rt f ^ H "^o^' IN MEMORIAM FLORIAN CAJORl fU^ I American Progressive Series, THE COMPLETE ARITHMETIC BY MILTON B. GOFF, A.M., M PROFESSOR OF MATHEMATICS IN THE WESTERN UNIVERSITY OF PENNSYLVANIA- PITTSBURGH : PUBLISHED BY H. I. GOURLEY, Agt 98 FOURTH AVENUE. Copyright, 1876. A. H. ENGLISH & CO. LANE S. HART, PRINTER AND BINDER, HARRISBURG, PA. CAJORI BY an experience of over twenty years in the class-room, the author has been convinced of these three things : 1. That both in our public schools and academies too many books hav2 been used - and too much time has been spent in the study of Arithmetic. 2. That while the study of Intellectual Arithmetic, as such, is doubtless benefi- cial, the greatest good results from its study in close connection with Written Arith- metic ; indeed, that the two cannot be separated. 3. That while it is impossible from any text-book, to teach a pupil all '* that he is to practice as a man," it is possible to comprise in less space than is usua"y done, and in a manner that will serve to strengthen the reasoning powers, many of the facts with which our youth will have to deal in after life. The aim, therefore, in the following pages has been to present in a form compact, but not obscure, all that is necessary for the instruction of youth in the science of Arithmetic, whether as a preparation for the ordinary vocations of life, or as a pre- liminary training for a course in Mathematics. The First Book in Arithmetic is intended for the younger pupils, and may be placed in their hands at a very early stage of advancement in their education. And it is safe to say, that those finishing the First Book will readily mastc" the second, or Complete Arithmetic. It is believed, however, that in the rural dis- tricts, as children do not enter school at as tender an age as in the cities, the present work will be found sufficiently elementary in the first part, even for beginners ; and although in the body of the work the matter is often as difficult, and quite as exten- sive, as in the Higher Arithmetic, the whole is arranged in so progressive a manner, that the natural and necessary development of the mind of the pupil will enabl© him to master every difficulty. We cannot see why Oral Arithmetic and Written Arithmetic should be sepa- rated ; for every process in the latter includes, more or less, the former. Yielding, therefore, to what seemed to be a necessity, we have made the " Oral " and " Writ- ten" alternate in every case where it was deemed judicious, presenting as great a variety of examples as is usually found in works devoted exclusively to the Intel- lectual, and better adapted, in our opinion, for teaching the pupil to think. At the same time, the " Oral" part can, if desired, be omitted without destroying connec- tion between subjects. The third object has been kept steadily in view. Examples and problems have been chosen with special reference to the wants of our American youth. And while we do not claim perfection in this matter, we do think we have taken a step in the right direction. We believe that no man whose son is to engage in any mechan- ical pursuit, will begrudge the space (249-257) given to the computation of carpenters', bricklayers', painters', and stonemasons' work. Nor will any one hav- ing to do with lands in the Western States, regard the time devoted to Government iii 918305 IV PREFACE. Lands (21 1^ 248) as time poorly spent. These articles we regard as being " for the greatest good of the greatest number"; since their position, somewhat early in the work, serves the purpose of giving to many who would never see them if placed on the last pages with Mensuration, an idea of what they shall have to labor at in after years. Besides, the rules may be laid by for future use and reference. In addition to the points named, we wish to call attention to our Outlines. It will be observed that we have a General Outline and for each chapter a Sub-outline. And although these may not be all that the heart could desire, we do think that every teacher may make them a valuable aid in the prosecution of his labors. We like them better than synopses placed at the close of the chapter ; for the mind, like the eye, takes in an object as a whole, then separates it into parts, then sub-divides each part, and so proceeds, until all the minutiae are, in turn, examined. This is the prin- cipal object of the Outlines ; though they also serve the purpose of a ready index. In the arrangements of subjects, deviation has been made from the " beaten track"; but only with great caution. The placing of Decimals immediately after Division of Integers was decided upon only after careful experiment by many of our most experienced educators. And last, though, perhaps, to some, not least, is the fact that although the publish- ers have spared neither pains nor expense (as the most casual examination of the books themselves shows) to make this series, consisting of the First Book in Arith- metic and The Complete Arithmetic, second to none, they propose to offer to their customers in these two volumes, what is usually found in from three to five, and at correspondingly low prices ; and thus relieve our '' humble poor," as well as those of moderate means, of much of the expense they can so ill afford in the procur- ing of text-books for their children. They do not expect, however, to change in a day the habits strengthened by years. There will be men who will differ from them and us in opinion, and who will think, despite what has been said, that two books are not enough ; that Intellectual and Written Arithmetic should be sepa^- rate works. With these the publishers have no quarrel, but, having acted in accord- ance with their best judgment, now submit to an intelligent public for their decision this question so important to all interested in the subject of education. One word to our fellow-teachers. We have made an earnest effort to present such a work as will meet with your approval and suit your wants. From comments of some of you on portions of the manuscript and proof-sheets that you have seen, we have reason to believe that you will be pleased with the result of our labors. But we cannot, and do not, expect you all to agree with us in all points. You have minds of your own ; and we are glad to know that you have not only the independence to think for yourselves, but that you possess that liberality of \iew, that grants others the privileges you claim for yourselves. Therefore, while we do not hope to escape your criticism, we do look for that honest, straightforward expression of opinion that becomes those who are engaged in a profession which, when prop- erly pursued, develops the noblest qualities of mind and heart. In the preparation of this work, we have been greatly aided and encouraged by many friends, whose kind suggestions have been thankfully received and freely used. To all we return our best thanks, as well as to our co-laborers, Messrs. J. M. Logan and H.I. Gourley, of the Pittsburgh Schools, to whose superior taste we are indebted for the neat arrangement of headings, plates, and outlines, and by whose care and vigilance many errors and crudities have been avoided. Wbstkrw University of Pa., June^ 1876. M. B. G. ''-F^^, PAGE Foggestions 7 CHAPTER I. INTEGKRS. O UTI.IJVE of Arithmetic 11 General Definitions 12 OUTLIA^E of Numeration and Notation 14 Arabic Numeration and Notation. . 15 Roman Numeration and Notation ... 21 <9 ^7J?'ZyJK^ of Addition 23 Addition 24 O UTLIJVE of Subtraction 87 Subtraction 38 O UTLIJ\rE of Multiplication .... 47 Multiplication 48 Review Problems 59 O UrZIJVE of Division 60 Division 61 Problems Under the Four Rules ... 73 CHAPTER II. DECIMALS. OUTm^E 79 Numeration and Notation 80 Addition 87 Subtraction 90 Multiplication 92 Division 95 FAGB United States Monet 101 Addition 103 Subtraction 105 Multiplication 107 Division 109 CHAPTER III. FRACTIONS. OUTLINE Ill Numeration and Notation , . 112 Reductions 117 Greatest Common DiviBor 119 Least Common Multiple 121 Cancellation 124 Addition 138 Subtraction , 142 Multiplication 145 Division 151 Review Problems 157 Converse Reductions 162 Aliquot Parts 164 Bills 167 CHAPTER IV. DENOMINATE NUMBERS. OUTLIJ^E 171 Definitions 173 Tables— Moneys 175 Weights 180 Measures .... 186 yi CONTENTS. PAOK Reductions 202 Addition 223 Subtraction 230 Multiplication 238 DiviBion 241 APPLICATIONS OF MEAS- URES. O UTLIJVB 244 Square Measure 245 Artificers' Work 249 Cubic Measure 254 E:xcavations and Embankments. 254 Masonry— Stone-work 255 Brick-work 257 Capacities 259 Time 262 CHAPTER V. RATIO AND PROPORTION. OUTLIJVB 267 Ratio 268 Variation 271 Proportion— Simple 274 Compound 280 Distributive 284 CHAPTER VI. PERCENTAGE. OUTLIJ^S 289, 290 Definitions 291 General Cases 293 APPLICATIONS. Profit and Loss 303 Insurance 315 Commission 318 Simple Interest 322 Partial Payments 339 Compound Interest 343 PAGE Discount— Bank 349 True 354 Stocks 357 Exchange 362 Taxes and Duties 367 Direct , 368 Indirect 369 Partnership 371 Bankruptcy 375 Average of Payments 376 ' CHAPTER VII. INVOLUTION AND EVOLU- TION. OVTLIJV^ 383 Involution 384 Evolution 386 Square Koot 387 Applications 390 Cube Root 893 Applications" 399 CHAPTER VIII. PROGRESSIONS. OUTLIJ\rJE 400 Arithmetical 401 Geometrical 404 Annuities 406 CHAPTER IX. MENSURATION. OUTLIJ\rE 409 Square Measure 410 Solids 415 Surfaces of Solids 417 Volumes of Solids 421 METRIC SYSTEM 42C LUMBERMEN'S RULES 42g MISCELLANEOUS PROBLEMS 42( Art. 1-19. The teacher must exercise discretion in use of definitions. Those in Section T. need not all be committed at once, as some of them are given in the body of the work as needed. 55. Addends^ although a comparatively new term, is not used without authority. 61 . In Mental Exercises, use the model best adapted to each pupil. If deemed best, give the younger pupils the mental problems as dictation exercises on the slate. 73. Refer to definitions (4 and 5), or explain fully concrete and abstract. 78. As early as possible, it is well to show that the placing of the subtrahend under the minuend, is a matter of convenience. 89. When multiplying we may regard both terms as abstract, and then attach to the product such name as the nature of the question demands. Thus, since 1 bbl. flour costs $8, 9 bbls. cost 9 times as much, or $72. 9x8 = 8x9= 72. The answer must be dollars. Therefore, $72, answer. 93. Prob. 10 may be contracted thus : 374781 1402 749562 525442962 First multiply by 2 ; then multiply this product by 7, placing the right-hand figure under 4, and the remaining figures in order to the left ; for 2 X 7 = 14. This is given merely as a sample. Once show the pupil how to lighten his labor, and generally he is not slow to take advantage of any short process ; and will soon reap great benefit. Try Prob. 21. Many operations in multiplication may be solved by such devices as this: 324 X 81=? 26244 324 X 18=? 324 X 108=? 2592 324 X 801 = ? 95. When ciphers are on the right of significant figures, either in the multipli- cand or multiplier, or both, they should not be considered until the significant fig ures are multiplied together, when the ciphers must be annexed to the product rii Vlll SUGGESTIONS. 102. Example Second. This may also be solved thus: If each of two per- sons receive $1, to divide $10 equally between them, each must receive as many times $1 as $2 is contained times in $10, or 5 times $1 = $5. This relieves us of the inconsistency of calling the divisor an abstract and the dividend a coticrete number. Or, since taking one-half oi %Vi is the same as multiplying $10 by J, we have the multiplicand and product of the same kind (89). 106-108. Show that the divisor may be written in any other convenient place as well as at the left or right of the dividend. Problems. The earnest teacher will not fail to supply the student with abun- dant examples. Pp. 73-78 are thought to afford a fair variety of such examples as the pupil needs to make him thoroughly familiar with the principles and operations already discussed. 112-119. Show the strong resemblance of a whole number, or Integer, and a decimal ; and exhibit in the strongest light the importance of the decimal point. 120. The Second Method of Numeration and Notation possesses such great advantages over the first method that we wonder at the limited use of the former. 129-1 38 . Too much care cannot be exercised in teaching " Multiplication of Decimals " and " Division of Decimals "; and no part of the Arithmetic will better repay this care ; for, pupils once thorough in these, will move along easily and rapidly. 138. Reducing dividend and di%'isor to the same denomination before divid- ing has the advantage of clearness, but is. sometimes inconvenient in practice. Thus : Divide 1728 by 1.2. Annexing .0 to 1728, we have 1728.0, dividend and divi- sor, both tenths^ and the quotient a whole number, 1440. Dividing .0001728 by 1.2 by this method, however plain it may be, is a clumsy performance. 167. In finding the factors of a number, ^& facts in this article may be greatly extended, as for example : Fourth. Any even number, the sum of whose digits is divisible by 3, has 6 for an exact divisor ; Fifth, Any number, whose two right- hand figures are divisible by 4, has 4: for an exact divisor, etc., etc. 183. Special Rules on p. 148 are given, that teacher and pupil may have both variety and choice. We prefer, however, the General Rule. 184. The same remarks apply to Rules on pp. 154, 155. 185. The pupil should be able at once to change a decimal into a common frac- tion, or a common fraction into a decimal. 186. 187» are eminently practical and should be thoroughly mastered. 192-232 embrace the tables used in Denominate Numbers, and are placed together as a matter of convenience for easy reference. The exercises which follow are placed under their appropriate headings, so that a table, or a convenient number of tables, with the exercises Belonging to each, may be readily assigned as a lesson. 193. The Table of Federal Money, together with definitions, are here in- serted to preserve the uniformity of the system. The representations of the coins of the United States are all that are coined at the present time. Of the coins of Great Britain, Germany, and France, only a lim- SUGGESTION'S. ix ited number are inserted. It is advisable, when possible, for the teacher to exhibit the actual coins. The same is true of all the weights and measures represented. 210, 211, 216. Show how square and cubic measure stand related to Long measure. 211. We see no good reason why a pupil should wait until he studies a treat- ise on surveying before he knows how a township is subdivided. By means of cut, p. 192, the teacher may suggest, and the pupil solve, a great number of interesting problems. By means of Section Maps of the Western States, the pupil may locate the principal cities, towns, etc. 214. The edge of a cube is one of its dimensions. The edge of any solid is a line formed by the meeting of two adjacent faces of that solid. 245. Although three methods are given, whether the pupil shall study them all at once should be determined by his stage of advancement. 246, 257, are entirely practical, and while intended for pupils in general, are especially useful for those who have no prospect of taking a course in Mathematics, or who cannot even find the time to go as far as Mensuration proper, as treated in the last pages of this work. Much is given here in a convenieni form that is not easy to find in any one book. 248. Nearly every pupil in the Western States will be interested in knowing the manner of dividing lands m his own county. 257. Prob. 4. Smce when the sides were 15 inches high, the wagon-bed held 50 43^1 cubic feet ; in order to hold 50 cubic feet, it must be -^^ of 15 inches high, or 17|t inches. Hence the depth is increased 2|? in. Probs. 13, 17 and 18 may be solved in the same manner. 258. This may be illustrated by the change apparent in a wptch, in going from San Francisco to New York, New York to Liverpool, Liverpool to Canton (China), Canton to San Francisco ; illustrating that a good time-keeper will lose a day in going round the world eastwardly ; and, in like manner, going westwardly will gain a day. 259. The word measure is here used in its general sense; for although 25 consists of a number of parts, yet, for the purpose of measuring, it is a unit. Fof example : How many centals in 2050 lbs. of wheat ? Here we divide 2050 by 100, the number of lbs. in one cental, and 100 lbs. is regarded as the measuring unit. 286. P. 285, Ex. Equimultiples of numbers are the products of those num- bers by a given number. Thus : 7 x 5 and 8x5 are equimultiples of 7 and 8. 291, 292. Note especially the difference between Rate per cent, and Rate. 305. In Higher Mathematics and Applications, Formulas are deemed invalu- able. Why not in Arithmetic ? 319. Solution of Ex. 1, in Review Problems: To make W^^ the selling price must be | of cost, or \ of 24c. = 30c. But 30c. is WS less than asking price. Since 1615^ = ^, « — J (= 5) of asking price, must equal selling price, and \ of asking price SUGGESTIONS. equals J of selling price ; therefore, | of asking price equals | of selling price. } of 30c. = 36c., asking price. 353. One method of computation well taught is worth more than all the others poorly taught ; and all special methods should be omitted with beginners. 396. Ex. 5 is solved by formula (g^ x 100^= Rj. Thus: B = 900, 2-4 5 P = $19.20, t ^ ih = i^ y- Then W|» x -\5^ x ii&ift^ = i 2% 8. 403-412. The five cases correspond to the five cases of simple interest. 426. Both Rules and Formulas are omitted in computations of Stocks and Bonds, because they are the simplest applications of percentage. Prob. 17. Another Solution : Dividing the given by the required rate % gives the rate. .75 x 500 = 375 ; or, $375 is the price. 427. Letters of Credit, that is, written orders on which partial payments are made at sight, are issued to travellers in all or nearly all civilized countries ; thus affording great security in the transportation of necessary funds for travelling ex- penses. The value of such letters, of course, is reckoned according to the principles governing the calculation of exchange. 436. The table on p. 366 will suggest to the ingenious teacher a great variety of interesting problems. 461. Pupils should also be exercised in solving Problems in Partnership and Bankruptcy by Distributive Proportion, pp. 285-288. 484. Prob. 1. Had Simpson paid the stipulated sums at the times agreed tipon, the party from whom he bought would have had at the end of 15 mo. the interest of $500 for 15 mo. ; of $600 for 9 mo. ; and of $700 for 3 mo. ; or, at 6^, he would have had $37.50 + $27 + $10.50 = $75 interest. Simpson then should pay him $2700 ($500 + $600 + $700 + $900) at such a time be/ore the end of 15 mo., that at the end of the 15 mo. its interest at 6$? would equal $75. In 12 mo. $2700, at 6^, will gain $162 ; and will gain $75 in ts\ of 12 mo. = 5| mo. Simpson, therefore, should pay $2700 in 15 mo. — 5| mo. = 9 J mo.^Ans. 485. Let the pupil work a number of examples by selecting both the first and the last dates as focal dates. 609. In the applications of Square and Cube Roots, no demonstrations are attempted. 531. Prob. 1 : As the first payment is made at the beginning ot the first year, and the tenth payment at the beginning oi the tenth year, the entire 10 payments are made within 9 yrs. and 1 da. We must, therefore, regard the annuity as having 10 yrs. to run. 567. It is a matter of regret that the Metric System has not come into general use in the United States. ^^^^5^ iniiis sf MrilJimBlic a H o H H H Oh g O 3. ITJVJT. CLASSIFICATION. J OPERATION. TERMS. SIGXS. CLASSIFICATION. ^ As to Object As to WJioleness of their Unit. As to their Na- ture. As to the Number ofJeindsofUn Us. j 4. Concrete. 5. Abstract. 6. Integral. 7. Fractional. 8. Mixed. Like. Unlike. 11. Simple. 12. Compound ■ 9. Numeration and Notation. Addition. Subtraction. Slultiplica tion. Division. Reduction. 13. Solution. 14. Problem. 15. Explanation, 16. Princiitle, 17. Example. 18. Analysis. 19. yjuie. Denominate Nttmbers. Percentage. Ratio and Proportion. Involution and Evolution. Progressions. Mensuration. CHAPTER I. SECTION I. — DEFINITIONS. 1. Arithmetic is the science of numbers and the art of computation. 3. A Wutuber is a unit, or a collection of units, and answers the question "How many ? " 3. A Unit is one, or a single thing; as, one, one dollar, OJie house, one bushel, one peck, one half. (5, jstote, and 150.) Numbers are classified as to dfyject into Concrete and Abstract. 4. A Concrete Number is one that is applied to some particular object; as, three books, four dollars, five miles. 5. An Abstract Number is one that is not applied to any object ; as, three, four, five. Numbers are classified as to their umi into Integral, Fractional, and Mixed. 6. An Integer is a whole number ; as, one, five, ten. 7. A Fraction is a number expressing one or more of the equal parts of an integer ; as, one half, five tenths. 8. A Mixed Number is composed of an integer and a fraction; as, six and one half, nine and five tenths. Numbers are classified as to their nature, into Like and Unlike. IS INTEGERS. 13 9. lATze Numbers are those which express the same kind of units ; as, two cents and four cents, six hats and one hat 10. Unlike Numbers are those which express differ- ent kinds of units; as, two cents and six hats, four cents and one hat. Numbers are classified as to the number of kinds of units, into Simple and Compound. 11. A Simple Number is a number expressing one kind of unit ; as, four pounds. 13. A Compound Number is a number expressing more than one kind of unit; as, four pounds five ounces. The terms used in computations in Arithmetic are Solution, Problem, Explanation, Principle, Example, Analysis, and Rule. 13. A Solution is a process of computation used to obtain a required result. 14. A Problem is a question for solution. 15. An Explanation is a statement of the reasons for the manner of solving a problem. 16. A Principle is a general truth upon wbich a process of computation is founded. 17. An Example is a problem used to illustrate a principle, or to explain a method of computation. 18. An Analysis is a statement of the successive steps in a solution. 19. A Mule is a direction for performing any compu- tation. OUTLINE OF NUMERATION AND NOTATION, ^ 23. ORIQiy. 24. CHARACTERS. 86. Significant. 25. Names. ■{ 27. Insignificant. i 29. Simple. 28. Values. ■{ \ 30. Xoco/. 33, Units or Ones. ^ 31. J'/oce 33. TeTW. or . 34. Hundreds. < TERMS. &c., l Q> III t^ E>< 5 ^1^ 400 040 004 444 440 of 404 This number is read 400 quadrillion 40 trillion 4 lUlion 4M million 440 thousand 404. The Periods above Quadrillions are Quintillions, SeTtillions, Septil- lions, Octillions, Nonillions, DecilUons, Undedllions, Duodecillions, Tredecillions, Quatuordecillions, Quindecillions, Sexdecillions, SeptendeciJf- lions, OctodeciUions, NovendeciUioTis, Vigintillions, &c. 20 IITTEGERS. PRINCIPLES OF NUMERATION AND NOTATION. 41 . First. Ten units of any order in a number are always equal to one unit of the next higher order. Second. Removing a figure one place to the left, increases its value tenfold. Removing a figure one place to the right, diminishes its value tenfold. Third. The name and value of the units represented hy a figure in a number are always those of its order in that number. Fourth. The absence of units in any order in a number is denoted "by RULES FOR NUMERATION AND NOTATION. 42. Numeration". — Beginning at the right, separate the numbers i?ito periods of three figures each. TJien read the number in the first period at the left, adding the na^ne of the period J do the same with each period in order toivard the right, omitting to name the units period and any period or periods composed only of ciphers. Notation. — Beginning at the left, write the hundreds, tens, and ones of each period in their proper order, filling all vacant orders and periods with ciphers. EXER CISES. Eead the following numbers : 1. 125 623 6. 423 152 391 11. 2 648 2. 400 119 7. 72 110 110 12. 9 999 3. 50 090 8. 95 486 000 13. 14 150 4. 62 124 9. 172 000 400 14. 40 000 5. 142 205 10. 909 412 613 15. 425 003 16. 144 044 014 005 009 21. 9 00 9 009 009 009 009 17. 75 000 000 000 075 22. 41 00 041 000 041 000 18. 48 370 490 563 23. 3 00 000 000 000 000 19. 49 123 436 000 000 24. 6 2 410 000 443 000 20. 11 010 001 936 127 25. 41 3 125 679 456 199 NUMEEATIOIf AND NOTATIOK. 21 Write the following numbers in figures : 1. Twenty-two thousand seven hundred sixty-five. 2. Eighty thousand two hundred one. 3. Thirty thousand thirty. 4. Four hundred ten thousand two hundred five. 5. Ninety thousand one. 6. Eight hundred thousand six hundred sixty-nine. 7. Nine hundred thousand one. 8. Five hundred thousand fifty. 9. One hundred million ten thousand one. 10. Ninety-one million seven thousand sixty. 11. Seventy million four. 12. Seven hundred million ten thousand one. 13. One billion one million forty. 14. Forty billion two hundred thousand five. 15. Seven hundred twenty-six billion fifty million one thousand two hundred forty-three. ROMAN NOTATIOIV. 43. In the Roman 3Iethod numbers are expressed by means of seven characters. 44. The Roman method of expressing numbers had its origin in Rome. 45. The Characters used in this system of expressing numbers are seven Letters, 46. The letters used are the following : . I, V, X, L, C, D, M. One, Five, Ten, Fifty, One hundred, Five hundred, One thousand. »55 INTEGERS. 47. When used alone each letter has a fixed value. 48. When a letter is repeated, its value is repeated. Thus, III are 3, XX are 20, COO are 300, MM are 2000. 49. When a letter folloivs one of greater value, their sum is the number expressed. Thus, VI are 6, LX are 60. 50. When a IqUqt precedes one of greater value, their difference is the number expressed. Thus, IV are 4, XL are 40. 51. When a letter is placed between two others ol greater value, the difference hetioeen it and their sum is the number expressed. Thus, XIX are 19, XIV are 14. 53. When a horizontal line is placed over a letter, its value is increased a thousand times. Thus V is 5000. 53. Uses, — Roman notation is not convenient for arith- metical computations, but is used principally in marking dials, numbering chapters, sections, &c., in books, and in Writing physicians' prescriptions. EX ER CISE8. Write the following numbers in letters : 1. 7 9. 88 17. 109 25. 482 33. 2300 2. 14 10. 94 18. 245 26. 375 34. 5050 3. 28 11. 73 19. 537 27. 160 35. 4875 4. 56 12. 59 20. 199 28. 543 36. 1876 5. 50 13. 60 21. 125 29. 215 37. 1250 6. 75 14. 85 22. 150 30. 312 38. 2275 7. 96 15. 63 23. 260 31. 219 39. 2116 8. 99 16. 89 24. 384 32. 420 40. 5000 OUTLINE OF ADDITION. TERMS. 55. Addends, or Parts, 56. Sum, or Amount, o 57. SIGNS. 61. TABLE. 58. Of Addition, 59. Of Equality. 60. Of Dollars. Q Q PRINCIPLES. CASES. 62. Like Numbers. 63. Like Orders, 64. L 65. IL When the sum of the units of each order is less than 10. Wh£n the svm of the units of any order equals or exceeds 10. 66. RULE. SECTION III )I)jI^JlQjM .^j,^ 54. Addition is the process of uniting two or more numbers to form one number. The Terms used in Addition are Addends, or Parts, and Sum, or Amount. 55. The Addends, or JParts, are the numbers to be added. 56. The Sum, or Amount, is the result obtained by Addition. 57. Signs are characters used for abbreviating expres- sions. 58. The Sign of Addition is an erect cross called plus ( + ), and when placed between numbers signifies that they are to be added. Thus, 3 + 2 means that 3 and 2 are to be added, and is read " 3 plus 2." The term plus, Latin, signifies more, or added to. « 59. The Sign of Equality is two short, horizontal, parallel lines (=), and when placed between numbers sig- nifies that they are equal. Thus, 3+2 = 5 means that 3 and 2 are equal to 5, and is read " 3 plus 2 equals 5." 60. The Sign of Dollars is a capital S with two vertical marks across it (I), and when placed before a num- ber expresses dollars. Thus, $120 means 120 dollars. ADDITION 25 61. ADDITION TABLE. 12 111 12 3 ONE. 3 4 5 6 7 8 9 1111111 4 5 6 7 8 910 SIX. 01234567 8 9 6666666666 6 7 8 910111213 1415 12 2 2 2 TWO. 3 4 5 6 7 8 9 2 2 2 2 2 2 2 SEVEN. 0123456789 7777777777 2 3 4 5 6 7 8 91011 7 8 910111213 1415 16 THREE. 0123456789 3333333333 EIGHT. 0123456789 8888888888 3 4 5 6 7 8 9101112 8 9 10111213 1415 16 17 12 4 4 4 FOUR. 3 4 5 6 7 8 9 4 4 4 4 4 4 4 NINE. 0123456789 9999999999 4 5 6 7 8 910 111213 910111213 1415 16 1718 12 5 5 5 FIVE. 3 4 5 6 7 8 9 5 5 5 5 5 5 5 TEN. 0123456789 10 10 10 10 10 10 10 10 10 10 5 6 7 8 9 10111213 14 101112131415 16 171819 26 INTEGEBS ;f ©i^at^xTGi^ci^eX Example. — James has two hats and John has 3. How many have both ? Solution. — Since 2 hats and 3 Tiats are 6 hats, they both have 5 hats. Or, 2 hats and 3 hats a/re 6 hats. TJierefore. they loth have 5 hats. Or, They both have 6 hais, because 2 fiats and 3 hois are 5 hats. 1. Willie found 4 marbles and bought 7. How many had he then ? 7 and 4 are how many ? 9 and 2 ? 2. Samuel has 5 walnuts and James gives him 4 more. How many has he then ? How many are 5 and 4 ? 4 and 5 ? 2 and 2 and 5? 3 and 2 and 4? 1+4+4=? 1 + 3 + 5=? 2 + 2 + 2 + 3 = ? 3. Charles bought 7 almonds at one store and 10 at another. How many did he buy in all ? 7 + 10 = ? 10 + 7 = ? 7 + 5 + 5 = ? 2 + 8 + 7 = ? 6 + 7+4 = ? 3+7 + 7 = ? 9 + 3 + 5 = ? 4. If a pound of sugar costs 9 cents and an orange 6 cents, what will both cost ? 6 cents and 9 cents are how many cents ? 5. Jennie's father paid 8 dollars for her muff, 7 dollars for her tippet. How much did he pay for both ? 7 + 8 = ? 6. If a house has 7 windows in the front and 6 in the rear, how many windows has it? 6 + 7 = ? 4 + 2 + 5 + 2 = ? 7. Mary paid 8 cents for an inkstand and 5 cents for a pencil. How much did she pay for both ? 5 cents + 8 cents = ? ADDITION. 27 How many are 8.10menand7men?,13.5 + 2 + 3? 9. 8 boys and 9 boys ? 14.5 + 3 + 1? 10. 6 girls and 8 girls ? 11. 5 tops and 9 tops ? 12. 3 pins and 8 pins ? 15.3 + 3 + 3? 16.4 + 4 + 4? 17.5 + 5 + 5? 18. 7 toys and 7 toys? 1 9. 6 miles and 8 miles ? 20. 8 cups and 4 cups ? 21. 7 cents and 5 cents ? 22. 6 figs and 5 figs ? 23. How many are 8 chairs and 4 dollars ? We cannot add 8 chairs and 4 dollars. For 8 chairs + 4 dollars = neither 12 chairs nor 13 dollars. Hence, 63. Only abstract numbers, or like concrete numbers, can be added. 24. How many are 3 tens and 9 thousands ? We cannot add 3 tens and 9 thousands. For 3 tens + 9 thousands = neither 12 tens, nor 12 thousands. Hence, 63. Only like orders of units can be added. fWfriiien ^^erci^e^-f CASE I. 64. When the sum of all the units of each order is less thaii 10> Example.— What is the sum of 11, 431, 2245, 3112 ? SOLUTION. 11 = 1 ten 1 one 431 = 4 hundreds 3 tens 1 one 2245 = 2 thousands 2 hundreds 4 tens 5 ones 3112 = 3 thousands 1 hundred 1 ten 2 ones Amount. 5799 = 5 thousands 7 hundreds 9 tens 9 ones Addends. . 28 IlS^TEGEKS. Explanation. — Since only like orders can be added, it is most con- venient to write like orders in the same column. We first add the ones, 1, 1, 5, and 3, and find tlieir sum to be 9, which we write in the ones' order of the amount. We then add the tens, 1, 3, 4, and 1, and find their sum to be 9, which we write in the tens' order of the amount. We then add the hundreds, 4, 3, and 1, and find their sum to be 7, which we write in the hundreds' order of the amount. We then add the thousands, 3 and 3, and find their sum to be 5, which we write in the thousands' order of the amount. The result, 5799, is the sum required. rJtOBLEMS. 123 12 132 .-67 (2.) 410 30 240 680 (3.) 432 201 222 855 (4.) 210 135 42 387 (5.) 1234 1211 4524 6969 (6.) 231 2213 6404 (7.) 210414 337052 142513 8848 689979 8. Add 6404, 2213, and 1231. A?is. 9848. 9. Add 3472, 4513, 1001, and 1000. Ans. 9986. 10. Add 10, 12, 21, 32, 11, and 3. Ans. 89. 11. Add 1321, 2312, 1241, and 3102. Ans. 7976. 12. Add 3131, 2112, 3421, and 1003. Ans. 9667. 13. Add 10102, 4030, 634, 11, and 2. Ans. 14779. 14. John gave 10 cents for a spelling-book, 12 cents for a writing-book, 15 cents for a slate, 20 cents for pens and ink, and 31 cents for an Arithmetic. How much did he give for all? Ans. 88 cents. 15. One horse cost 120 dollars, another 115 dollars, one harness 31 dollars, the other 30 dollars, and the carriage 200 dollars. How much did all cost? Ans. 496 dollars. 16. Gave 3216 dollars for land, 2150 for a house, 1502 for stock, and 2121 for a barn. How much did all cost ? Ans. 8989 dollars. ADDITIOi?^. f ©ral^xfGl^ciXGX Example. — How many are 16 and 9 ? (Written.) SOLUTION. EXPLANATION. ) 9 r6 ones and 9 ones are 15 ones = 1 ten 1st step. >■ 16 •< and 5 ones. ) 5 ( Write the 5 ones in ones' order of sum. ) 9 ( Add the 1 ten to tens' order of parts. 2d step. >• 16 •< 1 ten and 1 ten are 2 tens. ) 25 ( Write the 2 tens in tens' order of sum. Ans. 25. Example. — How many are 16 and 9 ? (Oral.) Explanation.— Add the right-hand figures, 9 and 6. Their sum is 15 ; retain the 5 in the mind, and Add the 1 to the left-hand figure, 1. Their sum is 3. Result, 25. PJBOBIjEMS. 1. How many are 18 and 9? 15 and 6? 17 and 5? 13 and 8? 16 and 4? 19 and 7? 2. How many are 24 and 7? 28 and 5 ? 26 and 9? 25 and 8 ? 29 and 5 ? 27 and 8 ? 3. How many are 36 and 6 ? 32 and 9 ? 38 and 5 ? 35 and 7? 37 and 4? 39 and 8? 4. How many are 47 and 8 ? 44 and 6 ? 49 and 7 ? 42 and 9 ? 46 and 5 ? 48 and 9 ? 5. How many are 57 and 4 ? 59 and 7 ? 55 and 6 ? 53 and 8? 57 and 9? 58 and 5? 6. How many are 62 and 9? 65 and 8 ? 67 and 7? 64 and 6? 68 and 9? 69 and 9? 30 INTEGERS. 7. How many are 75 and 7 ? 73 and 9 ? 77 and 5 ? 78 and 4 ? 79 and 6 ? 76 and 9 ? 8. How many are 83 and 9? 87 and 6? 85 and 8? 86 and 7 ? 88 and 7 ? 89 and 9 ? 9. How many are 91 and 9? 95 and 8? 98 and 6? 94 and 7 ? 99 and 5 ? 93 and 9 ? Example. — How many are 43 and 39 ? Explanation. — See tLe sum of right-hand figures (12), retain in the mind the ones of this sum (2). See the sum of left-hand figures (7), increase this by 1 (8), because the sum of the right-hand figures exceeded 9. Ans. 82. 10.44 + 37 = ? 45 + 36 = ? 42 + 39 = ? 48 + 34 = ? 11.51 + 19 = ? 56 + 18 = ? 57 + 15=? 59 + 13 = ? 12.62 + 18 = ? 65 + 13 = ? 64 + 25 = ? 69 + 15 = ? 13.71 + 18 = ? 75 + 25 = ? 74 + 16 = ? 79 + 11 = ? 14. 12 + 8 + 9 + 7 + 8 = ? 15 + 9 + 8 + 9 + 15 = ? 15. 14 + 7 + 9 + 12 + 18 = ? 17 + 3 + 40 + 20 + 9 = ? 16. 13 + 9 + 11 + 44 + 33 = ? 16 + 10 + 4 + 9 + 12 = ? 17.24 + 7 + 9 + 60 + 50 = ? 25 + 15 + 25 + 15 = ? 18. 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 = ? D. tWritten ^Xer eigesT-t CASE II. 65, When the sum of all the units of each order equals or exceeds 10, Example.— Find the sum of 314, 591, 856, 721, and 672. ADDITION. 31 SOLUTION. Explanation. — We write tlie numbers so that units q-i ^ of the same order stand in tlie same column, and thus form as many columns of figures as there are different 591 orders of units. 856 Then we add first the right-hand or units' column, 721 beginning at the top and adding downward. 672 4: + l + 6 + l+2 = 14:, OT 1 ten and 4 ones. Write the 4 ones directly under the ones, and add ^^^^ the 1 to the tens. In like manner we add the tens' column, 1 + 9 + 5 + 2 + 7 and the 1 ten which we obtained by adding the ones' column, making in all 35, or 3 hundreds and 5 tens. We write the 5 tens directly under the column of tens and add the 2 hundreds to the column of hundreds. The hundreds' column 3 + 5 + 8 + 7 + 6 and the 3 hundreds from the tens equal 31 hundreds, or 3 thousands and 1 hundred. Write the 1 hundred directly under the column of hundreds and the 3 thousands in the thousands' order of the result. This completes the work and gives us 3154 for the sum. We should obtain the same result by beginning at the bottom and adding upwards. It is mainly by repeated additions, in this manner, that we can prove the correctness of the result obtained. 66. EuLE. — Arrange the numbers to he added so that the same order of units shall stand in the same column. Find the sum of the ones'* column and place the right-hand figure of the sum directly under the ones and add the num- ber expressed hy the left-hand figure or figures, if any^ to the tens' column. Add in the same manner each column in order, always placing directly under the column added the right-hand figure of its sum, and adding the number expressed by the left-hand figure or figures, if any, to the next column, talcing care also to write at the left the whole of the last sum found. 32 IIJ^TEGERS. Pit OBI. JEMS '. (1.) (2.) (3.) (4.) (5.) (6.) (7.) 142 344 796 9 98 8 4501 177 792 438 976 472 69 2437 802 671 29 854 9 987 684 1121 1807 1263 1839 579 1064 7622 8. Add 873254, 289094, 627380, and 456987. Ans. 2246715. 9. Add 35709, 50804, 90006, and 47103. Ans, 223622. 10. Add 263, 5827, 39001, 70126, and 85. Ans. 115302. 11. Add 92, 924, 9857, 95064, and 572. Ans. 106509. 12. Add 3597, 82, 649, 30548, and 762. Ans. 35638. 13. Add 4689, 8694, 75, 64982. Ans. 78440. What is the sum 14. Of 5 + 15 + 25 + 35 + 45 + 55 + 65? Ans. 245. 15. Of 4+54+64 + 74 + 84 + 94 + 44? Ans. 418. 16. Of 23 + 33 + 43 + 53 + 63 + 73 + 83? Ans, 371. 17. Of 32 + 42 + 52 + 62 + 72 + 82 + 92? Ans. 434. (18.) 78243 8145 94067 72 21698 202225 313332 717782 (19.) 43285 16007 990 8622 73084 141988 265888 549864 (20.) 43 437 6908 97360 5033 109781 345678 565240 (21.) 9 89 789 6789 56789 •64465 89658 218588 (22.) 53102 8967 406 12 8 62495 95642 220632 ADDITION. 33 23. Add six hundred forty- two, three thousand one hun- dred twenty-four, seventy-nine thousand nine hundred six, eight hundred twenty-four, seven hundred five, and forty- seven thousand twenty-eight. Ans. 132229. 24. Find the sum of six million sixty thousand six, seven million nine hundred fifty thousand ninety-nine, ten million nine thousand eight hundred seven, and three hundred sixty-seven thousand forty-five. Ans, 24386957. 25. Add seventy thousand four hundred fifty-three, five million eight hundred six thousand twenty-eight, eighty million ninety-seven thousand nine, and twenty-five million seven hundred thousand. Ans. 111673490. 26. A man owns a shop worth $4000, a house and lot worth $7500, bank stock worth $1800, insurance stock worth $800, a carriage worth $450, and two horses worth $350. How much are all worth ? Ans. $14900. 27. Mr. B's State taxes are $91; County taxes, $321; School taxes, $421 ; Borough taxes, $375 ; and Poor tax, $39. What is the whole amount of taxes ? Ans. $1247. 28. A merchant deposits in bank on Jan. 1, $300; Jan. 3, $1716; Jan. 4, $5217; Jan. 9, $16714; Jan. 10, $516; and Jan. 11, $7060. How much in all has he deposited? Ans. $31523. 29. A drover bought 350 sheep from one farmer, 425 from a second, 750 from a third, 400 from a fourth, 1616 from a fifth. How many did ha buy from aU ? Ans. 3541. 30. From the savings bank drew out the following sums: $210, $35, $48, $16, $25, $45, $24, $16, $13, $75, $84, $7, and $16. How much did he draw out in all ? Ans. $614. 34 INTEGERS. 31. A merchant "checked out" of his hank $215, $3125, $21620, $17540, $2063, $216, $257, $301, $51, and $1617. How much in all did he check out ? Ans, $47005. 32. A man after " checking out " $947, found that he had remaining in bank $369. How much had he at first ? A71S. $1316. 33. After " checking out" $30, $40, $9, and $25, a grocer found that he had $711 remaining in bank. How much had he at first ? Ans. $824. 34. A college pays its president $2500 per year, its pro- fessors $16000, its assistant professors $4500, and its tutors $4800. How much does it pay annually for instruction ? Ans, $27800. 35. My income from rent of warehouse is $5750, from one dwelling-house $1200, from another $900, from bank stock $3500, from insurance stocks $1500, from my share in a steamboat $1643, from interest on money loaned $673. What is my entire income ? Ans. $15166. 36. G's expenses for 1 year were : rent $1200 ; taxes $225 ; insurance on furniture and household goods $65 ; household expenses $3652; James and Mary's school biUs $945 ; clothing for self and family $2250 ; traveling ex- penses and hotel bills $742. Find the sum. Ans. $9079. 37. A invests $2516 in a farm ; B $16525 in a mill ; C invests $21621 in a steamboat ; and D as much as A, B, and C together in a cotton factory. How much do they all invest? Ans, $81324. 38. A merchant bought of a jobber, prints amounting to $3520 ; muslins $3962 ; hnens $1733 ; silks $2375 ; woolen goods $7675. How much is the biU? Ans. $19265. ADDITIOIS^. 35 39. A farmer has 16 horses, 11 colts, 32 milch cows, 11 steers, 24 yearlings, 15 calves, 321 sheep, and 75 hogs. Of how many head does his stock consist ? Ans, 505 head. 40. A farmer has 50 acres planted in wheat, 75 in corn, 20 in barley, 14 in rye, 35 in oats, 16 in potatoes, and 7 in "garden truck." How .many acres in all has he planted ? Ans. 217 acres. 41. If in 10 years the expenses of a government are, respectively, $85279524, $74843279, $92573181, $68491183, $73111265, $79400295, $87765432, $95556443, $90200671, and $100785894, how much does the government spend in these 10 years? Ans. $848007167. 42. In January 1875, there were 31 days; in February 28 ; in March 31 ; in April 30 ; in May 31 ; in June 30 ; in July 31 ; in August 31 ; in September 30 ; in October 31 ; in November 30 ; and in December 31. How many days were there in the year 1875 ? Ans. 365 days. 43. By the P., F. W. and C. R. R. the distance from Pitts- burgh to Alliance is 83 miles ; from Alliance to Crestline, 106 miles ; from Crestline to Ft. Wayne, 131 miles ; from Ft. Wayne to Plymouth, 64 miles ; from Plymouth to Chi- cago, 85 miles. What is the distance from Pittsburgh to Chicago? A)is. 469 miles. 44. Add 24678, 423, 9846, 23, and 82764. 45. Add 4632, 8, 79, 8264, 38000, and 673289. 46. Add 79, 897, 438, 2986, 150, and 371436. 47. Add 873254, 289094, 627380, and 456987. 48. Add 35709, 50804, 90006, and 47103. 49. Add 263, 5827, 39001, 70126, and 85. 36 INTEGERS. 50. Add 92, 924, 9857, 95064, and 572. 51. Add 8, 28, 348, 4578, 90068, and 8. 52. Add 7, 77, 777, 7777, 7777, 777, and 77. 53. Add 66666, 6666, 666, 66, 6, 66, 666, and 6666. 64. 102 + 304 + 506 + 708 + 901 + 206 + 604=: ? 55. 2030 + 4050 + 6070 + 8090 + 1230 + 3110 = ? 56. 50709 + 2080 + 30604 + 8010 + 9076==:? 57. 864209 + 753186 + 999999 + 888888 = ? 53. 3207 + 10875 + 892 + 6075 + 124607 + 761 = ? 59. 456 + 654 + 789 + 897 + 231 + 123 + 875 = ? 60. 9 + 99 + 999 + 9999 + 99999 + 999999 + 25 = ? 61. A merchant paid $15796 for a lot and built a house upon it, which cost him $25650. How much did the lot and building cost ? Ans. $41446. 62. A man gave $2375 to his son, $1875 to his daughter, and donated $10984 to charitable purposes. How much did he give away ? Ans. $15234. 63. By selling a house and lot for $25875 I lost $987. What did both cost ? Ans. $26862. 64. A coal dealer shipped from Pittsburgh at one time 225675 bushels of coal ; at another time 186798 bushels ; at another time 75987 bushels. How many bushels did he ship in all ? Ans. 488460. 65. In 1870 the population of North America was 50,931,242, South America 26,677,419, Central America 2,690,635, and West Indies 4,065,945. What was their united population ? Ans. 84,365,241. OUTLINE OF SUBTRACTION. TERMS. 68. Minuend. 69. Subtrahend, 70. Difference, or JRetnainder, o M H < H PQ P • SIGNS, 72. TABLE. PRINCIPLES. CASES. 71. Of Subtraction, 59. Of Equality. 60. Of Dollars. r 73. ''Same Kind.** 74. ''Same Order,'' 76. "Same Number added" to both Minuend and Sub" trahend. 75. jT. 77. ZT. 78. RULE. 79. PROOF. SECTION IV. S; CBJ T^R; AC^'IOjM ^«S 67. Subtraction is the process of taking one number from another. The Terms used in Subtraction are Minuend, Subtrahend, and Difference, or Eemainder. 68. The Minuend is that number from which the other is to be taken. 69. The Subtrahend is that number which is to be taken from the other. 70. The Difference, or JRemainder, is the result obtained by Subtraction. 71. The Sign of Subtraction is a short horizontal line called minus ( — ), and when placed between two numbers signifies that the one after it is to be taken from the one before it. Thus, 5 — 2 means that 2 is to be taken from 5, and is read " 5 minus 2." The term minus, Latin, means lets, or diminished hy. Illustration. — In the expression 5—2 = 3, read 5 minus 2 equals 3, 5 is the rninuend, 2 the subtrahend, and 3 the difference, or remainder. SUBTRACTION. 39 72. SUBTRACTION TABLE 12 3 111 ONE. 4 5 6 7 8 910 1111111 SIX. 6 7 8 9 10 111213 1415 66 66666606 12 3 4 5 6 7 8 9 0123456 7 89 2 3 4 2 2 2 TWO. 5-6 7 8 91011 2 2 2 2 2 2 2 SEVEN. 7 8 910111213 1415 16 7777777777 12 3 4 5 6 7 8 9 0123456789 THREE. 3 4 5 6 7 8 9101112 3333333333 EIGHT. 8 91011121314151617 8888888888 12 3 4 5 6 7 8 9 0123456789 4 5 6 4 4 4 FOUR. 7 8 910 111213 4 4 4 4 4 4 4 NINE. 9 10 11 12 13 14 15 16 17 18 9999999 9 99 12 3 4 5 6 7 8 9 0123456789 5 6 7 5 5 5 FIVE. 8 91011121314 5 5 5 5 5 5 5 TEN. 10111213141516171819 10 10 10 10 10 10 10 10 10 10 12 3 4 5 6 7 8 9 0123456789 40 INTEGERS 0i«al^AfGiici^eX Example. — James had 9 apples and gaye his brother 4. How many had he left ? SoiiUTiON. — Since 4 apples from 9 apples leave 5 apples,, he had 5 apples left. Or, 4 apples from 9 apples leave 5 apples. Therefore, he had 5 apples left. Or, He had 5 apples left, because 4 apples from 9 apples leave 5 apples. 1. There are 10 books on the shelf. If I take away 8, how many will be left ? Take 8 from 10 and how many are left ? 6 from 8? 6 from 8? 1 from 10? 4 from 6? 6 from 7? 7 from 9? 3 from 10? 2. Susan has 9 cents and Mary has 6. How many more has Susan than Mary ? How many less has Mary than Susan. 9 less 6 are how many? 9 less 5? 9 less 4? 9 less 3? 9 less 2? 9 less 1? 9 less 9? How many are 3. 15 horses less 6 horses ? 4. 10 sheep less 4 sheep ? 5. 13 bushels less 7 bushels ? 6. 16 miles less 10 miles? 7. 12 dollars less 6 dollars? 8. 17 cents less 8 cents? 9. 14 birds less 5 birds ? 10. 15 marbles less 8 marbles? 11. 15 books less 9 books ? 12. 16 pens less 7 pens ? 13. 13 desks less 6 desks ? 14. 5 doors less 3 doors ? 15. 14 chairs less 8 chairs ? 16. 10 boys less 7 boys ? 17. 11 men less 8 men ? 18. 15 figs less 7 figs? 19. If you take 7 dollars from 14 dollars, how many remain ? SUBTRACTION-. 41 20. Nannie brought 6 pears to school, and gave 2 of them to her teacher. How many had she left ? 21. How many are 15 apples less 8 pears? We cannot take 8 pears from 15 apples. For 15 apples — 8 pears = neither 7 pears nor 7 apples. Hence, 73. Only abstract numbers, or like concrete numbers, can be subtracted. 22. From 9 tens -take 2 ones. We cannot take 2 ones from 9 tens. For 9 tens — 2 ones = neither 7 tens nor 7 ones. Hence, 74. Only like orders of units can be subtracted. + Written ^^vferei^esT-e CASE I. 75. When each figure in the Subtrahend rep- resents a less value than the correspond- ing figure of the Minuend. ■ Example.— From 68759 subtract 37422. SOLUTION. Explanation. — Since only like Minuend, Subtrahend, Kemainder, 68759 37422 31337 orders can be subtracted, for conve- nience we write the subtrahend so that its units stand under the units of the same order in the minuend. We now subtract each figure of the subtrahend from the figure above it in the minuend, writing each result in the remainder. 2 ones from 9 ones leave 7 ones ; 2 tens from 5 tens leave 8 tens; 4 hundreds from 7 hundreds leave 3 hundreds ; 7 thousands from 8 thousands leave 1 thousand ; 3 ten-thousands from 6 ten-thousands leave 3 ten-thousands. The result, 31337, is the difference or remainder. 42 INTEGERS. PB. O B LEJUS, 1. From 7395 take 2004. 2. From 36714 take 12511. 3. From 597263 take 84123. 4. 68764 — 63233 = ? 5. 190806 — 90603 = ? Ans. 5391. Ans. 24203. Ans. 513140. 6. 68754 — 5521 = ? 7. 190856 — 100203 = ? 8. Bought 30 shares of bank stock for 13789 and sokl them for $5999. How much did I gain ? 9. A owns 8 horses and B owns 4. How many more has A than B ? After each has bought 4 more, how many more has A than B ? Hence, 76. Wlien the same number is added to both minuend and subtrahend, it does not change their difference. -•>•- 4-lWi'itten^^Grci^eXt m^^^ CASE II. 77. Wlien a figure in the Subtrahend represents a greater value than the corresponding figure of the Minuend, Example. — From 54 subtract. 19. SOLUTION. Minuend, 54 Subtrahend, 19 Eemainder, 35 Explanation.— It is evident that we cannot subtract 9 ones from 4 ones. But by adding 10 ones to the minuend and 1 ten (10 ones) to the subtrahend (76), we can perform the subtraction very readily. 54 = 5 tens 4 ones ; add 10 ones and we have 5 tens 14 ones. 19 = 1 ten 9 ones ; add 1 ten and we have 2 tens 9 ones. And 2 tens 9 ones from 5 tens 14 ones = 3 tens 5 ones, or 35, the required difEereuce or remainder. S U B T R A C T I O K 43 Example. - SOLUTION. 924 376 548 -From 924 take 376. Explanation. — We write the numbers as in the margin and say mentally : 6 ones from 4 ones impossible ; but 6 ones from 14 ones leave 8 ones. Since we added 10 ones (1 ten) to the minuend, we must add 1 ten (10 ones) to the subtrahend. 7 tens and 1 ten are 8 tens ; 8 tens from 2 tens impossible ; but 8 tens from 12 tens leave 4 tens. Since we added 10 tens (1 hundred) to the minuend, we must add 1 hundred (10 tens) to the subtrahend. 3 hundreds and 1 hundred are 4 hun- dreds ; 4 hundreds from 9 hundreds leave 5 hundreds. Result, 548. 78. KuLE. — Write the suUraliend so that its units shall stand under the units of the same order in the minuend. Begin at the right-hand and subtract from each figure of the minuend the corresponding figure of the subtrahend and write the remainder underneath. If in the minuend any figure expresses a less number of ones than the corresponding figure of the subtrahend , add 10 to the former ; then subtract, and add 1 to the next figure of the subtrahend. Proceed thus till the work is complete. 79. Proof. — When the work is correct, the sum of the remainder and subtrahend equals the minuend. PI lOBLEM^ f. (1.) From 734 (2.) 652 (3.) 818 (4.) 963 (5.) 521 (6.) 520 Take 475 384 549 777 258 287 Ans. 259 From 233 7. 403 take 304. 11. 900 take 512. 8. 600 take 259. 12. 500 take 201. ^ 9. 800 take 468. 13. 400 take 102. 10. 700 take 237. 14. 4321 take 1234. 44 INTEGEKS, 15. From 8765 take 5678. 16. From 90807 take 6898. 17. From 60504 take 496. 18. From 30201 take 194. 19. How many are 100000 — 99999 ? Ans. 1. 20. How many are 100000 — 9999? Ans. 90001. 21. How many are 100000 — 999 ? Ans. 99001. 22. How many are 100000 — 99 ? Ans. 99901. 23. How many are 100000 — 9 ? Ans. 99991. 24. By the P. R. E. the distance from Pittsburgh to New York is 444 miles; from Pittsburgh to Philadelphia, 355 miles. What is the distance from Philadelphia to New York ? Ans. 89 miles. 25. By the same road, from Pittsburgh to Harrisburg is 246 miles. What is the distance of Harrisburg from Phila- delphia? From New York? 1st Ans. 109 miles. 2d Ans. 198 miles. 26. Washington was born A. d. 1732, and died'in A. d. 1799. How old was he at his death ? Ans. 611 years. 27. The sum of two numbers is 20560 ; the less is 3745. What is the greater? Ans. 16815. 28. A merchant sold on Saturday goods amounting to $2567 ; on Monday goods amounting to $1075. What is the difference in the two days' sales ? Ans. $1492. 29. If a man's property is valued at $15725, and his debts at $6837, what is he worth ? Ans. 88888. 30. If a man's property is valued at $20569, and his debts at $30880, what excess of debt has he ? Aiis. $10311. 31. A man sold a horse for $375, making $97. How much did the horse cost him ? Ans. $278. 32. A and B together bought property for $275315, and A furnished $98676 of the purchase money. How much did B furnish ? Ans. $176639. SUBTRACTION. 45 33. America was discovered by Columbus in 1492. How many years had elapsed in 1837 ? Ans. 345 years. 34. I deposited in the bank $1840, and drew out M75. How many dollars had I left ? Ans, $1365. 35. A man has property worth $10104, and owes debts to the amount of $7426. When his debts are paid, how much will he have left ? Ans. $2678. 36. A man having $100000, gave away $11. How many dollars had he left ? Ans. $99989. 37. A man bought a span of horses for $537 and a yoke of oxen for $297. How much more did he pay for the horses than for the oxen? Ans. $240. 38. A man having $8795 gave $2309 of it for a store. How much money had he left? Ans. $6486. 39. A merchant owing $35542 paid $22560. How much does he still owe ? Ans. $12982. 40. A merchant selling goods at $6789, gained $444. Whatdid they cost? Ans. $6345. 41. Chatham is 47 miles west of York, and Howe is 91 miles west of York. How far is it from Howe to Chatham ? Ans. 4A miles. 42. If the distance of the moon from us is 240000 miles, and that of the sun 95000000 miles, how much farther is it to the sun than to the moon ? Ans. 94760000 miles. 43. Washington died A. d. 1799, at the age of 67. In what year was he born ? Ans. 1732. 44. At an election the successful candidate received 5075 votes, and had a majority of 387 over the defeated candi- date. How many votes did the latter receive ? Ans. 4688 votes. 46 IKTEGERS. 45. How many years is it since the discovery of America by Columbus in 1492 ? 46. Find the difference between eight thousand seventeen, and seven thousand eighteen. 47. An army numbered before a battle thirty thousand five hundred ten men ; after battle twenty-six thousand seven hundred nineteen. What was the loss ? Ans. 3791 men. Find the difference 48. Between 85374 and 5805. 49. Between 4709 and G3489. 50. Between 75310 and 864. 51. Between 9811 and 98021. 52. Between 5678901 and 991112. 53. Between 82223 and 1359012. 54. Between 26590123 and 733334. 55. Between 6444445 and 924901123. 56. 4159012345 — 86665554 = ? 57. 65490123456 — 566666667 = ? 58. The difference of two numbers is 753, and the greater is 1031. What is the less ? Ans, 278. 59. What number added to 2056 makes 18000 ? 60. What number taken from 67890 leaves eGGQH ? 61. The thermometer stood at 19 degrees above zero in the morning, and 45 degrees above zero in the afternoon. How many degrees did it rise ? 62. A farmer raised 1220 bushels of wheat, and 700 bushels of com. He sold 875 bushels of wheat, and 225 bushels of corn. How many bushels of each had he left ? Ans. 345 of wheat ; 475, corn. OUTLINE OF MULTIPLICATION. 81. Product, 83. Multiplicand. TERMS. 82. Factors. (84. Multiplier. r 85. Of Mtatiplication. • o SIGNS. 69. Of Equality, 60. Of Dollars, M H . 95\ Parenthesis, —Vinculum < M 1-^ 86. TABLE. H ' 87. J^irs*. H 88. Second, PRINCIPLES. - 89. T/wrcl. 90. Fourth, g 6 X r 91. First. CASES. 92. Second. I 94. T/*irr Case JJ. 95. -For Case 2JJ. SECTION V. -^g — g^e^p-^aer- m mmiM'i'BiMi€ATi&>m I ^- ^,%<^^-^ 80. Multiplication is the process of repeating one riumber as many times as there are ones in another. Or, it is a short process of adding equal numbers. The Terms are Prodicct and Factors. 81. The Product is the result obtained by Multipli- cation. 82. The Factors are the numbers used to obtain the Product. "YYie factors are Multiplicand and Multiplier. 83. The Multiplicand is that factor which is to be repeated. 84. The Multiplier is that factor which shows how many times the Multiplicand is to be repeated. When it is said, 2 times 3 are 6, 6 is the Product and 3 and 3 are the Factors, of which 2 is the Multiplier and 3 the Multiplicand. 85. The Sign of Multiplication is an oblique cross ( X ), and when placed between two or more numbers signifies that they are to be multiplied together. Thus, 2x3 means that 2 and 3 are to be multiplied together, and is read "2 times 3." MULTIPLICATION. 49 86. MULTIPLICATION TABLE 1 1 1 ONE. 234567 8 9 11111111 6 SIX. 123456789 66666 6-666 1 23456789 6 12 18 24 30 36 42 48 54 1 2 2 TWO. 2345 6 789 22222222 7 SEVEN. 123456789 777777777 2 4 6 810 12141618 7 14 21 28 35 42 49 56 63 1 3 3 THREE. 23 456789 33333333 8 EIGHT. 123456789 888888888 3 6 9121518 2124 27 8 16 24 32 40 48 56 64 72 1 4 4 4 FOUR. 23456789 4 4_4_4_4_4jt_4 8 12 16 20 24 28 32 36 9 NINE. 123456789 999999999 9 18 27 36 45 54 63 72 81 1 5 5 5 FIVE. 23456789 55555555 TEN. 0123456789 10 10 10 10 10 10 10 10 10 10 10 15 20 25 30 35 40 45 10 20 30 40 50 60 70 80 90 50 INTEGERS. ..-f.- -f ©ral^ Example. — What will 4 apples cost at 3 cents each ? Solution.— Since 4 times 3 cents are 13 cents, 4 apples at 3 cents each will cost 12 cents. Or, 4 times 3 cents are 12 cents. Therefore, 4 apples. at 3 cents each will cost 12 cents. Or, 4 apples at 3 cents each will cost 12 cents, because 4 times 3 cents are 12 cents. Or, If 1 apple cost 3 cents, 4 apples will cost 4 times 3 cents or 12 cents. Therefore, 4 apples at 3 cents each will cost 12 cents. PBOBI^EMS What cost 1. 9 pencils at 6 cents each ? 2. 7 chairs at 8 dollars each ? 3. 5 hats at 4 dollars each ? 4. 3 lemons at 5 cents each ? 5. 4 tons of coal at $9 a ton ? 6. 8 harrels of flour at $8 a barrel ? 7. 6 yards of cloth at $7 a yard ? 8. 6 vests at 5 dollars each ? 9. 9 books at 3 dollars each ? How many are 11. 9x4cents? 4x9? 12. 7x5dollars? 5x7? 13. 5x9dollars? 9x5? 14. 3 X 8 cents ? 8 x 3 ? 15. 4x8dollars? 8x4? 16. 8x9dollars? 9x8? 17. 6x8 cents? 8x6? 18. 6x3dollars? 3x6? 19. 9x7dollars? 7x9? 20. 7x3dollars? 3x7? 10. 7 pairs of shoes at $4 a pair ? 21. If 8 pounds of beef will last a family one week, how many pounds will last them 5 weeks ? 2 weeks ? 7 weeks ? Example. — How many are 4 chairs x 3 dollars ? We cannot multiply 3 dollars by 4 chairs. For the product would neither be 12 chairs, nor 12 dollars. We can multiply 3 dollars by 4. For 4 times 3 dollars are ($3 + $3 + 13 + $3), or 12 dollars. Hence, MULTIPLICATION^. 51 87. The Multiplier is always an abstract number, 88. The Multiplicand may be an abstract, or concrete number. 89. The Product is always of the same kind as the true Multiplicand. Note. — In the explanation of the solution of problems which contain concrete numbers, the number used concretely is the true multiplicand. Thus, 4 chairs at |3 each, will cost 4 times "3 dollars" etc. In this explanation " 3 dollars" is the number used concretely, and is, there- fore, the true multiplicand. 90. The product is numerically the same in whatever order the factors are multiplied together. i'lWi«itten^xferci^eX'+ CASE I. 91. When the Multiplier does not exceed 9, Example. — How many are 564-f 564+564, or 3 x 564 ? Explanation.— In the Solution by Addition we find that the sum of 3 564's is 1692. But since 4 ones + 4 ones + 4 ones = 3x4 ones, and 6 tens + 6 tens + 6 tens = 3x6 tens, and 5 hundreds + 5 hundreds + 5 hundreds = 3 x 5 hundreds, the process can be very much shortened, as shown in the following Solution by Multiplication. Explanation.— In this Solution the Multiplicand 564 is written but once ; and as it is to be repeated 3 times, we write the Multiplier 3 under it, and commence at the right hand to mul. tiply. solution. By Addition. i 564 Parts, -j 564 ( 564 Sum, 1692 solution. By Multiplication. Multiplicand, 564 Multiplier, 3 Product, 1692 O'Z IKTEGERS. 3 times 4 ones are 12 ones = 1 ten and 2 ones. Write the 2 ones in ones' order of the product and reserve the 1 ten to add to the product of tens. 3 times 6 tens are 18 tens, plus 1 ten reserved, are 19 tens = 1 hundred and 9 tens. Write the 9 tens in tens' order, and reserve the 1 hundred to add to the product of hundreds. 3 times 5 hundreds are fifteen hundreds, plus 1 hundred reserved, are 16 hundreds = 1 thousand and 6 hundreds, which write in the thousands' and hun- dreds' orders. Hence the Product is 1692. equal to the Sum of the 1st Solution. PROBLEMS, Solve by both methods. 1. 4 times 83. 4. 6 times 344. 7. 2 times 476. 2. 5 times 126. 5. 7 times 519. 8. 8 times 155. 3. 3 times 693. 6. 4 times 195. 9. 9 times 619. 10. Multiply 372 by 3 ; by 4 ; by 5 ; by 6 ; by 7. 11. Multiply 5344 by 4 ; by 6 ; by 8 ; by 9 ; by 5. 12. Multiply 474372 by 5 ; by 7 ; by 9 ; by 6. 13. 67391 X 6 = ? 14. 36519 X 4 = ? 15. 62168 x7 = ? 16. 537354 x 9 = ? 17. 313763 X 8 = ? 18. 898767 x 5 = ? 19. What cost 456 barrels of flour, at $8 a barrel? Although $8 is the true Multiplicand, for convenience we use 8 for the Multiplier and 456 for the Multiplicand (SO, Note), but the Pro- duct is dollars. For if 1 barrel of flour cost $8, 456 barrels will cost 456 times $8 == $3643. 20. What cost 986 pounds of sugar, at 9 cents a pound ? 21. What is the cost of 7 lots of ground, at 15648 each ? 22. How much will 8 tons of hay cost, at $23 a ton ? Ans. $184. 23. How much will 4 horses cost, at 1225 each ? Ans. $900. 24. How much will a year's board cost, at $6 a week, there being 52 weeks in a year ? Ans. $312. MULTIPLICATION^. 63 25. There are 5280 feet in a mile. How many feet in 9 miles? Ans. 47520 feet. 26. There are 1760 yards in a mile. How many yards in 7 miles ? A7is. 12320 yards. 27. How far will a train of cars run in 8 hours, at the rate of 42 miles an hour ? Ans. 336 miles. 28. An iron manufacturer pays his hands $4269 a month. How much do their wages amount to in 6 months ? in 7 months? 1st Ans. $25614 ; 2d, $29883. 29. What will 8 farms cost at $12219 each? 30. How many square miles in 7 townships, if each con- tains 24375 square miles ? 31. How much will 328 yards of cloth cost, at $5 per yard? Ans. $1640. 32. What will 375 head of sheep cost, at 4 dollars a head ? Ans. $1500. 33. What will 2768 barrels of flour cost, at $7 a barrel? Ans. $19376. 34. There are 1728 cubic inches in one cubic foot. How many cubic inches in 8 cubic feet ? Ans. 13824 cubic inches. 35. In one square foot there are 144 square inches. How many square inches in 6 square feet. Ans. 864 square inches. 36. At $325 per acre what will a lot of 9 acres cost ? Ans. $2925. 37. What cost 675 melons at 8 cents each ? 38. What will 7 buggies cost at $225 apiece ? 39. A man travels 8 miles an hour. How far can he travel in 4897 hours? Ans. 39176 miles. 54 INTEGERS, ^-iVTritten^xfercisesr-f CASE 11^ 93. When the Multiplier exceeds 9. Example.— Multiply 583 by 47. 583 47 SOLUTION. Multiplicand, Multiplier, 1st Partial Product, 4081 2d Partial Product, 2332 Product, 27401 Explanation. — We first multiply 583 by 7, the num- ber of ones in the Multiplier as taught in Case I, and obtain 4081. We then in the same man- ner multiply 583 by 4, the number of tens in the Multi- plier and obtain 2332. But since 1 ten is 10 times as great as 1 one, 4 tens must be 10 times as great as 4 ones, and therefore each figure multiplied by 4 tens will produce a product ten times as great as the same figure multiplied by 4 ones. Therefore 583 multiplied by 4 tens, is ten times as great as when multiplied by 4 ones. But when multiplied by 4 ones the product is written so that the right-hand figure stands in the ones' place. Hence, to make it express ten times as much, it must be written so that the right-hand figure shall stand m the tens' place. Writing it thus and adding the two Partial Products, we have 27401, the required Product. 93. EuLE. — Write the Multiplier under the Multiplicand so that units of the same order shall stand in the same column. Multiply the entire Multiplicand hy each significant figure in the Multiplier, and write the several Partial Products ona under another so that the right-hand figure of MULTIPLICATION^. 55 each Partial Product will represent the same order as the figure of the Multiplier producing it. The sum of the Partial Products is the required Product. 1. 364805 X 92 = ? 2. 213254 X 607 := ? 3. 380097 X 504 = ? 4. 400562 X 904 = ? 5. 576348 x 746 = ? JPiJ O B LEMS. 6. 908070 x 908 =: ? 7. 174563 X 1385 = ? 8. 217572x4175 = ? 9. 764050 X 1731 = ? 10. 374781 X 1402 = ? These 10 Problems may be increased in number to 100 by multi- plying eacb Multiplicand by eacli of the 10 given Multipliers. 11. If a vessel sails 231 miles in 1 day, how far will she sail in 35 days ? in 72 days ? 1st Ans. 8085 miles. 12. If a man spends $25 a day for store expenses, how much will he spend in 730 days? Ans. $18250. 13. A man bought 16 lots at 1750 apiece. How much did they all cost him ? Ans. $12000. 14. A farmer raised 456 bushels of corn each year for 19 years. How much in all did he raise ? Ans. 8664 bushels. 15. A gentleman's expenses for 1 year were $3450. At the same rate what would they be for 25 years ? Ans. $86250. 16. In a school there are 19 teachers and 43 pupils for each teacher. How many pupils are there ? Ans. 817 pupils. 17. In a certain school district there are 351 pupils, and the cost of tuition per pupil is $14. What is the entire cost?" Ans. $4914. 18. What cost 450 tons of iron, at $45 per ton ? 66 INTEGEES. 19. What cost 37 tons of steel, at $63 per ton ? 20. What cost 372 yards of cloth, at $9 per yard ? 21. What cost 568 acres of ground, at $175 per acre ? 22. A railroad company bought 13 locomotives at 19500 apiece. What did they all cost ? Ans. $123500. 23. An institution spends in advertising $565 per year; at this rate how much would it spend in 15 years ? Ans. $8475. 24. Multiply three hundred forty-one million six hundred four thousand five hundred one by twenty-three thousand six hundred eight. Ans. 8064599059608. 25. Find the product of seventy-five million three hun- dred sixty-seven thousand four hundred fifty-one by seven thousand twenty-one. Ans. 529154873471. 4^ Wi itten ^JExTerei^eXt •'•^ — - CASE III. 94. When either factor has one or more ci" phers on the right. Example.— Multiply 325 by 10 ; by 100. Explanation. — Since 10 units of any order make 1 of the next higher order, the writing of a cipher on the right of a number, thus removing each of its figures one order to the left, must multiply it by 10. In like manner the writing of two ciphers on the right of a number multiplies it by too. Hence, to 325 x 1, which is 325, we annex one cipher, and have 10 times the number; annex two ciphers and we have 100 times the number. SOLUTION. Multiplicand, 325 Multiplier, 10 Product, 3250 MULTIPLICATION. 67 Example. — Multiply 325 by 50 ; by 500. SOLUTION. Explanation.— Since 50 is 10 times 5, 50 times 325 325 must be 10 times 5 times 325. -^ 5 times 325 is 1625, and 10 times as much, is 1625 with one cipher annexed, or 16250. 16250 In like manner the annexing of two ciphers gives the product when 325 is multiplied by 500. Example.— Multiply 3250 by 50. solution. Explanation.— 5 times 3250 is 5 times 325 with a 3250 cipher annexed, or 16250, and 50 times 3250 is KQ 16250 with one cipher annexed, or 162500, the re- quired product. 162500 95. Rule. — To the product of all the other figures annex as many ciphers as there are at the right of both factors. PR OBZE MS. 1. Multiply 375 by 100 ; by 420: 41900 by 9090. 2. Multiply 1450 by 3500; by 780: 3080 by 1450. 3. Multiply 6300 by 1700; by 4000: 7500 by 6300. 4. Multiply 245 by 36000; by 790: 1260 by 2450. . 5. Multiply 375000 by 360 ; by 42 : 1420 by 3750. 6. Multiply 5090 by 480; by 90 : 6030 by 5090. 7. Multiply 6203 by 9300; by 200: 4090 by 6200. 8. A barrel of flour weighs 196 pounds. How many pounds in 200 barrels ? Ans. 39200 pounds. 9. A barrel of fish weighs 200 pounds. How many pounds in 150 barrels ? Ans. 30000 pounds. 10. A barrel of salt weighs 280 pounds. How many pounds in 179 barrels ? Ans. 50120 pounds. 11. A barrel of lime weighs 240 pounds. How many pounds in 56 barrels ? Ans. 13440 pounds. 68 INTEGERS. 12. A bushel of coal weighs 76 pounds. How many pounds in 4800 bushels ? Ans. 364800 pounds. 13. How much will 20 city lots cost, at 14100 each ? Ans. $82000. 14. How much will 16 acres of ground cost, at $170 per acre? Ans. 12720. 15. How much will 450 head of cattle cost, at $56 a head ? Ans. $25200. 16. How much will 351 head of sheep cost, at $10 a head ? Ans. $3510. 17. How much will 402 head of horses cost, at $230 a head? A7is. $92460. 18. In printing 5000 copies of a book, 24 sheets of paper were used for each book. How much paper was used ? Ans. 120000 sheets. 19. In building a line of telegraph 370 miles long, the cost was $1004 per mile. What was the entire cost ? Ans, $371480. 20. At a plow factory 108 plows were made each day for 50 days. How many plows were made ? Ans. 5400 plows. 21. An agent travels 297 days at an average of 125 miles miles per day. Howmany miles does he travel in the year ? Ans. 37125 miles. 22. How many days in 8 years of 365 days each ? Ans. 2920 days. 95'. A Parenthesis ( ) is used to include such num- bers as are to be considered together. A Vinculum has the same meaning. Thus, 4 X (5 + 4), or 4 X 5 + 4 means that 4 is to be multiplied by 5 + 4 or 9. 4 multiplied by 9 = 36, Ans. Without the parenthesis, or vinculum, 4x5 + 4 means 4 times 5 and 4 added. 4 times 5 = 20 ; 20 + 4 = 24, Ans. MULTIPLICATION. 69 hlWi«itten?EXfepci*e^-f REVIEW PROBLEMS. 1. Multiply 325 — (15 x 15) by 47 x (1861 — 1814). Ans. 220900. 2. Multiply 801 — 169 + 533 by 624 — 25~^2r. Ans. 9801. 3. Multiply (40 x 50) — 1850 by (91 x 19) — 1579. Ans. 22500. 4. Multiply (75 x 24) — 849 by (63 x 38) - 1393. Ans. 951951. 5. Multiply 42 + (13 x 43) by 1761 - (38 x 23). Ans. 533087. 6. 75 + 8 X (25 + 38) x (89 — 72) x 7 — 100 = ? Alls. 59951. 7. 20 X (16 X 5 - 14 X 3) + (42 - 18 x 2) - 700 rzz ? Ans. QQ, 8. 15291 — 1342 + 863 — 14812 = ? 9. 1764 + 91426 - 92190 + HI = ? 10. What sum must be added to 271 x 300 to make the amount 131313 ? Ans. 50013. 11. What is the difference between 75011 — (671 x 102) and 463 x 201 — 86394 ? 12. (463 + 537—901) x 990 = ? Ans. 98010. 13. 96 + 8 X (35 + 40) by (56— 40x8)— 5 = ? Ans. 85608. 14. A speculator bought 800 acres of land at $20 an acre ; he sold at one time 350 acres at $25 an acre ; at another time 250 acres at $30 an acre. At what price must he sell the remainder in order to gain on purchase $8250. Ans. $8000. OUTLINE FOR DIVISION. O M H H 9^ TERMS. SIGNS. 97. Dividend, (Prodiict.) 98. Divisor, (Factor.) 99. Quotient, {Factor.) 100. O/ Division. 59. O/ Equality. 60. O/ Dollars. 95'. Parenthesis.— Vinculum, 101. TABLE. 102. OBJECT, OR OFFICE. PRINCIPLES. 103. l^insf. 104. Second. CASES. r 105. J. 107. II. L 109. JJJ. RULES. 106. Case I. 108. Case II. L no. Case III. SECTION VIo • g) ® (5 » ^ miWISiIO;M 96. Division is the process of finding how many times one number is contained in another. The Tertns in Division are Dividend, Divisor, and Quotient. 97. The Dividend is the number to be divided. 98. The Divisor is the number by which to divide. 99. The Quotient is the result obtained by Division. 1. When there is nothing left after division, the division is said to be exact ; and the divisor is called an exact divisor. 2. That part of the dividend which is left when the division is not exact is called the Remainder. 3 The remainder must always be of the same denomination as the dividend, because it is really a part of it. 4. A true remainder is always less than the divisor. 100. The Sign of Division is a short horizontal line placed between two dots (-^), and when placed between two numbers signifies that the one on the left is to be divided by the one on the right. Thus, Q -\-2 means that 6 is to be divided by 2, and is read " 6 divided by 2." Division is sometimes indicated by writinor the dividend above, and the divisor below a short horizontal line. Thus, f is read " 6 divided by 2." In writinqr numbers for solution, the Divisor and Dividend may be separated by a curved line. Thus, 2 ) C, or 6 ( 2 is read " G divid- ed by 2." " ~ 6X 62 INTEGERS. 101. DIVISION TABLE. ONE. 1)1 23456789 123456789 SEVEN. 7 V? 14 21 28 35 42 49 56 63 12 3 4 5 6 7~8~9 TWO. 2)^ 4 6 81012141618 123 4 56789 EIGHT. 8)81624 32 40 48 56 64 72 123456789 THREE. 3)3 6 91215 18 212427 NINE. 9 ) 9 18 27 36 45 54 63 72 81 123456789 123456789 FOUR. 4)4 81216 20 24 28 32 36 TEN. 10)10 20 30 40 50 60 70 80 90 123456789 123456789 FIVE, 5)5 1015 20 2530 35 40 45 ELEVEN. 11)11 22 33 44 55 66 77 88 99 123456789 123456789 SIX. 6)61218 24 30 36 42 48 54 T W EL VE. 12)12 24 36 48 60 72 84 96108 123456789 12345678 9 DIVISION. 63 10^. The Object of Division is twofold: First To find how many times one number is contained in another. Second. To find one of the equal parts of a number. First. Example. — A man has 10 dollars with which he wishes to buy books at 2 dollars each. How many books can he buy ? Solution. — Since 2 dollars a/re contained in 10 dollars 5 times, he can buy 5 hooks. Or, 2 dollars are contained in 10 dollars 5 times. Therefore, he can buy 5 looks. Second. Example. — If a man divides 10 dollars equally between 2 persons, how much does each receive ? Solution. — Since 2 persons receive 10 dollars, one person receives ONE-HALF of 10 dollars, or 5 dollars. Or, One-half of 10 dollars is 5 dollars. Therefore, each person receives 5 dollars. The preceding are representative examples of the two kinds of problems to which Division is applied. In the fird, the Divisor and Dividend are like numbers and the Quotient is an abstract number. Hence, 103. The Quotient is an abstract number, when the Divisor and Dividend are both abstract, or both concrete numbers. In the second example, the Divisor and Dividend are unlike numbers and the Quotient is the same kind as the dividend. Hence, 104. When the Divisor and Dividend are unlike num- bers the Quotient is alivays the same kind as the Dividend. The mode of reasoning in the solution of these two classes of exam- ples is somewhat different ; but in reality the operation is the same, viz.: to find how many times one number is contained in another, which is in harmony with the definition of Division (96). 64 INTEGERS 0i»at iExfGiicis%X -i ^^ How Many Tinies,'^ 1. How many pounds of rice, at 8 cents a pound, can be bought for 56 cents ? for 24 cents ? 2. When raisins are 9 cents a pound, how many pounds can be purchased for 72 cents ? for 36 cents ? 3. At 8 cents a bushel, how many bushels of coal can be bought for 64 cents ? for 48 cents ? 4. At $7 per ton, how much ice can be bought for $63 ? for 135 ? for 149 ? 5. How many pencils, at 6 cents each, can be bought for 48 cents ? for 54 cents ? 6. To how many boys can I give $30, giving each boy $5 ? giving each $6 ? 7. At 4 cents each, how many oranges can I buy for 36 cents ? for 16 cents ? 8. At 6 cents a quart, how many quarts of milk can be bought for 24 cents ? for 36 cents ? 9. A man paid 81 cents for beef, at 9 cents a pound. How many pounds did he buy ? 10. A lady gave 16 apples to some boys, giving 4 apples to each boy. How many boys were there ? 11. How many times can a bucket which holds 3 gallons be filled from a barrel which holds 27 gallons ? 12. To how many persons can I give $14, giving each person $2 ? 13. At 7 cents each how many oranges can I buy for 63 cents ? Pivisioiiir. 65 ©i*al ^.^or cisG^ '' Equal Parts.'^ When a number or tiling is divided into two equal parts, the parts are called halves ; into three, the parts are called thirds ; into four, fourths; into Jive, fifths; into ten, tenths, etc. A number is divided into two, three, four, five, ten, etc., equal parts ^y dividing by 2, 3, 4, 5, 10, etc., respectively. 1. What is a half of 8 ? Of 10 ? Of 18? Of 12? 2. What is a third of 9 ? Of 12 ? Of 24 ? Of 6 ? 3. What is a fourth of 10 ? Of 8 ? Of 36 ? Of 12 ? 4. What is a fifth of 20 ? • Of 45 ? Of 20? Of 35 ? 5. What is a sixth of 18 ? Of 54? Of 24 ? Of 36 ? 6. What is a seventh of 14 ? Of 28 ? Of 63 ? Of 35 ? 7. What is an eigh th of 40 ? Of 8 ? Of 48 ? Of 72 ? 8. What is a ninth of 27 ? Of 72 ? Of 18 ? Of 36 ? 9. If 7 pounds of rice cost 56 cents, what does 1 pound cost? 10. If 9 tons of ice cost $63, what does 1 ton cost ? 11. If 9 oranges cost 36 cents, what does 1 orange cost? 12. A lady gave 16 apples to 4 boys, giving to eaxih boy the same number. How many apples did each boy receive ? 13. If 4 quarts of milk cost 24 cents, how much does 1 quart cost ? 14. If 8 pencils cost 48 cents, what does 1 pencil cost? 15. If 8 bushels of coal cost 64 cents, what does 1 bushel cost? 16. A boy divides 84 cherries equally among 7 boys. How many does he give to each ? 66 INTEGERS. t^Wi^itten ^:?^erei CaUed Factors in MultipUcation. 824 1236 412 54384 13 54397 108. Rl'LE. — Write the divisor at the left or right of the dividend^ luith a curved line hettoeen them. Remainder. Dividend. Called Product in Multiplication. 70 INTEGERS. Separate mentally from the left of the dividend the figures expressing the smallest number that will contain the divisor. Divide this nuniber hy the divisor, and write the result as the first figure of the quotient. Multijjly the divisor hy this quotient figure, and subtract the product from that part of the dividend which was used. To the right of the reynainder, if there is any, annex the r.ext figure of the dividend, and divide the mimber thus formed by the divisor, ivriting the result as the second figure of the quotient. 'Proceed in this manner until all the figures of the dividend have been used. If any remainder with the next figure of the dividend annexed, does not contain the divisor, write a cipher in the quotient, annex to the remainder another figure of the divi- dend, and proceed as before. Proof. — Multiply the divisor and quotient together, and to the product add the remainder, if any. If the result is equal to the dividend, the ivorh is correct P ROBI.EM, 1. 67473 2. 22491 3. 44982 4. 179928 5. 112455 6. 157437 7. 202419 8. 292383 9. 314874 10. 472311 a 13 b 14 c 15 d 16 e 18 / ^1 9 24 h 27 i 32 j 34 Explanation.— Form 100 Problems by dividing each of the dividends 1, 2, 3, 4, 5, etc., by each of the divisors a, b, c, d, e, etc., as \a. 67473-7-13 = ? \g. 67473^24 = ? 6a. 157437-f-13 = ? 6d. 157437-^16 = ? DIVISIOK. 71 11. Divide three million twenty-five thousand six, by four thousand eighteen. Remainder, 3470. 12. In 638 days a government's revenue was $246356682. How much was that per day ? Ans. 1386139. 13. If a railroad earns $2600000 in 52 weeks, how much is that per week ? Ans. $50000. 14. A railroad 357 miles long cost $2502927. How much was that per mile ? Ans. $7011. 15. In one gallon of wine there are 231 cubic inches. How many gallons in 8547 cubic inches ? Ans. 37 gallons. 16. Sound travels 1119 feet in 1 second. How long would it take it to travel one mile, or 5280 feet ? Ans. 4 seconds, and 804 feet remainder. ■^ ^Written ^xfei'ci^e*^^ CASE III 109. WJien the Divisor is annexed. 1 with ciphers Example.— Divide 1462 by 100. Explanation. — We divide accord- ing to Explanation and Rule in 107 and 108, and obtain tlie quo- tient 14, and remainder 63. We can obtain tlie same result by a much shorter process, name'y, cutting off the two right-hand fig- ures of the dividend, thus 14/62, in wliich we have 14 for the quotient, and 62 for the remainder as before. solution. Divisor. Dividend. Quotient. 100 ) 1462 ( 14 100 462 400 62 Remainder. Example.— Divide 17563 by 1000. Solution.— 17563-5-1000 = 17, with 563 remainder. 73 INTEGERS. Explanation. — We cut off from tlie right, tliree figures (the num- ber of ciphers in the divisor) and have 14 as the quotient, and 563. the remainder. In any number all the figures to the left of ones, express tens ; to the left of tens, hundreds ; to the left of hundreds, thousands ; etc. Therefore, in the number 17563, 17 must express 17 thousands, and 563 must be the remainder. 110. EuLE. — Cut off from the right of tlie dividend as many figures as there are ciphers at the right of the divisor. The remaining figures form the quotient, and the figures cut off, the remainder. PROBLEMS. 1. Divide 48729 by 100 ; by 1000 ; by 10000. 2. Divide 637591 by 1000 ; by 100000 ; by 10. 3. Divide 1234567 by 1000000 ; by 10000. 111. When the divisor is any numler with ciphers annexed, the following Kule answers^ our purpose : KuLE. — Cut off from the right of the dividend as many figures as there are ciphers at the right of the divisor. Then divide the remaining figures in the dividend hy the divisor exclusive of its right-hand ciphers. To the remainder, if any after this division, annex the figures cut off from the dividend for a true remainder. Example.— Divide 2400 by 60. SOLUTION. 6|0 ) 240|0 40 Example.— Divide 13762 by 400. SOLUTION. 4!00)137|62 34 quotient, 162 remainder. PBOB Ex. 1. Divide 3472 by 40. 2. Divide 476375 by 70. 3. Divide 167478 by 90. ZEMS. 4. Divide 17560 by 400. 5. Divide 715723 by 700. 6. Divide 752562 by 340. SECTION VII. ,^» MWi'itten ^^erci^es^f MEVIEW PHOBIjEMS. 1. The addends of a number are 64, 92, 430, 7859, and 436177. What is the number ? Ans. 444628. 2. The parts of a number are 4876, 73, 23756, 400009, and 99999. What is the number? Ans. 528713. 3. The minuend is 48639 and the subtrahend is 29048. What is their difference? Ans. 19591. 4. The minuend is 93470 and the subtrahend is 39401. What is the remainder ? Ans. 54069. 5. The minuend is 47319 and the difference is 12499. What is the subtrahend ? Ans. 34820. 6. The minuend is 311409 and the remainder is 76234. What is the subtrahend ? Ans. 235175. 7. The subtrahend is 4008 and the difference is 17296. What is the minuend ? Ans. 21304. 8. The subtrahend is 93000 and the remainder is 639421. What is the minuend? Ans. 732421. 9. The multiplicand is 4215 and the multiplier 309. What is the product? Ans. 1302435. 10. The multiplicand is 937 and the product is 40291. What is the multiplier? Aiis. 43. 11. The multipher is 319 and the product is 320276. What is the multiplicand ? Ans. 1004. 74 INTEGERS. 12. The dividend is 382268 and the divisor is 421. What is the quotient ? Ans. 908. 13. The dividend is 387234 and the quotient is 909. What is the divisor? Ans. 426. 14. The divisor is 145 and the quotient is 4200. What is the dividend ? Ans. 609000. 15. The divisor is 43, the quotient, 64, and the remainder, 15. What is the dividend ? Ans. 2767. 16. The dividend is 2943 and the divisor is 66. What is the remainder ? Ans. 39. 17. The dividend is 9600 and the quotient is 95. What is the remainder? Ans. 5. 18. From the sum of 72 and 83 take the difference of 125 and 182. Ans. 98. 19. (127 + 468) — (1930 — 1678) = ? Ans. 343. 20. To the product of 45 by 23 add the quotient of 512 by 16. Ans. 1067. 21. ( 93 X 32 ) 4- ( 1462 -^ 43 ) = ? Ans. 3010. 22. Multiply the sum of 83 and 74 by their difference, and divide the product by 15. Ans. 94, 3 remainder. 23. (99 + 48) X (99 _ 48) - 7 H- 35 = ? Ans. 214. 24. Multiply the quotient of 3348 divided by 27 by the quotient of 150000 divided by 375. A?is. 49600. 25. (1287 -T- 99) X (30969 -^ 999 ) = ? Ans. 403. 26. Bought 160 acres of land at $5 an acre and sold it at $12 an acre. What did I gain ? Ans. $1120. 27. 140 X (16-7)=? (140 x 16)- (140 x 7)=? Ans. 1260. 28. Bought 375 barrels flour for 13000; sold them for $2260. How much per barrel did I lose ? Ans, $2. REVIEW PROBLEMS. 75 29. (3600 - 1350) -^ 450 = ? ( 3600 -^ 450) - (1350 — 450) = ? Ans. 5. 30. Bought 5 yards of cloth at $7 a yard, 3 yards of silk at $2 a yard, and 12 shirts at $3 apiece, and offered four $20 bills. What change was due me ? Ans. $3. 31. 176 - (9x7) + (5 X 8) + (14 x 5) = ? Ans. 3. 32. B's income one year was $2500; his expenses that year averaged $3 per day : how much did he save, the year having 365 days. Ans. $1405. 33. 10000 — (421 X 13) = ? Ans. 4527. 34. A farmer had $5, and sold his marketing for $28 ; he bought $4 worth of sugar, $3 worth of coffee, and $2 worth of spices. How much had he left? Ans. $24. 35. (1900 + 396) — (561 + 411 + 96) = ? Ans. 1228. 36. Mr. Hall, keeping his store open 310 days in 1 year, made $3100; and his family expenses for 365 days were 11095. How much did he make per working day more than he spent per spending day ? Ans. $7. 37. (20736 ^ 144) — (1728 -^ 144) = ? Ans. 132. 38. How many times can a 40-gallon barrel be filled from 5 casks, each holding 120 gallons ? Ans. 15 times. 39. 1700 X 400 -^ 34000 = ? Ans. 20. 40. Bought 300 barrels of apples at $3 per barrel ; 120 of them were stolen : at what price per barrel must sell the remainder so as to suffer no loss? Ans. $5. 41. (1900 X 20) -f- (1900 - 1140) = ? Ans. 50. 42. Five thousand six hundred seventy-eight, and another number, amount to twenty-five thousand four. What is the other number? Ans. 19326. 76 INTEGERS. 43. A public school pays its principal 11800 per annum; 2 assistant principals 1800 each; 6 medium teachers $600 each ; and 8 primary teachers $500 each. What is the cost of instruction ? Ans. $11000. 44. 2500 + (3 X 500) + (5 x 700) + (6 X 200) =? Ans. 8700. 45. If a mechanic receives $1400 a year for his labor, and his expenses are $840, in what time can he save enough to buy 32 acres of land at $140 an acre ? Ans. 8 years. 46. (280 X 64) -f- (2504 — 1384) = ? Ans. 16. 47. A's age is 35, and B's 27 years. What is the average of their ages ? Operation. -7- (35 years + 27 years) -4- 2 = 81 years. The average of two numbers is one-half their sum ; the average of three numbers, one-third their sura, &c. 48. Find the average of 2, 4, and 6. Ans. 4. 49. Find the average of 3, 5, and 10. Ans. 6. 50. Find the average of 6, 10, and 11. Ans. 9. 51. Five men are worth, respectively, $3000, $12000, $56000, $20000, and $35000. What are they worth on the average ? Ans. $25200. 52. A boy's earnings per month were as follows: $12, $10, $15, $13, $16, $18, $14, $8, $20, $17, $15, and $10. What were his average earnings per month that year? Ans. $14. 53. Farmer A's expenses and receipts one year were as follows : EXPENSES, RECEIPTS. ^01 ' Labor .... $315 For Wheat . . . . $364 (( Seed 64 " Corn .... . 140 it Stock . . . . 290 " Pork .... . 216 (( Fruit Trees . . 100 " Fruit. . . . . 79 u Family Expenses 590 "Hay ... . . 195 a A Reaper . . . 375 « Vegetables. . . 46 REVIEW PROBLEMS. Did he make or lose that year, and how much ? Ans. Lost $694. 54. Farmer B's expenses and receipts one year were as follows : EXPENSES. For Labor .... $516 " Seed 309 " Stock .... 50 " A Mower ... 190 " Family Expenses 478 Did he make or lose that year, and how much ? Ans. Made $194. 55. Farmer C's expenses and receipts one year were as follows : RECEIPTS. EXPENDITURES. RECEIPTS. For Grain . . " Wool . . " Beef . . . « Hay . . . " Vegetables $816 400 180 295 46 For Grain . " Stock . . . $4005 For Seed $145 . . 984 " Labor .... 875 Did he gain or lose money, and how much ? Ans. Gained $3969. 56. The sum of two numbers is 72, and their difference is 46. What are the numbers ? EXPLANATION. Since 72 is the sum of the numbers, if the difference 46 is subtracted from the sum 72, the remainder 26 will be twice the less number. SOLUTION. 72- 46 = 26. 26h- 2 = 13, Less No. 13 + 46 = 59, Greater No. PROOF. 59 -f 13 = 72, Sum. 59- 13 = 46, Difference. 72 + 46 = 118. 118 -^ 2 = 59, Greater No. 59- 46 = 13, Less No. 26 -i- 2 = 13, the less number, and 13 + 46 = 59, the greater number. Or, if the difference 46 is added to the sum 72, the amount 118 will be twice the greater number. 118 ^ 2 = 59, the greater number, and 59 — 46 = 13, the less number. 78 INTEGERS. 57. The sum of two numbers is 1196, and their difference is 504. What are the numbers ? A71S. Greater, 850; Less, 346. 58. The sum of two numbers is 8356, and their difference is 1792. What are the numbers ? Ans. Greater, 5074; Less, 3282. 59. A man paid 14200 for a house and lot, the lot being valued at $1950 more than the house. What was the value of each? Ans. Lot, $3075; House, 11125. 60. Two men are worth $43875, and one is worth $2965 more than the other. What is each man worth ? A71S. 1st man, $23420; 2d, $20455. The pupils will make original examples to illustrate the following problems : 61. Given the sum and all the parts but one, to fin 1 that one. 62. Given the greater of two numbers and their difference, to find the less. 63. Given the product and two of three factors, to find the otJier factor. 64. Given the product and multiplier, to find the multi- plicand. 65. Given the divisor, quotient, and remainder, to find the dividend. 66. Given the sum and difference of two numbers, to find the numbers. 67. Given the divisor and quotient, to find the dividend. 68. Given the product and one of two factors, to find the other factor. 69. The pupils will make two original examples in each of the fundamental rules. OUTLINE OF DECIMALS. 125 137 139 1st METHOD. < 113. Decimal Point. 114. Value. PLACES. 113. Point. 115. Tenths. 116. Ilundvedths. 117. Tlionsandths. &c., &c.t &c. 118 and 119. Bnles. 2d METHOD, ntiyciPLES. ( 120. X 121 ( 133. I 124. Decimal Point, and 122. Rules. First. Second. . ADDITION ...VZ6. Ttule. . .Sl7JJTiJ4CTJC»iV....128. JJuZe. . MTTLTIPLI- CATION. DIVISION, PRINCIPLES 132. JRule. PRIKCIPLES. ■1 130. 1st. 131. 2d. 133. 1st. 134. 2d. 135. Sd. 136. 4it'i«ion.. 7» CHAPTER II.. ~§fr(^ J0)EJCT[MA1L,§; II (jg-^v^ SECTION I '^-^ IMBMEIRJ^IIOM smm N©TA11®M (i^'^ «^s^ From the Outline of Arithmetic, pag^e 11, we learn that numbers are classified as to wholeness of their unit into Integral, Frac- tional, and Mixed Numbers. 8C NUMEEATION AKT> NOTATION. 81 We have already considered Numeration and Notation, Addition^ Subtraction, Multiplication, and Division of Integers, and are now prepared to proceed with Fractions. A Fraction ( 7 ) is a number expressing one or more of the equal parts of a unit ; dL^five tenths, one half. Fractions with reference to the mode of expressing them are of two kinds, namely. Decimal and Common. 112. Decimals are those fractions which express one or more of the parts of a unit which has been divided into tenths, hundredths, thousandths, &c. Decimals are expressed in accordance with the Principles of Arabic Numeration and Notation of Integers (41). The term decimal is derived from the Latin decern, which signifies 10. A number consisting of an Integer and Decimal is a Mixed Num- lev (8). By referring to the picture on page 80, you will notice a ball resting on the keystone (units) of the arch. This represents the Decimal Point, and is in decimals yfhaX ones of units are in inlegerSi—ihet First Place. 113. The Decimal Point is used to distinguish Decimals from Integers. When the point is placed between figures it is read and ; thus, 2.5 is read " 2 and 5 tenths." It may also be seen in the picture that places equally distant from units have corresponding names, with the exception of the decimal termination ths. Thus, tens, — tenths \ hundreds, — hundred^A*; &c 114. The Value which any figure in a decimal ex- presses is determined by the 'place it occupies. The Places are Point, TentTis, Hundredths, ThonsandtJis, Ten-thov^ sandths, &c. 6 5 DECIMALS. 115, Tenths is the Second Place or Order. .1 rzr 1 tenth. .4 = 4 tenths. .7 = 7 tenths. .2 = 2 tenths. .5 = 5 tenths. .8 = 8 tenths. .3 = 3 tenths. .Q = Q tenths. .9 = 9 tenths. The greatest number of tenths expressed ly one figure is 9. 10 tenths = 1. EXERCISES. 1. Read .6; 6.6; .9; 9.9; 18.7; 23.5. 2. Write eight tenths; 12 and 3 tenths; 745 and 4 tentha 116. hundredths is the Third Place or Order. .01 = 1 hundredth. .02 = 2 hundredths. .03 = 3 hundredths. .04 = 4 hundredths. .11 .32 .53 .84 11 hundredths. 32 hundredths. 53 hundredths. 84 hundredths. The greatest number of hundredths expressed by two figures is 99. 100 hundredths = 1. EXERCISES. 1. Read .24; 5.47; .88; 475.15. 2. Read 42.09; .75; 1876.93; .19. 3. Read 14.14; 23.23; 44.44; .07. 4. Read 111.11; 213.02; 400.04; 500.05. 5. Write one hundred and five hundredths. 6. Write fifty and forty-one hundredths. 7. Write seventy-six hundredths; 10 hundredths. 8. Write nine hundred ninety-nine and ninety-nine hun- dredths. 9. Write three thousand forty and forty thousand three h undred-thousandths. KUMERATIOX AND NOTATIOiq-. 83 117. Thousattdths is the Fourth Place or Order. .001 = 1 thousandth. I .098 = 98 thousandths. .014 = 14 thousandths. .942 = 942 thousandths. .144 = 144 thousandths. | .004 = 4 thousandths. 2%e greatest number of thousandths expressed by three figures is 999. 1000 thousandths = 1. ±:XEliCISES. 1. Read .359 ; 4000.004 ; 35.035. 2. Eead 485.485 ; 75.075 ; 19000.019. 3. Read .011; 408.008; 143.043. 4. Read 4279.279; 11.048; .099. 5. Write 325 thousandths; 4 thousandths. 6. Write 97 and 12 thousandths; 803 thousandths. 7. Write 4 and 4 thousandths; 47 thousandths. 8. Write 11245 and 45 thousandths; 9 thousandths. When the right-hand figure of a decimal is handredths ( 1 16), the decimal is read as hundredths; when the right-hand figure is thou- sandths (117) the decimal is read as thousandths ; &c. Hence, 118. Rule for Numeratoi?'. — Read the figures of a decimal the same as the figures of an integer , adding the name of the right-hand order of the decimal. 119. Rule for Notation. — Write the given decimal as a whole number. Prefix ciphers, if necessary, to fill the re- quired number of places. To the left write the decimal point. Example. — Read 123456.72345. By the preceding method we are required to perform three operar tions in order to read the example given : First.— To read the integer— if,? thousand J^56. Second. — To read the decimal as an integer — and 72 thousand 345. Third. — To ascertain the name of the right-hand decimal place by beginning at 7 and saying tenths, hundredths, thousandths, ten-thou- sandths, hundred-thousandths. 84 DECIMALS. SECOJfD METHOD. 120. The Decimal I^oint ( . ) occupies that place in a decimal number which gives narm to the decimal. Example. — Read 123456.72345. Integer. • Decimal. 123 45?i^34? Thous. Units. I23 45? = 123 thousand 4S6 and Thous. Cnits. .72 345 = 72 thousand 343 hundred-thou^ sandths. By the Second method the integer is read as in the First method. But in reading the decimal as an integer we notice that the point is in hundreds' order of thousands' period and know instantly that hundred- thousandths is the name of the decimal. Hence, 121. Rule for Numeration. — Read the figures of a decimal the sa7ne as the figures of an integer, adding the name of the "place in the decimal occupied hy the pointy count- ing from the right as in integers, 122. Rule for Notation. — Write the given decimal as a whole number, prefixing ciphers, if necessary, to cause the point to stand in the place luhich gives the name of the decimal. EXERCISES. Read the following: 1. .125 623 2. .400 119 3. .50 091 4. .62 124 5. .142 205 6. .423 152 391 7. .72 110 111 8. .00 095 486 9. .000 004 004 10. .400 092 11. 12. 13. 14. .125 .0 125 .00 125 .4 125 15. .111 222 NUMEEATIOi^ AND IS^OTATION. { 16. 421 234 .23 191 27. 1 004 .1 004 17. 75 629 .119 256 28. 93 093 .93 093 18. 40 000 .0 004 29. 175 175 .175 175 19. 914 .173 216 014 30. 42 420 .42 420 20. 400 000 .00 004 31. 750 000 .750 21. 50 000 .050 32. 400 000 .400 22. 10 000 .0 001 33. 27 516 .516 23. 19 019 .019 34. 44 550 .00 009 24. 128.000 000 128 35. 833 338 .000 009 25. 4.000 000 004 36. .00 000 000 009 26. 934 761 132 .4 37. .9 009 000 909 85 Write the following in figures: 1. Two hundred fifty-six ten-thousandths. 2. Nineteen thousand one millionths. 3. Fourteen hundred-thousandths. 4. Eight thousand ninety-nine billionths. 5. One million three ten-millionths. 6. Seven hundred forty thousand one millionths. 7. Ninety-eight and one hundred three thousand four ten-millionths. 8. Four thousand and four hundred-thousandths. 9. Four thousand four hundred-thousandths. 10. Four thousand and four hundred thousandths. 11. Six thousand and six hundred-thousandths. 12. Six thousand six hundred-thousandths. 13. Six thousand and six hundred thousandths. 14. Three hundred one billionths. 15. Ninety-three million and 93 millionths. 16. Eleven billion and eleven hundred-millionths. 17. Five tenths; Fifty hundredths; Five hundred thou- sandths; Five thousand ten-thousandths. 86 DECIMALS. 18. Seventy-five hundredths. 19. Twenty-eight thousandths. 20. One hundred nine ten-thousandths. 21. Fifteen and thirty-two hundred-thousandths. 22. Seven and three hundred five milUonths. 23. The decimal two thousand six ten-milhonths. 24. Two thousand six ten-milUonths. 25. Six thousand nine hundred- milUonths. 26. Six thousand 9 hundred-millionths. 27. Three hundred one billionths. 28. Three hundred and one billionth. 29. Two hundred and five ten-thousandths. 30. Forty thousand and thirty-four millionths. 31. Two thousand and four hundred-thousandths. 32. Six hundred and fifteen ten-millionths. 33. Six hundred units and fifteen ten -thousandths. 34. Fifteen and fifteen thousandths. 35. Three hundred thousand three hundred and three hundred-millionths. 36. Five million and eighty-five ten-millionths. 37. Twelve hundred-thousandths. 38. Four hundred units and four hundred sixty-five millionths. .5 = 5 tenths. .50 = 50 hundredths, or 5 tenths and hundredths. .500 = 500 thousandths, or 5 tenths hundredths and thousandths. From this we learn that .5, .50, and .500 are of the same value— 5 tenths. Hence, 1^3. Ciphers may he annexed to a decimal tvitJiout changing its value. 124. Ciphers may he omitted from the right of a deci- mal without changing its value. SECTION II, «^|Mli&il^aiS1gMtet^ 125. Example. — What is the sum of 43.249, 123.14, and 93.412? Explanation. — Since only like orders of units can be added, we write the nambers with like orders in the ^* same columns. 123.14 The decimal points then stand in a column. 93.412 Commence at the right and add as in integers, plac- 9^Q «ni ^°^ ^^^® decimal point in the sum directly under the decimal points in the parts. 126. KuLE. — Write the numbers so that the decimal points shall stand in a column. Add as in integers^ and place the point in the sum underneath the points in the addends, PnOBIs EMS. Find the sum 1. Of 6.09.5, 4.94, 8.573, and 2.6. Ans. 22.208. 2. Of 7.97, .0568, .04, and 5.6378. Ans. 13.7046. 3. Of .853, .0609, 9.7, and .00357. Ans. 10.61747. 4. Of .00328, .6, .88, 6.5, and .987654. Ans. 8.970934. 5. Of four and five tenths, forty-five hundredths, fifty- eight thousandths, eight thousand eight ten-thousandths, seven tenths, and ninety-six millionths. Ans. 6.508896. 6. Of 5 and 9 tenths, 4 and 58 hundredths; 6 and 29 thousandths, and 5708 millionths. A^is. 16.514708. 8S DECIMALS. 7. On the Pa. E. K. the distance from Pittsburgh to Wilkinsburg is 6.9 miles; from Wilkinsburg to Greensburg is 24.2 miles ; from Greensburg to Johnstown, 46.9 miles ; from Johnstown to Altoona, 38.7 miles; from Altoona to Harrisburg, 129.3. What is the distance from Pittsburgh to Harrisburg ? Ans. 246 miles. 8. On the P. W. & B. R K. the distance from Pittsburgh to Braddocks is 9.6 miles; from Braddocks to West Newton, 23.1 miles; from West Newton to Confluence, 51.5 miles; from Confluence to Cumberland, 65.3 miles. What is the distance from Pittsburgh to Cumberland ? Ans. 149.5 miles. 9. A farmer in buildiug a fence around his farm found that the sides measured respectively 28.17 rods, 30.485 rods, 25.672 rods, 68.875 rods, and 75.125 rods. How many rods of fence did he have to build ? Ans. 228.327 rods. 10. A dealer in Western lands sold to one party 325.75 acres ; to a second, 475.375 acres ; to a third, 37.565 acres ; and to a fourth, 379.598 acres. How many acres in all did he sell? Ans. 1218.288 acres. 11. A. bought from one dealer 36.75 tons of coal ; from a second, 25.3165 tons; from a third, 3.75605 tons; and from a fourth, 47.62 tons. How much did he purchase from all? A71S. 113.44255 tons. 12. At Pittsburgh, the rain-fall in Spring is 9.38 inches ; in Summer, 9.87 inches; in Autumn, 8.23 inches; and in Winter, 7.48 inches. What is the raiu-fall for the year? Ans. 34.96 inches. 13. In Washington City the rain-fall in the Spring is 10.45 inches ; in Summer, 10.53 inches ; in Autumn, 10.15 inches; in Winter, 10.07 inches. Kequired the rain-fall for the year. Ans. 41.20 inches. ADDITION. (14) 413.403 (15) 432.5931 (16) .48321 4.12356 96.845 .6973 4193.48 9.38 .831 95.631 .9 .952 4.1284 1.93 .7345 786.47 75.125 .12345 5497.23596 616.7731 3.82146 (17) 179.0009 (18) . 987.789 (19) 5.5 268.008 98.89 44.44 357.09 9.9 333.333 46,9 89.98 66.66 3. 789.987 7.7 853.9999 1976.546 457.633 89 20. What is the sum of three tenths, twenty-five hun- dredths, nine thousand, five millionths, forty-five and eighteen ten-thousandths ? Ans. 21. Each of the two sides of a rectangular field is 215.125 rods long ; and each of the ends is 124.75 rods wide. How many rods around the field ? A7is. 679.75 rods. 22. Add 75 ten-millionth s, 42 ten-thousandths, 37 hun- dredths, .427, 57.3004, and .044. Ans. 58.1456075. 23. What is the sum of 47 ten-thousandths, 15 hun- dredths, 8 tenths, 492 millionths, 130 ten-thousandths, and 5678 ten-millionths? Ans. .9687598. 24. What is the sum of 41.371 -f- 2.29 + 73.402 + 1.729? 25. What is the sum of 86.005 + 4.00003 + 2.000098 ? SECTION III mmmTmMmTn&m % 127. Example. — From 246.7235 take 27.8726. SOLUTION. 246.7235 27.8726 218.8509 liundredtlis ; Explanation. — We write tlie subtrahend so that its units stand under the units of the same order in the minuend. Then subtract exactly as in whole numbers. 6 ten-thousandths from 15 ten-thousandths — 9 ten- thousandths ; 3 thousandths from 3 thousandths = thousandths ; 7 hundredths from 12 hundredths = 5 9 tenths from 17 tenths = 8 tenths ; 8 ones from 16 ones = 8 ones ; 3 tens from 4 tens = 1 ten ; hundreds from 2 hundreds = 3 hundreds. Then we place the decimal point between the ones and tenths, or directly under the decimal point in the subtrahend. 128. EuLE. — Write ike suhtraliend so that its units shall stand under the units of the same order in the minuend. Subtract as in integers ; and ivrite the decimal point in the remainder directly under the decimal point in the sub- trahend. PRO BTj EMS. (1) 2.7536 2.4578 .2958 From (2) .3748 . 269 7 .1051 (3) 247.36 25.9734 (4) 34.4897 21.93 seventy-five thousandths, 90 221.3866 12.5597 seventy and five tenths take seven hundred Am. 69.725. SUBTRACTION. 91 6. From eighteen and forty-two ten-thousandths take eighteen hundred forty-two ten-thousandths. A71S. 17.8Ji. 7. From $.875 take $.3125. A7is. $.5625. 8. From five thousand seventeen millionths take .0000097. Ans. .0044073. 9. From Pittsburgh to Piiiladelphia by the P. R. R. the distance is 352.2 miles; from Pittsburgh to Altoona it is 116.7 miles. What is the distance from Altoona to Phila- delphia? Ans. 235.5 miles. 10. By the P., W. & B. R. R. the distance from Pittsburgh to Washington is 327 miles; from Pittsburgh to Cumber- land, 149.5 miles. What is the distance from Cumberland to Washington? Ans. 177.5 miles. 11. The average length of the school term in Pennsyl- vania in 1863 was 5.433 months, in 1869 it was 0.04 months. What was the increase ? Ans. 12. In 1863, the average cost per month for each pupil in the common schools of Pennsylvania was $.49 ; in 1869, it was $.97. What was the increase ? A7is. $.48. 13. At Fort Humboldt, California, the rain-fall in Winter is 15.03 inches; in Summer, 1.18 inches. How much greater is it in Winter than in Summer ? Ans. 13.85 inches. 14. Chicago is .0293317 days west of Washington, D. C, and the Allegheny Observatory is .0082252 days west of the same place. How far is Chicago w^est of the Allegheny Observatory ? A7is. .0211065 days. 15. One hectare of land equals 2.471 acres. How much land have I left after selling 1 hectare from 5.5 acres? Ans. 3.029 acres. 16. Subtract four hundred twenty-five ten -thousandths from four hundred and twenty-five ten-thousandths. Ans. 399.96. SECTION IV. ^V=: MUIL,TIi^IL,I(€JATIOjM ^:^ 139. Multiplication of Decimals is tlie same as Multipli- cation of Integers with the exception of locating the Decimal Point. Example. — Multiply 13.25 by 9.27. Explanation.— 13.25 multiplied by 7 =r 92.75 " A " (89). But as our multi- plier is .07, we have to divide this product by 100, which is done by removing the point two places to the left — " B." 13.25 multiplied by 2 = 20.50 "A." But as our multiplier is .2, we have to divide this product by 10, which is done by removing the point one place to the left-"B." 13.25 multiplied by 9 = 119.25 "A," " B." Notice First.— That with each stu;ceeding figure of the multiplier there is 1 figure less to cut off. So if each partial jwoduct is carried one place further to the left than the preceding it will bring the decimal point in it exactly under the one above, and consequently like orders will stand in the same columns in the partial products. Notice Second.— 7^ ''A" and " B" the 9 ones of the 3d' partial product is obtained by multiplying the ones of the multiplicand by the ones of the multiplier ; and the decimal point is placed at the right of ones. Hence, A B 13.25 13.25 927 9.27 92.75 .9275 2 6.50 2.650 119.2 5 119.25 122.8275 130. Tlie product of ones multiplied ly ones is ones. 92 M U L T I P L 1 C A T I O i^ 93 131. The number of decimal figures in a product must equal the number of decimal figures in the factors. -Multiply 3.04 by 3.2. Explanation. — The product of the 3 ones of the mul- tiplicand and the 3 ones of the multiplier is the 9 of the second partial product. Therefore, that figure is ones, and the 9 of the required product is ones also. Notice, that there are three decimals in the factors and Q « OQ three in the product (131). 133. Rule. — Write the multiplier under the multipli- cand. Multiply as in whole numbers, and point off from the right of the product as many places for decimals as there are decimal places in the factors. If there are not enough places in the product, make a sufficient number by prefixing ciphers. FROBT.EM8. (1) (3) (3) (4) (5) .075 .084 6.375 63.75 637.5 .07 .3 .8 .08 5.1000 .008 .00525 .0252 5.1000 5.1000 Multiply Ans. Multiply 6. .05 by 500. 25. 11. 7.94 by 4.5. 7. .35 by 225. 78.75. 12. 12.65 by .24. 8. .144 by 12. 1.728. 13. 48.125 bj ^ 1.08. 9. .0003 by 500. .15. 14. .625 by 64. 10. .75 by 1000. 750. 15. 61.76 by .0071. 16. Multiply 3.7267 by 10; by 100; by 1000. 94 DECIMALS. SOLUTION. Explanation. — Since moving 3 7267 X 10 37 2G7 ^ figure one place to the left 3*7267 X 100 - S72 67 i^^^^^^^s its value ten-fold, and d./.50/ X lUU_^/x5.0/ gij^p3 tl^g relative position of 3.7207 X 1000 = 3726.7 the figure and decimal point is the same whether we move the figure one place to the left or the decimal point one place to the right, we multiply 3.7267 by 10 by moving the decimal point one place to the right. By the same process of reasoning we multiply by 100 by removing the decimal point two places to the right, etc., etc. 17. What cost 17.35 tons of hay at $17,375 per ton ? 18. What cost 13.75 dozen eggs at 1.1875 per dozen ? 19. What cost 18.75 pounds of butter at $.1875 per pound ? 20. Multiply three hundred and three thousand one hun- dred forty-one ten-thousandths by twenty-four and three thousand five hundred sixty-one hundred-thousandths. Ans. 7218.232585101. 21. There are 16.5 feet in 1 rod. How many feet in 9.8 rods? Ans. 161.7 ft. 22. There are 5.5 yards in 1 rod. How many yards in 80 rods? Ans. 440 yards. 23. Bought 143.5 acres of land at $75.25 per acre. What did the whole cost? Ans. $10798.375. 24. AVhat is the product of 1.25 multiplied by 800 ? 25. What cost 4000.7 bushels com at $.75 a bushel? (26.) (27.) (28.) (29.) 91.24 40.083 .125 1.25 1.15 1.3 .125 52.1 30. (4.5 X 1.8) + (12.5 X .8) = ? Ans. 18.1. 31. (9.2 - .52) X (123 — 114.32) = ? Ans. 75.3424. 32. (683.21-1-316.79 - 999.83) x .17 = ? Ans. .0289. SECTION V. ■^ ©l^ISI^M I h^ ^ ^r 3 cents are contained in 6 cents 2 times. 3 ones are contained in 6 ones 2 times. 3 tenths are contained in (5 tenths 2 times. 3 hundredths are contained in 6 hundredths 2 times. 3 thousandths are contained in 6 thousandths 2 times. Hence, 133. When the divisor and dividend are like nnmhers, or are of the same order of units, the quotient is ones. Example.— Divide 37.18 by 1.43. SOLUTION. 1.43 ) 37.18 ( 26 28 6 8 58 8 58 Explanation. — The right-hand figures of the divisor and dividend are of the same order of units (hundredths) ; hence the quotient, 26. is an integer. P BOB LEMS. 1. Divide 5.85 by .65. 2. Divide 292.5 by 22.5. 3. Divide 59.163 by .481. 4. Divide 4.732 by .004. 5. Divide 59.994 by 3.333. ?. 9. Ans. 13. Ans, 123. Ans. 1183. Ans. 18. 134. The dividend must contain at least as many deci' mal figures as the divisor. DECIMALS, Example.— Divide 30 by 1.25. SOLUTION. 1.25 ) 30.00 ( 24 25 5 00 5 00 Explanation. — The right-hand figure of the divisor is hundredths. We annex two decimal ciphers to the dividend, thus making its right-hand fig- ure hundredths, and obtain ones for the quotient (134). PROBLEMS. 1. 279 -^ 4.5 = ? 2. 444 -^ 9.25 = ? Ans. 62. Ans. 48. 3. 21 -j- .375 = ? Ans. 56. 4. 91.2 -^ .95 = ? Ans. 96. 5. 5.1 -T- .075 = ? Ans, 68. One-half of 6 cents is 3 cents: 6 cents -^2 = ^ cents. One-half of 6 ones is 3 ones : 6 ones -j- 3 = 2 ones. One-half of 6 tenths is 3 tenths : 6 tenths -^3 = 2 tenths. One-half of 6 hundredtlis is 3 hundredths : .06 -4- 3 = .02. One-half of 6 thousandths is 3 thousandths. 3 ).006 .002 135. When the divisor is an integer, the quotient will be of the same orders of units as the dividend. Example.— Divide .4375 by 35. solution. 35 ) .4375 ( .0125 35 87 70_ 175 175 Explanation.— The divisor 35 is an integer. And since the dividend is tenths, hundredths, thousandths, and ten-thousandths, the quotient consists of the same orders of units (135). PBOBLEMS. 1. Divide .096 by 16. Ans. .006. 2. Divide .0162 by 18. Ans. .0009. Division. 97 3. Divide .0425 by 25. Ans. .0017. 4. Divide .00144 by 12. Ans. .00012. 5. Divide 1728.12 by 12. Ans, 144.01. 136. The quotient contains as many decimal figures as the tiumber of those in the dividend exceeds those in the divisor. Example.— Divide 4.6875 by 3.125. soLtrriON. Explanation. — ^The right-hand fig- 3.125 ) 4.68715 ( 1.5 ^^e of the divisor is thousandths, and -1 25 til© corresponding order of the divi- dend is 7 (thousandths). Hence, the 1 5625 quotient of 4.687 divided by 3.125 is 15625 an Luteger ( 1 33). Placing the decimal point after the quotient figure, 1, and completing the division, the quotient obtained is 1.5, which contains one decimal place (136). The dotted vertical line in the solution shows what figures of the dividend are used to obtain the integral part of the quotient. When decimal ciphers are annexed to form partial dividends, they are counted as decimal figures of the dividend. PBOBLEHrS. 1. Divide 30.688 by 22.4. Ans. 1.37. 2. Divide 5.7531 by 45.3. Ans. .127. 3. Divide 92.92261 by 9.119. Ans. 10.19. 4. Divide .0060858 by .483. Ans. .0126. 5. Divide 7.6875 by .041. Ans. 187.5. 6. Divide 64.86316 by 561.1. Ans. .1156. * 7. Divide .0000798 by .0042. Ans. .019. 8. Divide .6336 by .132. Ans. 4.8. 9. Divide 20.04 by .04. Ans. 501. 10. Divide 522,1054 by 42.31. Ans. 12.34 7 98 DECIMALS. 137. The right-hand figure of any remainder after divi- sion is always of the same order of units as the last figure of the dividend used to oMain it. Example. — Divide 15.5 by 1.3. SOLUTION. Explanation.— In the solution of the 1.3 ) 15.5 ( 11 following problems it is necessary that 23 the quotient shall consisi of an integer only. ^ ^ Since the 25 in the accompanying solu- 1 3 tion 19 tenths, the 12 is also tenths, and -1 o T> ' :, the true remainder is 1.2. !./« Remainder. 1. Into how many building lots, each containing .48 of an acre, can 8 acres of land be divided ? Ans. 16 lots, and .32 of an acre left. 2. How many ice-pitchers, each weighing 2.13 pounds, can a manufacturer make from 15 pounds of silver ? Ans, 7 pitchers, with .09 of a pound left. 138. EuLE. — Divide as in whole numbers, and point off from the right of the quotient as matiy places for decimals as the numher of decimal places in the dividend exceed the number in the divisor. If there are not so many places in the quotient, supply the deficiency hy prefixing ciphers. If in the course of division ciphers have been annexed to the dividend, regard each cipher annexed as a decimal place. PMOBZJBMS. 1. Divide .7 by 10. Ans. .07. 2. Divide .875' by 100. Ans. .00875. DIVISION. 3. Divide .56 'by .007. Ans. 80. 4. Divide 172.8 by .012. Ans. 14400. 5. Divide 11.7 by .0009. Ans. 13000. 6. Divide 14.4 by .00006. Ans. 240000. 7. Divide 5.85 by .65. Ans, 9 8. Divide 45.3 by .15. Ans. 302, 9. Divide 100 by .01. Ans. 10000. 10. Divide 72 by .18. Ans. 400. 11. Divide .72 by 18. A71S. .04. 12. Divide 96 by .016. Ans. 6000. 13. Divide .096 by 16. Ans. .006. 14. Divide .2 by 1.4. Ans. .142857 + . 99 Note. — The sign + after the answer to Problem 14 mdicates that the division is not exact. 15. Divide 63.75 by 100 ; by 1000 ; by 10000. SOLUTION. 100 =.6375 63.75-^ 1000 = .06375 63.75 63.75 10000 = .006375 Explanation. — Moving a fig- ure one place to the right or moving the decimal point one place to the left, divides by 10 ; two places divides by 100 ; &c., &c. Moving the decimal point of 63.75 two places to the left is the same as moving each figure two places to the right, and consequently divides each by 100, giving .6375. Moving the point three places to the left, divides by 1000, &c., &c. 16. Divide 5386.5 by 1000 17. Divide 8382.55 by 10000 18. Divide 8.75 by 100000 19. Divide 10.075 by 1000000 20. A merchant buys 10000 yards of calico for $645. What does it cost him per yard ? 21. Divide 5 tenths by 1 trilhon. Ans. .0000000000005. 100 DECIMALS. 22. Divide 56.285 by 6.3. Ans. 8.9341 + . 23. Divide 108.029 by 7.2. Ans. 15.004 + . 24. Divide 7500 by one thousand eight hundred seventy- five millionths. Ans. 4000000. 25. Divide .75 by 1875. Ans. .0004. 26. If 2.75 pounds of butter cost $1.03125, what is the price per pound ? Ans. 1.375. 27. If $8264.50 pay 137.19025 tax, what is the tax on one dollar? Ans. $.0045. 28. Gave $23.4375 for 75 pounds of coffee. What was the price per pound ? Ans. $.3126. 29. The mail train runs from Philadelphia to Pittsburgh, a distance of 355.2 miles, in 17.3 hours. What is the rate per hour? Ans. 20.53+ miles. 30. The " limited mail " runs the same distance in 10.3 hours; what is the rate per hour ? Ans. 34.48+ miles. 31. The fast mail runs from Pittsburgh to Chicago, a distance of 469 miles, in 13.084 hours. What is the rate per hour? Ans. 35.845+ miles. 32. If 2,3 yards of cloth make 1 coat, how many coats will 150 yards make ? Ans. 65 coats, .5 yards left. 33. How much muslin at $.165 a yard can be bought for $732.60? Ans. 4440 yards. 34. At $18.75 each, how many washstands can be bought for $506.25 ? Ans. 27 washstands. 35. A man walks 65.625 miles in 17.5 hours. How many miles, on an average, does he walk each hour ? Ans. 3.75 miles. 'Y^ SECTION VI UiNlliTE^Dj SJrAfTiES; MiOjNlEYr *-t5X5^(>'eM •"■ 139. United States Money, sometimes called Federal Money, consists of dollars, dimes, cents, and mills. 140. TABLE. 10 mills (m) = 1 cent . . . c. 1000 mills. 10 cents = 1 dime. . . d. %1 = 100 cents. 10 dimes = 1 dollar. . % 10 dimes. 141. The Unit of U. S. Money is the Dollar, Since the dollar is the unit of U. S. money, dimes, cents, and mills are respectively tenths, hundredths, and thmtsandth^ of the unit. 142. Dollars should be written as integers, with the sign (I) prefixed ; and dimes, cents, and mills as decimals. Thus, 9 dollars 2 dimes 5 cents 6 mills, are written $9,256. The denomination dime is not regarded in business operations, dimes being tens of cents. Thus, $9,256 is read 9 dollars S5 cents 6 mills, instead of 9 dollars ^ dimes 5 cents and 6 mills. 143. Since the two places, tenths and hundredths, are appropriated to cents, when the number of cents is less than 10 write a cipher in the place of tenths. Thus, 5 cents is written $.05. In business transactions, if the mills in the final result are 5 or more than 5, they are considered a cent ; if less than 5, they are not regarded. Thus, $3,186 would be called 13.19 ; $9,424, $9 42 ; $.375, $.38 ; &c. 101 102 DECIMALS. EXERCISES. 1. Read 14.12, $.18, $3.05, $10.10, $5,153. 2. Read $.04, $24,125, $125,003, $99,995. 3. Read $500.50, $13,033, $.007, $199,053. 4. Read $1,005, $100,009, $4,875, $164,664. 5. Read $1000.40, $89,091, $75.75, $77,777. 6. Write 15 cents ; 5 cents ; 9 mills. 7. Write 13 dollars 9 cents ; 5 dollars 80 cents. 8. Write 100 dollars 40 cents 3 mills. 9. Write 30 cents 6 mills ; 8 cents 7 mills. " 10. Write 5 dollars 50 cents ; 1000 dollars 7 cents. 144. Decimal parts of a dollar less than mills are expressed as decimals of a mill. $ .0003 is 2 tenths of a mill. $ .00025 is 25 hundredths of a mill. $ .000125 is 125 thousandths of a mill. $ .0052 is 5 and 2 tenth mills. $ .3552 is 35 cents 5 and 2 tenth mills. $4.1007 is 4 dollars 10 cents and 7 tenth mills. EXEItCI S E S , 11. Read $.054, $111.0006, $.49075, $55.5555. 12. Read $99.43753, $100.00001, $.7283, $99.0888. 13. Read $.40404, $.25025, $9.0999, $10.7558. 14. Write 45 cents 7 and 13 hundredth mills. 15. Write 3 dollars and 3 tenth mills. 16. Write 95 dollars 37 cents and 5 tenth mills. 17. Write 100 dollars 10 cents 4 and 9 tenth mills. 18. Write 11 cents 3 and 125 thousandth mills. 19. Write 5 cents and 3125 ten-thousandth miUs. 20. Write 44 dollars 44 cents 4 and 4 tenth mills. UKITED STATES MONEY 103 [^!i ^I>jI>)IT'IOjM '<^J^f^ 145. Since the dollar is the unit of TJ. S. Money (141), and cents, mills, and parts of a mill are decimals of a dollar, United States Money is added in the same manner as other decimals. Examples.— What is the sum of $100.50, $4,075, $.43125, and $900.90 ? SOLUTION. $100.50 4.075 .43125 900.90 $1005.90625 (1) $481.26 32.75 1.25 421.16 Explanation. — We write the numbers so that units of the same name stand in the same column ; as, cents in the column of cents, mills in the column of mills, etc. We then add the parts and place the decimal point in the sum, as in Addition of Decimals, not forgetting to prefix the sign ($ ) to the result. PR OBLEM, $936.42 (3) $593,715 26.43 5.421 18.62 $644,186 (3) $4.22 5.031 8.2004 9.136 $26.5874 (4) $.41 .092 .03 .468 $1. 5. Mr. B. pays $91.75 State taxes; $321.16 County taxes; $421.13 School taxes; $375.08 Borough taxes; and $39.17 Poor tax. What is the total amount of his taxes ? Ans. $1248.29. 104 DECIMALS. 6. A mercliant deposits in bank on Jan. 1, $300 ; Jan. 3, $1716.50; Jan. 4, $5217.45; Jan. 9, $16714.25; Jan. 10, $516.08; and Jan. 11, 17060.37. How much in all has he deposited? Ans. $31524.65. 7. A owes B the following sums: $3.75, $8.25, $9.62, $20.31, $17.42, $16.73, $24.40, $120.75, $16.35, $115.41, $25.64, $71.22, $41.78, $168.37, $415.93. How much does A owe B? Ans. $1075.93. 8. From the Savings' Bank, C drew out the following sums: $210, $35, $48, $16, $25, $45, $24, $16, $13, $75, $84, $7, and $16.13. How much did he draw out altogether? Ans. $614.13. 9. A merchant "checked out" of his bank $215, $3125.75, $21620, $17540, $2063.74, $216.14, $25750.75, $301.06, $51.22, and $1617.01. How much in all did he check out ? Ans. $72500.67. 10. A man after "checking out' $947, found that he had remaining in bank $369.89. How much had he at first? Ans. $1316.89. 11. After "checking out" $30.43, $49.34, $9, $25, a grocer found that he had $711.84 remaining in bank. How much had he at first ? Ans. $825.61. 12. A merchant bought of a jobber, prints amounting to $3520.50; mush ns, $3762.75; linens, $1733; silks, $2375.16; woolen goods, $7675.87. How much is the bill ? Ans. $19067.28. 13. A man bought a cow for 30 dollars, a horse for 104 dollars 60 cents, and a wagon for 85 dollars 40 cents. How much did all cost him ? Ans. $220. 14. Each pupil will prepare an original problem for class exercise. UNITED STATES MONEY. 105 mmmTRMmwiBm ii 146, United States Money is subtracted in the same manner as other decimals (145). Example.— From $98.45 take $49,375. SOLUTION. $98.45 49.375 $49,075 (1) $.36 .19 Explanation. — We write dollars under dollars, cents under cents, and mills under mills. We then subtract, and place the decimal point in tLie remainder as in Subtraction of Decimals, not for- getting to prefix the sign ( $ ) to the result. PROBLEMS. (2) $47.21 8.564 (3) $275,386 85.722 (4) $40,009 21.4304 $.17 $38,646 $189,664 $18.5786 5. A merchant sold on Saturday, goods amounting to $2565.75 ; on Monday, goods amounting to $1075.23. What was the dijfference in the two days' sale ? Ans. $1490.52. 6. If a man's property is valued at $15725.50, and his debts at $6837.37, what is he worth ? Ans. $8888.13. 7. If a man's property is valued at $20569, and his debts at $30880, what excess of debt has he ? Ans. $10311. 8. I deposited in bank $1840, and drew out $475.50. How much had I left ? Ans. $1364.50. 9. A man has property worth $10104, and owes debts to the amount of $7426.25. When he pays his debts, how much will be left ? Ans. $2677.75. 106 DECIMALS. 10. A man having $100000, gave away $1473.72. How much had he left ? Ans. $98526.28. 11. A merchant owing $35542.75, paid $22560.25. How much does he still owe? Ans. $12982.50. 12. A man having $8795.15, gave $2309.95 of it for a store. How much money had he remaining ? Ans. $6485.20. 13. A man bought a span of horses for $537, and a yoke of oxen for $297.50. How much more did he pay for the horses than for the oxen ? Ans. $239.50. 14. A man gave me his note for $240, and he has since paid all but $40.50 of it. How much has he paid on the note? Ans. $199.50. 15. A bank-clerk earned $150 a month, and his expenses were $67.43. How much did he save? Ans. $82.57. 16. I paid $25000 for a piece of property and afterward sold it for $19400.75. What was my loss ? Ans. $5699,25. 17. From $11.05 take $9,258. Ans. $1,792. 18. From $.001 take $.00095. Ans. $.00005. 19. From $10 take $5.75. Ans. $4.25. 20. From $5 take $1.85. Ans. $3.15. 21. I pay $5268 for property which I sell for $4319.67. What is my loss ? Ans. $948.33. 22. From 9 dollars subtract 8 dollars 99 cents 9 mills. Ans. $.001. 23. Subtract 99 dollars 99 cents 9 mills from $100,875. Ans. $.876. 24. From 1 dollar subtract $.0555. A^is. $.9445. 25. Each pupil will prepare an original problem for class exercise. UNITED STATES MON^EY 107 ^-5)«(5- m mmi/F'i^iji(^;MT'iom ^ 147. United States Money is multiplied in the same manner as other decimals (145). Example.— Multiply $13.75 by 4.5. SOLUTION. $13.75 4.5 6875 5500 $61,875 Explanation. — We write the multiplier under the multiplicand. We then multiply, and place the decimal point in the product, as in Multiplication of Decimals, not forgetting to prefix the sign ($) to the result. 1. Multiply $137.57 by 7. 2. Multiply $37,354 by 6. 3. Multiply $12,225 by 8. 4. Multiply $.347 by 9. 5. Multiply $2575 by .04. 6. Multiply $33.75 by .008. 7. Multiply $.40 by 500. 8. Multiply $57.27 by 504. 9. Multiply $.875 by 56. 10. Multiply $100 by .00001. 11. If a boy goes to the opera at evening, how much will it cost him for Ans. $962.99. Ans. $224,124. Ans, $97.80. A71S. $3,123. Ans. $103. Ans. $.27. Ans. $200. Ans, $28864.08. Ans. $49. Ans. $.001. a cost of $1.25 an 20 evenings ? Ans. $25. 108 DECIMALS. 12. A gentleman's expenses for one year were $3450.75. At the same rate what would they be for 25 years ? Ans. $86268.75. 13. What cost 450 tons of iron at $45.75 per ton? Ans. $20587.50. 14. What cost 37 tons of steel at $63.50 per ton ? Ans. $2349.50. 15. What cost 372 yards of muslin at 11 cents per yard ? Ans. $40.92. 16. What cost 568 yards of calico at 6 cents per yard ? Ans. $34.08. 17. A street-car conductor receives $1.75 per day. How much does he make in one year of 365 days ? Ans. $638.75. 18. An institution spends in advertising $565.34 per year. At this rate how much would it spend in 15 years ? Ans. $8480.10. 19. If a boy places in a Savings' Bank $1.15 per week, how much will he deposit in 1 year or 52 weeks ? Ans. $59.80. 20. A farmer sold 493 pounds of wool at $.875 per pound. How much did he receive for it ? Ans. $431.38. 21. A tinner worked 5.5 days at $1,375. per day. How much did he earn ? Ans. $7.56. 22. What cost 47 bushels of potatoes at $.655 per bushel ? Ans. $30.79. 23. What cost 35 yards of carpeting at $2,375 per yard ? Ans. $83.13. 24. What cost 77 bushels apples at $.875 per bushel ? At $.625 per bushel? Ans. 1st., $67.38; 2d., $48.13. 25. Each pupil will prepare an original problem for class exercise. UNITED STATES MONEY. 109 Jj^ MFWIMJi&jm 148. United States Money is divided in the same manner as other decimals (145). Example.— Divide $61,875 by 4.5. SOLUTION. 4.5 ) $61,875 ( $13.75 45^_ 16 8 13 5 3 37 315 225 225 Explanation.— We write the di- visor at the left of the dividend. We then divide, and place the decimal point in the quotient, as in Division of Decimals, not forgetting to prefix the sign ($) to the result, when the divisor is an abstract num- ber (102). PBOBIjEMS, 1. Divide $962.99 by 7. Ans, 1137.57. 2. Divide $224,124 by 6. Ans. $37,354. 3. Divide $97.80 by 8. Ans. $12,225. 4. Divide $3,123 by 9. Ans. $.347. 5. Divide $103 by .04. Ans. $2575. 6. Divide $.27 by .008. A^is. $33.75. 7. Divide $200 by 500. Ans. $.40. 8. Divide $28864.08 by 504. Ans. $57.27. 9. Divide $49 by 56. Ans. $.875. 10. Divide $.001 by .00001. Ans. $100. 110 DECIMALS. 11. Divide 13.123 by $.347. Ans. 9. 12. Divide $103 by $2575. Ans. .04. 13. Divide $.27 by $33.75. Ans, .008. 14. Divide $200 by $.40. A71S. 500. 15. Divide $28864.08 by $57.27. Ans. 504. 16. Divide $49 by $.875. Ans. 56. 17. Divide $.001 by $100. Ans. .00001. 18. If shovels are worth $1.15 apiece, how many can be had for $41.40? Atis. 36 shovels. 19. At $.875 per bushel, how many bushels of corn can be bought for $83,125 ? Ans. 95 bushels. 20. A merchant sold 2000 barrels of flour for $14750. How much did he get per barrel? Ans. $7,375. 21. A puddler's income one year of 313 working days was $2582.25. What was the average per day? Ans. $8.25. 22. A family of seven persons spent in one year of 365 days $1642.50. What was the average per day ? Ans. $4.50. 23. Bought 456 bushels of bituminous coal for $50.16. How much was that per bushel ? Ans. $.11. 24. At 6 cents a yard how many yards of calico can be purchased for $34.08 ? Ans. 568 yards. 25. If a street-car conductor earns $1.75 per day, how long will he be in earning $638.75 ? Ans, 365 days. 26. If a boy deposits $1.15 per week in a Savings' Bank, in how many weeks will he have deposited $59.80 ? Ans. 52 weeks. 27. A hvery man received $14208 for boarding 37 horses, 12 months. How much did he charge a month for each horse? Ans. $32. 28. Each pupil will prepare an original problem for class exercise. OUTLINE OF FRACTIONS. 150. FBACTIONAJL UNIT. p2 181. 182. 183. 184. 185. 186. I 187. 151. TERMS. CLASSIFICA- TION. PRINCIPLES. PREPARATORY. ^ j 152. ] 153. Denominator. Numerator. As to Value. As to Form. li As to Fractional Unit. 154. Proper. 155. Improper, 156. Simple. 157. Complex. 5 8. Compound. 159. Similar. 160. Dissimilar. 161. First.— Multiplied. 162. Second.— Divided. 163. Ttiird.—Not changed. « f 166. Z)mswsj 167. Prime. k^ b J or i^actor*. ( 168. Composite. 2t 1 169. Common Divisor or Factor. fe i. 170. Greatest Common Divisor. 1'71. Multi lulti-^ 1 pies. \ 1 72. Commwi Mvltiples. 73. Least Com. Mult. CASES. V 174. Cancellation. f 175. jr. Integers or Mixed Num- bers to Improper Fractions. 176. JTJ. Improper to Integers or Mixed Numbers. 177. III. Integers to Higher or Lotver Terms. 178. IV. To Lowest Terms. 179. V. Dissimilar to Similar. 180. VI. Dissimilar to Least Sim- ilar. ADDITION. SUBTRACTION. MULTIPLICATION. DIVISION. CONVERSE REDUCTION, ALIQUOT PARTS. BILLS. Ill 6 SIXTHS^ 5, seveMTHS^ 8 EIGHTHS^ iiiiiiilgj^^ iiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiniiiiit!! ' n!!!!," ""'"if CHAPTER III. SECTION I. (2i (5? -^®^ FB'ACT'10)MS 149. A Fraction is a number expressing one or more of the equal parts of a unit. 150. A Fractional Unit is one of the equal parts into which a unit is divided. 112 lifUMERATIOK AlfD NOTATIOlf. 113 The number of these parts indicates their name. Thus, One divided into two equal parts = 2 halves. One half is written |, One divided into three equal parts = 3 thirds. Two thirds are One divided mio four equal parts = A fourths. Three fourths" One divided into fioe equal parts = 5 fifths. Four fifths " One divided into six equal parts = 6 sixths. Five sixths " One divided into seven equal parts = 7 sevenths. Six sevenths " One divided into eight equal parts = 8 eighths. Seven eighths One divided into nin£, equal parts = 9 ninths. Eight ninths " One divided into ten equal parts = 10 tenths. Nine tenths " ^^^or .9. g When we express nine tenths thus, Jq, it is called a Common Frac- tion ; but when thus, .9, it is called a Decimal Fraction. Hence, With reference to the mode of expressing them. Fractions are of two kinds, namely : Common and Decimal. (See page 81.) A fraction is expressed by two numbers, one written under the other, with a horizontal line between them. 151. The Terms of a fraction are the two numbers used to express it, and are called the Denominator and the Numerator. 152. The Denominator is that term of the fraction which indicates the number of equal parts into which the unit is divided, and is written below the horizontal hne. 153. The Numerator is that term which indicates the number of equal parts taken, and is written above the horizontal line. Thus, in the fraction f , the 3 is the denomi- nator, and signifies that a unit is divided into three equal parts ; the 2 is the numerator, and signifies that two of these parts (thirds) are taken, or expressed. Fractions are read by first pronouncing the number in the numera- tor, and then the parts of the unit expressed by the denominator. Thus ^ is one half ; |, two thirds; ^, four twenty-firsts; ^, seven thirty-seconds. 8 114 FKACTIOKS. Read 2* 2f sy a> sf T? T> "!> i> Tf ity TT? A^ yf > 4o> Write 1. Four ninths. 2. Six twenty-firsts. 3. Five eighteenths. 4. Seven thirty-seconds. 5. Eight twelfths. 6. Nine twenty-fifths. 7. Two nineteenths. 8. Eight elevenths. 9. Eleven fifteenths. JEx^ncisjss. h h h h h h -V, m A, A, A H, «, «, I 10. Nine one-hundred-thirds. 11. Two three-hundred-fortieths. 12. Five two-hundredths. 13. One four-hundred-thirteenth. 14. Three three-hundred-firsts. 15. Seventy eightieths. 16. Nineteen ninetieths. 17. Four ten-thousandths. 18. Two hundred-thousandths. In reference to their value, fractions are of two kinds, namely; Proper and Improper. 154. A Proper Fraction is one whose numerator is less than its denominator ; as, ^, f . Such a fraction is called proper because its value is less than a unit, and is, therefore, properly expressed in & fractional form. 155. An Improper Fraction is one whose numer- ator equals or exceeds its denominator; as, f, ^. Such a fraction is called improper because its value equals or exceeds a unit, and, therefore, the proper expression for its value would be an integer or a mixed number. EXEJtClSES, Of the following fractions, name the class to which each belongs. T> f ? f > T> "^? T» TO? "TT"? ^^9 TU9 A» lT» "V* TtS* Tt> 4T> T*> M> TSTir* KUMERATIOK AND l^OTATION. 115 A Mixed Number is one expressed by an integer and a fraction ; as, In reading a mixed number, and belongs between the integer and fraction. Thus, 4| is read 4 and ^. In reference to their form, fractions are of three kinds, namely. Simple, Complex, and Compound. 156. A Simple Fraction is one whose terms are whole numbers ; as, |, f. 157. A Conipleoc Fraction is one which has a frac- tion in one or both of its terms ; as, |, -^. 158. A Compound Fraction is a fraction of a fraction ; as, \ of ^, which may also be written thus, | x J. In reference to their fractional unit, fractions are of two kinds, namely : Similar and Dissimilar. 159. Similar Fractions are those that have like fractional units; as, |, |^; -^^j, /j, v'V- 160. Dissimilaf* Fractions are those that have unlike fractional units ; as, f , J ; J, |, J. EXERCISES. Write in separate groups the similar fractions found in the following: T> 6' ^^ Tfy h T? iSf TT> T7' "8? T? IT' T> *, A, h A. I> f. I. «. I. «. i ?. 4. a, I, f. 4. A. A. ¥, V. tt> ¥. «. ^, ¥• If we multiply the numerator of f by 2, we obtain |. The frac- tional unit in | and | is the same, |. | has 3 ^imes as many fractional units as f . | ^ 2 _ 4 Again, If we divide the denominator of f by 2, we obtain f . The number of fractional units in | and f is the same, 2. But the value of the 116 FKACTIOi^S. fractional unit :^ is 3 times the value of the fractional unit \. f ^. 2 = f . Hence, 161. A fraction is multiplied hy multiplying its nu- merator or dividing its denominator, iy a number greater than unity. If we divide the numerator of f by 2, we obtain \. The fractional unit in f and \ is the same, \. \ has only one half as many fractional units as |. | -*- » = ^. Again, If we multiply the denominator of | by 2, we obtain |. The number of fractional units in f and | is the same. But the valae of the frac- tional unit \ is only one half the value of the fractional unit \. f X 2 = |. Hence, 162. A fraction is divided hy dividing its numerator or multiplying its denominator, ly a number greater than unity. If we multiply both terms of \ by 2, we obtain |. The number of fractional units in | is twice as many as in J, but the xoHue of each fractional unit is only one half as much. That is, the increase in the number of parts taken equals the decrease in the size of the parts. ixi=|. Again, If we divide both terms of | by 2 we obtain \. The number of frac- tional units in \ is only one half as many as in f , but the value of the fractit'-nal units is twice as much. That is, the decrease in the number of parts taken equals the increase in the size of the parts. 1 1: | = ^. Hence, 163. The value of a fraction is not changed hy either multiplying or dividing hoth terms hy the same number. Fractions primarily arise from performing the operation of division, when the division is not exact (99, 1). That of which the fraction expresses a part is called the Unit of the Fraction ; as in the expression $|, one dollar is the Unit of the Fraction. The Unit of the Fraction is not always a single thing ; it may be a collection of things taken as a whole ; as in the expression | of fifty men, fifty men is the unit of the fraction. SECTION II. I mmmmm'Enmm I -^--'**^®(^^'-->- 164. deduction of Fi'actions is changing them into other equivalent expressions. It also includes the changing of whole and mixed numbers to the form of a fraction. Before proceeding with Reduction of Fractions it is necessary to understand something of Measures and Multiples. 165. A Divisor^ or 3feasurej of a number is a number which will divide that number without a remain- der. Thus, 3 is a divisor, or measure, of 6. " Measure" is here used in the limited sense of exact divisor (99, 1). Any number may be used as the measure of any other number of the same kind (191, 259). 166. The Divisors or Factors of a number are the numbers which multiplied together will produce it. Thus, 2 and 3 are factors of 6 ; 2, 6, and 5, of 60. An Even Number is one that is exactly divisible by 2. The left- hand-page numbers of this book are even. An Odd Number is one that is not exactly divisible by 2. The right- hand-page numbers of this book are odd. A Prime Number is one that can not be separated into integral fac- tors ; as 2, 3. 5, 7, 11. 18. The number 1 is not regarded as a factor. 167. The I^Htne Factors or Frime Divisors of a number are the prime numbers which multiplied together 117 118 FKACTIONS. will produce it. Thus, 2, 6, and 5, or 2 and 30, or 5 and 12 are factors of 60 ; but 2, 2, 3, and 5 are the prime factors of 60, because they are the prime numbers whose product is 60. When the numbers are small, we can find the prime factors by in- spection ; but when they are large, we have generally to obtain them by trial. In making trial a knowledge of the following facta may be of ser- vice : First. — Any number ending in 0, 2, 4, 6, or 8 has 2 for an exact divisor. Second. — Any number the sum of ivhose digits is divisi- ble by 3, has 3 for a divisor. Third. — Any number ending in 5, or 0, has 5 for an exact divisor. Example. — Find the prime factors of 2310. SOLUTION. Explanation. — From the 1st Fact, we n N nqi rv see that 3 is an exact divisor of 2310. '- From the 2d Fact, we see that 3 is an 3 ) 1155 exact divisor of the quotient 1155. 5 \ 335 From the 3d Fact, we see that 5 is an . -, exact divisor of the quotient 385. f None of these Facts aid us in obtaining 11 the next factor ; but since 7 is a prime number and a divisor of 77, we divide by Ans, 2x3x5x7x 11. it and obtain 11, a prime number, for the quotient. The product of the divisors 2, 3, 5, 7, and the last quotient, 11, all of which are prime numbers, is 2310. Hence tli?y are the prime fac- tors of that number (167). KuLE. — Divide by any prime number except 1 that is an exact divisor ; divide the quotient in the same manner ; and so proceed until the quotient is a prime number. The several divisors and the last quotient are the prime factors. EEDUCTIOlf, 119 EXERCISES. Find the prime factors of 1. 6. Ans. 2, 3. 11. 32. Ans. 2, 2, 2, 2, 2. 2. 8. Ans. 2, 2, 2. 12. 34. Ans. 2, 17. 3. 12. Ans. 2, 2, 3. 13. 96. Ans. 2, 2, 2, 2, 2, 3. 4. 16. ^7^5. 2, 2, 2, 2. 14. 132. Ans. 2, 2, 3, 11, 5. 18. ^W5. 2, 3, 3. 15. 144. Ans. 2, 2, 2, 2, 3, 3. 6. 20. ^/i5. 2, 2, 5. 16. 385. Ans. 5, 7, 11. Z 24. ^/i5. 2, 2, 2, 3. 17. 1001. Ans. 7, 11, 13. 8. 27. ^ws. 3, 3, 3. 18. 4862. Ans. 2, 11, 13, 17. 9. 28. Ans. 2, 2, 7. 19. 8008. Ans. 2, 2, 2, 7, 11, 13, 10. 30. Ans, 2, 3, 5. 20. 2310. Ans. 2, 3, 5, 7, 11 A composite number is one tliat can be separated into inte- gral factors ; as 4, 6, 8, 9, 10, 13. 168. The Composite Factors or Divisors of a number are the composite numbers which multiphed together will produce it. Thus, 4 and 15 are the composite factors of 60, because they are the composite numbers whose product is 60. . If a number is an exact divisor of two or more numbers, it is said to be a divisor, or measure, common to them. 169. A Common Divisor or Measure of two or more numbers is an exact divisor of each of them. Thus, 2 is a divisor common to 2, 4, 6, 8, 10, &c., and is called a common divisor. Numbers are prime to each other when 1 is their only exact divisor. 170. The Greatest Common Divisor or Meas- ure of two or more numbers is the greatest number that is an exact divisor of each of them. Thus, 6 is the greatest 120 REACTIONS. number that will exactly divide 12 and 18, hence it is their greatest common divisor. Example. — Find the greatest common divisor of 105 and 175. FIRST METHOD. SOLUTION. Explanation. — Resolving 105 = 3 X 5 X 7 l^*h numbers into their prime ^^^5 = 5 X 5 X "7 factors, we find 5 and 7 to be TT.i c w n^ A tlie only factors common to In both are 5 X 7 = 35, Ans. . *i x. mr • -, . > " both numbers. Their product, 5x7, therefore, is the greatest common divisor. SECOJ^D METHOD. BOLunoN. Explanation. — 105 is the greatest 105) 175 (1 measure of itself. If it also measured -J ^- 175 it would be the greatest divisor of both numbers. 70) 105 (1 But it does not measure 175, because YO 70 remains. That is, 175 = 105 + 70. Now, as 70 is the greatest measure of Ans. 35 ) 70 ( 2 itself, if it also measures 105 it will 70 measure 175. But it does not measure 105, because 35 remains. TMfet is, if 175 = 105 + 70, then 175 = 35 + 70 + 70. Now, as 35 is the greatest measure of itself, if it measures 70 it will measure 105 (which = 35 + 70), and 175 (which = 35 + 70 + 70). Since 35 does measure 70, it will also measure 105 and 175. It must be the greatest common divisor, because it is tbe greatest num- ber contained in 105 and in the difference between 105 and 175. KuLE I. — Find the product of all the prime factors com- mon to the given numlers. Or, II. — Divide the greater number hy the less, and if there is a remainder, divide the preceding divisor ly it. So con- tinue dividing the last divisor hy the last remainder, till nothing remains. The last divisor will le the greatest com- mon divisor of the two numbers. EEDUCTIOK. 121 WJien iliere are more than two numbers, first find the greatest common divisor of tivo of them, then of that com- mon divisor and one of the other numbers, and so on till every number has been used. The last common divisor is the greatest common divisor of all the numbers. EX ERC I S JES, Find the greatest common divisor 1. Of 48 and 72. Ans. 24. 2. Of 36 and 54. Ans. 18. 3. Of 60 and 270. Ans. 30. 4. Of 180 and 300. Ans. 60. 5. Of 32 and 56. Ans. 8. 6. Of 91 and 143. Ans. 13. 7. Of 20, 50, and 70. Ans. 10. 8. Of 30, 42, and 63. Ans. 3. 9. Of 120, 210, and 345. Ans. 15. 10. Of 154, 210, and 287. Ans. 7. 11. Of 330,495, and 660. Ans. 165. 171. A Multiple of a number is any number which is divisible by it. Thus, 15 is a multiple of 5. 172. A Common Multiple of two or more numbers is a number which is divisible by each of them. Thus, 12 is a multiple common to 2, 3, 4, and 6. 173. The Least Common Multiple of two or more numbers is the least number that is divisible by each of them. Thus, 12 is the least common multiple of 2, 3, 4, and 6, because it is the least number that each will exactly divide. Example. — Find the least common multiple of 6, 9, and 12. I22 fEACTIOJS^S. FIRST METHOD. SOLUTION. Explanation. — Resolv- n 2 V 3 ^^^ ^® numbers into their prime factors, we see that ^ ^=^ ^ y^ *^ the answer must contain 12 = 3x3x3 2x2x3, because 12 = 2x2x3x3= 36, Ans. 3x2x3. - The answer must also contain 3x3, because 9 = 3x3. But, as there is already one 3 in 2 X 2 X 3, we need only use another 3 making 2x2x3x3. The answer must also contain 2x3, because 6 = 2x3. But in 2x2x3x3 we already have 2x3. Therefore 2x2x3x3 contains all the prime factors of all the numbers, and none of them more times than is necessary to contair/ every number. Hence 2x2x3x3 = 36 is the least commor multiple of 6, 9, and 12. Since in the above Example 6 is a factor of 12, it might have beei> omitted in the solution, and the same result obtained. Thus, 9 = 3 x 3 12 = 2 X 2 X 3 2 X 2 X 3 X 3 = 36, Ans. SECOJ^D METHOD. SOLUTION. Explanation. — We take the 2 ) 6, 9, 12 factor 2 out of 6 and 12, and then the numbers become 3, 9, and 6. ^ )_A_?i ? Again we take the factor 3 out, 13 2 ^^^ *^^y become 1, 3, and 2. Ov, Qv, Qv, o oa A^c. Now, we cannot take out any 2x0x3x2 = 36, Ans, ., c ^ r ^ o ^ other common factor from 1, 3, and 2. Hence we have rejected all factors which are not necessary to con- tain 6, 9, and 12. Those which are retained, namely, 2, 3, 3, and 2, must be necessary to contain those numbers. Hence, their product, namely, 2x3x3x2 = 36, must be the least number that will contain them. REDUCTION. • 123 Omitting the ' 6" as suggested in the First Method, and we have : 3) 0, 9, 12 3, 4, or 3 X 3 X 4 = 36, Ans. Rule I. — Resolve the numbers into their prime factors, and multiply together all those of the largest number, and such of the others as are not found in the largest number ; the product is the least common multiple. Rule II. — Write the numbers in a line, and divide by any prime number that is contained exactly in two or more of them. Write the quotients and undivided tiumbers in a line below. Divide these in the same manner. Continue thus till no number greater than 1 is exactly contained in any two of the numbers. TJien multi2oly all the divisors and remaining numbers together ; the product is the least common multiple. Wlien numbers are prime to each other, they have no com- mon factor to be rejected, and their least common multiple is their product. EXEItCISES. Find the least common multiple 1. Of 10, 15, and 20. Ans. 60. 2. Of 12, 18, and 24. Ans. 72. 3. Of 14, 35, and 56. Ans. 280. 4. Of 16, 24, 36, and 60. Ans. 720. 5. Of 20, 30, 50, and 75. Ans. ZOO. 6. Of 44, 66, 88, and 110. Ans. 1320. 7. Of 65, 78, 104, and 130. Ans. 1560. 8. Of 48, 80, 120, and 144. Ans. 720. 9. Of 8, 11, and 15. Ans. 1320. 10. Of 120, 180, 200, and 240. Ans. 3600. 11. Of 1, 2, 3, 4, 5, 6, 7, 8, and 9. Ans. 2520. 12. Of 7, 13, 23, and 31. Ans. 64883. 124 • FKACTIONS. 174. Cancellation is the process of ahlreviating oper- ations in division by rejecting equal factors from the divisor and dividend. The Sign of Cancellation is an oblique line drawn across a figure ; as 0, '^$, U$, &c. A factor is cancelled hy dividing both dividend and divisor hy that factor. Example. — Divide 4x5x6 by 2x4x5. SOLUTION. Explanation. — We write the 3 numbers that constitute tlie divi- ^ X ^ X P Q Ayig dend above a line, and those that ^ X 4 X ^ constitute the divisor below it. Observing 4 and 5 to be factors in both dividend and divisor, we cancel them in botli. Then observing that 2 is a factor in the 2 of the divisor and in the 6 of the dividend, we cancel it out of these, leaving in the dividend the factor 3, which is the required quotient. Since dividing a number by itself gives 1 for a quotient, if the factor or factors cancelled equal the number itself, the result will be One, which need not be written, except in the quotient where there are no other factors. EuLE. — Cancel all factors common to both dividend and divisor, and divide the product of the factors remaining in the dividend by the product of those remaining in the divisor. EXERCISES. 1. Divide 8 x 12 x 15 by 4 x 3 x 12. Ans. 1 0. 2. Divide 6 x 7 x 8 by 3 x 7 x 4. Ans. 4. 3. Divide 9x5x11 by 11x9x5. Ans. 1. 4. Divide 42 x 12 x 18 by 18 x 36 x 7. Ans. 2. 5. Divide 144x360 by 6x8x9x20. Ayis. 6. 6. Divide 8 xl2 x27 by 4 x3 x 12. Ans. 18. 7. Divide 20 x 32 x 35 by 4 x 5 x 16. Ans. 70. EEDUCTION. 125 MISC ELIjAN EOU S P It O B Tj E M S , 1. Find the prime factors of 325. Of 1872. Ansivers. 5, 5, 13 ; 2, 2, 2, 2, 3, 3, 13. 2. Find all the exact divisors of 24. Of 42. Answers. 2, 3, 4, 6, 8, 12, 24; 2, 3, 7, 6, 14, 21, 42. 3. Find the largest number that will divide without a remainder 720 and 960. Ans. 240. 4. Find the greatest common measure of 2145 and 3471. Ans. 39. 5. Find the least common multiple of 54, 378, 486, and 540. Ans. 34020. 6. Find the greatest common measure and the least com- mon multiple of 12, 36, 72, and 144. Ansivers, 12 ; 144. 7. What must be the width of carpeting to fit three rooms 18, 21, and 24 feet wide respectively ? Ans. 8. What is the least sum of money for which I can pur- chase sheep at 2, 3, 4, or 5 dollars each, and just expend the whole ? Ans. 9. How long is the longest measure that will accurately measure three pieces of cloth, respectively 27, 63, and 108 yards long ? Ans. 10. What is the l^st sum of money for which a man can buy cattle at 18, 30, or 36 dollars each ? How many can he buy at each price ? Ans. $180. 10 at $18; 6 at $30, or 5 at $36. 11. Find the value of the following expression by cancel- lation : 40 X 54 X 99 X 250 -j- 20 X 27 X 198 X 50. Ans. ] 2. A man bought 20 cases of muslin, each case contain- ing 50 pieces of 40 yards each, at 10 cents per yard, and paid in coffee, giving 80 bags of 200 pounds each ; what was the coffee per pound ? Ans. 25 cents. 126 FRACTIONS. CASE I. 175. Integers or Mixed Numbers^ to Im proper Fractions, -•>•- (i)ral'^:vfGrciXGX -^ -•-♦•• ^ Example. — How many thirds inl? 2? 4? 6? Solution. — In 1 there are 3 thirds ; therefore in 2 there are two times 3 thirds, which are 6 thirds. In 4 there are 4 times 3 thirds, which are 13 thirds, &c. PROBT.EMS, 1. How many fourths in 3? 8? 6? 5? 9? 2. How many fifths in 4 ? 7 ? 11 ? 12 ? 10 ? 3. How many sixths in 5? 9? 7? 8? 6? 4. How many sevenths in 8? 7? 6? 5? 4? 5. How many eighths in 7 ? 5 ? 4 ? 3 ? 1 ? In performing these examples it will be noticed, FiKST. That we change the form of the numbers, but not their rnlue. Second. That each whole number has now the form of a fraction. Example. — How many thirds in 3J^ Solution.— Since in 1 there are 3 thirds, in 3| there are 3 times 3 thirds + 1 third = ^/. 6. How many fifths in If ? 3| ? 6^ ? 7^ ? 7. How many eighths in 3 ? 2| ? 3^ ? 5| ? 8. How many sixths in 2 ? 4| ? 3^- ? 7| ? 9. How many fourths in 4 ? 4^ ? 8f ? lOJ ? 10. How many sevenths in 3 ? 5f ? 6^ ? 8-f- ? 11. How many ninths in 2 ? 2^ ? 5f ? 6| ? KEDUCTIOK. 127 Reduction of Fractions is changing their form without alter- ing their value (104). i-lVV^i'ittei^ ^xfercise^-f Example 1. — Eeduce 41 to ninths. BOLUTION. 9 41 369 9 ' Ans. Explanation. — Since in 1 there are 9 ninths, in 41 there are 41 times 9 ninths, or 369 ninths. The same answer may be obtained by multiplying the whole number, 41, by 9, and writing the result over the required denominator. Example 2. — Reduce 25^ to a fractional form. SOLUTION. 25H = 25 + H- 12 X 25 11 311 12 12 ~ 12 ' Ans. Explanation.— In 1 there are 12 twelfths; therefore in 25i^ there are 25 times 12 twelfths + 11 twelfths = ^. Hence the Rule. — Multiply the given denominator by the whole number. Add the given numerator , if any, to the product, and write the sum over the given denominator. PJROBLEMS 1. 4 to twelfths. 2. 16 to eighths. 3. 25 to thirtieths. 4. 121 to 140ths. 5. 132 to 75ths. Reduce 6. 7. 8. 9. 10. 167 to lOOths. 97 to 30-thirds. 125 to 25ths. 16 to 12tbs. Ill to 40-firsts. 128 PR ACTIONS. Reduce to Improper Fractions, 11. 37|. 12. 341yV 13. 542|i. 14. 742|if. 15. 2147,^. 16. 17571||. 17. 13562ff}-. 18. 26753ff. 19. 3672^^07. 20. 7im|. CASE II. 176. Improper Fractions to Integers or Mixed Numbers, 4- ©ral^:«fGiici8%X -^ Example 1. — Reduce ^ to an integer. Solution. — There are 3 thirds in 1. Since 3 thirds are contained in 12 thirds 4 times, 12 thirds = 4. Example 2. — Reduce ^ to a mixed number. Solution. — There s^re 4: fourths in 1. Since 4: fourths &re contained in 17 fourths 4 and 1 fourth times, 17 fourths = 4^. PROBLEMS. Reduce 1. ¥. ¥. ¥> ¥. ¥. and ¥ to integers. 2. -i^, ¥ -¥-' ¥? and ^ to mixed numbers. 3. ¥. ¥ ¥. ¥. ¥. and V to integers. 4" ¥? ¥> ¥> ¥j and ^ to mixed numbers. 5. In 14 o^ a foot, how many feet ? In ¥ ^^ an acre, how many acres ? In $^, how many dollars ? 6. In -2^ of a quart, how many quarts ? In 4^ of a peck, how many pecks ? In f of a bushel, how many bushels ? REDUCTION. 7. How many pints in ^ of a pint ? ^^ of a pint ? 8. How many days in -^ of a day ? ^ of a day 9. How many miles in -^^ of a mile ? ^ of 10. In ^ of a melon, how many melons ? 11. In ^ of a gallon, how many gallons ? 12. ^=?^ = ?^ = ?^=?^^=? 129 a mile? — l^. tlWiitten ^^erci^G^+ -♦•- Example 1. — Reduce -i^ to an integer. SOLUTION. Explanation.— Since 12 twelfths make 1, 12 ) 144 there are as many I's in 144 twelfths, as 12 is contained times in 144, which are 12 times. 12, Ans. Therefore, ^< = 12. Example 2. — Reduce f4 to a mixed number. ftOT TnPTO"V . ' Explanation.— Since there are \} in 1, 17 ) 9b ( 5ff, Ans. ^^^^^ ^re as many I's in f|, as 17 is con- 85 tained times in 96, which are 5|| times. Therefore, f« = 5H. 11 Rule. — Divide the numerator ly the denominator. Write the remainder, if any, over this denominator for the frac" tionalpart of the number. PROBLEMS. Reduce to integers 1. n- Ans, 8. 5. ^^2^. Ans. 91. 2. ^m- Ans. 12. 6. W- Ans. 24. 3. -WA Ans. 11. 7. W. Ans. 63 4. ¥f- Ans. 31. 8. m- Ans. 1. >0 FEACTIOKS. Keduce to mixed numbers 9. w. Ans. 7^. 14. 1^/1. ^ws. 312|f. 10. ^i*^. A71S. 52^. 15. 1^2,41.. Ans. 1974JI 11. H-}^' Ans. 205f?-. 16. i-y^K Jws. 754/y. 12. Wf- ^W5. 8^. 17. W. Ans. 13|4 13. WA^. -4^5. 27,%V 18. W. ^ws. 13H- CASE III. mi. Fractions to Higher or Lower Terms. A fraction is reduced to Higher Terms when the numerator and Denominator are expressed in larger numbers. Thus i = |, op ijs> or iV &c. A fraction is reduced to Lower Terms when the Numerator and Denominator are expressed in smaller numbers. Thus, ^ = f , or ^. •-♦-• — — 0i.ar3E:^(3 Example 1. — In ^ how many ninths ? In | ? In | ? Solution. — Since in 1 there are 9 ninths, in i of 1 there must be \ of 9 ninths, or 3 ninths; in f there must be f of 9 ninths^ or 6 ninths, &c. PJtOBLEMS. 1. In i how many 12ths ? Inf? f? f? |? V? J? 2. In|howmany24tbs? In|? |? |? |? V? ^? 3. In^J^^owmany 60ths? In^f? H? if? A? A? 4. In I how many 25ths ? In ^? |? |? -J? 5. In|howmany28ths? In^? f? 4? ^? J^? How may the answers to these questions be obtained ? Ans. By multiplying both terms of the fraction by an integer. Thus, \, |, &c.. REDUCTION. 131 are reduced to 12ths by multiplying both terms by 2 ; -^y* it»&c., are reduced to GOths by multiplying both terms by 5. Are the values of the fractions altered by this process? Why not? (163.) Example 2. — How many halves inf? f? -j^? ff? Solution.— Since in 1 there are |, in ^ there are ^ of |, or f . If there is | in |, in | there are as many halves as 2 is contained times in 4, which are 3 times. Therefore, | = |. In like manner f = |, 6. How many thirds in ^ ? In ^ ? if ? f f ? ff ? 7. How many sixths in if ? In H ? M ? «? if ? 8. How many eighths in ^ ? In \l ? f| ? U? H? How may the answers to questions 6—8 be obtained? What effect on the value of a fraction has dividing both terms bj the same number ? (163.) Example 1. — Reduce ^ to a fraction whose denominator is 72. SOLUTION. Explanation.— Since the denominator, 24, is 5 X 3 15 to be changed to 72, we must inquire by what 24 X 3 72 number 24 must be multiplied to produce 72. By dividing 72 by 24, we find the number to be 3. We multiply 34 by 3, and also the numerator by 3, so as not to chanj3re the value of the fraction (163), and obtain ||. Example 2.— Reduce ff to 24ths. SOLUTION. Explanation.— Since 96 is to be changed to 56 -> 4 _ 14 24, we divide 96 by 34 to find the divisor that will 96 -T- 4 24 effect the change. We find it to be 4. Dividing 96 by 4, we must also divide the numerator 56 by 4, in order that the value of the fraction may remain unchanged (163), and find that f f = ^f . 132 FEAGTIONS. Example S.—Reduce J to a fraction whose numerator is 15. SOLUTION. Explanation.— We first divide 15 hj 3 to 3 X 5 15 find the number by which we must multiply the 4x5 20 given numerator, 3, that the product may be 15. We find 6 to be the number, by which we mul- tiply the terms, and obtain f = ^f . Example 4. — Reduce ff to a fraction whose numerator is 8. Explanation. — We find by dividing 48 by 8 SOLUTION. that 6 is the number by which the given numer- 48-7-6 S^ ator must be divided to produce the required 96-1-6 16 numerator. Dividing both terms by 6 we have If = A- Rule. — Multiply or divide both terms of the fraction by such a number as will change the giveti denominator or numer- ator to the required one. PBOBL EMS, Reduce 1. 4 to 49ths. Ans. ff . 6. H to 3ds. Ans. f. 2. i to 72ds. Ans. ^. 7. 1^ to llths. Ans. ^. 3. ^ to 576ths. Ans. f^. 8. AV to 12ths. Ans. -^. 4. ^ to 1728ths. Ans.^^. 9. 1066 to 26ths. Ans. ^. 5. ^^to8784ths.^/i5.-^y. 10. ^A to 74ths. Ans. -f^. Reduce 11. J to numerator 24. Ans. M- 12. 1 to numerator 525. Ans. m- 13. -^ to numerator 438. Ans. fjf. 14. ^^ to numerator 273. Ans. m- 15. 3^5 to numerator 1243. Ans. ,WiV 16. f^ to numerator 5. Ans. A- EEDUCTION. 133 17. Reduce ^i^ to numerator 12. Ans, 3^. 18. Eediice y^^ to numerator 3. Ans. ^. 19. Reduce j-|| to numerator 13. Ans. \^. 20. Reduce f|^ to numerator 14. Ans, \4[, In the preceding problems, our multipliers and divisors are all in- tegers, but had they been fractions the application of the Itule would have been the same; for example, let it be re(iuired to reduce f to a fraction whose denominator is 10. We divide 10 by 4 to find the number by which we must multiply both terms, and obtain ^. ^^ of 4 are 10; ^^ times 3, or 3 times 1^, are -\a = 72 (176). Therefore, 7| f =v^, a complex fraction (157). In this Case we have seen that a fraction is reduced to lower terms by division of both its terms by a common divisor. It is plain, there- fore, that a fraction is reduced to its lowest terms by the division oi both its terms by their greatest common divisor. CASE IV. 178. Fractions to their Lowest Terms. ©rat^Exfoi^ciXoX ^^ Example. — Reduce Jf ho its lowest terms. Solution. — Since 6 is the greatest common divisor of 12 and 30, we divide both terms by 6 and have ^§ = f . Therefore, f is ^f reduced to its lowest terms. FltO BZ E3IS. 1. Reduce to lowest terms, f , ^, ^, i^, H, ^\, ^. 2. Reduce to lowest terms, ■^, ff , ||, ff , A, 1^. 3. Change H ^o its lowest terms, fj. if. ||-. f . 4. Change ^ to its lowest terms, -ff . ■^^. -Vf. ff- 5. Express with the smallest whole numbers the fractions wv, fi. m^ m> and if. A fraction is said to be in its Loivest Terms when the numerator and denominator are prime to each other (169). 134 FKACTIONS. ^-'^Wi'itten ^^erci^esr-i Example. — Reduce f j- to its lowest terms. FIRST SOLUTION. 45 -^9_5 54-^9 ~~6 SECOND SOLUTION. 45 54 15 18 THIRD SOLUTION. 5 5 ^ _5 6 04 ~ 6 Explanation. — In the 1st Solution, we first find the greatest com- mon divisor of 45 and 54 by (170), or by inspection, and divide by it both terms, and obtain f as the result. In the 2d Solution, we first divide both terms by the common factor 3. But as the terms are not yet prime to each other, we divide again in like manner by 3. The result is f , in which both terms are prime. In the 3d Solution, we simply canceled the factors common to both terms. Hence the Rules. — 1. Divide both terms ly their greatest common divisor; or, 2. Divide both terms by any common factor, and the re- sulting terms in the same manner, till they are prime ; or, 3. Cancel all factors common to both terms. 3. Jf- 5. if- 6.**. Ans. ^. Ans. |. Ans. f. Ans. f. Ans. -J. Ans. -J. PJt O B LBMS Reduce to lowest terms ITHT* 7. 9. in. ll' 26 2 5* Ans. -^. Ans. i. Ans. -j^r- Ans. i- Ans. A- Ans. i^. 13.m- 15. m- 16. iWr- !"• 2 32 3* Ans. -^. Ans. 41. Ans. ^\. Alls. -^. Ans. T^. Ans. A-. KEDUCTION- 135 CASE V. 179. Dissimilar to Similar Fractions, •4^- ^Wi^itteH^^rcises^ E iX AMPLE 1, SOLUTION. 2 3 X X 5 5 10 ~15 4 5 X X 3 3 12 ~15 -Reduce f and ^ to similar fractions. Explanation. — Thirds cannot be reduced to fifths, nor fifths to thirds, without resulting in complex fractions. (See 1st Note, page 133.) But since 3 times 5 = 15, we reduce f to fifteenths by multiplying its terms by 5 ; and since 5 times 3 = 15, we reduce | to fifteenths by multiplying its terms by 3. Hence, |, | = \h if. Example 2. — Reduce \, {, and f to similar fractions. solution. Explanation. — Since the product of the 1 V 4 X 6 24 denominators 2, 4, and 6 = 48, we may reduce the fraction to 48ths. This is done by multiplying the terms of each fraction 3x2x6 36 by the denominators of the other fractions. 4x2x6 48 ^^® terms of \ multiplied by 4 and 6 = f|; those of f , by 2 and 6 = ff ; and those of 5 X 2 X 4 _ 40 f , by 2 and 4 = |§. Hence, i, f , | = f |, 6x2x4 48 11,40. Fractions having the same denominator are said to have a common denominator. The common denominator of two or more fractions is a com-tnon multiple of their denominators. Reducing "dissimilar to similar fractions" is sometimes called reducing fraMions to equivalent fractions having a common denominator. Rule. — Multiply loth terms of each fraction hy all the denominators of the other fractions. 2x4x6 48 136 rUACTIOJ^S. 1* hobijEms, 1. Reduce |, f, and | to similar fractions. ■^ns. \i, 1^, if. 2. Reduce j, |, and | to similar fractions. ^^S' m, 1%, *. 3. Reduce | and ^ to similar fractions. Ans. ^, f|. 4. Reduce f and f to similar fractions. Ans. ||, f J. 5. Reduce | and -^ to similar fractions. ^ws. ||, |f. CASE VI. 180. Dissimilar to Least Similar JFractions, ^%ym{iG^;^^e (rcisi^ 1 £ 5 In examining the fractions f| and f f (answer to Prob. 3, Case V), ■we find that we may reduce them to lower terms, and still preserve a common denominator ; thus, by dividing both terms of f f and f^, by 4, we reduce them to ^ and ^ respectively. And since they cannot be reduced to any lower terms and at the same time have a common denominator, they are said to be reduced to least similar frac- tions. The least common denominator of two or more fractions is the least common multiple of their denominators. Reducing " dissimilar to least similar fractions " is sometimes called redttcing fractions to equivalent fractions hairing a least common denom- inator. Example. — Reduce ^^, -^, and -^ to least similar fractions. Explanation. — We BOLTJTION. A? A? A? *^® fractions. 20, 18, 12, their denominators. 180, the L. C. M. of the denominators 9, 10, 15, L. 0. M. -T- denominators. first find the L. C. M. of the denominators 20, 18, 12, to be 180 (173). We now find our mul- tipliers to be respective- ly 9, 10, 15 (page 131, Ex. 1^. EEDUCTION. 13? _3^ X 9 21 Multiplying both terms 20 X 9 " 180 of ^ by 9, ^\ by 10, and ^ ^ y^lQ 50 ^7 15, we have ^\\, ■^%, and 18 X 10 ^ 180 ^t . A A' ' ' Hence, to reduce dissiuu- 7 X 15 105 lar to least similar fractions, 12 X 15 ~" l80 we have this EuLE. — Multiply loth terms of each fraction ly the num- her of times its denominator is contained in the least common multiple of all the denominators. PROBLEMS. Eeduce to least similar fractions, 1. I, ^\, and H- ■^^«- tt 41. H' 2. A, ii, and fi-. Ans. -^, ^, «f 3. f, h I, tV ^^^- H' u> ii n 4. f, A, f, U' ^^s- A. «. M. 4i- 5. i i i i f ^'^^^ «. » li^ n^ u 6. I, 1, h ^. ^^^- *V, tli. m 7. I, -I, A, U' ^^'' lo-. H, f ^, U 8. V-. ii J- ^^^- -W. M, W 9. h I f. ^^5. ill. m> -ih 10. \S ^/, M. ^^^- W' W. -If 11. 4, i, f ^^5. u, if, if, 13- A A. A- ^^^- -^^^ A. U 13. Yjj, YW' "^* Ans. -ff^, 2^, -^^ 14. A, I, A. ^/2^. H, if, *■ 15. i «, if. Ans. If If, if, 16. A. i A- -4;?5. -^, If, ^, 17. i, A, A. ^^^. If, A, li- 18- i, h h h ^ns. ff , iA, ^, If 19. i ih-, A ^ns. A¥f, i¥A-. tMt 20. i, i, i ^/?5. A\, ^^ -tt SECTION III. < ** — -Mr — ^lJt ^j^^ 5 to be m Finally, we add ^" H" l^' together the two sums 38 and Or, 38 + If^ 1ft, and find our entire sum to Or 39M ^^ 39ft, the result required. Rule. — If the fractions have a common denominator, write over it the sum of all the numerators. If the fractions have not a common denominator, reduce them to a coiimion, or to their least common denominator, and then write the sum of the numerators over the common denominator. Add mixed numbers hy finding the sum of the integral and fractional parts separately, and then adding their sums. P B OBLEM8. 1. Find the sum of ^, If, and Jf Ans. 2^. 2. Find the sum of ff , 44? and ff- Ans. Iff. 3. Find the sum of ^, -Jf , \-^, and ^. Ans. 1-|-|. 4. Find the sura of -5%, |||, }^, and ■^. Ans. 2^. 6. Find the sum of -^4, -jW^, and ^^^. Ans. 1^^. 6. Find the sum of i, f , and \. Ans. 1^. 7. Find the sum of f , |, and ■^. Ans. 1\^\. 8. Find the sum of |, |f , and |. Ans. 2|4. 9. Find the sum of |, ^, and ^^. -4w5. Iff. 10. Find the sum of \, f , f , |, and |. ^W5. 3-|. 11. Find the sum of 7^, 8J, and 3. Ans, 12. Find the sum of 16|, 12|, and ^^. Ans. ADDITION". 141 13. Find the sum of 5|, 9^^, and llf Ans. 26^, 14. Find the sum of 7^, 3^, and 21|. A71S. 32^. 15. Find the sum of 5-gig^, 7^, and 9^^^^. Ans. 16. Find the sum of 21f , 18|, 4 and 26f Ans. 70fJ. 17. Find the sum of 17i 10, 14|, and 13|. Ans. 18. Find the sum off, 3jf 10|, and ^. Ans. 15^^. 19. What is the sum of 1246f pounds, 9464^^ pounds, and 204^2^ pounds ? Ans. 20. What is the sum of 1476 bushels, 46-^ bushels, and 978-1 bushels ? Ans. 21. Bought 4 pieces of muslin containing 90-J-, 40J^, 33^, and 61 J yards respectively. How many yards in all ? Ans. 225 J yards. 22. Four men weighed respectively 150J, 101, 185J, and 114| pounds. What was their entire weight. A7is. 23. My farm is divided into .5 fields : the first contains lOJ acres, the second 12| acres, the third 15| acres, the fourth 9^ acres, and the fifth 52Jf acres. How many acres in my farm? Ans. 100 acres. 24. A man has 3 sheep ; the first is worth |6f, the second t8|, and the third |9J. How many dollars are they all worth ? Ans. 25. John is 11^ years old; Mary is 2^ years older than John ; and Henry is 6-^ years older than Mary. How old is Henry ? Ans. 26. A man walked 25f miles on Monday, 27| miles on Tuesday, 30^ miles on Wednesday, 33| miles on Thursday. How far did he walk in all ? Ans. 117^|- miles. 27. Jones has 746-J^ acres of land, Wilson has 594- acres more than Jones, and Williams has as many acres as both. How many have all ? Ans. 3103^. SECTION IV. S: mm T'R' ACT rQ:IV 183. 0i*at^x:GitciXeX Example 1. — From ^ take 4. Solution.— Since the difference between 3 units and 1 unit is 3 units, the difference between 3 sevenths and 1 seventh must be 3 JPJJOBX^JIfS. 1. From A take ,s^; A; A; A; A; A- 2. From « take ^V; A; A; A; A; A- Example 2. — From | take }. Solution. — By reducing | to 8ths we have | ; then from | we take I, which leaves as many Sths as there are units in the difference be- tween 7 and 6, which is 1 unit. Therefore, if from | we take |, we have left |. 3. From i| take |; |; i; i; J; f 4. From fj- take A; ^s^; i; |; J; }; f Example 3. — From | take f. Solution. — The least common multiple of 8 and 7 is 56 ; f = ||, and f = If. 35 fifty-sixths less 24 fifty-sixths is 11 fifty-sixths. There- fore, if from f we take f, the remainder will be |^. 5. From f take |; i; |; f; ^; A- 6. From A take i; |; f ; -f^', A; A- Example 4.— From 10 take 3|. Solution.— 10 is equal to 9 + 8 eighths. From 8 eighthi take 5 142 SUBTRACTION. 143 eigJUhs, and we have 3 eighths left. From 9 take 8 and 6 remains. Therefore 10 less 3| = 6|. 7. How much is 11 - 2f ? 12 — 3| ? 13 - 7^ ? 8. How much is 15 — 3i ? 20 — 4| ? 24 — 5| ? Example 5. — How much is 4J — 2J ? Solution.— 4| ~ 4||, and 2| = 2||. Since we cannot subtract 28 thirty-dxths from 27 thirty-sixths, we reduce one of the 4 units to thirty-sixths, which added to 27 thirty-sixths gires 63 thirty-sixths. 63 thirty-sixtJis less 28 thirty-sixths are 35 thirty-sixths, and 2 from 3 leaves 1. Therefore 4| less 2| = 1||. 9. How much is 4J — 2^ ? 5^ — 2} ? 6^ — 2^ ? 10. How much is 5J — 3^ ? ^ — ^^ 21-^ - l^^^ ? ^•IWi'itten ^:^er ci^feX-i Example 1. 23 47 SOLUTION. 17 _ 23 — 17 47"" 47 47 What is if less H ? Explanation. — Since the dif- ference between 23 units and 17 units is 6 units, the difference be- tween 23 47ths and 17 47th8 must be 6 Ji.7th8, or :^, Ana. Example 2. — How much is ^ — ^^ ? Explanation.— We first reduce ^^ and ^^ to SJ^ths and find ^^ = 51 .?-^,n^i 1 1 sloths, 40 54C^A«, we have Example 3.— Subtract 3^^ from 20. SOLUTION. Explanation. — ^^ from 1 (|f), 20 ^::;19_j_l:^19_|_i6 leaves VV« 3 from 19 (amount loft o 7 QiJi Q_l_7, *^^^^ reducing 1 to 16ths) leaves ^tV — ^ + TT — ^ + T8- ig Annexing to this ^^, we have SOLUTION ^ X 17 _ _51 20 X 17 ~ 340 ^ X 20 _ ^ 17 X 20 ~" 340 144 FRACTIONS. Example 4. SOLUTION. ^= 4+ i= 4 + il From 16f subtract 4j: Explanation. — Since the fraction in the minu- end is greater than the one in the subtrahend, we take 1 from 16 and add to the f, making Y". Then we Ans. llfj- reduce the fractions to a common denominator, subtracting them as in Ex. 1 ; then subtract the integers, and prefix the latter remainder to the former. Rule 1.— Reduce, if necessary , the fractions to a common denominator ; then write the difference of the numerators over the common denominator, II. — If mixed numbers are to he subtracted, subtract the integers and fractions separately ; or reduce to improper fractions and proceed ly Rule I. rnoBL EMS. From 12. fj take |. Ans, i^, 13. 8| take 2|. Ans. 6^. 14. 9f take 5f. Ans. 4^. 15. lOf take 5}. Ans. 5^. 16. 7| take 2|. Ans. 4|. 17. 11 take 6|. Ans. 4|. 18. 15i take -J. Ans. l^, 19. 9 take 3f Ans, 5f 20. 13i take f . Ans, 12if. 21. 12^ take 7f . Ans, 4^^. 22. 16f takelOfI- Ans,5^. From 1. \^ take ^, 2. a take ^. 3. i take f 4. i take f . 5. f take f. 6. -f take f 7. -J take f 8. t take |. 9. ^take |. 10. ^ take f 11. H take J. 23. If a ton of hay costs $24f and a ton of coal $59J, what is the difference in their cost ? Ans. $35^. 24. A farmer having 300 acres of land, sold 149| acres. How many acres had he left ? Ans. 150^ acres. Ans. f Ans. A- Ans. i- Ans. A- Ans. T^- Ans. A- Ans. 1- Ans. |. Ans. U- Ans. A- Ans. A- SECTION V. m MULTIPLICATION r^ 183. Oral TK\:oi ri^oj^ -•♦•- Example 1. — If 1 pound of coffee costs J of a dollar, how much will 3 pounds cost ? 5 pounds ? 7 pounds ? Solution. — If 1 pound of coffee costs ^ of a dollar, 3 pounds wOl cost 3 times as much, or f of a dollar (161). 5 pounds will cost 5 times as much as 1 pound, or f = 1^ dollars. PR OBLEM8 . 1. At 1^ per pound, how much will 2 pounds of sugar cost? 3 pounds? 4 pounds? 5 pounds? 8 pounds? 2. At $f per gallon, what cost 3 gallons of molasses? 5 gallons ? 7 gallons ? 9 gallons ? Example 2.— If 1 barrel of flour costs $8, what will } of a barrel cost? J? -V-? i^? iy^? Solution.— If one barrel of flour costs $8, i of a barrel will cost \ of $8, or $2 ; and f of a barrel will cost 3 times |3, or $6. | of a bar- rel will cost 7 times ^ of $8, or $7 ; &c., &c. 3. If an acre of land is worth $2400, what is \ of it worth ? i? 4? J? i? i? ^^? JV? 10 145 146 FRACTIONS 4. If one yard of silk is worth $4.80, what is J of a yard worth? i? I? 1? A? A? A? Example 3. — A man owning f of a mill, sold f of his share. What part of the mill did he sell ? Solution. — If a man owning f of a mill, sells § of his share, he sells I of f of the mill. ^ of f of the mill is 3^^ of the mill, and f of f of the mill are 2 times -j'^, or /^ of the mill. Therefore, he sells /^ of the mill. 5. How much is f multiplied by f ? By f ? By |^ ? 6. ixf=? |xi=:? |xi = ? ixi = ? |x| = ? tlWi itteri ^Xjer ei;*e^ t Example 1. 5M = of, -Multiply H l>y 7. Explanation. — Since multi- plying the numerator hy a whole number, the denominator re- maining the same, multiplies the value of the fraction (161), we multiply 17 by 7 and write the product 119 over the denom- inator, and thus have y^ = 5^f = 5|. Or. Since dividing the denominator by a whole number, the numerator remaining the same, multiplies the fraction (161), we divide the denominator 21 by 7, and write the result under the numerator ; thus. 17 X 21 17 21-4- SOLUnON 7_n9_ — 21 ~ or 17 , = i = 'i- Example 2.— Multiply 42 by ^. SOLUTION. 42 X 13 _ 42 X 13 _ 546 17 ~ 17 ~ 17 or 42 X 13 _ 42 X 13 _ 546 17 "" 17 ~" 17 = 32-ft, = 32^. Explanation. — We are required to find the same part of 42 that |f is of 1. We can do this in two ways : 1st. We can find ^ MULTIPLICATION. 147 of 42 = ff, and then multiply yV of 42, or f f , by 13 to find ^| of 42, which equal 32^. Or, 2d. We can multiply 42 by 13 ; but 13 is seventeen times as great as our multiplier jf. Therefore to obtain the true product we must divide 42 x 13 by 17, and we have —-- — = 32j\, Ana. Example 3. — Multiply -j^ by |^. SOLUTION. Explanation. — ^We are required to find 5 11 55 ii of A- Is*- We can find ^V of iif ^7 T^ ^ 22 ^^ 156* multiplying the denominator by 12 (162), jf ^. Having found ^f ^ to be 3^, we find \^ by multiplying y|^ by 11 (161), obtain- ing for an answer, ^f^. Or, We may multiply ^^^ by 11 (161), giving us f|. But we were to have multiplied by \\, a number only yV as large as 11 ; therefore, we must divide our product by 12 (162). This will give the correct result, ^^^. Example 4.— Multiply 21J by 5. solution. 21f X 5 = 21 X 5 + } X 5 = 105 + 3f = 108J. Explanation. — We multiply 21 by 5, and | by 5, and add the two products. Example 5. — Multiply 4| by 7|. solution. Explanation. — We reduce both factors to improper fractions, and then multiply as in Example 3. 17 7 119 Since Example 1 can be solved thus, oT ^ 1 ~ "2r ' ^^^™P^® ^' tt„«, I X ^ = ^; and Example 4, thu., 31} x 5 = ^ x|=^; by changing the integers to a fractional form by writing 1 for the denominators, we can readily see by inspection, that Examples 1 to 5 may all be solved by the following 148 FKACTIOIfS. Rule. — Reduce each integer and mixed number to the form of a simple fraction ; then write the product of the numerators over the product of the denominators. To lessen labor, cancel all factors common to both terms, before multiplying. The following special rules can be deduced, and may be used, if thought best : To multiply a fraction by an integer, or an integer by a fraction, we deduce from Examples 1 and 3, Rule I. — Write the prodtLd of the integer and the numerator of the fraction over the denominator ; or, Write the quotient of the denominator of the fraction divided by the whole number under the numerator. To multiply a fraction by a fraction, from Example 3, we deduce. Rule II. — Write the product of the numerators over the proditet of the denominators. To multiply a mixed number by an integer, we have, from Exam- ple 4, Rule III. — Multiply the whole number and the fraction separately by the whole number , and add the products. To multiply mixed numbers, from Example 5, we have Rule IV. — Reduce mixed numbers to improper fractions and apply Rule II. Reducing Compound to Simple Fractions is the same as multiplying one fraction by another (158). PB o a z EM Multiply 1. ^ by 3. Ans, If 3. ^ by 4. Ans. 1-^. 3. m by 5. Ans. 3^. 4. tt by 6. Ans. l^V 5. ^ by 9. Ans. JJ. Multiply 6. -^ by 10. Ans. 2H 7. m by 7. Ans. 6ff| 8. fHbyS. Ans.'Hm 9. -^ by 9. Ans. 1^ 10. 4 by 20. Ans. K\i MULTIPLICATION. 149 Multiply Multiply 11. 28 by J. Ans. 21. 17. 165 by |. Ans. 103^. 12. 315 by |. Ans. 210. 18. 283 by f ^7i5. 220^. 13. 118 by f Ans. 59. 19. 292 by A- ^W5. 87f 14. 430 by f ^ws. 258. 20. 568 by A- Ans. 154^. 15. 348 by f Ans. 290. 21. 443 by t^j^. ^ws. 258t«^. 16. 756 by f ^ws 432. 22. 687 by H- Ans. 377H. 23. H by J. Ans. i>0 rRACTio]srs. 45. 9| tons of sugar, at $180 per ton ? Ans. $1728. 46. 15| yards of silk, at $5 per yard ? A71S. $7GJ = $76.88+. 47. 20| bushels rye, at $1.20 per bushel? Ans. $24| = $24.75. Multiply Ans. Multiply Ans. 48. 9^ by 5|. 54^. 52. 10-^ hy llf. 121A. 49. 18| by 4|. 80H. 53. 12J by 16f 203^. 60. 16| by 3i. 57H. 54. 131J- by 16f. 2187i 51. 17i by 12*. 230^. 55. 62J by 137i. 8593f. 66. What cost 19| cords of wood, at $3| per cord ? Ans. $66.23 + . 57. What cost 93^ bushels oats, at 33J cents per bushel? A71S. 68. What will 8f weeks board cost, at $6J^ per week ? Ans. 69. At 12^ cents each, what cost 12^ pints of peanuts ? Ans. $1.56i-. 60. What is the value of 204^^ acres of land at $72} an acre ? A ns, 61. What is the value of 580-|- pounds of sugar at 9^^ cents per pound ? Ans. 62. What will 85^ pounds of tea cost at $1.37^ per pound? Ans. $117.56^. 63. Bought 156 pounds of cheese at $.12^ per pound, 327 pounds of coffee at 26f cents per pound, and 17 barrels of apples at $2.87|^ per bbl. What was the cost of the whole ? Ans. $155.84|. 64. What is the value of 5| pieces of cloth, each piece containing 36} yards, at $.85 per yard? Ans. SECTION VI ^ ©I^riSI^M I 184. Example 1. — If I divide f of a dollar equally among 4 boys, what part will each boy receive ? Solution.— Each boy will receive J of f of a dollar, or f of a dollar (162). Pn OBZJEMS. 1. Divide ff by 2. By 3. By 4. By 6. By 9. 2. Divide if by 2. By 8. By 12. By 16. By 24. Example 2.— Divide 4 by 4. By 5. By 7. By 9. Solution. — | divided by 4 are -^ (162) ; f divided by 5 are ^; &c. 3. Divide fj by 3. By 4. By 9. By 6. By 7. 4. Divide f by 5. By 6. By 7. By 9. By 3. Example 3. — Divide 5 by J. 1st Solution. — Since 5 = *^, and since 20 units divided by 3 units equal ^, 20 fourths divided by S fourths must equal ^, or 6| (176). 2d Solution. — If we divide 5 by 3 we have f ; but our divisor is only ^ of 3 ; therefore to obtain the correct quotient we must multiply f by 4, giving us ^ = 6|. 161 152 FKACTION^S. 6. Divide 8 by}, f f i^. |. ^. ^V h A- 6. Divide 11 by f f. f i. f A- t^.- A- A- Example 4. — Divide | by ^, 1st Solution.— I = ||, and TT=lf- Since 77 units divided by 72 M7iif« = ^, 77 eighty-eighths divided by 72 eighty-eighths must equal il, or 1^. 2d Solution.— Dividing i'bjQ gives ^V (162) ; but our divisor is o^7 irr of nine, therefore, we mast multiply if\ by 11 to obtain the true result which is ^ = 1/^. 7. Divide A^yf |. h i' f f f I A- A- 8. Divide I by i. f. -J. f t^. A- f h f A- Example 5.— Divide 4| by 4. 5. 6. 7. 8. 9. 10. Solution. — 4 divided by 4 is 1 ; | divided by 4 is ^. Therefore 4| divided by 4 equal 1/^. 4| divided by 5 equal ^ -j- 5 = ff ; &c., &c. 9. Divide 7| by 6. 7. 10. 12. 13. 15. 16. 8. 9. 10. Divide 9^ by 2. 3. 4. 5. 6. 7. 8. 9. 11. 12. Example 6.— Divide 5| by 6f. 7J. 8^. Solution.— 5f = ^ = f| ; 6| = ^ = ff . 69 twelfths divided by 80 twelfths = If. 5| = ^ ; 7i = ^ ; and -^^ divided by »/ = || ; &c. 11. Divide 9| by 3f 2|. 3|-. 5|. 4^- H- 6^. 12. Divide 4^ by 2^. 3^ 6J. 5f 9f 3^. 8^. Example 1.— Divide ^ by 7. l8T solution. Explanation. -We may f2: "^ ' _£_ either divide the numerator 31 ~ 31 21 by 7 (162) ; or. DIVISION, 153 21 31 31 • 2d solution. Multiply the denominator 21 3 by 7 (162); or. X 7 ~ 217 ~ 31. We may reduce the divisor 3d solution. to 217 31st8, and then we have 21 217 21 3 21 Slsts divided by 217 SUts 31 • 31 ~ 217 " "31. = iii = -h- Example 2.— Divide 22 by f . 1st solution. 22 22 99. = 99. Explanation— In the first Solution, we first divide 22 by 2, which gives us 11 ; but in dividing by 2, we use a di- visor 9 times too large, and our quo- tient is only \ of what it should be. We therefore multiply it by 9 and ob- tain the correct result, 99. In the second Solution, we multiply 22 by 9 and divide the product by 2. This gives the same result as the first Solution, and is more convenient when the integer is not exactly divisible by the numerator. In the third process, we reduce both dividend and divisor to the same denominator, 9. Then 198 ninths divided by 2 ninths = ^^^ a 23 - 2d solution. 2 _ 22 X 9 9~ 2 3d solution 22 -^ 198 9 = 99. 5^ 16 1st solution. 5 5 X 4 = 5 r, _5^ 16 64' 64 2d solution. _35 35 "~16' 16 3d solution. 4_^^^ 7~112 • 112 7 = 35 64. Example 3.— Divide ^ by 4- Explanation.— Since we are 35 required to divide y^ by 4 sevenths^ g^ in the 1st Solution, we divide ^^ by 4, which gives us ^ (162). But as 4 is seven times as large as 4, ^ is only one seventh of the re- quired answer. Therefore multi- plying g\ by 7 will give the correct 35 result, which is ff. 64, In the 2d Solution, we first mul- tiply by 7 and then divide by 4, obtaining, of course, the same result. In the 3d Solution, we reduce -^-^ and | to ^y'^ and ^ respectively. Since 35 units divided by 64 units gives f|, -^ divided by ^ must give f f . 154 FEACTIONS. Example 4. — Divide 41J by 9. 1st solution. Explanation.— This Ex- 167 167 ., ample is performed as Ex- 41f -T- 9 = — -J- 9 = — = 4|f . ample 1, after the mixed o^ o^TTTmx^,.T number has been reduced 2d solution. 9 ^ 41 3 ^ ^ fraction. l IL It is sometimes prefer- 4|-| able, however, to solve it thus : 9 is contained in 41 f 4 times, with the remainder 5f = -\'-. 9 is contained in ^, |f of a time. Therefore,41| divided by 9 - 4f f. Example 5.— Divide 18| by 7|. solution. 65 18^ -;- 7f _ ^ . ^ _ ^ X ^^ _ ^ _ 2Aj. 19 Explanation. — We first reduce the mixed numbers to simple fractions, and then proceed as in Example 3, 1st Solution. Of course, by the 2d or 3d Solution the result would be the same. The Solution of Examples 1, 2, and 4 may, by supplying a denomi- 21 7 nator 1 to the integers in each, be efiected thus : Ex. 1. ^ -*-:§■ = ol X 21 1 3 „„ 22 2 22 9 „„„..,, 9 167 _x^ = gj Ex.2. ^^g=j-x- = 99. E^4. 41}^j = -3-x - = -^ = 4|f . By an inspection of these three Solutions and the let Solutions of Examples 3 and 5, we see that in every instance, after changing the integers and mixed numbers to fractions, we obtain our answer by multiplying the dividend by the divisor with its terms ivr verted, that is, with the numerator and denominator irUerchanged. The Reciprocal of a number is 1 divided by that number, so that the Reciprocal of a fraction is 'the same as a fraction with its terms inverted ; thus, f is the reciprocal of |. \ is the reciprocal of f or 2 ; |off; Aof¥;&c. Hence, in all cases of division, in which dividend, or divisor, or both, is a fraction, we can solve the problems by the following Rule. — Reduce integers and mixed numbers to the form vf simple fractions ; and then multiply the dividend hy the DIVISION. 155 divisor with its terms inverted ; or, by the reciprocal of the divisor. We may also have the following Special Rules : From Ex. 1, 1st, 2d, and 3d Solutions, we deduce for dividing a fraction by a whole number, this Rule I. — Divide the numerator by the whole number, and vyrite the quotient over the denominator. Or, Multiply the denominator by the whole number, and write the product under the numerator. Or, Reduce the fraction and whole number to a common denominator., and write the numerator of the latter under that of the former. To divide a whole number by a fraction, we deduce from Ex. 2, 1st., 2d, and 3d Solutions, the following Rule II. — Divide the whole number by the numerator, and multiply the quotient by the denominator. Or, Multiply the whole number by the denominator, and divide the product by the numerator. Or, Reduce the whole number and the fraction to a common denominator, and divide the numerator of the former by that of the latter. To divide a fraction by a fraction, we deduce from Ex. 3, 1st, 2d, and 3d Solutions, Rule III. — Divide the dividend by the numerator of the divisor, and multiply the quotient by the denominator of the divisor. Or, Multiply the dividend by the denominator of the divisor, and divide the product by the numerator of the divisor. Or, Reduce both fractions to a common denominator, and then divide the numerator of the dividend by the numerator of the divisor. Reducing Complex to Simple Frtictions is the same as dU f>iding one fraction by another (157). PROBLEM, Divide 2. 14^ by 20. 3. 14 by 54. 4. i|4 by 108. 5. « by 72. 6. ^ by 13. -AnS. yfy Ans. ^, Ans, yIy Ans. ^, Ans. yi^. Ans. ^, Divide 7. m by 96. 8. 1^ by 203. 9. ill by 213. 10. -^t by 24. 11. 'm by 144. 12. i,V/ by 91. 156 FRACTIONS. 13. 16 by f Ans, 20. 18. 483 by ^. ^m 536|. 14. 24 by f . Ans, 27. 19. 160 by ^. ^W5. 202f. 15. 36 by A. Ans. 66. 20. 272 by |f Ans. 335^?.. 16. 401 by f . ^n5. 467f 21. 256 by H. Jw5. 304if. 17. 453 by f|. ^715. 578f. 22. 365 by |f ^ws. 414^. 23. At $^5jj. per pound, how be bought for $25 ? 24. If one gallon of molasses can be purchased for $150 ? 25. At $Jf apiece, how many for $39 ? many pounds of cofiee can Ans. 100 pounds, cost $3^^, how many gallons Ans. 480. chickens can be purchased 93. Divide Divide 26. 5| by 8. Ans. yV 31. 24I3-V by 21. 27. 5i by 8. Ans. jr. 32. 7462J by 27. 28. 10| by 9. Ans. H- 33. 372JV by 27. 29. 72VV by 6. Ans. 12:rb. 34. 146741 by 140. 30. 143 A by 11. Ans. 13^. 35. 3146^1 by 324. 36. 8f by 24. Ans. 3i. 40. 8i by 3f . Ans. 2A. 37. 15| by 8^. Ans. If. 41.19i by 11 Ans. lOf. 38. 10| by 9f. Ans. If 42.73i by 9i. Ans. n- 39. 12J by 16|. Ans. }. 43. 54JI by 25f Ans. H. 44. If 7f yards of ribbon cost $4^|, what is the price per yard? Ans. $f 45. If 2} yards of cloth cost $17^^, what is the price per yard? Ans. $6f 46. If 3| gallons of oil cost $2^^^, what is the price per gallon? A71S. ^. 47. How many yards of carpet 2| ft. wide will be required to cover a floor 14^ feet in length and lOj^ feet in width ? Ans. 19^. DIVISION". Divide Divide 48. 1 by i. Ans. 2^. 53. li by a- 49. i by I ^7i5. f 54- it by -Wr- 50. A by f. ^ws. 1}. 55. IH by If. 51. A by H. ^/is. |. 56. m by H 52. H by if. Ans. 7^. 57. TftWrbyiftWr 157 58. Find the value of(f-J)-^(|- — f). ^?^s. If 59. Find the value of (i -^ %) + (i -^ -f^). Ans. 2fJ. 60. Find the value of (| x i) -r- ( J x |). Ans. f 61. Find the value of (^ x i) x (f x M). Ans. ^V 62. Find the value of |, or of ^ -^ f . Ans. fj. 63. Find the value of | of |, or of f x |. Ans, i. 64. Find the value of 4 + 1' orof (|-|)~(J+|). 65. Find the value of t^^> or of (| x H) "^ (i x i)- Ans, 11. ItEVIEW PROBIjEMS, Oi'^it 5^Xoi'cis^o^ 1. Jones owns J of a rolling mill, Markham f and Simp- son ^ ; how much do they all own ? 2. Smith sold -^ of his interest in his business to one of his clerks, and ^ to another. How much did he sell to both? 3. After paying out \ of my money, and depositing f of it in bank, what part had I left ? 4. Burnside sold at one time f of his farm, at another 168 FBACTIONS. ■^^, at another time -^; how much of his farm had he remaining ? 5. Hamilton owned f of a planing mill ; he sold f of his interest to Boyd. What part of the mill did he sell Boyd ? 6. James lost -f of his marbles. He then had 28. How many had he at first ? 7. A man purchased a steamboat, paying at the rate of $600 for ^ of ^ of it. How much did the boat cost him ? 8. A w^oman purchased 21 yards of silk for $75. What did she pay per yard ? 9. James bought 9 oranges for 58 cents. What did he pay apiece for them ? 10. A grocer sold 20^ pounds of lard for $4.10. What was that per pound ? 11. At 35^4 cents per pound, how many pounds of copperas can be bought for 66 cents ? 12. After selling 3^ acres to A, and 5-| acres to B, and 4J acres to 0, 1 find that I have | of my land left. How much had I at first ? 13. One bushel is the ^^ of ^ of my entire crop of oats. How many bushels of oats did I raise ? 14. A suit of clothes cost $50, and the price of a pair of boots was ^ of the cost of the suit. What was the price of the boots ? 15. If 7 men eat 6 loaves of bread in 1 day, how many loaves will 1 man eat in the same time ? 2 men ? 10 men ? 16. If^oi^ of the price of my farm equals the value of my dwelling, and my dwelling is worth $375, what is the price of my farm ? 17. If 7 yards of broadcloth cost $40 J, how much does 1 yard cost ? 18. What cost 7i pounds of sugar at 12 J cents per pound ? DIVISIOK. 159 19. If 95f cents be paid for 12J^ pounds of rice, what cost 1 pound? 20. If A can do J of a piece of work in 1 day, and B can do I of the same work in 1 day, what part of it will A and B together do in 1 day ? In how many days will they per- form the whole work ? 21. If a man can do a piece of work in 5 days, what part will he do in 1 day ? In 2 days ? In 3 days ? In 4 days ? 22. A can build a wall in 10 days, and B can build the same wall in 12 days. In what time will A and B together do the work ? 23. If C and D together can mow a field in 3J days, and D alone can mow the same field in 5^ days, in what time can alone mow the field ? 24. If I of I of a barrel of flour cost $1 J, what cost 1 bar- rel ? H barrels ? 25. Paid $3^ for a City Directory, and 2^ times as much for a Webster's Unabridged Dictionary ; what did I pay for both? 26. A market woman bought eggs at 3 for 5 cents, and sold them at 4 for 7 cents. How much did she gain on each egg? 27. A confectioner bought oranges at 30 cents for 12, and sold them again at the rate of 8 for 21 cents ; and then found that he had gained 64 cents. How many oranges did he buy ? 28. After selling 200 bananas at a loss of 6j cents each, I found that I could purchase lemons at 3 for 7 cents and sell them at 7J cents apiece. How many lemons must I buy and sell to make good my loss on the bananas ? 29. How many pencils at 7^. must be paid for 150 pens, at the rate of 1 J cents apiece ? 160 FRACTIONS. t'WVl:*ittei\^?&r-ci^GXt 1. A bought a farm for $3250, and sold ^ of it to B, and the remainder to C. What is the value of C^s property ? Ans. $2375. 2. A merchant bought 25 pieces of muslin at $3:^ per piece ; 125 pieces prints at $2| ; 35 pieces ticking at $7J. What was the amount of the bill ? Ans. $716.25. 3. A man owns a farm of 125^ acres, worth $62f per acre; another of 83} acres, worth $76^ per acre. How much are both farms worth ? Ans. $14240.144-. 4. A grocer bought 5 barrels of granulated sugar; the first barrel containing 225 J pounds, the second 221J^ pounds, the third, 23 1|^ pounds, the fourth, 229-J- pounds, and the fifth, 240 pounds, at 11 J cents per pound. He sold } of it at 12i cents per pound, and the remainder at 12} cents per pound. How much did he gain ? Ans. ^16.06 -{-. 5. From 3} acres, I sell to A f of an acre, to B, f of an acre, to 0, | of an acre, to D, f of an acre. How much is the remainder worth, at $2575-J per acre ? Ans. $2042^^. 6. Into how many house lots, each containing rf^ of an acre, can I divide 184 acres, and for how much apiece must I sell lots, to realize $95600? Ans. 150 lots; price, $637^. 7. A stove manufacturer purchased old metal at $^^ per hundred pounds, and makes out of it stoves weighing 350 pounds each, which he sells at $17^^ apiece. How much does he gain on each 100 pounds of old metal ? Ans. $4.13. 8. A man bought a farm for $17500, and sold | of it to Mr. B for $4500 ; \ of the remainder to Mr. for $4600; REVIEW PROBLEMS. 161 and the remainder to Mr. L at a profit of $1250. What was his whole gain ? A ns. $5975. 9. A huckster bought 125 bushels of apples at $J per bushel ; 35 bushels at $J per bushel ; and 25 bushels at ^ per bushel. At what price per bushel must he sell them to gain$30.61i? Ans.t.^Z 10. A hunter in pursuit of a fox that was ^\ miles ahead of him, ran 1^ hours, at the rate of 5 miles an hour. The fox advanced at the rate of 3f miles per hour, for 1^ hours. How far in advance of the hunter is the fox ? Ans, l^f miles. 11. Henry, William, and Jacob can remove a pile of wood in 4 hours ; and Henry and William together can remove it in 9|^ hours. In how many hours can Jacob alone remove it ? Ans. Q\\ hours. 12. Two men 115 miles apart start toward each other, and travel, one at the rate of 2||- miles per hour, and the other at the rate of 3J miles per hour. In how many hours will they meet ? Ans. 18|4 hours. 13. L, M, and N bought a drove of cattle : L paid for f of the drove, M, for -^ of it, and N, for the remainder. It was found that L paid $640 more than N. What did each pay, and what was the cost of the drove ? Ans. L, $3840 ; M and N, each 13200 ; $10240. 14. How many barrels of apples at $9 per barrel will pay for 95 cords of wood at $5J per cord ? Ans. 55^^. 15. A gentleman purchased a farm for $9000; he paid $95 for fencing and other improvements, and then sold it for $381^ less than he had expended. What did he receive for it ? Ans. $8713.75. '\Q. If a family of 5 persons consume f of a barrel of flour in one month, in how many months will they consume 9.1 barrels .? Ans. IZ^. 162 fractio:n^s. 17. If a man earns $112 per month and spends 177 in the same time, how long will it take him to save $941 ? Ans. 26-|^ months. 18. There are two outlets to a cistern ; by one 15 gdftlons, and by the other 20 gallons, can be emptied in one minute. How many gallons may be emptied by both in 17f min- utes? 19. The product of three numbers is 79. The first num- ber is 7, the second is 9|. What is the third ? Ans. l^J^. 185. Converse deductions. We have already learned that Fractions with reference to the mode ofexpremng them are either called ** Decimals" or ** Fractions.'* In the decimal form, the denomination, or fractional unit, is indicated by the position of the decimal point (119) ; but, in the fractional forin, the denomination, or fractional unit, is ex- pressed by the denominator' (152). We shall now proceed to learn how to change Decimals to Frac- tions and Fractions to Decimals under the heading Co^iverse He- duction. Example 1. — Reduce .875 to a common fraction. SOLUTION. Explanation. — .875 is read 875 thousandfJis. 375 7 Therefore our 72wwer«^c»r is 875, and our (fenow- .875 = TT^QQ = Q inator, 1000, and the fraction, yH^, which re- duced to its lowest terms by Rule (178) =|, required fractional form. Example 2.— Reduce .03-^ to a fraction. SOLUTION. Explanation. — We 31 10 10 1 write the number with- .03^ =^^— "3" -i- 10^ = goQ = 3Q out the decimal point, and express its fractional unit by the known denominator 100, which gives us the complex frac- tion ^~, which reduced to a simple fraction by Rule for Division of lUU Fractions = ^, required fractional fonn. CONVERSE REDUCTIONS. 163 Hence, to reduce a decimal to a fraction, we have the following EuLE. — Write the given numler of decimal units. Omit the decimal point, and express the fractional unit hy a de- nominator. Reduce the resulting fraction to its simplest form. Example 3.— Keduce | to a decimal. SOLUTION. 1st Explanation. — Annexing three ciphers to 7, 7 8 ) 7.000 reduces it to thousandths, and | of 7000 thousandths 8 ^^ 875 ~ ^'^^ thousandths or .875. 2d Explanation. — Since f expresses the quotient of 7 divided by 8, we annex decimal ciphers to 7, and divide by 8, as in Division of Decimals. Hence, to reduce a fraction to a dechnal, we have the fol- lowing KuLE. — Annex a decimal cipher or ciphers to the numera- tor and divide by the denominator. PROBLEMS, 1. Keduce .625 to a fraction. 3. Reduce .75 to a fraction. 5. What fraction = .4375 ? 7. What fraction = .008 ? 9. Reduce .28125 to a fraction. 11. Reduce .0375 to a fraction. 13. What fraction = .56 ? 15. What fraction = .096 ? 2. Reduce | to a decimal. 4. Reduce f to a decimal. 6. What decimal = T^g. ? 8. What decimal = T^ ? 10. Reduce -^ to a decimal. 12. Reduce ^% to a decimal. 14. What decimal = ^ ? 16. What decimal = ■^? 17. Reduce ^ to a decimal. Anf^. .3-J^or.33-J-or .333^ &c. Sometimes the decimal is interminable, and in such cases a fraction may be written after the decimal figures, the quotient being carried to any desired number of decimal places ; or the sign + may be written after the decimal figures, to show that the divisor is not exact. Decimal figures which continually repeat, are called a Repetend, 164 FRACTIONS. The value of a repetend is expressed by a fraction whose numer- ator is the repeating figures and whose denominator is as many nines as there are figures in the repetend. Thus, i = .333 &c. = f II = i ; I = .66 &c. = ff = | ; &c., &c. That this is correct is shown from the fact that \ produces .1111 &c.; therefore, f will produce .222 &c. ; |, .33 &c.; |, .555 &c. In like manner ^^ will produce .0101 &;c.; therefore, ^^ will produce .02 &c.,' &c., &c. 18. Eeduce the repetend .384615 to a fraction. 19. Reduce ^^ to a decimal. 20. Reduce the repetend .238095 to a fraction. 21. Reduce ^ to a decimal. 22. Reduce the repetend .142857 to a fraction. 23. Reduce ^ to a decimal. 24. Reduce the repetend .428571 to a fraction. 25. Reduce f to a decimal. 186. Multiplication and Division by Ali^ quot l^ai^ts. In common oral transactions, resort is often had to short methods of calculations, which in many, though not by any means in all cases, facilitate calculations. We shall learn these short processes under the heading Multijill' cation and Division by Aliquot I*arts, An Aliquot l^art of a number is any exact divisor of it. Thus, 6 is i of 12 ; 3 is i of 12 ; &c. Tlie Unit of an Aliquot Part is that number which is divided to obtain the part. The Aliquot Parts of $1 are as follows : 5 cents = ^V of $1. 16| cents = i of $1. ^ 61 " = A « " 20 " =i " " The Unit of the 81 " =tV '' " 25 " =i •' " > Aliquot Part LO " = tV " " 331- .* :^ ^ « u is 1. 3i " =i " " 50 " =i " " J ALIQUOT PABTS. 165 MULTIPLICATION. Example 1. — What cost 144 pencils, at $.06J a piece ? Solution. — Since 6^ cents = -^^ of a dollar, the cost of the pencils is j\ as many dollars as there are pencils. jV of 144 is 9. Therefore, l4i pencils cost $9. PROBLEMS. 1. What cost 32 yards of muslin, at 1.1 2|^ per yard? 2. What cost 36.6 yards sheeting, at 16f cents per yard? 3. What cost 24 yards of alpaca, at 33^ cents per yard ? 4. What cost 28 pounds of butter, at 20 cents per pound ? 5. What cost 64 wooden pails, at 25 cents per pail ? 6. What cost 19 pounds of tobacco, at 50 cents a pound? 7. What cost 120 oranges, at 10 cents each ? 8. What cost 24 slates, at 8-J- cents a piece ? 9. What cost 60 gum erasers, at 5 cents a piece ? 10. What cost 128 bottles of ink, at 6^ cents a bottle? In like manner we may multiply when the unit of the aliquot part is 100, 1000, &c. Example 2.— What cost 6 J shares of mining stock, at $144 per share ? Solution.— At $144 per share 100 shares cost $14400, but since 6^ is tV of 100, 6^ shares cost ^V of $14400, or $900. 11. What cost 12^ yards of alpaca, at 32 cents per yard ? 12. What cost 16| pounds of Java coffee, at 36.6 cents per pound ? 13. What cost 33| pounds of coffee, at 24 cents per pound ? 14. What cost 20 pounds of butter, at 28 cents per pound? 15. What cost 25 wooden pails, at 64 cents apiece ? 16. What cost 50 papers of tobacco, at 19 cents per paper? 166 FRACTIONS. 17. What cost 10 pairs of gloves, at $1.20 a pair? 18. What cost 8-J- pounds spices, at 24 cents per pound ? 19. What cost 5 quires paper, at 60 cents a quire ? 20. What cost 6^ gallons berries, at 11.20 per gallon ? DIVISIOIST. In a somewhat similar manner, we may perform certain examples in division. Example. — At 6J cents apiece, how many pencils can be bought for $9 ? Solution. — Since Q^ cents = ^^ of $1, 16 pencils could be bought for $1 ; and for $9, 9 times 16 pencils, or 144 pencils, could be bought. PM OBL EMS. 1. At 12-J^ cents per yard, how many yards of muslin can be had for $4? 2. At 16| cents per yard, how many yards of sheeting can be purchased for $6.10 ? 3. At 33-J^ cents per yard, how many yards of alpaca may be purchased with $8 ? 4 At 20 cents per pound, how many pounds of butter can I buy with $5.60? 5. For $16, how many wooden pails can I buy, at 25 cents a pail ? 6. For 19.50, how many pounds of tobacco, at 60 cents per pound, can I purchase ? 7. If I pay $9 for 144 pencils, how much is that apiece ? 8. The cost of 32 yards of muslin is $4 ; how much is the cost per yard ? 9. If I pay $8 for 24 yards of alpaca, how much is that per yard ? BILLS. 167 187. BILLS, A Sill is a detailed written statement of items, their prices, &c., which one party, called the Creditor, renders to another party, called the Dd)ior, The party purchasing, or owing, is the Debtor , and the party from whom he purchases, or whom he owes, is the Creditor. When the biL is paid, the person receiving payment writes at the bottom of the bill " Rec'd Payment," or " Paid," and signs the name or the names of the party, or parties, for whom he is acting, as an acknowledgment of the bill having been paid. The bill is then said to be receipted. EXERCISES. FIRST. Pittsburgh, Nov. 21, 1876. Messrs. B. H. Hlnchman & Sons, Bought of Pennsylvania Lead Co. 40 Kegs White Lead 100s. 4000 20 " " " 50s. 1000 80 " " « 25s. 2000 7000 lb.,@9|/. $691 25 Mr. Wm. Unger, SECOND. Pittsburgh, April 1, 1876. Bought of W. H. McClintock & Co. 54 yards Ingrain Carpet, @, $1.25 25 " Venetian Border, " $1. 68 " Carpet Lining, "12^^. Making and Laying 86 yards Ingrain and Border, @ 10!^. 20 yards Oil Cloth, @, 90<^. 15 " Binding, " 5^. Rec'd Payment, ~ W. H. McClintock & Co., Per James Sater. Jan. 28 it tt 29 *t 31 $67 50 25 8 50 8 60 18 75 $128 35 168 FKACTIOKS. Calculating by aliquot parts, the derk will carry out all the amounts of the Bill mentally. Mr. SmoN Martin, THIRD. Pittsburgh, April 1, 1876. To Murphy & Hamilton, Carpenters. JDr. Feb. 3 « « Mar. 5 " " u 31 t( « 120 feet of Flooring, @ 3/'. 3 days' time of Carpenter, @ $3 50. 1 day's do. do. " 3.50. Repairing Step Ladder, 250 feet Flooring, @, 3^^. 4 days' time of Carpenter, @ $3.50. Rec'd Payment, 50 $40 85 Mr. Martin Searle, Murphy & Hamilton. rOURTH. New York, May 3, 1876. To New York Tribune, Dr. Apr. 9 To Adv. 25 lines, 3 times, @ 25f . $ << 10 " " 3 lines, Local, @ $1. K 25 " " 14 '* @37|^. <( ayf or the interval of time between two successive passages of the sun 07er the same meridian ; but, as these intervals are of unequal length, the True Unit of Measure is a mean of those intervals, and is called a Mean Solar Day ; and is 24 hours long. 227. A Solar Year is the time in which the earth makes one revolution round the sun. 199 200 DE NOMINATE NUMBERS. 228. A Civil Day is in length the same as a mean solar day ; that is, 24 hours, and in most countries begins and ends at midnight. TABLE. 60 seconds (sec.) = 1 minute .... min. 60 minutes = 1 hour hr. 24 hours = 1 day da. 7 days = 1 week wk. 12 calendar months = 1 year yr. 365 days (or 52 wk. 1 da.) =1 common year. 366 days (or 52 wk. 2 da.) =1 leap year. 365 da. 5 hr. 48 min. 49.7 sec. = 1 solar year. 100 years = 1 century. By an inspection of the Table the pupil will see that a solar year is 5 hr. 48 min. 49.7 sec. longer than a common year ; and that 4 solar years are = 4 common years + 23 hr. 15 min. 18.8. sec. To make the Solar and Common year correspond, 1 day is added to every fourth common year, or to every year exactly divisible by 4 ; and that year ip called a leap year. But by adding 1 day every 4 years, too much is added by the difference between 24 hrs. and 23 hr. 15 min, 18.8 sec, or at the rate of 11 min. 10.3 sec. per year too much. In 100 yr , 18 hr. 37 min. 10 sec. too much will have been added. Hence every 100th year is called a common year ; but this would be dropping too much by the difference between 24 hr. and 18 hr. 37 min. 10 sec, or 5 hr. 23 min. 50 sec. every hundred years. By calling every 400th year a leap year, it would take 3840 common years to equal 3840 solar years and 1 day. Hence, 229. Every year divisible by Jf is a leap year, except those that have 100 and not JfiO for an exact divisor. The names and order of the calendar months are, Janu- ary, 1st mo. ; February, 2d mo. ; March, 3d mo. ; April, 4th mo. ; May, 5th mo. ; June, 6th mo. ; July, 7th mo. ; August, 8th mo. ; September, 9th mo. : October, 10th mo. ; Novem- ber, 11th mo.; December, 12th mo. TABLES. 201 January, March, May, July, August, October, and DecemTier have 31 days each. April, June, September, and November have 30 days each. February has 28 days, except in leap-year when it has 29. These numbers may be remembered by the aid of the following lines : Thirty days have September, April, June, and November, And all the rest have thirty-one. Save February, which alone Has twenty -eight ; and we assi^ To this, in leap-year, twenty-nine. MISCJELLAWIJOUS TABLES. 230. Counting, 12 things make 1 dozen. 12 dozBD, or 144 things, " 1 gross. 12 gross, or 1728 things, " 1 great gross. 2 things " 1 pair. 6 things " 1 set. 20 things " 1 score. 231. Paper. 24 sheets make 1 quire. 20 quires, or 480 sheets, " 1 ream. 2 reams " 1 bundle. 5 bundles « 1 bale. 232. Books. A folio book (fo) is made of sheets folded in 2 leaves. A quarto (4to) « " " " " " 4 (( An octavo (8vo) " " " " " " 8 a A duodecimo (12mo) " " " " « " 12 « Anl8mo " " « " " " 18 a A24mo « " " « 24 « A32mo " " '' " " " 32 ti The tervaa fdio, quarto, oetaeo, etc., denote the number of leaves into which a sheet of paper is folded in making books. 20$ DEI^OMINATE NUMBERS. 333. Copying. 72 words inake 1 folio, or sheet of common law. 90 ** " 1 " " *' " chancery. 234. Shoemakers^ Measure, Small Sizes.— No. 1. ^ inches. No. 2. 44 " 4. J = 44i in. No. 3. 4i " 4. i + J =: 4il in. &c., &c., &c. Large Sizes.— No. 1. %^ inches. No. 2. 8iJ " + i = m in. No. 3. 8ii " + J + J = 9i in. &c., &c., &c. t'-ii-^^^^^^ll^^m.^^S^ niS!ll-TJCT;IC|.lM i4 ^ 235. deduction of Denominate Numhers consists in changing their name without altering their value. MONEY. Oi«al iE.V Example 1.— How many dollars in 3 E. ? 5 E.? 10 E. ? Solution.— Since in 1 E. there are 10 dollars, in 3 E. there are 3 times 10 doUara = $30 ; &c., &c., &c BEDUCTION. 203 ritO BZJEMS, 1. How many dollars in 6 D. E. ? 20 D. E. ? 30 D. E. ? 2. How many farthings in 3 pence ? 2s. ? 3. How many pence in 3s. ? 12s. ? 4£ ? 4. How many s. in 6g.? 8 cr.? 16 florins? 5. How many millimes in 2 cL? 5 dc. ? 10 fr. ? 6. How many francs in 1 Napoleon ? 3 ? 5 ? 23 ? 7. How many c. in $1, Canada currency? In 110 ? In $20? 8. How many pfennigs in 1 mark ? In 10 marks ? In 24? In the foregoing problems we have increased the number of parts, diminished the size of each part and changed the name without chang- ing the value. This is called reducing to lower denominations, or Meduction Descending , which is performed bj multiplication. Example 2. — In 30 dollars how many E.? In $50 ? In 1100? Solution.— Since in $10 there is 1 E., in $30 there are 8 E. ; &c., &c PROBLEMS. 9. In $120 how many D. E. ? In $400 ? In $600 ? 10. In 14:4: far. how many pence? How many shillings? 11. In 2005. how many £ ? How many crowns ? 12. In 126s. how many g. ? How many florins ? 13. In 2000?w. how many centimes ? decimes ? francs ? 14. How many Napoleons in 20 francs ? In 200 ? In 500 ? In 280? In 640? 15. How many marks in 100 pfennigs ? In 1000 ? In 2400 ? In 2100 ? In 10000 ? In the last 7 problems we have diminished the number of parts, increased the size of each part and changed the name, without chang- ing the value of the expression. This is called reducing to a higher denomination, or Meduction Ascending, which is performed by division. d04 DENOMINATE NUMBERS, MONEY. ^IWiitton^Xorii.Hf^s f 236. Simple or Compound Denom^inate Numbers to Lower Denominate Numbers. Example 3.— Reduce £140 to farthings. SOLUTION. 205. 140 2800s. nd. 2800 336006?. A: far, 33600 134400 /an Example 4.- 80LUTI0N. 20s. 5 100s. + 8s. 108 1296^. + 5•- t Wi itteri ^Xfer c i^eX Reduce 210. 8 bu. 3 pk. to pt. 211. 9 bu. 1 pk. 5 qt. to pt. 212. 4 bu. 3 pk. 2 qt. 1 pt. to pt. 213. 3 qt. 1 pt. 3 gi. to gi. 214. 40 gal. 3 qt. 2 gi. to gi. 215. 5 bu. 7 qt. 1 pt. to pt. 216. 5 cong. to f 3 . 217. 8fl to f^L. 226. If 5 bushels of wheat many barrels can be made 4003 qt. ? Out of 8000 pt. ? Eeduce 218. 560 pt. to bu. 219. 602 pt. to bu. 220. 309 pt. to bu. 221. 31 gi. to qt. 222. 1306 gi. to gaL 223. 335 pt. to bu. 224. 640 f 3 to cong. 225. 3840 ni to f S . make a barrel of flour, how out of 1000 gal.? Out of 2U DEIS^OMINATE NUMBEBS. CIBCVLAB MEASURE. :i^1>Yi^ittei\^xrer c i^eX Eeduce 227. 45° 28' 55" to seconds. 228. 73° 58' to seconds. 229. 89° 20" to seconds. 230. 131° 3' 29" to seconds. Reduce 231. 163735" to degrees. 232. 266280" to degrees. 233. 320420" to degrees. 234. 471809" to degrees. TIME MEASURE. ^ ©i^at ^XGitciXeX 235. How many sec. in 3 min. ? 5 min. ? 7 min. ? 236. How many min. in 2 hr. ? 3 hr. ? 5 J hr.? 237. How many lir. in 3 da. ? 5 da. ? 7 da. ? 24 min. ? 238. How many da. in 3 wk. ? 10 wk. ? 48 hr. ? 96 hr. ? 239. How many hr. in 6 da. ? 8 da. ? 15 da. ? 1 wk. ? TIME 3IEASUBE. ^ ^ Wi ittcn ^Xeivi^l^^ Reduce 240. 15 hr. 25 min. to min. 241. 43 min. 54 sec. to sec. 242. 25 wk. 12 hr. to hr. 243. 4 da. 5 hr. 6 min. to min. 244. 1 cora. yr. to sec. 245. 1 solar yr. to sec. Reduce 246. 925 min. to hr. 247. 2634 sec. to min. 248. 4212 hr. to wk. 249. 6066 min. to da. 250. 31536000 sec. to da. 251. 31556929^!^ sec. to da. RED CrCTION 215 252. How many days from June 20 to Dec. 16 ? Ans. 179 da. 253. How many days from Jan. 16 to Sept. 24 (1876) ? Ans. 252 da. MISCELLANEOUS TABLES. 0i«al^\fei'.ciYi^ittei\^7cferci^eX'{; Example. — What cost 5 yards broadcloth, at £1 10s. 6dc per yard ? SOLUTION. £ s. d. 1 10 6 5 7 12 6 Explanation. — Since 1 yd. costs £1 10s. 6d., 5 yd. will cost 5 times as much. 5 times 6d. = 80d. = 2s. 6d. Write 6 in the pence place of the product, and reserve the 2s. to be added to the product of the shillings. 5 times 10s. = 50s., to which add the 2s., making 52s. = £2 12s. Write 12 in the shillings' place of the product, reserving the £2 to add to the product of pounds. 5 times £1 = £5, to which add the £2 reserved, making £7. The entire product, therefore, is £7 12s. 6d. EuLE. — Multiply the number represe7iting the denomina- tion at the right by the multiplier. Divide the product by the number of this denomination required to make 1 of the next higher. Write the remainder, if any, ifi the product under the number multiplied, and add the quotient to the product of the next higher de7iomination. Proceed in this manner till all the denominations have been multiplied. The result will be the answer sought. MULTIPLICATION. 239 PJtOBZTlMS. (1-) (2.) mi. fur. rd. ft. in. mi.fur. cli. p. 1. 5 7 13 10 8 2 6 7 3 15 5 4 29 4 28 3(i) 4 11 3 1 2 10 a)=6 29 4 28 3 10 (3.) (4.) A. R. P. ft. in. A. R. P. yd. ft. in. 8 2 16 138 100 7 3 27 15 6 50 2 3 17 33 4(J) 56 (|-)=108 17 33 5 20 23 3 2 16} 1 6 i=6 108 23 3 2 16 7 114 (5.) (6.) (7.) ou.yd.cu.ft. cu.in. 6 20 435 cd.cd.ft. 5 7 cu.ft. 13 cd. cu. ft. cu. in. 2 52 360 4 59 6 10 2 5 23 12 12 5 72 (8.) lb. oz.pwt.gr. 6 7 8 9 C. 0. 4 (9.) tl il 13 6 lb. 1 (10.) ! 3 3 gr. 10 7 1 10 6 9 7 39 8 10 6 5 3 11 6 13 4 4 1 10 11. What is the weight of 8 bales of hay, each weighing 5 cwt. 6 lb. ? Ans, 2 T. 48 lb. 240 DETS^OMINATE N^ UMBERS. 12. What is the weight of 10 packages of crockery, aver- aging in English Avoirdupois weight, 7 cwt. 26 lb. 7 oz. 9 dr. ? Ans. 3 T. 12 cwt. 40 lb. 11 oz. 10 dr. 13. Find the contents of 9 bbl., averaging 40 gal. 1 qt 1 pt. 1 gi. Ans. 363 gal. 2 qt. 1 pt. 1 gi. 14. How much castor-oil in 50 bottles averaging 3 f ^ 2 f 3 ? Ans. 1 Cong. 2 0. 2 f 1 4 f 3 . 15. How much wheat in 19 sacks, averaging 1 bu. 3 pk. 7 qt. 1 pt.? Ans. 37 bu. 2 pk. 6 qt. 1 pt. 16. Multiply 75° 45' 25" by 7. Ans. 1 0. 170° 17' 55". 17. Calling a solar year 365 da. 5 hr. 48 min. 50 sec, how much time is there in 18 solar years? Ans. 6574 da. 8 hr. 39 min. 18. How much molasses in 25 casks, if each cask contains 61 gal. 1 qt. 1 pt. ? Ans, 1534 gal. 1 qt. 1 pt. 19. If one silver spoon weighs 3 oz. 15 pwt. 13 gr., what is the weight of two dozen ? Ans. 7 lb. 6 oz. 13 pwt. 20. If a man travels at the rate of 22 m. 7 fur. 32 rd. 4 yd. per day, how far can he travel in 56 days ? Ans. 1286 m. 5 fur. 32 rd. 4 yd. 21. How many bushels of wheat will 12 acres produce, if one acre produces 30 bu. 3 pk. 7 qt. 1 pt. ? Ans. 371 bu. 3 pk. 2 qt. 22. How much cloth will it take for 16 suits of clothes, if it takes 8 yd. If qr. for one suit ? Ans. 135 yd. 23. If one load of wood measures 117 cu ft. 110 cu. in., how much will 40 loads of the same size measure ? Alls. 36 cd. 74 cu. ft. 944 cu. in. 24. If a quarter of beef weighs 216 lb. 7 oz., how mu(*h will 4 quarters weigh ? Ans. 865 lb. 12 oz. 25. If the daily motion of the moon is 13° 10' 35", how much is it in 15 days ? Ans. 197° 38' 45". ^ ^iv I>;l^IS;iO;]^ 245. t^\¥i?itteri ^x:Grci;^eX Example 1.— If 5 yd. of broadcloth cost £7 12s. 6d., what ^osts 1 yard ? 1st solution. £ s. d. I of £7 = 1 8 I of 12s. = 2 4f i of 6d. = H 1 10 6 1st Explanation. — Since 5 yd. cost £7 12s. 6d., 1 yard will cost | as much. ^ of £7 = £lf = £1 8s. ^ of 12s. = 2|s. = 2s. 4fd. iof6d. =lid. £1 8s.+2s. 4fd. + Ud.=£l 10s. 6d. 2d solution. £ s. d. 7 12 6 = £7.625 5 ) £7.62 5 £1.525 = £1 10s. 6d. 2d Explanation. — £7 12s. 6d.= £7.625. Dividing by 5, we have £1.525 = £1 10s. 6d. 3d solution. £ s. d. 7 12 6 = 1830d. 5 ) 1830d. 366d. = £1 10s. 6d. 16 3d Explanation.— £7 12s. 6d.= 1830d. Dividing by 5, we obtain 366d. = £1 10s. 6d. S41 242 DEKOMIHATE I^^UMBEES Example 2.— At £1 10s. 6d. per yd., how many yards of broadcloth can be bought for £7 12s. 6d. ? 1st solution. £ 8. d. £ s. d. 1 10 6)7 12 6(5 7 12 6 1st Explanation. — By inspec- tion we see that £7 12s. 6d. contains £1 10s. 6d. exactly 5 times. 3d solution. £ s. d. £ 7 12 6 = 7.625 1 10 6 = 1.525 £7.625 -r- £1.525 = 5 2d Explanation. — We reduce both dividend and divisor to £ and the decimal of a £, and divide as in division of decimals. 8d solution. £ s. d. d. 7 12 6 = 1830 1 10 6 = 366 1830d. -T- 366d. = 5 3d Explanation. • — We reduce both numbers tod., and then divide as in whole numbers. Hence, To find any required part of a simple or compound denominate number, we have KuLE I. — Find the required part of each simple denom- inate number composing the dividend, reducing these parts, if necessary, to compound denominate numbers. Then add as in addition of denominate numbers. Or, Reduce the dividend to any convenient denominate number, and proceed as in division of integers, or decimals. Then reduce the result, if desirable, to a denominate number. If we cannot by inspection divide one compound denominate number by another we apply KuLE II. — Reduce dividend and divisor to the same denomination, and divide as in integers, decimals, or frac- tions. DiYisioiq^. 243 PJt O B LEMS (1) (^) (3) mi. fur. rd. mi. fur. rd. A. R. P. 5)29 7 8 4)23 52 6 9 ) 35 1 1 2 3 3 28 5 7 33.6 rd. 6)75 (4) yd. 4 ft. in. 2 6 c 9 12 3 1 8 11 lb. )35 (7) oz. pwt. 2 3 10 3 2 7 14 5 19 li 1 (5) cu. yd .ft. in. 9)30 4 9 3 9 769 (6) cd. ft. in. 10 ) 93 12 632 9 39 1100 (8) lb. 1 3 3 gr. 3)11 1 6 1 17 3 8 4 2 12J (9) (10) T. cwt. qr. lb, oz. dr. gal. qt. pt. gi. 5)53 4 1 19 14 12 12 ) 47 1 1 2 10 12 3 13 15 12 3 3 1 2i 11. If 9 bales of hay weigh 2 T. 7 cwt. 25 lb., what is the average weight ? Ans. 5 cwt. 25 lb. 12. If 10 packages of crockery weigh in long ton weight, 3 T. 14 cwt. 3 qr. 12 lb. 11 oz. 10 dr., what is their average weight? Ans. 7 cwt. 1 qr. 26 lb. 7 oz. 9 dr. 13. If 9 barrels hold 363 gal. 3 qt. 1 pt. 2 gi., what is their average capacity ? Ans. 40 gal. 1 qt. 1 pt. 2 gi. 14. If 10 sacks contain 39 bu. 2 pk. 6 qt. of wheat, what is their average capacity ? Ans. 3 bu. 3 pk. 7 qt. 15. How many town lots containing 28 P. 30 yd. each, can be made of 3 A. 2 R 19 P. 25.25 yd. ? Ans. 20. 244 DENOMINATE NUMBERS, OUTLINE OF APPLICATIONS OF SQUARE, CUBIC, AND TIME MEASURES. O M H M -^ SQUARE. CUBIC. 247. Common, 248. Government Lands. 249. Lumber. C 250. Plasterers' Work. 251. Carpenters' Work. 252. Painters' Work. I 253. Paviors' Work. Artificers' Work, 254. Excavations and Embank- ments. Masonry, 255. Stone Work. 256. Brkk Work. ^ 257. Capacities, 258. TIME. APPLICATIOKS. 245 APPLICATIONS OF SQUARE, CUBIC, AND TIME MEASURES. 346. We propose in this chapter to give some such practical applications of Square, Cubic, and Time Measures, as the pupil is now prepared to understand ; omitting for the present, those that involve square root, cule root, &c. These latter will be treated of in another place. 347. SQUARE 31EASUBE. Example 1. — How many sq. yd. in the floor of a rect- angular room 5 yd. long and 4 yd. wide ? Solution. — Since a floor 5 yd. long and 1 yd. wide contains 5 sq. yd., a floor 5 yd. long and 4 yd. wide contains 4 times 5 sq. yd., or 20 sq. yd. Hence to find the area of a rectangular surface (204), we have this Rule. — Multiply the length by the width, expressed in the same denomination. F HOB LEMS. 1. How many sq. in. in a rectangular pane of • glass 16 X 18 in. ? How many sq. ft. ? What part of a sq. yd. ? 2. How many sq. ft. in a box of glass containing 60 panes, each 10 x 12 in. ? Ans. 50 sq. ft. 3. How many perches in a village lot 150 ft. long and 50 ft. wide? 27.55+ P. 4. How many acres in a square farm, eacli of whose sides is 40 rd. Ans. 10 A. 5. How many acres in a square piece of land measuring G miles on each side? Ans. 23040 A. 246 DENOMINATE NUMBERS. Example 2. — If a rectangular pane of glass 18 in. long, has a surface of 288 sq. in., how wide is it ? SOLUTION. Explanation. — Since a pane of glass 18 in. long and 18 ) 288 1 in- wide contains 18 sq. in., to contain 288 sq. in. it will Tn liave to be as many inches wide as 18 sq. in. are con- tained times in 288 sq. in., or 16 inches wida. Hence, to find one dimension of a rectangle (207), when the other dimension is given, we have this EuLE. — Divide the area hy the giveti dimension ; the quo^ tient is the other dime7ision, 6. A town lot contains 2400 sq. ft. It has 20 ft. front. What is its deptfi ? 7. A box of glass contains 50 sq. ft. How many panes 12 X 24 in. ? Ans. 25 panes. 8. How many rods in length is a rectangular farm con- taining 150 acres, if one end is 120 rd. in width ? 9. How many yards of carpet -| yd. wide, will cover a floor 15 feet square ? Ans. 40 yd. 10. A gentleman purchased Brussels carpet | yd. wide, at $1.75 per linear yard, to cover a room 22 feet long. The entire cost of carpet was $106.08f. What was the width of the room ? Ans. 15.5 ft. 11. *How many yards of oil-cloth will be required to cover a hall 25 feet long and 9J feet wide; and two recesses each 6 X 7 ft. ? 12. How much will it cost for blinds for 10 windows, each 7x3 ft., at 62i^ per yd. ? Ans. $14.58 + . 248. Government Zand Measure, 1. If in a Section there are 640 acres of land, how many acres are there in the E. | of S. |^ of Section 10 ? (See Dia- gram, page 191.) APPLICATION'S. 247 2. How many in the S. E. J of S. E. J of Sec. 10 ? 3. How many in the E. J of E. ^ of S. W. J of Sec. 21 ? 4. How many in the W. ^ of E. ^ of S. W. i of Sec. 36 ? 5. A farmer purchased the N. W. J of Sec. 1, Township 6, N., Range G E. of the oth Principal Meridian ; and sold 35 acres of it to his brother. How many acres had he left ? Ans, 125 A. 6. If he paid the usual Government pre-emption price ($1.25 per acre), and sold to his brother at the rate of 13.25 per acre, what did the part retained stand him ? Ans. $86.25. 7. If he afterwards exchanges his property for S. ^ of the S. W. J of Sec. 30, in the same Township, how much per acre does his new land cost him ? Ans. $1,078^. Show on the diagram the position of each purchase. 8. A speculator bought the K i of the E. |- of the S. W. i of Sec. 8 ; he afterwards purchased the N. ^ of the E. ^ of the N. E. i, the S. i of the W. J of the N. E. J, and the W. J of the S. i of the S. W. i of the same section. How much land did he purchase ? Ans. 160 acres. Make a diagram showing the shape of his land. 9. He afterwards made an even exchange of two of the pieces he had purchased, so that his property was in the form of a single square. Show in how many ways this may be done. 349. Lumber Measure, Lumber is reckoned by ioard measure; that is, it is regarded as cut into boards one inch thick. The unit of board measure is a square foot, one inch thick. If a piece of lumber 1 inch thick, contains a certain number of feet, a piece with the same dimensions ttvo inches thick, contains 248 DEKOMIKATE NUMBERS. twice as many feet ; three inches thick, three times as many feet, &c. The average width of a board is half the sum of the width of the ends. Example 1. — How many sq. ft. in a board 18 ft. long, 18 in. wide, and 1 in. thick ? Explanation. — We reduce the width SOLUTION. 18 in. to 1.5 ft. by dividing by 13; then 18 in. =: 1.5 linear ft. multiply 1.5 by J 8, and thus obtain 27, to -tn a a which we give the name sq. ft. As the board is 1 in. thick, the surface is the pro- 27 sq. ft. duct of the length and breadth, the true result being 27 sq. ft. Example 2. — How many feet in a plank 18 ft. long, 18 in. wide, and 1^ in. thick ? SOLUTION. Explanation. — We proceed, as .14 X 18 X IJ = 40|- sq. ft. in the last example, to find the area. Since the plank is 1^ times as thick as the board, it must contain 1\ times as many feet 1^ times 27=40|. Rule. — Multiply together the length and breadth (or aver- age Ireadth) in feet ; and this product ty the thickness in inches. PIIOBI.EMS, 1. How many feet in an inch board 10 ft. long and 15 in. broad ? Ans. l^ ft. 2. How many feet in a three-inch plank 10 ft. long and 15 in. broad ? Ans. 37^ ft. 3. In a log 12 ft. long and 15 in. square? Ayis. 225 ft. 4. In a joist 20 ft. long, 10 in. broad, and 3 in. thick? Ans. 50 ft. 5. In a scantling 16 ft. long and 4 in. square ? Ans. 21ift. APPLICATIONS. ^ 249 6. In a beam 30 ft. long, 8 in. broad, and 6 in. thick ? Ans. 120 ft. 7. What is the value, at 2J^ per foot, of a log 25 ft. long, 18 in. broad at one end, 14 at the other, and uniformly 14 in. thick? Ans. $11.67. 8. What is the value of a plank 30 ft. long, 2 ft. wide, H in. thick, at 3^^ per foot, board measure ? Ans. $3.15. 9. What cost 37 joists, each 20 ft. long, 9 in. broad, and 2^ in. thick, at $18 per 1000 feet ? Ans. $24.98. One thousand shingles 4 in. wide are estimated to cover one square of 100 sq. ft. 10. How many shingles will be required to cover a roof, the one side of which is 40 by 20 ft., and the other 40 by 18 ft. ? Ans. 15i M. ARTIFICERS' WORK. 250. Plasterers^ Work, Plastering is done by the square yard. Rules for Measuring. I. — Girt {measure) the walls and multiply hy the height from floor to ceiling. Note 1. — Deduct the half of all openings over two feet wide. 2. Measure all circular walls and ceilings twice. 3. Corners of chimneys and other external angles are charged extra if made with " gauged " mortar, otherwise they are included in the common measurement. 4. Stucco work is done by the square foot, when more than 13 in. wide. II. — Girt the cornice and multiply hy the length, 5. One foot in length of the cornice is added for each mitre oi' "return." Example 1. — At 30^ per yd., what will it cost to plaster a room 16^ ft. long, 13^ ft. broad, and 10 ft. high, having a 250 DEKOMIN^ATE NUMBERS. door 7J ft. by 3 ft., two windows each 6 ft. by 3, and a mantel 5 ft. by 4 ? SOLUTION. 16| + 16| + 13| + 13| = 60, the measurement around the room. 60 X 10 = 600, the number of sq. ft in the walls. 16^ X 131 = 222|, " " " " " •' ceiling. 822|, " " *• " « " room. 7^ X 3 = 22|, the number of sq. ft. in the door. 2 X 6 X 3 = 36 , " " " .< « « two windows. 5x4 = 20_, " " " " " " mantel. 78^, *' " " " " " openings. 78^ H- 2 = 391, the number of sq. ft. to be deducted (Note 1). 822f - 39^ = 783i sq. ft. = 87tV sq. yd. 87yV sq. yd, @ 30;^ = $26.12, Ans. Example 2. — What cost cornice 15 in. wide, about the same room, supposing the only mitres to be those at the corners, at 10^ per foot ? SOLUTION. Girth of room, 60 ft. ; 4 mitres, 4 ft. ; 60 ft. + 4 ft. = 64 ft. ; 15 in. = 1\ ft. ; 64 X 1^ = 80 sq. ft. 80 sq. ft. @, lOj^ = $8.00, Ans. PROBZEM8. 1. What cost the plastering of a circular wall 63 ft. in length and 13 ft. higb, at 25)* per yd. ? (See Note 2.) Ans. $45.50. 2. How many yd. of plastering will cover a ceiling 20 x 30 ft. ; and what will it cost at 15f per sq. yd. ? Ans. GQ^ yd. ; $10. 3. What cost the plastering of the walls and ceiling of a room 12 ft. 11 in. square and 9 ft. 6 in. high, at 20^ per sq. yd.? Ans, $14.62 +. APPLICATIONS. 251 4. What cost the cornice of a room 24 x 16 ft.; the room having two chimneys, breasts 1 ft. thick, and forming 4 extra mitres and four "returns; " the width of the cornice as measured by the line following all the indentations being 16 inches ; at 12J^^ per foot ? Ans. $16. 5. What cost the plastering of a house of 8 rooms, 4 of which are 18 x 15 ft. and 8^ ft. high ; 4 are 14 x 15 ft. and 9^ in height, and a hall 30 x 8 ft. and 9^ ft., allowance being made for 16 doors, 7^ x 3, and 18 windows 6^ x 3, at 15;^ per sq. yd. ? ^ws. $116.24. 251. Carpenters^ and Joiners^ Work. This work is done by the lineal or square foot, or by the square of 100 sq. ft. Framing the larger timbers used in building, such as sills, posts, and vertical timbers, principal rafters, &c., framing hips and valleys, chamfering posts and girders, raising plates, bond timbers, bridging joists, making mouldings, architraves, &c., are counted by the lineal foot. Planing posts and girders, making window-shutters, doors, wain- scoting, pews, shelving (sometimes lineal), ceilings, lattice-worR, cornice, &c., are reckoned by the sqitnre foot. Framing joists, small rafters, studding, putting on weather-boarding, shingling, sheathing, laying floors, making partitions, board fences, &c. , are charged for by the square. Note 1.— In measuring weather-boarding, deduct all openings over 4 ft. wide. 2. Window-sashes are estimated by the light. PBOBZEMS. 1. At $5 a square, what will it cost to shingle a roof 45 ft. long, each of its two slopes being 20 ft. broad ? Ans. $90. 252 DENOMINATE NUMBEES. 2. At 50^ a sq. yd., what costs a wainscot 2 J ft. high and 54 feet in length ? ^7i5. $7.50. 3. What will a floor 33 ft. long and 17 ft. broad cost, at 13.50 a square ? Ans. $19M. 4. What cost the framing of 4 sills, 6x8 in. and 20 ft. long, @ Sf per lineal foot ? Ans. $6.40. 5. What cost the making of 8 doors, 8x3 ft., 1 J in. thick, 6 panels, at 16^ per sq. ft.? Ans, 8 6. What cost the making of 20 shutters each 18 in. by 6 ft., at 15^ per superficial foot ? Ans. 127. 7. What will be the cost of a gravel roof 20 x 30 ft., at $5.50 per square ? Ans. $ 8. What costs the same roof of shingles @ $9.70. Of slate @ $11. Of tin at $16 per square ? A7is. to last, $96. 9. A square of slate roofing weighs 600 lb. What will be the weight of slate on the roof in the last problem ? Ans. 3600 lbs. 352. Painters^ Work, I^riming and glaring are done by the light. • Painting is done either by the lineal foot, or square yard, and is estimated as any other surface measure, hy the jjro- duct of the length and Ireadth, The principal thing is to know how to obtain these two dimensions; and for this purpose we give the following Directions for Measuring, In weather-boarding, measure not only the width of the board, but also the width of the exposed edge. For beaded partition boards, add to the width of the boards | inch for each bead. These measure- ments multiplied by the length of the boards give the area painted. For paling fences, measure the height, and to this add the width of each rail, to obtain one dimension, breadth. The other dimension, APPLICATIONS. 253 ^, is found by adding to the length of the fence, twice the width of each post, if rectangular, or one-half the girth, if cylindrical. [The measure of width of rails and posts is made at right angles with the fence.] To measure panel doors, shutters, &c., &c., find the length and breadth by pressing the measuring line into all the quirks and mouldings. For all weather- boarding and wall painting, take out half the openings. For shingle or board roofs, measure the butts or edges ; trellis or lattice work to be measured double, and Venetian shutters add one-half. To measure pilasters, commence at the wall and measure round to the bead at the end of the jamb casing, pressing the line into all the quirks. This gives one dimension ; the other is, of course, the length. Plain rosettes, in pilasters, add 1 foot to the length ; carved rosettes, in pilasters, add 3 feet to the length. Window jambs, cornice, &c., &c., are measured substantially in the same way, the object being to obtain the surface painted, whether it be plain or ornamental. mOBTj E MS , 1. What will the glazing cost in a house containing 17 windows of 6 lights each, at 25^ a light ? 2. What will it cost to paint both sides of a paling fence. 18 rd. long, and 5 feet high, the posts 4 x 4 in. and 2 to the rod, and rails 2 x 4in. at 21^ per sq. yd.? Ans. I85.06 + . 3. What cost the painting of 4 ft. high wainscoting of 6 in. wide beaded boards, of a room 15 x 17 ft., at 20^ per sq. yd.? Ans. $6.16 + . 253. Paviors^ Work. PliOBLEMS, 1. What will be the cost of flagging a walk 100 ft. long by 9 ft. wide, at 21f per sq. ft. ? Ans. $243. 2. At $6 per square, what will it cost to pave a yard 150 ft. long and 100 ft. broad ? Ans. $900. 254 DEN^OMIJS^ATE NUMBERS. 3. What will it cost to pave a street 30 ft. wide and 1560 ft. long, with Nicholson pavement, @ $.30 per sq. ft.? ^ Ans. $ 4. What cost the paving of an avenue 100 ft. wide and 3150 long with the Miller pavement at 44^ per sq. yd. ? Ans. $15400. CUBIC MUASUBE. 254. JExcavations and Enibankments, Excavations and Enihanknients of all kinds are reckoned by the cubic yard. Note. — Trenches for foimJations are counted double. Example. — How many yards of excavation in a cellar 40 ft. by 25 ft. and ^ ft. deep ? Solution. — Since the solid contents of a rectangular body are found by multiplying together its three dimensions (length, breadth, and thickness), we multiply 40 by 25, and this product by 4J^, obtain- ing as a result 4500 cu. ft. Dividing this by 27, the number of cu. ft. in 1 cu. yd., we have 166|, the number of cu. yd. of excavation. rjt OB1.EMS, 1. What cost the excavation of a trench for a foundation for a stone fence 500 ft. long, 2 ft. wide, and 1^ ft. deep, at 27/ per cu. yd. ? (See Note.) Ans. $30. 2. How many loads of 1 cu. yd. each will be required to fill a street 150 ft. long, 50 ft. wide, and 2^ ft. deep; and how much will the whole cost, at 18/ per cu. yd. ? Ans. 694f yds. ; $125. 3. A railroad company excavated a tunnel 1500 feet long, with a cross-section of 450 square feet. How many yards of earth were removed ? Ans. 25000 cu. yd. APPLICATIONS. 255 4. What costs the excavation for a cellar 5 ft. deep, for a dwelling house, the main building being 40 x 30 ft., and the L 18 X 14 ft., at 50)^ per cu. yd. ? Ans. $134.44. 5. What costs the excavation for a cellar under the main building of same dwelling-house, and an excavation H ft. deep and 1^^ ft. wide for the walls of the L (one of the short sides being attached to the main building), at 50^ per cu. yd. ? Ans, $115.03. MASOJVBT. 255. Stone Work. Stone Worh is done by the perch of 24| cu. ft. DIRECTIONS FOR MEASURING. Rough Stone. 1. Measure the outside girt of a wall. This measures the comers twice ; but as a corner is more diflBcult to build, this is a method of allowing for the extra work. 2. Add 9 in. in chimney breasts, pillars, &c., for each dressed and plambed corner. 3. Count openings less than 3 ft. wide as built up. Cut Stone. 1. In cut stone the measuring line follows the chisel, and the work is measured on the surface by the square foot. 2. Belt courses less than 1 ft. wide are measured by the linear foot. 3. Moulded and chamfered work and the whole face of the rise in moulded steps, is measured twice. 4. Circular moulded hoods, pediments, and lintels are measured twice. Example. — What is the cost of a foundation 36 x 20 ft, 4 ft. deep, and H ft. thick, built of rubble work with 11 in. belt course of cut stone, at $3.63 per perch for the rubble, and 25^ per foot lineal, for the belt course? 256 DENOMINATE NUMBERS. SOLUTION. EXPLANATIOX.— Mul- 36 4- 20 4- 36 + 20 = 112 ft., girth. tiplying the girth by the 112 X 4 X li =r 672 cu. ft. ^®P*^' ^°<$^?(i 359. HatiOf in Arithmetic, is the relation which ono number bears to another of the same kind. This relation is found by dividing one of the numbers by the other. The quotient is also the value of the ratio. The divisor may be regarded as the measure of the number divided. Example 1. — What is the relation of 5 to 25 ? Ans. 5 is -J- of 25. Example 2.— What is the relation of 18 to 6 ? Ans. 18 is 3 times 6. In Ex. 1, we notice that 25 is the measure of 5, and— is the value of the ratio. In Ex. 3, 6 is the measure of 18, and 3 the value of the ratio. 260. Any number and its measure, or any two numbers that are compared, are the Terms of a ratio. They are also a couplet. The Terms are named from the order of their position, Antecedent (going before), and Consequent (following). 261. The Antecedent is the first term, or the numler to he divided, or measured. 262. The Consequent is the second term, or the divisor, or 7neasure. RATIO AXD PROPORTIOiq^. 269 Katio is indicated in two ways, viz. : 363. First. — By a fraction, whose numerator is the antecedent, and whose denominator is the consequent. 264. Second. — By writing the consequent after the antecedent, with a colon between them. Thus, the ratio of 4 to 5 is written either f , or 4 : 5, and is read, " four is to five." We may change the former ex- pression into the latter, or tlie reverse. The terms of a ratio must not only express the same kind of quan- tity, but also the same denomination. Thus, the ratio of 3 quarts to 2 pecks is not | ; but if we reduce the 3 pk. to qt., the ration is y\. Oral'OExfGBciXGX Example. — What is the relation of 4 to 5 ? Solution. — Since we find the relation oi one number to another by dividing the former by the latter (259), the yeiAtiou of 4 to 5 is f ; or, 4 -r- 5 ; or, 4 : 5. PRO li L E3tS. 1. What is the relation of 3 to 6 ? 5 to 8 ? 7 to 9 ? 2. What is the -atio of 15 to 3 ? 8 to 6 ? 12 to 9 ? 3. 11 to 12? 7. 6^ to 121 ? 11. $20 to $5 ? 4. 13 to 26 ? 8. 2.5 to .5 ? 12. £16 to £4? 5. .9 to .27 ? 9. 100 to .25? 13. 10s. to 100s. ? 6. 50 to 25 ? 10. 20 to 25 ? 14. 16d. to 32d. ? Since the ratio of two numbers is expressed by a fraction whose numerator is the antecedent, and whose denominator is the consequent, it follows that 270 RATIO AKD PEOPORTION. 265. First. — Dividing the antecedent, or multiplying the consequent, divides the value of the ratio (163). 266. Secon"D. — Multiplying the antecedent or dividing the consequent, multiplies the ratio (161). 267. Third. — Multiplying or dividing both terms of the ratio by the same number does not affect the value of the ratio (163). In reference to the number of its terms, a ratio is either Simple or Compound. 268. A Simple Hatio is one that has only t^ o terms ; as4: 5; 6: 8; H; ft; &c. 269. A Compound Ratio is one which has two or more pairs of terms. Thus |xj; ^xfix4; (2:4)x (7 : 8) ; (3 : 5) X (7 : 10) x (5 : 13) are compound ratios. Example. — Reduce 3 : 5 and 10 : 13 to a simple ratio. SOLUTION. Explanation.— For 3:5=4-) 2 convenience, we ex- -I Q . -I Q 10 \ ^ ^ TF^^TT? ^^ " * 1'^' press the ratios in their fractional form and proceed as in multiplication of fractions, obtaining the simple ratio ^, or 6 : 13. Hence the Rule. — Write the given ratios in their fractional form, and multiply them together as in multiplication of fractions ; the product will be the simple ratio required. BATIO AND PKOPOKTIOK. 271 P J? OBLEMS. (1) Reduce the following to simple ratios (3) (3) (4) 3 : 15 2| : 3 20i : 7 7 : 41 17 : 20 7:1 5 : 11 16 : 41 16 : 19 5 : 16 9 : 23 9 : 54 3: 5 17 : 15 1 Which is the greater ■{ 7: 8, or 31 : 42 ^? { 9: 11 7 : 35 \ The ratio of the reciprocals of two numbers is called the reciprocal, or inverse, ratio of those numbers. Thus, the reciprocal of 2 is |^; of 3, 1 ; and the reciprocal ratio of 2 to 3 is the ratio of \ to \. The ratio of ^ to i is ^ -H 1 = I X f = f , or 5 ; 2. Hence, 270. The Reciprocal or Inverse Ratio of any two terms is the same two terms with their positions interchanged. 2*71. Variation. One thing varies directly as another, when it increases as the other increases, and decreases as the other decreases. Thus, if the rate of motion be uniform, the distances moved over will vary directly as the time ; that is, in twice the time, the distance moved over will be tivice as great, and in three times the time, it will be three times as great, &c. One thing varies inversely as another, when it increases as the other decreases, and decreases as the other increases. Thus, the time of moving over a given space varies inversely as the velocity ; that is, if the velocity is twice as great, the space will be moved over in one-half the time ; if the veloc- ity is three times as great the space will be moved over in one-third the time; &c. 272 RATIO AND PROPORTIONS'. JPBOPOBTIOm 273. Proportion is equality of ratios. 373. ui. proportion is a statement of the equality of ratios. Thus, f = f , which may also be expressed 6:3 = 8 : 4, or 6 : 3 : : 8 : 4, is a proportion. The last is read " 6 is to 3 as 8 is to 4." A7iy one of these expressions may he changed to either of the other two at pleasure. 274. The numbers which form a proportion are called proportionals, or tenns of the proportion. 275. Three numbers are proportionals when the first is to the second as the second is to the third. Thus, 2 : 4 :: 4 : 8. In this proportion the second term, 4, is called the mean proportional between the other two terms. 276. Four numbers are proportionals when the first is to the second as the third is to the fourth. Thus, 2 : 4 : : 6 : 12. 277. The Extremes of a proportion are the 1st and 4th terms. 278. The Means are the 2d and 3d terms. 279. The Antecedents are the 1st and 3d terms. 280. The Consequents are the 2d and 4th terms. 281. In every proportion, the product of the extremes is equal to the product of the means. Thus, in the proportions 3 : 6 :: 5 : 10, 3x10 = 6x5; and 7 : 2 :: 14 : 4,7x4 = 14x2. This Principle is called the test of the accuracy of the proportion. RATIO AND PROPORTION. 273 Tell which of the followiDg expressions are proportions: 1. 3 7 : : 5 9. 2. 6 8 : : 3 4. 3. 7 10 : : 14 20. 4. 6 8 : : 5 12. 5. 3 : 7 : 4 : 9. 6. 2 : 1 : 8 : 4. 7. 7 : 2 : 21 : 6. 8. 5 : 3 : 15 : 9. This Principle also affords, an easy method of finding any one term that may be wanting in a proportion. Example 1.— What is the 4th term in 9 : 18 : : 5 : ( ) ? SOLUTION. Explanation. — The product of the means 18 X 5 _ .^ . ^'^x 18 = 90. Q — 1^> -a-US. The product of the extremes must also be 90 (281). Since one of the extremes is 9, the other extreme must be 90 h- 9 = 10, Ana. Example 2. — What is the 2d term in 3 SOLUTION. 3 X 12 — 9, Ans. ( ) :: 4 : 12? Explanation. — The product of the ex- tremes is 3 X 12 = 36. The product of the means must also be 36 (281). Since one of the means is 4, the other mean must be 36 -5- 4 = 9, A71S. KuLE I. — Divide the product of the means hy the given extreme; the quotient is the other extreme. II. Divide the product of the extremes ly the given mean ; the quotient is the other mean. Cancelling common factors from the antecedents or consequents, or either couplet, wiU not destroy the proportion; because the results will in each case stand the test. rnO BLEM s. Find the unknown term in the following : 7. £( ) : £12 : : 15 : 3. 8. 12 : $5 : : 20 lb. : ( ) lb. 9. 8;^ : 24^ :: ( ) qt. : 6qt. 10. lOpwt. : ( )pwt. :: 15 : 6. 11. ( ) oz. : 12 oz. : : $5 : |6. 12. 7T. : 11 T.:: 14001b. : ( ). 1. 1 3::4:( ). 2. 4 • 2::10:( ). 3. 6 8 : : ( ) : 12. 4 9 : 5 :: ( ) : 10. 5. 3 ( )::8:6. 6. 5 : ( ) ::20:12. 274 RATIO AK1> PROPOETION. In reference to the number of terms in their ratios, proportions are Simple or Compound. 282. A Simple JProportion is a statement of the equality of two simple ratios. Thus, 5 : 8 : : 10 : 16, or A =z ^^, is a simple proportion. 283. A Compound I^roportion is a statement of the equality of a simple and a compound ratio, or of 2 • 3 two compound ratios. Thus, g ; g : : 4 : 6, or 2 x 6 : 3 X 9 : : 4 : 6, or = f, represents the equality of a compound and simple ratio; while q ! g -^ o • 4' ^^ 1 X 3 : 2 X 6 :: 5 X 2 : 10 X 4, or ^-^ = T^r^, rep- 2x6 10 X 4 ^ resents the equality of two compound ratios. SIMPLE PROPORTION. Simple Proportion is employed to find a fourth propor- tional, when three are given. 0raL 3E:Coi^ci^G^ •-♦•• Example. — If 9 lb. sugar cost 81)^, what cost 12 lb. ? Solution.— If 9 lb. cost 81;^., 12 lb. will cost more in the ratio of 12 to 9, or y- = f as much. I of 81/ = 108/. Therefore, if 9 lb. of sugar cost 81/, 12 lb. cost 108/, or $1.08. Pn OBLEMS. 1. If 6 yd. carpet cost $15, what do 10 yd. cost? 2. If a tree 20 ft. high casts a shadow 8 ft. long, how high is the tree whose shadow is 32 ft. long ? RATIO AND PROPORTIOI^. 275 3. 7 bbl. molasses cost $63. What cost 14 bbl. ? 4. 2 lb. coffee last 6 persons 8 days. How long will it last 24 persons ? 5. Paid $1.20 for 3 lb. butter. How many lb. can be bought for $3.60 ? 6. 5 horses eat a ton of hay in 8 months. How long would it take 2 horses to eat the same amount ? 7. If the boarding of 3 men for a week costs $21, how many men can be boarded the same length of time for $63 ? 8. If a 10^ loaf of bread weighs 12 oz. when flour is $8 a bbl., how much should it weigh when flour is $16 a bbl.? 9. If 3 lb. sugar cost 33^ what will 6 lb. cost ? 10. If 4 bu. apples cost $3, what do 8 bu. cost ? 11. If 5 T. iron cost $400, what will 8 T. cost? 12. If 6 bu. wheat cost $9, what do 5 bu. cost ? 13. If 11 lb. coffee cost $1.65, what do 12 lb. cost? 14. If 12 lb. rice cost 72)^, what do 11 lb. cost ? 15. If 15 bbl. beef contain 3000 lb., how many do 20 bbl. contain ? .> _ t'Wr^tten ^:^erciXeXt Example 1. — If 50 bu. wheat cost $130, what will 60 bu. cost at the same rate ? Explanation. — Since both terms of $ a ratio must be like numbers, we see that (156) we must compare 50 6?y. with 60 Mi.; and $130 with the requisite number of dol- lars in the result. Since proportion is an equality of ra- 50 ) 7800 tios, the ratio of 50 to 60, or |^, must be $156. Ans. equal to the ratio of $130 to $( ), or p^ ; 276 RATIO AND PROPORTION. or 50 : 60 : : $130 : $( ). Solving by 281, Rule I., we have f^ x $130 = $156. Example 2. — If 10 men can build a boat in 12 days, how long will it take 15 men to do the same work ? SOLUTION. Explanation.— If 10 men build the men men days days ^^^^ ^ ^^ days, 15 men will build it in 15 : 10 : : 12 : (8) less time ; in other words, the relation 12 X 10 , , of 15 to 10, or If , determines the relation ~"Y5 — ^ ^^' ^^^^' of 13 da. to the required time. 384. EuLE. — MaJce that number luMcli is of the same hind as the required answer, the third term. Then detertnine from the nature of the question whether the ansiver is to be greater, or less thari the third term. If greater, turite the less of the reinaining two terms as the first term of the proportion, and the greater as the second. If less, ivrite the greater of the two remaining terms as the first and the less as the second. Find the fourth term by 2S1, Rule I. As the processes are all multiplication and division, it is evident that the work may frequently be shortened by cancellation. This and all similar examples may be solved by Analysis. Thus, Solution 1. — If 10 men can build a boat in 12 da., it would take 1 man 10 times 12 da., or 120 da. And if 1 man will build the boat in 120 da., 15 men will build it in -^^ of 120 da., or 8 da., Ans. 2. If 10 men build a boat in 13 da., 15 men will build it in the same part of 13 da. that 10 is of 15, or i§ = | of 13 da., or 8 da.. Ana. mo B LJiJM s. 1. If 50 bu. wheat cost $75, what cost 40 bii.? Ans. $60. 2. If 20 yd. cloth cost $40, how much do 18 yd. cost ? A?is. $33. 3. If 12 lb. tea cost $10, how much would 25 lb. cost ? Ans. $20f . 4 If 7 caps cost $9, what do 10 cost ? A7is. $12f RATIO AND PROPORTIOK. 277 5. If $63 buy 21 hats, how many will $48 buy ? Ans. 16 hats. 6. If 30 bu. wheat make 6 bbl. flour, how many bu. will be needed to make 8 bbl.? Ans. 40 bu. 7. If 50 bu. wheat make 10 bbl. flour, how many bbl. of flour will 60 bu. make ? Ans. 12 bbl. 8. If a man walks 60 mi. in 4 da., how far would he walk in 6 da., at the same rate? Ans. 90 mi. 9. If a man travels 132 mi. in 6 da., how long will it take him to travel 462 mi., at the same rate ? Ans. 21 da. 10. If a man, at the rate of 3 mi. an hr., walks a certain distance in 12 hr., how long would it take him to walk the same distance, at 6 mi. an hr. ? Ans. 6 hr. 11. If a ship's provisions will last 36 men 216 da., how long would they last 60 men ? A7is. 129.6 da. 12. If 24 men can build a wall in 42 da., how long would it take 16 men ? • A7is. 63 da. 13. If 36 men can build a wall in 42 da., how many men would build it in 63 da. ? Ans. 24 men. 14. If a ship's provisions would last 48 men 162 da. , how many men would they last 324 da.? Ans. 24 men. 15. If 40 gal. syrup cost £6 10s. (130s.), what cost 50 gal. ? Ans. £8 2s. 6d. 16. If 5 lb. 8 oz. tea cost $8.80, how much do 3 lb. 2 oz. cost? Ans. $5. 17. If 5 bu. 4 qt. wheat cost $12.30, how much do 7 bu. 3pk. cost? A?is. $18.60. 18. If a ship has water enough to allow each of her crew of 50 men H pt. per day, till the end of her voyage, how much can be allowed if she takes on board 30 more men ? Ans. 3} gi. 278 EATIO AND PROPOETIOl^. 19. If $150 gain $12, what sum will gain $1200, in the same time, at the same rate ? Ans. $15000. Explanation. — Al- SOLUTION. though all the terms gain gain capital capital are dollars, the third $12 : $1200 : : $150 : $ (15000) term is a different /a;id 1200 X 150 12 from the 1st and 2d = $15000,^^25. terms. Therefore, we state the Problem as in the solution, comparing gain with gain, and capital with capital. 20. How long will oats last 12 horses, which would last 4 horses 24 wk. ? Ans. 8 wk. 21. How long will hay last 15 cows, which would last 8 cows 20 wk. ? Ajis. 10| wk. 22. How many men can do that work in 15 da., which would take 25 men 90 da. ? Ans. 150 men. 23. If 5 lb. sugar cost 80^, what cost | lb. ? Ans. 24. If i lb. tea cost 37^^, what cost 5 lb. ? A7is. $7.50. 25. If 6 T. iron cost £90 18s. 6d., what cost ^ T. ? 26. If -J bbl. flour costs $||, what costs | bbl. ? A71S. $6.75. 27. An upright stick 1 ft. 6 in. high, casts a shadow 3 ft., when a tree casts a shadow 192J ft. How high was the tree? Ans. 96| ft 28. If I borrow $150 for 6 mo., how long should the lender have $300 of my money to equalize favors ? Ans. 3 mo. 29. If I borrow $150 for 6 mo,, how many dollars should the lender have 5 mo. to equalize favors? Ans. $180. 30. If a set of men can lay the brick of a building in 18 days, working 10 hr. per day, in how many days could fchey lay them working 14 hr. per day? Ans. SOLUTION. sq. rd. sq. rd. rd. rd. 4 : 20 :: 1 : (5) RATIO AND PROPORTION^. 279 31. A rectangular lot containing 20 sq. rd. is 4 rd. wide. How long is it ? Explanation. — (See Figure, 209.) Four sq. rd. along one side are 1 rod wide. Therefore, 4 sq. rd. are to the 20 sq. rd., as the width of the 4 sq. rd. is to width of the 20 sq. rd. 32. If 2| mo. boarding cost $50f, what would 15^^ mo. boarding cost, at the same rate ? Ans. $314.65. 33. A lot 40 rd. long and 4 rd. wide contains an acre. How long must an acre lot be, if 8 rd. wide ? Ans. 20 rd. 34. If sound goes 6160 ft. in 5| sec, how far does it go in 16i sec. ? Ans. 18480 ft. 35. If the moon moves on the average 52° 42' 20" in 4 da., in what time would it make a complete circuit of the earth ? Ans. 27.32 da. + . 36. If a clock loses 1 min. 45 sec. in 36 hr., how much time will it lose in 2 wk. ? Ans, 16 min. 20 sec. 37. If $175 gain $21, what do $100 gain? Ans. $12. 38. If $350 lose $42, what do $250 lose ? Ans. $30. 39. If $800 pay $3 tax, what does $1 pay ? Ans. $.003}. 40. If $2500 earn $1000, what do $1000 earn ? Ans. $400. 41. What does $1 earn ? Ans. $.40. 42. If the 13^ loaf weigh 15 oz. when flour is $8 per bbl., what should it weigh at $12 per bbl. ? Ans. 10 oz. 43. What should it weigh at $6 per bbl.? Ans. 20 oz. 44. A 40 gal. barrel is filled with a defective gallon measure, and thus appears to hold 42| gal. How much short was the gallon measure ? Ans. 2 gi. 45. If I of a barrel of cider cost $1^, what is the cost of I of a barrel ? Ans. $lfi-. 280 RATIO AJ^D PROPORTION. Compound Proportion, Compound Proportion is used when the solution of a problem depends upon two or more pairs of conditions. -^i Otal ^xfoBci^eX Example.— If $50 gain $2 in 8 mo., how much will $25 gain in 10 mo. ? Solution.— If $50 gain $2 in 8 mo. , $25 will in tlie same time gain less, in the ratio of 25 to 50, or ff = | ; | of $2 = $1. If $25 gain $1 in 8 mo,, in 10 mo. it will gain more in the ratio of 10 to 8, or Jfl = f ; f of $1 = $li. Therefore, if $50 gain $2 in 8 mo., $25 will gain $1|^ in 10 mo. pn obIjEMS. 1. If 8 men mow 32 A. in 3 da., how many men can mow 48 A. in 12 da. ? 2. If 10 loaves are sufficient for 20 men 1 da., how many loaves will supply 40 men 2 da. ? 3. If 5 horses eat 10 tons of hay in 12 mo., how many tons will 8 horses consume in 6 mo. ? 4. If 3 boys earn $7 in 4 da., how much can 9 boys earn in 20 da.? 5. If 400 bu. coal are required to supply 8 fires 5 mo., how much coal would 2 fires consume in 10 mo. ? 6. If 5 weeks' boarding of 6 men cost $150, what will 7 wk. boarding of 5 men cost ? 7. If 10 men can build a wall 30 ft. long in 5 da., how long will it take 16 men to build a similar wall 96 ft. long? BATIO AND PROPORTIOiq-. 281 8. If 25 oranges are worth 50 lemons, and 6 lemons are worth 24^, how much are 15 oranges worth ? 9. If 10 men, by working 10 hr. per da., can build a boat in 8 da., how many da. would it take 12 men to do the work, if they wrought only 8 hr. per da. ? 10. If a 5^ loaf weighs 10 oz. when flour is 18 a bbl., how much should an 8^ loaf weigh, when flour is $5 a bbl. ? t^Wrdtten ^]^er ci^esTHr Example. — If 16 men in 20 da. 10 hr. long can build 5 mi. of railroad, in how many da. 12 hr. long can 64 men build 30 miles of railroad ? SOLUTION. 64 men : 16 men 12 hr. : 10 hr. :: 20 da. : ( ) da. 5 mi. : 30 mi. 64 X 12 X 5 : 16 X 10 X 30 :: 20 : (25) Explanation. — We write 20 da. for 3d tenn, because our answer, or 4th term, is required in da. Since 64 men will perform the work in less time than 16 men, our 4th term must be less than the 3d, and we therefore write 64 men for our 1st term and 16 men for the 2d. Since it would take a shorter time to do a piece of work by working 12 hr. per day than by working 10 hr. per day, our 4th term must be less than the 3d, and we therefore write 12 hr. for the 1st term and 10 hr. for the 2d Since it would take more time to build 30 mi. of road than 5, the 4th must be greater than the 3d term, and we therefore write 5 mi. for the 1st and 30 mi. for the 2d term. We then reduce the compound ratio to a simple one (269), and find the 4th term by Rule (284), or by (281, Rule I.) 282 RATIO AND PEOPORTION. 285. Rule. — Write as the tJiird term that number which is of the same kind as the required answer. Then, of the other terms, form a compound ratio, being careful to write the terms in pairs, and to state each pair of terms, as if the value of the 4th term depended on that pair alone. After which, find the fourth term in accordance with the Principle of Proportion (281). This and all similar examples may be solved by Analysis. Thus, Solution. — If 16 men can build 5 mi. (or any other numb.er of miles) of railroad in 20 da., 64 men will build the same amount in the same part of 20 da. that 16 is of 64, that is ^ of 20 da., or 5 da. ; and if they can build 5 mi. in 5 da., they can build 30 miles in f of 30 da. = 30 da. And if they can build it in 30 da., by working 10 hr. per da., they can, by working 12 hr. per day, build it in f|, or f of 30 da., or 25 da., Atis. This and similar examples may also be solved by Cause and Effect. Thus, SOLUTION. BECOND CAUSE. r 64 men 1 , j working I ' I 12 hr. per da., [ J I for( )da. J We have now to find the number of days in the Sec. cause. And since this is a proportion, the product of the extremes equals the product of the means, and we therefore can readily fill the blank ( ) by Rule I, or II. (281). PiJ OB LEMS. 1. If 6 men build a wall 3 rd. long in 5 da., how many men would build a wall 15 rd. long in 10 da. ? Ans. 2. If 24 cows eat 36 bu. com meal in 12 da., how many bu. will last 20 oows 60 da. ? Ans. 150 bu. FIRST CAITSB. r 1(5 men J working I lOhr.perda., [ for 20 da. EFFECT OF EFFECT OF 1st Cause. ] Zd Cause. 5 mi. 1 . 30mi. of 1 • of railroad - . railroad. RATIO AND PPtOPORTION. 283 3. If 675 gain $12 in 2 yr., what sum will gain $150 in 18 mo. ? Ans. $1250. 4. If 16 men earn $640 in 26 da., what will 48 men earn in 104 da. ? Ans. $7680. 5. If 3 men in 18 da. earn $162, how many dollars will 9 men earn in 12 da., with like wages? A7is. $324. 6. If 10 persons spend $600 in 8 mo., how much would 8 persons spend in 12 mo., at the same rate? Ans. $720. 7. If 7 persons spend $273 in 2 mo., in how many mo. would 5 persons spend $650? Ans. 6f mo. 8. If $250 gain $15 in 1 yr., in how many months will $750 gain $37.50 ? Ans. 10 mo. 9. If $300 gain $9 in 3 mo., what wiH gain $50 in 10 mo. Ans. $500. 10. If 8 men dig a ditch 6 rd. long in 10 da., how many men would dig 27 rd. long in 12 da. ? Ans. 30 men. 11. If 8 men dig a ditch 6 rd. long in 10 da., in what length of time would 30 men dig 27 rd. long? Ans. 12 da. 12. If 8 men dig a ditch 6 rd. long in 10 da., how far would 30 men dig in 12 da. ? Ans. 27 rd. 13. If 12 horses eat 18 bu. of oats in 6 da., how many bu. would last 20 horses 30 da.? Ans. 150 bu. 14. If 6 horses eat 4^ bu. oats in 3 da., how many days will 100 bu. last 10 horses? A7is. 40 da. 15. If 4 horses eat 12 bu. of oats in 6 da., how many horses will require 75 bu. in 30 da. ? A71S. 5 horses. 16. If 11 men mow 45 A. of grass in 6 da. of 10 hr. each, how many men will be required to mow 81 A. in 12 da. of 11 hr. each ? Ans. 9 men. 17. If 11 men mow 45 A. in 6 da. of 10 hr. each, how many hr. per da. must 9 men mow, to finish 81 A. in 12 da.? Ans. 284 RATIO AND PROPORTION 18. If 11 men mow 45 A. in 6 da. of 10 hr. each, how many A. will 9 men mow in 12 da. of 11 hr. each ? Ans. 81 A. 19. If 108 men build a fort in 24 J da. of 12|- hr. each, in how many da. would 84 men build it, working 10^ hr. per da. ? Ans. 37^ da. Distributive JPropoi^tion, 286. Distributive Proportion is, as its name implies, proportion applied to the distribution of a quantity into parts which have a given ratio. It is sometimes called Partitive Proportion. f ©pat^xfGitciXeX -^ -•'¥•- Example.— What parts of 20 are as 3 to 1. Solution. — Since 20 is to be divided into parts which are as 3 to 1, it must be divided into 3 + 1, or 4 equal parts, and 3 and 1 of these parts taken. 20 -f- 4 = 5, or 1 part ; three of these parts is 15, and one of them is 5. Therefore, the parts of 20 which are as 3 to 1, are 15 and 5. PMO BLE MS. What parts 1. Of 30 are as 2 to 3 ? 2. Of 63 are as 4 to 5 ? 3. Of 96 are as 3 to 5 ? 4. Of 84 are as 4 to 2 ? 5. Of 21 are as 2 to 1 ? 6. Of 49 are as 2 to 7 ? 7. Of 105 are as 7 to 8? 8. Of 110 are as 10 to 11? 9. Anna and Mary, aged respectively, 5 and 7 yr., found $1.20, and divided it in the ratio of their ages. How much did each receive ? 10. Jane, Ann, and Eliza contributed to the purchase of a present of an $18 clock to their teacher, in the ratio of 1, 2, and 3. How much did each contribute? EATIO AND PROPORTION. 285 11. E and F spent $29, F spending 6^ times as much as E. How much did each spend ? 12. Frank and James have 75^ ; Frank has J as much as James. How much has each ? 13. Fifty has two parts, one of which is ^ the other. What are the parts ? t Wi'itten ^?(ferci«e^-f Example. — Resolve 75 into two parts, of which the first shall be to the second as 7 to 8. SOLUTION. 7 : 8 , ratio of parts. 7 + 8 = 15, no. of parts. 75 -^ 15 = 5, one of the parts. 5 X 7 = 35, 1st Part, 5 X 8 = 40, 2d Part, i\ Ans. Explanation. — From the conditions of the Example, 75 is to be divided in the ratio of 7 to 8, or in such a manner that the sum of the same equimul- tiple of 7 and of 8 shall be equal to 75. And since 75 is to equal an equimultiple of 7 and 8 at the same time, it must be an equimultiple of their sum, 15. Now 75 contains 15, 5 times ; therefore, one part of 75 must con- tain 7, 5 times, and the other part must contain 8, 5 times. Hence the parts are 5 x 7 = 35, and 5 x 8 = 40. SOLUTION BY PROPORTION. 7 4-8 = 15,-15: 7:: 75: (35) 1st Part. 7 + 8 = 15, — 15 : 8 : : 75 : (40 ) 2d Part. 387. Rule. — Divide the given number by the sum of the terms of the ratio, and multiply the quotient by each term. Or, As the sum. of the terms of the ratio is to one of them, so is the number to be resolved, to its corresponding part. PR OBLJEMS What parts 1. Of48areasl, 2, and3? 2. Of 108 are as 2, 3, and 4 ? Ans. 8, 16, and 24. Ans. 24, 36, and 48. 286 RATIO AND PROPORTIOif. 3. Of 108 are as 5, 6, and 7 ? Ans. 30, 36, and 42. 4. Of 64 are as 1, 3, 5, and.7 ? Ans. 4, 12, 20, and 28. 5. Of 120 are as H, 2J, 3^, and ^ ? Ans. 15, 25, 35, and 45. 6. Of 95 are as 3J-, 4J, 5^-, and 6J? ^?i5. 16J, 21J-, 26J, and 31J. 7. A and B engaged in a speculation. A invested $300, and B $500. They gained $80. What was each man's share of the gain ? Ans. A% $30 ; B's, $50. 8. A, B, and bought a farm of 650 A., for which A paid $2000; B, $3000; and 0, $1500. How many A. should each have ? Ans. A, 200 A. ; B, 300 A. ; 0, 150 A. 9. Four men. A, B, C, and D, held a pasture for $54. A put in 3 cows ; B put in 5 ; put in 4 ; and D put in 6. How much should each pay ? A71S. A, $9; B, 115; 0, $12 ; D, $18. 10. Three farmers bought a threshing machine in part- nership for $330. The first gave $80; the second, $100; and the third, $150. What part of the time should each have the use of it ? Ans. The 1st, -^; the 2d, -J^; and the 3d, J|- of the time. 11. Eesolve 84 into two such parts that one shall be 6 times the other. Solution. — Since 84 is to be the same equimultiple of 1 and 6, it must be the equimuhiple of their sum, 7. If 84 is a multiple of 7, the factor by which 7 is multiplied must be | of 84, or 12. Therefore, the numbers are 12 times 1, or 12 ; and 12 times 6, or 72. 12. Divide 40 into two parts, so that one shall be 3 times the other. 13. Divide 144 into two such parts that one shall be eleven times the other. RATIO AND PROPORTION^. 28? 14. A and B divide 960 A. so that A has 5 times as many A. as B. How many has each ? Ans. A, 800; and B, 160 A. 15. M and N. together have $6300 ; and M has 1 J- times as much as N. How much has each ? Ans. M, $3600; N, $2700. 16. Kesolve 100 into two parts, one of which shall be 16 less than the other. SOLUTION. 100 — 16 = 84 84 -^ (1 + 1 =) 2 = 42, Less Number. ) 42 + 13 = 58, Greater Number. ) ' Explanation. — Since the less and the less plus 16 = 100, 16 plus twice the less = 100 ; and twice the less = 16 less than 100, or 84. Hence the less = | of 84, or 42 ; and the greater, 42 + 16 = 58. 17. Eesolve 150 into two parts, one of which is 40 less than the other. Ans. 55 and 95. 18. Kesolve 75 into two parts, one of which is 15 more than the other. Ans. 30 and 45. 19. H has $1500 more than I ; both have $25000. How much has each ? Ans. H, $13250 ; I, $11750. 20. M is 21 yr. younger than ; both ages equal 85 yr. How old is each ? Ans. M, 32 yr. ; 0, 53 yr. 21. What part of 400 is 100 more than twice the other part ? Ans. 300. The other part 100. 22. Two men built 2695 rods of fence, and the first built I as much as the second ; how many rods did each build ? Ans. 1st. 1225 ; 2d. 1470. 23. Divide 15400 into four parts that shall be to each other as i, -J, i, -^. Ans. 6000, 4000, 3000, 2400. 24. Resolve 132 into two such parts, that 4 times the first equals 7 times the second. SOLUTION. 4. X 1st = 7 X 2d; 1 X 1st = i X 2d. J: K^d -\-l X 2d = -.{i^l)x2d = ¥ X 2d = 132. i X 2d = 12; and i X 2d = 48, the second p^rt. i X ^^ = 7 X 12 : = 84, the first part. 288 RATIO AND PROPORTION. Explanation. — If 4 times the 1st = 7 times the 2d, then the 1st = :|^ of 7 times the 2d, or I of the 2d. Hence, | of the 2d + f of the 2d, or -^^i of the 2d = 132. Therefore, i of the 2d = ^ of 132, or 12 ; and | of the 2d = 4 X 12, or 48 ; and I of the 2d = 7 x 12, or 84, the first. 25. Eesolve 96 into two parts, of which 3 times one equals 5 times the other. Aiis. 60 and 36. 26. Eesolve 720 into two parts, of which 2 times one equals 7 times the other. Ans. 560 and 160. 27. Resolve 65 into two such parts that f of one shall equal 4" of the other. SOLUTION. EXPLANA- % X lst = 4t X 2d; i X lst = ^ X 2d; tion.-H f of ^Xlst = ^x 2d. ^^^ ^^^""^l' ^ of the other fx^^ + ^X^^=(f + ^-^)x^^ = 65; theniofth; and^- x2d = ^ = 5; ^ x2d=l! x5 = 35; 1st = ^ of f, and f X ^^ = 6 X 5 = 30, 1st part. or f of the 2d, and the whole of the 1st = 3 times f , or f of the 2d Hence, f + f , or i^ of the 2d = both parts, or 65 ; and -^ of 65, or 5, is | of the 2d, and 7 times 5 = 35, is the 2d part. And 65 — 35 = 30, is the 1st part. 28. Resolve 50 into two such parts that -J- of one equals J of the other. Ans. 20 and 30. 29. Resolve 25 into two such parts that twice one equals J of the other. Ans. 5 and 20. 30. If } of A's age is f of B's, and the sum of their ages is 51 yr., what are their ages ? Ans. A's, 24 yr. ; B's, 27 yr. 31. C and D own 400 A., and | of C's farm equals f of D's. How many A. has each ? Ans. C, 160 ; and D, 240 A. OUTLINE OF PERCENTAGE. Ph BEBC^Ifl^fJ^J^.-i^QO. 2d Meaning. T'TT'TJlVr^ i 291. RATE PER. CENT, U 2*er Cenf. 292. RATE. 294. AMOUNT. 295. D INFERENCE. 296. NUMERATIOJ^ AND NOTATION. ^ * ^^^ i 297. /. . . 298. RULES. 299. II, . 300. RULES. CASES, j 301. iZ/. . 302. RULES. 303. iF. . 304. RULES. 305. FORMULAE. ^,TKTT,M^c ^ 308. Co»«. 309. Rate %. TERMS.] 3jQ^ Profit and Loss. 311. Selling Trice, 312. -T. . . . 313. Rules, 314. JX. . . 315. Rules, 31G, 111. . . 317. 7;»t/e*. 318, IF. . . 319. Rules. ©OH CASES. H TERMS. ) 3J3J3^ Policy, Q I (327. Underwriters, 0<^KINDS. 15 O H O CASES. 323. J'/rR. 334. Marine. 325. Health. 336. ii/e. t 339. /. 328. J^ire anrf Marine. ■{ 330. /Z (33i./zr. 332. I>i/e jfn^Mrance, r334. Agent. 335. Commission Merchant. 336. Broker. TEli::^S.'{ 337. Brokerage. I 338. Consignment, 339. Consignor. *^340, Consignee. CASES i 341. Z. 342. TJ. TER3IS. 347. JLwtoMn*. KINDS. 346. Principal. 35V. Rate %. As to ntode of computation. {HI- f^^ As to Rate % charged. r352. I. 350. Legal. 351. lUegaL CASES • '<'< CC*-'cJ r 354. II. . \ 356. J/r. 19 . 353. Rules. . 355. Rules. . 357. Rules. ] 358. JF. . . 359. JJu/e*. 1360. V. . . . 361. JJt«?c*. <- 3 6 3 . Days of Grace. 364. MatuHty. rr ««»,„ -^65. The ''Promise.'" 368. i^aStra/« or 7Ve«. L Weight. (463. I*artners. 463. Company. < 464. Capital. 465. Bividend. { 466. Assessment. 467. Liabilities. zi«« ( 469. l>c6rocced» 461. PARTNERSHIP. " ( 473. Assignee. f 475. Equated Time. 476. J^oeai JDafe. 477. Term 0/ Credit. 478^ verage Term of Credit. KINBS.— 479. Simple. 480. Compound. 481. AN ACCOUNT.\lll ^'^^^^ce. COlif PUT A- {Simple Average 484. Rule. TIONS. S Compound Average . . 486. Rule, CHAPTER VI. SECTION I. wmmmM.mTA(^m -^. ^^. 288. I^er cent (contraction for per centum) means, by the hundred. Thus, 5 per cent of $1 is yfo of a dollar, or 5 cents. The term percentaffe, as commonly used, has two meanings. 389. 1. It is the name of a process of computing by hundredths. Thus, finding the value of 5 per cent of a dollar, is working an example in percentage. 390. 2. It is a name applied to any number of hun- dredths of fif?iy^7u'?2^ that is regarded as a basis of computation. Thus, in the example 5 per cent of $1 = 5c., 5c. is the percentage, $1 the basis, or hase. Per cent is frequently used in the same sense as the second meaning of Percentage. 291. Hate per Cent is the number of hundredths. Thus, in the expression 6 per cent of $1, is tlie rate per cent. In computations, we give the rate per cent a denom- inator 100, and thus obtain a common or decimal fraction called 292. The Mate, an expression denoting what part of 291 292 PERCEI^T AGE. the base equals the percentage. Thus, in 6 per cent of $1 = 6c., the rate is -^^s, -^, or .06. 293. The Base of percentage is that number ot which a certain per cent is computed. 394. The Amount is the sum of the base and per- centage. 295. The Difference, or Remainder , is the Base less the Percentage. 296. Bate per cent is usually written with the sign %, Thus 5 per cent may be written b%. The sign %, is always read ** per cent." Thus, 6% is read " Six per cent." Note. — The sign is generally used in business, but in calculations rate only is used, and is written as a decimal or common fraction. Thus, tlie rate per cent Q\ is the rate — ^, or yV> or .06^, or .0625 ; the rate % \%\ is the rate --^, or \, or .12^, or .125. One iwr cent of a number is ^^ of that numher ; less than one per cent of a numher is less than y^ of that number. One hundred per cent of a number is that number, because it is {^ of the number. Tioo hundred per cent of a number is twice that number, or f f§- of it ; etc. .01 .02 .03 4 100 _5_ 100 6 100 EXERCISES. Read as per cent the following rates : .oii .105 .75 .OOJ .02} .125 1.75 .005 .07^ .161 2.00 .0075 9i 100 87i 100 350 100 100 10 25 425 i 100 100 100 100 lOi 50 505 125 100 100 100 100 .001 2.1 2. 200 100 300 100 600 100 PERCENTAGE. 2 Write as rates the following rates per cent: >Seveii per cent. 100 per cent. 75 per cent Eight " " 1 " a 750 « " Nine " " 2 " a 3 " « Ten " « 200 " a 33^ " « Twenty " « 50 " a 1 a « Thirty-five " 500 " u 6i " « Sixty i" 6( 125 " " Eighty-five " n " (( 12J " « 293 Table showing what ^jar^ of anything a certain 7'ate % o/ it equals : 1^ = TiTr- 6,^ = fV. H% = A- 16f;? = f 3$?= W- 6i^ - A- 9^ = tItt- 25^ =i. 3^ = T^- 6|;^ = T^- 10^ = A. 33i$^ = i- 4$^= 1^- ^% =TiF- 13!^ = A- 66|^ = |. 5^= A. 8^ - A- 13i$g= i- K% =i. SECTION II. GENERAL CASES OF BEHCEj^TAGE. CASE I. 297. J5ase ancf l^a^e jjer cent given, to find Percentage* ©ral^ ■^♦•- ExAMPLE 1.-5^ of $10 is what part of it ? 10^ ? Solution.— 5 fc of $10 is yf^, or i^ of it. 10% of $10 is ^^VV or of it. 294 PERCENTAGE, Example 2.— How much is 5% of 110? 10^? Solution.— 5% of $iO is j^^, or ^V of $10 = 50;^. 10% of $10 ia i^oV or j\ of $10 = $1. Find 1. 1% of $100. 5. 2. 5% of 60 bu. 6. 3. Sfo of 125. 7. 4. 25^ of £64. 8. Find 9. 100^ of 3 cows. 10. 150% of 100 A. 11. 250% of 20 yd. 12. 40% of 1000 gr. I*JiOBIj:EMS, Find 7% of $100. i% of 1200. 2i% of 40 yd. 8. 37i^ of 32 mi. 13. A merchant gained in one year 28% of his capital. What part of his capital equals his gain ? 14. A man whose farm contained 280 acres, bought as many acres more as equaled 10% of what he already had. How many acres did he buy ? 15. In a school of 300 pupils, 40% are girls and the re- mainder are boys. How many of each ? 16. A gentleman bequeathed 20% of his estate to a college, 10% to an orphan asylum, and 10% to a public library, and the remainder he gave to his children. What % of his estate did the children receive ? 17. Suppose there were three, how many % would that be for each ; and what part of the estate would each receive ? 18. Suppose the estate worth $200000; how much would each child receive ? bWii{iim^:?Ce^ci^e^b Example. — Find 15% of 30 bu. of wheat. 1st solution. Explanation. — 15% of 30 ba. is the 30 bu. Base. same as .15 of 30 bu, and .15 of 30 bu. is .15 Rate. 4.5 bu. Therefore, 15% of 30 bu. is 4.5 bu. 4.5 bu. Percentage. PERCEKTAQE. 295 2d solution. Explanation.— Since 15% means 15 15^ _ 16_ _ _3_ hundredths, we take that part of 30 bu. 100% ~ 100 "~ 20 that 15% is of 100 /c, that is, ^fp^, or ^% /o of 30 bu. = ^ bu. «^ 20 bu. = 4| bu. Ans. Note. — We might solve the Example by Proportion, thus : 100% : 15%. : : 30 bu. : (4|-) bu. 298. Rules. — I. Multiply the Base hy the Rate. II. Take that part of the Base that the given rate % is of 200%. PROBTjEMS, 1. Find 25% of 64 horses. Ans. 16 horses. 2. Find 35% of $125. Ans. 143.75. 3. Find 45% of £13^. Ans. £5.88. 4. Find 16J% of 10 A. Ans. 1.65 A. 5. Find 21% of 18.5 bu. Ans. 3.885 bu. 6. Find 2|% of 17.25 mi. Ans. .474375 mi. 7. Find \% of $2.25. Ans. $.0028125. 8. Find 1% of 7.5s. Ans. 3.15 far. 9. Find 3^% of $51. Ans. $1.70. 10. Find 12J% of a ship. Ans. ^ of a ship. 11. Find 16|^ of an oil-well. Ans. -J- of an oil-well. 12. Find 33J% of an iron-mill. Ans. ^ of an iron-mill. 13. Find 133^% of $100. Ans. $133.33f 14. The distance from Pittsburgh to Philadelphia is 355.3 mi., and from Pittsburgh to New York is 25% more. What is the distance from Pittsburgh to New York ? Ans. 4:4A mi. 15. The distance from New York to Pittsburgh is 444 mi., and the distance from Philadelphia to Pittsburgh is 20% less. What is the distance between the two latter places ? Ans. 355.2 mi. 296 PERCENTAGE. 16. A merchant sold goods on Monday amounting to $275.37J; and on Tuesday, sold 37^% more. What was the amount of Tuesday's sales ? Ans. ^378.643^. 17. At Pittsburgh the spring rain-fall averages 9.38 in. ; in summer its average is 4.7^ more. What is the summer rain-fall ? Ans. 9.82 in. 18. The average cost of each pupil in the common schools of Pennsylvania in 1869 was $.97 per month ; in 1849, it was 49.4^ less. What was the cost in 1849 ? Ans. $.49. 19. The average length of the school term in 1863 was 5.433 mo.; in 1869 it was 11.2^ more. What was the increase Ans. .608 mo. CASE TI. 299. Base and Percentage given to find the Hate JPer Cent. €)i*ar^xfoiici• Qhal ^Xoitci^Gj^ ••♦>• ■ — Example. — 5 bu. are 20^ of how many bu. ? SoLtJTiON. — 20% of any number = ^ of that number. 6 bu. must be ^ of 5 times 5 bu., or 25 bu. Therefore, PERCENTAGE. 299 1. Is 10^ 10^ ? 2. Is 20 da., 15%? 3. Is 2i hr., b% ? 7. Is 8 fur., 3% ? 8. Is 3 mi., 33^% ? 9. Is 7i far., 15% ? Of what 4. Is 10 cii., 2% ? 5. Is £5, 10% ? 6. Is 8s., 5% ? 10. James had 18 marbles, which was 25% of what Thomas had. How many had Thomas? 11. 15 men, which was 15% of a company, deserted. What was the number of men in this company? 12. Out of $75, a lady gave 20% for a hat. Wh^t had she left? T ^r^tteii iE:?^erci5e^t EXAMPLE.- IST SOLUTION. 25 bu. -j- .05 = 500 bu 25 X 25 25 bu. are 5% of how many bu. ? Explanation. — Since the percentage is the product of the base and rate, used as factors, the base is the quotient obtained by dividing the percentage by the rate : and 25 bu. divided by .05 = 500 bu. Explanation.— If 5% of any number of bu. = 25 bu., 100% must equal if^ of 25 bu., or 20 times 25 bu. = 500 bu. Ans. Observe that we obtain this result, 1st, By multi- plying the 25 bu. by the ratio of 100% to 5% ; 2d, By dividing the 25 by the reciprocal of the same ratio, which is the rate % expressed as a decimal, or simply the rate. Hence, the 302. Rule. — I. Divide the given percentage hy the rate. Or, II. Multiply the percentage hy the ratio of 100% to the given rate %. 2d solution. ^=500. Or, 5% 5-^100 = 1 = ^0*^- Or- 5% : 100% : : 25 bu. : (500) bu 500 PEECENTAGE. PROBLEMS 1. Of what is 30 da., lb% ? Ans. 200 da. 2. Of what is 1 ch., 1^% ? Ans, 1 mi. 3. Of what is 1 pt., 12^^? Ans. 1 gal. 4. Of what is $8.50, 16f ^ ? Ans. $51. 5. Of what is $1000, Q^% ? Ans. $1600. 6. Of what is ^^ |^ ? Ans. $1.33^. 7. Of what is l^^, i% ? ^^s- ^18}. 8. Of what is £f , 337^^ ? Ans. 2|4s. 9. A farmer raised 40 bu. oats, which was 25^ of his crop of wheat. How much wheat did he raise ? Ans. 160 bu. 10. I sold for $325 J, 1^% of my interest in a flouring mill. What is the value of my remaining interest ? Ans. $4014.50. 11. From 1863 to 1869, the average length of the school term in Pennsylvania increased .608 mo., which was equal to 11.2^ of the average length of term in 1863. What was the length of that term ? Ans 5.43 mo. 12. Having sold 240 sheep at one time, and 210 at an- other, and 120 at another, I find that I have only b% of the original number left. How many sheep had I at first ? Ans. 600 sheep. 13. A man who owned 37^^ of a salt-works, sold 33-^^ of his interest for $756. 33J. At what price was the entire works valued? Ans. $6050.66. 14. A sold f of 62i^ of his U. S. Bonds, for $5000. What is the value of those he retained ? Ans. $5000. PERCENTAGE, 301 CASE IV. 303. Sum or Difference and Rate Per Cent given to find the Base, ©i^al ^E:^Grx^iXGX- '"^ Example 1. — What number is that which if increased by 20^ of itself, equals 48 ? Solution. — 20% of a number = \ o{ the number, whicli added to f or tbe number = f of the number, or 48. Since 48 is f of the number, \ of the number is \ of 48, or 8 ; and f , or the number, is 5 times 8, or 40. Therefore, 40 is that number to which if 20 % be added the sum is 48. Example 2. — What number is that which diminished by 12i^of itself, equals 56 ? Solution. — 12^% of a number = i of the number, which taken from f , or the number = | of the number, or 56. Since 56 is | of the number, | of the number is ^ of 56, or 8 ; and I, or the nimiber, is 8 times 8, or 64. Therefore, 64 is that number from which if 12|% be taken the difference is 56. PJtOBT^EMS. What number increased by 1. 20^ of itself = 24 ? 2. Vl\% of itself = 72 ? 3. 334^ of itself = 120? 4. 16f ^ of itself = 35 ? 5. 150^ of itself = 25 ? What number decreased by 6. 25^ of itself = 100 ? 7. 66|% of itself = 9^ ? 8. 4:% of itself = 24 ? 9. ^% of itself = 132 ? 10. 374^ of itself = 200 ? 11. My crop of beets this year was 42 bu., which was b% more than last year. What was it last year ? 12. I harvested this year 450 bu. of wheat, which is 12^^ more than my bins hold. What is their capacity ? 13. A man lost 20^ of his sheep, and had 120 left. How many sheep had he at first ? 302 PERCENTAGE. 14. Charles attended school 5 days in one week, which was 66f^ better than his brother James.. How many days did James attend ? 15. A man bought stock to the amount of 14000, which lacked 20^ of being all the money he had. How much money had he ? 16. A farmer purchased a piece of land, containing 45 A., which was just 16f ^ less than his farm. How many acres in the farm ? - 17. A merchant sold his bookkeeper an interest in hia business for $4000, and then found that his own interest was 250^ more than that of his bookkeeper. What was the yalue of his interest ? Of both their interests ? t Wi^itten ^x^erci^esTt Example 1. — What number increased by 29% of itsell equals 1677? 1st solution. Explanation.— Since the number 1 -|- .29 = 1.29 ^s increased by 29% or .29 of itself, 1fi77 _JL_ 1 9Q 1^00 "^^^^ ^^ 129% or 1.29 times tlie num- ber: and 1677 divided by 1.29 gives the quotient 1300, or the number required, or base. Explanation. — Represent- 2d solution. .^^ ^^^ g^g^ ^y ^QQ ^^ ^ ^g ^^^^ 100^ 4- 29^ = 129% ; the entire number, 1677 = 1677 -7- 129 = 13 100% + 29% = 129%. 13 X 100 = 1300, Ans, Now, if 129% = 1677, 1% must equal j^j of 1677, or 13, and 100% = 100 times 13, or Or, ^^^^ X 100 = ^ = 1300, Ans. 1300 129 1.29 Or, to change, by cancellation, the 129% : 100% : : 1677 : (1300 ) 129 to the form 1.29, and divide But, it is more convenient change, by ca 9 to the form 1 by this number. PERCE ]S^T AGE. 303 Example 2. — What number diminished by 31^ of itself equals 1587? Explanation.— Since the number is 1st solution. diminished by 31 f^ or .31 of itself, 1587 1 — .31 = .69 is 69% or .69 of the number; and 1587 X537 _:_ .69 = 2300 divided by .69 is 2300, or the number required, or base. Explanation.— 2d solution. - The Base, 100 /e, 100^ - 31^ = 6d%; -^^^' = ^^^^ = 1587 -^ 69 = 23 ; 23 X 100 = 2300, Ans. ^^^'^^^ ^^ ~ ^fj^ Or, 100% = 100 times 1587 1587 23, or 2300. Or, -gg- X 100 =^^^ = i^300, Ans. ^^ cancellation, we change 69 to .69 and ' divide, obtaining in 69^ : 100^ : : 1587 : ( 2300 ). both cases the same result, 2300. 304. Rule.— I. Divide the given number by 1 increased or diminished by the given rate. Or, II. Take such a part of the given number as 100% is of the sum or difference of 100% and the given rate %). PR O B L EMS. What number 1. Increased by 13^ of itself = 1582 ? Ans. 1400. 2. Increased by 17^^ of itself = 2350 ? Ans. 2000. 3. Increased by 22^ of itself = 366 ? Ans. 300. 4. Increased by 75^ of itself = 787.5 ? Ans. 450. 5. Increased by 125^ of itself = 833^ ? Ans. 370^^. 6. Increased by 325^ of itself = 2210 ? Ans. 520. 7. Increased by 22^ of itself is 1098? 304 PERCENTAGE. What number 8. Diminished by 16|^ of itself = 700 ? Ans. 840. 9. Diminished by 75 % of itself = 125 ? Ans. 500. 10. Diminished by 13^% of itself = 260 ? A7is. 300. 11. Diminished by 62^% of itself = 131.25 ? Ans. 350. 12. Diminished by 87^% of itself = 28^ ? Ans. 225. 13. Diminished by 93i^ of itself r= 12|- ? Ans, 200. 14. A drover bought 360 head of cattle, which was 25^ more than he already had. How many had he before the purchase ? Ans. 288 head. 15. The average school term in Pa., in 1869, was 6.04 mo., which was 11.2^ more than in 1863. What was the average term in 1863 ? Ans. 5.43 mo. 16. The average Pa. school term in 1863, was 5.43 mo., which was 10.05^ less than in 1869. What was the length in 1869 ? Ans. 6.04 mo. 17. The distance by the P. R. R. from Pittsburgh to Harrisburg is 246 miles, which is 125ff^ more than the distance by the same road from Harrisburg to Philadelphia. What is the latter distance ? Ans, 109.2 mi. 18. W. R. Ford bought a farm for a certain sum, ex- pended for stock 11^ of the price of the farm, and found that the whole cost was 17215. What was the cost of the farm alone ? Aris. $6500. 19. H. J. Gourley raised 800 bu. wheat, which was 25^ more than ^ of what Logan raised. How many bushels did Logan raise? A7is. 1280 bu. 20. A man dying bequeathed 10^ of his property to his son, 20;^ of the remainder to his daughter, 33^;^^ of this remainder to his wife, and the remainder, or $48000, to charitable purposes. Required the value of the estate. Ans, $100000. PERCENTAGE. 305 305. Formulas, The entire subject of Percentage may be summed up in the following formulas: Let r represent the rate. " S " " amount " D " « difference. Let B represent the base. " F " " percentage. " R « " rate %. Then the Kule in Case I. becomes = B ^r, or p = B x-^^. Case II. Case IIL r =: -j^ and ^ = ^ X i^^^. B =- and 5=^ X P. B ^ r R Case IV. J, S D B = -— - — , or 1 4- r' 1 — r A thorougli knowledge of the use of these formulas is about all the pupil needs to carry him through the Applications of Percent- age ; for by a simple substitution of the proper terms for B, P, R, &c. , they can be made applicable to all cases except Interest, in which the element time {t.) is introduced. 306. Applications of Percentage. Percentage is, because of its convenience, extensively used in business calculations. Its principal commercial applica- tions are in Profit and Loss, Insurance, Com- mission, Interest, Discount, Stocks, Exchange^ Taxes, Partnership and Bankruptcy. 20 SECTION III »MciFiT' Mmm iMmm '^^^•>0 307. Profit and Loss are terms used to express the gains or losses in business transactions. Gains and losses are computed at a certain rate per cent of the cost or the sum invested, and the operations involve the 2jri7icijples already explained in Percentage. 308. The Base is the cost or the sum invested. 309. The Kate % is the rate per cent of profit or loss. 310. The IPercentage is the profit or loss. 311 {a). The Amount is the sm7?2 of cos^ and j^rq/?^, or the selling price. 311 (5). The Difference is the co5^ Zess the Zo55, or the selling price. CASE I. 312. To find the sum gained, or lost, when the cost and rate % are given. ■■♦•- —{i ■\- €)i-alE^oicisG< Example 1.— A wagon that cost $150 was sold at a gain of 20,^. What was the gain ? Solution.— Since the gain was 20%, it was ^, or \ of $150, or $30. 316 PROFIT A KD LOSS. 307 Example 2. — A horse that cost 1300 was sold at a loss of (^f- What was the loss ? Solution.— Since the loss was 27%, it was ^^^ of $300, or $81. l*JtOBL]£MS. How much is gained or lost when the cost 1. Is $50, gain 20^ ? 6. Is $500, loss 12^? 2. Is $67, loss 10^ ? 3. Is $800, gain 30^ ? 4. Is $1000, loss 15^ ? 5. Is $30, gain 7. Is $800, gain 50^? 8. Is $600, loss 40% ? 9. Is $765, gain 100% ? 10. Is $250, gain 75% ? -.-♦-- t ^i^itten ^X^^er c i;^e^ i: SOLUTION. X ^ = $37.50, gain. 2 Or, $250 X .15 = $37.50, gain. Example. — What is the gain, when goods are bought for $250 and sold at a gain of 15% ? Explanation.— The gain is 15% (R, in Percentage), or ^J^ of the cost (B), which is found either by multiplying the cost by the number denoting the rate % , and dividing this pro- dact by 100 ; or by multiplying the cost by the rate (r), .15. In either case we obtain the same result, $37.50. 313. EuLES. — Take such part of the cost as the rate % is of 100%. (A X ^ = P.) Or, Multiply the cost hy the rate. {B x r ^ P.) I* ROBLEMS. 1. Sold goods which cost me $1250, at a gain of 12|^%. How much did I gain? Ans. $156.25. 308 PERCENTAGE. 2. How much do I lose by gelling a horse, which cost me $542, at a loss of 18^ ? Atis. $97.56. 3. I wish to sell my goods at a profit of 20%; how shall I mark calicoes that cost 10)^ ? Delaines that cost 40^ ? Towehng that cost 12^^ ? Silks that cost $3 ? Broadcloths that cost $4.50 ? Latins that cost $5 ? Ans. to last, $6. 4. I am closing out my stock, and willing to lose 12^^ on cost of my goods. How should I mark hose that cost 3'Z^ ? Collars that cost 8^ ? Cloaks that cost $32 ? Aiis. to last, 128. CASE II. 314. To find the rate % of gain or loss, when the cost and gain, or loss, are given. 4^ ©ral^:<^GFciXe, Example. — Bought a shawl for $16, and sold it at -an advance of $4. What % did I gain ? Solution. — My gain is $4 on $16, or y\ = ^ of cost. Since cost is 100 % , my gain is \ of 100 % , or 25 % . mo B LEMS. 1. Bought calico for 8^ per yard, and sold it so as to gain 2^ on each yard. What % did I gain ? 2. Bought shoes at $5, and sold them at $6. What was the gain % ? 3. Bought oranges at 3f apiece, and sold them at bf . What is the rate % gain ? 4. Bought apples at 3 for bf , and sold them at 4 for bf . What % was the loss ? 5. Bought a book for $2, and sold it for $1.50. What was the loss % ? PROFIT AND LOSS. 309 -•>. ^IWritten ^xTerei^er-i Example.— Bought a buggy for 1360, and sold it for $450. What was the rate % of gain ? SOLUTION. $450, selHng price. Explanation.— By subtraction we find 360, cost. the gain is $90. This is gained on $360 7 T~ (B). The rate % (R), therefore, must be $ 90, gain. ^5H,^ Qj. 1^ ^f -^QQ a/^ = 25 % . Or, by dividing -//jp of 100^ = 25^. the gain, $90 (P), by the cost, $360 (B), we Or, obtain the same result, .25 = 25%. $90-^1360=1.25 = 25^. 315. Rules. — Talce such a part of 100% as the gain or loss is of the cost. (b = -j^x 100%.\ Or, Divide gain or loss by cost. (P -r- B = r.) 1. Bought a house for $1750, and sold it for $500. What ^ is the loss ? Ans. 71-|^. 2. Bought a farm for $7650, and sold it for $9780. Re- quired the gain %- A7is. 27f|^. 3. Bought a horse for $160, and sold him for $25. What ^ did I lose? Ans. 84|^. 4. Bought delaines for 20^ a yard, and sold them at 37J^ a yard. Required the rate % gain. Ans. 87^-^. 5. Bought 100 doz. eggs at 15)^ a doz. ; but 20 doz. were broken in transportation. Sold the remainder at Slf a doz. What did I gain on the investment ? Ans. 65^%. 6. I buy a horse for $100, and after keeping him at a livery stable for 10 weeks at an expense of $5 a week, I sell him for $120. What is my rate % of loss ? Ans. 20^. 310 PEKCENTAGE. CASE III. 316. To find the cosf^ when the gain, or loss, and the rate % are given. Qved f^Xoi^cj^h K Example. — Sold my horse at an advance of 125, which was 20% of his cost. What was his cost ? Solution. — Since 20 %, or y%"|j, or |, of the cost is $25, f , or the cost, Is 5 times $25, or $125. PROBLEMS. 1. If $20 is 20% of the cost of 16 bbl. of flour, what is the entire cost ? 2. If I lose $16 on the price of an ox, my loss amounts to 12|^% of his value. What is his value ? 3. If $10 is 5% of the cost of my fare to San Francisco, what is the whole cost ? 4. A lost $450, which was 90% of all he had. How much did he have ? 5. A grocer sold some bacon for $250, which was equal to 12^% of the value of all he had. What was the entire value ? -•>•- t^ 1 itten ^?&i^ci;^es t. r- Example. — I sold my farm at a gain of $500, which was equal to 20% of its value. What was its value ? PROFIT AlfD LOSS. 311 Explanation. — The whole value SOLUTION. of the farm is 100^ of it. But 100% Yi^ittei\^:xrerei^esr^ Example. — By selling my farm for $4000, 1 gained 20^. What did my farm cost ? SOLUTION. 100^ + 20% = 120% = $4000. $4000 -T- 120 = $33^ = 1%. 1% X 100 = $33^ X 100 = $33334. Or, $4000 -J- (1 4- .20 = 1.20) $33334. Explanation. — We first add the gain, 20%, to the cost, 100%, to find what fo of the cost = the selling price, and ob- tain 120% of cost. PROFIT AND LOSS. 313 We then divide the $4000 by 120 to find l^o of the cost, or $33f This we multiply by J 00 to find 100 5^ of the cost, or the whole cost=|3333i-. Or, as we receive $1.20 for each $1 invested, we must have in- vested as many dollars as $1.20 is contained times in $4000. 319. KuLES. — Take stick part of the selling price as 100% is of 100% + the rate % of gain, or — the rate % of loss. Divide the selling price hy 1 increased hy the rate of gain, or diminished hy the rate of loss. I = B, or 47=-) 1 — r FBOBZEMS. What is the cost of goods which selling 1. At a loss of 20^ bring $30 ? Afis. $37.50. 2. At a gain of 25% bring $75 ? Ans, $60. 3. At a gain of 12|^ bring $.72 ? Ans. $.64. 4. At a loss of 8 J;;^ bring $132.11 ? Ans. $144.12. 5. Sold a watch at 33^^ gain, and with the money bought another, which I sold for $120, and lost 20^. Did I gain or lose, and how much ? Ans. $7.50 gain. 6. Sold a wagon at $250, and thereby gained 11^. What did the wagon cost ? Ans. $225.23, MEVIEW PROBLEMS, 1. How must goods be marked which cost 24/, so as to take off 16|^, and still make 25^ ? A7is. 36f . Note. — First find selling price at 25 % gain (Case I) ; then regard the selling price as cost and find asking price (Case IV), 2. Find the asking price when cost is $1.20, so as to fall 6J^, and still make 25^. Ans. ^l.QO. 314 PERCENTAGE. 3. Cost $1.50 ; from what price can I fall 10^, and still make 20^ ? Ans. 12. 4. Cost 9^ ; from what price can I fall 10^, and still make 50%? Ans.lbt 5. Cost 20)^ ; from what price can I fall 25%, and still make a profit of 20% ? Ans, 32^. 6. If goods selling at $2.40, realize 33-J% gain, what would be the rate % of gain at $2.10 ? Ans. 16|%. Note. — First find cost (Case IV.) ; then rate % of gain (Case II.) 7. If sales at $1.80 make 10% loss, what would be the rate % of gain at $2.20 ? Ans. 10%. 8. If property at $800 makes 20% loss, what would be the rate % loss at $750 ? Ans. 25%. 9. If a bank share at $62.50 is 25% gain, what would be the rate % loss at $48 ? Ans 4%. 10. If land at $56 an acre is 12% gain, what would be the rate % gain at $60 an acre ? Ans. 20%. 11. If nails at 6<^ a lb. make 4% loss, what would be the rate % gain at Hf a lb. Ans. 12%. 12. If a merchant realizes 20% gain on tea at 75^ a lb., what would he do at 56^^^ a lb. ? Ans. Lose 10%. 13. Selling price 15f ; gain 25% ; find rate % of gain at 16;^. Ans. 33-Jr%. 14. Selling price 16 f; loss 25%; find the rate % of gain at 25^. Ans. 25%. 15. A speculator sold 2 houses at $6000 each ; on one he gained 33 J% of his investment, and on the other he lost 33-J-% of his investment. Did he gain or lose, and how much? Ans. $1500 loss. 16. A man sold a house, which was presented to him, for $5000. What was his rate % of profit? Ans. ? ^^^^:^ SECTION IV 4 IMSTUMJtMC'E ^fc^ 320. Insurance is indemnity against pecuniary loss, pledged by one party to another. 321. The Premium is the sum paid for insurance, and is a certain per cent, of the sum insured. 322. The Policy is the contract between the parties. 323. Fire Insurance is indemnity for loss by fire. 324. Marine Insurance is indemnity for loss by navigation. 325. Health Insurance secures an allowance of money during the sickness or disability of the party in- sured. 326. Life Insurance secures a specified sum of money to a designated person, or to designated persons, or to their heirs, on the death of the party iiisured, or on his arriving at a certain age. 327. Underwriters or Insurers are the parties that insure against loss. 328. Fire and Marine Insurance, Fire and Marine Insurance are so completely covered by Cases I, 11, and III of Percentage, that it is thought unnecessary to do more than say that the Value Insured corresponds to the base . . . -B, Premium " " " percentage. P. Bate Per Cent, ** " '* rate % . . B, 815 316 PEBCENTAGE CASE I. 329. To find the I^remiuin^ when the Value and Kate % are given. l*lt O BLEM S. What is the Premium on 1. A house insured for $2000, at 1^%? A^is. $30. 2. A store iusured for $3500, at 2^? A71S, $70. 3. A barn insured for $900, at 1% ? Ans. $6.75. 4. Furniture insured for $1250, at |^ ? Ans. $7.81. .5. A church insured for $16000, at H% ? Ans. $300. 6. A factory insured for ^64000, at 2^^? Ans. $1440. 7. A ship insured for $15000, at 3^? Ans. $450. 8. A steamboat insured for $85000, at 2-J;^? Ans. $2125. 9. Freight, for $3780, at 11% ? Ans. $66.15. 10. A mill, for $12500, at ^% ? Ans. $312.50. 11. Clothing, for $875, at {% ? Ans. $7.66. CASE II. 330. To find the Rate %, when the Value and Premium are given. PMO B TjEMS. What is the Rate %, when the 1. Value insured is $500, premium $6 ? Ans. 1.%%. 2. Value insured is $1000, " $8? Ans. .H%. 3. Value insured is $800, " $3 ? Ans. %%. 4. Value insured is $3600, " $56 ? Ans. 1^%. 5. Value insured is $5000, " $125 ? Ans. 2^%. 6. Value insured is $8000, " $280 ? Ans. 3^%. 7. Value insured is $1800, " $24 ? Ans. 1^%. INSURANCE. 317 CASE III. 331. To find the Value Insured^ when the Rate % and J^remium are given. jPiJ O BLEMS. What is the value insured, when the 1. Premium is $6 and 1.2 the rate % ? Arts. $500. 2. Premium is 18 and .8 the rate % ? Ans. $1000. 3. Premium is $3 and | the rate % ? Ans, $800. 4. Premium is $24 and 1| the rate ^ ? Ans. $1800. 5. Premium is $56 and 1| the rate % ? Ans. $3600. 6. Premium is $125 and 2^ the rate ^ ? Ans. $5000. 7. Premium is $280 and ^ the rate % ? Ans. $8000. 332. Life Insurance. Life Insurance is of different kinds. The rates paid for life insurance depend upon the age of the party insured, and the kind of Policy taken out. For example, — it costs more per year to insure a man 50 yr. of age, than one 40 yr. of age. It also costs more per yr., to insure a man of any number of years of age, when the payments are to be made in a stated number of years, than when they are to be made every year he lives. It also requires a higher rate for any given man when the insur- ance is to be paid at the end of some stipulated period, if the man be living, or at his death should it previously occur. PR O B LEMS . 1. A man at the age of 53, paid at the rate of 5.38^ per year, for an insurance on his hfe. What was the annual premium on ^15000 ? Ans. $269. 2. After having paid an annual premium of 2.6^ on $10000 for 5 yr., the insured died. How much more than was paid was received by the heirs ? Ans. $8700. 3. A man paid for 40 years a premium of 2.53^ on $7500. How much did he pay in all ? Ans. $7590. Qy 6^ SECTION V. K ^Q>^ ^ CC>MMISSI©M I' ^ '^^^ . 333. Commission is the pay of an agent, collector, broker, &c., and is usually estimated on the amount of money he receives or expends. 334. An Af/ent or Factor is a person who transacts business for another. 335. A Commission Merchant is a merchant who buys and sells goods for others. 336. A broker is an agent who buys and sells stocks, notes, money, &c. 337. SroUerage is the commission of a Broker. 338. A Consignment is merchandise sent to an agent for sale. 339. A Consignor is a sender or maker of a con- signment. 340. A Consignee is an agent who receives a con- signment. Commission embraces the Fo^ir Cases of Percentage. Hence, the Bame Rules apply, and need not be repeated. The Proceeds of sales, or the moneys collected or invested correspond to the tase S, The Commission corresponds to the percentage . . P. The Rate % " « " rate % . . , . iJ. The Proceeds + Com. " '* " amount S* The Proceeds — Com, " " " difference . . . D, 318 COMMISSION. 319 CASE I. 341. To find Corn/mission. PROBLEMS. What is the commission on buying or selling 1. Flour for $7200, at 3^ ? Ans. $216. 2. Oats for $2800, at ^% ? Ans. $70. 3. Iron for $17500, at 1^% ? A7is. $262.50. 4. A sells B's consignment for $1985, at 3^ commission. How much does A remit B ? Ans. $1925.45. 5. buys for D merchandise worth $13000, at 2^^ com- mission. How much should D send C ? Ans. $13325. 6. E's broker buys for him $9000 worth of stocks, at 1% brokerage. What does E pay ? Ans. $9033. 75. 7. F's broker sells for him a note for $960, at ^% broker- age. How much does F get ? Ans. $957.60. 8. A consignee sells oil for $7250, at 3J^ commission. What does the consignor receive ? Ans, $6996.25. CASE II. 342. To find Bate %, P It O BIjEM s, 1. A merchant received $153 for selling $5100 worth of flonr. What % commission did he get ? Ans. 3%. 2. A real estate broker charged $53.75 for selling a fai-m for $2150. What rate % was his commission ? Ans. 2^%. 3. A broker charged me $218.75 for selHng $25000 worth of R R. stock. What % was the brokerage ? Ans. i%. 4. A factor charged $230,625 for selling 1025 bbl. apples at $5 a bbl., and $5.76 for selling 256 bu. potatoes at ^ per bu. What was the rate % commission ? Ans. A^%. 320 PERCENTAGE. CASE III. 343. To find Collections^ Purchases^ or Pro- ceeds of Sales 9 Commission and Kate % being given, PJt O BT.EMS, 1. Paid an agent 2^^ commission on the sale of lands. His bill was $524.50. What was the value of his sale ? Ans. $20980. 2. A real estate broker's income for 1875 was $10563. His commissions were 2.1^. What were his sales ? Ans. $503000. 3. My factor at Chicago sent me a bill on account of a consignment of grain to him, for $525, which included his commission at 3;^, and charges for freight, drayage, &c., $125. What was the value of the consignment ? Ans. $13333^. Note. — Deduct $125, charges, from the bill ; the remainder is the commission. 4. An auctioneer whose commission is 1%, charged $25 for selling some insurance stock. What was the value of the stock ? Ans. $2500. 5. A collector charged me $25 for collecting, at b%. How much money did I receive ? Ans. $475. CASE lY. 344 (a). To find Amount of Sale, or Collec tionSf Mate %, and Remittance being given; and, 344 (b). To find Investment, Bate %, and an Amount covering Investment and Commis- sion being given. Note.— Apply Rules in Case IV of Percentage. COMMISSION. 321 PR O B T.EMS, 1. My factor remits to me $3910, the proceeds of a con- signment of wheat, after deducting his commission of 3^. For what did he sell the wheat ? Ans. $3000. 2. 1 sent $5000 to a real estate broker in Chicago, to in- vest in lands, with instructions to deduct his commission at 4:%. What sum did he invest ? Ans. $4807.69. 3. A collector in Dubuque remitted to me $525, as due from a debt he had collected, after deducting h% for col- lecting. What was the debt ? Ans. $552.63. 4. A broker having intrusted to him 110000 to invest in stocks, performed the duty, charging ^% for his services. How much of the $10000 did he invest? Ans. $9913.26. 5. A commission merchant sold goods to the amount of $7560, charging 3^ commission. After paying $36.50 charges, he invested the balance in coffee, and charged 2J^ for the investment. If he paid 20^ per lb. for the coffee, how many lb. did he purchase ? Ans. 35593.65 lb. + 6. A commission merchant sold goods to amount of $6000, charging 1^% for his commission and $16 for freight. How- much did he remit ? Ans. $5894. 7. A broker having invested $3600 in stocks for me, sent me a bill for commission at \^% and $3 for telegraphing, &c. I sent him a draft for $1000, with instructions to take out his former bill, and invest the balance in bank stock. How much did he invest, after deducting a commission of 1% on last transaction ? Ans. $933.66 + . 8. I sent an agent $5922, with which to buy stocks, after deducting a commission of 6%. What was his commission, and how much did he invest ? Ans. Com. $282; invested, $5640. 21 SECTION VI ■^lA^ (5^ IMT;E:ItE;ST' ^i) 345. Interest is a compensation for the use of money. Interest has Five Cases ; and the operations do not differ from those in Percentage, except in respect to the new element, Time. In Interest, however, only money can be the hose; while in Percent- age the base may be anything. 346. The JPrincipal is the sum of money for the use of which compensation is made. 347. The Amount is the sum of the Principal and its Interest If I pay $6 for the use of $100 for 1 year, $6 is the interest; $100 the princvpal ; and $100 + $6, the amount. Interest as to methods of reckoning, is of two kinds. Simple and Compound. 348. Simple Interest is reckoned on the Principal alone at a certain % for some fixed period, generally 1 year. 349. Compound Interest is reckoned on the Prin- cipal for the first period, and after that on the Principal and Interest. Interest as to rate charged is of two kinds. Legal and lUegal. 350. Legal Interest is that established by law. 351. Illegal Interest, called Usury, is that which exceeds Legal Interest, and is not allowed by law. The taking of Usury is a violation of law, and usually has attached to it some penalty. 325 INTEREST. 323 351'. The Hate IPer Cent of Interest is the number denoting how many hundredths of the Principal equals the Interest. The laws regulating the rate per cent, and the forfeitures and pen- alties for usury are different in different States and Countries ; in the United States the most common legal rate per cent is 6 % . The following table will afford the pupil a general idea of the Legal Mates of Interest, Legal rate in England France Louisiana Dakota Georgia Kentucky Michigan Minnesota New Jersey New York South Carolina. Wisconsin Alabama , Texas... Arizona California... Colorado — Idaho Kansas Montana — Nebraska . . . Nevada Oregon Utah Washington. Wyoming T. .10^ .12^ All other States . . . Dist. of Columbia. Debts due U. S. . . . Canada Ireland Nova Scotia Rates allowed in Arkansas California Missis!?ippi ISIassachusetts.. Nevada Rhode Island. ■ . . . .Any rate, hy agreement. Dakota.... Nebraska 70 15^ Minnesota Oregon Texas Florida Illinois.. ., Michigan. . Missouri . . Tennessee Wisconsin. Louisiana . Ohio 10% Pennsylvania cial law few institutions lia, by spe- ) in cases of a v . . 7% itutions ) Kansas, do 324 PEKCEKTAGE. CASE I. 353. OChe Principal^ Bate % and Time given^ to find the Interest and Amount. 0i*al '^Xoi^ci^e'X Example 1. — At 6% per annum, what is the interest of 112 for 1 year ? Solution.— If the interest of $1 for 1 yr. is 6%, it is 6i^; and of $12, it is 13 times as much, or 13 times 6^ = 73^2^. Or, Since 6 % is yf ^ = /^^ of the principal, the interest of $13 for 1 year, at 6%, is /^ of $13, or $.73. Example 2. — What is the interest of $25 for 1 year, at 6% ? For 2 yr. ? For 3 yr. ? For 7 yr. ? Solution. — At 5 % the interest = y^, or -^^ of the principal ; ^ of $35 is $1^, or $1.35. For 3 yr. it is 3 times $1.35, or $3.50 : for 3 yr., 3 times $1.35, or $3.75 ; &c. I^MOBLBMS. 1. What is the interest of $50 for 3 yr., at 6^ ? 2. What is the interest of $75 for 10 yr., at b% ? At 6^ ? At 7^? At 8^? At 9^? 3. What is the interest of $150 for 5 yr., at b% ? At 6^ ? At 7^? At 8^? AtlO^^? At 12^? Example 3. — What is the interest of $100 for 1 yr. 3 mo., at 6^? At 8^? At 9^? At 10^ ? At 12^ ? Solution.— 1 yr. 3 mo. = 1^»^, or 1^, or f yr. The interest of $100 for 1 yr. at 6% is $6 ; therefore, the interest for f yr. at 6% = | of $6, or $7^. If the interest at 6% = $7|, at 8% it will be |, or | of $7i = $10; at 9%, I, or f of $7^ = $11^; &c., &c. 4. What is the interest of $200 for 2 yr. 9 mo., at 6^ ? At 8^? At 9^? At 10^? At 12^? 5. What is the interest of $500 for 1 yr. 3 mo., at 6^? At 7^? At 10^? At 15^? INTEREST. 325 Example 4.— What is the amount of $20 for 2 yr. 6 mo., at 6^? At 9^? At 12^? At 15^ ? 1st Solution.— At 6% for 2|, or f yr., the interest of $20 is $3; and $30 + $3 = $33, the amount. 3d Solution,— The interest of $1 for 2| or f yr., at 6%, is 15j^; and $1 + 15^ = $1.15, the amount. If $1 amounts to $1.15, $20 will amount to 30 times $1.15 = $33. The interest of $1 for f yr. at 9% is 32^^ ; and the amount, $1.22^ ; hence the amount of $30 is 20 times $1.33^ = $34.50 ; &c., &c. 6. What is the amount of $30 for 3 yr. 3 mo. at 8% ? At 12^? At 16^? 7. To what does $50 amount in 5 yr. 6 mo., at 6% ? At 10^? At 20^? Example 1.— What is the interest of $250 for 4 yr. at 6% ? 1st solution. Explanation. — We first find that part of the principal designated by the rate for 1 yr. to be $15, by mul- tiplying $350 by .06 ; we then multi- ply this interest for 1 yr. by 4, and find the interest = 4 x $15 = $60. Ans. Explanation. — We first mul- tiply the interest of $1 for 1 yr. by 4 to find the interest of $1 for 4 yr. This is equal to $.24. We then multiply the inter- est of $1 for 4 yr. by 350 to find the interest of $350, which is equal to $60, Ans, $250, .06, Principal. Eate. $15.00, 4 $60 Int. for 1 yr. Int. for 4 yr. $.06 = 4 2d solution. Int. of $1 for 1 yr. $.24 = 250, : Int. of $1 for 4 yr. Principal. 120 48 $60.00 = Int. of $250 for 4 yr. 326 PE RCEKTAGE, 353. EuLES. — I. Multiply the principal by the rate^ and that product hy the time. {B xr x t=F.) Or, II. Multiply the interest of $1 for the given time by the nu7nber of dollars in the principal. (rxtxB=P.) As the amount is the sum of the principal and interest, we have only to add these parts to find the amount. Find the interest 1. Of $150 for 2 yr., at 6^. Ans. $18. 2. Of $275 for 3 yr., at 7^. Ans. $57.75. 3. Of $128,121 for 8 yr., at b%. Ans. $51.25. 4. Of $650 for 5 yr. at %%. Aiis. $260. 5. Of $225 for 8 mo., at 12^. Ans. $18. 6. Of $580.75 for 6 mo., at 18^. Ans. $52.26f. Example 2. — What is the amount of $750 for 3 yr. at SOLUTION. $.08 = Int. of $1 for 1 yr. 3 $.24 = Int. of $1 for 3 yr. $1.00 $1.24 = Amt. of $1 for 3 yr. 750, Principal. 620 868 Explanation. — We first multiply the interest of $1 fori yr. at 8% by 3 to find the interest for 3 yr. To this we add $1 to find the amount of $1 for 3 yr., at 8%. This sum we multiply by 750 to find the amount of $750 for the same time and at the same rate. There- fore, to find the amount, we have this $930.00 = Amt. of $750 for 3 yr. Rule. — III. To $1 add its interest for the given time^ at given rate ; and multiply this sum by the number of dollars in the principal. (1 -\- r x t x B = S.) INTEREST. 327 What is the amount 7. Of $150 for 2 yr., at 6% ? Ans. I1G8. 8. Of $275 for 3 yr., at 7% ? A)is. $332.75. . 9. Of 1756.25 for 2 mo., at 15^ ? Ans. ^775.16. 10. Of $852.50 for 2.5 yr., at 9% ? Ans. $1044.31. 11. Of $116 for 8 yr., at 12^^ ? Ans. $232. 12. Of $120.50 for 7 yr., at 15^ ? Ans. $247.03. 13. Of $13520 for 3 yr., at 7^^ ? Ans. $16562. Example 3.— What is the interest of $200 for 3 yr. 9 mo. 18 da., at 7% per annum ? Solution by Rule I. $200, Principal. .07, Eate. $14.00, Int. for 1 yr. 3.8, Time in yr. (340.) 112 42 Solution by Rule n. $.07 = Int. of $1 for 1 yr. 3.8 Time in yr. $.266 = Int. of $1 for 3.8 yr. 200 $53.20, Int. of $200 for 3.8 yr. $53.20, Interest. What is the interest 14. Of $142.83 for 7 mo. 18 da., at 7% ? 15. Of $968.84 for 1 yr. 10 mo. 26 da., at 16. Of $6360 for 2 mo. 17 da., at 4J^? What is the amount 17. Of $630.37i for 9 mo. 15 da., at 5%? 18. Of $2831.20 for 6 mo. 9 da., at 9%? 19. Of $120 from Jan. 1, 1874, to May 16, 1876, at 8%? Note. — Find time between dates by Rule, p. 236. 20. Of $360 from Feb. 9, 1875, to May 18, 1876, at 9%? Ans. $401.31 Ans. $6.33. Ans. $61.21. Ans. $655.32. Ans. $2964.97. 328 PERCENTAGE, 21. Of $637.50 from Mar. 4, 1873, to Aug. 16, 1876, at 10^? Ans. $857.43. 22. Of $1184 from Apr. 29, 1873, to Dec. 5, 1876, at 6% ? Ans. $1397.12. 23. Of $1580 from May 20, 1872, to May 2, 1876, at 4^ ? Ans. $1829.64. What is the interest 24. Of $295.45 for 2 yr. 5 mo. 5 da., at 6^? Ans. S43.09. 25. Of $238.38 from Aug. 19, 1871, to Apr. 15, 1876, at 7^? Ans. $77.68. 26. Of $487,375 from June 15, 1875, to Sept. 30, 1875, at 1% a mouth ? Ans. $17.06. 27. Of $360 from Aug. 10, 1876, to Nov. 20, 1876, at i% a month ? Ans. $9* Example 4. — What is the interest of $150 for 5 yr. 9 mo. 12 da., at 11% per annum ? SoLtmoN BY Aliquots. $150 , Principal. .07 , Rate. $10.50 , Int. for 1 yr. 5 ALIQUOTS. $62.50 , Int. for 5 yr. 6 mo. = ^ yr. 5.25 , « *' 6 mo. 3 mo. = i yr. 2.625, « « 3 mo. 12 da. =^ yr. .35 , " « 12 da. $60,725, Int for 5 yr. 9 mo. 12 da. Explanation. — ^We first find the interest for 1 yr. ; then for 5 yr. For 6 mo.'s interest we take \ the interest for 1 yr. ; for 3 mo.'s, we take \ the interest for 1 yr. ; for 12 da,, we take ^ the interest for 1 yr. ; and the sum of the interest for 5 yr. + ^ yr. + i yr- + ^ yr. equals the interest for 5 yr. 9 mo. 13 da., or $60.73, correct An». INTEREST. 329 What is the interest 28. Of $75 for 2 yr. 8 mo. 12 da., at 7^ ? Ans, 114.18. 29. Of $120 from Jan. 1, 1873, to May 16, 1875, at S% ? Ans. 122.80. 30. Of $360 from Feb. 9, 1875, to May 18, 1876, at 9%? Ans, $41.31. 31. Of $637.50 from Mar. 4, 1873, to Aug. 16, 1876, a^ 10^? Ans. $219.93. 32. Of $1184 from Apr. 29, 1873, to Dec. 5, 1876, at 5% ? Ans. $213.12. Six Per Cent Method. In the calculation of interest we reckon only 360 days in a year. If then the interest of $1 for 1 yr. is 6 cents, or 60 mills, the interest for 1 day must be ^^^ of 60 mills, or | of a mill, and for 6 da. it must be 6 times ^ of a mill, or 1 mill ; for 60 da. or 2 mo., 60 times ^ of a mill, or 10 mills, or 1 cent. Therefore, If the interest is at 6% per annum, the interest of $1 is equal to one-half as many cents as there are months, or ^ as many mills as there are days. Having found the interest of $1, the interest of any number of dollars is readily found, by multiplying this interest by the number of dollars. And having found the interest of any number of dollars, at 6%, we can find it at any other rate % hy taking such a part of that at 6fof as the given rate % is of Ofo, Example 5. — What is the interest of $50, for 18 mo. 24 da., at S% ? SOLUTION. Int. of $1 for 18 mo. ® 6% = ^^ or $.09. Int. of $1 for 24 da. @ 6^ = -^ m., or $ 004. Int. of $1 for 18 mo. 24 da., (^ 6% = $.094. $.094 X 50 := $4.70, Int. of $50, @ 6^. $4.70 X f = $6.27, Int. of $50, @ 8%- 330 PERCENTAGE. What is the interest @ 6% 33. Of $200 for 200 mo. ? Ans. $200. 34. Of $150 for 18 mo. 12 da. ? Ans. $13.80. 35. Of $100 for 46 mo. 6 da. ? Ans. $23.10. 36. Of $80 for 62 mo. 15 da. ? Ans. $25. 37. Of $750 for 15 mo. 18 da. ? Ans. $58.50. 38. Of $425 for 4 yr. 4 mo. 9 da. ? Ans. $111.14. 39. Of $187.25 for 10 mo. 21 da. ? Ans. $10.02. 40. Of $250.63 for 9 mo. 27 da. ? Ans. $12.40. 41. Of $75 for 19 mo. ? Ans. $7.13. What is the amount 42. Of $500 for 1 yr. 6 mo. 12da., at 8% ? Ans. $561.33. 43. Of $400 for 2 yr. 2 mo. 18 da., at 7^? Ans. $462.07. 44. Of $200 for 2 yr. 6 mo. 24 da., at 6% ? Ans. $225.67. 45. Of $100 for 2 yr. 6 mo. 6 da., at d% ? A7is. $122.65. 46. Of $150 for 3 yr. 8 mo. 9 da., at 10^ ? Ans. $205.38. 47. Of $225 for 4 yr. 15 da., at 12^ ? Ans. $334.13. 48. Of $350 for 4 yr. 21 da., at 4^ ? ^/zs. $406.81. 49. Of $480 for 5 yr. 1 mo. 27 da., at d%? Ans. $554.28. 50. Of $645 for 6 yr. 3 mo. 20 da., at 7%? Ans. $929.69. Example 6.— What is the interest of $600 for 33 da., at e%? ^ ( M. ^ 5t No. of mills interest of $1. SOLUTION. \^ ^ J^^g ^ g^^ ^ ^3 3^^ ^^^ What is the interest 51. Of $450 for 63 da., at 7% ? Ans. $5.51. 52. Of $525 for 93 da., at S% ? Ans. $10.85. 53. Of $750 for 123 da. at 9%? Ans. $23.06. 54. Of $875 for 48 da., at 6% ? Ans. $7. IlfTEREST. 331 55. Of 1900 for 78 da., at 10^? Ans. $19.50. 56. Of $1000 for 33 da., at 12^ ? Ans. $11. 57. Of $75 for 63 da., at 6%? Ans. $.79. 58. Of $150 for 34 da., at 6% ? Ans. $.71. 59. Of $225 for 93 da., at 7^? Ans. $4.07. 60. Of $540 for 123 da., at 8% ? Ans. $14.76. All of our calculations in Interest, thus far, have proceeded on the assumption that there are only 360 da. in 1 yr., whereas there are 365. In some States, calculations of interest are made on the latter basis. In which case we need make no changes so far as entire years axe concerned ; but when there is a part of a year we have this Rule. — Calculate the interest by the methods already adopted, and subtract from the result -^^ of itself. The accuracy of this Rule is evident from the fact that 865 da. — 360 da. = 5 da, ; and 5 da. are ^f 5, or -^^^ of a year. So that 5, or any number of days in the latter method, contains tj^^ less time than the same number of days in the former ; and as a consequence the interest should be ^^ less. In accordance with this Rule find the true interest in Problems 51 and 60 inclusive. Example 7.— Find the interest of £45 12s. 6d. for 3 yr. 6 mo. 18 da., at b%. Explanation. — We first SOLUTION. find the value of £45 12s. £45 12s. 6d. = £45.625 6d. in pounds sterling. Rate .05 Multiplying this result by the given rate gives the in- Int. for 1 yr. ... £2.28125 terest for 1 year. Time in yr 3.55 Multiply the interest for 1 yr. by 3.55, the time in Int. for given time £8.0984375 = ^ears, and we have the in. £8 Is. ll|d., Ans. ' terest = £8.0984375 = £8 Is. llfd., Ans. The method by aliquot parts or any other method may be used. 332 PERCEI^TAGE. 61. What is the interest of £50 6s. 9d. for 4 yr. 1 mo. 15 da., at b% ? Ans, £10 7s. 7d. 2.825 far. 62. What is the amount of £75 4s. 6d. for 5 yr. 2 mo. 20 da., at ^% ? Ans. £92 18s. .69d. 63. What is the amount of £100 10s. 3d. for 8 yr., at ^% ? Ans. £148 15s. 2.04d. CASE II. 354. The !Principal^ Interest, and Time given, to find the Mate %. Example.— At what % will 150 yield $7, in 2 yr. 4 mo.? Solution. — 2 yr. 4 mo. = 2^ yr. If $50 yield $7 in 2|, or \ yr. , in 1 yr. it will yield $3 ; and if $50 yield $3 in 1 yr., $1 will yield ^^ of $3, or 6^. Therefore, $50 will yield $7 in 2 yr. 4 mo., at 6% . Or, Since the interest of $1 for 2 yr. 4 mo., or 2^ yr., at 1^ is 2^, or \f, the interest of $50 for the same time and at the same rate is 50 times If — %\. And if the interest at 1 % yields %%, in order to yield $7, the rate % must he as many as $| is contained times in $7, or 6 times. Therefore, $50 will yield $7 in 2 yr. 4 mo., at 6%. PB OBLEM8. At what % will $300 pay $45 in 3 yr. ? $450 pay $135 in 5 yr. ? $575 pay $184 in 4 yr.? $625 pay $50 in 2 yr.? 1. 2. 3. 4. 9. At what % will 1100 pay $10 in 6 mo. 10 da. ? 5. $300 pay $63 in3yr.? 6. $250 pay $45 in 2 yr. 3 mo. ? 7. $500 pay $70 in 2 yr. 4 mo.? 8. $10 pay $1.40 in 3 yr. 4 mo. ? INTEREST. 333 ,4.^ ^ IWieitteii ^Aler eisc Example.— At what % will $150 pay $20.25 in 2 yr. 3 mo.? Explanation. — We first SOLUTION. divide the given interest $20.25 -r- 150 = $.135 hy the given principal to ($.135 ^ 2}) X 100 = 6, the rate %. ^^^ *^« ^^^e for 2J years. ^ We next divide this rate ' by 2J to find the rate for ^^•^^ V 1 0O'/ — fi^ 1 y^- = 06, which multi- 150 X 2i "^ /' ~ /'• pUed by 100, gives the rate % , 6. 355. Rule. — Take such a part of 100% as the given interest is of the product of the principal and time. Divide the given interest hy the interest of the given prin- cipal for the given time at 1% ; the quotient is the rate %. Ult^B -"") (354.^DS0Lnxi0^.) Note the introduction of the new element, tiTne, which in the fop. mula is represented by " t." PBOB TjEMS. At what rate % will 1. $450 yield $52.50, in 2 yr. 4 mo. ? Ans. 5%, 2. $200 yield $28.60, in 2 yr. 4 mo. 18 da. ? Ans. 6%- 3. $150 yield $60.72, in 5 yr. 9 mo. 12 da. ? Ans. 7^. 4. $100 yield $31.65, in 3 yr. 6 mo. 6 da.? ^ns. 9%. 5. $150 yield $55,375, in 3 yr. 8 mo. 9 da.? Ans. 10%. 6. $225 yield $131,625, in 4 yr. 10 mo. 15 da. ? Ans. 12^. 334 PERCENTAGE. CASE III. 356. TJie Interest f Tinie^ and Rate to find the JPHncipal, given, ©l^at ^^Xfoi ci^fe ^ Example.— What principal will, in 3 yr., at 5%, gain $45 ? Solution. — Since $1 in 3 yr. at 5% will gain $.15; it will require as many dollars to gain $45, as $. 15 is contained times in $45, or 300 times. Therefore, $300 will gain $45 in 3 yr. at 5^. mOBIjEMS. What principal will produce 1. $80 in 2 yr. 6 mo., at 8^ ? 2. $90 in 3 yr. 9 mo., at 4^ ? 3. $105 in 4 yr. 8 mo., at 6^? 4. $150 in 20 yr., at b% ? 5. $52.50 in 5 yr. 3 mo., at h% ? 6. $200 in 2 yr. 6 mo., at S% ? 7. $200 in 5 yr., at 5^? 8. $20 in 2 yr., at b% ? 9. $50 in 5 mo., at 1%% ? 10. $100 in 2 yr. 6 mo., at 6^ ? 11. $75 in 10 yr., at %^% ? 12. $500 in 5 yr., at 20^? 13. $25 in 1 yr. 3 mo., at 12^ ? 14. $100 in 10 yr., at 10^ ? +lWi«itten ^^Grci.8es"-6 Example. — What principal will in 3 yr. 4 mo., at produce $100 ? 1st solution $100 500. $.06 X 3f times. Hence, our answer is $500. Explanation. — Since $1 in 3^ yr. at 6% will produce 31^ times 6^ = 20/^, it will require as many dollars to produce $100 as 20,«^ is contained times in $100, or 500 INTEREST. 335 Explanation.— Interest at 6% for 2d solution. 31 yr. = 20%, or i of the principal. % ^ *-,^^ _ dbKoo ^^^^^ ^^^^ ^ ^ ^^ *^^ principal, the 6^ X 34 ~ ®^^^^® principal must be 5 times $100, or $500. 357. Rules — I. Divide the given interest ly the interest of $1 for the given time at the given rate j the quotient is the principal I = ^.J II. Take such apart of the given interest, as 100% is of the given rate % multiplied ly the given time, (d— ^ x P = ^. ) What principal will gain 1. $135 in 5 yr., at 6^ ? Ans. $450 2. $184 in 4 yr., at 8^ ? Ans. $575 3. $262.50 in 6 yr., at ^% ? Ans, $625 4. $400 in 4 yr. 2 mo., at 10^ ? Ans. $960 5. £252 in 3 yr. 6 mo., at 9^? Ans. £800 6. £50 in 2 yr., at h% ? Ans. £500 CASE IV. 358. The Tinier Hate %, and Amount given, to find the JPrincipal* ©Fat ^:^GFci^G^ Example. — ^What principal will in 1 yr. 8 mo., at Q%, amount to $5.75 ? Solution. — Since the amount of $1 for 1 yr. 8 mo., at 9% = $1.15, it will require as many dollars to amount to $5.75, as $1.15 is contained times in $5.75, or 5 times. Therefore, $5 will, in 1 yr. S m«., at 9%, amount to $5.75. 336 PERCEIS^TAGE. PnOBT. EMS. What principal will amount 1. In 3 yr., at Q%, to $118 ? 2. In 4J yr., at 8^, to $68 ? 3. In 2^ yr., at 9^, to 1363 r 4. In 5iyr.,atl0^,to$152? 5. In 1 yr. 9 mo., at 8^, to $228 ? 6. In 3 yr. 3 mo., at 12^, to $278 ? 7. In 6 yr. 2 mo., at Q%, to $685 ? 8. In 1 yr. 8 mo., at 9^, to $230 ? +lWi«itten ^T^ercise^fc Example. — What principal will in 1 yr. 4 mo., at 6^, amount to $1188 ? SOLUTION. $1188 $1 + $.08 $1100. Explanation. — Since in 1 yr. 4 mo., at 6%, $1 will amount to $1.08, it will require as many dollars to amount to $1183 as $1.08 is contained times in $1188, or 1100 times. Therefore, $1100 wHl in 1 yr. 4 mo., at 6%, amount to $1188. 359. Rule. — Divide the given amount hy the amount of $lfor the given time at the given rate. \-t^ T ~ ^'1 mOB LEMS. What principal will amount 1. In 6 yr., at 7^, to $887.50 ? Ans. $625. 2. In 2 yr., at 6%, to $784 ? Ans. $700. 3. In 3 yr., at 7^, to $166.37^? 4. In 2 yr. 4 mo. 18 da., at 6^, to $228.60 ? Ans. $200. 5. In 1 yr. 6 mo. 12 da., at S%y to $561.33^ ? Ans. $500. 6. In 2 yr. 2 mo. 18 da., at 7^, to $462.06|? Ans. $400. 7. In 3 yr. 8 mo. 9 da., at 10^, to $205.37^? Ans. $150. INTEREST 337 CASE V. 360. The Interest, JPrincipal^ and Mate % given, to find the Time. ©tat 3E:^GFciXfeX Example. — In what time will $75 gain $9, at 6^ ? Solution.— At G%, $75 will in 1 yr. yield 75 x $.06, or $4.50. If $75 yield $4| in 1 yr., it will take as many years to gain $9 as $4| is contained times in $9, or 3 times. Therefore, it will take $75, at 6%, 2 yr. to gain $9. PROBLEMS. 1. In what time, at Q%, will $10 gain $1.20 ? $1.80 ? $2.40 ? $7.20 ? $8.70 ? $10 ? Yc, will $15 gain $1.05? $8.40? $15? ^, will $25 gain $1.25 ? $4.33i? 2. In what time, at 7 $4.20? $5.25? $7.35? 3. In what time, at 5 $.93| ? $6.25 ? $1.50 ? $3.15? $2.50? $.62|? ^lWl itf en ^xerei^e^sT I- ExAMPLE^In what time wiU $137.50 gain $28.87^, at 7^? SOLUTION. Explanation. — At 7%, $137.50 will $28.87^ gain in 1 yr. 137.50 x $.07 = $9,625; and 07 X $137 50 ~ ^* ^^ $137.50 gain $9,625 in 1 yr., to gain $38,875 will take as many yr. as $9,635 is contained times in $38,875, or 3 times. Therefore, at 7% , $137.50 will gain $38,875 in 3 yr. 361* Rule. — Divide the given interest ly the interest of the given principal at the given rate for one year ; the quo- tient is the time in years. I 7^ = t.\ 22 338 PERCENTAGE. fJROBLEMS , In what time will 1. 1256.25, at 5^, yield $51.25 ? Ans, 4 yr. 2. '$450, at 12^, yield $18 ? Ans, 4^o. 3. $540, at 8^, yield $14.88 ? ^/is. 124 da. 4. $480, at 3^, yield $74.28 ? ^t^s. 5 yr. 1 mo. 27 da. 5. $225, at 12^, yield $131,625 ? Ans. 4 yr. 10 mo. 15 da. 6. $400, at 7^, yield $62.06f ? Ans. 2 yr. 2 mo. 18 da. mOMISSOBY NOTES. 363. A JPromissory Note is a written promise to pay a specified sum of money, either on demand, or at some designated time after the date of the note. 363. When the time of payment is specified in the note it is customary to allow three more days, called Days of Grace. 364. A note is said to mature on the day on which it is due. 365. The promise is usually either to pay the bearer, or some person, mentioned in the note, or to Tiis order, 366. A Negotiable Note is one that can be trans- ferred, or sold. 367. A Joint Note is one signed by two or more persons. 368. The Face of a Note is the sum promised in the note. This sum should always be written in words in the body of the note. A promissory note should have in it the words " value received," or "for value received," as without these words, the holder may be required to prove that value was received. In some States it is customary also to insert ** without defalcation/* though for what object is not very plain. INTEREST. 339 369. The Maker, or Drawer, of a note is the per- son who signs it. 370. The Holder, or Payee, is the person to whom it is to be paid. 371. An Indorser is the person who writes his name upon a note, or other obligation, usually upon its back, thereby becoming responsible for its payment. 37^. A JProtest is a legal notice to the indorser of a note that the maker has failed to pay the note when due. As notes are not always paid when due, and holders agree to take a portion of the payment, and wait for the balance, receiving interest on the unpaid part, we have the term 373. A JPartial JPayment, which is a payment of part of a note, or any other legal obligation, such as a mort- gage^ article of agreement, &c. If stated on the back of the in- strument in writing by the holder, it is called an indorse^ meiit. JPABTIAL PAYMENTS. To compute interest when partial payments have been made. Settlements are made by the Courts of the United States, and by most of the individual States, by what is known as the United States Rule. Example. $3000. Pittsburgh, Sept. 1, 1874 On demand, for value received, I promise to pay to the order of J. M. Logan, three thousand dollars, without defal- cation, with interest from date. James Meredith. 340 PERCEN^TAGE. On this note were these indorsements: Received, Jan. 1/75. $500 ; March 1/75, $25 ; Apr. 7/75, $600; Oct. 13/75, $1000 ; May 19/76, $200. How much was due June 1/76 ? SOLUTION. Principal $3000 Int. from Sept. 1/74, to Jan. 1/75, 4 mo 60 Amt. of Prin. to date of 1st Payment $8060 1st Payment made Jan. 1/75 500 2d Principal $25^0 Int. of 2d Prin. from Jan. 1 to Apr. 7/75 40 .96 Amt. to Apr. 7/75 $2600.96 2d Payment (not equal to Int.) and 3d Payment . . . 625. 3d Principal $1975.96 Int. of 3d Prin. from Apr. 7 to Oct. 13/75 61.255 Amt. of 3d Prin. to Oct. 13/76 $2037.215 4th Payment 1000. 4th Principal $1037.215 Int. from Oct. 13/75 to May 19/76 37.340 Amt. of 4th Prin. to May 19/76 $1074.555 5th Payment 200. 5th Principal $ 874.555 Int. from May 19 to June 1/76 1.749 Sum due June 1/70 $876,304 374. Itule of the United States Courts. Find the amount of the 'principal up to the time of the first payment, if that payment equals or exceeds the interest ; hut if it does not, then find the amount up to the time when the sum of two or more payments equals or exceeds the interest. From this amount subtract the first payment, or, if neces- sary, the sum of the two or more payments. Use the remainder as a new principal, and proceed as before, until time of final settlement. INTEREST. 341 PROBIjEMS. 1. $2500. Philadelphia, Feb. 12, 1873. On demand we jointly and severally promise to pay Robert Mackay, or order, twenty-five hundred dollars, for value received, with interest from date. Benjamin Steinman. Lewis A. Morton. Indorsements.— June 12, 1873, $25; Dec. 12, 1873, $150; June 12, 1874, $750 ; Dec. 12, 1874, $350 ; Apr. 12, 1875, $250. What was due Apr. 12, 1876? Ans. $1331.61. 2. $7500. New Orleans, July 1, 1874. Six months after date, I promise to pay to the order of Samuel Hill and Co., seven thousand five hundred dollars, at the First National Bank of New Orleans, for value received. Due Jan. 1/4, 1875. B. Simpson. Indorsements.— Jan. 4, 1875, $150; July 4, 1875, $1000; Aug. 19, 1875, $1000 ; Sept. 4. 1875, $3000. Find balance due Jan. 4, 1876, at h%. Ans. $2629.30. 3. $1500. New York, Aug. 26, 1873. Four months after date I promise to pay J. Bos well, or order, fifteen hundred dollars, for value received. Due Dec. 26/29, 1873. John Tweddle. Indorsements.— June 29, 1874, $50 ; Dec. 29, 1874, $350 ; Mar. 29, 1875, $20 ; Sept. 29, 1875, $370 ; Dec. 29, 1875, $200. What was due June 29, 1876, at 1% ? Ans. $717.91. 4. $1750^. St. Louis, Oct. 25, 1874. Four months after date, I promise to pay to the order of J. B. Jones, one thousand seven hundred fifty dollars and twenty-five cents, at the Merchants' Bank, value received. Due Feb. 25/28, 1875. G. J. Luckey. 342 PERCEN^TAGE. Credits on this Note.— Feb. 28, 1875, $50 25; June 14, 1875, $25 ; Oct. 20, 1875, $350 ; May 8, 1876, $575. Find balance due Sept. 8, 1876, at Q%. A^is. $878.86. Many business men settle notes and accounts on short time according to the following, often called 375. The Commercial or Merchants^ Hale, Find the amount of the principal at the time of settlement. Then find the amount of each payment from the time it loas made until settlement, and suhtract the sum of the amounts of the payments from the amount of the principal. Example. $950. Columbus, 0., Feb. 3, 1876. On demand, I promise to pay to the order of H. I. Gourley, nine hundred fifty dollars, without defalcation, value received. Edwaed M. JoHNSTOiq^. Indorsements.— Mar. 1, '76, $150 ; June 3, '76, $96 ; July 8, 76, $300 ; Dec. 20, '76, $250. How much was due Jan. 17, 1877? SOLUTION. Amount of $950, from Feb. 3, '76, to Jan. 17, '77, $1004466f. Amount of $150, from Mar. 1, '76, to " " " $157.90 Amount of $ 96, from June 3, '76, to " " *' $ 99.584 Amount of $300, from July 8, 76, to " *' " $309.45 Amount of $250, from Dec. 20, '76, to " " " $251,125 i 818.059. Am. $ 186.41. The following notes bear interest from date : 6. Face $1000; date Apr. 16, 1876. Payments.— $400, Aug. 16, 1876 ; $360, Oct. 20, 1876. Balance due Feb. 25, 1877, at 6^ ? Ans. $271.40. 6. Face, $1500 ; date, Aug. 16, '75 ; rate %, 10. INTEREST. 343 Payments.— $300, Nov. 1, 75 ; $200, Feb. 26, 76 ; $250, Apr. 22, 76 ; $400, June 1, 76. Balance due Aug. 16, 1876 ? Ans, $450.56. 376. In Connecticut, the Supreme Court has adopted the following principles in the calculation of interest : First. — Payments made when Merest has run a year or more, and those less than the interest, are treated as in the U. S. rule. Secoi^d. — A payment made within a year from the begin- ning of any interest, draws interest for the rest of that year, if that year does not extend heyond settlement; and its amount must ie taken from the aynount of the principal for that year. But if the year does extend leyond settlement, the amounts are computed for hotJi principal and payment, to settlement. The difference of these amounts is the balance due. In some States, as in Vermont and New Hampshire, what is called ^^ Annual Interest,'^ is allowed; that is, if interest is not paid when due, it will bear simple interest. Example. — $100 is to be paid in 3 yr. with annual interest. At the end of the 1st year $6 int. is due. If these $6 are not paid until the end of the three years, two years simple interest ($6 x .13 = $.72) must be added to the interest. In like manner, if the $6 int. due at the end of the second yr. is not paid until the end of the 3d year, one year's simple interest ($6 x .06 = $.36) must be added. This will make the sum paid at the end of the three years = Principal, $100 + Ist year's int., $6 + 2yr.'s int. on 1st year's int., $.72 + 2d yr.'s int., $6 + 1 yr.'s int. on 2d yr.'s int., $.36 + 3d yr.'s int., $6 = $119.08, or $1.08 more than by TJ. S. rule. COMPOUND IlfTEBEST. In Compound Interest, as in Simple Interest, there are 5 cases. It is, howerver, thought best to discuss the one following, only : 344 PERCE KTAGE. 377. TJie rrincipal, Mate % and Time given, to find the Amount, and Interest, Example.— Find the amount and interest of $200 for 2 yr. 3 mo., at 5^. SOLUTION. Principal $200. Amt. at end of Ist year, $200 x 1.05 -- $210. , Amt. at end of 2d year, $210 x 1.05 = $220.50 Amt. at end of 2^ years, $220.50 x 1.01^ = $223.26 Subtract the Principal $200.00 And there remains the interest $ 23.26 378. Rule. — Find tlie amount for the 1st period of time, and malce it the principal for the 2d period; mahe the amount of the 2d the principal for the 3d, and so on. The last amount will he the one sought. SuUract the 1st Principal from the last amount ; the remainder is the compound interest. T n OB L EMS. At compound interest find the amount and interest 1. Of $300 for 2 yr. 4 mo., at 6%. Ans. $343.82; $43.82. 2. Of $500 for 5 yr., at 7^. Ans. $701.27; $201.27. 3. Of $350 for 4 yr. 6 mo., at 3^. Ans. $399.84; $49.84. 4. Of $600 for 7 yr., at 8^. Ans. $1028.29; $428.29. 5. Of $125 for 8 yi\, at 4^. Ans. $171.07; $46.07. 6. Of $375 for 6 yr., at 5^. Ans. $502.54 ; $127.54. Note.— Observe that the amount of any principal for a given num- ber of years is equal to the product of 1 plus the rate used as a factor as many times as there are units in the number of years, multiplied hy the principal. Thus, Ex. 2 is, 1.07 x 1.07 x 1.07 x 1.07 x 1.07 x $500 = $701.27 + . II^TEREST. 345 The labor of computing compound interest is greatly lessened by the use of the following 379. TABLE, Showing the amount of 1 unit of money at Compound Interest, at from dfo to 8% for periods not greater than 35 yr. Yr. 1 3 per cent. 4 per cent. 5 per cent. 6 per cent. 7 per cent. 8 per cent. 1.03 1.04 1.05 1.06 1.07 1.08 2 1.0609 1.0816 1.1025 1.1236 1.1449 1.1664 3 1.092727 1.124864 1.157625 1.191016 1.225043 1.259712 4 1.125509 1.169859 1.215506 1.262477 1.310796 1.360488 5 1.159274 1.216653 1.276282 1.338226 1.402551 1.469328 6 1.194052 1.965.319 1.340096 1.418519 1.500730 1.586874 7 1.229874 1.315932 1.407100 1.503630 1.605781 1.713824 8 1.266770 1.368569 1.477455 1.593848 1.718186 1.850930 9 1.304773 1.423312 1.551328 1.689479 1.838459 1.999004 10 1.343916 1.480244 1.628895 1.790848 1.967151 2.158924 11 1.384234 1.539454 1.710339 1.898299 2.104851 2.a31638 12 1.425761 1.601032 1.795856 2.012196 2.252191 2.518170 13 1.468534 1.665074 1.885649 2.13i928 2.409845 2.719623 14 1.512590 1.731676 1.979932 2.260904 2.578534 2.937193 15 1.557967 1.800944 2.078.928 2.396558 2.759031 3.172169 16 1.604706 1.872981 2.182875 2 540352 2.952163 3.425942 17 1.652848 1.947901 2.292018 2.69*773 3.158815 3.700013 18 1.70M33 2.025817 2.406619 2.854a39 3.379932 3.996019 19 1.753506 2.106849 2.526950 3.025600 3.616527 4.315701 20 1.806111 2.191123 2.653298 3.207135 3.8696&4 4.660957 21 1.860295 2.278768 2.785963 3.399564 4.140562 5.033833 22 1.916103 2.369919 2.925261 3.603537 4.430401 5.4.36540 23 1.973587 2.464716 3.071324 3,819750 4.740529 5.871463 24 2.032794 2.56:3304 3.22510'J 4.0489:35 5.072366 6.341180 25 2.093778 2.665836 3.386355 4.301871 5.427432 6.848475 Example 2. — What is the amount and interest of S250 for 9 yr. 6 mo., at 7% ? SOLUTION. Ami of $1 for 9 yr., @ 7^ = $1.838459 250 Amt. of $250 for 9 yr., @ 7^ = 1459.615 Amt. of $1 for 6 mo., @ 7% = 1.03^- Amt. of $250 for 9 yr., 6 mo. %1% = $475.70 Principal $250 Int. of $250 for 9 yr. 6 mo. %1% = $225.70 Explanation. —We first find in the Table opposite 9 in the 1st 346 PERCENTAGE. column, and under the column headed 7%, that the amount of $1 for 9 yr. @ 7% is $1.838459. Multiplying this by 250, we find the amount of $250 for the same period = $459,615. This result multiplied by the amount of $1 for 6 mo., at 7%, gives us $475.70, the amount of $250 for 9 yr. ^ mo., at 7%. Subtracting the original principal from the last amt. leaves the interest = $225.70. 380. Rule. — For a whole number of years. Multiply the amount of one unit of money hy the number of such units in the principal. The result is the amount. From this sub- tract the principal The result is the compound interest. For a final fraction of a year. Use the amount of the whole number of years as a principal for the fractional time. PROBLEMS, 1. What is the compound interest of $300 for 18 yr. 9 mo., at 4^, payable annually? Ans. $325.98. 2. What is the compound interest of $400 for 20 yr. 4 mo. ;24 da., at b%, payable annually ? Ans. $682,55. 3. What is the compound interest of $1000 for 25 yr. 8 mo., at 6^, payable annually ? Ans. $3463.55. 4. What is the interest of $1600 for 8 yr. 6 mo., at 6^, payable semi-annually ? Ans. $1044.56. Note.— Observe that this is the same as 17 yr., at 3%. 5. What is the interest of $1000 for 8 yr. 6 mo., at 8^, payable semi-annually ? Ans. $947.90. 6. What is the interest of $100 for 42 yr., at S% ? Ans. $2433.94. Note. — First find amount for 25 yr. ; then using this amount as a new principal, find its amount for 17 yr. SECTION VII > IDIS3COWMT' -^ ^*<^£(£>^ 381. Discount is a deduction from a price, or from a debt. Discount is of three kinds, namely: Trade, Bank, and True or Equitable, Discount. 383. Trade Discount is a deduction from a price, or from cost. Thus, a seller is said to discount 7iis goods, when he deducts a part of the price. An article is some- times said to be sold at a discount, when it is sold for less than it cost. Trade Discount is usually reckoned as a percentage of the price, or cost. 383. IBank Discount is a deduction made from the face or amount of a note, or other obligation, due at some future time. 384. A Hank is an institution legally established for the purpose of deahng in money. 385. A Bank of Deposit is a bank legally em- powered to receive and take charge of the money of others. 386. A Deposit is money, or its equivalent, intrusted to the keeping of a bank. 387. A Depositor is a person who makes a deposit. 388. A Check is a written order on a bank for money. 389. A JBanIc of Issue is a bank legally empowered to pay out its own promissory notes as money. These notes are frequently called bank notes, or bank bills. ^47 348 PEKCEKTAGE. 390. A Sank of Discount is a bank legally em- powered to loan money. Banks of Deposit, of Issue, and of Discount are often combined in one institution. The affairs of a bank are usually managed by a Board of Directors, chosen by the Stockholders. The board of directors select A President, who presides over their meetings, and in a bank of issue, signs all bills, or notes issued ; A Cashier, who has charge of the accounts of the bank, and who, also, signs bills, or notes ; One or more Tellers, who are called receiving and paying tellers, to designate their position. A bank is said to discount a note when it purchases the note for a sum less than the note will be worth when it becomes due. The sum which is paid for a note is called the avails, or proceeds, of the note. In bank discount three days of grace are always considered as part of the time a note has to run. Hence, a note for 30 days is always considered 33 days ; one for 60 days, 63 days, &c. ; and the discount is reckoned also, for this additional time, 391. True^ or Equitable jyiseoiint^ is also a de- duction made from the face or amount of a note, or other obligation due at some future time ; but it differs from Bank Discount in this respect, — 393. In Bank Discount, the face of the note, if luitJiout interest, or the amount of the face, if with interest, is taken as the base of calculation, and the discount is calcu- lated exactly as Simple Interest, Case I ; while in True Discou7it, not the face of the note, but a sum which, in a given time, will amount to the face or amount of the note, is the base. For Example.— The bank discount of $106 for 1 yr., at 6% (no days of grace) = $6.36 ; but the true discount of $106 is -^ of $100 = $6 : for $100 is the sum which in 1 yr., at 6% , will amount to $106. DISCOUNT. 349 Bank Discount. Bank Discount has Five Cases, which correspond to the Five Cases in Simple Interest. CASE I. 393. Tlie Bate %, Time, and Face or Amount of Note given, to find the Discount. Example. — What is the bank discount of $350, due 3 mo. hence, at %% discount ? What are the proceeds ? SOLUTION Explanation.— Adding 90 da. + 3 da. = 93 da.. Time. I ^^ ^'^'^'^.^ ^^^ ^^ * ^ da., time of Dis. = AV or $350 X .08 X tV. = i7.23J, Dis. ^a^/ y^. The interest of $350 — $7.23i = $342.77, Pro. $350, the face of the note, for xV?r yr» at 8 fc = $7.23^, or discount. The proceeds = $350 - $7.33^ = $342.77. The avails, or proceeds, may be found by subtracting the discoumt from the face, or amount, of the note. 394. KuLES. — Compute the interest on the face of the note {or, if it hears interest, on its amount when due) at the given rate % for three days more than the given time ; the result is the bank discount. (B x r x t = P.) Or, Multiply the interest of Si, for the given time, at given rate, ly the numher of dollars to he discounted, (r xt x B= P.) PROBLEMS. 1. What is the discount of $250, discounted for 4 mo., at 9% ? What are the proceeds ? Ans. $7.69 ; $242.31. 2. What are the discount and proceeds of a note for $75Q discounted for 90 da., at S% ? Ans. $15.50 ; $734.50. 35Q PERCENTAGE. 3. What are the discount and avails of a note for $3250, discounted for 60 da., at 1% ? A7is. 139.81 ; $3210.19. 4. A note for $650 dated Jan. 1/76, due in 3 mo., was discounted Mar. 1/76, at Q%. What were the discount and proceeds? ^ns. $3.68 ; $646.32. Note. — When a note is made for a given number of months, its maturity is reckoned by calendar months, but its Discount by the actual number of days. CASE II. 395. The Discount^ Time, and Face, or Amount given, to find the Hate %, Example. — The discount of a 3 mo. note for $100, was $1.55. What was the rate % discount ? SOLUTION. Explanation. — The int. 90 da. + 3 da. = 93 da. ^^ ^1^^ ^«^ ^^ ^^^ «^ xWyr-, $100 X .01 X AV = ^-2581, Dis. ^! 1^^ ''I ^-^.^^^ = discount. *-^^ *.^^r.: « /.^ Smce the ffiven discount is' $1.55 - $.3584 = 6, or 6^. ^^ ,._ ^^^ ^^^ ^^ ^^^^ ^ as many times 1 ^ as $.258^ are contained times in $1.55, or 6 times, or 6%. 396. KuLES. — Divide the given discount hy the discount of the face {or amount) for the time at 1%. The quotient is the rate %. (^--^^ = e) Or, Take such U part of 100% as the given discount is of the product of the time and face (or amount) of the note. (_^^XiO.^=iJ.) PBOBLEMS. Find the rate % at which a note is discounted when 1. The face is $400 ; discount, $4.20 ; time, 60 da. Ans. 6^ DISCOUNT. 351 2. The face is $150 ; discount, $4.10 ; time, 4 mo. What is the rate % ? Ans. 8%. 3. The face is $250 ; discount, $5.81 ; time, 3 mo. What rate % is charged ? Ans. 9^. 4. The face is $800 ; discount, $14.47 ; time, 3 mo. Re- quired the rate %. Ans. 7%- 5. The face is $900; discount, $19.20 ; time, 61 da. What is the rate % of discount ? Ans. 12^. CASE III. 397. The Discount, Time^ and Hate % given, to find the Face. Example. — The discount of a 4 mo. note, at 8%, was $4.10. What was the face ? Explanation. — The int. SOLUTION. or discount of $1 for 123 da., 120 da. 4- 3 da. = 123 da. or ^ jr., at 8% is $.0273^. II X .08 X tV^t = $.0273^, Dis. Since the given dis. is $4.10, $4.10 - $.0273J = 150, or *^" ^^"^ ^/. *^" ^f " ^^,to? 41 KA T? ^® many times $1 as $.0273^ $150, h ace. ^^^ contained times in $4.10, or 150 times, or $150, Face. 398. EuLE. — Divide the given discount hy the discount of$l for the given time, at given rate ; the quotient is the face, \r x t I ntOBLEMS . 1. Discount is $11.96 ; time, 4 mo. ; rate %, 10. What is the face of note ? Ans. $350. 2. Discount is $2.75 ; time, 1 mo. ; rate %, 6. Find face of the note. Ans. $500. 352 PEKCEKTAGE. 3. Discount is $4.20; time, 60 da.; rate %, 6. Eequired the face of note. Ans. $400. 4. Discount is $14.47 ; time, 3 mo. ; rate %, 7. Face of note required. Ans, 1800. 5. Discount is $13.06 ; time, 91 da. ; rate %, 5. What is the face of note ? Ans, $1000. CASE lY. 399. The Tinie^ Mate %^ and Proceeds given^ to find the Face, Example. — What must be the face of a note which, dis- counted for 4 mo., at 8^, will yield $145.90 ? SOLUTION. Explanation. — The int. 130 da. + 3 da. = 123 da. o^ dis. of $1 for 123 da., or $1 X .08 X ^ = $.02731, Dis. ^Viy yr., at 8% is $.0273,^. $1 - $.0273i = $.97261, f>ro. ^.^0273^^2;, lincf tlTe $145.90 - $.9726f = 150, or ^^,^ proceeds are $145.90, $150, Face. the face of the note must be as many times $1 as $.9726| are contained times in $145.90, or 150 times, or $150, Face. 400. KuLE. — Divide the given proceeds hy the proceeds of $1 for the time, at rate given ; the quotient is the face of the note. (^ — 7 -r = B.) PBOBZEMS, What must be the face of a note payable in 1. Thirty days, rate %,'6, to draw $100 ? Ans. $100.55. 2. Ninety days, rate %, 6, to draw $295.35 ? Ans. $300. 3. Three months, rate %, 9, to draw $244.19 ? Ans. $250. 4. Ninety days, rate %y 15, to draw $1000 ? Ans. $1040.31. DISCOUNT. 353 5. Time, four months; rate %, 8; proceeds, $2165.45. What is the face of note ? Ans. $2226.30. 6. Time, four months ; rate %, 8 ; proceeds, $7350. Re- quired the face of note. Ans. $7556.54. 7. Time, five months ; rate %, 10 ; proceeds, $800. Face of note required. ^?^5. $835.51. CASE V. 401. The Face, Mate % and Discount or Proceeds given, to find the Time. Example. — Paid $4.10 for discounting a note of $150, at 8^. For what time was it discounted ? SOLUTION. Explanation. — The int. $150 X .08 = $12, Dis. or dis. of $150 for 1 yr., at $4.10 -r- $12 = ^glgpo. or ^ yr. 8% = $12. Since the given tV^t yr. = 4 mo. 3 da. ^^' ^ ^^•^^' *^^ *^« ^^ ^ "^ be that part of a yr. repre- sented by the quotient of $4.10 -4- $12, or yVjT* o^" xVff TT. = 4 mo. 3 da.. Time. 403. KuLE. — Divide the given discount hy the discount of the face at the given rate for one year ; the quotient is the time, ( 73 = t. ) \B X r I PBOBJLEMS. 1. The face of a note is $300 ; discount, $4.65 ; rate %, 6. How long has it to run ? Ans. 93 da. 2. The face of a note is $500 ; discount, $2.83 ; rate %, 6. In what time will it be due ? Ans. 34 da. 3. The face of a note is $250 ; discount, $5.81 ; rate %, 9. How long before it matures ? Ans. S mo. 3 da, 4. The face of a note is $350 ; discount, $11.96 ; rate %, 10. How long has it to run ? A7is. 4 mo. 3 da. 354 PERCEKTAGE. 6. Its face is $800 ; discount, 814.47 ; rate ^, 7 ? A71S. 3 mo. 3 da. 6. Its face is $1000 ; discount, $13.06 ; rate %,5? Arts. 94 da. 7. Its face is $100 ; discount, $1.55 ; rate ^, 6 ? Ans. 3 mo. 3 da. OOrue or JEquitahle Discotmt. True Discount^ like Banh Discount, has Five Cases, corresponding, with but httle variation, to the Five Cases in Simple Interest. The attentive pupil -will have noticed that the party who sells his note at a bank pays a discount not only on the mcmey he recei'oes (the proceeds), but also on what he allows, or pays, the hank for the v-se of the money. If the discount were for a long time, the injustice of this would be quite evident. For Example.— Suppose a man wished to sell his note of $1000, running 10 years, at 10^ discount. How much would the note on this supposition yield him ? Ans. Nothing. For at 10% for 10 yr. the dis- count of $1 = $1, and the discount of $1000 = $1000, and $1000 (Face) - $1000 (Discount) = 0: Banks, however, do not discount notes for long periods. In True Discount, discount is reckoned only on the »um paid for the note, mortgage, or other obligation. . 403. Case 1.— Present Worth, Bate %, and Titne given, to find the Discount, or the Face* 404. EuLE. — Multiply the present worth ty the rate, and faat product hy the time, {B x r x t = P,) (353. I.) To find the Fa4^, add to the present worth, the discount. 405. Case II. — Present Worth, Time, and Discount given^ to find tlie Rate %• DISCOUNT. 355 406. EuLES. — Take such a part of 100% as the given discount is of the product of the present worth and time. Divide the given discount hy the interest of the present toorth for the given time at 1% ; the quotient is the rate %. in — ^? — i = ^l (355.) \B X 1% xt J ^ ' 407. Case 111.— Discount^ Hate %, and Time given, to find Present Worth. 408. Rule. — Divide the given discount hy the interest of SI for the given time and rate; the quotient is the present worth. [^^ = b) (357. L) 409. Case lY.—Face or Ainounty Hate %, and Thne given, to find Present Worth, 410. Eule. — Divide the face, or amount, hy the amount of $1 for the given time and rate ; the quotient is the present worth. (jT^x7 = ^') (3S9-: In importance, this Case would come first ; but in order that the Cases of Interest, Bank Discount, and True Discount may be uniform, it has been thought best to place them in the present order. Of course, the Present Worth having been found, the Discount may be obtained by subtracting the Present Worth from the Face. 411. Case Y. — The Present Worth, Hate %, and Discount given, to find the Time. 412. Rule. — Divide the given discount by the interest of the Present Worth, at the given rate, for 1 ye^r ; the quo- tient is the time, i ^ = t.\ (361.) 356 PERCENTAGE. PR O It T.EMS. Case I. — Find the discount and face of a note whose 1. Present Worth is $150 ; rate %, 6 ; time, 4 mo. Ans. $3; $153. 2. Present Worth is $1200 ; rate %, 7 ; time, 6 mo. Ans. $42; $1242. 3. Present Worth is $600 ; rate %, 4 ; time, 1 mo. Ans. $2; $602. 4. Present Worth is $3802.281; rate %, 7; time, 4 yr. 6 mo. Ans. $1197.719 ; $5000. Case II. — Find the rate % discount of a note whose 5. Present Worth is $400 ; discount, $5 ; time, 3 mo. Ans. 6%. 6. Present Worth is $325 ; discount, $6.50 ; time, 2 mo. 12 da. Ans. 10%. 7. Present Worth, £800; discount, £12 8s.; time, 3 mo. 3 da. Ans. Q%- 8. Present Worth, $3802.281 ; discount, $1197.719 ; time, 4 yr. 6 mo. A7is. '7%. Case III. — Find the present worth of a note whose 9. Discount is $3; time, 4 mo.; rate %, 6. Ans. $150. 10. Discount is $2.60 ; time, 4 mo. 10 da. ; rate %, 9. Ans. $80. 11. Discount is $452.85 ; time, 3 yr. 9 mo. ; rate %, 7. Ans. $1725.15. 12. Discount is $91.07 ; time, 2 yr. 8 mo. ; rate %, 6. Ans. $569.19. Case IV. — Find the present worth of a note whose 13. Face is $602 ; time, 1 mo. ; rate %, 4. Ans. $600. 14. Face is $2178; time, 3 yr. 9 mo. ; rate %, 7. Ans, $1725.15. DISCOUNT. 357 15. Face of a note is $5000 ; time, 4 yr. 6 mo. ; rate %, 7. What is the present worth ? Ans, $3802.281. 16. Face of a note is $5560; time, 1 yr. 6 mo.; rate %, 7. Find the present worth. Ans. $5031.67. Case V.— To find time. 17. Present Worth of a note is $1200; discount, $42; rate %, 7. What is the time ? Ans.Q ma 18. Present Worth of a note is $820; discount, $12.71; rate %, 6. How long does it run ? Ans. 3 mo. 3 da. 19. Present Worth of a note is $3802.281; discount, $1197.719; rate %, 7. Find fche time. Ans. 4 yr. 6 ma 20. Present Worth of a note is $3000 ; discount, $540 ; rate %, 5. What is the time ? Ans. Z yr. 7 mo. 6 da. SECTION VIII. ■ST'OCKS i ^ 413. StocJcs are the funds, or capital, of Corporations. 414. A Corporatiofi is a company authorized hy a charter, or law, to act and be considered as a single indi- vidual, 415. A Charter is a legal instrument incorporating certain persons, and defining the powers and duties of the corporation. 416. Stocks are usually represented by a certain num^ ber of equal parts called Shares, which are usually $100 each, though they may be of any value agreed upon. 358 PEKCENTAGE. 417. Certificates of Stocky called Scrij)^ are state- ments showing the number of shares which an individual owns. 418. Stockholders are the owners of the shares of stock. 419. Stocks are at par when the shares sell for their face. 420. Stocks are above par, or at a premiiun, when the shares sell for more than their face. 431. Stocks are heloiv pwr, or at a discount, when the shares sell for less than their face. 422. The Market Value of stocks is the price at which they sell. 423. Dividends are portions of the earnings divided among the stockholders in proportion to the stock they own. 424. Assessments are contributions made by stock- holders to supply the deficiency, when from any cause the expenses are in excess of the income. 425. I^onds are written obligations by which parties bind themselves to pay specified sums, at or before certain times specified in the bonds. Bonds are of various kinds : United States Bonds, State, County, City, Borough, Eailroad, Express, Bridge, Mining Co. Bonds, &c. Of these, the XJ. S. Bonds are probably the most impor- tant. The principal ones are 1. " 6's of '81 ," bearing 6% interest, and payable in 1881. 2. "5-20's," bearing 6% interest, and payable in not less than 5 nor more than 20 yr. from their date, at the pleasure of the govern- ment. Interest paid semi-annually in gold. STOCKS. 359 3. "10-40's," bearing 5% interest, redeemable after 10 yr. from their date ; interest semi-annually in gold. 4. "5's of '81," bearing ofo interest, redeemable after 1881 ; inter- est paid quarterly in gold. 5. "l|'s of '86," bearing 41% interest, redeemable after 1886; interest paid quarterly in gold. 6. "4's of 1901," bearing 4% interest; the principal payable after 1901 ; the interest paid quarterly in gold. 436. In Coni2)utati07is of Stocks and Bonds it is important to remember 1st. That premium and discount are reckoned as a percentage of the par value, and not the market value. 2d. That Par Value is the same as Base lq Percentage. That Rate Per Cent is the same as llate Per Cent in Percentage. That Premium, or Discount, is the same as Percentage in Per- centage. That Market Value is the same as Sum or Difference, in Per- centage. Hence, in Stocks and Bonds, we have simply to apply the rules in Percentage. PRO B L EMS. 1. What cost 30 shares bank stock, par value $100, at \^% premium, brokerage ^% ? Solution.— |100 x 30 x 1.09 + (3000 x .00^, brokerage) = $3277.50, Ans. 2. What cost 25 shares railroad stock, par value 150, at 12|^^ discount ? Solution.— $50 x 25 x .87| = $1093.75, Ans. 3. What cost 100 shares P. R. R. stock, par value $50, at 9^% premium; brokerage, ^% ? ^ns. $5500.25. 4. What cost 25 6% U. S. Bonds, par value $100, at 116| ? Ans. $2903.13. 5. What cost 30 $500 10-40's, at 16|^ premium, broker- age!^? Ans. $17550. 360 PEECEITTAGE. 6. What cost 50 IIOOO U. S. 6's of 1881, at 21 J^ pre- mium ? Ans. 7. What cost 30 shares Michigan Southern R R, par value $50, at 11% premium, brokerage 1^^? Ans. 11681.88. 8. Invested $12000 in stock whose par value was $10000. What % premium did I pay ? Solution.— $13000 — $10000 = $2000. J^^ x 100% = 20%, Ans. $10000 9. Paid $800 for 20 shares R R stock, par value $50. What was the rate % discount ? Solution.— $50 x20=$1000. $1000— $800= $200. J^ x 100% = 30%, ^715. 10. Paid a premium of $100 on 20 shares of mining stock whose par value was $50 per share. What was the rate % ? Ans. 10%. 11. Having purchased 25 shares of a quicksilver-mine, at an advance of $20 per share for $1750, what %, premium did I pay ? Ans. A0%. 12. How many $100 10-40's, at a premium of 16^, bro- kerage i%, can I purchase for $4998. 75 ? Solution.— At a premium of 16%, a $100 bond is worth $116; to this add \%,ot $.35 for brokerage, and the entire cost is $116.25. For $4§98.75, I can purchase as many $100 bonds as $116.35 is con- tained times in $4998.75, which are 43 times. Therefore, 43 is the number of bonds. 13. How many dollars in gold, at a premium of 12^^, brokerage ^^, can be bought for $1130 in greenbacks? Ans. $1000. STOCKS. 361 14. How many dollars in silver, at a discount of 2^, bro- kerage ifc, can be purchased for 19812.50 ? Ans. $10000. 15. How many shares P. K. E. stock, par value $50, can I purchase for $6875, when they are at a premium of 9|-^, and brokerage costs me i% ? A7is. 125 shares. 16. How many $500 bonds of '81, at a premium of 16-^^, brokerage 1^, can be purchased for $58250 ? A71S. 100 bonds. 17. At what price will a 6% bond of $500 yield 8% ? Solution. — 6% of $500 = $30. At 8% it will require as many dollars to yield $30, as $.08 is contained times in $30, that is, 375 times ; or, at $375, a 6% $500 bond will yield Sfo. At what price will a 18. 10^ bank stock, par value $100, yield S% interest ? Ans. $125. 19. S% raiboad bond of $1000, yield 6% ? Ans, $1333J. 20. 10-40 U. S. $500 bond, yield 6^%, when gold is at a premium of 12 J^^ ? Ans. $432.69. 21. 6% mining stock, par value $50, pay S% ? Ans. $31.25. 22. Which is the better investment, 5-20*s at 122, or 10-40's at 116 ? Ans. The former. 23. Which is the better investment, " 4|'s of '86 " at par, or"4'sofl901"atlH^dis. ? 24. Which is the better investment 'aO-40's" at 112, or " 4i's " at par ? Ans, The latter. SECTION IX jj (T^^S) ® (f) ^i:^^ ^ ]BX.C'H^Mrafty or Tret^ is an allowance for waste in handling. 459. Gross Weight is the weight of the goods and the thing which contains them. 460. Wet Weight is weight remaining after the de- ductions. In the collection of specific duties, after all allowances are made, the problem reduces to one of simple multiplication. In calculating ad valorem duties, after the deductions are made the problem becomes one in percentage. Pit OBLEMS, 1. What is the duty on 150 boxes raisins, 16 lb. per box, at %f per lb. ; tare 25^ ? Solution.— 150 boxes of 16 lb. each — 2400 lb. The tare, 25% of 2400 lb. = 600 lb. Subtracting the tare from 2400 lb , there remain 2400 lb. — 600 lb. =: 1800 lb., on which to collect the duty of 2^ per lb. 2i^ x 1800 = 3600)* = $36, Arts. 2. What is the duty on 250 T. steel, at $45 per T.? Ans. $11250. 3. What is the duty on 50 bbl. sugar of 250 lb. each, tare 10^, atlj^? Jt^s. $168.75. 4. What is the duty on carpeting invoiced at £560 5s., at Solution.— £560 5s. = £560.25. $4.8665 x 560.25 = $2-726.4566}, which at 35% gives $954.26, Ans. 5. What is the duty on 50 crates of crockery, Invoiced at $9500 ; breakage, 10^ ; at 30^ ad valorem ? Ans. $2565. TAXES AND DUTIES 371 6. What is the duty on 400 doz. bottles of Porter, in- voiced at $350 ; breakage, 10^ ; at 25^ ad valorem ? Ans. 178.75. 7. What is the duty on leather invoiced at $9740 ; dam- ages, 40^; at 15^ ad valorem ? Ans. $876.60. 8. A manufacturer paid to the U. S. Government $2500 duties; excise taxes, $3500; income tax, $450; and used revenue stamps to the value of $1750, and postage-stamps worth $963. How much did he contribute to the support of the National Government ? Ans. $9163. SECTION XI ^ WAMTmmmmmiW' ■ ■|! io- 461. Partnership is the association of two or more persons who unite their money and labor in the transaction of business. Partnersliip is of two kinds. Simple and Compound. Simple Partnersliip is that in which the capital of each partner is employed for the same period of time. Compound Partnership is that in which the capital of each partner is employed for uneqvM periods of time. 462. Partners are persons associated in business. 463. A business association is called a Company ^ Firnif or Souse. 464. Capital, or Stock, is the property employed in business. 372 PERCENTAGE. 465. A Dividend is that which is divided among partners. 466. An Assessment is a demand for money to be paid by each partner, for his share of the expenses, or losses. 46T. The Liabilities of a company are its debts. PJt O BL EMS, 1. A and B enter into partnership. xVs capital is $5000; B's, $3000. They gain $2400. What is each partner's share ? Explanation. — We SOLUTION. find the amount of cap- A's capital = $5000 ital invested by both B's " = $3000 partners to be — the entire gain is 5^2400. A and B's capital = $8000 we find the rate % A and B's gain = $2400 gain by Case II. in M« X 100^ = 30^ gain (399). Xar^iX $5000 X .30 = $1500, A's gain (29')'). Case I. of Percentage. $3000 X .30 = 900, B's " ^^^^^ ^1^° Examples 7, ^ 8, and 9, page 286.) 2. Two partners, M and N, having gained $2700, divided it with reference to the capital employed by each. How much will they each receive, provided M's capital is $9000 and N's $18000 ? Aiis. M, $900, and IST, $1800. 3. Messrs. Jones, Smith & Simpson lost in business $7500. What was the loss of each, on the supposition that Jones' capital was $15000, Smith's, $20000, and Simpson's, $25000 ? ^wg. Jones, $1875 ; Smith, $2500 ; Simpson, $3125. PARTKERSHIP. 373 4. Brown and Long shared £7284 12s. profits of specula- tion. B's stock was £5525, L's £7735. Find the share of each. Ans. B, £3035 5s.; L, £4249 7s. 5. The firm of Dayton & Co. dissolved, dividing £7896 18s. 6d. D's stock was £4000, and each of his two partners had £1000. Find the share of each. Ans. D, £5264 12s. 4d.; the others each, £1316 3s. Id. 6. In a partnership between A, B and 0, A invested $3000 for 6 mo.; B, $5000 for 8 mo.; and C, $6000 for 7 mo. They gained $5000. What was each partner's share ? Explanation. — $3000 SOLUTION. employed for 6 mo. = 6 x $3000 X 6 = $18000, A'S for 1 mo. $3000, or $18000 employed $5000 X 8 r= $40000, B's " " for 1 mo. ; $5000 for 8 mo. $6000 X 7 = $42000, C's " " = ^ ^ ^^^^^' ^' ^^^^ ^^^ 1 mo. ; and $6000 for 7 mo. $100000. = $42000 for 12mo. Hence, ,, the combined capital of A, /^' B, and C was equivalent $18000 X .05 = $900, A's gain, to $100000 for 1 mo. On $40000 X .05 = $2000, B's " H' ^^T!^ ^^l^ r"" ' ^ ' $5000, or 5%. Therefore, $42000 X .05 = $2100, C's " each gained 5% on what was equivalent to his stock for 1 mo. ; and A receives 5% of 6 x $3000 ; B, 5% of 8 x §5000; and C, 5% of 7 X $6000. 7. Taft and Rodgers divided $15000 July 1, 1875. T be- gan Oct. 1, 1874, with $24000, and took in R, Feb. 1, 1875, with $16800. Find their shares. A71S. T, $10800 ; R, $4200. 8. M, N, and P hired a pasture for $68.40. M pastured 1 cow 30 weeks; N, 1 cow 30 weeks, and another 20 weeks; and P, 2 cows 30 weeks, and 1 cow 12 weeks. What should eacli pay ? Ans. M, $13.50 ; N, $22.50 ; P, $32 40. t'oUU X 100^ 374 PERCENTAGE. 9. Siemon and Bright began business Apr. 1, 1873, each with a capital of $10000. They took in Williams Jan. 1, 1874, with $5000. They dissolved partnership July 1, 1875, sharing $31500 profit. Find the share of each ? Ans. S and B, each $13500 ; W, $4500. 10. Macnum & Seward began business Dec. 1, 1873 ; M, with $7500 ; and Seward, with $6000. July 1, 1874, M took out $1500. They dissolved partnership June 1, 1875, paying $5662.50 debt. What had each to pay ? Ans. M, $2962.50; S, $2700. 11. A school had an attendance one term equal to that of 1 pupil 2300 days. Mr. Caldwell had sent 2 pupils each 90 days, and 1 pupil 60 days. How much of $149.50, the cost of the school, should he bear, if pay was according to attendance? Ans. $15.60. 12. Cook and Little formed a partnership for 1 year. C had I of the capital and L | ; they agreed to pay Shai-p J of the profits for managing the business. At the end of the year they divided $12000. What was the share of each ? Ans. S, $3000; C, $5400; L, $3600. 13. Sill, Eea and Collins shared $30000 profit on the fol- lowing conditions : for every $4 of S's stock there were $3 of R's and $2 of C's, and for every 3 mo. of S's time there were 2 mo. of R's and 1 of C's. What was the share of each ? Ans. S, $18000 ; R, $9000; C, $3000. 14. At the end of a year, A, B and C divided profits amounting to $7500, A receiving $3600; B, $2400; C, $1500. A had been engaged in the partnership 9 mo. ; B, 8 mo., and C, 3 mo. What was the ratio of their capital ? Ans. As 4, 3 and 5. SECTION XII. BJAMKR-CJfP'TCY '^^ (5^ 4i'QS. Ba^ihrtiptcy^ or Insolvency , is iDability to pay debts, through lack of property. • A person unable to pay his debts is called a bankrupt^ or an insol- vent. The liabilities of a bankrupt are his debts. 469. Debtors are persons who owe. 470. Creditors are persons to whom debts are owed. 4'71. Assets are such portions of property as can be appropriated to paying debts. 472. N'et JProceeds is the value remaining after all necessary deductions have been made. 473. An Assignee is a person selected to take charge of the assets of an insolvent, or bankrupt, and apply them to the payment of the creditors. PB. O B r. EMS. 1. Samuel Jackson placed in the hands of an assignee for the benefit of his creditors, a farm which sold for $5600; a mill property which brought $17000; bank stock worth 18000 ; insurance stock worth $5000. His entire liabilities were $69600. The assignee's charges were $10680, and other expenses were $1720. What % of his indebtedness did he pay ? A?is. 33^%. 2. How much did S. Thompson receive whose claim was $5000? Ans. $1666|. 375 376 P E R C E I^- T A G E , 3. How niiich did Amos Miller get whose claim was 118000? Ans. 16000. 4. How much did James Horner obtain whose claim was $600? Ans. $200. SECTION XIII. ^■¥EiKA(3.E ®F PiAYMEMTSJ 474. Average of JPayments is a process of finding the time for paying in a single payment two or more debts which are due at different times, so that neither debtor nor creditor may suffer loss of interest. 475. The Averaf/e, or Equated Tinier is the time determined by this process. 476. A Focal Date is any assumed date with which other dates are compared. 477. The Term of Credit is the time elapsing be- tween the contracting of a debt and its maturity. 478. The Average Term of Credit is the average time between the contracting of several debts and their maturities. Average of Payments is of two kinds, Simple and Compound. 479. In SlTTiple Average, all the items belong to one side of an account, that is, all are debts, or all are credits. 480. In Compound Average, there are items be- longing to both sides of an account. 481. A7i Account is a formal statement made by one party of his commercial transactions with another. AVERAGE AND PAYMEN^TS. 377 482. An account has two sides, one the debtor side, marked Dr.^ showing items of debt ; the other, the credit side, marked Cr., showing the items of credit. 483. The JBalance of an account is the difference between the amounts of the Dr. and Or. sides. Simple Average, Example. — A owes B $300, to be paid in 4 mo., and $500 to be paid in 6 mo. When could 1800 be paid by A to B, so as to cause no loss to either party ? SOLUTION. Explanation.— In this Exam- 1300 X .02 = $6.00. P^^ ^® regard A as having the $500 X .03 = $15.00. r «'' ^^V ' "" "r ^''' for 6 mo. Now, our question is, for how many months would the use of their sum, $800, be an equiv- alent. At 6% (we may use any rate % ), the interest of $300 for 4 mo. equals $6; and the interest of $500 at the same rate % for 6 mo. equals $15; together, $6 + $15 = $21. The interest of $800 ($300 + $500) for 1 yr. at 6 % is $48. Therefore, instead of the use of $300 for 4 mo. and $500 for 6 mo., the use of $800 for |i yr,,or5^ mo., (361) is an equivalent. All problems in Average of Payments may be solved by the fore- going method ; but when Interest tables are not at hand, the method generally adopted is exhibited in the following solution. Explanation.— The use of $300 $300 X 4 = $1200 ^'^^ ^ ^^^' ^^ equivalent to the use of tfj^nn ^ (K — *qonn ^ ^ ^^^^' ^'^ ^^^^^' ^^'' ^ '^''' ' *^^ $21. $800 X .06 = = $48. (353.) «yr. = = H mo. (361.) $800 ) $4200 to the use of 6 x $500, or $3000, 7j for 1 mo. ; and the sum of these, 5^ mo. ^^,^QQ ^ ^gO^^ ^^ ^^2QQ^ .g ^^^^ alent to the use of $4200 for 1 mo. But we wish to know for how 378 PEKCENTAGE. many months we can use $800, so that its use shall be equivalent to the use of $4200 for 1 mo., which is, as we have just shown, equivalent to the use of $300 for 4 mo., and $500 for 6 mo. This we find by dividing $4200 by $800. Our quotient, 5^, gives the number of months. Hence, the 484. Rule. — Multiply each item iy its time, and divide the sum of the products by the sum of the item^. PROBLEMS. 1. John Simpson bought a farm, for which he was to pay $500 cash ; $600 in 6 mo. ; $700 in 1 yr., and $900 in 15 mo. He afterward agreed to pay it all at one time. Find the time. Ans. 9|^ mo. 2. Bought goods to the amount of $5000, of which $1000 was to be paid in 1 mo., $1000 in 2 mo., $2000 in 3 mo., and the balance in 6 mo. If I gave a note at once for the whole amount, how long should the note run ? Ans. 3 mo. 3. Bought of Joseph Moon & Co. goods as follows : Jan. 3, 1876, $300, on 1 mo. credit; Jan. 4, 1876, $400, on 2 mo. credit ; Jan. 7, 1876, $500, on 3 mo. credit. At what time will $1200 ( =z $300 -f ^400 + $500) cancel the debts ? SOLUTION. Explanation. — We first Focal Date, Feb. 3, 1876. add the terms of credit to each Feb. 3/76, $300 x = date to find when each item is due. Mar. 4/76, $400 X 30 = 12000 We then fix upon Feb. 3 Apl. 7/76,. $500 X 64 = 32000 as the focal date (any other ~7ZZZ ^ TTITT date would answer our pur- 1300 ) 44000 ^^^ ^„a fi^a j,„^ J^„y 37 days intervene between Feb. Feb. 3 + 37 = Mar. 11/76, Ans. ^ ^"^ *^^ ^^^^ «^ ^^^^ «*^«' item becoming due. After which we proceed exactly as in the Example. That is, we multiply each item by its time from the focal date, and divide the sum of these AVERAGE or PAYMENTS. 379 products by the sum of the items. We find by this process the quo- tient to be 37, which is the number of days after Feb. 3/76, when the sum of the items is due. Adding 37 da. to Feb. 3/76, gives Mar. 11/76, Ans. Note. — Always use the exact number of days as found in Table, page 237. Find the average time of payments of the following purchases : (4.) Jan. 6, $450, (3 mo. cr.) Mar. 24, $600, (3 mo. cr.) Feb. 18, $800, (3 mo. cr.) Apr. 25, $700, (3 mo. cr.) Ans. June 7. (5.) Apr. 27, $900, (6 mo. cr.) June 29, $1000, (4 mo. cr.) May 16, $700, (5 mo. cr.) July 14, $1500, (3 mo. cr.) Ans. Oct. 21. (6.) July 31, $1200, (5 mo. cr.) Sept. 5, $8000, (3 mo. cr.) Aug. 15, $5000, (4 mo. cr.) Oct. 10, $6000, (2 mo. cr.) Ans. Dec. 11. (7.) Jan. 2, mdse. $1500, (4 mo. cr.) Feb. 2, mdse. $2000, (4 mo. cr.) Jan. 19, mdse. $2500, (4 mo. cr.) Mar. 2, cash $3000. Ans. Apr. 23. (8.) B owes A, as follows : Mar. 1, 1875, for cash, $400 ; Apr. 1, mdse. $300, (3 mo. cr.) ; June 1, mdse. $500, (3 mo. cr.). Mnd the average time for B's debts. Ans. June 16. (9.) I buy goods, Jan. 6, (3 mo. cr.) for $700 ; Feb. 18, (4 mo. cr.) for $800; Mar. 7, (3 mo. cr.) for $9000; Apr. 7, (1 mo. cr.) for $1600. Find the average time of payment. Ans. May 31. 380 PERCENTAGE. Compound Average, Example. — Find the equated time for paying the bal- ance of the following account : Dr. James Morrow in acct. with John Swan. Or. 1875. Dec. 6. 1876. Jan. 7. Feb. 2. 1876. To Mdse, 4 mo. 500 Feb. 1. Mar. 3. " " 3 mo. 600 Apr. 8.. " ** 3 mo. 400 By Mdse, 4 mo. 600 " " 5 mo. 300 " Cash, 100 Solution. Focal Date, Apr. 6, 1876. Apr. 6/76, $500 x = June 1/76, $600 x 56 = 33600 Apr. 7/76, $600 x 1 = 600 Aug. 3/76, $300 x 119 = 35700 May 2/76, $400 x 26 = 10400 Apr. 8/76, $100 x 2 = 200 1500 11000 1000 69500 1500 1000 69500 11000 500 ) 58500 Apr. 6/76 — 117 da 117 Dec. 11/75, Ana. Explanation. We first select the earliest date on either side as the focal date, and then proceed as in Problem 3, Simple Average, to find the product of each item by its time from this focal date. We next find the sum of these products on the Dr. and Or. sides, respectively. In this Ex. we find the sum of the products on the Dr. side 11000, and on the Cr. side 69500 ; and the difference of these sums 58500 We interpret the result in this way, — the Dr. side is entitled to the interest of $1 for 11000 days ; and the Gr. side, to the interest of $1 for 69500 days ; that is, the Cr. side is entitled to the interest of $1 for 58500 (69500 — 11000) more days than the Dr. side. The balance of items on the Dr. side is $500 ($1500 — $1000). Therefore, this $500 should have been AVERAGE OF PAYME1n[TS. 381 paid enough earlier than the focal date, to balance the interest of $1 for 58500 days on the Cr. side ; or 58500 -j- 500, or 117 days before the focal date, which gives Dec. 11/75, as the equated time for paying the balance of the account. We have used the earliest as the focal date ; the latest would be just as convenient. If an account is settled at the equated time, the balance of items is the true balance of the account ; if hefore the equated time, the bal- ance of items is discounted for the intervening time; if after the equated time, interest is allowed on the balance of items from that time to settlement. Bank Discount is used without days of grace. 485. EuLE. — Multiply each debt hy the units of time he- tween its maturity and the focal date, and divide the balance of products hy the balance of items. Then add the number of units of time i7i the quotient to the focal date, when both balances are on the same side of the account, but subtract from the focal date, when balances are on different sides of the account. Note —If the last date is selected as the focal date, we subtract from the focal date t?ie number of units of time in the quotient, when both balances are on the same side of the account ; and add th^m when balances are on different sides of the account. PBOB LEM S. De. (1.) James Johnston". Or. 1875. 1875. Apr. 1. To Cash, 1200 Feb. 8. By Mdse, 4 mo. 2300 " 21. " Mdse,4mo. 4000 Mar. 8. « (( « 1500 May 30. « <( <( 3000 " 28. U *€ tt 1000 Find the face of the note, and the date from which it bears interest, to settle the above account. Ans. Face, $3400; Date, Oct. 22/75. 382 PEliCEli! TAGE. (3.) Equate the following acct.: De. Andeew Kichey in acct. with Alex. Maetin^. Ce. 1875. Apr. 35. May 16. June 15. 1875. To Mdse, 3 mo. 1700 May 10. U (( it 2500 June 3. " •* <♦ 5000 1 " 27. By Mdse, 3 mo. " Casli, 2000 4000 8000 Ans. Martin owes $4800, and int. from May 14/75. (3.) Balance the following acct. at 6%, Sept. 1, 1875 : De. Chas. Atwell in acct. with M. F. Millee. Ce. 1875. 1875. May 13. To Mdse, 4 mo. 3560 40 June 24 By Cash, 2500 June 5. U (C t( 1775 25 ! July 14 " Mdse, 4 mo. 3335 50 July 19. (t It €( 2188 80 Aug. 1. (( ti (< 4516 10 Ans. Miller owes $2827.15 -$34.87 (dis. from Nov. 14) = $2792.28. Instead of using Notes and Rtjle on page 381, some accountants find the value of each item at the required date. Thus, in Ex. 3, the worth on September 1. 1875, at 6% (interest or discount depending on the date, and using Bank Discount without grace) Of .S3560.40 is $3553.28. Of $1775.25 is $1765.19. Of $2188.80 is $2159 .98. $7478.45. Of $2500.00 is $2528.75. Of $3335.50 is $3294.36. Of $4516.10 is $4447 .61. $10270.72. And $10270.72 - $7478.45 = $2792.27, amount due from Miller, Sept. 1. 1875, a result slightly different from that found by the other method. Solve in like manner Ex. 2. OUTLINE OF POWERS AND ROOTS. 487. A POWER. < 490. 1st Power. 4:91. 2d " yor Square. 492. Sd ** , or Cube. 493. Index, or Exponent. 494. Principle. 495. Rule. 488. INVOLVINQ A NUMBER. 489. A ROOT OF A NUMBER. 497. Square Root. ROOTS. J 498. Cube Root. 499. Rational Roots, 500. Swrda. 501. THE SIGN. k5 '502. Extracting. 503. Principle. 504. JBuie /or Integerg. 505. Bute /or Fractions. ^ > * 506. Triangle. SQUARE ROOT, M 507. iJiflr/t<- Angled ^ { Hypotenuse, 508. •< Perpendicular 509. Pnnciple. g Triangles. 510. i?«fe«. *1 511. J.r(f &c., to infinity. Ans, f 5. A man bought a house containing 24 windows, at 3 mills for the first window, 6 for the second, 12 for the third, &c. What did he pay for the house ? ^/^s. $50331.645. ANNUITIES. 529. An Annuity is a sum of money payable annually, or at regular periods of time. 530. The Afnoiint of an annuity is the sum of the payments and their interest for the specified time. To find the amount of an Annuity at Com^ pound Interest. Example. — To what will an annuity of $10 amount in 3yr., at6^? SOLUTION. Explanation. — We first $10 X 1.062 = $11,236 find the amount of $10 for $10 X 1.06 = $10.60 ^ ^^- ' *^^^ ^^ ^^^ ^^^ 1 y^- ' $10 = $10.00 !,f ^^^"^ f ^^^ ^^' ^" ''""'• The sum of these amounts, $31,836. $31.83(5, is the value of the p. annuity at the end of the 3d ' yr. Or, we may regard the $10 X 1.06^ $10 _ 13-j^ ggg^ annuity as the 1st term of 1.06 — 1 a Geometrical Series ; the amount of %l for 1 yr., at tJie given rate as the ratio ; and the number of years, the number of terms, and perform the example by (527, 528). ANNUITIES 407 To facilitate computations, the following table answers a very good purpose. It gives the amount of any unit for any number of periods from 1 to 40, at 5, 6, and 1%. TABLE Of amounts of $1 or £1 annuity per annum, at compound interest. £ 03 1 5 per cent. 6 per cent. 7 per cent. 1 5 per cent. 6 per cent. 7 per cent. 1.000 000 1.000 000 1.000 000 21 35.719 252 39.992 727 44.865 177 2 2.050 000 2.060 000 2.070 000 22 38.505 214 43.392 290 49.005 739 3 3.152 500 3.183 600 3.214 900 23 41.430 475 46.995 828 53.436 141 4 4.310 125 4.374 616 4.439 943 24 44.501 999 50.815 577 58.176 671 5 5.525 631 5.637 093 5.750 739 25 47.727 099 54.864 512 &3.249 030 6 6.801 913 6.975 319 7.153 291 26 51.113 454 59.156 383 68.676 470 7 8.142 008 8.393 833 8.6;>i 021 ^ 54.669 126 63.705 766 74.483 823 8 9.549 1C9 9.897 468 10.259 803 28 58.402 583 68.528 112 80.697 691 9 11.026 564 11.491 316 11.977 989 29 62.322 712 73.639 798 87.346 529 10 12.577 893 13.180 795 13.816 448 30 66.438 848 79.058 186 94.460 786 11 14.206 787 14.971 643 15.78;3 599 31 70.760 790 84.801 677 102.073 041 12 15.917 127 16.869 941 17.888 451 39 75.298 829 90.889 778 110.218 154 13 17.712 983 18.882 138 20 140 643 a3 80.063 771 97343 165 118.933 425 14 19 598 632 21.015 066 22.550 488 34 85.066 959 104183 755 128 258 765 15 21.578 564 23.275 970 25.129 022 35 90.320 307 111.434 780 138.236 878 16 23.657 492 25.670 528 27.838 054 m 95.836 323 119.120 867 148.913 460 17 25.840 366 28.212 880 30.840 217 37 101.628 139 127.268 119 160.337 400 18 28.132 385 30.905 653 a3.999 033 38 107.709 546 135.904 206 172.561 020 19 30.539 004 33.759 992 37.378 965 39 114.095 023 145.058 458 185.640 292 20 33.065 954 36.785 591 40.995 492 40 120.799 774 154.761 966 199.635 112 To use this table we have the following 531. EuLE. — Find the amount of 1 unit for the given time and rate, and multiply this amount hy the number of units. Example. — To what sum will an annuity of S225 amount in 12 yr., at 7^ ? SOLUTION. Explanation. — We find $17.888451 X 225 = $4024.901. in the table, under the head- ing 7%, and opposite 12 in the left-hand column, the number 17.888451, which is the number of 408 PROGEESSIONS. dollars to which an annuity of $1 will amount in 12 yr,, at 7%. Mul- tiplying this by the given number of dollars, 225, we have $4024901, the value of an annuity of $225 running 12 yr., at 7%. PROBLEMS . 1. A gentleman deposited for his son $150 for 10 consecu- tive years in a savings bank that paid b% compound in- terest. To what sum did the annuity amount immediately after the 10th payment ? Ans, $1886.084. 2. A gentleman paid $400 a year rent, for 15 yr. To what sum did his 15 yr. rent amount, at 7^ ? Ans. $10051.61. 3. A merchant at the age of 30 insured his life for $10000, paying a premium of $23.30 on $1000. He died just after making the 30th annual payment. How much more money would his family have received had this premium been improved at 6^ compound interest ? Ans, $8420.56. 4 A father desired to deposit annually in a savings bank such a sum as would amount in 10 payments, at 5^, to $1886.684. What must be the annual deposit ? Solution. — Since in 10 yr., at 5 % , $1 annuity would amount to $12.5779 + , it will require as many dollars annually to amount to $1886.684, as $12.5779 + is contained times in $1886.684, or 150 times. Therefore, he must deposit annually $150, Ans. 5. A dealer in real estate sells $600 lots for $100 cash, and the balance in 10 equal annual payments, which pay both principal and interest, at 7^. How much is each pay- ment? Solution.— In 10 years, at 7%, compound interest, $500 will amount to $983.5755 (378). An annuity of $1 for 10 yr., at 7% , pro- duces $13.81645 (see Table). It will, therefore, take an annuity, or yearly payment of as many dollars as $13.81645 is contained times in $983.5755 to yield $983.5955 in 10 yr. $983.5755 -^ $13.81645 = 71. 189. The annual payment is, therefore, $71,189, Ana. 6. What would have been the annual payments at &% ? Ans. $67,934. OUTLINE OF MENSURATION. t O TERMS. AREAS. OF LINES ■ V. 533. ^ Surface. 534. An Area. 535. A Triangle. 536. Tlie Base. 537. The Altitude. 538. Tlie Diagonal. 539. To find area of any Parallelogram. 540. To find area of any Triangle {A). 541. To find area of any Trapezoid. 542. To find area of any Triangle (li). 543. To find area of any Plane Figure. 544. To find area of any Regular Figure. 545. 548. To find area of any Circle. 546. To finil Circumference. 47, To find Uiameter. 549. A Solid. '550. Convex Surface. 551. Entire Surface. SURFACE, -i 552. To find surface of Prism. 553. To find surface of Pyramid. 554. To find surface of Frustum. 555. A Cylinder. 557. A Cone. 559. A Frustum. 561. ^ Sphere. 563. 656. To find Surface. 658. To find Surface. 660. To find Surface. 562. To find Surface. VOLUME. 40Q To find volume of Prism or Cylin- der. 564. To find volume of Pyramid or Cone, 565. To find volume of Frustum. . 566. To find volume of Sphere. CHAPTER IX A MEM SJXIR^TiId^ 532. Mensuration treats of the measurement . of magnitudes. In reference to the kinds of magnitude, mensuration is of four kinds, namely, Lines, Angles, Suj'faces, and Solids. A Straight Line is one that does not change its direction. Lines and Angles, and Surfaces and Solids in part, have been some- what fully discussed in Arts, 202-224, and also 247-258. It is proposed in this Chapter simply to complete the subject of mensura- tion, so far as may be proper in a work of this kind, by adding some Definitions, Explanations, and Rules, which the pupil is now prepared to comprehend, with reference to surfaces and solids. SQUABE MEASUHB. 533. A Surface is that which has length and breadth without thickness (204). 534. An Area is a definite amount of surface. 535. A Triangle is a plane surface bounded by three straight lines. (Fig. 1.) Triangles are equilateral, isosceles, and scalene, when the three sides are equal, when two sides are equal, and when no two sides are equal, respectively. They are also right-angled, obtuse-angled, acute-angled, and equi- angular, when they have one right-angle, when they have OTie obtuse- angle, when all the angles are acute, and when all the angles are eqvM, respectively. 41U MEKSURATIOiq- 411 Fig ABC, Fig. 1, is a right-angled trian- gle, and also a scalene triangle. ADC, Fig. 2, and D C B, Fig. 4, are obtuse-angled-triangles, and also scalene tri- FiG. 2. Fig. 3. Fig. 4. A Quadrilateral is a surface bounded by four straight lines, Figs. 2, 3, and 4. Quadrilaterals are subdivided into par- allelograms, trapezoids, and trapeziums, that is, four-sided figures having their opposite sides parallel, ABEC, Fig. 3; having two sides parallel and two in- clined to each other, Fig. 4; having no two sides parallel. Fig. 2, respec- tively. Lines are said to be parallel when they will not meet, however far they may be produced, as C E and A B, Fig.,3 ; and D C and A B, Fig. 4. Parallelograms include rectangu- lar and square fig- ures (207, 208); the rhombus, all of whose sides are equal and angles oblique ; and the rhomboid, whose opposite sides are equal and angles oblique, ABEC, Fig. 3. A surface bounded by five sides is called a pen- tagon ; one B by six sides. 412 MENSURATION^. a hexagon ; by seten, a heptagon ; by eight, an octagon ; by ^«;i, a deca- gon ; by eleven, an undecagon; by twelve, a dodecagon. 536. The JBase of a plane figure is the side on which it is supposed to rest, as B 0, Fig. 1 ; A B, Figs. 2, 3, and 4. 537. The Altitude is the perpendicular distance be- tween the base and the vertex of the opposite angle, or between the base and the opposite side ; as A B, Fig. 1, and C D, Fig. 3, and D E, Fig. 4. 538. The Diagonal of a plane figure is the straight line joining the vertices of two angles not connected by one of the sides; as A 0, Fig. 2, and D B, Fig. 4. In Fig. 3, if we cut off the part ADC and place it to the right of the figure C D B E, so that A C will coincide with B E, we see that A B E C will be changed to a rectangle (207), which is equivalent to A B E C, and whose base is equal to A B, and altitude C D ; and we find the area by multiplying together the two dimensions A B and C D (209). Therefore, to find the area of any parallelogram we have this 539. EuLE. — Multiply the base hy the altitude. And since a triangle is one-half of the parallelogram having the same base and altitude, (See ABC, Fig. 3,) we find its area by this 540. EuLE. — A. Multiply the lase hy one-half the altitude. And since a trapezoid is only two -triangles, BCD and BAD, Fig. 4, and they have the same altitude, D E, and the base of one is A B, and D C may be regarded as the base of the other, we may take the sum of the areas of the two triangles as the area of the trapezoid. And as we have the area of one of the triangles equal to one-half its base multiplied by its altitude, and the area of the other triangle equal to one-half its base by the same altitude, we must have the area of the trapezoid equal to 54:1. One-half the su7n of its two sides, mMiiMed hy its altitude. MEI^SURATION. 41S The area of a triangle is also found by this 542. EuLE. — B. From the half sum of the three sides subtract each side separately. Theti extract the square root of the continued product of these remainders and the half sum of the sides.. The area of any plane figure may be found 543. By dividing it i^ito triangles, calculating the area of each triangle separately, and then finding the sum of these areas. If a figure is regular, that is, has its sides and angles equal, each to each, its area may be found 544. By multiplying the distance around it {the peri- meter), hy one-half the perpendicular distance from its center to one of its sides. If a regular figure has its sides exceedingly small and an exceed- ingly large number of them, the sides taken together form the circum- ference (223) of a circle, (222), and the perpendicular distance froip the centre to one of its sides is the radius (223) ; and, therefore, we 9ay that 545. Hie area of a circle equals the product of the cir- cumference, hy one-half the radius. It is proven in Geometry, that the ratio of the circumference of a circle to its diameter is 3.14159, nearly. By means of this truth we deduce various rules for determining different parts of a circle when other parts are known. Thus, since the circumference is 3.14159 times the diameter. To find the circumference when the diameter is known, 546. Rule. — Multiply the diameter, or twice the radius, by 3.U159. To find the diameter when the circumference is known, 547. Rule. — 1. Divide the circumference by SJJf.159, 414 MENSURATION". Or, 2. Multiply the circumference hy.81831 ( = j. From the foregoing rules we may readily deduce the following additional rules for finding the area of a circle. 548. Multiply the square of the diameter hy .786 J^, Multiply the square of the radius hy 3.1j^159. Multiply the square of the circumference hy ,079577, PROBLEMS. Find the area of a triangle whose 1. Base is 25 ft, altitude 6. Ans. 75 sq. ft. 2. Base is 100 rd., altitude 50. Ans. 2500 sq. id. 3. Base is 16 yards, altitude 16. Ans. 128 sq. yd. 4. Base is 37^ ft., altitude 25. Ans. 468| sq. ft. 5. Sides are 6, 10, and 12 rods. Ans. 29.933 sq. rd. 6. Sides are 9, 15, and 20 yd. Ans, 63.277 sq. yd. 7. Sides are 13, 26, and 35 ft. Ans, 139.77 sq. ft. Find the area of a parallelogram whose » 8. Base is 13, altitude 20 ft. Ans. 260 sq. ft. 9. Base is 250, altitude 10 rd. Ans. 2500 sq. rd. .10. Base is l^, altitude 16 yd. Ans. 200 sq. yd. 11. Base is 37J in., altitude 6 ft. A7is. 2700 sq. in. Find the area of a trapezoid whose bases 12. Are 6 and 10 ft., altitude 5 ft. Ans. 40 sq. ft. 13. Are 25 and 35 yd., altitude 9 ft. Ans. 90 sq. yd. 14. Are 25 and 35 ft., altitude 9 yd. Ans. 90 sq. yd. 15. Are 75 ft. and 17 yd., altitude 10 ft. Ans. 70 sq. yd. Find the area of a trapezuim whose diagonal 16. Is 21, perpendiculars 3 and 5 ft. A7is. 84 sq. ft. 17. Is 36, perpendiculars 21 and 30 ft. Ans, 918 sq. ft. 18. Is 31, perpendiculars 27 and 16 yd. Ans. 666|^ sq. yd. 19. Is 15, perpendiculars 8 and 10 rd. Ans, 135 sq. rd. MENSURATION^. 415 Find the area of a regular figure of 20. Five sides, each 10 ft; per. 6.88 ft. Ans. 172 sq. tt. 21. Six sides, each 10 ft. ; per. 8. 66 ft. Ans. 259.8 sq. ft 22. Eight sides, each 20 ft; per. 24.14 ft. A71S. 1931.2 sq. ft. Find the circumference of 23. A circle whose diameter is 10. Ans. 31.4159. 24. A circle whose diameter is 25. A7is. 78.54. 25. A circle whose radius is 10. Ans. 6^.8318. 26. A circle whose radius is 100. Ans. 628.318. Find the diameter of 27. A circle whose circumference is 36. Ans. 11.45916. 28. A circle whose circumference is 25. Ans. 7.958 Find the area of a circle whose 29. Circumf. is 20 ft ; radius, 3.1831. Ans. 31.831. 30. Circumf. is 25 ft ; radius, 3.97887. Ans. 49.7359. 31. Diameter is 13 rd. 32. Diameter is 16 yd. 33. Eadius is 35 ft 34. Circumference 13.5 ft. 35. Circumference 12.5 rd. Ans. 132.7326 sq. rd. Ans. 201.0624 sq. yd. A71S. 3848.46 sq. ft Ans. 14.5 sq. ft Ans. 12.434 sq. rods. SOLIDS. 549. A Solid f or Body^ is that which has length, breadth and thick- ness, or three dimensions. A JPrism is a solid, bounded by plane sur- faces, two of which, the ends, or bases of the prism, are equal and similar plane figures ; and the sides, or faces, parallelograms ; as Figs. 1, 2, and 3, and (215, Fig. 3.) 416 MENSURATION. Fig 3. Fig. 4. Both plane fibres and solids are often re- garded as having two bases, an upper and a lower base. In Fig. 1, A B C is the lower, and D E F, the upper base. In Fig. 2, A B C D, the lower, and E F G H the upper base. Prisms are named from the form of their bases. Fig. 1 represents a triangular prism, because its bases ABC and D E F are tri- angles ; Fig. 2, a quadrangular prism, be- cause its bases A B C D and E F G H are quadrilaterals; Fig. 3, a pentagonal prism, because its bases ABODE and FGHIJ are pentagons. The edges of a prism are the lines in which the bounding surfaces meet. Thus, in Fig. 1, A D, C B and B E are edges ; in Fig. 2, A E, B F, &c. are edges. A JRlght Prism is one whose faces are perpendicular to the base. Fig. 3 is a right prism, because each of its faces, as A B G F, is perpendicular to the base A B C D E. A Pyramid is a solid having for its bases any polygon, and for the rest of its surface, plane triangles. Fig. 4. It terminates in a point called its vertex, S, Fig. 4. Pyramids are named from their bases. The one represented in the margin is a pentan- gular pyramid, because its base is a pentagon. The altitude of a pyramid is the perpen- dicular distance from its vertex to its base. If the pyramid is a rigM pyramid, the perpen- dicular will fall on the middle of the base ; as, S 0, Fig. 4. The slant height is the length of a line drawn from the vertex perpendicular to one side of the base ; as, S M, Fig. 4. MENSURATION. 417 Fig. 5. The frustum of a pyramid is the part that remains after the top is cut off by a plane parallel to the base, Fig. 5. A B C D E repre- sents the lower, and fghij the upper base. They are named from their bases ; the one represented in the margin being pentagonal. The altitude of a frustum is the perpendicular distance between its bases. The slant height of a frustum is the perpendicular distance be- tween the two parallel sides of one of its faces ; as, L K, Fig. 5. SURFACES OF SOLIDS, 550. The Convex Surface of a prism, or pyramid, or frustum, is all of its surface except the base, or bases. 551. The Entire Surface is the convex surface and the surface of the base or bases. By insi)ection we can readily see that in order to obtain the surface of solids we have simply to apply the rules already learned. For example — to find the convex surface of the prism. Fig. 1, we have only to find the area of the three parallelograms A B E D, B E F C, and A C F D, and then find their sum, which is just the same as multiply- ing the sum of A B, B C, and C A by the altitude A D. If we wish the entire surface, we add the area of the two triangles, ABC, and D E F. Hence, to find the surface of a prism we have this 552. Rule. — Multiply the perimeter of the base hy the altitude. To this result, add the area of the bases for the entire surface. In like manner, to find the convex surface of a pyramid, we find the sum of the areas of the triangles that form its surface ; the slant 27 418 MENSUEATIOis. height of the pyramid being the altitude of the triangles. From which we deduce this 553. EuLE. — Multiply the perimeter of the base by one- half the slant height. To this add the area of base, for the entire surface. Again, since the surface of the frustum of a pyramid is composed of trapezoids, we see that the surface can be obtained by the following : 554. EuLE. — Multiply the sum of the perimeters of the parallel bases, by one-half the slant height. Add the areas of the bases, for the entire surface. P ROBIjEMS. What is the convex surface of a prism whose 1. Altitude is 10 ft. and perimeter of base 16 ft. ? Ans. 160 ft. 2. Altitude is 13 yd. and perimeter of base 14 yd. ? Ans. 182 sq. yd. 3. Altitude is 6J in. and perimeter of base 14 in. ? Ans. 87^ in. 4. What is the entire surface of a square prism whose altitude is 25 ft. and each side of the base 7 ft. ? Ans. 798 sq. ft. 5. What is the entire surface of a triangular prism whose altitude is 5 yd. and each side of the base, 6 yd. ? Ans. 121.176 sq. yd. 6. What is the convex surface of a pentangular pyramid whose slant height is 30 ft. and each side of its base is 3 ft. ? Ans. 225 sq. ft. 7. What is the entire surface of a quadrangular pyramid whose slant height is 25 ft. and each side of the base is 6 ft. ? Ans. 336 sq. ft. MENSURATIOif. 419 8. What is the entire surface of a triangular pyramid whose slant height is 35 in. and each side of its base is 6 in. ? Ans. 330.588 sq. in. 9. What is the convex surface of the frustum of an octa- gonal pyramid whose slant height is 20 ft. and each side of whose lower base is 6 ft. and upper base 4 ft. ? Fig. 6. Ans. 800 sq. ft. 555. A Cylinder is a solid whose ends are equal parallel circles, Fig. 6. When the line 'that joins the centers of the bases is perpendicular to the bases, the cylinder is called a Right Cylinder ; the line, MN, Fig. 6, is the axis and also the altitude of the cyl- inder ; and since the cylinder is a prism with a great number of faces, to find its convex and entire surface, we have only to write " circum- ference " for 'perimeter in the rule for finding the surface of a prism, and obtain this 556. Rule. — Multiply the circumference of the base ly the altiticde. To this result add the areas of the bases, for the entire surface. 5511.. A Cone is a solid having a circle for its base and tapering to a point called its vertex, S, Fig. 7. The altitude is the perpendicular distance from the vertex to the base. The slant height is the shortest distance from the vertex to the circumference of the base, AS, Fig. 7. A cone is a pyramid with a great number of faces ; and hence to find its surface we use the 55S. Rule. — Multiply the circum- ference of the base by one-half the slant height. To this add the area of base, for the entire surface. 420 MEKSURATIOK Fig. 8. 559. The JFrustufii of a Cone is the part left after the top has been cut off by a plane parallel to the base, Fig. 8. The artist has done his work so well that we need not define altitude (line representing the axis), slant height and radius. To find the surface we have the 560t EuLE. — Multiply the sum of the circumferences of the two hases hy half the slant height, and add the areas of hoses for the entire surface. Fig. 9. 561. A Sphere is a volume bounded by a uniformly curved surface, every point of which is equally distant from a point within called the center, Fig. 9. The Diameter of a sphere is a straight line passing through the center and terminating at both ends with the surface, MN, Fig. 9. The Radius is the distance from the center to the surface, OS, Fig. 9. The Circumference is the greatest distance around the sphere. The relations of a diameter, radius, and circumference of a sphere are the same as those that the diameter, radium, and circumference of a circle bear to each other. To find the surface of a sphere we use this 563. EuLE. — Multiply the diameter hy the circumfer- ence. Or, Multiply the square of the diameter hy S.lJflG, P110B1.EMS. Find the convex surface of a 1. Cylinder, alt. 10 ft. ; circum. of base 31.83. Ans. 318.3 sq. ft MENSURATION". 421 2. Cylinder, alt. 25 ft. ; radius of base 10 ft. Ans. 1570.8 sq. ft. 3. Cone, slant height 20 ft. ; radius of base 5 ft, Ans. 314.16 sq. ft. 4. Frustum of a cone whose slant height is 36 ft. ; radius of upper base 10 ft., of lower base 15 ft. Ans. 2827.44 sq. ft 5. Find entire surface of frustum of a cone, slant height 80 ft. ; radius of upper base 5 ft., and lower base 2^ ft. Ans. 569.415 sq. ft. 6. Find the surface of a sphere whose diameter is 25. A71S. 25 X 3.1416 X 25 r= 1963.5. 7. Find the surface of a sphere whose radius is 7 rd. Ans. 615.7536 sq. rd. VOLUMES. To find the volume of a prism or cylinder, 563. Multiply the area of the base by the altitude. To find the volume of a pyramid or cone, 564. Multiply the area of the base by ^ the altitude. To find the volume of the frusttim of a pyramid or cone, 5G5. To the sum of the areas of the two bases, add the square root of their jjroduct, and multiply this result by ^ of the altitude. To find the volume of a sphere, 5GG, Mtiltiply the cube of the diameter by .5236, Or, Multiply the cube of the radius by Jf.l888. PR on LJEM S. 1. Find the capacity of a cylindrical measure 18-|- inches in diameter, and 8 in. deep. Ans. 1 bu. 2. Of a cylindrical bucket 5 inches in diameter and 7 in. deep (wine measure). Ans. 2.38 qt. 433 MENSUKATIOK. 3. Find the volume of a triangular bar 15 ft. long, and whose sides are 1 in., 1 in., and 1.414 in. Ans. 90 cu. in. 4. How many conical glasses, each 3 inches in diameter and 4.902 in. deep, can be filled from a qt. bottle? Ans. 5. 5. What is the volume of a quadrangular pyramid, eacli side of the base being 100 ft. and the altitude 75 ft. ? Ans. 250000 cu. ft. 6. What is the volume of the frustum of a pyramid 24 ft. high, 6 ft. square at one end and 4 ft. square at the other ? Ans. 608 cu. ft. 7. How many bbl. will a cistern 12 ft. deep, Avith upper diameter 9 ft. and lower diameter 3 ft., hold ? Ans. 87.3 bbl. 8. How many cubic feet of iron in a cannon ball 36 in. in diameter? Ans. 14.1372. 9. How many cubic ft. of granite in a cylindrical monu- ment 36 ft. high, with a base whose radius is 2 ft. ? 10. How many square feet of sheet-iron J in. thick can be made of a cylindrical shaft 20 ft. long and 4 inches in diameter ? Ans. 83.776 sq. ft. 11. What are the relative values of a ball of gold 4 inches in diameter and a cylinder of same diameter and altitude ? Ans. 2 : 3. 12. What is the capacity of an oval lime-kiln, whose top, bottom, and greatest diameters are 6, 2, and 7 ft. ; the dis- tances of the top and bottom diameters from the greatest diameter being 5 and 15 ft. ? A7is. 370.26 bu. « Multiply the sum of the square of the top diameter and twice the square of the greatest diameter by the distance between these diam- eters. Do the same with the bottom and greatest diameter and their distance. Then multiply the sum of these two results by .2G18. [(36 + 2 X 49) X 5 + (4 + 3 X 49) x 15] x .2618 = 575.96 cu. ft. THE METRIC SYSTEM. 423 THE METRIC SYSTE3I. 567. The Metric System is a decimal system of Weights and Measures^ founded on a certain unit of length called the metre = 39.37079 inches. The system is an exceedingly valuable one, bat its limited use in this country does not justify our treating the subject as fuUy as its importance deserves. The Metric System adopts a standard unit of measure, and then forms lower denominations by the decimal snh-measures of this unit; and higher denomina~ tions by the decimal multiples. Thus, taking the metre as the unit, and using the Latin prefixes, deci (tenth), centi (hundredth), milli (thousandth), we have Deci-mQtYQ (dess'-e-meet-ur) =.1 metre (meet'ur). Centi-TCLQivQ (cent'-e-meet-ur) = .01 metre. Milli-m.QtrQ (mil'-e-meet-ur) = .001 metre. And by using the Greek prefixes deca (ten), lieda (hun- dred), kilo (thousand), myria (ten-thousand), we have Deca-TCieivQ (dek'-a-meet-ur) = 10 metres. Ifecta-metre (hec'-ta-meet-ur) = 100 metres. /u7o-metre (kil'-o-meet-ur) = 1000 metres. Myria-metre (mir'-ee-a-meet-ur) = 10000 metres. In like manner taking the litre as the unit of meas- urCf we have milU-litre, centi-litTe, deci-litTe, deca-litre, &c. The Units of Measure are 568. The Metre , the Linear Unit, nearly 1 ten-mil- lionth of the distance from the equator to either pole. 569. The Are (air), which is the Unit of Surface Meas- ure, and is a square whose side is equal to 10 metres, or 1 decametre. 570. The Stere (stair), which is the Unit of Solid Measure, and is a cube whose edge is 1 metre. 424 THE METRIC SYSTE 571. The Litre (leet-ur), which is the Unit of Measures of Capacity, and is a cube whose edge is .1 metre, or 1 decimetre. 573. Gratnme (gram), which is the Unit of Weight, This is a cube of pure water whose edge is .01 metre, or 1 centimetre, weighed in a vacuum, at a temperature, when its density is greatest, 39.2° Fahrenheit's Thermometer. JsrUMBBATION AND NOTATION. 573. Each standard unit is represented by the initial letter of its name. Thus, M, for metre ; A, for are ; S, for stere, &c. A multiple is represented by the first letter of its prefix followed by the first letter of the standard unit ; and a sub-multiple by the first letter of its prefix followed by the initial letter of the standard unit written in small letters. Thus, MM. stands ioi myriametres, mm. for millimetres ; DM. stands for decametres, dm. for decimetres, &c., &c., though in general, since all de- nominations are decimal, the name of the standard unit is the only name that is necessary to express a number. Thus, 21.632 M., or M. 21.632, OT2\y^Q^2\^iG2i.^ twenty-one and six hundred thirty-two thousandths metres ; 32.3 L., or L. 32.3 is read thirty-two and three-tenths litres, TABLES. Long and Linear Measure. .1 I I I i 1 I I 31476.3 4 1 Read, thirty-one thousand four hun- dred seventy-six and three hundred forty-one thousandths metres. Note. — Observe the resemblance between M. 31476.341, and $31476.341. Instead of placing the separatrix between the metres and decimetres, which we do in measuring cloth and ordinary distances, we may place it between the kilometres and hectometres, which we do when measur- ing long distances, as the length of rivers, &c. The number should then be read 31 and 476341 hundred-thousandths kllotnetres. By omitting the separatrix, the number maybe read 31476341 mil' Ihnetres. 10 mm. 1 Centimetre. 10 cm. 1 Decimetre. 10 dc. 1 Metre. 10 M. 1 Decametre. 10 DM. 1 Hectometre. 10 HM. 1 Kilometre. 10 KM. 1 Myriametre. THE METRIC SYSTEM. 425 Surface or Square Measure. This measure is used in measuring land and other sur- faces. The Unit of Measure is an are = 100 sq. M. = 119.60 sq. yd., or a hectare = 2.471 A. i . I ■.A. I 2 34.42 HA. : Two hundred thirty-four and forty-two hundredths ares. The denomination centare is sometimes written centim'6. It is customary also to express the area of ordinary sur- faces by the square of the linear metre, called the square meter = 1550 sq. in. TABLE, 100 cen tares = 1 are 100 ares = 1 hectare . TABLE. 100 sq. cm. = 1 sq, dm. 100 sq. dm. = 1 sq. m. «5 B a ^ ^ « 347.3017 is read three hundred forty-seven and two thousand sevens- teen ten-thousandths square me- tres. Solid or Cubic Measures. The denominations decastere and dedstere are sometimes used when applied to the measure of the volunie of fire- wood and building-timber, but ordinarily the stere = 35.316 cu. ft. TABLE. 10 decistere = 1 stere. 10 stere = 1 decastere. 37489.004 is read 37489 and 4 thousandths steres. Volumes, such as excavations, embankments, &c., are ex- pressed by the cube of some denomination of linear meas- ure; thus. 426 THE METRIC SYSTEM TABLE. 1000 cu. millim. = 1 cu. centim. 1000 cu. centim. = 1 cu. decim. 1000 cu. decim. = 1 cu. metre = 1 stere. £ J I I ^ ^ J s 15.113601603 is read 15 and 113001003 hUllonths cubic metres. Measure of Capacity, This corresponds to our Liquid and Dry Measures. The Litre = .908 qt. Dry Measure, or 1.0567 qt. Liquid Meas- ure (^1 kilogramme, or 2.2046 lb. Av.), is the Unit in Liquid Measure, and the Hectolitre in Dry Measure. TABLE, 10 millilitres 10 centilitres 10 decilitres 10 litres 10 decalitres 10 hectolitres = 1 centilitre. = 1 decilitre. = 1 Litre. z= 1 decalitre. = 1 Hectolitre. = 1 kilolitre. 3 2 1 . 4 7 5 6 is read 321 and 4756 ten-thousandths Li- tres = 3.214.756 Hectolitres. Weight. This corresponds to our Troy, Apoth dupois Weights. The IJiiit of Measure = 15.432 grains. TABLE. 10 Milligrammes =1 Centigramme. 10 Centigrammes =1 Decigramme. 10 Decigrammes =1 Gramme. 10 Grammes =1 Decagramme. 10 Decagrammes =1 Hectogramme. 10 Hectogrammes =1 Kilogramme, 10 Kilogrammes =1 Myriagramme. 10 Myriagrammes=l Quintal, 10 Quintals =1 Tonneau. ecaries, and Avoir- is the Gra^nme ^ g. s'S :S