,** 
 "*'-. 
 
 

 LIBRARY 
 
 UNIVSf>5TY Of 
 
 r 
 
PIKE'S 
 
 8 
 
 DESIGNED TO FACILITATE 
 THE STUDY OF THE SCIENCE OF NUMBERS. 
 
 COMPREHENDING 
 \ * 
 THE MOST PERSPICUOUS AND ACCURATE RULES. 
 
 ILLUSTRATED BY USEFUL EXAMPLES. 
 
 TO WHICH ARE ADDED 
 APPROPRIATE QUESTIONS, 
 
 FOR THE EXAMINATION OF SCHOLARS; 
 
 AND 
 
 Stiotrt Sfistr w of 
 
 BY DUDLEY LEAVITT, 
 
 Teacher of Mathematicks and Natural Philosophy. 
 
 -** 
 
 CONCORD : 
 
 PUBLISHED BY JACOB B. MOORE- 
 
 1826. 
 
DISTRICT OP NEW-HAMPSHIRE, to wii 
 
 f District Clerk's Offkt. 
 
 ?^^JTE IT REMEMBERED, &at on<he eighth day of Apiil 4 A. D. J826,andin the fif- 
 5 L. S.> *J tieth year of the Independence of the United States of America, JACOB B. MOORE, 
 *^v% of said j)i str ict, hath deposited in this Office the title of a book, the right whereof 
 he claims as proprietor, in the words following, to wit : 
 
 " Pike's System of Arithmetick Abridged : designed t facilitate the study of the science of 
 numbers. Comprehending the most perspicuous and accurate rulet, Illustrated by useful exam- 
 ples. To which are added appropriate questions for the examination of Scholars ; and a short 
 system of Book-Keeping. By DUDLEY LEAVITT, Teacher of Mathematicks and Natural 
 Philosophy." 
 
 In conformity to the act of the Congress of the United States, entitled, "An act for the encour- 
 agement of learning, by securing the copies of Maps, Charts and Books, tr'-^e Authors and Pro- 
 prietors of such copies, during the time therein mentioned ;"' and also to an act, entitled, 
 " An act supplementary to an act, entitled, an act far the encouragement of learning, by secur- 
 ing the copies of Maps, Charts and Books to the Authors and Proprietors of such .-.opK s, during 
 the times therein mentioned, and extending the benefits thereof to the arts of aligning, engrav- 
 ing and etching historical and other prints." 
 
 CHARLES W. CUTTER, 
 Clerk of the District of New-Hampshire, 
 A trne copy of record : 
 
 Attest, CHARLES W, CUTTER, Clerk 
 
 GIFS 
 
I PIKE'S ARITHMETICK is universally acknowledged to be the most 
 complete system ever published in the United States. It is 
 the source, indeed, from which most of the common arithme- 
 ticks have been compiled. Several years of laborious study 
 were devoted to the work by its author, and on its first appear- 
 ance, it obtained a very high reputation. Others have built 
 upon his foundation, and their works have been deservedly 
 popular ; but still, in the fulness and correctness of his rules, 
 and the simplicity of their illustration, PIKE STANDS PRE- 
 EMINENT.] 
 
 DARTMOUTH UNIVERSITY, 1786. 
 
 AT the request of Nicolas Pike, Esq. we have inspected his System of Arithniet;ck, 
 whicii we cheerfully recommend to the publirk, as easy, accurate, and complete. And 
 we apprehend there is no treatise of the kind extant, from which so great utility may 
 afise to Schools. 
 
 B. WOODWARD, Math, and Phil. Prof. 
 JOHN SMITH, Prof, of the Learned Languages. 
 I do sincerely concur in the preceding recommendation. 
 
 J. WHEELOCK, President of the University. 
 
 PROVIDENCE, RHODE ISLAND, 1735. 
 
 WHOEVER may have the perusal of this treatise on Arithmetick may naturally conr 
 dude I might have spared myself the trouble of giving it this recommendation, as the 
 work will speak more for itself than the most elaborate recommendation from my pen 
 can speak for it : But as I have always been delighted with the contemplation of mathe- 
 matical subjects, and at the same time fully sensible of the utility of a work of this 
 nature, I 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 
 sheets 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 aathor, in treating on numbers, has done it with so much perspicuity 
 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 tyro into the 
 first principles ol numbers, are novelties 1 have notmet with in any book I have seen. 
 "VVingate, Hutton, Ward, Hill, and many other authors whose names might be adduced, 
 if necessary, have claimed a considerable share of merit ; but when brought into a 
 comparative point of view with this treatise, they are inadequate and defective. This 
 volume contains, besides what is useful and necessary in the comrnou affairs of life, a 
 great fund for amusement aud entertainment. The JMechanick will fiud in it much more 
 than he may have occasion 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. 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 of this work. 
 
 BENJAMIN WEST. 
 
 269 
 
iv RECOMMENDATIONS. 
 
 UNIVERSITY inr CAMBRIDGE, 1786. 
 
 HAVING, by the desire of Nicolas Pike, Esq. inspected the. foiii.wing volume in 
 manuscript, we beg leave to acquaint the pubhck, that in our opinioi. it is a wcik \\ell 
 executed, and contains a complete sjstean of Arithmetick. The jules are plain, :md 
 the cemonstrations perspicuous and satisfactory ; and we esteem i: the best calculated, 
 of any single piece we hve me f with, to eact youth, hy natural and easy gradations, 
 into a methodical find thorough acquaintance with the science of figures Peisc ns cf all 
 desc iptious may find in it every thing, respecting numbers, m cessary to their business ; 
 and not only so, but it they have a speculative turn, and mathematical taste, may meet 
 with much for their entertainment at a leisure hour. 
 
 We heartily recommend it to schools, and to the community at large, and wish that 
 the industry and skill of the Author may be rewaided. for so beneficial a work, by 
 meeting with the general approbation and encouragement of the publick. 
 
 JOKPH WILLARD, D D. Presiil. ni of the University. 
 
 E. WIGGLES\VORTH. S. T. P. Hollis. 
 
 S. WILLIAMS, LL.D. Math, et Phil. Nat- Prof. Hollis. 
 
 YALE COLLEGE. 1786. 
 
 Uroir examining Mr. Pike's system of Arithmetick, in manuscript, I find it to be a 
 work of such mathematical ingenuity, that I esteem myself honored in join in? with the 
 Rev. President Willard. and other learned gentlemen, in recommending it to the pub- 
 lick as a production of genius, interspersed with originality in this part of learning, and 
 as a book, suitable to be taught in schools: of utility to the merchant, and well adapted 
 even for University instruction. I consider it of such merit as that it will probably gain 
 a very general reception and use throughout the republick of letters. 
 
 EZRA STILES, President. 
 
 SCHENECTADY, OCT 16, 1822. 
 
 I HAVE for many years been fully acquainted with Pike's System uf Aritluretick, 
 and am fully persuaded of its excellence; I do not know of any treatise of n ; ore prac- 
 tical utility ; the arrangement of its pans is natural, its rules are plain and easih un~ 
 rtcrtooJ am 1 Hpplied, and it contains all that is of any importance to the Meicantile or 
 Scientific Arithmetician. 
 
 T. M'AULEY. S. T. D. 
 Late Professor of Mathematicks, Union College. 
 
 AMHERST, MASS. FEB. 9, 1822. 
 
 I HAVE long been acquainted with Pike's Arithmetick, and think it the best of any 
 extant, for those who wish to acquire a thorough knowledge of Arithmetiek as a science 
 and an art. I cheerfully recommend the work to the patronage of the Instructors of 
 youth in Academies and Schools, as combining more excellencies than any other Arith- 
 metick now ia use. 
 
 ZEPH. SWIFT MOORE, 
 President of the Collegiate Institution, at Amherst, Mass. 
 
 Extract of a Letter from My. Benedict, Tutor of Williams College, dated 
 
 WILLIAMS COLLEGE. JANUARY 2, 1822. 
 
 ' FROM the experience which I have had in instructing youth, I have had occasion 
 to acquaint myseli with many, if not most of the Systems of Arithmetick new in use in 
 this country. I can therefore speak with some more confidence than I otherwise should, 
 from having proved their excellencies and defects hy actual trial of them. It is most 
 certain that as a complete System on this important part of education, the work under 
 consideration stands pre-eminent. It is impossible that Arithniet : ck should be so treated 
 of. as not to leave much to be done by the instructor. Still, as I think, Pike's System 
 wil! enable tlie teacher to benefit his scholars, to give tliem sound theoretical ard prac- 
 tical knowledge in this branch, fo induce them to think and leason rlosely, and increase 
 their power <f arithmetical invention, far more than any other within the compass of 
 my knowledge. 
 
 GEORGE W. BENEDICT. 
 
'a lioticr. 
 
 THE SYSTEM OF ARITHMETICS, of which this work is a new 
 and careful abridgment, has been so long an inmate of our acad- 
 emies and higher seminaries of learning, that its merits are fa- 
 miliar to all. Though numerous treatises on the subject have 
 successively appeared, since the work of PIKE was first publish- 
 ed,* few have been able to sustain even an ephemeral reputation, 
 excepting such as were built upon the labours of our author, yet 
 simplified in some of the more intricate parts. Excellent as that 
 work is acknowledged to have been, it had its defects ; espe- 
 cially in its want of conformity to the federal notation, and of sim- 
 plicity, and attraction to the scholar, in a few of the rules. An 
 abridgment was some years since published, in which little else 
 was done than to change the notation ; and for want of that con- 
 ciseness in the fundamental rules, which some treatises of less 
 real merit possessed, the book was superseded, and has been ne- 
 glected in this section of the country, even by those who avow 
 their preference for Pike's as a complete system. 
 
 At the suggestion of several experienced teachers, the Pub- 
 lisher was induced to put to press a NEW ABRIDGMENT of Pike's 
 Arithmetick, embracing all such portions of the original w r ork as 
 should be necessary for the use of common schools, or of private 
 learners ; containing also, many new and practical illustrations of 
 the more important rules. For this purpose, several improve- 
 ments suggested by instructors of schools and academies have been 
 incorporated in the work ; and the whole has undergone the pa- 
 tient and careful revision of a gentleman well known to the pub- 
 lick, as a teacher of great merit both in Mathematicks and Nat- 
 ural Philosophy, and who has devoted many years to the instruction 
 of youth of both sexes. 
 
 *Inl7S8. NICOLAS PIKE was a native of Somersworth in New-Hampshire ; was 
 born Oct. 6, 1743; graduated at Harvard College, 1766; was employed a great portion 
 of his life in instructing youth, and died at Newburyprrt. Ms. Dec. 9, 1819. 
 
vi PUBLISHER'S NOTICE. 
 
 One great object in this abridgment has been to simplify the 
 general rules, by placing before the scholar their constituent parts 
 illustrated by plain and easy examples. In executing the work, 
 nothing superfluous has been added, and nothing omitted that 
 would contribute to perfect its design, and render it serviceable 
 to youth. Those, however, who are in the habit of teaching su- 
 perficially, with a view of flattering the pupil and the parent with 
 the mistaken idea of extraordinary progress, may probably raise 
 objections against the work, as containing too many things to be 
 committed to memory that they will burden and confuse the 
 mind of the scholar. Such persons have yet to learn the capa- 
 city of the young mind. For, though it may be true, that a mass 
 of complex ideas crowded into the mind of a scholar would em- 
 barrass and perplex him it is equally true, that in proportion to 
 the number of simple propositions impressed upon the memory of 
 a child, will be the progress of his understanding in strength and 
 capacity. 
 
 In the arrangement of the present work, regard has been paid 
 to the natural dependence of the several parts upon each other. 
 Though in some instances, it differs from the common method, it 
 is believed to be the more correct and useful. Some of the old 
 and obsolete rules of Tare and Trett, &c, have been omitted, and 
 the Duties and Custom-House Allowances of our own country 
 substituted. Several rules, such as Position, Alligation, Permu- 
 tation, &c. are inserted in this work, more for the purpose of grat- 
 ifying the curiosity and exercising the mind of the scholar, than 
 forthe-'r utility in business. The PRACTICAL parts are those upon 
 which the greatest attention has been and should be bestowed ; 
 and it is upon the improvements 4n these generally, as well as 
 upon the established character of the author, that the publisher 
 rests his belief of the merits of this edition. 
 Jpril 14, 1826. 
 
TABLE OF CONTENTS, 
 
 fage.\ 
 Numeration, 9 
 Table, 10 
 Notation by Roman Letters., 12 
 Addition, 13 
 
 Tnhlr 1/1 
 
 page, 
 General Method of Making Taxes, 140 
 Rule of Three Direct, 142 
 ,. _ . 1 1~ verse I'i3 
 
 The Double Rule of Three, 143 
 Conjoined Proportion, 145 
 Practice, 147 
 Single Fellowship, 153 
 Double Fellowship, 155 
 Custom-House Allowances, 158 
 American Duties, 159 
 Discount, 160 
 Barter, 162 
 Loss and Gain, ItJl 
 Equation of Payments, 165 
 Involution, 167 
 Evolution, 167 
 TaWe of Powers, 168 
 Extraction of the Square Root, 169 
 Extraction of the Cube Root, 177 
 Rule for extracting the roots of all 
 powers, 183 
 Alligation Medial. 184 
 Alligation Alternate, 185 
 Single Position, ISO 
 Double Position, 191 
 Permutation, 193 
 Gauging, 194 
 vlecnanical PowersOf the Lever, 196 
 Of the Wheel and Axle, 198 
 Of the Screw, 197 
 Useful an! Diverting Questions, 198 
 
 Subtraction, 17 
 Multiplication, 20 
 Table 21 
 Division, 27 
 
 ,, Tnhlr 5R 
 
 Federal Money, 33 
 Table of American Coins, 33 
 Do. of Foreign Coins, 39 
 Compound Addition, 40 
 Subtraction, 52 
 Problems, resulting from a comparison of 
 the preceding rules, 56 
 Reduction, 59 
 Vulgar Factions, 70 
 Decimal Fractions, 89 
 Compound Multiplication, 99 
 Division, 105 
 Simple Interest, 111 
 Commission, 
 Buying or Selling Stocks, 118 
 Insurance, 119 
 Simple Interest by Decimals, 119 
 Compound Interest 121 
 by Decimals, 123 
 Duodecimal, or Cross Multiplication, 124 
 The Single Rule of Three, 130 
 
 INDEX TO T 
 
 Addition, 14 
 Ale or Beer Measure, 50 
 Aliquot parts of money weight, 
 fcc. 112, 147, 148 
 American Coins, 
 Apothecaries' Weight, 44 
 Avoirdupois do. 43 
 Cloth Measure, 44 
 Decimals, 89 
 Decimal parts of a year, 120 
 Discount per cent. 148 
 Division, 28 
 Dry Measure, 51 
 Duodecimals, 124 
 English Money, 41 
 Federal Money, 33 
 Foreign Coins, 39 
 
 0^7** Beside the tables above enumeratec 
 BOOK, published in connexion with this v 
 jevral not found in common arithmetic*;*, 
 
 HE TABLES. 
 
 Interest by decimals, 119, 123 
 Land or Square Measure, 48 
 Long do. 45, 4G 
 Money, 33, 41 
 Motion, 47 
 Multiplication, 21 
 Notation, 12 
 Numeration, 10 
 IVnoe and shillings, 41 
 Powers, 168 
 ^olid Measure, 49 
 Square do. 48 
 Taxes, for making, 141 
 Time, 46 
 Troy Weight, 42 
 Wine Measure, 49 
 
 , the scholar will find in the CIPHERING- 
 alume. a variety of rary nseful tables, and 
 
EXPLANATION OF ARITHMETICAL SIGNS. 
 
 Signs. 
 
 = Two parallel horizontal lines are the sign of equality. It 
 shows that the number before, is equal to the number after it. 
 Example, ] dollars 100 cents, is read thus, 1 dollar is equal 
 
 to 100 cents. 
 
 -f- Two short lines crossing each other at right angles, are the 
 sign of Addition. It shows that numbers with this sign be- 
 tween them, are to be added together. Example, 5+7=12, 
 is read thus, 5 added to 7, or 5 plus 1, is equal to 12. 
 
 A short horizontal line is a sign of subtraction. It shows that 
 the number after it, is to be taken irom the number before it. 
 Example, 12 7^=5, is read thus, 12 less 7, or 12 minus 7, is 
 equal to 5. 
 
 X Two short lines crossing each other in the form of an X, are 
 the sign of multiplication. It shows that the number before 
 it, is to be multiplied by the number after it. Example, 
 6X5=30, is read thus, 6 multiplied by 5, is equal to 30. 
 
 -7- A short horizontal line between two points, is the sign of divis- 
 ion. It shows that the number before it, is to be divided by 
 the number after it Example, 30-~6==5, is read thus, 30 
 divided by 6 is equal to 5. 
 
 : : : : Four points or colons are the sign of proportion ; and to show 
 that numbers are proportional, they are written thus, 
 2 : 4 : : 9 : 16, which are read, 2 is to 4 as 8 is to 16. 
 
 I 3 This sign signifies the second power or square. 
 I 3 This sign signifies the third power or cube. 
 
 2 This sign prefixed to any number shows that the square root of 
 V the number is required. 
 
 s This sign, prefixed to any number, shows that the cube root of 
 V the number is required. 
 
 CORRECTION'S AND REMARKS, 
 
 Though unwearied pnins have been taken to have the typographical execution of this work 
 correct, it is not expected that it is so, in all respects and the publisher wiil thank those teach- 
 ers, or other persons who discover errors, to point them out that they may be corrected in fu- 
 ture editions. The following are the only errors we have discovered : 
 
 Page 142, Quest. 1. For the divisor 9, read 3. 
 ' 144, Quest 1. Though correctly stated and wrought by one method, is not according to 
 
 the rule there givtn. 
 Page 168. Tabk of Power f. Square cube of 3, for 728, read 729. 
 
ARITHMETICS 
 
 AHITHMETICK is the science of Numbers ;* and com- 
 prises the following principal rules, viz. I. NOTATION 
 or NUMERATION ; II. ADDITION ; III. SUBTRACTION ; IV. 
 MULTIPLICATION ; and V. DIVISION. The four last are 
 called simple, when the numbers are all of one denom- 
 ination ; compound, when the numbers are of different 
 denominations. 
 
 These five rules are called principal or fundamental, because 
 the whole art of arithraetick is comprehended in their various 
 operations. 
 
 I. NUMERATION. 
 
 1. NUMERATION teaches us to read or write any sum or number, 
 by means of the following ten characters, called figures :f 
 
 Cipher. One. Two. Three. Four. Five. Six. Seven. Eight. Nine: 
 
 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. 
 
 2. The value of a figure, when alone, is called its simple value, 
 and is invariable. Figures have also a local value, which varies 
 
 * Number is either an unit, or a collection of unit?. A single thing is an unit, or 
 one. One and one are two. One and two are three. And thus, by constant addition, 
 all numbers are generated. 
 
 f These figures, which are of Arabic or Indian origin, were introduced Into Europe by 
 the Moors, about the year 1150: they were formerly all called ciphers, whence it came 
 to pass that the art of arithmetick was called ciphering. 
 
 1 . What in Arithmetick 1 2. What are called itsfundamental rules 1 3. What 
 
 does Numeration teach ? 4. How many figures are used to represent numbers ?-~ 
 
 5. How do you determine their value ? 
 
 B 
 
|0 NUMERATION. 
 
 according to the place in which they stand. In a combination of 
 figures, reckoning from the right to the left, the figure at the 
 right, or in the first place, represents its simple value ; that in 
 the second place, ten times the simple value ; that in the third 
 place, ten times the value of that in the second place ; and so on, 
 in a tenfold increase. 
 
 EXAMPLE. Write down the sum 4444. The first figure at the right, 
 in the place of units, has its simple value, or the same as if standing 
 alone -four. The second, in the place of tens, signifies four tens, or 
 forty. The third figure, in the place of hundreds, signifies four hun- 
 dred, or ten times its value in the place of tens. The fourth figure 
 is in the place of thousands^ bearing ten times its value in the place 
 immediately preceding. 
 
 3. A cipher 0, though of no signification itself, when placed on 
 the right hand of figures, in whole numbers, increases their value 
 in the same tenfold propoition. Thus 9, signifies only nine, itf 
 simple value. Place a cipher on the right, (90) it becomes nine- 
 ty ; and by placing two ciphers at the right, thus (900) it be- 
 comes nine hundred. 
 
 ZSTUMEH.A.TION TABLE, 
 
 3. 
 
 3 2 
 3 2 I 
 
 1. Period of units. 
 
 2. Do. of thousands, 
 
 3. Do. of millions. 
 
 nine units, or nine. 
 
 nine tens, and eight units, or ninety-eight. 
 
 nine hundreds, eight tens, and seven units, or 
 
 nine thousand, 876. [nine hund. eighty-seven. 
 
 ninety-eight thousand, 765. 
 
 nine hundred eighty-seven thousand, 654. 
 
 nine millions, 876 thousand, 543. 
 
 ninety-eight millions, 765 thousand, 432. [321. 
 
 nine hundred eighty -seven millions, 654 thous. 
 
 JVbfe. Six places of figures, beginning on the right, are called a 
 period ; but they are commonly divided into half periods of three 
 
 6. In a combination of figures, hoiv do you ascertain their value ? 7. What is 1h* 
 nature of the, cipher ? 8. What is a period in Numeration ? 
 
NUMERATION. 1 1 
 
 figures each. This division enables us to read any number of figures 
 as easily as we can read the first period. 
 
 RULE. Commit the words at the head of the Table, viz. units, 
 tens, hundreds, &c. to memory ; then, to the simple value of each 
 figure, join the name of its place, beginning at the left hand and 
 reading towards the right. More particularly 1. Place a dot under 
 the right hand figure of the 2d, 4th, 6th, 8th, &c. half periods, and 
 the figure over such dot will, universally, have the name of thousands. 
 2. Place the figures 1, 2, 3, 4, &c. as indices, over the 2d, 3d, 4th, 
 &c. period : These indices will then show the number of times the 
 millions are involved the figure under 1 bearing the name of mil- 
 lions, that under 2, the name of billions, (or millions of millions) .; 
 that under 3, trillions, (or millions of millions of millions.) 
 
 EXAMPLE. 
 
 Sextillions. Quintillions. Quatrillions. Trillions. Billions. Millions, Units, 
 
 tii. un. th. un. th. un. th. un. th. un. th. un. c.x.t. c.x.u 
 
 654321 
 $13,208.000,341.620,057.219,356. 819,379.120,406. 129,763 
 
 H H H H H H H 
 
 cr D- er cr cr sr sr 
 
 o o o o o o o 
 
 I I 1 I I I I 
 & & g- & & & s- 
 
 Note. Billions is substituted for millions of millions ; Trillions, for 
 millions of millions of millions ; Quatrillions, for millions of millions 
 of millions of millions. Quintillions, Sextil lions, Septillions, Octil- 
 lions, Nonillions, Decillions, Undecillions, Duodecillions, fee. answer 
 to millions so often involved as their indices respectively denote. 
 
 The right hand figure of each half period has the place of units, 
 of that half period ; the middle one, that of tens, and the left hand 
 one, that of hundreds, 
 
 APPLICATION. Let the scholar now read, or write down in words at 
 length, the following numbers : 
 
 8 437 709.040 3.476.194 7.184.397.647 
 
 17 3.010 879.096 84.094.007 49.163.189.186 
 
 129 76.506 4.091.875 690.748.591 500.098.422.700 
 
 Write down, in proper figures, the following numbers : 
 Fifteen 
 
 Two hundred and seventy-nine 
 
 9. How are numbers commonly divided ? 10. Of what use is this division ? -li 
 
 What M the rule for Numeration ? 
 
12 
 
 ROMAN NOTATION, 
 
 Three thousand four hundred and three 
 
 Thirty-seven thousand jive hundred and sixty- ? 
 seven i 
 
 Four hundred one thousand and twenty -eight 
 
 Nine millions seventy-two thousand and two 
 hundred 
 
 Fifty-five millions three hundred nine thou- 
 sand and nine 
 
 Eight hundred millions forty-four thousand 
 and fifty-five i 
 
 Two thousand five hundred and forty-three) 
 mil IT mis four hundred and thirty one thou-\ 
 sand seven hundred and two \ 
 
 NOTATION BY ROMAN LETTERS, 
 
 ROMAN NOTATION is the method of representing numbers by 
 Letters ; and is now chiefly used to number the chapters of 
 books, &c. Seven letters are used for this purpose, viz. I, V, 
 X, L, C, D, and M. /, signifies 1 ; V, 5 ; X, 10 ; L, 50 ; C, 
 100 ; >, 500 ; and M, 1000. 
 
 TABLE. 
 
 I, 
 
 One. 
 
 XV, Fifteen. ,CC, Two Hundred. 
 
 ", 
 
 Two. 
 
 XVI, Sixteen. CCG, Three Hundred. 
 
 in, 
 
 Three. 
 
 XV 7 II, Seventeen. CCCC, Four Hundred. 
 
 IV, 
 
 Four. 
 
 iXVUl, Eighteen. D, or IQ, Five Hundred. 
 
 v, 
 
 Five. 
 
 X(X, Nineteen. DC, Six Hundred. 
 
 vi, 
 
 Six, 
 
 XX, Twenty. 
 
 DCC, Seven Hundred. 
 
 VII, 
 
 ^even. 
 
 XXX, Thirty. 
 
 DCCC, Eight Hundred. 
 
 VIII, 
 
 fcight. 
 
 'XL, Forty. 
 
 DCCCC, Nine Hundred. 
 
 IX, 
 
 Nine. 
 
 L, Fifty. 
 
 M, or CIo, One Thousand. 
 
 X, 
 
 Ten, 
 
 LX, Sixty. HJQ, Five Thousand. 
 
 XI, 
 
 Eleven. 
 
 LXX. Seventy. j^OOOi Fifty Thousand.* 
 
 Xri, 
 
 Tvvehe. 
 
 LXXX, Eighty. looo^OOD^ Five Hund- Thous. 
 
 Xlil, 
 
 Tn'rieen. 
 
 XC, Ninety MDCCCXXV, One Thous. Eight 
 
 XIV, 
 
 Fourteen. 
 
 C, Hundred. [Hund. and Twenty-five. 
 
 * Sometimes thousands are represented by drawing a line over the top of the numeral 
 letttr.- Mas. V represents five thousand, L fty thousand, CC two hundred thousand. 
 
 12 Wliat is Roman Notation ? 13. What is its use ? 14. How many letters 
 
 are used for this purpose ? 15. What number does each represent ? 
 
SIMPLE ADDITION. 13 
 
 RULES. 
 
 1. The number of a letter is doubled as often as it is repeated ; 
 thus, 1, represents one ; II, two ; X, ten ; XX, twenty ; XXX, 
 thirty. 
 
 2." A less literal number placed after a greater, augments the 
 value of the greater ; if put before, it diminishes it. Thus, VI 5 
 is 6 ; IV, is 4 ; XI, is 11 ; IX, is 9, &c. 
 
 II. ADDITION. 
 
 ADDITION is the putting together of two or more numbers, or 
 sums, to make them one total, or whole sum. 
 
 SIMPLE ADDITION 
 
 Is the adding of several numbers together, which are all of one 
 sort, or kind ; as, 7 pounds, 12 pounds, and 20 pounds, being added 
 together, make a sum total, or aggregate, of 39 pounds. 
 
 RULE. 
 
 Place units under units, tens under tens, &c. : draw a line un- 
 derneath, and begin with the units : After adding up every figure 
 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 
 manner through every column or row, and set down the whole 
 amount of the last row.* 
 
 PROOF. 
 
 Begin at the top of each column, and add the figures down- 
 wards, in the same manner as they were added upwards, and, if 
 it be right, this aggregate will be equal to the first amount. Or, 
 cut off the upper line of figures, and find the amount of the rest ; 
 then if this amount and upper line, when added together, be 
 equal to the sum total, the work is supposed to be right. 
 
 * This rule is founded on the known axiom, that il the whole ',s equal to the sum of all 
 its parts." The method of placing the numbers, and carrying for tens, is evident from 
 the nature of notation ; for any other disposition of the numbers would alter their value; 
 and carrying 1 for every 10, from an infeiior to a superior denomination, is evidently 
 right ; because 1 unit in the lattercase is equal to the value of 10 units in the former. 
 
 16. How are other numbers represented 1 17. What is Addition? 18. IVJiat 
 
 is Simple Addition ? 19. Repeat the rule. 20. Why do you carry for ten, m arftf- 
 
 ing simple numbers 1 21. What is your method of proof ? 
 
14 
 
 SIMPLE ADDITION. 
 
 ADDITION TABLE. 
 
 [It is not necessary that this Table be cammitted to memory, so as to repeat it 
 
 wholly out of the book. This would be indeed a tedious task. When the pupil 
 
 can read the two first columns, viz. 2 and 6, 2 and 3, 2 and 7, &c., and cover the 
 
 third, viz. 8, 5, 9, &c., and recite it readily, it will be sufficiently committed.] 
 
 2 and 6 are 8 
 
 5 and 9 are 14 
 
 Sand 11 are 19 
 
 11 and6are!7 
 
 2 3 5 
 
 5 13 18 
 
 8 7 15 
 
 1 3 14 
 
 279 
 
 5 7 12 
 
 8 13 21 
 
 1 8 19 
 
 2 4 .6 
 
 538 
 
 8311 
 
 1 4 15 
 
 2 9 11 
 
 5 5 10 
 
 8 8 16 
 
 1 12 23 
 
 257 
 
 5 11 16 
 
 8 5 13 
 
 1 T 18 
 
 2 8 10 
 
 5 8 13 
 
 8 10 18 
 
 1 2 13 
 
 2 12 14 
 
 5 6 11 
 
 8 6 14 
 
 1 9 20 
 
 224 
 
 527 
 
 8 2 10 
 
 1 13 24 
 
 2 10 12 
 
 5 12 17 
 
 8 12 20 
 
 1 5 16 
 
 2 11 13 
 
 549 
 
 8917 
 
 1 11 22 
 
 2 13 15 
 
 5 10 15 
 
 8 4 12 
 
 1 10 21 
 
 3 and 9 are 1 2 
 
 6 and 13 are 19 
 
 9 and 6 are 1 5 
 
 12andl4are26 
 
 3 7 10 
 
 6 7 13 
 
 9 3 12 
 
 12 6 18 
 
 358 
 
 639 
 
 9716 
 
 12 3 15 
 
 3811 
 
 6 8 14 
 
 9 13 22 
 
 12 ' 10 22 
 
 369 
 
 6511 
 
 9 10 19 
 
 12 7 19 
 
 347 
 
 6 12 18 
 
 9 7 16 
 
 12 9 21 
 
 325 
 
 6 4 10 
 
 9 5 14 
 
 12 5 17 
 
 3 10 13 
 
 6 10 16 
 
 9211 
 
 12 2 14 
 
 336 
 
 628 
 
 9 12 21 
 
 12 8 20 
 
 3 12 15 
 
 6 6 12 
 
 9 8 17 
 
 12 13 25 
 
 3 11 14 
 
 6 9 15 
 
 9 4 13 
 
 12 4 16 
 
 3 13 16 
 
 6 11 17 
 
 9 11 20 
 
 12 11 23 
 
 4 and 13 are 17 
 
 7 and 9 are 16 
 
 10 and 9 are 19 
 
 13and2are 15 
 
 459 
 
 7 13 20 
 
 10 13 23 
 
 13 7 20 
 
 4 11 15 
 
 7 3 10 
 
 10 T 17 
 
 13 10 23 
 
 4 8 12 
 
 7 7 14 
 
 10 11 21 
 
 13 3 16 
 
 426 
 
 7 11 18 
 
 10 3 13 
 
 13 8 21 
 
 4 6 10 
 
 7 5 12 
 
 10 10 20 
 
 13 5 18 
 
 437 
 
 7 8 15 
 
 10 4 14 
 
 13 13 26 
 
 4 12 16 
 
 729 
 
 10 16 
 
 13 9 22 
 
 4711 
 
 7 6 13 
 
 10 12 22 
 
 13 4 17 
 
 4 10 14 
 
 7 12 19 
 
 10 5 15 
 
 13 12 25 
 
 4 9 13 
 
 7411 
 
 10 2 12 
 
 13 6 19 
 
 448 
 
 7 10 17 
 
 10 8 18 
 
 13 11 24 
 
SIMPLE ADDITION. 15 
 
 EXAMPLES. 
 . What is the amount of 3406, 7980, 345, and 27 ? 
 
 H H 55 H cj 
 
 rt JT C TI B 
 
 If.?? $^7" Here we begin by writing 
 
 [ g. down the several numbers, units 
 
 under units, tens under tens, &c. 
 
 s 3.4 6 Then draw a line under them. 
 
 | We now commence adding at the 
 
 7.980 foot of the right hand column, 
 345 and say, 7 and 5 are 12, and 6 
 2 7 are 18. This exceeding ten, we 
 write down the right hand figure 
 
 Total amount, 11.7 58 8 under the column of units, and 
 
 Amount with tht carry 1 to the next column ; and 
 
 Upper line cut off, 8.352 say, 1 and 2 are 3, and 4 are 7, 
 
 and 8 are 15. We wrile down 5 
 
 Proof, 11.7 58 at the foot of the column, and 
 
 proceed to the next ; 1 and 3 are 
 
 4, and 9 are 13, and 4 are 17. We write down in the same way the 
 right hand figure, 7, under its column ; and carrying 1 to the next, 
 say 1 and 7 are 8, and 3 are 11. This being the last column, we 
 write down the whole amount 11, and find the sum total to be 11.758. 
 The method of proof is sufficiently explained. 
 
 2. 3. 4. 
 
 389261 2136784 37696941 
 
 789794 8297698 49760823 
 
 849798 8297694 45697615 
 
 487697 4876695 82132435 
 
 999996 1234697 48769206 
 
 9 4 8 2 1 7092032 48769209 
 
 5. 6. 7. 
 
 37856 378269 141 
 
 975 402607 3672 
 
 1234 702 82971 
 
 14 1246 34676 
 
 5612 2132 1459 
 
 2075 45178 427 
 
 16287 10276 12 
 
16 
 
 SIMPLE ADDITION. 
 
 8. What is the amount of three 
 hundred and sixty-five, eight hun- 
 dred and seven, five hundred arid 
 sixty, twenty-five, thirty-seven, and 
 one hundred one ? Ans. 1895. 
 
 9. The hind quarters of a cow 
 weigh one hundred and three 
 pounds each ; the fore quarters 
 weigh ninety-seven pounds each ; 
 the hide sixty-thiee pounds, and 
 the tallow fifty-six pounds ; what 
 is the whole weight of the cow ? 
 
 Ans. 519 pounds. 
 
 10. A man has four farms ; the 
 first is worth two thousand seven 
 hundred and twenty-five dollars ; 
 the second is worth three thousand 
 eight hundred and nineteen dol- 
 lars ; the third is worth one thou- 
 sand six hundred and ten dollars ; 
 the fourth is worth five hundred 
 and twelve dollars ; what are they 
 all worth ? 
 
 Ans. 8.666 dollars. 
 
 11. A man possesses a tract of 
 land, which contains forty-nine 
 thousand eight hundred and thir- 
 ty-five acres ; suppose he had six 
 tracts of equal dimensions, how 
 many acres would the whole con- 
 tain ? Ans. 299.010. 
 
 12. What is the amount of three 
 hundred, seventy-five, two, forty- 
 seven, thirty-three, nine thousand 
 seven hundred and eighty-four, 
 twenty thousand one hundred and 
 fifty, seven hundred and sixty-five 
 thousand and ninety one, and one 
 million seventy-five thousand and 
 forty-seven? Ans. 1.870.529. 
 
 13. Add seventy-five millions 
 nine hundred and sixty thousand 
 eight hundred, two hundred and 
 twenty-five thousand, and one hun- 
 dred and forty together. 
 
 Ans. 76.185.940. 
 
 14. W T hat is the sum of four 
 thousand and twenty five, seventy- 
 five thousand six hundred, eight 
 hundred thousand four hundred 
 and fifty, five millions three hun- 
 dred and ten thousand, thirty mil- 
 lions seven hundred and twenty, 
 aiid nine hundred fifty millions ? 
 Ans. 936.190.795. 
 
 1 5. What is the sum of one mi 11- 
 ion five hundred thousand, three 
 hundred and eleven thousand, nine- 
 ty thousand six hundred and ten, 
 fifty thousand and forty- five. 
 
 Ans. 1.951.655. 
 
 REMARK, As it is of great consequence in business to perform 
 addition readily and exactly, the learner ought to practise it till 
 it become quite familiar. If the learner can readily add any two 
 digits, he will soon add a digit to a higher number with equal 
 ease. It is only to add the unit place of that number to the digit, 
 and if it exceed ten, it raises the amount accordingly. Thus, be- 
 cause 8 and 6 are 14, 48 and 6 are 54. It will be proper to mark- 
 down under the sums of each column, in a small hand, the figure 
 that is carried to the next column. This prevents the trouble of 
 
SIMFLE SUBTRACTION. 17 
 
 going over the whole operation again, in case of interruption or 
 mistake. If you want to keep the account clean, mark down the 
 sum and figure you carry on a separate paper, and after revising 
 them, transcribe the sum only. After some practice, we ought 
 to acquire the habit of adding two or more figures at one glance. 
 This is particularly useful when two figures which amount to 10, 
 as 6 and 4> or 7 and 3, stand together in the column. Every 
 operation in arithmetick ought to be revised to prevent mistakes ; 
 and as one is apt to fall into the same mistake if he revise it in 
 the same manner he performed it, it is proper either to alter the 
 order, or else to trace back the steps by which the operation 
 advanced, which will lead us at last to the number we be- 
 gan with. 
 
 -**- 
 
 III. SUBTRACTION. 
 
 N teaches to take a less number from a greater, to 
 find a third, shewing the inequality, excess or difference between 
 the 'given numbers. The greater number is called the Minuend ; 
 the lesser number is called the Subtrahend. The difference be- 
 tween them, or what is left after subtraction is made, is called 
 the Remainder. 
 
 SIMPLE SUBTRACTION 
 
 Teaches to find the difference between any two numbers, which 
 ire of a like kind. 
 
 RULE. 
 
 Place the larger number uppermost, and the less underneath, 
 so that units may stand under unis, tens under tens, &c ; then 
 drawing a line underneath, begin with the units, and subtract the 
 lower from the upper figure, and set down the remainder ; but 
 if the lower figure be greater than the upper, add ten to the 
 upper, and subtract the lower figure therefrom : To this differ- 
 
 22. What is Subtraction 1 23. What is the m 'nutnd ? 24. What is the sub- 
 trahend ? 25. What the remainder 1 2tf. How d you proceed in subtracting 
 
 tima's numbers* 
 
 c 
 
 
18 SIMPLE SUBTRACTION. 
 
 ence add the upper figure, which being set down, you must add 
 one to the lower figure of the next column, for that which you 
 borrowed ; and thus proceed through the whole. 
 
 PROOF. 
 
 jt 
 
 Add the remainder and the less number together ; if the work 
 be right, the amount will be equal to the greater number : Or, 
 subtract the remainder from the greater sum, and the difference 
 will be equal to the less. 
 
 EXAMPLES. 
 
 From 3724 Minuend O^T The operation of this example is 
 
 Take 2583 Subtrahend very plain. The two sums being written 
 
 tlown according to the rule, we draw a 
 1141 Remainder line underneath, and beginning at the right 
 
 hand figure, say 3 from 4 leaves 1, whi^h 
 -Proof 3724 we set down in. the next column, the sub- 
 
 trahend (8) being greater than the minuend, 
 
 we add 10 to the upper figure, making it 12, and say, 8 eight from 12, 
 there remain 4, which is written down. We now carry 1 to the next 
 column, for that which we just borrowed, and say, 1 to 5 is 6, and 6 
 from 7 leaves 1, which we put down ; and in the next column, tak- 
 ing 2 from 3, leaves I, which we write down, and the work is 
 done. The method of proof will be extremely easy to the learner. 
 
 3. 4. 5. 
 
 Feet. * Cwt., Doll*. 
 
 879647 9187641 10000 
 
 164348 91843 9999 
 
 6. 7. 
 
 1 000200340000 1 00200300400400600700S009CO 
 
 3189910304 980760540320nu23045067089. 
 
 What is the melhod of proof ? 
 
SIMPLE SUBTRACTION. 
 
 8. The Arabian or Indian meth- 
 od of notation was first known in 
 England in 1150$ how long is it 
 to the present time ? 
 
 9. What number must be added 
 to 6892, so that the sum shall be 
 8265 ? Am. 1373. 
 
 10. A gentleman purchased fif- 
 teen thousand eight hundred and 
 forty acres of land ; he sold two 
 thousand three hundred and fifty 
 to one man ; four thousand five 
 hundred to another ; and three 
 thousand two hundred and twenty- 
 five to a third ; how many acres 
 
 had he left ? 
 
 Am. 5.765. 
 
 11. A man died, leaving an 
 estate of 12.650 dollars, which he 
 bequeathed as follows : 2.500 dol- 
 lars to each of his three daughters, 
 and the remainder to his son ; 
 what was the son's share ? 
 
 Ans. 5.150 dollars. 
 
 12. If the greater of two num- 
 bers be seven hundred and fifty 
 millions, and the less five hundred 
 and forty thousand ; what will be 
 the remainder or difference ? 
 
 Ans. 749.460.000. 
 
 13. A man's property is worth 
 10.650 dollars ; but he owe* A 
 1.800 dollars, B 1.260 dollars, C 
 750 dollars, and D 500 dollars ; 
 what will remain after his debts 
 are paid ? Ans. 6.340 dolls. 
 
 REMARK. At every stage of the scholar's progress, it would 
 be well for him to review his previous studies to fix in his 
 memory the most important rules ; and endeavour to under- 
 stand thoroughly the reasonableness and accuracy of the principles 
 he is taught. The simple rules of arithmetick are easily ac- 
 quired ; but if the scholar hurries over them, as is too often the 
 case, without fully understanding their meaning and application, 
 he will find his subsequent studies as much retarded, as he would 
 find them aided, were he to become MASTER OF HIS SUBJECT as 
 he goes along. The instructor should question his pupil on every 
 rule, and explain the principles of the rules, in every case, where 
 fae scholar desires it 
 
Jiff SIMPLE MULTIPLICATION. 
 
 IV. MULTIPLICATION. 
 
 MULTIPLICATION teaches how to increase the greater of tw 
 numbers given, as often as there are units in the less ; performs 
 the work of many additions in the most compendious manner ; 
 brings numbers of great denominations into small, as pounds into 
 shillings, pence, or ifarthiugs, &c. ; and, by knowing the value of 
 one thing, we fin^the value of many. 
 
 There are three parts in Multiplication, viz. 
 The sum to be multiplied is called the Multiplicand. 
 The sum by which you multiply is called the Multiplier. 
 The result of the operation is called the Product 
 The Multiplicand and Multiplier are likewise called both to- 
 gether/actors, or that by which the operation is performed. 
 
 SIMPLE MULTIPLICATION 
 
 Is the multiplying of any two numbers together, without having 
 regard to theii; signification ; as, 7 times 8 are 56, &c. 
 
 RULE. 
 
 1. Place the Multiplier under the Multiplicand, so that units 
 stand under units, tens under tens, &c., and draw a line under 
 them. 
 
 2. Beginning at the right hand, multiply each figure in the 
 multiplicand by each in the multiplier, placing the first figure of 
 every line directly under its respective multiplier, and to the 
 product of the next figure carry one for every ten, as in addition. 
 
 3. Add the several products together, and their sum will be 
 the total product requiied. 
 
 PROOF. Multiply the multiplier by the multiplicand.* 
 
 It is indispensably necessary that the Multiplication Table be 
 comm tied perfectly to memory , before the scholar proceeds to the exam- 
 ples under the rules. 
 
 * Tne better way of proving Multiplication, is by Division. It may also be proved by 
 casting out the 9's , but as the work will sometimes prove by that method, when in fact 
 wrong, the rule is omitted. 
 
 28. Whnt is JWultip lication ? --- 29. Whai are the terms used in multiplication'! - - 
 30 What is th( multiplicand? --- 31. W hat the multiplier * -- 32. And what is tt-e 
 product? S3. What are ih, two first sometimes called 1 - 34. What is Simple 
 Multiplication 1 - 35. W 'hat is the general rvltfor multiplying' simple numbers 1 - 
 ">$. How do you prove your sum to be right ? -- 37. Repeat the Multiplication 
 
SIMPLE MULTIPLICATION. 
 
 21 
 
 MUL 
 
 2 times 1 are 2 
 2 4 
 3 6 
 4 8 
 5 10 
 6 12 
 7 14 
 8 16 
 9 18 
 10 20 
 11 22 
 12 24 
 3 times 1 are 3 
 2 6 
 3 9 
 4 12 
 5 1.3 
 6 18 
 7 21 
 8 24 
 9 27 
 10 30 
 11 33 
 1 2 36 
 
 TIPLICATIOW TA 
 
 5 times 9 are 45 
 10 50 
 11 55 
 12 60 
 
 BLB. 
 
 9 times 5 are 45 
 6 54 
 7 63 
 8 72 
 9 81 
 10 90 
 11 -99 
 12 108 
 
 6 tunes 1 are 
 2 12 
 3 18 
 4 24 
 5 30 
 6 3G 
 7 42 
 8 48 
 9 54 
 10 60 
 11 66 
 12 72 
 
 10 umes 1 are 10 
 2 20 
 3 30 
 4 40 
 5 50 
 6 60 
 7 70 
 8 80 
 9 90 
 10 100 
 11 110 
 12 120 
 
 7 times 1 are 7 
 2 14 
 8 21 
 4 28 
 5 35 
 6 42 
 7 49 
 8 56 
 S 63 
 10 70 
 11 77 
 12 84 
 
 1 1 times 1 are 11 
 2 22 
 3 33 
 4 4 1 
 5 55 
 6 66 
 7 77 , 
 
 80 o 
 oo 
 
 9 ( f9 
 10 110 
 11 121 
 12 132 
 
 4 times 1 are 4 
 2 8 
 3 12 
 4 16 
 5 20 
 6 24 
 
 7O 
 rf%O 
 
 8 32 
 9 36 
 10 40 
 11 -44 
 12 48 
 
 8 times 1 are 8 
 2 16 
 3 24 
 4 32 
 5 40 
 48 
 7 56 
 8 64 
 9 
 10 80 
 1 I 88 
 12 96 
 9 times 1 "are 9 
 2 18 
 3 27 
 4 36 
 
 12 times 1 are 12 
 
 C) <*A 
 
 3 36 
 4 43 
 r,o 
 72 
 7 84 
 8 
 9 108 
 10 120 
 11 
 
 5 times 1 are 5 
 
 2 : 10 
 
 3 15 
 
 4 C 2* 
 5 25 
 6 30 
 7 36 
 8 40 
 
 
22 SIMPLE MULTIPLICATION. 
 
 EXAMPLES. 
 
 1. What is the product of 90631, multiplied by 8 ? 
 
 Multiplicand, 90631 (jr Here, in the first place, we write 
 Multiplier, 8 down the multiplicand ; then we write the 
 
 multiplier under ths unit figure of the mul- 
 
 Product, 725.048 tiplicand, and draw a line underneath We 
 
 then say, 8 times 1 are 8 ; as the product 
 
 does not exceed nine, we write it underneath ; we then say, 8 times 
 3 are 24, we write down the right hand figure 4, and reserve the 
 left hand figure 2, to be added to the product of the next figure ; we 
 then say, 8 times 6 are 48 and two are 50, we write down the right 
 hand figure 0, and reserve the left hand figure 5, to he added to the 
 product of the next figure ; we then say, 8 times is and 5 are 5. 
 and write down 5 ; we then say, 8 times 9 are 72 : this being the last 
 figure, we write down the whole product 72, and find the answer to 
 fee 725.048. 
 
 2. 3. 4. 
 
 769308 49-80076 763896 
 
 3 4 5 
 
 Product, 2307924 
 
 5. What will be the product of 25.375, multiplied by 10 ?- 
 
 6. Multiply 750.005 by 11. 
 
 7. Multiply 900.301.399 by 12. 
 
 8. Multiply 989.000 by 9. 
 9- Multiply 568.098 by 8. 
 
 10. Multiply 639.408 by 10. 
 
 11. Multiply 789.795 by 7. 
 
 CASE I. 
 
 When the Multiplier consists of several figures. 
 
 RULE. Write down the multiplicand, and under it the mul- 
 tiplier, units under units, tens under tens, &c. Then multiply 
 by each significant figure in the multiplier separately, beginning 
 with units, and writing the first figure of each different product 
 directly under the figure/ by which you are multiplying. Add the 
 several products together, and you have the sum total. 
 
 38, What is yowr rule, when the multiplier c9nsists of several figures ? 
 
SIMPLE MULTIPLICATION, 23 
 
 EXAMPLES. 
 
 I. Multiply 7654 by 543. 
 
 Proof by Multiplication.* 
 
 7654 643 
 
 543 7654 
 
 22962=3 times the multiplicand. 
 306 IS =40 times do. 
 
 38270 =500 times do. 
 
 4.156.122=543 times do. 
 
 4.156.122 Products 
 
 When the multiplier is a number consisting of several figures, 
 after we have found the product of the multiplicand by the first 
 figure of the multiplier, we suppose the multiplier to be divided in- 
 to parts, and after the same manner, find the product of the multi- 
 plicand by the second figure of the multiplier ; but as the figure, by 
 which we are multiplying, stands in the place of tens, the product 
 must be ten times its number ; and, therefore, the first figure in this 
 product must be written in the place of tens, or, which is the same, 
 directly under the figure by which we are multiplying. And pro- 
 ceeding in the same manner with all the figures of the multiplier, 
 separately, it is evident, we shall multiply all the parts of the multi- 
 plicand by all the parts of the multiplier ; therefore, the sum of these 
 several products will be the whole product required. 
 
 CO 7 " If there are ciphers between the significant figures of the 
 multiplier, write them in a line with the product of the next signi- 
 ficant figure, directly under their places in the multiplier. As, in 
 Example 2d. 
 
 2. 3. 4. 
 
 647906 46293845 91861284 
 4003 106 6875 
 
 1943718 
 259162400 
 
 2593567718 
 
 * The reason efthe method of proof by multiplication, depends apoa this propositioc, 
 that if two numbers aie to be multiplied together, either of them may be made the multi- 
 plier or the multiplicand, and the product will be the same, JRwmplt. X4~24> and 
 4X6 -24. 
 
M SIMPLE MULTIPLICATION 
 
 CASE H. 
 
 When there are ciphers on the right hand of either the multiplicand 
 or multiplier, or both. 
 
 RULE. Neglect those ciphers ; then place the significant 
 figures under one another, and multiply by them only ; add them 
 together, as before directed, and place to the right hand as many 
 ciphers as there are in both the factors. 
 
 EXAMPLES. 
 
 1. 2. 3. 
 
 67910 956700 $30137000 
 
 5600 320 9500 
 
 Froduct, 380.296.000 
 
 4. What is the value of a farm of 600 acres, at 20 dollars an 
 acre ? 
 
 Ans. 12.000 dollars 
 5. Multiply 50.750.000 by 75.000. 
 
 Ans. 3.806.250.000.00*. 
 
 39. How do you proceed when any of the right hand figures of the multiplicand c> 
 multiplier are ciphers ? 
 
SIMPLE MULTIPLICATION. 2g 
 
 CASE III. 
 
 To multiply by 10, 100, 1000, #c. 
 
 RULE. Set down the mu tiplicand underneath, and join tire 
 ciphers in the multiplier to the right hand of them.* 
 
 EXAMPLES. 
 
 3. 4. 
 
 613975 8473965 
 
 1000 10000 
 
 Prod. 579350 
 
 CASE IV. 
 
 Wlien the multiplier is a composite number, (or exactly equal to the 
 product of any two figures in the multiplication table ^ 
 
 RULE. Multiply first by one of those figures, and that product 
 by the other, the last product will be the answer.f 
 
 EXAMPLES. 
 
 1. 2. 3. 
 
 Multiply 59375 by 35 39187 by 48 91632 by 56 
 
 Product, 2078126 
 
 * This is evident from the nature of numbers, since every cipher annexed to the right 
 of a number increases the value of that number in a tenfold proportion. 
 
 T The reason of this rule is obvious; for any number multiplied by the component 
 parts of another, must give the same product as if it were multiplied by that numbel 
 at once. 
 
 40. What is the rule when the multiplier is 10, 109, 1QW, #c. ? 1. When tft 
 
 multiplier is a composite number, what is your ruU ? 
 
 D 
 
20 SIMPLE MULTIPLICATION, 
 
 CASE v. 
 
 To multiply by 9, 99, 999, fyc. 
 
 RULE. Annex as many ciphers to the right of the multiplicand 
 as there are figures in your multiplier, and from the number thus 
 produced, subtract the given multiplicand, and the remainder 
 will be the product. 
 
 EXAMPLES. 
 
 I. Multiply 5384976 by 9999. 
 
 53849760000 O^r There being four 9s in the given multiplier, 
 
 5384976 add four ciphers (0000) at the right hand ; then 
 
 write the multiplicand underneath, and subtract ac* 
 
 53844375024 cording to the rule. 
 
 Multiply 371967 by 999. 
 
 Ans. 371595033. 
 
 PROMISCUOUS EXERCISES. 
 
 1. What is the product 
 237956, multiplied by 3728 ? 
 Ms. 887.099.968. 
 
 of 
 
 2. If 4 bushels of wheat make 
 1 barrefc of flour, and the price of 
 wheat bti 1 dollar a bushel, what 
 will 225 barrels of flour cost ? 
 
 3. Multiply 308879 by twenty 
 thousand five hundred and three. 
 Ans. 6.332.946.137. 
 
 4. Multiply 876956 by 990000. 
 Ans. 868.186.440.000. 
 
 5. If a man rise an hour earlier 
 every day, how much useful time 
 will he gain for study or labour in 
 20 years, there being 365 days in 
 a year ? 
 
 6. WbaJt will be the total pro- 
 duct of ninety-eight millions seven 
 hundred sixty-three thousand five 
 hundred and forty-two, multiplied 
 by the same sum ? 
 
 Ana. 9.734.237.228.385.764. 
 
 42. What if the rvkfor multiplying by 9, 99, 
 
SIMPLE DIVISION. 27 
 
 V. DIVISION. 
 
 DIVISION is the method of finding how many times a less num- 
 ber is contained in a greater ; or dividing a quantity given, into 
 any number of parts assigned ; and is a concise way of perform- 
 ing several subtractions. 
 
 There are four parts to be noted in Division : 
 
 1. The Dividend, or number given to be divided. 
 
 2. The Divisor, or number given to divide by. 
 
 3. The Quotient, or answer to the question, which shows how 
 often the divisor is contained in the dividend. 
 
 4. The Remainder (which is always less than the divisor, and 
 of the same name with the dividend) is very uncertain, as there 
 is sometimes a remainder, and sometimes none. 
 
 SIMPLE DIVISION 
 
 Is the dividing of one number by another, without regard to 
 their values ; as 56, divided by 8, produces 7 in the quotient : 
 That is, 8 is contained 7 times in 56. 
 
 GENERAL RULE. 
 
 1. Draw a curve line on each side of the dividend, and write 
 the divisor at the left hand. 
 
 2. Take the same number of the first left hand figures of the 
 dividend that there are in the divisor, if they be equal to, or 
 greater than the divisor ; but if they be less than the divisor, 
 take one more ; find the number of times the divisor is contained 
 in them, and write a figure representing the number at the right 
 hand of the dividend, which will be the first figure of the 
 quotient. 
 
 3. Multiply the divisor by this quotient figure, and write the 
 product under that part of the dividend taken. 
 
 4. Subtract this product from that part of the dividend taken, 
 and bring down the next figure of the dividend, and place it at the 
 right hand of the remainder : then find a quotient figure, multi- 
 ply and subtract as before directed ; proceed in the same man- 
 ner until all the figures in the dividend are brought down and 
 divided. 
 
 5. When a figure has been brought down and placed f at the 
 right hand of the remainder, if the number be less than the 
 divisor, write a cipher in the quotient, and bring down another 
 figure. 
 
 43. What is Division ? 44. How many parts are there in Division, and what 
 
 fire they 1 45. Wh*t is Svnpk Division '! 40. What is your rule ? 
 
28 
 
 SIMPLE DIVlSIOiY. 
 
 PROOF. Multiply the divisor raid quotient together, and add 
 the remainder, if there be any, to the product : If the work be 
 right, that sum will be equal to the dividend. 
 
 DIVISION TABLE. 
 
 1| 2| 3 ; 4 
 
 51 6 7 8 9 
 
 1 I 
 
 1C 
 
 11 
 
 12 
 
 To USE THE TABLE 
 
 Look for the divisor or 
 number by which you wish 
 :o divide in the left hand 
 3erpendicular column. 
 Then trace the horizon- 
 tal column in which the 
 
 2 1 4 6 j 81 
 3' 1 9*121 
 
 OJ12 : 1416 
 
 5'l82124 
 
 ! i 
 
 n 
 
 20 
 30 
 
 22 
 
 i 
 
 33 
 44 
 
 24 
 
 36 
 
 4 162 
 
 QJ242S3236 
 
 -i i j 
 330;3540'45 
 
 40 
 
 48 
 
 5 . 2 
 
 50 
 
 55 
 
 66 
 
 60 
 
 ;he dividend or numb.er 
 into wliiol; you wish to di- 
 vide, then trace that col- 
 umn to the top and you 
 will find the product or 
 number of times the divi- 
 sor is contained in the div- 
 
 6 
 
 i 36424854 
 
 60 
 
 72 
 
 7 
 
 4956 
 
 w 
 
 70 
 
 77 
 
 84 
 96 
 
 8 
 
 '64 
 
 72 
 
 80 
 
 88 
 
 9 
 
 81 
 
 90 
 
 99!l08 
 
 the exact number into 
 which you wish to divide 
 in the table, look for the 
 next less one, and the dif- 
 ference between them 
 will be what is over 
 
 10 
 
 100 
 
 110 
 
 120 
 
 11 
 
 121 
 
 132 
 
 12 
 
 144 
 
 EXAMPLES. 
 
 1. How many times is 3 contained in 175817 
 
 Div sor. Dividend. Quotient. 
 
 3) 1 7 5 8 1 7 (58605 
 1 5 
 
 2 5 
 
 2 4 
 
 1 8 
 1 8 
 
 1 7 
 1 5 
 
 2 Remainder. 
 
 0^7=- Here we first write down 
 the dividend, and making a curve 
 on each side, place the divisor (3) 
 at the left hand. In this example, 
 we see, that 3, the divisor, cannot 
 be contained in 1, the first figure 
 of the. dividend ; therefore we take 
 two figures, (17) and inquire how 
 often 3 is contained therein, which 
 finding to be 5 times, we place the 5 
 in the quotient, and multiply the 
 divisor by it, setting the first figure 
 of the multiplication under the 7 in 
 
 the 
 
 47. How do you prove your work to be right 
 
SIMPLE DIVISION. 29 
 
 PROOF. the dividend, &c. We then subtract 
 
 58605 Quotient. 15 from 17, and find a remainder of 
 
 X3 Divisor-f-2 2, to Ihe right hand of which we 
 
 - -- bring down the next figure of the 
 
 175817 dividend, viz. 5 ; then, we inquire 
 
 how often the divisor 3 is contained 
 
 Observe that the remainder 2, 1 in 25, and finding it to be 8 times, 
 is here added in multiplying > we multiply by 8, and proceed as be- 
 by 3. j fore, till we bring down the 1, when 
 
 finding we cannot have the divisor in. 
 
 1, we place Oin the quotient, and bring down 7 to the 1, and proceed 
 as at the first. 
 
 JVbfe. When there is no remainder to a division, the quotient istUe absolute and per- 
 fect answei to the question ; but when there is a remainder, it may be observed, that it 
 goes so much towards anrther tune as it approaches the divisor ; thus, if the remainder 
 be halfthe divisor, itwillgo half of a time more, and so on ; in order, therefore, to com- 
 plete the quotient, put the last remainder to the end of it, above a line, and the divisor 
 below it, as in Example 2. Hence the origin of vulgar fractions, which will be treated 
 of hereafter. 
 
 The reason of the proof is plain -, for, since the quotient is the number of times the 
 dividend contains the divisor, the product of the quotient and divisor must, evidently, 
 be equal to the dividend. 
 
 As the quotient and divisor are always multiplied during the operation, a simple method 
 of proof is, by adding the several products and remainder (if any) together as they stand. 
 Thus in the above example 1 prod. 15 .... 
 
 2 prod. .24 ... 
 
 3 prod. ..18. . 
 
 4 prod. ____ 15 
 Remainder .... 2 
 
 175817 Equal to dividend* 
 
 2. 3. 
 
 29) 1 53598(5296^ 493)9 1 876375( 1 41 50 
 
 145 
 
 85 
 58 
 
 261 
 
 188 
 174 
 
 14 
 
 48. When thert is no remainder, what is the quotient 1 49. When there is a 
 
 > h<l * its Mtw* ? 50. What reason have yon for yow method of 
 
30 SIMPLE DIVISION. 
 
 CASE I. 
 
 SHORT DIVISION, or when the divisor does not exceed 1. 
 
 RULE 1. Find how often the divisor is contained in the first 
 figure, or figures of the dividend, setting it under the dividend, 
 and carrying the remainder, if any, to the next figure, as so 
 many tens. 
 
 2. Find how often the divisor is contained in this dividend, and 
 set it down as before, continuing so to do, till all the figures ia 
 the dividend are used. 
 
 Note. The work in Short Division is done mentally, that is, divid- 
 ed in the mind, and the result only written down ; whereas in Long 
 Division the operation is written at large. 
 
 EXAMPLES. 
 
 In this example, we write down the dividend, 
 and draw a curve line at the left, and a straight 
 line underneath. We then take the first figure of 
 the dividend 9, and find that 4 is contained in 9, 2 
 times ; we write 2 under the 9, and then, mentally, 
 multiply the divisor by it, and the product is 8, which we subtract 
 from 9 and the remainder is 1 ; we then take the remainder 1, and 
 the next figure of the dividend 2, and the number is 12 ; we find 4 
 is contained in 12, 3 times ; we write 3 under the 2 and multiply the 
 divisor by it, and the product is 12, which we subtract from 12 and 
 nothing remains ; we then take the next figure of the dividend 4, 
 and find that 4 is contained in 4 once ; we write 1 under the 4, and 
 multiply the divisor by it, and the product is .4, which we subtract 
 from 4, and nothing remains. All the figures of the dividend having 
 been divided, we find that 4 is contained ia 924, 231 times. 
 
 12)1196437847536 
 Quotient, 35967 1 remainder. 
 
 CASE II. 
 
 When tJiere is one cipher r more at the right hand of the divisor. 
 
 RULE. It or they must be cut off ; also, cut off the same 
 iiumber of figures from the dividend, and then proceed as in case 
 first : But the figures which were cut off from the dividend 
 must be placed at the right hand of the remainder. 
 
 51. What is Short Division? 5*2. How is it performed 1 53. WhentherS 
 
 $rr ciphers at thtright hand oftht divisor, how c?o you proceed ? 
 
SIMPLE DIVISION. 31 
 
 EXAMPLES. 
 1. 2. 
 
 65100)3794326175(58374 5193|000)8937643|893( 
 
 325 
 
 544 
 
 520 
 
 243 
 195 
 
 482 
 455 
 
 276 
 26Q 
 
 1675 Remainder. 
 
 . To divide by 10, 100, 1000, &c., cut off as many figured 
 from the right hand of the dividend, as there are ciphers in the 
 divisor ; the left hand figures will be the quotient, and the right hand 
 figures cut off will be the remainder. 
 
 CASE III. 
 
 When the divisor is such a number as any two or more figures in the 
 
 table multiplied will make. 
 
 RULE. Divide the dividend by one of these figures, and the 
 quotient by the other ; the last quotient will be the answer. 
 
 EXAMPLE. 
 
 1. What is the quotient of 196473 divided by 72 ? 
 
 1st method. 2d method. 
 
 9)196473 8)196473 
 
 8)21830 9)24559 
 
 Quotient, 272857 Rem. 2728 57 
 
 54. What is your rule, vehen the divisor is the product of emu twofigwrts in ihe mtff- 1 
 fe6/e? 
 
-32 SIMPLE DIVISION. 
 
 CASE IV. 
 
 To divide by fractions, or parts of an unit. 
 RULE. If the numerator, or upper figure, is an unit, multiply 
 the given number by the denominator, or under figure, and the 
 product will be the answer : But if the numerator is more than 
 an unit, multiply the given number by the denominator, and 
 divide the product by the numerator. 
 
 EXAMPLES. 
 
 1. Divide 848 by J. 2. Divide 496 by f . 
 
 4=denom. 8=denom. 
 
 Ans. 3392 Numer. 3)3968 
 
 1322 Ans. 
 When the divisor is a whole number and a fraction 
 
 Multiply the whole number of the divisor by the denominator 
 of the fractional part, and add the numera-tor to the product for a 
 new divisor ; then, multiply the dividend by the denominator of 
 the fraction for a new dividend : lastly, divide the new dividend 
 by the new divisor, and the quotient will be the answer. 
 
 EXAMPLES. 
 
 1. Divide 693 by 24f 
 
 24f J693 O^r We multiply 24, the whole number of 
 
 4 4 the divisor, by 4, the denominator of the frac- 
 
 tional part, and add the numerator 3 to the 
 
 99)2772(28 Ans. product, and the sum is 99 ; then we multiply 
 
 198 the dividend 693 by the denominator of th$ 
 
 fraction, and the product is 2772 ; lastly, we 
 
 792 divide 2772 by 99, and the quotient is 28 ; con- 
 
 792 sequently 24f*are contained in 693, 28 times. 
 
 2. Divide 6375 by 5J. Ans. 1159^- 
 
 3. In one rod there are 5J yards ; how many rods are there 
 in 1760 yards? Ans. 320. 
 
 4. What is the quotient of 10142, divided by 3 ? 
 
 Ans. 2766. 
 
 55. How do you divide byfractins ? - 56- Hw, wAw- the divisor u <* wtolt nwn* 
 btr and fraction 
 
FEDERAL MONEY. 33 
 
 FEDERAL MONEY. 
 
 FEDERAL MONHY is the coin of the United States, established 
 by Congress in 1786. The Gold Coins are the Eagle, Half 
 Eagle, and Quarter Eagle. The Silver Coins are the Dollar, 
 Half Dollar, Quarter Dollar, Dime, and Half Dime. The Cop- 
 per Coins are the Cent, and Half Cent. Mill is only imaginary, 
 there being no coin less than a half cent. The denominations of 
 Federal Money are Eagles, Dollars, Dimes, Cents and Mills. 
 
 TABZiE.* 
 
 10 Mills ] fCent c.l 10= 1 cent 
 
 10 Cents 1 ,3 I Dime d. I 100= 10= 1 dime 
 
 10 Dimes j 1 Dollar $ or D. [ 1000= 100= 10= 1 dollar 
 
 10 Dollars J [Eagle E.J 10000=1000= 100=10=eagle 
 
 In keeping accounts, and in reading Federal Money, eagles and 
 dimes are not named ; eagles being read tens, &c. of dollars ; 
 and dimes, tens of cents. Dollars are separated from cents by a 
 point or comma ; all the figures at the left hand of the point are 
 dollars the two first at the right hand are cents, and the third 
 is mills. When there is a fourth figure, it represents tenths of 
 mills the fifth, hundredths of mills, &c. Thus, $25,7525 are 
 read twenty-five dollars, seventy-five cents, two mills, and five 
 tenths of a mill. 
 
 * By an act of Congress it was resolved, that there should be 
 
 Pure. Standard. 
 
 Two Gold < 1. The Eagle . $10 weighing 247,5 grs. 270 grs. 
 
 coins : TJZ. { 2. The half Eagle - 5 123,75 135 
 
 fl. The Dollar . - 1 371,25 41ft 
 
 | 2. Half Dollar . ,50 ct$. 185,625 208 
 
 Six Silver ! 3. Quarter Dollar ,25 92,8125 104 
 
 coins ; viz. l 4. Double Dime = ,20 74,25 83,2 
 
 I 5. Dime ... ,10 37,125 41,K 
 
 (6. HalfDiroa . = ,05 18,5625 20,8 
 
 Two Copper 41. The cent . ,10 mills. 208 20B 
 
 coins ; viz. \ 2. The half cent ~ ,05 mills. 104 104 
 
 Any sum in Federal Money may be read either in the lowest denomination, or partly 
 in the kigher, and partly in the lowest ; thus, $54,321 may be read 54321 mills, or 5432 
 cents 1 mill, or 543 dimes 2 cents 1 mill, or 5 eagles 4 dollars 3 dimes 2 cents 1 mill ; all 
 which denominations may be easily distinguished by the decimal point, thus - 
 Eag. dot. di. cent. mill. 
 
 5,4,3,2,1. 
 
 The method best adapted to practical purposes, ani! which has been sanctioned by a 
 law of the United States, is the decimal form of expression by a decimal point, making 
 the dollar the money unit. Dollars, therefore, will occupy the place of units, and the less 
 denominations will be decimal parts of a dollar, and distinguished by the decimal point. 
 
 57. What is Federal Money ? 58. What are the different denominations 1 
 
 08. What denominations are named in keeping accounts 1 60. How are dollars 
 
 separated from ctnts ? 61. What are the figurts *t th< left, and nt the right oftht 
 
 E 
 
34 FEDERAL MONE. 
 
 Note I. To write any number of cents less than ten Write a 
 figure representing the number of cents, then write a cipher at the 
 left hand in the ten's place. Five cents are written thus ,05. 
 
 2. To write any number of mills less than ten, when there are no 
 cents in the sum Write a figure representing the number of mills, 
 then write two ciphers at the left hand in the places ot cents. Five 
 mills are written thus ,005. 
 
 3. Parts or fractions of a cent are sometimes used instead of mills 
 and tenths of a mill j as ,12J cents for ,125. ,06J cents for ,0625. 
 
 As Federal Money increases in a tenfold proportion, the 
 rules for addition, subtraction, multiplication and division of sim- 
 ple numbers, are applied to Federal Money. The rules for the 
 operations in Decimals are also strictly applicable. [See Deci- 
 mal Fractions.'] 
 
 REDUCTION OF FEDERAL MONEY. 
 
 The method of changing or reducing the denominations of Fed- 
 eral Money is very simple, and needs perhaps no illustration, 
 were the scholar always presumed to comprehend the preceding 
 observations. In any case, the following concise rules will be 
 easily understood. 
 
 RULES. EXAMPLES. 
 
 Given No. Reduced. 
 
 \. To reduce dollars to cents : Add two 
 
 c ^nn 
 . , .-i i ) xjouu cents. 
 
 ciphers to the given number. } ' p 
 
 8. To reduce dollars to milk: Annex ) |2g=25000 mills 
 three ciphers to the given number. } v 
 
 3 To reduce any mm of several fgure, 
 to the lowest denomination mentioned : Read 
 
 ., , , .-I i , i /3t).*i*: - ' 
 
 the whole sum m the lowest denomma- ( ip 135 125 mills 
 tion. ) 
 
 4. To reduce cents to dollars : Cut off 
 
 the two right hand figures of the given Cents. c. 
 
 number, which will be cents those re- 
 maining on the left are dollars. 
 
 5. To reduce mills to dollars : Cut off 1 
 
 the three right hand figures ; those at f Mills. c.m, 
 
 the left are dollars the two first on the f 7515=$75,125 
 right are cents, the third mills. } 
 
 62. How do you write a number of cents less than 10 1 - 63. How is a less number 
 of mills than. 10 written 1 - 64. How are dollars reduced to cents and to mills 1 - 65 
 How do you reduce cents or mills to dollars f 
 
FEDERAL MONEY. 35 
 
 REDUCTION OF OTHER CURRENCIES TO FEDERAL MONEY, &c. 
 CASE I. 
 
 To reduce Lawful Money, or the Currency of New-England, Vir- 
 ginia, Kentucky, and Tennessee, to Federal Money the dollar 
 be ing 6 shillings. 
 
 RULE. As the value of a dollar is equal to three-tenths of a 
 pound, when pounds are given to be changed, annex three ciphers 
 to the sum, and divide the whole by 3 ; the quotient is the answer 
 in cents. 
 
 EXAMPLES. 
 
 1. Change 523 to Federal Money. 
 
 3)523000 
 
 1 74333 J cents. Am. 1743 dolls. 33 J cts. 
 
 2. Change 29 to Federal Money. Ans. $96 66 cts. 
 
 When pounds and shillings are given, to the pounds annex half 
 the number of shillings and two ciphers, if the number of shillings 
 in the given sum be even ; but if the number be odd, annex half 
 the number, and then 5 and one cipher, and divide by 3 ; the 
 quotient is the answer in cents. 
 
 EXAMPLES. 
 
 1 . Change 59 1 8*. to Federal Money, 
 
 3)59900 
 
 199GG| cts, Am. 199 dolls. 66f cts. 
 
 2. Change 93 13*.. to Federal Money. 
 
 3)93650 
 
 31216 cts. Ans. 312 dolls. 16f cts. 
 
 Change the following sums, viz. 
 
 . *. dolls, cts. 
 
 3. - 129 13 - Am. 432 16 
 
 4. 63 15 - 212 50 
 
 When there are shillings, pence, &c. in the given sum, annex 
 for the shillings as before directed, and to these add the farthings 
 
 re 
 
 3. How is lawful money changed into the federal cut rency ? -67. How, when then 
 different denomination*, as pounds, shillings, pence, 8fC, ? 
 
3G FEDERAL MONEY. 
 
 in the given pence and farthings, observing to increase their 
 number by one when they exceed 12, and by two when they ex- 
 ceed 37, and divide as before. 
 
 EXAMPLES. 
 
 1. Change 21 85. 4jd. to Federal Money. 
 
 3)21419 4 is annexed to the pounds for half the 
 
 shilling?, and 19 for the farthings in 4JJ, 
 7139cts. and excess f 12. 
 
 Ans. 71 dolls. 59 J cts. 
 
 2. Change 721 85. lljd. to Federal Money. 
 
 3)721497 In this example, 4 is annexed to the pounds for 
 
 half the even shillings, and 47 for the farthings 
 
 240499 cts. in lljd, and excess of 37, and then 5 is added to 
 
 the figure next to half the shillings, making it 9 
 
 in place of 4 for the odd shilling. 
 
 Am. 2404 dolls. 99 cts. 
 
 3. Change 24 11*. 7fd. to Federal Money. 
 
 Ans. $81 94. 
 
 4. Change 2001 1*. 3Jd. to Federal Money. 
 
 Ans. $6670 21|. 
 
 CASE II. 
 
 To reduce New- York and North- Carolina Currency to Federal 
 Monty the dollar being 8 shillings. 
 
 RULE. Add a cipher, and divide by 4.* 
 
 Note. If there are shillings, pene, &c. given in any case, they 
 must always be reduced to the decimal of a pound, and annexed to 
 the given pounds before dividing by 3 ; and in all such cases three 
 figures must be cut *ff at the right hand for decimals of a dollar. 
 
 EXAMPLES. 
 
 1. Reduce 44 New-York currency to Federal Money. 
 
 4)440 
 
 110 Ans. $110. 
 
 2. Reduce 24 12s. to Federal Money. Am. $61 5f. 
 
 * As a dollar is equal to four-tenths of a pound, dividing pounds, or pounds and deci- 
 mals of a pound, by 4, must, evidently, give dollars, cents and mills. 
 
 68. What is your rule for changing JVew- York currency, fyc. to Federal mtnfl/ ? 
 
FEDERAL MONEY. 37 
 
 CASE III. 
 
 To reduce Pennsylvania, New- Jersey, Delaware, Maryland and 
 
 Ohio Currency to Federal Money the dollar being Is. 6d. 
 RULE. Multiply by 8, and divide the product by 3, the quo- 
 tient will be the answer.* 
 
 EXAMPLE. 
 
 Reduce 243 New-Jersey to Federal Money. 
 8 
 
 3)1944 
 
 648 federal money. AM. $848. 
 
 CASE IV. 
 
 To reduce South- Carolina and Georgia Currency to Federal Money 
 
 the dollar being 4s. Sd. 
 
 RULE. Multiply by 30, and divide the product by 7, the que- 
 tieat will be the answer.f 
 
 EXAMPLE. 
 
 Reduce 300 South-Carolina and Georgia to Federal Money. 
 300 
 30 
 
 7)9000 
 1285,71,4- =federal. Ans. $1285,71,4*. 
 
 CASE V. 
 
 To change Sterling to Lawful Money the dollar being 4$. 6d 
 RULE. Add J to the Sterling the sum will be Lawful. 
 
 * As a dollar is equal to three-eighths of a pound, dividing by , or wkich is the same, 
 multiplying by 8 and dividing the quotient by 3, must give dollars, cents and mills. 
 
 f As a dollar is equal to seven-thirtieths of a pound, dividing hy _Z, or which i* the 
 same, multiplying by SO and dividing the quotient by 7, must give dollars, cents and 
 mills. 
 
 69. What is the rule for changing Pennsylvania, Sfc. to the same currency ? 70. 
 
 How is Georgia and South-Carolina, currency changed ? 71. How is sterling' re- 
 duced to laitful money ? 
 
38 FEDERAL MONEY 
 
 EXAMPLE. 
 
 In 3 17 Sterling, how much Lawful ? 
 3)347=rsterling. 
 115 13 4 added. 
 
 462 13 4=lawful. Ans. 462 13 4. 
 
 CASE VI. 
 
 To change Lawful to Sterling Money. 
 
 RULE. From the Lawful subtract J, the remainder will be 
 Sterling. 
 
 EXAMPLE. 
 
 Reduce 462 13 4 Lawful to Sterling. 
 4)462 13 4=lawful. 
 115 13 4 subtract. 
 
 347 00 0=sterling. Ans. 347. 
 
 CASE VII. 
 
 To change Canada and Nova- Scotia Currency to Federal Money 
 the dollar being 5 shillings. 
 
 RULE. As the value of a dollar is equal to one-fourth of a 
 pound, multiply the sum, when in pounds, by 4, for dollars. 
 
 When there are shillings, &c, reduce the given sum to pence, 
 annex two ciphers, and divide by 60, for cents. 
 
 EXAMPLES. 
 
 1. Change 36 Canada Currency to Federal Money. 
 
 36 
 4 
 
 Jlns. 144 dolls. 
 
 2. Change 528 12s. 6d. Canada Currency to Federal Money. 
 
 20 Or thus, 528 
 
 4 
 
 10572 
 
 12 2112 
 
 10 shill.= 2 
 
 6|0)1268700jO 2*. 6d.= 50 
 
 211450cts. 2114 50 
 
 Jlns. 2114 dolls. 50 cts. 
 
 72. Lawful to sterling ? 73. How is the currency of Canada and Nova-Scoti* 
 
 changed to federal money ? 
 
FEDERAL MONEY. 39 
 
 CASE VIII. 
 
 To reduce English or Sterling Money to Federal Money. 
 
 RULE. Multiply English money by 40, divide the product 
 by 9, and the quotient will be dollars, or dollars, cents and mills.* 
 
 EXAMPLE. 
 
 Reduce 100 English to Federal money. 
 100 
 
 9)4000 
 
 44,44? Ans. 
 
 -*- 
 
 Shewing the rates at which the following foreign coins and currencies 
 are estimated at the Custom- Houses of the United States. 
 
 Pound sterling of Great-Britain, - - 4,44 
 
 Pound of Ireland, - * - 4,10 
 
 Livre tournois of France, ,18J 
 
 Franc, do. , ,18| 
 
 Silver Rouble of Russia, ,75 
 
 Florin or Guilder of the United Netherlands, - ,40 
 
 Mark Banco of Hamburg, - ,33J 
 
 Rix Dollar of Sweden and Denmark, l,OcT 
 
 Real of Plate of Spain, - - ,10 
 
 Real of Veilon of do. ,05 
 
 Milrea of Portugal, - 1,24 
 
 Tale of China, ... 1,48 
 
 Pagoda of India, 1,84 
 
 Rupee of Bengal, ... ,50 
 
 NOTE. Any number of either of the above coins or currencies 
 may be easily reduced to Federal Money, by multiplying the value 
 of one in cents by the given number, and the product will be cents. 
 
 * As a dollar is equal to nine fortieths of a pound, therefore multiplying by 40, and 
 dividing the product by 9, must give dollars, cents and mills. 
 
 74. Hoio is sterling money reduced to federal 1 75. What is the value in federal 
 
 money of the pound sterling of Gr cat- Britain ? 76. Of the Irish pound 1 77. Of 
 
 the Livre tournois of France 1 78. Of the silver rouble of Russia ? 79. Of the 
 
 t*U of China ? 80. And of the rupee of Bengal ? 
 
40 COMPOUND ADDITION. 
 
 COMPOUND ADDITION. 
 
 ND ADD ITU; J 1 : is the adding together of several mini- 
 ig different denominations, as, pounds, shillings, pence* 
 
 hnnrlrPfls. miartfrj frr 
 
 ii _ ; . ; i iA. i \^ MI v: ^, n iu i' ic ciui-io 
 
 5, hundreds, quarters, &c. 
 RULE. 
 
 the cumbers, so that those of the same denomination 
 
 cr. veetly ijj.der each other. 
 
 t coi.imn or deriomiaation together as in whole 
 nun -'!ivicie the sum by as ma.-iy of the same denomin- 
 
 of the iie-if greater, setting down the remain- 
 der u 'ucid added ; carry the quotient to the next 
 sir/ * n r ccnnnuing the same to the last column, 
 add) lion.* 
 
 PROOF. ii simple addition. 
 
 * . MONEY. 
 
 'See page 33.) 
 
 2. 
 
 cts. m, 
 
 375 
 
 29 18 
 
 7 12 5 
 199 18 7 
 
 30 01 
 
 24 1 1 3 
 
 * Tl-e rcr-: > rom what has been said in addition of simple 
 
 a umbel's is 1 in the pence is - ~maj to 4 in the farthings; 1 
 
 in tie. sl.i h .' ii. tnn jjuunc's.to 20 in the shillings; there- 
 
 fore, C^M Oionej ;:'!.., '' >m each column, properly, in 
 
 hold good iu the ? Idition of 
 pound w) . \ t ,. 
 
 ?'.. n'L ' -S2. M 7 /!,;/ ;* Me rwie ? 83. Wftai thi 
 
 method of ff; , :. are rter tn an ffg-/e ? 
 
COMPOUND ADDITION. 41 
 
 Table of English Money. 
 
 4 Farthings ) C Penny, marked, qrs. d. 
 
 12 Pence > MAKE ONE < Shilling, , s. 
 
 20 Shillings ) ( Pound, . 
 
 NOTE $pz\ farthing, or a quarter of any thing. 
 
 zr2 faithings, or half of any thing. 
 
 $3 farthings, or three quarters of anything. 
 
 FENCE AND SHILLING TABLE. 
 
 20 pence make 5.1 
 30 -.-- 2 
 40 - - - - 3 
 
 i8?100 - - 
 4*120 - - 
 
 - -5.8 
 
 - - 9 
 - - 10 
 
 dAt 
 
 60 - 
 70 - 
 80 - 
 
 - - -3 
 
 ... 3 
 ... 4 
 
 ,.0! 
 10 
 
 erf 
 
 50 
 
 .... 4 
 
 2^240 .... 
 
 20 
 
 Ol 
 
 90 - 
 
 ... 4 
 
 10 
 
 60 
 
 .... 5 
 
 
 
 9 
 
 20 shill. 
 
 makel 
 
 5.0 
 
 100 - 
 
 ... 5 
 
 
 
 70 
 
 .... 5 
 
 10 
 
 \ 
 
 30 - - 
 
 _ 
 
 1 
 
 105 
 
 110 - 
 
 ... 5 
 
 10 
 
 80 
 
 - - - - 6 
 
 8 
 
 \ 
 
 40 n - 
 
 - - 
 
 2 
 
 i 
 
 120 - 
 
 - - - 6 
 
 
 
 90 
 
 .... 7 
 
 6 
 
 
 50 - - 
 
 . 
 
 2 
 
 10' 
 
 130 - 
 
 ... 6 
 
 10 
 
 
 
 
 \ 
 
 
 
 
 i 
 
 
 
 
 EXAMPLES. 
 1. 
 
 . 5. d. 
 
 47 17 li CO" In this example, we begin by placing 
 the numbers of one denomination under 
 
 24 13 9 each other, that is to say, pounds under 
 
 36 10 6 pounds, shillings under shillings, and pence 
 
 21 15 7 under pence. We then begin with the least 
 
 17 14 6 denomination, which is pence, and find the 
 
 15 11 5 amount to be 46, which we bring into shil- 
 
 10 10 2 lings by dividing them by 12, we write 
 
 the remainder 10 under the coJumn, and 
 
 Ans. 174 13 10 carry the quotient 3 to the column of shil- 
 
 lings ; we then find the amount of the 
 
 126 15 11 column of shillings to be 93, which we 
 
 bring into pounds by dividing them by 20 ; 
 
 Proof, 174 13 J.O we write the remainder 13 under the col- 
 
 umn of shillings, and carry the quotient 4 
 
 to the pounds ; we then find the amount 
 
 of the column of pounds to be 174, which we write under the column 
 of pounds, and find the answer to be 174. 13s. lOd. 
 
 85. What is a farthing ? 86. Repeat the table of English Money. 87. How 
 
 nany pence and farthings are there in a pound ? 
 
42 
 
 COMPOUND ADDITION. 
 
 2. 
 
 
 
 s. 
 
 d. 
 
 qrs. 
 
 847 
 
 11 
 
 11 
 
 2 
 
 491 
 
 19 
 
 6 
 
 I 
 
 59 
 
 6 
 
 10 
 
 G 
 
 747 
 
 16 
 
 1 
 
 2 
 
 
 
 
 
 s. 
 
 d. 
 
 qrs. 
 
 915 
 
 10 
 
 10 
 
 2 
 
 64 
 
 8 
 
 9 
 
 1 
 
 419 
 
 2 
 
 10 
 
 2 
 
 491 
 
 19 
 
 11 
 
 # 
 
 762 
 
 17 
 
 6 
 
 1 
 
 5 
 
 16 
 
 11 
 
 3 
 
 
 
 
 2. TROY WEIGHT.* 
 
 By this weight are weighed Gold, Silver, Jewels, Electuaries^ 
 and all Liquors. 
 
 24 Grains 
 
 20 Pennyweights 
 
 12 Ounces 
 
 Table. 
 
 grs."\ f Pennyweight, marked 
 
 >MAKE ONE< Ounce, 02. 
 
 ) r Pound, ft or l 
 
 1. 
 
 ft. oz. pwt. 
 
 767 10 17 
 
 39 6 9 
 
 417 11 16 
 
 935 9 8 
 
 ft. 
 
 649 
 32 
 841 
 473 
 
 EXAMPLES. 
 
 2. 
 
 oz. pwt. gr. 
 11 19 20 
 965 
 10 11 20 
 949 
 
 
 
 ft. 
 
 859 
 
 3. 
 
 oz. pwt. gr. 
 9 15 20 
 
 437 10 17 22 
 642 3 7 
 738 9 4 23 
 
 * The original of all weights used in England, was a grain or kernel of wheat, gath- 
 ered out ot the middle of the par, and, being \vo'l dried, 32 of them were to make one 
 peunywej^ht, 20 pennyweights one ounce, ar-u 12 ounces one pound. But in later time* 
 it was thought sufficient to divide the same pennyweight into 24 equal parts, still called 
 grains, being the least weight n^w in common use ; and from thence the' rest are com- 
 puted. 
 
 88, WhuiitTroy 
 
 What are its denominations ? 
 
COMPOUND ADDITION. 
 
 . Gold is tried by fire, and reckoned in carats, or the 24th 
 part of any quantity. Such gold as will abide the fire without loss ia 
 accounted 24 carats fine ; if it lose 2 carats, it is called 22 carats fine, 
 &t,i A pound of silv iT * wh-ch loses nothing in trial, is 12 ounces fine. 
 JWmi is the base metal naix ui with gold or silver, which abates ita 
 fineness. O^ 8 * 175 Troy oz. are equal to 192 Avoirdupois oz., and 
 17o Iba. Troy are eq ;a! to 144 Ibs. Avoirdupois. 1 Ib. Troy = 5760 
 grains, and t Ib. Avoirdupois = 7000 grains. 
 
 3. AVOIRDUPOIS WEIGHT. 
 
 By Avoirdu.pt is are weighed all coarse and drossy goods, gro- 
 ceries, bread, tallow, hay, leather, and all metals, except gold and 
 silver. 
 
 16 Drams, 
 16 Ounces 
 28 Pounds 
 
 4 Quarters 
 20 Hundred Weight 
 
 Table. 
 
 dr. -\ f Onnce, marked 02. 
 
 \ Pound, Ib. 
 
 MAKE ONE< Quarter of a hund. wt. qr. 
 ) Hund. weight, 112lbs. Cwt. 
 ( Ton, T. 
 
 EXAMPLES. 
 
 1, 
 
 Ib. oz. dr. 
 
 19 13 12 
 
 21 9 6 
 4 15 15 
 
 22 10 5 
 
 2. 
 
 T. cwt. qrs. Ib. 
 
 59 13 2 17 
 
 6 17 1 21 
 
 45 11 3 25 
 
 57 16 2 19 
 
 3. 
 
 T. cwt. qrs. Ib. oz. dr. 
 
 91 17 2 25 13 15 
 
 19 9 17 10 Ifc 
 
 14 13 2 9 11 
 
 47 11 3 19 14 
 
 NOTE. In Avoirdupois Weight, several other denominations are 
 used in particular goods, viz. 
 
 A bbl. of Pot Ashes, 
 A bbl. of Pork, 
 A bbl. of Beef, 
 A quintal of Fish, 
 12 particular things, 
 12 dozen, 
 
 200 Ibs. 
 220 Ibs. 
 = 220 Ibs. 
 = 112 Ibs. 
 1 dozen. 
 1 gross. 
 
 < 144 dozen, = 
 $ t 20 particular things, 
 \ 5 do. do 
 5 24 sheets of paper, 
 20 quires, ^: 
 
 great gvoss. 
 score, 
 taliy. 
 
 quire, 
 ream. 
 
 90. W!:nt is a Carat 1 } 91. What is the difference between Troy and Avoirdu- 
 pois ? 92. What is the use of Avoirdupois ? 93. Repeat the Table 94. 
 
 ethtr dttiominations are used in Avoirdupois ? 
 
COMPOUND ADDITION. 
 4. APOTHECARIES' WEIGHT. 
 
 By this weight, apothecaries and physicians mix their medicines 
 but they buy and sell their commodities by avoirdupois. 
 
 20 Grains, 
 3 iScrupies, 
 8 Drams, 
 
 12 Ounces r 
 
 grt. 
 
 Tabk. 
 
 MAKE ONE 
 
 Scraple, 
 Dram, 
 Ounce, 
 Poundj 
 
 marked, 
 
 EXAMPLES. 
 
 |. 
 
 3 B gr- 
 
 9 1 17 
 
 3 2 14 
 
 6 1 17 
 
 4 
 8 1 
 
 5 2 
 
 2. 
 
 3 3 B gr. 
 10 7 2 19 
 6 3 12 
 17 
 12 
 
 7 6 1 
 952 
 
 8 3 1 
 740 
 
 11 
 
 4 
 
 3. 
 
 ib. 3 3 B g*. 
 
 12 11 6 1 15 
 
 4 9 1 12 
 
 91 10 7 2 16 
 
 4812 19 
 
 9 6 3 17 
 
 3452 9 
 
 5. CLOTH MEASURE. 
 
 Table. 
 
 2 Inches, in. 
 
 4 ^Hi. l r? ; or 9 Inches, 
 
 4 CjiuTt- f&, or 36 Inches, 
 
 3 Quartero, or 27 Inches, 
 b Quarters, or 45 Inches, 
 6 Quarters, or 54 Inches, 
 
 4 tt'j-cii-tftrs, 1 In. & 1 5th, or 
 37 laches aud one 5th, 
 
 3 Quarters and two thirds, 
 
 MAKE ONE 
 
 ' Nail, marked, n. 
 
 or no. 
 
 Quarter of a Yard, 
 
 (jr. 
 
 Yard, 
 
 Yd. 
 
 Ell Flemish, 
 
 E. Fl. 
 
 Ell English, 
 
 jri jjt 
 Jlj* FJ- 
 
 Ell French, 
 
 E.Fr. 
 
 EH Scotch, 
 
 E. Sc. 
 
 , Spanish Var. 
 
 
 95, What is the use of Apothecaries' Weight? 96. What are its denomiruttiaris? 
 
 -97. W hat are the varieties in Cloth Measure f 
 
COMPOUND ADDITION. 
 
 45 
 
 EXAMPLES. 
 
 ft. 
 
 Yd qr. n. 
 
 76 2 3 
 
 3 3 1 
 
 42 3 3 
 
 57 2 2 
 
 2. 
 
 E.E- qr. n. 
 
 91 3 2 
 
 49 4 3 
 
 623 
 
 84 4 1 
 
 3. 
 
 E.FI. qr. n. 
 
 75 2 i 
 7 1 3 
 
 84 2 
 
 76 2 3 
 
 Note. The Scotch allow one English yard in every score yards. 
 All Scotch and Irish linens are bought by the English or American 
 yard, which are the same, and all Dutch linens by the Ell Flemish : 
 out are all sold in America by the American yard. 
 
 6. LONG MEASURE. 
 
 The use of Long Measure is to measure the distances of places*, 
 or any thing, where length is considered without regard to breadth, 
 
 Table. 
 
 3 Barley-corns, bar.'} 
 
 12 Inches, 
 
 3 Feet, 
 
 5J Yards, or 16 J feet, 
 40 Rods, Perches or Poles, 
 or 132 Paces, 
 
 8 Furlongs, 
 
 60 Geographicaljor 69 J stat- 
 ute miles 
 360 Degrees, 
 
 MAKE ONE 
 
 Inch, marked, 
 
 Foot, 
 
 Yard, 
 
 Rod, Perch, or Poj e , 
 
 in., 
 
 ft- 
 
 yd. 
 pol. 
 
 fur. 
 mils. 
 
 Furlong, 
 Mile, 
 
 Degree, deg. or * 
 
 .Great Circle of the Earth 
 
 98. What is Long Measure? 99. Repeat the table 1 
 
46 
 
 COMPOUND ADDITION. 
 
 EXAMPLES. 
 
 1. 2. 3. 
 
 Pol. ft. in. Mil. for. pol. Deg. mil. fur. pol. ft. in. bar. 
 
 12 n 10 
 
 9 10 9 
 
 8 12 11 
 7 15 6 
 
 9 7 36 
 
 7 3 19 
 
 4 1 24 
 
 6 5 12 
 
 759 56 6 29 15 10 2 
 
 217 39 1 36 11 61 
 
 497 63 7 24 9 81 
 
 562 17 11 13 00 
 
 NOTE. Distances are also measured with a chain, 4 rods long, con- 
 taining 100 links. 
 
 finches 
 
 25 links 
 links 
 1 chains 
 
 8 furlongs 
 4 inches 
 fi points 
 
 ~ 1 link, * 12 lines 1 inch. 
 
 1 ple. \ 12 inches 1 foot. 
 
 1 chain. j 6 feet ~ 1 fathom.f 
 
 1 furlong, j 3 miles 1 league. 
 r= 1 mile. \ 5 feet 1 geometrical pace, 
 rr 1 hand.* ; 66 feet rz 1 gunter's chain. 
 
 1 line. 
 
 
 7. TIME. 
 
 
 
 TabU. 
 
 
 60 Seconds 
 
 1 ( 
 
 Minule, n 
 
 60 Minutes 
 
 
 Hour, 
 
 24 Hours 
 
 
 Day, 
 
 7 Days 
 
 > MAKE ONE { 
 
 Week, 
 
 4 Weeks 
 
 I | 
 
 'Month, 
 
 1 3 Months 1 day & 6 hours, 
 or 365 days & 6 hours, 
 
 J I 
 
 Julian Year, 
 
 marked 
 
 s. m. 
 h. 
 d. 
 
 w. 
 mo. 
 
 Y. 
 
 * Used in measuring the height of horses. 
 
 f Or French toise, used in measuring the depth of water and cordage, and is equal to 
 6 feet 4 inches English measure. 
 
 100. In what other way are distances measured? 101. Whatis a chain ? 102. 
 
 What afurlongl 103. A fathom? 104. What is a league! 105. What 
 
 is a geometrical pace? 106. What are the divisions of time? 
 
COMPOUND ADDITION. 47 
 
 EXAMPLES. 
 
 1. 2. 3. 
 
 W. d. b. m. s. Y. mo. d. Y. mo, w. d. h. m. s. 
 
 3 6 22 57 42 19 10 19 57 11 36 23 29 55 
 
 1 5 19 31 28 7 9 27 4 8 1 I 19 45 38 
 
 2 3 17 9 15 4 8 16 29 9 2 3 17 18 19 
 
 3 9 17 58 1 11 14 46 10 2 5 11 50 13 
 
 NOTE. The civil Solar year of 365 days being short of the true, 
 by bk. 48w. 57s. occasioned the beginning of the year to run forward 
 through the season nearly one day in four years ; on this account, 
 Julius Caesar ordained that one day should be added to February every 
 fourth year, by causing the 24th day to be reckoned twice ; and be- 
 cause this 24th day was the sixth, (sextilis) before the kalends of 
 March, there were in this year two of these sextiles, which gave the 
 name of Bissextile to this year, which being thus corrected, was, 
 from thence, called the Julian year. 
 
 8. MOTION, OR CIRCULAR MEASURE. 
 
 This Measure is used in reckoning latitude and longitude ; also 
 in computing the revolution of the earth and other planets round 
 the sun. 
 
 Table. 
 
 60 Seconds (//) } ("Minute, - - 
 
 60 Minutes f 1 Degree, - 
 
 30 Degrees MAKE ONE )Sign, - ,, 
 
 12 Signs, or 360 degrees,) ( Circle of the Zodiack.* 
 
 * The Zodiuck is the great circle of the Sphere, containing the 12 signs, through 
 which the Sun passes. 
 
 107. What is the origin of the JulianYear ? 108. What is a degrtc ? 109. 
 
 Wk*t is the use of the tabU of Motion ? 
 
48 COMPOUND ADDITION 
 
 EXAMPLES. 
 
 1 ct o 
 
 I *.. O. 
 
 ' " sig-. f tf gig. ' /f 
 
 17 55 48 10 5 37 42. 104 7 52 16 
 
 1 37 51 102 7 25 72 648 5 27 24 
 29 19 45 14 8 26 11 393 6 17 13 
 19 19 37 72 4 32 ]7 136 7 38 24 
 
 107 6 47 498 5 42 19 
 
 9. LAND OR SQUARE MEASURE. 
 
 By this are measured thipgs that have both length and breadth, 
 
 Tabk. 
 
 144 Inches make one Square Foot. 
 
 9 Feet --- Yard. 
 
 Yards, or ) 
 
 --- Pole - 
 
 40 Poles - - - - = - Rood. 
 
 4 . Roods, or 160 Rods, or > . 
 
 4840 Yards, $ 
 
 640 Acres - - --- Mile. 
 
 EXAMPLES. 
 
 1. 2. 3. 
 
 Pole. feet. inch. Yard. feet. inch. Acre. rood. pole. feet, inch, 
 
 36 17 137 28 7 119 756 3 37 245 128 
 
 19 248 119 93 75 29 1 29 93 25 
 
 12 96 75 29 6 120 416 3 31 128 119 
 
 18 110 122 4 8 12 37 1 19 218 '20 
 
 110. What is a pok in Square Measure 111. What a f*oi 1 112. /few - 
 
 ny yards in enacrt? 
 
0OMPOUND ADDITION. 
 10. SOLID MEASURE. 
 
 49 
 
 This Measure is used to measure things that have length, breadth 
 and thickness. 
 
 Table. 
 
 1728 Inches make one 
 
 27 Feet - 
 
 40 Feet of round timber, or > 
 50 Feet of hewn timber, $ 
 128 Solid feet, i. e. 8 in length, 4 in 
 breadth^ and 4 in height, 
 
 Foot. 
 Yard. 
 
 Ton or Load. 
 Cord of Wood. 
 
 1. 
 
 Ton. ft. in. 
 29 46 1229 
 
 12 19 
 18 11 
 
 64 
 917 
 
 19 8 1002 
 
 EXAMPLES. 
 
 
 2. 
 
 
 Yard. 
 
 ft. 
 
 in. 
 
 75 
 
 22 
 
 1412 
 
 9 
 
 26 
 
 195 
 
 3 
 
 19 
 
 1091 
 
 28 
 
 15 
 
 1722 
 
 Cord. 
 37 
 9 
 
 48 
 8 
 
 3. 
 
 ft. in. 
 119 1015 
 110 159 
 127 1017 
 956 
 
 NOTE. 16 solid feet are called a foot in measuring wood, and 8 
 feet a cord. Also, 24f solid feet are called a perch in measuring 1 
 stone. 
 
 11. WINE MEASURE. 
 
 All brandies, spirits, perry, cider, mead, vinegar, honey and 
 oil, are measured by wine measure. 
 
 4 gills, (gi.) 
 2 pints, 
 4 quarts, 
 31J gallons, 
 63 gallons, 
 2 hogsheads, 
 2 pipes, or 4 hogsheads 
 
 Table. 
 
 MAKE ONE 
 
 pint 
 
 ~1 7 
 
 gallon, 
 barrel, 
 hogshead, 
 pipe, or butt, 
 tun, 
 
 ft. 
 
 qt 
 
 gal. 
 
 bar. 
 
 hhd. 
 
 P. or B. 
 
 T. 
 
 113. What is afoot in Solid Measure ? 114. What is a cord ofwood^-^ 1 15- 
 
 How many feet of timber in a ton 1 11G. What is a, perch? 117. What fire the 
 
 de-nominations of Wine Measure ? 
 
 G 
 
COMPOUND ADDITION. 
 
 I. 
 
 Tie. gal. qt. pt. 
 
 37 39 3 1 
 
 9 17 2 I 
 
 4 28 
 
 32 19 1 1 
 
 EXAMPLES. 
 
 2. 
 
 Hhd gal, qt. qt. 
 
 51 53 1 1 
 
 21 39 3 
 
 9 18 1 
 
 0921 
 
 Tun. hhd. gal. qk 
 
 37 2 37 2 
 
 19 ' 1 59 1 
 
 28 2 00 
 
 19 47 1 
 
 NOTE. The wine gallon contains 231 cubick inches. 
 
 12. ALE OR BEER MEASURE. 
 
 2 Pints, pt. 
 
 4 Quarts, 
 
 8 Gallons, 
 8J Gallons, 
 
 9 Gallons, 
 2 Firkins, 
 
 2 Kilderkins, 
 
 1 J Barrel, or 45 Galloes, 
 
 2 barrels, 
 
 3 Barrels, or 2 Hhds. 
 
 Table. 
 
 - MAKE ONE 
 
 Quart, <jf. 
 
 Gallon, gal. 
 
 Firkin of Ale in Lond. A,fir. 
 Firkin of Ale or Beer., 
 Firkin of Beer in Lon. BJir. 
 Kilderkin, kil. 
 
 Barrel, bar, 
 
 Hogshead of Beer, hhd. 
 Puncheon, pun. 
 
 Butt butt. 
 
 J. 
 
 A.B. fir. gal. 
 
 49 3 7 
 
 26 2 3 
 
 904 
 
 17 3 
 
 EXAMPLES. 
 
 2. 
 
 B B. fir. gal. 
 
 29 1 8 
 
 19 3 5 
 
 16 3 
 
 9 1 8 
 
 3. 
 
 Hhd. gal. qt. 
 
 379 63 3 
 
 79 2 
 
 121 57 2 
 
 467 19 1 
 
 NOTE. The beer gallon contains 282 cubick inches. 
 
 118. How many cubick inches are there in a gallon of wine ; find how many in a gat- 
 Ion of beer ? 119. What are the divisions in Ale or Beer Measure ? 
 
COMPOUND ADDITION. 1 
 
 13. DRY MEASURE. 
 
 This measure is applied to all Dry Goods, such as Corn, Seeds, 
 Fruit, Roots, Salt, Sand, Coals, &c. 
 
 Table. 
 
 2 Pints 
 
 2 qunrts, 
 
 2 Potties, 
 
 2 Gallons, 
 
 4 pecks, 
 
 2 Bushels 
 
 2 Strikes, 
 
 2 Cooms, 
 
 4 Quarters, 
 4$ Quarters, 
 
 5 Quarters, 
 2 Weys, 
 
 1. 
 
 Qr. bu. pk. qt. 
 64 7 3 7 
 
 9415 
 19 6 2 I 
 
 4020 
 
 MAKE ONE 
 
 Quart, 
 Pottle, 
 Gallon, 
 Peck, 
 Bushel, 
 Strike, 
 Coom, 
 Quarter, 
 Chaldron, 
 Chaldron in 
 Wey, 
 Last, 
 
 ft. 
 
 pot* 
 gal. 
 pk. 
 bu. 
 sir. 
 
 CO. 
 
 qf 
 
 ch. 
 London 
 wey. 
 last. 
 
 EXAMPLES. 
 
 2. 
 
 Bu. pk. qt. pt. 
 
 37 2 5 1 
 
 19 3 7 
 
 16 2 1 
 
 5161 
 
 3. 
 
 Chal. bu. pk. qt. 
 
 37 27 3 5 
 
 6' 29 1 7 
 
 15 30 
 
 4 11 3 
 
 NOTE. A Winchester bushel is 1 8 J inches in diameter, 8 inches 
 deep, and contains 2,150^ cubick inches. 
 
 \ barrel of Mackerel and other barrelled fish, by law in New- 
 Hurc: - ' ;<?, s to contain not less than 29, nor more than 30 gallons. 
 In jL^s.ichaaetts. it is 30 gallons. In Connecticut and New- York, 
 the Shad and Salmon barrel, must contain 200 Ibs. 
 
 120. What is the use e/ Dry Measure 1 121. Repeat the table 1 
 
52 COMPOUND SUBTRACTION. 
 
 COMPOUND SUBTRACTION. 
 
 COMPOUND SUBTRACTION teaches to find the difference or ex- 
 cess between any two sums of divers denominations. 
 
 RULE. 
 
 Place those numbers under each other,which are of the same de- 
 nomination, the less being below the greater ; begin with the least 
 denomination, and if it exceed the figure over it, borrow as many 
 units as make one of the 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 
 borrowed.* 
 
 PKOOF. The same as in simple subtraction. 
 1. FEDERAL MONEY. 
 
 1. 
 
 D. c. m. 
 
 From 39 15 5 
 
 Take 28 17 2 
 
 EXAMPLES. 
 
 2. 
 E. D. c. r 
 
 21 
 10 
 
 5 
 
 3. 
 
 D. c. m, 
 
 100 00 
 
 48 87 5 
 
 
 D. 
 
 c. 
 
 ml 
 
 Lent 
 
 7159 
 
 12 
 
 8 
 
 Received f 
 
 245 
 
 37 
 
 5) First add 1 
 
 at sev- 1 
 
 31 H 
 
 15 
 
 7f received, 
 
 eral \ 
 
 2000 
 
 00 
 
 ? from the s 
 
 times. ( 
 
 1092 
 
 92 
 
 o) 
 
 Rec'd in all 
 
 
 
 
 Remains due 
 
 
 
 
 * The reason of this rule will readily appear, from what was said in subtraotion of 
 simple numbers ; for the borrowing depends upon the same principle, and is only differ- 
 ent as the numbers to be subtracted are of different denominations. 
 
 122. Whtst is Compound Subtraction 1 123. What is the rule 1 124. And the 
 
 method of proof? 125. What is the difference between simple and compel nd subtrac- 
 tion ? 
 
COMPOUND SUBTRACTION. 53 
 
 2. ENGLISH MONEY. 
 1. 2. 
 
 8. p. qrs. s. p. qrs. 
 
 Borrowed 349 15 6 1 Lent 791 9 8 1 
 Paid 195 11 8 1 Received 107 16 4 2 
 
 3. TROY WEIGHT. 
 
 1. 2. 3. 
 
 ft. oz. pwt. gr. ft. oz. pwt. gr. ft. oz. pwt. gr. 
 
 749 5 13 16 189 8 12 10 543 3 9 13 
 
 91 9 19 13 143 4 11 19 179 1 15 18 
 
 4. AVOIRDUPOIS WEIGHT. 
 
 lb. oz. dr. Cwt. qrs. Ib. T. cwt. qrs. Ib, oz. dr. 
 
 7 9 12 8 2 13 9 11 3 17 5 Id 
 
 3 12 9 4 1 15 3 12 1 19 10 9 
 
 5. APOTHECARIES' WEIGHT. 
 
 1. 2. 3. 
 
 lb. 3 39 gr. lb. 3 39 gr. lb. S 3 9 gr 
 
 71 9 3 1 13 65 10 6 2 I 84 1 1 1 1 
 
 37 8 4 1 16 31 8 4 2 65 9 3 1 17 
 
54 COMPOUND SUBTRACTION*. 
 
 6. CLOTH MEASURE. 
 1. 2. 3. 4. 
 
 Yd. qr. n. E E. qr. n. E-F1. qr. n. E-Fr. qr. n. 
 
 35 1 2 467 3 1 765 13 549 4 2 
 19 1 3 291 3 2 149 21 197 4 3 
 
 7. LONG MEASURE. 
 
 1. 2. 3. 
 
 Yd. ft. in. Mil. fur. pol. Deg. mil, fur. pol. vd. ft. in. bar. 
 
 8 2 10 76 3 11 38 41 3 29 2 1 7 2 
 
 17 2 11 21 3 21 19 35 5 31 3 1 9 1 
 
 8. TIME. 
 
 1. 2. 3. 
 
 W. d. h. m. s. Y mo. d. Y. mo. w. d. h. m. s. 
 
 6 17 13 27 19 7 3 13 48 9 2 5 19 21 31 
 
 1 21 16 41 35 4 4 19 19 9 3 4 20 19 49 
 
 9. MOTION. 
 
 1. 2. 3. 
 
 or ff Si o / rt g, o f o 
 
 79 21 31 6 11 17 48 4 19 41 22 
 
 41 41 52 * 3 8 39 29 1 22 19 45 
 
COMPOUND SUBTRACTION. 55 
 
 10. LAND OR SQUARE MEASURE. 
 
 1. 2. 3. 
 
 . rood. pole. Acre. rood. pole. Acre. rood. pole, feet inch. 
 
 29 1 10 29 3 19 56 2 16 27 110 
 
 24 1 25 17 1 36 29 21 210 129 
 
 11. SOLID MEASURE. 
 
 1. 2. 3. 
 
 Ton. ft. in. Yds. ft. in. Cord. ft. in. 
 
 49 19 1100 79 11 917 349 97 1250 
 
 38 36 1296 17 25 10R5 192 127 1349 
 
 12. WINE MEASURE. 
 
 1. 2. 3. 
 
 Hhd. gal. qt. pt. Tier. gal. qt. Tun. hhd. gal. 
 
 79 21 2 1 19 17 1 532 1 19 
 
 38 61 3 1 12 29 2 197 1 47 
 
 13. ALE AND BEER MEASURE. 
 
 1. 2. 3. 
 
 A.B. fir. gal. qt. B.B. fir. gal. qt pt. Hhd. gal. qt. 
 
 39 1 2 1 21 3 5 2 769 17 1 
 
 24 3 6 2 14 1 7 2 1 391 42 3 
 
56 COMPOUND SUBTRACTION. 
 
 14. DRY MEASURE. 
 
 1. 2. 3. 
 
 Qr. bu. pk. qt. Bu. pk. qt. pt. Chal. bu. pk. qt. 
 
 56 2 2 I 91 1 3 2 39 ^2 2 1 
 
 39 3 1 2 29 2 I 1 24 25 3 2 
 
 PROBLEMS, 
 
 RESULTING FROM A COMPARISON OF THE PRECEDING RULES. 
 
 By a due consideration and application of these Problems, many questions may 
 be solved in a short and easy manner, although some of them are generally supposed 
 to belong to higher rules.] 
 
 PROBLEM 1 . Having the sum of two numbers, and one of them 
 given, to find the other. 
 
 Rule. Subtract the given number from the given sum, and the 
 remainder will be the sum required. 
 
 Let 288 be the sum of two numbers, one of which is 115, the 
 ether is 'required. 
 
 From 288 the sum, 
 
 Take 115 the given number. 
 
 Remains 173 the other. 
 
 PROB. 2. Having the greater of two numbers, and the differ- 
 ence between that and the less given, to find the less. 
 
 Rule. Subtract the one from the other. 
 
 Let the greater number be 325, and the difference between that 
 and the other 198 : What is the other ? 
 From 325 the greater, 
 Take 198 the difference. 
 
 Remainder 127 the less. 
 
PROBLEMS. 57 
 
 PROB. 3. Having the lesser of two numbers given, and the 
 difference between that and a greater, to find the greater. 
 
 Rule. Add them together. 
 
 Given $ 127 the less nnmber 
 Ulven ) 198 the difference. 
 
 Sum 325 the greater number required. 
 
 PROB. 4. Having the sum and difference of two numbers giv- 
 en, to find those numbers. 
 
 Rule. To half the sum add half the difference, and the sum 
 is the greater ; and from half the sum take half the difference, 
 and the remainder is the less. Or, from the sum take the differ- 
 ence, and half the remainder is the least ; to the least add the 
 given difference, and the sum is the greatest. 
 
 What are those two numbers, whose sum is 48, and difference 14 ? 
 2)48 2)14 
 
 Half sum =24 Half diff 7 
 
 24+7=31 the greater ; and 24 7 = 17 the less. 
 
 Or, 48 14+2=17 ; and 17+14=31. 
 
 PROB. 5. Having the sum of two numbers and the difference 
 of their squares* given, to find those numbers. 
 
 Rule. Divide the difference of their squares by the sum of the 
 numbers, and the quotient will be their difference : You will then 
 have their sum and difference, to find the numbers by Problem 4. 
 
 What two numbers are those, whose sum is 32, and the difference 
 of whose squares is 256 ? 
 
 Half sum 16 
 Half diff. 4 
 32)256(8 difference. 
 
 256 Greater 20 
 
 Less 12 
 
 PROB. 6. Having the difference of two numbers, and the dif-' 
 ference of their squares given, to find those numbers. 
 
 * The square of a number is the product of it multiplied into itsetf, 
 
 H 
 
58 PROBLEMS. 
 
 Rule. Divide the difference of their squares by the differ- 
 ence of the numbers, and the quotient will be their sum ; then 
 proceed by Problem 4. 
 
 What are those two numbers, whose difference is 20, and the dif- 
 ference of whose squares is 2000 ? 
 
 20)2000(100 sum. 50+1060, the greater ; and 
 501040 the less. 
 
 PROB. 7. Having the product of two numbers, and one of 
 them given, to find the other. 
 
 Rule. Divide the product by the given number, and the quo- 
 tient will be the number required. 
 
 Let the product of two numbers be 288, and one of them 8 ; we 
 demand the other. 8)288 
 
 Ans. 3G 
 
 PROB. 8. Having the dividend and quotient, to find the divisor. 
 Rule. Divide the dividend by the quotient. 
 
 Hence we get another method of proving division. 
 
 $ 288 the dividend, | 36)288(8 divisor. 
 
 1 { 36 the quotient. | 288 
 
 Required the divisor. 
 
 PROB. 9. Having the divisor and quotient given, to find the 
 dividend. 
 
 Rule. Multiply them together. 
 
 1 
 
 8 the divisor, 36 
 
 36 the quotient. 8 
 
 Required the dividend. 
 
 288 the dividend. 
 
 REMARK. The scholar, having now surveyed the ground- 
 work of his studies, begins to see their application to the com- 
 mon concerns of life ; and it is important, while proceeding in 
 the higher rules, that his memory be strengthened by repeated 
 examinations in the previous studies. The instructor is advised, 
 therefore, to state questions of his own, promiscuously under 
 the several rules, that the good scholar may have an opportunity 
 of proving to his teacher and friends, by prompt and ready an- 
 swers to difficult questions, that he thoroughly understands the 
 subject before him. This hint, improved now, may be of essen- 
 tial service hereafter. 
 
REDUCTION. 59 
 
 REDUCTION. 
 
 REDUCTION is the method of changing numbers of one denom- 
 ination into others of different denominations, without altering 
 their value. 
 
 It is of two sorts, viz. Descending and Ascending. 
 
 REDUCTION DESCENDING 
 
 Teaches to change numbers from a higher denomination to & 
 lower ; and is performed by multiplication. 
 
 RULE. 
 
 Multiply the highest denomiaation given, by so many of the next 
 less, as make one of that greater, and thus continue, till you 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 II, of Simple Multiplication, it ap- 
 pears, that FEDERAL MONEY is reduced from higher to lower denom- 
 inations, by annexing as many ciphers ns there are places from the 
 denomination given to that required ; or, if the given sum be of dif- 
 ferent denominations, by annexing the several figures of all the de- 
 nominations in their order, and continuing with ciphers, if necessa- 
 ry, to the denomination required ; or, what amounts to the same 
 thing, by reading the whole number from the left to the required de- 
 nomination, as one number in the required denomination. 
 
 EXAMPLES. 
 
 I. In 3 eagles 2 dollars, how many mills ? 
 
 Ans. 32000m. 
 
 *The reason of this rule is obvious ; for pounds are brought into shillings by multiply- 
 ing them by 20 ; shillings into pence by multiplying them by 12 ; and pence into far 
 things by raulti plying them by 4 ; and this will be true in the reduction of numbers con- 
 sisting of any denomination whatever. The reason of the rule in Reduction Ascend- 
 ing, is equally evident ; for farthings are brought into pence by dividing them by 4 ; 
 pence into shillings by dividing them by 12 ; and shillings into pounds by dividing them 
 by 20. 
 
 126. What is Reduction ? 127. Of how many parts does it consist?- 128. 
 
 W]vit is Reduction Descending 1 129. What is the i-vlefor changing any number 
 
 from a greater to a less denomination ? 130. W hut is the reason of this rule ? 
 
66 REDUCTION. 
 
 2. In 91 dollars 75 cents, how many cents ? 
 
 Ans. 9175C. 
 
 3. In 50 eagles, how many dollars ? Ans. $500. 
 
 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 27 15s. 9d. 2qrs. how many farthings ? 
 
 s. d. qrs, {/= In multiplying hy 20, 
 
 27 15 92 we add in the 15s ; by 12^ 
 
 Multiplied by 20=shillings in a pound, the 9d. and by 4, the 2qrs. 
 
 - - which must always be done 
 
 555=shilliugs. in like cases. 
 - by J 2=pence in a shilling. 
 
 6669=pence. 
 . by 4=farthings in a penny. 
 
 Ans.=2C678 
 
 PROOF. 
 
 4)26678 O^r This is in fact, in the present 
 
 case, simply changing the order of the 
 
 12)6669 2qrs. question, thus : In 26678 farthings, how 
 
 many pounds ? 
 2jO)55|5 9d. 
 
 Ans. 27 15 9 2 
 
 8. In 36 12s. lOd. Iq. how many farthings ? Ans. 35177. 
 
 9. In 719 9s. lid. how many halfpence ? Ans. 345358. 
 
 10. In 37 pistoles, at 22s. how many shillings, pence and far- 
 things ? Aas- 814<sA;//, 
 
 9768 pence. 
 39072 farlk. 
 
 11. In 53 moidores, at 36s., how many shillings, pence and far- 
 things? An*. 1908s. 
 
REDUCTION fel 
 
 12. Reduce 47 guineas, (at 28s. each,) and one fourth of a guin- 
 ea into shillings, six-pences, groats, (4c/.) three-pences, two-pences, 
 pences, and farthings. 
 
 Ans. 1323 shillings, 2646 sixpences, 3969 groats, 5292 three-pen- 
 s, 7938 two-pences, 15876ckand 63504qrs. 
 
 ces 
 
 REDUCTION ASCENDING, 
 
 Teaches how to change numbers from a lower to a higher de 
 nomination ; and 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 denomination, 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 
 higher denominations by cutting off as many places as the given de- 
 nomination stands to the right of that required ; the figures cut off 
 belonging to their respective denominations. 
 
 EXAMPLES. 
 
 1. How many eagles in 32000 mills ? Ans. 3E. 2D. 
 
 2. In 9175 cents, how many dollars? Ans. 9 ID. 75C. 
 
 3. In 500 dollars, how many eagles ? Ans. 50 
 
 4. In 4414 mills, how many dimes? 
 
 i 
 
 5. In 9317 mills, how many dollars? 
 
 6. In 39050 mills, how many dollars? 
 
 7. In 547325 farthings, how many pence, shillings, and pounds ? 
 
 131. How do you change numbers of a lower denomination into a higher ; or, in other 
 how iviil you reduce farthings to pounds, or mills to eagles 1 
 
62 REDUCTION. 
 
 Farthings in a penny 4)547325 
 
 Pence in a shilling 12)136831 Iqr, 
 
 Shillings in a pound 2lO)1140|2 7d 
 
 570 2s. 7d. Iqr. 
 Ans. 136831d. ; 11402s.; 570. 
 
 NOTE. The remainder is always of the same denomination with 
 the dividend. 
 
 8. Bring 35177 farthings into pounds. 
 
 9. Bring 365358 half pence into pence, shillings and pounds. 
 
 10. In 39072 farthings, how many pistoles, at 22s. ? 
 
 11. In 63504 farthings, how many pence, twopences, threepen^ 
 ces, groats, sixpences, shillings, and guineas ? 
 
 NOTE. These questions may serve as proofs to some of those in 
 Deduction Descending. 
 
 REDUCTION ASCENDING AND DESCENDING. 
 
 1. MONEY. 
 
 1. la 97, how many pence, and English or French Crowns, at 
 6s. 8d.? jjjw. 2323Gd. & 291 crowns. 
 
 2. In 735 French crowns, how many shillings and French guin- 
 eas,' at 26s. 8d. ? Ans. 4900*.,. & 183 gum. 24s. 
 
 NOTE. When it is required to know how many sorts of coin, of dif- 
 ferent values, and of equal numbers, are contained in any number of an- 
 other kind ; reduce the several sorts of coin into the lowest denom- 
 ination mentioned, and add them together for a divisor ; then reduce 
 the money given into the same denomination for a dividend, and the 
 quotient arising from the division will be the number required. 
 
 0^7= Observe the same direction in weights and measures. 
 
 132. / wish to ascertain hmo many moidores, guineas, pistoles, dollars, shillings and 
 sixpences, each of the like number, are contained in 275 half joes : how shall I proceed? 
 
REDUCTION. 63 
 
 3. In 275 half Johannes, how many moidores, guineas, pistoles, 
 dollars, shillings, and six-pences, of each the like number? 
 
 A moid, is 36s.=72 six-pences. 275 half joes. 
 
 A guin. is 28s.=56 do. 48s. in a johan. 
 
 A pistole is 22s.=44 do. 
 
 A dollar is 6s. =12 do. 2300 
 
 One shilling is =2 do. 1100 
 1 do. 
 
 13200 shillings. 
 
 Divisor 187 six-pences. 2 six-p. in ashill. 
 
 Dividend=26400 six-pences. 
 
 187)26400(141 of each, and 33 six-pences, or 16s. 6d. over, the 
 answer. 
 
 4. A gentleman distributed 371. 10s. between 4 persons in the 
 following manner, viz. that as often as the first had 20s. the second 
 should have 15s. the third 10s. and the fourth 5s. What did each 
 person receive ? 
 
 (The first man 15. 
 a )The second, ll 6s. 
 
 ) The. third, l 10s. 
 (The fourth, 315s. 
 
 2. TROY WEIGHT. 
 
 3. How many grains in a silver bowl, that weighs 3 Ibs. 10 ez. 12 
 pwt. ? 
 
 ife oz. pwt. 
 3 10 12 
 12 ounces in a pound. 
 
 46 ounces. 
 
 20 pennyweights in an ounce. 
 
 932 pennyweights. 
 24 grains in one pennyweight, 
 
 3728 
 1864 
 
 Proof. 24)22368 grains, Ans. 
 2|0)93|2 
 12)46 12 pwt. 
 3fe lOoz. 
 
64 REDUCTION. 
 
 2. In 13 ingots of gold, each weighing oz v ,5 pwt. how many 
 grains ? Ans. 57720 grains. 
 
 3. In 97397 grains, how many pounds ? 
 
 Ans. IGlbs. 10 oz. 18 pwt. 5gr. 
 
 4. How many rings, each weighing 5 pwt. 7 gr. may he made of 
 3 Ibs. 5 oz. 16 pwt. 2 gr. of gold ? Ans. 158. 
 
 3. AVOIRDUPOIS WEIGHT. 
 
 Cwt. qrs. fe. oz. 
 
 ]. In 91 3 17 14 how many ounces ? 
 4 
 
 307 quarters. Proof. 
 
 28 16)164702 
 
 2943 28)10293 14 oz. 
 
 735 
 
 4)367 17 ft. 
 10293 pounds. 
 
 16 Cwt. 91 3 qrs. 
 
 61762 
 10293 
 
 164702 ounces. 
 
 2. In 12 tons, 15 cwt 1 qr. 19 Ibs. 6 oz. 12 dr. how many 
 drams ? As. 7323500 drams. 
 
 3. In 44800 pounds, how many drams and tons ? 
 
 Ans. 11468800 dr. and 20 tons. 
 
REDUCTION. 65 
 
 4. In 47 Ibs. 9oz. 13 pwt. 17 grs. Troy, how many pounds 
 Avoirdupois ? 
 
 Ibs. oz. pwt. gr. 
 47 9 13 17 
 12 
 
 573 
 
 20 
 
 11473 
 
 24 
 
 
 
 45899 
 22947 
 
 7iOOO) 2751369(39 Ibs. 
 21 
 
 65 
 63 
 
 2369 
 16 
 
 14214 
 2369 
 
 71000)371904(5 oz. 
 35 
 
 5904 
 16 
 
 17424 
 2904 
 
 A A ft A 
 
 TOOO)461464(64o drams. 
 42 
 
 4464 
 
66 
 
 REDUCTION. 
 
 4. LONG MEASURE. 
 
 1 . How many barley-corns will 
 Boston, it being 43 miles ? 
 
 Proof. 
 
 Miles. 
 
 43 
 
 8 
 
 344 furlongs. 
 40 
 
 13760 rods. 
 
 68800 
 6880 
 
 72680 yards. 
 3 
 
 227040 feet. 
 12 
 
 2724480 inches. 
 3 
 
 3173440 Ans. 
 
 3)8173440 
 
 12)2724480 
 
 3)227040 
 
 11)75680 
 
 6880 
 2 
 
 410)137610 
 8)344 
 43 
 
 reach from Newburyport to 
 
 Here we divide by 11, 
 and multiply the quotient 
 by 2, because twice 5J is 
 11 ; or we might have 
 first multiplied by 2, and 
 then have divided the 
 product by 11. 
 
 2. How many barley corns will reach round the globe, it being 
 360 degrees ? 
 
 Ans. 4755801600. 
 
 3. How many inches from Newburyport to London, it being 2700 
 miles'? Ans. 17107200Q. 
 
 4. How many yards, feet and inches from Concord, N. H. to the 
 City of Washington, it being 498 miles ? 
 
REDUCTION. 
 5. TIME. 
 
 1 , In 20 years, how many seconds ? 
 
 d. h. 
 
 365 6 in a year. 
 24 
 
 1466 
 730 
 
 8766 hours in one year. 
 
 20 
 
 175320 hours in 20 years. 
 60 
 
 10510200 minutes in do. 
 60 
 
 Proof. 
 6|0)631 15200(0 
 
 6|0)1051920|0 
 
 2|0)17532|0 
 
 4x6)8766 
 
 4)1461 
 
 365d. 6h, 
 
 631152000 seconds in do. 
 
 2. Suppose your ag to be 15y. 19d. llh. 37m. 45s., how many 
 conds are there in it, allowing 365 days 6 hours to the year ? 
 
 Ans. 475047465. 
 
 3. The war of the Revolution commenced April 19, 1775, and In- 
 dependence was declared July 4, 1776 : how many seconds between? 
 
 6. LAND OR SQUARE MEASURE. 
 
 1. In 29 acres, 3 roods, 19 poles, how many roods and perches' 
 29 3 19 410)47719 
 
 119 roods. 
 40 
 
 Ans. 4779 perches. 
 
 4)119 19p. 
 
68 REDUCTION. 
 
 2. ID 1997 poles, how many acres ? 
 
 Ans. 12a. Ir. 37p 
 
 3. How many square feet, square yards, and square poles, IB 
 a square mile ? 
 
 Ans. 27878400 feet. 
 3097600 yards. 
 102400 poles. 
 
 7. SOLID MEASURE. 
 
 1. In fifteen tons of hewn timber, how many solid inches ? 
 
 15 tons. PROOF. 
 
 50 510 
 
 1728)1296000(7510 
 
 750 feet. 12096 
 
 1728 15 tons. 
 
 8640 
 6000 8640 
 
 1500 
 
 5250 
 750 
 
 1296000 inches.. ..<flns. 
 
 In 25 cords of wood, how many inches ? 
 
 Ans. 5529600. 
 
 GRINDSTONES ARE USUALLY SOLD BIT THE SOLID FOOT* 
 
 The contents may be found by the following 
 
 RULE. Multiply the sum of the whole diameter by the half 
 diambter, and this product by the thickness, and you have the 
 product in cubick inches. 
 
 133. How are grindstones measured tmd sod ? 
 
REDUCTION. 69 
 
 3. What is the solid content of a grindstone 32 inches in diam- 
 eter, and 3 inches thick ? 
 
 32 diameter. 1728)2304(1 foot. 
 
 16 half diameter. 1728 
 
 48 576 
 
 16 3 
 
 ) 1728(1 third. 
 3 1728 
 
 2304 solid inches. Ans. 1J feet. 
 
 8. WINE MEASURE. 
 
 1. In 9 hhd. 15 gal. 3 qts. of wine, how many quarts /? 
 
 9 15 3 PROOF. 
 
 63 4)2331 
 
 32 63)582 3 qts. 
 55 
 9 hhd. 15 gal 
 
 582 
 4 
 
 2331 quarts.. ..Ans. 
 
 , In twelve pipes of wine, how many pints ? 
 
 Ans. 12096. 
 
 REMARK. It is thought unnecessary to give examples of re- 
 duction under all the tables of Weights, Measures, &c. ; as the 
 attentive scholar will readily understand the correct process, 
 from the exercises he has already had. The learner may, how- 
 ever, find it useful to turn to those tables under which no exam- 
 ple is here given state questions of his own accord and, hav- 
 ing found their true answers, write them down with the other 
 examples in his CIPHERING-BOOK. 
 
70 FRACTIONS. 
 
 FRACTIONS.* 
 
 FRACTIONS are parts of an unit, or whole number. When ^ 
 whole is expressed by figures, the number is called an integer : 
 but when a part, or some parts, of a thing are denoted by figures, 
 as, i one fourth, f two thirds, T \ three tenths, &c. of a thing, these 
 figures are called fractions. 
 
 Fractions are divided into two kinds, VULGAR and DECIMAL. 
 VULGAR FRACTIONS. 
 
 A VULGAR FRACTION is that which can have any denominator ; 
 and is expressed by two numbers written one above the other, 
 thus f , with a line between. 
 
 The figure above the line is called the numerator, ( 5 
 
 The figure below the line is called the denominator, \ 8 
 
 The denominator (which is the divisor in division) shows how 
 many parts the integer is divided into ; and the numerator 
 (which is the remainder after division) shows hew many of those 
 parts are meant by the fraction. 
 
 Fractions are either proper, improper, single, compound, or 
 mixed. Any whole number may be made an improper fraction, 
 by drawing a line under it, and putting unity, or 1 , for a denomi- 
 nator ; as, 9 may be expressed, fractionwise, thus f, and 12 
 thus, V, &c. 
 
 1. A single, or simple fraction, is a fraction expressed in a simple 
 form ; as, , , T \, &c. 
 
 2. A compound fraction is a fraction expressed in a compound form, 
 being a fraction of a fraction ; or, two or more fractions connected 
 
 * The term, FRACTION, signifies a broken part or parts of any thing or number ; and 
 these parts can bo represented by figures, as well as whole things or numbers. It was 
 shewn on page 29, that fractions arise from the operations of division ; and hence we 
 may see the necessity of understandiug something of the arithmetick of Vulgar Frac- 
 tions, even though it be in some respects " a tedious and intricate rule." 
 
 134. What are fractions'! 135. What is a Vulgar Fraction 1 136. How is 
 
 it written ? 137. What is the figure above the line, called 1 138. W hot is the one 
 
 below the line called !? 139. What is the meaning, or the use, of these terms ? 140. 
 
 Are there different kinds of Vulgar Fractions ? 141. How can a whole number l/t 
 
 made an improper fraction! 142. Whtft is a simple fraction 7 143. Whatiia- 
 
 compound fraction ? 
 
VULGAR FRACTIONS. 7 1 
 
 together ; as, of f, f of f T of i ; which are read thus one half 
 of three fourths, two sevenths of five elevenths of nineteen twen- 
 tieths, &c. 
 
 3. A proper fraction is a fraction, whose numerator is less than its 
 denominator ; as, J, f, , &c. 
 
 4. An improper fraction is a fraction, whose numerator is larger 
 than its denominator ; as, f , , f &c. 
 
 5. A mixed number h composed of a whole number and a fraction ; 
 as, 7|, 35/3, &- c -5 * na * i s > sevea aQ d three fifths, &,c. 
 
 6. The common measure of two or more numbers, is that number, 
 which will divide each of them without a remainder: Thus, 5 is the 
 common measure (or divisor) of 10, 20, and 30; and the greatest 
 number which will do this is called the greatest common measure. 
 
 7. A number, which can be measured by two, or more, numbers, 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, 
 GO, 80, 100, are multiples of 4 and 5 ; but their least common multi- 
 ple is 20: 
 
 8. A prime number is that, which can only be measured (that is, 
 divided) by itself, or an unit ; as, 5, &c. 
 
 9. A perfect number is equal to the sum of all its aliquot parts. 
 
 PROBLEM I. 
 To find the greatest common measure of two or more numbers. 
 
 RUL 1. If there be two numbers only, divide the greater by 
 the less, and this divisor by the remainder, and so on, always di- 
 viding the last divisor by the last remainder, till nothing remain ; 
 then will the last divisor be the greatest common measure re- 
 quired. 
 
 2. When there are more than two numbers, find the greatest 
 common measure of two of them, as before ; then of that com- 
 mon measure and one of the other numbers, and so on, through 
 all the numbers, to the last ; then will the greatest common 
 measure, last found, be the answer. 
 
 144. TV hat is a proper fraction? 145. What is an improper fraction? 146. 
 
 W hat is a mixed number ? 147. What is a common measure of two or more num 
 
 bers 1 148. What is a. common multiple ? 149. What is the difference between a 
 
 prime and a perfect number 1 150. How da ^ou .find ff>e %rfa i ?*f common m.en*ure of 
 
 r more numbers ? 
 
72 VULGAR FRACTIONS. 
 
 3. If 1 happens to be the common measure, the given numbers 
 are prime to each other, and found to be incommensurable, or in 
 their lowest terms.* 
 
 EXAMPLES. 
 
 1. What is the greatest common measure of 36 and 96_? 
 36)96(2 
 
 72 $3r It is evident, that 
 
 12 is the greatest number 
 24)36(1 that will divide 36 and 96 
 
 24 without a remainder ; 
 
 therefore, it is the great- 
 
 Greatest common measure = 12)24(2 est common measure, or 
 
 24 answer. 
 
 2. What is the greatest common measure of 1224 and 1080? 
 
 Ans. 72, 
 
 PROBLEM II. 
 
 To find the least common multiple of two or more numbers. 
 
 RULE 1. Divide by any number that will divide two or more 
 of the given numbers without a remainder, and set the quotients, 
 together with the undivided numbers, in a line beneath. 
 
 2. Divide the second line as before, and so on, till there are 
 no two numbers that can be divided ; then the continued product 
 of the divisors and quotients will give the multiple required. f 
 
 * The truth of this rale may be shown from the first example ; for since 12 divides 
 24, it also divides 24+12, or 36. Again, since 12 divides 24 and 36, it also divides 
 36X2+24, or 96. 
 
 f The reason of this rule may be shown from the first example, thus : It is evident, 
 that 3X5X8X10 1200 may be divided by 3, 5, 8, and 10, without a remainder; bat 
 10 is a multiple of 5 ; therefore, 3x5x8x2240 is also divisible by 3, 5, 8, and 10. 
 Also, 8 is a multiple of 2 ; therefore, 3x5x4x2=120 is also divisible by 3, 5, 8, and 
 10 ; and is evidently the least number Vhat can be so divided. 
 
 151. W hat is the rule for finding the greatest common multiple of two or more nwM* 
 bers ? 
 
REDUCTION OF VULGAR FRACTIONS. 73 
 
 EXAMPLES. 
 
 1. What is the least common multiple of 6, 10, 16, and 20 
 
 *5)6 10 J.6 20 O^T We survey the given numbers, 
 
 -- and find that 5 will divide two of them, 
 
 *2)6 216 4 viz. 10 aud 20, which we divide by 5, 
 
 --- bringing into a line with the quotients, 
 
 *2)3 182 the numbers, which 5 will not measure : 
 
 --- again, we view the numbers in the sec- 
 
 *3 1 *4 1 ond line, and find 2 will measure them 
 
 all, and we get 3, 1, 8, 2, in the third 
 
 line, and find that 2 will measure 8 and 
 
 ***** 2, and in the fourth line get 3, 1, 4, 1 all 
 
 5X2X2X3X4 240 Ans. prime ; we then multiply the prime 
 
 numbers and the divisors continually into 
 each other, for the number sought, and 
 find it to be 240. 
 
 2: What is the least number, which can be divided by the 9 digits 
 separately, without a remainder ? Ans. 2520. 
 
 REDUCTION OF VULGAR FRACTIONS, 
 
 Is the bringing them out of one form into another, in order to 
 prepare them for the operations of Addition, Subtraction, &c. 
 
 CASE I. 
 
 To abbreviate or reduce fractions to their lowest terms.* 
 
 * A fraction is in its lowest terms, when represented by the least numbers possible : 
 thus ^ when reduced to its lowest terms is . 
 
 W f \at is reduction of vulgar fractions 1 153. When is a fraction in its 
 
 lowest terms ? 
 
 K 
 
74 REDUCTION OF VULGAR FRACTIONS. 
 
 RULE. 
 
 Divide the terras of the given fraction by any number, which 
 will divide them without a remainder, and the quotients again in 
 the same manner ; and so on, till it appears there is no number 
 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 will be the terms of the frac- 
 tion required. 
 
 NOTE. That dividing both the terms, that is, both numerator and 
 denominator of the fractions, equally by any number, whatever, will 
 give another fraction, equal to the former, is evident ; and if those di- 
 visions be performed as often as can be done, or the common divisor 
 be the greatest possible, the terms of the resulting fraction must be 
 the least possible. 
 
 1 . Any number ending with an even number or cipher is divisi- 
 ble by 2. 
 
 2. Any number ending with 5, or 0, is divisible by 5. 
 
 3. If tbe right hand figure of any number be 0, the whole is di- 
 visible by 10. 
 
 4. If the two right hand figures of any number be divisible by 4, 
 the whole is divisible by 4. 
 
 5. If the sum of the digits constituting any number, be divisible 
 by 3, or 9, the whole is divisible by 3 or 9. 
 
 6. All prime numbers, except 2 and 5, have 1, 3, 7, or 9, in the 
 place of units, and consequently all other numbers are composite and 
 capable of being divided. 
 
 EXAMPLES. 
 
 1. Reduce fff to its lowest terms. 
 
 8> (4) (3) 
 5 *H=f *=&=* the answer. 
 
 154. \V Hat are the rules for reducing fractions to their lowest terms ? 
 
REDUCTION OF VULGAR FRACTIONS. 75 
 
 Or thus : 
 
 288)480(1 Therefore 96 is the greatest com- 
 
 288 mon measure ; 
 
 192)288(1 and 96)ff=} the same as before. 
 
 192 
 
 Common measure 96)192(2 
 192 
 
 2. Reduce Jfc to its lowest terms. Ans. J 
 
 3. Reduce T 4 /4 to its lowest terms. Ans. J 
 
 4. Reduce Jf ff to its lowest terms. Ans. j- 
 
 CASE II. 
 
 To reduce a mixed number to its equivalent improper fraction. 
 
 RULE.* Multiply the whole number by the denominator of the 
 fraction, and add the numerator of the fraction to the product ; 
 under which subjoin the denominator, and it will form the fraction 
 required. 
 
 * All fractions represent a division of a numerator by the denominator, and are taken 
 altogether as proper and adequate expressions of the quotient. Thus the quotient of 3 f 
 divided by 4, is f. 
 
 1&5. How do you reduce a mixed number to its equivalent improper fraction ? 
 
76 REDUCTION OF VULGAR FRACTIONS. 
 
 EXAMPLES. 
 
 1. Reduce 36f to its equivalent improper fraction. 
 
 36 We multiply 36 by 8, and adding the nil- 
 
 X8_|_5 merator 5 to the product, as we multiply, the 
 
 sum, 293, is the numerator of tKe fraction 
 
 Answer, 293 sought, and 8 the denominator; so that 
 
 the improper fraction, equai to 36|. 
 8 
 
 36X8+5 293 
 
 Or, Answer as before. 
 
 8 $ 
 
 2. Reduce 127 T \ to its equivalent improper fraction. 
 
 Ans. 
 
 3. Reduce 653 T 3 ^ to its equivalent improper fraction. 
 
 Ans. 
 
 CASE III. 
 
 To reduce an improper fraction to its equivalent whole or mixed 
 number. 
 
 RULE. Divide the numerator by the denominator, the quo- 
 tient will be the whole number, and the remainder, if any, writ- 
 ten over the given denominator, will form the fractional part. 
 
 NOTE. This rule is, evidently, the reverse of the rule in Case II, 
 and has its reason in the nature of common division. 
 
 156. Whatis the rule for redwing- an improper fraction to Us equivalent whole or 
 mixed number ? 
 
REDUCTION OF VULGAR FRACTIONS. 
 
 77 
 
 EXAMPLES. 
 
 1. Reduce 2 f 3 to its equivalent whole or mixed number. 
 
 8)293(36f Ans. 
 
 24 
 
 53 
 
 48 
 
 Or, 2|3-293-^8=36f as before. 
 5 
 
 2. Reduce 2 \%* to its equivalent whole or mixed number. 
 
 Ans. 127 T V 
 
 3. Reduce y to its equivalent whole number. 
 
 Ans. 
 
 CASE IV. 
 
 To reduce a compound fraction to an equivalent simple one. 
 
 RULE. Multiply all the numerators continually together for a 
 new numerator, and all the denominators, for a new denominator, 
 and they will form the simple fraction required. 
 
 NOTE. If any part of the compound fraction be a whole or mixed 
 number, it must first be reduced to au inproper fraction. 
 
 That a compound fraction may be represented by a single one, is 
 evident, since a part ef a part must be equal to some part of the 
 whole. The truth of the rale for this reduction may he shewn as fol- 
 lows, 
 
 Let the compound fraction to be reduced be of 4. Then J of | 
 4~r3z=:^ 4 T , and consequently of 4 X 2= 2 8 T the same as by the rule, 
 and the like will be found true in all cases. 
 
 If the compound fraction consist of more numbers than two, the 
 two first may be reduced to one, and that one and the third will be 
 the same as a fraction of two numbers, and so on. 
 
 157. How do you reduce a compound, to a simple fraction ? 
 
78 REDUCTION OF VULGAR FRACTIONS, 
 
 EXAMPLES. 
 
 1. Reduce J of f of of f to a simple fraction. 
 
 1X2X3X4 
 
 = T Vo=i the Ans. 
 
 2X3X4X5 
 
 Or, by expunging the equal numerators and denominators, it will 
 give i as before. 
 
 2. Reduce of of of |i to a simple fraction. 
 3X4X5X11 
 
 4X5X6X12 
 
 Or, by expunging the equal numerators and denominators, it will 
 3X11 
 
 be -- =4f =W as before. 
 6X12 
 
 3. Reduce J of off of 12J to a simple fraction. 
 
 CASE V. 
 
 To reduce fractions of different denominators to equivalent fractions, 
 having a common denominator. 
 
 RULE. Multiply each numerator into all the denominators, 
 except its own, for a new numerator, and all the denominators 
 into each other continually, for a common denominator. 
 
 158. What is the rule for reducing fractions of different denominators to 
 alent fractions, having a common denominator ? 
 
REDUCTION OF VULGAR FRACTIONS. 79 
 
 EXAMPLES. 
 
 1. Reduce J, f, and f to equivalent fractions, having a common 
 denominator. 
 
 1X5X8= 40 the new numerator for J, 
 2X4X8= 64 ditto f. 
 
 5X4X5=100 ditto f. 
 
 4X5X8=160 the common denominator. 
 
 Therefore, the new equivalent fractions are T 4 F o- 7 -rVu anc * Teib tne 
 answer. 
 
 2. Reduce J, f , j, , and , to fractions, having a common de- 
 nominator. 
 
 NOTE. When there are whole numhers, mixed numbers, or com- 
 pound fractions, given in the question, they must first be reduced t 
 their simple forms. 
 
 CASE VI. 
 
 To reduce any given fractions to others, which shall have the least 
 common denominator. 
 
 RULE. Find the least common multiple (by problem II t 
 page 72,) of all the denominators of the given fractions, and it 
 will be the common denominator required ; then divide the com- 
 mon denominator by the denominator of each fraction, and multi- 
 ply the quotient by the numerator for a new numerator ; the new 
 numerators written over the common denominator will form the 
 fractions required in their lowest terms. 
 
 159. What is the rule for reducing any given fractions to others which shall have 
 the bast common denominator ? 
 
80 REDUCTION OF VULGAR FRACTIONS. 
 
 NOTE. The common denominator is a multiple of all the denomi- 
 nators, and coQsequontly will divide by any of them ; therefore, 
 proper parts may be taken for all the numerators as required by 
 the rule. 
 
 EXAMPLES. 
 
 1. Reduce J, f, and J to fractions having the least common de- 
 nominator possible. 
 
 4)3 4 8 
 
 4X3X2r=24=lcast common 
 3 1 2 denominator. 
 
 243X1=8 the first numerator ; 244X3=18 the second numera- 
 tor ; 24-7-8X7=21 the third numerator. 
 
 Whence the required fractions are ^ , JJ, |J. 
 
 2. Reduce J, , f , and f to fractions having the least common de- 
 nominator. 
 
 Ans. fS, ||, j. f , f |. 
 
 *- x CASE VII. 
 
 To reduce afractiVnjf one denomination to an equivalent fraction of 
 
 denomination. 
 
 RULE. Reduce the given fraction to a cjompound one by com- 
 paring it with all the denominations between it and that denom- 
 ination you would reduce it to ; lastly, reduce this compound 
 fraction to a single one, by Case V, and you will have a fraction 
 of the required denomination, equal in value to the given fraction. 
 
 NOTE. The reason of the rule m^y be shown in the following 1 man- 
 ner : As there are 12 pence in a shilling, four-fifths of one penny 
 can be only a twelfth part as much of 12 pence or a shilling-, as it is 
 of one penny. Hence, to reduce four-fifths of a penny to the frac- 
 tion of a shilling, the given fraction must be diminished 12 times, or 
 one-twelfth of it will be the equivalent fraction oi a shilling. And in 
 general, the fraction of one denomination must be as much diminish- 
 ed to be an equivalent fraction of a higher denomination, as is indi- 
 
 160. How do you reduce a fraction of one denomination to an equivalent fraction 
 of a higher ? 
 
REDUCTION OF VULGAR FRACTIONS. 8 1 
 
 cated by the number of parts of the given denomination which make 
 one of the higher denomination. 
 
 EXAMPLES. 
 
 1. Reduce 4 of a cent to the fraction of a dollar. 
 
 By comparing it, it become* 4 of T ' 7 of T V, which, reduced^ Case 
 V. will be 4X1 XI = 4 
 
 js do11 - Ans. 
 
 and 7X10X10= 700 
 
 2. Reduce f of a penny to the .fraction of a pound. 
 
 Ans. t - 
 
 3. Reduce 4 of an ounce to the fraction of a Ib. Avoirdupois. 
 
 Ans. 
 
 4. Reduce J of a pennyweight to the fraction of a Ib. Troy 
 
 Ans. 
 
 CASE VllT. 
 
 To reduce a fraction of one denomination to the fraction of another, 
 but less, retaining the same value. 
 
 RULE. Multiply the given numerator by the parts of the de- 
 nominations between it and that denomination you would reduce 
 it to, for a new numerator, which place over the given denom- 
 inator : Or, only invert the parts contained in the integer, and 
 make of them a compound fraction as before ; then reduce it to 
 a simple one. 
 
 . This rule is the reverse of the preceding, and the reason 
 of it may be shown in a similar manner. The fraction of a higher 
 denomination is obviously less than the equivalent fraction of a low- 
 er denomination ; for example, f of a pound are 6 / shillings, or 12 
 shillings. Whence the value of the trac'ion must be increased, to 
 render it an equivalent Jrnction of a lower denomination, so many 
 times as there are parts of the less denomination in the higher. 
 
 EXAMPLES. 
 
 1. Reduce T i T of a dollar to the fraction of a cent. 
 
 161. W hat is the rule for reducing a fraction of one denomination to knottier <iw, 
 retaining the same value 1 
 
 L 
 
82 REDUCTION OF VULGAR FRACTIONS. 
 
 By comparing the fraction it will be jj of V of J T j then 
 1 X 10 X 10 100 *4 
 
 Ans. d. 
 
 175 X 1 XI 175 7 
 
 / 
 2. Reduce ? i of a pound to the Action of a penny. 
 
 Reduce ^V of a Ib. Avoirdupois to the fraction of an ounce. 
 
 Ans. ^ oz. 
 
 4. Reduce T /^ of a Ib. Troy to the fraction of a pwt. 
 
 Ans. J pwt. 
 
 CASE IX. 
 
 To find the qalue of a fraction in the known parts of the integer, as 
 of coin, weight, measure, fyc. 
 
 RULE. Multiply the numerator by the parts in the next in- 
 ferior denomination, and divide the product by the denominator ; 
 and if any thing remain, multiply it by the next inferior denom- 
 ination, and divide by the denominator as before, and so on, as f?r 
 as necessary ; and the quotients placed after one another, in 
 their order, will be the answer required. Or, reduce the nu- 
 merator, as if it were a whole number, to the lowest denomina- 
 tion, and divide the result by the denominator ; the quotient will 
 be the number of the lowest denomination, (which must be 
 brought into higher denominations as far as it will go,) and the re- 
 mainder will be a numerator, to be placed over the given denom- 
 inator for a fraction of the lowest denomination. 
 
 NOTE. From this rule, in connexion with what has been said of 
 Reduction of Federal Money, it appears, that, annexing to the given 
 numerator as many ciphers, as will fill all the places to the lowest 
 denomination, and dividing the numbrr so form* d by the denomina- 
 tor, the quotient will be the answer in the several denominations, 
 and the remainder a numerator to be placed over th^ given denom- 
 inator, forming a fraction of the lowest denomination. 
 
 162. JHoiv do you find the value of a fraction in tht ktiuwn ^arts of the integtr ? 
 
REDUCTION OF VULGAR FRACTIONS. 33 
 
 EXAMPLES. 
 
 i. What is the value of g of a dollar ? 
 
 By the general rule. By the note. 
 
 5 jj. d. C. m. 
 
 10 8)5 
 
 8J50( 2 625 
 
 $ 6 10 
 
 Ans. 6d. 2c. 5m. 
 8)20( 4 or 62c. 5m. 
 
 c. 2 10 
 
 8)10 
 
 m. 5 
 
 Or thus. 
 g5 . 5000m. and so 00 m.=:825ni,=62c. 5m. answer, as before. 
 
 2. What is the value of JJ of a dollar ? 
 
 $. d. c. m. 
 
 64)17 (2d. 6c. 5|m, 
 
 128 
 
 or, 26c. 5f ra. 
 420 
 384 Or, $17.:=i7000m. And 
 
 360 i VV> o m.=:26 5f m 
 
 320 
 
 26c. 5|m. answer as before. 
 40 
 
 64 
 
 3. What is the value of 4 of a pound ? 
 
 Ans. 14s. 3d. l|qr. 
 
 4. What is the value of | of a Ib. Avoirdupois ? 
 
 Ans, 12oz. 12|dr. 
 
U4 REDUCTION OF VULGAR FRACTIONS. 
 
 CASE X.* 
 
 To reduce any given quantity to the fraction of any greater denom- 
 ination of the same kind. 
 
 RULE. Reduce the given quantity to the lowest term men- 
 tioned, for a numerator ; then reduce the integral part to the 
 same term for a denominator ; which will be the fraction re- 
 quired. 
 
 NOTE It appears from this rule, and what has been said before, 
 that, in Federal Money, where the given quantity contains no frac- 
 tion of its lowest denomination, the annexing of as many ciphers to 
 1 of the required denomination, as will extend to the lowest denom- 
 ination in the given quantity, will form a denominator, which, placed 
 under the given quantity used as one number for a numerator, will 
 make the answer, which may he reduced to its lowest terms. Or, 
 if there be a fraction of the lowest denomination, multiply the giv- 
 en whole numbers hy its denominator, adding its numerator, for a 
 numerator ; and let the denominator itself, at the left of as many 
 ciphers as were mentioned above, be a denominator ; the fraction 
 so formed will be the answer ; which may be reduced to its lowest 
 terms. 
 
 EXAMPLES. 
 
 1. Reduce 6d. 2c. 5m to the fraction of a dollar. 
 By the general rule. 
 
 6d. lOd. int. part. 
 
 X10+2 10 
 
 100 By the note. 
 
 10 $.' d. c. m. 
 
 625 
 
 625 1000 :|g. 
 
 1 000 
 And T 6 o 2 o 5 o !$ Ans. Ans. as before. 
 
 2. Reduce 26c. 5m. to the fraction of a dollar* 
 
 * This ease is the reverse of the former, therefore proves it. 
 
 0^/~ If there he a fraction given with the said quantity, it must be farther reduced to 
 the denominative parts thereof, adding thereto the numerator. 
 
 163. What if your rule for reducing any givtn quantity to the fraction of a greater 
 denomination of the same kind ? 
 
ADDITION OF VULGAR FRACTIONS. 85 
 
 By the general rule. 
 
 2Cc. lOOc. int. part. 
 
 XIO+5 10 
 
 265 1000 
 
 8 
 
 8000 
 
 By the note. 
 
 $. d. c. iru 
 
 2t55Xb-f5=2 1 2 5 
 
 . IK 
 b 41 
 
 And Jl,X8c=8 000 
 
 Ans. as before. 
 
 3. Reduce 14s. 3jd. 4 to the fraction of a pound. 
 
 Ans. = 
 
 4. Reduce 12cz. 12idr. to the fraction of a Ib. Avoirdupois. 
 
 Ans. |lb. 
 
 ADDITION OF VULGAR FRACTIONS. 
 
 RULE. 
 
 REDUCE compound fractions to single ones ; mixed numbers to 
 improper fractions ; fractions of different 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 he 
 the sum of the fractions required. 
 
 NOTE. Fractions, before they are reduced to a common denomina- 
 tor, are entirely dissimilar, and therefore cannot be incorporated 
 with one another ; but when they are reduced to a common denom- 
 inator, and made parts of the same thinaf, their sum, or difference, 
 may then be as properly expressed by the sum or difference of the 
 numerators, as the sum or difference of a^y two quantities whatever, 
 by the sum or difference of their individuals ; whence the reason of 
 the rule?, both for Addition and Subtraction, is manifest. 
 
 164. What is the rule for addition tf vulgar fractions 1 
 
36 SUBTRACTION OF VULGAR FRACTIONS. 
 
 EXAMPLES. 
 
 1. Add j, off, 4, and 5J together. 
 First J of ~| ; 4-i ; an d5j=V. 
 Then the fractions are f , f, A, V. 
 
 2 )S f f y. 
 
 y X.3 x i x 1X212 least common denominator. 
 
 124 3X 3 
 12 b 2 
 
 24 = 3X 3 9 j 
 2-6- 2X 2- 4f 
 2 1 12X 418? 
 2-2- bX 11-66) 
 
 new numerators. 
 
 127 sum of the numerators. 
 
 12 least common denominator. 
 Ans. y/ =10 Jj- 
 
 2. What is the sum of T \ of 4|, f of -J, and 9 ? 
 
 Ans. 
 
 3. Add together }$. fc. ^-c. and-|m. 
 
 Ans. 20c. 9m. 
 
 4. Add 1. -|s. and |d. together. 
 
 Ans. 2s. 8 T y 5 d. 
 
 5. Add J of a week, J of a day, ^ of an hour, and 2 of a minute 
 together. 
 
 Ans. 2 days, 2 hours, 30 minutes, 45 seconds. 
 
 SUBTRACTION OF VULGAR FRACTIONS. 
 
 RULE. 
 
 PREPARE the fractions as in Addition, and the difference of the 
 numerators, written above the common denominator, will give 
 the difference of the fractions required.* 
 
 * In subtracting mixed nurabprs, when the fractions have a common denominator, and 
 the numeratoi in the subtrahend is less than that in the minuend, the difference of the 
 whole numbers will be a 'vhole number, and the difference of the numeratms a numera- 
 tor to be placed over the given denominator ; this whole number and the fraction thus 
 formed will be the remainder ; but, when the numerator in the subtrahend is giea er 
 than that in the minuend, subtract the numerator in the subtrahend from the common 
 denominator, adding tne numerator in the minuend, and carrying one to the integer of 
 the subtrahend. Hence, 
 
 165. What is your rult for the subtraction of vulgar fractions ? 
 
MULTIPLICATION OF VULGAR FRACTIONS. 87 
 
 EXAMPLES. 
 
 1. From J take f of ;. 
 
 f of g=i|=L.-/ T . Then the fractions are f and / ff . 
 
 5X 2 4= *0 h=AV ^ *=Wn therefore, 
 
 4X28=112 com. den. ) rYs i-j viV^i remainder. 
 
 2. , From -. take Ans. if J. 
 
 3. Take 3jc, from | of 2jg. Ans. 43J-C. 
 
 4. From 1. take f^s. Ans. 4s. l}d. 
 
 5. From 5 weeks take 1 9f days. 
 
 Ans. 15da. 4ho. 48min. 
 
 MULTIPLICATION OF VULGAR FRACTIONS. 
 
 RULE. 
 
 REDUCE compound fractions to Simple ones, and mixed numbers 
 to improper fractions ; then the product of the numerators will 
 be the numerator, and the product of the denominators, the de- 
 nominator of the product required. 
 
 NOTE. Where several fractions are to he multiplied, if the nu- 
 merator of one fraction be equal to the denominator of another, their 
 equal numerators and denominators may be omitted. 
 
 Hence, a fraction is subtracted from a whole number, bv taking the numerator of the 
 fraction from its denominator, aud placing the remainder over the denominator, them 
 taking one from the whole number. 
 
 EXAMPLES. 
 
 From 12 m 12 
 
 Take^Tf 7^ ^J 
 
 Rem. 51 4| llf 
 
 166. What is the rule for multipkjir.gvutffar fractions ? 
 
88 DIVISION OF VULGAR FRACTIONS. 
 
 EXAMPLES. 
 
 1. What is the continued product of 4|, j, J of J, and 6 ? 
 
 1X7 
 
 4J=V, J of J= =ft, and 6=f . 
 4X8 
 
 13X1X 7X6 
 
 Then uxiXJ_Xf~ -- - |if m the answer. 
 3X5X32X1 
 
 2. Multiply T \ by ^V Ans. &. 
 
 3. Multiply 5| by J-. Ans. J. 
 
 4. Multiply J of 5 by f off Ans. T \. 
 
 DIVISION OF VULGAR FRACTIONS. 
 
 RULE. 
 
 PREPARE the fractions as before ; then invert the divisor, and 
 proceed exactly as in Multiplication : The products will be the 
 quotient required. 
 
 NOTE. To multiply a fraction by an integer, divide the denom- 
 inator, or multiply the numerator by it ; and to divide by an integer, 
 divide the numerator, or multiply the denominator by it. 
 
 EXAMPLES. 
 
 1. Divide of 17 b of. 
 
 1X17 
 of 17=J of y - ~y, and f of |=|f=i ; therefore, 
 
 3X1 
 17X2 
 
 y-^Jn -- =yz=llj, the quotient required. 
 3X1 
 
 2. Divide 4 by f . Ans. 1 
 
 3. Divide 5J by 7 f . Ans. | 
 
 4. Divide | by 9. Ans. ^ 
 
 5. Divide J of J of bv off. Ans. |. 
 
 167. What is the rule for the division of vulgar fractions 1 
 
DECIMAL FRACTIONS. 89 
 
 DECIMAL FRACTIONS. 
 
 -The denominations of Federal Money are purely decimal ; dollars being 
 units o- \viiole lumbers, ilimss tenths of a dollar, cents huiidredt'is of a dollar, .ind 
 mills thousandths of a dolltr ; consequently, Federal Money and Decimal Fractions 
 are subject to ti:e same method* of operation.] 
 
 A DECIMAL FRACTION is a fraction whose denominator is a unit 
 with so many ciphers annexed as the numerator has places of fig- 
 ures. 
 
 As the denominator of a decimal fraction is always 10, 100, 
 1000, &c. the denominators need not be expressed : For the 
 numerator only may be mude to express the true value. For 
 this purpose it is only required to write the numerator with a 
 point before it at the left hand, to distinguish it from a whole 
 number, when it consists of so many figures, as the denominator 
 Wh ciphers annexed to unity, or i : So T 5 o is written -5 ; j 3 oi 
 33 ; T VVV, '735, &c. 
 
 The point prefixed is called a Separatrix. 
 
 But if the numerator has not so many places as the denominator 
 has ciphers, put so many ciphers before it, viz. at the left hand, 
 as will make up the defect : So write T 7 thus, .05 ; and -,-/.-. 7 
 thus, -006, &c. And thus do these fractions receive the form of 
 whole numbers. 
 
 We may consider unity as a fixed point, from whence whole 
 numbers proceed infinitely increasing toward the left hand, and 
 decimals infinitely decreasing toward the right hand to 0, as in 
 the following 
 
 TAB.LE. 
 
 1, r < 
 
 ai C 
 
 **" 03 ^ 
 
 r- 5 % a " e 
 
 ~ 3 -co 
 
 t^ O 
 
 o^= 
 "9 
 
 . ns c 2 *o ! -C 
 
 55555555-55555555 
 
 0)Q)a)wQ?v4}C ojQjttJtyaiciOo 
 <joooo-ou ooyoocuo 
 
 ccsren--^^ rcr;^~r3^s 
 
 C-~. cu 3. ft. . Q. 
 -c.cjr^rjr^"^ *-> 
 
 .- +-. s-< t- C tfi 
 
 Whole numbois. Decimal parts. 
 
 168. What are Decimal Fractions? 169. Explain the principles on which deci- 
 mals are founded 1 170. Wkai is said of unity 1 171. Explain lite Table*. 
 
 % 
 
90 DECIMAL FRACTIONS. 
 
 All the figures at the left hand of the decimal point are whole num- 
 bers. The 5 in the 1st place at the right hand of the point, repre- 
 sents 5 tenths ; 5 in the 2nd place, 5 hundredths ; ft in the 3rd place, 
 5 thousandths ; 5 in the 4th place, 5 ten-thousandths ; 5 in the 5th 
 place, 5 hundred thousandths ; 5 in the 6th place, 5 millionths; 5 in 
 the 7th place, 5 ten-millionths ; 5 in the 8th place, 5 hundred-mil- 
 lionths ; and all of them taken together, are read thus : Fifty-five 
 million, five hundred fifty-five thousand, five hundred and fifty-five 
 hundred millionths.* 
 
 Ciphers, placed at the right hand of a decimal fraction, do not al- 
 ter its value, since every significant figure continues to possess the 
 same place : So -5, -50, and '500, are all of the same value, and each 
 equal to J. 
 
 But ciphers placed at the left hand of a decimal, do alter its value, 
 every cipher depressing it to T ' of the value it had before, by re- 
 moving every significant figure one place further from the place of 
 units. So -5, -05, -005, all express different decimals, viz. -5, T 5 ff ; -05, 
 TO ? 005, -nfop 
 
 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 ciphers as make them 
 equal in places with that which has the most. So these decimals, -5, 
 06, -455, may be reduced to the decimals -500, -060, -455, which 
 have all 1000 for their denominator. 
 
 Of decimals, that is the greatest, whose highest figure is greatest, 
 whether they consist of an equal or unequal number of places : Thus, 
 5 is greater than -459, for if it be reduced to the same denominator 
 with -459, it will be -500. 
 
 * It is evident from the table, that since the deci nal parts decrease in a tenfold propor- 
 tion from the place of units towards the right hand, they must increase in a tenfold pro- 
 portion from tiie right, hand towards the left, which proves that decimal fractions are 
 suhject to the same, law of notation, and consequently of operation, as whole numbers. 
 
 To read decimal figures. - Begin at the left hand, and read them in the same manner as 
 we read whole nuinDers, and add to the whole the name of the place of the right ham! 
 j&gure. 
 
 EXAMPLES. 
 
 24 is read 24 hundredths ; 
 035 is read 35 thousandths ; 
 0050 is read 50 ten thousandths ; 
 000750 is read 750 millionths ; 
 00000009 is read 99 hundred millionths. 
 
 172 Whit are all figures at the left hand of the point, or separatrix ? 173. What 
 
 does the first figure at the right of the decimal point represent 1 174. What the 
 
 second, third, fourth, 8fc ? 175. *What effect have ciphers placed on either hand of a 
 
 decimal 1 176 f-ow arc decimal fractions of unequal denominators reduced to a com- 
 
 mo/i denominator 7 
 
ADDITION OF DECIMALS. 91 
 
 A mixed number, viz. a whole number witb a decimal annexed, is 
 equal to an improper fraction, whose numerator is all the figures of 
 the mixed number, taken as one whole number, and the denominator 
 that of the decimal part. So 45-309 is equal to YoVoS as is evident 
 from the method given to reduce a mixed number to an improper 
 fraction: Thus, 45xlOO(H-309- 4 T 5 o 3 /e 9 as above. 
 
 ADDITION OF DECIMALS. 
 
 'v 
 
 RULE. 
 
 1. Place the numbers, whether mixed, or pure decimals, un- 
 der each other, according to the value of their places. 
 
 2. Find their sum, as in whole numbers, and point off so many 
 places for decimals, as are equal to the greatest number of deci- 
 mal p aces in any of the given numbers. 
 
 EXAMPLES. 
 
 1. Find the sum of 1 9-073+2-3597+223+-01 9758 1+3478 -1+ 
 12358. 
 
 19-073 
 2-3597 
 223. 
 
 0197581 
 3478-1 
 12-358 
 
 3734-9104581 the sum. 
 
 2. Required the sum of 429+2 1-37+355-003+1 -07+1 -7 ? 
 
 Ans. 808-143. 
 
 3. Required the sum of 973+1 9+1 -75+93-7 164+-9501 ? 
 
 Ans. 1088-4165. 
 
 4. Required the sum of -3+1 1-973+49+-9+1-731 4+34 3 ? 
 
 Ans. 103-2044. 
 
 177. What is the rule for addition of decimals ? 
 
92 MULTIPLICATION OF DECIMALS, 
 
 SUBTRACTION OF DECIMALS. 
 RULE. 
 
 Place the numbers according to their value : Then subtract 
 as in whole numbers, and point "off the decimals as in Addition. 
 
 EXAMPLES. 
 
 1. Find the difference of 1^93-13 and 817-05693. 
 
 From 1793-13 
 Take 8l7-05.i93 
 
 Remainder 976-07307 
 
 2. From 171-195 take 125-9176. Ans. 45-2774. 
 
 3. From 219-1384 take 195-91. Ans. 23-2^84. 
 
 4. From 480 take 245-0075. Ans. 234-9925. 
 
 MULTIPLICATION OF DECIMALS. 
 
 RULE. 
 
 1. Whether they be mixed numbers, or pure decimals, place 
 the factors and multiply them as in whole numbers. 
 
 2. Point off so many figures from the product as there are 
 decimal places in both the factors ; and if there be not so many 
 places in the product, supply the defect by prefixing ciphers. 
 
 NOTE The reason of th1*rule for pointing 1 off the figures for dec- 
 imals, is evident from the Rotation of decimals. Thus, -5X-5=-25; 
 for 5X i 5 or once-five tenths. But as the multiplier is less than uni- 
 ty, or tenths, multiplying is only taking tenths of tenths, and so many 
 tenths of tenth? 5 , are evidently so many hundredths. So al^o, tenths 
 of hundredths would be thousandths; hundredths of hundredths would 
 b<* ten thousandths; and so on. 
 
 173 Whit is */'<- rulr.for subtracting decimal fractions 1 17tf. What is the rule 
 
 for multiplication of decimals 1 
 
DIVISION OF DECIMALS. 93 
 
 EXAMPLES. 
 
 i. Multiply -09345 
 by -00163 
 
 7036 
 14070 
 2345 
 
 0000^82235 the product. 
 
 2. Multiply 25-238 by 12-17. Ans. 307-14646. 
 
 3. Multiply -3759 by -945. Ans. -3552255. 
 
 4. Multiply -84179 by -0385. Ans. -032408915. 
 
 To multiply by 10, 100, 1000, &c remove the separating point so 
 many places to the right hand, as the multiplier has ciphers. 
 
 (3-45 
 1 34- 
 
 So -345 multiplied by 100 V makes 34-5 
 flOOO.) (345, 
 
 For -345X 10 is 3-450, &c. 
 
 DIVISION OF DECIMALS. 
 
 RULE. 
 
 1. The places of decimal parts in the divisor and quotient, 
 counted together, must always be equal to those in the dividend ; 
 therefore, divide as in whole numbers, and, from the right hand 
 of the quotient, point off so many places for decimals, as the dec- 
 imal places in the dividend exceed those in the divisor. 
 
 2. If the places of the quotient be not so many as the rule re- 
 quires, supply the defect by prefixing ciphers to the left hand. 
 
 3. If at any time there be a remainder, or the decimal places 
 in the divisor be more than those in the dividend, ciphers may 
 be annexed to the dividend or to the remainder, and the quotient 
 carried on to any degree of exactness. 
 
 189, What is the rule for division of decimals 1 
 
94 
 
 DIVISION OF DECIMALS. 
 
 EXAMPLES. 
 
 1- 2. 
 
 1 17841075(-000538087, &c. -3719)38OOOO(102-178, 
 
 1U95 0^7= In Example 1st, the 3719 
 
 divisor having no decimals, the -- 
 834 quotient must have so many as 8100 
 657 there are in the dividend. In 7438 
 Example 2d, the dividend be- - 
 1771 ing an integer, must have at 6620 
 1752 least so many ciphers annexed, 3719 
 -- as there are decimals in the 
 1907 divisor, and so far the quotient 29010 
 1752 will be whole numbers ; then 26C33 
 annexing more ciphers, the --- 
 
 1555 remaining figures in the quo 
 1533 tient will be decimals, accord- 
 - ing to the rule. 
 
 22 
 
 29770 
 29752 
 
 3. 133)5737(43-1353+* 
 
 4. 23-7)65321(2756-16+ 
 
 5. t72)918-217(12753+ 
 
 6. 125-17)315-6293(1253+ 
 
 7. 1-317)29-417(92+ 
 
 When decimals or whole numbers are to be divided by 10, 100 3 
 1000, <kc. [viz. unity with ciphers] it is performed by removing the 
 separatrix, in the dividend, so many places toward the left hand, as 
 there are ciphers in the divisor. 
 
 dividing 7654, the quotient is 
 
 765-4 
 76-54 
 
 7654 
 
 * This character, placed here, shews that the quotient is not complete, ami that it 
 may he lurthei continued by annexing ciphers to the remainder. 
 } These questions are le/t unpointed in the quotient, to exercise the learner. 
 
 181 When decimals or vthole numbers are to be divided by 100, 1000, fyc., how do 
 you proceed ? 
 
REDUCTION OF DECIMALS. 95 
 
 REDUCTION OF DECIMALS. 
 
 CASE I. 
 
 * 
 To reduce a vulgar fraction to its equivalent decimal. 
 
 RULE. 
 
 Divide the numerator by the denominator, as in Division of 
 Decimals, and the quotient will be the decimal required. Or, 
 so many ciphers as you annex to the given numerator, so many 
 places must be pointed off in the quotient, and if there be not so 
 many places of figures in the quotient, the deficiency must be 
 supplied by prefixing so many ciphers before the quotient figures. 
 
 NOTE. By annexing one, two, three, &c. ciphers to the numera- 
 tor, the value of the fraction is increased ten, a hundred, &c. times. 
 After dividing, the quotient will of course be ten, a hundred, &c. 
 limes too much ; the quotient must therefore be divided by ten, a 
 hundred, &c to give the true quotient or fraction. In the first exam- 
 ple, J is made 'yo^ 1 -?- 5 , which, divided by 1000, is T ' Vf =,125 3 
 and is the rule. 
 
 EXAMPLES. 
 
 1. Reduce J to a decimal. 8)1-000 
 
 125 Ans. 
 
 2. Reduce f , f , and to decimals. 
 
 Ans. -375, -625, -666+, 
 
 3. Reduce J, J, J, J, f , f , and J to decimals. 
 
 Ans. -25, -5, -75, -333+, -8, -833+, -875. 
 
 1. Reduce T 5 , f, T VV, and ^. to decimals. 
 
 Ans. -263+, -692+, -025, -25. 
 
 CASE II. 
 
 To reduce numbers of different denominations, as of Money, Weigh^ 
 and Measure, to their equivalent decimal fractions . 
 
 RULE. 
 
 1. Write the given numbers perpendicularly under eachother 3 
 for dividends, proceeding orderly from the least to the greatest. 
 
 182. How do you proceed to reduce a vulgar fraction to its equivalent decimal ? 
 
 183. Wknt is your rule for reducing numbers of different denominations to their 
 ftjuiculent decimai value ? 
 
96 REDUCTION OF DECIMALS. 
 
 2 Opposite to each dividend, on the left hand, place such a 
 number, for a divisor, as will bring it to the next superior de- 
 nomination, and draw a line perpendicularly between them 
 
 3 Begin with the highest, and write the quotient of each 
 division as decimal parts on the right hand of the dividend next 
 below it, and so on, until tney are all used, and the last quotient 
 will be the decimal sought. 
 
 NOTE. 'The reason of the rule may be explained from the first 
 example : thus, three faj things are ^ of a penny, which, brought to a 
 decimal, is 75 ; consequently, 8-Jd. may bo expressed, 8-75d. but 
 3-75 is -J.J4. of a p^nny , 8 2 7 5 g- of a shilling, which f>rou$nt to a d -ci- 
 mal, is '7^l66s.+ "in like manner, 1 7-729 1 6t>3.+ are 
 ^|JUJLe -3 ti 6458-i- as by the rule. 
 
 EXAMPLES. 
 
 1. Reduce 17s. 8jd. to the decimal of a pound, 
 
 3- 
 
 8-75 
 
 17 729166, &,c. 
 
 8864o8, &,c. the decimal required. 
 
 e, in dividing 3 by 4, we suppose two ciphers to^be an- 
 aexed to the 3, which make it 3-00, and -75 is the quotient, which 
 we write against 8 in the next line ; this quotient, viz. 8-75, hoi .g 
 pfiice and decimal parts of a penny, we divide th m by 12, which 
 brings them to shillings and decimal parts ; we therefore divide by 
 20, and (there being no whole number) the quotient is decimal parts 
 of a pound. 
 
 2. Reduce 1, 2, 3, 4, and so on to 19 shillings, to decimals. 
 
 Shillings. 
 
 i 
 
 2 3 
 
 4 
 
 5 
 
 6 
 
 Answers. 
 
 05, 
 
 1, -15, 
 
 2, 
 
 25, 
 
 3, 
 
 Shillings. 
 
 7 
 
 8 9 
 
 10 
 
 11 
 
 12 
 
 Answers. 
 
 -35, 
 
 4, -45, 
 
 5, 
 
 55, 
 
 e* 
 
 Shillings 
 
 13 14 
 
 15 16 
 
 17 
 
 18 
 
 19 
 
 Answers. 
 
 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 cipher to the shillings, and tken halving them, you will 
 have their decimal expression. 
 
1 
 
 004 16+, 
 
 23 4 
 0083>, -0126, -01666-f, 
 
 5 
 
 0208+, 
 
 6 78 
 025, -02916-h, -0333+, 
 
 9 
 0375, 
 
 10 11 
 
 0416+, -04583-K 
 
 REDUCTION OF DECIMALS. 97 
 
 3. Reduce 1, 2, 3, and so on to !t pence, to the decimals of a 
 shilling-* 
 
 Pence. 123456 
 
 Answers. -083+, -166, -25, -333,+, -4 16+, -5, 
 
 Pence. 7 8 9 10 11 
 
 Answers. -583+, -666+, -75, -833+, -916+. 
 
 4. Reduce 1, 2, 3, &c. to 1 1 pence, to the decimals of a pound. 
 
 Pence. 
 
 Answers. 
 
 Pence. 
 
 Answers. 
 
 Pence. 
 
 Answers. 
 
 ASE III. 
 
 Tojind the decimal of any number of*shillingSj pence, and farthings, 
 by inspection. 
 
 RULE. 
 
 1. Write half the greatest even number of shillings for the 
 first decimal figure. 
 
 2. Let the farthings in the given pence and farthings, possess 
 the second 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 exceed 36. 
 
 NOTE. If there be no shillings, or only one shilling in the given 
 sum, a cipher must be written in the first place, or place of tenths. 
 
 If the farthings in the given pence and farthings, do not exceed 9, 
 a cipher must be written in the second place, or place of hundredths. 
 
 The invention of the rule is as follows: As shillings are so many 
 20ths of a pound, half of them must be so many tenths, and conse- 
 quently take the place of tenths in the decimals; but when they are 
 odd, their half will always consist of two figures, the first of which 
 will be half of the even number, next less, and the second a 5 : Again, 
 farthings are 960ths of a pound, and had it happened that 1000, in- 
 stead of 960, had made a pound, it is plain any number of farthings 
 would have made so many thousandths, and might have taken their 
 place in the decimal accordingly. But 960 increased by ^ part of 
 
 * The answers to this question are the same as the decimal parts of afoot. 
 
 184. What is the rule for finding the decimal tf any number of shillings, pence, Sfc, 
 by inspection ? 
 
 N 
 
98 REDUCTION OF DECIMALS. 
 
 itself, is =1000, consequently any number of farthings, increased by 
 their ^ part, will be n exact decimal expression for them : Whence, 
 if the number of farthings be more than 12, ^V P art is greater than 
 Jqr. and, therefore, 1 must be added ; and whnn the number of far- 
 things is more than 36, ^ partis greater than IJqr. for which 2 must 
 be added. 
 
 EXAMPLES. 
 
 1. Find the decimal of 13s. 9|d. by inspection. 
 
 6. . =Jof 12s. 
 
 5 for the odd shilling. 
 39 farthings in 9f d. 
 Add 2 for the excess of 36. 
 
 691 rrdecimal required. 
 
 2. Find by inspection the decimal expressions of \ 8s. 3Jd. and 
 17s. 8Jd. Ans. -914 and -885. 
 
 CASE IV. 
 
 To find the value of any given decimal in the terms of the integer. 
 
 RULE. 1. Multiply the decimal by the number of parts in the 
 next less denomination, and cut off so many places for a remain- 
 der, to the right hand as there are places in the given decimal. 
 
 2. Multip'y the remainder by the next inferior denomination, 
 and cut off a remainder as before 
 
 3. Proceed in this manner through all the parts of the integer 
 and the several denominations, standing on the left hand, make 
 the answer. 
 
 EXAMPLES. 
 
 1. Find the value of -73968 of a pound. 
 20 
 
 14-79360 
 12 
 
 2-09280 Ans. 14s. 9Jd. 
 
 185. What is the rule for fating the value of any g-tven decimal in the terms of 
 file integer ? 
 
COMPOUND MULTIPLICATION. 99 
 
 2. What is the value of -679 of a shilling ? 
 
 Ans. 8-148-d. 
 
 3. What is the value of -617 of a cwt. ? 
 
 Ans. 2qrs. 13lb. loz. 
 
 4. What is the value of -397 of a yard ? 
 
 Ans. Iqr. 2-352n. 
 
 5. What is the value of -8469 of a degree ? 
 
 Ans. 58m. 6fur. 35po. Oft. 11 in. 
 
 6. What is the value of -569 of a year ? 
 
 Ans. 207d. 16h. 26m. 24sec. 
 
 COMPOUND MULTIPLICATION. 
 
 COMPOUND MULTIPLICATION is the multiplying of sums of dif- 
 ferent denominations ; it is useful in finding the value of Goods, &Q. 
 And, as in Compound Addition, we carry from the lowest de- 
 nomination to the next higher, so we begin and carry in Com- 
 pound Multiplication ; one general rule being to multiply the 
 price by the quantity. 
 
 NOTE. The product of a number, consisting of several parts or 
 denominations, by any simple number whatever, will be expressed by 
 taking the product of that simple number, and each part by itself, ag 
 so many distinct questions : Thus, 331. 15s. 9d. multiplied by 5, wiil 
 be 1651. 75s. 45d. (by taking the shillings from the pence, and the 
 pounds from the shillings, and placing them in the shillings and pounds 
 respectively,) 1681. 18s. 9d. and this will be true when the multipli- 
 cand is any compound number whatever. 
 
 CASE I. 
 
 When the multiplier or quantity does not exceed l&i 
 
 Multiply the price of one yard, pound, &c. by the whole quan- 
 tity or number of yards, pounds, &c. the product will be the 
 
 answer. 
 
 186. What is Compound Multipli 
 tiptier does not exceed 12 ? 
 
 ication ? 187, What is the rule when the mvl- 
 
JOO COMPOUND MULTIPLICATION 
 
 EXAMPLES. 
 
 1. What will 5 yards of broadcloth amount to, at l. 12$. 4$d. a 
 yard ? 
 
 . s. d. (r In this example, we write down l. 12s. 
 
 i 1 2 4 J 4Jd. the price of one yard, and then write 5, the 
 
 5 number of yards, under the least denomination. 
 
 We n-uitiply 2 farthings by 5, and the product 
 
 8 1 10 is 10 farthings, which we bring into pence by 
 
 dividing them by 4 ; we write down the remain- 
 
 ing 2 farthings and reserve the quotient, 2 pence, 
 
 to be added to the product of the pence. We then multipK 4 pence 
 
 by 5 and the product is 90 pence, and 2 pence which we reserved 
 
 are 22 pence, which we bring into shillings by dividing them by 12 ; 
 
 we write down the remainder 10 pence, and reserve the quotient, 
 
 1 shilling, to be added to the product of the shillings. We then mul- 
 
 ti(/iy 12 shillings by 5, and the product is 60 shillings, and 1 shilling 
 
 thai we reserved are 61 shillings, which we bring into pounds by 
 
 div d:i ( g them by 20 ; we write down the remainder 1 shilling, and 
 
 res. rv? the quotient, 3 pounds, to be added to the product of the 
 
 pou ;us. WP then multiply 1 pound by 5, and the product is 5 
 
 pounds, and 3 pounds which we reserved are 8 pounds ; this being 
 
 the highest denomination, wfr write down the whole amount 8 pounds 
 
 and tind the product or answer to be 8. is. 
 
 2. Multiply 4 13s. 4%d. by 10. 
 
 3. Multiply 8 15. llfd. by II. 
 
 4. Multiply 13 12s. lid. by 7. 
 
 5. Multiply 14 17*. 8Jd. by 9. 
 
 Ans. 133. 19*. 
 
 REMARK. The facility of reckoning in Federal Money, com- 
 pared with pounds, shillings, &c. may be seen from the examples 
 given in this and the following cases. 
 
 The general rule is 
 
 Multiply as in simple multiplication, and from the product point 
 off so u.any places for cents and mills, as there are places of cents 
 and mills in the price. 
 
COMPOUND MULTIPLICATION. 
 
 101 
 
 In Pounds, Shillings, drc. 
 
 !. What will 9 yards of cloth 
 amount to, at 5s. 4d. per yard ? 
 
 Price per yard, 054 
 Number yards, 9 
 
 280 
 2. Six yards at 9s. lOd. per 
 
 yard? 
 
 Ans. 2 19s. 
 
 In Dollars, Cents, and Mills. 
 
 1. What will 9 yards of cloth 
 amount to, at 88c. 9m. per yard ? 
 
 Price, 
 
 88c. 9m. 
 9 
 
 $8 -00 1 
 
 2. Six yards at gl-22c. 9m. per 
 
 yard ? 
 
 Ans. $7-37c. 4m. 
 
 CASE II. 
 
 Where the multiplier, that is, the quantity, is above 12. 
 
 Multiply by two such numbers, as, when multiplied together, 
 will produce the given quantity, or multiplier. 
 
 EXAMPLES. 
 
 In Pounds, Shillings, grc, 
 
 1. What will 144 yards of cloth 
 cost, at 3s. 5'Jd. per yard ? 
 
 t. d. 
 
 3 5J price 1 yd. 
 Mult, by 12 
 
 Produces 2 
 Mult, by 
 
 1 6 price 12 yds. 
 12 
 
 24 18 price 144 yds. 
 
 2. 24 yards at 6s. 3jd. ? 
 
 Ans. 7 11s. 6d. 
 
 3. 27yds. at 9s. 10d.*? 
 
 In Dollars, Cents and Mills. 
 
 1. What will 144 yards of cloth 
 cost, at 57c. 6|m. per yard ? 
 
 c. m. 
 
 5764 Or, -57.34 
 144 12 
 
 23056 
 23056 
 5764 
 
 $83-0016 
 
 69168 
 12 
 
 $83-0016 
 
 2. 24 yards at $1, 5c. 2m. ? 
 
 Ans. $25, 24c, 8m. 
 
 3. 27 yds. at $1, 22c. 9m. ? 
 
 * New- York currency, 
 
 188. What is the rule when the multiplier exceeds 12 ? 
 
102 
 
 COMPOUND MULTIPLICATION. 
 CASE III. 
 
 When the quantity is such a number, that no two numbers in the table 
 will produce it exactly. 
 
 Multiply by two such numbers as come the nearest to it ; and 
 for the number wanting, multiply the given price of 1 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 sub- 
 tract from the last product, and the remainder will be the an- 
 swer. 
 
 EXAMPLES* 
 
 In Pounds, Shillings, &e. 
 
 1. What will 47 yards of cloth, 
 at 17s. 9J. per yard, amount to ? 
 
 . s. d. 
 
 17 9 price of lyd. 
 Mult, by 5 
 
 Produces 489 price of 5 yds. 
 Mult, by 9 
 
 In Dollars, Cents, ar.d Millt. 
 
 1. What will 47 yards of cloth, 
 at $2. 95c. 8m. per yard, amount 
 to? 
 
 $2-958 
 47 
 
 20706 
 11832 
 
 Ans. $139-026 
 
 Produces 39 18 9 price of 45yds. 
 Add 1 15 6 price ot 2yds. 
 
 Ans.41 14 3 price of 47 yds. I 
 
 NOTE. 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 
 
 2. 67 J- yards at 16s. 4d.*? 
 
 Ans. 55 2s. 6d. 
 
 3. 59 yards at lOs.j? 
 
 Ans. 29 10s. 
 
 2. 67 yards at $2 4c. ? 
 
 Ans. $137 81c. 
 
 3. 59 yards at $1 33jc.? 
 
 Ans. $78 66c. 6m. 
 
 * N. Y. currency. 
 
 f Pennsylvania currency; $1 7s. 63. 
 
 189. Ifow do you proceed wh.cn the multiplier or quantity is such a number as no tics. 
 lumbers in the multiplication tabU will produce exactly 1 
 
COMPOUND MULTIPLICATION. ]03 
 
 CASE IV. 
 
 When the quantity is any number above the Multiplication Table. 
 
 Multiply the price of 1 yard by 10, which will produce the 
 price of 10 yards : This product, multiplied by 10, will give th& 
 price of 100 yards ; then you must multiply the price of 100 by 
 the number of hundreds in your question ; the price of ten by 
 the number of tens ; and the price of unity, or .1, by the number 
 of units : lastly, add these several products together, and the sum 
 will be the answer. 
 
 EXAMPLES. 
 
 1. What will 359 yards of cloth, at 4s. 7d. per yard, amouut to ? 
 
 5 . d. 
 
 4 7^ price of 1 yard. 
 l5 
 
 2 6 3 price of 10 yards. 
 
 to 
 
 2 6 price of 100 yards, 
 
 69 7 6 price of 300 yards. 
 
 5 times the price of 10 yds. 11 11 3 price of 50 yards. 
 
 9 times the price of 1 yd. 2 1 7J price of 9 yards. 
 
 Ans. 83 4 price of 359 yards. 
 
 2. 297 yards at 17s. 8d. per yard ? 
 
 Ans. 256 15s. 7jd. 
 
 3. 297 yards at $2 88c. 2m. per yard ? 
 
 Ans. $855 95c 4i, 
 
 CASE V. 
 
 To find the, value of a hundredweight or 112 pounds, having the 
 price of one pound given. 
 
 Multiply the price of 1 pound by 7, the product will be the 
 price of 7 pounds ; multiply the price of 7 pounds by 4, the pro- 
 
 190. When the quantity is a number above the multiplication table, how do you pro- 
 ceed 1 191. What is the method of finding the value of o cwt. or 112 Lbs. the price of 
 
 1 pound being given ? 
 
104 COMPOUND MULTIPLICATION. 
 
 duct will be the price of 28 pounds or 1 qr. of a cwt. ; then 
 multiply the price of 28 pounds by 4, and the product will be 
 the price of 112 pounds, or a c\vt. 
 
 EXAMPLES. 
 1 , What will 1 cwt. of lead come to, at 6Jd. per pound ? 
 
 6Jd. price of 1 Ib. 
 7 
 
 3 9 price of 7 Ibs. 
 
 15 2 price of 28 Ibs. or cwt. 
 4 
 
 3 8 price of 1 cwt. 
 
 2. 1 cwt. of tin at 2jd. per Ib. 
 
 Ans. l Is. Od. 
 
 3. 1 cwt. of chalk at 1 Jd per Ib. ? 
 
 Ans. 14s. 
 
 4. 1 cwt. of chalk at 2c. 1m. per Ib. ? 
 
 Ans. $2 35c. 2m. 
 
 PRACTICAL QUESTIONS IN WEIGHTS AND MEASURES. 
 
 1. What is the weight of 4 hogsheads of sugar, each weighing 7 
 cwt. 3qrs. 19lb. ? Ans. 31cwt. 2qrs. 20lb. 
 
 2. What is the weight of 6 chests of tea, each weighing \3cwt. 2 
 qrs. 91b. ? Ans. 21 cwt. Iqr. 261b. 
 
 3 If I am possessed of 1 J dozen of silver spoons, each weighing 
 3oz. 5pwt. 2 dozen of tea spoons, each weighing Ibpwt. 14gr. 3 
 silver cans, each 9oz. 7pwt 2 silver tankards, each 21 oz. 15pwt. 
 and 6 silver porringers, each lloz. 18pwt. ; what is the weight of 
 the whole? Ans. 181b. 4oz; 3pwt, 
 
 4. In 35 pieces of cloth, each measuring 27J yards, how many 
 yards? Ans. 97 1| yards. 
 
 5. How much brandy in 9 casks, each containing 45 gallons, 3qt. 
 Ipt? Ans. 412gal. 3qt. Ipt. 
 
 6. If I have 9 fields, each of which contains 12 acres, 2 roods, and 
 25 poles ; how many acres are there in the whole ? 
 
 Ans. 11 Sac. 3r. 25p. 
 
COMPOUND DIVISION. 
 
 COMPOUND DIVISION. 
 
 COMPOUND DIVISION is the dividing of numbers of different 
 denominations : in doing which, always begin at the highest, and 
 when you have divided that, if any thing remain, reduce it to 
 the next lower denomination, and so on, till you have divided 
 the whole, taking care to set down your quotient figures under 
 their respective denominations.* 
 
 INTRODUCTORY EXAMPLES. 
 
 S. d. 
 
 1. Divide 549 17 9 by 5. O^Hn this example, having di- 
 
 vided the pounds, the 4 which re- 
 
 109 19 6J mains is 4, which are equal to 
 
 80s and 17 in the shillings, make 
 
 '37s. ; we then find that 5 is contained 19 times in 97, and 2 over : we 
 set down the 19 under the shillings, and reduce the 2s. that remain 
 into pence, and they make 24, which, added to the 9d. in the ques- 
 tion, make 33d. ; then how often 5 in 33 ? 6 times, and 3 over. 
 We set down G under the pence, and reduce the 3d. which remain to 
 farthings they make 12: 5 is contained in 12 twice ; we therefore 
 set down Jd. and the 2 which remains, being f of a farthing, is dis- 
 regarded. 
 
 2. 3. 
 
 T. cwt. qrs. Ib. oz. dr. Ib. oz. pwt. gr. 
 
 3)29 13 2 25 1 13 10)849 11 12 14 
 
 4. Divide 581. 19s. lljd. by 11. 
 
 5. Divide 371. 17s. 4fd. by 7, 
 
 *To divide a number, consisting of several denominations by itsimple number, is, evir 
 dently, the same as dividing all the parts of which it is composed by the same simple 
 number. And this will be true, when any of the parts are not an exact multiple of the 
 divisor; for if the excess of that multiple has its proper value in the next less denomina- 
 tion, the dividend will still be divided into paits, and the true quotient will be found as 
 before. Thus, 81. Is. lO^d. divided by 5, will be the same as 51. 60s. 20d. lOqr. divid- 
 ed by 5, which is equal to 11. 12s. 4d. as by the rule. 
 
 192. What is Compound Divmvn ? 
 
 o 
 
J06 COMPOUND DIVISION. 
 
 CASE I.* 
 
 Having the price of any number of yards, fyc. within the pence 
 table^ to find the price of 1 yard, fyc. 
 
 If the quantity do not exceed 12, proceed by setting down 
 the price, and dividing it by the quantity ; which quotient will 
 be the price of one yard, required ; but if the quantity exceed 
 12, then divide by two such numbers as, when multiplied together, 
 will produce the quantity, and the last quotient will be the 
 value of 1 yard, required. 
 
 EXAMPLES. 
 
 In pounds, shillings, &c. In dollars, cents, and mills. 
 
 1. If 9 yards of cloth cost 4 
 3s. 7 Jd. what is it per yard ? 
 
 *. d. 
 
 9)4 3 7J 
 
 9 3J Ans. 
 
 2. If 84 cows cost 253 13s. 
 what is the price of each ? 
 
 Ans. 3 Os. 4|d. 
 
 1. If 9 yards of cloth cost 13 
 93f c. what is it per yard ? 
 
 9)13-9375 
 
 1-5486-f-Ans. 
 
 2. If 84- cows cost $845 50c. 
 what is the price of each ? 
 
 Ans. $10 6Jc, 
 
 NOTE. If there be a remainder after the division by one of the 
 parts of a composite number before the last, that remainder must be 
 divided according to the rule for division of fractions. Thus, 
 
 If 35 yards cost 37 1 Is. what is the price of 1 yard ? 
 
 * 
 35=5X7 5)37 11 
 
 7)7 10 2 Iqfr. 
 1 1 5 Iffqr. Ans. 
 
 CASE II. 
 
 Having the, price of Icwt. or Il2lbs. to find the price of one pound, 
 Divide the price of 1 cwt. by 8, that quotient by 7, and this 
 
 *This case proves the 1st and 2d cases in Compound Multiplication. 
 
 193. Having the price of several yards, how do you find the price of one ? 194. In 
 case of a remainder, how do you proceed*! 195. Having the price of Icwt. or more, 
 how do you find that of a pound ? 
 
COMPOUND DIVISION. 107 
 
 quotient by 2, and the last quotient will be the price of one pound, 
 required. The reason is, 8 X 7 X & = 112 Ibs. or 1 cwt. 
 
 EXAMPLES. 
 
 1. If I cwt. of flax cost 2 7s, 8d. what is that per Ib. ? 
 
 *. d. 
 8)2 7 8 
 
 7)0 5 11 J 
 
 2)0 10 Ofqr. 
 
 5^ d. price of 1 pound. 
 
 2. At $156 per cwt. what is the price per Ib. ? 
 
 Ans. $1 39f c, 
 
 3. At $3 33c. 3m. per cwt. what is the cost of 1 Ib. ? 
 
 Ans. 2c. 9 T y 7 m. 
 
 4. If 1 cwt. of sugar cost 3 7s. 6d. what is that per Ib.? 
 
 Ans. 7jd. 
 
 CASE III.* 
 
 Having the price of several hundred weight, to find the price per 
 
 pound. 
 
 Divide the whole price by the number of hundreds, which will 
 give the price per cwt. and then proceed as in the last case. 
 
 EXAMPLES. 
 
 1. If 5 cwt. sugar cost 13 8s. 4d. what is that per Ib. ? 
 
 s. d. 
 5)13 6 4 
 
 8) 2 13 8 price of Icwt. 
 
 7) 6 8J price of 14lb. or J cwt. 
 
 2) 11 price of 21b. or -fa cwt. 
 
 5| price of lib. 
 *Tbis a proves the 6th in^Com pound Multiplication. 
 
108 COMPOUND DIVISION. 
 
 5, If 3cwt. of raisins cost $50 52c. what cost 1 lb. 
 
 * 
 3)50-52 
 
 8)16-84 
 7)2-105 
 
 1 5c. T \m. Aus. 
 
 3. If Sjcwt. cost 3.64, what cost lib. ? 
 
 Ans. 1C. 
 
 4. If If cwt. cotton wool cost 6 10s. 8d. what is that per lb.2 
 
 5. If 3| cwt. cost $11 5Cc. what is the cost of 1 lb. ? 
 
 6. If 11 J cwt. cost $87 33c. what cost 1 lb. ? 
 
 CASE IV.* 
 
 Having the price of any number of yards, fyc. to find the price of 
 
 I yard. 
 
 Divide the price by the quantity, beginning at the highest de- 
 nomination, and if any thing remain, reduce it into the next, and 
 every inferior denomination, and at each reduction divide as be- 
 fore, remembering each time to add the odd shillings, pence, &c. 
 if there be any, and you will have the value of unity required. 
 
 NOTE. If there be J, J, or f of a yard, pound, &/C. multiply both 
 the price and quantity by 4, and then proceed as above directed ; or, 
 in Federal Money, work by decimals. 
 
 EXAMPLES. 
 
 1 . If 95Jlb. of figs cost $ 6 what are they per lb.? 
 
 * This case proves the 3d and 4th cases in Reduction. 
 
 196. Having the price of any number of yank, how do yovjind (he price ef one ? 
 
COMPOUND DIVISION. i<)9 
 
 Quantity = 95jlb. Price = l6 13 6| 
 
 Mult by 4" 4 
 
 Produces 382 for a divisor. Product 66 14 3 for a di- 
 
 [vidend 4 . 
 
 382)66 14 3(0 3 5|||| per Ib. 
 
 20 
 _ _ D.c.m.dec.c. m. 
 
 382)1334(3 95|-95.5)55.59375(.58 2-f-Ans. 
 1146 4776 
 
 188 7843 
 
 12 7640 
 
 382)2259(6 2037 
 
 1910* 1910 
 
 349 
 4 
 
 382)1396(3 
 1146 
 
 250 
 
 . Bought 33J yds. cloth for $8i 63c. 2m. ; what did I pay 
 
 Ans. $2 57c. 5m. 
 
 . If 33J yds. baize cost 25 13s. 9d. what is it per yard? 
 
 Ans. 15s. 6jd. ^. 
 
 PRACTICAL QUESTIONS. 
 
 1. Divide J ^ 9n * ^ ^ I among 5 men and 4 womcD, and 
 give the men twice as much as the women. 
 
 Men. Worn. 
 
 5 and 4 
 . by 2 
 
 10 shares. 
 Add 4 women's shares. 
 
 1 4 number of equal shares ia the whole = Divisor, 
 
110 COMPOUND DIVISION. 
 
 s. d. s. d. 
 
 Divide by 14)273 9 4( 19 10 8=1 woman's share. 
 14 4 women. 
 
 133 
 
 126 
 
 7 
 20 
 
 14)149(10 
 14 
 
 9 
 12 
 
 14)112(8 
 112 
 
 78 
 
 2 
 
 8=women\s share. 
 
 19 
 
 10 
 
 8 
 2 
 
 39 
 
 1 
 
 4=1 man's share. 
 5 men. 
 
 195 
 
 78 
 
 6 
 
 2 
 
 8=mea's share. 
 8=women's share. 
 
 273 
 
 9 
 
 4 Proof. 
 
 
 
 
 $ 
 
 14)911-555(65'111+=1 woman's share. 
 84 4 women. 
 
 71 260-444 women's share. 
 70 
 
 15 65-111+ 
 
 14 2 
 
 15 130-222+ =1 man's share. 
 
 14 5 
 
 15 651-111+ = men's share. 
 
 14 260-444+ = women's share. 
 
 1 911-555+ Proof. 
 
 2. A man purchased a farm containing 1 ISA. 3r. 25p. and he solo 
 one ninth part of it ; how many acres did he sell ? 
 
 Ans. 12A. 2r. 25p. 
 
 3. Divide 21 har. 30 gals. 2 qts. by 4. 
 
 Ans. 5 bar. 15 gals. 2 qts. 
 
 4. A merchant purchased 9 hogsheads of rum, containing 1-0&9 
 gallons 2 qts. ; what number of gallons did each hogshead contain ? 
 
 Ans. 1 1 5 gals, 2 qts. 
 
 5. Divide 45 tons of round timber by 8. 
 
 Ans. 5T. 25ft. 
 
SIMPLE INTEREST. HI 
 
 6. A man purchased a freight of wood containing 93 cords, and 
 piled it into six ranges of equal dimensions; what was the number of 
 cords in each range 1 
 
 Ans. 15C. 64ft. 
 
 7. If the earth perform 5 revolutions round the sun in l-826d. 5h. 
 4m. 45s., in what time will it perform one revolution ? 
 
 Ans. 365d. 5h. 48m. 57sec, 
 
 3. Divide 60 2' by 6. 
 
 Ans. 10 0' 20"- 
 
 INTEREST. 
 
 INTEREST is a premium allewed for the loan of money. 
 
 It is estimated at a certain number of dollars, pounds, &c. for 
 each hundred dollars, pounds, &c. for one year ; and in the same 
 proportion for a greater or less sum, or for a longer or shorter 
 time. Hence, interest is said to be so much per cent, or per 
 centum, per annum. 
 
 The terms used in interest are principal, rate, and amount. 
 
 The principal is the sum of money lent. 
 
 The rate is the sum allowed for the use of one hundred dollars 
 or pounds for one year.* 
 
 The amount is the sum of the principal and interest. 
 
 Interest is of two kinds, Simple and Compound. 
 
 SIMPLE INTEREST. 
 
 SIMPLE INTEREST is when a premium is allowed for the prin- 
 cipal only. 
 
 . Commission, Insurance, purchasing Stocks, or any thing 
 else, rated at so much per cent, are calculated by the Rules for Sim- 
 ple Interest. 
 
 * Lawful or legal interest is that which is allowed by law. In England, the rate is 
 five per cent. In the New England States, is is six; and in the State of New York, it 
 is seven per cent. 
 
 197. What is Interest ! 7198. How is it estimated? 199. What terms are used in, 
 interest 7200. Whit is the principal! SQL What the rate ; and amount? 202. 
 What it fli-mpls Interest? 
 
112 
 
 SIMPLE INTEREST. 
 
 GENERAL EULE. 
 
 1 Multiply the principal by the rate, and divide the product 
 by 100, the quotient will be the interest for one year. 
 
 2, To find the interest for more years than one ; multiply the 
 interest of one year by the given number of years, and the pro- 
 duct will be the interest for that number of years. 
 
 3. To find the interest for parts of a year, as months and days ; 
 for the months, take aliquot parts of the interest for one year ; 
 and ibr the days, take aliquot parts of the interest for one month, 
 allowing 30 days to the month. 
 
 TABLE OF ALIQUOT PARTS. 
 
 Parts of a year. 
 6 months J 
 
 4 =| 
 
 s = i 
 
 5? 4 
 
 i* ," =i 
 
 i = T V 
 
 Parts of a month. 
 15 days = 
 10. = 
 
 6 
 
 5 
 
 3 
 
 =1 
 
 NOTE. 
 When the rate 
 per cent, per an- ' 
 num is 
 
 <9 1 
 8 
 
 6 
 4 
 3 
 .2, 
 
 multiply ! f 
 the prin- j f 
 cipal bv 
 
 If] 
 
 of the given number of months, 
 and you will have the interest 
 for the given period of time.* 
 
 EXAMPLES. 
 
 1. What is the interest of 347 dollars 50 cents, at 6 per cent, for 
 
 ane year 
 
 $347-50 
 6 
 
 20-8500 
 
 Ans. $20 85c. 
 
 *The reason of this rule is, that as 9 is | of 12 months, if the rate be 9 per cent, 
 multiply the principal by | of the months; if the rate be 8 per cent, by | of the num- 
 ber of months ; and if 6, by hall the months, &c. 
 
 203.. What w the, general rule for computing Simple Merest ? 
 
SIMPLE INTEREST. 
 
 113 
 
 2. What is the interest of 365 Hcts. 6 mills, for three years, 7 
 months, and 6 days ? 
 
 $365-146 principal. 
 6 rate. 
 
 6 months J)21-90 876 interest for 1 year. 
 3 
 
 65-72 628 interest for 3 years. 
 1 month 1)10-95 438 interest for 6 months. 
 6 days i) 1-82 573 interest for 1 month. 
 36 514 interest for 6 days. 
 
 $78-87 153 interest for 3 years, 7 months, and 6 
 days ; that is $78 87c. 1 j^m. 
 
 NOTE. Because 7 months are not an even part of a year, take two 
 such numbers as are even parts, and which added together will make 
 7 (6 and 1) 6 months are j of a year, therefore for 6 months, divide 
 the interest of one year by 2 ; again, 1 month is | of 6 months, there- 
 fore for 1 month, divide the interest of 6 months by 6. For the 
 days, because 6 days are } of a month, or of 30 days, therefore for 6 
 days, divide the interest of 1 month by 5. Lastly, add the interest of 
 all the parts of the time together ; the sum is the answer, 
 
 3. What is the interest of 329 17s. 6d. for 3 years, 7 months, 
 and 12 days, at 5 per cent, per annum ? 
 
 
 
 325 
 
 8. d. 
 17 6J 
 5 
 
 6 months J) 
 
 1 month....i) 
 10 days J) 
 2 days ) 
 
 
 
 Then 16 
 
 s. d. 
 9 10J- 
 3 
 
 interest 
 
 do. 
 do. 
 do. 
 : do. 
 do. 
 
 of 1 year. 
 
 of 3 y*ars. ' 
 of 6 months, 
 of 1 month, 
 of 10 days, 
 of 2 days. 
 
 16-49 
 20 
 
 9-87 
 12 
 
 7 8J 
 
 49 
 8 
 1 
 
 9 7^ 
 4 11; 
 7 5j 
 9 \l 
 1 9' 
 
 10-52 
 4 
 
 59 
 
 13 
 
 Answer. 
 
 
 2-10 
 
 4. What is the interest of $500 for 3 years, 9 months, and 25 days, 
 at 6 per cent.? 
 
 Ans. 
 
 5. What will be the interest of $60-625 for 2 years, 10 months, 
 and 20 days, at 6 per cent. ? 
 
 ADS. J10-5083-K 
 
114 SIMPLE INTEREST. 
 
 6. What is the interest of $124 for 3 months and 10 days, at 6 per 
 cent. ? Ans. $2-066f . 
 
 7. What is the interest of $215-65 for 5 years, 5 months, and 6 
 days, at 6 per cent. ? 
 
 Ans. $70-3019. 
 
 8. What will be the interest of -75 cents for 4 months, at 6 per 
 cent. ? 
 
 Ans. -015. 
 
 To find the interest of any sum at 6 per cent, per annum, for any 
 number of months. 
 
 RULE. Multiply the principal by half the number of months, 
 and divide the product by 100 ; the quotient will be the interest 
 for the given time.* 
 
 EXAMPLES. 
 
 1. What is the interest of $225 for 4 months, at 6 per cent. ? 
 
 $225 principal. 
 
 2 half the number of months. 
 
 $4-50 Ans. 
 
 2. What is the interest of 1575 dollars for 10 months, at 6 per 
 cent. ? 
 
 To find the interest of any sum^ for any number of days. 
 
 RULE 1 . Multiply the principal by the given number of days, 
 and that product by the rate, then divide the last product by 
 365 X 100, and the quotient will be the interest. 
 
 *The reason of the rule. When the time is months, multiplying by the rate for the 
 time, gives the answer. This rate is found by multiplying the time by the given rate 
 per cent, for a year, and dividing the product by 12; the quotient is the rate required, 
 and is always equal to half the number of months, when the yearly rate is 6 per cent. 
 
 204. What is the rule for finding the interest of any sum at 6 per cent- per annum, 
 for any number of months 1 205. What is the reason of the rule ? 
 
SIMPLE INTEREST. 
 
 RULE 2. Multiply the principal by the given number of days, 
 and divide the product 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 will be the interest. 
 
 EXAMPLES. 
 
 1. What will be the interest of 320 dollars for 146 clays, at 6 per 
 cent. ? 
 
 $ days. rate. 
 
 320X ( 46X6 280320 ; and 365X 1 00=36500, 
 Then 280320-^36500 =$7-68 the interest. 
 
 2. What is the interest of $226-50 for 292 days, at 6 per cent.? 
 
 Ans. $10-824. 
 
 3. What will be the interest of $225-50 for 22 days, at 6 per 
 cent. ? 
 
 Ans. $10-824|ff. 
 
 4. What is the interest of 146 dollars for 146 days, at 5 per cent. ? 
 
 Ans. $2-92. 
 
 To find the interest on bonds, notes of hand , <^c. when partial pay- 
 ments have been made, or endorsed on them. 
 
 RULE 1 . Find the interest of the given sum to the first pay- 
 ment, which either alone, or with any preceding payment, if any, 
 exceeds the interest due at that time, and add that interest to the 
 given sum. 
 
 RULE 2. From this amount subtract the payment made at 
 that time with the preceding payments, if any, and the remain- 
 der will form a new principal ; the interest of which find and 
 subtract as of the first sum, and so on till the last payment. 
 
 NOTE. This mode of computing interest, is established by law in 
 Massachusetts for making up judgments on securities for money draw- 
 ing interest, and on which partial payments are endorsed. This mode 
 is the most equitable, because the payments are applied to keep down 
 the interest, no part of which, in this method of computation, forms 
 any part of the principal, drawing interest. 
 
 20G. What are the rules for finding the interest of any sum for any number of days ? 
 
 -207 What is the rule for computing interest upon bonds, notes, fyc, upon which 
 
 partial payments have been made 1 
 
J16 SIMPLE INTEREST. 
 
 EXAMPLES. 
 
 1. A. gave a note to B. dated Jan. 1, 1320, for $1000, payable on 
 demand, with interest ; on which were the following endorsements*. 
 
 
 
 
 
 
 yeart. 
 
 months. 
 
 March 1 
 
 1821, 
 
 received 
 
 75 
 
 dollars 
 
 1 
 
 2 
 
 July 1 
 
 1821, 
 
 (C 
 
 20 
 
 dollars 
 
 
 4 
 
 Sept. 1 
 
 1822, 
 
 u 
 
 20 
 
 dollars 
 
 1 
 
 4 
 
 Nov. 1 
 
 1822, 
 
 u 
 
 750 
 
 dollars 
 
 
 4 
 
 March 1 
 
 1823, 
 
 u 
 
 100 
 
 dollars 
 
 
 4 
 
 What was the balance due July 1, 1823, interest being computed at 
 6 per cent. ? 
 
 Calculation* 
 
 1000 Principal. 
 
 70 Interest to March 1, 1821, 1 yr. 2 month*. 
 
 1070 Amount. 
 
 75 First payment deducted. 
 
 995 New principal March 1, 1821. 
 19-90 Interest to July 1, 1821, 4 months. 
 
 1014-90 Amount. 
 20 00 Second payment deducted. 
 
 994 90 New principal July 1, 1821. 
 79-592 Interest to Nov. 1, 1822, 1 yr. 4 month?. 
 
 1074-492 Amount. 
 
 20-000 Third payment Sept. 1, 1822. 
 750-000 Fourth payment Nov. 1, 1822. 
 
 770-000 Sum of 3d and 4th payments deducted, 
 
 304-492 New principal Nov. 1, 1822. 
 
 6 089 Interest to March I, 1823, 4 months. 
 
 310-581 Amount. 
 
 100 000 Fifth payment deducted. 
 
 210-581 New principal March 1, 1823. 
 4-212 Interest to July 1, 1823, 4 months. 
 
 Ans. $214-793 Balance due July 1, 1823. 
 
COMMISSION. 
 
 117 
 
 A simple method of operation, is, first to set down against 
 each payment, (as in Ex. I.) the time for which the interest is to be 
 cast : then set down the sums, interest, payments, &c. in columns, as 
 follows : 
 
 
 Principal, 
 
 Time. 
 
 Interest 
 
 Payments. 
 
 Excess. 
 
 
 dolls, mills. 
 
 months. 
 
 dolls, iri.lls. 
 
 dolls, mil Is. 
 
 dolls, milts. 
 
 1 
 
 1000-000 
 
 14 
 
 70-000 
 
 75-000 
 
 5-000 
 
 
 5-000 
 
 
 
 
 
 2. 
 
 995-000 
 
 4 
 
 19-900 
 
 20000 
 
 100 
 
 
 100 
 
 
 
 
 
 3. 
 
 994-900 
 
 16 
 
 79-592 
 
 20-000 
 
 
 4. 
 
 u u 
 
 4 
 
 6-089 
 
 750-000 
 
 
 
 
 
 85-681 
 
 770-000 
 
 684-319 
 
 
 684-319 
 
 
 
 
 
 5. 
 
 310-581 
 
 4 
 
 4-212 
 
 100-000 
 
 95-788 
 
 
 95-788 
 
 
 
 
 
 
 $214-793 Balance due July 1, 1823. 
 
 2. Supposing a note of 867 dollars 33 cents, dated January 6, 1814. 
 upon which the following payments should be made, viz. 
 
 1. April 16, 1817, - - g!36-44 
 
 2. April 16, 1819, - - 319- 
 
 3. Jan. 1, 1820, - - 518-68. 
 
 What would be due July 11, 1821 ? 
 
 Ans. $215-103. 
 
 COMMISSION. 
 
 COMMISSION is a premium, at so much per cent, allowed a per- 
 son called a correspondent, factor or broker, for assisting mer- 
 chants, and others, in purchasing and selling goods. 
 
 RULE. Multiply the given sum by the rate per cent., and cut 
 off the two right hand figures, as in Simple Interest. 
 
 208. What is Commission 1 209. What is the rule for calculating commission* 
 
1 18 BUYING AND SELLING STOCKS. 
 
 EXAMPLES. 
 
 1. What is the commission on the purchase of goods, the invoice 
 f which amounts to 1250 dollars, at 2J per cent? 
 
 Ans. $31-25. 
 
 2. What must I allow my correspondent for selling goods to the 
 amount of 4325-75 at a commission of 5 per cent.? 
 
 Ans. $216-2875. 
 
 BUYING AND SELLING STOCKS. 
 
 STOCK is a general name for the capitals of trading companies^, 
 or of a fund established by government, the value of which is va- 
 riable. 
 
 RULE. If the stock be above par, that is, when 100 dollars of 
 stock are worth more than 100 dollars, multiply the given sum 
 by its value per cent over 100 per cent, and divide the product 
 by 100 ; then add the quotient and given sum together, the 
 amount will be the value required. 
 
 If the stock be below par, that is, when 100 dollars of stock 
 are worth less than 100 dollars, multiply the given sum by its 
 value per cent, less than 100 per cent, and divide the product by 
 100 ; then subtract the quotient from the given sum, the remain- 
 der will be the value required. 
 
 EXAMPLES. 
 
 1. What is the value of $7500 United States Bank Stock, at 112J 
 per cent. ? 
 
 Ans. $8437-50. 
 
 2. What is the value of $5400 of Stock, at 97 per cent. ? 
 
 Ans. $5238. 
 
 310. What are Stocks 1 211. What is the rule for finding the value of any kindof 
 
 stock, the value per cent, being given 1 
 
INSURANCE....SIMPLE INTEREST BY DECIMALS. 119 
 INSURANCE. 
 
 INSURANCE is a security by contract, to indemnify for a speci- 
 fied sum, the insured for such loss or damage as may happen to 
 the property, for a limited time. 
 
 The premium is the sum paid by the insured for the insurance 
 of his property, and is generally at so much per cent. 
 
 RULE. Multiply the sum to be insured by the rate per cent, 
 and divide the product by 100 ; the quotient will be the premi- 
 
 um or answer. 
 
 EXAMPLE. 
 
 What will be the annual premium for insuring a house against loss 
 by fire, valued at $5600, at J per cent. ? Ans. $42. 
 
 SIMPLE INTEREST BY DECIMALS. 
 
 A TABLE OF RATIOS FROM ONE DOLLAR, &C. TO TEN DOLLARS. 
 
 Ratio is the simple interest of $1 or 1 for one year, at the rate per cent, 
 agreed on, and is found by dividing the rate by 100, and reducing it 
 to a decimal. Thus T f T = -06, and T | = '05. 
 
 I Rate per cent, ratios. | Rate per cent, ratios. | Kate per cent. ratios. j 
 
 1 
 
 01 
 
 4 
 
 04 
 
 7 
 
 07 
 
 14 
 
 0125 
 
 44 
 
 0425 
 
 74 
 
 0725 
 
 ij 
 
 015 
 
 
 045 
 
 
 075 
 
 -f 
 
 0175 
 
 4f 
 
 0475 
 
 7J 
 
 0775 
 
 2 
 
 02 
 
 5 
 
 05 
 
 8 
 
 08 
 
 24 
 
 0225 
 
 54 
 
 0525 
 
 84 
 
 0825 
 
 2 i 
 
 025 
 
 
 055 
 
 
 085 
 
 2| 
 
 0275 
 
 5? 
 
 0575 
 
 8 S 
 
 0875 
 
 3 
 
 03 
 
 6 
 
 06 
 
 9 
 
 09 
 
 34 
 
 0325 
 
 6* 
 
 0825 
 
 94 
 
 0925 
 
 
 035 
 
 
 065 
 
 
 095 
 
 3| 
 
 0375 
 
 3 
 
 0675 
 
 9| 
 
 0975 
 
 
 
 10 
 
 1 
 
 212. Wkat is Insurance'* 213. Whn f is the premium ?- 
 
 for finding theprtrtiiwn, the rate per cent, being' given ? 
 
 -214. What is the rale 
 
120 
 
 SIMPLE INTEREST BY DECIMALS. 
 
 A TABLE for the ready finding of the decimal parts of a year, equal to 
 any number of months and days, allowing 12 months, or 365 days to 
 the year. 
 
 Days. 
 
 decimal parts. 
 
 l>ays 
 
 decimal parts. 
 
 Months. 
 
 decimal parts. 
 
 1 
 
 00274 
 
 30 
 
 082192 
 
 1 
 
 083333 
 
 2 
 
 005479 
 
 40 
 
 109589 
 
 2 
 
 166666 
 
 3 
 
 .008219 
 
 50 
 
 136986 
 
 3 
 
 25 
 
 4 
 
 010959 
 
 60 
 
 '164383 
 
 4 
 
 333333 
 
 5 
 
 013699 
 
 70 
 
 191781 
 
 5 
 
 416666 
 
 r 
 O 
 
 016438 
 
 80 
 
 219178 
 
 6 
 
 5 
 
 7 
 
 019178 
 
 90 
 
 246575 
 
 7 
 
 583333 
 
 8 
 
 021918 
 
 100 
 
 273973 
 
 8 
 
 666666 
 
 9 
 
 024657 
 
 200 
 
 547945 
 
 9 
 
 75 
 
 10 
 
 027397 
 
 300 
 
 -821918 
 
 10 
 
 833333 
 
 20 
 
 054794 
 
 365 
 
 1-000000 
 
 11 
 
 -916666 
 
 RULE. 
 
 1 . Multiply the principal by the ratio, the product will be the 
 interest for one year. 
 
 2. To find the interest for more years than one, multiply the 
 interest of one year by the given number of years, aud the pro- 
 duct will be the interest for that number of years. 
 
 3. To find the interest for months, or months and days ; multi- 
 ply the interest of one year by the decimal parts which are equal 
 to the given number of months, or months and days, and the pro- 
 duct will be the interest for that time.* 
 
 EXAMPLES. 
 
 1. What is the interest and amount of $475-50 for 4 years, 9 
 months, 25 days, at 6 per cent. ? 
 
 *This rule is a contraction of the general rule for simple interest. 
 215. fniflf iy the rule for computing simple interest by decimals ? 
 
COMPOUND INTEREST. 121 
 
 $ cts. 
 
 9 mo. = -75 475 50 Principal. 
 
 20 da = -054794 -06 Ratio. 
 
 5 da. =-013699 
 
 28-5300 Int. for one year. 
 Fwf, 4-818498 4*818493 Time. 
 
 855900 
 2567700 
 1141200 
 2282400 
 285300 
 2282400 
 1141200 
 
 Interest, 137-4716052900 
 Principal, 475-50 
 
 $612-9716 Amount, or answer. 
 
 2. If a factor sell a cargo for me, to the amount of $1750, on com- 
 mission at 2J per cent, and purchase me another cargo of the value of 
 $2500, on commission.at If per cent, j what will his commission on 
 both cargoes amount to ? 
 
 ADS. $83-125. 
 
 COMPOUND INTEREST. 
 
 COMPOUND INTEREST is that which arises from the interest be- 
 ing added to the principal at the end of each year, and making the 
 amount the principal for the next succeeding year. 
 
 RULE. Find the interest of the given sum for one year, and add 
 it to the principal, the amount is the principal for the second year ; 
 then find the interest of that amount, and add as before, and thus 
 proceed for any number of years required. 
 
 216 What is Compound Interest? 217. What is the rule for c*kultiin$ com'* 
 
 poiind interest ? 
 
 Q 
 
122 COMPOUND INTEREST, 
 
 Subtract the first principal from the last amount, and the re- 
 mainder will be the compound interest for the whole time.* 
 
 EXAMPLES. 
 
 1. What is the compound interest of $680 for 4 years, at 6 per 
 cent? 
 
 $ 
 680 given sum or first principal. 
 
 40-80 interest of do. for one year. 
 
 720 80 amount or principal for 2nd year. 
 43-248 interest of do. 
 
 764-048 amount or principal for 3rd year. 
 
 45.843 interest of do. 
 
 t 
 
 809 891 amount or principal for 4th year. 
 48-593 interest of do. 
 
 858-484 amount for 4 years. 
 680-000 first principal subtracted. 
 
 Ans. 178-484 compound interest for 4 years. 
 
 2. What is the compound interest of 400, for 3 years, at 5 per 
 nt.? Ans. 63 Is. 
 
 cent. 
 
 3. What will be the amount and compound interest of 1000 dollars 
 for 5 years, at 6 per cent. ? 
 
 A ( $1 338-226 amount. 
 ns ' I % 338-226 comp. int. 
 
 * It is not usually necessary to carry the work beyond mills ; therefore, when the fig- 
 ure next beyond mills, at the right is 5 or more than 5, increase the number of mills 1 ; 
 when it is less than 5, it may be omitted, and the result will be exact enough for commcm 
 purposes. 
 
COMPOUND INTEREST BY DECIMALS. 
 
 123 
 
 COMPOUND INTEREST BY DECIMALS. 
 
 TABLE showing the amount o/$l, or 1, at 5 and 6 per cent, per an- 
 num, compound interest, from I year to 20 years. 
 
 Years. 
 
 5 per cent 
 
 6 per ct nt. 
 
 Years. 
 
 5 per cent. 
 
 6 percent. 
 
 1 
 
 1-050000 
 
 060000 
 
 11 
 
 1-710339 
 
 1-898298 
 
 2 
 
 i- 102500 
 
 123600 
 
 12 
 
 1-795856 
 
 2-012J96 
 
 3 
 
 157625 
 
 191016 
 
 13 
 
 1-883649 
 
 2-132928 
 
 4 
 
 215506 
 
 262479 
 
 14 
 
 1-979931 
 
 2-260903 
 
 5 
 
 276281 
 
 338225 
 
 15 
 
 2-078928 
 
 2-396558 
 
 6 
 
 340095 
 
 418519 
 
 16 
 
 2-182874 
 
 2-547271 
 
 7 
 
 407100 
 
 503630 
 
 17 
 
 2-292018 
 
 2-692772 
 
 8 
 
 477455 
 
 5U3848 
 
 18 
 
 2-406619 
 
 2 854339 
 
 9 
 
 551328 
 
 689478 
 
 19 
 
 2-526950 
 
 3-025599 
 
 to 
 
 628894 
 
 790847 
 
 20 
 
 2-653297 
 
 3-207135 
 
 RULE 1. Multiply the given principal continually by the 
 amount of one dollar, or one pound, for one year, at the rate per 
 cent, given, until the number of multiplications be equal to the 
 given number of years the last product will be the amount for the 
 whole time ; from which subtract the given principal, and the re- 
 mainder will be the compound interest. Or, 
 
 RULE 2. Multiply the amount of one dollar, or one pound> 
 for the given number of years, by the given principal, the product 
 will be the amount required ; from which subtract the given prin- 
 cipal, and the remainder will be the compound interest. 
 
 EXAMPLES. 
 
 1. What will be the amount and compound interest of $500 for 3 
 years, at 6 per cent. ? 
 
 By rule 1. $500 X 1-06 X 1-06 X 1-06 = $595-508 Amt. 
 $595-508 500 = 95-508 Comp. Int. 
 
 By rule 2. Amount of one dollar for 3 years = 1-191016 
 
 principal = 503 
 
 Amount $595-508000 
 500 
 
 Comp. Int. $95-508000 
 
 218. What is the method of computing compound interest by decimals 1 
 
124 DUODECIMALS. 
 
 2. What will be the compound interest of 320 for 4 years, at 6 
 per cent. ? 
 
 Ans. 68 19s. 2d. 3-52 qrs. Rule I. 
 
 3. What will $50 amount to in 20 years, at 6 per cent, compound 
 interest ? 
 
 Ans. $160-35675. Rule 2. 
 
 4 What will be the compound interest of $1500 for 15 years, at 
 6 per cent. ? 
 
 Ans. $2094-837. Rule 2. 
 
 DUODECIMALS ; * 
 
 OR 
 
 CROSS MULTIPLICATION. 
 
 THE rule of Duodecimals is particularly useful to Workmen 
 and Artificers, in casting up the contents of their work 
 
 The denominations are feet, inches or primes, seconds, thirds, 
 fourths j fifths, &c there being no limit to the division. 
 
 \Z fifths, (marked '"") are 1 fourth, ("") 
 
 12 fourths -- 1 third, "('") 
 
 12 thirds --- 1 second, (" ) 
 
 12 seconds 1 inch or prime, (') 
 
 12 inches or primes - - - 1 foot (ft.) 
 
 Glaziers' and Masons' work is measured by the foot. 
 Painting, plastering, and paving are done by the yard. 
 Partitioning, flooring, slating, rough boarding, by the square of 
 100 feet. 
 
 Stone and brick work by the rod of 16J feet, whose square is 
 Bricks also are laid by the thousand. 
 
 * Duodecimals are a species of compound numbers, decreasing in an uniform ratio of 
 12. fiom a greater denomination to a less; hence their name. 
 
 219. What are [)u<,dr.nmalsl 220. What is the use of duodecimals 1 221. 
 
 What are the denominations used in duodecimals ? 
 
DUODECIMALS. 125 
 
 RULE. 
 
 1. Under the multiplicand write the corresponding denomina- 
 tions of the multiplier. 
 
 2. Multiply each term in the multiplicand, beginning at the 
 lowest, by the feet in the multiplier, and write the result of each 
 under its respective term, observing to carry an unit for every 12, 
 from each lower denomination to its superior. 
 
 3. In the same manner multiply the multiplicand, by the inches 
 in the multiplier, and write the result of each term in the multi- 
 plicand, thus multiplied, one place to the right hand in the product. 
 
 4. Proceed in the same manner with the other parts in the mul- 
 tiplier, which if seconds, write the result two places to the right 
 hand ; if thirds, three places, fyc. and their sum will be the answer 
 required. 
 
 N OTE> Feet multiplied by feet give feet Feet multiplied by inch- 
 es give inches Feet multiplied by seconds give seconds Inches 
 multiplied by inches give seconds Inches multiplied by seconds give 
 thirds Seconds multiplied by seconds give fourths. 
 
 EXAMPLE. 
 
 Multiply 7 feet, 3 inches, 2 seconds, by 1 foot, 7 inches and 
 seconds. 
 
 F. i. " 
 
 0^f"Here we multiply the 7f 
 Sin. 2" by the If. in the multiplier, 
 which gives seconds, inches and 
 Prod, by the feet, 7 3 2 "' feet. We next multiply the same 
 
 do. by primes, 4 2 10 2 "" 7f. Sin. 2" by the 7in. saying 7 times 
 do. by seconds, 1996 2 are 14 which is once 12 and 2 
 
 over, which (2) we set down one 
 
 11 7 9116 place to the right hand, that is, in 
 the place of thirds, and carry one 
 to the next place, and proceed in 
 
 the same manner with the other terms. Lastly, we multiply the mul- 
 tiplicand by the 3" saying 3 times 2 are 6, which we set down two 
 places to the right hand and so proceed with the other terms of the 
 multiplicand. The sum of all the products is the answer. 
 
 222, What is the rule ? 
 
126 DUODECIMALS. 
 
 APPLICATION AND USE OF DUODECIMALS. 
 
 /. To find the superficial contents of boards, $c. where length and 
 breadth only are considered. 
 
 RULE. Multiply the length by the breadth, and the product 
 will be the superficial content. 
 
 NOTE- If the board or plank is tapering, add the width of both 
 ends together, and take half the sum for the mean width, which mul- 
 tiplied by the length, will give the contents. 
 
 EXAMPLES. 
 
 1. How many feet in a beard 10 feet 7 inches long, and 9 incites 
 wide? 
 
 Ans. 7ft. 11' 3" 
 
 2. What number of feet are there in a floor 16 feet 6 inches loug. ; 
 and 12 feet 8 inches wide ? 
 
 Ans. 209 feet. 
 
 //. To find the solid content of timber, stone, bales, trunks, fyc. 
 
 RULE. Multiply the length by the breadth, and the .product 
 by the depth or thickness ; the last product will be the content 
 in solid or cubick feet, and parts of a foot. 
 
 EXAMPLES. 
 
 1. How many cubick feet in a stick of timber 12 feet 10 inches- 
 long, 1 foot 7 inches wide, and 1 foot 9 inches thick ? 
 
 Ans. 35ft. 6.' 8." 6.'" 
 
 2. Ho'vr many cubick feet, and perches of 24| feet, are there in a 
 cellar wall, 130 feet 8 inches long, 8 feet high, and 2 feet 9 inches 
 thick ? 
 
 Ans. 2874ft 8.'= 116 ^ perches. 
 
 NOTE. Bricklayers value their work at the rate of a brick and a 
 half thick; and if the thickness of the wall is more or less, it must 
 be reduced to that thickness, which may be done by the following 
 
 223. What is the rnh for meawring boards by duodecimals ? 224. What is tht 
 f /ule for fading the solid content 0f timber, stone, fyc. ? 
 
DUODECIMALS. 127 
 
 RULE.- Multiply the area of the wall by the number of the 
 lialf bricks in the thickness of the wall, the product divided by 3 
 will give the area. 
 
 JIL To measure drains, vaults, dikes, cellars, fyc. 
 
 RULE. Multiply the length, width and depth in feet together, 
 and divide by 216. 
 
 NOTE --Diggers work by the square of 6 feet long, wide and deep, 
 making 216 cubic feet to a square. 
 
 EXAMPLES. 
 
 1. The re is a drain 200 feet long, 3J feet wide, and 5} feet deep ; 
 how raany squares does it contain ? 
 
 Ans. 18-3- squares. 
 
 2. How many squares are in a vault 8 feet square, and OJ feet 
 deep? 
 
 Ans. 2f f squares. 
 
 IV. To measure wood. 
 
 RULE. Multiply the length by the width, and the product by 
 the height, the last product will be the content in cubick feet, 
 and parts of a foot, which are brought into cords by dividing them 
 by 128, or into cord wood feet by dividing by 16. 
 
 NOTE. A cord of wood is a pile 8 feet long, 4 feet wide, and 4 feet 
 high, containing 128 cubick feet, or 8 feet of cord wood. A foot of 
 cord wood is a pile 4 feet long, 4 feet wide, and 1 foot high, contain- 
 ing 16 cubick feet. 
 
 EXAMPLES. 
 
 I. How many cords areHhere in a pile of wood 176 feet long, 
 feet 9 inches wide, and 4 feet 3 inches high? 
 
 Ans. 21]i| cords. 
 
 225. What is the rule for measuring drains, vwlfs, &>c, 1 226. What is the rule 
 
 Jor measuring wood duodecimally ? 
 
128 
 
 DUODECIMALS. 
 
 2. How many cord wood feet are there in 41 load of wood 8 fee? 
 long, 3 feet 10 inches wide, arid 5 feet 4 inches high? 
 
 Ans. lOf feet. 
 
 3. How many cord wood feet are there in a pile 16 feet long, 5 
 feet 3 inches wide, and 4 feet 1 inch high ? 
 
 NOTE 2. As the superficial contents of one end of a con! of wood 
 are exactly double to the number of cord wood feet, therefore, in or- 
 der to find the number of feet of cord wood, in any load, multiply the 
 height by the width, duodecimally, and divide the product by 2, the 
 quotient will be the number of cord wood feet, and parts of a foot, 
 which the load contains. 
 
 EXAMPLES. 
 
 1. How many feet of wood in a 
 load 3ft. Gin. high, and 3ft. Gin. 
 wide ? 
 
 F. ' 
 
 3 6 
 3 6 
 
 10 6 
 1 9 
 
 2)12 3 
 65 Ans. 
 
 2. How many feet of wood in a 
 load 4ft. Sin. high, and 3ft. 5in, 
 wide ? 
 
 2)14 6 3 
 7J Ans. 
 
 these examples, each load i considered as consisting of two tiers, each of which 
 is supposed to be cut four feet long, according to law. 
 
 NOTE 3. After having multiplied the height and width of any load 
 of wood together, the figui/es which occupy the place of inches* in the 
 product, are not twelfth parts of a foot, because, as they are to be di- 
 vided by two* they are only twenty-fourths of a foot; therefore. 3 is-J, 
 4 is , 8 is J, 8 is J and 9 /'is . When the figures in the place of inch- 
 es happen to be 5, 7, 10, /or 11, as these figures are not even parts of 
 24, we call 5, J- ; 7, ; 1/0, f ; 11, f ; or J, as the case may be; that 
 is, if the figure in the tnbd place be less than 6, we call 11, f ; but, 
 if it be more than 6, theiji we call 1 1, . 
 
 The figures in the tlilird place are so inconsiderable, that they are 
 sot reckoned into the Contents of a load. 
 
DUODECIMALS. 
 
 129 
 
 3. How many feet of cord 
 wood, in a load 3ft. 9in. high, 
 and 2ft. lOin. wide 2 
 
 F. ' 
 
 3 9 
 
 2 10 
 
 3)10 7 6 
 5 Ans. 
 
 4. How many feet of cord 
 wood, in a load 4ft. 7in. high, and 
 3ft. Sin. wide ? 
 
 F. ' 
 
 4 7 
 3 8 
 
 2)16 9 8 
 8f Ans. 
 
 In the above examples, the 7, which occupies the place of inches 
 In the one, I call J of a foot, although it is in reality one twenty-fourth 
 part of a foot more than a quarter. ' The 9, which occupies the place 
 of inches, in the other example, is exactly f of a foot. 
 
 The 6, which possesses the third place, in one example, and the 
 8, in the other, are not reckoned into the quantity. 
 
 5. How many feet of cord wood in a load 4ft. 4in. high, and 3ft. 
 tin. wide? 
 
 In this example 13 divided by 2 
 are 6J, and 4, which occupies the 
 place of inches, is i of a foot; 
 therefore J which is |, being ad- 
 ded to J, produces f , which is ex- 
 actly f. 
 
 2)13 4 4 
 6 Ans. 
 
 6. How many feet of cord wood are there in a load which is 3f 
 ilin. high, and 3 ft. lOin wide ? 
 
 Ans. 7J. 
 
 NOTE 4. When wood is cut less, or more, than 4ft. long, find the 
 contents of the load by the foregoing examples ; then deduct or add, 
 as the case may require, so many forty-eighths of a foot, as the num- 
 ber of feet in the load will produce when multiplied by the number 
 of inches it falls short, or overruns. 
 
 7. How many feet of cord wood are there in a load 4ft. high, and 
 3ft. wide, and cut only 3ft. 9in. long ? 
 
 227. When wood is cut more or less than 4ftet long, how doyoujind the contents of a 
 
130 SINGLE RULE OF THREE. 
 
 F. ' 
 
 In this example, the contents of 
 
 3 the load, in case it were cut 4ff, 
 
 long, is 6ft. Euf as it lacks Sin. 
 
 2)12 of 4ft. multiply 3 by 6, and the 
 
 product is 18, which is eighteenfor- 
 
 6 ty-cighths ; and || = , which be- 
 
 f deduct. ing deducted from 6, leaves 5f , 
 
 the real quantity of the load. 
 5f Ans. 
 
 8. How many feet of cord wood in a load, 4ft. Sin. high, and 3ft. 
 Gin. wide, and cut 4ft. 7in. long ? 
 
 Ans. 8f . 
 
 9. How many feet of cord wood in a load, 3ft. 7in. high, and 3ft. 
 Sin. wide, and cut 3ft. Gin. long ? 
 
 Ans. 5|. 
 
 10. How many feet of cord wood in a load, 4ft. Sin. high, and 3ft 
 Sin. wide ? 
 
 Ans. 71-. 
 
 THE SINGLE RULE OF THREE. 
 
 THE SINGLE RULE OF THREE teaches how to find a fourth num- 
 ber, proportional to three numbers given ; for which reason it is 
 sometimes called the RULE OF PROPORTION. It is called the 
 Rule of Three, because three terms or numbers are given, to find 
 a fourth. And because of its great and extensive usefulness, it is 
 often called the GOLDEN RULE. 
 
 This Rule is usually considered as of two kinds, viz. Direct and 
 Inverse : a distinction, however, which is totally useless, and which 
 has been avoided by some of the best writers It may not be 
 ar: i-s, however, to exp.ain the difference usually understood be- 
 tween the two parts of this rule. 
 
 The Rule of Three Direct is that in which more requires more, 
 or less requires less. As in this ; if 3 men dig 21 yards of trench 
 in a certain time, how much will 6 men dig in the same time ? 
 Here more requires more, that is, 6 men, which are more than 3 
 
 218. What are w? to understand by the Single Rule of Three? 229. Why is it 
 
 called the Rule of Three, or the Golden Rule 1 230. ft it of more kinds than one * 
 
SINGLE RULE OF THREE. 131 
 
 men, will also perform more work in the same time. Or, when 
 it is thus : if 6 men dig 42 yards, how much will 3 men dig in the 
 same time ? Here then less requires less, or 3 men will perform 
 proportionally less work than 6 men, in the same time. In both 
 these cases then, the Rule, or the Proportion, is Direct ; and the 
 stating must be 
 
 thus, As 3 : 21 : : 6 : 42, 
 or thus, As 6 : 42 : : 3 : 21. 
 
 But the Rule of Three Inverse, is when more requires less, or 
 less requires more. As in this ; if 3 men di^ a certain quantity 
 of trench in 14 hours, in how many hours will 6 men dig the like 
 quantity ? Here it is evident that 6 men, being more than 3, will 
 perform an equal quantity of work in less time or fewer hours. 
 Or thus : if 6 men perform a certain quantity of work in 7 hours, 
 in how many hours wiii 3 men perform the same ? Here less re- 
 quires more, for 3 men will take more hours than 6 to r>e form the 
 same work. In both these cases then the Rule, or the Proportion, 
 is Inverse ; and the stating must be 
 
 thus, As 6 : 14 : : 3 : 7, 
 or thus, As 3 : 7 : : 6 : 14. 
 
 And in all these statings, the fourth term is found, by multiply- 
 ing the 2d and 3d terms together, and dividing the product by the 
 1st term. 
 
 Of the three given numbers, two of them contain the supposi- 
 tion, and the third a demand. And for stating and working ques- 
 tions of these kinds, observe the following 
 
 GENERAL RULE. 
 
 1. Write that number which is of the same name or kind with 
 the answer or number required, for the second term. 
 
 2. Then consider whether the answer must be greater or less 
 than the second term. If the answer must be greater than the 
 second term, write the greater of the two remaining numbers on 
 the right for the third term, and the other on the left for the first 
 term ; but if the answer must be less than the second term, 
 write the less of the two remaining numbers on the right for the 
 third term, and the other on the left for the first term. 
 
 3. Multiply the second and third terms together, divide by the 
 first, and the quotient will be the answer to the question, which, 
 
 231. What is the nature of these useless distinctions ? 232. What is the general 
 
 rult for stating and performing all questions in simple proportion 1 
 
132 SINGLE RULE OF THREE. 
 
 (as also the remainder) will be in the same denomination in whiefe 
 you left the second term, and may be brought into any other de- 
 nomination required.* 
 
 NOTE. The chief difficulty that occurs in the Rule of Three, is the 
 right placing of the numbers, or stating of the question; this being 
 accomplished, there is nothing to do, but to multiply and divide, and 
 the work is done. 
 
 To this end the nature of every question must be considered, and 
 the circumstances on which the proportion depends, observed, and 
 common sense will direct this ii the terms of the question be under- 
 stood. 
 
 1. The fourth number is always found in the same name in which 
 the second is given, or reduced to ; which, if it be not the highest 
 denomination of its kind, reduce to the highest, when it can be done. 
 
 2. When the second number is of divers denominations, bring it to 
 the lowest mentioned, and the fourth will be found in the same name 
 
 * This rule is founded on the obvious principle, that the magnitude or quantity of any 
 effect, varies constancy in proportion to the varying pait of the cause : Thus, the quan- 
 tity of goods bought, is in proportion to the money laid out; the space gone over by nn 
 uniform motion, is in proportion to the tiiiK., &c. 
 
 It has been shewn in Multiplication of Money, that the price of one multiplied by the 
 quantity, is the price of the whole; and in Division, that the price of the whole divided 
 by the quantity, is the price of one. Now, in all cases of valuing goods, &c. where one 
 is the first term of the proportion, it is plain, that Jie answer fount! by this rule, will be 
 the same as that found by Multiplication of Money; and where one is the last term of 
 the proportion, it will he the same as that found by Division of Money. 
 
 In like manner, if the first term be any number whatever, it is plain that the product 
 of the second and third terms will be greater than the true answer required, by as much 
 as the price in the second term exceeds the price of one, or as the first term exceeds an 
 unit; consequently this product, divided by the first term, will give the true answer re-> 
 quired. 
 
 NOTE 1 When it can be done, multiply and divide as in Compound Multiplication, 
 and Compound Division. 
 
 2. If the first term, and either the second or thiid can be divided by any number with- 
 out a remainder, let them be divided, and the quotient used instead of them. 
 
 Q^T" The following methods of operation, whtn they can be used, perform the work in a 
 much shorter manner than the general rule. 
 
 1. Divide the second term by the first; multiply the quotient into the third, and the 
 product will be the answer. 
 
 2. Divide the thud terra by the first; multiply the quotient into the seoond, and the 
 product will be the answer. 
 
 3. Divide the first term by the second, and the third by that quotient, and the last quo- 
 tient will be the answer. 
 
 4 Divide the first term by the third, and ihe second by that quotient, and the last quo- 
 tient will be the answer. 
 
 2:3. What is the reason of this rule ; or why will multiplying' the second 
 terms together, fyc. produce tin trve answer'? 
 
SINGLE RULE OF THREE. ' 133 
 
 to which the second is reduced, which reduce back to the highest 
 possible. 
 
 3 If the first and third be of different names, or one or both of 
 divers denominations, reduce them both to the lowest denomination 
 mentioned in either. 
 
 4. When the product of the second and third is divided by the 
 first; if there be a remainder after the division, and the quotient be 
 not the least denomination of its kind ; then multiply the remainder 
 by that number, which one of the saim* denomination with the quo- 
 dent contains of the next less, and divide this product again by the 
 first number ; and thus proceed till the least denomination be found, 
 or till nothing remain, 
 
 5. If the first number be greater than the product of the second 
 and third; then bring the second to a lower denomination. 
 
 6. When any number of barrels, bales, or other packages or pie- 
 ces are given, each containing an equal quantity, let the content of 
 one be reduced to the lowest name, and then multiplied by the ghee 
 number of packages or pieces. 
 
 7. If the given barrels, hiiiee, pieces, &,c. be of unequal contents^ 
 (as it most generally happens) put the separate content of each prop- 
 erly under one another, then add them together, and you will have 
 the whole quantity. 
 
 8. When any of the terms are given in Federal Money, the opera- 
 tion is conducted ia all respects as in simple numbers, observing only 
 to place the point or separatrix between dollars and cents, to point 
 off the results according to what has been taught already in Fedeial 
 Money, and Decimal Fractions. 
 
 PROOF. The method of proof is by inverting the question. 
 That is, put the fourth number, or answer, in the first place ; the 
 third in the second ; and the second in the third ; work as before 
 directed, and the quotient will be the first number. 
 
 EXAMPLES. 
 
 1. If 4 yards of cloth cost 16 dollars, what will 12 yards come to, 
 at the same rate ? 
 
 yds. $ yds, 
 
 4 : 16 :: 12 (7- In this question the answer 
 
 12 must be money, therefore we write 
 
 16 dollars for the second term. As 
 
 4)192 12 yards must cost more than 4 
 
 yards, we write 12 yards on the 
 
 $48 Ans. right of 16 dollars for the third 
 
 term, and 4 yards on the left for 
 the first term. 
 
 234 -. hen. tl>e first, second, or third terms are of different denominations. hw d* 
 yojt proceed I 235. What is the method of proff? 
 
134 SINGLE RULE OF THREE. 
 
 2. Proof of Example 1. If 48 dollars will pay for 12 yards of 
 cloth, how many yards can he bought for 16 dollars? 
 
 $ yds. $ 
 
 (r In this question the answer 
 
 12 must be yards ; therefore, we write 
 
 12 yards for the second term. As 
 
 48)192(4 yds. Ans. 16 dollars will pay for a less num- 
 
 192 ber of yards than 48 dollars, we 
 
 write 16 dollars on the right of 
 
 the 12 yards for the third term, and 48 dollars on the left for the first 
 term. 
 
 3. If 5 yds. of cloth cost $10, what will 20 yds, come to ? 
 
 yds. $ yds. 
 
 As 5 : 10 : : 20 
 
 H 
 
 $40 Ans. 
 
 ere we divide the 2d term by the Ibt, and multiply the quotient 
 into the 3d, for the answer. 
 
 yds. $ yds. 
 As 5 : 10 : : 20 
 
 40 Ans. 
 
 Here we divide the 3d term by the 1st, and multiply the quotient 
 into the 2d, for the answer. 
 
 4. If 20 yds. cost 120, how many yards may we have for 30 ? 
 
 $ yds. $ 
 As 120 : 20 : : 30 
 
 120-r-20=6 quot. and 30-^6=5 yards, Answer. 
 Here we divide the 1st term by the 2d, and then, the 3d term by 
 the quotient for the answer. 
 
 $ yds. $ 
 Again, As 120 : 20 : : 30 
 120_:_30^4 quot. and, 20-^-4=5 yards, Ans. 
 
 Here we divide the 1st term by the 3d, and then, the 2d term by 
 that quotient for the answer. 
 
SINGLE RULE OF THREE. 
 
 135 
 
 5. If 1 cwt. of tobacco cost 5 1 2s. 9 Jd. ; what will 8 cwt. do. cost ? 
 
 cwt. s. d. cwt. 
 As 1 : 5 12 9 : : 8 
 8 
 
 Ans. 45 2 4 
 
 Here there is no need of reducing the middle term, because it 
 can be performed by compound multiplication, the first term being 
 an unit. 
 
 6. If 8 cwt. of tobacco cost 45 2s. 4d. ; what is that per cwt? 
 
 9. d. 
 8)45 2 4 
 
 Ans. 5 12 
 
 Here there is no need of reducing the middle term, because it may 
 be performed by compound division only, the 3d term being an unit. 
 
 7. If 2 cwt. 1 qr. 14 Ibs. of sugar be worth g21-75, what will be 
 the value of 42 cwt. 3 qrs. at the same rate? 
 Cwt. qr. Ib. $ c. Cwt. 
 
 2 1 14 : 21-75 : : 42 
 
 4 4 
 
 qrs. 
 
 3 
 
 9 qrs. 
 
 76 
 19 
 
 266 Ibs. 
 
 4788 Ibs. 
 2175 
 
 171 qrs. 
 
 O^In this example, the 
 first and third terras are com- 
 pound numbers, therefore, we 
 reduce them both to the least 
 denomination mentioned in ei- 
 ther, viz. pounds. The se- 
 cond term is reduced to the 
 least denomination mentioned 
 in it, by calling it all cents. 
 We then multiply the third 
 term by the second, and divide 
 the product by the first term, 
 
 $ c. and the quotient is 39 150 cents, 
 
 266)10413900(391-50 Ans. which we reduce to dollars by 
 
 798 pointing off the two right 
 
 hand figures. 
 
 2433 
 
 2394 
 
 23940 
 33516 
 4788 
 9576 
 
 399 
 266 
 
 1330 
 1330 
 
136 SINGLE RULE OF THREE. 
 
 8. If 9 cwt, 3 qrs. of sugar cost 21 17s. 6d., what will 2 cwt. lor. 
 1 Ib. cost ? 
 
 Cwt- q- . d. Cwt. qr. Ih. 
 
 93 27 17 6 211 
 
 4 20 4 
 
 39 557 "9 
 
 28 12 28 
 
 312 6690 73 
 
 78 19 
 
 2092 263 
 
 Ib. d. Ib. 
 
 As 1092 : 6690 : : 263 
 263 
 
 1092)1759470( 1611 
 1092 - 
 
 -- 2lO)13|4 3d. 
 6674 -- 
 
 6552 6 14s. 3d. Answer. 
 
 1227 
 1092 
 
 1350 
 1098 
 
 258 
 4 
 
 1092)1 032(0 qr. 
 
 CfHere it will be seen, that the first and third terms are of the 
 same kind, but of different denominations, and therefore are reduced 
 to the same name or denomination, and that the demand of the ques- 
 tion lies on the 3d term. 
 
 2 That the middle term, being 1 given in pounds, shillings and 
 pence, is reduced to pence. But 
 
 3. If the second term were in federal money it would be sufficient 
 <3 proceed according to decimals. 
 
SINGLE RULE OF THREE. 137 
 
 % 
 
 3. It my income be 109 guineas per annum, I desire to know what 
 1 may spend per day, so that I may lay up 45 at the year's end? 
 
 Ans. o 5s. IGjd. ^ per day. 
 
 NOTE 1. You must subtract 45 from the value of 109 guineas. 
 
 2. There being 365 days in a year, your question must next be 
 stated thus : 
 
 D. Gmn. D. . d. qrs. 
 As 365 : 10945 : : 1 : 5 10 3^ the Ans. 
 
 10. If my salary be .43 12s. 5d, per annum, what does it amount 
 to per week ? 
 
 Ans. 16s. 
 
 The Stating. NOTE. As there are 52 weeks 
 
 W. s, d. W. and 1 day in a year, you will get 
 
 As 52 : 43 12 5 : : 1: the Ans. the true answer to the above ques- 
 
 tion by the following ratio. 
 
 D s. d. D 
 As 365 : 43 12 5 : : 7 : 16s. Sfffd. 
 
 11. Suppose my income to be 16s. 8|ffd. per week, what is it per 
 annum. 
 
 Ans. 43 13s. 
 
 NOTE 1. You must first reduce the middle term to pence. 
 
 2. You must multiply by 365 (the denominator of the fraction) and 
 add to the product the 283 which remains ; and remember always to 
 do so in similar cases. 
 
 3. You must divide by 7, the first terra and the quotient will be the 
 answer in 365ths of a penny, which (in all similar cases) must be first 
 divided by the denominator, and then brought into pounds. 
 
 12. How many yards of cloth may be bought for $195 75c. of 
 which 9J yds. cost $11 2c. ? 
 
 Ana. 168 yds. 3 qrs. 
 
 13. A merchant, failing in trade, owes in all 3475, and has in mon- 
 ey and effects but 2316 13s. 4<1. : Now, supposing his effects are 
 delivered up, pray, what will each creditor receive on the pound ? 
 
 s. d. 
 
 As 3475 : 2316 13 4 :: 1 : 13s. 4d, Ans 
 
138 SINGLE RULE OF THREE. 
 
 14. If the distance from Newburyport to York be 31 miles; I de- 
 mand how many times a wheel, whose circumference is 15J feet will 
 turn round in performing the journey ? 
 
 Ans. 10560 times. 
 
 15. What will 37J gross of buttons come to at 13 cents per dozen ? 
 
 Ans. $58 50c. 
 
 16. If T 7 F of a ship cost $1163 what is the whole worth ? 
 
 Ans. $2658 28c. 5m. 
 
 1 7. A merchant bought 9 packages of cloth, at 3 guineas for 7 yards : 
 each package contained 8 parcels, each parcel 12 pieces, and each 
 piece 20 jards ; how many dollars c&me the whole to, and how many 
 per yard ? 
 
 Yds guin pack. $ 
 As 7 : 3 : : 9 : 34560 Ans. for the whole cost. 
 
 yds. guin. yd, $ 
 As 7 : 3 : : 1 : 2 Ans. per yard. 
 
 18. A merchant bought 49 tuns of wine for $9 10; freight cost $90 ; 
 duties $10; cellar $31 67c. ; other charges $50 and he would gain 
 $185 by the bargain ; what must I give him for 23 tuns ? 
 
 Tuns. $ $ $ $ c. $ $ Tuns. $ 
 As 49 : 910+904-40+31 67+50+185 : : 23 : 613 33c. Ans. 
 
 19. 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 43d 
 parallel of latitude ; the degree of longitude, in this latitude, being 
 about 51 statute miles? 
 
 H. D. M. M. M. 
 
 As 24 : 360 X 51 :: 1 : 12f Ans. 
 
 20. If a staff, 4 feet long, cast a shade (on level ground) 7 feet, 
 what is the height of that steeple, whose shade, at the same time 
 measures 198 feet?* 
 
 F sh. F. hei. F. sh. F. hei. 
 As 7 : 4 :: 198 : 113| Ans. 
 
 *As the rays oflight from the sun may be considered parallel, line lengths of the shad- 
 ows must bo proportioned to the heights of the objects. Hence the reason of the state- 
 ment of this question. 
 
SINGLE RULE OF THREE. 139 
 
 21. Said Harry to Dick, my purse and money are worth 3J guin- 
 eas, but the money is w6rth eleven times as much as the purse ; pray, 
 how much money is there in it ? 
 
 Ans. 4 3s. 5d. 
 
 22. If f of a yard cost of a , what will T 7 T of a yard cost ?* 
 
 : : : T 7 T : |X T V4=ttf Answer. 
 Or, | : f :: T 7 T : |XiX T y^fi = l 7s. IJJd. 
 
 23. There is a cistern, having four cocks ; the first will empty it in 
 ten minutes ; the second in 25 minutes ; the third in 40, and the fourth 
 in 80 minutes j in what time will all four, running together empty it ? 
 
 Cist. Min. (o ) Cist. Min. Cist. Min. 
 
 : 1 : : 60 : HA As llj : 60 : : 1 : 5J Ans. 
 /|V 45 60X4 
 
 ^ * } thatis : 60 : : 1 : -b\. 
 
 4 45 
 
 11 J Cist. 
 
 24. If the earth revolves 366 times in 365 days, in what time does 
 it perform one revolution ? 
 
 Ans. 23h. 56' 3" 56'" -f 1 Sidereal day.t 
 
 25. If the earth makes one complete revolution in 23h. 56' 3"+, 
 in what time does it pass through one degree ? 
 
 Ans. 3' 55" 20'". 
 
 26. If the earth performs its diurnal revolution in a solar day,J or 
 24 hours ; in what time does it move one degree ? 
 
 Ans. 4'. 
 
 * If the first term of the statement be a Vulgar Fraction, whether the other terms are 
 or not, after the first and third terms are reduced to the same denomination, invert the 
 first term as in division of Vulgar Fractions, and the product of the three terms will of 
 course be the answer. 
 
 The student should work the questions in Vulgar, or Decimal Fractions, according as 
 the rules for fractions require. 
 
 f A sidereal day is the space of time which happens between the departure of a star 
 from, and its return to the same meridian again. 
 
 \ The solar day is that space of time which intervenes between the sun's departing from 
 any one meridian, and its return to the same again. 
 
140 METHOD OF MAKING TAXES. 
 
 27. If f yd. cost gj what will 40jyds cpme to ? 
 
 An?. $59 6c. 
 
 28. At $3f per cwt. what will 9|lb. come to ? 
 
 Ans. 31 c. 3m, 
 
 29 A sonduit has a cock, which will fill a cistern in -2 of an hour? 
 this cistern has 3 cocks; the first wil! empty it in 1-25 hour, the se- 
 cond in -625 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. 
 
 30. In a certain school, ^th of the pupils study Greek, j\ study 
 Latin. | study Arithmetick, ^ read and write, and 20 attend to other 
 things ; what is the. number of pupils ? 
 
 =T 8 7n then 2U=fr and ft : 20 : : - : JOO Ans. 
 
 GENERAL METHOD OF MAKING TAXES. 
 
 RULE.- In the first place an inventory of the value of all the es- 
 tates, both real and personal, and the number of polls for which 
 e::ch person is rateab e, 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 : 2dlv, make 
 a tabie, by multiplying the value on the dollar by 1,2, 3, 4, 
 &c. 3dly, From the inventory take the real and personal estates 
 of each man, and find them separately in the tabie, which will 
 shew you each man's proportional share of the tax for real and 
 personal estates. 
 
 NOTE. If any part of the tax is averaged on the polls, or otherwise, 
 before stating- 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 av- 
 erage you nrust have a separate column, as well as for the real and 
 personal estates. 
 
 236. What method would y<M& pursue were you to undertake the assessing <ff tmvn 
 
METHOD OF MAKING TAXES. 
 
 141 
 
 EXAMPLES. 
 
 1. Suppose the General Court should grant a tax of $500000, of 
 which the town of Portsmouth is to pay $5312 50c. and, of which the 
 polls, being 1550, are to pay $1 25c. each: The town's inventory 
 amounts to 460000, what will it be on the dollar, and what is A'a 
 tax, whose estate (as by the inventory) is as follows, viz. real $1376, 
 personal $1149, and he has 3 polls ? 
 
 Pol. $ c. Pol. $ c. 
 
 First, As 1 : 1 25 :: 1550 : 1937 50 the average part of the tax 
 to be deducted from $5312 50c. and there will remain $3375. 
 
 $ $ $ 
 
 Secondly, As 450000 : 3375 : : 1 
 
 7jm. on, the dollar. 
 
 $ $ c. m. 
 
 1 is 7J 
 
 2 1 5 
 30 2 2 J 
 
 4 3 
 
 5 _ o 3 7 J 
 
 6 4 5 
 70 5 2 
 80 6 0" 
 9 6 74 
 
 10 7 5 
 
 TABLE. 
 
 $ $ c. m. 
 
 20 is 15 
 
 300 22 5 
 
 40 
 
 30 
 
 500 37 5 
 
 600 45 
 
 700 52 5 
 
 80 ' 60 
 
 90 67 5 
 
 1000 75 
 
 $ $ c. 
 
 200 is 1 50 
 
 300 2 25 
 
 400 3 00 
 
 500 3 75 
 
 600 4 50 
 
 700 5 25 
 
 800 6 00 
 
 900 6 75 
 
 1000 7 50 
 
 Now to find what A's rate will be, 
 His real estate being $1370, we. 
 
 Real. 
 $c, m. 
 
 Personal. 
 $ c. m. 
 
 Polls. 
 $c. m. 
 
 Total. 
 $ c. m. 
 
 find by this table, that $1000 
 
 jq $7 r,0r 
 
 10320 
 
 8 bl 7J 
 
 375 
 
 22687J 
 
 that $300 is - - - - 2 25 
 that $70 is - - - - 52 5m. 
 and that $6 is - - - 45 
 
 
 
 
 
 For his real estate $10 32 
 
 
 
 i 
 
 
 In like manner we find his tax } 
 foi 1 personal estate to be $ 
 
 His 3 polls, at $1 25c. each are 
 
 61 7 
 
 i* 
 
 3 75 $ 
 
 +$1032=$2268c. 7ji 
 or, $22 69c. Ans. 
 
 2. Suppose a tax of $755 be laid on a town, and the inventory of 
 all the estates in town amounts to $9345, what must A. pay, whose 
 estate is $149? 
 
 Ans. $12*127. 
 
142 RULE OF THREE DIRECT. 
 
 RULE OF THREE DIRECT. 
 
 THE RULE OF THREE DIRECT teaches, by having three num- 
 bers given, to find a fourth, which shall have the same ratio to the 
 second, as the third has to the first. 
 
 RULE. 
 
 1. State the question by making that number, which asks the 
 question,* the third term ; that which is of the same name or 
 kind, the first term ; and that which is of the same name or kind 
 with the answer, the second term. 
 
 2, Multiply the second and third terms together, and divide the 
 product by the first term, and the quotient will be the fourth 
 term, or answer. 
 
 The notes under the general rule are applicable to this rule. 
 
 EXAMPLES. 
 
 1. If 3 bushels of corn be worth $1-80, what is the value of 1 2 
 bushels ? 
 
 bu. $ c. bu. 
 3 : 1-80 : : 12 
 12 
 
 9)2160 
 
 #7-20 Ans. 
 
 In this example, 12 bushels asks the question, and is made (he 
 third term; 3 bushels being of the same name, is made the first term ; 
 and $1-80 being of the same name with the answer, is made the se- 
 cond term. Here the third term is greater than the first, and the 
 question evidently requires the fourth term or answer to be greater 
 than the second ; therefore, the question belongs to the Rule of Three 
 Direct. 
 
 2. If 6 Ibs. sugar cost 10s. what will 33 Ibs. cost at the same rate ? 
 
 Ans. 2 15s. 
 
 *The term which asks or moves the question has generally some words like these be- 
 fore it, viz. What will? What cost? How many ? How far ? How much ? How 
 long ? &c. 
 
 237. What is the method / operation in the Rule of Three Direct ? 
 
DOUBLE RULE OF THREE. 143 
 
 RULE OF THREE INVERSE ; 
 
 OR, 
 
 RECIPROCAL PROPORTION. 
 
 THE RULE OF THREE INVERSE teaches, by having three num- 
 bers given, to find a fourth, which shall have the same ratio to the 
 second, as the first has to the third. 
 
 RULE. State and reduce the terms as in the Rule of Three 
 Direct ; then multiply the first and second terms together, and 
 divide the product by the third, the quotient will be the fourth 
 term, or answer. 
 
 EXAMPLES. 
 
 1. If 6 men can do a piece of work in 18 days, in what time can 
 12 men do it ? 
 
 m, d. m. 
 
 6 : 18 : : 12 In this example, the third term is greater than 
 6 the first, yet it is evident, that the question re- 
 
 quires the fourth term or answer to be less than 
 
 12)108 the second; therefore, this question belongs to 
 
 the Rule of Three Inverse. 
 9 days Ans. 
 
 2. If a man perform a journey in 15 days, when the day is 12 hours 
 long, in how many days will he do it, when the day is but 10 hours ? 
 
 Ans. 18 days. 
 
 THE DOUBLE RULE OF THREE. 
 
 THE DOUBLE RULE,OF THREE, or COMPOUND PROPORTION, is 
 the method of resolving such questions as require two or more op- 
 erations by Simple Proportion. 
 
 It always consists of an odd number of terms given, as five, sev- 
 en^ &c, These terms are distinguished into terms of supposition, 
 and terms of demand ; the number of the former always exceeding 
 that of the latter by one, which is of the same name or kind with 
 the answer or term required. 
 
 238. How do you proceed in the Rule of Three. Inverse ? 239, What is Compound 
 
 Proportion, or the Double Rule of Three ? 240. How is it distinguished from Sim 
 
 pie Proportion ? 
 
144 DOUBLE RULE OF THREE. 
 
 RULE. 
 
 1 . State the question, by placing the three conditional terms 
 in such order that that number which is the cause of gain, loss, or 
 action, may possess the first place ; that which denotes space of 
 time, or distance of place, the second ; and that which is the gain, 
 loss, or action, the third. 
 
 2. Place the other two terms, which move the question, under 
 those of the same name. 
 
 3. Then, if the blank place, or term sought, fall under the 
 third place, the proportion is direct, therefore, multiply the 
 three last terms together, for a dividend, and the other two for a 
 divisor ; then the quotient will be the answer. 
 
 4. But if the blank fall under the first or second place, the 
 proportion is inverse, wherefore multiply tlie first, second and 
 last terms together, for a dividend, and the other two, for a divi- 
 sor ; the quotient will be the answer.* 
 
 EXAMPLES. 
 
 1. If a principal of $100 gain #6 interest in one year, what will 
 $ 500 gain in 8 months ? 
 
 Statement & operation. 
 $P $lni. $P. 
 
 100 : 6 : : 500 In this question, the answer required is 
 
 12 8 interest, therefore, $6 must be the middle 
 
 term. As $500 will gain more interest in 
 
 1200 4000 the same time than 1 100 ; $500 must be 
 
 6 placed on the right for the third term ; and 
 
 $100 on the left for the first term. Anil as 
 
 12100)240100 the same sum will gain more interest in 12 
 
 months than in 8 months, the 8 must be pla- 
 
 $20 Ans. ced under the third term, and the 12 under 
 the first term. The operation is obvious on inspecting it. 
 
 2. If $100 in 12 months will gain $6 interest; in what time will 
 
 i 750 gain $30 interest ? 
 
 Ans. 8 months. 
 
 The reasoH of this rule for stating, and of the method of operation, is evident from 
 the nature of Simple Proportion ; for every line in this case is a particular statement in 
 that rule. If, then, all the separate dividends be collected into one dividend, and all the 
 divisors into one divisor, their quotient must he the answer. 
 
 241. Repeat the rule ? 
 
CONJOINED PROPORTION. 145 
 
 3. If 7 men can reap 84 acres of wheat in 24 days ; how many men 
 Can reap 100 acres in 10 days ? 
 
 Ans. 20 men. 
 
 4. If 12 men can earn 3 8s. in 8 days ; what will 21 men earn in 
 15 days? 
 
 Acs. 27 11s. 3d, 
 
 5. If a family of 9 persons spend 450 dollars in 5 months; how 
 much would be sufficient to maintain them 8 months, if 5 persons 
 more were added to the family ? 
 
 Ans. 11 20 dolls. 
 
 CONJOINED PROPORTION. 
 
 CONJOINED PROPORTION is when the coins, weights, or meas- 
 ures 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 antecedents have to their consequents, the 
 proportion between the first antecedent and last consequent is 
 discovered, as well as the proportion between the others in their 
 several respects. 
 
 NOTE. This rule may generally be abridged by cancelling equal 
 quantities or numbers that happen to be the same in both columns ; 
 and it may be proved by as many stating? in the Single Rule of 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 antece- 
 dents at the left hand, and the consequents at the right, and write 
 the last number on the left hand ; then multiply all the numbers 
 in the left hand column continually together for a dividend ; and 
 all the numbers in the right hand column for a divisor ; divide, 
 and the quotient will be the answer. 
 
 242. What is the meaning of Conjoined Proportion 7 243. When you icish to 
 
 know how many of the first sort of coin, fyc. mentioned in any qvcst^on are equal to a giv- 
 en quantity in the last, how do you proceed ? 
 
 T 
 
146 CONJOINED PROPORTION. 
 
 EXAMPLES. 
 
 1. If 12 Ibs. of Boston be equal to 10 Ibs. of Amsterdam, and 10 Ibt, 
 of Amsterdam be equal to 12 Ibs. of Paris ; how man}' pounds of Bos- 
 ton are equal to 80 Ibs. of Paris ? 
 
 Antecedents. Consequents. 
 
 12 of Boston=:10 of Amster. , Then 12X10X80=9600 dividend; 
 10 of Amster.=:l2 of Paris. I and 10X12=120 the divisor; 
 80 of Paris ? I Therefore 9600120-80 Ibs. Ans. 
 
 2. If 140 braces of Venice be equal to 150 braces of Leghorn, and 
 7 braces of Leghorn be equal to 4 American yards ; how many Veni- 
 tian braces are equal to 32 American yards ? 
 
 Ans. 52^ braces. 
 
 CASE II. 
 
 When it is required to find how many of the last sort of coin, weight, or 
 measure mentioned in the question, are equal to a given quantity of 
 thefirst. 
 
 RULE. Place the numbers alternately, beginning at the left 
 hand, and write the last number on the right hand ; then multiply 
 all the numbers in the right hand column continually together for 
 a dividend, and all the numbers in the left hand column for a di- 
 visor ; divide, and the quotient will be the answer. 
 
 EXAMPLES. 
 
 1. If 12 Ibs. of Boston be equal to 10 Ibs. of Amsterdam, and 100 
 Ibs. of Amsterdam be equal to 1 20 Ibs. of Paris ; how many pounds of 
 Paris are equal to 80 Ibs. of Boston ? 
 
 Antecedents. Consequents. 
 
 12 of BostonzrlO of Amsterdam. I 10X120X80=96000 dividend ; 
 
 100 of Amsterdam=l 20 oi Paris. and 12X100=1200 divisor; 
 
 80 of Boston ?' Then 96000-^1200=80 Ibs. Ans. 
 
 2. If 140 braces of Venice be equal to 150 braces of Leghorn, and 7 
 braces of Leghorn be equal to 4 American yards; how many Ameri- 
 can yards are equal to 52 T 4 5 Venetian braces ? 
 
 Anfl. 32 yards. 
 
 244 How. when you wish to ascertain how many of the last are equal to a certain 
 quantity named in tliejirst ? 
 
PRACTICE. 
 
 147 
 
 PRACTICE.* 
 
 PRACTICE is a contraction of the Rule of Three Direct, when 
 the first term happens to be an unit or one ; it has its name from 
 its daily use among Merchants and Tradesmen, heing an easy and 
 concise method of working most questions which occur in trade 
 and business. 
 
 PROOF. By the Single Rule of Three, Compound Multiplica- 
 tion, or by varying the parts 
 
 Before any advances are made in this rule, the learner should 
 commit to memory the following 
 
 TABLES OP ALIQUOT, OR EVEN PARTS. 
 
 Mquot, or even pa ts of Money. 
 
 I'lrf. of a 
 
 shil. 
 
 oi a . 
 
 Parts of a Pound. 
 
 Parts of 
 
 a Dollar. 
 
 d. 
 
 s. 
 
 
 
 s. d. 
 
 c. 
 
 $ 
 
 6 - 
 
 4 
 
 ?V 
 
 10 = 4 
 
 50 
 
 = 4 
 
 4 = 
 
 4 
 
 = eV 
 
 68=4 
 
 33J 
 
 
 3 = 
 
 i 
 
 
 5 = i 
 
 25 
 
 = 4 
 
 2 
 
 
 = T "o 
 
 4 = i 
 
 20 
 
 
 1 A- 
 
 J. 
 
 Te o- 
 
 S 4 *~ 
 
 168- 
 
 JL 
 
 1 - 
 
 TV 
 
 _i_ 
 
 2 6 = I 
 
 12f 
 
 = 1 
 
 3 
 
 rV 
 
 "slo" 
 
 1 8 = iV 
 
 H 
 
 A 
 
 4 
 
 *'T 
 
 ? |^- 
 
 1 4 T 5 
 
 6^ 
 
 
 I = 
 
 iV 
 
 !__ 
 
 1 3 = A 
 
 5 
 
 aV 
 
 Parts 
 
 of 2 
 
 Shili. 
 
 1 = A 
 
 4 
 
 zr - 1 - 
 
 d. 
 
 
 2s. 
 
 10 = 2T 
 
 f,* 
 
 = ^ 
 
 1 
 
 
 
 A 
 
 8 = aV 
 
 2 
 
 
 1J 
 
 ~ 
 
 TV 
 
 5 = TV 
 
 1 
 
 = VT 
 
 2 
 
 
 
 JL 
 
 1 2 
 
 4 = A 
 
 
 
 3 
 
 
 
 i 
 
 3 = tV 
 
 
 
 4 
 
 
 
 i 
 
 02 = 7^0 
 
 
 
 6 
 
 = 
 
 1 
 
 
 8 
 
 *""* 
 
 J 
 
 
 * Perhaps no method can be more simple and concise to find the value of goods in 
 Federal Money, than thv, general rule of multiplying the price by the quantity, as given 
 in Multiplication of Federal Money or Decimals ; therefore, the application of this rule 
 to Federal Money is almost useless. Yet, as English merchants, trading with Americans, 
 make imt the invoices of their goods in sterling money, an acquaintance with this excel- 
 lent rule is necessary to every one, employed in mercantile pursuits. 
 
 245. What is Practice ? 246. Explain to me the use of the tablet 7 247. What 
 
 are aliguot parts of any quantity ? 
 
148 
 
 PRACTICE. 
 
 M'quot, or even parts of Weight. 
 
 Parts of a Cwt. Parts of Cvvt. 
 
 Q,rs. Ib. Cvvt.i Ib. & Cwt. 
 
 Parts of Cwt 
 Ib. I Cwt. 
 
 Parts of a Ton. 
 Cwt. qr. T. 
 
 2 - J 128 - 
 
 14 - J 
 
 10 - J 
 
 i o = 4 
 
 14 = i 
 
 ^ = i 
 
 5 = 1 
 
 16 = | 
 
 8 ^ 4 
 
 4 = T 
 
 4 - i 
 
 14 - ] 
 
 ^ - 4 
 
 O ] 
 
 TT 
 
 22 = ^ 
 
 8 - J- 
 
 4 = TV 
 
 
 2 = T -S 
 
 7 - 1 - 
 
 
 
 11 1 
 
 __ 4 = A 
 
 
 
 T 
 
 1 J^ 
 
 of Discount per cent . 
 
 s.d.^ 
 i^ percent. 3 
 
 O 
 
 s.d.^ 
 8fpercent.=l 9 
 
 10 c) o 
 
 c 
 
 p 
 
 s. d. ^ 
 22J per cent.=4 6 
 
 QPj~ f> O 
 
 
 
 j>3 r\ Q 
 
 o> 
 
 101 p. 
 
 n 
 
 SO ft A 
 
 
 
 c; i A 
 
 T! 
 
 1 F, ^ |) 
 
 *c 
 
 <2c; 7 n 
 
 
 fil 1 ^ 
 
 
 17 1 ^ fi 
 
 
 -in s A 
 
 *TD 
 o 
 
 7J =1 6J 
 
 C. 
 
 20 ^ 0, 
 
 ?- 
 
 45 =9 
 c;n in o 
 
 c 
 
 3 
 CL, 
 
 
 
 
 
 
 
 Practice admits of a great variety of rules, the multipli- 
 cation of which serves little else than to confuse the mind of the 
 scholar. The most useful are here given, and all that are actu- 
 ally necessary for the man of business ; but the scholar who wish- 
 es to pursue his studies beyond this elementary treatise, will find 
 the subject amply treated of in the large Arithmetick of PIKE. 
 
 CASE I. 
 
 To fold the value of any number of pounds, fyc. ichen the price of one 
 pound is an aliquot part of a dollar. 
 
 RULE. As the value of any number of pounds at one dollar a 
 pound, will be just so many dollars as there are pounds : there- 
 fore, call the pounds dollars, and divide them by such an aliquot 
 part as the price is of a dollar, the quotient will be the answer in 
 dollars. 
 
 248 How do you find the value of any number of pounds, fyc. when 'the price per lb, it 
 vn ali(juot part of a foliar ? 
 
PRACTICE. 
 
 149 
 
 EXAMPLES. 
 
 1. What will be the value of 360 
 yards of cloth, at *25 cents a 
 yard ? 
 
 cts. 
 
 g90 Ans. 
 
 3. What will be the value of 
 125 pounds of butter, at -12J cents 
 a pound ? 
 
 Ans. $ 15-625. 
 
 2. What will be the value of 75 
 pounds of loaf sugar, at -20 cents 
 a pound ? 
 
 cts. $ 
 
 20 = 1)75 
 
 $15 Ans. 
 
 4. What will 1250 pounds of 
 cheese come to, at -06J cents a 
 pound ? 
 
 Ans. $78-125. 
 
 CASE II. 
 
 To find the value of any number of yards, fyc. when the price of one 
 yard is an aliquot part of a pound. 
 
 RULE. Call the number of yards so many pounds, and divide 
 them by such an aliquot part as the price is of a pound, the quo- 
 tient will be the answer in pounds. 
 
 EXAMPLES. 
 
 1. What will 63 gallons of rum 
 come to, at 6s. 8d. a gallon ? 
 
 s. d. 
 6 8 = 
 
 21 Ans. 
 
 3. What is the value of 500 
 bushels of potatoes, at 2s. 6d. a 
 bushel ? 
 
 Ans. 62 10s. 
 
 5. What will be the value of 
 150 bushels of corn at 3s. 4d. a 
 bushel ? 
 
 Ans. 25. 
 
 2. What is the value of 125 
 yards of linen, at 5s. a yard? 
 
 s. 
 5=1)125 
 
 31 5s. Ans. 
 
 4. What will 37J yards of 
 sheeting come to, at Is. 8d. ;> 
 yard? 
 
 Ans. 3 2s. 6d. 
 
 G. What will 120 gallons of 
 wine come to, at 10 shillings a 
 
 gallon ? 
 
 Ans. 00. 
 
 249. Jfow do youftnd the value of a certain number of yards, when the price of on? 
 is an aliquot part of a pound ? 
 
150 
 
 PRACTICE. 
 CASE III. 
 
 To find the value of any number of pounds, fyc. when the price of one 
 pound is an aliquot part of a shilling. 
 
 RULE. Call the given number of pounds so many shillings, and 
 divide them by such an aliquot part as the price is of one shilling, 
 the quotient will be the answer in shillings. 
 
 EXAMPLES. 
 
 1. What will be the value of 
 650 pounds of cheese, at 6 pence 
 a pound ? 
 
 d. s. 
 
 6 - )650 
 
 2lO)3 c 2j5 
 16 5s. Ans. 
 
 3. What is the value of 37 J yds. 
 ef ribband, at 3 pence a yard ? 
 Aus. 9s. 
 
 2. What will 1364 pounds of pork 
 come to, at 4 pence a pound? 
 d. s. 
 
 4 = J)1864 
 
 2jO)62|l 4d. 
 
 31 Is. 4d. Ans. 
 
 4. What will 75J yards of tape 
 come to, at Ijd. a yard ? 
 
 Ans. 9s. 5jd. 
 
 CASE IV. 
 
 To find the value of any number of yards, fyc- when the price of on& 
 yard is shillings, or sJilllngs, pence and farthings, and not an ali- 
 quot part of a pound. 
 
 RULE. Write down the given number of yards, &c. and call 
 them so many pounds. Divide the given price into aliquot parts, 
 either of a pound, or of one another ; then divide by the several 
 aliquot parts, and the sum of the quotients will be the answer in 
 pounds. 
 
 EXAMPLES. 
 
 1. What will he the value of 36 
 yards of silk vesting, at 7s. 6d. a 
 vard ? 
 
 s. 
 5 = J)36 
 
 2. What will 72 yards of Casf 
 mere come to, at 15 shillings 
 yard ? 
 s. 
 
 2s.6d.= )9 
 4 10s. 
 
 5 = ^)36 
 18 
 
 
 13 10s. Ans. 
 
 54 Ans. 
 
 
 250. How, when the price of a pound is an aliquot part of a shilling ? 
 
PRACTICE. 
 
 151 
 
 3. What is the value of 139 gal- 
 lons of wine, at 9s. lOd. a gallon ? 
 Ans. 68 6s. lOd. 
 
 5. What will 146 yards of broad- 
 cloth come to, at 14s. 9d. a yard ? 
 Ans. 107 13s. "Gd. 
 
 4. What will 49 J pounds of tea 
 come to, at 3s. lld. a pound ? 
 Ans. 9 15s. lljd. 
 
 6. What will 120 barrel* of ap- 
 ples come to, at 11s. 3d. a barrel? 
 Ans. 67 10s. 
 
 CASE V. 
 
 To find the value of any number of pounds, fyc. when the price of one 
 pound is pence, or pence and farthings, and not an aliquot part of a 
 shilling. 
 
 ]|pLK. Write down the given number of pounds, &c. and call 
 them shillings. Divide the given price into aliquot parts, either 
 of a shilling, or of one another ; then divide by the several aliquot 
 parts, and the sum of the quotients will be the answer in shillings. 
 
 EXAMPLES. 
 
 1. What will be the value of 56 
 pounds of raisins, at 8 pence a 
 pound ? 
 
 d. s. 
 
 2 = 1)28 
 
 9 4 
 
 2iO)3j7 4 
 
 1 17s. 4d. Ans. 
 
 3. What is the value of 541 
 yards of cotton shirting, at 9 Jd. a 
 yard ? 
 
 Ans. 20 17s. Ojd. 
 
 2. What will 48J yards of ri 
 band come to, at 7J pence 
 yard ? 
 
 d. s. d. 
 
 G = )48 6 
 
 11 = 4)24 3 
 6 0| 
 
 2jO)3'jO 3| 
 
 1 10s. 3|d. Ans. 
 
 4. What will 672 pounds of but- 
 ter come to at IJf pence a 
 pound ? 
 
 Ans. 32 1 8s. 
 
 2 r 'l . 7/otp do you proceed, when the price is not an aliiuot part of a pound, or of a 
 shilling ? 
 
152 
 
 PRACTICE. 
 
 CASE VI. 
 
 To find the value of any quantity consisting of several denominations, 
 wke*i the price of one hundredweight, fyc. is given in sec eral denom- 
 inations. 
 
 RULE. Multiply the price of a cwt. &c by the number of hun- 
 dreds. Divide the remainder of the quantity into aliquot parts, 
 either of a cwt or of one another ; then divide by the several ali- 
 quot parts, and the sum of the product and quotients will be the 
 answer. 
 
 EXAMPLES. 
 
 1. What will be the value of 5 
 
 2. What will 12 cwt 
 
 Iqr. 161bs, 
 
 cwt. 3 qrs. 14 Ibs. of 
 
 sugar at 
 
 of hay come to, at 4 
 
 shillings 6 
 
 9 56 a cwt. ? 
 
 
 pence a cwt. ? 
 
 
 qrs. g cts. 
 
 
 
 s. 
 
 d. 
 
 2 = J' 
 
 9-56 
 
 
 i qr- = 4 
 
 4 
 
 6 
 
 
 5 
 
 
 
 
 12 
 
 
 47-80 
 
 
 
 2 14 
 
 
 
 iqr.= i 
 
 4-78 
 
 
 14 Ibs. = J 
 
 1 
 
 4 
 
 141bs = \ 
 
 2-39 
 
 
 2 Ibs. = 4 
 
 
 
 6| 
 
 
 M95 
 
 
 
 
 Off 
 
 $56-165 
 
 Ans. 
 
 2 15 
 
 9 f Ans, 
 
 3. What cost 12 cwt. qrs 
 
 4. What is the value 
 
 of 14 cwt, 
 
 7 Ibs. at $6-34 a cwt. ? 
 
 Ans. $76-476. 
 
 3 qrs. 7 Ibs. at 13s . 8d. a cwt. ? 
 Ans. 10 2s. 5jd. 
 
 5. What is the value of a piece 
 of broadcloth measuring 36 yards 
 3 qrs. at $7-75 a yard ? 
 
 Ans. $284-8125. 
 
 6. What is the value of a farm 
 containing 144 acres, 3 roods, 25 
 poles, at Q 10s. an acre ? 
 
 Ans. 941 17s. 9fd, 
 
 252. How will you gf-t the value of i quantity of several denominations, when thf 
 price of one cwt, fyc is given in different denominations ? 
 
SINGLE FELLOWSHIP. 353 
 
 FELLOWSHIP. 
 
 FELLOWSHIP is a rule, by which the accounts of several mer- 
 chants or other persons, trading in partnership, are so adjusted, 
 that each may have his share of the gain, or sustain his share of 
 the loss, in proportion to his share of the joint stock and the time 
 of its continuance in trade. 
 
 NOTE. By this rule, a bankrupt's estate may be divided among his 
 creditors. 
 
 SINGLE FELLOWSHIP. 
 
 SINGLE FELLOWSHIP is, when the stock of each of the several 
 'partners is continued in trade the same length of time. 
 
 RULE. 
 
 As the whole amount of stock, is to the whole gain or loss, so is 
 each man's stock, to his share of the gain or loss.* 
 
 PROOF. Add all the shares of the gain or loss together; and 
 if the work be right, the sum will be equal to the whole gain or 
 loss. 
 
 EXAMPLES. 
 
 1. A, B, and C traded in partnership. A put into the stock 400 
 dollars, B put in 300 dollars, and C put in 200 dollars ; they gained 
 270 dollars : what was each man's shape of the gain in proportion to 
 what he put in ? 
 
 * That their gaia or loss, in this Rule, is in proportion to their stocks, is evident : For, 
 is the times, in which the stocks are in trade, are equal, if I put in one half of the whole 
 stock, 1 ought to have doe haif of the gain; if my part of the stock be one fourth, my 
 share of the gain or loss ought to be one fourth also. And generally the same ratio that 
 the whoie stock has to the whole gain or loss, must each person's particular stock have to 
 his respective gain or loss. 
 
 253. What is Fellowship? -254. What is Single Fdloioship'! 255. What is 
 
 We rule; and upon what principle is it founded 1 256. What is the method of proof 1 
 
 u 
 
354 SINGLE FELLOWSHIP. 
 
 dfa df f# r 
 
 400 A's stock. As 900 : 270 : : 200 
 300 B % 9 stock. 200 
 
 200 C's stock. 
 
 $ $ 9|00}540|00 
 
 As $900 : 270 : : 400 
 
 400 $60 C's gain. 
 
 9JOO)1080|00 
 
 C$120 A's gain. 
 
 $120 A's gain, Ans.? 90 B's gain. 
 
 f 60 C's gain. 
 
 270 Proof. 
 
 $ $ $ 
 
 As 900 : 270 :: 300 
 
 300 
 9|00)8 10)00 
 
 $90 B's gain. 
 
 2. Divide the number 360 into four equal parts, which shall be to 
 each other, as 3, 4, 5 and 6. 
 
 As 3+4+5+6 : 360 : <J? '' \ Answer. 
 
 S3 : 60\ 
 4: 8of 
 5 : lOOf 
 6 : 120) 
 
 360 Proof. 
 
 3. A gentleman dying, left two sons and a daughter, to whom he 
 bequeathed the following sums, viz. To the first son he gave 1200 
 dollars, to the second 1000 dollars, and to the daughter, 800 dollars : 
 but it was found, that his whole estate amounted only to 750 dollars : 
 what must each child receive of the estate, in proportion to the 
 legacies ? 
 
 ( 300 first son's portion. 
 Ans.< 250 second son's portion. 
 ( 200 daughter's portion. 
 
 4. A, B, and C traded in partnership, A put in 385 dollars, 50 
 cents; B put in 297 dollars, 75 cents; and C put in 175 dollars, 25 
 cents ; they gained 343 dollars, 40 cents : what was each one's share 
 of the gain ? 
 
 ($154-20 A's gain. 
 Ans.< 119-10 B's 
 f 70-10 C's 
 
DOUBLE FELLOWSHIP. 155 
 
 DOUBLE FELLOWSHIP. 
 
 DOUBLE FELLOWSHIP, or Fellowship with time, is when the 
 stocks of partners are continued unequal times. 
 
 RULE. Multiply each man's stock by the time it was contin- 
 ued in trade. Then, As the whole sum of the products is to the 
 whole gain or loss, so is each man's particular product to his par- 
 ticular share of the loss or gain.* 
 
 EXAMPLES. 
 
 1. A, B, and C traded in company. A put in 400 dollars, for 
 months ; B put in 300 dollars, for 6 months ; C put in 200 dollars for 
 5 months : they gained 320 dollars j what was each man's share of 
 
 the gain ? 
 
 
 
 
 
 
 
 $ 
 
 
 $ 
 
 $ 
 
 $ 
 
 $ 
 
 $ 
 
 400 
 
 
 300 
 
 200 
 
 As 6400 
 
 ; 320 : : 
 
 1800 
 
 9 
 
 
 6 
 
 5 
 
 
 1800 
 
 
 $3600 
 
 A's pr. 
 
 1800 
 
 1000 
 
 
 256 
 
 
 1800 
 
 B's pr. 
 
 
 
 
 32 
 
 
 1000 
 
 C's pr. 
 
 
 
 
 
 
 
 
 
 i 
 
 * 
 
 64|00)5760!00($QO B'i 
 
 As 6400 
 
 : 3*0 : 
 
 : 3600 
 
 
 
 576 
 
 
 
 3600 
 
 
 
 
 
 
 
 
 
 t $ ' $ 
 
 As 6400 : 320 : : 1000 
 1000 
 
 64|00)11520|00($180 A's gain. 
 
 64 64)00)32()0|00($50 C's gain. 
 
 320 
 
 512 
 
 512 
 
 ($180 A's gain. 
 
 Ans. < 90 B's gain. 
 
 ( 50 C's gain. 
 
 g320 Proof. 
 
 * When the stock of each partner is employed in trade the same length of time, the 
 share of the gain or loss is, evidently, in proportion to the stock; and when the stock of 
 each partner is the same, but employed in trade a different length of time, the share of the 
 gain or loss is in proportion to the time ; biu when neither are the same, the share of the 
 gain or loss of each partner must, evidently, be in proportion to the product of the stock 
 multiplied by the time. 
 
 257. What is Double Fellowship ? 258 . What is the rule ? 259. Explain the 
 
 difference between the two kinds of Fellowship, 
 
156 CUSTOM HOUSE ALLOWA&CEb. 
 
 2. Tv/o merchants entered into partnership for 18 months; K to? 
 put into stock $ U)0, and at the end of 8 months he put in S^OO more . 
 L, at tirst, put in $1100, and at the end of 4 months took out $280. 
 Now at the expiration of the time, they found they had gained il052 : 
 What is each man's just share ? 
 
 Ans. K, $385-90. L, $666-10. 
 
 3 A, B, and C, hold a pasture in common, for which they pay 20 
 per annum. In this pasture, A had 40 oxen for 76 days ; B had 36 
 oxen for 50 days, and C had 50 oxen for 90 days. I demand what 
 part each of these tenants ought to pay of the 20. 
 
 s. d. q. 
 
 CQ 10 2 Iff |f A's part. ' 
 Ans. -?3 17 1 Of ||i B'spart. 
 (9 12 8 24$ C'spar? 
 
 CUSTOM HOUSE ALLOWANCES. 
 
 THE only allowances which are made in the weight of goods f 
 at the Custom-Houses of the United States, are /are, and draft or 
 scalage. 
 
 Tare is an allowance made for the weight of the box, bag, 
 hogshead, cask, &c. which contains the goods, and it is either at 
 so much per cent, or at so much per box, <fcc. or the real or ac- 
 tual tare. *-, 
 
 Draft or scalage is an allowance of J per cent on the whole 
 gross weight of all goods, except tea and sugar, which is to be 
 deducted before casting the tare. 
 
 On tea, no draft is allowed. On sugar, there is a deduction to 
 be made for draft or scalage, before casting the tare, of 7 pounds 
 on every hogshead, 4 pounds on every tierce, 2 pounds on every 
 barrel, and 4 pounds on every Havanna box. 
 
 Grots weight is the whole weight of any kind of goods, together 
 with the box, bag, or cask, &c. which contains them. 
 
 Net weight is the weight of the goods after all allowances have 
 been deducted. 
 
 260. What allowances are made at the American custom houses 1 261. What is 
 
 tare 1 draft or scalage ? gross weight 1 and net weight ? 
 
CUSTOM HOUSE ALLOWANCES. 157 
 
 CASE I. 
 
 To find the net weight of goods, when the tare is at so much per cent., 
 with the allowance for draft or scalage. 
 
 RULE. 
 
 1. When the scalage is J per cent, divide the whole gross 
 weight by 200, and the quotient will be the scalage. 
 
 2. When the scalage is FO much per hogshead, box, &c multi- 
 ply the scalage of one by the number of hogsheads, boxes, &c. 
 and the product will be the whole scalage. 
 
 3. Subtract the scalage from the gross weight, then multiply 
 the remainder by the tare per cent, and divide the product by 
 100, the quotient will be the whole tare, which subtract from 
 what remained after the scalage was deducted, and the remain- 
 der will be the net weight.* 
 
 EXAMPLES. 
 
 1. A merchant bought 15 boxes of brown Havanna sugar, weighing 
 6740 Ibs. orross, tare 15 per cent, scalage 4 Ibs. per box; what was 
 the net weight of the sugar, and what was its value at $10-25 per 
 hundred weight ? 
 
 C 5G78 lbs.-50 cwt. 2 qrs. 22 Ibs. net weight. 
 $519-63811 value. 
 
 2. What is the pet weight of 4 bags of coffee, weighing 480 Ibs. 
 gross, allowing 2fper cent, for tare, and J per cent, for scalage ; and 
 what is the value of the net weight, at '18 cents per pound ? 
 
 A $ 438 Ibs. net weight. 
 ' $84-24 value. 
 
 CASE II. 
 
 To find the net weight of goods, when the tare is at so much per box, 
 ccHfft, chest, fyc. either with or without the allowance for draft or 
 scalage. 
 
 * Tn casting tare and scalage, remainders, which do not esneed half a pound, are not 
 usually reckoned, but if they exceed half a pound, then another pound is to be reckoned. 
 
 262. What is the rule to find the net weight of goods, when the tare is at so much per 
 cent, with an allowance for draffor scalage ? 
 
158 CUSTOM HOUSE ALLOWANCES. 
 
 RULE. 
 
 1. If draft or scalage be allowed, cast and deduct it as directecf 
 in Case 1 st. 
 
 2. Multiply the tare of one by the number of boxes, chests,, 
 &c. the product will be the whole tare, which subtract from what 
 remained after the scalage was deducted, or from the whole gross 
 weight, when no scalage is allowed, and the remainder will be 
 the net weight. 
 
 EXAMPLE. 
 
 1. What is the net weight and value of 4 casks of raisins weighing 
 520 Ibs. gross, tare 12 Ibs. per cask, allowing J percent, for draft or 
 scalage, at 7J cents per pound ? 
 
 . $ 469 Ibs. net weight. 
 ADS> $35-175 value. 
 
 CASE III. 
 
 To find the net weight of goods, when the real or actual tare is allowed. 
 
 RULE. 
 
 Subtract the whole amount of tare from the whole gross weight, 
 and the remainder will be the net weight. 
 
 EXAMPLES. 
 
 1. What is the net weight and value of 50 boxes of figs, weighing 
 1250 Ibs. gross, whole amount of tare 175 Ibs. with an allowance of 
 per cent, for scalage, at -10 cents per pound ? 
 
 . ^ 1069 Ibs. net weight. 
 '' $106-90 value. 
 
 2. What is the net weight and value of 4 firkins of butter, weigh- 
 ing as follows : 
 
 No. 1, gross 75 Ibs. tare 12 Ibs. 
 
 2, do. 83 Ibs. do. 14 Ibs. 
 
 3, do. 98 Ibs. do. 17 Ibs. 
 
 4, do. 120 Ibs. do. 21 Ibs. 
 at '15 cents per pound ? 
 
 . $312 Ibs. net weight. 
 '' \ $46-80 value. 
 
 263. What is the rule, when the tare is at so much per box, 8ft. either with or without 
 allowance ? 264. How do youjind the weight of goods, when the real tare is allowed 1 
 
AMERICAN DUTIES. 159 
 
 AMERICAN DUTIES. 
 
 DUTIES are imposed by law on goods, wares and merchandise, 
 imported into the United States, at certain rates per centum ad 
 valorem, or at certain rates per ton, hundred, pound, gallon, &c. 
 
 The ad valorem rates of duty upon goods, wares and merchan- 
 dise, are estimated by adding 20 per cent, to the actual cost there- 
 of, if imported from the Cape of Good Hope, or from any place 
 beyond it ; and 10 per cent, if imported from any other place or 
 country, including all charges ; commissions, outside packages, 
 and insurance excepted. 
 
 CASE I. 
 
 To find the duty on any amount of goods, wares or merchandise, at 
 any given rate per centum ad valorem. 
 
 RULE. Reduce the cost of the goods to Federal Money, add 
 20 per cent, if imported from or beyond the Cape of Good Hope, 
 or 10 per cent, if imported from any other place, then multiply 
 the amount by the given rate per cent, and divide the product by 
 100, the quotient will be the duty required. 
 
 EXAMPLES. 
 
 1. What will be the duty on an invoice of woollen goods, imported 
 from London, which cost 1250 10s. sterling, at 30 per cent, ad 
 valorem ? 
 
 1250-5 sterling cost. 
 $444=.l sterling.* 
 
 50020 
 50020 
 50020 
 
 $5552-220 actual cost in Federal Money. 
 555-222 tea per cent, added. 
 
 go 107-442 amount. 
 30 
 
 Ans. $1 832-23260 duty required. 
 
 * The rates at which foreign coins and currencies are estimated at the Custom Houses 
 of the United States may be found on page 39. 
 
 265 How are duties imposed upon, imported goods 1 266. In what manner are the 
 
 ad valorem rates estimated ? 267. What is the rule for finding the dvty on any 
 
 amourt, at a. given rate per cent, ad valorem ? 
 
160 DISCOUNT. 
 
 2. What will be the duty on an invoice of silk goods, imported 
 from France, which cost 2650 Francs, at 20 percent, ad valorem ? 
 
 Ans. J109-31J. 
 
 3. What will be tho duty on an invoice cf silk and cotton goods, im- 
 ported from India, which cost 25000 rupees, at 25 prr cent, ad 
 valorem ? 
 
 Ans. g3750. 
 
 CASE II. 
 
 To find the duty on any amount of goods, wares or merchandise, at 
 any given rate per pound , gallon, fyc. 
 
 RULE. Multiply the given rate per pound, gallon, &c, by the 
 number of pounds, gallons, &c. and the product will be the duty 
 required. 
 
 EXAMPLES. 
 
 1. What will be the duty on 150 chests of Hyson tea imported di- 
 rect from China in a vessel of the United States, weighing gross 
 11250 !bs. tare 20 Ibs. per chest, at -40 cents per pound ? 
 
 Ans. $3300. 
 
 2. What will be the duty on 20 pipes of French brandy, 4th proof, 
 containing 2520 gallons, at -48 cents per gallon ? 
 
 Ans. $1 209-60. 
 
 3. What will be the duty on 25 hogsheads of brown su2far, weigh- 
 ing 43750 Ibs. gross, allowing 12 per cent. tare, and 7 pounds per 
 hogshead for draft or scalage, at -03 cents per pound ? 
 
 Ans. $11 50-38. 
 
 DISCOUNT. 
 
 DISCOUNT is an allowance made for the payment of any sum of 
 money before it becomes due ; or it is the difference between 
 any sum due at some future time, and its present worth. 
 
 The present worth of any sum, or debt, due at any future time, 
 is such a sum, as, if put at interest, would in that time and at the 
 rate per cent, for which the discount is to be made, amount to the 
 given sum or debt. 
 
 268. How do you find the. duty, at any given rate, per pound, fc ? 2tJ9. What is 
 
 Discount! 270. What is the present worth of a sum due at a future time ? 
 
DISCOUNT. 161 
 
 RULE. 
 
 1. As the amount of $100 or 100 at the given rate and time, 
 Is to the interest of $100 or > 100 at the same rate and time, so is 
 the given sum to the discount. 
 
 Subtract the discount from the given sum, and the remainder is 
 the present worth. 
 
 2. As the amount of $100 or 100 at the given rate and time, 
 is to $100 or 100, so is the given sum to the present worth. 
 
 Subtract the present worth from the given sum, and the re- 
 mainder is the discount.* 
 
 EXAMPLES. 
 
 1. What is the discount and present worth of $500 payable in 8 
 months, allowing discount at 6 per cent, per annum ? 
 
 By Rule 1. By Rule 2. 
 
 $ $ 
 
 100 100 
 
 Int. for 8 mo. 4 Int. for 8 mo. 4 
 
 Amount, 104 Amount, 104 
 
 $ $ $ $ $ $ 
 
 1U4 : 4 : : 500 104 : 100 :: 500 
 
 500 500 
 
 r 
 
 104)2000( 19-23 T V discount. 104)50000(480-76ff present worth. 
 104 -- 416 ---- 
 
 -- 480-76jf present worth. -- 19-23 T ' T discount. 
 
 960 840 * 
 
 936 832 
 
 2400 8000 
 
 208 728 
 
 320 720 
 
 312 624 
 
 * That an allowance ought to be made for the payment of money before it becomes 
 due, which is supposed to bear no interest until after it is due, is very reasonable; and 
 this allowance ought to be such a sum, a c , being put at interest until the debt becomes due, 
 would amount to the interest of the debt for tiie same time. 
 
 The reason of the rules is evident from the nature of simple interest ; for since the 
 debt may be consideied as the amount of some principal (called here the present worth) 
 at a certain rate percent, for the given time, that amount must he in the same proportion 
 either to its principal or interest, ap the amount of any other sum at the same rate and 
 for the same time, is to its principal or interest. 
 
 271. What are the rides of Discount! 272. Jlndwhat are the reasons vpon which 
 
 they nre founded ? 
 
 w 
 
]62 BARTER. 
 
 2. What is the present worth of 150 payable in 3 months, allow- 
 ing discount at 5 per cent, per annum ? 
 
 Aus. 148 2s. lid. 2|qrs. 
 
 NOTE. The only correct principle of calculating discount is by 
 the preceding rule ; yet banks and commercial persons, in discount- 
 ing notes, &c. deduct the interest for the time (including the days of 
 grace) for the discount. When discount is made for present payment 
 without any regard to time, the interest of the sum for one year i? 
 the discount. 
 
 BARTER. 
 
 BARTER is the exchanging of one commodity for another, and 
 teaches merchants so to proportion their quantities, that neither 
 shall sustain loss.* 
 
 PROOF. By changing the order of the question. 
 
 RULE. 
 
 1 . When the quantity of one commodity is given with its value or the 
 value of its integer, <? also the value of the integer of some other com- 
 modity to be exchanged for it, to find the quantity of this commodity : 
 Find the value of the commodity of which the quantity is given, 
 then find how much of the other commodity at the rate proposed 
 may be had for that sum. 
 
 2- When the quantities of two commodities are given, and the rate 
 of selling them, to find, in case of inequality , how much of some other 
 commodity must be given : -Find the separate values of the two giv- 
 en commodities ; subtract the less from the greater, and the dif- 
 ference will be the amount of the third commodity, whose quality 
 and rate may be easily found. 
 
 * The rules of Barter arc merely applications of thr Rule of Three, and are easily un- 
 derstood. 
 
 273. What is Barter ? 274. Explain to me the given rules, if you understand 
 
 them 1 
 
BARTER. 163 
 
 S. When, in bartering, one commodity is reckoned above the ready 
 money price, to findtlie bartering price of the other : Say, as the 
 ready money price of the one, is to its bartering price ; so is that 
 of the other, to its bartering price : Next, find the quantity re- 
 quired, according to either the bartering or ready money price. 
 
 EXAMPLES. 
 
 1. How much tea at 9s. 6d. per Ib. must be given in barter for 156 
 gallons of wine, at 12s. 3Ad. per gallon ? 
 
 Galls, 
 
 9s. 6d.= 
 
 23010 
 
 d. Ib. d. Ib. oz. 
 
 As 114 : I : : 23010 : : 201 13/ T \ Ans. 
 
 price, quan. price, quan. 
 s. d. gals. s. d. Ib. oz. 
 Or, As 12 3J : *156 : 9 6 : 201 13 T 5 -& Ans. as before. 
 
 2. A and B would barter ; A has 150 bushels of wheat, at $l-25c. 
 per bushel, for which B gives 65 bushels of barley, worth -62Jc. per 
 bushel, and the balance in oats at -37jc. per bushel; What quantity 
 of oats must A receive from B ? 
 
 Ans. 39 If bushels. 
 
 3. A has linen cloth, at -30c. per yard, ready money, in barter -36c. ; 
 B has 3610 yards of ribband, at -22o. 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 must A give B ? 
 
 Ans. The rate of ribband is -26c. 4m. per yard, and B must re- 
 ceive 1 980| yards of linen, and $200 in cash. 
 
164 LOSS AND GAIN. 
 
 LOSS AND GAIN. 
 
 Loss AND GAIN is a rule by which merchants and traders find 
 what they gain or lose by trading, and at what rate per cent. : 
 It also teaches them to find the price for which any kind of goods 
 must be sold, in order to gain or lose any given rate per cent 
 The different cases are only particular applications of the Rule 
 of Three. 
 
 CASE I. 
 
 When goods are bought at one price, and sold at another, to find what 
 is gained or lost, and the gain or loss per cent. 
 
 RULE. Find the gain or loss by subtraction ; then, As the 
 price the goods cost, is to the gain or loss, so is $100 or 100 to 
 the gain or loss per cent. 
 
 EXAMPLES. 
 
 1. If I buy cloth for $2 a yard, and sell it for g2-50 a yard ; what 
 do 1 gain per yard, and what do 1 gain per cent, or by laying out 
 100 dollars ? 
 
 jjc. % c. $ 
 
 Sold for 2-50 As 2 : -50 : : 100 
 
 Cost 2-00 100 
 
 Gained -50 per yard. 2)5000. 
 
 Gained $25-00 per cent. 
 
 NOTE. When goods are bought or sold on credit, the present 
 worth of the value of the goods for the time, must be found, in order 
 to find the true gain or loss. 
 
 2. If I buy cloth at $3-50 per yard for cash, and sell it at $4-12 per 
 yard on a credit of six months, what do 1 gain per cent, allowing dis- 
 count at 6 per cent, a year on the selling price ? 
 
 Ans. gl 4-284 per cent. 
 
 3. Bought 12 cwt. 2 qrs. of sugar at $10-20 per cwt. on a credit oi 
 4 months, and sold the same at $10-25 per cwt. for cash ; what waa 
 the whole gain, and the gain per cent, allowing discount at 6 percent. 
 a year on the purchase price ? 
 
 <$3-12j whole gain. 
 \ 2-50 gain per cent. 
 
 275. What is Loss and Gain ? - 273. What is the rule for finding loss or gain in 
 business ? - 277. How do you proceed, when goods are bovgM or sold on credit 1 
 
EQUATION OF PAYMENTS. 165 
 
 CASE II. 
 
 To find the price for which any kind of goods must le sold, in order 
 to gain or lose any given rate per cent. 
 
 RULE. As $100 or 100 is to the purchase price, so is $100 
 or 100 with the profit per cent, added, or loss per cent, sub- 
 tracted, to the selling price. 
 
 EXAMPLES. 
 
 1 . Bought linen at -60 cents per yard, how must it be sold per yard, 
 in order to gain 25 per cent. ? 
 
 $ c. t c. 
 
 As 100 : -60 : : 125 : -75 Ans. 
 
 2. Bought a piece of cloth at $2-75 per yard, and sold it at a loss 
 of 15 percent. : what was it sold for per yard ? 
 
 Ans. $2-3375. 
 
 3. Bought 50 gallons of brandy, at -75 cents per gallon, but by ac- 
 cident, 10 gallons leaked out : At what rate must I sell the remainder 
 per gallon, to gain upon the whole prime Cost, at the rate of 10 per 
 cent. ? 
 
 Ans. $l-03c. IJm. 
 
 EQUATION OF PAYMENTS. 
 
 EQUATION OF PAYMENTS is the finding of a time to pay at once, 
 several debts due at different times, so that neither party shall 
 sustain loss. 
 
 RULE. Multiply each payment by the time at which it is due ; 
 then divide the sum of the products by the sum of the payments, 
 and the quotient will be the equated time.* 
 
 *This rule is founded on a supposition, that the sum of the interests of the several 
 debts which arc payable before the equated time, from their terms to that time, is equal 
 to the sum of the interests of the debts payable after the equated time, from that to their 
 terms; but this is not correct for by keeping a debt unpaid after it is due, the interest of 
 it is gained for that time ; but by paying a debt before it is due, the payer does not lose 
 the interest for that time, but the discount only, which is less than the interest ; there- 
 fore, the. rule is not accurately true ; however, in most questions which occur in busi- 
 ness, the errour is so trifling, that it will generally be made use of as the most eligible 
 method. 
 
 278. How do you ascertain at what price you must sell an article in order to gain so 
 
 muchper cent. ? 279 What is Equation of Payments ? 280. What is the rule , 
 
 and on what is it founded 1 
 
166 EQUATION OF PAYMENTS. 
 
 EXAMPLES. 
 
 1. A owesB g380to be paid as follows, viz. glOOin 6 months, $120 
 in 7 months, and $160 in 10 months: What is the equated time lor 
 the payment of the whole debt? 
 
 100X (5= 600 
 
 12UX 1 840 
 
 160X101600 
 
 100+ 120+160=380)3040(8 months, Ans. 
 3040 
 
 C 2. A owes B 104 15s. to be paid in 4J months, 161 to be paid 
 in 3i months, and 152 5s. to be paid in 5 months : What is the 
 equaled time for the payment of the whole ? 
 
 Ans. 4 months and 8 days. 
 
 3. There is owing to a merchant 998, to be paid 178 ready 
 money, 200 at 3 months, and 320 in 8 months ; I demand the in- 
 different time tor the payment of the whole ? 
 
 Ans. 4 months. 
 
 4. The sum of $164 16c. 6m. is to be paid, Jin 6 months, J in 8 
 months, and in 12 months: what is the mean time for the payment 
 of the whole ? 
 
 Ans. 7 months. 
 
 5. A merchant has 360 due him, to be paid at 6 months, but the 
 debtor agrees to pay J at the present time, and at 4 months ; 1 de- 
 mand the time he must have to pay the remainder, at simple interest, 
 so that neither party may have the advantage of the other ? 
 
 J=180 paid down. 
 J=120 paid at 4 months. 
 60 unpaid. 
 
 Now as he pays 180 dollars 6 months, and 120 dollars 2 months be- 
 iore they are respectively due, say, as the interest of 60 dollars for 
 1 month, is to 1 month, so is the sum of the interest of 180 dollars for 6 
 months, and of 120 dollars for 2 months, to a fourth number, which 
 added to the 6 months, will give the time for which the 60 dollar? 
 ought to be retained. 
 
 Ans. 28 months. 
 
1NVOLUTION....EVOLUTION. 1 67 
 
 INVOLUTION. 
 
 INVOLUTION is the method of finding the powers of numbers. 
 
 Powers of numbers are the products arising from the continu- 
 al multiplication of numbers into themselves. 
 
 Any number may itself be called the root or first power. If 
 the first power be multiplied by itself, the product is called the 
 second power, or the square ; if the square be multiplied by the 
 first power, the product is called the third power, or the cube ; 
 if the cube be multiplied by the first power, the product is call- 
 ed the fourth power, or the biquadrate, &c. 
 
 Thus 4 is the root or 1st power of 4. 
 4X4=16 is the 3d power, or the square of 4. =4 3 
 4X4X4=64 is the 3d power, or the cube of 4. =4 3 
 4 X4X 4X4=256 is the 4th power, 01 the biquadrate of 4, &c. =4* 
 
 The small figure points out the order of the power, and is call- 
 ed the Index or Exponent, 
 
 Rule for finding the powers of numbers. 
 
 Multiply the given number, or first power continually by itself, 
 till the number of multiplications be one less than the index of 
 the power to be found, and the last product will be the power 
 required. 
 
 NOTE. The powers of vulgar fractions are found by raising each 
 of their terms lo the power required. If the power of a mixed 
 number be required, either reduce it to an improper fraction, or re- 
 duce the vulgar fraction to a decimal. 
 
 EVOLUTION. 
 
 EVOLUTION, or the extraction of roots, is the operation by 
 which we find any root of any given number. 
 
 The root is a number whose continual multiplication into itself 
 produces the power, and is denominated the square, cube, biqua- 
 drate, or 2d, 3d, 4th root, &c. accordingly as it is, when raised 
 to the 2d, 3d, 4th, &c. power, equal to that power. Thus, 4 is 
 the square root of 16, because 4x4=16. 4 also is the cube 
 root of 64, becajise 4x4x4=64 ; and 3 is the square root of 9, 
 and 12 is the square root of 144, and the cube root of 1728, be- 
 cause 12x12x12 = 1728, and so on. 
 
 2S1. What is I'iyolut : on? 232. What are powers tf numbers ? 283. Now 
 
 do you find these powers 1 284. What is Evolution ? 285. What is a root ? 
 
163 
 
 EVOLUTION. 
 
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SQUARE ROOT. 169 
 
 There is no number of which we cannot find any power ex- 
 actly ; but there are many numbers, of which the exact roots 
 can never be obtained. Yet by the heJp of decimals, we can 
 obtain these roots to any necessary degree of exactness. 
 
 Those roots which cannot be exactly obtained, are called surd 
 roots ; and those which can be found exactly, are called rational 
 roots. 
 
 Roots are sometimes denoted by writing this character \/ be- 
 fore the power, with the index of the power over it ; thus the 
 
 cube root of 64 is expressed \/ 64, and the square root of 64 is 
 expressed \/ 64, the index 2 being omitted when the square root 
 is required. 
 
 EXTRACTION OF THE SQUARE ROOT. 
 
 THE EXTRACTION OF THE SQUARE HOOT is the method of find- 
 ing a number, which, being multiplied by itself, shall produce the 
 given number. 
 
 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 show the number of figures 
 the root will consist of. 
 
 2. Find the greatest square 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 oi 
 the root, and the square number under the period, then subtract 
 it therefrom, and to the remainder bring down the next period for 
 a dividend. 
 
 3. Place 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, 
 (except 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 
 divisor ; 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 r dividend. 
 
 28G. Can youjindthe jiowr aim' root of any number'? 287. What is the distinc- 
 tion in the roots ? 283. What is the extraction of the Square Root ? 289. What is 
 
 the rule ? 
 
 X 
 
170 SQUARE ROOT. 
 
 5. Double the figures already found in the root, for a new di- 
 visor, (or bring down your last divisor for a new one, doubling the 
 right hand figure of it,) and from these, find the next figure of the 
 root as last directed, and continue the operation in the same man- 
 ner, till you have brought down all the periods.* 
 
 * The rule for the extraction of the square root may be illustrated hy attending to th 
 process by which any number is raised to the square. The several products of the mul- 
 tiplication are to be kept separate, as in the proof of the rule for multiplication of simple 
 numbers. Let 37 be the number to he raised to th" square. 
 
 37X37=136 
 37 
 
 (37 (3<j+7=37 
 
 2X3)42 =2X3X7 
 49=73 
 
 Now it is evident that 9, in the place of hundredth!, is the greatest square in this 
 product ; put i's root. 3. in the qm-tiei t and 900 is taken from the product. The next 
 products are 21+21=2X3X7 for a dividend. Do-thle the root already found, and it is 
 2X3, fora divisor, which gives 7 for the quotien* which annexed to the divisor, and the 
 whole then multiplied by it, gives 2X3X?(=42 +7X7{=49) which, placed in their 
 proper places, completely exhausts the remainder of the square. The same may be. 
 shown in any other case, and the rule becomes obvious. 
 
 Perhaps the following method may he considered more simple and plain. Let 
 37=30+7. be multiplied as :n the demonsnation of multiplication of simple number?, 
 <and the products kept separate. 
 
 30+7 
 30+7 
 
 900+30X7 
 
 30X7+49 
 
 900+2x30X7+49=1316, the sum and square 
 900 30+7 
 
 2X30+7X7)2X30X7+49 
 2X30X7+49 
 
 The root of 900 is 30 an<l leaves the two other terms, which are exhausted by a dmsm 
 formed and multiplied as directed in the rule. 
 
 290. Explain to me the nature of this rule- 
 
SQUARE ROOT. J7I 
 
 NOTE 1. If when the given power is pointed off as the power re- 
 quires, the left hand period should be deficient, it must nevertheless 
 stand as the first period. 
 
 2. If there be decimals in the given number, it must be pointed 
 both ways from the place of units: If when there are integers, the 
 first period in the decimals he deficient it may be completed by an- 
 nexing so many ciphers 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 operation may be continued at 
 pleasure by annexing ciphers. 
 
 EXAMPLES* 
 
 1. Required the square root of 729 ? 
 
 729(27 the root. The given number being distinguished irr- 
 4 to periods, we seek the greatest square num- 
 
 ber in the left hand period (7) which is 4, 
 
 47)329 of which the root (2) being placed to the 
 
 329 right hand of the given number, aftor the 
 
 manner of a quotient, and the square number 
 000 (4) subtracted from the period (7) to the re- 
 
 mainder (3) we bring down the next period 
 PROOF. (29) making for a dividend, 329. Then the 
 
 27 double of the root (4) being placed to the left 
 
 27 hand fora divisor, we say how often 4 in 32? 
 
 (excepting 9 the right hand figure) the answer 
 189 is 7, which we place in the root for the se- 
 
 54 cond figure of it, and also to the right hand 
 
 of the divisor ; then multiplying the divisor 
 729 thus increased by the figure (7) last obtained 
 
 in the root, we place the product underneath the dividend, and sub- 
 tract it thereirom, and the work is done. 
 
 DEMONSTRATION OF THE REASON AND NATURE OF THE RULE. 
 
 The superficial content of any thing, that is to say, the number of 
 square feet, inches, &c. in the surface of a field, a floor, &c. is found 
 by multiplying the length into the breadth. Thus, if a piece of land 
 be 10 rods in length, and 10 in width, it is a square, and the measure 
 of one of its sides is the root, of which the superficial content of the 
 piece of land is the 2d power. Or, supposing you have a piece of 
 cloth 1 yard wide, and 225 yards in length, and you wish to know 
 how many square yards it will cover, you must so arrange the 
 of the whole that they may be in a square form. 
 
172 
 
 SQUARE ROOT. 
 
 Now, suppose yon have 144 square pieces of wood, and wish to 
 know how many pieces would be on a side, were the whole arranged 
 into a square form. To determine this, you must extract the square 
 root 144 ; the first step of which is to point off the number into periods 
 of two figures each, This shows how many figures the root will con- 
 sist of, and is done on this principle, that the pioduct of any two num- 
 bers can have, at most, but as many places of figures, as are in both the 
 factors, and at least but one less. 
 
 144(1 
 1 
 
 44 
 Fig. 1. 
 
 
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 1 
 
 
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 The left hand period being 1. the 
 square of it will be 1. and likewise the 
 root will be 1 . But as we have nothing 
 to do at present with the right hand pe- 
 riod, we will omit it, and consider only 
 the left hnnd period, which being in the 
 place of hundreds, must be called 100; 
 hence the operation, at present, will be 
 to find the square root of 100. The 
 root of 1, is 1, but as there are two pe- 
 riods in 100, there will be two figures 
 in its root, and as the figure already 
 obtained in the root is equal to its peri- 
 od, there is nothing remaining for the 
 next period ; and as the next period con- 
 sists wholly of ciphers, the next figure 
 of the root will be a cipher, so that the 
 root of 100 is 10. By this process we 
 
 have disposed of 100 of the pieces into the form represented by Fig. 
 
 1, viz. 10 pieces on a side. 
 
 The reason for placing the square number underneath the period, 
 and subtracting it from the period, as directed in the rule, is as fol- 
 lows. When we have obtained the root of the left hand period, we 
 have disposed of as many pieces as the greatest square of the left hand 
 period represents, and by subtracting the square of the root from 
 its period, we make it smaller by as many as the square of the root 
 represents; thus in the example given, 1 in the quotient represents 
 10, the square of which is 100, which 1, under the left hand period, 
 represents. This, subtracted from the left hand period, leaves 44 ; 
 so that 100 pieces have been disposed of as represented by Fig. 1, and 
 44 pieces are now to be added to it, in such manner that the square 
 form will be preserved. To do this, the rule directs to " place tke 
 double of the root already found on the left hand of the dividend for n 
 divisor*" 
 
SQUARE ROOT. 
 
 173 
 
 144(12 
 
 22)44 Now the first figure of the 
 
 44 roof shows the number of pie- 
 
 ces there arc on a side of Fig. 1, 
 viz. 10 In order to preserve 
 
 Fig- 2. the square form, the additions 
 
 A. must be made on two adjoin- 
 ing sides of the square, as in 
 Fig. 2. Now it is evident 
 that if there were just 20 pie- 
 ces left, after disposing of 100, 
 there would be just enongh to 
 make a row on two sides of 
 Fig. 1, and if there were 40 
 pieces left, they would make 
 two rows, on two sides, as rep- 
 resented by the rows a, e, and 
 o, n, Fig. 2. Hence the reason 
 
 j of placing the double of the 
 
 \o\n~\ root on the left of the dividend 
 for a divisor. In making the 
 
 additions , c, and o, n, you will observe there is a deficiency, A. 
 which is not filled. To fill this deficiency, the rule directs to " except 
 the right hand figure^ and likewise to "place the quotient figure on tke 
 right hand of the di-visor" Now the deficiency A. must be limited by 
 the additions a, e, and o, n, consequently the figure, expressing the 
 width of these additions, expresses the root of this deficiency, which, 
 multiplied into itself, gives the superficial contents of the deficiency 
 Thus Fig. 2 shows the disposition of 144 pieces iato a square form. 
 
 a 
 
 
 
 
 
 
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 I 
 
 
 
 
 
 
 
 
 
 
 
 
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 1 
 
 
 
 
 
 
 
 
 
 
 
 
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 I 
 
 
 
 1 
 
 
 
 
 
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 1 
 
 | 
 
 
 
 
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 1 
 
 1 
 
 2. Required the square root of 575-5 ? 
 
 575-50(23-98+, root. 
 4 
 
 43)175 
 
 129 
 
 469)4650 
 4221 
 
 4708)42900 
 38304 
 
 4596 remainder. 
 
J74 SQUARE ROOT. 
 
 3. What is the square root of 10342656 ? 
 
 Ans. 3216. 
 
 1. What is the square root of 964-5192360241 ? 
 
 Ans. 31-05671. 
 
 5. What is the square root of -0000316969 ? 
 
 Ans. 00563. 
 
 Rules for extracting the Square Root of Vulgar Fractions and Mixed 
 
 Numbers. 
 
 1. Reduce the fraction to its lowest terms for this and all other 
 
 roots. 
 
 2. Extract the root of the numerator for a new numerator, and' 
 the root of the denominator for a new denominator. 
 
 3. Or reduce the vulgar fraction to a decimal, and extract its 
 root. 
 
 4. Mixed numbers may be reduced to improper fractions, and 
 the root of the numerator and denominator extracted, or the vulgar 
 fraction may be reduced to a decimal, and annexed to the whole 
 number, and the root of the whole extracted, 
 
 EXAMPLES. 
 1. What is the square root of T i|| ? 
 
 T i| T 16(4 root of the numerator. 
 
 16 
 
 1681(41 root of the denominator. 
 16 
 
 Gl)81 Therefore T 4 T is the root of. 
 
 81 [the given fraction. 
 
 Or, 1681)16-(-00951S1439-f. And V 
 
 2, What is the square root of f-iai ? 
 
 Ans. 
 
 3. What is the square root of 42J ? 
 
 Ans. 
 
 291. IV hat is the rule for extracting the Square Root of Vulgar Fractions and mixed 
 numbers ? 
 
SQUARE ROOT, 175 
 
 APPLICATION AND USE OF THE SQUARE ROOT. 
 
 PROBLEM I. To find a mean proportional between two numbers. 
 
 Multiply one of the given numbers by the other, and* 
 extract the square root of the product, and the root will be the 
 mean proportional required. 
 
 NOTE. When the first number is as many times greater than the 
 second, as the second is times greater than the third, the second 
 number is called a mean proportional between the other two, 
 
 EXAMPLE. 
 
 1. What is the mean proportional between 36 and 144 ? 
 
 3GX 144-5184, and */ 5184=72 Ans. 
 
 PROBLEM II. To find the side of a square equal in area to any 
 given superficies whatever. 
 
 RULE. Find the area, and the square root is the side of the 
 square sought.* 
 
 EXAMPLES. 
 
 1. If the area of a triangle be 160, what is the side of a square equal 
 in area thereto ? 
 
 VI 60=1 2-649+ Ans. 
 
 PROBLEM III. A certain general has an army of 5625 men : pray, 
 how many must he place in rank and file, to form them into a square ? 
 
 V 5625=75 Ans.t 
 
 PROBLEM IV. If a pipe 6 inches bore, will be 4 hours in running 
 off a certain quantity of water, in what time will 3 pipes, each 4 in- 
 ches 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. Ans. 
 
 * A square is a figure of four equal sides, each pair meeting perpendicularly, or a fig- 
 ure whose length and breadth are equal. As the area, or number of square feet, inches, 
 fcc. in a square, is equal to the product of two sides which are equal, the second power 
 is called the square. Hence the rule of PROBLEM II. is evident. 
 
 flf you would have the number of men be double, triple, or quadruple. &c. as many 
 in rank -rsin file, extract the square root of , J. &c. of the given number of men, and 
 thatwil 1 he the number of men m file, which double, triple, quadruple, &c. and the pro- 
 duct will te the number in rank. 
 
 292. How do you find a mean proportional between two numbers ? 
 
170 SQUARE ROOT. 
 
 PROBLEM V. A line 36 yards long will exactly reach tVorn 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 24X2-1=576. Then, 1296576=720, and 
 /v /71>0=26-83+yards, the Answer. 
 
 PROBLEM VI. The height of a tree growing in the centre of a cir- 
 cular island 44 foot in diameter, is 75 feet, and a line stretched from 
 the top of it over to the hither edge of the water, is 256 feet. 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, 655365625=59911 
 andv'59911=244-76-hmd 244-76 V~ 222 ' 76 feet ? Answer. 
 
 PROBLEM VII. Suppose a ladder 00 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? 
 
 Ans. 102-64 feet, 
 
 PROBLEM VIII. 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, 
 
 EXAMPLES. 
 
 1. The difference of two numbers is 20, and the difference of their 
 squares is 2000; what are the numbers ? 
 
 Ans. 60 the greater. 40 the less. 
 
 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. 
 
 . $ Harry's share 12. 
 Ans> ) Jack's share 6. 
 
CUBE! ROOT. 177 
 
 EXTRACTION OF THE CUBE ROOT. 
 
 A CUBE is any number multiplied by its square. To extract 
 the cube root, is to find a number which, being multiplied into its 
 square, shall produce the given number. 
 
 RULE. 
 
 1 . Separate the given number into periods of three figures each, 
 by putting a point over the unit figure and every third figure be- 
 ; yond the place of units. 
 
 2. Find the greatest cube in the left hand period, and put its 
 root in the quotient. 
 
 3. Subtract the cube thus found, from the said period, and to 
 the remainder bring down the next period, and call this the divi- 
 dend. 
 
 4. Multiply 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. 
 
 6. Multiply the triple square by the last quotient figure, and 
 write the product under this 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. 
 
 7. 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 till the whole be finished.* 
 
 * The process for extracting the cube ro;>t may be illustrated in the same manner as 
 chat for the square root. Take the same number 37, ami multiply as before, collecting 
 the twice 21 into one sum, as they belong to the same place, and the operation will be 
 simplified, 37s 50653. 
 
 C 49=72 4H 72 
 
 372 J 42 =2X3X7 4202x30X7 
 
 900=302 
 
 37 the multiplier. 37 
 
 293. Wind is a Cube ? 294. What is the method of extracting the cube root of a 
 
 given numkfr 1 295. I wish you to illustrate the process under this ride, by one of 
 
 the examples given. 
 
 Y 
 
178 CUBE ROOT. 
 
 NOTE. The same rule must be observed for continuing the ope- 
 ration and pointing for decimals, as in the square root. 
 
 EXAMPLES. 
 
 1. Required the cube root of 436036824287. 
 
 436036824287(7583 the root 
 343 
 
 1st Divisor= 1491 0)93036=1 st Dividend. 
 
 73500 
 
 5250 
 
 125 
 
 78875=1 st Subtrahend. 
 
 343=73 343=73 
 
 294 =2X3X72 
 
 f> 7 o _ / 63 ' = 32 X 7 6300=303 x 7 
 
 - \ 147 =3X72 1470=30X73 
 
 1 26 =;? X 32 X 7 1260foz2 X 302 x< 
 
 27- -=3s 27000=303 
 
 27 (37 27000 (,30+7 
 
 3x32 ) 189 -=SX32X7 3X302 ) 18900=3X302X7 
 
 3X3 ) 441 -=3X^X73 3X30 ) 4410=3X30X72 
 
 As 27 or 27000 is the greatest cube, its root is 3 or 39, and that part of the cube is ex- 
 hausted by this extraction. Collect those terms which belong to the same places, and we 
 have 33X7=63. and 2X32X7=126, and 63+ 126=3X32 X7=189 ; and X3X?2 = 
 294and3X72=147,aad 294+147=441=3x3X72 fora dividend, which divided by 
 the divisor, formed according to the rule, the quotient is 7, for 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 subtrahend, according to the rule. 
 The same may be shown in any other case, and the universa.ity of the rule hence inferred- 
 The other method of illustration, employed in the square root, is equally applicable iu 
 this case. 
 
 3?=30-}-7, and 30+72=3024-2X30X7+73 
 
 30 +7 the mutiplier. 
 
 303+2X302 X 7+30X72 
 
 302 x 7+2X 30X 73+73 
 
 373=50653=303+3x302 X7+3X30X7 
 303 
 
 Divisor 3X302 +3X30 )3X30sX 7+3X30X72 +7 3 div. 
 
 3X302X7+3X30X72+73 subtrahend. 
 
 It is evident that 303 is the greatest cube. When its root is extracted, the next three 
 terms constitute the dividend ; aud the several products formed by means of the quotient 
 or second figure in the root, are precisely equal to the remaining parts of the power 
 whose root was to be found. 
 
CUBE ROOT. 179 
 
 Divisor=1689750)14161 824=2d Dividend, 
 
 13500000 
 
 144000 
 
 512 
 
 1 364451 2=2d Subtrahend. 
 
 3d Divisor 172391940)517312287=3d Dividend. 
 
 317107600 
 
 204660 
 
 27 
 
 517312287=3 Subtrahend. 
 
 The Method of Operation. 
 7X7X300 = 14700 1st Triple square. 
 7X30 210 = 1st Triple quotient. 
 
 14910 = 1st Divisor. 
 
 14700X5 = 73500 
 5X5X210 = 5250 
 5X5X5 = 125 
 
 78875 = 1st Subtrahend. 
 
 75X75X300 = 1687500 = 2d Triple square. 
 75X30 2250 = 2d Triple quotient 
 
 1689750 = 2d Divisor, 
 
 1687500X8 = 13500000 
 2250XCX8 = 144000 
 8X8X8 = 512 
 
 13644512 = 2d Subtrahend. 
 
 758X758X300 =172369200 = 3d Triple square. 
 758X30 = 22740 = 3d Triple quotient 
 
 172391940 = 3d Divisor. 
 
 172369200X3 =517107600 
 22740X3X3 = 204660 
 3X3X3 = 27 
 
 517312287 = 3d Subtrahend. 
 
180 CUBE ROOT. 
 
 DEMONSTRATION OF THE REASON AND NATURE OF THE RULE 
 
 A block of wood, or any solid body, having six equal sides, all ex- 
 actly square, is a CUBE; the root of which is the measure in length of 
 one of its sides. To gain a distinct understanding of the subject, let 
 the scholar provide himself with little blocks of wood, and build 
 them up into a cubick form, according to the rule. 
 
 First make a cubick block of any given size, and mark it with the 
 letter A Then make three other blocks of a square form, of an in- 
 definite thickness, but all equal to each other, each of which will just 
 cover one side of the block A. and mark them B. C. and D. Place 
 these blocks on three adjoining sides of the block A. when there will 
 be deficiencies at the three points where the blocks B. C. and D. 
 meet. These deficiencies .must be filled with three other blocks, 
 each of which must be just equal in length to one side of the block 
 A. and mark these blocks with E. r F and G. When the blocks E. F. 
 G are put in their places, there will be a deficiency at the place 
 where the ends of these blocks meet. Thi? deficiency must be filled 
 with another block, which mark H, To illustrate the rule, take the 
 following number. 
 
 10648(2 
 8 
 
 2 
 
 In the first place 1 seek the greatest cube in the left hand period, 
 and place its root, 2, in the quotient. The cube of 2 is 8, which I 
 place under the left hand period, aad subtract it therefrom, which 
 leaves a remainder of 2. Now as there are two periods in the given 
 number, there must be 2 figures in the root, consequently, 2, in the 
 quotient, does not express 2, merely, but 20 ; and the cube of 20 is 
 8000, which 8, under the period 10, represents; thus 8000 of the 
 parts of 10648, are disposed of into a cubick body, the length of each 
 side of which is equal to 20 of those parts, and to render the explan- 
 ation more plain, we will consider these parts as cubick feet, so that 
 each side of this body is 20 feet square, and this body we will have 
 represented by the block A. Now as each side of this block is 20 
 feet square, there are 400 feet OB each side of it. Now 8000 feet 
 are disposed of in this block, consequently there are 2648 cubick feet 
 to be added to the block A. in such a manner, that its cubick form will 
 be preserved. To do this, the additions must be made to three sides 
 of the block, and these additions are represented by the blocks, B. C. 
 and D. each of which containing 400 leet, the sum of the whole is 
 1200. Thus it is evident, that if there were 1200 feet more, there 
 would be just enough to cover three sides of the block A. ; and it is 
 to find the contents of these three sides, that the rule directs to u mul- 
 tiply the square of the quotient by 300." The square of the quotient 
 shows the superficial contents of one side of the block A. viz. 400, 
 for 2 in the quotient is in reality 20, and 20X20=400, and it is be^ 
 cause the cipher is not annexed to the quotient figure, that we are 
 
CUBE ROOT. 181 
 
 directed to multiply the square of the quotient by 300 instead of 3, as 
 in the following work. 
 
 4 10648(2 1200 60 
 
 300 824 
 
 1200 triple square. 1260)2648 2400 240 
 
 2400 
 
 2 240 
 
 30 8 cube of the root. 
 
 60 triple quotient. 2648 subtrahend. 
 1200 
 
 1260 divisor. 
 
 The rule directs to " multiply the quotient by 30." This is to ob- 
 tain the contents of the blocks E. F. and G. and the sum of these is 
 taken for a divisor, because the number of times the dividend con- 
 tains the divisor, will be equal to the number of feet, the additions to 
 the block, A. are in thickness. 
 
 The rule next directs to " multiply the triple square by the last quo- 
 tient figure." Now the triple square represents the superficial con- 
 tents of the three blocks B. C. and D. and the last quotient figure 
 shows the thickness of those blocks, consequently, multiplying the 
 triple square by the last quotient figure gives the cubick contents of 
 those blocks. Then the rule directs to kt multiply the square of the 
 last quotient figure by the triple quotient." Now the triple quotient is 
 the length of the blocks E. F. and G. and the quotient figure shows 
 their breadth, and their thickness ; hence multiplying the square of 
 the last quotient figure by the triple quotient, gives the cubick con- 
 tents of these blocks. Next we are directed to cube the last quotient 
 figure. This cube shows the cubick contents of the block H. and the 
 sum of these is equal to the cubick contents of the blocks B. C. and D.; 
 and E. F. and G. 
 
 2. What is the cube root of 34965783 ? 
 
 Ans. 327. 
 
 3. What is the cube root of 84-604519 ? 
 
 Ans. 4-39, 
 
 4. What is the cube root of -008649 ? 
 
 Ans. -2052-K 
 
 What is the cube root of iff ? 
 
 Ans. f . 
 
182 CUBE ROOT. 
 
 APPLICATION AND USE OF THE CUBE ROOT. 
 
 PROBLEM 1. To find two mean proportionals between any two 
 given numbers. 
 
 RULE 1. Divide the greater number by the less, and extract the 
 cube root of the quotient. 
 
 2. Multiply the root, so found, by the least of the given numbers, 
 and the product will be the less. 
 
 3. Multiply this product by the same root, and it will give the greater. 
 
 EXAMPLES. 
 
 1. What are the two mean proportionals between 6 and 750? 
 
 750-^-6=125, and \ri2b=5. Then 5X6=30 the less, and 
 
 30X5=1.50 the greater. Ans. 30 and 150. 
 
 2. What are the two mean proportionals between 56 and 12096 ? 
 
 Ans. 336 and 2016. 
 
 PROBLEM 2. Thesideofacubebemggiven^tojlndthesideofactibe 
 which shall be any number of times greater or less than the given cube.* 
 
 RULE. Cube the given side, and if the required cube be greater 
 than the given one, multiply the cube of the given side by the given 
 proportion, and the cube root of the product will be the side of the 
 cube required. But if the required cube be less, divide the cube of 
 the given side by the given proportion, and the cube root of the quo- 
 tient will be the side of the required cube. 
 
 EXAMPLES. 
 
 1. There is a cubick box whose side is 18 inches ; what is the side 
 of a box that will contain 8 times as much ? 
 
 18X18X18=5832, and 5832X8=46656. Then V46&56=36 in. Ans. 
 
 2. There is a cubick box whose side is 24 inches ; what is the side 
 of a box that will contain one sixty-fourth part as much ? 
 
 Ans. 6 inches. 
 
 * The solid, called a cube, has its length and breadth and heightall equal As the num- 
 ber of solid fret, inches, fee. in a cube are found by multiplying the height and length and 
 breadth together ; that is, by multiplying one side into itself twice, the third power of a 
 number is called the cube of that number. The solid contents of similar figures are in 
 proportion to each other, as the cubes of their similar sides or diameters. 
 
CUBE ROOT. 183 
 
 PROBLEM 3. The diameter of a globe or ball being given, to find 
 ihe diameter of a globe or ball that shall be any number of times great- 
 er or less than the given one. 
 
 RULE. Cube the given diameter, and multiply or divide it by the 
 given proportion as the question may require ; and the cube root of 
 the product, or quotient, will be the diameter of the globe required. 
 
 EXAMPLES. 
 
 1. If the diameter of a globe be 12 inches, what will be the diame- 
 ter of a globe one eighth part as large ? Ans. 6 inches. 
 
 2. If the diameter of a globe be 6 inches, what will be the diame- 
 ter of a globe 64 times as large ? Ans. 24 inches. 
 
 A GENERAL RULE FOR EXTRACTING THE ROOTS 
 OF ALL POWERS.* 
 
 1. Prepare the given number for extraction, by pointing off 
 from the units' place, as the required root directs. 
 
 2. Find the first figure of the root by trial, or by inspection 
 into a table of powers, and subtract its power from the left hand 
 period of the given number. 
 
 3. To the remainder bring down the first figure in the next pe- 
 riod, and call it the dividend. 
 
 4. Involve the root to the next inferior power to that which is 
 given, and multiply it by the number denoting the given power, 
 for a divisor. 
 
 5. Find how many times the divisor may be had in the divi- 
 dend, and the quotient will be another figure of the root. 
 
 6. Involve the whole root to the given power, and subtract it 
 always from as many periods of the given number as you have 
 found periods in the root. 
 
 * The roots of most powers may be found by the square and cube root only , therefore 
 when any even powei is g'wen, Uic easiest method will be (especially in a very high 
 power,) to extract the square root of it, which reduces it to half the giv>n power ; then 
 the square root of that power reduces it to half the same power ; and so on till you come 
 to a square or cube. 
 
 For example ; suppose a twelfth power be given ; extracting the square root of the 
 twelfth power reduces it to a sixth power ; and extracting the square root of the sixth 
 power reduces it to a cube. 
 
 296. \Vhat is the general method of extracting roots of all powers '? 
 
184 ALLIGATION MEDIAL. 
 
 7. Bring down the first figure of the next period to the re 
 mainder for a new dividend, to which find a new divisor, as be- 
 fore, and in like manner proceed till the whole be finished. 
 
 EXAMPLES. 
 
 1. What is the cube root of 135796744? 
 
 135796744(514 the root, or answer. 
 5X5X5=125= 1st subtrahend. 
 
 5X5X3=75 )107 = 1st dividend. 
 
 51X51X51=132651 = 2nd subtrahend. 
 51X51X3=7803)31457 = 2nd dividend. 
 514X514X514=135796741 = 3d subtrahend. 
 
 2. What is the biquadrate or fourth root of 19987173376 ? 
 
 Ans. 376. 
 
 ALLIGATION. 
 
 ALLIGATION is the method of mixing two or more simples of dif- 
 ferent, qualities, so that the composition may be of a mean or mid- 
 dle quality ; it consists of two kinds, viz. Alligation Medial and 
 Alligation Alternate. 
 
 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, is to 
 the whole value ; so is any part of the composition to its mean 
 price or value. 
 
 297. What is Alligation 1 298. Whatis Alligation Medial! 299. What is 
 
 the rule in Alligation Medial ? 
 
ALLIGATION ALTERNATE. 185 
 
 EXAMPLES. 
 
 1. A grocer mixed 4 cwt. of sugar at $10 a cwt. with 5 cwt. at 
 ^9-50 a cwt. and 3 cwt. at $3-75 a cwt. what was the value of one 
 cu't. of this mixture ? 
 
 owt. $ $ c. cwt. 
 
 4 at 10 =40-00 12 : 
 
 5 " 9-50=47-60 
 3 " 8-75^26-25 
 
 12)113*75 
 cwt. 12 $113-75 
 
 $9-479-1- Ans. 
 
 2. If a bushel of Indian corn at -75 cents a bushel, be mixed with 5 
 bushels of rye at -80 cents a bushel, and 15 bushels of oats at 30 cents 
 a bushel; what will be the value of a bushel of the mixture? 
 
 Ans. '53^ cents. 
 
 3. A wine merchant mixes 12 gallons of wine at -75 cents a gal- 
 lon, with 24 gallons at -90 cents, and 10 gallons at $MO; what is a 
 gallon of this composition worth ? c . m 
 
 Ans. 
 
 4. A goldsmith melted together 8oz. of gold of 22 carats fine,* 1 Ib. 
 oz. of 21 carats fine, and 10 oz. of 18 carats fiue ; what is the fineness 
 of the composition? 
 
 Ans. 20 T 8 ^ carats fine. 
 
 ALLIGATION ALTERNATE 
 
 Is the method of finding what quantity of each of the ingredi- 
 ents, whose rates are given, will compose a mixture of a given 
 rate : So that it is the reverse of Alligation Medial, and may be 
 proved by it. 
 
 * If an dunce or any other quantity of pure gold he divided into 24 equal parts, these 
 parts are called carats, but gold iso<ten mixed with some baser metal, which is called the 
 alloy, and the mixture is said to be so many carats fine, according to the proportion of 
 pure gold contained in it; thus, if 22 carats of pure gold and 12 of alloy be mixed to- 
 gether it is said to be 22 carats fine. See page. 43. 
 
 300. What is Mligaiion Alternate ? 
 
 z 
 
186 ALLIGATION ALTERNATE, 
 
 CASE I. 
 
 When the mean rate of the whole mixture, and the rates of all the m- 
 gredients are given, and the quantity not limited. 
 
 RULE. 
 
 1. Place the several rales or prices of the simples, being re- 
 duced to one denomination, in a column under each other, and the 
 mean price reduced to the same denomination, at the left hand. 
 
 2. Connect with a continued line the price of each simple or 
 ingredient, which is less than that of the mean rate, with one, or 
 any number of those, which are greater than the mean rate ; and 
 each greater rate or price with one, or any 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 are 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. a Ib. some at 2s. some af 
 4*. and some at 5s. a Ib, : How much of each sort must he mix, that 
 he may sell the mixture at 3s. 4d. a Ib. ? 
 
 * By connecting the less rate to the greater, and placing the difference between them 
 and the mean rate alternately, the quantifies resulting are such, that there is precisely 
 as much gained hy 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 ivS always a less with a greater than the mean price, there 
 will be an equal balance of ios* and gain between every two, and consequently an equal 
 balance on the whole. 
 
 If any of the simples be of little or no value with respect to the rest, its rate is supposed 
 to be nothing; as water mixed with wine, aijd alloy with gold or silver. 
 
 301. "How do you find the quantify of each ingredient, when the mean rate of the wholt 
 'mixture, and the rates of all the ingredients are given, and the quantity not limited ? 
 
ALLIGATION ALTERNATE. 
 
 Ib s. d. 
 
 20 at 1 6 
 
 8 " 2 
 
 16 " 4 
 
 22 " 5 
 
 Ib. s. d. 
 
 8 at 1 6 
 
 20 " 2 
 
 22 "40 
 
 
 d. 
 
 
 !18 ^ 
 
 Mean rate 40d. 
 
 24-> 
 48-' 
 
 
 60 ' 
 
 
 d. 
 
 
 ( 18 ^ 
 
 Mean rate 40d. 
 
 t 
 
 148--^ 
 (60 
 
 Ans. 
 
 Ans. 
 
 16 
 
 
 
 NOTE. Questions in this case admit of as many answers, as there 
 are various ways of connecting the rates of the ingredients together. 
 
 2. A grocer would mix the following qualities of sugar ; viz. at 10 
 cents, 13 cents, and 16 cents a pound; what quantity of each sort 
 must be taken to make a mixture worth 12 cents a pound ? 
 
 ib Ibs. at 10 cts. 
 
 Ans. ?2 Ibs. at 13 cts. 
 
 f 2 Ibs. at 18 cts. 
 
 3. It is required to mix rum at 80 cents, and at 70 cents a gallon 
 with water, that the mixture may be worth 75 cents a gallon ; what 
 quantity of each must be taken ? 
 
 80 galls, at 80 cents. 
 Ans. < 5 galls, at 70 cents. 
 ( 5 galls, of water. 
 
 4. A goldsmith would mix gold of 19 carats fine, with some of 16, 
 18, 23, and 24 carats fine, so that the compound may be 21 carats 
 fine ; what quantity of each must he take ? 
 
 5 oz. of 16 carats fine. 
 5 oz. of 18 carats fine. 
 Ans. <( 5 oz. of 19 carats fine. 
 10 oz of 23 carats fine. 
 10 oz. of 24 carats fine. 
 
J88T ALLIGATION ALTERNATE. 
 
 CASE II. 
 
 When the rates of all the ingredients, the quantity of but one of them, 
 and the mean rate of the whole mixture are given, to find the several 
 quantities of the rest, in proportion to the quantity given, 
 
 RULE. 
 
 Take the difference between each price, and the mean rate, 
 and place them alternately as in Case I Tlfen, as the difference 
 standing against that simple, whose quantity is given, is to that, 
 quantity, so is each of the other differences, severally, to the sev- 
 eral quantities required. 
 
 EXAMPLES. 
 
 1. A merchant has 20 Ibs. of tea at $1 04 a pound, whiob he would 
 mix with some at 98 cents, some at 92 cents, and s- me at 80 cents a 
 pound ; how much of each sort must he take to mix with the 2O 
 pounds, that he may sell the mixture at 96 cents a pound ? 
 
 4 stands against the given quantity. 
 16 
 8 
 2 
 
 Ibs. Ihs. cts. 
 (16 : 80 at 98) 
 
 As 4 : 20 :: 1 8 40 " 9V An?. 
 f 2 10 " 80S 
 
 2. Bought a pipe of brandy containing 120 gallons, at $1'30 a gal- 
 lon ; how much water must be mixed with it to reduce the first cost 
 to $MO a gallon 1 
 
 Ans. 21 T T galls. 
 
 3. How much gold of 1 .">, 20 and 24 carats fine, and how much al- 
 loy, must be mixed with 10 oz. of 18 carats fine, that the compositioa 
 may be 22 carats fine ? 
 
 f 10 oz. of 16 carats fine. 
 A \ 10 oz. of 20 carats fine. 
 ' \ 170 oz. of 24 carats fine. 
 * 10 oz. of alloy. 
 
 302. When the rates of all the ingredients, the quantity of but one of them, and thf 
 mean rate of the whole mixture ire given ; what is the rule for finding the several quan 
 tities of the rest, in proportion to the given quantity ? 
 
ALLIGATION ALTERNATE....FOSITION, ]89 
 
 CASE III. 
 
 When the rates of the several ingredients, the quantiir to le - 
 ded, and the mean rate of the whole mixture are given* to j~> .- 
 much of each sort will make up the quantity. 
 
 RULE. 
 
 Write the difference between the mean rate, and the serf: 
 prices alternately, as in Case I ; then, as tl^e si m of the qu 
 ties or differences thus determined, is to the given qiutnt?!y 
 whole composition ; so is the difference of each rate, to tlie ic- 
 quired quantity of each rate. 
 
 EXAMPLES. 
 
 1. A merchant having sugais at 12 dollars, 10 dollars, and 8 dol- 
 lars a cwt. would make a mixture of 30 cwt. worth 9 dollars a cwt. ; 
 what quantity of each must be taken ? 
 
 cwt. dols. 
 
 1 C[ : 5 at 12 ) 
 
 1 As G : 30 : : <1 : 5 " JO VAtfc. 
 1+3=4 (4 : 20 " 8 ) 
 
 sum=6 30 
 
 2. A goldsmith has several sorts of gold; viz. of 15, 17, 20 and ' 
 carats fine, and would melt together of all these sorts, so much as 
 may make a mass of 40 oz. of 18 carats fine ; how much of each sort 
 is required? 
 
 16 oz. of 15 carats fine. 
 . . 8 oz. of 17 carats fine. 
 k 4 oz of 20 carats fine. 
 12 oz. of 22 carats fine. 
 
 POSITION. 
 
 POSITION is a rule, by which any true or required numbers are 
 found by means of false or supposed numbers. It is of two kinds, 
 Single and Double. 
 
 303. \Vhen the rates of the several ingredients the quantity to be compounded, and the 
 mean rate of the whole m'>r.iurp. are given ; what is the rule for finding how much of each 
 sort will make up the quantity ? 304. What is Position ? 
 
190 SINGLE POSITION. 
 
 SINGLE POSITION 
 
 Is the working of one supposed number, as if it were the true 
 one, to find the true number. 
 
 RULE. 
 
 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 errors 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 
 agree with the sum, it is right. 
 
 EXAMPLES. 
 
 1. A school master, being asked how many scholars he had, said, 
 if 1 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 ? 
 
 Suppose he had 80 As 290 ; 435 : : 80 
 
 As many = 80 80 
 
 2 as many =60 120 
 
 | as many = 40 29|0)3480jO(120 Ans. 120 
 
 J as many = 20 29 90 
 
 J as many =10 60 
 
 58 30 
 
 290 58 15 
 
 435 Proof. 
 
 2. A person lent his friend a sum of money unknown, to receive 
 interest for the same at 6 percent, per annum. Sin pie interest, and at 
 the end of 12 years, received for principal and interest $860 : What 
 was the sum lent ? 
 
 Ans. $500. 
 
 *The operations contained in the question being performed upon the answer or num- 
 ber to be found, will give the result contained in the question. The same operations, per- 
 formed on any other number, will give a certain result. When the results are propor- 
 tional to their supposed numbers, it is manifest that the result of the operations performed 
 on the supposed number, must be to the supposed number, as the result in the question is 
 to the true number or answer. In any case, when the results are not proportional to 
 their supposed numbers the answer cannot be foun^ by this rule. 
 
 305. What is Single Position! 306. What is the rule for working questions in 
 
 Single Position ? 
 
DOUBLE POSITION. 191 
 
 3. A, B and C joined their stocks, and gained $ 9 >53 12jc. of which 
 A took up a certain sum, B took up four times so much as A, and C, 
 live times so much as B : What share of the gain had each ? 
 
 i$lA 12|c. A's share. 
 
 Ans. ? 56 50 B's share. 
 
 (282 50 C's share 
 
 4. A and B, talking of their ages, B said his age was once and an 
 half the age of A ; C said his was twice and one tenth the age of both, 
 and that the'sum of their ages was 93 : what was the age of each ? 
 
 Ans. A's 12, B's 18, and C's 63 years. 
 
 5. Seven eights of a certain number exceeds four fifths by 6 : what 
 is the number ? Ans. 80. 
 
 6. What number is that, which, being increased by f , |, and of 
 itself, the sum will be 234J ? Ans. 90. 
 
 DOUBLE POSITION. 
 
 DOUBLE POSITION teaches to resolve questions by making two 
 suppositions of false numbers. 
 
 Those questions, in which the results are not proportional, to 
 their positions, belong to this rule : such are those, in which the 
 number sought is increased or diminished by some given number, 
 which is no known part of the number required. 
 
 RULE.* 
 
 1. Take any two convenient numbers, and proceed with each 
 according to the conditions of the question. 
 
 2. Place the result or errours against their positions or suppos- 
 
 Pos. Err. 
 30 12 
 ed numbers, thus, ~\T and if the errour be too great, mark it 
 
 20^ L 6 
 with + > an d if too small, with . 
 
 * The rule is founded on this supposition, that the first errour is to the second, as the 
 difference between the true and first supposed numher is to the difference between the 
 true and second supposed number : When that is not the case, the exact answer to the 
 question cannot he found by this rule. 
 
 307. What is Double Position ? 308. And what is the rule ? 
 
HOUCLE POSITION". 
 
 : ; that is, the first position by the 
 Lib i errouTj :.ui-:i ihe la$tj:>osition by the first errour. 
 
 4 If the errours be alike, that is. both too small or both too 
 gr;.r3.t, divide the difference of the products by the difference of 
 the errours, and the quotient will be the answer. 
 
 5. If the errours be unlike ; that is, o^e too small, and the 
 other too ^reat, divide the sum of the products by the sum of the 
 errours, and the quotient will be the answer. 
 
 NOTE. When the errours are the same in quantity, and unlike in 
 quality, half the sum of the suppositions is the number sought. 
 
 EXAMPLES. 
 
 ] . A lady bought damRsk for a gown, at 8s. per yard, and lining for 
 it at 3*. per yard; the sjown and lining contained 15 yards, and the 
 price of the whole was ,3 I Os. : How many yards were there of each ? 
 
 Suppose 6 yards damask, value 48s. 
 Then she must have 9 yards lining, value 27s. 
 
 Sum ot their values =: 75s. 
 So that the first errouris 5 too much, or + & 
 
 Again, suppose she had 4 yards of damask, value 32s. 
 Then she must have 11 yards of lining, value 33s. 
 
 Sum of their values = 65s. 
 So that the second errour is 5 too little, or 5s. 
 
 65+ s. d. 
 
 Then "V" 5 yards at 8s. = 2 00 
 
 4^^5__ 10 yards at 3s. = ] 10 
 
 20 30 3 10 proof. 
 
 20 
 
 Sum of errors 5-}- 5 = 10) 50 
 
 Ans. 5 yds, damask, and 15 5=10 yds. 
 Or, 6 + 44-2=5 as before. [lining. 
 
PERMUTATION. 193 
 
 2. A labourer was hired for 60 days upon this condition, that for 
 every day he wrought, he should receive 75c. ; and tor everyday he 
 was idle, should forfeit 37jc. ; at the expiration of the time he re- 
 ceived 18: How many days did he work, and how many was he 
 idle ? 
 
 Ans. He was employed 36 days, and was idle 24. 
 
 3. 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 ? 
 
 Ans. 80 feet. 
 
 4. A farmer, having driven his cattle to market, received for them 
 all $320, 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, 5 cows, and 20 calves. 
 
 PERMUTATION. 
 
 PERMUTATION is the method of finding how many different ways 
 the order or position of any given number of things may be chang- 
 ed or varied. 
 
 To find the number of permutations or changes that can be made of 
 any given number of things, all different from each other. 
 
 RULE. 
 
 Multiply all the terms of the natural series of numbers, from 
 one up to the given number, continually together, and the last 
 product will be the answer required.* 
 
 *The reason of this rule may be shown thus, any one thing a is capable of one posi- 
 tion only as, a. 
 
 Any two things a and b are capable of two variations only ; as ab t ba ,- whose number 
 is expressed by 1X2. 
 
 If there be three things a, b, and c , then any two of them, leaving oul the third, will 
 have J X2 variations ; and consequently, when the third is taken in, there will be 1X2X3 
 variations ; and so on as far as you please. 
 
 309. What is Permutation ? 310. What is the rule for finding the number of 
 
 changes that can be made of any given number of things, all differentfrom each other ? 
 
 AA 
 
194 GAUGING. 
 
 EXAMPLES. 
 
 1. .How many changes or variations can be made of the three first 
 letters of the alphabet ? 
 
 a b c 
 c b 
 
 1X2X3=6 Ans. Changes or varia- 
 
 tions at large. 
 
 2. Christ church, in Boston, has 3 bells : how many changes may 
 be rung on them ? 
 
 Ans. 40320. 
 
 3. Nine gentlemen met at an inn, and were so well pleased with 
 their host, and with each other, that in a lrotick,they agreed to tarry 
 so long as they, together with their host, could sit every day in a dif- 
 ferent position at dinner : Pray how long, had they kept their 
 agreement, would their irolick have lasted ? 
 
 Ans. 3628800 days,=9935 T 5 / T years. 
 
 GAUGING. 
 
 GAUGING is the art of measuring all kinds of casks or vessels 
 used for liquor, and of determining the quantity they will contain. 
 
 The instruments used in gauging are the gauging rod, callipers, 
 the sliding rule, and Gunter's scale. 
 
 RULE. 
 
 Take the dimensions of the cask in inches, viz. the diameter at 
 the bung and head, and the length of the cask ; subtract the head 
 diameter from the bung diameter, and note the difference. 
 
 If the staves of the cask be much curved or bulging between 
 the bung and head, multiply the difference between the bung and 
 head diameter by -7 ; if not quite so much curved, by 65 ; if 
 they curve yet less, by -6 ; and if they are aimost or quite straight, 
 by '55, and add the product to the head diameter, the sum will 
 be a mean diameter, by which the cask is reduced to a cylinder. 
 
 311 What is Gauging 1 312. What are the rules for measuring casks, and find- 
 ing tkeir contents in gallons ? 
 
GAUGING. 195 
 
 Square the mean diameter, thus found, then multiply it by the 
 length : divide the product by 294 for wine, or by 359 for ale or 
 beer, and the quotient will be the content in gallons. 
 
 NOTE 1. These divisors are found by dividing 231 and 282 by 
 
 7854. If the square of the mean diameter be multiplied by -7854 
 
 and the product multiplied by the length, the last product will be the 
 
 content in cubick inches, which being divided by 231 for wine, or by 
 
 282 for ale or beer, the quotient will be the content in gallons. 
 
 NOTE 2. The length and head diameter are usually taken by cal- 
 lipers, allowing, for the thickness of both heads, 1 inch, Ij inch, or 
 2 inches, according to the size of the cask. 
 
 The head diameter must be taken close to its outside, and for small 
 casks, add 3 tenths of an inch ; for casks of 30, 40, or 50 gallons, 4 
 tenths; and for larger casks 5 or 6 tenths ; and the sum will be very 
 nearly the head diameter within. 
 
 The bung diameter is usually taken by the gauging rod, and in ta- 
 king it, observe by moving the rod backward and ,foi ward, whether 
 the stave, opposite the bung, be thicker or thinner than the rest, and 
 if it be, make allowance accordingly. 
 
 EXAMPLE. 
 
 What is the content in wine, and ale or beer gallons, of a cask, 
 whose bung diameter is 35 inches, head diameter 27 inches, and 
 length 45 inches ? 
 
 Bung diaraeter=35 Square of the diameter^- 1062-76 
 
 Head diameter=27 Length= 45 
 
 Difference=8 531380 
 
 7 425104 
 
 5-6 47824-20 
 
 Add the head diam. 27 
 
 47824-20294=162-69+ 
 
 Wine galls. f . 
 47824-20-7-359=1 33-21+ r 
 Ale or Beer galls. 
 
 Squared 1062-76 
 
 313. How do you obtain the divisors, named in the rule ? 314. How are the differ* 
 
 rut diameters of the cask to be ascertained ? 
 
196 MECHANICAL POWERS. 
 
 MECHANICAL POWERS. 
 
 OF THE LEVER OR STEELYARD. 
 
 It is a principle in mechanicks, that the power is to the 
 weight, as the velocity of the weight, is 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 160 Ibs. 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 dis- 
 tance from the prop to the power is 10 1=9 feet; therefore, 
 as 1 foot : 9 feet : : 160 Ibs. : 1440 Ibs. Ans. 
 
 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 sufficient incumbrance for the 
 horse : Now supposing two passengers to weigh 336 pounds, and the 
 body of the chaise 84 pounds more ; what will the beast in both these 
 cases bear, more than his harness ? 
 
 Weight of the passengers and chaise = 420 Ibs. and 9 feet = 108 
 inches ; 
 
 in. 11*. 
 
 Then, as 108 in. : 420 Ibs. j j^ ! ^fi Ans - 
 
 OP THE WHEEL AND AXLE. 
 
 The proportion for the wheel and axle (in which the 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 sus- 
 pended by the axle. 
 
 315. What is a general principle in mechanicks 1 - =-316. What is the power of the 
 
MECHANICAL POWERS. }97 
 
 If the diameter of the axle be 6 inches, and (he diameter of the 
 wheel, 60 inches ; .what weight suspended from the axle will be bal- 
 anced by a power of 1 Ib, applied to the wheel ? 
 
 in. in. Ib. Ibs. 
 As 6 : GO : : 1 : 10 Ans. 
 
 OP THE SCREW. 
 
 The power is to the weight which is to be raised, as the dis- 
 tance between two threads of the screw, is to the circumference 
 of a circle described by the power applied to the end of the 
 lever. 
 
 RULE. 
 
 Find the circumference of the circle described by the end of 
 the lever ; then, as that circumference is to the distance be- 
 tween 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 proportional to the quantity of surface, but to the weight of 
 the incumbent part ; and, at a medium, J part of the effect of the 
 machine is destroyed by it, sometimes more and stmetimes less. 
 
 There is a screw whose threads are an inch asunder; the lever by 
 which it is turned 30 inches long, and the weight to be raised a ton, 
 or 2240 Ibs. : What power or force 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-diameter of the circle, the diameter is 
 60 inches; then, as 113 : 355 : : 60 : 188-496 inches nearly, the 
 circumference. 
 
 in. in. Ibs. Ibs. 
 Therefore, 188-496 : 1 : : 2240 : 11-88+ Ans. 
 
 Let the weight be 2240lb. the power ll-881b. and the lever 30 inch- 
 es : Required the distance between the threads? 
 
 Ibs. Ibs. in. in. 
 
 As 2240 : 11-88 : : 288-496 : 1 nearly, Ans. 
 
 Let the power be ll-881b., the weight 22401b., and the threads an 
 inch asunder, to find the length, of the lever. 
 Ibs. Ibs. in. in. 
 
 As 11-88 : 2240 : : 1 : 188-5; then, as 355 : 113 : : 188-5 : 60 
 inches nearly, the diameter, and 60-7-230 inches, Ans. 
 
 318. What is the power of the screw 1 319. What is the rule for finding the power 
 
 'which must be applied to the end of the lever, sufficient to turn the screw, that is, to raise 
 
 any given weight ? 320. When required, lioiv do you Jind the distance between the 
 
 spiral threads of the screw ,- or the length of the lever ? 
 
( 198 ) 
 
 USEFUL AND DIVERTING* EXERCISES. 
 
 1. Whaf difference is there between twice 5 and twenty, and twice 
 twenty live ? Ans. 20. 
 
 2. In an orchard of fruit trees J of them bear apples, J pears, 
 plums, 60 of them peaches, and 40 cherries; how many trees does 
 the orchard contain ? Ans. 1200. 
 
 3. A merchant begins the world with $500, and finds that by his 
 distillery he clear* $5009, in 6 years ; by his navigation, $5000 in 7J 
 years ; and that he spends in gambling $5000 in 3 years ; how long 
 will his estate last ? Ans. 30 years. 
 
 4. A can do a piece of work alone in 7 days, and B in 12 ; in what 
 4ime will both, working together, finish it ? Ans. 4 T 8 days. 
 
 5. A. and B. are on opposite sides of a circular field 268 poles 
 about ; they being to go round it, both the same way at the same in- 
 stant of time ; A. goes 22 rods in 2 minutes, and B. 34 rods in 3 min- 
 utes; how many times will they go round the field before the swifter 
 
 overtakes the slower? ^ A 16^ times. 
 
 - 
 
 6. A water tub holds ^3 gallons ; the pipe, which conveys the water 
 to it usually admits 7 gallons in 5 minutes ; and the tap discharges 20 
 gallons in 17 minutes. Now, supposing both these to be carelessly 
 left oppn, 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 filled ? 
 
 Ans. 32 min. 58f[| sec. 
 
 7. A hare starts 12 rods 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 her at the rate of 
 16 miles an hour ; how long will the course hold, and what space will 
 be run over, from the spot where the dog started ? 
 
 Ans. 2288 ft. 97 seconds. 
 
 8. Required the number, from which, if 7 be subtracted, and the 
 remainder be divided by 8, and the quotient be multiplied by 5, and 
 4 added to the product, the square root of the sum extracted, and 
 three fourths of that root cubed, the cube, divided by 9, the last quo- 
 tient may be 2 1 ? Ans. 103. 
 
 9. Suppose a lighthouse built on the top of a rock ; the distance 
 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, 840- yards; and fiom the top of the light house 900 
 yards ; the height of the light house is required ? 
 
 Ans. 76-77+ yds. 
 
 10. Sound, uninterrupted, moves at the rate of 1142 feet per se- 
 cbnd; if the time between the lightning and thunder be one minute, 
 at what distance was the explosion ? Ans. 12-977+ mile* 
 
USEFUL AND DIVERTING EXERCISES. 199 
 
 11. If the earth be 7911 miles in diameter, and the moon 2180 
 miles; how many moons will it take to make one earth ? 
 
 Ans. 47-788. 
 
 12. A father left his estate of $1300 per annum to his only sen, but 
 he being only 14 years of age, his guardian was to pay 100 per an- 
 num for board, education, &c., and the surplus was to be put out to 
 interest for his benefit, at 6 percent, compound interest ; now allow- 
 ing no loss, what sum had his guardian to pay him when he was of 
 age ? Ans. 1115 dols 33 cts. 5 m. 
 
 13. Hiero, king of Sicily, ordered his jeweller to make a crown, 
 containing 63 ounces of gold. The workman thought that substituting 
 part silver was only a proper perquisite, which taking air, Archimedes 
 was appointed to examine it, who on putting it into a vessel of water 
 found it raised the fluid 8-2245 cubick inches, and having discovered 
 that the inch of gold weighed 10-36 ounces, and that of silver bat 
 5-85 ounces, he found what part of the king's gold had been changed. 
 Repeat the process and inform us what part of it was gold, and what 
 silver? A , $ 28-8038 oz. silver. 
 
 u I 34- 1962 oz. gold. 
 
 14. A person having driven a stock of cattle to market, received 
 for them all 450 dollars, he was paid at 50 dollars for each horse, 20 
 dollars for each cow, and 4 dollars for each sheep ; the number of 
 cows was double the number of horses ; and there were three times 
 as many sheep as cows : what did he receive for the horses, what for 
 the cows, and what for the sheep, and how many of each sort were 
 there (24 Sheep $ 96 
 
 Ans. J 8 Cows $160 
 f 4 Horses $200 
 
 15. A gate-keeper is to receive 6 cents for every wagon, 4 cents 
 for every gig, 2 cents for every horseman, and 1 cent for every footman 
 that passes the gate ; at the year's end he found that 3150 gigs had 
 passed, and that 7 gigs passed when 5 wagons did, and 4 horsemen 
 passed when 6 footmen did, and 5 footmen passed when 3 gigs did, 
 what number of wagons, horsemen, and footmen passed, and how 
 much did the gate-keeper receive ? 
 
 Gigs 3150 
 
 Wagons 2250 
 
 Ans. ^ Footmen 
 
 Horsemen 3500 
 
 Amount of toll $383-50 
 
 16. If 1000 bricks lie 6 inches from each other in a straight line, 
 and a person be employed to gather them up one by one, and place 
 them on a pile which is one foot from the first brick, how far will he 
 have walked when he shall have placed the last brick on the pile ? 
 
 Ans. 91M. 7fur. 186yds 2ft. 
 
 17. A man dying left his wife in expectation that a child would be 
 afterwards added to the surviving family ; and making his will, order- 
 ^d, that, if f he child were a son, of his estate should belong to him, 
 
200 USEFUL AND DIVERTING EXERCISES. 
 
 and the remainder to Ids mother; but if it were a daughter, he ap- 
 pointed the mother g, and the child the remainder. But it happened, 
 that the addition was both a son and a daughter, by which the widow 
 lost in equity, $2400 more than if there had been only a girl. What 
 would have been her dowry had she had only a son ? 
 
 Ans. $2100. 
 
 18. When first the marriage knot was tied 
 Betwixt my wife and me, 
 My age with hers did so agree, 
 As nineteen does with eight and three ; 
 But after ten and half ten years, 
 We man and wife had been, 
 Her age came up so near to mine, 
 As two times three to nine. 
 
 What were our ages at marriage ? 
 
 Ans. 57 and 33. 
 
 19. Three jealous husbands with their wives, being ready to pass 
 by night over a river, do find at the water side a boat which can car- 
 ry but two persons at once, and for want of a waterman, they are ne- 
 cessitated to row themselves over the river at several times : The 
 question is, how those six persons shall pass by 2 and 2, so that none of 
 the three wives may be found in the company of one or two men, 
 unless her husband be present ? 
 
 20. As I was going to St. Ives, 
 I met seven wives, 
 Every wife had seven sacks, 
 Every sack had seven cats, 
 Every cat had seven kits, 
 Kits, cats, sacks and wives, 
 How many were going to St. Ives ? 
 
 21. A countryman having a Fox, a Goose, and a peck of corn, in his 
 journey, came to a river, where it so happened that he could carry 
 hut one over at a time. Now as no two were to be left together that 
 might destroy each other ; so he was at his wit's end how to dispose of 
 them ; for, says he, tho' the corn can't eat the goose, nor the goose 
 eat the fox ; yet the fox can eat the goose, and the goose eat the corn. 
 The question is, how he must carry them over that they may not de- 
 Tour each other ? 
 
 22. A man driving his geese to market, was met by another, who 
 said, Good morrow master, with your hundred geese ; says he, 1 have 
 not an hundred, but if I had half as many as 1 now have, and two geese 
 and a half beside the number I now have already, I should have an 
 hundred. How many had he ? 
 
 23. Two men were driving sheep to market, says one to the other, 
 give me one of yours and I shall have as many as you ; the other says. 
 give me one of yours and 1 shall have as many again as you. How 
 many had each ? 
 
[After the scholar, male or female, has acquired a competent knowledge of Arithme- 
 t!ck, or of its fundamental rules, instruction in the mode of keeping accounts should be 
 attended to. By this it is not meant to recommend that the son or daughter of every 
 farmer, mechanick, or shop keeper should enter deeply into the science as practised by 
 the merchant, engaged in extensive business, for such study would engross a great portion 
 of time which might be more usefully employed in acquiring a proper knowledge of a 
 trade or other employment. 
 
 Persons employed in the commoa business of life, who do not keep regular accounts, 
 are subjected to many losses and inconveniences ; to avoid which, the following simple 
 and correct plan, is recommended for their adoption. 
 
 Let a small book be made, or a few sheets of paper sewed together, and ruled after the 
 examples given in this system. In the book, termed the Day Book, are duly to be entered, 
 daily, all the transactions of the master or mistress of the family, which require a charge 
 to be made, or a credit to be given to any person. No article thus subject to be entered, 
 should on any consideration, be deferred till another day. Great attention should be giv- 
 en to write the transaction in a plain hand ; the entry should mention all the particulars 
 necessary to make it fully understood with the time when they took place ; and if an ar- 
 ticle be delivered, the name of the person to whom delivered is to be mentioned. No 
 scratching out may be suffered ; because it is sometimes done for dishonest purposes, and 
 will weaken or destroy the authority of your accounts. But if, through mistake, any 
 transaction should be wrongly entered, the error must be rectified, by a new entry ; and 
 the wrong one may be cancelled by writing the word .Error, in the margin. 
 
 A book, thus fairly kept, will at all times show the exact state of a person's affairs, and 
 have great weight, should there at any time be a necessity of producing it in a court of 
 Justice. 
 
 The instructor, who feels a parental solicitude for the permanent welfare of his pupils, 
 cannot in any way so mucU contribute to their success in life, with so little trouble, as to 
 teach them to understand this abridged, complete and simple system of Book-Keeping, 
 It contains all the important principles of extended and expensive works on the science; 
 all, in fact, that is necessary to be known by the Farmer, Mechanick, and Shopkeeper, 
 relating to accounts ; and yet with very little explanation and repeated copying and bal- 
 ancing the accounts, will be so fully understood and deeply impressed on the memory of 
 scholars of common mind, as never to be forgotten ; while their knowledge of common 
 arithmetick and practical penmanship will thereby be greatly improved.] 
 
 BOOK-KEEPING is the art of recording mercantile transactions 
 in a regular and systematick manner. 
 
 Book-Keeping by SINGLE ENTRY chiefly records the transactions 
 on creciit, and for this purpose two books are necessary, called 
 the Day-Book, and the Leger* 
 
 THE DAY-BOOK. 
 
 Each page ot the Day-Book should be ruled with two columns 
 on the right hand for dollars and cents, and one column on the 
 left for inserting the page or folio of the Leger on which the ac- 
 count is posted. 
 
 B* 
 
BOOK-KEEPING. 
 
 The Day-Book begins with an account of the owner's proper* 
 ty, debts, &c. ; then follows a detail of the occurrences of trade, 
 set down in the order of time in which they take place. 
 
 The name of the person, or customer, is first written with the 
 term Dr. or Cr. annexed, according as he becomes debtor or cred- 
 itor by the transaction ; and this may be distinguished by the 
 following general nth. 
 
 The person who receives is Debtor, and the person who gives 
 or parts with any thing is Creditor. 
 
 Thus, if I sell goods on credit, I enter A. B. the buyer, Dr. to 
 the goods, specifying their quantity and value. 
 
 If I buy goods on credit, I enter C. D. the seller, Cr. by the 
 goods, specifying their quantity and value. 
 
 By the same rule, if I pay money, the person to whom I pay it, 
 is made Dr. to cash for the amount ; and if I receive money, 
 the person from whom I receive it, is made Cr. by cash for the 
 amount. 
 
 And if debts be contracted or discharged by any other means, 
 the same rule is to be observed ; the person who becomes in- 
 debted to me, is entered Dr. and the person to whom I become 
 indebted, Cr. Also, the person whose debt I discharge, is made 
 Dr. and he that discharges a debt due by me, is Cr. 
 
 TEIE LEG-SB. 
 
 Each page of the Leger should be ruled with two columns on 
 the right hand for dollars and cents, and with one column on 
 the left, for entering the date of the transaction. 
 
 In the Leger, the dispersed accounts of each person in the 
 Day-Book, are collected together, and the Drs. and Crs. are 
 placed upon opposite sides of the same page, or book. 
 
 N. B. It is usual in commencing business, to mark the first Day- 
 Book and Leger with the letter A, and succeeding books, with 
 B, C, &c. 
 
 The person's name is written in large^characters as a title; on 
 the left hand page, he is styled Dr. and on the opposite, or right 
 hand page Cr. On these pages the transactions are entered as 
 they stand Drs. or Crs. in the Day-Book. For instance, A. B. is 
 debited for whatever he has bought of me ; and on the opposite 
 page, he is credited for the payments he has made. In short 
 whatever I have given him is on the Dr. side, and what he has 
 given me on the Cr.; and the difference between the Dr. andCr. 
 sides is called the balance. 
 
 The Leger should have an index, in which the titles of the accounts should be arran- 
 ged under their initial letters, with the number of the page or folio in the Leger, on which 
 tiie account is posted, 
 
FORM OF A DAY-BOOK. 
 
 DAY-BOOK.. ..A. 
 
 Concord, JV. H. Jan. 1, 1826. 
 
 Inventory of all my properly and debts, taken 
 this day by me, Charles Hastings, of Concord. 
 
 $ cte. 
 
 Money on hand, 500-00 
 
 Merchandize, 1500-00 
 
 Furniture, - - 200-00 
 
 Real Estate, 150000 
 
 Bank Stock, 500-00 
 
 Horse, chaise and harness, - - 300 00 
 Cows, hogs, farming tools, &c, - 75-00 
 Thomas Jackman owes me on ac- 
 count, 125-00 
 
 Sundry promissory notes, - - - 375 00 
 
 5075-00 
 1 owe I. & F, Williams on account, 75-00 
 
 My net property, $5000-00 
 
 Q^T" At the commencement of every year, if no oftener, 
 the trader, farmer ormechanick should make an inventory 
 of his property, either in the general form above, or by en- 
 tering the articles individually ; also of his debts, securities, 
 &c. ; in order that he may know the exact state of his af- 
 airs. 
 
 Entered. 
 1 
 
 Entered. 
 
 2 
 
 Entered. 
 3 
 
 Entered. 
 
 Concord. N. H.Jan. 1, 1826. 
 
 Thomas Jackman, 
 
 To balance of former account, 
 
 Dr. 
 
 Isaac F. & W. Williams, Or. 
 
 By balance of former account, - - - 
 
 Robert Childs, 
 
 Dr. 
 
 To 3 yds. broadcloth, at $3-50 per yd. $10-50 
 " 7 do. linen, at -60 4-20 
 
 Feb. 28. 
 
 Isaac F. & W. Williams, 
 
 To cash paid their order, - - - 
 
 Dr. 
 
 125 
 
 75 
 
 00 
 
 00 
 
 1470 
 
 75 
 
 00 
 
FORM OF A DAY-BOOK. 
 
 Concord, .V. H. Feb. 28, 1826. 
 
 Entered. 
 6 
 
 Entered. Robert Childs, Cr. 
 
 3 By cash, 
 
 March 3. 
 
 Thomas Jackraan, Cr. 
 
 Entered. 
 
 ' By his note of this date, payable in 30 days, to 
 
 balance his account, ...._- .^5 
 4th. 
 
 Samuel A. Morrison, Dr. 
 
 To cash paid him, 2-00 
 
 Entered. To order on J C. West, 3-00 
 
 4 To 10 yds. shirting at 20 cents, - - - 2-00 
 
 i. By his bill of work to this date - - - 
 
 Jacob B. Moore, Cr 
 
 Entered. By 1 dozen Marshall's Spellings, - -1-50 
 J dozen Pike's Arithmetick, - - - 5-00 
 " 1 Walker's Dictionary. .... 1-00 
 u 1 dozen Murray's Reader, - - -4-50 
 
 1 Writing Book, -12J 
 
 " 1 Slate, -25 
 
 u 2 quires paper, -42 
 
 10th. 
 
 Entered. Dudley Leavitt, Dr 
 
 6 To cash paid his order this day, - - 
 
 14th. 
 
 Entered. Samuel A. Morrison, Dr 
 
 To cash, - 
 
 Jacob B. Moore, Dr 
 
 rf To 3 barrels cider, at $1 00 3-00 
 
 " 6 bushels corn, -75 4-50 
 
 " 2 do. rye, " -67 1-3 
 
 To cash balance, ---.... 3-95 
 
 Dudley Leavitt, 
 By his bill for instructing my son, 
 
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BOOK-KEEPING. 
 
 It is the practice of some accountants in single entry, to have a 
 column ruled on the page of the Leger, for the purpose of marking 
 against each entry the page of the Day-Book whereon it is originally 
 made. But this, in common business, is wholly unnecessary the date, 
 in most cases, being a sufficient guide to any entries in the Duy-Book. 
 
 When all the transactions are correctly posted into the Leger, each 
 account is balanced by subtracting the amount of the less side from 
 the amount of the greater, and entering the remainder or balance on 
 the less side, by which the two sides are made equal. 
 
 When the place assigned in the Leger for any person's account is 
 filled with items, the person's name must not be entered the second 
 time, but may be transferred to another page in the following man- 
 ner, viz. Add up the columns on both sides, and against the sum 
 write, " Amount transferred to page " inserting the number of the 
 page where the new account is opened. After titling the new ac- 
 count and entering the number of the page in the index, write on the 
 Dr. side of the new account, " To amount brought from page " 
 inserting the number of the page from which the old account was 
 brought, and on the Cr. side, " By amount brought from page ," in- 
 serting also the page where the old account was ; and place the sums 
 in the proper columns. 
 
 When the first Leger is rilled up, a new one may be opened as fol- 
 lows, viz. At the end of the preceding Leger, draw out a balance ac- 
 count, entering the debits and credits on their respective Dr. and Cr. 
 side, and transfer each unbalanced account to its respective Dr. or Cr. 
 side to the new Leger. The first Leger may be marked A, the se- 
 cond B, and so on in alphabetical order. 
 
 To find iha profit and loss, for any given time, an inventory of the 
 property and debts must be taken, and the net amount of which, when 
 compared wila the net amount of property at commencing business, 
 will show the profit nnd loss. 
 
 In Book-Keeping by Double Entry, a variety of books are made 
 use of, some of which, though not absolutely necessary to the me- 
 chanick or farmer, would be convenient and useful to him on many 
 accounts. 
 
 1. The Cash Book. This should be made of a sufficient quantity 
 of paper ruled in the form of a leger, with columns on each page for 
 entries of cash received and cask paid. The practice of entering" sums 
 as received, and all sums as paid out, may serve on some occasions to 
 correct misjmderstandiDgs on settlement of accounts, besides the op- 
 portunity it affords of ascertaining at any time the state of a man's 
 money concerns. 
 
 2. Account of Expenses. This may be kept in a separate book, 
 ruled like a Day-Book, or it may bo onterad as an account in the Leg- 
 er. The prudent farmer or mechanick will ever feel anxious to 
 know the amount of his contingent expenses, the better to ascertain 
 the exact stute of his income. 
 
 3. Letter BooL As bad consequences frequently result from a neg- 
 lect to preserve copies of letters sent on business, a book should be 
 kept for the purpose of entering and preserving a copy of every let- 
 ter or memorandum concerning business which is sent awav bv an in- 
 dividual. J J 
 
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RETURN TO the circulation desk of any 
 University of California Library 
 or to the 
 
 NORTHERN REGIONAL LIBRARY FACILITY 
 Bldg. 400, Richmond Field Station 
 University of California 
 Richmond, CA 94804-4698 
 
 ALL BOOKS MAY BE RECALLED AFTER 7 DAYS 
 2-month loans may be renewed by calling 
 
 (415)642-6233 
 1-year loans may be recharged by bringing books 
 
 to NRLF 
 Renewals and recharges may be made 4 days 
 
 prior to due date 
 
 DUE AS STAMPED BELOW 
 
58479 .1 
 
 14 DAY USE 
 
 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 
 
 SUMMER SESSIONS 
 
 DEC b 1965 
 
 JM31 !96s 
 
 J*h. )<?C f 
 
 Jfl*l.. 
 
 & ' 
 
 *L/ T *&(> 
 
 
 
 
 a^y~- 
 
 
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 JAY i 1966 
 Ivipy' i 1966 
 
 
 / K ftti 
 
 
 7* / f 99 
 
 ' 'i 
 
 
 NO/5 1^65 
 
 
 MAS 7 locjr 
 
 
 Ji/^lM / ' '' 
 
 
 r V JUt' i'O KtU -10 AM 
 
 ^V^X 8 ?,^ 8 U-EggSnU. 
 

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