have -esg&DG 1247436 1871134 1871154 1871154 • 1663248 1663248 185 035508376=prc3. worth of the reversion. CASE 11. When {he present worth of the reversion y rate and time are giverij to find the annuity. Rule 1. Multiply that power of the ratio signified by the time of reversion, by the present worth, and the product will be the amount of the present worth for the time before the annuity com- mences. 2. Multiply that power of the ratio signified by the time of con- tinuance plus 1, by the last product. 3. Multiply that power of the ratio, signified by the time, by the aforesaid product, and this last product, divided by that power of the ratio denoted by the time minus unity, will give the annuity. Or, Divide the continual product of the present worth, ihat power of the ratio denoted by the time of continuance, that j)ower of it denoted by the time of reversion, and the ratio minus 1, by that power of the ratio denoted by the tinie of continuance minus J, and the quotient will be the annuity. Examples. 1. What annuity, to be entered upon 2 years hence, and then to continue 4 years, may be purchased for gl85 03^899, at 6 f>er ct. ? First Method. 1 06x1 06= l-l236==2d. power ofthe ratio. Multiply by 185 Q36=pre*ent wortii. , 67416 ■ 33708 5.61800 89888 11236 207-9064496 amount for the lime of reversion. AT COMPOUND INTEREST. 329 3rou^ht up. 207-9064496 amount for the time of reversion. Multiply by I 33822 =:3th power of the ratio. 415812 4th power of the ratio= 1-26247 415812 Multiply by 207-906 1663248 623718 757482 623718 11362230 207906 883729 2524940 From 278-22396732 Take 262-47508782 26247508782 Divide by l-06|4--l = -26247)15-7488750(60 the annuity required. Or, 185-036xM236=207-906 207 •906X 1 -33822— 207- 906x 1 -26247 Then, 1 •26247— 1 "^^^^ ^*"^- Second Method. Jl85036=present worth of the reversion. l-26247=4th power of the ratio. 1295252 Or by Table 4th, divide the 740144 present worth of the reversion 370072 by the difference between the 1110216 present worth of j^l for the time 370072 both in being and reversion, and 185036 the time in being, and the quo- tient will be the annuity. 233-6024 l-1236=2d. power of the ratio. 1401614-1 oi7'^9— 5 P**' ^^^^^ ^H^ ^^ ^^^ 7008072 ^'^J^^^— ^ time in being k revsn. 4672048 _. ^ present worth of gl for 2336024 1 ^^^-^^ — ^ the time in being 2336024 3'O8402)185'04!2(60 Ans. 262-47565664 •06=ratio— 1. -;.-0tJi4_i=:.o6247)15 7485393984(60. 185-036x1 -26247X1 •1236x1-06—1 ^^' r26247i:i =='^^' 2. The present worth of a lease of a house is iG431 15s. 7d. 2 7819qrs. taken in reversion for 20 years ; but not to commence till the end of 8 years, allowing £6 per cent, to the purchaser : What is the yearly rent? Anj5. £60. ^ s <5aU PURCHASING FREEHOLD ESTATES PURCmSljVG AJVJVUITIES FOREVER, OR FREEHOLD ES TAXES, AT COMPOUND IJVTEREST. CASE I. JVhen the annuity^ or yearly rent, and the rate are giveUf to find ike present worth or price. Rule.* As the rate per cent, is to £100 or $100 so is the yearly rent, to the Talue required. Or, Divide the yearly rent by the ratio less 1, and the quotient will be the value required. Examples. 1. What is the worth of a freehold estate of £60 pel: annum, al- lowing 61. per cent, to the purchaser ? £ £ £ 6 : 100 :: CO Or, I'OG— 1=:'06)60'00 60 ■ - — 1000 6)6000 • • £1000 Ans. 2. An estate brings in yearly $15 : What will it sell for, allow- ing the purchaser 6 per cent, compound interest? Ans. ^1500. CASE n. When the price, or present worth, and rate are given, to find the an- nuity y or yearly rent. Rule. As£100 or glOOis to the rate go is the present worth to its rent. Or, Multiply the present worth by the ratio less 1, and the pro- duct will be the yearly rent. Examples. 1. If a freehold estate be bought for £1000 allowing £6 per cent, to the purchaser : What is the yearly rent ? £ £ £ 400 : 6 :: 1000 6 Or, 1000 X -06= £60. 100)6000(£60 Ans. 600 2. If an estate be sold for gl500 and 5 per cent, allowed (o (he buyer ; what is the yearly rent ? Al)s. $15. * The reason of this rule is obvious ; for since a year's interest of the prico» •which is given for it, is the annuity, there can neither more nor less be made oi that price, than of the aimuity, whether it be employed at simple or compound interest. It has also been proved under Case I. of the Present Worth of Annu- ities &:c. at Compoond Interest. Case II. and HI. follow dirc6tly from the rule for Case I. and their rules are hence manifest. AT COMPOUND INTEREST. 33i CASE HI. When the present worthy or price, and yearly rent, are given, to find ike rale. Rule. As the present worth is to the rent ; so is £100 or ^100 to the u'ate. Or, Divide the rent by the present worth ; add 1 to the quo- tient, and the sum will be the ratio of the rate per cent. Or, Divide the sum of the present worth and rent by the pres- ent worth, and the quotient will be the ratio. Examples. 1. If an estate of £60 per annum be bought for £1000 what rate of interest was allowed the purchaser for his money ? £ £ £ 1000 : 60 :: 100 Or, 1 000)60 •00(-06-l-l==l '06 100 6000 1000)6000(£6 Ans. Or, to 1000=present worth. Add 60=rent. 1000)1060(1-06 1000 6000 6000 2. An estate of §75 per annum was purchased for ^1500 what rate of interest bad the buyer for his money ? Ans, 5 per cent. To find at how many years' purchase an estate may be bought. CASE I. JVhen the rate of interest is given, to find the number of years. Rule. Divide £100 or g 100 by the rate, and the quotient will be the |v years. Examples. 1. How many years' purchase should a gentleman offer for the I purchase of an estate, to have 6 per cent, for his money ? 6)100 16 666-f = 16| years. 2. How many years' purchase is ah estate worth, allowing 5 per ;ent. to the purchaser ? Ans. 20 years^ 332 PURCHASING FREEHOLD ESTATfcs CASE n. When the number of years* purchase, at -which an estate is bought. Or sold, is given, to find the rate of interest. Rule. Divide £lOO or ^100 by the number of years, and the quotient will be the rate. Examples. 1. A gentleman gives 16| years' purchase for a farm ; what in- terest is he allowed ? 16|=16-666-|-)100000(6 per cent. Ans, 2. A gentleman gives 20 years' purchase for an estate ; what in- terest has he ? Ans. 5 per cent. PURCHASIJVG FREEHOLD ESTJITES IJV REVERSION. CASE I. The rate and rent of a freehold estate being given, to find the present worth of reversion. Rule.* 1. Find the present worth of the annuity or rent, (by Case 1, of purchasing Freehold Estates, page 330,) as though it were to be entered on immediately. 2. Divide the last present worth by that power of the ratio dct noted by the time of reversion (by Discount by Compound Inter- est) and the quotient will be the answer required. Or, 1. Having found the present value of the estate, supposing it to be immediate : Multiply the annuity, or rent, by the present worth of II. or $1 corresponding with the time of reversion and rate in Table 4th, and the product will be the present worth of the annuity, or rent, for the time of reversion ; or the value of the present possession. 2. Subtract the value of the possession from the value of the estate, and the remainder will be the value of reversion. Examples. 1. Suppose a freehold estate of 601. per annum to commence 2 years hence, be put up to sale ; what is its value, allowing the purchaser 61. per cent. ? Fii^st Method. l-Oe— l='06)60'00=rent per annum. 1000=present worth, if entered on immediately. * By the first step, the present worth is found for the present time ; but as the estate is not to be entered on for a certain time, discount for that time must be allowed at Compound Interest. This is the second step, and the jDropriety of j the rule is manifest. Case 11. needs no illustration. IN REVERSION, 533 F06|^ = l-1236)1000'000(889996=£889 19s. lid.— present worlh of 10001. for 2 years, or the whole present worth required. Second Method. 1.06—1= •06)60-00 1000=present worth, for immediate possession. In Table 4th. we have l-33339=value of II. for 2 years. Multiply by 60=rent. 110'00340=value of possession. From 1000-OpOO Subtract 110-0034 889*9966=value required. 2. Suppose an estate of §75 per annum, to commence 10 years hence, were to be sold, allowing the purchaser 5 per cent ; what is its worth ? Ans. $920 87c. Im. (nearly.) CASE II. The value of a Reversion y the Time prior to its Commencemeyit, and rate of Interest given, to find the Annuity or Rent. Rule. 1. Multiply the price of the reversion by that power of the amount of 11. or gl for 1 year, denoted by the time of reversion, and the product will be its amount, (by Case 1 of Compound In- terest.) 2i Find the interest of the amount (by Case 1st Simple Interest) and it will be the annuity, or yearly rent. Examples. 1. A freehold estate is bought for £889*9966 which does not commence till the end of 2 years ; the buyer being allowed €1. per cent, for his money ; 1 desire to know the yearly income ? 889-9966=price of the reversion. Multiply by Toel^ = 1-1236 denoted by the time of reveriioH. 63399796 26699898 17799932 8899966 8899966 1000'00017976=amount of the reversioa. •06 Ans. £60-00 2. If a freehold estate, to commence 10 years hence, be sold iov ^920 87c. Im. allowing the purchaser T) per cent. ; what »746215 •725587 •70496 •666042 7 •759918 •734828 •710681 •687869 •665057 •622749 8 •730690 •703185 •676839 652125 •627412 •582009 9 .702587 672904: .644609 618253 •591898 •543933 10 11 675564 643928 •613913 •586153 •558394 •508349 •649581 •616199 •584679 •573733 •562787 •475092 12 •624597 •589664 •556837 •526903 •49696<) •444012 13 •600574 •564271 •530321 •49958 •468839 414964 14 •:-77475 •539973 •505068 •473684 •442301 .387817 15 116 •555264 •516720 481017 •449141 417265 •362446< •533908 •494469 •458311 •425979 393647 338734 17 •513373 •473176. •436297 •40383 •371364 •316574 18 •493628 •4528 •415521 •382932 350343 •295864 !l9 •474642 •433302 •395734 •363123 •330513 •276508 J20 •456387 •414643 •376889 •344346 •311804 •258419 •438833 •396787 •358942 •326568 •294155 241513 |22 •421955 •379701 •34185 •309677 •277505 •225713 23 •405726 •36335 •325571 .293684 •261797 •210947 24 •390121 •347703 •310068 •278523 •246978 •197146 [26 26 375117 •332731 •305303 •26915 •232998 •184249 •360689 •318402 •281241 •250525 •21981 •172195 27 •340816 •304691 •267848 •237608 •207368 •160930 128 •333477 •^91571 •255094 •225362 •19563 •1*0402 l29 •320651 •279015 •242946 •213715 •184556 •14056^ |30 131 •308309 •267 •231377 •202743 •17411 •131367 290460 •255502 220359 •192307 •164255 •122773 j32 •285058 •2445 •209866 .•182411 •154957 •114741 33 •274094 •233971 •199872 •173029 •146186 • 107234* 34 •263552 •223896 •190355 164133 •137912 •100219 35 36 254415 •214251 •18129 •155692 •147399 •130105 •093663 •243669 •205028 •172057 •122741 •087535 37 •234297 •196299 •164436 •140114 •115795 •081808 3tr -^25285 •18775 •156605 •132893 •109182 •076456 39 .216671 •^179605 •149148 •126075 •103002 071455 40i -208289 •171929 •142046 •119608 •09717 066780 TABLES. TABLE III. S37 KEWIIfG THE AMOUNT OF £ 1 OR $1 ANNUITY FOR ANT NUMBER OF YEARS, FROM 1 TO 49. 1 4 pr. cent. 4^ pr. cent. 5 pr. cent. 5h pr. cent, 1- 6 pr. cent. 7 pr. cent. 1- 1- 1- 1- 1- 2 2-04 2-045 2-05 2-055 £-06 2-07 3 3-1216 3-137025 3-1525 3-16802 3-1836 3-2149 4 4-246464 4-278191 4-31012U 4-34226 4-374616 4-43994 5 6 5-416322 6-632975 5-47071 5-525631 6-801913 5-58109 5-637093 5-75073 7-15329 6-716892 6-888051 6-975318 7 7-898294 > 8-0 19 152 8-14200!i 8-266894 8-393837 8-65402 8 9-214266 9-380014 9-549109 9-721573 9-897467 10-2598 9 10-582795 10-802114 11-026564 1 1-256259 11-491315 11-9799 10 12.006107 12-2882 12-577892 12-875354 13-180794 13-8164 I] 13-486351 13-841179 14-206787 14-583498 14-971642 15-7836 12 15-025805 15-464032 15-917126 16-38559 16-86994 17-8884 13 16-626838 17-159913 17-712983 18-286798 18-882132 30-1406 \4 18-291911 18-932109 19-59863'i 20-292572 21015064 22-5504 15 20-02358{^ 20-784054 21-578563 22-408663 23-275968 25-1290 16 21-824531 22-719337 23-657492 24-64114 25-672527 27-8880 17 23-697512 24-741707 25-840366 26-996402 28-212879 30-8402 18 25-645413 26-855084 28-132385 29-481205 30-905652 33-9990 19 27-671229 29-063562 30-529004 32-102671 33-759991 37-3789 20 29-778078 31-371423 33-065954 34-868318 36-78559 40-9954 21 31-969202 33-783137 35-719252 37-786075 39-992725 44-8651 22 34-24797 36-303378 38-505214 40-864309 43-392289 49-0057 23 36-617888 38-93703 41-430475 44-111846 46-995826 53-4361 24 39-082604 41-689196 44-501999 47-537998 50-815576 58-1766 25 41-645908 44-56521 47-727099 51-152588 54-86451 63-2490 — — __. — 26 44-311745 47-570645 51-113454 54-96598 59-156381 68-2490 27 47-084214 50-711324 54-669126 58-989109 63-705763 74-4838 , 28 49-967583 53-993333 58-402583 63-23351 68-528109 80-6976 29 52-966286 57-423033 62-3227 12 67-711353 73-639796 87-3465 30 56-084938 61-007067 66-438847 72-435478 79-058183 94-4607 31 59-328335 64-752388 70-76079 77-419429 84-801674 102-073 32 62-701469 68-666245 75-298829 82-677498 90-889775 110-218 33 66-209527 72*756226 80-063771 88-22476 97-343161 118-933 34 69-857908 77-030256 85-066959 94-077122 104-183751 128-258 35 73-652225 81-496618 90-320307 100-251363 111-434776 138-236 36 77-598314 86-163966 95-836323 106-765188 119-120863 148-913 37 81-702246 91-041314 101-628139 113-637274 127-268114 160-337 38 85-970.336 96- 138205 107-709546 120-887324 135-904201 172-561 130 .90-40915 101-464424 114-095025 128-5361 27 145-053453 185-640 Uo 95-025516 ! 07-030323 120-799774 136-6056141 1.^4-761961 199-635 T t TABLLS. [lEWING Tilt MIT. -LA TABLE IV. AVollllI OF £\ OR ${ AXNUIIY, TOR AJfY KtTMBfiE OF YFARS FROM 1 TO 40. ys,|4 per ceut.l 4^j per ct, |.> per ceiit..( 5J. per ct. |6 per cent.| 7 per ct. j 1 0,il(il54 0,95694 0,95238 0,94786 0,94339| 0,9346! o 1,88609 1,87267 1,85941 1 8463 1,83339! 1,8080 3 2,77509 2,74896 2,72326 2,6979 2,67301 2,6243 4 3,62989 3,58752 3,64696 3,49862 3,4651 3,3872 i~6 4,45182 6,24214 4,38997 ~5^57"87 4,32948 "'5767"669 4.26759 4,21236 4,92732 4,1001 4,7666 4,97699 1 7 6,00205 6,8927 5,78637 5,66888 6,68238 6,3892 8 6,73274 6,59589 6,46321 6,30522 6,20979 6,9712 p 7,43533 7,26879 7,10782 6,91786 6,80169 6,5152 10 n 8,11089 7,91272 8,52892 7,72173 7,49866 7,36008 7,0235 8,76048 8,30641 8,04898 7,88687 7,4986 !l2 9.38607 9,11858 8,86326 8,6707 8,38384 7,9426 13 9,98566 9,68285 9,39367 9,06522 8,86268 8,3576 14 10,56312 10,22282 9,89864 9,63396 9,29498 8,7454 jl5 m 11,11839 10,73954 10,37966 9,97824 9,71225 9,1079 11,65229 11,23401 10,83777 10,39936 10,10589 9,4466 17 12,16567 11,70719 11,27407 10,79852 10,47726 9,7632 18 12,65929 12,15099 11,68958 11,17687 10,8276 10,059 19 13,13394 12,69329 12,08632 11,63549 11,16811 10,335 20 ¥l 13,59032 13,00793 12,46221 11,87641 11,46992 11,76407 10,694 14,02916 13,40472 12,82115 12,1976 10,835 22 14,45111 13,79442 13,163 12,50299 12,04158 11,061 23 14,85684 14,14777 13,48807 12,79246 12,30338 11,272 24 15,24696 14,49648 13,79864 13.06682 12,65035 11,469 25 15,62208 14.82821 14,09394 13.3688 12,78336 11,653 26 15,98277 15,14661 14,37518 13,67338 13,00316 11,826 27 16,32959 1^,4513 14,64303 13,80702 13,21053 11,986 28 16,66306 15,74287114,89813 14,02848 13,40616 12,137 29 16,98371 16,02189 15,14107114,23838 13,69072 12,277 30 37 1 7,20202 16,28889 15.37246 T5^92¥l 14,43733 T4T6259~ 13,76483 13,92908 12,409 17,58349 12,631 32 17,87355 16,78889 15,80268 14,80463 14,08398 12,646 33 18,14764 17,02286 16,00256 14,97404 14,22917 12,753 34 18,4112 17,24676 16,1929 15,13461 14,36613 12,864 35 ,36 18,66461 18»,9082T; 17,46101 17,66604 16,37419 15,2868 14,49533 12,947 16,54086 15,43106 14,61722 13,035 37 19J4258 17,86224 16,71129 16.66779 14,73211 13,117 38 19,36787 18,04999 1G,86789| 16,6974 14,84048 13,193 ,30 19,58448 18,22965 17,01^704 16,82024 14,9427 13,264 |40 19.79277 18,40158 17.16909 15.93667 15.03913 13,331 TABLES. 339 TABLE V. THE ANJfUITY WHICH £l OR $1 WILL PURCHASE FOR ANY NUMBER OF YEARS TO C03IE, FROM 1 TO 40. 1 4 per ct. 1,04 4^ per ct. 1,045 5 per ct. 1,05 54 per ct. 1,055 6 per ct. 1,06 7 per ct^ 1,07 2 ,5302 ,534 ,5378 ,54162 ,54544 ,55309 3 ,36035 ,36377 ,36721 ,37065 - ,37411 ,38105 4 ,27549 ,27874 ,28201 ,28582 ,28859 ,29523 5 6 ,22463 ,19076 ,22779 ,23097 ,19702 ,23487 ,20092 ,23739 ,24389 ,19388 ,20336 ,20977 7 ,16661 ,1697 ,17282 ,17671 ,17913 ,18556 8 ,14853 ,15161 ,15473 ,15859 ,16103 ,16747 9 ,13449 ,13757 ,14069 ,14455 ,14702 ,15349 10 n! ,12329 ,11415 ,12638 ,1295 ,12039 ,13334 ,12424 ,13587 ,12679 ,14238 ,11725 ,13336 12 ,10655 ,10967 ,11282 ,11667 ,11927 ,12592 13 ,10014 ,10327 ,10645 .11031 ,11296 ,11965 14 ,09467 ,09782 ,10102 ,10489 ,10758 ,11435 15 ,08994 ,09311 ,09624 ,10022 ,10296 ,10979 16 ,08582 ,08901 ,09227 ,0962 ,09895 ,10586 17 ,0822 ,08542 ,0887 ,0926 ,09544 ,10243 18 ,07899 ,08224 ,08555 ,08947 ,09235 ,09941 19 ,07614 ,07941 ,08274 ,08699 ,08962 ,09676 20 21 ,07359 ,07128 ,07688 ,08024 ,08427 ,08718 ,09439 ,0746 ,078 ,08198 ,085 ,09230 22 ,0692 ,07254 ,07597 ,07998 ,08303 ,09041; 23 ,06731 ,07068 ,07414 ,07825 ,08128 ,08880* 24 ,06559 ,06899 ,07247 ,07653 ,07968 ,08719' 25 2G ,06401 ,06744 ,07095 ,07503 ,07367 ,07823 ,0769 ,0858 Ij ,06257 ,06602 ,06956 ,08457} 27 ,06124 ,06472 ,06829 ,07242 ,0757 ,083431 28 ,06001 ,06352 ,06712 ,07128 ,07459 ,082391 29 ,05888 ,06241 ,06604 ,07023 ,07358 ,08145; 30 31 ,05783 ,06139 ,06506 ,06926 ,07272 ,08058 ,05685 ,06044 ,06413 ,06837 ,07179 ,07980 32 ,05595 ,05956 ,06328 ,06754 ,071 ,07908 ,07841 33 ,0551 ,05874 ,06249 ,06678 ,07027 34 ,05431 ,05798 ,06175 ,06607 ,06959 ,07779 35 ,05358 ,05727 ,06107 ,06541 ,06899 ,07724 36 ,05289 ,0566 ,06043 ,0648 ,06839 ,07672 37 ,05224 ,05598 ,05984 ,06423 ,06785 ,076241 38 ,05163 ,0554 ,05928 ,0637 ,06735 ,07579 39 ,05106 ,05485 ,05876 ,06321 ,06689 ,07539 ,07501 40 ,05052 ,05434 ,05828 ,06274 ,06646 o4tf TABLEfe-. TABLE VL VALUE OF AN ANNUITY OF j£l OR $1, AT DIFFERESTT RATES PER CENT. PAYABLE YEARLY, HALF YEARLY, aUARTERLY, DAILY OR MOMENTLY, rOR EVER. Rate per cent, 3 H 4 4i 5" 6 ^^ 7 Perpetuity payable yearly. Half yearly. Quar- terly. Daily. Perpetuity payable momently. 33,3333 28,5714 25,0000 22 2*^22 2o!oOOO 18,1818 16,6666 15,5556 14,2857 33,6022 28,7886 25,2525 22,4699 20,2429 18,4298 16,9147 15,6299 14,5349 33,6927 28,9083 25,3807 22,5938 20,3583 18,5529 17,0380 15,7550 14.6575 33,8238 29,0841 25,4858 22,7175 20,4763 18,6630 17,0554 15,7743 14,7694 33,8308 29,2908 25,4992 22,7244 20,4959 18,6812 17,1635 15,7902 14,7800 J\'ote. This Table is calculated by the Rule, Case I. of basing Annuities for ever." Pur^ TABLES. 341 TABLE VII. VALUE OE AN ANNUITY OF £l OR $1, FOR A SINGLE LIFE. 3 per 3^ per 4 per 4i per 5 j)er 6 per Age. cent. cent. cent. cent. cent. cent. 10 19,87 18,27 16,88 15,67 14,60 12,80 12 19,60 18,05 16,69 15,61 14,47 12,70 16 19,19 17,71 16,41 15,27 14,27 12,55 18 18,76 17,33 16,10 15,01 14,05 12,48 20 18,46 17,09 15,89 14,83 13,89 12,30 22 18,15 16,83 15,67 14,64 13,72 12,15 25 17,66 16,42 15,31 14,34 13,46 12,00 28 17,16 15,98 14,94 14,02 13,18 11,75 30 16,80 15,68 14,68 13,79 12,99 11,60 33 16,25 15,21 14,27 13,43 12,67 11,35 35 15.86 14,89 13.98 13,17 12,45 11.15 38 15,29 14,34 13,52 12,77 12,09 10,90 40 14,84 13,98 13,20 12,48 11,83 10,70 43 14,19 13,40 12,68 12.02 11,43 10,35 45 13,7S 12,99 12,30 11,70 11,14 10,10 48 13,01 12,36 11,74 11,19 10,68 9,75 50 12,51 11,92 11,34 10,82 10,35 9,45 53 11,73 11,20 10,70 10,24 9,82 9,00 55 11,18 10,69 10,24 9,82 9,44 8,70 50 10,32 9,91 9,52 9,16 8,83 8,20 60 9,73 9,36 9,01 9,69 8,39 7,80 63 8,79 8,49 8,20 7,94 7,68 7,20 65 8,13 7,88 7,63 7,39 7,18 6,75 6P 7,10 6,91 6,75 6,54 6,36 6,00 .70 6,38 6,22 6,06 5,92 5,77 5,50 73 5,25 5,14 5,02 4,92 4,82 4,60 75 4,45 4,38 4,29 4,22 4,14 4,00 77 3,63 3,57 3,52 3,47 3,41 3,30 79 2,78 2,74 2,70 2,67 2,64 2,25 80 2,34 2,31 2,28 2,26 2,23 2,15 This Table is formed by ascertaining from Bills of mortality ths mean Itngth of the lives of persons of a certain age, and then cal- culating the value of the annuity for the number of years they may thus be expected to live. This mean is called the Expectation ot" Life, at any given age, which is exhibited in the following table. :42 CIRCULATING DECIMALS. TABLE VIIL ^EXPECTATION OF LIFE AT SEVERAL AGES. Age, 1 Expec. Age. 20 Expec. Age. 42 Expec. Age. 62 Expec. Age. 82 Expec. 29,80 33,62 21,65 11,68 3,31 2 37,92 22 32,46 45 20,10 65 10,20 85 2,64 3 40,18 25 30,83 47 19,07 67 9,20 87 2,40 4 41,32 27 29,74 50 17,55 70 7,80 90 2,12 6 41,77 30 28,12 52 16,55 72 7,00 92 1,60 8 41,34 32 27,04 55 15,10 75 5,84 93 1,14 10 40,25 35 25,34 57 14,12 77 6,15 94 0,90 12 38,57 37 24,30 59 13,15 79 4,40 96 0,66 16 36,64 40 22,82 60 12,66 80 4,00 96 0,50 18 34,77 i CIRCULATING DECIMALS ARE produced from Vulgar Fractions, whose denominators do not measure their numerators, and are distinguished by the con- tinual repetition of the same figures. 1. The circulating figures are called rrpetends ; and, if one figure only repeats, it is called a sn?«-/e rcpetend : As 1111, &c. '6666, &c. 2. A compound repeicnd has the same figures circulating alter- nately : As -OlOtOi, &c. -379379379, &c. 3. if other figures arise before those which circulate, the deci- mal is called a mixed repetend ; thus, '375565, &c. is a mixed single repeiend^ and '378123123, he. a mixed compound repetend. 4. A single repetend is expressed by writing only the circula ting figure with a point over it; thus, -1111, &c. is denoted by •1, and -6666, &c. by -6. 5. Compound repetends are distinguished by putting a point over the first and last repeating figures ; thus, -010101, &c. is writtea •01, and -379379379, &c. thus, -379. 6. Similar circulating decimals are such as consist of the same number of figures, and begin at the same place, either beture ur after the decimal point ; thus, '3 and '5 are similar circulates ; as are also 3-54 and 7*36, &:c. 7. Dissimilar repetends consist of an unequal number of figures, and begin at different places. C. Similar and conterminous circulates are such as begin and end at the same place ; as 4734576, 9w3528 and -05463, kc. REDUCTION OF CIRCULATING DECIMALS. 345 REDUCTION OF CIRCULATING DECIMALS. CASE I. To reduce a simple Repeiend to its equivalent Vidgar Fraction. Rule.* 1. Make the given decimal the numerator, and let the denomi- nator be a number, consisting of so many nines as there are re- curring^ places in the repetend. 2. If there be integral figures in the circulate, so many cyphers must be annexed to the numerator as the highest place of the repetend is distant from the decimal point. Examples. 1. Required the least vulgar fractions equal to -3 and -324. • •3=1=1 ; and .324=A|4.==.r.. Ans. -i- and if. ?, Reduce -7 to its equivalent vulgar fraction. Ans. ^. 3. Reduce 2.37 to its equivalent vulgar fraction. Ans. VsV* 4. Required the least vulgar fraction equal to -384615. Ans. y^. CASE H. To reduce a mixed Repetend to its equivalent Vulgar Fraction. RULE.J 1. To so many nines as there are figures in the repetend, annex «o many cyphers as there are finite places, (that is, as there are decimal places before the repeiend) for a denominator. "'■ If unily, with cyphers annexed, be divided by 9 ad infinitum^ the quotient v'ill be 1 continually ; that is, if i be reduced to a decimal, it will produce the circulate •!, and since • 1 is the decimal equivalent to ^, -2 vnW =?., 3=3, and so on till 'O^-g^l . Therefore every single repetend is equal to a vulgar fraction, whose numerator is llie repeating figure and denominator 9. Again, J.- or -1- being reduced to decimals, make -010101, &c. and -001001001, k ^'c. ad inJmitumr=z-0\ and -001 ; tliat is, 9V="01i and ^ly =-001, consequently /,,-~-02, r;\— -03, kc. and g fg =-002, g fg- =:-003, &c. and the same will hold universally. t In like manner for a mixed circulate ; consider it as divisible i»to its finite and circulating parts, and the same principle will be seen to run through them also; thus the mixed circulate -13 is divisible into the finite decimal -1, and the repetend -03 : but '1— jVi ^^<^ *03 would be equal to f provided the circulation began immediately after the place of units ; but as it begins after the place of tenths, it is | of _.i_— JL, and so the vulg;ar fraction='13 is _.>_-|-_3_=r .»„-|-_a. =-j}|-, and id the same as by the rul'?. 344 REDUCTION OF CIRCULATING DECIMALS. 2. Multiply the nines in the saifJ denominator by the finite pail, and add the repeating decimals to the product for the numerator. 3. If the repetend begins in some integral place, the finite value of the circulating part must be added to the finite part. Examples. 1. What is the vulgar fraction equivalent to '153 ? There being 1 figure in the repetend, and 2 finite places, I an- nex 2 cyphers to 9 for a denominator, viz. 900; then 1 multiply the 9 in the denominator by the two figures in the finite part, and add the repeating figure for a numerator; thus, 9xl5-f 3=138 numerator. Therefore, 'IGS— if^==y2_;L the Ans. 2. What is the least vulgar fraction equal to -4123 ? Ans. ||^|. 3. Required the finite number equivalent to 45-78 ? Ans. 45||,i CASE HI. To' make any number of dissimilar repctends similar and coniermi nous ; that is, of an equal number of places. Rule.* Change them into other repetend.e, which shall each consist of so many figures, as the least common multiple of the sums of the several numbers of places, found in allthe repetends, contains units. Examples. 1. Make 6 317 ; 3-45 ; 62-3 ; 191-03 ; -057 ; 5-3 and 1-359 sim- ilar and conterminous. Here, in the first repetend, there are three places, in the se- cond, one, in the third, none, in the fourth, two, in the fifth, three, iM the sixth, one, and in the seventh, one. Now find the least common multiple of these several sum«, thus : 3\.3, 1, 2, 3, 1, 1 j „ ^^^ 2x3=6 units ; thereforp, the similar and conterminous repetends must contain 6 place.^.j =* All)'' gjivcn repetend whatever, whether sing^le, compoun'l, pure, or inixed, itiuy be triinsformcd into another repefend, which shall coir ii^t of an equal ur gi-enlri-^iiiiiil-ci- of figures at plcatiure ; thus, •.> may be transformed into '33, or -333, ^c. also -TOr-'TinO-r-v-TOT, and so on. t The learner may observe that the similar and contermiuous repetends be- ^j-in just so far from unity, as is the farthest among; the dissimilar repetends ; and il. is £0 in all cases. REDUCTION OF CIRCULATING DECIMALS, 345 Dissimilar made similar and conterminous, 6-317= 631731731 3<45 = 3 46655555 52-3 = 62 30000000 191-03 —191 03030303 •057= •05705705 ' 5-3 = 5-33333333 1-369= 1-3599999:9 }. 2. Make '531, -7348, -07 and 0603 similar and conterminous. CASE IV. To find whether the decimal fraction, equal to a given vulgar one, be finite or infinity, and how many places the repeiend will con- sist of. Rule* 1. Reduce the given fraction to its least terms, and divide the denominator by 2, 5 or 10, as often as possible. 2. Divide 9999, &;c. by the former result, till nothing remain, and the. number of 9s used will show the number of places in the repetend ; whicii will begin after so many places of figures as there were 10s, 2s, or 5s, divided by. If the whole denominator vanish in dividing by 2, 5 or 10, the decimal will be finite, and will consist of so many places as you perform divisions. * In dividing 1-000, &c. by any prime number whatever, except 2 or 5, the figures in the quotient will begin to repeat over again as soon as the remainder is 1 : and since 999, &:c. is less than 1000, &:c. by 1, therefore 999, &c. divided by any number whatever, will, when the repeating figures are at their period, leave for a remainder. Now, whatever number of repeating figures we have, wdien the dividend is 1, there will be exactly the same number, when the dividend is any other number whatever. Thus, let -390539053905, &c. be a circulate, whose repeating part is 3905. Now, every repetend (3905,) being equally multiplied, must give the same pro- duct : For although these products will consist of more places, yet the overplus in each, being alike, will be carried to the next, by which means, ea^ product will be equally increased, and consequently every four places will continue alike. And the same will hold for any other number. Now from hence it appears that the dividend may be altered at pleasure, and the number of places in' the repetend will still be the same ; thus, -J_".-09 ; and fi ^^ -]\-X4;--3C, whence ihc number of pl.irc:, in each arc rililtc r .1 346 ADDITION OF CIRCULATING DECIMALS. Examples. 1. Required to find whether the decimal equal to 2W0 ^^^ finite or infinite, and if" infinite, how many places that repetend will con sist of. 475 19 \ (2) (2) (2) :56=28=14=7. \ 475 19 \ ( First 25 ) ^ 2 ]\n=l /2800 112 / 7)999999 Then, Taq^^t '^ therefore, because the denominator 112 did not vanish in dividing^ by 2, the decimal is infinite, and, as six 9s were used, the circulate consists of 6 places, beginning at the fifth place, because four 2s were used in dividing. 2. Let ji be the fraction proposed. 3. Let f be the fraction proposed. ADDITION OF CIRCULATING DECIMALS. Rule. 1. Make the repetends similar and conterminous, and find their sum as in common addition. 2. Divide this sum (of the repetends only) by so many nines as there are places in the repetend, and the remainder is the repe- tend of their sum ; which must be set under the figures added, with cyphers on the left hand, when it has not so many places as the repetends. 3. Carry the quotient of this division to the next column, and proceed with the rest as infinite decimals. Examples. 1. Let 5'3+59'435C+397 6+519+-39+217'5 be added together, 5 3 = 6-3333333 59'4366= 59-435G356 397-6 =:397-360GGG6 519- —519-0000000 •39 ~ -3939393 ^n7-5 .=217-6555555 1199-3851303 1199 3851305 the sum. in this question, the sum of the repetends is 2851303, which divided by 999999, gives 2 to carry to the next column 5,3,0, &c. and the remainder is 851305. xMULTIPLlCATION OF CIRCULATING DECIMALS. 3 n 2. Let 3275-3194-36-454-123-19+5-3173-f n2-3513+lM31 + •125+29-10053 be added together. Ans. 3593-00042. SUBTRACTION OF CIRCULATING DECIMALS. Rule. Make the repetends similar and conterminous, and subtract as Tisual, observing, that if the repetend of the number to be sub- iracted be greater than (he repetend of the number it is to be ta- ken from, then the right hand of the remainder must be loss by inity than it would be if tlie expressions were finite. Examples. 1. From 5703 take 29-73587 5703 =5703030 23-73587=29-73587 27.29442 ^he difference. 2. From 325 17 take 137-5819. Ans. 1 87-5957. MULTIPLICATION OF CIRCULATING DECIMALS, Rule. 1. Turn both the terms into their equivalent vulgar fractions, and find the product of those fractions as usual. 2. Turn the vulgar fraction expressing the product, into an equivalent decimal one, and it will be the product required. Examples. 1. Multiply -54 l)v -15.. 'b\=l^^j\ and •15=ii=J7_ ■ixXi'E—ih—'^^^ ^he product. 2. Multiply 378-5 by 236. Ans. 8959-148. DIVISION OF CIRCULATING DECIMALS. y Rule. 1.' Change both the divisor and dividend into their equivalent vulgar fractions, and find their quotient as usual. 2. Turn the vulgar fraction expressing the quotient, into its equivalent decimal, and it will be the quotient required. 348 ALLIGATION. Examples. 1. Divide '54 by '15. •B4=5A=:=:_6^ an^ •13=11= "L it-^4V=AXV=Vt =3ff=:3'606493 the quotient. 2. Divide 3458 by 6. Ans. 618 S3. ALLIGATION IS the method of mixing two or more simples of different quali- ties, so that the composition may be of a mean or middle quahty ; It consists of two kinds, viz. Alligation Medial, and Alligation Al- ternate. ALLIGATION MEDIAL Is, when the quantities and prices of several things are given, to find the mean price of the mixture compounded of those things. Rule. As the sum of the quantities, or the whole composition, ^s to their total value ; so is any part of the composition to its mean price or value. Examples. 1. A Tobacconist would mix 60lfe of tobacco, at 6d. per ft with 50ife at Is. 40ife at Is. 6d. and 30ife at 2s. per ft: What is 1ft of this mixture worth ? ft s. d. £ s. ft £ ft ■ 60 at 6 is 1 10 As 180 : 10 : 1 60 -_ 1 — 2 10 1 40 — 1 6 — 3 — 30—2 — 30 Sum of the ? io^t « i i i.. simples, M80 Total value 10 10 20 180)200(13. 180 20 12 180)240(lid. s. d. 180 Ans. 1 CO ipr.ft 2. A farmer would mix 20 bushels of wheat at gl per bushel, IG bushels of rye at 75c. per bushel, 12 bushels of barley at 50c. ALLIGATION ALTERNATE. 349 per bushel, and 8 bushels of oats at 40c. per bushel ; What is the value of one bushel of this mixture ? Ans. 73c. 5|m. 3. A wine merchant mixes 12 gallons of wine, at 75c. per gal Ion, with 24 gallons at 90c. and IG gallons at §1 10c. : What is a gallon of this composition worth ? An?. 92c. 6m. 4. A goldsmith melted together 8oz. of gold of 22 carats fine, 1^ life 8oz. of 21 carats fine, and lOoz. of 18 carats fine : Pray what I is the quality, or fineness of the composition ? ^^2+Yi^ J^lj carats fine, Ans. 8+20+10 5. A refiner meits 51b of gold of 20 carats fine ^ith 8ft of 1^ carats fine : How much alloy must be put to it, to make it 22 ca- rats fine ? r=r . 22— 6X20+8X1 S^i-S+S^Sy^^. Answer. It is not fine enough by o^\ carats, so that no alloy i^ must be added, but more gold. ALLIGATIOjY ALTERjYATE'^ Is the method of finding ivhat quantity of each of (he ingredi- ents, whose rales are given, will compose a mixture of a given rate : So that it is the reveise of Alligation Medial, and may he proved by it. CASE I. The whole work of this case consists in linking the extremes truly together and taking the differences between them and tho mean price, which differences are the quantities sought. Rule. 1. Place the several prices of the simples, being reduced t© one denomination, in a column under each other, the least upper- most, and so gradually downward, as they increase with a line oi' ■ Demon. By connecting the less rate willi the greater, and placing the dif- Jcreace between them and the mean rate alternately, or one after the other ui turn, the quantities resulting nre such, that there is precisely as much gained by one quantity as is lost by the other, and therefore the gain and loss, upon the whole, are equal, and are exactly the proposed rate.- In like manner, let the number of simples be what it may, and with how many soever, each one is linked, since it is always a less with a greater than the mcau price, there will be an equal balance of loss and giiin between every two, anc' consequently an equal balance on the whole. It is obvious from the rule, that questions of this sort admit of a great variety of answers ; for having found one answer, we may find as many more as we please, by only multiplying or dividing each of tlie quantities found, by 2, 3, 4, &c. the reason of which is evident ; for if two quantities of two simples make a balance of loss and gain with respect to the mean price, so must also the dou- ble or triple, the half or third part, or any other ratio of these quantities, and so on ad infinitum. If any one of the simples be of little or no value with respect to the rest, ifs rate is supposed to be nothing, as water mix^d with wine, and alloy with gold }ind silver. obO ALLIGATION ALTERNATE. connection at the left hand, and the mean price at the icl't hand of all. 2. Connect, with a continued line, the price of each simple, or ingredient, which is less than that of the compound, with one or any number of those which are greater than the compound, and each greater rate or price with one or an}' number of the less. 3. Place the difference, between the mean price (or mixture rate) and that of each of the simples, opposite to the rates with which they ai^ connected. 4. Then, if only one difference stand against any rate, it will be the quantity belonging to that rate ; but if there be several, their sum will be the quantity. . Examples. 1. A merchant has spices, some at Is. 6d. per Jfe, some at 2s. some at 4s. and some at 5s. per fe : How much of each sort must he mix that he may sell the mixture at 3s. 4d. per 3fe '{ Mean rate40d 28 38 16 04-20 .20 22 22-fl6 oo 28 at J 6 28 — 2 38 — 4 0, J8 — 5 Note. I'lie-e seven answers arise from as many various ways of linking the ralt'« of the ingredients together. 2. *A fnerchant has Canary wine, at 3s. per gallon, Sherry, at 23. Id. and Claret at Is. 5d. per gallon : How much of each sort must he take, to sell it at 2s. 4d. per gallon ? Meanratc28d. ^36'*. 3f n 14 at 3 0) ]d. { 25^) 8 8 2 1) f 17-^ 8 8 15) per gallon. * Note, the 2d and 3(1 qnctiom admit but of one way of linking;, and po but of one answer ; yet all numbers in the same proportion between tliemselvc.-:, as the numbers which compose the answer, will likewise satisfy the condition of the question. ALLIGATION ALTERNATE. o. How ranch barley at 40c. rye at 60c. and wheat at 80c. per bushel, must be mixed together, that the compound may be worth 62iG. per bushel? Ans. 17| busfiels of barley, Ilk ofr^e, and 25 of wheat. 4. A gohlsmilh woukl mis gohl o(^l9 carats fine, witli some of 16, 18, 23 and 24 carats tine, so that the compound may be 21 ca- rats fine : What quantity of each must he take ? Ans. 5oz. of 16 carats tine, 3oz. of 18, 3oz. of t^, lOoz. of 23, and lOoz. of 24 carats line. 5, It is required to mix several sorts of wine, at 60c. 90c. and ^1 I5c. per gallon, with water, that the mixture may be worth Idc. per gallon : Of how much of each sort must the composition con- sist ? Ans. 40gall8. of water, 15galls. of wine, at 60c. ISgalls. do. at 90c. and 75galls. do. at gl 15c. (?ASE II. IVhen the rates of all the ingredients, the quantity of hut one of them, and the mean rate of the zvhole mixture arc given, to find the sev- eral quantities of the rest, in proportion to the quantity given. Rule. Take the differences between each price, andahe mean rate, and place them alternately, as in Case 1. Then, as the differenco standing against that simple, whose quantity is given, is to that quantity, so is each of the other differences, severally, to the seve- ral quantities required. Examples. 1. A merchant has 40lfe of tea, at Gs. per ft, which he would mix with some at 5s. 8d. some at 5s. 2(1. and some at 4s. 6d. : How much of each sort must he take, to mix with the 40Ife, that he may sell the mixture at 5s. 5d. per ft t ft li) 10 14 14 stands against the given quantitv^ ft s. d. 10 : 28y^j at 4 6 ) 10 : 28/^— -5 2) per ft 14 : 40 —5 8) 2. A farmer being determined to mix 20 bushels of oats, at GOc. per bushel, with barley, at 75c. rye, at gl, and wheat, at ^1 25c. per bushel ; I demand the quantity of each, which must be mixed with the 20 bnt;heis of oats, that the whole qijantity may be worth 90c. per busliel ? • Ans. 70 of barley, 60 of rye, and 30 of wheat, (or 20 of each.) 3. How much gold of 16, 20 and 24 carats line, and how much illoy, must be mixed with lOoz. of 18 carats line, that the compo- Mtion maybe 22 carats fine. Ans. lOoz. of 16 carats tine, 10 of 20, 170 of 24, and 10 of alloy, As 14 : 40 :o2 ALIJGATiON ALTERNATE. . ALTERNATION TOTAL.* CASE IIL rVheji the rates of the several ingredients, the quantifi) to be compound- ed, and the mean rate of the rvhole mixture are given, to find how much of each sort will make up the qnanlity. Rule. Place the difiercnccs befweni the mean ralp, atid (lie several prices alte^natel3^ as in (.'a?e I ; tlien, as the sorn of the qnanti- ues, or differences thus deteitnined, is to the given (|uantity, or whole composition ; so is the difference of each rate, to the re- qiured quantity of each rate. Ea'amplfs. 1. Suppose I have 4 soi'ts of currant?, at TJd. 12d. Vui\. and 22d. per Ye ; the worst will not sell, and the he?t are too dear; I there- fore conclude to mix 1^20Jfe and so much of each port as to sell them at 16(1. per Ife ; how much of each sort must 1 take ? d. Ife mm' r 8*< 6 ' fi) : 3G at 8d.T 56c]. \ ^^JJ,^ ^^ oo : 120 :: ") 4 : 24 - 18d. [P"' ^ [22-^ 8 [g : 48 — 22d. j ' Sum— 20 120 2. A goldsmith has several sorts of gold ; viz. of 15, 17, 20 antl 22 carats tine, and would melt toi^ether, ofall these sorts, so much as may make a mass of 40oz. 18 carats fine ;' how much of each sort is required ? Ans. ]6oz. 15" carats line, 8oz. 17, 4oz. 20, and 12oz. of 22 ca- rats line. * I'o llvi:^ Case Lelong'- iJuil cnrionr, qiiostion concornin^ king' Hiero's crown. Tliero, king of Sja-actise, gave orders for a crown \o ttc? made entirely of pin r ;?;oid ; but suspecting the "workmen ]iad debased i I, by mixing with it silver oi copper, he recemmended the discovery of f he fraud Iq th^ famous Archiinedci, .uid desired to know the exact quantity of alloy in the croivn,* Archimedes, in order to detect the imi>osition, procured two other itiasi^cr, one of pure gold, and the other of silver, or copper, and eadi of the same ^^eight with the former ; and by j5utting each separately hito a, vessel At// of Aval or, the quantity of water expelled by tliem, determined their .^pociiic bulks ; from v/hich, and their given weights, it is easier to determijic the quaiitities of gold and alloy in the crown by this case oi Alhgatiou, than by an Algebr.iic process. Suppose the weight of each mass to have been [Ah. the Aveight of the water expelled by the alloy, 23oz. by the gold, 13oz. and by the crown lOoz. tliat is, !bat their specifick bulks Avere as 23, 13, "and 16 ; then, Avhat Avcre the quanti- ties of gold and alloy respectively in the crown ? Here, the rates of the simples are 23 and 13, and of the compound 16, whence, < 13 \7 of gold > Antl the 'sum of these is 7-f-3=10, Avhich should have ( 23_ /3 of alloy \ been but 5, Avhence, by the rule, ' ... - ( 7 : ?A\\). of gold ) ,, . ^^•''■- ^3:U]b.ufaHoyi^^^^"^^^''"^- 16 ALLIGATION ALTERNATE. vJ5r> o. A merchant would mix 4 sorts of wine, of several prices, viz. 3t 75c. gl 25c. $1 60c. and $1 62-}c. per gallon ; of these he wonld have a mixture of 72 gallons, worth gl 37^0. per gallon ; what quantity of each sort must he have ? Ans. 8 at 75c. 16 at $1 25c. 40 at gl 50c. and 8 at gl 62^Q. Or, 16 at 75c. 8 at $1 25c. 8 at $1 50c. and 40 at $ I 62ic. 4. How many gallons of water of no value, must be mixed with wine, at 4s. per gallon, so as to fill a vessel of 80 gallons, that may be afforded at 2s. 9d. per gallon ? Gal. Gal. Gal. Gal. i 0)15 I 48^33 33 Sum 48 A i.imiiea simple^ ^ ^^ gallons, at 90 = 10 80c. $ 20 16 80 Gal. $ c. Gal. c. As 20 : 16 80 :: 1 : 84 per g-alion. Now, having found the rate of the limited simples, the question may stand thus : How much wine, at 80c. and 87ic. per gallon, must be mixed with 20 gallons at 84c. per gallon, that the mixture may be worth 82Jc. per gallon ? 80 ^ m-5 82A / 84 87i-^ 2-1 61 gallons, at 80c. 2 L 84 As n JgJ ::20: J^^ !!!!!!!!:i! 8^' £!!!!!!!" * J Answer. * The tbree last Cases need no demonstration, as the 2d and 3d evidently result from the fir>t, an-l the last from AUi^alion Medial, and the second Case in Altcnrate. W w \ POSITION. Proof. 52 gallons at 80c. ~ CO - - 871 = 8 - - 75 = 12 . . 90 = ^41 60c. 17 50 6 10 80 92 - - 821 = 75 90 2. How much gold, of 14 and \6 carats fine, must be mixed with 6oz. of 19, and 12 of 22 carats fine, that the composition may be 20 carats fine ? Ans. 1 ^''joz. of each sort. POSITION. POSITION is a rule, which, by false or supposed numbers, ta- ken at pleasure, discovers the true ones required. It is divided into two parts ; single and double. SIJVGLE POSITIOjY. Single Position teaches to resolve those questions, whose results are proportional to their suppositions : such are those which re- quire the multiplication or division of the number sought by any proposed number ; or when it is to be increased or diminished by itself a certain proposed number of times. 1. Take any number, and perform the same operations with it as are described to be performed in the question. 2. Then say, as the sum of the errours is to the given sum, so is the supposed number, to the true one required. Proof. Add the several parts of the sum together, and if it agrees with the sum, it is right. Examples. 1. A school master, being asked how many scholars be had, said, If I had as many more as I now have, three quarters as many, half as many, one fourth and one eighth as many, I should then have 435: Of what number did his school consist? * The operations contained in the question heia* performed upon tlie answer or number to be fovmd, will give the result contained in the question. The same operations, performed on any other number, will give a certain result. When the results are proportional to their supposed numbers, it is manifest that one result must be to the result in the question, as the supposed 7iwnhcr is to the true (Hie or answer. In any cases, when the results are not proportional to their sup- positious, tlie answer cannot be found by tliis rule. SINGLE POSITION. Suppose he had 80 As many = 80 ^ as many = 60 A as many = 40 1 as many ~ 20 I- as many — 10 290 As 290 : 435 :: 80 80 120 120 90 60- 30 13 29|0)3480|0(120 Ans. 29 58 68 435 Proof. 2. A person lent his friend a sum of money unknown, to receive interest for the same at 6 per cent, per annum, simple interest, and at the end of 12 years, received for principal and interest J260 : What was the sum lent ? Ans. =. s. d. Then >^ 5 yards at 8s. = 2 4^^^5— 10 yards at 3s. = 1 10 20 30 3 10 proof. 20 Sum of errours — 5-i- 5=^10)50 Ans. 5yds. damask, and 15 — 5= 10yds. lining. Or, 6+4—2=5 as before. 2. A and B have the same income ; A saves } of his ; but B, by spending £30 per annum more than A, at the end of 8 years finds himself £40 in debt; what is their income, and what doc- each spend per annum ? C 80_^120-[- Ans. Their income is £200 per ann. Suppose { V ( 160-*- ■*- 40-1- Also, A spends £175 and B £205 pci annum. Then, 80 — 10=70 A's expense per annum, and 70-|-3(» = 100, B's expense per annum. Then 100x8—80x8=160, which should have been 40 ; therefore, IGO — 40=120 more than it should be, for the first errour. In like manner proceed for the second errour. 3. A and B laid out equal sums of money, in trade : A gained a sum equal to i of his stock, and B lost §225, then A's money was double that of B : What did'earh lay out ? Ans. §600. %. .>5B DOUBLE POSITION. 4. A labourer was hired for 60 days upon this condition, thatlbi every day he wrought, he should receive 75c. ; and for every day he was idle, should forfeit 37ic. ; at (he expiration of the time he received $18 : How many days did he work, and how man^was he idle ? Ans. He was employed 36 days, and was idle 24. 5. A gentleman has two horses of considerable value, and a carriage worth £100 ; now if the first horse be harnessed in it, he and the carriage together will be triple the value of the se- cond ; but if the second be put in they will be 7 times the value of the first : What is the value of each horse ? Ans. One £20 and the other £40. 6. There is a fish, whose head is 10 feet long ; his tail is as long as his head and half the length of his body, and his body as long as the head and tail : What is the whole length of the fish ? Head=l0 First, suppose the body 20—. ^ lO— Tail =30 V Body=40 2d. suppose it 30-^^ 5— " — Ans. 80 feet. 7. What number is that, which, being increased by its i, its ], and 5 more, will be doubled? Ans. 20. 8. A farmer, having driven his cattle to market, received for them all }J320, being paid at the rate of $24 per ox, $16 per cow, and $6 per calf; there were as many oxen as cows, and 4 times as many calves as cows : How many were there of each sort? Ans. 5 oxen, 6 cows, and 20 calves. 9. A, B, and C built a ship, which cost them $5000, of which A paid a certain sum, B paid $500 more than A, and C $500 more than both ; having finished her, they fixed her for sea, with a car- go worth twice the value of the ship : The outfits and charges of the voyage, amounted to -} of the ship ; upon the return of which, they found their clear gain to be § off of the vessel, cargo and expenses : Please to inform me what the ship cost them, several- ly ; what share each had in her, and what, upon the final adjust- njcnt of their accompts, they had severally gained ? 600.^^1500— Suppose it cost A ^l^ 1000-^^500-f Ans. A owned -^\ of the ship, which cost him $875, and his share of the gain, was $1093 75c. B owned ^1, which cost $1375, and his gain was $1718 75c. C owned ^J, which cost $2750, and his gain was $3437 50c. PERMUTATIONS AND COMBINATIONS. 3o9 PERMUTATIONS AND COMBINATIONS. THE Premutation of Quantities is the shewing^ how many dif- ferent ways any given number of things may be changed. This is also called variation, alternation or changes ; and the only thing to be regarded here is the order they stand in ; for no two parcels are to have all their quantities placed in the same situation. The Combination of quantities is the shewing how often a less number of things can be taken out of a greater, and combined together/ without considering their places, or the order they stand in. This is sometimes called election, or choice ; and here every parcel must be different from all the rest, and no two are to have precisely the same quantities, or things. The Composition of Quantities is the taking of a given number of quantities out of as many equal rows of different quantities, one out of every row, and combining them together. Here no regard is had to their places ; and it differs from Com- bination only as that admits of but one row of things. Combinations of the same form are those, in which there are the same number of quantities, and the same repetitions ; thus, a6cc, bhad, deef, &c. are of the same form ; but abbcy abbb, aacc are of different forms. Problem I. To find the number of permutations^ or changes, that can be made of any given number of things all different from each other. KULE.* Multiply all the terms of the natural series of numbers, from 1 up to the given number, continually together, and the last product wiil be the answer required. Examples. 1. Christ church, in Boston, has 8 bells: How many changes may be rung on them ? 1x2x3x4x5x6x7x8=40320 Ans. 2. Nine gentlemen n^et at an inn, and were so pleased with their host, and with each other, that in a frolick, they agreed to tarry so long as they, together with their host, could sit every day in a different position at dinner: Pray how long, had they kept their agreement, would their frolick have lasted ? Ans. 9941 If 4 years. 3. How many changes, or variations, will the alphabet admit of? Ans. G20448401 733239439360000. *= The reasoa of this lule may hs shewn thus, any one thing; a is capable of one position only, us a. Any two things a and b are capable of two variations only ; as o6, ba ; who've number is expressed by 1 X2. If there be three things a, b and c ; then any two of them, leaving out tlie third, w"l have 1X2 variations ; and consequently when the third is taken in, there will be 1 X2-X^ variations ; and so on, a? iuv as you ploa*e. 360 PERMUTATIONS AND COMBINATIONS. Problem II. Jny number of different things being given^ to find how many changt^ can be made out of them by taking any given number of qudniitie? at a time. Rule.* Take a series of numbers, beginning at the number of thing? given, and decreasing by 1, as many terms as the number of quan- tities to be taken at a time ; the product of all the terms will be the answer required. Examples. 1 . How many changes may be rung with 4 bells out of 8 ? 8 7 66 6 -— Or, 8X7X6X5 (=4 terms) =1680 Ads. 336 1680 2. How many words can be made with 6 letters of the alphabet^ admitting a number of consonants may make a word ? 24x23x22x21x20x19 (6 terms) =96909120 Ans. Problem III. 'Any number of things being given, whereof there are several things of one sort, several of another, 4'C. to find how many changes may be made out of them all. RuLE.f 1. Take the series 1x2x3x4, &c. up to the number of thing? given, and find the product of all the terms. * This Rule, expressed in term?, is as follows ; mXni — 1 X w — 2Xni — rj, &c. to n terms ; whence m = number of things given, and n =■ quantities to be ta- ken at a time. '4.mu-T> 1 • 1- . *i 1X2X3X4X5,&C. tow. T This Rule IS expressed m terms thus ; ~ — —7- ^ lX2X3,&c.tojtfXiX2x3,Acc.to^,&:c- wlience m = number of things given, p =z number of things of the first sort, q = number of things of the second sort, &;c. Any 2 quantities, a, 6, both different, admit of 2 clianges ; but if the quanti- ties are the same, or ab becomes aa, thei'e will be only one alteration, which may 1X2 be expressed by =1 . '^ 1X2 Any 3 quantities, «, b, c, all different from each other, admit of G variations -, but if the quantities are all alike, or, a be become aaa, then the 6 variations 1 X2X>3 w^ill be reduced to 1, which may be expressed by —- -=:i. Asain, if twf 1X2X3 o ' . quantities out of three are alike, or abc become aac; then the 6 variations will 1X2x3 be reduced to these 3,rtar, caa, aca,' which may be expressed by — •■-— = 3. IX-*- and so of any greater number. PERMUTATIONS AND COMBINATIONS. 361 2. Take the series 1x2x3x4, &c. up to the number of the giv- en things of the first sort, and the series, 1x2x3x4, &c. up to the number of the given things of the second sort, &lc. 3. Divide the product of all the terms of the first series by the joint product of all the terms of the remaining ones, and the quo- tient will be the answer required. Examples. 1. How many vatialions may be made of the letters in the word Zaphnathpaaneah ? 1X2X3X4X5X6X7X8X9X10X11X12X13X14X16 (== number of letters in the word) =1307674368000. 1X2X3X4X6 (= number of as^ = 120 1X2 (= number of ps) = 2 1 (= number of ^s) = 1 1x2x3 (== nuiiiber of As) = 6 1X2 (= number of ns) = 2 2X6X1X2X120=2880)1307674368000(454033600 Ans. 2. How many different numbers can be made Qi the following figures, 1223334444? Ans. 12600. Problem IV. To find the number of comhinations of any given number of things, all different from one another, taken any given number at a time. Rule.* 1. Take the series 1, 2, 3, 4^ &,c. up to the number to be taken at a time, and find the product of all the terms. 2. Take a series of as many terms, decreasing by 1, from the given number, out of which the election is to be made, and find the product of all the terms- 3. Divide the last product by the former, ard the quotient will be the number sought. Examples. 1. How many combinations may be made of 7 letters out of 12 ? 1X2X3X4X5X6X7 (= the number to be taken at a time)=5040. 12X11X10X9X8X7X6(= same number from I2)=3991680. 5040)3091680(792 Ans. 2. How many combinations can be made of 6 letters out of the 24 letters of the alphabet ? Ans. 134596. m m — 1 m — 2 m — 3 * This Rule, expressed algebraically, is — x X X , fee. to n 12 3 4 terms ; where m is the number of given quantities, and n those to be taken at a time. Note. In any given number of quantities, the number of Combinations in- vrcases gradually till you come about the even numbers, and then gradually decreases. If the number of quantities be ci'en, half the number of places will -]iew the greatest number of Combinations, that can be made of those quanti- Jes ; but if odd, then those two numbers which are the middle, and whose sum s equal to the givcu number of quantities, v.iU show the :>;re.itest. niimler of •Jcmbiautions, X T 362 TERMUTATIONS AND COMBINATIONS. 1 1 -6 5 _^ * ' 1 1 34"* 3. A general was asked by his king- what reward he should cori' fer on him for his services ; the general only required a penny lor cverv file, of 10 men in a file, which he could make out of a com- iiany of 90 men : What did it amount to ? Ans. £23836022841 4. A farmer bargained with a gentleman for a dozeo sheep, (at 2 dollars per head) which were to be picked out of 2 dozen ; but being long in choosing them, the gentleman told him that if he would give»him a cent for every different dozen which might be chosen out of the two dozen, he should have the whole, to which the farmer readily agreed : Pray what did they cost him ? Ans. §27041 56c. 5. How many locks, whose wards differ, may be unlocked with a key of 6 several wards ? Ans. 63 : 6 of which may have one single ward, 15 double wards, 20 triple wards, 15 four wards, 6 five wards, and 1 lock 6 ward* Ward s. Locks w ard s. Locks. T ' 61 T " 5' 2 15 2 10 - 3 4 > in 6 = - 20 15 ► 3 4 ► in 5 = ' 10 5 5 L6J 6 . 1- Problem V 5 1 To find the numlcr of combinations of any given number of things, by taking any given number at a time ; in which there are several tilings of one sort, several of another, <^'C. Rule. Fipd the number of different fornix, which the things, to be ta- ken at a time, will admit of, in the following manner : 1. Place the things so that the greatest indices may be first, and the r<»st in order. 2. Begin with the first letter and join it to the second, third, fourth, &c. to the last. 3.' Join the second letter to the third, fourth, &c. to the last j and so on till they are all done, always rejecting such combinations as have occurred before ; and this will give the combinations of all the twos. 4. Join the first letter to every one of the twos ; then join the second^ third, &c. as before ; and it will give the combinations of all the threes. '5. Proceed in the same manner to get the combinations of all the fours, fives, &c. and you will at last get all the several ferms of combination, and the number in each form. 6. Having found the number of combinations in each form, add them all together, and the sum will be the number required. PERMUTATIONS AND COMBINATIONS. obo Example. Let the things proposed be aaabbc : It is required to find the number of combinations of every 2, of every 3, and of every 4 of these quantities. Combinations at large. Forms. Combinations in each form aa,aayab,ab,ac a2,62 2 aa,abyab,ac ab^acjjc 3 ab,ab,ac — bbM 6=sum of the twos. be a3 1 aaa,aabfaabyaac a^b,a^c,b^ a,62c4 aab,aab^aac abc 1 abby(tbc .— bbc 6=sum of the threes. daab,aaab,aa(tc a^b.a^c o aabb^aabc aH^ 1 abbe a^bCfb^ac 2 Ans. 5 combinations of every 2 ; 4 quantities. 5=9um of the fours, of every 3, and 6 of every Problem VI. To find the changes of any given number of things^ taken a given, number at a time ; in which there are several given things of one sorty several of another^ 4*c. Rule. 1. Find all the different forms of combination of all the given things, taken, as many at a time, as in the question, by Problem 6. 2. Find the number of changes in any form, (by Problem 3,) and multiply it by the number of combinations in that form. 3. Do the same for every distinct form, and the sum of all the products will give the whole number of changes required. Example. How many changes can be m.ide of every 4 letters out of these 6, aaabbc 1 No. of forms. Comb. a^b^a^c a"bcJ)"ac !! 1X2X3X4=24 1X2X3 = 6 1X2X3X4=24 1X2X1X2= 4 1X2X3X4=24 Changes. Qy I = 12 364 MISCELLANEOUS MATTERS (2X 4= I, rix 6= Therefore, posite sides of a circular field 268 poles about ; they begiti to go round it, both (ue same way, at the same instant of time ; A goes 22 rods in 2 minutes, and B 34 rods in 3 minutes : How many times will they go round the field, before the swifler overtakes the slower? min. po. mitj. ])o. 2 : 22 ^ - ( 1 1 A goes in a minute. 3 : 34^ ■• ^ • ^ llAB do. do. therefore, B gains 11] — 1 1=^ of a pole of A every minute. And, as ipo. : Imin. ;: -f^po. (=^half round the field) : 402min. (=the time in which B will overtake A.) Then, min. ])o. min. uo. And, VeV — ^^y times round the field, A travels ; and VeV^l"^ times round the field, B travels. 18. If 15 men can perform a piece of work in 1 1 days, how many men will accomplish another piece of work, four times as large in a fifth part of the time ? Ans. 300 men. 19. If A can do a piece of work alone in 7 days, and B in 12 ; let them both go about it together: In what time will they finish it? Days. work.day works, work. work. work, work.day.work.day. . \ ? S Thpn J-4--i-=.lll As 15. • I" 1 • 4_8^ An 's the power which does the work three times, for each agent is united with others three times. Then f^4X3=yJ|=5i5i days, the time in which all together will do the work. Then rVra—o^jrVVz"^'^ power, by taking A, B, and C's from the sum of the whole power to do the work once. Then YaV ^^y^^ ~ ^'^ ^''^^- '" ^^^^ ^^^^ way, is found V^^d. = A's time. Vt¥^- — ^'^ ^'*"^' ^"^ Vri^- ~ ^'^ 'io^e, Ans. 23. A cistern, for water, has 2 cocks to supply it ; by the first, it may be filled in 45 minutes, and by the second, in 55 minutes; it has likewise a dischargmg cock, by which it may, when full, be emptied in 30 minutes ; Now, if these three cocks he all left open, when the water comes in, in what time will the cistern be filled ? Cist. Hour. Cist. h. m. s. As -4242 : 1 :: 1 : 2 21 20 1 Ans. Or, by vulgar fractions, more accu- rately, 2h. 21m. 2543. Ans. Gains in an hour '4242 of a cistern. 24. A water tub holds 73 gallons ; the pipe, which conveys the water to it, usually admits 7 gallons in 5 minutes ; and the tap dis- charges 20 gallons in 17 minutes : Now, suppof.ing these both to be carelessly left open, and the water to be turned on at 4 o'clock in the morning; a servant, at 6, finding the water running, puts in the tap ; in what time, after this accident, will the tub be full ? Ans. The tub will be full at 32m. 58j J-fs. after 6. 25. A has a chest of tea, weighing 3icwt. the prime cost of which is .£60: Now, allowing interest at 6 per cent, per annum, how must he rate it per Ife to B, so that, by taking his note of hand, payable at G months, he may clear §50 by the bargain ? Interest £2 5s. Then as 3icwt. : £60 -f £15-1- £2 5s. :;aife : 3s. llffd. Am. 26. Suppose tltBt^merican continental debt to be 18 millions, what annuity, at 6 per cent, per annum, will discharge it in 25 years ? By Table 5, of annuities, page 339, -07823 is the annuity which £1 will purchase in 25 years, then, -07823x18000000— £1408140 An? The annual interest of the deht= 1080000 3Iin. Cist. Min. Cist. 45 : 1 : : GO : ; 1-3333 5^ : 1 : : GO : ; 10909 2-4242 30 : 1 : : GO : ; 2- Therefore, there must be a sinking fund of £328140 per ann. 27. The hour and minute hands of a watch arc exactly together at 12 o'clock; When are they next (ogether? B 372 MISCELLANEOUS QUESTIONS. The velocities of the two hands of a watch, or clock, are to each other, as 12 to 1 ; therefore, the difference of velocities is 12—1 kc. Ans. 1 5 27fV) 2 10 54/j-) 3 16 2\j\) 28. A hare istarts 12 lods before a hound; but is not perceived by him till she has been up 45 seconds ; she scuds away at the rate of 10 miles an hour, and the dog, on view, makes after at the rate of 16 miles an hour : How lono: will the course hold, and what space will be run over, from the spot wheie the dog started? 2288 feet = the ground run over by the dog. 9'7|sec. Ans. 29. In a series of proportional numbers, the first is 4, the third 12, and the product of the second and third is 112-B: What is the difference of the second and fourth ? Ans. 18-8. 3.0. A fellow said that when he counted his luits; two by two, three by three, four by four, five by live, and six by six, there was still an odd one ; but when he told them seven by seven, they came out even : How many had he ? 2X3X4X3X6=720, and 720+T->7=103 even, Ans. 721. 31. There is an island 50 miles in circumference, and three men start together to travel the same way about it : A goes 7 miles per day, B 8, and C 9: When will they all come together again, and how far will each travel ? 50x7-f50x8-f-50x9-~7+8-|-9'=50 days. A 350 miles, B 400, and C 450, Ans. 32. Suppose A leaves Newburyport at 6 o'clock on Monday morning, and travels towards Providence, at the rate of 4 miles per hour without intermission ; and that, at 3 in the afternoon, B sets out from Providence for Newburyport, and travels constantly at the rate of 7 miles an hour: Now suppose the distance between the two towns to be 90 miles ; whereabout on the road will they meet ? 6-f 3=9 hours, and 9x4=36 mile;i, the time and distance A had travelled before B started. Then 90—36=54 miles remain to be travelled by both ; now, as both togfether lessen the distance 7+4 = 11 miles an hour, therefore y\ of 54-|-36=55j\ miles from Newburyport ; which is near Ames's, at Dedham. 33. If, during ebb tide, a wherry should set out from Haverhill to come down the river, and at the same time, another should set out from Newburyport, to go up the river, allowing the distance to be 18 miles ; suppose the current forwards one and retards the other H mile per hour; the boats are equally laden, the rowers equally good, and, in the common way of working in still water, would proceed at the rate of 4 miles per hour : VVhere, in the ri- ver will the two boats meet ? Ans. I2fm. from Haverhill, and 5|m. from Newburyport. .34. A gentleman making his addresses in a lady's family who had live daughters ; she told him that their father had made a will, "vhich imported that the first four of the girls' fortunes were, to- MISCELLANEOUS QUESTIONS. 373 gether, to make gSOOOO ; the last four, $66000; the three last with the first, g60000; the three first with the last, 56000 ; and the two first with the two last, $64000, which, if he would unravel, and make it appear, what each was to have, as he appeared to have a partiality for Harriet, her third daughter, he should be welcome to her: Pray, what was Miss Harriet's fortune ? A-fB+C-f-D =30000 B4.CH-D+E=66000 A +C-fD+E=60000 A-fB +D-|-E=64000 296000 Then, 296000—4 the number of combinations=74000 the sum of their fortunes. A+B+C +£=56000 ^ Then, AH-B-1-C4-D4-E=74000 And A + B +D-1-E=64000 Ans. Harriet's fortune =§10000 35. Three persons purchase a vessel in company, towards the payment whereof A advanced f , B ^, and C, $900: What did A and B pay, each, and what part of the vessel had C ? ^'^^ C's part of the vessel. $2100 A advanced. $2250 B ad^ vanced. 36. A and B cleared, by an adventure at sea, 45 guineas, which was £35 per cent, upon the money advanced, and with which they agreed.to purchase a genteel horse and carriage, whereof they were to have the use in proportion to the sums adventured, which was found to be 11 to A, as often as 8 to B : What money did each adventure ? As JC35 : 100 :: 45 guineas : jei80=the whole adventure. - As 11+8 • 180 -r^ '-^^^"^ 4s.2i|d. A's. 37. A, B and C are to share £100 in the proportion of i, ] and } respectively ; but C dying, it is required to divide the whole gum properlv, between the other two? £ 9. d. 67 2 lOf A's share in all, } . 42 17 14 B's share in ail, I ^"^• Proof 100 38. A, B and C have among them 135 guineas ; A's-j-B's are to B's+C's, as 5 to 7, and C's — B's to C'sH-B's as 1 to 7 : How many had each ? A+B. B+C. Suppose A'sH-B's=50 ; then, as 5 : 7 :: 50 : 70 ; as 7 : 1 :: 70 : iO=C's— B's; then 70—10=60, and 60-ir2=30=B's ; 50—30= 20=A's, and 30+10=40=C's, by the supposition : Now 20+30+ 40=90, which should have been 135, therefore, C 20 : 30=A's. As 90 : 135 :: {30 : 45=B's. (40 : 60=C's. Snm=135 proof. 39. There are three horses, belonging to different men, employ- ed in a team to draw a l©«d of salt froui Newburyport to Boston, i^ 374 MISCELLANEOUS qUESTiOKS. for £2 10s. : A and B are supposed to do -j^ of the work ; A and C y5„ and B and C j\ of it ; they are to be paid proportionally Can vou divide it as it should be? Ans. A's=:19|f|s. B's=: 9i||s. C's=21/^|s. Proof 60s.=the sum. 40. I would put 20 hogsheads of London beer into 10 wine pipes, and desire to know what the cask must contain, which will receive the difference, 231 solid inches being the wine gallon, and 282 that of beer. Beer hhd. = 54 gall, and 54x282x20=304560 solid inches. Wine pipe =126 gall, and 126x231x10=291060 solid inches, and 304560—291060 - — '■ ^2 =47f4 beer gallons, Ans. 41. Being about to plant 5292 trees equally distant in rows, the length of the grove is to be three tfmes the breadth : How many of the shorter rows will there be ? Ans. viz. i of the trees are to form an exact square, the side whereof being 42, shews how many come into a short row. 42. A general, disposing his army into a square battalion, found he had 231 over and above, but increasing each side with one sol- dier, he wanted 44 to fill up the square : How many men did his army consist of? 231+44=275, and 275^-^-2=137, then "137x137+231 = 19000 Ans. Proof, 138x138=19044. 43. 1 want the length of a shoar, the bottom of which, beiwg set 9 feet from the perpendicular side of a house, will support a weak place in the wall, 22i feet from the ground ? Ans. 24 feet, 2| inches. 44. A line 35 yards long will exactly reach from the top of a fort, standing on the brink of a river, known to be 27 yards broad, to the opposite bank : What is the height of the wall ? Ans. 22 yards, 9| inches, nearly. 45. Suppose a light house built on the top of a rock; the dis- tance between the place of observation and that part of the rock level with the eye 620 yards ; the distance from the top of the rock to the place of observation, 846 yards, and from the top of the light-house 900 yards : the height of the light-house is required ? f~— "-*~ ' f^ " V 900 X 900— 620x620— V 846x846—620x620=76 •77yds. Ans. 46. Tfie sum and difference of the squares of two numbers given, to find those numbers. Rule. From the sum take the dilTerence, and half the remainder is the jquare of the less, %vhich, taken from the sum of the squares; wift give the square of the greater. 1 MISCELLANEOUS QUESTIONS. o75 A and B have between them a number of guinea?, which are to be so divided, that the sum of their squares may be 208, and the difference of their squares 80; supposing A's the greaternumber, how many has he more than B ? 208—80—2=64 the square of B's, and 208—64=144 the square of A's; therefore ^144— -^64=4, Ans. 47. Having the sum of two numbers ^ and the sum of their squares given, to find those numbers. Rule. From the square of their sum take the sum of their squares : then from the sum of their squares take this remainder, and the square root of the difference will be the difference of the two num- bers. To half their sum add their difference, and the sum will be the greater. From half the sum take half their difference, and the remainder will be the less. A and B have 50 guineas between them, which are to be so di- vided, as that the sum of the squares of the two numbers shall be 1300 : How many had each, supposing A to have the greater num- ber ? 30x50—1300=1200 ; then, ^1 300— 1200=10 difference. Now 50—2+ 10-^2=30= A's. And 50-t^— 10~-2=20=B's, Ans. 48. Having the difference of two numbers^ and the sum of their squares given, to find those numbers. Rule. From the sum of their squares take the square of their differ- ence : to the sum of the squares add the remainder, and the square root of thisj sum will be the sum of the required numbers ; then, with the half sum and half difference proceed as in the last ques- tion. A number of guineas are to be divided between A and B, in such a manner that A may have 50 more than B, and that the sum of the squares of the respective shares may be 12500: What number had each ? 12500— 50x50— 10000, and >/l2500-f-10000=150=sum of their shares. Then, Tbii^Z-\Eo^^=\00 A's ; and 150-r-2---'50^i= 50 B's, Ans. 49. Having the sum of the squares of tm'o numbers, and the square of their half sum given, to find thoss numbers. Rule. From the sum of the squares take twice tbe square of the haU sum, and the square root of half the remainder will be their halt' ^liffercnuo. with which and the half sum proceei' as before dirt^cted. :m MISCELLANEOUS qUESTIONS. « Let tlie sum of the squares of two numbers be 3161, and Ihe square of their half gum 1560-25 : Required those numbers ? -3161—1560 2 5x2^4 5 40 5-t-2--20-25 and v^20^5=:4-6=i difference, and -v/1560-25=:39-5=i sum ; then, 39 5+4-5—44 the greater, and 39*5— 4 6=35 the less, Ans. 50. — 1. If the quantity of matter^ (or weights) of any two bodies, put in motion, be equal, ihe force by which they are moved will be in .'proportion to their velocities, or swiftness of motion. 2. If the velocities of these bodies be equal, their forces will be di- rectly as the quantities of matter contained in them, that is, as their weights- 3. If both the quantities of matter, and the velocities be unequal, the forces with which the bodies are moved, will be in a proportion compounded of their quantities of matter and velocities. Suppose the battering ram of Vespasian weighed 60000ife ; that it was moved at the rate of 24 feet in one second, and that this was sufficient to demolish the walls of Jerusalem : With what ve- locity must a cannon ball, which weighs 42lfe be moved, to do the same execution ? The velocity of the ram being 24, and the weight of the ball 42, compounded, will make a fraction— f|— 4, and 4x60000=34285f feet in a second, Ans. 61. A body weighing 30ife is impelled by such a force as to send it 20 rods in a second : VVitli what velocity would a body weighing 12lfe move, if it were impelled by the same force ? 30X20 — Y77— =50 rods in a second, Ans OF GRAVITY. 52, The gravity of bodies above the surface of the earth decreases in a duplicate ratio (or as the squares of their distances) in semidiam- ciers of the earth, from the eartlis centre. Supposing a body to weigh 400ife at 2000 mile ■»»iibove the earth's surface : What would it weigh at the surface, estimating the earth's semidiameter at 4000 miles ? From the centre to the given height being IJ- semidiameters ; multiply the square of 1| by the weight, and the f)roduct will be the answer. l-5xl'5X400=900ife Ans. 53. If a body weigh 900ife at the surface of the earth, what will it weigh at 2000 miles above the surface ? This being the reverse of the last, therefore, H- -5= 1-5 and 900 -^l-5Xl'5'=400ifs Ans. 54. A certain body on the surface of the earth, weighs ISOlfe -. JIovv high must it be carried to weigh but 20115? v^l 80-^20=3, Ans. 3 semidiameters from the earth's centre, that is 8000 above its surface. I MISCELLANEOUS QUESTIONS. 3T7 bb. To what height must a ball be raised to lose half its weight ? As 1 : 398206X3982-06 :: 2 : 31713603-6872, and -v/3 17 13603-6872 —5631-48 : and 5631-48— 3982'06=1649-42 miles, Ans. 56. Jit ZDhat distance from the earth zn'Oiild a balloon be suspended be- tween the earth and moon ? Rule, As the Slim of the square roots of their quantities of matter is to the distance of their centres, so is the square root of the quantity of matter in the earth, to the distance from the earth's centre. The proportional quantity of matter in the earth being to that in the moon as 4124 to 1: and the distance of their centres 240000-|- 398206 + 1090: therefore, as>/41-24+\/l:240000-f-3982U6-f 1090 :: ^/41 24 : 212051-49. And 21205149— 3982 06— 208069-43 miles from the earth's surface, Ans. 57. — 1. // the diameters of two globes be equal, and their densities different^ the weight of a body on their surfaces will be as their densities. 2. If their densities be equals and diameters diff'erent, the weight will be as their diameters. 3. If their diameters and densities be both different, the weight will be as the product of their diameters and densities. If a stone weigh lOOife at the surface of the earth, required its weight at the surfaces of the sun and the several planets, whose densities are known respectively ? Sun. Jupiter. Saturn. Earth. Moon. TheiT densities 100 78 5 36 392-5 464 ^ Diameters in > 883217 58. 89170-81. 79042-35. 7964-12 2180 Eng. miles, y ^ As 7964512x392-5 : 100:; 8 83217-58 X100 : 2825 •461b. at the Sun. 89170-8} x7»-5 : 220-411b. at Jupiter. 79042-35'x3d : 91061b. at Saturn, L 2180 X464 : 32-35lb. at the Moon. what will be the height of a tide raised by the earth on the sur- face of the moon under similar circumstances? The attraction of one of those bodies on the other's surface i** directly as its quantity of matter, and inversely as its diameter ; therefore, as 2180x2180x2180x464 : 5 :: 7964 x 7%4x7964x392v']^ ■ 206-22 directly. And as 2180 : 206-22 :: 7964 : 56448 inverse- ly, Ans. OF THE FALL OF BODIES. 59. Heavy bodies near the surface of the earth, fall one foot the iirst quarter of a second ; three feet the second quarter ; five feet in the third, and seven feet in the fourth quarter ^ that is, 16 ieet In the first second.* ■'- Tiic exact velocil'*' ;u rr(n(o i« IG ] ia ^];p ^omnd • hv^ iv ^hp p'l f^ ivjll he, i78 MISCELLANEOUS QUESTIONS. The velocities, acquired by bodies in falling, are in proportion to the squares of the times in which they fall ; for instance, Let go three bullets together ; stop the first at one second, and it will have fallen 16 feet. Stop the next at the end of the second second, and it will have fallen (2x2= 1) four times 16, or 64 feet ; and stop the last at the end of the third second, and the distance fallen will be (3x3=9) nine times 16, or 144 feet, and so on. Or, which is the same, the space fallen through (in feet) is al- ways equal to the square of the time in 4ths of a second. Or, by multiplying 16 feet by so many of the odd numbers, be- ginning at unity, as there are seconds in any given time ; viz. by I for the first second, by 3 for the second, by 5 for the third, and so on, these several products will give the spaces fallen through, in each of the several seconds, and their sum will be the whole dis- tance fallen. The velocity given, to find the space fallen through* KULE. 1. The square root of the feet, in the space fallen through, will ever be equal to one eighth of the velocity acquired at the end of the fall ; therefore, 2. Divide the velocity by 8, and the square of the quotient will be the distance fallen through, to acquire that velocity. Suppose the velocity of a cannon ball to be about } of a mile, or 660 feet per second : From what height must a body fall, to ac- quire the same velocity per second ? 660—8=82-5 and 82-6x82'6=6806i feet,=lj\\ mile, Ans. 60. Tlie time given, to find the space fallen through. KuLE. 1. The square root of the feet, in the space fallen through, wrll ever be equal to four times the number of seconds the body has been falling; therefore, 2. Multiply the time by 4, and the square of the products will be the space fallen through in the given time. How many feet will a body fall in 5 seconds ? 5X4=20, and 20x20=400 feet, Ans. 61. A bullet is dropped from the top of a building, and found tc» reach the ground in If seconds : Required its height? 1 -75x4=7, and 7x7=49 feet, Ans. Or, If =7qrs. and 7x7=49. Or, l-75xl-75xl6=49feet, Ans. 62. What is the difference between the depths of two wells, into each of which should a stone be droj)})ed in the same instant, one would reach the bottom in 5 seconds, and the other in 3 ? 5x4=20, and 20x20=400 feet, 3xl = 12,.and 12x12=144 feet. Ans. 256 fett. 63. Ascending bodies are rclardcd in the same ratio that de- scending bodies are accelerated ; therefore, if a ball, discharged from a guu, return to Ihc earth in 12 seconds : Uo»v high did it as- cend ? MISCELLANEOUS QUESTIONS. 37H The ball being half of the time, or 6 seconds, in its ascentjthere- fore, 6x4=24, and 24x24=576 feet, Ans. 64. The velocity per second given, to Jind the lime.. RULK. 1. Four times the number of seconds, in which a body has been falling, is equal to one eighth of thn velocity, in feet, per second, acquired at the end of the fail ; therefore, 2. Divide the given velocity by 8, and one fourth part of the quotient will be the answer. How long must a bullet be falling, to acquire a velocity of 160 feet per second ? 160-f-8— 20, and 20—4=5 seconds, Ans. 65. The space through which a body has fallen, given, to find the time- it has been falling. RuLK. 1. Four times the number of seconds, in which the body has been falling, will ever be equal to the square root of the space, in feet, through which it has fallen ; therefore, 2. Divide the square root of the space fallen through by 4, and the quotient will be the time, in which it was falling. In how manv seconds will a bullet fall tlirough a space of 10125 feet ? Vl0125=100-6, and 100 6^-4=25-15 seconds=25" 9"' AbS. 66. In what time will a musket ball, dropped from the top of a steeple, 484 feet high, come to the ground ? V'484=22, and 22-^4=51 seconds, Ans. 67. To fin(Jl the velocity, per second, with which a heavy body will begin to descend, at any distance from the eartli's surface. Rule. As the square of the earth's semidiameter is to 16 feet, so is the square of any other distance from the earth's centre, inverse- ly, to the velocity with which it begins to descend per second. With what velocity, per second, will an iron ball begin to de- scend if raised 3000 miles above the earth's surface ? As 4000X4000 : 16 :: 4000+3000x4000+3000 : 5.22449 feet, Ans, 68. How high must a ball be raised above (he earth's surface, to begin to descend with a velocity of 522449 feet per second ? As 16 : 4000X4000 :: 5-22449 : 49000000, and V49000()6o"=7000. Wherefore, 7000—4000=3000 miles, Ans. 69. To find the mean velocity of a falling body. Rule. Divide the space fallen through by the number of seconds it was falling, and the quotient will be the mean velocity. A musket ball dropped from the top of a steeple 484 feet high in 51 seconds : Required its mean velocity ? 484-~5-5— 88 feet per seroi^l. Am. 380 MISCELLANEOUS QUESTIONS. 70. To find the velocity acquired by a falling body^ per second, (or by a stream of water, having the perpendicular descent given) at the end of any given period of time. Rule. 1. The velocity acquired at the end of any period is equal to twice the mean velocity, with which it passed during that period. Or, 2. Multiply the perpendicular space fallen through by 64, and the square root of the product is the velocity required. If a ball fall through a space of 484 feet in 5i seconds, with what velocity will it strike ? By the former part of the rule. By the latter part, with- 484-4-5*5—88, and out regarding the time. 88x2=176, Ans. ^484x64=176, Ans. 71. There is a sluice, or flume, one end of which is 2i feet low- er than the other : What is the velocity of the stream per second ? 2-6x64=160, and ^160=12-649 feet, Ana. 72. The velocity, zcith which a falling body strikes, given, to find the space fallen through. Rule. Divide the square of the velocity by 64, and the quotient will be the height required. If a ball strike the ground with a velocity of 56 feet per second, from what height did it fall ? 5t)X56-r.64=49 feet, Ans. 73. The mean velocity of a fluid, or stream, is 1^-649 feet per second : What is the perpendicular fall of the stream ? 12-649xl2-649-~64=2i feet, Ans. 74. The weight of a body, and the space fallen through, given, to find the force with which it will strike. Rule. The momentum, or force, with which a falling body stiikes, is equal to its weight multiplied by its velocity ; tiierefore, find the velocity, by Problem 70, and multiply it by the weight, which ^vill produce the force required. If the rammer, used for driving ihe piles of Charlestown bridge, weighed 2^ tons, or 4500lfe and fell through a space of U) feet, with what force did it strike the pile ■ V'10x64=:±25 3=velocity, and 26-3x4500=^ 1138501fe momen- tum, Answer. 75. The weight and momentum, or /striking force, given, to find the space fallen through. Rule. Divide the momentum by the weight, and Ibe quotient will be Uie velocity ; then divide the square of the velocity by 64, and the quotient will be the ?pace fallen through. MISCELLANEOUS QUESTIONS. 381 If the aforementioned rammer weighed 4500ife and struck with a force of 1138501fe : From what heig^ht did it fall ? n3850~4500=25-3, and 25^3x2 5~3~6 4 — -[0 feet, Ans. 76. If it were required to know with what quantity of motion, momentum or force, a fluid, moving with a given velocity, strikes upon a fixed obstacle, Rule. By Problem 72 find the fall, which will produce the given velo- city ; multiply that height by 62-51fe Avoird. for clean' river water, by 63}fe for dirty water, and by 64 for sea water. Suppose a stream of clear water to move at the rate of 5 feet per second, and to meet with a fixed obstacle (or bulk head) 15 feet wide and 4 feet high : What is the momentary, instantaneous pressure of the stream ? 5X5~~64=ff and 26-T-64="39 of a foot, for the perpendicular fall of the water. Now 62-5X'39=24-375lfe the pressure upon each square foot, which, multiplied by 60, (the number of square feet in the obstacle) gives 14625lfe going with the given velocity of 5 feet per second ; therefore, 1462'5x 5=7312-5ife Ans.* 77. The velocity of water, spouting through a sluice, or aper- ture in a reservoir, or bulk head, is the same that a body would acquire by falling through a perpendicular space equal to that be- tween the top of the water in the reservoir and the aperture. What is the velocity of water issuing from a head of 5 feet deep ? By Problem 70lh 64x5=320, and ^/:32(T=18 feet nearly, Ans. 78. If the velocity of a stream issuing through the bulk head of a mill, be 16 feet per second, what head of water is there. l6xl6"-^64=4 feet, Ans. 79. The quantity of water discharged from a hole in a vessel, is as the square root of the height of water above the aperture. A miller has a head of water 4 feet above the sluice : How high must the water be raised above the opening, so that half as much again water may be discharged from the sluice in the same time ? -v/4=2, and half as much again as 2, is 2-1-1^=3, for the square root of the required depth ; therefore, 3x3=9 feet high, Ans. OF PENDULUMS. 80. The time of a vibration, in a cycloid, is to the time of a heavy body's de.-cent through half its length, as the circumference of a circle to its diameter, that is, as 3- 14 16 to 1 : therefore, (as a body descends freely, by gravity, through about 193'5 inches in the first second) to find the length of a pendulum vibrating se- conds. Rule. As 31416x31416 : 1x1 :: 193 5 : 196 inches, the half length, and 19-6X2=39-2 inches, the length. * Water being a yielding substance, loses two thipTs of it? power in produc- m2: effects. 382 MISCELLANEOUS (QUESTIONS. 81. To find the length of a pendulum^ that "will swing any given time. Rule. Multiply the square of the second? in any given lime by 39-2 and the product will be the length required, in inches. Required the lengths of several pendulums, which will respec- tively swing i seconds, | seconds, seconds, niinutes and hours ? •25X-25X39-2=24o inches for ^ seconds. •5X-5X39-2=9-8 inches for A seconds, ixl x39'2=39-2 inches for seconds, as above; 60X60x39-2=the inches in 2 miles and 1200 feet, for minutes; and 1 hour=3C00 seconds, therefore 3600X3600X392 :=^the inches in 8018 miles and 96 feet, for hours, Ans. 82. What is the difference between the length of a pendulum, which vibrates half seconds and one which swings three seconds ? 3X3X39^— •5X-5X39"^=:343 inches=28/^ feet, Ans. 83. To find the time which a pendulum of any given length will swing. Rule. Divide the given length by 39-2, and the quotient will be the square of the time in seconds. Or, as 6-261 (the square root of 39-2) is to the square root of the given length, so is 1 second to the time of 1 oscillation : that is, di- vide the square root of the given length by 6*261, and the quo- tient will be the time of one vibration of that pendulum. How often will a pendulum of 9 8 inches vibrate in a second ? By the former part of the rule, 9-8^39-2=*25 of a second, and ^/'25=^■5 of a second, the time of one vibration, that is, it vibrates balfseconds, or 60~-5=120 limes in a minute. By the latter part. V9^=3-13, and v'39-2=6-261, therefore, 3;13-r6-26]= o of a second. 84. I observed, that while a stone was falling from a precipice, a string, (with a bullet at the end) which measured 25 inches, (to the middle of the ball,) had made 5 vibrations : What was the iieight of the precipice ? 25-r39-2=:-6377+, and >/6377=-7986— of a second, the time of one vibration, and -7986X5=1 seconds, nearly, the time of the ftone's descent ; then 4X4=16, and 16x16=256 feet, Ans. c5. To find the true depth of a welly by dropping a stone into ity also the time of the stone's descent and of (he sound's ascent. ^ Rule. 1. Take a line of any length, and by the last Problem find the ti'me from the dropping of the stone till you hear it strike the bot- tom. 2. Multiply 73088 (=16X4X1 142; 1 142 feet being the distance, which sound moves jn a second) by the number of seconds till you hear the stone strike the bottom. 3. To this product add 1304164 (=the square of 1142) and from the square root of the sum take 1142. 4. Divide the square of the remainder by 64 (=16X4) and the quotient will be the depth of the well in feet. MISCELLANEOUS QUESTIONS. 383 o. Divide the depth by 1142, and the quotient will be the time of the sound's ascent, which, being taken from the whole time, wili leave the time of the stone's descent in seconds. Suppose I drop a stone into a well, and a string with a plummet, tvhich measured to the mitdle of the ball, 25 inches, made 5 vi- brations before I heard the stone strike the bottom : Required the depth, time of the stone's descent, and of the sound's ascent : 25-^39-2=-6377, and v''6377=-7986, and -7986X5—4 seconds to the hearing of it strike ; then, -v/73088X44. 1304164— 1142 = 121-53; and 121 53X 121-53-f-64=230-77 feet, the depth, and 23077-r-1142=-2 of a second, the time of the sound's ascent, and 4 — •2=3-8 seconds, the time of the stone's descent. OF THE LEVER OR STEELYARD. 86. It is a principle in mechanicks, that the power is to the weight, as the velocity of the weight, to the velocity of the power. Therefore, to find what weight may be raised or balanced by any given power, say ; As the distance between the body to be raised or balanced, and the fulcrum or prop, is to the distance between the prop and the point where the power is applied ; so is the power to the weight which it will balance. If a man, weighing 160ife, rest on the end of a lever 10 feet long, what weight will he balance on the other end, supposing the prop one foot from the weight ? The distance between the weight and prop being 1 foot, the distance from the prop to the power is 10 — 1=9 feet ; there- fore, as 1 foot : 9 feet :: 160ife : 1440ife, Ans. 87. If a weight of 1440lfe were to be raised with a lever 10 feet long, and the prop fixed one foot t>om the weight, what power or weight, applied to the other end of the lever would balance it ? As 9 : 1 :: 1440 : 1601fe, Ans. 88. If a weight of 1440Ife be placed 1 foot from the prop, at what distance from the prop must a power of 160lfe be applied, to balance it? As 160 : 1440 :: 1 : 9 feet, Ans. 89. At what distance from a weight of 1440ife, must a prop be placed, so as that a power of 1601fe, applied 9 feet from the prop may balance it? As 1440 : 160 :; 9 : 1 foot, Ans. 90. In giving directions for making a chaise, the length of the shafts between the axletree and backhand, being settled at 9 feet, a dispute arose whereabout on the shafts the centre of the body should be fixed. The chaise maker advised to place it 30 inches before the axletree ; others supposed 20 inches would be a sulTi- cient incumbrance for the horse : Now supposing two passengers to weigh 3cwt. and the body of the chaise |cwt. more : VVhat will the beast in both these cases bear, more than his harness ? Weight of the chaise and passengers 3|cwt.--420ifc, and 9 f^et- tOB inches. , „ In. ft *"• ''* i u) • nG?-> Then, as 1C8 : 420 ;: >o ' iti| J Ans. 384 MISCELLANEOUS QUESTIONS. OF THE WHEEL AND AXLE. 01. The proportion for ihe wheel and axle (in which ihe power is applied to the circumference of the wheel, and the weight is raised by a rope, which coils about the axle as the wheel turns round) is, as the diameter of the axle is to the diameter of the wheel, so is the power applied to the wheel, to the weight suspended by the axle. A mecbanick would make a windlass in such a manner, as that life applied to the wheel, should be equal to lOlfe suspended from the axle ; now, supposing the axle to be six inches diameter, re- quired the diameter of the vvheel ? 11). in. lb. in. As 10 : 6 :: 1 : 60 inversely, the diameter required. 92. Suppose the diameter of the wheel to be CO inches: Re- quired the diameter of the axle, so as that life on the wheel may balance lOlfe on the axle ? lb. in. lb. in. Inversely, as 1 : 60 :: 10 : 6 diamett.r required. 93. Suppose the diameter of the axle 6 inches, and that of the wheel 60 inches, what power at the wheel will balance lOIfe at the axle? in. lb. in. lb. Inversely, 6 : 10 :: 60 : 1 Ans. 94. Suppose the diameter of the wheel 60 inches, and that of tbft axle 6 inches ; what weight at the axle will balance life at fhe wheel ? in. lb. in. lb. Inversely, as 60 : 1 :: 6 : 10 Ans. OF THE SCREW. 95. The power is to the weight, which is to be raised, as the distance between two threads of the screw, is to the circumference of a circle described by the power applied at the end of thelever. Rule. Find the circumference of the circle described by the end of the lever; then, as that circumference is to the distance between the spiral threads of the screw ; so is the weight to be raised, to the power which will raise it, abating the friction, which is not propor- tional to the quantity of surface ; but to the weight of the iiicijrn- bent part; and, at ci medium, -} part of the effect of the macliine is destroyed by it, sometimes more and sometimes less. There is a screw, whose threads are an inch asunder ; the level* by which it is turned 30 inches long, and the weight to be raised a ton, or 22408i : What power or Ibrce must be applied to the end of the lever, sufficient to turn the screw — that is, to raise the weight. The lever being the semi 982 1 99-57 ( 3222-48 X 3222-48 X .i222-4B=93- 12 —^ ; 7687-85X7687-85X7687-85= Ml ^- \ 4189-69x4189-69x4189-69= 6-86 ' 2180X2180 X2180 =48-74 7964- 12X7964- 12x7964-12— < IVon.^nC, Von/^nf. .Von.^nZ ^.o^ > tess x; N.*B. The above diameters and mean distances in English miles answer to the same in geographical miles, as they were deduced from observations on the transits of Venus over the Sun in 1761 and 1769. 106. Suppose the density of the Moon 464, and that of the Earth 392-5 : Required the proportion between the quantity of matter In the Earth and in that of the Moon, allowing the Earth's diameter to be 796412, and the Moon's 2180 miles, and supposing the Earth a complete sphere, which, however, it is not? 7964-12X7964' 12X7964- 12X392-6 There is -g^- X2180 X^O X464 ^^^'^^ ^^"^^« *^^ quantity of matter in the Earth that there is in the Moon ; or, the Earth's weight is so many limes that of the Moon. 107. The mean diameter of the Earth's orbit, (or annual path round the Sun)supposing it a circle, is in English miles 190437141'7: Required its mean niolion, (or the space through which it moves in its orbit,) per miimte ? 19043714I'7X311i6— 598'n7324-36 miles in circumference; then, Days. As 366-25-: 598277324-36 :: 1' : 1 137-49 miles, Ans. N. B. The Earth's diurnal motion round its axis is ll^ miles per minute, at the equator. OF THE SPECIFICK GRAVITIES OF BODIES. The specifick gravities of bodies are as their densities, or weights^ bulk for bulk ; thus, p. body is said to have two or three times the speciiick gravity of another, when it contains two or three times as much matter in the same space. A body, immersed in a fluid, will sink, if it be heavier than its bulk of the Huid. If it be suspended therein, it will lose so much of what it weighed in the air, as its bulk of the fluid weighs. Hence, all bodies of equal bulk, which will sink in fluids, lose equal weights when suspended therein, and unequal bodies lose in pro- portion to their bulks. The hydrostaiick balance differs very litde from a common bal- ance that is nicely made ; only it has a hook at the bottom of each scale, on which small weights may be hung by horse hairs, so that a body suspended by the hair, may be immersed in water wi-thou' wetting the scales. MISCELLANEOUS QUESTIONS. 33: How to find the Specifick Gravities of Bodies. if the body, thus suspended under the scale, at one end df the balance, be first counterpoised in air by weights in the opposite scale, and then imnnersed in water, the equilibrium will be imme- diately destroyed ; then, if as much weight be put into the scale, to which the body is suspentled, as will restore the equilibrium, (without altering the weights in the opposite scale) that weight, which restores the equilibrium, will be equal to a quantity of water as big as the immersed body ; and if the weight of the body in air be divided by what it loses in water, the quotient will shew how much that body is heavier than its bulk of water. Thus, if a guin- ea suspended in air, be counterbalanced by 129 grains in the oppo- site scale, and then, upon being immersed in water, it becomes so much lighter as to require 7} grains to be put into the scale over it, to restore the equilibrium, it shews that a quantity of water, of equal bulk with the guinea, weighs 7 25 grains ; by which divide 129 (the weight of the»guinea in air) and the quotient will be 17-793 ; which shews that the guinea is 17*793 tiines as heavy as its bulk of water. Thus may any piece of gold be tried, by weighing it first in air, and then in water; and if, upon dividing the weight in air by the loss in water, the quotient comes out 17*793, the gold is good : If the quotient be 18, or between 18 and 19, the gold is very fine : but if it be less than 17, the gold is too much alloyed by being mix- ed with some other metal. ; If silver be tried in this manner and found to be 11 times as hea- vy as water, it is very fine : It it be 10^ times as heavy, it is stand- ard ; but if it be of any less weight compared with water, it is mix- ed with some lighter metal, such as tin, k,c. If a piece of brass, glass, lead, or silver, be immersed and sus- pended in different sorts of fluids, the different losses of weight therein will shew how much heavier it is than its bulk of the fluid ; that fluid being lightest, in which the immersed body loses least of its aerial weight. Common clear water, for common uses, is generally made a standard for comparing bodies by, whose gravity may be represent- ed by unity, or 1, or, in case great accuaracy be required, by 1*000, where 3 cyphers are annexed to give room to express the ratios of other gravities in larger numbeis in the table. In doiuj-^ this there is a twofold advantage ; the first is, that, by this mean the soecifick gravities of bodies may be expressed to a much greater degree of accuracy. The second is, that the numbers of the Ta- ble, considered as whole numbers, do also express the ounces Avoirdupois contained in a cubick foot of every sort of matter there- in specified; because a cubick foot of common water, is found by experiment to weigh very nearly 1000 ounces Avoirdupoi'5, or 62i. pounds. 388 MISCELLANEOUS QUESTIONS. A TABLE OE THE SPECIFICK GRAVMTIES OF SEVERAL SOLID AND FLUID BODIES ; WHERE THE SECOND COLUMN CONTAINS THEIR ABSOLUTE WEIGHT, AND THE THIRD THEIR RELATIVE WEIGHT, IN AVOIRDUPOIS OUNCES. A Cubick Foot of jAbso, I Rela. \ wt. I wt. A Cubick Foot of i Abso.l Rela. I wt. I wt. Platina rendered mal- ) leable and hammered ^ Very fine Gold - - Standard G old - - Guinea Gold - - _ Moidore Gold - - - Quicksilver - - - - Lead Fine Silver - _ _ Standard Silver - - Rose Copper - - - Copper - - - - - Plate Brass - - - - Steel Cast Brass - - - - Iron ------ Block Tin - - - - Cast Iron - - - - Lead Ore - - - - Copper Ore - - - t)iamond - - - - Crystal Glass - - - White Marble - - - Black Marble - - - Rock Crystal - - - Gi-een Glass - - _ Clear Glass - ^ - 20170 20-170 fFlint - - c^ Pavinof - St^^^i Cornelian (^Free - - 19637 18888 17793 17140 13600 11325 11087 10535 9000 8843 8000 785 7850 7645 7321 7135 6800 3776 3400 3150 2707 2704 2658 2620 2600 2582 2570 2568 2352 19-637 18-888 17-793 17-140 13-600 11-325 11-087 10-535 9-000 8-843 8-000 7-852 7-850 7-645 7-321 7-135 6-800 3-775 3-400 3-150 2-707 2-704 2-658 2-620 2-600 2-582 2-570 2-56 2-352 Brick Liver Sulphur - - Nitre Alabaster - - - - Dry Ivory - - - - Brimstone - - - Solid subs, of Gun Pow, Alum ----- Ebony - - - - Human Blood - - - Amber - - - - Cow's Milk - - - Sea Water - - - Pure Water - - - Red Wine - - - Oil of Amber - - - Ptoof Spirits - - - Dry Oak - - - - Olive Oil - - - - Loose Gun Powder - Spirits of Turpentine x\lcohol or Pure Spirit Elm and Ash - - - Oil of Turpentine - - Dry Crab Tree - - ^ther White Pine - - - Sassafras Wood - - Cork - - - - - Common Air - - - Inflammable Air - - 2000 2000 1900 1875 1825 1800 1745 1714 1117 1054 1030 1030 1030 lOOO 993 978 925 925 913 872 864 850 800 772 765 732 569 482 240 2-000 2-000 1-900 1-875 1-825 1-800 1-745 1-714 1-117 1-054 1-030 1-030 1-030 I -000 0-993 0-978 0n)25 0-925 0-913 0-872 0-864 0-850 0-800 0-772 0-765 0-732 0-569 0-482 0-240 0-00125 0-00012 The use of the Table of Specifick Gravities will best appear by several Examples. How to discover the quantity of adulteration in metals. Suppose a body be compounded of gohl and silver, and it be reqriied to find the quantity of each nietaj in the compound. First, find the Specifick gravity of the compound, by weighing- it in air and in water, and divivling its aerial weight by what it loses thereof in water, and the quotient will shew its specifick gravity, or how many times heavier it is than its bulk of water. Then, sub- tract the specifick gravity of silver (found in the Table) from that of the compound, and the specifick gravity of the compound from that of the gold : the first remainder will j.hew the b-.jlk of gold, and the latter, the bulk of silver in the whole compound; and if these remainders be multiplied by the respective specifick gravi- ties, the products will shew the proportional weights of each me*.« al in the body. MISCELLANEOUS QUESTIONS. 389 Suppose the specifick gravity of the confipounded body be 14 ; that of standard silver (by the Table) is 10 535, and that of standard gold 18-888 ; therefore, 10-635 from 14, remains 3465, the pro- portional bulk of the gold in the compound ; and 14 from 18-888, remains 4-888, the proportional iw/A; of silver in the compound: then, 18-888, the specifick gravity of gold, multiplied by the first remainder 3 465, produces 65-447 for the proportional weight of gold ; and 10 535, the specifick gravity of silver, multiplied by the last remainder, produces 51-495 for the proportional weight of sil- ver in the vv^hole body : So that for every 65 447 ounces or pounds of gold, there are 51-495 ounces or pounds of silver in the body. Hence it is easy to know whether any suspected metal be genu- ine, or alloyed or counterfeit, by finding how much heavier it is than its bulkof water, and comparing the same with the Table ; if they agree, the metal is good ; if they differ, it is alloyed or counterfeited. How to try Sjiirilous Liquors. A cubick inch of good brandy, rum, or other proofspirits, weighs 234 grains ; therefore if a true inch cube of any metal weighs 234 grains less in spirits than in air, it shews the spirits are proof: If it lose less of its aerial wei 850/75 V equal measures. 109. What proportional weight of rectified spirits of wine and water must be mixed, to make proof spirit, the specifick gravities as before ? 1000 20 Ans. = — , or as 20 to 17. 850 17 110. What is the specifick gravity of best French brandy, con- gisting of 5 parts, measure, of rectified spirits of wine, and 3 parts watpr ? 390 MISCELLANEOUS QUESTIONS. 850x5=4250 1000x3=3000 64-3= 8 ) 7250 906'25=specirick gravity. 111. A retailer has 30 gallons of rum, whose speciiick gravity is 900 t How much water must he add to reduce it to standard proof? fjp- ( 1000 "NSB } ' g. rum. g. wat. g. rum. g. wat. "^^, 900/75^ As 75 : 25 :: 30 ; 10 to be added. 112. The cubick inch of common glass weighs about l-36oz. Troy : ditto of salt water •5427oz. ditto of brandy •48927oz. Sup- pose then, a seaman has a gallon of brandy in a bottle, which weighs 4it{> Troy, out of water, and to conceal it, throws it over- board into salt water : Pray, will it sink or swim, and by how much is it heavier or lighter than the same bulk of salt water? 54 4i}fe=54oz.— weight of bottle — - =39-7059 cub. in. in the bottle. 1 36 Add 231" =do. in the brandy. 270.7059=ditto in both. Then, 270 7059X'5427=146 912oz.=weight of salt water occu- pied by the bottle and brandy. And -48927 (=weight of a cu- bick inch of brandy) x231=l 1302+oz and 11302+54=16702oz. =weight of the bottle and brandy. From this take the weight of the salt water, viz. 146192oz. Ans. Supposing the bottle full, it is 20II0Z. heavier than the same bulk of salt water, and therefore will sink. Given the weight to be raised by a balloon^ to find its diameter. Rule. 1. As the sprcific difference between common and inflammable air, is to one oubick foot : so is any weight to be raised, to the cu- bick feet contained in the balloon. 2. Divide the cubick feet by '5236, and the cube root of the quotient will be the diameter required, to balance it with common air; but, to raise it, the diameter must be somewhat greater, or the weight somewhat less. 113. I would construct a spherical balloon, of sufficient capacity io ascend with 4 persons, weighing, one with another, 160ife, and the balloon and a bag of sand weighing 601fe : Required the diame- ter of the balloon ? By the Table of Specifick Gfravities, page 388, I find a cubick foot of common air weighs 1 25 ounces Avoirdupois, and a cubick foot of inflammable air '12 of an ounce Avoirdupois ; therefore. MISCELLANEOUS QUESTIONS. 391 Ife ife ft oz. 1 25— IS^MSoz. difference. And 160x4+60=700=11200. oz. cub. foot. oz. cub. feet 3 9911-5044 As M3 : 1 ;: 11200 : 9911-5044. And V —=26. 65 feet, •5236 [Ans. Given the diameter of a balloon^ to find what weight it is capable of raising. Rule. 1. Multiply the cube of the diameter by -5236, and the product will be the content in cubic feet. 2. As one cubick foot is to the specifick difference between com- mon and inflammable air; so is the content of the balloon to the weight it will raise. 114. The diameter of a balloon is 26 65 feet: What weight is it capable of raising ? 26-65x26-65X26-65X-5236i=9911-4-f cubick Ceet. And cub. foot. oz. cub. feet. oz. As 1 : 1-13 :: 9911 4-|- : 1 1199882=7001fe nearly. If the magnitude of any body be multiplied by its specifick gra- vity, the product will be its abi^olute weight. 115. What weight of lead will cover a house, the area of whose roof is 6000 feet, and the thickness of the lead yi^ of a foot ? 6000Xj^jf=50 cub. feet, and its specifick gravity 11325X60= tons. cwt. qrs. lbs. oz. 566250 ounces=15 15 3 26 10 Ans. To find the magnitude of any thing Tvhen the weight is known. Divide the weight by the specifick gravity in the Table, and the quotient will be the magnitude sought. 116. What is the magnitude of several fragments of clear glass, whose weight is 13 ounces ? 13-r-2600=-005 of a cubick foot, and -005X1728=8.640 cubick ! inches, Ans. Hating the magnitude and weight of any body given^ to find its spe- cifick gravity. Divide the weight by the magnitude, and the quotient will be the specifick gravity. 117. Suppose a piece of marble contains 8 cubick feet, and weighs 1353ilfe or 21656 ounces : What is the specifick gravity ? i 21656~-8=2707 the specifick gravity required, as by the Table, ^, To find the quantity of pressure against the sluice or bank, which pens ^' water. Multiply the area of the sluice, under water, by the depth of the :entre of gravity, (which is equal to half the depth of the water) iu feet; and that product again by H.^} ftho number of pounds Avoir- 392 MISCELLANEOUS qUESTlONS. dupois in a cubick foot of fresh water) or by 6 4-4lfe (the'Avoiruu- pois weight of a cuhlck foot of salt water) and the pr'oduct will be th*e number of pounds required. 118. Suppose the length of a sluice or floom be 30 feet, the width at bottom 4 feet, and the depth of the water 4 feet; what is the pressure against the side of the sluice ? 30X4=120 feet the area of the bottom, and 120X2 (the depth of the centre of gravit}') gives 240 cubick feet, and 240X62-5= 15000ife=6T. 13cwt. 3qrs. 20Jfe Ans. The perpendicular pressure ofjluids on the bottoms of vessels is estimat- ed by the area of the bottom multiplied by the altitude of the fluid. 119. Suppose a vessel 3 feet wide, 6 feet long, and 4 feet high, what is the pressure on the bottom, it being filled with water to the brim? 3X3=15 square feet, the area of the bottom, and 16X4=60 cubick feet, and 60X62-5=3750Ifei=t33 cwt. 1 qr. 261b. THE USE OF THE BAROMETER. The Barometer is so formed, that a column of quicksilver is sup- ported within it to such a height as to counterbalance the weight of a column of air, of an equal diameter, extending from the barome- ter to the top of the atmosphere. 120. At the surface of the earth, the height of this column of quicksilver is, at an average, almost 30 inches ; when the barom- eter is at that height; what is the pressure of atmosphere on a square foot, and on the surface of a man's body, estimated at 14 square feet? As the cubick foot of quicksilver is 13600 ounces. Avoirdupois, and as the height in the barometer, is 2'5 feet, therefore 13600X25 =34000 ounces, =2125 pounds on a square foot ; and 2125X14=^ 29750 pounds on a man's body. 121. If the mercury in a barometer, at the bottom of a tower, be observed to stand at 30 inches, and, on being carried to the lop of it) be observed at 29*9 inches : What is the height of the tower ? Divide 13600, the specifick gravity of quicksilver, by 1*25, the specitick gravity of air, and the quotient will be ti)e height of the tower, in tenths of an inch. 13600 10880 « =10880 tenths, and =1088 inch.=90| feet Ans, 1-25 10 The number of feet, in height, of the atmosphere, corresponding with yV of an inch on the barometer is variable, depending oo the temperature and density of the atmosphere. The variation, depending on the temperature, is shewn in the fol- lowing Table, calculated for every 5 degrees, from 32 to 80, Fahren- heit's Thermometer, from whence it may be easily calculated, for the intermediate degree*) by allowing f „^^ of a foot/or each degree. MISCELLANEOUS QUESTIONS. 393 TABLE. Thermo. Feet. 32" 86- 8G 35 87^49 40 88-54 45 89-60 50 90-66 65 91-72 60 92-771 65 93-82 70 94-88 76 , 95 93 80 96-99 The altitude, thus found, will be to the altitude cor- rected for the density of the air, inversely, as the mean height of the barometer, at the two stations, is to 30 inches ; therefore, Rule. — Multiply the mean height corresponding to the mean temperature ef the two barometers (found in the Table) by the tenths of an inch in the difference of the two barometers, and this product by 30 ; divide this last product by the mean height of the two barometers, and the quotient will be the an- swer, or height required, with the errour of a few feet only, if the height be less than a mile.* 122. At the first station, suppose the barometer to stand at 29, and the thermometer at 60 ; at the second station, the barometer at 28, and the thermometer at 40 : What is the height of the 2d station or the distance between the two places of observation ? Barometer. First station =29 Second station =28 Add i sum==28'5=mean height of the two barometers. 29 28 Difference= 1 = 10 tenths of an inch. Thermometer. First station =60 Second station =40 1)100 60=mean height of the two thermometers, against which, in the Table you will find 90-66, the mean temperature of the two barometers. Now, according to the rule 90 66x 10x30~ 28'5==954-3feet, the Answer, nearly. * Let^ = mean height of the barometer at its two stations, (or of two barom- eters, one at each station) in inches ; d = difference of the two barometers in tenths of an inch ; and n =z number from the Table answering to the mean tem- 30dn perature of the two thermometers accompanying.the barometer, then = the h. altitude required nearly. B 3 394 OF THE VALUE OF COINS. The Act of Congress of April 29, 1816, regulating the cnrrency within the United States of the gold coins of Great Britain, France, ^c. enacted, That, of the gold coins of Great Britain and Portugal, 27 grs.=:100 cents, or 1 pwt.—80f cents ; Of France, 27igrs.= do. do. =87^ do. Spain, 28igrs.= do. do. =84 do. Crowns of France, weighing 449grs. = irOcenls, or loz. = 117c. Five franc pieces, weighing 386grs. =93-3 cts. or loz.=ir6cts. The Spanish dollar, weighing not less than 416grs.==100 cents. The Federal dollar is to be of the weight and purity of the Spanish dollar; but, it is to weigh 416grs. and to contain 371^grs. of pure silver. The pound Sterling of England is ^4*44 in the United States. The dollar is reckoned, therefore, at four shillings and six pence sterling; but in 1820, four shillings and six pence of English silver coin contained only 363^grs. of pure silver, being 8grs. less than ia contained in the Federal dollar. One pound Troy weight of Standard Gold in England, contains 6280grs. of pure gold, and is coined into £4Q 14s. 6d. or 11214 pence. Hence a pound sterling contains 1 13 0016grs. of pure gold. 'J'he Eagle contains 247-5grs. of pure gold, and hence the pound sterling is worth in gold ^4 56-572, and hence the dollar is worth in English gold 4s. 4-5656d. . One pound Troy of Standard silver in England contains 6328grs. of pure silver, and is coined into 66 shillings or 792 pence. As the dollar contains 37Jigrs. of pure silver, the dollar is worth in Eng- lish silver 4s. 71858d. In a pound sterling there is 1614 545grs. of pure silver, which is equal in silver to §4-34 8943. Taking the mean of the values of the dollar and the pound ster- ling in gold and silver, The value of the dollar is 4s 5-8757d. sterling. And the value of the pound sterling is g4-45-7331. This mean value of the dollar and the pound sterling is very near the values at which they are commonly estimated. [See ^' Report of the Secretary of Slate upon Weights and Meas- ures," appendix C. to Congress, Feb. 1821.] The standard price of gold in England, is £3 17s. 10|d. an oz. and of silver 5s. 2d. an oz. The standard weight of the English jguinea is 5pwt. 9|grs. ; but it usually weighs 5pwt. 8grs. The standard coin of France is to contain one tenth of alloy, and the standard value of gold to silver is 15 to 1. TABLES. 395 The following Tables are calculated according to the value of Foreign Gold established by act of Congress, April 29, 1816. TABLE I. TABLE II. V^alue of LCngl] sh and 1- ortuguese Gold, in dollars, cents, and mills, throug:hout the United States. | Gr. Ctg, m. Pwts. $ cts. 1 3 7 1 881 2 7 4 2 1 77f 3 11 1 3 2 66| 4 14 8 4 3 55 5 181 5 4 44 6 22 2 6 5 331 7 25 9 7 6 32 8 29 6 8 7 11 9 331. 9 8 10 37' 10 8 89 11 40 7 11 9 77| 12 44 4 12 10 G6 13 48 1 13 11 551 14 51 8 14 12 44 15 55 5 15 13 331 16 591 16 14 22 17 63^ 17 15 11 18 GGs. 18 16 19 70' 4 19 16 89 20 74 loz. 17 77| 21 771 J^ote. 88 8. cents, 22 8lt the value of 1 pen- 23 85^ 2 ny-weigh and Port it of Eng. ugs. Gold. Value of French Gold, in dolls, cents, and mills, in the United States. Gr. Cts. m. Pwt. $ Cts. 1 3 63. 1 87i 2 7 3 2 1 741 3 10 9 3 2 613 4 14 5/- 4 3 49 5 18 2 ' 5 4 361 6 21 8 6 5 231 7 25 4-6- 7 6 10| 8 29 r^ 8 6 98 9 32 7_3_ 9 7 851 10 36 3_7. 10 8 721 11 40 11 9 59| 12 43 6-^j 12 10 47 13 47 3 13 11 341 14 50 9 14 12 211 15 54 51 15 13 083 16 58 2 16 13 96 17 61 8 17 14 831 18 65 4-6^ 18 15 701 19 69 l'' 19 16 57| 20 72 7 loz. 17 45 OJ 76 3 ^ 22 80 Q Mte. 87i cents is the V alue of 1 23 83 6A pwt. of ] ^re. Gold. K) K) ^« .^ ^_ ^^ ^_. /^ so Ococo->3050T*^03^s»-ooc^<^0:OT*-oo^o^-7f C0--a*»0i-*Oi»-*cr'OCnOCr|M M|M 00|Ol I0|- Mj— *-|— r o CD *-» ^^ H- CO CO <{ oi en o N *^ 03tsi»— ococ5— ^ 12 60 13 44 14 28 15 12 15 96 16 80 ^1 Ci OOCOCO3coo^oo^cna)Coo toco4^oa>^^co►^oo5J^co4-J* 396 TABLES OF EXCHANGE. N. Hamp- New Jer- shire, New York sey, Penn- South Caro- Canada Federal Massachu- and sylvania, lina and and Nova- French. Money. setts,R.Isl- and, Con. North Ca- rolina. Delaware, and Mary- Georgia. scotia. feVirginia. land. *; c. X, s. d. Jj. s. d. £ s. d. £ s. d. £ £. d. ^ Sou. 00 1 ¥s II T^o if 1 iii 00 2 m m lA h% n 3 2oV 22.2. ^25 2t\ n^ H ■^ 3^^. 00 4 2|2 m 3t\ 2/. 22 4^ 0-0 6 3| 4 f 4fV 2| 3 5 I 0-0 6 4^ m ^j\ 3/j ^ 6^ 00 7 ^oV m 6f^. m H ^.^ 0-0 8 ^M m 7^ m ^ s ! 0-0 r m m ^tV Ni 5! 9^\ 1 ^ i ^1 9 ^ f 6 1^ i • , 1 ■ 1 2 2 1 n 2 41 1 6 1 -4 1 1 10 2 1 1 1 0-3 1 9| 2 3 1 6 1 H i 0-4 2 4 4 3 2l 3 2 2 2 0-5 C 3 ' 4 O' 3 9 2 4 ' 2 G 2 12 1 Ob 3 7 1 4 9| 4 6 2 9^ 3 3 3 07 4 2 f 6 7;l 5 3 3 3 i 3 6 3 13 X 4 4 8 C 4 9 1 6 4f 6 3 8 i 4 2 1 4 8 4 \ 0-9 6 4 A 7 2| 6 9 4 6 4 14 1 \ 10 6 8 7 6 5 5 5 2-0 3-0 4-0 12 16 15 9 4 10 10 10 18 1 4 12 6 14 15 15 15 14 1 12 1 10 18 8 1 21 rrO C 1 10 2 1 17 6 1 3 4 1 8 1 5 26 5 6-0 1 16 2 8 2 5 1 10 31 10 7-0 C 2 2 2 16 2 12 6 1 12 8 1 15 36 15 8-0 2 8 3 4 3 1 17 4 2 42 9-0 2 14 3 12 3 7 6 2 2 2 5.0 47 5 10-0 3 4 3 15 2 6 8 . 2 10 52 10 20-0 6 8 7 10 4 13 4 5 105 30-0 9 0.0 12 11 5 7 7 10 157 10 40-0 12 16 15 9 6 8 10 210 50-0 GOO 15 18 20 18 15 11 13 4 14 12 10 262 10 24 22 10 15 315 70-0 ( 21 28 2G 5 16 6 8 17 10 367 10 r { bou. Tour. J 111 1 0-01/, H n i f 0* ""^""^ 02^ 2| 21 H H n\ '3 ' 0-041 4 3| n n 4 1 4 05| S-^ 5 31 3 -' i 6 006|i 4 Qi H 3=^ ^l 6 0-081 8 n H 41 ^ 1 7 009-1-1 9i ^ H H lOi^ 8 O-lli lOf 10 H 6 11 % 9 0121 1 l^l 7 63 3 13 i 10 o-isf 1 1^ 101 n ^■i- 11 0-15.{ 1 2-1 1 1^ n ^1 12 1 0-I6| 1 4 1 3 n 9' 24 17 -J 20 0-33I 2 8 2 6 1 6^ 1 6 2 1 15 30 0-60 4 3 9 2 4 23 2 12 -I 40 0-661 5 4 50 3 n 30 2 3 10 60 83i 6 8 6 3 3 lO-a 3 9 4 7 1 60 1-00 8 7 6 4 8 4 6 2 5 5 7 M6.! 94 8 9 5 H 53 6 2 1 8 l-33i 10 8 10 6 2i 60 7 9 1-50 12 11 3 7 o' 6 9 7 17 ^ 100 1-66^ 13 4 12 6 7 n 7 6 8 15 " 1 '3-33^, 1 6 8 1 50 !5 H 15 17 10 200 , 6 -661 2 13 4 2 10 1 11 n 1 10 35 300 10-00 4 3 15 2 6 8 2 50 52 10 400 13-331 5 6 8 5 00 3 2 25 3 00 70 500 16-66| 6 13 4 6 50 3 17 n 3 15 87 10 6 00 2000 8 7 10 4 13 4 4 10 105 7 00 23 331 9 6 8 8 15 5 8 106 5 50 122 10 8 00 26-662 10 13 4 10 6 4 H 6 00 140 900 3000' 12 1150 7 0' 6 15 157 15 1 10 33-331 13 6 8 12 10 7 15 6| 7 10 175 20 -G6G61 26 13 4 25 15 11 H 15 00 350 30 100 00 40 37 10 23 6 8 22 100 525 ; 40 133-331 53 6 8 50 31 2 Q6 30 00 700 50 166661 GG 13 4 62 10 38 17 ^ 37 10 875 CO 200 00 80 75 46 13 4 45 1050 j 70 233-331 93 6 8 87 10 54 8 10^ 52 10 1225 80 266 66| 106 13 4 100 00 62 4 H 60 00 1400 ] 90 300-00"' 120 00 112 100 70 67 10 1575 ! 100 333-33J 133 6 8 125 77 15 6^1 75 1750 ;200 C 666 66-;;- 266 13 4 1250 155 11 H Ii50 00 3500 ;300 100000 400 375 233 6 8 1225 00 5250 iOO C 1333-33 533 6 8 500 311 2 25 300 00 7000 500 C 1666-66 666 13 4 1625 '388 17 9i^ 375 8760 598 TABLES. TABLE ©F THE VALUE OP SEVERAL PIECES OF COIN, IN THE FEDERAL COIN, AND THE SEVERAL CURRENCIES OF THE UNITED STATES. N. Hamp. N. Jersey, Federal Mass. New York Pennsylva- S. Carolina Coin. R. Island, and North nia, Dela- and Con. and Carolina. ware and Georgia. Virginia. Maryland. £ s. d. Cents. £ s. d. £ s. d. J^ s. d. J^ofa Dollar 0-06 1 n 6 5f ^ iofa Pistareen 0-10 Vir. 8 9| 9 H iofa Dollar 0-11 1 8 10| 10 6f i of ditto 012 1 9 1 iH 7 A Pistareen 020 1 2| 1 6 1 6 Hi An Eng. Shilling 022 1 1 4 1 7f 1 8 1 Of 1 of a Dollar 0-25 1 6 2 1 101 1 2 Half ditto 0-50 3 4 3 9' 2 4 A Dollar 100 6 8 7 6 4 8 French Crown 110 6 7 8 9-} 8 3 5 U pwt. gr. ! fr. Guineas 6 4-64 I I 7 31 1 16 n 1 14 1 1 21 En.Guinea 6 6 4-66 f 1 8 1 17 4 1 15 1 1 31 1 Johann. 9 8-00 2 8 3 4 3 1 17 4 Pistole 4 5 3-53 111 1 2i 1 8 3 1 6 6 16 5 Moidore 6 18 600 1 16 2 8 2 5 1 8 Donbloonl7 14 66 1 4 8 5 16 5 12 3 10 Blanks 24 == 1 Perrot. 480 = 20 == 1 Mite. 9600 — 400= 20 = 1 Gram. Table of Refiner's Weights. J>foie. What they denominate a carat is the ^'^ of a !fe an oz. or any other weight. Dutch Weights for Gold and Silver. Note, 32aces=l engel, 20 engels=l ounce, 8 ounces=lmark, for jrross gold. Also, 24 parts=l grain, 12 grain8=l carat, 24 farats=l mark, for fine gold. The mark weights are 1 per cent, lighter than our Troy weight. TABLES. A TABLE OF COMMISSION OR BROKERAGE. 399 Goods or stock sold Shut. 1 2 . . 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 \Pounds 1 9 10 20 30 40 60 60 70 80 90 100 200 300 400 500 600 700 800 900 1000 at i per cent. s. d. 0, 04 04 Oi Of oi oi ol oi oi oh 01 Of Of 0| 1 81 91 lOf 1 2 3 4 5 6 7 8 9 10 1 1 10 2 2 10 3 3 10 4 4 10 5 at 1 per | c ent. Oi 04 0^ 0^ 01 0| 1 1 n u M u 11 11 2 2 2A n 2h 5 7 9i 1 1 2i 1 41 1 7 1 9^ 2 4 6 8 10 12 14 16 18 1 2 3 4 5 6 7 8 9 10 at H per cent. 1, 04 Oh Oi Oj 1 u u H 1* II 2 24 2i 2^ 2^ 3 3 3i 34 7i lOf at 2 per cent. 12 18 1 1 1 4 1 7 1 10 3 4 10 6 7 10 9 10 10 12 13 10 15 1 1 1 1 1 2 4 6 8 10 12 14 16 18 20 04 0^ Of 1 14 u 1| 2 24 2* 3 3i 3h 31 4 4i 4i 4| 9^ 1 1 2 2 2 3 3 4 8 12 16 4 8 12 16 at^ li per 1 cent. j 04 0^ 0| 1 n u 2 24 2h 3 3^ 3i 31 4 4i 4i 5 54 5i 06 1 5 7 10 12 15 17 20 22 25 1 6 2 26 30 36 40 46 50 10 15 00 5 10 15 00 5 10 00 10 00 10 00 10 00 10 00 at 3 pe" cent. 1 1 1 2 2 2 3 6 9 12 15 18 21 24 27 30 Oh 1 l4 II 2 24 2| 3 3^ 31 44 4i 5 54 5I 6 64 6^ 7 1 2-2 1 9h 2 41 3 3 7 4 21 4 91 6 4i 6 0*^ 12 18 40 10 16 2 80 14 00 00 00 00 00 00 00 00 00 00 400 TABLES. A TABLE / THE RETURNS OF THE NEAT PROCEEDS OF AN ACCOrNT OF SALES /FROM A FACTOR TO HIS EMPLOYER, RESERVING HIS COMMISSIONS FOR / REMITTANCE. Sum to be Sum to be Sum tobel Sum to be! NeatP re- remitted, remitted, Neat pro ceeds. remitted, remitted, ceedi: reserving reserving " reserving 2i|reservmg 5j 2i per ct. 5 per ceni per ct. com - per ct. com- commisn. commisn. mission. mission. £ s. el. JC s. d. £ s. d. £ s.d. £ s. d. £ s. d. 3 3 21 6 5 17 Of 5 14 5i- 4 4 31 7 6 16 7 6 13 4 j 5 5 4f 8 7 16 H 7 12 4i 6 51 51 9 8 15 7i 8 11 5i 7 61 6i 10 9 15 U 9 10 51 3 8 71 7i 20 19 10 3 19 lU 1 9 81 8i 30 29 5 4^ 28 11 5i \ 10 91 9^ 40 39 5^ 38 1 m \ 11 101 101 50 48 15 7i 47 12 4i 1 111 lU 60 58 10 8| 57 2 lOi 2 1 lU 1 101 70 68 5 10 66 13 4 3 2 iH 2 m 80 78 Ul 76 3 9| 4 3 10| 3" 91 90 87 16 1 85 14 3i 5 4 1O.5 4 9^ 100 97 11 21 95 4 9 6 3 104 5 8i 200 195 2 51 190 9 6i 7 6 10 6 8 300 292 13 8 285 14 3i 8 7 91 7 7i 400 390 4 10^ 380 19 Oi 9 8 9i 8 6| 500 487 16 H 476 3 94 10 9 9 9 6i 600 585 7 31 571 8 6| 1 19 6i 19 0^ 700 682 18 4| 666 13 4 2 1 19 Oi 1 18 H 800 780 9 9 761 18 1 ^ 2 18 6'^ 2 17 1| 900 878 111 857 2 10 4 3 18 Oi 3 16 2i 1000 975 12 2i 952 7 7i 5 4 17 6^ 4 15 21 Suppose I have the neat proceeds, or balance of an account of sales 3251. 17s. 9d. in my hands and would make remittance to my employer, reserving my commission at 2i per cent. What sum must be remitted, so that my employer's account may be closed ? Against. £ s. d. fSOO 0~] 20 5 10 7 9 £ c. d. 292 13 8 19 10 3 4 17 61 9 9 6 10 8| S stands. To be remitted £317 18 9i Answer. TABLES. A TABLE, 40i WiEWliVG THE arUMBER OF DAYS FROM ANT DAT IN ANY MONTH TO THE S13IE DAY ly ANY OTHER MONTH THROUGH THE YEAR. From Jan Feb Mar Ap May Juu Jul Aug Sep Oot Nov Dec| To Jan. 365 "3! 59 "90 120 131 181 212 243 273 304 334 334 365 T8 '59 Tl' 120 150 181 212 242 273 303 306 337 365 3] 61 92 122 153 184 214 245 275 306 334 365 30 61 9\ ^ 153 183 214 24^ 245 276 304 335 365 Tl 6l '92 123 153 214 214 24f: 304 334 365 30 ~6i — 92 122 153 183 18. 215 243 274 .304 335 36^ 31 ~62 ~92 123 153 153 184 ,212 243 273 304 334 3^:5 31 61 T2 122 122 \53 iFi 212 242 273 303 334 365 30 61 91 9,i l23 151 182 2r2 243 273 304 335 365 31 61 6i 92 120 151 181 2\2 242 273 304 334 365 30 31 62 "90 1211 151 182 2r2 243 274 304 335 365 Feb. Mar, Apr. May. June. July. Aug. Sept. Oct. Nov. Dec. The Use of the preceding Table of number of days, will easily ippear from the following examples. Suppose the number of da^s between the first, or 10th, or 30th, &c. of January, and the 1st, or 10th or 30th, &c. of October, were required : Look in the column under January for October, and I against that month you will find 2^73, which is the number of days ^ between the said times; and so for the days between any other two months. If the given days be different, it is only adding or subtracting their inequality to or from the tabular number. How many days from the 6th of April to the 12th of January ? From the 6th of April to the 6th of January is 275, and adding the 6 overplus days, it makes 281 days. And from the 6th of June to the 1st of February is 240 days. Note. After February 31 (in leap years) increase each num- ber with an unit or 1. rj o, U)^ TABLES. A Table of the Measure of Length of the principal places in Europe, compared txith the American yard. 100 Aunes or Ells of England, - , - - = 125 100 of HoUand or Amsterdam, Haerlem, Leyden^ the Hague, Rotterdam, Nuremberg, and > = 75 other cities of Holland, 7 100 ~ of Barbant, or Antwerp, - - - =-76 100 of France and Oznaburg, - - - = 1284- 100 ofHamburg, Frankfort, Leipsick, Bern, & Basil, = 62^^ 100 ofBreslau, =60" 100 ofDanlzick, ,....= 66f 100 of Bergen and Drontheim, , . _ = 68i 100 of Sweden and Stockholm, . - - = 65| 100 of St. Gall, for linens, . - - - = 87| 100 of ditto for cloths, - - - . = 67 100 of Geneva, =124| 100 .Canes of Marseilles and Montpelter, - - =214^ 100 ofThoulouse and High Languedoc, - - = 200" 100 of Genoa, of 9 palms, -.-.=: 245i 100 of Home, - = 227i 100 Varas of Spain, - - - - . - = 93i 100 of Portugal, - - , . - - =123 100 Cavidos of Portugal, =75 100 Brasses of Venice, ._«--= 73^ 100 of Bergamo, - - - - - = 7li 100 of Florence and Leghorn, - " ' =64 100 of Milan, ..-..-= 58^ The use of the follomng Table , directiitg^ how to buy and sell by the hundred. If you buy or sell any thing by the great hundred (llSlfe) and desire to know, .by the fe, what the hundred is valued at, observe the following examples. 1. If you buy sugar at GU. per ife, look for 6ld. in the left hand column of the Table, against it in the second column, you will find £3 3s. which is the value of Icwt. at that rate. 2. If Icwt. (112Jfe) cost £9 4s. 4d. to know how much it is pet* ife, look £9 4s. 4d. in the tburth column, and against it in the next^ -left hand column, you will find Is. 7ld. which is the price per S5. Again, If you buy one hundred weight of goods for 91. 4s. 4d.| and retail it at Is. 9|d. per ife, it comes at that rate to 101. 3s. ; thei^ take 91. 4s. 4d. from 101. .3s. am!, by the remainder, you will findj that you have gained 18s. 8d. And in this manner you may, with ease, cilculate any quantityl by the following Table. TABLES. 40? A ^ABLE DIRECTmc HOW TO BUY AND SELI, BY THE HUNDRED. d. £ s. d. s. d. i; s. d. s. d. £ s. d. i 2 4 1 Oi 6 14 4 2 01 116 4 ^ 4 8 1 01 5 )6 8 2 01 11 8 8 f 7 1 0| 5 19 2 01 11 11 1 H 9 4 1 1 6 1 4 2 1 2 11 11 13 4 11 8 1 ^i 6 3 ( 11 15 8 n 14 1 1^ 6 6 2 ]i 11 18 H IG 4 1 If 6 8 4 2 H 12 4 2 18 8 1 9 6 10 8 2 2 12 2 8 ^i 1 1 iii G 13 2 21 12 5 n 3 4 1 21 6 15 4 2 21 12 7 4 21 5 8 1 21 6 17 8 2 21 12 9 8 3 8 1 3 7 2 3 12 12 H 10 4 ' 31 7 2 4 2 31 12 14 4 31 12 8 1 31 7 4 8 2 3 . 12 16 8 31 ^ 15 1 31 7 7 2 3| 12 19 4 1 17 4 .1 4 7 9 4 2 4 13 1 4 44 19 8 1 44 7 11 8 2 44 13 3 8 4i 2 2 1 4i 7 14 2 41 13 6 4f 2 4 4 1 4^ 7 16 4 2 41 13 8 4 5 2 6 8 1 5 7 18 8 2 6 13 10 8 ■ ^} 2 9 1 r.! 8 1 2 51 13 13 5h 2 n 4 1 «^4 8 3 4 2 5l 13 15 4 5f 2 13 .8 1 51 8 5 8 2 5| 13 17 8 6 2 16 1 G 8 8 2 6 14 6i 2 18 4 1 C)i 8 10 4 2 61 14 2 4 4 3 8 1 61 8 12 8 2 61 14 4 8 6l 3 3 . 1 6l 8 15 2 61 14 7 7' 3 6 4 1 7 8 17 41 2 7 14 9 4 3 7 8 1 7i 8 19 8 2 71 ^^14 11 8 3 10 1 71 9 2 2 71 14 14 7_^ 3 12 4 1 71 9 4 4 2 71- 14 16 4 8* 3 14 8 1 8 9 6 8 2 8 14 18 8 ^T 3 17 1 ^4 9 9 2 81 15 1 8i 3 19 4 1 8l 9 114 2 81 15 3 4 S| 4 1 8 1 sl 9 13 8 2 8| 15 5 8 9* 4 4 1 9 9 16 2 9 15 8 ^T 4 t) 4 1 91 9 18 '1 2 91 15 10 4 91 4 8 8 ^ 91 10 8 2^ 91 15 12 8 9-? 4 11 1 9l 10- 3 2 9^ 15 15 10 4 13 4 1 10 10 5 4 2 10 15 17 4J ^^i 4 15 8 I 10^ 10 ^ l^ - 10 j 15 Id 8 lOi 4 18 1 101 10 10 2 lOJ 10 2 C lOf 5 4 1 10^ 10 12 A 2 101 ]G 4 4 11 5 2 8 1 11 10 14 C 2 11 2 11^ 16 6 q ^H 5 5 1 Hi 10 17 16 9 111 5 7 4 1 ni 10 19 4 2 111 16 11 4 111 5 9 8 1 111 il 18 2 111 ]6 13 8 12' 5 12 2 114 3 16 16 404 TABLES. A Comparison of the American Foot with the Feet of other Cininiries. The American Foot bei inches, the feet of several America, - - - - London, - - - - Antwerp, - - - - Bologna, - - - - Bremen, - - - - Cologne, - - - - . Copenhagen, - - - Amsterdam, - - - Dantzick, - - - - Dort, - - - - - Frankfort on the Main, The Greek, - - - Lorrain, - - • - - Mantua, Meckliri, - - - - . Middleburg, - - - France, - - - - - Prague, - . - . Rhyneland or Leyden, Kiga, ....-, Koman, Old Roman, - - - Scotch, - - - - . Strasburgh, - - . . Toledo, - - . - Turin, Venice, - . - - ng divided into 1000 parts, or into other countries will be as follow. , 12 Parts. 1000 1000 946 1204 964 954 965 942 944 1184 948 1007 958 1569 919 991 1066 1026 1033 1831 967 970 1005 920 899 1062 1162 t- - Inch. Dec. 12- 12' 11-352 14-448 11 568 11 448 11-580 11 304 ll-32a 14-208 11-376 12084 11-496 18828 11028 11-892 12 792 12-312 12-39G 21-972 1 1 -604 11-640 12 060 11040 10788 12-744 13-944 A Table representing the Conformity of the weights of the principal trading Cities of Europe with those of America. lb. 100 of England, Scotland and Ireland, 100 of Amsterdam, Faris, Bordeaux, k.c. loo of Antwerp, or Brabant, 100 of Rouen, the Viscount}', . - 100 of Lyons, the city, 100 of Hochelle, 100 of Thoulouse, and Upper Languedoc, 100 of Marseilles and Provence, 100 of Geneva, - - . - 100 of Hamburg, - . , . 100 of Frankfort, 100 of Leipsick, - - . - - - 100 of Bremen, - ' - - 100 of Russia, - - " 100 of Vienna and Trieste, of America. lOOlb. Ogz, 109 8 104 ^ 113 14 94 3 110 9 92 G 88 11 123 107 4 111 11 104 5 110 88 4 123 TABLES. 405.. 'A Table representwg the Conformity of the Weights of the -principai trading Cities of Europe with those of .America. lb, ' cf America. 100 of Genoa, . - - - - . = 73 100 of Leghorn, - - - - . = 77 4 100 of Milan, .,---- = G3 3 100 of Venice, - - - = 65 11 100 of Naples, . - - - - = ei 10 100 of Seville, Cadiz, &c. - - . = 103 2 100 of Portugal, = 77 4 100 of Liege, . - - - - = 104 loo of Spain, ---..- = 97 Note, The Spanish Arrobe is 25 Spanish pounds, = 24 4 A Table to cast up wages, orex-| A Table to iiud wages or expense--: penses , for a year, at so much| for a month, week or day, at ec' per da y, week, or month. week.^bi/ month j hy'ytar d. much by the year. > Dai/.\b]/ by vyrj by month. | by week. 1 A?.' ■ • j s. d. £ s. d. jC. s. d. jt:. s. £. JC. s. d. £. s. d. f , 1 7 2 4 1 10 5 1 1 6.i 4' ■c c? : 2 1 2 4 8 3 10 2 3 01 C;-) i^ ! 3 1 9 7 4 11 3 '7^, 3 4 74 1 u i 1 4 2 4 9 4 6 1 8 4 6 n 1 C4, :H\ 5 2 11 11 8 7 12 1 ? 5 6 7 8 "12^; 1 ii )) .0 3:r 1 6 3 6 14 9 2 6 9 2 tjA vj 4 7 4 1 16 4 10 12 n 5" 7 10 9 2 8^, 4^ 8 4 8 18 8 12 3 4 8 12 3J, 3 0^ 5i 9 5 3 1 1 13 13 9 (T> 13 91 3 5^ 6 10 5 10 1 3 4 15 4 2 "7 10 11 15 4 m 3 10 7 -eh n 11 G 5 1 5 8 16 14 16 4 n 1 7 1 8 18 5 12 18 5 4 7i 8 2 14 2 16 36 10 -T 13 19 114 4 IH 8i 3 1 1 4 4 54 15 rr 14 1 1 51 5 41 94 4 I 8 5 12 73 D 15 1 3 0^ 5 9 J 9| 5 1 15 7 91 5 1 4 6 l.t \Gh 6 2 2 8 8 109 10 ■D 17 1 C 1 6 6i Hi 7 2 9 9 16 127 15 c s 18 1 7 7>^ 6 101 llf 8 2 16 114 146 19 1 9 11 7 St^ 1 0^ 3 3 12 12 164 5 "0 5- 20 30 40 1 10 2 6 8^ 0:1 r. 7 11 8 6 "4' 1 H 1 7| 10 U 3 3 10 17 14 15 8 182 10 200 15 3 1 4^ 15 2 2i 12 I 4 16 16 219 3 50 3 16 8^ 19 2^ 2 9 13 4 11 18 4 237 5 ■-< 60 4 12 0^ 3 0^ 3 3i 14 i 18 19 12 21 255 10 CO 70 80 5 7 C 2 41 9 6 10 10^ 8:1 3 10 4 4i 15 I) 5 5 273 15 10 5 5 12 19 22 8 23 16 t 292 310 5 03 90 Too 6 18 1 14 91 4 1li 17 7 13 5 \\i 4.i 5 5-1 18 r> 6 25 4 328 10 200 15 10 J 3 16 3i 10 lUi 19 6 13 2Q 12 346 15 300 23 3i 5 15 0:1 16 6U 20 7 28 365 c 400 30 13 8A 7 13 5 I 1 11 I 500 38 7 1-^ 9 11 9i 1 7 41 1 |1000|76 14 3 19 3 Gli2 14 9i] 40G PERPETUj\L almanack:. COCOCOCOCO. CCiCOCOCOCOCOCOC»COODC5COCDCOCO-3-J-3-^--->->^ D -, ^ ^ CO o -. g r. 3 S g^:H ^-^ ft) to ^ "^ ^"^ C — - UJ O (TJ P Ix! > ^ 3* < o cr w o ^ '^ ^ sq cr rs CD 03 3- 3 ^~ o etj s -» g ^' rs n -• o :i- ^- c- = o_ .^ — . as CT- C ^^ 3 e:^ CU- r^ ta 3 ^ D- S en t^ g- o CC 03 C O g 3 .' J^ (f rt) ^^^ CD to to CT CO -^ IN 2 ^ . ^ Oi to (~* ^ u 05 CO >:0 'II _ r*'^ * W to ^ ■^ -1 o oc 1 to ^ — 3 s (JX CO >— • k;^. c- a 5^ - . ^ N'"^ k^ fc.^ ^>f=^ ^ TO CO to 0- 1^ s * t3^ to to l-i ^^ <» o OC' a. 3 i CO Oj :« ^ KT' to S 1 CO 4i^ 1 H > CO C? CO CO OO CC CO CO CO 00 CO CO CO CO CO 4- CO CO u* c^ u, CO U) w U) CJ ^s io *a t3 O <^ CO --1 O tn tJ-^ Oi tS >— O CO CO ~^ O) t4 ^ f _ K •^ "!" "^ CO CO CO CO CO *o >^ *o W <0 H- t? S5 J1 TABLES, 40^ AT ABLE for reducing Troy wt. a' rABLE for reducing Avoirdudois wt, i to Avoirdupois. into Troy. TVT Av. Troyl Avoirdupois. Jv ! Troy, Jlvoir. 1 Troy. gr. drm. oz. lb. oz. dr. cH ib. oz. pw. gr. ib. lb. oz. pw. gr. 1 •04 1 1 1-55 i 13-67 1 1 2 11 16! 2 •07 2 2 3-11 2 3 20-51 2 2 5 3 8| -3 4 5 •11 •15 •18 3 4 5 3 4-66 4 6-22 ♦ 5 7-77 4 1 2 1 334 2 6 68 o 4 3 4 7 15 10 6 16 6 •22 6 6 9-32 3 3 10 02 5 6 18 8 7 •26 7 7 10-89 4 5 1336 6 7 3 10 8 •29 8 8 12-44 5 6 16-7 7 8 6 1 16 9 10 •33 •36 9 10 9 14 10 15-56 6 7 20 04 8 9 8 13 8 11 •40 11 22 1-09 7 8 23-38 9 10 11 5 12 •44 lb. 8 9 272 10 12 1 16 16 13 •47 1 13 2-65 9 10 606 . 20 24 3 13 8 14 15 16 •51 '55 •58 2 3 4 1 10 5-3 2 7 8 3 4 10-6 10 U 11 9-4 12 12 74 30 40 36 48 6 10 7 6 16 17 •62 5 4 1 13-25 12 13 1508 50 60 9 3 8 18 'G6 6 4 14 15-9 13 14 19-42 60 72 11 10 •69 7 5 12 2-56 14 15 22 76 70 85 16 16 20 21 •73 •77 8 9 6 9 5-21 7 6 7-86 15 17 2 1 80 97 2 13 8 22 •80 10 8 3 10-52 oz 90 109 4 10 23 •84 20 16 7 5-03 1 18 65 100 121 6 6 16 pw. 30 24 10 15-54 2 1 16 11 200 243 13 8 1 3 0-88 1-75 2.63 40 50 60 32 14 10-05 41 2 4-57 49 5 15-08 3 4 2 14 16-5 3 12 22 300 400 364 486 7 1 6 16 4 3-51 70 57 9 9-6 5 4 11 3 5 500 607 7 13 8 5 4-39 80 65 13 4-11 ■ 6 5 9 9 600 729 2 c 5-27 90 74 13-62 7 6 7 14-5 700 850 8 6 16 7 8 6-14 7-02 100 200 82 4 9-15 164 9 2-28 8 7 5 20 800 972 2 13 8 9 7-9 300 246 13 11-42 ' 9 8 4 1-5 900 1093 9 10 8-78 400 329 2 4-57 10 9 2 7 1000 1215 3 6 16 11 9-65 500 411 6 13-71 11 10 12 5 2000 2430 6 13 8 12 13 14 10-53 11-41 12-29 600 700 800 493 11 6-85 576 658 4 9-14 12 13 10 18 18 11 16 235 3000 4000 3645 4861 10 1 6 16 15 13-16 900 740 9 2-28 14 I 15 6 5000 6076 4 13 8 16 14-04 1000 822 13 11-42 15 I 1 13 10-5 16000 7291 8 17 14-92 2000 1645 11 6-84 I • 18 15-79 3000 2528 9 2-26 19 16-67 4000 |3291 6 13-68 iO« TABLES. I. TABLE OF THE MONEY OF COMxMERCIAL COUx^TRIES, WITH THEIR VALUE IN STERLING AND FEDERAL MONEY. MONEY is of two kinds, Real, and Imaginary, or coin and money of account. Real Money is (he coin ofa country, ivhose value is established at a certain amount. Lnagi'nary money is merely a denomination to express a certain sum» which is not represeiited by any coin, as a pound, a livre, a mill. The denominations of imaginary money are employed with or without those of real money in keeping accounts. UNITED STATES. Sterling:. £. s. d. Dolls. Cts: Eagle, gold coin, 2 6 0* 10 00 Half Eagle, do. 1 2 6 5 00 Quarter do. do. 11 3 2 50 Dollar, silver, 4 6 1 00 Half dollar, do. 2 3 50 Quarter doll. do. 1 H 25 Dime, do. ^ 10 Halfdime, do. 2tV 05 Cent, copper. OH 01 Half cent, do. Ot^^ 00-^ ENGLAND AND SCOTLAND. £• s. d. $ Cts. Dec. Pound} =20 shillings, 1 Q 4 44 -441- Slulling=12 pence, 1 2222f Fenny = 1 farthings, 1 1 01 -So/^ 9 pence, 9 16-661 6 pence, 6 11.11^ 4 pence or 1 groat, 4 07 40^^ 3 pence, 3 05 55| 2 pence, 2 03 70t} 6^ pence, h 12 50 3} pencef, H 06-25 Crown, silver. 5 1 11. n^ Guinea, gold. 1 1 4 66'66?r he English crown passe s in the U.S. at, |l."stcrling. t The pound was anciently a pound Troy of silver ; it now contains only one - third as much. The pound sterling in the time of William the Conqueror or 1066, A. D. was to the present pound sterling as 31 to 10. Articles were then about ten times cheaper than at present. So that his revenue of ^400,000 sterling, was equal to about J£ 12,400,000 sterling at present. TABLES. 40S IRELAND. Denominations are the same as in England, but sterling is to Irish mpne^ as 13 to 12, or £12 sterling=£13 Irish. £. s. d. D. c. dec. Pound=20 shillings, 18 5f^ 4 10-25f»JL Shiliing=12 pence, 11-j-V 20'6U— 13 pence, 10 22 22f 22s. 9 pence=Eng. guinea, 110 4 66-66^ 4s. IGJd. 4 6 1 00 PRANCE. £. s. d. D. c. dec. Livre=20 sols, K) 18-51|| Sol=12 deniers, Oi 0O-92if Denier, 0^\ 00 07y%-{- Crown or Ecu=:6 livres, 5 1 IMl^ Pistole=*lo livres, 8 4 1851814 Louis d'or=24 do. 10 4 44-44f Ecu of exchange=60 sols, 2 6 55.65 The livre or livre tournois is estinoated in the ? ^ ini United States, at 5 " ^"^ '^' NEW COINS. £. s. d. D. c. dec. Franc=fi livre tournois, lOj 18-75— Deciin=yV franc, l^V 01-871— Centim=y^7 franc, Oj\\ 00 18|— Five franc piece, silver, 4 2| 9375 — 2 francs, ' - 1 81 37-50— Crown of exchang=3 livres, 2 6 65 55| In the United States the five franc pieces are } ^ 93.30 estimated at y And the franc at 18-731- There are silver coins of the value of 1 franc, and off, i, and \ franc. The franc is to weigh 18yYoV g^^. composed of— of pure silver, and yV alloy. The gold coins are also to contain j\ of alloy. £. s. d. D. c. 20 francs, gold, 16 lOi 3 75 40 do. do. 1 13 9 7 60 SPAIN. Accounts are kept in money ofvellon or current dollars, and mo- aey of plate, or hard or plate dollars. £. s. d. , D. c. Dollar, plate=10 rials plate, 4 6 1 00 Doll, current or piastre=8 do. 3 7} 80 Pistanne=2 rials plate, lOf 20 Rial, plate=34 maravedies, 0| 10 Quarto=4 do. 5fj 01-1714. D 3 4iO TABLES Sterling, Ma rave di, 100 rials plate=188y\ rials vel. 100 rials vellon=53| plate, Rial vellon, 32 rials vellon= 17 rials plate. Ducat of Excbange=375 mar. Pistole— 36 rials plate, In the United States, the rial vellon is estimated at 5 cents, or half a rial plate, and equal to 20 to a Spanish Dollar. The lower denominations are different in some parts of Spain. Thus, accounts are sometimes kept in the following denominations at £ s. ^ a. $ c. 0/t\ 00-29^^ 2 6 10 0000 1 3 m 5 31-25 ^u 05-311 4 IHI 1 10-29,^ 16 . 2|. 3 60- £ s. d. $ c. Plate dollar=20 rials vellon, 4 6 1 00 Rial veUon=8^ quartos, 2tV 05 Quarto, OH 00-58|f BARCELONA, VALENCIA, SARAGOSSA. £ s. d. $ c. Plate dollar=37^ sols, 4 6 1 00 Livre=:20 sols, " 2 H 53i Sol=12 deniers, m 02| Denier, O/i OOf BILBOA. £ s. d. $ c. Plate dollar=i20 rials vellon, 4 6 1 0000 Rial vellon=-34 maravedies, ^h 05-00 Maravedie, OfiV 0014jf. ST. LUCAR. £ s. d. $ e. Plate dollar=ilO rials plate, 4 6 1 0000 Rial plate=»16 quartos, Q 5A 1000 Q,uarto, on 00-62^ PORTUGAL. £ s. d. $ c. Millrea=1000 reas==10testoons,0 5 n 1 26-00 Rea, OtV« 00-12,1 Vintin=20 reas, h\ 02-60'' Testoon=5 vintins=100reas, 6f 1260 Crusade of exchange =4 testoons, 2 3 60-00 New crusade=4| do. 2 H 60-00 Moidore=48 testoons, 1 7 6 00 -00 Joanese or I Johan.=64 testoons, 1 16 8 00-00 The millrea is 126 cents in the United States. TABLES, 411 HOLLAND. Sterling. Guilder or Flojln=20 stivers, Stiver=2 grotes, Grote=8 pennlngs, Penning, Shilling=6 stivers=12 grotes Pound Flemish=20s =6 guilders,0 10 Kix dolIar=2i guilders, 4 Ducat, 1 16 Sovereign, 1 7 ^ The florin is 40 cents in the United States. d. ^1 6 c. 4000 0200 01-00 00- 12 J 12-00 40-00 0000 00-00 00-00 HAMBURGH. £ s. Pound=20 shillings Flemish, Oil Fl. shil.=12 grotes or pence Fl. Grote orpenny=6 deniers, Mark banco=32 grotes or 2f shil. 1 ' Rix donar=3 marks, 4 Stiver or shilling lubs,- Ducat=6-* marks, 9 The Bank money of Hamburgh is superior to the currency ; the agiOf or rate, varies, and is sometimes 20 per cent, or more in favor of the Bank money. d. 3 6^ 6 6 H $ c. 2 60 12-600 01^ 33i 1 00 02^V 2 16| BREMEN. £ Rix dollar=72 grotes=2i marks, Grote, d. 0^ $ c. 76 01^,- ANTWERP. £ s. d. Guilder=3i shillings=:40 grotes, 1 9f Shilling=12 grotes, 6^f Grote, 0|^ Stiver=^2 grotes, l/« $ c. 40 12 01 02 VIENNA AND TRIESTE. Florin=60 cruitzers, Cruitzer=i=4 fenings, Batzen=4 cruitzers, Rix dollar: Livre=20 soldi, Specie dollar=30 batzen, Ducat=60 batzen, li florins, d. 4 HI H 61-85/^ $ 00-866— 03-46 1 2 m 09-87 03 7 07-4 The value of these denominations is the same generally through Austria, Swabia, Frauconia, Bohemia, Silesia, and Hungary. 412 TABLES. HANOVER AND SAXONY. Sterling. £s. d. $ c. Rix dolIar=24 groshen, 3 6 77j Grosh=12 fening, If 03if Gould orguilder=16groshen, 2 4 61-85^^ Ducat=4 guilders or goulds, 9 4 2 07-40 Specie dollar or hard dollar, 4 8 1 03-7 POLAND AND PRUSSIA. £ s. d. $ c. His dollar=90 groshen, 3 6 77f Florin =30 do. 0. 1 2 25^ Grosh=3 shelons, Oj\ 00-866- Ducat=8 florins, 9 4 2 07-4 Frederic d'or=5 rix dollars, 17 6 3 88f LIVONIA. £ s. d. $ c. Rix dollar==90 groshen, 3 6 77f Florin =30 do. 12 25ff Marc = 6 do. 2^ 05/^ Grosh=6 blackens, 0/j 00-866"- RUSSIA. JE s. d. $ c. Ruble=100 copec5, 04 6 100 Poltin= 60 do. 2 3 60 Copec=4 poluscas, 0|4 01 2ervenitz=2 rubles, 9 2 00 DENMARK AND NORWAY. £ s, a. $ c. Rix dollar=6 marks, 4 6 1 00 Crown =4 do. 3 66| Marc=16skilling8, 9 16f Skilling, OyO^ 01 «f Ducat=^ll marcs, , 8 3 1 83| SWEDEN AND LAPLAND. £ s. d. $ c. Rix dollar=3 silver dollars, 4 8 1 03^^ Silver doUar=3 copper dollars, 1 6| 34|f Copper dollar=4 copper marcs, 6| 1 l-}f | Copper marc=8 runstics, ]| 02|-ff Ducat=2 rix dollars, 9 4 2 07^-} SWITZERLAND. £ s. d. $ c. Rix dollar=108 cruitzers, 4 6 l 00 CruitzeT=4 feaingf, 0} 00S>2i^ TABLES. 413 Sterling. £s, d. $ c. Fening=3 rap, Oi oo-ss^v Sol=12f€nings, li 02-771 Livre or gulden=20 sols, 2 6 65f GENOA, CORSICA, &c. £ s. d. $ 6 3 1 16| Pistole=13 lires. 16 3 3 61i Louis d'or=16 lires * 1 00 4 44f ROME AND CIVITA VFXCHIA. riurrent crown=10 julios 6 1 ni Julio=8bayocs 6 11* Bayoc=5 quatrini 03 oiVv Stamped Julio=10 1 bayocs 71 13| Stamped crown=12 julios 6 1 33* Pistole=31 julios 15 6 3 34^ FLORENCE AND LEGHORN. Piafitre=6 lires 4 © 88| Lire =20 soldi 8 14-81-i Soldi=12 denari Of 01-23 Ducat=7i lires 5 1 Hi Pistole =22 lires NAPLES. 14 S 3 25-93 Pucat=100 grains 3 4 74-07|i Orain=3 quatrini 0/^ 00-74 Carlin=10 grains. 4 07-40 Ounce=3 ducats 10 2 22-225^ Pistole =4-6 carl ins VENICE. 15 4 £ s. d. 3 33-33^ Ducat=24 gros, 4 4 96/,. Gros or grosi=5J- soldi, 2J- 04J^ SQldi=12 denari, O^f 00-77? Lire=20 soldi, 8if 15-53^ 4J4 TABLES. In Leghorn, a piastre= 20 soldi of Genoa. Naples, a ducat = 86 " ** Milan, a crown = 80 " " Sicily, a crown =127| " " PALERMO IN SICILY. Sterling. £ s. d. Once or Onge=30 tarins, 10 Tarin=20 grains, Grain, Dollar of Sicily=240grs.=12 tarin, 4 Spanish dollar=252 grains, 4 9f 0-2JL ^125 *^2 5 6 $ C. 2 40 08 00*4 96 1 00 TURKEY. Piastre=80 aspers, Asper=4 inangars, Parac=3 aspers, Ostic: Caragrouch=100 aspers, Xerifi=10 solotas=200 aspers, 10 do.=i solota, £ s. 4 Of If 6 6 10 88f 1 2 03'33i IMli. mix 22-22| Piastre=100 aspers or 40 paras, Asper, 136 Paras=Spanisb dollar, Para, 10 Spanish dollars=34 piastres. SMYRNA. £ ARABIA. Piastre=60 comashees=80 caveers, Comasbee=7 carrets, Larin— 80 do. SeqMin=100 comashees, Tomond=B0 Larins, d. 3H on $ c. 29'41y\ 00-29 1 00 00 00-73/^ 10 00 d. 6 0-^- '^1 lOf 6 6 $ c. 1 00 01 '661 1905 1 66 66f 15 00 EGYPT. PiastFe==80 aspers, Asper, Medin=3 aspers, Dollar=^30 medins, Ecu or crown=96 aspers, Sultanin=2 ecu, ?argo dollar=70 medini, s. 4 4 5 10 10 d. If a 6 OlllJ^ 03-33/^ 00 11 22-22f 33-33 m TABLES. 415 ALGIERS, TUNIS AND TRIPOLI. Sterling. £ s. d. $ c. Dollar=:4 doubles. 4 6 1 00 DoubIe=2 rials, 1 n 25 Rial=10 aspers, 6| 12.', Pistole=^16 doubles. 16 10^ 2 75 Zequin=180 aspers, 10 1^ 2 25 Pataca Chica, lU 21/. Sultanin=8i patacas chicas, 8 If 1 81 Piastre=3 " " 2 lOi 64 MOROCCO, FEZ, TANGIERS, AND SALLEE. £ s. d. $ c. Dollar=28 blanquils, 4 6 1 00 Blanquil=24 fluces, 1-H 03-57! Ounce=4 blanquils, 7f 14-28-i Q,uarto=14 '* 2 3 50 Zequin=56 " 9 2 00 Pistole =100 " 16 04 PERSIA. 3 57| £ s. d. $ c. Bovello=12 abashees, 16 3 55-56f Abashee=4 shahees, 1 4 29 63— Sbahee=10 coz, 4 07-41-~ Larin =25 do. 10 18-51 Or=5 abashees, 6 8 BOMBAY. 1 4314 £ s. d. $ c. Rupee=4 quarters, 2 3 50 Quarter— 20 pices=100 reas , 6f m- Rea, o^v^ 00-125 Pagoda=14 quarters, 7 lOi 1 75 Rupee, gold=GO do. 1 13 9 7 50 Current niohuss=15 current rupees, 1 1 1| 4 68f Current rupee=50 pice. 1 41 31{ MADRAS AND PONDICHERRY. £ s. d. $ c. Pagoda=S6 fanauis, 8 3/^ 1 84 Fanam=8pice=80 cash, 2f 05i Pice=2 viz=lO cash, 014 00 64— Rupee = 10 fanams, 2 3f 51J- Crown=2 rupees. 4 7^ 1 02f- Rupee,* gold=4 pagodas, 1 13 1|4- 7 36 Bengal, or new Sicca rupee, 2 3 50 * Lack of rupees is 100000, and 340 Sicca rupees pas3 currently f6r 100 Star pagodas, 416 TABLES. Pagoda is 184 cents in the United States. Sea shells, called cowries are used for change ; their value va- rigs with their quality. CALCUTTA AND CALLICUT. Sterling. £ s. d. $ c. Rnpee = 16 annas, 2 3 60 Anna=12 pices, 1|| 03-|- Fanam=4 do. 6f 12| Crown or Ecu=2 rupees, 04 6 1 00 Pagoda=56 annas, 7 10^ 1 75 100 Sicca rupe€S=116 current rupees. PEGU, JAVA, SUMATRA, fcc. £ s. d. $ c, Dollar=900 fettees, 4 6 1 00 Fettee=10 cori, 0/^ OOJ- Tical=500 fettees, 2 6 55f Rial, crown or ecu=2 ticals, 05 1 1^ CHINA. JE s. d. $ c. Tale=10 maces, 6 8 1 48^^ Mace=10 candereens, 8 14"81 Candereen=10 cash, 0| 01-48 The Spanish dollar passes at 72 candereens, which makes the candereen worth, 01-38| But the tale is reckoned in the United States, 1 4800 JAPAN. £ s. d. $ c. Tale=10 mace=rix dollar, 3 4i 75 Mace=10 candereens. 4^V 07i- Candereen=2 pitis, MANILLA. 2,V 03-7^ £ s. d. $ c. Dollar~8 reals. 4 6 1 00 Real=12 quartos, 6^ m quarto, BATAVIA. Oy\ oC\ £ s. d. $ c. Spanish dollar=64 stivers. 4 6 1 00 Stiver, op OlVk Uix dollar=48 stivers 3 4f 75 Ducatoon— 80 do. 5 7X 1 25 TABLES. 417 COLOMBO IN CEYLON. je s. d. $ c. Spanish dolIar=64i stivers, 4 6 1 00 Sliver, Of-« 01-55 Rupee=30 stivers, 2 I/3 46'63i Rix doIIaf=48 do.=8 shillings, 3 4^\ 74-45 ENGLISH WEST INDIES. The principal difference is in the number of shillings in the Spanish dollar, while a pound is 20 shilling's, and a shilling is 12 pence, at Jamaica and Bermudas, the Spanish dollar is 6 shillings and 8 pence. Hence the £ s. d. $ c. Pound=20 shillings=3 dollars, 13 6 3 00 Shining=12 pence, Sj\ 15 Penny, 0|| 01-25 At Barbadoes, the Spanish dollar is 6 shillings and 3 pence. Hence a Pound=20 shillings 14 4f 3 20 Shilling=12 pence 8i| IG Penny Off 01^ FRENCH WEST INDIES. In some ofthe Islands the Spanish dollar passes for 8 livres and 6 sols, and in others for 9 livres. In the former, the Livre=20 sols Sol=12 deniers 8 livres 5 sols In the latter, the Livre=20 sols Sol=12 deniers 9 livres IN MARTINICO, TOBAGO AND ST. CHRISTOPHERS, The English inhabitants keep their accounts in the denomina- tions of English money, and the French, in those of France. But the round dollar passes for 9 shillings, and the current dollar at 8 shillings and three pence, or, the round is to the current dollar as 12 to 11. So that 99 livres=ll round dollars or =12 current dollars. s. d. 6A OOif 4 6 D. c. 123V 00-60§f 1 00 6 4 6 lli 00-55f 1 00 SPANISH WEST INDIES. Dollar==8 reals Real £. s. d. 4 6 6£ $ c. 1 00 12i. E 3 410 TABLES. TABLE II. OF THE MONEY OF THE JEWS, GREEKS, AND ROMANS, "WITH THE VALUE IN STERLING AND FEDERAL MONEY.* \' Talenl=60 manch Manch, or Hebrew mina Shekel=2 becah Becah=10 gerah Gerah Sextula Siclus aureus Talent of Gold Iti this estimate the n to 1. OF THE JEWS. je. s. d. $ c. 342 3 9 1620 83|- 60 shekels 6 16 m 30 41| 2 3f 50|f 1 lU 26f|- Hh 1/^ 12 ^ 2 67/^ 1 16 6 8 111 5475 25533 33i • value of gold is to that of silver nearly as OF THE GREEKS. Drachma— 1| tetrobolom Tetrobolum=2 diobolum DioboIum=2 obolus Oboliis=2 hemiobolum Hemiobolum=4 chalcus Chalcus=7 septon Didrachmon=2 drachma Tetrard statu=2 didrachma Mina=100 drachmae Talent=60 minaa . IOC Talents Statu aureus=25 drachma Statu daricus— 50 do. according to Josephus, s. d. 71 £ 3 193 15 19375 16 13 1 12 31 2tV lA OOff 1 31 2 7 4 7 $ c. 14i| 09ff 04if I 02i|l Olifl OOlneariy 281^ 5711 14 35-V 861 111 86111 111 3 58f^ 7 17if OF THE ROMANS. Denarius~2 quinarii Q,uinariu8=2 sestertii Sestertius=2^ asor libella =4 teruncii Teruncios Sestertium=1000 sestertii * Auihora differ respecting the precise value of ancient money, estimate is here g^iven, which is, at least sufficiently near the truth. s. d. 7f 3| $ c. 14if 07J^ Hf 03lfl 8 ^i 1 51 00 j% 35 87 nearlj= The commop TABLES. 419 d. D. 8072 18 4 35879 63 Decius sestertium=1000 ses- tertia, Centies sestertium, or centies HS. was 10000 sestertia, or 80729 3 4 Millies HS. was 100000 ses- tertia, or 807291 13 4 And the millies centies HS. was the sum of the last two, or 888020 16 8 Aureus=2i3 denarii 16 If This ratio of the aureus to the denarius is that mentioned by Tacitus. 3 68H TABLE III. OF MEASURES OF LENGTH AND CAPACITY AND WEIGHTS OF VARIOUS COUNTRIES. THE table of English and American measures has been given under compound addition. Compared with the French measure, the English inch is 0-253994053958632382 12354. of a French metre. A Foot English Yard=3 feet Rod or pole=5i yards Mile=320 rods League=3 miles Ell Englishes quarters of a yard Ell Flemish=3 do. Ell French=6 do. Wine gallon=231 cubic inches English Ale do.=282 do. Gallon, dry measure— 268-8 do. Bushel=32 quarts=2150-42 do. Wine quart=57*75 do. Ale do.=70-5 do. Dry do.=67-2 do. Wine Hogshead=63 gallons French metres. 0-30479286-1- 0-914378594* 5-02908227— 1609-30632588-h 4827-919977644- M42973 0-685784 1-371568 French litres. 3-3735 4-6208 4.4043 .35-2343 0-9463 M552 11011 238-4509 SCOTLAND. 3 Feet make 1 ell, 1 Mile=5760 feet, 30 Scotch ells are equal to 30 Scotch miles, English. 37-2 inches. 5952 feet. 31 yards. 31 miles. 426 TABLES. I Fall=6 ells, 2234 inches 48 Scotch acres are very nearly 61 acres- For the measure of Wheats Peas, Beans, Rye, and While Salt, 100 Bolls equal 409 bushels, Winchester measure. "For Barley, Oats, and Malt. 100 Bolls equal 596 bushels, Winchester measure. f{ote. The Boll varies in d^^rent parts of Scotland. IRELAND. The Irish and English foot and yard are equaL The Irish mile, =2240 yards. II miles Irish, 14 miles. 121 acres " 196 acres. The Irish bushel contains 1740-8 cubic inches. Metre, Deca-metre=10 metres, Hecto-metre=100 " Kilo-metre=1000 " Myria-metre= 10000 " Deci-metre=y^j metre, Centimetre=T Milli-metre=- i 'O J 10 FRANCE.* French feet. 3078444 30-78444 307 8444 3078-444 30784-44 French inches. 3-6941328 0-36941328 0-036941328 French lines. 443-295936 English feet- 3-2809167 32-809167 328-09167 3280-9167 32809-167 English inches. 3-93710004 0-393710004 0-0393710004 English lines. 472-4520048 Metre, Quadrant of the Meridian, 100 French degrees, 90 English degrees. * France.is the only nation, which has established an invariable standard of measure. The linear unit of the French measure is the metre. By accurate observations and calculations the length of the meridian from the Equator to the pole, v.'hich passes through the city of Paris, was ascertained to be 5130740 toises of six feet each of the ancient French measure. This number of toises is equal to 30784440 French feet, or 32fl091G7 English feet. The metre is one ten millionth part of this arc of the meridian, or 3-078444 feet, which is 3 feet J 1 lines, and _2iL?_93_s_ of a line of tlie former French measure. All other 10 measures are derived from the metre. In England, it has been proposed to make the length of the pendulum to vi- brate seconds at London, the standard of measure. At the level of the sea, and when the temperature is 62d. Far. and in lat. 51d. ?iV 8-4" N. the length of the pendulum to vibrate in a second is 30-1386 inches, English, as very accurately •.ietermined by Capt. Katar. According to Capt. Kater's measure, the French metre is 39-37076 inches English, at the same temperature, and may" be taken with sufficient accuracy to be 39-371 inches. Thc^o measures will vary a little according to the scale on which they are estimated. U Troughton's scale of 36 inches be taken as the standard, General Roy^s Scale is 36-00036 inches, and Bird's Parliamentary standard of 1758, is 36-000^3 inches. And if the scale of 1758 be the standard, Troughton's scale is 35-99077 inches. According to Mr 'l£as?lcr, the French metre is 3-23168733 feet on Troughton's Scale. French feet. English feet. 6 6-3946266-h 3 63f 3-87824. 1- 1 0657711+* Ot\ 0'0888I42-1-. TABLES. 421 French feet. English feet, Degree=54 min. Eng. 100 min. Fr. 307844-4 328091-67 Minute==32'4 sec. Eng. lOOsec. Fr. 3078-444 3280-9167 The are is the square of the deca-metre, and is the unit for square measure. French square feet. English square feet. Are, 947-681746113 107644143923 Square metre or Centiare, 9-47681746113 10-7644143923 The litre is the cube of the decimetre, dind is the unit for dry and liquid measure. French cubic inches. Eng. cubic inches. Litre, 60-412416 61028 Hecto!itre=13 veltes 3i pints, « 6102 8 Paris pint is yVoWoVo ^^^^^ ^^ 46-95 cubic inches. The stere is a cubic metre, and is the unit for Cubic or Solid Measure. French cubic feet. English cubic feet. Stere, 20-17385 35 317 OF THE OLD FRENCH MEASURES. The Toise, Aune or Ell, Foot=12 inches, Inch=12 lines, A French League is nearly 2J English miles, or about j^ of an English League. By a decree of 1812, the Toise, Aune, Foot, &c. are allowed to be the denominations of measure for the common people of France, in the following ratios to the metre. Toise=2 metres, 6 56 English feet nearly. Foot—} metre, 05468 lnch=yV metre, 2734 Aune or*'ElI=li metre, 3-9371 Bushel^i Hectolitre, 762 854 cubic inches. The old litron=40'393455i|^ French cubic inches, by statute, but the common iitron is 48 8224 English cubic inches. Weight. . English. Paris pound or 16oz.=:2 marcs, 7560grs. Ounce, 4725 16 pounds are 211bs. Troy. 63oz. 64oz. Troy. 100 pounds 1081bs. Avoirdupois, 92if do. 100 do. 25 do. 27 do. 1 Kilogramme 45-35 do. 1 Hectogramme 453.50 do. '■ The ratio of the French to the Eu2;lish foot here assigned is very little cui- I'orcnt from 1 to 1 jl^g, which v/as furaierly considered as the true ratio. 422 TABLES. HAMBURGH. English. 100 Ells, 62^ yards. 16 do. 10 do. 1 German mile, 4 miles. 100 lbs. 1074 lbs. 93iM> 100 do. 1 Shippound, 280 do. SWEDEN. 1026275 feet, 1000 feet. 1000 do. 974-397 do. 1 Can or Kann, 159-864 cub. inch, 877-7125 Victualievigt, lOOOlbs. Troy. 1333-4935 Skulpounds, 1000 do. Av. HOLLAND. 100 Ells, 75yds. 1331 do. 100 do. 1 Dutch mile. 3i miles. 1 ife=2marcs=16oz. 5775grs. 100 lbs. 109ilbs. 9I2V9 <^o- 100 do. 219 do. 200 do. ANTWERP. 100 Brabant Ells, 74yds. 135/^ do. 100 do. 100 lbs. 104ilbs. , 96 do. 100 do. Q,uintal=10 myriogrammes, 2041 do. 100 pots of Brabant, 361 gallons. BREMEN. 100 fe, 110ft. 90if do. 100 do. 1 Last, 80 bushels. DENMARK. 96 lbs. lOOlbs. 100 lbs. 1041 do. 1 Shippound=320]bs. or 20 } lispounds, J 33.31 do/ K^tyt^^ vlV/k 1000 Feet, 1049 RUSSIA. 1 Arsheen, 28 inches. 9 • do. 7 yards. TABLES. 423 100 lbs. English. SBiibs. 1 Pood=40lbs. 33,Vdo. 1 Borquit=10 poods, 333 do. Russian Verst, f mile. SPAIN. Vara or Barra=i| of French ell, 2-7471 feet. Vara, of Catalonia=li do. 5-8173 do. lOOlfe, 97ft. Arobe=25lbs. 241 do. Spanish league, 3| miles. BILBOA. 100 Varas, 108yds. 100 ft, 106ilbs. 100 ft of iron, 100 ft. 32 velts. 66 gallons. 100 fanagues, 152 bushels. PORTUGAL. Cavedo=26| Eng. in. or accurately. 26-6933 inches. Vara 43| do. do. 44-66 32ft=l arobe, nearly 33fe. Q.uintal=4 arobes or 128fe 132ft. 60 Alquiers or 1 Moy, 24 bushels. Fanga=4 alquiers, If do. Canado=4 quarteels. 3 pints. Almude— 12 canados, 41 wine gallons NAPLES. Cane, =7 feet, or 84 inches. Pound of silk, 12oz. 1001 bs. 64|lbs. Cantar=100 rotolos, 196lbs. LEGHORN. 145 lbs. n2ft. 100 brasses, 64yds. Cane=4 brasses. 2ifyds. 116 sacks, 100 quarters, 800 bushels. TRIESTE. Brace, 27 inches, or f yard. 100 ft of Vienna, 123 ft Av, 31 Staros, 1 quarter. , 8 bushels. Staro, 2f ■ do. Barrel of wine, 18 gallons. 124 TABLES. PALERMO IN SICILY. Englisli. Cantar, 1761bs. Av. Rottoli=Yj,^ canlar, lllfe nearly. Po'Mui r;eight, fife. Salm. 485 fe. Caffis, 3i gallons. Caular .of oil, 25 do. Barrel, 9 do. SMYRNA. Pike, 1 yd. nearly 45 Okes, nearly 123| ft Av. 40| do. 112 do. BOxMBAY. Maud, iCwt. Surat maud, 37ilbs. Surat Candy^21 S. mauds, 784 do. Bombay Candv=21 B. mauds, 588 do. The ife is the English 3fe. MADRAS. Picul=100 cattas, 133ilbs. Calta, Hife. Maud, 25fe Troy, Candy=20 mauds, 500lbs. do. CALCUTTA. Cavid, ^l\ Bazar maud, fCwt. Factory do. 75 ife. 1 Maud=40 Seer?, and 1 Seer=16 chittacks. CHINA. Covid, Uj% inches. Picul=100 cattas, 1331 lbs. Catta=^16 tales, I3- *• BATAVIA. Picul = 100 cattas = 125lfe Dutch = 1335ife. MANILLA. lOOfe, 101%. Arobe or 25ife, 26 do. Picul=5i arobes, 143 do. ACCOUNT OF THE GREGORIAN STYLE, &c. 425 JAPAN. English. Hichey or Ichan, 3| feet. Catta=16 mace, 13^fe. IVIace = 10 tales, | do. Balec, 16i gallons. FRENCH WEST INDIES. 104ife, 112 lbs. 100ft, lOTy^ibs. AN" ACCOUNT OF THK OUKGORIAN OR NEW STYLE, TOGETHER WITH SOME CHRONOLOGICAL PROBLEMS, FOR FINDING THE EPACT, GOLDEN NUMBER, moon's AGE, &:c. POPE GREGORY the XIMth made a reformation of the cal- endar. The Julian calendar, or old style, had, before that time, been in general use all over Europe. The year, according to the Julian calendar, consists of three hundred and sixty five days and six hours ; which six hours being one fourth part of a day, the common years consisted of three hundred and sixty five days, and every fourth year, one day was added to the month of February, which made each of those.jyears three hundred and sixty six days, which are usually called leap years. This computation, though near the truth, is more than the solar year by eleven minutes, which. In one hundred and thirty one years, amounts to a whole day. By which the Vernal iKquinox was anticipated ten days, from the time of the general council of Nice, held in the year 325 of the Christian iEra, to the time of Pope Gregory ; who therefore caused ten days to be taken out of the month of October in 1582, to make the /Equinox fi\ll on the 21st of March, as it did at the time of that council. And, to pre- vent the like variation for the future, he ordered that three days should be abated in every four hundred years, by reducing thQ leap year at the close of each century, for three successive cen- turies, to common years, and retaining the leap year at the close of each fourth century only. This was at that time esteemed as exactly conformable to the true solar year; but Dr. Halley makes the solar year to be three hundred and sixty five days, five hours, forty eight minutes, fifty four seconds, forty one thirds, twenty seven fourths, and thirty one fifths: According to which, in four hundred years, the Julian year of three hundred and sixty live days and six hours will exceed the solar by three days, one hour and i\hy five minutes, v/hich is near two hours, so that in fifty centuries it will amount to a day. Though the Gregorian calendar, or new style, had long been used throughout the greatest part of Europe, it did not take place in Great Britain and America till the first of January, 1752; and in September following, the eleven days were adjusted by calling the third day of that month the fourteenth, and continuing the rest in their order. F 3 126 CHRONOLOGICAL PROBLEMS. CHRONOLOGICAL PROBLEMS. Problem I. As there are three leap years to be abated in every four centuries : tQ shew how to find in which century the last year is to be a leap year^ and in which it is not. Rule. Cut off two cyphers, and divide the rettiaining figures by 4 ; if nothing remain, the year is a leap year. ExAMP. 1. The year 18|00. Examp. 3. The year 20|00. 4)18(4 4)20(5 16 20 2 Examp. 2. The year 19iOO. Examp. 4. The year 40|00. , 4)19(4 4)40(10 16 40 3 The first and second examples, having remainders, shew the years to be common years of three hqndred and sixty five days ; but the third and fourth, having no remainders, are leap years of three hundred and sixty six days. Problem II. Jo find, with regard to any other years, whether any given year be leap year, and the contrary. Rule. Divide the proposed year by 4, and if there be no remainder, after the division, it is leap year ; but if 1, 2 or 3 remain, it is the first, second or third after leap year. Examp. 1. For the year 1784, Examp. 2. For the year 178G. 4)1784(446 4)1786(446 16 16 18 18 16 16 24 26 24 24 ~j^ "^ ^ second after \ leap year. Problem III. To find the Dominical Letter for any year, according to the Julian method of calculation. Rule, Add to the year its fourth part and 4, and divide that sum by 7 : if nothing remain, the Dominical Letter is G ; but if there be any ^ CHRONOLOaiCAL PROBLEMS. 427 remainder, it shews the letter in a retrograde order from G, be- ginning the reckoning with F ; or, if it be subtracted from 7, you will have the index of the letter from A, accounting as follows: A B C D E F G 1 2 3 4 5 C 7 ExAMP. For the year 1786. Given year=1786 Add ^ Its fourth = 446 And 4 7)2236(319 21 13 7 66 63 And 7 — 3=4=t), reckoning from A. Problem IV. To find the Dominical Letter for any year according to the Gregori- an computation. Rule. Divide the year and its fourth part, less 1 (for the present cen- tury) by 7 ; subtract the remainder after the division, from 7, and this remainder will be the index of the Dominical Letter, as be- fore : if nothing remain it is G. ExAMP. 1. For the year 1810. Examp. 2. For the year 1812.* .,, 5Givenyear=1810 1812 -^^^ i Its fourth = 452 453 2262 2265 Subtract 1 1 7)2261(323 7)2264(323 21 21 16 16 14 14 21 24 21 21 And 7— 0=7=G. And 7— 3=4=0. * Here it is to be observed, that every leap year has two Dominical Letters ; that, found by this rule, is the Dominical Letter from the twenty fifth day ©i ^^ 42S CHRONOLOGICAL PROBLEMS. Problem V. To find the PrimCy or Golden JVumher. Rule. Add 1 to the given year; divide the sum by 19, and the remain- der, after the division, will be the Prime ; if nothing remain, then 19 will be the Golden Number. ExAMP. F'or the year 1786. To the given year 1786 Add I 19)1787(94 H 171 77 76 1 Golden Number. The Golden Number, or Lunar Cycle, is a period of 19 years, invented by Meton, an Athenian, and from him called the Metonick, Cycle. The use of this cycle is to find the change of the moon ; because after 19 years, the changes of the moon fall on the same days of the month as in the former 19 years ; though not at the same time of the day, there being an anticipation of one hour, twenty seven minutes, forty one seconds, and thirty two thirds ; Tvhich, in 312 years, amount to a whole day. Hence, the Golden Number will not show the true change of the moon for more than three hundred and twelve years, without being varied. But the golden number is not so well adapted to the Gregorian, as the Ju- lian calendar : The epact being more certain in the new style, to . find which, the golden number is of use. Problem VL To find the Julian Epact. Rule. First find the Golden Number, which multiply by 11, and the product, if less than 30, will be the number required ; if the pro- duct exceed 30, then divide it by 30, and the remainder is the epact. ExAMP. 1. For the year 1786. February to the end of the year ; and the next in tlie order of the alphabet serves from the first of January to the twenty fourth of February. In the 2d Example, D is the Dominical Letter for the year; but E, the next in the order of the alphabet, is th^ Dominical Letter for January and February. From this interruption of the Dominical Letter every fourth year, it is twenty eight years before the Dominical Letter returns to the same order, which, were it not for the leap years, would return to the same every seven years. This Cycle of twenty ei^ht years is called the Cycle of the &im. CHRONOLOGICAL PROBLEMS. ^39 To the given year 1786 Add 1 19)1787(94 171 77 76 Golden Number=l and 1x11 = 11 t^ie Julian Epact, ExAMp. 2. For the vear 1791. 1791 1 i9)1792(94 171 82 76 6=Golden Number, and 6x11=66, therefore 30)66(2 60 Problem VII. Epact 6 To find the Gregorian Epact. Rule. Subtract 11 from the JuHan Epact: If the subtraction cannot be made, add 30 to the Julian Epact ; then subtract, and the remainder will be the Gregorian Epact: if nothing remain, the Epact is 29. Or, take 1 from the Golden Number, and divide the remainder by 3 ; if 1 remain, add 10 to the dividend, which sum will be the Epact; if 2 remain, add 20 to the dividend ; but if nothing remain, the dividend is the Epact. ExAMP. 1. For the year 1786. Examp. 3. For the year 1791. The Julian Epact being 11 The Julian Epact being but 6 Subtract 11 Add to it 30 Because nothing remains, the Epact is 29. Or, Examp. 2. For the year 1786. The Golden number being I Take from it 1 Divide by 3)0(0 There being bo remainder, the Epact is 29, as before. Subtract 1 1 Gregorian Epact=25 Or, Examp. 4. For the year 1791. The Golden number being 6 Take from it 1 3)5(1 3 430 CHRONOLOGICAL PROBLEM^. Therefore, as 2 remains, add 20 to the dividend, and it gives the Epact 25, as before. A general Rule for finding the Gregorian Epact forever. Divide the centuries of any year of the Christian Era by 4, (re- jecting the subsequent numbers ;) multiply the remainder by 17, and to this product add ihe quotient multiplied by 43 ; divide this sura plus 86 by 26, multiplying the Golden Number by 11, from which subtract the last quotient, and rejecting the thirties, the remainder will be the Epact. ExAMP. For the year 1786. Rejecting the subsequent numbers 86, it will be 17. 4)17(4 16 Golden Number= 1 1 Multiply by 1 1 Multiply by 17 — — 11 1 7 Subtract the last quotienl= 1 1 Add 4X43=172 — ■ • 00 189 Therefore, as nothing remains, Add 86 the Epact is 29, as before. 23)275(11 25 A TABLE OF THE IvINETEEN EPACTS FOB, THE JULIAN AND GREGO- RIAN ACCOCTNTS, BY THE GOLDEN NUMBER. J ulian Grsg. Julian Gre^. Julian Greg. G. N. Epact. Epact. G. N. Epact. Epact. G.N. Epact. Epact. 1 11 29 7 17 6 13 23 12 2 22 11 8 28 17 14 4 23 3 3 i 22 i 9 9 28 15 15 4 4 14 3 10 20 9 16 26 15 5 25 14 11 1 20 17 7 26 6 6 25 12 12 1 18 19 18 29 7 18 Problem VIH. To calculate the Moon's age on any given day. Rule. To the given day of the month, add the Epact and num- ber of the month : If the sum be less than 30, it is the Moon's age, but if it exceed 30, then take 30 from it, and the remainder will bo the Moon's r.ge. ** JVote. The numbers to be added to the following months, are SI follow : CHRONOLOGICAL PROBLEMS. 431 'January 'July r 5- February 2 August 6 Tu- March 1 '2' September, 8 April October ] 8 May 3 Nove(i)ber 10 .June 4 .December LlO. Example. For January 25th, 1786. (Given (lay =25 Add < Epact =29 ( No. of the month =00 64 Subtract 30 24=Moon's age Problem IX. To find the times of the New and Full Moon, and the first and last Quarters. Rule. Find the Moon's age on the given day, then, if it be 15, the Moon will be full on that day, and by counting 7^ days back- ward and forward you will have the first and last quarters, and by counting backward and forward 15 days, you will have the times of the last and next change ; but if the age of the Moon be greater than 15, take 15 from it, and the remainder will shew how many days have passed since the last full moon, and counting these back- ward, you will have the day the last full moon happened on, and by knowing that, we can find the change, or either of the quarters, as before. Again, if the age of the moon, on the assumed day, be less than 15, then take that from 15, and the remainder will shew how many days are to run till the next full moon, which you will have by ad- ding the remainder to the assumed day ; and proceeding as before, you will have the days of the change, and either quarter as above. ExAMp. For January 25th, 1786. C Assumed day =25 Add I Epact =29 f Number of the month=00 54 Subtract 30 Moon's age=24 Subtract 15 Take the days since the last full moon= 9 From the assumed day=25 To the day of the fall moon=16th Add 15 ^i> CHROxNOLOGICAL PROBLEMS. New Mooou 5 is: From the full Moon 16 Take 7^ First quarter 9th To the fullMoon=16 Add l-l Last quarter==23 Problem X. The time of ihe Moon's coming to the South, after the Sun ^ being given, tc find the a. " .. — ^.-^ _ * '. 'F passes in any part of the line, as at c ; extend the compass- es at pleasure, unless a dis- tance be assigned, and de- scribe an arch 6; with the same extent in some other ^ T B part of the line AB, as at e, describe the arch a ; lay a ruler to the extremities of the arches, and draw the line E F, which will be parallel to the line A B. Problem VI. To make an Angle equal to any number of degrees* It is required to lay off an acute Angle of 35° on a given line AB. Take 60 degrees frorp the line of chords in the compasses, set one foot of the compasses in the point A, describe an arch CD, at pleas- ure ; then set one foot of the com- passes in the brass centre, in the beginning of the line of chords, and bring the other to 35 on the line ; with this extent set one foot in C, with the other intersect the arch CD, in a, and through a draw the line AE, so will EAB be an angle of 35 degrees. If the angle had been obtuse, suppose 125°, then take 90° from the line of chords ; set one foot in C, and intersect the arch in b ; then take 35° from the same line of chords, and set them from b to di a line drawn from A through d to F will make an angle, FAB, of 125°. To measure an angle by the line of chords, is only to take the distance on the arch between the lines AB and AE, or AB and AF, and lay it on the line of chords. Problem VII. To make a Triangle^ whose sides shall be equal to three §iven lines, provided any two of them be longer than the third. Let A,B,C, be the three given lines ; draw a line AB, at pleasure ; take the line C in the compasses, set one foot in A, and with the other make a mark at B ; then take the given line B in the compasses, and setting one foot in A, draw the arch C ; then take the line A in the compasses, and intersect the arch C in C ; lastly, draw the lines AC and BC, and the friangle will be completed. H r^ 442 GEOMETRICAL PROBLEMS. to am; Proelem VIIL To make a Square^ having equal sides, equal given line. Let A be the given line ; draw a line a AB equal to the given line ; from P raise — — — a perpendicular to C pqual to AB, with _ ■' the same extent, set one foot inCandde scribe the arch D ; also with the same ex- tent, set one foot in A and intersect the arch D ; lastly, draw the lines AD and CD, and the square will be completed. In like manner may a Parallelogram be constructed, only attending to the differ- ence between the length and breadth. A Problem IX. To describe a Circle, which shall pass through any three given Points, which are not in a straight line. Let the three given points be A,B,C, through which the circle is to pass. Join the points AB and BC with right lines, and bisect these lines; the point D, where the bisecting lines cross each oth- er, will be the centre of the circle required. Therefore, place one point of the compasses in D, extending the other to cither of the given points, and the circle, described by that radius, will pas? through all the points. Hence, it will be ea«>y to find the cen- tre of any given circle ; for, if any three points are taken in the circumference of the given circle, the centre will be rea- dily tbnnd as above. 'J'he same may al- so be observed, when only a part of the circumference is given. Problem X. To describe an Ellipsis or Oval mechanically. Dravv two parallel lines, as L and M, at a moderate distance, by Prob. 5 ; then draw two others at the same distance, across the for- mer, as N and O ; by the crossing of these lines will be made a fig- ure ABCD, of four sides ; extend the compasses at pleasure, and set- ting one foot in D, describe the arch cde ; with the same extent, set one foot in B, and describe the arch fgh ; then set one foot in C, and contract them so as to reach the point c, and describe the arch Im ; with the same extent, and one foot in A, describe the arch ik, and the oval will be completed. In the same manner, with a greater or less extent of the compass- es, may a greater or less oval be made by the same four sided fig- ure A BCD. MENSURATION OF SUPERFICIES, Lc, 443 MENSURATION OF SUPERFICIES AND SOLIDS. Section I. Of Superficies. SUPERFICIES, or surfaces, are measured by the superficial inch, foot, yard, &c. according to the measures peculiar to differ- ent artists. The superficial inch, foot, &c. is one inch, foot, &c. in length and breadth; and, because 12 inches make one foot of Long Measure, therefore 12x12=144 inches make 1 superficial foot, 3x3=9 feet, a yard, &c. The superficial content of every surface is found by the proper rule of its figure, whether square, triangle, polygon, or circle. Article 1. To measure a Square y having equal sides. Rule. Multiply the side of the square into itself, and the product will be the area or superficial content, of the same name with the de- nomination taken, either in inches, feet, or yards, respectively. Let ABCD represent a square, whose side is 12 a inches or 12 feet. Multiply the side 12 by itself, ' thus, 12 inches. 12 feet. 12 inches. 12 feet. Area=144 inches. 144 feet. Bij the Sliding Rule. Set 1 to the length on B, then, find the breadth on A, and oppo site to this on B, you will have the content. By Guntcr^s Scale. Extend the dividers from 1, on the line of numbers, to the length ; that distance, laid the same way from the breadth, will point out the answer. Art. 2. To measure a Parallelogram or long Square. Rule. Multiply the length by the breadth, and the product will be the area, or superficial content* Let ABCD represent a parallelogram, whose A B length is 5 feet, and breadth, 4 feet. Multiply 5 by 4. Length 5 Breadth 4 _ D Area 20 Ans. * If the parallelogram be divided into squares by drawings lines as in the figure, it is obvious on inspection, that the number of squares must always be equal to the product of the length and breadth. The same may be shown ou the square also. The area of a Rhombus or Rhomboides is equal to that of a T)4iraUeIogram of the same base and altitude. 444 MENSURATION OF SUPERFICIES The content of this figure is found on the sliding rule and scale, as the former. Art. 3. When the breadth of a Superficies is given^ to find how much in length will make a square foot^ yardy ^c. Rule. As the breadth is to a foot, yard, &c. so is a foot, yard, &c. to the length required to make a foot, yard, &,c. Or divide 144 by the breadth, and the quotient will be the length required. How much, in length, of a board 21 feet wide, will make a square foot? In. br. In. leng. In. br. In. leng. As 30 : 12 :: 12 : 48 12 30)144(4-8 inches, length required. 120 240 240 In. Breadth=30)144(4-8 inches, Ans. Art. 4. To measure a Rhombus. JJefmiiion, A rhombus is a figure with four equal sides, in the form of a diamond on cards, having two angles greater and two less than the angles of a square : the former are called obtuse angles, and the latter, acutey or sharp, angles. Rule. Multiply the side by the length of a perpendicular, let fall from one of the obtuse angles to the side opposite such angle. Let ABCD represent a rhombus, each of whose sides is 16 feet: AAr ^^ perpendicular let fall from the obtuse \ angle, at B, on the side DC, will in- \ tersect it in the point E, so will BE \ be 12 feet ; and this being multiplied \ into the given side, the product will \ be the area of the rhombus. T> V y V r ^ ^V Side=16 By the Sliding Rule. Fer.=12 Set 1 on A to the length on B; find the perpendicular height on A, against which on 192 area. B is the content. By Gunter. The extent from 1 to the perpendicular height will reach from the length to the content. AND SOLIDS. 44c Art. 5. To find the Area of a Rhomhoides. Defirtntion. A rhomboides is a figure, whose opposite sides and opposite angles are equal. Rule. Multiply one of the longest sides by the perpendicular let fal: from one of the obtuse angles on one of the longest sides. Let ABCD represent a rhomboides ; the longest sides AB and CD being /; /B 16-5 feet, and the perpendicular AE, ■"^•7 feet. Side=:16-5 jO __ Perp. 9-7 ^ The content is found on the sliding rule, and scale, as in the last figurp. Ans. 16005 feet. Art. 6. To measure a Triangle.* Rule. If it be a right angled triangle, multiply the base by half the perpendicular, or half the base by the perpendicular, and the pro- duct wiH be the area: but if it be an oblique angled triangle, (whether obtuse, or acute,) multiply half the base by the length of the perpendicular let fall on the base from the angle opposite to it, and the product will be the area. The longest side of a irian- gle is usually called the base, except in a right angled triangle, where the longest of the two legs, which include the right angle, is called the base. In the right angled triangle ABC right angled at C ; the base AC is 18-8 feet, and the perpendicular BC=12-6. Base =18-8 Or, Perp.=12-6 jPerp.= 6-3 Base 664 1128 9-4 504 1134 11844 area. 118-44 area The oblique angled triangle ABC being given, let fall a perpendicular from the angle at B on the base AC, and that perpendicular is the height of the triangle. The base AC being i 15-6, and the perpendicular BD=9, ■ to find the area. * A triangle is half a parallelogram of the same base and altitude ; hence th« rule. In a right angled triangle, the longest side is called the hypotenuse ; the next longest, the oase ; and the shortest side, the perpendiculjxr. 446 MENSURATION OF SUPERFiCIEt 78=half the base. 9=height of the angle. 70-2=area. By the Sliding Rule. Set 1 on A to the length of the base on B, and opposite to half the length of the perpendicular, on A, you will have the content on B. By Gunter. The extent from 1 to half the length of the perpendicular will reach from the length of the base to the content. In this place it may be proper to instruct the learner in one of the properties of a right angled triangle : viz That the j'quare of the longest side of a right angled triangle, usually called the hy- potenuse, is equal to the sum of the squares of the two other sides, usually called the legs ; which is of great use, for by this mean, any two sides of a right angled triangle being given, the other may be found by common Arithmetick. Thus, in the right angled tri- angle ABC, the base AC and perpendicular BC being given, the hypotenuse AB may be found by extracting the square root of the sum of the squares of the ba^e and perpendicular. Base 18 8 Perp. 12 6 353-44=square of the base. 18-8 12 6 158-76=squareoftheperp. 1504 1504 188 756 252 126 61 2-20(22'63 hypotenuse 4 353-44 158-76 42)112 84 446)2820 2676 • 4523)14400 13569 831 And, if the hypotenuse and one of the legs be given, the other may be found by subtracting the square of the given leg from the square of tiie hypotenuse. There are some numbers, the sum of whose squares make a per- fect square, of which sort are 3 and 4, whose squares, being added together, make 25, which is Ih^ square of 5 ; therefore, if the base of a triangle be 4, and the perpendicular 3, the hypotenuse will be 5 ; and if any of these numbers be multiplied by any other number, those products will be the j^ides of right angled triangles, as 6, 8, 10, and 15, 20, S5, &c Thus artiticers, when they set off the corner of a building, usually measure 6 feet on one side, and 8 feet on the other, then laying a 10 feet pole across, it makes the corner a true right angle. AND SOLIDS. 44' Art. 7 I'here is another method of finding the area of triangles, the three sides being given. Rule. Add the three sides together, then take the half of that sum, and out of it subtract each side severally ; and muhiply ihe half of the sum and these remainders continually, and the square root of this product will be the area of the triangle. In the oblique triangle ABC, the base AC is given loG, the side AB is 10 4, and the side BC is 9-2, to find the area. 15 6 17-6 17-6 17-6 o .. 10-4 — 16-6 —10 4 — 92 8-4 35 2 sum ^C 17-6=half the sum. . 17-6 2 35-2 7*2 704 2464 253-44 84 101376 202752 2128-896 2128-8960(46'139=area. 16 86)528 516 921)1289 921 9223)36860 27669 92269)919100 830421 88679 Art. 8. To measure a Trapezium. Definition. A trapezium is an irregular figure of four unequatl sides, and uwequal angles. Rule. Draw a diagonal line from one of the angles to the oppo- site angle, as AC, and then will the trapezium be divided into two triangles, of which the diagonal is the common base : then, letting fall perpendiculars from the other opposite angles on the diagonal^ add those perpendicuUrs tof^ether, and multiply half that sum into the diagonal, or half of the diagonal into the sum of the perpendic- ulars, and that product will be the area of the trapezium. In the trapezium ABCD, the cliagonal AC is 24, the perpen- dicular DE 0, and the perpendi- cular EF 10. The sum of the perpendiculars is 16, whose half is 8, which being multiplied in* to 24, will give the area. ^41 MENSURATION OF SUPERFICIES 24 8 192=area. By the sliding Rule. Set 1 on A to I the sum of the perpendiculars on B, and opposite the length of the diagonal on A, you will have the area on B. By Gunter. The extent from 1 to ^\he sum of the perpendiculars will reach from the length of the diagonal to the area. Art. 9. To measure any irregular figure. Rule. Divide the figure into triangles, by drawing diagonals from one angle to another; then measure all the triangles by ei- ther of the rules, already taught, at Article 6 or 7, and th.e sum of the several areas of all the triangles will be the area of the given fiorure. The irregular figure ABCDEF being given, divide it into triangles by the diagonals FB, EB, and DB : then may the triangles be meas- ured by letting fall perpendiculars on their respective bases, as Ba, B6, Dc, ¥d, and multiplying those perpendiculars by half their res- pective bases. In the triangle AFB the base FA is loo, and the perpendicular Ba49 ; in the triangle FBE the base BE is 9:2, and the perpendi- cular Yd 52 ; Hi the triangle EBD, the base BE is the same as be- fore, and the perpendicular Y)c 44; and in the triangle DCB, the base DC is 80, and the perpendicular B6 38 ; by which the area of each may be found by Art. 6, as follows. 60=half AF. 49=perp. aB, 46=ha!f BE. 52=perp. Yd, 2450=area of AFB. 46=halfBE. 44=perp. Dc. 184 184 9'. 230 2392=area of FBE. 38=perp. B&. 40=halfDC. 2450 2024 2392 1520 838G=area of the figure ABCDEF. 2024=area of EBD. 1520=area of DCB. AND SOLIDS. 44& In dividing any irregular figure into triangles, the triangles will be less, by two, and the diagonals less, by three, than the numbey of the sides of the figure. If there be a long, irregular figure like the following, the mean breadth may be found very nearly, by measuring the breadth at certain equal distan- ces along AB, and dividing the sum of the breadths by their number. Let the length, AB, be 16 rods, the 1st breadth AC 3 9 rods, the 2d 4 rods, the 3d 3 93 rods, the 4th 4-3 rods, the 5th 4-25 rods, the 6th 4-5 rods, the 7th 48 rods, and the 8th 4 9 rods ; what is the area ? 3^9-f4--f.3-96+4-3+4-254-4-6+4-8-{-4 9 34-6 _ ~ __ jjjg mean 34-6 breadth. Then —^Xl6=69-2 rods, Ans. Art. 10. 7o measure a Trapezoid. Definition. A trapezoid is the segment of a triangle, cut by a line parallel to the base. Rule. Add the parallel sides together, and multiply half that sum by the perpendicular breadth. In the trapezoid ABCD,thesideAD 18 24, the side BC is 16, and the perpen- dicular breadth Ba is 10, to find the 24= AD 16=BC 40=sum. -Ad 20= 10: ■ 1 "2 Ba. sum. area by adding the sides BC and AD and multiplying half 200=area. their sum by the perpendicular breadth Ba. By the Sliding Rule. Set 1 on A to the equated length on B, and against the breadth on A you will have the area on B. By Gunter. The extent from 1 to the breadth will reach from the equated length to the area. Art. 11. To measure any regular Polygon. Definition. A regular polygon is a figure whose sides and an- gles are all equal ; they are usually denominated from the number of their sides. 1 3 *■ 450 MENSURATION OF SUPERFICIES Thus, A figure having equal sides and angles is a 'Trigon. Tetragon. Pentagon. Hexagon. Heptagon* Ocfagon. Enneagon. Decagon. Endecagon. Dodecagon. Rule. Blultiply the length of one of the sides by the number of sides ; then, this product by the half of a perpendicular let fall from the centre of the figure to the middle of one of the sides, and the product will be the area of the polygon. In the pentagon ABODE, each side is 95, and the perpendicular FG 65 36, to find th,e area. 95=lei>gth of a side. 6=number of sides. '^* 475=:sum of the sides. 32-68— i of the perpendicular. 15623-00=area of the pentagon. By ike Sliding Rule. Set 1 on A to J the perpendicular on B, and against the sum of the sides on A you will have the area on B. By Gunter. The extent from I to half the length of the perpendicular, will reach from the sum of the sides to the content. But for the more ready measuring regular polygons, the follow- ing Table, containing multipliers for all regular figures from the triangle to the dodecagon, will be of use to the learner. Number •of sides. Names. Multipliers. Number of sides. Name?. Multipliers. 3 4 5 6 7 Trigon. Tetragon. Pentagon. Hexagon. Heptagon •433oiar 1- 1-720477 2-589076 3-633959 8 9 10 11 12 Octagon. Enneagon. Decagon. Endecagon. Dodecagon. 4-828427 6-181827 7-694209 9-361 11196 If the square of the sidr of a polygon be multiplied by the mul- tipli«r of the like figure, the product will be the area of the figure isought. AND SOLIDS. 45i To measure a Circle and its Paris. - In the annexed circle ABCD, the arch line ABCD is called the periphe- ryy the length of which is called the circumference : Any line, as DB or AC, passing through the centre E, cuts the circle into two equal parts, called se- micircles, or half circles ; and such lines are called diameters of the cir- cle : If two diameters be drawn through a circle, at right angles to each other, then, the four equal divisions of the oircle are called quadrants : half the diameter as EB, is called the radiuSf or semidiameter. Art. 12. The Diameter of a Circle being given, to find the Circum- ference.'^ Rule. This may be done by either of the following propor- tions in whole numbers, as 7 is to 22, or more exactly, as 113 is to 355 ; or in decimals, as 1 is to 3- 14 159 ; so is the diameter of a circle to the circumference. * J^ote. 1. If the diameter of any circle be \ "^"jtipjied ) y^ < 3-14159, the product ) -^ ^^ eircuinference. ^divided S"yX -31831, the quotient S 2. If the diameter of any circle be \ «»«}tiplied ) . ( -886227, the product ) . ^^ ^-^^^ ^f ^,-, equal square, divided 5^^1-128379, the quotients 3. If the diameter of any circle be \ n^'^l^lied \ , ( -866024, the product ) is the side of the equilateral \ divided SI '1547, the quotient S triangle inscribed. 4. If the diameter of any circle be \ "multiplied \ , S -707016, the product \ is the^ide of the square I divided \ "^ \ 1-414213, the quoUent \ inscribed. 5. If the square of the diameter of any circle C multiplied ) . \ -785398, the product ^ • ,, ^^ Idivided I ^y ] 1-273241; the quotient \ '' ^^"^ ^''''- 6 . If the circumference of any ci rcle 7. It the circumference of any circle , ^ $ multiplied ) , J -282094, the product ) is the side of tlie ' Idivided I ^ ( 3-544907, the quotient \ square equal. 8. If the circumference of any circle ^ S multiplied ) , \ -2756646, the product ) is the side of the equilateral ''^ idivided \ ^^ } 3-6275939, the quotient S triangle inscribed. 9. If the circumference of any circle j^ C multiplied 7 ^ C -225079, the product 7 is the side of the '^ idivided 5 ^ 14-442877, the quotient J square insci'ibed. 10. If the square of the circumference of any circle , C multiplied ) , S -079577525, the product ( ■ ,,^ ^^„^ ''' idivided \ ^y \ 12-56636217, the quotient \ '' ^^'^ ^""^- n. If the area of any circle ) multiplied ) , J 1-273241, the product ) is the square of ^^ I divided \ ^ ( -785398, the quotient \ the diameter. 452i MENSURATION OF SUPERFICIES ExAMP. A circle whose diameter is 12, to find the circumference. As 7 : 22 :: 12 As 113 : 356 :: 12 As 1 : 3-14159 :: 12 12 !2 12 7)264(37-71 =^ cir- > 113)4260(37-699 cir. 37-69908 cir. 21 cumference. S 339 64 870 49 791 5C 790 49 678 10 1120 7 1017 3 103 J^ote. 3-14159 may be contracted to 31416 without any sensible difference. Art. 13. The Circumference of a Circle being given, to find the Di- ameter. RuLft. As 22 is to 7; or 355 to 113; or as 1 to -31831, so is the circumference of a circle to the diameter. ExAMP. The circumference of a circle being 326, to find the diameter. 1 2. If the area of any circle , J multiplied?, J 12-56636217, the product ? is the square of the I divided S ^ f -079577525, the quotient I circumference. 13. When the diameter of 1 circle is 1, and the diameter of another is 2, th« circumference of the first is equal to the area of the second,=3' 141592. 14. If the circumference be 4, the diameter and area are equal,= 1-273241. 15. If the diameter be 4, the circumference and area are equal,= 12-566368. Hence, because circles are the most capacious of all figures, if ihe/burth part of a circle be squared^ it will not be equal to the area of that circle (as is common- ly supposed) although the four sides added together are equal to the circumfa- rence of that circle. In a circle whose diameter is 24, circumference 75*4, and area 452-4, the fourth. part of the circumference is 18'85, the square of which is only 355*3225, that i?, 97-0775 less than the truth : and the larger the circle is, the greater will the er- rour be. For further proof of this matter ; If a cylindrical pint, beer measure, whose content is 35-25 cubicle inches, be beaten into a perfectly square form, it will con- tain only 28-902 cubick inches, which is less than the truth by 6-3484+ ; the area of the circle is 8-7615859288, and the area ofthe square only 6-8813320653076624. Hence appears the reason, why taking the fourth part of the girth in measur- ing a cylinder (or a round stick of timber) is false. 16. If the diameter of one circle be doable to that of another, the area ofthe first circle will be four times the area of the second, because the areas of circles are as the squares of their diameters ; see Art. 15. ^ AND SOLIDS. 453 22 : 7 :: 326 365:113:: 326 1 : -31831 :: 326 7 326 •72 diam. 678 326 22)2282(103 190986 22 226 63662 82 339 95493 66 355)36838(103-76 diam. 103-76906 = di- , 355 ameter. This 160 . proportion is 154 1338 the most accu- 1065 rate. 60 44 2730 — 2485 16 245 Art. 14. To find the Area of a Circle. Rule. Multiply half the diameter by half the circumference and the product is the area. If the diameter be given, find the circumference by Art. 12. If the circumference be given, find the diameter by Art. 13. ExAMP. A circle whose diameter is 12, and circumference is 37-7, given, to find the area? 18-85=half the circumference. 6=half the diameter. 113-10=ar€a of the given circle. JVote. A circular ring is the figure contained between the peri- pheries of two concentric circles. Hence, the area of a circular ring must be the difference of the areas of the two circles. Art. 15. The Diameter being given to find the Area of a Circle without finding the Circumference. RuLF. Multiply the square of the diameter by -7854,* and the product will be the area of the circle, whose diameter was given. ExAMP. The diameter of a circle being 12, to find the area ? •7854 12x12= 144 31416 31416 7854 1130976~area. * When the diameter is 1, the area is found to be -7854, and as the areas of ■circles are as the squares of their diameter?, the rule is evident. # 454 MENSURATION OF SUPERFICIES By the Sliding Rule. Set 1 on A to the diameter on B, then find -7854 (which ex- presses the area of a circle whose diameter is 1) on A, against which on B is a 4th number, then find this 4th number on A, against which on B is the area. By Gunter. The extent from 1 to the length of the diameter reaches from •7864 to a 4th number, and from that 4th number to the area. Art. 16. The Circumference of a Circle being given, to find the Area without finding the Diameter. Rule. Multiply the square of the circumference by -07958, and the product will be the area of the circle. ExAMP. The circumference of a circle being 37*7, to find the area. 1421-29 37'7 -07958 37-7 1137032 710G45 1279161 994903 2639 2639 1131 1421'29™square. 1131062582=area of the circle. Art. 17. The Dimensions of any of the parts of a Circle being given, to iind the side of a Square equal- to the Circle. Rule. If the area of the circle be given, extract the square root of the area, which will be the side of a square equal to the circle : If the diameter or circumference be given, find the area by Art. 15 or 16, and then extract the square root, as before. And this is a general rule to find the side of a square equal to any superficial fig- ure, regular or irregular : for the square root of the area of any figure whatever, is the side of a square equal to the given figure. But with regard to circles, if the diameter be given ; multiply it by -886, and the product will be the side of an equal square : or, as 1*3'545 is to 12, or 1354 to 1200: so is the diameter of a circle to the side of a square equal to the given circle. And, if the cir- cumference be given, multiply it by -282 for the side of an equal square. Or, divide it by 3-515, and the quotient »vill be the side of an equal square. ExAiiP. 1. Exam p. 2. Let the diameter of a circle be The circumference being 37 7 12, to find the side of a square to find the side of an equal equal to the circle ? square ? •886xl2=10-632=side of the 37-7X'282=10-63I =side of spuare. the square. Or, as 13-545: 12:: 12: 10 631 Or, 37'7-~3-545=10-63i, —the side. AND SOLIDS. 453 Art. 18. The Area of a Circle being given, iojind the Diameter. KuLE. Multiply the given area by 1-2732, and the product will be the square of the diameter ; then, extracting the square root of the pr(Kiuct, you will have the diameter.* ExAMP. The area of a circle being 113 09, to find the diameter. 1-2732 11309 143-986188(11*999=^ 1 114588 381960 12732 12732 21)43 2^1 229)2298 2061 143-986188 2389)23761 21501 23989)226088 215901 10187 remainder. Art. 19. Tlie Area of a Circle being given, to find the circumference. Rule. Multiply the given area by 12-566, and extract the square root of the product, which root will be the circumference required. ExAMP. The area of a circle being 113 03 to find the circumfe- rence. 12-566 11303 1420'3349(37-68=circumferencc, 9 37698 376980 12566 12566 67)520 469 746)5133 4476 1420-33498 7528)65749 60224 5525 remainder. Art. 20. The Side of a Square being given, to find the Diameter of a circle equal to the Square, whose Side is given. Rule. Multiply the given side by 1128, and the product will be the diameter of a circle, whose area is equal to the area of the * As the area of a circle, whose diameter is 1, is -7854, the area divided by •7{{54 must give the square of the diameter ; but a? l-273'i2 is the reciprocal ©f '7854, the rule is evident. 436 MENSURATION OF SUPERFICIES given square. Or, if the side of the square be divided by -886, the quotient will be the diameter. Or, as 12 to 13-34, so is the side of any square to the diameter of an equal circle. ExAMP. The side of a square being 10*635, to find the diameter «f a circle equal to that square ? 10-636XM28=12 nearly. Or, 10-635-~-886=12=diameter. Or, as 12 : 13-54 :: 10 635 : 12 nearly* Art. 21. The Side of a Square being given, to find the circumference of a Circle equal to the given Squaie. Rule. Multiply the given side by 3-645 and the product will be the circumference required. Or, divide it by 282, and the quo- tient will be the circumference. ExAMp. The side of a square being 10-631, to find the circum- ference of a circle equal to that square. 10-631x3 645=37'686=circum. Or, -282) 10-631 (37-698 circum. Art. 22, To find the Area of a Semicircle, the Diameter being given. Rule. Find the area of the circle by Art. 15, and take the half of it. In the same manner may the area of a quadrant, or a quarter of a circle, be found, by taking a fourth part of the area of the whole circle. But with regard to measuring a sector, or a segment of a circle, it will be necessary first to show how to find the length of the arch line of a sector, and the diameter of the circle to a given segment. Art. 23. A Segment of a Circle being given, to find the length of the Arch Line. Rule. Divide the segment into two equal parts ; then measure the chord of the half arch, from the double of which subtract the chord of the whole segment ; and one third of that difference, be- ing sdded to the double of the chord of the half arch, will give the length of the arch line. ExAMP. In the segment ABCD, jj the whole chord ADC is 216, and ^x-'^t!^^^^^'^'^^^^'*^*^ the chord AB or BC 126, to find y^^;f j ^N^\ the arch line ABC. /^^^ ' ^^*^^\ 126=chord AB or BC. A^:L _J JS^ 252=double. 252=double of AB. 216=ADC, to be subtracted. 12=^ difference added. 3)36=difrerence. 264=Ienglh of the arch ABC. 12x=^ difference . AND SOLIDS. 457 Art. 24. The Chord and versed Sine of a Segment being given, to find the Diameter of a Circle. Rule. Multiply half the chord by itself, and divide the product by the versed sine ; then add the quotient to the veHed sine, and the sum will be the diameter of the circle. B Example. In the segment ABCD, the chord AC is 1869-5, and the versed sine 13D 4235, to find the diameter. { half the 934-75 934-75 I chord AC 467375 664325 373900 280425 841275 423 "•5)873757-5625(2063 1 = 8470 423-6= =DE =BD, "E 26757 25410 2486-6=diameter BDF. 13475 12705 1 7706 4235 3471 Art. 25. To measure n Sector. Definition. A sector is a part of a circle, contained between an arch line, and two radii or semidiameters of the circle. ^""''^S >'^'< Rule. Find the length of half the arch by Art. 23 : Then mul- tiply this by the radius or semidiameter, and the product will be the area. ExAMP. 1. In the sector ABCD, given the radius AD or DC 72 feet, the chord AC=126 feet, and the chord AB or BC=70, to find the area of the sector. First. 70=chord AB or BG. 2 —~ Carried over. K 3 458 MENSURATION OF SUPERFICIES 140 Brought over. 126^AC, subtract. 3)14 4'GC 140 Secondly. 72-33=halfthe arch. 72=railius. 14466 60631 144'16=length of the arch 5207-76=area. [ABC,by Art. 23. 72-33 E.YAMP. 2. In the sector ABCD, greater than a semicircle, given the radius AE or ED=112, the chord BD (of half the arch ABD)==204, and the chord BC (of half the arch BCD)=120, to find the area of the sector. 120=::BC. ! 2 240 204 subtract. 3)36 12 240 Add. 252=half the arch ABD. 112=radius. S04 252 262 Che arch Art. 23. 28224=area of the sector. ot^s— ^i-engthof -^^~ I BCD, by . Art. 26. To find the Area of a Segment of a Circle. Definition. A segment of a circle is any part of a circle cut off by a right line drawn across the circle, which does not pass through the centre, and is always greater or less than a semicircle. ExAMP. 1. To find the area of the segment ABC, whose chorrt AC is 172, tlie chord of half the arch ABC, viz. BC=104, and tlit> versed sine BD=68-48. RuL£. By Art. 23, find the length of the arch line ABC, and by Art. 24, the diameter FB ; (hen multiply half the a chord of the arch ABC by half the diam- eter, and the product will be the area of the sector ABCE : then find the area of the triangle AEC, whose base AC is 172, and perpendicular heiglU 34, found by subtracting the versed sine BD from half the diameter; and the area of Ibe trian- gle AEC, being subtracted from the area of the sector ABCE, will leave the area of the segment ABC AND SOLIDS. 4i9 104=BC. 2 208 172= AC, subtract. 3)36 86=halfADC. 86 516 688 58-48)7396-00(126-47=DEF 5848 ' 58'48=BD, add. 12 208 add> 220=arch line ABC. 110=^halfarch. 92*475=radiu8. 110 1 5480 184-95=diameter BF. 11696 37840 I diameter. 35088 924750 92475 27520 23392 41280 40936 10172-25=areaorthe sector. 344 86=half the base=AD. 101 72 25=area of the sector. 34=perpendicular DE. 2924 =area of the triangle. 7248 25=area of the segment. 2924=area of the triangle. ExAMP. 2. In the segment ABCD greater than a semicircle, giv- en the chord of the whole segment AD= 136, the chord AC of half the arch ACD=146, the chord AB or BC one fourth of the arch ACD =86, and the radius AE or ED= 80, to find the area of the segment ABCD. First find the area of the sector ABCDE, by Art. 25, at the second Example ; then find the area of the triangle AED, by Art. 6, and, ad- ding the area of the triangle to the area of the sector, you will have the area of the segment. 86=chord AB. 172 146=chord AC, subtract. 736 8-66G 172 =double of AB, add. 180-666=arch line ABC. 80=radius. 3)20 IM53»280=area of the sector Carried over* 460 MENSURATION OP .SUPERFFCIES Brought over. 68=half the base AD. 42=perpendicular E 136. 2856=area of the triangle AED. 14453'28=area of the sector, [add. 17309'28=area of the segment. 136 272 Note 1. The area of a Lune or Crescent, is calculated by the preceding rule. A Lune is a figure made by two circular arcg, which intersect each other, as ACBD. The area of the Lune is the difference of the two segments, which are contain- ed by the arcs and the chord. Thus the difference of the segments ACBE and ADBE is the area of the crescent ACBD. Note 2> A Circular Zone is a figure con- tained between two parallel chords. If the chords be equal, it is called a middle zone, as A ABCD. The area of a zone is evidently the difference between the area of the circle and D the areas of the two segments. Art. 27. To find the Area of an Ellipsis, Definition. An ellipsis, or oval, is a curve which returns into it- self like a circle, but has two diameters, one longer than the oth- er, the longest of which is called the transverse, and the shortest the conjugate diameter. Rule. Multiply the two diameters of the ellipsis together ; then multiplying the product by -7854, this last product will be the area of the ellipsis. ExAMP. In the ellipsis ABCD, the transverse diameter AC is 88, and the conjugate diameter BD is 72, to find the area, 88 72 176 C16 C336 7854 25344 31680 50688 44352 The conteftt is found by the sliding rule and Gunter, in the same way as the circle, only using the product of the two diam- eters as the square of the diameter of a circle. 4976-2944=area. Mensuration of Superficies is easily applied to Surveying : lhu«, take the angles of the plot with a good compass, then Dpkeasure the AND SOUDS. 461 sides with Gunter's chain, which note down in links (or chains and links, which is done by separating the two right hand figures of your links by a comma, your chain being 100 links) then cast up the con- tents, according to the rule of the figure, cutting off the five right hand figures of the product, and those at the left hand, if any, are acres ; then multiply the five figures cut off, by 4, by 40, and by 272i, cutting off as before, and those at the left hand, will be roods, poles, and feet, respectively. Section II. Of Solids. Solids are measured by the solid inch, foot, or yard, kc. 1728 of these inches, that is 12x12x12, make one cubick or solid foot. The solid content of every body is found by rules adapted to their particular figures. Art. 28. To measure a Cube.* Definition. A cube is a solid of six equal sides, each of which is an exact square. * Here follows a Table of the Proportions, which the following Solids hav;-^ to the Cube and Cylinder, having the same Base and Altitude. Solid Inches. 1. A Cube whose side is 12 inches, contains 1728 2. A Prisw, having an equilateral triangle, whose side is 12 ) j.'q^.oi im;hes from its Base^ and its Altitude 12 inches, contains \ is 'i-1 of the above cube, and contains \ 6. A Sphere or Globe, whose axis or diameter is 12 inches, equal ) to the side of the cube, is 11 of it, and contains > 7. A Cone, whose base and altitude are each 12 inches, equal ) to the side of the cube, is JL. of it, and contains ^ 8. A ParaJboHck Conoid, -whose diameter at the base and height, ) p'~?.ro«5 are each 12 inches, being ^ its circumscribing cylinder, contains s, ' ^'^^"^ 9. A Hyperbolick Conoid, whose height, and diameter at the ) base, are each 12 inches, is 5 of its circumscribing cylinder, and V 565'49 contains ) 10. A Parabolick Spindlc,-whose height and middle diameter are } ^-o'^.oo^ each 12 inches, is JL of its circumscribing cylinder, and contains <> '""^ ^* Hence arises a different method of finding their contents. General Rule. If the base of the solid, whose contents you would find, I e rectilinear, consider it as Parallehpipedon ; if curved, as a Cylinder, and find the content accordingly : then take such a part of the content, thus found, as is specified in the preceding Table, which if the parts be taken in incher-, will bn the solid content of the given Bgure, in inches, wliich, divided by 1728, will givr the cubick feet. ExAMP. 1. There is a triangular prism, the side of whose base is 48 inche-, and whose perpendicular height is 108 inches : what is its solid content? The base being right lined, I consider it as a paraUelopipedon, the side cf whose base it 4o inches, and whose length is 108 inches, and as 784*24 is con - 249-41:^. 1357-17 904-78 452-38829 462 MENSURATION OF SUPERFICIES The solid foot is composed of 1728 inches ; for a 8011(1, that is 1 foot, or 12 inches every way, that is 12x12x12, contains 172S inches. Rule. Multiply the side by itself and that product by the same side, and this last product will be the solid content of the cube.t ExABip. The side of a cube AB, being 18 inches, or 1 foot and 6 inches, to find the content ? 1 foot 6 inches=l'5 foot. 18 inches. 15 18 75 144 15 18 Carried over. 2-25 324 tained 2'20340712 times in a cubick foot ; 2-20340712 is a divisor, to divide the content of the parallelopipedon by; therefore 48X48X108^2*203407 12= 112930-56 solid inches=65-353 solid feet. Had the dimensions been given in feet, it would have been 4x4X9-t- 2-203407 12=65-353 feet. ExAMT. 2. There is a square pyramid, whose height is 12 feet, and the side of whose base is 3-5 feet ; what is its content ? 3-5 X 3-5 X 12-^3=49 feet, Ans. ExAMP. 3. There is a triangular pyramid, whose height is 15 feet, and the side of whose base is 5 feet : what is its content ? 5 X5 X 15-^-7=53-57 feet, Ans. ExAMP. 4. There is a cylinder whose diameter is 2-5 feet, and whose length is 24 feet ; what is its content ? Here, the diameter is to be considered as the side of the base of a parallelo- pipedon. Therefore, 2-5 X2-5X24Xll-M4=l 17-857 feet, Ans. Ex AMP. 5. There is a spherical balloon, whose diameter is 50 feet ; how many eubick feet of air does it contain ? Here, the diameter is to be considered as the side of a cube. Therefore, 50x60X50x11-^21=65476-19 feet, Ans. ExAMP. 6. There is a cone, whose height is 15 feet, and the diameter of whose base is 5 feet ; what is its content? Here, the diameter of tlie base is to be considered as the side of the base of a parallelopipedon, and its height, as the length. Therefore, 5X5X15X5-^19=98^84 feet, Ans. ExAMP. 7. There is a parabolick conoid, whose diameter at the base is 2-9 feet, and whose height is 6 feet ; what is the content ? This solid being ^ of a cylinder ; we must first find the content as of that of a cylinder, and then halve it. Therefore, 2-9 X2-9 X6 X 11-^14=39-647, and 39-647-^2=19-823, Ans. ExAMP. 8. There is a hyperbolick conoid, whose diameter at the base is 2-9 feet, and whose height is 6 feet ; what is the content ? Fii-st, find the 14375 240528760=area of the base. If the square of the diameter of a cylinder be multiplied by •7854, and the solidity divided by that product, the quotient will be the length, and. if the content be divided by the length, the quotient will be the area of the end, from which the iiameter \$ found by Art, IR. L 3 iuo x^JENSURATlON OF SUPERFICIES 'j'he learner may, for his practice, reduce all the dimensions to inches, and find the solid content in inches, which being divided by 1728, the quotient will be the solid content in feet : or, if he finds the area at the end in inches, and multiplies that by the length in feet, and divides by 144 ; the quotient will be feet. - This is a general rule for finding the content of any straight solid body, of equal bigness from end to end, of whatever form the bases are : for, if the area of the base be multiplied by the length, the product will be the solid content. By the SUditig Rule. Set 13 5, the square root of 183-34 (which is a guage point arising from the division of 144 by -7854) found on D, to the diam- eter found on C, and opposite to the length, on D, you will find the content on C. Or, as 42 64 is to the circumference ; so is the length in feet to a fourth number, and so is that fourth number to the answer. Note. The superficial content of a cylinder is found by multiply- ing the circumference of one of the bases into the length, and to the product adding the areas of the two bases, or ends. When the diameter is given in inches, to find what length will make a solid foot. Rule. As the given diameter is to 13 631 : so is 12 to a fourth number, and so is that fourth number to the required length. If the diameter be given in foot measure : Rule, as the given diame- ter is to 1*128 : so is 1 to a fourth number, and so is that fourth number to the required length. Or, divide 1728 by the area at the end in inches, and the quotient will be the required length. To find how much a Cylindrick or round Tree, that is equally thick from end to end, will hew to, when made square. Rule. Multiply twice the square of its semidiameter by the length, then divide the product by 144, and the quotient will be the answer. If the diameter of a round slick of timber be 24 inches from end to end, and its length 20 feet : how many solid feet will it contain, ivhen hewn square ; and what will be the content of the slabs which reduce it to a square ? 12X12X2X20 —- =40 feet, the solidity when hewn square. 24X24X-7854X20 jj^ =62 8 feet, or 2x2X*7854x20=62-8 the total solidity, whence 62-8—40=22 8 feet, the solidity of the slabs. Note. The rule of workmen for measuring round timber is to multiply the square of the quarter girt or one fourth of the cir- cumlerence, by the length. This rule allows about one fifth, for the bark, waste in hewing, &c. The example above, in which the diameter of the cylinder is 1 foot 9 inches, and the length 12 fee* AND SOLIDS, 4&r 6 inche?, will give the quarter girt 1 3744 feet, and Hie solid con- tent is l-3744^xl2-5=23*61 feet, which is nearly four Mha of 30C625, the content hy the accurate rule. A rule, nearly correct, is to multiply twice the square of one fifth of the circumference by the length. Thus, in the example, } of the circumference is 10995, and 2 x .1 -0995^x12-5= 30-22 feet. Art. 31. To measure a Prism. Definition. A prism is a body with two equal or parallel ends, either square, triangular, or polygonal, and three or more sides, which meet in parallel lines, running from the several angles at one end, to those of the other. Rule. Prisms of all kinds, whether square, triangular or poly- gonal, are measured by one general rule, viz. Find the superficial content, or area at the base (or end) by the proper rule of Sect. K and this multiplied by the length, or height of the prism, will give the solid content. ExAMp. The side of a stick of timber, AB, hewn three square, is 10 inches, and the length, AC, is 12 feet, to find the content ? Side= 10 inches. I Perpendicular=4-33 inches. 43-3=area at the end. 12 feet=length. 144)519G(3-6 feet, content. 432 87{> 864 12 Note. The superficial content is found by adding the areas of the several quadrilateral and triangular figures which compose it. Art. 32. To measure a Pyramid. Definition. Solids, which decrease gradually from the base till they come to a point, arc generally called pyramids, and are of dif- ferent kmds, according to the figure of their bases ; thus, if it has a square base, it is called a square pyramid ; if a triangular base, a triangular pyramid : If the base be a circle, a circular pyramid, or simply a cone. The point, in which the top of a pyramid ends, is called a Vertex, and a line drawn trom the vertex, perpendicular to the base, is called the height of the pyramid. Rule. Find the area of the base, whether triangular, square, polygonal or circular, by the rules in superficial measure : then, multiply this area by one third of tiie height, and the product will be the solid content of the ryr^vmil 468 MENSURATION OF SUPERFICIES ExAMP. !• Id a triangular pyramid, the height BE, being 48, and each side of the base 13 : the base being a triangle, let the perpendicular height DE be 1 1 ; to find the content. 5-5=half ED. 13=base AC, 165 55 7]'5=area ef the base. 16=1 of the height EB. 4290 715 n44'0=content ExAMP. 2. In a quadrangular pyramid, the height JBE being 48, and each side of the base 13, to find the content. 13 13 39 13 169=area of the base. 16=J- of the height £B. 1014 169 2704==coDtent. ExAMP. 3. To measure a Cone^ — The diameter AC being 13, and the height BD 48, to find the content, 13 13 39 13 169 •7854 676 845 i 1352 1183 132'7326=area of the base. AND SOLIDS. 4«9 Brought up. 132-7326 16=i of the height. 7963956 1327326 2123-7216=content. Note. The superficial content of all pyramids is fouad by tak- ing the sum of the several areas, which compose them. That of a cone, by multiplying the circumference of the base into half the line joining the vertex and any point in that circumference, and adding the area of the base to the product. Art. 33. To measure the Frustum of a Pyramid. Definition, The frustum of a pyramid is what remains after the top is cut off by a plane parallel to the base, and is in the form of a log greater at one end than the other, whether round, or hewn three or four square, &c. Rule, If it be the frustum of a square pyramid, multiply the side of the greater base by the side of the less ; to this product add one third of the square of the difference of the sides, and the sum will be the mean area between the bases ; but if the base be any other regular figure, multiply this sum by the proper multipli- er of its figure in the Table, Art. 11. and the product will be the mean area between the bases : lastly, multiply this by the height, and it will give the height of the frustum. ExAMP. 1. In the frustum of a square pyramid the q side of the greater base AD=16, the side of the less, ^ ^ BC=6, and the height KF=40, to find the content. ^ ^ 15=AD. 15 6=BC. 6 Prod.=;:90 9=difference. Add 27 9 117 3)81?=square of the difference. A.< X 40 — 27=1 of the square. 4680=content. Or, if it be a tapering square stick of timber, take the girth of it in the middle ; square ^ of the girth (or multiply it by itself ia inches) then say, as 144 (inches) to that product ; so is the length, taken in feet, to the content in feet. ExAMP. 2. What is the content of a tapering square stick of timber, whose side of the largest end is 12 inches, of the least end, 8, and whose length is thirty feet. One fourth of the girdj in the middle=10, and 10x10=100, the area in the middle ; then, as 144 ; 100 :; 30 feet : 20 03 feet the content. 4T0 MENSURATION OF SUPERFICIjt:^ By the Sliding Rule. Set 12 on D to -} of the circumference on C, and against the length on D is the answer on C. By Gttnter. The extent from 12 to i of the circumference doubled, or twice turned oyer, will reach from the length to the content. ExAMP. 3. In the frustum of a tiiangular pyramid, the side of the greater base AC;=15, as before, the side of the leys BD=6, and the height EF=40, to find the content. 15=AC. 6=:BD. 9=difference of the sides. 9 i)81=:square of the diiference. 27=i of the square. 15 e 90 Add 27 117 '433 multiplier. 351 351 468 50-661 =mean area. 40=height. 2026-440=content. Or, if it be a tapering three square stick of timber, you may find the area midway from end to end, then, as 144 is to that area, so is the iength, taken in feet, to the content in feet. Ex AMP. 4. To ineamre the Frustum of a Cone. Rule. Multiply the diameters of the two bases together, and to the product add one third of the square of the difference of the diameters : then multiplying this sum by -7854, it will be the mean area between the two bases, which being mullipiied by the length of the frustum, will give the solid content. Or, to the areas of the top and bottom add the square root of the product of those areas, and the sum, multiplied by one third of the height of the frustum, will give the solidity. When fjijfures run uniformly taper ; but not to a point (they be- ing considered as portions of the cone or pyramid) we may find the solidity by supplying what is wanting to complete the figure, and then deducting the part (iut ofi'. A general rule for completing every straight sided solid, nshose ends are parallel and similar. As the difference of the top and bottom diameters is to the per- pendicular height, (or depth which is the same :) so is the long^'Si diameter to the altitude of the whole cene or pyramid. AND SOLIDS. 471 ExAMp. 1. The former cone in Art. 32, Exanip. 3, being cut off in the middle, the greater diameter AC is 13, the less BD 6i, and height EF 24, to fmd the content of the frustum. AC=13 inches. ' 13 BD=6-5 inches. 6-5 65 78 84'6 Add 14083 98-583 •7854 394332 492915 788664 690081 6 5=difference. 6-5 325 390 i4-083=i of the square. 144)1858'248(12-9045 feet content 144 77-427|0882=mean area. 24 feet=length. 418 288 309708 154854 1302 1296 1858-248=content. 648 576 720 720 ExAMP. 2. What number of barrels, each 32 gallons of Ale measure, is contained in a cistern whose largest diameter is 6 feet, and smallest diameter 5 feet, and whose depth is 8 feet ? 6x5=30 12 i 6--5I :1= 01- 3 ^J 30i mean diameter. •7854 23-5620 •26ia 23-8238 mean area. 8 190-5904 content in feet. 1728 inches in a solid foot. 329340 2112 cubic inches, which divided bj- 9024, the cubic mches in a barrel or 32 gallons, gives 36-5 barrels nearly, Ans. 472 MENSURATION OF SUPERFICIES If the answer had been required in Beer Measure, where the barrel contains 36 gallons, the answer would have been 32*4 bar- rels. Note. If when the end diameters of a conical cistern are giv- en, it is required to find the length of the cistern to contain a cer- tain number of barrels; divide the cubic fieet contained in the number of barrels by the mean area, and the quotient will be the height. Let the mean area be as in the last Ex. to find the length of the cistern to contain 60 barrels of 32 gallons of Ale measure. 26111 111 &c.=cubic feet in 50 barrels, which divided by i23 8238, the mean area, gives 1095 feet, for the length of the cistern, Ans. To find the diameters of the cistern, when the content, and length, and difference of the diameters, are given, see Art. 53. Art. 34. To measure a Sphere or Globe, Definition. A sphere or globe is a round solid body, in the mid- dle of which is a point, from which all lines drawn to the surface are equal. Rule. Multiply the cube of the diameter by -5236, and the product will be the solid content. Or, multiply the circumference by the diameter, which will give the superficial content ; then multiply the surface by one sixth of the diameter, and it will give the solidity. Or, multiply the cube of the diameter by 11, and the product divided by 21, will give the solidity. ExAMP. The diameter, AB, of a globe, is 4-5 feet ; to find the solid content. 4-5 4-5 225 180 20-26 4-5 10125 8100 91-125 •523G 546750 273375 182250 455G25 47-7130500 AND SOLIDS. 473 ' Note. If the circumference, or greatest circle of the sphere, be given, multiply the cube of it by 016887 for the content. 'I'he surface of the globe may be found by multiplying the square of the diameter by 3' 1416 ; or by multiplying the area of its greatest circle by 4, or the square of the circumference by •3183. When the solidity of a globe is given, the diameter maybe found by dividing the solidity by -5236, and extracting the cube root of the quotient. Or, if the circumference be required, divide the solidity by •016887, and the cube root of the quotient will give it. Art. 35, To measure the Solidity of a Frustum or Segment of aGlobe^ Definition. The frustum of a globe is any part cut off by a plane. Rule. To three times the square of the seraidiaraeter of the base, add the square of the height; then multiplying that sum by the height, and the product by -5236, you will have the solid content. ExAMP. The height BL) beirig 9 inches, and the diameter of the base AC 24 inches : to find the content. 12=8emidiam6ter. 4617 jy 12 -5236 144=square. X 3 27702 13851 A 9234 23085 432 . f. Add 9X9= 81= J f?"^^.^^, — ^ ^^'^ ^^•S^^-2417-4612=solid content. 613 X 9=height. 4617 To measure the Surface of a Frustum or Segment of a Globe. Rule. Find the diameter of the globe by Art. 24, and the sur- face of the whole globe, by Art. 34 ; then, as the diameter of the globe is to the height of the frustum ; so is the surface of the globe to the surface of the frustum ; then, by Art. 15,fifjd the area of the base ; add these two together, and the sum will be the whole surface of the frustwni. Art. 36. To measure the middle Zone of a Globe. Definition. This part of a globe is somewhat like a cask, two equal segments being wanting, one on each side of the axis. KuLE. To twice the square of the middle diameter, add the square of the end diameter; multiply that sum by -7854, and that product, multiplied by one third of the length, will give the go* lidity. M 3 474 MENSURATION OF SUPERFICIES Or, to four times the square of the middle diameter add twicr the square of the end diameter ; that sum multiplied by '7854, and that product by one sixth of the length, will give the solidity. Note. This rule is applicable to the frustum of a cone or py- ramid. If the middle diameter of a zone be 20 inches, the end diame- ters each 16 inches, and length 12 inches : Required its solidity ? 20x20x2+ 16xl5x-7854x4=3317-6296, Ans. Art. 37. To measure a Spheroid. Definition. A spheroid is a solid body like an egg^ only both it§ ends are the same. RuLEi. Multiply the square of the diameter of the greatest circle, viz. the diameter of the middle (DB in the fig- ure) by the length AC, and that pro- , duct by -5236, and you will have the -A, solidity. ExAMP. Tke diameter BD being 20, and the length AC 30, to find the content. 20x20x30x-6236=6283-2, Ans. Art. 38. To measure the middle Frustum of the Spheroid, Definition. This is a cask like solid, wanting two equal seg- ments to complete the spheroid. Rule. The same as in Article 36. If the middle and end diameters of the middle frustivm of a sphe- roid be 40 and 30 inches, and its length 50 ; what is its solidity ? 50-^-3— 16 6, then40x40x2l|-30x30X-7854xl6-6=53454-324,Ans. Art. 39. To measure a Segment^ or Frustum of a Spheroid. Definition. This is a part of a spheroid made by a plane, par- allel to its greatest circular diameter. Rule. 'To four times the square of the middle diameter add the square of the base diameter, then multiply that sum by -7854, and the pro 27729 tons. Depth in the hold, 30) As a merchant ship, 29188*6 ts QUESTIONS IN MENSURATION. 4?; (QUESTIONS IN MENSURATION. 1. THE largfest of the Egyptian pyramids is square at the base, and m easures 693 feet on a side : how much ground does it cover ? 696 X393 1764 •-^=^— = 1764 poles, and 7^=11 acres and 4 poles, Ans. 2. What difference is there between a floor 20 feet square, and two others, each 10 feet squa re ? 20X20—10X10+ 10X10=200 feet, Ans. 3. There is a square of 2500 yards in area : what is each side of the square, and the breadth of a walk along one side and one end, which may take up just one half of the square ? 2500 V2500=50 yards, each side. V"^— ^35-35, and 50— 33-35 =:14'65 yard?, breadth of the walk, Ans. 4. A pine plank is 16 feet and 5 inches long, and I would have just a square yard slit off: at what distance from the edge must the line bo drawn ? A square yard=1296 inches, and 16 feet 5 inches=197 inches. 1 296 Therefore, -r^=6-J-f^ inches, Ans. 5. If the area of a triangle be ,900 yards, and the perpendicular 40 yards: required the length of the base ? 900X2 ■ ■ =45 yards, Ans. 6. If the three sides of a plane triangle be 24, 16, and 12 perch- es : required its area ? 24+16+12 2 =2^5 26—24—2; 26—16=^10; 26—12=14, and 2 V26X 14X10X2=85-32 perches, =area. Again, as 24: 16+12 :: 16 — 12 : 4-6+, the difference of the segments of the base ; then, 4-6+ ' 12— ~--=9'6, and >/ 12X12— 9-6x9-6=7-ll the perpendicular on the longestside; whence 24^2X7- 11 =85*32, area as above. 7. Required the area of a circular garden, whose diameter is 12 rods? 12X12X'7854 = 113 0976 poles, Ans. 8. The wheel of a perambulator turns just once and a half in a rod : what is its diameter ? 16 5X| = ll circumference, and l]X-31831=3i feet, Ans. 9. Agreed for a platform to the curb of a round well,dt7id. per cquare foot : the inward part, round the mouth of the well, is 36 inches diameter, and the l>reail(h of (he platform was to be 15| inches : wiiat will it como (o .' 484 QUESTIONS IN MENSURATION. 36-|-l5-5x2=67the greatest cIiam.;67x67X-7864—.36x36X'7854 2507-8722 s= ~ — =17'4157 square feet, at 7|d. per foot, =10s. 10,\(1. [Aug. 10. Required the difterence between the area of a circle, whose radius (or semidiameter) is 60 yards, and its greatest inscribed square ? 60X2=100 the diameter, and lOOXlOOX-7854=:7864 the area of the circle ; then, 60X50X2=:6000 the area of the greatest in- scribed square, and 7864 — 5000=2854, Ans. 11. There is a section of a tree 25 inches over ; I demand the difference of the areas of the inscribed and circumscribed squares, and how far they differ from the area of the section ? ______ 26X25— 12 5xT2-5X2=312'5 the difference of the squares. 25X25 — 26X26X-7854=134125 the circumscribed square, more than the section, and 25X25X-7854~"12-5X 1^-6 X 2=178-375 inscribed square, less than the area of the section. 12. Four men bought a grindstone of 60 inches diameter : how much of its diameter must each grind off, to have an equal share of the stone, if one first grind his share, and then another, till the stone is ground awaj, making no allowance for the eye ? Rule. Divide the square of the diameter by the number of men, subtract the quotient from tho square, and extract the square rgot of the remainder, which is th«i lengtii of the diameter after the first man has ground his share ; this work being repeated by subtract- ing the same quotient from the remainder, for every man, to the last ; extract the square root of the remainders, and subtract those roots from the diameters, one after another ; the several remain- ders will be the answers. 60 - From 60 60 Take 61-9615 4)3G00 Remains 8 0385— 1st share Quot.=900 From 519615 Take 424264 From 3600 Take 900 Rem. 9 6361~=2d share. v'2700=551-9615, to be taken from 60. Subt. 900 From 42 4264 V'1800=42-4264, from 51-0616. Take 30' Subt. 900 — ' ^/900=30, from 42-4264. Rem. 12 4264=3d share. And 30 inches=1th share. QUESTIONS IN MENSURATION. 485 13. If a cubick foot of iron were hammered, or drawn, inlo a square bar, an inch about, that is, ^ of an inch square : required its length, supposing there is no waste of metal ? 12X12X12 •2qX25X 4~''*^^^ inches,=576 feet, Ans. 14. Required the axis of a globe, whose solidity may be just equal to the area of its surface ? -7854X4 '5^36 '^^^ inches, Ans. ■ 15. A joist is 71 inches wide, and 2i thick ; but I want one just twice as large, which shall be 3f inches thick : what will be the breadth ? 75X2 25x2 —— =9 inches, Ans. 16. I have a square stick of timber 18 inches by 14 ; but one of a third part of the timber in it, provided it be 8 inches deep, will serve : how wide will it be ? 18X14 — - — -~-8=]0i inches, Ans. 17. A had a beam of oak timber, 18 inches square throughout, and 25 feet long, which he bartered with B, for an equilateral tri- angular beam of the same length, each side 24 inches : required the balance at Is. 4d. per foot? 18x18x25 — — =56-95, solidity of the square beam. The perpendicular let fall on one of the sides of the triangu- lar beam is 20*7846 inches, and the half perp. = 10-3923 ; then ■10-3923x24 — —jj- =1732 foot, area at the end, and 1-732x25=43 3 feet, solidity of the triangular beam ; therefore 56-25--43-3=12-95 feet, at U. 4d. per foot=17s. 3-2d. balance due to A, Ans. 18. What is the difference between a solid half foot, and half a foot solid ? 12x12x6 g^g^^ =4, therefore, one is but -| of the other. 19. A lent B a solid stack of hay, measuring 20 feet every way ; sometime afterward, B returned a quantity measuring every way 10 feet : vy^hat proportion of the hay remains due ? 20X20X20—10X10X10=7000 feet=l, Ans. 20. A ship's hold is 75^ feet long, 18J wide, and 7A deep : how many bales of goods 3i ieet long, 2i deep, and 2f wide, may be slowed therein, leaving a gang way the whole length, of 3i feet wide ? 75-5X18-5'x7-25— 75-5X7-25X3-2"5 on- .. , , ~ =38i5-44 bales, Ans. 3-5X2-25X2-75 21. If a stick of timber be 20-1 feet long, 16 inches broad, and 8 inches thick, and 3i solid feet be sawed off one end : how long will the stick then be ? 1728x3o 20-5 — '^xlT'^'^^ ^"^^^ ^v inches, Ans, 4ati QUESTIONS IN MENSURATION. 22. The solid content of a square stone is found to be 1361 feet j its length is 9^ feet : what is the area of one end ? and if the breadth be 3 feet 11 inches, what is the depth ? 136-5X1728 2069-0526 9-5X12 —^^^'^ 20G9C52G inches, and ■;pj = 44022 inches, Ans. 23. I wotild hnve a cnbick box made capable of receiving just 50 bushels, the bushel containing 2150-425 solid inches : what will be the length of the side ? 3 V2150-4X56=47'55 inches. 24. A statute bushel is (0 be made 8 inches high, and 18i inches diameter, to contain 2176 cubick inches ; (though the content of the dimensions is but 2160 425 inches) 1 demand what the diame- ter of the bushel must be, the height being 8 inches; and what the height, the diameter being I81 inches, to contain 2176 cubick inches ? Solidity. Height— 8)2176 and ^/272X1 •273=186 diameter. 18-6X18-5 X 7854=268 80315=area, and the solidity Area= 272 2176-i-2.68-8=8-0956 inches, height. 25. There is a garden rolling stone 66 inches in circumference, and 31 cubick feet are to be cut off from one end, perpendicular to the axis : where must the section be made ? 1728x3-5 A — , 10 r — =14-65 inches from one end, Ans. Area= 4125 ' 26. I would have a syringe of ]i inch diameter in the bore, to bold a quart, wine measure : what must be the length of the pis- ton, sufficient to make an injection with ? 1 5X1'5X-7854=1-76715, and 231 --4=57 75 the cubick inches 57-75 in a quart, then j.„g^j;=32 679 inches, Ans. 27. If a round pillar, 9 inches diameter, contain 5 feet: of what diameter is that column, of equal length, which measures 10 times as much ? As 5 : 9X9 :: 5X10 : 810, and -v/810=28-46 inches, Ans. 28 There is a square pyramid, each side of whose base is 30 inches, and whose perpendicular height is 120 inches, to be divid- ed by sections parallel to its base into 3 equal parts : required the perpendicular height of each part? 30X30X40=36000 the solidity in inches, now | thereof is 24000, 3»n(l 1 is 12000. Therefore, ^;.3e000:,.0X,20X,.0::jf^O^°J;'SoS'r'>- Vi 152000=104-8 Also, V'576000=83-2. Then, 120— 104-8 ^152 length of the thickest part, and 104-8— 83 2=21-6 length of the middle part, consequently 83 2 is the length of the top part. 29. Suppose the diameter of the base of a conical ingot of gold la be 3 inches, and its height 9 inches ; what length of wire may QUESTIONS IN MENSURATION. 487 be expected from it, without loss of metal, the diameter of the wire being one hundredth part of an inch ? 3x3x-7854X3=:21-2038 the solidity of the cone. 21-2058 •01 X Ol 'x'^7S^~^^^^^^ inch. =4 miles, and 460 yards, Ans. 30. Suppose a pole to stand on a horizontal plane 75 feet in height above the surface : at what height from the ground must it be cut ofl', so as that the top of it may fall on a point 55 feet from the bottom of the pole, the end, where it was cut off, resting on the stump, or upright part ? As the whole length of the pole is equal to the sum of the hy- potenuse and perpendicular of a triangle, (the 55 feet on the ground being the base) this, as well as the following question, may be solved by this Rule. From the square of the length of the pole (that is, of the sum of the hypotenuse and perpendicular) take the square of the base ; divide the remainder by twice the length of the pole, and the quotient will be the perpendicular, or height at which it must be cut off. 75X2 31. Suppose a ship sails from latitude 43°, north, between north and east, till her departure from the meridian be 45 leagues, and the sum of her distance and difference of latitude to be 1 35 leagues : I demand her distance sailed, and latitude come to? 135X135— 45X4*5 ^^ , . ^„^„ ,„„ ., ^ , =60 leagues, and 60X3=180 miles=3 de- 135X2 grees the difference of latitude, 135 — 60=75 leagues the distance. Now as the vessel is sailing from the equator, and consequently Ihe latitude is increasing : Therefore, To the latitude saijed from 43'',00' N. Add the difference of latitude 3 ,00 And the sum is the latitude come to=46 ,00 N; BOOK KEEPING BOOK KEEPING is a systematic record of mercantile trans- actions. Every mercantile transaction consists in giving one thing for another. This change of property requires a distinct record in the books prepared for the purpose, so as to enable the man of business to know the true state of his affairs, and of his accounts with an individual. The importance of a correct knowledge of Book keeping, to the man of business is obvious. His books should exhibit the result of each transaction, and the general result of the whole. Book keeping may be performed either by Single or Double Entry. The method of book keeping by single entry is the most simple, and is sufficient for the generality of Mechanics, Farmers, Retail Merchants, &c. The method by Double Entry is more systematic in its principles, and more certain in its conclusions, and is much to be preferred for wholesale or any extensive business. in Single Entry only, persons are entered as debtor and creditor. BOOK KEEPING BY SINGLE ENTRY. In the practice of single entry, two principal books, the Day Book or Waste Book, and the Leger, and one auxiliary book, the Cash Book, are necessary. ]. THE DAY BOOK OR WASTE BOOK. The Day Book sliould begin with an account of all the property ^ debts, &c. of the person, and be followed by a distinct record o* all the transactions of the trade in the order of time in which they occur. Some accountants use also a Blotter, in which tlie changes of property are recorded, and the Day Book is only a copy of the Blotter, written in a more- fair and plain manner Each page of the Day Book should be ruled with three columns on the right side for pounds, shillings, and pence, or with two co- lumns for dollars and cents, as the accounts are to be kept in one or the other of these denominations of money. The following is the order observed in making an account in the Day Book : First, the date ; next, the name of the person with the abbreviation Dr. or Cr. at the right hand, as he is debtor or credi- tor by the transaction; and then, the article or articles with the price annexed, unless the article be money, and the value carried BOOK KEEPING, &c. 489 out in the ruled columns, with the sum of the whole placed direct- ly under, when there is more than one charged. Thus, for ex- ample, January 1st, 1820. David Bradley, To 2 yards of broadcloth at 42s. per To lOlbs. loaf sugar at 2s. 2d. Dr. yard, £ 4 1 5 a. 4 1 5 8 8 January 2d. Simon Jones, By 3 bushels of wheat at 9s. Cd. Cr. I 8 6 The following rule shows whether a person is to be entered as Dr. orCr. on the Day Book. The person who receives any thing from me is Dr. to me, and the person from whom I receive is Cr. Or, The person, who becomes indebted to me, whether by re- ceiving goods or money or by my paying his debts, must be enter- ed Dr. ; and the person to whom I become indebted, whether by receiving from him goods or money, or by the payment of my debts, must be entered Cr. The following general direction is to be observed in keeping the Day Book, viz. Enter on the Day Book every case of debt or credit relating to the business in the order of time in which it takes place, and in language so explicit as not to be mistaken. This rule is most important, because the Day Book is the deci- sive book of reference in case of any supposed mistake or error in the accounts in the Leger. 2. THE LEGER. The various accounts of each person are collected from the Day Book, and placed or posted under his name, and on two opposite pages of the Leger, as they are Dr. or Cr. The name of the person is to be written in large and fair characters as a title, and the ac- counts in which he is Dr. are to be written on the left hand page, and those in which he is Cr. on the right hand page of the same folio. If the name be written only on the Dr. page, the title of the other page is to be, Contra or Ca. Cr. The Leger is ruled with a margin for the date of each transaction, or with a column for the page of the Day Book which contains the account, or with both. It must also be ruled with two or three columns on the right of each page for the denominations of money, as they may be Dolls, and Cents, or£ s. and d. If the Leger be a wide Folio, the ac- counts of Dr. and Cr, may be placed on tho same page,, as in the following example^ " 490 BOOK KEEPING Dr. Jan.l 5 11 Jan.l 6 James '. Fowler, Cr. $ c. $ c. Iron 3Cwt. 15 60 Jan.2 4 Wheat 12 bushels, 18 50 Kiim 10 galls. 10 20 6 5 Corn 7 do. 3 71 Wine 3 galls. 36 10 7 Cash, 7 84 Lot Ford, Dr. Contra Cr. Broadcloth 2 yds. 9 50 Jan.3 4 Beef UOlbs. 4 40 Wine 1 gall. 2 12 7 6 Wood 3 cords, 6 00 In the preceding example, the two columns on the left both of the Dr. and Cr. accounts contain the date of the transaction and the page of the Day Book on which the original account is to be found. Next follows the article and its quantity, which should be written in few words, and then its amount in the money columns. Either the date or the page of the Day Book which contains the account, is anjply sufficient in the Leger, and the latter is to be preferred. The Leger exhibits at one view the accounts with an individual, as it contains on the Dr. siiie whatever he has received, and on the Cr. side whatever he has paid. The difference between the sums of Dr. and Cr. called the Balance, shows the state of the trade in this instance. An Index must accompany the Leger, in which the names are arranged alphahetically, with the page of the Leger on which each account is to be found. See the Index to the Leger for Ex. 2, The following general directions are to be observed in forming the Leger. Let each account be posted from the Day Book in its proper place in the Leger. If a mistahe be maile, let it be cor- rected by an account in the Day Book, clearly stating the correc- tion, and then let this account be posted in its proper place in the Leger, that no blot or erasure may disfigure its pages. THE CASH BOOK. la this book are recorded the daily receipt and payment of men ey. For this purpose there are two polumns, one for money re- ceived, and the other for money paid, in which should be recorded merely the date, to or by whom paid, and the sum. The Cash Book is convenient, but not absolutely necessary. By some ac- countants other auxiliary books are used, which are found to be useful or imj)ortant in some particular business. These the ac- countant will readily form for himself, as circumstances may render nef:es-ary. Note 1. As several of the preceding books may be necessary in the progress of business, they should be distinguished by letter- ing them in the following manner. Day Book A, Day Book B, &c. Leger A. Leger B. &c. And in posting accounts into the Leger, there must be a reference to Day Book A. or B. kc. as the ac- count is Ibuod in the one or the other. See Example 2. BY SINGLE ENTII.Y 4in Note 2. In the following example the barter of any article, as well as the sale of an article for cash, is entered into the Day Book, although such accounts are not to be posted into the Leger. This is not generally practiced, but the accountant will often find a material benefit in recording even these changes of property. EXAMI'LE L SINGLE ENTRY. DAY BOOK. January 1, 1820. My whole property is a debt of 400 dollars due me from Samuel Richards, the balance of my in- heritance. $ c. Jan.l, 1820. Samuel Richards, Dr. To balance Irom the estate, 400 ^ 3rd. Samuel Richards, Cr. By broadcloth 105 yards at 3 dolls per yard, 315 192 4th. John Higgins, Dr. To 55 yards of broadcloth, at $3 50c. per yard, 50 5th. Exchanged 40yds. of broadcloth, for 24Cwt. Iron at g5. 6th. John Higgins, Cr. By cash, on account, 180 7th. Sold S. M.20Cwt. of Iron at gSiperCwt. forcash. 105 Note. The preceding example contains so few accounts, that the formation of the Cash Book is unnecessary. It is sufficient, however, to illustrate the principles of Single Entry ; while it is so short that the whole may be easily comprehended by the pupil. The Leger, in which this Day Book is posted, is on the following page. 492 BOOK KEEPING EXAMPLE I. SINGLE ENTRY. LEGER. Day Book 1 Dr. Samuel Richards Balance fromestate, John Higgins, Dr, Broadcloth, 55yds. i 400 192 c. 60 Day Book 1 1 Contra Cr. Broadcloth 105yds. Balance, Contra Cash, Balance, Cr. $ c. 315 85 400 180 12 50 — 192 50 To balance the accounts, place the difference of the several ac- counts under the sQ^alier side. Thus in the account with Samuel Richards, §85 is the Cr. to balance ; and, in the account with John Higgins, $\^ 50c. is the balance on the same side. It is obvious that I have gained by the trade. Were not the gain evident, on inspection, it would be made so by the following inventory from the preceding Leger. January 8, 1820. Due from Samuel Richards, John Higgins, . - . On hand 10yds. broadcloth at jj3 per yard, Cash from broadcloth and iron, 4Cwt. of iron at g5 - My property Jan. 1, Amount, i c. 85 00 12 50 30 00 285 00 20 00 432 50 400 00 Gain, 32 50 BY SINGLE ENTRY. 19: EXAMPLE II. SINGLE ENTRY DAY BOOK A. January 1, 1821. Inventory -of all my property and debts, taken this day by me, A. B. £ s. d. Ready money, 300 House and Furniture, 500 Williams Farm, 600 Merchandise, 555 Produce, 45 £ S' d. I owe, on accounts, ' To Henry Hardy, £15 10s. To Thomas Howe, 30 2000 45 10 . My net estate. 1954 10 January 1, 1821. Henry Hardy, By former account, balance Cr. 15 30 9 9 18 1 2 3 1 14 16 18 15 15 6 "2 5 16 6 Thomas Howe, By former account, balance Cr. Salmon Rogers, To 20 bushels of wheat, at 9s. pe To 6 yards of broadcloth, at 33s. Dr. ;r bushel, per yd. ~0 John Wheat, To 20 gallons of Rum, at 6s. 9d. Dr. per gall. John Taylor, To 61bs. loaf sugar, at 2s. 7d. pe; Igall. of Rum, Dr. ^ ife 6 9 "3 2d. Simon Pond, To 6 bushels of wheat, at 9s. per Dr. bushel. John Wheat, By cash, on account 60s. Cr. 3d. Henry Hardy, To 3 gallons of wine, at 12s. per 30 bushels of wheat, at 9s. 6d Dr. gall, per bush. ~0 2) BOOK KEEPING ) January 3d, 1821. Titus Coale, Dr. To 20 gallons of Rum, at 7s. per gall. ^cwt. of Havanna sugar, at 60s. per cwt. 7 15 7 1 15 13 4th. John Wheat, Cr. By SOlbs. of nails, at 8d. per ife 4 Simon Pond, Cr. By cash, on account 25s. 1 3 3 6 16 14 6th. Peter Owen, Dr. To goods dehvered C Paige, by your order, 8 Dixon Ferry, Dr. To 66 yards cotton cloth, at Is. 4d. per yd. 8 8th. Henry Hardy, Dr. To 2 yards blue broadcloth, at 36s. per yd. 3 12 Salmon Rogers, Cr. By cash to balance, 18 20 18 6 Peter Pindar, Dr. To 30 galls, wine, at 13s. 6d. per gall. 9th. Titus Coale, Cr. By 25 bushels wheat, at 8s, 6d. per bush. 10 12' 6 Hervey Brown, Dr. To 6yds. hrowfi linen, at Is. 9d. per yd. 1 15 6 9 8 14 ~3 16 4 9 lOvh. John Merrill, Dr. To lOIfe nails, at lOd. per ft 16Jfe brown sugar, at lid. 4 8 ~0 Thomas Howe, Dr. To cash, on account. 4 Titus Coale, Dr. To 16galls. Hum, at 7s. 3d. 12th. Peter Owen, Cr. By 20J- bushels wheat, at 9s. 6 BY SINGLE ENTRY. 49i Jannaiy 13th, 1821. Simon Pond, Dr. To 3 gallons Rum, at 7s. l-^gall. wine, at 12s. 4d. 1 1 18 (3 6 1 19 6 14th. John Wheat, Cr. By cash in full, 4U. 8d. 2 J 8 17th. Thomas Howe, Dr. To 20gaHs. rum, at 7s. 3d. per gall. 5 do. wine, at 12s. 4d. 40lbs. nails, at 9d. 7 3 1 11 6 1 10 16 8 8 18th. Charles Gray, Dr. To 3 gaUons French Brandy, at 12s. per gall. lOf lbs. loaf sugar, at 2s. 4d. 1 1 16 5 I 3 1 1 Dixon Ferry, Cr. By 1 mahogany table, 72s. 1 wash stand, 17s. 3 12 17 4 9 19th. Simon Pond, Cr. By cash, on account 43s. 6d. 2 3 6 20th. Peter Pindar, Cr. By 40 bushels of wheat at 95. cash, on account 45s. 18 6 20 1 6 2 23d. John Taylor, Cr. By 20|Ife butter, at Is- Id. 3 Henry Hardy, Dr. To 30 bushels of wheat, at 9s. 66. 14 2 2 5 8 15 3 Is 26th. Hervey Brown, Cr. By cash, on account 8s. 9d. 9 Titus Coale, Cr. By 6i bushels wheat, at 8s. 6d. cash, on account 3s. 3d. 3 ~6 4} BOOK KEEPING January '29th, 1821. Peter Owen, To Icvvt. white Havanna sugar, j lOgalls. Rum, at 8s. By cash, on account 35s. 2(1, at 63s. Cr. Dr. 3 4 7 3 ~3 "o 1 15 2 Dixon Ferry, To 20lbs. brown sugar, at 1 hi. Dr. 18 4 John Merrill, By cash in full 23s. Cr. 1 3 3 3 3 12 31st. Thomas Howe, To Icwt. Havanna sugar, at G3s. Dr. Charles Gray, By ^ bushels wheat, at 8s. Gd. Cr. 3 Samuel Lyman, To 100 bushels of wheat, at 9s. By cash, on account £40 10s. Cr. Dr. 45 40 3 3 Joshua Noble, To Icwt. Havanna sugar, at 63s. Dr. SINGLE ENTRY 497 LEGER INDEX, TORMZB FOR THE FOLLOWING LEGER, INTO WHICH THE PRECEPING DAY BOOK IS POSTED. A. B. C. Brown, Hervey p. 2. Coale, Titus p. 2. E. F. Ferry, Dixon 2. G. H. Gray, Charles 3. Hardy, Henry 1. Howe, Thomas 1. K. L. M. Lyman, Samuel 3. Merrill, John 2. Noble, Joshua 3. Owen , Peter 2. Pindar, Peter 2. Pond, Simon 2. % Rogers, R. Salmon 1. 3. T. Taylor, John 1, u. v> W. Wheat, John 1. X. y. Note. In the following Leger, both the date of the transaction and its page on the Day Book are given, merely to exercise the learner, as only one of them is essential. P:; i9(} 0) BOOK KEEPING Example H. LEGER A. 1821. P. D.B. A. Jan. 3 1 8 Q 23 3 Jan. 10 2 17 3 31 4 Jan. 1 1 Jan. 1 1 Jan. 1 1 Henrj Hardy, Wine 3 gallons at 12s.* Wheat 30 bushels at 9s. 6(1. Blue broadcloth 2yds. at 36s. Wheat 30 bushels at 9s* 6d. Thomas Howe, Cash, Rum 20 gallons at 7s. 3d. Wine 6 do. ISs. 4d. Nails 40lbs. 9d. Havana Sugar iCwt. Salmon Rogers, Wheat 20 bushels, Broadcloth 6yds. John Wheat, Rum 20 gallons, John Tajlor, Loaf Sugar 6ibs. liurn 1 gallon, Dr. Dr. Dr. Dr. Dr £ s. 1 16 14 5 3 12 14 6 33 18 15 7 5 3 1 1 10 3 3 30 16 16 * In posting accounts into the Leger, some accountants write tlie article, quan- l.i(y,Hnd price, as is here done ; others omit the price in the Leger, as the col- BY SINGLE ENTRY. Example II. LEGER A. 499 (1) 1821. Jan, 1 Jan. 1 Jan. 8 Jan. 2 4 14 Jan. 23 P. D.B A. I Contra Balance of former account, Balance, Contra Cash, Contra Cash, Nails 50lbs. Oash, , Contra 3 Butter 20iib, Cr. Contra Cr. Balance of former account. Ur. Cr. Cr. 10 ^)^ li 30 18 3 1 13 2 1 6 15 rectness of each account is to be ascertained in theDay Book. The latter meth- od is sufficient, and is generally followedl m, this Exampl«. 5d(; (2) BOOK KEEPING Example H. LEGER A. 1821. Jan. 2 13 P. D.B. A. 1 3 2 2 2 4 2 4 2 2 2 Simon Pond, Dr. Vheat 5 bushels, Rum 3 galls.=21s. and Winel| gall.= IBs. 6d. 2 1 4 7 5 1 s. 6 19 -4 16 16 6 6 Jan. 3 10 Titus Coale, Dr. Rum 20galls.=140s. and Sugar |Cvvt.= 16s. Rum 16 galls. 13 11 Jan. 6 29 Peter Owen, Dr. Goods per your order to C. Paige, Havanna Sugar, lcwt.=63s. and Rum 10 galls. =80s. 3 7 10 16 3 8 ~8 Jan. 6 29 Dixon Ferrj, Dr. Cotton Cloth 66yds. Brown Sugar 201bs. 3 „ , ,. 4 20 14 18 Ts 5 8 4 Jan. 8 Peter Pindar, Dr. Wine 30 gallons, Jan. 9 Hervey Bi^own, Dr. Brown Linen, ^yds. 8 9 Jan. 10 John Merrill, Dr. Nails lOlb. Brown Sugar 161b. 8 14 4 8 1 1 o\ ,nkt , BY SINGLE ENTRY. Example II. LEGER A. sei <2) 1821. Jan. 4 19 P. D. B. A. 2 3 Contra Cr. Cash, Cash, Balance^ 1 2 ■ 4 10 2 13 s. 5 3 16 "4 12 15 3 IT 6 ^> Jan. 9 26 2 3 2 4 3 3 Contra Cr. Wheat 25 bushels, Wheat 61 do. Cash, 6 3 3 ~0 Jan. 12 29 Contra Cr. Wheat 20i bushe!s, Cash, 9 1 4 15 6 2 10 IS Jan. 18 Contra Cr. 1 Mahogany Table 72s. and 1 Wash Stand 17s. Balance, 4 4 9 4 13 "0 Jan. 20 Contra Cr. Wheat 40 bushels=£i8 and cash £2 5s. 20 5 Jan, 26 3 4 Contra Cr. Cash, 1 3 9 Jan. 29 ' Contra Cr. Cash, • 502 (3) BOOK KEEPING Example II. LEGER A. 1821. Jan. 18 P. D. B. A. 3 Jan. 31 4 4 Jan. 31 Charles Gray, French Brandy, 3 gallons, Loaf Sugar, lO^lbs. Balance, Joshua Noble, Havanna Sugar, Icwt. Dr. Samuel Ljman, Dr. Wheat 100 bushels, Dr. d. 45 3 BY SINGLE ENTRY. Example II. LEGER A. 503 (3) 1821. Jan. 31 P. D. B. A. 4 Contra Cr. Wheat 8J bushels, 3 40 4 45 3 s. 12 10 10 "o 3 d. 3 Jan. 31 4 Contra Cr. Cash, Balance, ~0 Contra Cr. Balance, CASH BOOK, FORMED FROM THE DAY BOOK, EXAMPLE II. 'Jan. 2. 4. 8. 14. 19. . 20. ^26. 29. 31. Received. Of John Wheat, 3 S. Pond, S. Rogers, J. Wheat, S. Pond, P. Pindar, H. Brown, T. Coale, P. Owen, J. Merrill, S. Lyman, 1 18 2 o s. 5 18 1 3 5 8 3 15 3 40 10 73 13 24 4 4 8 49 8 8 300 ~0 349 8 8 ^ Paid. £ s. d. Jan. 10, Thomas Howe, 15 4 To this add my ex- pences for the month, being 9 4 4 Amount paidl^ 24 4 8 Amount received, do. paid. Excess, Cash, Jan. 1. Do. Feb. 1. In balancing the Leger, as in Example 2, draw two licavy black lines under the accounts in which the sums of Dr. and Cr. are equal, to show that the account is settled, 'Where the sums of Dr. and Cr. are unequal, place the sum to balance, under the smaller m4 BOOK KEEPING account, nriting against it the word, Balance. As there is to be no line drawn under these accounts, and as there is no reference HI the marginal columns to the Day Book, it will be obvious on in- spection, that such accounts are not settled. TO FIND THE PROFIT OR LOSS, An Inventory of the property and of the debts must be taken, as follows, from Example 2. February 1, 1821. Inventory of all my property, and of the sums doe to me, or owed by me, taken this day by me A. B. £ 3. d. Ready Money, 349 8 8 House and Furniture, 604 9 Williams Farm, 600 Merchandise, 480 Produce, 10 1943 17 8 £ s. d. Due from Henry Hardv, 18 8 S. Pond, 16 D. Ferry, H 4 S. Lyman, 4 10 J. Noble, 3 3 27 1 i owe Charles Gray, 11 3 26 9 Difference, 26 Feb. 1. 9 9 9 Net amount of property, 1970 6 8 do. Jan. 1. 1954 10 Profit in the month, 15 IS 8 GENERAL REMARK ON SINGLE ENTRY. Book Keeping by Single Entry, shows clearly the state of ac- counts with individuals, but it does not exhibit the true state of his affairs to the book keeper himself. For this purpose, he must take an Inventory of- all his property and debts, to ascertain the quan- tity of goods unsold, and the net amount of his property, and thence the profit or loss of trade, in the manner just taught. This is a work of much difficulty and trouble, if the business be exten- sive. It is for this reason, thut book keepit>g by Double Entry is preferred in extensive trade. BY SINGLE ENTRY« 505 SHORTER METHOD OF POSTING ACCOUNTS IN SINGLE ENTRY. The method a-lready giren, of posting accounts from the Day Book into the I.eger, is g«?neraily considered the most correct. The following shorter method is perhaps more commonly used. The Leger is ruled as before, and merely the amount of an ac- count is posted into the Leger, preceded by the page and letter of th6 Day Book on which the account is found. The first two ac- counts of the preceding Leger are here posted from the Day Book for ah example of this method. LEGER. SHORTER METHOD. Dr. Heiirj A 2. £16 Is. and £3 12 A 4. £14 5s. £ 19 14 33 15 11 _3 30 s. 13 5 To 16 3 d. 4 81 ~0 Hardj, A 1.£15 10s. Balance, Cr. £ 15 18 33 s. 10 8 T8 d. Dr. Thomas A 3. £1^5 Os. 4d. and £11 16s. 8d. A 4. £3 3s. Howe, lA l.£30. Cr. 30 In this Leger, A 1, A 2, &c. means that on page 1, 2, &c. of Day Book marked A, that particular account is to be found. The manner of balancing the Books, and of ascertaining the Profit or Lo?s is the same as before taught. To make the subject familiar the learner should be directed to form a Day Book for himself, and to carry the accounts through the several books, ac- cording to the preceding principles. SHORTEST METHOt) OF KEEPING ACCOUNTS. Only one Account Book is necessary in the practice of this meth- od. It is formed precisely like the Leger in Single Entry, except that there is no column of reference to any other book. The transactions of trade are entered under the names against the date on v\rhich they take place. An alphabet for the arrangement of the names is found convenient for reference to the various ac- counts. This Account Book is designed to answer the double purpose of Day Book and Leger. if the person be careful to enter every instance of debt and credit at the time it occurs, he will be able to ascertain at any time the state of his accounts in a particular case. This is the great object of this metho?!; 'vhich is exhibited io the feliowinpr example, Q3 6Qi^ BOOK KEEPING ACCOUNT BOOK. i Linden, Jan. 1, 1820. — =-i 182G. John Wilson j Dr. $ c. Jan. 1 To 3 cords of wood, ^1 50 per cord, 4 60 6 To li days work, by hired man, at 66 cents, 1 00 9 To 6i bushels rye, at 60 cents, 2 75 13 To 3 bushels wheat, at $1 76, 5 26 Feb. 1 To 5 cords wood, at gf 60, ^ 7 50 March 9 To 7 bushels oats, at 31 cents. 2 17 16 To work with hired man and horses, one day, 1 66 24 83 18 Cash to balance, 34 25 17 1820. Peter Pajwell, Dr. July 1 To life hyson tea, 1 66 9 To lOlbs. brown sugar, at 19 cents, 1 90 26 To 3 gallons Rum, at $1 17, 3 61 Sept, 9 To 91bs. blister steel, at 10 cents, 90 11 To 6|yds. calico, at 54 cents. 3 51 Oct 3 To 3vds. cotton cloth, at 18 cents, 54 13 To 2lbs. loaf sugar, at 31 cents, 62 1821. Jan. 1 To life hyson tea, 1 46 3 To goods delivered by your order to E, T. 3 72 V7 72 BY SINGLE ENTRY. SHORTEST METHOD. -607 1820. Jan. 1 5 13 Feb. 1 March 7 Linden, Jan. 1, 1820. John Wilson, Cr. By 1216 shingle nails, at 10 cents, By 25ilbs. cheese, at 7 cents, By your order on John Gibbs, for goods, By 20ilbs. Butter, at 18 cents, By licwt. iron, at ^6 60 per cwt. 77P c. 20 78 75 69 75 17 1820. Contra. Cr. Aug. 4 By 121bs. butter, at 121 cents, 1 50 7 By work 2 days by your man. 1 33 Oct. 3 By 661bs. chees€, at 7 cents. 3 92 10 By cash, 3 00 1821. Jan. 1 By 12 bushels rye, at 50 cents, 6 00 3 By cash to balance, 1 97 17 72 m^ BOOK KKEPINQ BOOK KEEPING BY DOUBLE ENTRY * THIS npethod of Book Keeping differs from that by Single Entry in t\^o important respects, viz, things, as well as persons, are en- tered as Dr. and Cv. and as Dr. and Cr. to each other, and each account is entered twice in the Leger. For the latter particular it is called Double Entry. In the practice of Double Entry, three Principal Books, and four Auxiliary Books are neceseary. PRINCIPAL BOOKS. The^eare the Day Book or Waste Book, the Journal, and the Leger. ^ L THE DAY BOOK OR WASTE BOOK. IpjfP' The Day Book beginj^, as in Single Entry, with an Inventory oX ^ '" all the property and (iebts of the merchant, and is followed by a regular account of Ib^ transactions in business, in the order of time in which they occur, stated in language so explicit and ful8 that there can be no miited by Exs^mple I. of Single Entry. (1) EXAMPLE I. DOUBLE ENTRY. DAY OR WASTE BQOK. January 1, 1821. My whole property is a debt of ^400 due me from Samuel Richards, the balance of my inheritance. Samuel Richards owes me the balance of my inher- ifance, 400 c. ord. Bought of Samuel Richards 105 yards of broadcloth at 3 dolls, per yard. 315 4th. Sold John Higgins 55yds. of broadcloth at $3 50c. per yard. 192 50 5lh. Bartered 40yds of broadcloth for 24cwt. of Iron, at 5 dolls. |)<^r Cwt. 120 6th, Received of John Hi^jjins in part. 180 7th. Sold 20cwt. of Iron to S. M. for Cash at 51 dolls, per Cwt. 105 * The general principles of this system of Book Keeping are taken from the Sys- tem in Rees' Cyclopedia, which is generally adopted by the merchants of London. BY DOUBLE ENTRY. 500 Note. In recording transactions in the Day Book, the above order is generally to be preserved, viz. 1st the date ; 2d the kimi of transaction in the active voice, as owes, s61d, bought, exchanged, &c. ; 3d the name of the person ; 4th the article and quantity ; 6th the price ; and 6th the amount in the columns of money. II. THE JOURNAL. The object of the Journal is to prepare the accounts for thf Leger. To effect this, the Dr. and Cr. of every article contained in the Day or Waste Book, is ascertained and expressed in the Journal. The Journal is ruled v^^ith two blank columns on the left, viz, one for the date and the other for the page in the Leger, and with the proper columns for money on the right, as in the following journal of the preceding Day Book. EXAMPLE I. DOUBLE ENTRY. JOURNAL. 0^ ^ - . Jan. 1 p L. yamuel Kictiards Dr. to :5tock §400 OU For the balance of my inheritance, 400 3 Broadcloth Dr. to Samuel Richards 315 00 For 105yds. broadcloth at 3 dolls, per yd. 315 4 John H!£,-irins Dr. to broadcloth 192 50 For 55ydsr broadcloth at $3 50 per yd. 192 50 6 Iron Dr. to broadcloth 120 00 To 40yds broadcloth at 3 dolls, per yard ) for 24cwt. of iron, ^ 120 ^- 6 'Cash Dr. to John Higgins 180 00 Received of him oh account, 180 7 Cash Dr. to Iron ^ 105 00 Received for 20cwt. of iron at J5 25 per Cwt. 105 To understand the method of forming the Journal, the following distinctions must be attendecVto. Accounts are distinguished in- to personal, real, and fictitious. Personal accounts are those in which a person is entered as Dr. or Cr. and are the same as in Single Entry. Thus, in the preceding Journal,. Samuel Jiichards is Dr in one account. Real accounts are those of property of any kind, as cash, hous- es, cloth, furniture, adventure, 5fc. In the preceding example Broadcloth is Dr. to Samuel Richards, andiron Dr. to broadcloth, kc. Fictitious accounts are those of stock, and profU and loss. Stoclg is used for the ovvner of tjie bouks. In the preceding^ Example, 5ia BOOK KEEPING Samuel Richards is Dr. to stock, i. e. to the owner of the books. Profit and loss is used for either gain or loss in the course of trade. This account does not appear in the Journal but in the Leger. If two or more persons or things are included in the same account in the Journal, they are expressed by the term, Sundries. Rules for distinguishing Dr. and Cr. in the Journal, are the fol- lowing. The person to whom or for whom I pay, or whom I enable to pay, is Debtor. The person for whom or from whom I receive, or by whom I am enabled to pay, is Creditor. Whatever comes into my possession or under my direction, is Dr. Whatever passes out of my possession or from under my contrqul, is Cr. The phrases, In debtor, and Out creditor, briefly express the points in these rules. Thus, in the preceding Journal, Broadcloth, EXAMPLE I. DOUBLE ENTRY. LEGER. Date 1820. P. J. Stock, Dr. P. L. $ 400 315 27 342 c. 50 50 50 Jan. 1 3 1 I Samuel Richards, Dr. To Stock, Broadcloth, Dr. To Samuel Richards, 105yds. at ^3 per yard, Profit and loss. 4 5 1 1 John Higgins, Dr. To broadcloth, SSyds. at $S 50, 192 Iron^ Dr. To broadcloth, 24cwt. at g3. Profit and loss, 120 5 125 7 8 1 Cash, Dr. Po John Higgins, Iron, 20cwt. at $5\, 180 105 285 Profit and Loss, Dr. m^ BY DOUBLE ENTRY. 511 Ithich come« into my possession, is Dr. to Samuel Richards by whom it is paid in part for stock in bis hands ; John Higgins is Dr. to broadcloth ; Iron is Dr. to broadcloth, and so on. III. THE LEGER. The object of the Leger is the same as in Single Entry. But as things, as well as persons, are introduced in the Journal, they must have separate accounts in the Leger also, where the respective Drs. and Crs. are to be arranged under their respective heads. The Legerjis ruled with columns for the denominations of money on the right side, immediately before which is a column for refer- ence to the page of the Leger in which the corresponding account is found ; and on the left side, is a column for dates, and another for the page of the Journal in which the account may be found. The following Leger for the preceding Example is formed on this plan. See foot of this and the preceding page. EXAMPLE I. DOUBLE ENTRY. LEGER. Date. P. 1820. J. Jan. 1 1 Contra Cr. By Samuel Richards, P. L. $ 400 c. 3 1 Contra Cr. By Broadcloth, 1 1 315 4 6 1 1 Contra Cr. By John Higgins, 55yds. at $3 50, Iron, 40yds. at $3 192 120 50 6 b 1 1 Contra Cr. By cash, 1 180 105 — Contra Cr. By cash, for 20cwt. at $5 25, Contra Cr. Contra Cr. By cloth, Iron, 27 6 32 50 50 12 BOOK KEEP li^G In polling the Journal to form Iho preceding Leger, Samut'! Richards is posted Dr. tu Stock J5400, and Stock is posted Cr. by Samuel Richards for the same sum ; Broadcloth is then entered Dr. ill the next account to Samuel Richards, and Samuel Richards Cr. by broadcloth ; John Higgins is posted Dr. to broadcloth, and Broadcloth Cr. by J Higgins, and so on. It is obvious that each Dr. must have a Cr. and each Cr. a Dr. and that every transaction relating to any one account, whatever may be its place in one ac- count in the Loger, niust be posted also on the proper side under the head to which it belongs. Thus, while Cash is Dr. to J. Hig- gins in the 6th account for ^180, J. Higgins is Cr. by cash for the same sum in 4th account; and while Cash is Dr. also to iron, Iron is Cr. by cash to the same amount, in the 6th account. On inspecting the preceding Leger it is evident, that in the per- sonal accounts, as those of S. Richards and J. Higgins, all the arr tides for which they are indebted are posted on the Dr. side, and all the articles they pay are on the Cr. side of the account ; in the real accounts, as those of broadcloth, and iron^ the quantity bought is posted on the Dr. side, and the quantity soid on the Cr side so that the quantity unsold and the profit or loss may be readily as- certamed. In the fictitious account of Profit and Loss, the loss is to be posted on the Dr. side, and the profit on the Cr. side, so that the difference must show the net gain or loss, by which the stock has been increased or diminished in the course of trade. Having ascertained that the accounts have been correctly posted, the next step is to balance the Leger. This is to be don# in th^ following matiner. To show this method more clearljs the pre- ceding Leger is repeated, and the several steps m balancing the accounts are subjoined. I'd the preceding Leger subjoin a new account. Balance Dr. and Cr. Begin with the next account to Stock, and place the balance of Dr. and Gr under the smaller, to make them equal, viz. ^85 on the Cr. side. Put this sum on the Dr. side of the account, Bal- ance. For if S. Richards is Cr. by Balance 85 dolls, then Balance must be Dr. to S. Richards for the same sum. Next, lo balance the Broadcloth account, the quantity on the two sides mu5t first be made equa}, and the value of the unsold cloth, at first cost, be placed \inder the amount sold, viz. 10 yards at jJS, amoufiting to ^oO. Thp. whole sum, viz. |J342 50 must be equal to the amount bought afid the profit on that sold. Then «J30 must be placed ou the Dr. side of Balance, for if Broadcloth be Cr. for the balance unsold, then Balance must be Dr. for the same sum. Proceed in this manner, through all the personal and real account?. BY DOUBLE ENTRY. 513 Proof.* Having balanced all the accounts except those of Stock, Profit and Loss, and Balance, in the first place, close the account of Prof- it and Loss, by making it Dr to the stock gained, viz. ^32 50, and then make Stock Cr. by the same sum. Next, let Stock be bal- anced by the necessary sura, viz. ^432 50, and Balance be made Cr. by the same sum. If the sides of the account, Balance, are now equal, the work is right. This proof is complete from this consideration. By this method, the cash in hand, the debts due, and the goods unsold, are contain- ed on one side, and what is owed, is contained on the other side. Another method of proof is to add the profit to, or subtract the loss from, the original stock, and the sum or difference placed to the Cr. of Balance, will be equal to the &um on the Dr. side of Balance. General Remark. The Journal should be kept up with the Day Book, and the ac- counts should be regularly posted into the Leger, that the books may be as nearly even as possible. And at the end of every month, the balance should be made, the Journal and Leger having been carefully examined to see that all the records of the Day Book have been carefully transferred into the Journal, and correctly posted from the Journal into the Leger. The following Example is sufficient to exhibit the principles of Double Entry. It was designed to.be so short that the student might have the whole before him at one view. It is too short, however, to render any of the auxiliary books necessary. In ex- tensive business, however, these books are essential. In the prac- tice too of Double Entry, the work will be shortened by forming the Journal, in the manner shewn in the next Example. An ac- count of the auxiliary books will afterward be given, and a speci- men of each one, as connected with the next Example of Double Xntry. * See pages 514 and 615. R 514 BOOK KEEPING EXAMPLE I. DOUBLE ENTRY. LEGER. Date. 1820. Jan. 1 J. Stock, Dr. By Balance, my net estate, P. L. 1 1 1 1 $ 432 400 315 27 342 192 120 5 c. 50 50 50 50 Samuel Richards, Dr. To Stock, 3 4 5 ■ Broadcloth, Dr. To Samuel Kichards, 105yds. at ^3 per yard, Profit and loss, John Higgins, Dr. To broadcloth, 55yds. at ^3 50, Iron, Dr. To broadcloth, 24cwt. at ^5, Profit and loss, 1 1 125 180 105 285 7 i ! Cash, Dr. To John Higgins, Iron, 20cwt. at g5i, Profit and Loss, Dr. To stock gained. i 32 50 Balance, Dr. By Samuel Richards, Broadcloth, unsold, John Higgins, Iron, unsold. Cash, 85 30 12 20 205 43!:^ 50 50 IBY DOUBLE ENTRY, 515 EXAMPLE I. DOUBLE ENTRY. LEGER. t Date. 1820. Jan. 1 3 P. J. 1 1 Contra Cr. By Samuel Richards, Profit and loss, P. L. 1 1 $ 400 32 432 315 85 400 c. 50 50 50 50 Contra Cr. By Broadcloth, 105 yards, at g3, Balance, 1 4 6 6 1 1 Contra Cr. By John Higgins, 55yds. at $3 50, Iron, 40yds. at $3 Balance, 10yds. unsold, at $3 lU5yds. 1 1 1 1 1 1 192 120 30 342 Contra Cr. By cash. Balance, 180 12 192 50 50 7 1 1 Contra Cr. By cash, for 20cvvf. at $5 25, Balance, 4cwt. at ^5, 24owt. 105 20 125 285 27 5 -32 50 50 Contra Cr. By balance, Contra Cr. By cloth, Iron, Contra Cr. By slock, my net estate, 1 432 50 1 516 BOOK KEEPING AUXILIARY BOOKS. ^^^|^ These are 1, the Cash Book ; 2, the Bill Book j 3, the Invoice 5ook ; and 4, the Sales Book. These books are important to the accountant, as a record of particular transactions referred to in the Day Book, and as origin- al and particular records of those transactions. They aid especial- ly in posting accounts into the Leger. They may be considered as a kind of Day Book, in aid of the general Day Book, and it is obvious, that if all the particular accounts were arranged under general heads in separate baoks, the common Day or Waste Book would be unnecessary, except as exhibiting a general history of the changes of property. I. THE CASH BOOK. The Cash Book is a record of all money paid or received. It i« referred to in the Day Book, by the initials C. B. It is formed like the Leger, with a Dr. and Cr. side, the Dr. side containing all money received, and the Cr. all money paid. The transactions are to be regularly entered into the Cash Book as they take place, with the dates, names, and all necessary particulars. The man of business will find it convenient to have separate columns for some transactions, as of money accounts at a Bank, Brokers, kc. and for some small incidental expenses, as well as for money lent and repaid immediately. The money accounts should be transferred to the Leger at least every/ month. When the cash account is entered into the Journal, it is written, Cash Dr. to Sundries, for money received, and Sun- dries Dr. to Cash, for money paid, mentioning all the necessary particulars. The following Caj^h Book shows the manner in which this book js formed and kept. It belongs to Lxample 2, of Double Entry, and U the Cash Book referred to in the Day or Waste Book of that Example, The same remark applies to the auxiliary books which follow the Caeh Book- Topost the Cash Book into the Leger, Make Cash Dr. to Sundries for the afriount received, and Cr. by Sundries for the amount paid Then make each account Dr. to (-ash, for the respective sums paid, and Cr. by Cash for the respective sums received. See Example 2, Leger. i BY DOUBLE ENTRY. 517 CASH BOOK. JANUARY, 1820. Jan. Dr. $ c. Jan. Cr. ■ $ c. 3 To interest for discount- ing Wm. Burr's Bill 2 By charges on merchan- dise, per the Venus No. 13. € 39 for Naples, 23 14 Bills receivable, No. 11, 3 Bills payable, No. 13, Williams & Co. 1600 Wm. Burr, 1440 23 Ship Fhebus, received 7 Charges on mer. per the for freight, 400 87 Dolphin, for Bilboa, 32 5© 25 Farm in Cambridge, re- 12 Bills payable, No. 11, ceived for produce, lao 35 George Myers, 2222 26 Bills receivable, No. 12, 21 Charges for sales, per the George Murray, 1300 Betsy, pd. customs, &;c. 439 24 31 Interest, half year's div- 23 Bills payable. No. 12, idend at the bank. 300 John Howe, 600 31 Funded property, sold 25 Ship Phebus, paid for re- 9000 at 811, 7272 pairs. 130 2S 31 Debentures, received, 300 27 Charges on mer. per the Henry for Jamaica, 51 50 11339 61 31 Expeuses of House, 150 52 5089)01 II. THE BILL BOOK. The Bill Book is a record of all Bills of Exchange receivable or payable. The reference in the Day Book, is by the initials B. ^. or, by B. R. for bills receivable, and B. P. for bills payable. Bills Receivable are tho^e paid or to be paid to the merchant. Bills Payable are those drawn on the merchant or to be paid by him. The particulars of each kind of bills are entered in the Bill Book under their separate beads of B. R. or B. P. The records of the Bill Book are entered into the Journal, un- • n"^ tM'O. pj P CD " S-TJ O P S V4 o (D O >-< o o OR •tj (t) V- td o & P? Pr8 d a"? » > n 3 c r *-i 2 '' O 9. w ■T-1 S^ f-!?r 8 n •^ >— c & D '.^ t-i-; p C • ■ p J c P t^ T p r :- r- ? '/"• O] g CT> p » ■:»^ n i^ r ffl 3 tS vi 3 3 3- J£. r/) 'Jt^ c 3 1 Ship Phebus, 400 87 lb 1 Farm in Cambridge^ 160 35 11 2 Debentures, 300 31 1 Funded property. 7272 11339 67 BY DOUBLE ENTRY. 527 EXAMPLE 11. JOURNAL. JANUARY 1820. (2) Ja. P. L. Sundries Dr. to Cash, For sums paid this month as per C. B. $ c. 2 1 Charges on merchandise, viz. per Venus, 23 00 7 Dninhiri S^ ^D 9^ Rrf-^pv 4SQ ^1 X 1 97 Hrnrv 51 50 ^ i .546 24 3 2 Bills payable, No. 13, 1440 00 xJ^U 12 11, 2222 00 oq lo f^an nn ^O 426'!> 25 1 Ship Phebus, 130 25 31 3 House expenses. 150 52 5089 ^ Bills Receivable Dr. to Sundries, For bills received this month, as per B. K. 2 2 To Cyrus Coate, No. 1, due March 1, 411 22 2 Lemuel Rogers, No. 2, Feb. 15, 2500 2911 Sundries Dr. to Bills Payable, For bills accepted by me this month, as per B. P. 15 20 3 To Gilson & Co. No. 1, due Feb. 1, Joseph Lockwood No. 2, 3, 800 3600 24 Smith & Son, No. 3, March 1, 2150 6550 George Parish Dr. to Sundries, For amount of Invoice of Sugar, per the Venus, for Naples, as per I. B. and W. B. 2 1 1 2 To Merchandise, 395 43 Charges, 23 00 Commission, 13 46 Insurance, 8 40 440 29 Samuel Rogers Dr. to Sundries, Amount of invoice per Dolphin, for Bilboa, as per VV. B. p. 2. 1 1 To Merchandise,^ 2000 00 Charges, 32 50 1 Commission, 60 75 2 Insurance, 34 15 2127 40 BOOK KEEPING EXAMPLE II. JOURNAL. JANUARY 182) J a. 16 31 26 27 26 3 Merchandise Dr. to Jos. Lockwood, To amount of goods bought of him, as per bills of parcels. Insurance Dr. to Globe Insnr. Co. For amount of Insurance, as per I. B. Per Venus, for Naples, $3 40 Per Dolphin, for Bilboa, 34 15 Per Henry, for Jamaica, 206 23 Debentures Dr. to Merchandise, For drawbacks received this month. Sundries Dr. to Sales per the Betsey, For amount of 8 pipes of Port Wine, on account of James Greaves, as per S. B. To Smith & Son, 3 pipes at 1 month, Joseph Lockwood, 5 pipe-^ at 2 months, Sundries Dr. to Sundries, For amount of Invoices per the Henry, for Jamai- ca, as per W. B. p. 3. Insur. Spring & Jones, George Black, Adventure to Ja maica, inCo with J. Broker, my half, James Broker, his half, Sums, $ c. 10000 87 247 200 Merchan. 1120 00 Charg. 12 50 Commis. 35 75 1800 00 15 00 56 50 2500 00 24 00 90 00 5420 00 51 50 U!2 25 37 25 62 50 105 48 '205 23 Sales per the Betsey Dr. to Sundries, To charges on merchandise, 439 24 Commission, 33 94 Interest, 4 03 James Greaves, for net proceeds of 8 pipes of Port Wine, as per S. B. 756 93 462 772 120 193 1359 1359 78 50 14 60 74 74 5858 98 1234 14 -\ BY DOUBLE ENTRY. 529 ALPHABETICAL INDEX TO THE LEGER. A. Adventure to Jam. 3, D. Debentures, G. Gilson & Co. Globe Ins. Co. Greaves. James B. Balance, 3 Bills payable, 2 Bills receivable, 2 Black, Geo. 2 Broker, James 3 E. H. House, House Expenses, C. Cash, Charges on Mer. Coate, Cyrus Commission, F. Farm in Camb. 1 Funded Property, 1 Furniture, 1 I. Insurance, Interest, K. N. L. M. Lockwood, Jos. 3 Merchandise, 1 O. R. T. Rogers, Lemuel U. P. Parish, Geo. 3 Profit & Loss, 3 Sales per Betsey, 3 Ship Phebus, 1 Smith & Son, 2 Spring & Jones, 2 Stock, 1 W. (Cr Before attempting to balance the Leger, it must be ascertained whether the Journal has been correctly formed from the Day Book and Auxiliary Books, and whether the journal has been correctly posted into the Leger. In examin- ing the books for this purpose, a point or some mark should be placed against the several accounts found to be correctly entered in the Journal and Leger, and this pointing or marking continued through all the accounts. It is then common to make a trial balance, on a separate piece of paper, before forming the account called Balance, in the Leger. If the books can be thus balanced, the several balances are then placed under Balance, and the work is finished. Note. The examples here given are sufficient to illustrate the method of Double Entry. The teacher should not suffer the pupil to pass over any point, until it is well understood. If more examples should be desired, he can direct the learner to take Ex. 2. of Single Entry, and form from it the several books in Double Entry. After this has been done, the pupil should form, for himself a larger Day or Waste Book for Double Entry, and carsy Iho account throvigh all the form", according to the principles taught in this sv-lpm, T 3 650 BOOK KEEPING (1) EXAMPLE II. LEGER. JANUARY 1820. i).IJ. P. 1 Dr. Stock, To Sundries, Balance, Dr. Cash, To Stock, Sundries, Dr. t unded To Stock 12000a80i Profit and Loss, Or. h arm in To Stock, Profit and Loss, I *. i^ou>4:j in ihe To Stock, Di\ Furnitur,', Po stock. Dr. JShip To Stock, Cash, Profit and Loss, Dr. Merchandise, To Stock, Jos. Lockwood, Dr. Charges on To Cash, Dr.<'oinxnis!*ion, To Profit and Loss, $ 11504 47049 5r.5h3 13500 ll"339 '24SrM^ 9660 27 61 961j: 4500 160 .35 20501 1200 9500 130 270 G900 6400 10000 l8553 5089 19750 24839 7272 2415 9687 160 4500 4660 2050 1200 400 9500 9900 395 43 2000 5420 200 50 8384 94 61 87 16400 37 23J 32I5O 50 51 539 546 13 60 182 33 94 290 40 BY DOUBLE EXAMPLE II. LEGER. Eb'TRV. JANUAR-Y 1820. 53i Ja J. P. 3 1 2 i 1 1 1 2 1 2 2 3 3 3 Dr. Insurance, To Globe Insuraace Company, L. 3 1 1 3 1 1 1 1 $ 247 c . 78 50 D. Ja 2 7 27 1 31 26 2 22 1 1 1 J. P. 2 2 3 1 1 2 1 3 o 2 1 1 1 Ca. Cr. By George Parish, Lemuel Rogers, Sundries, L. o i 1 3 1 1 'J * < 3-1 205 40 15 23 247 2900 2911 7S L>r. Bills To Stock, Sundries, 2900 2911 Keceiva.ble, Cr. By Cash, Bejance, .5811 5811 1 Dr. Bills To Cash, Bakmce, 4262 6550 Payable, Cr. By Stock, Sundries, 4262 6550 10812 108 li^ 1 31 Dr. Debentures, To Stock, Merchandise, 1300 200 Ca. Cr. By Cash, Balance, 300 1200 1500 50 1500 1700 50 35 40 50 ) Dr. James To Stock, Greaves, Cr. To Sales per B«tsey, Balance, 3 3 2 2 3 1 _ 1 3 756 943 03 171X) 411 1149 1560 1 Dy. Cyms To Stock, 1560 Coale, Cr. By Bills Receivable, Balance, jj A5 1 Dr. Lenauel To Stock, Sundries, 1175 2127 Kogfer^*, Cr. By Bills Receivable, Balance, 2500 802 3302 2640 40 3302 2200 440 40 29 40 ] 2 Dr. George To Stock, Sundries, ] 2 3 3 Parish, Cr. By JBalanc*, 20 2640 29 50 17 67 24 2G i)r, 8milh &L To Bills Payable, Sales per iietsey, 2150 462 !Son, Cr. By Stock, Balance, 21.50 462 2612 2612 27 Dr. bpnng it To Sundries, Balance, 1205 461 1666 Jones, Cr. By Stock, IGGQ 67 27 Dr. George To Sundrie?, 1934 hiack, Cr. By Stock, Balance, 1175 759 ' 1934 53:^ T. p. RETURN TO DESK FROM WHICH BORROWED EDUCATION - PSYCHOLOGY LIBRARY This book is due on the last date stamped below, or on the date to which renewed. Renewed books are subject to immediate recall. 7 DAY USE DURING W1AY13 196S k--'^ ff AUG 8 RECD'9 WTZJ ?^ MAY BE RE AFTER $^P15 1 3 8 :^ SUMMER SESSIONS AM ml LP 21A-15m-4.'63 (U64VlliUJ4 ' 'ib G< neral Library ■ Uaivcn sity c^ California Berkeley House, Furniture, Ship PLebus, Merchandise, Bills Receivable, Debentures, James Greaves, < yyrus Coate, Lemuel Rog'ers, Gpcr;^c Parish, Smith k. Son, Geor-'C Black, Advcii. to Jamai. 1 1 2050 1 1 1 1200 1 ] 9500 1 8384 94 i> 2911 2 1200 50 2 043 07 2 1149 35 i 802 4C i 2 2640 29 i 9 462 1 2 751) 1 ! 3 1 135:-|7.:|! 60027 1 81 Jos. Lockwood, Globe Insur. Co. Stock, 1450 .l:?59 0000 il232 800 306 4 310 247 150 27 160 270 290 310 •058 6550 > 461 \ 90 5628 247 47049 58 21 42 78 52 60027 89 f y cl5796' 882173 THE UNIVERSITY OF CALIFORNIA UBRARY -a"V • • -v^i/iSJ''