SA MP L E COPY. New Practical Arithmetic J PRICE. Pop Intpoduotion, - - - - -• - - 5«^- cts. Allowance for old book in use of similap gpade, when given in exchange, - - - - - - _AO <3|^g_ Booka ordered for introduction will he delivered at above named prices in any part of the United States. Sample copies for examination, with a view to introduction, will be sent by mail to teachen or school officers on receipt of the introduction price. Address, Clark & Maynard, ' 5 Barclay St., ITew York. (P. O. Box 1619.) THOMSON'S NEW GRADtD StRlES. NEW PRACTICAL AEITHMETIC: FOR GRAMMAR DEPARTMENTS By JAMRS B. THOMSOS, LL. D., ^UTHOB OF DAT & THOMSON S ARITHMETICAL SERIES ; EDITOR OF DaT'S SCUOO'4 ALGEBRA, LSGENDRS'S GEOHETRT, ETC. TfllETY-FIFTH EDITION. NEW YORK: CLARK & MAYNARD, PUBLISHERS. 5 barclay street. Chicago: 46 Madisoj^ Street. -••■': .•ipifois.ok^s vMathematical Series I. A Graded Series of Arithmetics, in three Books, viz. : New Illustrated Table Book, or Juvenile Arithmetic. With oral and slate exercises. (For beginners.) 128 pp. New Rudiments of Arithmetic. Combining Mental with Written Arithmetic. (For Intermediate Classes.) 224 pp. New Practical Arithmetic. Adapted to a complete business education. (For Grammar Departments.) 384 pp. II. Independent Books. Key to New Practical Arithmetic. Containing many valuable sug- gestions. (For teachers only.) 168 pp. New Mental Arithmetic. Containing the Simple and Compound Tables. (For Primary Schools.) 144 pp. Complete Intellectual Arithmetic. Specially adapted to Classes in Grammar Schools and Academies. 168 pp. III. Supplementary Course, New Practical Algebra. Adapted to High Schools and Academies. 312 pp. Key to New Practical Algebra. With full solutions. (For teachers only.) 224 pp. New Collegiate Algebra. Complete Higher Arithmetic. (In preparation.) \* Each book of tJie Series is complete in itself. Copyright, 1872, by James B. Thomson. EDUCATION DEPT PREFACE THE New Practical Arithmetic now offered to the public, is the third and last of the works which constitute the author's " New Graded Series." The old "Practical" was issued in 1845, and revised in 1853. Since that time important changes have taken lilace in the commercial world. These changes necessarily affect business calculations, and demand corresponding modifications in text books. To meet this demand, the "New Graded Series" was undertaken. Each part of the series has been reinvesti- gated and rewritten; — the whole being readjusted upon the graded plan, and brought down to the present wants. Among the objects aimed at in the present work are the following : I. To make the definitions clear, concise, and com,' 2. To present the principles of the science in a series of distinct and consecutive propositions. 3. To lead the mind of the pupil, through the analysis of the examples immediately following the respective propositions, to discover the principles by w"hich all similar examples are solved, and enable him to sum up the prin- ciples thus developed, into a brief, comprehensive rule. 4. The "how" and the "why" are fully explained. 5. Great pains have been taken to ascertain the Standard Weights and Measures authorized by the Government; and to discard from the Tables such denominations as are ohsolete, or not used in this country.* * Laws of Congress ; Profs, Haspler, Bache and Egleston ; Reports of Supts. Harrison and Calkins ; also of the Committee on Weights and Measures of the Paris Exposition, 1867. ivi34:y349 PREFA CE. 6. The Metric System is accompanied with brief and appropriate explanations for reducing it to practice. Its simplicity and comprehensiveness have secured its use in the natural sciences and commerce to such an extent in this and foreign countries, that no student can be said to have a finished education, without a knowledge of it. 7. Particular attention has been paid to the develop- ment of Analysis, the grand common sense rule which business men intuitively adopt as they enter upon practical life. 8. The examples are new and abundant ; — being drawn from the various industrial arts, commerce, science, etc. 9. The arrangement of the matter upon the page, and the typogi-aphy, have also received due attention. Teachers who deal much with figures, will be pleased with the adoption of the Eranklin type. The ease with which these figures are read, is sufficiently attested by their use in all recent Mathematical Tables. Finally, it has been the cardinal object to adapt the science of numbers to the present wants of the farm, the household, the workshop, and the counting-room ; — in a word, to incorporate as much information pertaining to business forms, and matters of science, as the limits of the book would perm-it. In this respect it is believed the work is unrivaled. While it puts forth no claim to mathematical paradoxes, it is believed teachers will find that something worthy of their attention is gained, in nearly every Article. In conclusion, the author tenders his most cordial thanks to teachers and the public for the very liberal patronage bestowed upon his former Arithmetics, known as " Day and Thomson's Series." It is hoped the " New Graded Series" will be found worthy of continued favor. James B. Thomson. I5[ew York, Jvly., 187^ CONTENTS. PAfiB N"umber, 9 Rotation, lo Arabic Notation, lo Koman Notation, 15 To Express Numbers by Letters, - - - -17 JSTumerationy 17 French Numeration, 18 English Numeration, 20 Addition^ 21 When the Sum of each Column is less than 10, - 23 When the Sum of a Column is 10 or more, - - 24 Carrying Illustrated, 24 Drill Columns, 29 Subtraction, 31 When each Figure in the Subtrahend is less than that above it, 32 When a Figure in the Subtrahend is greater than that above it, 34 Borrowing Illustrated, 34 Questions for Review, _ . _ - - 38 Multiplication, 40 When the Multiplier has but one Figure, - - 43 When the Multiplier has more than one Figure, - 45 To find the Excess of 9s, 47 Contractions, -------49 VI COKTEN^TS. PASB Division^ 53 The Two Problems of Division, - - - - 55 Short Division, - - -. - - - '57 Long Division, - - - - - -- 61 Contractions, 65 Questions for Review, - - - - - 69 General Principles of Division, - - - -71 Problems and Formulas in the Fundamental Eules, 7 2 Analysis, - 77 Classification and Properties of Numbers, - - 80 The Complement of Numbers, - - - - 82 Divisibility of Numbers, S^ Factor Ing, - - - - - - - 85 Prime Factors, 86 Cancellation^ 88 Greatest Common Divisor, - - - - - 91 Least Common Multiple, 96 Fractions, 99 To find a Fractional Part of a Number, - - 102 General Principles of Fractions, - - - - 103 Reduction of Fractions, - - - - - 104 A Common Denominator, iii The Least Common Denominator, - - - 112 Addition of Fractions, 114 Subtraction of Fractions, 117 Multiplication of Fractions, - - - - - 120 General Rule for multiplying Fractions, - - 125 Division of Fractions, 126 General Rule for dividing Fractions, - - 131 Questions for Review, 132 Fractional Relations of Numbers, - - - 134 J)ecinial Fractions, 139 Eeduction of Decimals, 144 Addition of Decimals, - 146 Subtraction of Decimals, - - - - - 148 Multiplication of Decimals, - - - - - 149 Piyision of Decimals, - - - • - 152 C K T E K T S . Vil PAGB United States Money, i54 Addition of U. S. Money, 15S Subtraction of U. S. Money, - - - "' - 15 9 Multiplication of U. S. Money, - - - - 160 Division of IT. S. Money, - - - - -161 Counting-room Exercises, 163 Making out Bills, ^164 Business Methods, ^^^ Cofnpound Nmnhers, 171 Money, ^7^ Weights, 175 Measures of Extension, i77 Measures of Capacity, 182 Circular Measure, - 184 Measurement of Time, - - - - - - 186 Reduction, 1^9 Application of Weights and Measures, - - - i94 Artificers' Work, _----- 196 Measurement of Lumber, ^9^ Denominate Fractions, ^^^ Metric Weights and Measures, - - - 207 Apphcation of Metric Weights and Measures, - 214 Compound Addition, - 216 Compound Subtraction, 219 Compound Multiplication, 224 Compound Division, 226 Comparison of Time and Longitude, - - - 227 Percentage^ ^2° Notation of Per Cent, 230 Five Problems of Percentage, - - - - 233 Applications of Percentage, - - - 241 Commission and Brokerage, ^ ' ' ' ^^l Account of Sales, -----•"4 Profit and Loss, ^"^l Interest, ------ ^J Preliminary Principles, - - - " 55 Six Per Cent Method, ------ 25 mi CONTENTS. PAoa Method by Aliquot Parts, - - - - -259 Method by Days, 260 Partial Payments, - - - - - -267 Compound Interest, - - - - - -273 Discount, - -276 Banks and Bank Discount, - - - - 278 Stock Investments, - - - - - -280 Government Bonds, 281 Exchange, .-.---- 285 Insurance, ------- 292 Taxes, --------- 295 Duties, -------- 298 Internal Kevenue, ------ 300 Equation of Payments, - - - - - 301 Averaging Accounts, -.---- 304 Ratio, ---..--- 307 Proportion^ ---.--- 309 Simple Proportion, - - - - - -311 Simple Proportion by Analysis, - - - - Z'^Z Compound Proportion, - - - - -316 Partitive Proportion, - - - - - -319 Partnership, - - * - - - - - 320 Bankruptcy, 323 Alligation, - - - - - - -324 TnroliitioTif 330 Formation of Squares, - - - - - 332 UroJution, -- - - - - - -333 Extraction of the Square Eoot, - - - 335 Applications of Square Root, - - - - 33^ Formation of Cubes, 343 Extraction of the Cube Root, - - - - 345 Applications of Cube Eoot, - - - - 349 Arithmetical JProgression, - - - - 350 Geometrical I*rogression, - - - ^^^ Mensuration, 355 Miscellaneous Examples, ----- 360 ARITHMETIC Art. 1. Arithtnetic is the science of numbers. Aritlimetic is sometimes said to be both a science and an art : a science when it treats of the theory and properties of numbers ; an art when it treats of their applications. Notes. — i. The term arithmetic, is from the Greek arithmetike, the art of reckoning. 2. The term science, from the Latin scientia, literally signifies knowledge. In a more restricted sense, it denotes an oi'derly ar- rangement of the facts and principles of a particular branch of knowledge. 2. Number is a unit,or a collection of nnits. A Unit is any single thing, called 07ie. One and one more are called tivo ; two and one more are called three ; three and one more, four, etc. The terms one, two, three, four, are properly the names of numbers, but are often used for numbers themselves. Note. — The term unit is fr9m the Latin unus, signifying one. 3. The TJnM One is the standard by which all num- bers are measured. It may also be considered the hase or element of number. For, all icliole numbers greater than one are composed of ones. Thus, tivo is composed of one and one. Three is one more than two ; but two, we have seen, is composed of ones ; hence, three is, and so on. Questions. — i. What is arithmetic? . What else is it sometimes said to be? When a science ? An art ? 2. Number ? A unit ? 3. The tcandard by which numbers are measured ? The base or element of number * 1.0 . KOTATIOK. 4. Numbers *are either abstract or concrete. An Abstract dumber is one that is not applied to any object; as, three, five, ten. A Concrete Number is one that is applied to some object; as, five peaches, ten books. NOTATION. 5. JS'otatioii is the art of expressing numbers by figures, letters, or other numeral characters. The two principal methods in use are the Arabic and the Roman. Note. — ^Numbers are also expressed bj words or common lan- guage ; but this, strictly speaking, is not Notation. ARABIC NOTATION. 6. The Arabic Notation is the method of express- ing numbers by certain characters called figures. It is so called, because it was introduced into Europe from Arabia. 7. The Arabic figures are the following ten, viz : 1, 2, 3, 4, 5, 6, 7, 8, 9, o. oiic, two, three, four, five, six, seven, eight, nine, naught. The first nine are called significant figures, or digits ; the last one, naught, zero, or cipher. Notes. — i. Tbe first nine are called aignijicant figures, because each always expresses a number. 2. The term digit is from the Latin digitus, a finger, and was applied to these characters because they were employed as a substi- 4. What is an abstract number? Concrete? 5. What is notation? The principal methods in use ? 6. Arabic notation ? Why so called ? 7. How many figures does it employ ? What are the £rst nine called ? The last one ? Note. Why called significant figures ? Why digits ? Meaning of digitus ? Why is the ki^t called naught ? Meaning of zero ? Of cipher ? NOTATION". 11 tute for the ^w^ers upon wliicli the ancients used to reckon. The term originally included the cipher, but is now generally restricted to i\xQ first nine. 3. The last one is called naught ; because, when standing alone, it has no value, and when connected with significant figures, it denotes the absence of the order in whose place it stands. 4. Zero is an Italian word, signifying nothing. The term cipher is from the Arabic sifr or sifreen, empty, vacant. Subsequently the term was applied to all the Arabic figures indis, crimiuately ; hence, calculations by them were called ciphering. 8. Each of the first nine numbers is expressed by a single figure, — each figure denoting the number indicated by its name. These numbers are called units of the first order, or simply units. 8, a, Nine is the greatest tiumber expressed by one figure. Numbers larger than nine are expressed thus : Te7i (i more than 9) is expressed by an ingenious de- vice, which groups ten single things or ones together, and considers the collection a 7iew or second order of units, called ten. Hence, ten is expressed by writing the figure i in the second place with a ciplier on the right; as, 10. The numbers from ten to nineteen inclusive are ex- pressed by writing i in the second place, and the figure denoting the units in the first ; as, II, 12, 13, 14, 15, 16, 17, 18, 19. eleven, twelve, thirteen, fourteen, fifteen, sixteen, seventeen, eighteen, nineteen. Twenty (2 tens) is expressed by writing the figure 2 in the second place with a ciplier on the right; as, 20. TJiirty (3 tens) by writing 3 in the second place with a cipher on the right ; and so on to ninety inclusive ; as, 20, 30, 40, 50, 60, 70, 80, 90. twenty, thirty, forty, fifty, sixty, seventy, eighty, ninety. 8. How are the first nine numbers expressed ? What are they called ? What is the greatest number expressed by one figure ? How is ten expressed ? Twenty ? Thirty, etc. ? The numbers between lo and 20 ? From 20 to 99 inclusive ? 12 l^OTATION". The numbers from twenty to thirty, and so on to ninety^ nine (99) inclusive, are expressed by writing the tens in the second place, and the units in the first ; as, 21, 2 2, 23, 34, 35, 46, 57, 99. twenty- one. twenty- two, twenty- three. thirty- four, thirty- five, forty- six, fifty- seven, ninety- nine. 8, h. Ninety-nine is the greatest number that can be expressed by tivo figures. A hundred (i more than 99) is expressed by grouping ten units of the second order together, and forming a new or third order of units, called a hundred. Thus, a hun- dred is expressed by writing 1 in the third place with two ciphers on the right; as, 100. In like manner, the numbers from one hundred to nine hundred and ninety-nine inclusive, are expressed by writ- ing the hundreds in the third place, the tens in the second, and the units in the ^r5^. Thus, one hundred and thirty- five (i hundred, 3 tens, and 5 units,) is expressed by 135. 8, c, Nine hundred and ninety-nine is the greatest number that can be expressed by three figures. TJiousands, and larger numbers, are expressed by form- ing other netv orders, called the fourth, fifth, etc., orders ; as, tens of thousands, hundreds of thousands, millions, etc. Notes. — i. The names of tlie first ten numbers, one, two, tliree, etc., are primitive words. The terms eleven and ticelve are from the Saxon endlefen and twelif, meaning one and ten, two and ten. Thir- teen is from thir and teen, which mean three and ten, and so on, 2. Twenty is from the Saxon tweentig, tween, two, and ty, tens; i. e., two tens. Thirty is from thir and ty, three tens, and so on. 3. The terms hundred, thousand, and million are primitive words, having no perceptible analogy to the numbers they express.- From the foregoing illustrations we derive the following principle : 8. How is a hundred expresped ? ThousandB, and larger numbers ? KOTATION". It 9. The Orders of Units increase hy the scale of ten That is, ten single units are one teri ; ten tens one liun dred ; ten hundreds one thousand ; and, universally. Ten of any lower order make a unit of the nexi higher. Notes. — i. If the term unit denotes one, how, it may be asked, can ten things or ones be a unit, ten tens another unit, etc. And how can the figures i, 2, 3, etc., sometimes denote single things or ones ; at others, tens of ones, and so on. The answer is, units are of two kinds, simple and collective. A simple unit is a single thing or one. A collective unit denotes a group of ones, regarded as a whole. 2. To ilfustrate these units, suppose a basket of pebbles is before us. Counting them out one by one, each pebble is a simple unit. Again, counting out ten single pebbles and putting them togethei ill a group, tliis group forms a unit of the second order, called ten. Counting out ten such groups, and putting them together in one pile, this collection forms a unit of the third order, called hundred. In like manner, a group of ten hundreds forms a unit of the fourth order, called thousand, and so on. These different ^ww^s of ten are called units on the same principle that a group of ten cents forms a unit, called a dime ; or, a group of ten dimes, a unit, called a dollar. Thus, it will be seen that the figures I, 2, 3, 4, etc., always mean one, two, three, four units as their name indicates ; but the value of these units depends upon the place the figure occupies. 10. From the preceding illustrations, it will be seen that the Aratic Notation is founded upon the following principles : ist. Numbers are divided into groups called units, of theirs/, second, third, etc., orders. 2d. To express these different orders of units, a simple and a local value are assigned to the significant figures, according to the place they occupy. 3d. If any order is wanting, its place is supplied by 3. cipher. 9. How do the orders of units increase? Units make a ten ? Tens a hundred* 10. Name the principles upon which the Arabic Notation is founded ? 14 KOTATIOK. 11. The Simple Value of a figure is the number of units it expresses when it stands alone, or in the riglit- hand place. The Local Value is the number it expresses when connected with other figures, and is determined by the jJilace it occupies, counting from the riqlit. 12. It is a general law of the Arabic Notation that the value of a figure is increased tenfold for every place it is moved from the riglit to the left ; and, conversely, The value of each figure is diminished tenfold for every place it is moved from the left to the right. Thus, 2 in the first place denotes two simple units ; in the second place, ten times two, or twenty ; in the third place, ten times as much as in the second place, or two hundreds, and so on. (Art. 8.) 13. The number denoting the scale by which the orders of units increase is called the radix. The radix of the Arabic Notation is ^e^ ; hence it is often called the decimal notation. KoTES. — I. The term radix, Latin, signifies root, or tase. The term decimal is from the Latin decern, ten. 2. The decimal radix was doubtless suggested by the number of fingers (digiti) on both hands. (Art. 7, Note.) Hence, 1^. To Express Numbers by Figures, Begin at the left hand and write the figures of the give?i orders in the successive places toward the right. If any intermediate orders are omitted, supijly their places with ciphers. II. The simple value of a figure ? Local ? 12. What is the law aa to movWg a figure to the right or left? 13. What is the radix of a system of notation? Tho radix of the Arabic ? What else is the Arabic system called ? Why ? NoU. Meaning of radix? Decimal? What Bu^refted the decimal radix? 14. Rule for expressing numbers by figures ? l^^OTATIOiT. EXERCISES. Express Uie following numbers by figures: 1. Three hundred and forty-five. 2. Four hundred and sixty. 3. Eight hundred and four. 4. Two thousand three hundred and ten. 5. Thirty thousand and nineteen. 6. Sixty- three thousand and two hundred. 7. One hundred and ten thousand two hundred and twelve. 8. Four hundred and sixty thousand nine hundred and thirty. 9. Six hundred and five thousand eight hundred and forty-two. 10. Two millions sixty thousand and seventy-five. ROMAN NOTATION. 15. The Mofnan dotation is the method of ex- pressing numbers by certain letters. It is so called because it was employed by the Romans. The letters used are the following seven, viz. : I, V, X, L, C, D and M. The letter I denotes one; Y,five; lL,ten; lu, fifty ; Q, one hundred; J), five hundred; M, one thousand. Interven- ing and larger numbers are expressed by the repetition and combination of these letters. 16. The Eoman system is based upon the following general principles : ist. It proceeds according to the scale of ten as far as a thousand, the unit of each order being denoted by 15. Roman notation? Why so called ? Letters employed? The letter I de- note? V? X? L? C? D? M? 16. Name the first principle upon which it Is based. What is the effect of repeating a letter ? Of placing a letter of less ralne before one of greater value? If placed after? If a line is placed over a tetter? 16 N^OTATIOJs". a single letter. Thus, I denotes one ; X, ten; C, one hun- dred ; M, one thousand. 2d. Repeating a letter, repeats its value. Thus, I denotes one ; II, two ; III, three ; X, ten ; XX, twenty, etc. 3d. Placing a letter of less value before one of greater value, diminishes the value of the greater by that of the less ; placing the less after the greater, increases the value of the greater by that of the less. Thus, Y denotes five, but IV denotes only four, and VI six. 4th. Placing a horizontal line over a letter increases its value a thousand times. Thus, I denotes a thousand ; X, ten thousand ; C, a hundred thousand ; M, a million. TABLE. I, denotes one. XXX, denotes thirty. n, " two. XT., forty. III, " three. L, fifty. IV, " four. LX, sixty. V, « five. LXX, seventy. VI, " six. LXXX, eighty. VII, " seven. xc. ninety. VIII, " eight. c, one hundred. IX, " nine. oc, two hundred. X, i> ten. ccc. three hundred. XI, " eleven. cccc, four hundred. XII, " twelve. D, five hundred. XIII, « thirteen. DC, six hundred. XIV, " fourteen. Dec, seven hundred. XV, (( fifteen. DCCC, eight hundred. XVI, " sixteen. DCCCC nine hundred. XVII, " seventeen. M, one thousand. XVIII, « eighteen. MM, two thousand. XIX, " nineteen, MDCCCLXXI, denotes one thous- XX, it twenty. and eight hundred and seventy-one. Notes. — i. Four was formerly denoted by IIII ; nine by Villi ; forty by XXXX ; ninety by LXXXX ; jive hundred by 10 ; and a thousand by CIO. KOTATIOJ^. 17 2. Annexing to 10 (five hundred) increases its value ten times. Thus, loo denotes five thousand ; lOOO, fifty thousand. Prefixing- a C and annexing a to CIO (a thousand) increases its value te7i times. Thus, CCIOO denotes ten thousand, etc. Hence, 17. To Express Numbers by Letters, Begin at the left hand or highest order, and write the letters denoting the given number of each order in successioti. Note. — The Roman Notation is seldom used, except in denoting chapters, sections, heads of discourses, etc. EXERCISES. Express the following numbers by letters : I. Fourteen, 2. Twenty-nine, 3. Thirty-four, 4. Sixty-six, 5. Forty-nine, 6. Seventy-three, 7. Eighty-eight, 8. Ninety-four, 9. Ninety-nine, 10. 107, II. 212, 12. 498, 13. 613, 14. 507, 15. 608. 16. 724, 17. 829, 18. 928, 19. 1004, 20. 1209, 21. 1363, 22. 1417, 23. 1614, 24. 1671, 25. 1748, 26. 1803, 27. 1876. NUMERATION. 18. Numeration is the art of reading numbers ex- pressed by figures, letters, or other numeral characters. Note. — The learner should be careful not to confound JV^umeration with Notation. The distinction between them is the same as that between reading and writing. 19. There are two methods of reading numbers, the French and the English. 17. Rule for expressing numbers by letters? 18. Numeration? Note. Dis- tinction between Numeration and Notation ? 19. How many methods of reading numbers ? 18 NUMERATIOJS". FRENCH NUMERATION. 20. The French divide numbers into periods of thre^ ■figures each, and then subdivide each period into units, tens, and hundreds, as in the following TABLE. o . tj SS ^S ;=1«^ :3»3 _2g ■53 h| s§ s| h| ■ol 2 -Sig "sa =53 -Sj ^ . rS^-^'^ nj'^S r^W. •73'^iffererice, when a figure in the Subtra- hend is greater than the corresponding figure of the Minuend, Ex. I. What is the difference between 8074 and 4869 ? ist Analysis. — Let tlie numbers be set down as in ist method. tlie margin. Beginning at the right, we proceed 8074 liin. thus: 9 units cannot be taken from 4 units; we 4860 Sub therefore take i ten from the 7 tens in the upper number, and add it to 4 units, making 14. Sub- •720^ Rem, tracting g units from 14 units leaves 5 units, which we place under the figure subtracted. Since we have taken i ten from 7 tens, there are but 6 tens left, and 6 tens from 6 tens leaves tens. Next, 8 hundreds cannot be taken from o hundreds; hence, we take i thousand from the 8 thousand, and adding it to the o, we have 10 hundreds. Taking 8 hundreds from 10 hundreds leaves 2 hundreds. Finally, 4 thousand from 7 thousand (8 — i) leave 3 thousand. The remainder is 3205. 2d Analysis. — Adding 10 to 4, the upper figure, 2d method. makes 14, and 9 from 14 leaves 5. Now, to balance 8074 Min. the ten added to the upper figure, we add i to 4860 Sub. the next higher order in the lower number. Adding 1 to 6 makes 7, and 7 from 7 leaves o. Next, since ^20=; Rem we cannot take 8 from o, we add 10 to the o, and 8 from 10 leaves 2. Finally, adding i to 4 makes 5, and 5 from 8 leaves 3. The remainder is 3205, the same as before. Note. — The chief diflerence between the two methods is this : In the first, we subtract i from the next figure in the ujpper number ; in the second, we add 1 to the next figure in the lower number. The former is perhaps the more philosophical ; but the latter is mora convenient, and therefore generally practiced. By inspecting this illustration, it will be seen, If a figure in the lower number is larger than that above it, we add 10 to the upper figure, then subtract, and add i to the next figure in the lower number. Rem —Instead of adding 10 to the upper figure, many prefer to take the lower figure directly from 10, and to tlie remainder add the upper figure. Thus, 9 from 10 leaves i, and 4 make 5, etc. 37. Adding 10 to the upper figure is called borrowing ; and adding i to the next figure in the lower number pays it SUBTRACTION. 35 I^oT£S. — I. The first method of borrowing depends upon the obvi- ous principle that the value of a number is not altered by transfer- ring a unit from a higher order to the next lower. 2. The reason that the second method of horroicing does not affect the difference between the two numbers, is because they are equally increased , and when two numbers are equally increased, their dif- ference is not altered. 3. The reasor for borrowing 10, instead of 5, 8, 12, or any other number, is because ten of a lower order are equal to one of the next higher. If numbers increased by the scale of 5, we should add 5 to the upper figure ; if by the scale of 8, we should add 8, etc. (Art. g.) 4. We begin to subtract at the right, because when we borrow we must pay before subtracting the next figure. (2-) (3-) (4-) (5-) (6.) From 784 842 6704 8042 9147 Take 438 695 3598 5439 8638 38. The preceding principles may be summed up in the following GENERAL RULE. I. Place the less number under tlie greater, units under units, etc. II. Begin at the right, and subtract each figure in tJia lower number from the one above it, setting the remainder under the figure subtracted. (Art. 35.) III. If a figure in the lower number is larger than the one above it, add 10 to the upper figure ; then subtract, and add I to the next figure in the loiver number. (Art. s^.) 39. Peoof. — Add the remainder to the subtrahend; if the sum is equal to the minuend, the work is right. 38. How write numbers for subtraction? How proceed when a figure in the lower number is greater tlian the one above it? 35. Note. Why write the less number under the greater, etc.? Why subtract the figures separately? Why set the remainder under the figure subtracted ? 37. What is meant by borrow- ing ? What is meant by paying or carrying ? Note. Upon what principle does the first method of borrowing depend ? Why does not the second method of borrowing aff"ect the diff'erence between the two numbers ? Why borrow 10 instead of 5, 8, 12, etc. ? Why begin to subtract at the right band? ^9. How prove subtraction ? Note. Upon what does this proof depend ? 3G SUBTEACTIOK. Notes, — i. This proof depends upon the Axiom, that the wJiole u equal to the sum of all its parts. 2. As soon as the class understand the rule, they should learn to omit all words but the results. Thus, .^n Ex. 2, instead of saying 9 from 4 you can't, 9 from 14 leaves 5, etc., B&y Jive, naught, three, etc. EXAMPLES. Find the difference between 84065 and 30428. (2.) (3-) (4.) (5.) 824 rods, 4523 pounds, 6841 years, 735^ acres, 519 rods. 4456 pounds. 3062 years. S^3^ acres. From Take (6.)- From 23941 Take 1 2367 (.0.) From 638024 Take 403015 (7.) 46083 23724 (II.) 7423614 2414605 (8.) 52300 25121 (12.) 8605240 3062431 (9.) 483672 216030 (I3-) 9042849 6304120 14. From 85269 pounds, take 33280 pounds. 15. From 412685 tons, take 103068 tons. 16. From 840005 acres, take 630651 acres. 17. What is the difference between 95301 and 60358? 18. What is the difference between 1675236 and 439243 ? 19. Subtract 2036573 from 5670378. 20. Subtract 35000384 from 68230460. 21. Subtract 250600325 from 600230021. 22. A man bought a house and lot for $36250, and paid $17260 down : how much does he still owe ? 23. A man bought a farm for I19200, and sold it for $17285 : what was his loss? 24. A merchant bought a cargo of tea for 1 1265 235, and Bold it for I1680261 : what was his gain? 25. A's income is $645275, and B's $845280: what is the difference in their incomes ? SUBTRACTIOK^. 3? 26. A bankrupt's assets are $569257, and his debts $849236 : how much will his creditors lose ? 27. Subtract 9999999 from iiiniio. 28. Subtract 666666666 from 7000000000. 29. Subtract 8888888888 from loooooooooo. 30. Take 200350043010 from 490103060756. 31. Take 53100060573604 from 80130645002120. 32. Take 675000000364906 from 901638000241036. ;^;^, From two millions and five, take ten thousand. 34. From one million, take forty-five hundred. 35. From sixty-five thousand and sixty-five, take five hundred and one. 36. From one billion and one thousand, take one million. 37. Our national independence was declared in 1776: how many years since ? 38. Our forefathers landed at Plymouth in 1620: how many years since ? 39. A father having 3265 acres of land, gave his son 1030 acres : how many acres has he left ? 40. The Earth is 95300000* miles from the sun, and Venus 68770000 miles: required the difierence. 41. Washington died in 1799, at the age of 67 years: in what year was he born ? 42. Dr. Franklin was born in 1706, and died in 1790: at what ago did he die ? 43. Sir Isaac Newton died in 1727, at the age of 85 years : in what year was he born ? 44. Jupiter is 496000000 miles from the sun, and Saturn 909000000 miles : what is the difierence ? 45. In 1865, the sales of A. T. Stewart & Co., by ofiicial returns, were $39391688; those of H. B. Claflin & Co., $42506715 : what was the difference in their sales? 46. The population of the United States in i860 was 31445080; in 1850 it was 23 191 87 6: what was the increase ? * Kiddle's Astronomy. SUBTRACTlOlf. 38 QUESTIONS FOR REVIEW. 1. A man's gross income was Si 5 65 a month for two successive months, and his outgoes for the same time 8965 : what was his net profit ? 2. A man paid I2635 for his farm, and $758 for stock; he then sold them for $4500 : what was his gain ? 3. A flour dealer has 3560 barrels of flour; after selling 1380 barrels to one customer, and 985 to another, how many barrels will he have left ? 4. If I deposit in bank $6530, and give three checks of $733 each, how much shall I have left ? 5. What is the difference between 3658 + 256 + 4236 and 2430 + 1249? 6. What is the difference between 6035+560 + 75 and 5003 + 360 + 28? 7. What is the difference between 891+306 + 5007 and 40 + 601 + 1703 + 89? 8. What is the difference between 900130 + 23040 and 19004+ 100607 ? 9. A man having I16250, gained $3245 by specula- tion, and spent I5203 in traveling: how much had he left? 10. A farmer bought 3 horses, for which he gave $275, $320, and $418 respectively, and paid $50 down: how much does he still owe for them ? 11. A young man received three legacies of $3263, I5490, and $7205 respectively; he lost I4795 minus I1360 by gambling : how much was he then worth ? 12. What is the difference between 6286 + 850 and 6286-850? 13. What is the difference between 11 325 — 2361 and 8030 — 3500? 14. What number added to 103256 will make 215378? 15. What number added to 573020 will make 700700 ? SUBTEACTION". 39 i6. What number subtracted from 230375 will leaye 121487 ? 17. What number subtracted from 317250 will leaye 190300 ? 18. If the greater of two numbers is 59253, and the difference is 21 231, what is the less number? 19. If the difference between two numbers is 1363, and the greater is 45261, what is the less number? 20. The sum of two numbers is 63270, and one of them is 29385 : what is the other? 21. What number increased by 2343 — 131, will become 3265-291 ? 22. What number increased by 3520+ 1060, will become 6539-279? 23. What number subtracted from 5009, will become 23404-471? 24. Agreed to pay a carpenter $5260 for building a house; $3520 for the masonry, and $1950 for painting: how much shall I owe him after paying $6000 ? 25. If a man earns $150 a month, and it costs him $6;^ a month to support his family, how much will he accumulate in 6 months ? 26. A's income-tax is $1165, B's is $163 less than A's; and C's is as much as A and B^s together, minus $365 : what is C's tax ? 27. A is worth $15230, B is worth $1260 less than A; and is worth as mucli as both, wanting $1760: what is B worth, and what C ? 28. The sum of 4 numbers is 45260; the first is 21000, the second 8200 less than the first, the third 7013 less than the second : what is the fourth ? 29. A sailor boastingly said ; If I could save I263 more, I should have ^1000: how much had he? 30. The difference between A and B's estates is $1525 ; B, who has the least, is worth $17250: what is A worth ? MULTIPLICATION. 40. Multiplication is finding the amount of a number taken or added to itself, a given number of times. Tlie Multiplicand is tlie number to be multiplied. The Multiplier is the number by which we mul- tiply. The Product is the number found by multiplici*tion. Thus, when it is said that 3 times 6 are 18, 6 is the mul- tiplicand, 3 the multiplier, and 18 the product. 41. The multiplier shows liow many times the multi- plicand is to be taken. Thus, Multiplying by i is taking the multiplicand once ; Multiplying by 2 is taking the multiplicand twice ; and Multiplying by any whole number is taking the multi- plicand as many times as there are units in the multiplier. Notes. — i. The term wvltiplication, from the Latin multipUco, multus, many, and plico, to fold, primarily signifies to increase by regular accessions. 2. Multiplicand, from the same root, signifies that which is to he tnul- tiphed ; the termination nd, having the force of to he. (Art. 33, n.) 42. The multiplier and multiplicand are also called Factors; because they make or produce the product Thus, 2 and 7 are the factors of 14. Notes. — i. The term/ac^or, is from the Latin /aa^?, to produce. 2. When the multiplicand contains only one denomination, the operation is called Simple Multiplication. 40. What ie multiplication? What 1e the numher multiplied called? The mimher to multiply hy ? The result ? When it is said, 3 times 4 are 12, which is the multiplicand? The multiplier? The product? 41. What does the multi- plier show ? What is it to multiply by i ? By 2 ? 42. What else are the nnmhers to be multiplied together called ? WTiy ? NoU. Meaninsr of the term factor ? What is the operation called when the multiplicand contains only one denomi- nation ? 43. The sign of multiplication. MULTIPLICATION, 41 43. The Sign of Multiplication is an oblique cross ( X ), placed between the factors. Thus, 7x5 means that 7 and 5 are to be multiplied together, and is read " 7 times 5," " 7 multiplied by 5," or " 7 into 5." 44. Numbers subject to the same operation are placed within ?i parenthesis ( ), or under a line called a vinculum ( ). Thus (4 + 5) X 3, or 4 + 5 X 3, shows that the sum of 4 and 5 is to be multiplied by 3. MULTIPLICATION TABLE. 2 times 1 are 2 2 " 4 3 " 6 4 « 8 5 6 7 8 9 10 II 12 3 times 1 are 3 2 " 6 3 " 9 4 " 12 5 " 15 6 " 18 7 " 21 8 " 24 9 " 27 " 30 1 " 33 2 " 7,6 times are 4 a 8 a 12 a 16 a 20 a 24 a 28 (i 32 a 36 a 40 a 44 a 48 5 times 1 are 5 2 " 10 a 15 20 25 30 35 40 45 50 55 60 6 times I are 6 2 3 4 5 6 7 8 9 10 II 12 12 18 24 30 36 42 48 54 60 I 66;i; 72 \i: 7 times 1 are 7 2 " 14 21 28 35 42 49 56 ^Z 70 77 ^4 8 times 9 times 10 times II times 12 times I are 8 I are Q I are 10 I are 11 I are 12 2 " 16 2 " 18 2 ' 20 2 " 22 2 " 24 3 " 24 3 " 27 3 ' 30 3 " Z2> 3 " 36 4 " 32 4 " 36 4 ' 40 4 " 44 4 " 48 5 " 40 5 " 45 5 ^ 50 5 " 55 5 " 60 6 " 48 6 " 54 6 ^ 60 6 « 66 6 - 72 7 " S^ 7 " 63 7 ^ 70 7 " 77 7 " 84 8 " 64 8 " 72 8 ' 80 8 " 88 8 " 96 9 " 72 9 " 81 9 ^ 90 9 " 99 9 " 108 10 " 80 10 " 90 10 ' 100 10 " no 10 " 120 II " 88 II « 90 II ' 1 10 II "121 II " 132 12 « 96 12 " 108 12 ' 120 12 " 132 12 « 144 Tlie pupil will observe tliat the 'products by 5, terminate iq 5 and o, alternately. Tbus, 5 times i are 5 ; 5 times 2 arc 10. ^ ^ Si m ^ ^ » m i5 m 4:2 MULTIPLICATION. The products by lo consist of the figure multiplied and a cipher.- Thus, 10 times i are lo ; ten times 2 are 20, and so on. The first nine products by 11 are formed by repeating the figure multiplied. Thus, 11 times i are 11 ; 11 times 2 are 22, and so on. The first figure of the first nine products by 9 increases, and the second decreases regularly by i ; while the sum of the digits in each product is 9. Thus, 9 times 2 are 18, 9 times 3 are 27, and so on. 45. When the factors are abstract numbers, it is immaterial in what order they are multi- plied. Thus, the product of 3 times 4 is equal to 4 times 3. For, taking. 4 units 3 times, is the same as taking 3 units 4 times ; that is, 4 x 3 = 3 x 4. Notes. — i. It is more convenient and therefore customary, to take the larger of the two given numbers for the multiplicand. Thus, it is better to multiply 5276 by 8, than 8 by 5276. 2. Multiplication is the same in principle as Addition; and is sometimes said to be «- short method of adding a number to itself a given number of times. Thus: 4 stars + 4 stars + 4 stars=3 times 4 or 12 stars. 46. The iwoduct is the same name or hind as the multi- plicand. For, repealing a number does not change its nature. Thus, if we repeat dollars, they are still dollars, etc. 47. The multiplier must be an abstract number, or con- sidered as such for the time being. For, the multiplier shows Jiow many times the multiplicand is to be taken ; and to say that one quantity is taken as many times as another \s heavy — is ait urd. Note. — When it is asked what 25 cts. multiplied by 25 cts., or 2S. 6d. by 2s. 6d , will produce, the questions, if taken literally, are nonsense. For, 2s. 6d. cannot be repeated 2s. 6d. times, nor 25 cts. 25 cts. times ; but we can multiply 25 cts. by the number 25, and 2S. 6d. by 2\. In like manner we can multiply the price of i yard by the number of yards in an article, and the product will be the cost. 45. When the factors are abstract, docs it make any difference with the pro- duct which is taken for the multiplicand ? Note. Which is commonly taken f Why ? What is Multiplication the same as ? 46. What denomination is the pra duct ? Why ? 47. "What must the multiplier be ? Why ? MULTIPLICATION. 43 CASE I. 48. To find the Product of two numbers, when the Multiplier has but one figure. I. What is the cost of 3 horses, at $286 apiece ? Analysis. — 3 horses will cost 3 times as mucli as opebation. 1 horse. Let the numbers be set down as in the $286 Multd. margin. Beginning at the right, we proceed thus : ^ Mult. 3 times 6 units are 18 units. We set the 8 in units' place because it is units, and carry the i ten to the $858 Prod, product of the tens, as in Addition. (Art. 29.) Next, 3 times 8 tens are 24 tens and i (to carry) make 25 tens, or 2 hun- dred and 5 tens. We set the 5 in tens' place because it is tens, and carry the 2 hundred to the product of hmidreds. Finally, 3 times 2 hundred arc 6 hundred, and 2 (to carry) are 8 himdred. We sot the 8 in hundreds' place because it is hundreds. Therefore the 3 horses cost $858. All similar examples are solved in like manner. By inspecting the preceding analysis, the learner will discover the following principle : Each figure of the multiplicand is multiplied hj the multiplier, heginning at the rigM, and the result is set do2vn as in Addition. (Art. 29.) Notes. — i. The reason for setting the multiplier under the multi- plicand is simply for convenience in muliiplyiug. 2. The reasons for- beginning to multiply at the right hand, as well as for setting down the units and carrying the tens, are tho same as those in Addition. (Art. 29, n.) 3. In reciting the following examples, the pupil should care- fully analyze each ; then give the results of the operations re- quired. 49. Units multiplied hy miits, it should be observed, produce imits ; tens by units, produce tens ; hundreds by units, produce hundreds ; and, universally, If the multiplying figure is units, the product will be the same order as the figure multiplied. Again, if the figure multiplied is units, the product will 40. What do units multiplied by units produce ? Tens by units ? Hundreds V units ? When the multiplying fl^re is units, what is the product J 44 MULTIPLICATION^, be the same order as the multiplying figure; for the product is the same, whichever factor is taken for the multiplier. (Art. 45.) (^•) (3-) (4-) (5-) Multiply 2563 13278 2648203 48033265 By 4 5 6 7 6. What will 8 carriages cost, at $750 apiece? 7. What cost 9 village lots, at $1375 a lot? 8. At $3500 apiece, what will 10 houses cost? 9. At $865 a hundred, how much will 11 hundred- weight of opium come to ? 10. If a steamship sail 358 miles per day, how far will she sail in 1 2 days ? 11. Multiply 86504 by 5. 12. Multiply 803127 by 7. 13. Multiply 440210 by 6. 14. Multiply 920032 by 8. 15. Multiply 603050 by 9. 16. Multiply 810305 by 10. 17. Multiply 753825 by II. 18. Multiply 954635 by 12. 19. What cost 4236 barrels of apples, at $7 a barrel ? 20. What cost 5167 melons, at 11 cents apiece? 21. At 6 shillings apiece, what will 1595 arithmetics cost? 22. At $12 a barrel, what will be the cost of 1350 bar- rels of flour ? 23. Wliat will 1735 boxes of soap cost, at $9 a box ? 24. When peaches are 1 2 shillings a basket, what will 2363 baskets cost ? 25. If a man travels 8 miles an hour, how far will he travel in 3260 hours ? 26. A builder sold 10 houses for $4560 apiece: how much did he receive for them all ? 27. What will it cost to construct 11 miles of railroad, at $12250 per mile? 28. What will be the expense of building 12 ferry-boats, at I23250 apiece ? MULTIPLICATIOK. 45 CASE II. 50. To find the Product of two numbers, when the Multiplier has two or more figures. I. A manufacturer bought 204 bales of cotton, at $176 a bale: what did he pay for the cotton ? Analysis. — Write the numbers as in the mar- $176 Multd. gin. Beginning at the right : 4 times 6 units are 204 Mult 24 units, or 2 tens and 4 units. Set the 4 in units' place, and carry the 2 to the next product. 4 7^4 times 7 tens are 28 tens and 2 are 30 tens, or 3 3^2 hundred and o tens. Set the o in tens' place and carry the 3 to the next product. 4 times i hun- $35904 Ans. are 4 hun. and 3 are 7 hun. Set the 7 in hundreds' place . The product by o tens is o ; we therefore omit it. Again, 2 hundred times 6 units are 12 hun. units, equal to i thousand and 2 hundred. Set the 2 in hundreds' place and carry the i to the next product. 2 hundred times 7 tens are 14 hundreds of tens, equal to 14 thousand, and i are 15 thousand. Set the 5 in thousands' place, and carry the i to the next product, and so on. Adding these results, the sum is the cost. Remakk. — The results which arise from multiplying the multipli- cand by the separate figures of the multiplier, are called partial pro- ducts ; because they are parts of the whole product. By inspecting this analysis, the learner will discover, 1. Tlie multiplicand is multiplied hy eacli figure of the multiplier, hegin^iing at the right, and the result set down as in Addition. 2. The first figure of each partial product is placed under the multiplying figure, and the sum of the partial products is the true answer. Notes. — i. When there are ciphers between the significant figures of the multiplier, omit them, and multiply by the next sig- nificant figure. 2. We multiply by each figure of the multiplier separately, when it exceeds 12, for the obvious reason, that it is not convenient to multi- ply by the whole of a large number at once. 3. The first figure of each partial product is placed under the multiplying figure; because it is the same order as that figure. 4. The several partial prof\ucts are added together, because the whole product is equal to the sam of aU its parts. 46 MULTIPLICATION. (^■) (3.) (4.) (5.) Multiply 426 5^3 1248 2506 By 24 35 52 304 51. The preceding principles may be summed up in the following GENERAL RULE I. Place the muUijMer under tlie multiplicand, units under units, etc. II. When the multiplier has hut one figure, heginning at the right, multiply each figure of the multiiMca^id iy it, and set down the result as in Addition. III. If the multiplier has two or more figures, multiply the multiplicand hy each figure of the multiplier separately, and set the first figure of each partial product under the multiilying figure. Finally, the sum of the partial products will he the ansiuer required. Note. — The pupil should early learn to abbreviate the several steps in multiplying as in Addition. (Art. 31, 7i.) PROOF. 52. By Multiplication. — Multiply the multiplier hy the multiplicand ; if this result agrees ivith the first, the ivork is right. Note. — This proof is based upon the principle that the result will be the same, whichever number is taken as the multiplicand. 53. By excess of 9s. — Find the excess of gs in each factor separately ; then multiply these excesses together, and reject the gs from the result ; if this excess agrees with the excess jf 9s in the answer, the work is right. 51. How write numbers for multiplication ? When the multiplier has hut one figure, how proceed ? When it haa two or more ? Wliat ie finally done with the partial products ? 4S. Note. Why write the multiplier under the multiplicand, units under units ? Why begin at the right hand ? 50. W^hat are partial products ? Note. Why multiply by each figure separately ? Why set the first figure of each Tinder the multiplying figure ? Why add the several partial products together ? 52. How prove multiplicetion by multiplication ? Note. Upon what ie this proof based ? 53. How prove it by excess of 98 ? MULT IPLICATIOS-, 47 Note. — This method of proof, if deemed advisable, may be omitted till review. It is placed here for convenience of reference. Though depending on a peculiar property of numbers, it is easily applied, and is confessedly the most expediitous method of proving multipli- cation yet devised. 54. To find the Excess of 98 in a number. Beginning at the left hand, add the figures together, and as soon as the sum is 9 or more, reject 9 and add the remainder to the next figure, and so on. Let it be required to find the excess of 9s in 7548467. Adding 7 to 5, the smn is 12. Rejecting 9 from 12, leaves 3 ; and 3 added to 4 are 7, and 8 are 15. Rejecting 9 from 15, leaves 6 ; and 6 added to 4 are 10. Rejecting 9 from 10, leaves i ; and i added to 6 are 7, and 7 are 14. Finally, rejecting 9 from 14 leaves 5, the excess required. Notes. — i. It will be observed that the excess of 9s in any two digits is always equal to the sum, or the excess in the sum, of those digits. Thus, in 15 the excess is 6, and 1 + 5=6; so in 51 it is 6, and 5 + 1=6. In 56 the sum is 11, the excess 2. 2. The operation of finding the excess of 9s in a number is called casting out the gs. EXAMPLES. I. Wliat is the product of 746 multiplied by 475 ? OPKBATION. 746 475 3730 5222 2984 Proof by Excess of gs. Excess of 9s in multd. is 8 " 9 s in mult. " 7 Now 8 X 7 = 56 The excess of 9s in 56 is 2 The excess of 9s in prod, is 2 And 2 = 2 Proof by Mult. 475 746 2850 1900 3325 Ans. ' 354350 ■^ns. 354350 (2.) (3.) (4.) (5.) Multiply 5645 18934 48367 231456 By 43 65 75 87 54. How And the excess of 98 ? 48 MULTIPLICATION. (6.) (7.) (8.) (9.) Multiply 1421673 2342678 4392460 5230648 By 234 402 347 526 10. Mult. 640231 by 205. II. Mult. 520608 by 675. 12. Mult. 431220 by 1234. 13. Mult. 623075 by 2650. 14. Mult. 730650 by 2167. 15. Mult. 593287 by 6007. r6. Mult. 843700 by 3465, 17. Mult. 748643 by 2100. 18. Mult. 9000401 by 50001. 19. Mult. 82030405 by 23456. 20. How many pounds m 1375 chests of tea, each chest containing 6;^ pounds ? 21. What cost 738 carts, at I75 apiece? 22. At 43 bushels per acre, how many bushels of wheat will 520 acres produce? 23. At $163 apiece, what will be the cost of 1368 covered buggies ? 24. If a man travel 215 miles per day, how far can he travel in 365 days ? 25. There are 5280 feet in a mile: how many feet in 256 miles? 26. What cost 21 15 revolvers, at I23 apiece? 27. Bought 1978 barrels of pickles, at $17 ; how much did they come to ? 28. If railroad cars are $4735 apiece, what will be the expense of 500 ? 29. How far will 2163 spools of thread extend, each containing 25 yards? 30. Bought 15265 ambulances, at $117 apiece: what was the amount of the bill ? 31. What will 3563 tons of railroad iron cost, at 168 per ton ? 32. How far will a man skate in 6 days, allowing he skates 8 hours a day, and goes y miles an hour? MULTIPLICATION^. 4^ CONTRACTIONS. 55. A Composite NtiTnher is the product of two or more factors, each of which is greater than i. Thus 15=3 X 5, and 42=^2 x 3 x 7, are composite numbers. Notes. — i. The product of a number multiplied into itself is called a power. Thus, 9=3 x 3, is a power. 2. Composite is from tlie Latin compono, to place together. 3. The term^ factors oxi^ parts must not be confounded with each other. The former are multiplied together to produce a number ; r'lie latter are added. Thus, 3 and 5 are tlxe factors of 15 ; but 5 and 10, 6 and 9, 7 and 8, etc., are the parts of 15. 1. What are the factors of 45 ? Ans. s and 9. 2. What are the factors of 24 ? Of 27? Of 28 ? Of 30? 3. What are the factors of 32? Of 35 ? Of 42? Of48.f' 4. What are the factors of 5 4 ? Of 63 ? Of 7 2 ? Of 84 ? CASE I. ^ 56. To multiply by a Cofnposite Number. I. A farmer sold 15 boxes of butter, each weighing 20 pounds : how many pounds did he sell ? Analysis. — 15 = 5 times 3; hence, 15 boxes will weigh 5 times as much as 3 boxes. Now, if I box weighs 20 pounds, 3 boxes will weigh OPEEATIOIT. 20 pounds. 3 times 20, or 60 pounds. Agam, if 3 boxes ^ 60 weigh 60 pounds, 5 times 3 boxes will weigh 5 times 60, or 300 pounds. He therefore sold 300 T>ounds. In the operation, we first multiply by. 5 the factor 3, and the product thus arising by the -other factor 5. Hence, the Ans. 300 pounds, EuLE. — Multiply the multiplicand by one of the factors of the multiplier, then this product hy another, and so on, till all the factors have been used. The last product will he the answer, 35. A composite number? Note. The difference between fiactors ano parts! ;6 Hew multiply by a compoaiie number ? z 50 MULTIPLICATION. Notes. — i. This rule is based upon the principle that it is immaterial in what order two factors are multiplied. (Art. 45.) The samt, illustration may be extended to three or more numbers. For, the product of two of the factors may be considered as one Qumber, and this may be used before or after a third factor, etc. 2. The process of multiplying three or more factors together, is called Continued Multiplication ; the result, the continued product. 2. Multiply $568 by 35. Ans. $19880. 3. Multiply 2604 by 25. 4. Multiply 6052 by 48. 5. Multiply 8091 by 63. 6. Multiply 45321 by 72. 7. In I cubic foot there are 1728 cubic inches: how many cubic inches are there in 84 cubic feet ? 8. If a ton of copper ore is worth $5268, what is the worth of 56 tons? 9. What cost 125 houses, at $1580 apiece? CASE II. 57. To multiply by lO, 100, lOOO, etc. 10. What will 100 cows cost, at $31 apiece? Analysis. — Annexing a cipher to a number moves each figure one place to the left ; but moving a figure one place to the left increases ioS value ten times ; therefore annexing a cipher to a number multiplies it by 10. In like manner, annexing two ciphers, multi- plies it by a hundred, etc. (Art, 12.) Therefore, $31 x ioo=$3ioa Hence, the Rule. — Annex as mamj ciphers to the multiplicand as there are ciphers i?i the multijMer. Note. — The term annex, -from the Latin ad and ne^to, to join tc^ eignifies to place after. 11. What cost 1000 horses, at I356 apiece? 12. Multiply 40530 by 1000. 13. Multiply 9850685 by loooo. 14. Multiply 84050071 by 1 00000. * 15. Multiply 360753429 by loooooo. .57. How maltiply by to, too, 1000, etc. T MULTIPLICATION. 51 CASE III. 58. To multiply, when there are Ciphers on the right of either or both Factors. 20. What is the product of 87000 multiplied by 230 ? Analysis. — The factors of the ruuhiplicand are 87000 87 and 1000; the factors of the multiplier are 23 2^0 and 10. We first multiply the factoi*s consisting — r of significant figures ; then multiply this product ^^ by the other two factors (1000 x 10), or loooo, ^74 by annexing 4 ciphers to it. Hence, the -4 ?^5. 20010000 EuLE. — Multiply the significant figures together ; and to the result annex as many ciphers as are found on the right of doth factors. Note. — This rule is based upon the two preceding cases; for, by supposition, one or both the given numbers are composite ; and one of the factors of this composite number is 10, 100, etc. (21.) (22.) (23.) Multiply 2130 64000 83046 By 700 52 2000 24. In I barrel of pork there are 200 pounds: how many pounds in 3700 barrels? 25. What will 2300 liead of cattle cost, at $80 per head? 26. In $1 there are 100 cts. : how many cts. in I26000? 27. The salary of the President is I50000 a year: how much will it amount to in 21 years ? 28. If a clock ticks 86400 times in i day, how many times will it tick in 7000 days? 29. If one ship costs $150000, what will 49 cost? 30. 670103700x60030040? 31. 800021000x80002100? 32. 570305000x40000620? 58. When one or both factors have ciphers on the right ? Note. Upon what is this rule based ? 53 MULTIPLICATION. ^^. 467234630X27000000? 34. 890000000x350741237? 35. 9400000027 X 28000000 ? 36. Multiply 39 millions and 200 thousand by 530 thousand. 37. Multiply 102 times 700 thousand and i hundred by 601 thousand and twenty. 38. Multiply 74 millions and 21 thousand by 5 millions and 5 thousand. 39. Multiply 31 millions 31 thousand and 31 by 21 thousand and twenty-one. 40. Multiply 2 billions, 2 millions, 2 thousand and 2 hundred by 200 thousand and 2 hundred. CASE IV. 59. To multiply 13, 14, 15, op I with a Significant Figure annexed. 41. If one city lot costs I3245, what will 17 lots cost? Analysis. — 17 lots will cost 17 times as much as 3245 x 17 I lot. Placing the multiplier on the right, we multiply the multiplicand by the 7 units, set each figure one place to the right of the figure multi- plied, and add the partial product to the multi- plicand. The result is $55165. Hence, the EuLE. — I. Multiply the multiplicand hy the iinits^ figure of the multiplier, and set each figure of the partial product one place to the right of the figure multiplied. II. Add this partial product to the multipflicandyand the result tvill be the true product. Note.— This contraction depends upon the principle that as the tens' figure of the multiplier is i, the multiplicand is the second par- tial jjroduct ; hence its first figure must stand in tejis' place. 42. Multiply 1368 by 13. 43. Multiply 2106 by 14. 44. Multiply 3065 by 15. 45, Multiply 6742 by 16. 46. Multiply 25269 by 18. 47. Multiply 83467 by 19. t;o. How mnltiply hy ii, 14, n;, or i with a significant figure annexed? Note, Upon what is this contraction haeed T PIVISION. 60. Division is finding how many times one numh«=i is contained in another. The Dividend is the number to be divided. The Divisor is the number by which we divide. The Qiiotierit is the number found by division, and shows how many times the divisor is contained in the dividend. The Manainder is a part of the dividend left after division. Thus, when it is said, 5 is contained in 1 7, 3 times and 2 over, 17 is the dividond, 5 the divisor, 3 the q utient, and 2 the remainder. Notes. — i. The term dicision, is from tlie Latin divido, to part, or divide. The term dididsnd, from the same root, signifies that which is to he divided ; the termination nd, having the force of to he. (Art. 33, n.) Quotient is from the Latin quoties, signifying how often or how many times. 2. The remairide?' is always the same denomination as the divi- dend ; for it is an undivided part of it. 3. A proper remainder is always less than the divisor. 61. An obvious way to find how many times the divisor is contained in the dividend, is to subtract the divisor from the dividend continually, till the latter is exhausted, or till the remainder is less than the divisor; the numler of subtractions will be the quotient. Thus, taking any two numbers, as 4 and 12, we hav3 12 — 4=8; 8 — 4 = 4; and 4—4 = 0. Here are three subtractions; therefore, 4 is contained in 12, 3 times. Hence, Division is sometimes said to be a sliort method of con^ tinned subtraction. 60. What is diviBioTi ? The number to he divided called ? To divide by! The remit? The part left? JV?)fe. Meanirifr of the term division ? Dividend? Quo- tient ? Wliat denomination is the remainder ? Why ? 54 DIVISIOK. But there is a shorter and more direct way of obtaining the quotient. For we know by the multiplication table, that 3 times 4, or 4 taken 3 times, are 1 2 ; hence, 4 is contained in 12, 3 times. 62. Division is the reverse of multiplication. In mul- tipIicatioQ both factors are given, and it is required to find the iwoduct ; in division, one factor and the ])roduct (which answers to the dividend) are given, and it is required to find the other factor, which answers to the quotient. Hence, division may be said to hQ finding a quotient which, WMltiplied into the divisor, will jjrodnce the dividend. Note. — When tlie dividend contains only one denomination, tlio operation is called Simple Dicidon. DIVISION TABLE. I is in 2 is in 3 is in 4 is in 5 is in 6 is in I, once. 2, once. 3, once. 4, once. 5, once. 6, once. 2, 2 4, 2 6, 8, 2 ,10, 2 12, 2 3, 3 6, 3 9, 3 12, ^ 15,. ^ 18, 3 4, 4 8, 4 12, /j 16, A J20, ^ ^24, 4 5, 5 10, 5 15, 5 20, 5 |25, q 30, 5 6, 6 12, 6 18, 6 24, 6 I30, e 36, 6 7, 7 14, 7 21, 7 28, 7 35, 7 42, 7 8, 8 16, 8 24, 8 32, 8 40, 8 48, 8 9. 9 18, 9 27, 9 2>^, 9 45, 9 54, 9 10, IC 20, 10 30, 10 40, 10 j5o,__ 10 |6o, 10 7 is5 in 8 is in 9 is in 10 is inl II is in 12 is in 7, once. 8, once. 9, once. 10, once. 1 ii,once.| 1 2 once. 14, 2 16, 2 18, 2 20, 2 22, 2 24, 2 21, 3 24, 3 27, 3 30, 3 2>2>, 3 36, 3 28, 4 32, 4 36, 4 40, 4 44, 4 48, 4 35, 5 40, 5 45, 5 50, 5 55, 5 60, 5 42, 6 48, 6 54, 6 60, 6 66, 6 72, 6 49, 7 56, 7 63, 7 70, 7 77, 7 84, 7 56, 8 64, 8 72, 8 80, 8 88, 8 96, 8 63, 9 72, 9 81, 9 90, 9 99, 9 ro8, 9 70, 10 80, 10 90, 10 100, 10 no, 10 120, 10 61. What is an obvions way to find how many times the divisor is contained In tlie dividend ? WTiat is division sometimes called ? DIVISIOIS. DD OBJECTS OF DIVISION. 63. The object or office of Division is twofold : ist. To find how many times one 7iumber is contained in another. 2d, To divide a niimher itito equal yarts. 63, «• To find how ^uantj fimes one number is contained in another. 1. A man has 15 dollars to lay out in books, which are 3 dollars apiece : how many can he buy ? Analysis. — In this problem the object is to find how many times 3 dols. are contained in 15 dols. Let the 15 dols. be represented by 15 counters, or unit marks. Separating these into groups of 3 each, there are 5 groups. There- fore, he can buy 5 books. <'^o|<>oo|<><><>|<><><>|<><^<^ 63, l>» To divide a number into equal parts. 2. If a man divides 15 dollars equally among 3 persons, how many dollars will each receive ? Analysis. — The object here is to divide 15 dols. into 3 equal parts. Let the 15 dols. be represented by 15 counters. If we form 3 j^roups, first putting i counter in each, then another, till the counters are ex- hausted, each group will have 5 counters. Therefore, each person will receive 5 dollars. <>0^0<>|<><>0<>|00<><>^ Remark. — The preceding are representative examples of the two classes of problems to which Division is applied. In the first, the divisor and dividend are the same denomination, and the quotient is times, or an abstract number. In the second, the divisor and dividend are different denominations, and the quotient is the same denomination as the dividend. Hence, 64. When the divisor and dividend are the same denom- ination, the quotient is always an abstract number. 62. Of what is division the reverse ? What is given in multiplication ? What required ? What is given in division ? What required ? Note. When the divi- dend contains but one denomination, what is the operation called ? 63. What is the object or oflace of division ? 63, a. What is the object in the first problem? 63, b. What in the second ? Rem. What is said of first two problems ? j: ly isi ON. When the divisor and dividend are different denomina- tions, the quotient is always the sa7ne denomination as the dividend. Notes. — i. The process of separating a number into equal parts, as required in tlie second class of problems, gave rise to the name " Division." It is also tlie origin of Fractions. (Art. 134.) 2. The mode of reasoning in the solution of these two classes cf examples is somewhat different ; but the practical operation is the same, viz. : to find how many times one number is contained in another, which accords with the definition of Division. 65. When a number or thing is divided into tivo equal parts, the parts are called halves; into three, the parts are called thirds ; into four, they are called fourths ; etc. The numher of parts is indicated by their name. 66. A number is divided into two, three, four, five, etc., equal parts by dividing it by 2, 3, 4, 5, etc., respectively. 3. What is a half of 12 ? A third of 15 ? A fourth of 20 ? A fifth of 35 ? A seventh of 28 ? 4. What is a sixth of 42 ? A seventh of 56 ? A ninth of (iT^ ? An eighth of 72 ? A twelfth of 108 ? 67. The Sign of Divisioii is a short horizontal line between two dots (-^), placed before the divisor. Thus, the expression 28-^7, shows that 28 is to be divided by 7, and is read, "28 aivided by 7." 68. Division is also denoted by loriting the divisor under the dividend, with a short line between them. Thus, -2^ is equivalent to 28-^-7. It is read, "28 divided by 7," or "28 sevenths." 69. Division is commonly distinguished as Sho7i Divis- ion and Long Division. 64. When the divisor and dividend are the same denomination, what is the quotient ? When different denominations, what ? 65. When a number is divider] into two equal parts, what are the parts called ? Into three ? Four ? 66. IIow divide a number into 2, 3, 4, etc., equal parts ? 67. What is the sisrn of division ? What does the expression 35+7 show? 68. How else is division denoted? SHOET DIYISIO]^. 70. Short Division is the method of dividhig^ when the results of the several steps are carried in tlie mind, and the quotient only is set down. 71. To divide by Sliort Division. Ex. I. If apples are $3 a barrel, how many barrels can you buy for I693 ? Analysis.— Since $3 will buy i barrel, $693 operatio. will buy as many barrels as $3 are contained ^-?)^6o*^ times in $693. Let the numbers be set down as ^ in the margin. Beginning at the left, we proceed H^^^- 231 Mr. thus : 3 is contained in 6 hundred, 2 hundred times. Set the 2 in hundreds' place, under the figure divided, becavse it is hundreds. Next, 3 is contained in 9 tens, 3 tens times. Set the 3 in tens' place under the figure divided. Finally, 3 is contained in 3 units, I time. Set the i in units' place. Ans. 231 barrels. Rem. — This problem belongs to the ist class, the object being to find how many times one number is contained in another, (Art. 63, a.) Solve the following examples in tlie same manner : (2.) (3.) (4.) (5.) 2)4468 3 )3696 4)4848 5) 5555 6. A man having 27543 pounds of grapes, packed them for market in boxes containing 5 pounds each : how many boxes did he fill, and how many pounds over? Analysis. — He used as many boxes as there operation. are times 5 pounds in 27543 pounds. Let the 5)27543 pounds, numbers be set down as in the margin. Since Quot i;qo8"boxes the divisor 5 is not contained in the first figure ^^^^j p Q^gj, of the dividend, we find how many times it is contained in the first two figures, which is 5 times and 2 over. We set the quotient figure 5 under the right hand figure divided, because it is the same order as that figure, and prefix the remainder 2, mentally to the next figure of the dividend, making 25. Now 5 is in 25, 5 times, and no remainder. Again, 5 is not contained in 4, the next figure of the dividend ; we therefore place a cipher in the ^o DIVISIOIS". quotient, and prefix the 4 mentally to the next figure of the dividend ^ as if it were a remainder, making 43. Finally, 5 is in 43, 8 times^ and 3 over. He therefore filled 5508 boxes, and had 3 pounds remainder. Hence, the EuLE.— I. Place the divisor on the left of the dividend, and heginning at the left, divide each figure hy it, setting the result under the figure divided. II. If the divisor is not contained in a figure of the div- idend, put a cipher in the quotient, and find hoio many times it is contained in this and the next figure, setting the remit under the right hand figure divided. III. If a remainder arise from any figure before the last, prefix it mentally to the next figure, and divide as before. If from the last figure, place it over the divisor, and annex it to the quotient. Notes. — i. In the operation, the divisor is placed on the left of tne dividend, and the quotient iinder it, as a matter of convenience. When division is simply represented, the divisor is either placed under the dividend, or on the right, with the sign (-5-) be- fore it. 2 The reason for beginning to divide at the left hand is, that in dividing a higher order there may be a remainder, which must be prefixed to the next lower order, as we proceed in tlie operation. 3. We place the quotient figure under the figure divided ; because the former is the same order as the latter. (Art. 71.) 4. When the divisor is not contained in a figure of the dividend, we place a cipher in the quotient, to show that the quotient has no units corresponding with the order of this figure. It also preserves the local value of the subsequent figures of the quotient. 5. The final remainder shows that a part of the dividend is not divided. It is placed over the divisor and annexed to the quotient to complete the division. 70. What is Short Division? 71 How write nnmtrers for short division? The next step ? When the divisor is not contained in a figure of the dividend, how proceed ? When there is a remainder after dividing a figure, how ? If there is a remainder after dividing the last figure, what? Note. Why place the divisor on the left of the dividend ? Why begin to divide at the left hand ? Why place each quotient figure under the figure divided? Wliy place a cipher in the quotient, whin the divisor is not contained in a figure of the dividend ? What does the final remainder show? Why place it over the divisor and annex it to the Quotient? DIVISION. 59 The pupil should early learn to abbreviate the language used In tlie process of dividing. Thus, in the next example, instead of Baying 5 is contained in 7 once, and 2 over, let him pronounce the quotient figures only ; as, one, four, jive, seven. I. A man divided 7285 acres of land equally among his 5 sons : what part, and how much, did each receive ? Solution. — i is i fifth of 5 ; hence each had i fifth part. (Art. 65.) ^gain, 7285 A-7-5 = i457 A; hence each received 1457 A. (Art. 66.) (2.) (3.) (4.) (5.) O43678 4 3)560346 4) 689034 5 )7482 39 (6.) (7.) (8.) (9.) ^ )3972647 7 )480 6108 8)7390464 9)8306729 (10.) (II.) (12.) T 0)57623140 1 1)6673 01451 1 2)81602 5 2397 13. At $2 apiece, liow many hats can be bought for $16486 ? (Art 48, Note 3.) 14. At $4 a head, how many sheep can be bought for $844? 15. How many times are 3 rods contained in 26936 rods ? 16. A man having $42684, divided it equally among his 4 children : how much did each receive ? 17. If a quantity of muslin containing 366 yards is di- vided into 3 equal parts, how many yards will each part contain ? 18. If 84844 pounds of bread are divided equally among 6 regiments, how many pounds will each regiment re- ceive ? 19. Eight men found a purse containing $64968, which they shared equally: how much did each receive? CO DIVISIOIS^. 20. Divide 4268410 by 4. 21. Divide 5601234 by 6. 22. Divide 6403021 by 5. 23. Divide 7008134 by 7. 24. Divide 8210042 by 11. 25. Divide 9603048 by 8. 26. Divide 23468420 by 10. 27. Divide 32064258 by 9. 28. Divide 46785142 by 8. 29. Divide 59130628 by 7. 30. Divide 653000638 by 11. 31. Divide 774230029 by 12 32. In 7 days there is i week: bow many wrecks in 26^6^ days? 33. If $38472 are divided equally among 6 persons, how much will each receive ? 34. How many tons of coal, at 87 a ton, can be pur- chased for I63456? 35. At $9 a barrel, how much flour can be bought for $47239? 36. In 12 months there is i year: how many years in 41260 months? 37. A merchant laid out $45285 in cloths, at $7 a yard: how many yards did he buy ? 38. At 8 shillings to a dollar, how many dollars are there in 75240 shillings? 39. In 9 square feet there is i square yard: how many square yards are there in 52308 square feet? 40. If a person travel 10 miles an hour, how long will it take him to travel 25000 miles? 41. A market woman having 845280 eggs, wished to pack them in baskets holding i dozen each : how many baskets did it take ? 42. If a prize of |i 16248 is divided equally among 8 men, what will be each one's share ? 43. If in 7 townships there are 2346281 acres, how many acres are there to a tow^nship ? 44. At $8 a barrel, how many barrels of sugar can be bought for $111364? 45. At $1 1 each, how many cows can be had for $88990 ? LOITG DIYISIOH". 72. Long jyivision is the method of dividing, when the results of the several steps and the quotierit are both set down. 73. To divide by Long Division. I. A speculator paid $31097 for 15 city lots: what did the lots cost apiece ? Analysis.— Since 15 lots cost $31097, i operation. lot will cost as many dollars as 15 is con- I5)$3i097($2073y^ tained times in 31097. Let the divisor and 30 ' " dividend be set down as in the margin. — Beginning at the left, the first step is to jog find how many times the divisor 15, is 105 contained in 31 (which is 2 times), and set the 2 on the right of the dividend. ^y Second, multiply the divisor by the ac quotient figure, and set the product 30, under the figures divided. Third, sub- 2 tract this product from the figures divided. Fourth^ bring down the next figure of the dividend, and place it on the right of the remainder, making 10 for the next partial dividend, and proceed as before. But 15 is not con- tained in 10 ; we therefore place a cijjher in the quotient, and bring down the next figure of the dividend, making 109. Now 15 is in 109, 7 times. Set the quotient figure 7 on the right, mul- tiply the divisor by it, subtract the product from the partial divi- dend, and to the right of the remainder, bring down the succeeding figure for the next partial dividend, precisely as before. Now 1 5 is in 47, 3 times. Setting the 3 in the quotient, multiplying, and sub- tracting, as above, the final remainder is 2. We place this remainder 72. What is Long Division ? 73. How write the numbers ? What is the first stop? The second? The third? The fourth? If the divisor is not contained in a partial dividend, how proceed ? What is to be done with the last reinain-ler ? Note. What is the difference between short and long division ? Of what order is the quotient figure ? 62 DIVISION. over the divisor and annex it to the quotient. The divisor and divi- dend being different denominations, the quotient is the same as the dividend (Art. 64). Therefore the lots cost $2073-,% apiece. Hence, the EuLE.— I. Find liow many times the divisor is contained in the fewest figures on the left of the dividend, that will contain it, and set the quotient on the right. II. Multiply the divisor hy this quotient figure, and sul- tract the product from the figures divided. III. To the right of the remainder, Iring down the next figure of the dividend, and divide as before. IV. If the divisor is 7iot contained in a partial dividend, place a cipher in the quotient, bring down another figure, and continue the operation till all the figures are divided. If there is a remainder after dividi7ig the last figure, set it over the divisor, and annex it to the quotie7it. Notes. — i. The parts into which the dividend is separated in finding the quotient figure, are called partial dividends, because they are parts of the whole dividend, 2. Short and Long Division, it will be seen, are the same in prin- ciple. The only difference is, that in one the results of the several steps are carried in the mind, in the other they are set down. Short Division is the more expeditious, and should be employed when the divisor does not exceed 12. The reasons for the arrangement of the parts, and for beginning to divide at the left hand, are the same as in Short Division. 3. The quotient figure in Long as well as in Short Division, is always of the same order as that of the right hand figure of the par- tial dividend. 4. To prevent mistakes, it is customary to place a mark under the several figures of the dividend as they are brought down. 5. After the first quotient figure is obtained, for each succeeding figure of the dividend, either a significant figure or a cipher must be put in the quotient. 6. If the product of the divisor into the figure placed in the quo- Uent is greater than the partial dividend, it is plain the quotient figure is too large, and therefore must be diminished. If the remainder is equal to or greater than the divisor, the quo- tient figure is too sm-cUl, and must be increased. DIVISION. 63 PROOF. 74. By Multiplication. — Multiply tlie divisor and quo^ tient together, and to the product add the remainder. If the result is equal to the dividend, the work is right. Note. — This proof depends upon the principle, that Division is the reverse of Multiplication ; the dividend answering to the pro- duct, the divisor to one of the factors, and the quotient to the other. (Art. 62.) 75. By excess of 9s. — Multiply the excess of 95 in the divisor hy that in the quotient, and to the product add the remainder. If the excess of gs i?i this sum is equal to that in the dividend, the work is right. Hvide 1 8 1403 by 67, and prove the operation. A?is. 270711. Proof. — By Multiplication.— 2707 x 67 = 181369, and 181369 + 34 the rem. = 181403 the dividend. By excess of 9s. — The excess of 9s in the divisor is 4, and the excess in the quotient is 7. Now 4x7=28, and 28 + 34=62; the excess of 93 in 62 is 8. The excess of 9s in the dividend is also 8. 3. Divido 34685 by 15. 4. Divide 65456 by 16. 5. Divide 41534 by 20. 6. Divide 52663 by 25. 7. Divide 420345 by 39. 8. Divide 506394 by 47. 9. Divide 673406 by 69. 10. Divide 789408 by 77. II. Divide 4375023 by 86. 12. Divide 5700429 by 93. 13. Divide 6004531 by 59. 14. Divide 8430905 by 78. 15. Divide 7895432 by 89. 16. Divide 9307108 by 98. 17. How many acres of land at $75 per acre, can I buy for I18246? 18. At $83 apiece, how many ambulances can be bought for $37682? 73. N^ate. If the product of the divisor into the figure placed in the quotient, is greater than the partial dividend, what does it show f If the remainder is equal to or greater than the divisor, what ? How is Division proved ? 64 Divisioif. 76. To find the Quotient Figure, when the Divisor is large. Take i\\Q first figure of the divisor for a trial divisor, and find hovv' many times it is contained in the first or first tivo figures of the dividend, making due allowance for carrying the tens of the product of the sec07id figure of the divisor into the quotient figure. 19. Divide 18046 by 673. Analysis. — Taking 6 for a trial divisor, it is 673)18046(26 contained in 18, 3 times. But in multiplying 7 by i ^45 3, there are 2 to carry, and 2 added to 3 times 6, . make 20. But 20 is larger than the partial divi 4^86 dend 18 ; therefore, 3 is too large for the quotient 40 -^jS figure. Hence, we place 2 in the quotient, and proceed as before. (Art. 73, n.) e ,g 20. Divide 3784123 by 127. 21. Divide 4361729 by 219. 22. Divide 8953046 by 378. 23. Divide 9073219 by 738. 24. How many shawls at I95, can be bought for $42750 ? 25. In 144 square inches there is i square foot: how many square feet are there in 59264 square inches ? 26. A quartermaster paid 829328 for 312 cavalry horses: how much was that apiece ? 27. If 128 cubic feet of wood make i cord, how many cords are in 69240 cubic feet ? 28. If a purse of $150648 is divided equally among 250 sailors, how much will each receive ? 29. In 1728 cubic inches there is i cubic foot: how many cubic feet are there in 250342 cubic inches ? 30. If 560245 pounds of bread are divided equally among 11 200 soldiers, how much will each receive? 31. Div. 36942536 by 4204. 32. Div."573oo652by5i29. 33. Div. 629348206 by 52312. 34. Div. 730500429 by 61^73. 35. Divide 7300400029 by 236421. 7,6. Divide 8230124037 by 463205. 37. Divide 843000329058 by 203963428. DIVISION. C5 38. A stock company having $5000000, wa^ divided into 1250 shares: what was the value of each share? 39. A raih'oad 478 miles in length cost $18120000 : what was the cost per mile ? 40. A company of 942 men purchased a tract of land containing 272090 acres, which they shared equally: what was each man's share ? 41. A tax of $42368200 was assessed equally upon 5263 towns : what sum did each town pay ? 42. The government distributed $9900000 bounty equally among 36000 volunteers: how much did each receive? 43. Since there is i year in 525600 minutes, how many years are there in 105 192000 minutes. CONTRACTIONS. CASE I. 77. To divide by a Composite Number, Ex. I. A farmer having 300 pounds of butter, packed it in boxes of 15 pounds each : it is required to find how many boxes he had, using the factors of the divisor. Analysis. — 15 = 5 times 3. Now if he puts 5 pounds into a box, it is plain he would use as many boxes as there are 5s in 300 or 60 operation. jive-pound boxes. But 3 five-pound boxes 5)300 lbs, make i fifteen-pound box; hence, 60 five- 3)60 51b. boxes. pound boxes will make as many 15 pound Ji^ig, 20 boxes. boxes, as 3 is contained times in 60, which is 20. In the operation, we first divide by one of the factors of 15, and the quotient by the other. Hence, the KuLE. — Divide the dividend hy one of the factors of the divisor, and the quotient thence arising by another factor, and so on, until all the factors have ieeji used. The last quotie7it ivill he the one required. 77. When the divisor Is a composite number, how proceed ? OG DIVISIO-N". Notes. — i. This contraction is the reverse of multiplying by a composite number. Hence, dividing the dividend (which answers to the product) by the several factors of one of the numbers which pro- duced it, will evidently give the quotient, the other factor of which the dividend is composed, (Art. 56.) 2. When the divisor can be resolved into different sets of factors, the result will be the same, whichever set is taken, and whatever the order in which they are employed. The pupil is therefore at liberty to select the set and the order most convenient. 2. Divide 357 by 21, using the factors. 3. If 532 oranges are divided equally among 28 boys, what part, and how many will each receive ? 4. A dairyman packed 805 pounds of butter in 35 jars : how many pounds did he put in a jar 2 5. How many companies in a regiment containing 756 soldiers, allowing 61, soldiers to a company? 6. Divide 204 by 12, using different sets of factors. 7. Divide 368 by 16, using different sets of factors. 8. Divide 780 by 30, using different sets of factors. 78. To find the True Memainder, when Factors of the divisor are used. Ex. 9. — A lad picked 2425 pints of chestnuts, which he wished to put into bags containing 64 pints each: it is required to find the number of bags he could fill, and the number of pints over, or the true remainder. Analysis. — The divisor 64 equals the factors 2x8x4, Di- .,. . , 1 ,7 OPERATION. vidmg 2425 pmts by 2, the quo- 2)2425 tient is 1212 and i remainder. ~ But the wm^« of the quotient 1212 8)1212 — 1 Ipt.istr. are 2 times as large as pints, 4)^5^ — 4j 4^2= 8pt. adr. which, for the sake of distinction, oy -? ' 3 x 8 X 2 =:=48pt. 3d r. we will call quarts ; and the re- raainder i, is a pint, the same as ^ns. 37 bags, and 5 7 pt. ovei; the dividend. (Art. 60, n) Next, dividing 1212 quarts by 8, the quotient is 151, and 4 remainder. But the units of the 78 How is the trae remainder found ? Note. To what is it equal ? What phonld he done with it ? The object in multiplying each partial remainder by all th3 preceding divisors except its own ? . DIVISIOif. 6? quotient 151 are 8 times as large as quarts; call them pecks ; the remainder 4, wliich denotes quarts, must be multiplied by the preceding divisor 2, to reduce it back to pints. Finally, dividing 151 pecks by 4, the quotient is 37, and 3 remainder. But the units of tlie vjuotient 37 have four times the value of pecks ; call them hags ; and tne remainder 3, vi^hich denotes pecks, must be multiplied by the last divisor 8, to reduce them back to quarts, and be multiplied by 2 to reduce the quarts to pints. Having filled 37 bags, we have three partial remainders, i pint, 4 quarts, and 3 pecks. The licxt step is to find the true remainder. We have seen that a unit 0/ tlie 2d remainder is twice as large as those of the given dividend, which are pints ; hence, 4 quarts = 8 pints. Again, each unit of tl.e 3d remainder is 8 times as large as those of the preceding dividend, vv hich are quarts , therefore, 3 pecks = 3 x 8 or 24 quarts ; and 24 ql J. -^ 24 x 2 or 48 pts. The sum of these partial remainders, I pt. + 8 pt. + 48 pt. = 57 pts., is the true remainder. Hence, the EuLB. — Multiply each partial remainder ly all the divisors preceding its own ; the sum of these results added to the first J ivill be the true reynainder. Notes. — i. Multiplying each remainder by all the preceding di- visors except its own, reduces them to units of the same denomina- tion as the given dividend. Hence, the true remainder is equal to the sum of the partial remainders reduced to the same denomina- tion as the dividend. 2. When found, it should be placed over the given divisor, and be annexed to the quotient. 10. Divide 43271 by 45. 11. Divide 502378 by 67,. 12. Divide 710302 by 72. 13. Divide 3005263 by 84. 14. Divide 634005 11 by 96. 15. Divide 216300265 by 144. CASE II. 79. To divide by 10, 100, 1000, etc. Ex. 16. Divide 2615 by 100. Analysis. — Annexing a cipher to a figure, we have seen, multiplies it by 10 ; conversely, re- opeeatioh. moving a cipher from the right of a number l[oo)26 15 must diminish its value 10 times, or divide it Ans. 26 15 Rem. by 10 ; for, each figure ' in the number is re- paoved one place to the right. (Art. 12,) In like manner, cutting G8 DIVISIO^-. off two figures from the right, divides it bj loo ; cutting off three, divides it by looo, etc. In the operation, as the divisor is loo, v^^e simply cut off two figures on the right of the dividend ; the number left, viz., 26, is the quotient; and the 15 cut off, the remainder. Hence, the Rule. — From the right of the dividend, cut off as many figures as there are ciphers in the divisor. The figures left will he the quotient ; those cut off, the remainder. 17. Divide 75236 by 100. 20. 9820341 by looooo. 18. Divide 245065 by 1000. 21. 9526401 by loooooo. 7:9. Divide 8052 11 by loooo. 22. 80043264 by looooooo. CASE III. 80. To divide, when the Divisor has Ciphers on the right, 2^. At $30 a barrel, how many barrels of beef can be bought for I4273? Analysis. — The divisor 30 is composed of the factors 3 and 10. Hence, in the operation, operatioti. we first divide by 10, by cutting off the right- 3|Q ;4 7 j3 hand figure of the dividend ; then dividing the Quot. 142, i Eem. remaining figures of the dividend by 3, the A71S. 142] ^ Us. quotient is 142, and i remainder. Prefixing the I remainder to the 3 which was cut off, we have 13 for the true remainder, which being placed over the given divisor 30, and an- nexed to the quotient, gives 142 3^ bis. for the answer. Hence, the Rule. — I. Cut off the ciphers on the iHght of the divisor and as many figures on the right of the dividend. II. Divide the remaining part of the dividend ly the remaining part of the divisor for the quotient. III. Annex the figures cut off to the remainder, and the result will he the true remainder. (Art. 78.) Note.— This contraction is based upon the last two Cases. The true remainder should be placed over the ichole divisor, and be an- nexed to the quotient. 70. ITow proceed when the divisor is 10, 100, etc. ? 80. How when there ar/ ciphers on the ri^ht of the divisor? Note. Upon what is this contraction based' DiYisiOiq^. G9 (4. Divide 45678 by 20. 25. 81386 by 200. 26. Divide 603245 by 3400. 27. 74032 1 by 6500. 28. Divide 7341264 by 87000. 29. 8004367 by 93000. 30. Divide 61273203 by 125000. 31. 416043271 by 67000a 32. In 100 cents tbere is i dollar: how many dollars in 37300 cents ? S3. At $200 apiece, how many horses will $45800 buy? 34. If $75360 were equally distributed among 1000 men, how much would each receive ? 35. At I4800 a lot, how many can be had for I25200 ? 36. How many bales, weighing 450 pounds each, can be made of 27000 pounds of cotton ? QUESTIONS FOR REVIEW, INVOLVING THE PRECEDING RULES. 1. William, who has 219 marbles, has 73 more than James: how many has James ? How many have both ? 2. A farmer having $6S sheep, wishes to increase his flock to 775 : how many must he buy? 3. The diflFerence of two persons' ages is 19 years, and the younger is 5 7 years : what is the age of the elder ? 4. What number must be added to 1368 to make 3147 ? 5. What number subtracted from 41 18 leaves 1025 ? 6. What number multiplied by 95 will produce 7905 ? 7. The product of the length into the breadth of a field is 2967 rods, and the length is 69 rods: what is the breadth? 8. A man having 5263 bushels of grain, sold all but 145 bushels: how much did he sell? 9. What number must be divided by 87, that the quotient may be 99 ? 10. If the quotient is 217, and the dividend 7595, what must be the divisor ? 11. If the divisor is 341 and the quotient 589, what must be the dividend ? 70 DIVISION. 12. A merchant bought 516 barrels of flour at $g a bar- rel, and sold it for $5275 : how much did he gain or lose ? 13. How long can 250 men subsist on a quantity of food sufficient to last i man 7550 days? 14. How many pounds of sugar, at 11 cents, must be given for 629 pounds of coffee, at 17 cents ? 15. At $13 a barrel, how many barrels of flour must be given for 530 barrels of potatoes worth $3 a barrel ? 16. A man having $15260, deducted $4500 for personal use, and divided the balance equally among his 7 sons : how much did each son receive ? 17. A man earns 12 shillings a day, and his son 8 shil- lings: how long will it take both to earn 1200 shilhngs? 18. If a man earns $19 a week, and pays I2 a week fo^ boarding each of his 3 sons at school, how much will he lay up in 1 2 weeks ? 19. A farmer sold 6 cows at $2;^, 150 bushels of wheat at $2, and 75 barrels of apples at $4, and laid out his money in cloth at $7 a yard: how many yards did he have ? 20. If I buy 1 36 1 barrels of flour at $7, and sell the whole for $12249, how much shall I make per barrel ? 21. Tlie earnings of a man and his two sons amount to $3560 a year; their expenses are $754. If the balance is divided equally, what will each have ? 22. A man having $23268, owed $1733, and divided the rest aniong four charities : how much did each receive ? 23. How many sheep at $5 a head, must be given for 30 cows at $42 apiece ? 24. A father bought a suit of clothes for each of his 3 sons, at $123 a suit, and agreed to pay 17 tons of hay at $12 a ton, and the rest in potatoes at $4 a barrel: how many barrels of potatoes did it take ? 25. If you add $435 to $567, divide the sum by $334, multiply the quotient by 217, and divide the product by 59, what will be the result? Diyisioi^^. 71 26. If from 1530 you take 319, add 793 to the remainder, multiply the sum by 44, and divide the product by 37, what will be the result ? 27. A man's annual income is $4250; if he spends $1365 for house rent, $1439 ^r other expenses, and the balance in books, at $3 apiece, how many books can he buy ? 28. A farmer having $3038, bought 15 tons of hay at $1 1, 3 yoke of oxen at $155, 375 sheep at $5, and spent the rest for cows at $41 a head : how many cows did he buy ? GENERAL PRINCIPLES OF DIVISION. 81. From the nature of Division, the absolute value of the quotient depends both upon the divisor and the dividend. The relative value of the quotient ; that is, its value com- pared with the dividend, depends upon the divisor. Thus, 1. If the divisor is equal to the dividend, the quotient is i. 2. If the divisor is greater than the dividend, the quotient is less than i. 3. If the divisor is less than the dividend, the quotient i^ greater than i. 4. If the divisor is i, the quotient is equal to the dividend. 5. If the divisor is greater than i, the quotient is les? than the dividend. 6. If the divisor is less than i, the quotient is greater than the dividend. 82. The relation of the divisor, dividend, and quotient is such that the divisor remaining the same. Multiplying the dividend by any number, multiplies the quotient by that number. Thus, 12-^2=6; and (12x2) -7-2=6x2. Conversely, 81. Upon what does the absolute vahie of the quotient depend? Its relative Talue ? If the divisor is equal to the dividend, what is true oi the quotient ? If greater? If less? If the divisor is t, what is the quotient? If £;reater than i ? If less than i f 82. What is the effect of multiplying the dividend ? 83. Of dirid- Incrit? TZ DIVISION, 83. Dividing tlie dividend l>y any number, divides the quotient by that number. Thus, (12-^2)-:- 2 = 6h- 2. 84. Multiplying the divisor by any number, divides the quotient by that number. Thus, 24-^-6=4; and 24-r (6x2) =4~ 2. Conversely, 85. Dividi7ig the divisor by any number, multiplies the quotient by that number. Thus, 24-^ (6-^2) =4 x 2. 86. Multi2)lying or dividing both the divisor and d'^V^- Je/it? by the same number, does not 6j//er the quotient. Thus, 48-^8 = 6; so (48x2)4-(8x2)=:6; and (48-^2)-t- (8-^-2)1:^6. Note. — Multiplying or dividing tlie dividend, produces a like effect on the quotient ; but multiplying or dividing the divisor, pro- duces the opposite effect on the quotient. 86, «• If a number is both multiplied and divided by the same number, its value is not altered. Thus, (7 x 6) -=-6 is equal to 7. PROBLEMS AND FORMULAS PERTAINING TO THE FUNDAMENTAL RULES. 87. A I^rohlem is something to he done, or a question to be solved. 88. A Formula is a specific rule by which problems are solved, and may be expressed by common language, or by signs. 89. The four great Problems of Arithmetic have already been illustrated. They are — ist. When two or more numbers are given, to find their dum, or amount. (Art. 29.) 2d. When two numbers are given, to find their difference, 84. What is the effect of multiplying the divipor ? Of dividing it ? 86. WTial Ib the eflFect of multiplying or dividing both the divisor and dividend ? 87. What id a problem? 8S. A formula? 89. The four great problems ot arithmetic 1 PROBLEMS AN"D FORMULAS. 73 3d. When two factors are giyen, to find tlieir product (Art. 50.) 4tb. When two members are given, to find how man^ times one is contained in the other. (Art. 73.) 90. These problems constitute the four fundamental rules of Arithmetic, called Addition, Subtraction, Multi- plication, and Division. Notes. — i. These rules are called fundamental, because upon them are based all arithmetical operations. 2. As multij)licatioji is an abbrevated form of addition, and division of subtraction, it follows that every change made upon the value of a number, must increase or diminish it. Hence, strictly speaking, there are but two fundamental operations, viz. : aggregation and diminution, or increase and decrease. 3. The following problems, though subordinate, are so closely con- nected with the preceding, that a passing notice of them may not be improper, in this connection. 91. To find the greater of two numbers, the less and their difference being given. 1. A planter raised two successive crops of cotton, the smaller of which amounted to 4168 bales, and the dif- ference between them was 11 23 bales: what was the greater crop ? Analysis. — If the difference between two numbers be added to the less, it is obvious the sum must be equal to the greater. There- fore, 4168 bales + 1 123 bales or 5291 bales must be the greater crop. Hence, the Rule. — To the less add the difference, and the sum will be tlie greater. (Art. 39.) 2. One of the two candidates at a certain election, re- ceived 746 votes, and was defeated by a majority of 411: how many votes did the successful candidate receive ? 90. What do these constitute? Note. Why so called? What is given and ■what required In addition ? In subtraction ? In multiplication ? In division ? Where begin the operation in addition, subtraction, and multiplication ? Where tn division ? WTiat is the difference between addition and subtraction ? Between addition and multiplication ? Between subtraction and division ? Between multiplication and division ? 74 PEOBLEMS AKD FORMULAS. 92. To find the less of two numbers, the greater and their difference being given. 3. The greater of two cargoes of flour is 526.7 barrels, and their difference is 1348 barrels: how many barrels does the smaller contain ? Analysis. — The difference added to the less number equals the greater ; therefore, the greater diminished by the difference, must be equal to the less ; and 5267 barrels minus 1348 barrels leaves 3919 barrels, the smaller cargo. Hence, the EuLE. — From the greater subtract the difference, and the result ivill he the less. (Art. 38.) 4. At a certain election one of the two candidates re- ceived 1366 votes, and was elected by a majority of 219 votes : how many votes did the other candidate receive ? 93. The Product and one Factor being given, to find the other Factor. 5. A drover being asked how many animals he had, replied that he had 67 oxen, and if his oxen were multi- plied by his number of sheep, the product would be 37520: how many sheep had he? Analysis. — Since 3752013 Si product, and 67 one of its factors, the other factor must be as many as there are 67s in 37520; and 37520-^67=560. (Art. 93.) Therefore, he had 560 sheep. Hence, the EuLE. — Divide the product hy the given factor, and the quotient will he the factor required. (Art. 62.) 6. The length of a certain park is 320 rods, and the product of its length and breadth is 51200 rods: what is its breadth ? 94. The Product of three or more Factors and all the Factors but one being given, to find that Factor. 7. The product of the length, breadth, and height of a certain mound is 62730 feet; its length is 45 feet, and its breadth 41 feet: what is its height? 91. How find the greater of two numbers, the less and difference being given? Q*, How find the less, the greater and difference being given ? PROBLEMS Al^^D FORMULAS. 75 Analysis. — The contents of solid bodies are found by multiplying their length, breadth, and thickness together. Now as the length is 45 feet, and the breadth 41 feet, the product of which is 1845, the height must be 62730 feet divided by 1845, or 34 feet, Hence, the EuLE. — Divide the given product by the product of the given factors, arid the quotient will le the factor re- quired. (Art, 93.) 8. The continued product of tlie distances wh' h. 4 men traveled is 1944630 miles; one traveled 45, nother 41, and another 34 miles: how far did the fourth travel? 95. To find the Dividend, the Divisor and Quotient being given. 9. If the quotient is 7071, and the divisor 556, what is the dividend ? Analysis. — Since the quotient shows how many times the divisor is contained in the dividend, it follows, that the product of the divisor and quotient must be equal to the dividend. Now 7071 x 556 = 3931476 the dividend. Hence, the Rule. — Multiply the divisor ly the quotient, and the result luill he the dividend. (Art. 74.) 10. What number of dollars divided among 135 persons will give them $168 apiece? 96. To find the Divisor, the Dividend and Quotient being given. 11. What must 8640 be divided by that the quotient may be 144? Analysis. — Since the quotient shows how many times the divisor is contained in the dividend, it follows that if the dividend is divided by the quotient the result must be the divisor, and 8640^144=60. Therefore, the divisor is 60. Hence, the Rule. — Divide the dividend by th^ quotient, and the result will be the divisor. (Arts. 62, 93.) 93. The product and one factor being given, how find the other fiactort 76 PHOBLEMS AND FORMULAS. 12. If the dividend is 7620, and the quotient 127, what must be the divisor ? 97. The Sum and Difference of two numbers being given, to find the Nmnbers. 13. The sum of two numbers is 65, and their difference 15 : what are the numbers ? Analysis. — The sum 65 is equal to the greater number increased by the less; and the greater diminished by the difference 15, is equal to the less (Art. 38). Hence, if 15 is taken from 65, the re- mainder 50, must be twice the less. But 50-7-2—25 the less num- ber ; and 25 + 15=40 the greater number. Proof. — 404-25 = 65 the given sum. Hence, the Rule. — From tlie sum take .the difference, and half the remainder ivill be the less number. To the less add the difference, and the result ivill be the greater. (Art. 39.) 14. A merchant made $5368 in two years, and the dif- ference in his annual gain was $976: what was his profit each year ? 15. The whole number of votes cast for the two candi- dates at a certain election was 5564, and the successful candidate was elected by a majority 708 : how many votes did each receive ? 16. A lady paid S250 for her watch and chain; the former being valued $42 higher than the latter : what was the price of each ? 17. Two pupils A and B, solved 75 examples; B solving 15 less than A: how many did each solve? 1 3. A and B found a pocket-book, and returning it to the owner, received a reward of $500, of which A took $38 more than B : what was the share of each ? 94. When the product of three or more factors, and all hut one are given, how find that one ? 95. IIow find the dividend, the divisor and quotient being given? 96. How find the divisor, the dividend and quotient being given ? ANALYSIS. 98. Analysis primarily denotes the seioaration of an object into its elements. 99. Analysis, in AviihrnQiic,!^ the process of tracing the relation of the conditions of a problem to each other, and thence deducing the successive stejjs necessary for its solution. Notes, — i. The application of Analysis to aritlimetic is of recent date, and to this source the late improvements in the mode of teaching the subject are chi-efly due. Previous to this, Arithmetic to most pupils, was a hidden mystery, regarded as beyond the reach of all but the favored few. 2. The pupil has already learned to analyze particular examples, and from them to deduce specific rules by which similar examples may be solved ; but Arithmetical Analysis has a much wider range. It is applied with advantage to those classes of examples commonly placed under the heads of Barter, Percentage, Profit and Loss, Simple and Compound Proportion, Partnership, etc. In a word, it is the grand Common- Sense Rule by which business men perform the great majority of commercial calculations. 100. 1^0 specific dtrectio7is can be prescribed for analyti- cal solutions. The following suggestions may, howeveic be serviceable to beginners : ist. In general, we reason from the given value of one, to the value of two or more of the same kind. Or, 2d. From the given value of two or more, to that of one. In the first instance we reason from a part to the whole ; in the second, from the ichole to a yart. 3d. Sometimes the result of certain combinations ia given, to find the original number or lase. 98. What is the primary meaning of analysis ? 99. What is arithmetical analysis? Note. What is said as to its utility in arithmetical and business calculations ? 100. Can specific rules be given for analytical solutions t What general directions can you mention ? 78 Al^ALTSIS. In such cases, it is generally best to begin with the result, and reverse each operation in succession, till the original number is reached. That is, to reason from the result to its origin, or from effect to cause. 1. A farmer bought 150 sheep at $2 a head, and paid for them in cows at $20 a head : how many cows did it take to pay for the sheep ? 2. How much tea, at 85 cents a pound, must be given for 425 pounds of rice, at 10 cents a pound ? 3. How much sugar, at 1 2 cents a pound, must be given for 288 pounds of raisins, at 18 cents a pound? 4. How much corn, at 80 cents a bushel, must be given for 160 pounds of tobacco, at 30 cents a pound? 5. How much butter, at 40 cents a pound, must be given for 62 yards of cahco, at 20 cents a yard ? 6. Bought 189 yards of linen, at 84 cents a yard, and paid for it in oats, at 42 cents a bushel : how many bushels did it take ? 7. Paid 18 barrels of flour for 30 yards of cloth, worth $6 a yard : what was the flour a barrel ? 8. A farmer gave 15 loads of hay for 45 tons of coal, worth %6 a ton : what did he receive a load for his hay ? 9. A sold B 35 hundred pounds of hops, at I27 a hun- dred, and took 45 sacks of coffee, at $14 a sack, and the balance in money : how much money did he receive ? 10. James bought 96 apples, at the rate of 4 for 3 cents, and exchanged them for pears at 4 cents apiece: how many pears did he receive ? 11. A farmer being asked how many acres he had, replied, if you subtract 20 from the number, divide the remainder by 8, add 15 to the quotient, and multiply by 5, the product will be 125 : how many had he ? Analysis, — Taking 125 as the base, and reversing the several operations, beginning with the last, we have 125 -f- 5 = 25, the miraber before the multiplication. Again, subtracting 15 from 25. we have 25 — 15 = 10, the number before the addition. A2^ALYSIS. 79 Next, multiplying lo by 8, we liave lo x 8=:8o, the number before tlie division. Finally, adding 20 to 80, we have 80 + 20=100, the number required. 12. What number is that, to which, if 25 be added, and the sum multiplied by 9, the product will be 504? 13. What number, if diminished by 40, and the remain- der divided by 8, the quotient will be 58 ? 14. A man being asked how many children he had, answered, if you multiply the number by 11, add 23 to the product, and divide the sum by 9, the quotient will be 16 : how many children had he ? 15. The greater of two numbers is 3 times the less, and the sum of the numbers is s^ : what are the numbers ? Analysis.— The smaller number is i part, and the larger 3 parts ; hence, the sum of the two is 4 parts, which by the conditions is 36. Now, if 4 parts of a nijm.ber are 36, i part is equal to as many units as there are 4s in 36, or 9. Therefore 9 is the smaller number, and 3 times 9 or 27, the greater. 16. The sum of two numbers is 72, and the greater is 5 times the less • what are the numbers ? 17. Divide 472 into three such parts, that the second shall be twice the first, and the third 3 times the second plus 13. Analysis. — Calling the first i part, the second will be 2 parts, and the tliird 6 parts plus 13 ; hence the sum of the three, in the terms of the first, is 9 parts plus 13, which by the conditions is 472, Taking 13 from 472 leaves 459, and we have 9 parts equal to 459. Now if 9 parts equal 459, i part is equal to as many units as 9 is contained times in 459, or 51. Therefore the first is 51, the second 2 times 51 or 102, the third 3 times 102 plus 13 or 319. 18. A and B counting their money, found that both had $473, and that A had 3 times as much as B plus $25 ; how much had each ? 19. The sum of two numbers is 243, the second is three times the first minus 25 : what are the numbers ? 20. What number is that to which if 315 be added, the sum will be 250 less than 2683 ? (For further applications of Analysis, see subsequent pages.) CLASSIFICATION AND PROPERTIES OF NUMBERS. 101. Numbers are divided into abstract and concrete, simple and compound, prime and comp)osite, odd and even, integral, fractional, and mixed, Tcnown and unhnoivn, similar and dissimilar, commensurable and incommen- surable, rational and irrational or surds. Def. — I. Abstract Numbers are those which are not applied to things ; as, one, two, three. 2. Concrete Numbers are those which are applied to things ; as, two caps, three pencils, six yards. 3. Simple Numbers are those which contain only one denomination, and may be either abstract or concrete; as, 13, 11 pounds. 4. A Compound Number is one containing two or rn^o^e denamina- tions, which have the same base or nature ; as, 3 shillings and 6 pence ; 4 yards 2 feet and 6 inches. 5. A Prime Number is one which cannot be produced by multi plication of any two or more numbers, except a unit and itself. All prime numbers except 2 and 5, end in i, 3, 7, or 9. 6. A Composite Number is the product of two or -move factors, each of which is greater than i ; as, 15 (5 x 3), 24 (2 x 3 x 4), etc. (Art. 5£,.) A prime number diflfers from a composite number in two respects : First, In its origin ; Second, In its divisibility ; the former being divisible only by a unit and itself; the latter by each of the factors which produce it. 7. Two numbers are prime to each other or relatively prime, when the only nimiber by which both can be divided without a remainder, is a unit or i ; as, 5 and 6. 8. An Even Number is one wliich can be divided by 2 without a remainder; as, 4, 6, 10. 9. An Odd Number is one which cannot be divided by 2 without a remainder; as, 3, 5, 7, 9, 11. Name the kind of each of the following numbers, and why : 3, 10, 17, 21, 28, 31, 56, 63, 72, 81, 44, 39, 91, 67, 51, 84, 99, 100. 10. An Integer is a number which contains one or more entire units only; as, i, 3, 7, 10, 50, 100. loi. How are numbers divided ? An abstract number ? Concrete ? Simple ? Compound ? Prime ? Composite ? When are two numbers prime to each other ! Eren ? Odd ? An integer ? A fraction ? A mixed number ? PROPERTIES OF ]SrUMBERS. 81 11. A Fraction is one or more of the equal parts into which a unit is divided ; as, i-half, 2-thirds, 3-fourths, etc. 12. A Mixed Number is an integer and a fraction expressed together ; as, 5^^, iif, etc. 13. Like or similar numbers are those which express units of the same kind or denomination ; as, 3 shillings and 5 shillings, four and seven, etc. 14. Unlike Numbers are those which express units of different kinds or denominations ; as, 2 apples and 3 oranges. 15. Commensurable Numbers are tliose which can be divided by the same number, without a remainder ; as, 9 and 12, each of which can be divided by 3. 16. Incommensurable Numbers are those which cannot be divided by the same number without a remainder. Thus, 3 and 7 are incom- mensurable. 17. A given number is one whose value is expressed. A number is also said to be given when it can be easily inferred from something else which is given. Thus, if two numbers are given, their sum and difference are given. 18. An Unknown Number is one whose value is not given. 102. A Factor of a number is one of the numbers, which multiplied together, produce that number. 103. An Exact Divisor of a number is one which will divide it without a remainder. Thus 2 is an exact divisor of 6, 3 of 15, etc. Notes. — i. An exact divisor of a number is always & factor of that number ; and, conversely, a factor of a number is always an exact divisor of it. For, the dividend is the product of the didsor and quotient, and therefore is divisible by each of the numbers that pro- duce it. (Art. 62.) 2. The terms divisor and factor are here restricted to integral numbers. 104. A 3Icasii7^e of a number is an exact divisor of that number. It is so called because the cor)iparath)0 magnitude of the number divided, is determined by this standard. Like numbers ? Unlike ? Commeneurable ? Incommensurable ? What is a given number ? An unknown 7 A factor ? An exact divisor ? A measure ? 82 PROPERTIES OF NUMBERS. 105. An aliquot part of a number is a factor or an ixact divisor of that number. Note. — The terms factors, measures, exact divisors, and aliquot •parts, are different names of the same thing, and are often iised as synonymous. Thus, 3 and 5 are respectively the factors, measures, exact divisors, and aliquot parts of 15. 106. The reciprocal of a nurriber is i divided by that number. Thus, the reciprocal of 4 is 1-7-4, or |. COMPLEMENT OF NUMBERS. 107. The Complement of a Nurriber is the dif- ference between the number and a unit of the next higher order. Thus, the complement of 7 is 3; for 10 — 7 = 3; the complement of 85 is 15 ; for 100 — 85 = 15. 108. To find the Conipleinent of a Number. Subtract the given number from 1 with as many ciphers annexed, as there are iyitegral figures in the given number. Or, begin at the left hand, and subtract each figure of the given number from 9, except the last significant figure on tlhe right, which must be tahen from 10. Note. — The second method is based upon the principle that when we borrow, the next upper figure must be considered i less than it is. (Art. 37, n.) Find the complement of the following numbers : I. 328. 6. 6072. II. 56239. 16. 73245- 2. 567. 7. 8256. 12. 64123. 17- 1234567- 3. 604. 8. 9061. 13. 102345. 18. 2301206. 4. 891. 9. 13926. 14. 261436. 19. 3021238. 5. 4638. 10. 23184. 15. 40061. 20. 7830426. lo-;. An aliquot part? Note. What is paid as to the use of these fonr terms? 106. The reciprocal of a number? 107. The complement? 108. How find th« •omplement of a number ? PEOPERTiES OF :n^umbers. 83 DIVISIBILITY OF NUMBERS. 109. One number is said to be divisible by anotlier, when there is no remainder. The division is then com- When there is a remainder, the division is incomplete; and the dividend is said to be indivisible by the divisor. . Notes. — i. In treating of the divisibility of numbers, the term divisor is commonly used for exact dimsor. 2. Every integral number is divisible by the unit i, and by itself. It is not customary, however, to consider the unit i, or the number itself, as & factor. (Art. 102.) 110. In determining the divisibility of numbers the ioWoyfmg pro2:)erties ov facts are useful: Prop. i. Any number is divisible by 2, which ends with o, 2, 4, 6, or 8. 2. Any number is divisible by 3, if the sum of its digits is divisible by 3. Thus, 147 the sum of whose digits is 1+4 + 7=12, is divisible by 3. 3. Any number is divisible by 4, if its ti'^o right hand figures are divisible by 4; as, 2256, 15368. 19384. 4. Any number is divisible by 5, which ends with o or 5 ; as, 130, 675. 5. Any even number is divisible by 6, which is divisible by 3. Thus, 1344-7-3=448 ; and 1344-^-6=224. 6. Any number is divisible by 8, if its three right hand figures are divisible by 8 ; as, 1840, 1688, 25320. 7. Any number is divisible by 10, which ends with o ; by 100, if it ends with 00 ; by 1000, if it ends with 000, etc. 8. Any number is divisible by 12 which is divisible by 3 and 4. 9. Any number divided by 9 will leave the same remainder as the warn of its digits divided by 9. Hence, 10. Any number is divisible by 9 if the sum of its digits is divisible by 9. Thus, the sum of the digits of 54378 is 5 + 4 + 3 + 7 + 8, or 27. Now, as 27 is divisible by 9, we may infer that 54378 is divisible br9- 10-,. When is one number divisible by a.iotber ? When indivisible ? no. Whei Is a number divisible by 2 ? By 3? By 4? By 5? By 6? By 8? By 10? 84 PKOPEKTIES OF NUMBERS. 11. As 9 is a multiple of 3, any number divisible by 9, is also divisible by 3. 12. Any number divided by 11 will leave tbe same remainder 2ja the sum of its digits in the even places, taken from the sum of those in the odd places, counting from the right, the latter being increased or diminished by 11 or a multiple of 11. Thus, the sum of the digits in the even places of 314567, viz., 6 + 4 + 3, ^^ '^3> is equal to the sum of the digits in the odd places, viz., 7 + 5 + 1, or 13 ; there- fore, 314567 is divisible by 11. Hence, 13. Any number is divisible by 11 when the sum of its digits in. the even places is equal to the sum of those in the odd places, or when their difference is divisible by 1 1. (For a demonstration of the properties of 9 and 11, see Higher Arithmetic.) 14. If one number is a divisor of another, the former is also a divisor of any multiple of the latter. (Art. 103, n.) Thus, 2 is a divisor of 6 ; it is also a divisor of the product of 3 times 6, of 5 x 6, and of any whole number of times 6. 15. If a number is an exact divisor of each of two numbers, it is also an exact divisor of their sum^ their difference, and their product. Thus, 3 is a divisor of 9 and 15 respectively ; it is also a divisor ot 9 + 15, or 24; of 15—9, or 6; of 15 X9, or 135. 16. A composite number is divisible by each of its prime factors, by the product of any two or more of them, and by no other number. Thus, the prime factors of 30 are 2x3x5. Now 30 is divisible by by 2, 3 and 5, by 2 x 3, by 2 x 5, by 3 x 5, and by 2 x 3 x 5, and by no other number. Hence, 17. The least divisor of a composite number, is a prime number. 18. An odd number cannot be divided by an even number ?i divisor of them. (Art. 123.) Hence, the Ans. 2. Rule. — Resolve each of the given numbers into two factors, one of which is common to them all. Find common divisors of the following numbers : 2. 8, 16, 20, and 24. Ans. 2 and 4. 3. 12, 15, 18, and 30. 6. 21, 28, 35, 49, d^. 4. 36, 48, 96, and 108. 7. 20, 30, 70, and 100. 5. 42, 54, dd, and 132. 8. 60, 75, 120, and 240. 9. 16, 24, 40, 64, 116, 120, 144, 168, 264, 1728. GREATEST COMMON DIVISOR. 125. The Greatest Common Divisor of two or more numbers, is the greatest number that will divide each of them without a remainder. Thus, 6 is the greatest common divisor of 18, 24, and 30. Notes. — i. A common divisor of two or more numbers is the same as a common factor of those numbers ; and the greatest common divisor of them is their greatest common factor. Hence, 2. If any two numbers have not a common factor, they cannot have a common dinsor greater than i. (Art. 112, n.) 3. A common divisor is often called a common measure, and the greatest common divisor, the greatest common measure. 123. A common divisor? 125. Greatest common divisor? Note. A common fllvisor is the same as wliat ? Often called what ? C0MM0 2^^ DIVISOES. FIRST METHOD. 126. To find the Greatest Common Divisor by continued Divisions. I. What is the greatest common divisor of 28 and 40 ? Analysis. — If we divide the greater number by the less, the quotient is i, and 12 remainder. operation. Next, dividing the first divisor 28, by the first 28)40(1 remainder 12, the quotient is 2, and 4 remainder. 28 Again, dividing the cecond divisor by the second remainder 4, the quotient is 3, and o rem. The )28(: 12 26(2 last divisor 4, is the greatest common divisor. —3 Demonstration. — We wish to prove two points : 4)^2(3 1st, That 4 is a common divisor of the given num- 1 2 bers. 2d, That it is their greatest common divisor. First. We are to prove that 4 is a common divisor of 28 and 40. By the last division, 4 is contained in 12, 3 times. Now, as 4 is a divisor of 12 ; it is also a divisor of i\ie product of 12 into 2, or 24. (Art. no, Prop. 14.) Next, since 4 is a divisor of itself and 24, it must be a divisor of the sum of 4 + 24, or 28, which is the smaller number. (Prop. 15.) For the same reason, since 4 is a divisor of 12 and 28, it must also be a divisor of the sum of 12 + 28, or 40, which is the larger number. Hence, 4 is a common divisor of 28 and 40. Second. We are now to prove that 4 is the greatest common divisor of 28 and 40. If the greatest common divisor of these numbers is not 4, it must be either greater or less than 4. But we have shown that 4 is a common divisor of the given numbers; therefore, no number less than 4 can be the greatest common divisor of them. The assumed number must therefore be greater than 4. By supposi- tion, this assumed number is a divisor of 28 and 40; hence, it must be a divisor of their difference 40—28 or 12. And as it is a divisor of 12, it must also divide the product of 12 into 2 or 24. Again, since the assumed number is a divisor of 28 and 24, it must also be a di- visor of their difference, which is 4 ; that is, a greater number will divide a less without a remainder, which is impossible. (Art. 81, Prin. 2.) Therefore, 4 must be the greatest common divisor of 28 and 40. Hence, the 126. Explain Ex. i. Prove that 4 is the greatest common divisor of 28 aad 40. Eule ? Note. If there are more than two numbers, how proceed ? If th< Mst divisor is i, then what? What is the greatest common divisor of two or m«ra prime numbers, or numbers prime to each other ? COMMOiq^ DIVISORS. 93 EuLE. — Divide the greater nuniber ly the less; then divide the first divisor ly the first remainder, the second divisor by the second remainder, and so on, until nothing remains; the last divisor will he the greatest common divisor. Notes. — i. If there are more tlian two mimbers, begin witli the smaller, and find the greatest common divisor of two of them ; then of this divisor and a third number, and so on, until all the numbers have been taken. 2. The greatest common divisor of two or more prime numbers, or numbers prime to each other, is i. (Art. 112, n.) 3, If the last divisor is i, the numbers are prime, or prime to each other ; therefore their greatest common divisor is i. Such numbers are said to be incommensurable. (Art. loi, Def 16.) 2. What is the greatest com, divisor of 48, 72, and 108 ? 3. What is the greatest common divisor of 72 and 120? SECOND METHOD. 127. To find the greatest Common IHvisor by Prime Factors. 4. What is the greatest common divisor of 28 and 40 ? Analysis. — Resolving the given numbers into operation factors common to both, we have 28 = 2x2x7, ^ 2x2 x*- and 40=2x2x10, Now the product of these ~~ ' factors, viz., 2 into 2, gives 4 for the greatest 4° 2 X 2 X 10 common divisor, tlio same as by the first method. Ans. 2 X 2^4. 5. What is the greatest com. div. of 30, 45, and 105 ? Analysts. — Setting the numbers in a hori- opekatiok. zontal line, we divide by any prime nimaber, as 3)30, 45, 105 3, that will divide each without a remainder, cVio iT ^ and set the quotients under the corresponding numbers. Again, dividing each of these quo- ' ^' ' tients by the prime number 5, the new quotients 3 -^ 5^^^S> ^^^« 2, 3, and 7, are prime, and have no common factor. Therefore, the product of the common divisors 3 into 5, or 15, is the giaatest common divisor. (An. no, Prop. 16) Hence, the 127. What is the rule for the second method? Woie. What is the object o/ placing ths numbers in a horizontal line ? 94 COMMOK DIVISORS. Rule. — I. Write the numbers in a horizontal line, and divide ly any prime number that tvill divide each without a remainder, setting the quotients in a line heloiv. II. Divide these quotients as before, and thus jiroceed till •[10 number can be found that loill divide all the quotients ivithout a remainder. The product of all the divisors will be the greatest common divisor. Notes. — i. This rule is the same as resolving the given numbers into prime factors, and multiplying together all that are common. 2. The advantage of placing the numbers in a horizontal line, is that the prime factors that are common, may be seen at a glance. 3. When required to find the greatest common divisor of three op more numbers, the second method is generally more expeditious, and therefore preferable. 4. When the given numbers have only one common factor, that factor is their greatest common divisor. 5. Two or more numbers may have several common divisors ; but they can have only one greatest common divisor. Find the greatest common divisor of the following: 6. 67, and 42. 14. 10, 28, 40, 64, 90, 32. 7. 135 and 105. 15. 12, 2>^, 60, 108, 132. 8. 24, 36, 72. 16. 16, 28, 64, 56, 160, 250. 9- 60, 75, 12. 17- 576 and 960. 10. 75, 125, 250. 18. 1225 and 592. II. 42, 54, 60, 84. 19. 703 and 1369. 12. 72, 100, 168, 136. 20. 1492 and 1866. 13. 60, 84, 132, 108. 21. 2040 and 4080. 22. A merchant had 180 yards of silk, and 234 yards of poplin, which he wished to cut into equal dress pat- terns, each containing the greatest possible number of yards : how many yards would each contain ? 23. Two lads had 42 and 63 apples respectively : how many can they put in a pile, that the piles shall be equal, and each pile have the greatest possible number ? 24. A man had farms of 56, 72, and 88 acres respectively, which he fenced into the largest possible fields of the sam^ number of acres : how many acres did he put in each ? MULTIPLES. 128. A 31ultiple is a number which can he divided by another number, without a remainder. Thus, 12 is a multiple of 4. Remark. — The Term Multiple is also used in the sense of product : as when it is said, " if one number is a di- visor of another, the former is also a divisor of any mul- tiple (product) of the latter." Thus, 3 is a divisor of 6 ; it is also a divisor of 7 times 6, or 42. Note. — Multiple is from the Latin multiplex, having many folds, or taken many times ; hence, a product. But every product is didsiUe by Ms factors ; hence, the term came to denote a dividend. The former signification is derived from the formation of the number ; the latter, from its dimsihility. 129. A Common Multiple is any number that can be divided by two or more numbers without a re- mainder. Thus, 1 8 is a common multiple of 2, 3, 6, and 9. 130. A common multiple of two or more numbers may be found by multiplying them together. That is, the pro- duct of two numbers, or any entire number of times their product, is a common multifile of them. (Art. 128, ni) Notes. — i. The factors or divisors of a multiple are sometimes called sijh-muttiples. 2. A number may have an unlimited number of multiples. For, according to the second definition, every number is a multiple of itself; and if multiplied by 2, the product will be a second multiple ; if multiplied by 3, the product will be a third multiple ; and univer- sally, its product into any ichole number will be a multiple of that number. (Art. 128.) Find a common multiple of the following numbers: 1. 2, 3 and 5. 4. 2, 8 and 10. 7. 13, 7 and 22. 2. 3, 7 and II. 5. 7, 6 and 17. 8. 19, 2 and 40. 3. 5, 7 and 13. 6. 11, 21 and 31. 9. 17, 10 and 34. 128. Meaning of the term multiple? 129. What is a common multiple? 130. How found? Not6. What are sub-multiples? How many multiples has a number ? 96 MULTIPLES LEAST COMMON MULTIPLE. 131. The Least Common Multiple of tvv^o or more numbers, is the least number that can be divided by each of them without a remainder. Thus, 15 is the least common multiple of 3 and 5. 132. A Multiple, according to the first definition, is a coraijosite number. But a composite number contains all the prime factors of each of the numbers which produce it. Hence, we derive the following Principles : Prin. I. That a multiple of a number must contain all the prime factors of that number. 2, A common multiple of two or more numbers must contain all the prime factors of each of the given numbers. 3. The least common multiple of two or more numbers is the least number which contains all their prime factors, each factor being tat en as many times only, as it occurs in either of the given numbers, and no more. 133. To find the Least Common Multiple of two or more numbers.. I. What is the least common multiple of 18, 21 and 66 ? ist Method. — We write the num- ist operation. bers in a horizontal line, with a comma 2)18, 21 66 between them. Since 2 is a prime fac- TT ~ "~ tor of one or more of the given num- -— ^-^ bers, it must be a factor of the least 3> 7> ^ ^ common multiple. (Art. 132, Prin. i.) 2x3x3x7x11 = 1386 We therefore divide by it, setting the quotients and undivided numbers in a line below. In like manner, we divide these quotients and undivided numbers by the prime number 3, and set the results in another line, as before. Now, as the numbers in the third line are prime, we can carry the division no further ; for, they have no common divisor greater than i, (Art. loi.) Hence, the divisors 2 and 3, with the numbers in the last line, 3. 7 and II, are all the prime factors contained in the given numbers, and each is taken as many times as it occurs in either of them. 131. What is the leaBt common multipls? 132. What principles arc deriTcd? MULTIPLES. 97 Therefore, the continued product of these factors, 2x3x3x7x11, or 1386, is the least common multiple required. 2d Method. — Resolving each num- 2d operation. ber into its prime factors, we have l8 = 2X3X 3 18=2x3x3, 21 = 3x7, and 66=2x3 21 = 3x7 XII. Now, as the least common mul- 66 = 2x3x11 tiple must contain all the prime fac- tors of the given number, it must con- 2x3x11x7x3 = 13 86 tain those of 66, viz., 2x3x11; we therefore retain these factors. Again, it must contain the prime factors of 21, which are 3x7, and of 18, which are 2x3x3, each being taken as many times as it is found in either of the given numbers. But 2 is already retained, and 3 has been taken once ; we therefore cancel the 2 and one of the 3s, and retain the other 3 and the 7. The continued product of these factors, viz., 2x3x11x7x3, is 1386, the same as before. Hence, the Rule. — I. Write the numbers in a horizontal line, and divide ly any 'prime number that tvill divide tivo or more of them without a remainder, placing the quotients and numbers U7idivided in a line below. II. Divide this line as before, and thus proceed till no two numbers are divisible by any number greater than i. Tlie continued product of the divisors and numbers in the last line will be the answer. Or, resolve the mimbers into their prime factors ; mid- iiply these factors together, talcing each the greatest number of times it occurs in either of the given numbers. The product will be the ansiuer. Notes. — i. These two methods are based upon the same principle, viz. : that the least common multiple of two or more numbers is the least number which contains all their prime factors, each factor being taken as many times only, as it occurs in either of the given numbers. (Art. 132, Prin. 3.) 2. The reason we employ prime numbers as divisors is because the given numbers are to be resolved into prime factors, or factors ^rm6 133. How find the least common multiple ? Note. Upon what principle are these two methods hased ? Why employ prime numbers for divisors in the first method ? Why write the numbers in a horizontal line ? What is the second method ? Advantage of it ? How may the operation be shortened ? When the given numbers are prime, or prime to each other, what is to be done ? 98 MULTIPLES. to each other. If the divisors were composite numbers, they would be liable to contaiu factors common to some of the quotients, or numbers in the last line ; and if so, their continued product would not be tlie least common multiple. 3. The object of placing the numbers in a horizontal line, is to resolve all the numbers into prime factors at the same time. 4. The chief advantage of the second method lies in the distinct- ness with which the prime factors are presented. The former is the more expeditious and less liable to mistakes. 5. The operation may often be shortened by cancelling any num- ber which is a factor of another number in the same line. (Ex. 2.) When the given numbers are prime, or prime to each other, they have no common factors to be rejected ; consequently, their con- tinued product will be the least common multiple. Thus, the least common multiple of 3 and 5 is 15 ; of 4 and 9 is 36. 2. Find the least common multiple of 5, 6, 9, 10, 21. Analysis. — In the first line 5 2)5, 6, 9, 10, 21 is a factor of 10, and is therefore ^w I ~ cancelled ; in the second, 3 is a — ^—^ — lactor of 9, and is also cancelled. 3? 5? 7 The product of 2X3X3X5X 7=630, the answer required. Find the least common multiple of the following : 10. 2>^, 48, 72, 96. 42, 6^^ 84, 108. 120, 144, 168, 216. 96, 108, 60, 204. 126, 154, 280, 560. i44> 256, 72,300. 250, 500, 1000. 17. Investing the same amount in each, what is the smallest sum with which I can buy a whole number of pears at 4 cents, lemons at 6, and oranges at 10 cents ? 18. A can hoe 16 rows of corn in a day, B 18, C 20, and D 24 rows : what is the smallest number of rows that will keep each employed an exact number of days ? 19. A grocer has a 4 pound, 5 pound, 6 pound, and a T2 pound weight: what is the smallest tub of butter that can be weighed by each without a remainder ? 3. 8, 12, 16, 24. 10. 4. 14, 28, 21, 42. II. 5 36, 24, 48, 60. 12. 6. 25? 40, 75. loo- 13. 7- 16, 24, ^2, 40. 14. 8. 22, 2>3, SS, 66. 15. 9- 30, 40, 60, 80. 16. FRACTIONS. 134. A Fraction is one or more of the equal parts into which a unit is divided. The numher of these parts indicates their name. Thus, when a unit is divided into tiuo equal parts, the parts are CdXl^di halves ; when into three, thej are called thirds; when into four, fourths, etc. Hence, 135. A half is one of the iivo equal parts of a unit ; a third is one of the three equal parts of a unit; a fourth, one of the four equal parts, etc. Note. — The term fraction is from the Latin frango, to Ireak. Hence, Fractions are often called Broken Numbers. 136. The value of these equal parts depends, First. Upon the magnitude of the unit divided. Second. The numher of parts into wiiich it is divided. Illustration. — ist. If a large and & small apple are each divide J into two, three, four, etc., equal parts, it is plain that the parts of the former will be larger than the corresponding parts of the latter. 2d. If one of two equal apples is divided into two equal parts, and the other mto four, the parts of the first will be twice as large as those of the second ; if one is divided into two equal parts, the other into six, one part of the first will be equal to three of the second, etc Hence, Note. — A Jialf is twice as large as ?i fourth, three times as large as a sixth, four tim.es as large as an eighth, etc. ; and generally, The greater the number of equal parts into which the unit is di- vided, the less wi!l be the value of each part. Conversely, The less the number of equal parts, the greater will be the value of each part. 134. What is a fraction ? What does the number of parts indicate ? 135, What is a half ? A third ? A fourth ? A tenth ? A hundredth ? 136. Upon what does the size of these parts depend ? 99 100 FBACTIONS. 137. Fractions are divided into two classes, common and decimal. A Common Fraction is one in which the unit is divided into any number of equal parts. 138. Common fractions are expressed by figures writ- ten above and below a line, called the numerator and denominator; as -|, f, ^. 139. The Denominator is written lelow the Hue, and shows into lioio many equal parts the unit is divided. It is so called because it names the parts ; as halves, thirds, fourths, tenths, etc. The Numerator is written above the line, and shows Uoiv many parts are expressed by the fraction. It is so called because it numlers the parts taken. Thus, in the fraction f, four is the denominator, and show^s that the unit is divided vnio four equal parts; tliree is the nu- merator, and shows that tliree of the parts are taken. 140. The Terms of a fraction are the numerator and denominator. Note. — Fractions primarily arise from dividin^^ a single unit into equal parts They are also used to express a part of a collection of units, and apart of a. fraction itself; as i half of 6 pears, i third of 3 fourths, etc. But that from which they arise, is always regarded as a iDhole, and is called the Unit or Base of the fraction. 141. Common fractions are usually divided into proper, improper, simple, compound, complex, and mixed numbers. 1. A Proper Fraction is one whose numerator is less than the denominator; as ^, f, |. 2. An Imjyroj^er Fraction is one w^hose numerator equals or exceeds the denominator; as f, f. 3. A Sitnple Fraction is one having but one 137. Into how many elapses are fractions divided? A common fraction? 138. How expressed ? 139. What does the denominator show ? Why so called? The numerator ? Why so called ? 140. What are the terms of a fraction ? FRACTIOI^S. 10] numerator and one denominator, eacli'of whdcli is d^'xvkole number, and may be proper or improper ; as f , f . 4. A Compound Fraction is a fraction of a frac- tion ; as J of f . 5. A Co^nple'x Fraction is one which has a frac- I 2i tional numerator, and an integral denominator; as, -, — . 4 5 Remark. — Those abnormal expressions, haring fractional denoyn- inators, commonly called complex fractions, do not strictly come under the definition of a fraction. For, a fraction is one or more of the equal parts into which a unit is divided. But it cannot properly be said, that a unit is divided into Sjtlis, 45dths, etc. ; that is, into 3^ equal parts, 4I equal parts, etc. They are expressions denoting divi- sion, having o, fractional divisor, and should be treated as such. 6. A llixed dumber is a whole number and a frac- tion expressed together ; as, 5 1, 34^^. Note. — The primary idea of a fraction is a part of a unit. Hence, a fraction of less value than a unit, is called a proper fraction. All other fractions are called improper, because, being equal to or greater than a unit, they cannot be said to be a pa7't of a unit. Express the following fractions by figures : 1. Two fifths. 6. Thirteen twenty-firsts. 2. Four sevenths. 7. Fifteen ninths. 3. Three eighths. 8. Twenty-three tenths. 4. Five twelfths. 9. Thirty-one forty-fifths. 5. Eleven fifteenths. 10. Sixty-nine hundredths. 11. 112 two hundred and fourths. 12. 256 five hundred and twenty-seconds. 13. Explain the fraction -J. Analysis. — \ denotes i of i ; that is, one such part as is obtained by dividing a unit into 4 equal parts. The denominator names the parts, and the numerator numbers the pirts taken. It is common because the unit is divided into any number of parts taken at ran- dom ; proper, because its value is less than i ; and simple, because it has but one numerator and one denominator, each of which is a whole number. 141 . Into what are common fractions divided ? A proper fraction ? Improper ? Simple ? Compound ? Complex ? A mixed number ? 15- *. 20. ¥• 1 6. ioff. 21. fl- 17- «• 22. If. 1 8. fofif. 23- If. 19. 4i 8* 24. 10 102 FRACTIOI^S. 14/ Explain the fraction f. Analysis. — } denotes f of i, or ^ of 3. For, if 3 equal lines are each divided into 4 equal parts, 3 of these parts will be equal to J of I line, or ^ of the 3 lines. It is common, etc. (Ex. 13.) Read and explain the following fractions : 25. 1 6 J. 26. 28fL 27. J off of |. 28. f of f of 2j. 29. — ^. Note. — The 21st and 22d should be read, " 68 twenty-Jlrsts,*' " 85 thhty-seconds," and not 68 twenty-ones, 85 thirty-twos. 142. Fractions, we haye seen, arise from division, the mcmerator being the dividend, and the denorninator the divisor. (Art. 64, ?2.) Hence, The value of a fraction is the quotient of the numerator divided by the denominator. Thus, the value of i fourth is I -i- 4, or J; of 6 halves is 6-^-2, or 3; of 3 thirds is 3-^3, or I. 143. To find a Fractional Part of a number. I. What is I half of 12 dimes? ANx\lysis. — If 12 dimes are divided into 2 equal parts, i of these parts will contain 6 dimes. Therefore, 2 of 12 dimes is 6 dimes. Note. — The solution of this and similar examples is an illustra- tion of the second object or office of Division. (Art. 63, &.) 4. What is J of 36 ? -I- of 45 ? | of 60 ? | of 6^ ? 5. What is I of 48 ? ^ of 63? -^ of 190? ^^ of 132? 6. What is J of 12 dimes? Analysis. — 2 thirds are 2 times as much as 3. But i third of 12 is 4. Therefore, § of 12 dimes are 2 times 4, or 8 dimes. Hence, the 142. From what do fractions arise ? Which part is the dividend ? Which the divisor ? What is the value of a fraction ? FRACTIOJ^^S. 103 Rule. — Divide the give^i number hy the denominator, and multiply the quotient by the numerator To find J, divide the number bj 2. To find J, divide the number by 3. To find J, divide by 3, and multiply by 2, etc. 7. What is J of 56 ? Ans. 42. 8. What is f of 30 apples ? Of 45 ? Of 60 ? 9. What is f of 42 ? 1^ of 56 ? f of 72 ? 10. What is I of 45 ? A of 50? ^^ of 66 ? ^'^ of 108? GENERAL PRINCIPLES OF FRACTIONS. 144. Since the numerator and denominator have the mme relation to each other as the dividend and divisor, it follows that the general principles established in division are applicable to fractions. (Art. 81.) That is, 1. If the numerator is equal to the denominator, the value of the fraction is i. 2. If the denominator is greater than the numerator, the value is less than i. 3. If the denominator is less than the numerator, the value is greater than i. 4. If the denominator is i, the value is the numerator. 5. Multiplying the numerator, multiplies the fraction. 6. Dividing the numerator, divides the fraction. 7. Multiplying the denominator, divides the fraction. 8. Dividing the denominator, multiplies the fraction. 9. Multiplying or dividing both the numerator a7id de- nominator by the same number does not alter the value of the fraction. 144. If the numerator is equal to the denominator, what is the value of the fraction? If the denominator is the jjreater, what? If less, what? If the de- nominator is r, what ? What is the effect of multiplying the numerator ? Divid- ing the numerator ? Multiplying the denominator ? Dividing the denominator ? 104 FEACTIOIfS. REDUCTION OF FRACTIONS. 145. deduction of fractions is changing their terms, Tvdthout altering the value of the fractions. (Art. 142.) CASE I. 146. To Reduce a Fraction to its Lowest Terms. Der — The Lowest Terms of a fraction are the smallest numbers in which its numerator and denominator can be expressed. (Art. 10 1, Def. 7.) Ex, I. Eednce fl to its lowest terms. Analysis. — Dividing both terms of a fraction ^st method. by the same number does not alter its value. .24 12 (Art. 144, Prin. 9.) Hence, we divide the given ^^^ 16 terms by 2, and the terms of the new fraction j 2 -> by 4 ; the result is |. Bat the terms of the frac- 4)77 — ~ A7IS. tion I are prime to each other ; therefore, they 16 4 Or, if we divide both terms by their greatest 2d method. common divisor, which is 8, we shall obtain the qx24 3 same result. (Art. 126.) Hence, the ^32 ~4^^^* Rule. — Divide tli^ numerator and denominator C07i- tinualltf hy any numher that will divide loth without a re- mainder, until no numher greater than i will divide them. Or, divide hoih terms of the fraction hy their greatest common divisor. (Art. 126.) Notes. — i. It follows, conversely, that a fraction is reduced to higher terms, by multiplying the numerator and denominator by a common multiplier. Thus, §=i^, both terms being multiplied by 8. 2. These rules depend upon the principle, that dividing both terms of a fraction by the same number does not alter its value. When the terms are large, the second method is preferable. 145. What is Reduction of Fractions? 146. What are the lowest terms of a fraction ? How reduce a fraction to its lowest terms t KEDUCTIOK or FEACTIOKS. 105 Reduce the following fractions to their lowest terms 2. if- 9- a- 16. ilf. 23- -fo%- 3-U- 10. m- 17. ih 24. m-i- 4. H- II. m- 18. in- 25- -m- 5- if- 12. Hi- 19- i¥oV 26. #^A- 6.U- 13- m- 20. m- 27. mf- 1-U- 14. t¥^- 21- t¥A. 28. ilU- s.ki- 15- «• CASE 22. i¥^V II. 29. AV 7. To re#uce an Im proper Fraction to a Whole or M'lQced Number, I. Reduce ^^- to a whole or mixed number. Analysis. — Since 4 fourths make a unit or i, 45 fourths will make as many units as 4 is contained times in 45, which is ii;^. Hence, the 4)45 iiiAns. Rule. — Divide the numerator hy tlie denominator. Notes. — i. This rule, in effect, divides both terms of the fraction by the same number ; for, removing the denominator cancels it, and cancelling the denominator divides it by itself. (Art. 121.) 2. If fractions occur in the answer, they should be reduced to the loijcest terms. Reduce the following to whole or mixed numbers : . iji. 8. ^F. 14. hS'^- 20. mi^- . -^. 9- w. 15- ¥t¥- 21. ift&r- ^. ^i^. 10. %^-. 16. 10000. 22. ^S:V/^. . W. II. nf. 17. ¥A\^- 23- 4fVW/. . i^. 12. w. 18. mi^. 24. ^t"-.^- . ^^^ 13- fM. 19. mi^- 25- 'fih'^- 26. Ii\ ^|4?^'*- of a pound, how many pounds ? 27. In ^^^^-i^ of a dollar, how many dollars ? 28. In ^^ j|g^ " of 9^ y^ar, bow many years? How reduce an imp.'"oper fraction to a wLole or mixed number f 106 KEDUCTIOiT OF FRACTIOKS. CASE III. 148. To reduce a Mixed Number to an I^nproper FracUoru I. Reduce 8f to an improper fraction. Analysis. — Since there are 7 sevenths in a unit, there 03 must be 7 times as many sevenths in a number as there ,. "^ are units, and 7 times 8 are 56 and 3 sevenths make ^7^. j~ Therefore, 8^=^7^. ^' ^^5- Or thus : Since i=f, 8 = 8 times ^ or ^7^, and f make ^7^. In the operation we multiply the integer by the given denominator, and to the product add the numerator. Hence, the EuLE. — Multiply the whole number ly the given denomi- nator ; to the product add the numerator, and place the mm over the denominator. Eemark. — A wJiole number may be reduced to an improper fraction by making i its denominator. Thus, 4=^. (Art. 81.) Reduce the following to improper fractions : 2- i5f 6. HSii- 10. i573f 14. 478^- 3. i8f. 7- l8^^. II. 2564. 15. 57AV 4- 35i 8. 295:^V- 12. 3640!. 16. 8|«. 5. SiU- 9- 806^?^. 13- 8624A. 17. 9^'A. 18. In 263y\ pounds how many sixteenths ? 19. Change 641^^ mile to fortieths of a mile. CASE IV. 149. To reduce a Compound Fraction to a Simple one, I. Reduce J of f to a simple fraction. Analysis. — .3 fourths of | = 3 times i fourth of f. ist method. Now I fourth of | is -,% ; for multiplying the de- 3 2 6 nominator divides the fraction; and 3 fourths of 4 7~~72 1 = 3 times -,% or ,^^ ; for multiplying the numerator multiplies the fraction. (Art. 144.) Reduced to its J^« — -zn- lowest terms 1^2=^. '12 2 148. How reduce a mixed number to an improper fraction? Eem. A whole number to an improper fraction ? EEDUCTIOK OF FRACTIOKS. 10? Or, since numerators are dividends and 3d method. A /nominators divisors, the factors 2 and 3, i, S 2, 1 which are common to both, may be cancelled. 2 4 ^ ^^^ Ans. (Art. 144, Prin. 9.) The result is |. Hence, the ' Rule. — Cancel the common factors, and 2:)lace the product of the factors remaining in the numerators over the product of those remaining in the denominators. Notes. — i. Whole and mixed numbers must be reduced to irriproper fractions, before multiplying the terms, or cancelling the factors. 2. Cancelling the common factors reduces the result to the lowest terms, and therefore should be employed, whenever practicable. 3. The numerators being dividends, may be placed on the right of a perpendicular line, and the denominators on the left, if pre- ferred. (Art. 122, Rem.) 4. The reason of the rule is this: multiplying the numerator of one fraction by the numerator of another, multiplies the value of the former fraction by as many units as are contained in the numerator of the latter ; consequently the result is as many times too large as there are units in the denominator of the latter. (Art. 144, Prin. 5.) This error is corrected by multiplying the two denominators to- gether, (Art. 144, Prin, 7.) 2. Reduce f of J of ^V of 2 J to a simple fraction. Ans. j|. Reduce the following to simple fractions : 3. 1 of ^\. 9. 1 of 2V of 60. 15. ^3 of f I of iJ. 4. f of f 10. ^ of li of if. 16. J of f of I of f 5. f of M-. 1 1 . H of li of f J. 17. ii of IJ of 65 J. 5. f of if- 12. If of -I of If. 18. fi of If of 84i 7. f of if. 13. M of /y of 45. 19. I of ^ of ig. 8. f of tV- 14. f of I of f J. 20. f of i of 163 J. 21. Reduce ^f of f^ of f-J of 9! to a simple frac- tion. 22. To what is I of fj of 3^ bushels equal? 2T,. To what is f of 4 J of ^ of a yard equal ? -49, How reduce a compound fraction to a simple one ? Note. What must bo TTT? TS^ DS' J, I 5 3 T^32I46 4- ^? 2(1' T2^- 12. y, ^, ^, y, y. 5'hhil 13. 9h Hi f of 40. 6. f, f, 4i 14. T^ of 13, j\^, /2V 7. A. A. f of i2i 15. 7|, A of 1 7, il 8. A. h f of 10. 16. ,\% m uU' n 5 14 ^I C3 T^ IIS 2ftQ T447 Tr. 9nf82 7r.f^r, tS 15 2 6 5 17 2 8 10. Y?7 01 o?, -g^ or 40. 15. -yj^, :fjjQ, T^T5"« 114 ADDITION OF FRACTIOi^S. ADDITION OF FRACTIONS. 155. Addition of Fractions embraces two classes of examples, viz. : those which have a common denominator and those which have different denominators. 156. When tivo or more fractions have a common de- nominator , and refer to the same kind of unit or base, their numerators are like parts of that unit or base, and therefore are like numbers. Hence, they may be added, suitracted, and divided in the same manner as whole num- bers. Thus, I and ^ are 12 eighths, just as 5 yards and 7 yards are 12 yards. (Art. lor, Def. 13.) 157. When two or more fractions have different denom- inators, their numerators are unlike parts, and therefore cannot be added to or subtracted from each other directly, any more than yards and dollars. (Art. 10 1, Def. 14.) 158. To add Fractions which have a Common Denom.- inator. 1. What is the sum of f yard, | yard, and | yard ? Analysis.— 3 eigliths and 5 eighths operation. yard are 8 eighths, and 7 are 15 eighths, f + 1 + -g-— ¥> ^r i| y. equal to i^ yard. (Art. 147.) For, since the given fractions have a common denominator, their numerators are like numbers. Hence, the EuLE. — Add the numerators, and jflace the sum over the common denorninator. Note. — The answers should be reduced to the lowest terms; and If improper fractions, to whole or mixed numbers. Add the following fractions . 2. -f-, f, and f 5. ^^, ^, T^, and «*. 3- A, A, A, and «. 6. ^^i^, ^6^, «$, and.||f. 4. ^, A, ^, and H- 7. Ml. Hh lif. and i|l 158. How add fractioUB that have a com^ion denominator? ADDITIOi^ OF FU ACTIONS. 115 159. To add Fractions which have Different Denomina- tors, 8. Find the sum of J of a pound, f of a pound, and | of a pound. Analysis.— As these fractions operation. have different denominators, 4 x 5 X 8 = 160, com. d, their numerators cannot be added 1x5x8= 40, ist nu. in their present form, (Art. 157.) 2x4x8= 64, 2d " We therefore reduce them to a 5x4x5 = 100, 3d " common denominator, which is g^^ni of nu., 204. Hence, 160. Then \=-,^^, |=-,Vo, and ^q 5^ ioo_204 § = -\M. Adding the numerators, — 7" + "7~ + "~7~ — "^7 ' , or I ^^ " ''^' . ^,^ ^, 160 160 160 160 and placing the sum over the common denominator, we have }^^, or i^^, the answer required. For, reducing fractions to a common denominator does not alter their value ; and when reduced to a common denominator, the nu merators are like numbers. (Arts. 153, 156.) Or, we may reduce them to the least common denominator, which is 40, and then add the numerators. (Art. 154.) Hence, the KuLE. — Reduce the fractions to a common denominator, and place the sum of the numerators over it. Or, reduce the fractions to the least common denomina- tor, and over this place the sum of the numerators. Remarks. — i. The fractional and integral parts of mixed num- bers should be added separately, and the results be united. Or, wJiole and mixed numbers may be reduced to improper frac- tions, then be added by the rule. (Art. 148.) 2. Compound and complex fractions must be reduced to simple ones ; then proceed according to the rule. (Ex. 20, 28.) (For Addition of Denominate Fractions, see Art. 315.) 9. What is the sum of 5, 2 J, and io| ? Analysis, — Reducing these fractions to a common 5 =5 denominator 12, we have, 5 = 5 ; 2i=::2-,4,; and io|= 24 = 2-1^7 iOi^2, Adding the numerators, 4. twelfths and 9 twelfths are if— i-i\. i and 10 are 11 and 2 are 13 and 5 are 18. Ans. iSff. Or, 5=f or ^; 2^=:M; and io|= J-i^a Now ft + f I + W= W, or i8f,- Ans. 159, How when they have not ? Note. How add whole and mixed numbers ? Compound and complex fractions ? T2 116 .6 ADDITION OF feactio:ms • Add the following : (10.) (11.) (12.) (I3-) (14.? 2l 3A 4l Si "1 27A H 8A U 461 li (IS-) I9i 47i 68f (i6.) 68A (17.) (18.) 2o7i i7Sl 62|- 207 49A 368J ('9-) 45° 67^ 3715 20. What is the sum of J of J, f of 7, and ^ of 9I ? 21. What is the sum of f of f, f of |, and f of 7 ? 22. What is the sum of f of 4 J, f of 3, and f of 10 J ? 23. A beggar received $if from one person, $2^ from another, and $3 J from another : how much did he receive from all ? 24. If a man lays up $43! a month, and his son I27I, how much will both save ? 25. What is the sum of f , |, -J J, and f pound ? 26. A shopkeeper sold i j^ yards of muslin to one cus- tomer, 8 J yards to another, 2 5 -J to another: how many \ yards did he sell to all ? \ „ 27. A farmer paid |i8J for hay, $45t\ for a cow, $150! for a horse, and $275 for a buggy: how much did he give for all? 28. What is the sum of — ^, — , and t 5 3 4 Analysis. — Rt lucing the complex fractions to simple ones, we have, -4i=|-5=:,% ; ^i = \^^3=i ; and ^=l-^4=-h, or i Now, 53 4 fp l-l + i-t^ + \¥ + ^^ = 2a^l^«- 48A6 »72I «2 140 A3 29. Add '^, ^, and ^-^A 30. Add ^, =f^, and ^. ^23 7 573 SUBTRACTION OF FRACTIONS. 117 SUBTRACTION OF FRACTIONS. 160. Subtraction of Fractions embraces tut classes of examples, viz.: those which have a common denominatory and those which have different denominators. 161. To subtract Fractions which have a Common Denominator, I. What is the difference between |f and \^ ? Analysis. — 13 sixteenths from 15 sixteentlis - . 'Ti .1 • J T-. OPERATION. leave 2 sixteenths, the answer required. For, T C T •J 2 since the given fractions have a common denomi- _:?. •£ ^^ .^ nator, their numerators, we have seen, are like 16 16 16 numbers. (Art. 156.) Hence, the EuLE. — Take the less numerator from the greater', a7id place the difference over the common denominator. 2 From if take \%. 5. From J-Ji take -Jf J 3. From fl take f f. 6. From |4| take %%%. 4. From \l\ take ^V 7- From ^^/o take ^«oV 162. To subtract Fractions which have Different Denominators, 8. From f of a pound, subtract j of a pound. 6 _ 24 , Analysis. — Since these fractions have different 7 28 denominators, their numerators cannot be subtracted 3 2 1 in their present form. We therefore reduce them to a a ~ 28 common denominator, which is 28 ; then subtracting as above, have 2^8. the answer required. (Art. 153.) Ans. — Hence, the 28 Rule. — Reduce the fractions to a common denominator, and over it place the difference of the numerators. 161. How subtract fractions that have a common denominator? 162. How when they have different denominators ? Bern. How subtract mixed numbers \ Compound and complex fractions ? A proper fraction from a whole number? 118 SUBTRACTIOi^ OF FRACTIOIn^S. Remarks. — i. HhQ fractional and integral parts of mixed numbers should be subtracted separatelj, and the results be united. Or, they may be reduced to improper fractions, then apply the rule. Compound and complex fractions should be reduced to simple ones, and all fractions to their lowest terms. 2. A. proper fraction maybe subtracted from a wliole number by taking it from a unit; then annex the remainder to the whole number minus 1. Or, the whole numher may be reduced to a fraction of the same denominator as that of the given fraction ; then subtract according to the rule. (Art. 151.) 3. The operation may often be shortened by finding the least com mon denominator of the given fractions. (Art. 154.) (For subtraction of Denominate Fractions, see Art. 317.) 9. What is the difference between 1 2} pounds and 5^ pounds ? AiTALYSTS. — The minuend I2^=i2f. Now i2|— 5|=6|, Ans. Or, 12^-^5;^ or^ii; and 5! = ^^. Now^4^— \^=^/, which reduced \o a mixed number, equals 6|, the same as before. (10.) (II.) (12.) (13.) (14.) ^ . From I . 5 1 7i- 23I . ^ U-^ Take t "^ z\ 5 1 ^ ^Sf li (15.) (16.) (17.) (18.) (19.) From 5*1 7ii ^\% 8^ 9t| Take 3if 4H 3lf SM 7t'A « 20. From a box containing 56^ pounds of sugar, a grocer took out 23 J pounds : how many pounds were left in the box? '" , 21. From a farm containing 165^^^ acres, the owner sold * 78 J acres: how much land had he left? 22. From 13 subtract f. Analysis. — ist. Reducing 13 to sevenths, we have i3=-f, and 3,^—5 = 87^, or I2f Ans. Or, 2d. Borrowing i from 13, and reducing it to 7th s, we have 13 = 12?^, and 12^— ^=i2f, An». SUBTRACTION OF FRACTIONS. 119 23. From 46 take 7 J. 24. From 58 take 2o|. 25. From 84f take 41. "/ 2S. From 150I take 8^. l^ 27. From no take 7-AV 28. From 1000 take 999f J 29. Subtract J of J of 3 from ^ of f of i^. Analysis.— Reducing i 4 j_i 4 i.3_3_£^ the compound fractions ^ ^ t ^ ^^~2.^ K 2.~ K ~20 to simple ones, then i^Ojj IlSi"? a common denominator, - of - of 3 = - of - of - = - == ^ — - the subtrahend becomes 3 4 ^4^4 ^^ is, the minuend \l. . 7^ Nowi§-A = /o. Ans. ^-^^*- 20* (30.) (31') (32') (33*) From foff |ofii Aof4i ift of 28 Take iof| f of A f of 3 | of 4! 34. From ^ of 62^ subtract ^ of 16 J. 35. A's farm contains 256! acres, B's 43ixV acres: what is the difference in the size of their farms ? 36. If from a yessel containing 230^^ tons of coal, 119^ tons are taken, how much will there be left ? 37. A man bought a cask of syrup containing 58^ gallons; on reaching home he found it contained only 17I gallons : how many gallons had leaked out ? 38. A lady having $100, paid $8 J for a pocket hand- kerchief, $15^ for a dress hat, $46! for a cloak: how much had she left ? 39. From ^^ subtract — . '^^ 6 5 Analysis.— Reducing the complex 5I. j,.^ jj 5- fractions to simple ones, then to a ~^^^^"~^^'~T2—'5% common denominator, the minuend j becomes fi the subtrahend VS. (Arts. -^=^^^=^-^=z I-6 152, 153.) Now, f^-U = H, or 5 \l,Ans. Ans.^—2^ 40. From --^ take t- 4i. From ^ take ^i ^ t: 8 ^22 120 MULTIPLICATION OF FRACTIOXS. MULTIPLICATION OF FRACTIONS. CASE I. 163. To multiply a Fraction by a Whole Kuniher, Ex, I. What will 4 pounds of tea cost, at | of a dollar 8 pound ? 1st Method. — Since i pound costs $|, 4 pounds ist operation. wiU cost 4 times as much, and 4 times $| are $|-X4=$-^ \tt— I2L, whicli is the answer required. For, and $^^=$2^ multiplying the numerator, multiplies the frac- tion. (Art. 144, Prin. 5.) 2d Method. — If we divide the given de- 2d operation. nominator by the given number of x>ounds, $|-r-4 = $|, or $2^ we have 1-^-4= $f, or $22, which is the true answer. For, dividing the denominator, multiplies the fraction. (Art. 144, Prin. 8.) Hence, the EuLE. — Multiply the numerator hy the whole nurriber. Or, divide the denominator hy it. Remabks. — I. When the multiplicand is a mixed number, the fractional and integral parts should be multiplied separately, and the results be united. Or, the mixed number may be reduced to an improper fraction, and then be multiplied as above. 2. K a fraction is multiplied by its denominator, the product will be equal to its numerator. For, the numerator is both multiplied and divided by the same number. (Art. 86, a.) Hence, 3. A fraction is multiplied by a number equal to its denominator by cancelling the denominator. Thus, f x 8 = 5. (Art. 121.) 4. In like manner, a fraction is multiplied by any factor of its de- nominator by cancelling that factor. 2. Multiply 2 7f by 6. Analysis. — Multiplying the fraction and integer operation. separately, we have 6 times f=-S^, or 3I ; and 6 27^ times 27=162. Now 162 + 31=1651, the answer. ^ Or,thus:27|=H^;andH^x6=«F^ori65l, ^ns. Ans. 165I 163, How multiply a fraction by a whole number ? Upon what does the first method depend? The second? Rem. How multiply a mixed number by a whole one ? How multiply a fraction by a number equal to its denominator f How by any factor in its denominator ? Ml [JLTIP LICATIi ON OF ERACTIONB. (3-) (4.) (5.) (6.) (7.) Mult. If a 23J 35i 48i By _7 _9 8 10 12 (8.) (9.) (10.) (II.) (12.) Mult. n U Ui 98A ssifV By 48 100 78 26 48 12i 13. Multiply lif by 239. 14. What cost 8 barrels of cider, at 1;^ a barrel? 15. At $25!: each, what will 9 chests of tea cost? 16. What will 25 cows come to, at I48J apiece? 17. What cost 27 tons of hay, at $29^ a ton? 18. What cost 35 acres of land, at $45 J per acre ? 19. At $34 J apiece, what cost 50 hogsheads of sugar. 20. At $45f apiece, what will 100 coats come to ? 21. What cost 6 dozen muffs, at $7 5 J apiece ? CASE II. 165. To Multiply a JVhole Kimiher by a Fraction. Def. Multiplying by a Fraction is taking a sertain part of the multiplicand as many times as there are UTce parts of a unit in the multiplier. But to find a given part of a number, we divide it into as many equal parts as there are units in the denominator, and then take as many of these parts as are indicated by the numerator. (Art. 143.) That is, To multiply a number by ^, divide it by 2. To multiply a number by ^, divide it by 3. To multiply a number by J, divide it by 4 for J, and multiply this quotient by 3 for f, etc. Hence, Remarks. — i. Multiplying a whole number by a fraction ig the same as finding a fractional part of a number, which the pupil should here review with care. (Art. 143.) I)3f. What is it to multiply by a fraction ? How multiply by i ? By i ? By 1 1 122 MULTIPLICATION?^ OF FEACTIONS. 2. When the multiplier is i, the product is equal to the multipli- cand. When the multiplier is greater than i, the product ia greater than tlie multiplicand. When the multiplier is less than i, the product is less than the multiplicand. 1. What will J of a ton of iron cost, at $55 a ton ? ist Method. — Since i ton costs $55, | of a ton opekatiok. will cost f times $55, or f of $55. But f of $55 H)S5 = 3 times i of $55. Now \ of $55==$55-^4, or $133 $13! ; and 3 times $i3l=$4i4, the answer required. 3 For, by definition, multiplying by a fraction is a g ^71^ taking a part of the multiplicand as many times as there are like parts of a unit in the multiplier. Now dividing the whole number by 4, takes i fourth of the multiplicand once ; and multiplying this result by 3, repeats this part 3 times, as indicated by the multiplier. (Art. 143.) 2d Method.— I of $55=^ of 3 times I55. But $55 $55 X 3 = $165, and §i65-T-4— $414, the same as 2 before. For, i of 3 times a number is the same as /i')76c^ 3 times 4- of it. Hence, the . - . ., - ^ * Ans. $4ii Rule. — Divide the whole numher hy the denominator of the fraction, and multiply hy the numerator. Or, Multiply the whole numher hy the numerator of the fraction, and divide hy the denomi?iator. Remarks. — i. The fraction may be taken for the multiplicand, and the whole number for the multiplier, at pleasure, without affect- ing the result. (Art. 45.) 2. When the multiplier is a mixed number, multiply by the frac- tional and integral parts separately, and unite the results. Or, reduce it to an improper frax^tion ; then proceed according to the rule. (Ex. 2.) 2. What will 8f tons of copper ore cost, at $365 a ton ? Analystb.— 8| =^^ ; now ^/ tons will come to ^i- times $365 ; and $365 X Y=$3i39- Ans. 165. How is a whole number multiplied by a fraction? Eem. WbaX, is the operation lilce ? When the multiplier is i, what ia the product ? When the mul- tiplier is greater than i, what ? When lesa, what? Re.n IIow multiply a whole by a mixed number? MULTIPLICATIOIf OF FRACTIONS. 123 (3-) (4.) (5-) {6., (7.) Mnlt. 65 89 96 87 100 By i ^ 3i: 4j 5f 8. What cost ^ of an acre of land, at $38 an acre ? 9. At $29 a ton, what cost 8| tons of hay ? 10. "What cost 2 8f tons of lead, at $223 per ton ? 11. What is fjof 845? 15. Multiply 79 by yf. 12. W^hat isy'^ of 1876 ? 16. Multiply 103 by gf. 13. What is If of 1000? 17. Multiply 1 00 1 by2iyV 14. What is J-J§ of 2010 ? 18. Multiply 1864 by 37-^. CASE III. 166. To multiply a Fraction by a Fraction. I. What will f of a pound of tea cost, at $^ a pound ? Analysis. — \ of a pound will cost i sixth operation. as much as i pound; and ^ of $,'0- is $o%- "HT >< |— ts"? ^^^ %i Again, ^ of a pound will cost 5 times as by cancellation, much as ^; and 5 times %ia are $6B=$4, 3> ^ ^ -' "^^^ ^3 the answer required. 2, IBt 5, 2 Or, having indicated the multiplication, canrd the common fac- tors 3 and 5, and then multiply the numerators together and the de- nominators ; the result is $i^, the same as before The reason is this: Multiplying the denominator of the multipli- cand 1^0 by 6 the denominator of the multiplier, \S\q result, $,^„, is 5 times too small ; for, we have multiplied by ^ instead of ^ of a unit, the true multiplier. To correct this, we must multiply this re- sult by 5, which is done by multiplying its numerator by 5. (Art. 144.) The object of cancelling common factors is twofold: it shortens the operation, and gives the answer in the lowest terms. Hence, the KuLE. — Cancel the common factors ; then multiply the numerators together for the neio numerator, and the dejiom- inators for the new denominator. (Art. 144, Prin- 9.) 166. How multiply a fraction hy a fraction ? Object of cancelling ? E^m. ITow multiply compound fractloau ? Mixed numljers ! Coinplex fractions ? 124 MULTIPLICATIOK OF FBaCTIOI^^S. Remarks. — i. Mixed numbers should be reduced to improper fractions, and complex fractions to simple ones ; then apply the rule. 2. Multiplying by a fraction, reducing a compound fraction to a simple one, and finding a fractional part of a number, are identical operations. (Arts. 143, 149, 166.) 2. Multiply f of J of f by i of 4. SoLtmON. — It is immaterial as to the result, whether 3i the fractions are arranged in a horizontal or in a per- i^ pendicular line, with the numerators on the right and 6 the denominators on the left. (Art. 122, Mem.) 2. 5=^ Perform the following multiplications : 3- 7- ifx 4. Ax A 8. iix 5- nx^ 9- ^x 6. «x f 10. fix TT II- tX ^X y fy 12. |Xt2_x f TT I3* TXT X -g^XT X -4:^ TT 14- T^XyyX^-j 15. Multiply f of f of 25 by I of -§^ of j. 16. Multiply I of tV of 33i by | of if of i8f. 17. If I quart of cherries costs J of J of 45 cents, what will f of T^ of a quart cost ? 1 8. What will 5^ barrels of chestnuts cost, at 6 J dollars ft barrel? Ans. $$6. (19.) (20.) (21.) (22.) Mult. 24I 37-1 62J 165-J By _8|- _^ jH _92| 23. Wbat is product of ~ multiplied by — . 5^ 21 . 21 2} 8 8 Analysis. — -^= — =-6 = — , and -^ — — ~3 = ~. 64 24 3 3 9 7,21 8. 7 7 Cancelling common factors, we have i^r; X - =: =^-,Ans, 3.^^ ^.3 3x3 9 24. Multiply ^ by ^t 25. Multiply ^5 by ^. MULTIPLICATIOiq^ OF FRACTIONS. 125 167. The preceding principles may be reduced to the following GENERAL RULE. I. Reduce tvhole and mixed numlers to improper frac- tions, and coynplex fractions to simple 07ies. II. Cancel the common factors, and place the product of the numerators over the product of the denominators. EXAMPLES. 1. What cost I: of a yard of calico, at $J a yard ? 2. What cost f of a pound of nutmegs, at $f a pound ? 3. At $f| a gallon, what will J of a gallon of mokssca cost? 4. Multiply 23 J by 6}. 7. Multiply pi-f by 43^. 5. Multiply 42-^ by 8f . ' 8. Multiply i64f by 75xV 6. Multiply 65I by <)\. 9. Multiply 2oof by 86^^ 10. What is the product of f of | of f of f into f of ^^ ofA? 11. What is the product of f of 3I into \ of 19J ? 12. When freight is $i| per hundred, what will it cog* to transport 27 hundredweight of goods from New Yorl' to Chicago ? 313. What will 16 J quarts of strawberries come to, at 26J cents per quart ? 14. At $3f a barrel, what will be the cost of 47! barrels of cider ? 15. Bells are composed of J tin and f copper: how many pounds 0. each metal docs a church-bell contain which weighs 3750 pounds ? jL^ 16. What will 45 men earn in 15} days, if each earns ''$2j per day? 17. How many feet of boards will a fence 603^ rods long require, allowing 74 J feet of boards to a rod ? 167. What is the general rule for multlpljdng fractions ? 12G DIVISION or FRACTIOKS. DIVISION OF FRACTIONS. CASE I. 168. To divide a Fraction by a Wliole Number, 1. If 4 yards of flannel cost If, what will i yard cost? ist Method. — i yard will cost i fourth as much as 4 yards ; and dividing the numerator ^®* method. by 4, we have $|h-4=^$|, the answer required. ^~A- __ ^2 ^^5; For, dividing the numerator by any number, 9 divides the traction by that number. 2d Method. — Multiplying the denominator 2d method. by the divisor 4 (the number of yards), the 8 8 result is .^^ = |, the same as before. For, 9x4 36 multiplying the denominator by any number, g divides the fraction by that number. (Art. 144, Ans. — = $^ Prin. 7.) Hence, the ^ EuLE. — Divide the numerator hy the whole number. Or, multiply the denominator hy it. Remaek. — When the dividend is a mixed number, it should be reduced to an improper fraction ; then apply the rule. 2. Divide %\ by 9. Ans. $f -^9=Vt- Perform the following divisions : /, 1 6 • * *T I50_i_.or' TT 4060 fsr\ 4. -u C 49 5- 1^ r, 5 268 • A>j x'7 52326_j!_Tr» «T ^ 350_:_^p, T-? 60045 • t C 6. |f--33- . 10. flf-8i. 14. fH-m-91. 15. A man having J of a barrel of flour, divided it equally among 5 persons : what part of a barrel did each receive ? 16. If 12 oranges cost IxVtt' what will i orange cost? 17. If 7 writing-books cost $f, what will that be apiece ? 18. If 6 barrels of flour cost I45I, what will i barrel cost? 168. How divide a fraction by a whole number? Upon what does the flrpt method depend? The second? Which is preferable? Eem. How is a mised number divided by a whole one ? DIVISIO:?^ OF FRACTIONS. 127 Perform the following divisions : 19. 8ox^3:~l2. 23. 865^-82. 27. 1000^^^50. 20. 762V^i4. 24. 490/0-^40. 28. 46841^68. 21. 28f-^20. 25. 758if-48. 29. 78961-^-25. 22. 511-^27. 26. 97511-^63. 30. 9684^^84. 31. If 5 yards of cloth cost $42|, what will i yard cost. 32. If 6 horses cost $7561, what will i horse cost? ^^. If 45 lbs. of wool cost $5 4 J, what will i lb. cost ? CASE II. 168. «• To divide a Whole Kinnher by a Fraction. I. At S| a yard, how many yards of cashmere can you buy for 820 ? Analysis.— At %\ & yard, $20 will operatiok. buy as many yards as there are 20x4=180 y. at %\. fourths in $20, and 4 times 20 are 80. 8o^3=:26f y. at %%. But the price $J, is 3 times as much Or, 20 X |=26|- y. at $J. as $^; therefore, at $f, you can buy only ^ as many yards as at %\ ; and ^ of 80 yards is 26| yards, the answer required. For, multiplying the whole number by the denominator 4, reduces it to fourths, which is the same denomi- nator as the given divisor. But when fractions have a common denominator, their numerators are like niimhers ; therefore, ona may be divided hy the other, as whole numbers. (Art. 156.) But multiplying the whole number 20, by the denominator 4, and dividing the product by the numerator 3, is the same as inverting the fractional divisor, and then multiplying the dividend by it. Hence, the Rule. — Multiply the luliole numher hy the fraction in- verted. (Art. 165.) Remarks. — i. When the divisor is a mixed number it should be reduced to an improper fraction ; then divide by the rule. (Ex. 16.) 2 A fraction is inverted, when its terms are made to exchange places. Thus, f inverted becomes f . 3 After the denominator is inverted, the common factors shouIJ be cancelled. (Art. 149, n.) 168, a. How divide a whole number by a fraction ? How does it appear that t'cvis process will give the true answer ? 128 DIYISIOJf OF fractio:n"S. Perform the following divisions : 2. 95-1- 5- i75^A- 8. 576^^. 3. i68-^tV 6. 26I-^^7. 9. 1236-^^. 4- 245 -^iiF- 7- 34S-^f 10. 624o-^AV II. How many yards of muslin, at $| a yard, can be bought for $19 ? Analysis. — At $^ a yard, $19 will buy as many yards as there Bxe thirds in 19, and 19 x ^=57. Therefore $19 will buy 57 yards. 12. 27=how many times ^ ? 14. 38=rhow many times -f? 13. 5 3= how many times ^ ? 15. 67 = how many times ^ ? 16. How many cloaks will 45 yards of cloth make, each containing 4^ yards ? AiSTALYSis. — Keducing the divisor to an improper fraction, we have 41^=^, and 45 yds. x $=^9^, or 10 cloaks., Ans. 17. Divide 88 by lof. 19. Divide 785 by 62}. 18. Divide 100 by 12 J. 20. Divide 1000 by Sj-}. CASE III. 169. To divide a Fraction by a Fraction, when they have a Common I>enorninator. Eemark. — This case embraces two classes of examples: First, those in which the fractions have a common denominator. Second, those in which they have different denominators. 21. At$f apiece, how many melons can be bought for $J? Analysis. — Since $| will buy i melon, %l will buy operation. fts many as $1 are contained times in $| ; and 3 l~^f=^^i eighths are in 7 eighths, 2 times and i over, or 2^ times, Ans, 2J m. (Art. 156.) Hence, the EuLE. — Divide the numerator of the dividend })y that of the divisor. Note. — When two fractions have a common denominator, their numerators are like numbers, and the quotient is the same as if they were whole numbers. (Arts, 64, 156.) 22. Divide f| by ^. 24. Divide f| by J J. 23. Di vide fl^yji' 25. Divide j| by jf. 169. IIow divide a fraction by a fraction when they have a common denominator? DIVISION OF FEACTIOJSTS. 12S 170. To divide a Fraction by a Fraction, when they have Different Denominators, 1. At 8J a pound, how much tea can be bought for $f ? 1ST Analysis. — $f will buy as many pounds -|=;|-=^9^ as $1 are contained times in $i $t = $t. Re- f = A ducing f and | to a common denominator, they 9 _^ g -— q .^l. g become -i%- and 1%, and their numerators like mim- "^ ' J^ ^ ' bers. Hence, A-r-,%=9-r- 8, or li ^m. li pound. 9-^^— ^^^"^ Remarks. — i. In reducing two fractions to a common denominO' tor, we multiply the numerator of each into the denominator of the. other, and the two denominators together. But in dividing, no use is made of the common denominator ; hence, in practice, multiply- ing the denominators together may be omitted. (Art. 153.) 2D Analysis.— $f=$!l. At $| a pound, $1 will buy as many pounds as $| are contained times in $1. Now i-i-^=3.-r-f, and ^-r-|=J, or f pounds, the quotient being the divisor inverted. Again, if $1 will buy | pounds, $| will buy f of f pounds ; and | of i=f, or 1 1 pounds, the same as before. Remarks. — 2. In this analysis, it will be seen, by inspection, that the numerator oi each fraction is also multiplied into the denomi- nator of the other. Both of these solutions, therefore, bring the same combinations of terms, and the same result, as inverting the di- visor, and multiplying the dividend by it. (Art. 166.) Hence, the EuLE. — Reduce tlie fractions to a com. denominator, and divide the numerator of the divideyid hy that of the divisor. Or, multiply the dividend by the divisor inverted. Notes. — i. The first method is based upon the principle that lumerators of fractions having a com. denom. are like numbers. The second is evident from the fact that it brings the same com- binations as reducing the fractions to a common denominator. The divisor is inverted for convenience in multiplying, 2. After the divisor is inverted, the common factors should be cnnceiled, before the multiplication is performed. 3. Mixed numbers should be reduced to improper fractions, com- pound and complex fractions to simj)le ones. 4. Those expressions which have fractional denominators, are tq- redncedtosimple fractions hy the above rule. Ex.14, (^^t 141, Rem.) 170. How when they have not? Note. Upon what principle is the first method based ? The eecond ? Show this coincidence. What is done with mixed nana bers, compound, and complex fractions. lO. TTn;-^ A'tt. 11. m~ ■ffS- 12. f-ii- ■tWj- 13- t¥A -T¥/y 130 DIVISION OF FRACTIOKS- Perform the following divisions : .,4_i_3 »tIS_-_9 3- TT • TT- /• -3T • ^y 4- A- A- 8. i-i^U- 14. Eeduce -^J to a simple fraction. Solution. — The gWen expression is equivalent to 27-^ -f- to \, and is reduced to a simple fraction by performing the division indicated. A71S. I or 2|. (Art. 141, Rem.) A I T C ^ 15. Eeduce-^ 17. Reduce-^ T T - T '7o ^ 16. Rednce — f- 18. Reduce -^^ 19. Divide 777-1^ dollars by 129!^ dollars. 20. How many rods in 23 20 J feet, at 16^ feet to a rod ? 21. How many times 30J sq. yards in 320yi3- sq. yards? 22. What is the quotient of f of J of 4^ divided by f of |? 2 23 2 5 o Solution. — - of - of 4A divided by - of „ = ^ 3 4 7^4 - X X 1- X ^ X - = -^ — 124. Ans. S 4 2 2 5 5 ^ ^17 ^^ When the perpendicular form is adopted, the 5 dicisor must b3 inverted, before its terms are arranged. 5 1^3 — ^ 2-| 23. Di^ddefoff of|by|ofi-. 24. Divide ^ of J of f by i of J. 25. Divide f of f of 4I by -| of J of 4^. .1 g 26. What is the quotient of | divided by -^ ? 3 3? Analysts.— The dividend | rrr i x 4 ; the divisor -_-=:- X -^. f 3i I 16 I S I lis I Invertingthe divisor, etc., ^^ ^ x ^ X - X y =— ^^^^ ^^^^ ^ ^^ 27. What is quotient of -^ divided by -^? DITISION" OF FRACTION'S, 131 171. The preceding principles may be reduced to the following GENERAL RULE. Reduce whole arid mixed numbers to improper fractions, compound and complex fractions to simple ones, and mul- tiply the dividend hy the divisor inverted. Note. — After the divisor is inverted, the common factors should be cancelled. EXAMPLES. 1. If a young man spends $2| a month for tobacco, in what time will he spend 813 J ? (Art. 48, Note 3.) 2. If a family use 5-J: pounds of butter a week, how long will 45 ^- pounds last them ? 3. If $1 will buy I yard of gingham, how much will 4. How many tons of coal, at $7} a ton, can be bought for $1251? 5. How many times will a keg containing 13I gallons of molasses fill a measure that holds |^ of a gallon ? 6. What is the quotient of 2^^^ divided by 8f ? 7. What is the quotient of f| divided by If? 8. What is the quotient of \l% divided by J| ? 9. What is the quotient of 45 J divided by 25 J ? 10. How much tea, at $i| a pound, can be bought for$75l? 11. How many acres can be sowed with 57! bushels of oats, allowing if bushel to an acre ? 12. A man having 57I acres of land, wished to fence it into lots of 5f acres : how^ many lots could he make ? 13. How many yards of cloth, at $6f, can you buy for $268f ? 14. What is the quotient of | of f of Jf H-f of | of 2 J ? 15. What is the quotient of f of J of 8f -^4f | ? [71. What is the general rnle for dividing: fractious f 132 DIVISION OF FKACTIOI^S. QUESTIONS FOR REVIEW. 1. A book-keeper adding a column of figures, made the T3sulfc $563!; proving his work, he found the true amount to be I607J: how much was the error? 2. A merchant paid two bills, one I278I, the other $34of, calling the amount I638J : what should he have paid ? What the error ? 3. What is the sum of 35I and 23I minus 8|? 4. A speculator bought two lots of land, one containing 47x77 acres, the other 6:^f acres: after selling 78^ acres, how many had he left ? 5. What is the sum of ^-J plus \ ? 6. What is the sum -- plus - plus — ,- ? 4 ' 12 ^ 5j 7. A man owning f|^ of a ship worth I48064, sold J of his share. What part of the ship did he sell ; what part does he still own, and what is it worth ? 8. A farmer owning 75! acres, sold 31 J acres, and after- ward bought 42 J acres : how many acres did he then have ? 3 ^5 9. What is the difference between | and ~ ? 6 k1 10. What is the difference between -^ and ~ ? si 6 11. What is the difference between --— and — J? H 12^ 12. If it requires i J bushels of wheat to sow an acre, how many bushels will be required to sow 28| acres? 13. How many feet in 148I- rods, allowing 16 J feet to a rod? 14. If a pedestrian can walk 45^ miles in i day, how far can he walk in 18J days ? 15. What will 37 J barrels of apples come to, at $2| per barrel ? QUESTIONS FOR REYIEW. 133 1 6. The sum of two numbers is 68 |f, and the difference between them is i3f : what are the numbers ? 17. What is the product of 4 i^to ^f ? 18. What is the product of ^ into ^^ ? 6f 12 19. What is the product of — f into -|? 1 2 1- 2 J 20. How many days' work will 100 men perform in || of a day ? 21. A man owning y\ of a section of land, sold -} of his share for $i2f : what is the whole section worth, at that rate? 22. How many times is -^jj of f of 5 J contained in 23I ? 23. Divide f of i8| by f of f ^ of f of 3i|. 24. If a gang of hands can do ^ of a job in 5-^ days, I ^hat part of it can they do in i day ? A 25. If f of a yard of satin will make i vest, how many vests can be made from 31^ yards ? 26. How many oil-cans, each containing if gallon, can be filled from a tank of 6i| gallons ? 27. If a man walks 3 J miles an hour, how long w^ill it take him to walk 45 ^^ miles ? 28. By what must JJ be multiplied to produce 15!? 29. How many bushels of apples, at f of a dol., are required to pay for 6 pair of boots, at $6 J ? 30. A farmer sold 330^^ pounds of maple sugar, at i6| cents a pound, and took his pay in muslin, at 22 J cents a yard: how many yards did he receive? 31. Divide the quotient of 12^ divided by 3I by the quotient of 6J-f- 3I ? 32. What is the quotient of -- — -4- ? T TT 7,\ 12 33. What is the quotient of 12^ times -?■ 4- ~j? 134 FRAGTIOKAL RELATION FRACTIONAL RELATION OF NUMBERS. 172. That Numbers may be compared with each other frac^ iionahy, they must be so far of the same nature that one may prop- erly be said to be a part of the other. Thus, an inch may be com- pared with 2ifoot ; for one is a twelfth part of the other. But it can- not be said that 9, foot is any part of an hour ; therefore the former cannot be compared with the latter. 173. To find what part one number is of another. 1. What part of 4 is i ? Analysis. — If 4 is divided into 4 equal parts, one of those parts is called I fourth. Therefore, i is -^ part of 4. 2. What part of 6 is 4 ? Analysis. — i is \ of 6, and 4 is 4 times \, or % of 6. But |=| (Art. 146) ; therefore, 4 is § of 6. Hence, the Rule. — Make the number denoting the part the numera- tor, and that ivith which it is compared the denominator. Note. — i. This rule embraces /<9wr classes of questions : ist. What part one wJiole number is of another, 2d. What part a fraction is of a whole number. 3d. What part a whole number is of a fraction. 4th. What part one fraction is of another. 2. When complex fractions occur, they should be reduced to wimple ones, and all answers to the lowest terms. (Art. 146.) 3. What part of 75 is 15 ? Of 84 is 30 ? 4. Of 91 is 63? 6. Of 81 is 18? 8. Of 256 is 72? 5. Of 48 is 72 ? . 7. Of 100 is 75 ? 9. Of 375 is 425 ? 10. What part of i week is 5 days? 11. A man gave a bushel of chestnuts to 17 boys: what part did 5 boys receive ? 12. At $13 a ton, how much coal can be bought for $10 ? 13. A father is 51 years old, and his son's age is 17: what part of the father's age is the son's ? OF K^ UMBERS. 135 14. If 8 pears cost 35 cents, what will 5 pears cost ? Analysis. — 5 pears are f of 8 pears ; hence, if 8 pears cost 35 cents, 5 pears will cost f of 35 cents. Now ^ of 35 cents is 4I cents, And 5 eighths are 5 times 4^ cents, which are 21^ cents. 15. If 5 bar. of flour cost $45, what w^ill 28 bar. cost? 16. If 50 yds. of cloth cost $175, what will 17 cost? 17. If 25 bu. of apples cost $30, what will no bu. cost? t8. What part of 5 is f ? Analysis. — Making the fraction which denotes the part the nu- merator, and the whole number the denominator, we have a fraction to be divided by a whole number. For, all denominators may be con- sidered as divisors. Thus, ^-i-s=-^o, Ans. (Art. 142.) 19. What part of 25 is J? 21. What part of 30 is -^^ ? 20. What part of 35 is -^^^- ? 22. What part of 40 is ^ ? 23. If 5 acres of land cost $100, what will J acre cost ? Analysis. — i acre is ^ of 5 acres, and f of an acre is | of i, or -^^ of 5 acres. Hence, ^ of an acre will cost 2^0 of $100. NoWa^,- of $100 is $5 ; and 3 twentieths are 3 times 5 or $15. 24. When coal is $95 for 15 tons, what will f ton cost? 25. If 19 yards of silk cost $60, what will | yard cost? 26. What part of f is 2 ? Analysis. — Making the wJiole number which denotes the part, ths numerator, and the fraction the denominator, we have a whole num- ber to be divided by a fraction. Thus, 2^|— 2 x §— \^, Ans. 27. What part of f is 8 ? 29. What part of f is 11? 28. What part of | is 12 ? 30. What part of ^V is 20 ? 31. What part off is f? Analysis. — Making the fraction denoting the part the numerator, and the other the denominator, we have a fraction to be divided by a fraction. Thus, f^|=| x f =f, ^/^.s. (Art. 170.) 32. What part of | is f ? 34. Wliat part of |f is ^^ ? SS. What part of f| is fj ? 35. What part of i| is f | ? $6. 6J is what part of 25 ? Analysis. — Reducing the mixed number to an improper fraction, we have 6^=*/, and \^-T-25 = i-, Ans. 136 FRACTIONAL RELATIOi^. 37. What part of 100 is 12^? 40. Of 100 is 62J ? 38. What part of 100 is $si^ 4i- C)f 100 is iSJ? 39. What part of 100 is i6f ? 42. Of 100 is 87^? 43. 12 J is what part of 18J ? Analysis. — Reducing the mixed numbers to improper fractions we have, 12^=^", and i8|=^4^. Now ^/-f-i/=f, Ans. 44. What part of 62^ is i8| ? 45. Of 874 is 31^ ? 46. At til ^ pound, how much tea will $f buy ? 47. At $JJ per foot, liow niany feet of land can be bought for $ii ? 48. A lad spent iSJ cents for candy, which was 62-J cents a pound : how much did he buy ? 49. A can do a certain job in 8 days, and B in 6 days * what part will both do in i day ? 50. What part of 4 times 20 is 9 times 16 ? 51. What part of 75 x 18 is Io5-^25 ? 52. What part of {6S — 24) x 14 is 168 -^ 12 ? 174. To find a Number, a Fractional Part of it being giver*. Ex. I. 9 is I of what number ? Analysis. — Since 9 is i third, 3 thirds or the whole number must be 3 times 9 or 27. Therefore, g is a third of 27. Or, thus : q is ^ of 3 times 9, and 3 x 9=27. Therefore, etc. 2. 21 is J of what number? Analysis. — Since f of a certain number is operation. 21 units, I or the whole number must be as many 2l-r-|:— 21 X^ units as | are contained times in 21 ; and 2i-;-|=: 21x^=2 8. 21 X ^=28, the answer required. For, a whole number is divided by a fraction by multiplying the former by the latter inverted. (Art. 168, a.) Or, thus : Since 21 is | of a certain number, i fourth of it is i third of 21, or 7. Now as 7 is I fourth of the number, 4 fourths must be 4 times 7 or 28, the same as before. Hence, the Rule. — Divide the number denoting the part hy the fraction. Or, Find one part as indicated hy the numerator of th^ fraction^ and multiply this hy the denominator. OF LUMBERS. 137 Note. — The learner Bhould observe the difference between finding ^ of a number, when f or the whole number is given, and when only f or a part of it is given. In the former, we divide by the denomi- nator of the fraction ; in the latter, by the numerator, as in the second analysis. If he is at a loss which to take for the divisor, let Mm substitute the word parts for the denominator. 3. 56 is f of what ? 7. 436 is f of what ? 4. 68 is J of what? 8. 456 is f of what? 5. 85 is f of what ? 9. 685 is -^ of what ? 6. 1 15 is f of what ? 10. 999 is y^ ^f what ? 1 1. A market man being asked how many eggs he had, replied that 126 was equal to -^ of them : how many had he ^ 12. If y\ of a ship is worth $8280, what is the whole worth ? 13. A commander lost f of his forces in a battle, and had 9500 men left: how many had he at first? 14. fj is :! of what number? Analysis.— f^ is i of 4 times f^' ; and 4 times fr=2|. i5« fi is y of what number ? Analysis. — Since ii^? of a certain number, \ of that number must be \ of H, and ^ of fl^/s, or \. Now if \ of the number=:;^, ^ must equal 7 times \=\, or i|, Ans. 7, 21 % 7 Or, dividing |i by f we have -- x . =-, or i J, Ans. 4, 2J.O. A 4 16. If is I of what number ? 17- If is f of what number? 18. 18J IS f of what number? Analysis.— 1 8|=^4^ Since \^ = 1, 1 = '^^-, and ^=^ja or 30, Ans^ Or, \'^-H-^=^/ X «=fi^,i=3o, the same as before. ^9' 372 is f of what number? 20. 66f is f of what number ? 21. 48 is f of f of what number ? Analysis. — f of ^=^. The question now is, 48 is f of what number ? Ans. 48 x f , or 108. 133 FRA.CTIOKAL RELATIONS. 2 2. 1 1 2 is I of f of what number ? 23. In i of 120 how many times 15 ? Analysis. — h of 120 ie 13^; and I are 7 times 13^ or 93^. I^ow 93i-i5=^l^-i5=\¥ or 6|, ^tis. 24. How many yards of brocatelle, at I9 a yard, can be bought for -J of $100 ? 25. A man paid ^ of $280 for 84 arm-chairs : what was tliat apiece ? 26. 90 is f of how many times 17 ? Analysis. — As 90 is & of a certain number, i is ^ of 90, which is 15 ; and t are 7 times 15 or 105. Now 17 is in 105, 6-^- times. Therefore, etc. 27. 125 is f of how many times 20 ? 28. A man paid 60 cents for his lunch, which was ^ of his money, and spent the remainder for cigars, which were 5 cents each : how much money had he ; and how many cigars did he buy ? 29. ^ of no is f of what number ? Analysis. — i^n of no is 11, and 1%, 9 times 11 or 99. Now, since 99 is t of a number, y of it must be ^ of 99, which is 11, and ^ must be 7 times 11 or 77. Therefore, etc. 30. 1^ of 126 is J of what number ? 31. f of 90 is I of how many times 11 ? Analysis. — f of 90 is 50. Now as 50 is f of a number, ^r is i of 50 or 10; I is 8 times 10 or 80. Finally, 11 is contained in 80, 7i^ times. Therefore ^ of 90 is § of 7 1\ times 11. 32. -I of 96 is 3^ of how many times 20 ? ^^, f of 120 is 1^ of how many sevenths of 56 ? Analysis. — g of 120 is 100. If 100 is | of a number, ^ is \ of the number; now \ of 100 is 25, and | is 9 times 25 or 225. Finally, f of 56 is 8, and 8 is contained in 225, 28^ times. Therefore, ^ of 120 is f of 28^ times | of 56. 34. f of 35 is I of liow many tenths of 120? "DECIMAL FRACTIONS. 175. Decimal Fractions are those in which the unit is divided into tenths, hundredths, thousandths, etc. They arise from continued divisions by lo. If a unit is divided into ten equal parts, the parts are called tenths. JSTow, if one of these tenths is subdivided into ten other equal parts, each of these parts will be one- tentli of a tenth, or a hundredth. Thus, -f^— lo or ^ of ^—^^. Again, if one of these hundredths is subdivided into ten equal parts, each of these parts will be one-tenth of a hundredth, or a thousandtli. Thus, ttJ-jj-t- ^o—-^-}^^, etc. NOTATION OF DECIMALS. 176. If we multiply the unit i by lo continually, it produces a series of whole numbers which increase regu- larly by the scale of lo ; as, I, lo, loo, looo, loooo, looooo, loooooo, etc. Now if we divide the highest term in this series by lo continually, the several quotients will form an inverted series, which decreases regularly by ten, and extends from the highest term to i, and from i to xV? ttottj tx^> ^^^ so on, indefinitely ; as, looooo, loooo, looo, loo, 10, I, ^, yj^, ToW^ ^tc. 177. By inspecting this series, the learner wiU perceive that ^'^ fractions thus obtained, regularly decrease toward the right by the scale of lo. If we apply to this class of fractions the great law of Arabic Notation, which assigns different values to figures, 175. What are decimal fractions? How do they arise? Explain this upon Uie blackboard. 177. By what law do decimals decrease ? 140 DECIMAL FRACTIOI^S. according to the place they occupy, it follows that a figure standing in the first place on the rigid of units, denotes tenths, or i tenth as much as when it stands in units' place ; when standing in the second place, it denotes hun- dredths, or I tenth as much as in the first place ; when standing in the third place, it denotes thousandths, etc., each succeeding order below units being one tenth the value of the preceding. Hence, 178. Fractions which decrease by the scale of ten, may be expressed like whole numbers ; the value of each figure in the decreasing scale being determined by the place it occupies on the right of units. Thus, 3 and 5 tenths may be expressed by 3.5 ; 3 and 5 hundredths by 3.05 ; 3 and 5 thousandths by 3.005, etc. 178, tthtut etc., are common or decimal fractions. All fractions whose denominator is vyritten under the numerator, fulfil the conditions of common fractions, and may be treated as such. But fractions which arise from dividing a unit into 10, 100, 1000, etc., equal parts, ansicer to tJie definition of decimals, whether the denom- inator is expressed, or understood. i33, a. How do decimals differ from coiuuiou fraction..-^ ? 14:4 DECIMAL FRACTIONS. REDUCTION OF DECIMALS. 184. To Reduce Decimals to a Common Denom^inator, 1. Reduce .06, 2.3, and .007 to a common denominator. Analysis — Decimals containing the same number .06 = 0.060 of figures, have a common denominator. (Art. 179.) « •3 = 2 -200 By annexing ciphers, the number of decimal fig- ares in each may be made the same, without altering * ' * ' their value. (Art. 181.) Hence, the Rule. — Make the number of decimal figures the same in eachf hy annexing ciphers. (Art. 181.) 2. Reduce .48 and .0003 to a common denominator. 3. Reduce 2 to tenths ; 3 to hundredths ; and .5 to thousandths. Ans. 2.0 or f^; 3.00 or f^f; .500. 185. To Reduce Decimals to Common Fractions. 1. Reduce .42 to a common fraction. Analysis, — The denominator of a decimal is i, with as many ciphers annexed as there are figures in its numerator; therefore the denominator of .42 is 100. (Art. 179.) Ans, .42=-j^o%. Hence, the Rule. — Erase the demmal pointy and place the denomi- nator under the numerator. (Art. 179.) 2. Reduce .65 to a common fraction; then to its lowest terms. Ans. .65 = A^ir^ ^^^ Twu=ii- Reduce the following decimals to common fractions : 3. .128 7. .05 II. .0007 15. .200684 4. .256 8. .003 12. .04056 16. .0000008 5. .375 9. .0008 13. .00364 17. .12400625 6. .863 10. .0605 14. .00005 18. .24801264 186. To Reduce Com^moti Fractions to Decimals. I. Reduce | to a decimal fraction. Analysis — | equals i of 3. Since 3 cannot be divided operation. fcy 8 ; we annex a cipher to reduce it to tenths. Now \ 8)3.000 of 30 tenths is 3 tenths, and 6 tenths over. 6 tenths AnsT^vfk reauced to hundredths =60 hundredths, and |^ of 60 hundredths =7 hundredths and 4 hundredths over. 4 hundred tha reduced to thousandths =^0 thousandths, and \ of 40 thousandtha rr5 thousandtlis. Therefore, ?l = .375. Hence, the t DECIMAL FRACTIOi^S 145 Rule — Annex ciphers to the numerator and divide by the denominator. Finally, point off as many decimal figures in the result as there are ciphers amiexed to the numerator. Note. — If the number of figures in the quotient is less than the number of ciphers amiexed to the numerator, supply the deficiency by prefixing ciphers. Demonstration. — A fraction indicates division, and its value is the numerator divided by the denominator. (Arts. 134, 142.) Now, an- nexing one cipher to the numerator multiplies the fraction by 10 ; annexing two ciphers, by 100, etc. Hence, dividing the numerator with one, two, or more ciphers annexed, gives a quotient, 10, 100, etc., times too large. To correct this error the quotient is divided by 10, 100, 1000, etc. But dividing by 10, 100, etc., is the same as pointing off an equal number of decimal figures. (Art. 181 .) Reduce the following fractions to decimals : 2. i 6. I 10. ^ 14. ^^ 3. f 7. A 1 1- it 15.7!^ 4- J 8. I 12. ^V -16. ^l-Q 1 8. Reduce f to the form of a decimal. Analysis. — Annexing ciphers to the numerator operation. and dividing by the denominator, as before, the quo- 3)2.000 tient consists of 6 repeated to infinity, and the re- "^6666 etc. mainder is always 2. Therefore | cannot be exactly expressed by decimals. 19. Reduce /y to the form of a decimal. Analysis. — Having obtained th^ee quotient operation. figures 135, the remainder is 5, the same as the 37(5.000000 original numerator; consequently, by annex- .135135? ^tc. ing ciphers to it, and continuing the division, we obtain the same set of figures as before, repeated to infinity. Thcsrefore ^ cannot be exactly expressed by decimals. 187. When the numerator, with ciphers annexed, is exactly divisible by the denominator, the decimal is called a terminate decimal. .J85. How reduce a decimal to a common fraction ? 186. How reduce a com- mon fraction to a decimal ? Note. If the number of figujes in the quotient is less t^an that in the numerator, what le to be done ? Explain the reason for pointing off the quotient. 116 ADDITION OF DECIMALS. When it is not exactly divisible^ and the same figure o set of figures continually recurs in the quotient, the deci nial is called an interminate or circulating decimal. T\\Q figure or set of figures repeated is called the repetend Thus, the decimals obtained in the last two examples ar< interminate, because the division, if continued forever, wil; leave a remainder. The repetend of the i8th is 6; thai of the 19th is 135. Notes. — i. After the quotient has been carried as far as desirably the sign ( + ) is annexed to it to indicate there is still a remainder. 2. If the remainder is such that the next quotient jBgure would be 5, or more, the last figure obtained is sometimes increased by i, and the sign (— ) annexed to show that the decimal is too large. 3. Again, the remainder is sometimes placed over the divisor and annexed to the quotient, forming a mixed fraction. (Art. i8c, n^ Thus if I is reduced to the decimal form, the result may be expressed by .6666+ ; by .6667— ; or by .6666 .|. (For the further consideration of Circulating Decimals, the student is referred to Higher Arithmetic.) Eeduce the following to four decimal places : 20. i 22. f 24. -i\ 26. Jf 21. f 23. $ 25. -3-\ 27. 4i Eeduce the following to the decimal form : 28. 75f 30- 26ii|- 32. 465^^ 34. lAo-a^ 29- 136! 31- 346H 2>Z' 523^3- 35- 956^ ADDITION OF DECIMALS. 188. Since decimals increase and decrease regularly by the scale of 10, it is plain they may be added, subtracted, multiplied, and divided like whole numbers. Or, they may be reduced to a common denominator, then be added, subtracted, and divided like Common Fractions. (Arts. 156, 184.) 187. When the numerator with ciphers annexed is exactly divisible by tho denominator, what is the decimal called ? If not exactly divisible, what ? What Is the figure or set of figures repeated called? ADDITION OF DECIMALS. 147 189. To find the Amount of two or more Decimals. I. Add 360.1252, 1. 91, 12.643, and 152.8413. Ai^ALYSis. — Since units of tlie same order or operation. like number's only can be added to eacli other, we 3^^'^ ^5^ reduce tlie decimals to a common denominator by 1.9 100 annexing ciphers; or, which is the same, by 1 2.6430 writing the decimals one under another, so that 152.8413 the decimal points shall be in a perpendicular ^715.527.5195 line. (Arts. 28, n. 156.) Beginning at the right, we add each column, and set down the result as in whole numbers, and for the same reasons. (Art. 29, 7i.) Finally, we place the deci- mal point in the amount directly under those in the numbers added. (Art. 178, a) Hence, the Rule. — I. Write the numUrs so that the decimal points shall stajid one under another, with tenths under tenths, etc. II. Beginning at the right, add as in whole numbers, and place the decimal poitit in the amount under those in the numbers added. Note.— Placing tenths under tenths, hundredths under hundredths, etc., in efiect, reduces the decimals to a com. denominator; hence th« ciphers on the right may be omitted. (Arts. 181, 184.) (2.) (3.) (4.) (5.) 41.3602 416.378 36.81045 4.83907 4.213 85.1 .203 .293 61.46 .4681 5-3078 .40 375.265 4.3S 87.69043 5.1067 482.2982 Ans. 375.2956 9.25 3.75039 6. What is the sum of 41.371 +2.29 + 73.4024- 1.729? 7. What is the sum of 823.37 + 7.375 + 61. i +.843? 8. What is the sum of .3925 + .64 + .462 + .7 +.56781 ? 9. What is the sum of 86 005 + 4.0003 + 2.00007 ? 10. What is the sum of 1.7 13 + 2.30 + 6. 400 + 27.004? 11. Add together 7 tenths; 312 thousandths; 46 hun- dredths; 9 tenths; and 228 ten-thousandths. 12. Add together 23 ten-thousandtks ; 23 hundred- thousandths; 23 thousandths; 23 hundredths; and 23. i?9. How are decimals added ? How point oflf tlie amount f 148 SUBTRACTIOK OF DECIMALS. 13. Add together five hundred seventy-five and seven- tenths; two hundred fifty-nine ten-thousandths; five- miUionths; three hundred twenty hundred-thousandths. 14. A farmer gathered lyf bushels of apples from one tree ; 8^ bushels from another ; loj bushels from another ; and 16 J bushels from another. Required the number of bushels he had, expressed decimally. 15. A grocer sold 7^ pounds of sugar to one customer; 11.37 pounds to another; lof pounds to another; 25^ pounds to another; and 21.^75 pounds to another: how many pounds did he sell to all ? SUBTRACTION OF DECIMALS. 190. To find the Difference between two Decimals. I. What is the difference between 2.607 ^^^ -7235 ? Analysis. — We reduce the decimals to a common operation. denominator, by annexing ciphers, or by writing the 2.607 same orders one under another. (Art. 189, n.) For, -7235 units of the same order or like numbers only, can Ans. 1.8835 be subtracted one from the other. Beginning at the right, we perceive that 5 ten-thousandths can not be taken from o ; we therefore borrow ten, and then proceed in all respects as in whole numbers, (Art. 38.) Hence, the Rule. — I. Write tlie less number under the greater, so That the decimal points shall stand o?ie under the other, with tenths tinder tenths, etc. II. Bcginniiig at the right hand, proceed as in subtract- ing luhole numbers, and place the decimal point in the remainder under that in the subtrahend. Note. — Writing the same orders one under another, in effect reduces the decimals to a. common denominator. (Art. 189, n.) (2-) / (3-) (4.) (5-) From 13.051 7*0392 20.41 S5.3004 rake 5.22 * .43671 3 -0425 67.35246 190. How are decimals Bubtracted ? How point off the remainder? MULTIPLICATION^ OFDECIMALS. 149 Perform the following subtractions : 6. 13.051 minus 5.22. 12. 10 minus 9.1030245. 7. 7.0392 minus 0.43671. 13. 100 minus 994503067, 8. 20.41 minus 3.0425. 14. i minus .123456789. 9. 85.3004 minus 7.35246. 15. i minus .98764321. 10. 93.38 minus 14.810034. 16. o.i minus .001. 11. 3 minus 0.103784. 17. o.oi minus .00001. 18. From 100 take i thousandth. 19. From 45 take 45 ten-thousandths. 20. From I ten- thousandth take 2 millionths. 21. A man having $673,875, paid $230.05 for a horse: how much had he left ? 22. A father having 504.03 acres of land, gave 100.45 acres to one son, 263.75 acres to another: how much had he left ? 23. What is the difference between 203.007 and 302.07 ? ^ 24. Two men starting from the same place, traveled in and is read " 7 and ',V,j dollars." Mills also are sometimes expressed by a common fraction. Thus, 12 cts. and 5 mills are written $0,125, or $0.12^ ; 18 cis. and 7.2 mills are written $0.1875, or $0.18]-, etc. 4. In business calculations, if the mills in the result are 5 or mor«, they are considered a cent ; if less than 5, they are omitted. 1. Write forty dollars and forty cents. 2. Write five dollars, five cents and five mills. 3. Write fifty dollars, sixty cents, and three mills. 4. Write one hundred dollars, seven cents, five mills. 5. Write two thousand and one dollars, eight and a half cents. 6. Write 7 hundred and 5 dollars and one cent. 7. Write 84 dollars and 12} cents. 8. Write 5 and a half cents: 6 and a fourth cents; 11 aiid three-ft)urth8 cents, decimally. 156 UNITED STATES MOKEY. 9. Write 7 dollars, 31 and a fourth cents^ decimally. 10. Write 19 dollars, 31:1 cents, decimally. 11. Write 14 eagles and 8 dimes, decimally. 12. Write 5 eagles, 5 dollars, 5 cents, and 5 mills. 198. To read United States money, expressed decimally. Rule. — Gall the figures on the left cf the decimal point, doU lars ; those in the first two places on the right, cents ; the next figure, mills ; the others, decimals of a mill. The expression $37.52748 is read 37 dollars, 52 cts., 7 mills, and 48 hundredths of a mill. Note. — Gents and mills are sometimes read as decimals oi a do?lar. Thus, $7,225 may be read 7 and 225 thousandths dollars. Read the following : 1. $204.30 5. $78,104 9. $1100.001 2. $360.05 6. $90,007 10. $7.3615 3. $500.19 7. $1001.10 II. $8.0043 4. $61,035 8. $1010.01 12. $10.00175 REDUCTION OF UNITED STATES MONEY. CASE I. 199. To reduce Dollars to Cents and Mills. 1. In $67 how many cents? Analysis. — As there are 100 cents in every opebation. dollar, there must be 100 times as many cents 67 X 100 = 6700 as dollars in the given sum. But to multiply Ans. 6700 cts. by 100 wo annex two ciphers. (Art. 57.) 2. In $84, how many mills? Analysis.— For a like reason, there are 84 X 1000 = 84000 1000 times as many mills as dollars, or 10 Ans. 84000 millji. times as many mills as cents. Hence, the Rule.— :7b reduce dollars to cents, multiply them ly 100. To reduce dollars to mills, multiply them hy 1000. To reduce cents to mills, multiply them hy 10. 198. How read United States money ? UN^ITED STATES MOl^EY. 157 Note. — Dollars and cents are reduced to cents ; also dollars, cents, and mills, to mills, by erasing the sign of dollars ($), and the deci- mal point. 2. Reduce $135 to cents. 6. Reduce 97 cents to mills. 3. Reduce $368 to mills. 7. Reduce $356.25 to cents. 4. Reduce $100 to mills. 8. Reduce $780,375 to mills. 5. Reduce $1680 to cents. 9. Reduce I800.60 to mills. CASE II. 200. To reduce Cents and Mills to Dollars. I. In 6837 cents how many dollars? Analysis. — Since 100 cents make i dollar, . opeeation. 6837 cents will make as many dollars as 100 is l)68|37 contained times in 6837, and 6837^100=68, $68.37 Ans. and 37 cents over. (Art. 79.) In like manner any number of mills will make as many dollars aa 1000 is contained times in that number. Hence, the Rule. — To reduce cents to dollars, divide them ly 100. To reduce mills to dollars, divide them ly 1000. To reduce mills to cents, divide them ly 10. (Art. 194.) Note. — The first two figures cut off on the right are cents, the next one mills. 2. Reduce 1625 cts. to dols. 6. Change 89567 cts. to dels. 3. Reduce 8126 m's to dols. 7. Change 94283 m's to dols. 4. Reduce loooo m's to dols. 8. Change 85600 m's to cents. 5. Reduce 9265 m's to cents. 9. Change 263475 m's to dols. 10. A farmer sold 763 apples, at a cent apiece : how many dollars did they come to ? n. A market woman sold 5 hundred eggs, at 2 cents each: how many dollars did she receive for them ? 12. A fruit dealer sold 675 watermelons at 1000 mills apiece : how many dollars did he receive for them ? 199. How reduce dollars to cents ? To mills ? How cents to mills ? Notg. How dollars to cents and mills? 300. How reduce cents and mills to dollars? NoU. VVhot are the figures cut off? 158 U 1^11 ED STATES MOKEY. ADDITION OF UNITED STATES MONEY. 201. United States Money, we have seen, is founded upon the decimal notation; hence, all its operations are precisely the same as the corresponding operations in Decimal Fractions, 202. To find the Amount of two or more Sums of Money, 1. Whatis the sum of $45,625 ; $109.07; and $450,137 ? Analysis. — Units of the same order only can be operation. added together. For convenience in adding, we there- $45,625 fore write dollars under dollars, cents under cents, etc., 109,07 with the decimal^ points in a perpendicular line. Begin- 45 o. 1 3 7 ning at the right, we add the columns separately, $604.8-^ placing the decimal point in the amount under the points in the mmibers added, to distinguish the dollars from cents and mills. (Art. 197.) Hence, the KuLE. — Write dollars under dollars, cents under cents, etc., and proceed as in Addition of Decimals. Note. — If any of the given numbers have no cents, their place ehould be supplied by ciphers. 2. A man paid $13. 62^ for a barrel of flour, $25.25 for butter, $9.75 for coal: what did he pay for all ? 3. A farmer sold a span of horses for $457.50, a yoke of oxen for $235, and a cow for $87.75 : how much did he receive for all ? 4. What is the sum of $97.87-1-; $82.09; $2o.i2|? 5. What is the sum of $81.06; $69.18; $67.16; $7.13? 6. What is the sum of $101.101 ; $2io.io|; $450.27^? 7. Add. $7 and 3 cents; $10; 6\ cents; i8| cents. 8. Add $68 and 5 mills ; 87 J cents; 31 J cents. 9. A man paid $8520.75 for his farm, $1860.45 for his stock, $1650.45 for his house, and $1100.07 for his furni- ture : what was the cost of the whole ? 201. What is said of operations in United States Money? 202. How ndd TTnited States Money? Note. If any of the numbere hav* no centD, how proceeds Uiq'ITED STATES MOis^ET. 1^\) 10. A lady paid $31 J for a dress, $15!- for trimmings, and $7-J for making: what was the cost of her dress ? 11. A grocer sold goods to one customer amounting to $17.50, to another $3o.i8f, to another $2i.o6:J, and to another $5 1.73 : what amount did he sell to all? 12. A young man paid $31.58 for a coat; $11.63 f'^^ ^ vest, $14.11 for pants, $10.50 for boots, $7 J for a hat, and $1 J for gloves : what did his suit cost him ? SUBTRACTION OF UNITED STATES MONEY. 203. To find the jyifference between two Sums of Money. 1. A man having $1343.87!, gave $750.69 to the Patriot Orphan Home : hoAV much had he left ? Analysis. — Since the same orders only can be sub- opiration, tracted one from the other, for convenience we write $1343.875 dollars under dollars, cents under cents, etc. Begin- 750.69 ning at the right, we subtract each figure separately, $CQ'?.i8^ and place the decimal point in the remainder under that in the subtrahend, for the same reason as in subtracting deci- mals. (Art. 190.) Hence, the Rule. — Write the less number under the greater, dollars under dollars, cents under cents, etc., and 'proceed as in Subtraction of Decimals. (Art. 190.) Note. — If only one of the given numbers has cents, their place in the other should be supplied by ciphers. 2. A man having $861.73, lost $328,625 in gambling: how much had he left ? 3. If a man's income is $1750, and his expenses $1145.3 7 J, how much does he lay up ? 4o A gentleman paid $1500 for his horses, and $975 J for his carriage : what was the difference in the cost ? 5. A merchant paid $25 73 J for a quantity of tea, and sold it for $3158^ : how much did he make ? 203. How 8u1)tract United States money? JVoie. If either number has no cents, what is to be done ? 160 U]S^ITED STATES MONEY. 6. If I pay $5268 for a farm, and sell it for $4319.67, how much shall I lose by the operation ? 7. From 673 dols. 6 J cents, take 501 dols. and 10 cents. 8. From ion dols. 12^ cents, take 600 dols. and 5 cents. 9. From I dollar and i cent, subtract 5 cents 5 mills. 10. From 7J dols. subtract 7J cents. 11. From 500 dols. subtract 5 dols. 5 cents and 5 mills. 12. A 3^oung lady bought a shawl for $35 J, a dress for ^231, a hat for $iof, a pair of gloves for |ij, and gave the clerk a hundred dollar bill : how much change ought she to receive ? MULTIPLICATION OF UNITED STATES MONEY. 204. To multiply United States Money. 1. What will 9 J yards of velvet cost, at $18 J per yard ? Analysis. — 9.^ yards will cost 9^ times as mucli as $18.2 c; I yard. Now g\ yds.=9.5 yds., and $i84-=$i8.25. „_- We multiply in tlie usual way, and since there are tJiree decimal figures in both factors, we point off three " ^ in the product for the same reason as in multiplying ^ 5 ^ decimals. (Art. 191.) Hence, the ^173-375 Rule. — Multiply, and jmint off the product, as in Multi- plication of Decimals. (Art. 191.) Notes. — i. In United States Money, as in gimple numbers, the multiplier must be considered an abfitract number. 2. If either of the given factors contains a common, fraction, it is generally more convenient to change it to a demnal 2. What will 65 barrels of flour cost, at $11.50 a barrel? 3. What will 145.3 pounds of wool cost, at $1.08 a pound? 4. What cost 75 pair of skates, at $3.87!- a pair ? 5. What cost 6;^ gallons of petroleum, at 95^ cents a gallon ? 6. At $4.17^ a barrel, what will no barrels of apples cost? 204. How cniltiply United States money ? Kote. What must the multiplier be r UKITED STATES MON^EY. 161 Perform the following multiplications : 7. $511x1.9!-. II. $765,401x6.05. 8. $6.07-!- X 2.3 J. 12. $.07 X. 008. 9. $10.05 X 6^. 13. $.005 X 1000. 10. $100,031x3.105. 14. $1,011 X. 001. 15. What cost 35 pounds of raisins, at iSJ cents a pound ? 16. What cost 51 pounds of tea, at $1.15 J a pound? 17. What cost 150 gallons of milk, at 37 J cts. a gallon ? 18. What cost 12 dozen penknives, at 31 J cents apiece? 19. What cost 13 boxes of butter, each containing 16 J pounds, at 37^ cents a pound? 20. A merchant sold 12 pieces of cloth, each containing 35 yards, at $4! a yard : what did it come to ? 21. What is the value of 21 bags of coffee, each weigh- ing 55 pounds, at 47^ cents a pound? 22. What cost 55 boxes of lemons, at $3 J a box ? 2;^. The proprietor of a livery stable took 37 horses to board, at $32 a month: how much did he receive in 12 months ? DIVISION OF UNITED STATES MONEY. 205. Division of United States money, like simple divi- sion, embraces two classes of problems : First. Those in wliicli both the divisor and dividend are money. Second. Those in which the dividend is money and the divisor is an abstract number, or regarded as such. In the former, a given sum is to be divided into parts, the value of each part being equal to the divisor; and the object is to ascertain the number of parts. Hence, the quotient is times or an abstract number, (Art. 63, a.) In the latter, a given sum is to be divided into a given number of equal parts indic&ted by the divisor ; and the object \s to ascertain the value or number of dollars in each part. Hence, the quotient is money, the same as the dividend. (Art. 63, 6.) ao5. What does DiviBion of U. S. money embrace ? The first class ? The second f 162 UNITED STATES MOls'EY. 206. To Divide Money by Money, or by an Abstract Number 1. How many barrels of flour, at $9.56 a barrel, can be bought for $262.90 ? Analysis.— At $9.56 a barrel, $262.90 will operation. 6uy as many barrels as $956 are contained $9.56)$262.9o(27.5 times in $262.90, or 27.5 barrels. As the 191 2 divisor and dividend both contain cents, T^^io tliey are the same denomination ; therefore, 6692 the quotient 27, is a whole number. Annex- 3— .1 1 .1 1 47 oO mg a cipher to the remamder, the next quo- ' o tient figure is tenths. (Art. 193, Rem) -zZ Hence, the Rule. — Divide, and point off the quotient, as in Division of Decimals. (Arts. 64, 193.) Notes. — i. In business matters it is rarely necessary to carry the quotient beyond mills. 2. If there is a remainder after all the figures of the dividend have been divided, annex ciphers and continue the division as far as de- sirable, considering the ciphers annexed as decimals of the dividend. 2. If 67^ ca,ps cost 1 1 23.75, what will i cap cost? 3. If 165 lemons cost $8.25, w^hatwill i lemon cost? ^4. Paid I852.50 for 310 sheep : what Tfas that apiece? L 5. If 356 bridles cost $1040, what will i bridle cost? Find the results of the following divisions : 6. $17.50-:- 1. 1 75 10. $.oo5^$.o5 7. $365.07 -^ $1.01 II. $ioi-^$i.oi 8. $1000-7-25 cts. 12. $5oo-^$.o5 ^^ 9. $.25H-$25 13. $l200-^$.002 ^ 14. If 410 chairs cost $1216, what will i chair cost? 15. At $9j a ton, how much coal will $8560 buy? 16. When potash is $120.35 per ton, how much can be bought for $35267.28? 17. If 2516 oranges cost $157.25, what will one cost? t8. Paid $273.58 for 5000 acres of land: what was that per acre ? 2o6. How divide money ? NoU. If there is a remainder, how proceed ? UNITED STATES MOifEY. 163 COUNTING-ROOM EXERCISES. 207. The Ledger is the principal look of account? kept by business men. It contains a Irief record of their monetary transactions. All the items of the Bay Booh are transferred to it in a condensed form, for reference and preservation, the debits (marked Dr.) being placed on the left, and the credits (marked Or.) on the right side. 208. Balaneing an Aecotcnt is finding the differ- ence between the debits and credits. Balance the following Ledger Accounts : (!•) (2.) Dr. Cr. Dr. Or. $8645.23 $2347.19 $76421.26 $43261.47 160.03 141.07 2406.71 728.23 2731.40 2137.21 724.05 6243.41 4242.25 3401.70 86025.21 75.69 324-31 2217.49 9307.60 53268.75 3313-17 168.03 685.17 3102.84 429.18 329.17 61.21 456.61 4536.20 2334-67 7824.28 32921.70 641.46 4506.41 60708.19 6242.09 182.78 239.06 1764.85 20374-34 2634.29 2067.12 32846.39 290.25 3727-34 675-89 385-72 4536.68 840.68 1431.07 23.64 42937-74 6219.77 4804.31 6072.77 819.35 4727.91 6536.48 50641.39 30769.27 23-45 720.34 2062.40 8506.35 650.80 36.45 301-53 97030.48 3267.03 7050.63 26.47 876.20 680.47 904.38 805.03 89030.50 64.38 78.05 2403.67 384.06 8350.60 307-63 10708.79 70207.48 207, What is the ledger ? What does it contain ? IIow balance an account ? 164 BILLS, MAKING OUT BILLS. 209. A Bill is a tvriUen statement of goods sold, ser- vices rendered, etc., and should always include the price of each item, the date, and the place of the transaction. A Bill is Beceiptedf when the person to whom it is due, or his agent, writes on it the words "Received payment," and his name. 209, a, A Statement of Account is a copy of the items of its cleMts and credits. Notes. — i. The abbreviation Dr. denotes debit or debtor ; Or., credit or creditor ; per, by ; the character @, stands for at. Thus, 5 books, @ 3 shillings, signifies, 5 books the price of which is 3 shil- lings apiece. 2. The learner should carefully observe the form of Bills, tho l^lace where the date and the names of the buyer and seller are placed, the arrangement of the items, etc. Copy and find the amount due on the following Bills : i\\ Thtladsi.fbia, March zst, 18 j2. Hon. Heney Barkard, To J. B. Lippii^coTT & Co., Dr. For 6 Webster's Dictionaries, 4to., @ $12.50 " 8 reams paper, @ $3-75 " 36 slates, @ $0.27 $75 30 9 00 00 72 $114 $84 $30 72 Credit. By 10 School Architecture, (S $5.45 " 8 Journals of Education, @ $3.75 Balance, $54 50 GO 50 209, What is a bill ? 209, a. A statement of account ? Where is the date placed? (See form.) The name of buyer? Seller? How is the payment of a bill shown ? JVote. What does Dr. denote ? Or. ? (2.) BILLS. 165 New Yobk, April 2d, 1871. Mrs. W. C. Gabfield, Bought of A. T. Stewaet. 28 yds. silk, 35 yds. table linen, 6 pair gloves, 43? yds. muslin, I do3. pr. cotton hose. Amount. @ $3-50 @, $2.12^ @$i.75 @ $0.33 @, $0.80 Eeceived Payment, A. T. Stewart. \3v New Orleans, May 5th, 1872. George Peabody, Esq., Bought of Jacob Barker. Feb. I .^7 Marc I13 25 April L30 75 bis. pork, 160 bis. flour. 500 gals, molasses, 75 boxes raisins, 256 gals, kerosene. @ $25.00 @ $8.75 % $0.93 @ $5,371- @ $0.87^ Amount, Received Payment, J. Barker, By John Howard. fA\ Boston, May 23343-75 EuLE. — To the number of articles, add \, |, J of itself, as the case may he; the sum will he their cost. 17. At $1.25 per acre, what will 168 acres of land cost? 18. At $1.50 apiece, what will 365 chairs cost ? 19. What cost 512 caps, at %i^ apiece ? 20. At $i.i6| apiece, what cost 12 dozen fans? 21. A man sold 200 overcoats at a profit of $1.20 apiece : how much did he make ? 213. To find the Number of articles, the Cost being given, and the Price of one an Aliquot part of $1. 22. How many spellers, at 12^ cents apiece, can he bought for $25 ? Analysis. — 12^ cents=$|; therefore $25 will $.i2j=$|^ buy as many spellers as there are eighths in $25, $25-^$-|-=20o and 25-f-i=200. Ans. 200 spellers. Hence, the Rule. — Divide the cost of the whole hy the aliquot part of%i, expressing the price of one. 23. How man}^ yards of flannel, at 50 cents, can be bought for $6.83 ? 24. How many pounds of candy, at 33^ centig, can be purchased with $375 ? 25. Paid $450 for cocoa-nuts which were 25 cents each: how many Avere bought ? 26. At 20 vients each, how many pine-apples can be pur- chased for $538 ? 211. How find the cos."" of articles by aliquot parts of $i ? 212. How when the price is $1 plus an aliquot part of $1? 213. How find the number of articleb, when cost is given, and the price of one is an aliquot part of $1 ? 8 170 COUNTIIf G-EOOM EXERCISES- 214. To find the Cost of articles, sold by the 100, or 1000. 1. What will 1765 oranges cost, at $6,125 a hundred? Analysis. — At $6,125 for each orauge, 1765 operation. oranges would cost 1765 times $6 125, and $6,125 x $6,125 1765 = $10810.625. But the price is $6,125 for a 1765 hundred ; therefore this product is 100 times too $108.10621; large. To correct it, we divide by 100, or remove the decimal point two places to the left. (Art. 181.) In like manner if the price is given by the 1000, we multiply the 'price and number of articles together, and remove the decimal point in the product three places to the left, which divides it by 1000. Hence, the Rule. — Multij^tly the price and niimher of articles together, and divide the product ly 100 or 1000, as the case may require, {xi rt, 181 .) Note. — In business transactions, the letter C is sometimes puv for hundred; and M for thousand. 2. What will 4532 bricks cost, at I17.25 per M. ? 3. What cost 1925 pounds of maple sugar, at I12.50 pev hundred? 4. What cost 25268 feet of boards, at $31.25 per thous- and? 5. At $5f per hundred, how much will 20345 pounds of flour come to ? 6. At I6.25 per hundred, what will 19263 pounds of codilsh come to ? 7. What cost 10250 envelopes, at I3.95 per thousand? 8. What cost 1275 oysters, at $1.75 per hundred? 9. What cost 13456 shingles, at $7.45 per M. ? 10. What cost 82 rails, at $5 J a hundred ? 11. What cost 93 pine apples, at $15.25 a hundred? 12. What cost 355 feet of lumber, at $45 per thousand^ 214. How find the cost of articles sold by the 100 or 1000 ? Note. What does C stand for? WhatM? COMPOUND NUMBERS. 215. Slmjyle J^umbers are those which contain units of one denomination only; as, three, live, 2 oranges, 4 feet, etc. (Art. loi.) 216. Compound JV^umbers are those which con- tain units of two or more denominations of the sat7ie nature; as, 5 pounds and 8 ounces; 3 yards, 2 feet and 4 inches, etc. But the expression 2 feet and 4 pounds, is not a com- pound number; for, its units are of unliJce nature. Notes. — i. Compound Numhers are restricted to the divisions of Money, Weights, and Measures, and are often called Denominate Nunibers. 2. For convenience of reference, tlie Compound Tables are placed togrether. If the teacher wishes to give exercises upon them as they are recited, he will find examples arranged in groups i responding with the order of the Tables in Arts, 276, 279. MOIfEY. 217. MJonej/ is the measure or standard of value. It is often called currency, or circulating medium, and is of two kinds, metallic and paper. Metallic Money consists o^ stamped pieces of metal, called coins. It is also called specie, or specie currency. I^aper Money consists of notes or hills issued by the Government and Banks, redeemable in coin. It is often called paper currency. U15. What are simple numbere ? 216. Compound? Give an example of each. Nd^e. To what s.-e compound numbers restricted? 217, What is money? Me- tallic money ? Paper money ? 172 COMPOUND NUMBERS. UNITED STATES MONEY. 218. United States Money is the national cur- realty of the United States, and is often called Federal Money. Its denominations are eagles, dollars, dimes, cents, and mills.. TABLE. lo mills (m) are i cent, ct. lo cents " I dime, . _ . - - e?. lo dimes " i dollar, _ _ - - dol. or $. lo dollars " i eagle E. 219. The Metallic Currency of the U7iited States consists 0^ gold and silver coins, and the minor coins.* 1. The gold coins are tlie double eagle, eagle, half eagle, quarter eagle, three dollar piece, and dollar. The dollar, at the standard weight, is the unit oi value. 2. The silver coins are the dollar, "trade" dollar, hMf dollar, quarter dollar, and dime. 3. The minor coins are the s-<^ent and 2>-cent pieces, and the cent. 220. The weight and purity of the coins of the United States are regulated by the laws of Congi-ess. 1. The standard weight of the gold dollar is 25.8 gr. ; of the quar- ter eagle, 64 5 gr. ; of the 3-dollar piece, 77.4 gr. ; of the half eagle, 129 gr. ; of the eagle, 258 gr. ; of the double eagle, 516 gr. 2. 'W\Qice'ght of the dollar is 412^ grains, Troy; the "trade" dollar, 420 grains ; the half dollar, \i\ grams ; the quarter dollar and dime, one-half and one-fifth the weight of the half dollar. 3. The weight of the 5-ccnt piece is 77. 16 grains, or 5 grams ; of the 3-cent piece, 30 grains ; of the cent, 48 grains. 4. The standard purity of the gold and silver coins is nine-tenths pure metal, and one-tenth alloy. The alloy oi gold coins is silver AxiA. copper ; the silver, by law, is not to exceed one-tenth of the whole alloy. The alloy of silver coins is pure copper. 219. Of what does the metallic currency of the United States consist? What are the gold coins ? The silver ? The minor coins ? 220. How is the weight and purity of United States coins regulated ? * Act of Congrass, March 3d, 1873. COMPOUl!^D NUMBERS. 173 5. The five-cent and three-cent pieces are composed of one-fourth nickel and three-fourths copper ; the cent, of 95 parts copper and 5 parts of tin and zinc. They are known as nickel and bronze coins. Notes. — i. The Trade dollar is so called, because of its intended use for commercial purposes among the great Eastern nations. 2. The gold coins are a legal tender in all payments; the siher coins, for any amount not exceeding $5 in any one payment ; the minor coins, for any amount not exceeding 25 cents in any one payment. 3. The diameter of the nickel 5-cent piece is two centimeters, and its weight 5 grams. These magnitudes present a simple relation of the Metric weights and measures to our own. 4. The silver 5-cent and 3-cent pieces, the bronze 2-cent piece, the old copper cent and half -cent are no longer issued. Mills were never coined. 221. The I*aper Currency of the United States consists of Treasury-notes issued by the Government known as Greenbacks, and Banh-notes issued by Banks. Note. — Treasury notes less than $1, are called Fractional Cur- rency. ENGLISH MONEY. 222. English 3Ioney is the national currency of Great Britain, and is often called Sterling Money. The denominations sltq poimds, shillinys, pence, and fa? thinys. TABLE. 4 farthings (gr. or /ar.) are I penny, - - d. 12 pence " i shilling, - s. 20 shillings " I pound or sovereign, - - £ 21 shillings " i guinea, g. Notes. — i. The gold coins are the sovereign and lialf sovereign. The pound sterling was never coined. It is a bank note, and is represented oj the sovereign. Its legal value as fixed by Congress is $4.8665. This is its intrinsic value, as estimated at the U. S. Mint. What is the alloy of gold coins ? Of silver ? 221. Of what does U. S. paper carrency cousist? 222. English money? The denominations? The Table? Jfote. Is the pound a coin ? How represented ? What is the value of a pound ? 174 COMPOUIS^D NUMBERS. 2. The silver coins are the crown (5s.) ; the half-crown (28. 6d.) . the florin (2s.) ; the shilling (i2d.) ; the six-penny, four-penny, and three-penny pieces. 3. The copper coins are the penny, half-penny, and farthing. 4. Farthings are commonly expressed asfractions of a penny, as y^d. 5. The oblique mark (/) sometimes placed between shillings and pence, is a modification of the long/. CANADA MONEY. 223. Canada 3Io7iey is the legal currency of the Dominion of Canada. Its denominations are dollars, cents, and mills, which have the same value as the cor- responding denominations of IT. S. money. Hence, all tlie oj)erations in it are the same as those in U. S. money. Note. — The present system was established in 1858. FRENCH MONEY. 224. French Money is the national currency of France. The denominations are the franc, the decime, and centime, TABLE. 10 centimes . _ . _ are i decTme. 10 decimes - - - - " i franc. Notes. — i. The system is founded upon the decimal notation ; hence, all the operations in it are the same as those in U. S. money, 2. The franc is the unit ; decimes are tentJis of a franc, and centimes hundredths. 3. Centimes by contraction are commonly called cents. 4. Decimes, like our dimes, are not used in business calculations; they are expressed by tens of centimes. Thus, 5 decimes are ex- jressed by 50 centimes; 63 francs, 5 decimes, and 4 centimes are written, 63.54 francs. 5. The legal value of i\\Q franc in estimating duties, is 19.3 cents, its intrinsic value being the same. Shilling ? Florin ? Crown ? How are farthings often written ? 223. What is Canada money? Its denominations ? Their value? 224. French money ? lis denomiuationa ? The Table ? mte. The unit ? The value of a franc ? WEIGHTS. 225. WeigJit is a measure of the force called gravity^ by which all bodies iend toward the center of the earth. 226. Net Weight is the weight of goods without the bag, cask, or box which contains them. Gross Weight is the weight of goods with the bag, cask, or box in which they are contained. The weights in use are of three kinds, yiz : Troy, Avoir- dupois, and Apothecaries^ Weight. TROY WEIGHT. 227. Troy Weight is employed in weighing gold, silver, und jewels. The denominations ixyq pounds, ounces* pennyioeights, and grains. TABLE. 24 grains {gr) are i pennyweiglit, - - - pwt, 20 pennyweights " i ounce, oz. 12 ounces " i pound, lb. Note. — The unit commonly employed in weighing diamonds, pearls, and otheT jewels, is the carat, which is equal to 4 grains. 228. The Standard Unit of weight in the IJnited States, is the Troy pound, which is equal to 22.794377 cubic inches of distilled water, at its maximum density (39.83° Fahrenheit),* the barometer standing at 30 inches. It is exactly equal to the Imperial Troy pound of England, the former being copied from the latter by Captain Kater.f Note. — The original element of weight is a grain of wheat taken from the middle of the ear or head. Hence the name grain as a unit of weight. 225. What is weight ? 226. Troy weight ? The denominations ? The table ? 227. The standard nnit of weight ? Note. The original element of weight ? 2»8. Avoirdupois weight ? The denominations ? The Table? * Hassler. t Professor A. D. Bache. 176 COMPOUND KUMBEES. AVOIRDUPOIS WEIGHT. 229. Avoirdupois Weight is used in weighing all coarse articles ; as, hay, cotton, meat, groceries, etc., and all metals, except ^o?r 243. How is the area of rectangular surfaces found ? 245. Cubic measure ? The denominations? Table? ^ote. Describe a cord of wood? A cord foot? A register ton ? Shipping ton ? 346. What is a cube ? A cubic inch ? Yard ? 182 COMPOUND IfUMBERS. 247. -^ rectangular body is one bounded by six rectangular sides, eack opposite pair being equal and parallel; as, boxes of goods, blocks of hewn stone, etc. Wtien all the sides are equal, it is called a cuhe ; when the opposite gides only are equal, it is called a parallelopiped. 248. The contents or solidity of a body is the quantity of matter or 5j??ace it contains. 249 The contents of a rectangular solid are found, by multijjlying the length, hreadth, and thichness together. MEASUEES OF OAPAOITT. 250. The capacity of a vessel is the quantity of space included within its limits. Pleasures of Capacity are divided into two classes, iry and liquid measures. DRY MEASURE. 251. l^ry JMeasure is used in measuring grain, fruit, salt, etc. The denominations are chaldrons, bushels, jJGcks, quarts, and^JiVz^^. TABLE. 2 pints {pt.) - - are I quart, (fi. 8 quarts - . - " i peck, ph. 4 pecks, or 32 qts., " i bushel, hu. 36 bushels - - - " I chaldron, .... cli. 252. The Standard Unit of Dry Measure is the bushel, which contains 2150.4 cubic inches, or 77.6274 lbs. avoirdupois of distilled water, at its maximum densit}'. 247. What is a rectangular body ? When all the sides are equal, what called ? When the opposite sides only are equal? 248. What are the contents of a solid holy? 24Q. How find the contents of a rectangular solid? 2co. The caparitv of a vessel? 2i?i. Dry measure? The denominations ? The Table? 252. The standard unit of dry measure? COMPOUND NUMBERS. 183 It is a cylinder i8J in. in diameter, and 8 in. deep, the same as the old Winchester bushel of England.* The British Imperial Bushel contains 2218.192 cu. inches. Notes. — i. The dry quart is equal to i^ liquid quart nearly. 2. Tlie chaldron is used for measuring coke and bituminous coal. 2E3. The Standard Bushel of different kinds of grain, seeds, etc., acc-o^ding to the laws of New York, is equal to the following number of pounds : 32 lbs. = I bu. of oats. 58 lbs. = I bu. of corn. 44 lbs. = I " Timothy seed. ( wheat, peas, „ ,- ,M buckwheat, or 60 lbs. = I " •< potatoes, or 48 lbs. = I "1 barley. ( clover-seed. 55 lbs. E= I " flax-seed. 56 lbs. = I " rye. 62 lbs. 100 lbs. _ u J beans, or blue- "" ( grass seed. = I cental of grain. Notes. — i. The cental is a standard recently recommended by the Boards of Trade in New York, Cincinnati, Chicago, and other large cities, for estimating grain, seeds, etc. Were this standard gen- erally adopted, the discrepancies of the present system of grain deal- ing would be avoided. 2. Bushels are changed to centals, by multiplying them by the nufnber of pounds in one bushel, and dividing the product by 100. The remainder will be hundredths of a cental. LIQUID MEASURE. 254. Liquid Pleasure is used in measuring milk, wine, vinegar, molasses, etc., and is often called Wine Measure. The denominations are liogslieads, larrels, gal- lons, quarts, innts, and gills, TABLE. 4 gills {gi^ are i pint, pt. 2 pints " I quart, qt. 4 quarts " i gallon. ... - . - gal. 31.^ gallons " I barrel, bar. or bbl 63 gallons " I hogshead, - - - - Jihd. 254. Liquid measure ? The denominations ? Table ? * Professor A. D. Bache. 184 COMPOUJS^D LUMBERS. 255. The Standard Unit of Liquid Measure is tha gallon, which contains 231 cubic inches, or 8.338 lbs. avoirdupois of distilled water, at its maximum densit}^. The British Imperial Gallon contains 277.274 cu. inches. Notes. — i. The barrel and hogshead, as units of measure, are chiefly used in estimating the contents of cisterns, reservoirs, etc 2. Beer Measure is practically obsolete in this country. The old beer gallon contained 282 cubic inches. CIRCULAR MEASURE. 256. Circular Pleasure is used in measuring angles, land, latitude and longitude, the motion of the heavenly bodies, etc. It is often called Angular Measure. The denominations are signs, degrees, minutes, and seconds. TABLE. 60 seconds (") are i minute, - - - ' 60 minutes " i degree, - - - o , or deg. 30 degrees " i sign, - - - - s. 12 signs, or 360° " i circumference, - dr. Note. — Signs are used in Astronomy as a measure of the Zodiac 257. A Circle is a plane figure bounded by a curve line, every part of which is equally distant from a point within called the center. The Circumference of a circle is the curve line by which it is bounded. The Diameter is a straight line drawn through the center, ter- minating at each end in the circu7nference. The Radius is a straight line drawn from the center to the circumference, and is equal to half the diameter. An Arc is any part of the circumfer- ence. In the adjacent figure, A D E B F is the circumference ; C the center ; A B the diameter ; C A, C D, C E, etc., are radii ; A D, D E, etc., are arcs. 255. The standard of Liquid meaeure ? Note. What of Beer measure ? 256. In what is Circular measure used ? The denominations ? The Table ? 357. WM 111 a circle ? The circumference ? Diameter ? Radius ? An arc f COMPOUND i^UMBEKS. 18^ 258. A Plane Angle is the quantity of divergence of two straight lines starting from the same point. ^ The Lines wliich form the angle are called the sides, and the point from which they start, the vertex. Thus, A is the ver- tex of the angle B A C, A B and A C the .-^ sides. A ^ ^"- 259. A JPerpendicular is a straight line which meets another straight line so as to make the two adjacent angles equal to '\ each other, as A B C, A B D. J Each of the two lines thus meeting is | perpendicular to the other. ^ CUD 260- A Rif/ht Angle is one of the two equal angles formed by the meeting of two straight lines which are perpendicular to each other. All other angles are called oblique. 260. a. The Measure of an angle is the arc of a circle included between its two sides, as the arc D E, in Fig,, Art. 258. 261. A Degree is one 360th part of the circumference of a circle. It is divided into 60 equal parts, called minutes ^ the minute is divided into 60 seconds, etc. Hence, the length of a degree, minute, etc., varies according to the magnitude of different circles. The length of a degree of longitude at the equator, also the aver- age length of a degree of latitude, adopted by the U. S, Coast Sur- vey, is 69.16 statute miles. At the latitude of 30° it is 59.81 miles, at 60° it is 34.53 miles, and at 90° it is nothing.* 262. A Semi -cir cum fei'ence is half a circumference, or 180°. A Quadrant is onefourth of a circumference, or 90°. 263. If two diameters are drawn perpendicular to each other, they will form four right angles at the center, and divide the cir- cumference into four equal parts. Hence, A right angle contains 90° ; for the quadrant, which measures it, is an arc of 90°. 258. A plane angle? The sides ? The vertex? 259. A perpendicular ? 260. A right angle ? 260, a. What is the measure of an angle ? 261. What is a degree? Upon what does the length of a degree depend ? What its length at the equator ? 262. What is a semi-circumference ? How many degrees does it contain ? A quadrant? 263. If two diameters are drawn perpendicular to each other, what is the result ? How many degrees in a right angle ? * Encyclopedia Britannica. 18G COMPOUND NUMBERS. MEASUREMENT OF TIME. 264. Time is a portion of duration. It is divided into centuries, years, months, iveelcs, days, hours, minutes* and seconds. TABLE. 60 seconds {sec.) are I minute - . - - m. 60 minutes " I hour - . - h. 24 hours " I day - - - d. 7 days I week . - - w. 365 days, or ) ,< 52 w. and I d. i I common year - c.p 366 days " I leap year - - I.y- 12 calendar months {mo.) " I civil year - - y- 100 years I century „ - c. WOTE. — In most business transactions 30 days are considered a monih. Four weeks are sometimes called a lunar month. 265. A Civil Year is the year adopted by govern- ment for the computation of time, and includes both com- mon and leap years as they occur. It is divided into 12 calendar months, as follows : January (Jan.) ist mo., 31 d. r'ebruary (Feb.) 2d " 28 d. March (Mar.) 3d " 31 d. April (Apr.) 4th " 30 d. May (May) 5th " 31 d. June (Jime) 6th " 30 d. July (July) 7th mo., 31 d. Aug-ust (Aug.) 8th " 31 d. September (Sep.) 9th " 30 d. October (Oct.) loth " 31 d. November (Nov.) nth " 30 d. December (Dec.) 12th " 31 d. Notes. — i. The following couplet will aid the learner in remem^ bering the months that have 30 days each : " Thirty days hath September, April, June, aud November." All the other months have 31 days, except February, which in common years has 28 days ; in leap years, 29. 266. Time is naturally divided into days and years. The former are measured by the revolution of the earth on its axis ; the latter by its revolution around the sun. 264. What is time ? The denominations ? The Table f COMPOUifD NUMBERS. 187 267. A Solar year is the time in which the earth, starting from one of the tropics or equinoctial points, re- volves around the sun, and returns to the same point. It is thence called the tropical or equinoctial year, and is equal to z^S^- sh. 48m. 49.7 sec* Note. — i. The excess of tlie solar above tlie common year is 6 hours ot \ of a day, nearly ; hence, in 4 years it amounts to about i day. To provide for this excess, i day is added to every 4th year, which is called Leap year or Bissextile. This additional day is given to February, because it is the shortest month. 2. Leap year is caused by the excess of a solar above a comm&ti year, and is so called because it leaps over the limit, or runs on i day more than a common year. 3. Every year that is exactly divisible by 4, except centenrJal years, is a Leap year ; the others are common years. Thus, 1868, '72, etc., are leap years ; 1869, '70, '71, are common. Every centennial year exactly divisible by 400 is a leap year ; the other centennir.l years are common. Thus, 1600 and 2000 are leap years; 1700, 1800, and 1900 are common. 268. An Apparent Solar Day is the time hetwccn the apparent departure of the sun from a given meridian and his return to it, and is shown by sun dials. A True or Mean solar day is the average length of apparent solar days, and is divided into 24 equal parts, called hours, as shown by a perfect clock. 269. A Civil Day is the day adopted by govern- ments for business purposes, and corresponds with the mean solar day. In most countries it begins and ends at midnight, and is divided into two parts of 12 hours each; the former being designated A. m.; the latter, p. m. Note. A meridian is an imaginary circle on the surface of the earth, passing through the poles, perpendicular to the equator. A. M. is an abbreviation of ante meridies, before midday ; p. m., of post meridies, after midday. 266- How is time natarally divided ? How is the former caused ? The latter ? * Laplace, Somerville, Baily's Tables. 188 COMPOUND l^UMBEES. MISCELLANEOUS TABLES 12 tilings are i dozen. 12 doz. " I gross. £4 sheets are i quire of paper. 20 quires " i ream. 2 leaves are i folio. 4 leaves " i quarto or 4to. 8 leaves " i octavo or 8vo. 12 gross are i great gross. 20 things " I score. 2 reams are i bundle. 5 bundles " i bale. 12 leaves are i duodecimo or i2ma 1 8 leaves " i eighteen mo. 24 leaves " i twenty-four mo. Note. — The terms folio, quarto, octavo, etc., denote the number ©f leaves into which a sheet of paper is folded in making books. 270. Aliquot Parts of a Dollar, or 100 cents. 12^ cents = $^ 10 cents = $iV 8^ cents = $iV 50 cents = %\ 33;^ cents = $;^ 25 cents = $1" 20 cents ■= %s i6| cents = $^ 6^ cents = $1^^ 5 cents == liV 271. Aliquot Parts of a Pound Sterling, loshil. =£,\ 6s. 8d. =-- £:V 5 shil. = £L 4shil. =£^ 3S. 4d. 2S. 6d. IS. 8d. = £ ,V 272. Aliquot Parts of a Pound Avoirdupois. 12 ounces = f pound. 8 " =i " 5^ " -^ " 4 ounces \ pound. 2 " i " I " h " 273. Aliquot Parts of a Year. 9 months = \ year. 8 " = f " 6 " =i " 4 " =^ " 3 months 2 " = t I " = ,J i year. 274. Aliquot Parts of a Month. 5 days = \ mentis. 3 " =-h " 2 " =,V " I " =^cr " 20 days = § month. 15 " =i " 10 " = i " 6 " = ^ " REDUCTION. 275= Hediiction is changing a number from one denomination to another, without altering its value. It is either descending or ascending. Reduction Descending is changing higher de- nominations to loiver ; as, yards to feet, etc. Reduction Ascending is changing lower denomi- nations to higher; as, feet to yards, etc. 276. To reduce Higher Denominations to Lower, 1. How many farthings are there in £23, 7s. sJd. ? Analysis.— Since there arc 20s. in a operation. pound, there must be 20 times as many £23, 78. 5d.. I far. shillings as pounds, p^i^s the given shil- 20 lings. Now 20 times 23 are 460, and "~T~ 460s. + 78. =4673. Again, since there are t I2d. in a shilling, there must be 12 times as many pence as shillings, p?w« the S^'^9 d. given pence. But 12 times 467 are 5604, 4 and '56o4d. + 5d, = 56o9d. Finally, since 22437 f^^* ^ns. there are 4 farthings in a penny, there must be 4 times as many farthings as pence, plus the given farthings. Now 4 times 5609 are 22436, and 22436 far. + 1 far. = 22437 far Therefore, in £23, 7s. s^d. there are 22437 far. Hence, the EuLE. — Multiply the highest denomination hy the numher required of the next lower to make a unit of the higher, and to the product add the lower denorni nation. Proceed in this manner tvith the successive denomina- tions, till the one required is reached. 2. How many pence in £8, los. 7d. ? Ans. 2047d. 3. How many farthings in 12s. gd. 2 far.? 4. How many farthings in £41, 5s. 4icL.? 275. What is Reduction? How many kinds? Descending? Ascendinc? 276. How are higlier denominations reduced to lower? Explain Ex. i from th« blackboard ? 190 REDUCTION. 277. To reduce Lower Denominations to Higher, 5. In 22437 farthings, how many pounds, BhiUin^s, pence, and farthings ? Analysis. — Since in 4 farthings there operation. is I penny, in 22437 farthings there are 4)22437 far. as many pence as 4 farthings are con- i2V^6oQd. I far. tained times in 22437 farthings, or "wT ^ 5609 pence and i farthing over. Again, }^ ' 5 • since in 12 pence there is i shilling, in -£23, 7§. 5609 pence thert^ are as many shillings A71S. £23, 7s. 5 d. I far. as 12 pence are contained times in 5609 pence, or 467 shillings and 5 pence over. Finally, since in 20 shillings there is i pound, in 467 shillings there are as many pounds as 20 shillings are contained times in 467 shillings, or 23 pounds and 7 shillings over. Therefore, in 22437 farthings there are £23, 7s. sd. i far. Hence, the EuLE. — Divide the given denomination by the mimier re- quired of this denomination to mahe a unit of the next higher. Proceed in this manner with the successive denomina- tions, till the one required is reached. Tlie last quotient^ with the several remainders annexed, ivill he the ansiver. Note. — The remainders, it should be observed, are the same denomination as the respective dividends from which they arise. 278. Proof. — Eeduction Ascending and Descending prove each other ; for, one is the reverse of the other. 6. In 2047 pence how many pounds, shilhngs, and pence ? 7. In 614 farthings, how many shillings, pence, etc.? 8. How many pounds, shillings, etc., in 39610 farthings ? 9. An importer paid £27, 13s. 8d. duty on a package of English ginghams, which was 4d. a yard: how many yards did the package contain ? 10. A railroad company employs 1000 men, paying each 4s. 6d. per day : what is the daily pay-roll of the company ? 11. Reduce 18 lbs. 6 ounces troy, to pennyweights. 277. How are lower denominations reduced to higher? 273. Proof? Explain Ex. 5 from the blackboard. EEDUCTlOi^. 191 12. Reduce 32 lbs. 9 oz. 5 pwt. to pennyweights. 13. How many pounds and ounces in 967 ounces troy? 14. How many pounds, etc., in 41250 grains? 15. How many rings, each weighing 3 pwt., can bo made of i lb. 10 oz. 9 pwt. of gold ? 16. What is the worth of a silver cup weighing 10 oz. 16 pwfc., at i2-| cents a pennyweight? 17. Eeduce 165 lbs. 13 oz. Avoirdupois to ounces. 18. Reduce 210 tons 121 pounds 8 ounces to ounces. 19. In 4725 lbs. how many tons and pounds? 20. In 370268 ounces, how many tons, etc.? 21. What will 875 lbs. 1 1 oz. of snuff come to, at 5 cents an ounce ? 22. A grocer bought i ton of maple sugar, at 10 cents a pound, and sold it at 6 cents a cake, each cake weighing 4 oz. : what was his profit ? 23. In 250 lbs. 6 oz. Apothecaries' weight, how many drams ? 24. In 2165 scruples, how many pounds ? 25. How many feet in 1250 rods, 4 yards, and 2 feet? For the metliod of multiplying by 5^ or i6|, see Art. 165. Ans. 20639 feet. 26. How many feet in 365 miles 78 rods and 8 feet? 27. In 5242 feet, how many rods, yards and feet? Remark. — In order to divide the second •5)^242 ft. dividend 1747 yards by 5K the number jv -, ni of yards in a rod, we reduce both the ^^\ '^' ^ divisor and dividend to halves; then divide one by the other. Thus, 5^ = 11 I 03494 half yds, halves ; and 1747=3494 halves ; and 11 is ^/^5. 317 r. 3^ yd. I ft. contained in 3494, 317 times and 7 over. But the remainder 7, is lialxies ; for the dividend was halves; and 7 half yards=3^- yards. (Art. 168, a, 169.) 2^. Reduce 38265 feet to miles, etc. 29. How many rods in 461 leagues, 2 miles, 6 furlongs? 192 REDUCTION. 30. Reduce 7 m. 6 fur. 23 r. 5 yds. 8 in. to inches, and prove the operiition. 31. Reduce i m. 7 fur. 39 r. 5 yds. i ft. 6 in. to inches, and prove the operation. ^32. If a farm is 2 J miles in circumference, what will it cost to enclose it with a stone wall, at $1.85 a rod? ^^. At 6J cents a mile, what will be the cost of a trip round the world, allowing it to be 8299.2 leagues ? 34. How many eighths of a yard in a piece of cloth 57 yards long? 35. How many sixteenths of a yard in 163 yards ? ^6. In 578 fourths, how many yards? 37. In 1978 sixteenths, how many yards? 38. How many vests will 16J yards of satin make, allow- ing J yard to a vest ? 39. A shopkeeper paid $2.50 for 18J yards ribbon, and making it into temperance badges of ^ yard each, sold them at 12 J cents apiece : what was his profit ? 40. How many square rods in 43816 square feet? 41. How many acres, etc., in 25430 square yards ? 42. Reduce 160 acres, 25 sq. rods, 8 sq. ft. to sq. feet. 43. Reduce 100 sq. miles to square rods? 44. A man having 64 acres 8 sq. rods of land, divided it into 6 equal pastures: how much land was there in each pasture ? 45. My neighbor bought a tract of land containing 15J acres, at I500 an acre, and dividing it into building lots of 20 square rods each, sold them at $250 apiece: how much did he make ? 46. Reduce 85 cubic yards, 10 cu. feet to cu. inches. 47. Reduce 250 cords of wood to cu. feet. 48. Reduce 18265 cu. inches to cu. feet, etc. 49. Reduce 8278 cu. feet to cords. 50. Reduce 164 bu. i pk. 3 qts. to quarts. 51. Reduce 375 bu. 3 pks. i qt. i pi to pints. KEDUCTI02 result is 3 qts. and .57500 qt. In like " ' manner, we multiply the decimal .57500 — . qt. by 2 to reduce it to pints, and the .60006 gl. result is i pt. and .15000 pt. Finally, we Ans. 3 qt. i pt. 0.6 ^i multiply the decimal .15000 pt. by 4 to reduce it to gills, and the result is .6 gill. Therefore, .89375 gal. equals 3 qt. i pt. 0.6 gi. Hence, the Rule. — Multiply the denominate decimal ly the mi^nher required of the next loiver denomi7iation to make one of the given denomination^ and point off the product as in multi^ plication of decimals. (Arts. 191, 276.) Proceed in this manner ivith the decimal part of the suc- cessive ^products, as far as required. The integral part of the several products ivill he the a^iswer. 34. What is the value of £.125445 in shillings, etc. ? 35. "What is the value of .91225 of a Troy pound ? 36. Reduce .35 mile to rods, etc. 37. How many quarts and pints in .625 of a gallon ? 2,2>. How many minutes and seconds in .651 degree? / 39. How many days, hours, etc., in .241 week ? 4c. What is the value of .25256 ton ? 41. What is the value of .003 of a Troy pound? 42. What is the value of £5.62542 ? 204. How are denominate decimals rednred from higher denominations to whole numbers of a lower denomination ? Explain Ex. 33 upon the blackbcfiro 206 DEK0 3IIKATE DECIMALS. CASE VI. 295. To reduce a Compound Number from lower to a Doiiominate Decimal of a higher denomination. 43. Eeduce 3 pk. 2 qt. i j)t. to the decimal of a bushel Analysis. — 2 pints make i quart ; hence operation. ipt. 2^t. 3-3125 pk. tliere is \ as many quarts as pints, and i of 2 I qt. is .5 qt., wliicli we write as a decimal on o the right of the given quarts. In like manner there is \ as many pecks as quarts, and \ of 4 2.5 is .3125 pk., which we place on the right Ans. .8281*25 bu. of the given pecks. Finally, there is | as many bushels as pecks, and \ of 3.3125 is .828125 bu. Therefore, 3 pk. 2 qt. I pt. equals .828125 bushel. Hence, the EuLE. — Write the numbers in a column, inlacing the lowest denomination at the top. Beginning with the lowest, divide it hy the numler re- quired of this denomination to make a unit of the next higher, and annex the quotient to the next denomination. Proceed in this manner with the successive denomina- tions, till the one required is reached. 44. Eeduce 6 oz. 10 pwt. 4 gr. to the decimal of a pound, 45. Eeduce 10 lb. to the decimal of a ton. 46. Eeduce 3 fur. 25 r. 4 yd. to the decimal of a mile. 47. Eeduce 9.6 pwt. to the decimal of a Troy pound. What decimal part of a barrel are 15 gal. 3 qt. 49. What decimal part of 2 rods are 2\ fathoms ? 50. What decimal part of £3, 3s. 6d. are 15s. io.5d. ; 51. What part of a barrel are 94.08 lbs. of flour? 52. Eeduce 7.92 yds. to the decimal of a rod. 53. Eeduce i day 4 hr. 10 sec. to the decimal of a w< 54. Eeduce 45 sq. rods 25 sq. ft. to the decimal of an a 55. Eeduce 53!^ cu. feet to the decimal of a cord. \)^ 295. How are compound numbers reduced from lower denominationa denominate decimals of a higher? Explain Ex. 43 upon the blackboard? METEIO WEIGHTS AND MEASURES, 296. Metric Weights and Pleasures are founded upon the decimal notation, and are so called because their primary unit or base is the 3Ieier.* 297. The Meter is the miit of le^igth, and is equal to one ten-millionth part of the distance on the earth's surface from the equator to the pole, or 39.37 inches nearly. Notes. — i. The term metei' is from the Greek metron, a measure. 2. The standard meter is a har of platinum deposited in the archives of Paris. 298. The Metric System employs /z^e different units or denominations and seven prefixes. The units are the me'ter, U'ter, gram, ar, and ster.^ 299. The names of the higher denominations are formed by prefixing to the unit the Greek numerals, deJc'a 10, hek'to 100, JciTo 1000, and myr'ia loooo ; as, deh'a-me'ter, 10 meters; hek'to-me'ter, 100 meters, etc. The names of the lower denominations, or divisions of the unit, are formed by prefixing to the unit the Latin numerals, dec'i (des'-ee) .1, cen'ti (cent'-ee) .01, and mil'li (mil'-lee) .001; as, dec'i-me'ter, y\j meter; cen'ti-me'ter, yj^ meter ; mil'li-me'ter, -j^^^ meter. Note. — The numeral prefixes are the key to the whole system ; their meaning therefore should be thoroughly understood. 296. Upon what are the Metric Weights and Measures founded ? Why so called ? 297. What is the meter ? 299. How are the names of the lower denom- inations formed ? The higher ? * This system had its origin in France, near the close of the last century. Its simplicity and comprehensiveness have secured its adoption in nearly all the countries of Europe and South America. Its use was legalized in Great Britain in 1864 ; and in the United States, by Act of Congress, 1866. t The spelling, pronunciation, and abbreviation of metric terms in this work, are the same as adopted by the American Metric Bureau, Boston, and the Metrological Society, New York. lo cen'ti-me'ters lo decl-me ters lo meters lo dek'a-me'ters lo hek'to-me ters 10 kilo-meters 208 METKIC OR DECIMAL LINEAR MEASURE. 300. The TJnit of Length is the Meter, whicli is equal to 39.37 inclies.* The denominations are the mil'li-rne'terf cen'ti-rnetery dec'i-me'ter, meter, deh'a-meter, hek' to-me' ter , kil'o-me'tery and myria-mdter. TABLE. ID mil'li-me'ters {mm.) make i cen'ti-me'ter - - mi. decimeter - - dm. 3IE'TEll - - m. dek'a-meter - - Dm. hek'to-me'ter - - Hm. Ml' o-me' ter - - Km. myr'ia-me'ter - Mm. Notes. — i. The Accent of each unit and jirejix is on the first syllable, and remains so in the compound words. To abbreviate the compounds, pronounce only the prefix and the first letterof the unit ; as, centim, millim, centil, decig, hektug, etc. 2, The meter, like our yard, is used in measuring cloths, laces, moderate distances, etc. For long distances the kilometer (3280 ft. 10 in.) is used ; and for rninute measurements, the centimeter or millimeter. 3. Decimeters, dekameters, hektometers, like our dimes and eagles, are seldom used. SQUARE MEASURE. 301. The Unit for measuring surfaces is the Square Meter, which is equal to 1550 sq. in. The denominations are the sq. cen'ti-yne'ter, sq. dec'i- me'ter, and sq. me' ter. 100 sq. cen'tim. make i sq. dec'i-me'ter - - sq. dm. 100 sq. dec'im. *' i Sq. 3Ie'ter - - sq. m. Note. — The square meter is used in measuring floorings, ceilings, etc. ; square deci-meters and centi-meters, for minute surfaces. 300. What is the unit of Linear Measure ? Its denominations ? Recite the Table. 301. What is the unit for measuring surfaces ? Recite the table. * Authorizetl by Act of Con^rcs.^^, 1866. WEIGHTS AND MEASURES. 209 302. The JJtiit for measuring land is the Ar, which is equal to a square dekameter, or 119.6 sq. yards. The only suMivision of the Ar is the cen'tar ; and the only multiple is the heJc'tar. Thus, 100 cent'ars (ca.) make i Ar - - - a. 100 ars " I hek'tar - - Ha. Notes. — i. The term ar is from the Latin area, a surface. 2. In Square Measure, it takes 100 units of a /oip^r denomination to make one in the next higher ; hence each denomination must have two places of figures. In this respect centars correspond to cents. CUBIC MEASURE. 303. The Unit for measuring solids is the CuMc Meter, which is equal to 35.316 cu. ft. TABLE. 1000 cu. mil'lim. make i cu. cen'ti me'ter - cu. cm. 1000 cu. cen'tim. " i cu. dec'i me'ter - eu. dm. 1000 cu. dec'im. " i Cu. Me'ter - cu. m. Notes. — i. The cubic meter is used in measuring embankments, excavations, etc. ; cubic centimeters and millimeters ^iov minute bodies. 2. Since it takes 1000 units of a lower denomination in cubic measure to make one of the next higher, it is plain that, like mills, each denomination requires three places of figures. 304. In measuring wood, the Ster, which is equal to a cuhic meter, is sometimes used. The only subdivision of the ster is the dec'i-ster ; and the only multiple, the deh'a-ster. Thus, 10 dec'i-sters make i Ster - - - 8. 10 sters " I dek'a-ster - Bs. Notes. — i. The term ster is from the Greek stereos, a solid. 2. In France, fireicood is commonly measured in a cubical box, whose length, breadth, and height are each i meter. 3. As the ster is applied only to wood, and probably will never come into general use, its divisions and multiples may be omitted. Note. How many places of figure? does each denomination occupy ? Wl y ? 302. What is the unit for measuring Iniid? Its divisions? Its miil(i)]es? 303. What are the units for measuring solids ? Recite the Table. How many places does each denomination in cubic measure occupy ? Why ? 210 METRIC OR DECIMAL DRY AND LIQUID MEASURE. 305. The Unit of Dry and Liquid 3l€a8iire is the Li'ter {lee'ter), which is equal to a cuMc decim, or t.0567 liquid quart or 0.908 dry quart. The denominations are the mil'li-U'ter, cen'ti-li'ter, dec'i-li'ter, li'ter, deh'a-li'ter, hek'to-U'ter, kil'o-liter. 10 mil'li-li'ters {mi) make i cen'ti-li'ter 10 cen'ti-li'ters 10 dec'i-li'ters 10 li'ters 10 dek'a-li'ters 10 hek to-liter 10 kilo li ters d. dL I. DL HI. KL ML dec'i-li'ter - LVter - - dek'a-liter - hek'to-li'ter kil'o-li'ter - myria-li'ter Notes. — i. The liter is used in measuring milk, wine, etc. For small quantities, the centiliter and milliliter are employed ; and for large quantities, the dekaliter. 2. For measuring grain, etc., the heJctoliter, which is equal to 2.8375 bushels, is commonly used. 3. The term liter is from the Greek litra, a pound. WEIGHT. 306. The Unit of Weight is the Gram, which is equal to 15.432 grains. The denominations .are the mil'li-gram, ceji'ti-gram, dec'i-gram, gram, deh'a-gram, Tiehto-gram, kW o-gram, ynyr'ia-gram, and ton'neau or ton. 10 milligrams {mg.) make i cen'ti-gram - eg. 10 centigrams I decigram - - dg. 10 dec'i-grams I Gram - - - g- 10 grams I dek'a-gram - - Bg. 10 dek'a-grams 1 hek'to-gram - - Hg. 10 hek to-grams I KiUo-(fr In days ! 222 COMPOUls^D SUBTE ACTION. 2. What is the number of days between Xov. loth, 1869, and March 3d, 1870 ? 3. A person started on a journey Aug. 19th, 1869, and returned Nov. ist, 1869: how long was he absent? ; 4. A note dated Jan. 31st, 1870, was paid June 30th, 1870: how many days did it run ? 5. How many days from May 21st, 1868, to Dec. 31st, following ? 6. Tiie building of a school-house was commenced April I st, and completed on the loth of July following : how long was it in building ? 320. To find the difference of Latitude or Longitude. Def. t. — Latitude is distance from the Equator. It is reckoned in degrees, minutes, etc., and is called North or South latitude, ac- cording as it is north or south of the equator. 2. Longitiide is the distance on the equator between a convene tional or fixed meridian and the meridian of a given place. It is reckoned in degrees, minutes, etc., and is called East and West longi- tude, according as the place is east or icest of the fixed meridian, lentil 180^ or half the circumference of the earth is reached. 7. The latitude of JSTew York is 40° 42' 43" N. ; that of !New Orleans, 29° 57' 30" N.: how much further north is New York than New Orleans ? * Analysis. — Placing the lower latitude [f^. Y. 40° 42' 43'' under the higher, and subtracting as in the ^. Q. 29° 57' 30" preceding rule, the Ans. is 10° 45' 13". J^^g^ 'io°'45'"i3^ 8. The latitude of Cape Horn is 55° 59' S., and that of Cape Cod 42° i' 57.1" N. : what is the difference of lati- tude between them ? Analysis. — As one of these capes is north of the equator and the other south, it is plain that the difference of their latitude is the sum of the two distances from the equator. We therefore add the two latitudes together, and the result is 98° o' 57 i'. 320. Z>«/". What 19 latitude? Longitude? How find the difference of latitude or lono;itucle between two })laces ? * The latitude and longitude of the places in the United States here given ar« taken from the American Almanac, 1854; those of places in foreign lands, mostly £ro;ii Bo'^ditch's Navigator. COMPOUND SUBTRACTION. 223 9. The longitude of Paris is 2° 20' East from Green- wich; that of Dublin is 6" 20' 30" West: what is the dif- ference in their longitude ? Analysis. — As tlie longitude of one of tliese places 2° 20' 00'' is East from Greenwich the standard meridian, and 6° 20' 30" that of the other West, the difference of their longi- '^'aq'^-^q'' tude must be the sum of the two distances from the '^ standard meridian. Ans. 8 ' 40' 30". Hence, the Rule. — I. If hoth places are the same side of the equator, or the standard meridian, subtract the less latitude or longitude from the greater. II. If the places are on different sides of the equator, or the standard meridian, add the two latitudes or longitudes together, and the sum loill he the ansicer. ro. The longitude of Berlin is 13° 24' E. ;* that of Xew Haven, Ct., 72° 55' 24'' W. : what is the difference ? 11. The longitude of Cambridge, Mass., is 71° 7' 22"; that of Charlottesville, Va., 78° 31' 29": what is the difference ? 12. The latitude of St. Petersburg is 59° 56' north, that of Rome 41° 54' north : what is their difference ? 13. The latitude of Albany is 4?° 2>9' ^"y ^^^t of Rich- mond 37° 32' 17": what is the difference? 14. The latitude of St. Augustine, Flor., is 29° 48' 30"; that of St. Paul, Min., is'44° 52' 46": what is the dif- ference ? 15. The longitude of Edinburgh is 3° 12' W. ; that of Vienna 16° 23' E.f : required their difference ? 16. The latitude of Valparaiso is 2>2>'' 2' S. ; that of Ha- vana is 23° 9' jST.: what is the difference? 17. The latitude of Cape of Good Hope is 34° 22' S. ; that of Gibraltar, 36° 7' N. : what is their difference? 18. The longitude of St. Louis is 90° 15' 16"; that of Charleston, S. C, 79° 55' 38": what is the difference? * Encyc. Brit. ■*■ Encyc. Amer. COMPOUND MULTIPLICATION. 321. To multiply Compound Numbers. I. A farmer raised 6 acres of wheat, which yielded 15 hn. 3 pk. I qt. per acre: how much wheat had he ? Analysis. — 6 acres will produce 6 times as opekatioic. much as i acre. Beginning at tlie right, 6 times ^^- P^- ^*- 1 qt. are 6 qts. As 6 quarts are less than a ^'^ "^ a peck, the next higher denomination, we set the 1 6 under the term multiplied. 6 times 3 pk. are A'^S. 94 2 6 18 pk. Since 4 pk. = i bu., 18 pecks=4 bu. and 2 pk. over. Setting the remainder 2 under the term multiplied, and carrying the 4 bu. to the next product, we have 6 times 15 bu.-go bu., and 4 bu. make 94 bu. Therefore, etc. Hence, the KuLE. — I. Write the multiplier under the lowest denomi- nation of the multiplicand, and, leginning at the rights multiply each term in succession. II. If the product of any term is less than a itnit of the next higher denomination, set it under the term multiplied. III. If equul to one or more units of the next higher de- nomination, carry these units to that denomination, and write the excess under the term multiplied. Notes. — i. If the multiplier is a composite number, multiply by one of the factors, then this partial product by another, and so on. 2. If a fraction occurs in the product of any denomination except the lowest, it should be reduced to lower deiiominations, and be united to those of the same name as in Compound Addition. (Art. 3 1 4) (2.) (3.) Mult. 12 T. 7 cwt. 16 lb. £21, 13s. 8Jd. By 8 7_ |. What is the weight of 10 silver spoons, each weigh- ing- 3 oz. 7 pwt. 13 gr.? 321. How are compound numbers multiplied? Nfce. If the multiplier is a composite number, how proceed ? When a fraction occurs In ku; denoninatiop except the last, how ? COMPOUND MULTITLICATIOX. 226 (5-) (6-) Mult. 9 oz. 13 pwt. 7 gr. by i8. Mult. lo r. i yd. i ft. by 7. 9 oz. 13 pwt. 7 gr. 10 r. I yd. i ft. o in. 3 7 2 lb. 4 19 21 71 3^ I 6 Jycl.= I 6 14 lb. 5 oz. 19 pwt. 6 gr. ^?^5. ^?i5. 71 r. 3 yd. 2 ft. 6 in. 7. If a family use 27 gal. 2 qt. i pt. of milk in a montli, how much will they use in a year ? 8. If a man chops 2 cords 67 cu. feet of wood per day, how much will he chop in 9 days ? 9. What cost 27 yards of silk, at 17s. 7jd. sterling per yard? 10. If a railroad train goes at the rate of 23 m. 3 fur. 2 1 r. an hour, how far will it go in 24 hours ? 11. How much corn will 6^ acres of land produce, at 30 bu. 3 pk. per acre ? 12. How many cords of wood in 17 loads, each contain- ing I cord 41 cu. ft. ? 13. How much hay in 12 stacks of 5 tons, 237 lbs. each ? 14. How much paper is required to print 20 editions of a book, requiring 65 reams, 7 quires, and 10 sheets each ? 15. If a meteor moves through 5° 2;^' 15'' in a second, how far will it move in 30 seconds ? 16. If the daily session of a school is 5 h. 45 min., how many school hours in a term of 15 weeks of 5 days each ? 17. A man has 11 village lots, each containing 12 sq. r. 4 sq. yd. 6 sq. ft. : how much do all contain ? 18. If I load of coal weighs i T. 48^ lb., what will 72 loads weigh ? 19. How many bushels of corn in 12 bins, each con- taining 130 bu. 3 pk. and 7 qt. ? 20. A grocer bought 35 casks of molasses, each contain* ing 55 gal. 2 qt. i pt. : how much did they all contain? COMPOU]S"D DIYISIOK 322. Division of Compound Nuinbers, like Simple Division, embraces tico classes of problems : First. — Those in wliicli the dividend is a compound number, and the divisor is an abstract number. Second. — Those in which both the divisor and dimdend are com- pound numbers. In the former the quotient is a compound number^ In the latter, it is times, or an abstract number. (Art. 64.) 323. To divide one Compound Number by another, or by an Abstract number. Ex. I. A dairy-woman packed 94 lb. 2 oz. of butter in 6 equal jars: liow much did each jar contain ?• Analysis. — The number of parts is given, opeeation. to find the value of each part. Since 6 jars 6)94 1^- 2 OZ. c:)ntain 94 lbs. 2 oz., i jar must contain \ Ans. iq lb. 11 OZ. of 94 lbs. 2 oz. Now \ of 94 lbs. is 15 lbs. and 4 lbs. remainder. Reducing the remainder to oz., and adding the 2 oz., we have 66 oz. Now ^ of 66 oz. is 11 oz. (Art. 63, b) 2. A dairy-woman packed 94 lbs. 2 oz. of butter in jars of 15 lbs. II oz. each : how many jars did she have? Analysts. — Here the size of each 94 lb. 2 oz. = 1506 oz. part is given, to find the number of 1 5 lb. 1 1 oz. — 251 oz. parts in q4 lb. 2 oz. Reduce both ^ OZ.)ic;o6 oz numbers to oz., and divide as in sitn- ^ c. • ' pie numbers. (Art. 63, a.) Hence, the ^^^- 6 jars. fiuLE. — I. When the divisor is an abstract number, Begiyming at the left, divide each de}iommation in suc- cessio7i, and set the quotient under the term divided. If there is a remainder, reduce it to the next lower de- nomination, and, adding it to the given units of this de- nomination, divide as before. 11. When the divisor is a compound number, Reduce the divisor and dividend to the loivest denomina- tion contained in either, and divide as in simple numbers. 322. How many classes of examples does division of compound numbers em bi-ace? The first? The second? 323. What i? the rule? TIME AIN'D LONGITUDE. 227 Note. — If the divisor is a composite number, we may diride by ite factors, as in simple numbers. (Art. 77.) 3. Divide 29 fur. 19 r. 2 yd. i ft by 7. 4. Divide 54 gal/ 3 qt. i pt. 3 gi. by 8. 5. A miller stored 450 bu. 3 pks. of grain in 18 equal bins : how much did he put in a bin ? 6. A farm of 360 A. 42 sq. r. is divided into 23 equal pastures: how much land does each contain? 7. How many spoons, each weighing 2 oz. 10 pwt., can be made out of 5 lb. 6 oz. of silver ? 8. How many iron rails, 1 8 ft. long, are required for a railroad track 15 miles in length? 9. How many times does a car-wheel 15 ft. 6 in. in cir- cumference turn round in 3 m. 25 r. 10 ft. ? 10. How many books, at 4s. 6Jd. apiece, can you buy for £2, 14s. 3d.? 11. If 6 men mow ^6 A. 64 sq. rods in 6 days, how much will I man mow in i day ? 12. A farmer gathered 150 bu. 3 pk. of apples from 24 trees : what was the average per tree ? COMPARISON OF TIME AND LONGITUDE. 324. The Earth makes a revolution on its axis once in 24 hours ; hence ^V part of its circumference must pass under the sun in I hour. But the circumference of every circle is divided into 360^, and -/4 of 360'' is 15°. It follows, therefore, that 15° of longi- tude make a difference of i hour in time. Again, since 15° of longitude make a difference of i hour in time, 15' of longitude {-^c^ of 15") will make a difference of i minute {-^^ of an hour) in time. In like manner, 15" of longitude (^Hr of 15'), will make a difference of I second in time. Hence, the following TABLE. 15° of longitude are equivalent to i hour of time. 15 " " " I minute " i.«5" '* " " I second 228 TIME AXD LONGITUDE. CASE I. 325. To find the Difference of Time between two^ places, the difference of Longitude being given. Ex. I. Tlie difference of longitude between 'New York and London is 73° 54' 3": what is the difference of time? Analysis. — Since 15° of Ion. are equiv- Operation. alent to i hour of time, the difference of 15 ) 73^ 54 3'' time must be -rs part as many hours, min- . T „ y- utes, and seconds as there are degrees, -4 00 • 3 • etc., in the dif. of Ion. ; and 73^ 54' 3"-t-i5=:4 h. 55' 36.2" Hence, the Rule. — Divide the difference of longitude by 15, and the degrees, minutes, and seconds of the quotient will le the diff'erence of time in hours, minutes, and secojids. (Art. 323.) 2. The difference of longitude between Savannah, Ga., and Portland, Me., is 10° 53' 2": what is the difference in time? 3. The longitude of Boston is 71° 3' 30'^ W., that of Detroit is 83° 2' 30" W.: when it is noon in Boston what is the time at Detroit ? 4. The longitude of Philadelphia is 75° 9' 54", that of Cincinnati 84° 27': when it is noon at Cincinnati what is the time at Philadelphia ? 5. The Ion. of Louisville, Ky., is 8s° 30', that of Bur- lington, Vt, 73° 10' : what is the difference in time ? 6. When it is noon at Washington, wliat is the time of day at all places 22° 30' east of it? What, at all places 22° 30' west of it ? 7. How much earlier does the sun rise in New York, Ion. 74° 3", than at Chicago, Ion. 87° 35'? 8. How much later does the sun set at St. Louis, whose longitude is 90° 15' 16" W^, than at Nashville, Tenn., whose longitude is 86° 49' 3" ? 325. How find the diflFerence of time between two places, the difference ol longritnde hein^f i^iven ? I tlME AJfD LONGITUDE. ?>39 CASE II. 326. To find the Difference of Longitude between two places, the difference of Time being given. 9. A whaleman wrecked on an Island in the Pacific, found that the difference of time between the Island and San Francisco was 2 hr. 27 min. 54! sec: how many de- grees of longitude was he from San Francisco? Analysts. — Since 15° of Ion. are equira Operatiok. lent to I hour of time, 15' of Ion. to i min. 2 h. 27 m. 54.6 seC. of time, and 15" to i sec. of time, there must j r be 15 times as many de^rrees, minutes, and A /COO''/ seconds in the difference of longitude as -^^•^* 3" 5^ 39 there are hours, minutes, and seconds in the difference of time ; and 2 h. 27 m. 54.6 s. x 15=36° 58' 39". Hence, the 'Rule.— Multiply the difference of time hij 15, ^nd the hours, minutes, and seconds of the product tvill he the difference of longitude in degrees, minutes, and seconds. TO. The difference of time between Eichmond, Ya., and JSTewport, E. I., is 24 min. 2>^ sec. : what is the difference of longitude ? 1 1. The difference of time between Mobile and Galveston is 27 rain. | sec: what is the difference of longitude? 12. The difference of time between Washington and San Francisco is 3 hr. i min. 39 sec: what is the dif- ference in longitude ? 13. The distance from Albany, N. Y., to Milwaukee, is nearly 625 miles, and a degree of longitude at these places is about 44 miles : how much faster is the time at Albany than at Milwaukee ? 14. The distance from Trenton, N. J, to Columbus, 0., is nearly 400 miles, and a degree of longitude at these places is about 46 miles: when it is noon at Columbus what is the time at Trenton ? 526. How find the difference of longitude, when the difference of lime 's glvwfl" PERCENTAGE. 327 JPer Cent and Unte Per Cent denote hun- dredths. Thu^, i per cent of a number is yjo part of that number ; 3 per cent, y^, &c. 328. I^ercentage U the resuU obtained by finding a certain per cent of a number. Note. — The term per cent, is from the Latin per, by and centum, hundred. NOTATION OF PER CENT. 329. The Sign of JPer Cent is an oUiqiie line between two ciphers {%) ; as 3^, 15^. Note. — The sign (%), is a modification of the sign division (-r-), the denominator 100 being miderstood. Thus 5 % =-1^0 = 5-^100. 330. Since per cent denotes a certain 7;ar^ of a hundred, it may obviously be expressed either by a common fraction, wliose denominator is 100, or by decimals, as seen in the following TABLE. 3 per cent is written .03 7 per cent " .07 10 per cent " .10 25 per cent " .25 50 per cent *' .50 100 per cent *' i.oo 125 per cent *' 1.25 300 per cent " 3.00 \ per cent is written .ocf^ \ per cent " ,0025 f per cent " -0075 f per cent " .006 2^ per cent " .025 7 1 per cent " .074 31^ per cent " -3125 112^ per cent " 1.125 Notes. — i. Since hundredtJis occupy two decimal places, it follows that every per cent requires, at least, two decimal figures. Hence, 327. What do the terms per cent and rate per cent denote ? 328. What is per- centagre? Note. From what, are the terms per cent and percentage derived? 320. What is the sign of per cent? 330. How may per cent he expressed? liote. How many decimnl places does it require? Why? If the given per cent \8 less than 10, what is to be done ? Wliat is 100 per cent of a numher ? PERCEN-TAGE. 231 \f the given per cent is less than lo, a cipher must be prefixed to the figure denoting it. Thus, 2 ^ is written .02 ; 6 %, .06, etc. 2. A hundred per cent of a number is equal to the number itself', for |§iT is equal to i. Hence, 100 per cent is commonly written i.oo. If the given per cent is 100 or over, it may be expressed by an integer, a mixed number, or an improper fraction. Thus, 125 pei cent is written 125 %, 1.25, or |§f,. Hence, 331. To express Per Cent, Decimally, Write the figures denoting tlie per cent in the first two places on the right of the decimal point ; and those denoting parts of I jyer cent, in the succeeding places toward the right. Notes. — i. When a given part of i per cent cannot be exactly expressed by one or two decimal figures, it is generally written as a common fraction, and annexed to the figures expressing the integral per cent. Thus, i\\fo is written .04^, instead of .043333 + . 2. In exi^ressing per cent, when the decimal point is used, the words per cent and the sign (%)must be omitted, and mee versa. Thus. .05 denotes 5 per cent, and is eq-ual to too or 2^1 \ t>ut .05 per cent or .05 and is equal to tijuott or 2inro. Express the following per cents, decimally: 1. 2%, 6%, Z% 14;^, 20^, 35^, 60^, 72^. 2. Zo%, \o\%, 104%, 150^, 210^, 300^. 3. life, 4l% H%, H%, loifc 332. To read any given Per Cent, expressed Decimally. Read the first two decimal figures as per cent; and those on the light as decimal parts of i per cent. Note. — Parts of i per cent, when easily reduced to a common fraction, are often read as such. Thus .105 is read 10 and a half per cent ; .0125 is read one and a quarter per cent. Read the folio v/ing as rates per cent : 4. .05; .07; .09; .045; .0625; .1875; -125; .165; .27. 5. .10; .17; .0825; .05125; .ssl; .i6|; .75375- 6. i.co; 1.06; 2.50; 3.00; 1. 125; 1.0725; 1.83J. 331. How express per cent, decimally? Note. When a part of i per cent, can- not be exactly expressed by one or two decimal figures, how is it commonly written ? 332. How read a given per cent, expressed decimally ? S'32 PERCENTAGE. 333. To change a given Per Cent from a Decimal to a Common Fraction. 7. Change 5^ to a common fraction. Ans. xSry^ A- 8. Change .045 to a common fraction. A?is. -^q. Rule. — Erase the decimal point or sign of i^er ceyit (%), and supply the required denominator. (Art. 179.) Note. — When a decimal per cent is reduced to a common fraction, then to its lowest terms, this fraction, it should be observed, will express an equivalent rate, but not the rate per cent. Change the following per cents to common fractions : 9. 5 percent; 10^; 4^; 207^; 25^; 50^; 75>t. 10. 6i per cent; 12!^; ^%^, S3i%'y 62^^. / \^ i pel' cent; i%y \%\ ifc, i%] Wo', -hi\ 25^. 334. To change a common Fraction to an equivalent Per Cent. 12. To what per cent is \ of a number equal ? Analysis. — Per cent dienoie^ hundredths. The J. m 1.00-^3 question then is, how is ^ reduced to hundredths? i.oo— -j -= - •j ^ Annexing ciphers to the numerator, and dividing by the denominator, we have ^ = 1.00-4-3 or .33.^. Hence, the Rule. — An7iex two ciphers to the numerator, aiid divide ly the denominator. (Art. 186.) 13. To what ^ is i equal ? J? f? i? |? |? f? 14. To what % is ^\ equal? tV? ^V ^V? 2V? 2V? A? 15. To whatsis f equal? i? i? |? i? tV? «? 335. In calculations of Percentage, four elements 01 'parts are to be considered, viz. : the hase, the rate per cent, {he. percentage, and the amount. 1. The base is the number on which the percentage is calculated. 2. Tlie ra^^ per ccn^ is the number which shows how many hundredths of the base are to be taken. 331. How change a given per cent from a decimal to a common fraction ? 334. How change a common fraction to an equal per cent ? 335. How many parlf &re to be considered in calculptions by percenta;;e? PERCE XT AGE. 333 3. The percentage is the number obtained by taking that portion of the base indicated by the rate per cent. 4. The amount is the base plus, or minus the percentage. The relation between these parts is such, that if any two of them are given, the other two may be found. Notes. — i. The term amount, it will be observed, is here employed in a modified or enlarged sense, as in algebra and other departments of mathematics. This avoids the necessity of an extra rule to meet the cases in which the final result is less than the base. 2. The conditions of the question show whether the percentage ia to be added to, or subtracted from the base to form the amount. 3. The learner should be careful to observe the distinction between percentage and per cent, or rate per cent. Percentage is properly a product, of which the given per cent or rate per cent, is one of i'he factors, and the base the other. This cai^ is the more necessary as these terms are often used indiscriminately. 4. The terms |7er cent, rate per cent, and rate, are commonly used as synonymous, unless otherwise mentioned. PROBLEM I. 336. To find the Percentage, the Base and Rate being given. Ex. I. What is 5 per cent of $600 ? Analysis. — 5 per ceift is .05 ; therefore 5 per $600 B. cent of a number is the same as .05 times that .05 K. number. Multiplying the base, $600, by the rate J[^g^ ~|^^o!oo P. .05, and pointing off the product as in multipli- cation of decimals, the result is $30. (Art. 191.) Hence, the Rule. — Multiply the base ly the rate, exjjressed deci- mally. Formula. Percentage — Base x Pate. Notes. — i. When the rate is an aliquot Y>SiTt of 100, the percentage may be found by taking a like part of the base. Thus, for 20;^, take i ; for 25%, take \, etc. (Arts. 105, 270.) 2. When the base is a compound number, the lower denominations should be reduced to a decimal of the highest ; or the higher to the lowest denomination mentioned ; then apply the rule. 3. Finding a per cent of a number is the same as finding ^frac^ tionaZ part of it, etc. The pupil is recommended to review with care. Arts. 143, 165, 191. Explain them. What is the relation of these parts ? The diflFerence between per- centage and per cent ? 234 PERCENTAGE. 3. s% of I807 ? II. sifc of 1000 men ? 4. 5^ of 216 bushels? 12. loj^ of 1428 meters? 5. 8^ of 282.5 yds. ? 13. 50$ of $1715.57? 6. 4^ of 216 oxen? 14. ^^ of £21.2 ? 7. 5}^ of 150 yards ? 15. J;^ of 500 liters? 8. 16% of $72.40? 16. i% of 230 kilograms? 9. 1 2% of 840 lbs. ? 1 7. 100;^ of 840 pounds ? 10. 14^^ of 451 tons? 18. 200^ of $500? 19. A farmer raised 875 bu. of corn, and sold g^ of it : how many bushels did he sell ? 7 ^ . V ^ 4hv 20. The gold used for coinage contains 10^^ of alloy: how much alloy is there in 3 1 pounds of standard gold ? > ^ ^ 21. A man having a hogshead of cider, lost 15^5^ of it by leakage: how many gallons did he lose? ^ 22. A garrison containing 4000 soldiers lost 21^ of them by sickness and desertion : what was the number lost ? 23. A grocer having 1925 pounds of sugar, sold 12J per cent of it : how many pounds did he sell ? ANAiiYSls.— 12^ fo is i of 1005, and 100% of operation. a number is equal to the number itself ; there- 8)1925 Ibs. fore iih per cent of a number is equal to ^ of ^^5 240.62c; that number, and g^ of 1925 IbR. is 240^ lbs. In the operation we take ^ of the base. Solve the next 9 examples by aliquot parts: 24. Find 25^ of $86c. 26. i2-|^ of 258 meters. 25. 10;^ of 1572 pounds. 27. 20;^ of 580 liters. 28. A drover taking 2320 sheep to market, lost 25^ of them by a railroad accident : how many did he lose ? 29. A farmer raised 468 bu. of corn, and ssi% as many oats as corn : how many bushels of oats did he raise ? 30. A young man having a salary of $1850 a year, spent 50 per cent of it: what were his annual expenses? 31. What is 33i-% of 1728 cu. feet of wood ? 32. What is 12 J per cent of £16, 8s.? •^^"S. How find the percentafre when the ba!»e and rate are giv(»n ? When tho rate is an aliquot part of loo, how proceed? When a compound number? PERCENT AGE. 235 PROBLEM II. 337. To find the Amount^ the Base and Rate being given, 1. A commenced business with $1500 capital, and laid up S% the first year : what amount was he then worth ? Analysis. — Since he laid up Sfc, he was opekation. worth his capital, $1500, plus 8% of itself. 1 1 500 13. But his capital is 100% or i time iteslf; and. i-o8? i H- R. 100;^ +8;^ =108% or 1.08; therefore he was 120.00 worth 1.08 times 1 1 500. Now $1500 x 1.08= 1500 $1620. We multiply the base by i plus the ^1620 00 Am't given rate, expressed decimally. (Art. 191.) 2. B commenced business with $1800 capital, and squander id 6% the first year: what was he then worth ? Analysis. — As B squandered 6%, he was $1800 B. worth his capital $1800, minus 6% of itself. .q^ J^. But his capital is 100^ or i time itself; and .~ ioo%— 6%r=94% or .94. Therefore he had ,94 Tfloo times $1800 : and $1800 x .g4=:$i692. Here we ^ . , multiply the base by i minus the given rate, $1692.00 Am t expressed decimally. (Art. 191.) Hence, the Rule. — Multiply the base by 1 plus or minus the rate, as the case may require. The result will be the amount. Formula. Amount = Base x (i ± Bate), Note. — i. The character ( ± ) is called the double or ambiguous 8'gn. Thus, the expression $5 ± $3 signifies that $3 is to be added to or subtracted from $5, as the case may require, and is read, " $5 plus or minus $3." 2. The rule is based upon the axiom that the whole is equal to trie sum of all its parts. 3. Wlien, by the conditions of the question, the amount is to be greater than the base, the multiplier is i plus the rate ; when the amount is to be less than the base, the multiplier is i minus the rate. 337. How find the amount when the base and rate are given ? 338. How else Ss the amount fouad, when the base and rate are given ? UA 236 PERCE:NrTAGE. 338. When the hase and rate are given, the amount rirv also be obtained by fiv^tfindrngthepercentageytlien adding it to or subtracting it from tJie base. (Art. t,z^') 3. and D have 1000 sheep apiece; if adds i$% to his flock, and D sells 12% of his, how manj^ sheep will each have ? 4. A merchant having I2 150.38 in bank, deposited ']% more : what amount had he then in bank ? 4 ^ % 5. If you have $3000 in railroad stock, and sell 5^ of it, what amount of stock will you then have ? ^^ ' C. 5 ^ 6. The cotton crop of a planter last year was 450 bales; this year it is 12 per cent more: what is his present crop ? 7. An oil well producing 2375 gallons a day, loses 15^ of it by leakage : what amount per day is saved ? 8. A gardener having 1640 melons in his field, lost 20^^ of them in a single night: what number did he have left ? 9. A man paid I420 for his horses, and 12;^ more for his carriage : what was the amount paid for the carriage ? 10. A man being asked how many geese and turkeys he had, replied that he had 150 geese; and the number 01 turkeys was 14^ less: how many turkeys had he? 11. A fruit grower having sent 2500 baskets of peaches to New York, found g% of them had decayed, and sold the balance for 62 cts. a basket: what did he receive for hts peaches ? 12. A Floridian having 4560 oranges, bought 25^ mor-, and sold the whole at 4 cts*, each : what did he receive for them? ^1^^ 13. If a man's inc6tne is $7235 a year, and he spends ZZ\7o of it, what amount will he lay up ? • 14. A man bought a house for $8500, and sold it for 2oi more than he gave : what did he receive for it ? - - '^ 15. A merchant bought a bill of goods for $10000, and sold them at a loss of 2l%\ what did he receive ? PERCENTAGE. 237 PROBLEM III. 339. To find the Mate, the Base and Percentage being given; Or, to find what Per Cent one number is of another. I. A clerk's salary, being 1 1500 a year, was raised $25c', what rate was the increase ? Analysis. — lu this example $1500 is the operation. base, and $250 the percentage. The question 1 500)^2^0^00 P. then is this: $250 is what per cant of $1500? ^^5. i6|-^^ E. iSTow $250 is iWo of $1500; and $250^11500 = .16666, etc., or i6|^. The first two decimal figures denote the per cent ; the others, parts of i^. (Arts, 331, 2.) Hence, the EuLE. — Divide the percentage hy the base. Formula. Bate = Percentage ~ Base. Notes.— I. This prob. is the same as finding what part one number is of another, then changing the common fraction to hundredths. (Arts. 173, 186, 334-) It is based upon the principle that percentage is a product o! which the base is a factor, and that dividing a product by one of its factors will give the other factor. (Art. 93.) 2. The number denoting the base is always jireceded by the word of, which distinguishes it from the percentage. 3. The given numbers must bo reduced to the same denomination ■, and if there is a remainder after two decimal figures are obtained, place it over the divisor and annex it to the quotient. 2. What ^ of 15 is 2 ? 6. What % of £3 are 1 5s. ? 3. What % of $20 are $5 ? 7. What % of 56 gals, are 7 qts.? 4. What % of 48 is 16 ? 8. What % are 5 dimes of $5 ? 5. W^hat ^of $5 are 75 cts.? 9. What ^ of | ton is I ton ? 10. The standard for gold and silver coin in the IT. S. is 9 parts pure metal and i part alloy : what % is the alloy? *| ' II. From a hogshead of molasses 15 gals, leaked out; f-^\-^t what per cent was the leakage ? ^'^ J4 12. A grocer having 560 bbls. of flour, sold f of it. what per cent of his flour did he sell ? _ 13. A horse and buggy are worth f>475 ; the buggy is worth $110; what % is that of the value of the horse ? 3^Q. How find the rate, when the base and percentage are given ? To what id tills problem equivalent ? A'ow. Upon what is it based ? 238 P E Tv C E is" T A G E , PROBLEM IV. 340. To find the Base, the Percentage and Rate being given. 1. A father gave his son $30 as a birthday present, which was 6% of the sum he gave his daughter: how much did he give his daughter? Analysis. — The percentage $30 is the product of opekation. the base into .06 the rate; therefore $30-f-.o6 is the .o6)$30.oo other factor or has3 ; and $30^. 06 =$500, the sum he Jiris7%<^oo gave his daughter. (Art. 193, n.) Or, since $30 is 6 % of a number, i % of that number must be I of $30, which is $5 ; and 100 fc is 100 times $5 or $500. It. is more concise, and therefore preferable, to divide the per- centage by the rate expressed decimally ; then point off the quotient as in division of decimals. (Art. 193.) Hence, the EuLE. — Divide the percentage hy the rate, exp^^essed decimally. Formula. Base — Percentage -^ Bate. Notes. — i. This problem is the same as finding a number when a given per cent or a fractional part of it is given. (Arts. 174, 334.) 2. The rule, like the preceding, is based upon the principle that percentage is a product, and the rate one of it^ factors. (Art. 335, n.) 3. Since the percentage is the same part of the base as the rate is of 100, when the rate is an aliquot part of 100, the operation will \^ shortened by using this aliquot part as the divisor. 2. 40 is 12^^ of what number ? Solution. — i2y/c=^ and 40-7-^=40x8=320. Ans. 3. 20 = 5^ of what number? Ans. 400. 4. 15 bushels =6^ of what number ? 5. $29 = 8;^ of what? 6. 45 tons = 25^ of what ? 7. £150 = 33^^ of what? ^8. 3 7.5 = 6 Jf^ of what? '9. 45 francs = 1 2^% of what ? 10. 40=^^ of what ? 11. 50 cts.=:-}^ of what ? 12. $100=1;^ of what? 13. $35.20=1;:^ of what? 14. 68 yds.= i25^of what ? 340. How find the base, when the percentage and rate are given ? Note. Upon what does the rule depend ? When the rate is an aliqnot part of loo, how proceed ? PEECENTAGE. 23j 15. 2;"^ of $150 is 6^ of Avliat sum ? 16. 12^ of 500 is 60^ of what number? 17. A paid a school tax of $50, which was i^ on the valuation of his property : what was the valuation ? , 18. B saves ^i^fo of his income, and lays up $600: what is his income ? . 19. A general lost 16}^ of his army, 315 killed, no prisoners, and 70 deserted; how many men had he? 20. According to the bills of mortality, a city loses 450 persons a month, and the number of deaths a year is 1^% of its population : what is its population ? PROBLEM V. 341. To find the Base, the Amount and Rate being given. 1. A manufacturer sold a carriage for $633, which wasJ 5^^ more than it cost him : what was the cost ? Analysis. — The amount received $633, is operation. equal to the cost or base plus 5 [7% of itself. 1.055)1^6 33.00 Now the cost is 100^ or i time itself, and Ans. $600 100% +5.J %=i.o5^ ; hence $633 equals 1.05^ times the cost of the carriage. The question now is; 633 is 105?^ or 1.05^ times what number? If 633 is 1.05I times a certain num- ber, once that number is equal to as many units as i.osi is contained times in 633 ; and 633-7-1.055=1600. Therefore the cost was $600. 2. A lady sold her piano for $628.25, which was 12-J^^, less than it cost her : w hat was the cost ? Analysis. — There being a loss in this case, operation. the amount received, $628.25, equals the cost .875)1)628.250 or base wmw« 12^% of itself. But the cost is ji^ns. I718 100% or I time itself, and 100 f c— 12)1%— .^'jV, ; hence $628.25 equals .87^ times the cost. Now if $628.25 equals .87^ times the cost, once the cost must be as many dollars as .87.} is contained times in $628.25, or $718. Hence, the KuLE. — Divide tJie amount hy i plus or miiius the rate,. as the case may require. Formula. Base — Amount -f- (i ± Rate). 341. How find tlie l.;ise the amount and rate being given ? Note. Upon wiiai 24:0 PERCEiTTAGE. Notes. — i. This problem is the same as finding a number wliich is a gioen per cent greater or less than a given number. 2. The rule depends upon the principle that the amount is a vro- duct of which the base is one of the factors, and i plus or miuwi the rate, the other. 3. The nature of the question shows whether i is to be increased or diminished by the rate, to form the divisor. 4. When the rate is an aliquot part of 100, the operation is often shortened by expressing it as a common fT&ction. Thus 25%=^; and I or | + ^=f, etc. 3. What number is 8^ of itself less than 351?^. 325. 4. What number is 5 1^ of itself more than 378 ? A. 400. 5. What number diminished 33^^ of itself will equal 539-3- ? 6. 2275 is 25^ more than what number? 7. I is 12^^ more than what number? Analysis.— i2i^=i ; and f -f-iir=f-j-|=f§ or f. Ans. 8. f is 10^ less than what number ? 9. A owns I of a ship, which is i6f^ less than B's part : what part does B own ? 10. A garrison which had lost 28^ of its men, had 3726 left : how many had it at first ? 11. A merchant drew a check for $4560, which was 25^ more than he had in bank : how much had he on deposit ? 12. The population of a certain place is 8250, which is 20^ more than it was 5 years ago : how much was it then ? 13. A man lays up $2010, which is 40^ less than his in- 'come : what is his income ? 14. A drover lost 10^ of his sheep by disease, 1$% were stolen, and he had 171 left : how many had he at first ? 15. The attendance of 9- certain school is 370, and 7^^ of the pupils are absent : what is the number on register ? 16. An array having lost 10^ in battle, now contains 5220 men : what was its original force ? docs this rule depend? How determine whether i is to be increased or dimin- ished by the rate ? When the rate is an aliquot part, how proceed? To what 19 this problem equivalent ? APPLIOATIOS'S OF PEECEI^TAGE. 342. The JPrincij^les of Percentage are applied to two important classes of problems: First. Those in which time is one of the elements 0/ calculation ; as, Interest, Discount, etc. Second: Those which are independent of time ; as. Com- mission, Brokerage, and Profit or Loss. COMMISSION AND BROKERAGE. 343. Cormnission is an allowance made to agents, collectors, brokers, etc., for the transaction of business. I^roUer^age is Commission paid a broker. Notes. — i. An Agent is one who transacts business for another, and is often called a Commission Merchant, Factor, or Correspondent. 2. A Collector is one who collects debts, taxes, duties, etc. 3. A Broker is one who buys and sells gold, stocks, bills of ex- change, etc. Brokers are commonly designated by the department of business in which they are engaged ; as, Stock-brokers, Exchange- brokers, Note-brokers, Merchandise-brokers, Real-estate-brokers, etc. 4. Goods sent to an agent to sell, are called a consignment ; tho person to whom they are sent, the consignee ; and the person send- ing them the consignor. 344. Commission and Brolcerage are computed at a certain per cent of the amount of business transacted. Hence, the operations are precisely the same as those in Percentage. That is. The sales of an agent, the sum collected or invested by him, are the base. The per cent for services, the rate. The commission, the percentage. The sales, etc., plus or minus the commission, the amount. 342. To what two classes of problems are the principles of percentage applied? 343. What is commission? Brokerage? Note. An agent? What called? A collector ? Broker ? 344. How are commission and brokerage computed ? What Is the base? The rate? The percentage? The amount? iVoie. The net proceeds ? 242 COMMISSION AND BEOKERAGE. Notes. — i. The rate of commission and brokerage varies. Com* mission merchants usually charge about 2\ per cent for seUi'n{j goods, and 2i- per cent additional for guaranteeing the payment, Btock-brckers usually charge ^ per cent on the par value of stocks, without regard to their market value. 2. The net proceeds of a business transaction, are the gross amount of sales, etc., minus the commission and other charges. 345, To find the Conmiissionf the Sales and the Rate being given. Multiply the sales ly the rate. (Problem I, Percentage.) Notes. — i. When the amount of sales, etc., and the commission are known, the net proceeds are found by subtracting the commission, from the amount of sales. Conversely, 2. When the net proceeds and commission are known, the amount of sales, etc., is found by adding the commission to the 7iet proceeds. 3. When both the amount of saies, etc., and the net proceeds are known, the commission is found by subtracting the net proceeds from the amount of sales. 4. In the examples relating to stocks, a share is considered $100, unless otherwise mentioned. (Ex. 2.) (For methods of analysis and of dedvicing the rules in Commission, Profit or Loss, etc., the learner is referred to the corresponding Problems in Percentage.) \j I. A merchant sold a consignment of cloths for $358: what was his commission at 2 J per cent ? Solution. — Commission =$358 (sales) x .025 (rate)=$8.95. Ans. 2. A broker sold 39 shares of bank stock : what was his brokerage, at J per cent ? Solution. — 39 shares =$3900; and $3900 x .005 = $19,500. Ans. 3. Sold a consignment of tobacco for $958.25 : what was ny commission at 2,i% ? 4. A man collected bills amounting to $11268.45, and sharged 2>i%'- what was his commission; and how much did he pay his employer ? 345. How find the commleeion or brokerage, when the sales and the rate arc ilwn ? iV»V, How find the net nroceeds ? COMMISSIOIT Ai^D BROKERAGE. 5. A commission merchant sold a consignment of goods for $4561, and cliarged 2^% commission, and 3^ for guaranteeing the payment: what were the net proceeds? 6. An agent sold 1530 lbs. of maple sugar at i6f cts, for which he received 2\% commission: what were the net proceeds, allowing $7.50 for freight, and $3,10 for storage ? 346. To find the Rate, the Sales and the Commission being given. Divide the commission hy the sales. (Prob. III. Percentage.) 7. An auctioneer sold goods amounting to $2240, for which he charged $53.20 commission : what per cent was that ? Solution. — Per cent = $53.20 (com.) -^ $2240 (sales) = .02375, or 2% per cent. (Art. 339, 331, n.) 8. A broker charged $19 for selling $3800 railroad stock: what per cent was the brokerage ? 9. Received $350 for selling a consignment of hops amounting to $7000: what per cent was my commis- sion ? 10. An administrator received $118.05 ^^^ settling an estate of $19675 : what per cent was his commission? 347. To find the Sales, the Commission and the Rate being given. Divide the commission ly the rate. (Prob. lY, Per ct.) 11. An agent charged 2% for selling a quantity of muslins, and received $93.50 commission : what was the amount of his sales ? Solution.— Sales=:$93. 50 (com.)-^.o2 (rate)=$4675. An8. 346. How find the per cent commission, when the sales and the commission are given ? 347. How find the amount of sales, when the commission and the rate arc ^von ? 244: COMMISSION AND BROKERAGE 12. Received $45 brokerage for selling stocks, which was I-/0 of what was sold: what was the amount of stocks S3ld? 13. A commission merchant charging 2 1^ commission, and 2 ^/o for guaranteeing the payment, received $2 10.60 for selling a cargo of grain : what were the amount of sales, and the net proceeds ? 14. A district collector received $67.50 for collecting a school tax, which was 4^% commission : how much did he collect, and how much pay the treasurer ? 15. An auctioneer received $135 for selling a house, which was 1'^%: for what did tlie house sell; and how much did the owner receive ? 348. To find the Sales, the net proceeds and per cent commission being given. Divide the net proceeds hy i minus the rate. (Prob. V. ) 16. An agent sold a consignment of goods at 2\% com- mission, and the net proceeds remitted the owner were $3381.30 : what was the amount of sales ? Solution.— Sales=$338i. 30 (net p.)-^-975 (i— rate) =$3468. Ans. 17. A tax receiver charged 5^ commission, and paid $4845 net proceeds into the town treasury: what was the am3unt collected? Note. — In this and similar examples, the pupil should observe that the base or sam on which commission is to be computed is the sum collected, aaJ not the sunt jjaid ocer. If it were the latter, the agent would have to collect his own commission, at his own expense, and his rate of commission would not be ytV, but j-^^. In the cot lection of $100,000, this would cause an error of more than $350, 18. After retaining 2}% for selling a consignment of flour, my agent paid me $6664 : required the amount of yaljs, and his commission. 348. How find the sales, etc., the net proceeds and the per cent commission being given ? COMMISSIOif A]S^D BEOKERAGE. 245 19. After deducting ij^ for brokerage, and $45.28 for advertising a house, a broker sent the owner $15250: for what did the house sell ? 20. An administrator of an estate paid the heirs $25686, charging 2^% commission, and $350 for other expenses: what was the gross amount collected ? 349. To find the Sum Invesfedf the sum remitted and the per cent commission being given. 21. A manufacturer sent his agent $3502 to invest hi wool, after deducting his cemmission of 3^ : what sum dit L lie invest ? AifALYSis. — The sum remitted $3502, includes both the sum in- vested and the commission. But the sum invested is 100% of itself, and 100% +3% (the commission) =103%. The question now is : $3502 is 103% of what number? $3502-7-1.03 — $3400, the sum in- vested. (Art. 340, n.) Hence, the Rule. — Divide the sum remitted ly 1 plus the per cent commission. (Prob. V, Per ct.) Note. — The learner will observe that the base in this and sim- ilar examples is the sum incested, and not the sum remitted. If it were the latter, the agent would receive commission on his commis- sion, which is manifestly unjust. „ 22. A teacher remitted to an agent $3131.18 to be laid out in philosophical apparatus, after deducting 4% com- mission : how much did the agent lay out in appa- ratus ? 23. If I remit my agent $2516 to purchase books, after deducting 4% commission, how much does he lay out in books ? 24. Remitted $50000 to a broker to be invested in city property, after deducting 1^% for his services: how much did he invest, and what was his commission ? 340. How find the sum invested, the sum remitted and the per cent commis" sion being given? 346 ACCOUKT or SALES. ACCOUNT OF SALES. 350. An Account of Sales is a written statement, made by a commission merchant to a consignor, contain- ing the prices of the goods sold, the expenses, and the net proceeds. The usual form is the following: Sales of Grain on accH of E. D. Barker, Esq., Chicago. DATE. BUTEK. DESCRIPTION. BUSHELS. PRICE. . EXTENSION. I87I. April 3 J. Hoyt, A. Woodruff, Hecker & Co., Gross am Winter wheat. Spring " Corn, :)unt. 565 870 1610 @ $2.10 @ 1.95 @ 1.05 $1186.50 1696.50 1690.50 $4573-50 Charges. Freight on 3045 bu., at 20 cts., Cartage " " $15 -30, Storage " 38-75. Commission, ^ifo. $609.00 1530 38.75 114.34 Net proceeds, _$777:39 $3796.11 New York, July sth, 1S71 J. Henderson & Co. Bx. 25. Make out an Account of Sales of the following: James Penfield, of Philadelphia, sold on account of J. Hamilton, of Cincinnati, 300 bbls. of pork to W. Gerard & Co., at $27; 1150 hams, at I1.75, to J. Eamsey; 875 kegs of lard, each containing 50 lb., at 8 cts., to Henry Parker, and 750 lb. of cheese, at 10 cts., to Thomas Young. Paid freight, $65.30; cartage, 1 15.25 ; insurance, $6.45; commission, at 2%. \¥hat were the net proceeds ? 26. Samuel Barret, of New Orleans, sold on account of James Field, of St. Louis, 85 bales cotton, at I96.50; 6s barrels of sugar, at $48.25 ; 37 bis. molasses, at $35. Paid freight, I45.50; insurance, I15 ; storage, $35.50; •jomraisRi^'m, 2^%. What were the net proceeds? 350. What is an accoant of Bales ? PROFIT AND LOSS. 247 PROFIT AND LOSS. 351. Profit and Loss are the sums gained or lost in business transactions. They are computed at a certain per cent of the cost or sum invested, and the operations are the same as those in Percentage and Commission. The Cost or sum invested is the Base ; The Per cent profit or loss, the Rate ; The Profit or Loss, the Percentage ; The Selling Price, that is, the cost plus or minus thu profit or loss, the Amount. 352. To find the Profit or Loss, the Cost and the Per Cent Profit or Loss being given. Multiply the cost hy the rate. (Problem 1, Per ct.) Note.— When the 'per cent is an aliquot part of loo, it is gener ally shorter, and therefore preferable to use the fraction. (Art. 336, n.) I. A man paid 1 2 50 for a horse, and sold it at 15^ profit: how much did he gain? Ans. $37.50. 2. A man paid I450 for a building lot, and sold at a loss of 11^: how much did he lose ? Ans. I49.50. 3. Paid $185 for a buggy, and sold it 12^ less than cost: what was the loss ? 4. Paid |i 10 for a pair of oxen, and sold them at 20^ advance : what was. the profit ? 5. A lad gave 87^ cts. for a knife, and sold it at 10% below cost: how much did he lose? 6. Bought a watch for I83I, and sold it at a loss of 20^: what was the loss ? 7. Bought a pair of skates for I4.20, and sold them at Zd>i% advance : required the gain ? 351. What are profit and loss? How reckoned? To what does the cost or sum invested answer? The per cent profit or loss? The profit or loss? The selling price ? 352. How find the profit or loss, when the cost and per cent ar« given ? Note. When the per cent is an aliquot part of 100, how proceed ? 248 PROFIT AKD LOSS. 353. To find the Selling Frice, the Cost and Per Cent Profit or Loss being given. Multiple/ the cost by i plus or minus the per cent. (Prob. II. Percentage.) Note. — When the cost and per cent profit or loss are given, the aeUing price may also be found by first finding the profit or loss; then add it to or subtract it from the cost. (Art. 338.) 8. A man paid $300 for a house lot : for what must he sell it to gain 20^? Analysis. — To gain 20%, he must sell it for the cost plus 20;^. That is, selling pr.=$30o (cost) x 1.20 (i + 20%) =$360. Ans. 9. A farmer paid I250 for a pair of oxen : for how much must he sell them to lose 15^? Selling pr.=$25o (cost) x .85 (i — i5%)=$2i2.5o. Ans. 10. A and B commenced business with $2500 apiece. A adds ij% to his capital during the first six months, and B loses 17^ of his: what amount is each then worth ? 11. A merchant paid $378 for a lot of silks, and sold them at 20^ profit: what did he get for the goods ? 12. If a man pays $2750 for a house, for how much must he sell it to gain y% ? 13. If a man starts in business with a capital of I8000, and makes igfo clear, how much will he have at the close of the year ? 14. If a merchant pays 15 cts. a yard for muslin, how must he sell it to lose 25^? 15. Bought gloyes at $15 a dozen: how must I sell them a pair, to lose 20^ ? 16. Paid $25 per dozen for pock'et handkerchiefs: for what must I sell them apiece to make ssi per cent? 353. How find the Belling price, when the cost and per cent profit or loss ara given. Note. How else may the selling price be found ? PROFIT AND LOSS. 249 17. Paid $196 for a piece of silk containing 50 yds.: how must I sell it per }'ard to gain 25^? 18. Bought a house for $3850: how must I sell it to make i2|c^? 19. A speculator invested $14000 in flour, and sold at a loss of 8-}%: what did he receive for his flour? 354. To find the Fer Cent, the Cost and the Profit oi« Loss being given. Divide the profit or loss hy the cost. (Prob. Ill, Per ct.) Note. — When the cost and selling price are given, first find the profit or loss, then the 2)er cent. (Art. 339.) 20. If I buy an acre of land for $320, and sell it for $80 more than it cost me, what is the per cent profit ? Solution.— Per cent=$8o (gain)-H$32o (cost)==.25 or 25%. Arts. 21. A jockey paid $875 for a fast horse, and sold it so as to lose $250 ; what per cent was his loss ? 22. If I pay 22 J cts. a pound for lard, and sell it at 2 J cts. advance, what per cent is the profit ? 22,. If a newsboy pays 2^ cts. for papers, and sells them at I -J cent advance, what per oput is his profit? 24. If a speculator buys apples at $2.12} a barrel^ an(i sells them at $2.87 J, what is his per cent profit? Analysis. — $2.87^ — $2. 12^=$. 75 profit per bl. Therefore, $.75 (gain)-4-$2.i2i- (cost)=.35,V or 3Sri%- (Art. 339, n.) 25. If I sell an article at double the cost, what per cent is my gain ? 26. If I sell an article at half the cost, what per cent is my loss ? 27. If I buy hats at $3, and sell at $5, what is the per cent profit ? 28. If I buy hats at $5, and sell at $3, what is the per cent loss ? 354. How find the per cent profit or loss, when the cost and profit or lose ara given ? 250 PROFIT Al^D LOSS. 29. If a man's debts are I3560, and he pays only $1780, what per cent is the loss of his creditors ? 30. If f of an article be sold for ^ its cost, what is the per cent loss ? Analysis. — If f are sold for ^ its cost, ^ must be sold for ^ of \, or Jg the cost, and | for f or f the cost. Hence, the loss is ^ the cost -, and i-f-i=.333 or sskfc 31. If you sell ^ of an article for | the cost of the whole, what is the gain per cent ? 32. If I sell I of a barrel of flour for the cost of a barrel, what is the per cent profit ? ^:^. If a milkman sells 3 quarts of milk for tlie price he pays for a gallon, what per cent does he make ? 34. Bought 3 hhd. of molasses at 85 cts. per gallon, and sold one hhd. at 75 cts., the other two at $1 a gallon* required the whole profit and the per cent profit? 355. To find the Cost^ the Profit or Loss and the Per Cent Profit or Loss being given. Divide the profit or loss hy tlie given rate. (Problem IV, Percentage.) Note. — When the per cent profit or loss is an aliquot part of 100, the operation may often be abbreviated by using this aliquot part as the divisor. (Art. 341, n) 35. A grocer sold a chest of tea at 25^ profit, by which he made I22J: what was the cost ? Solution.— Cost= $22.50 (profit)-T-. 25 (rate)=$90, Ans. Or, cost=$22.5o-r-^=$go. Ans. (Art, 340, n.) 36. A speculator lost $1950 on a lot of flour, wiiich was 20;^ of the cost: required the cost? 37. Lost 65 cents a yard on cloths, which was 13^ of the cost: required the cost and selling price ? Analysis.— The cost=65cts.-^.i3=$5; and I5 — $.65=^4.35 the selling price. (Art. 340, n.) 355. How find the cost, when the profit or loes and the per cent profit or loss are ijiven. PROFIT AND LOSS. 251 38. Gai 110(1 $2 J per barrel on a cargo of flour, which was 20;^: requirod the cost and selling price per barrel? Analysis. — 20%=:^^ and $2.5o-i-^=$i2.5o cost, and $12.50 + 12.50= $15, selling price. 39. If I sell coffee at lo^ profit I make lo cts. a pound: what was the cost ? 40. A man sold a house at a profit of 33 J/^, and thereby gained $7500: required the cost, and selling price? 41. If I make 20;^ profit on goods, what sum must I lay out to clear $3500, and what will my sales amount to ? 42. A and B each gained $1500, which was 12^^ of A*s and 16^ of B's stock : what was the investment of each ? 43. If a merchant sells goods at 10^ profit, what must be the amount of his sales to clear $25000 ? 44. A market man makes ^ a cent on every egg he sells, which is 25^ profit: what do they cost him, and how sell them? 356. To find the Cosf^ the Selling Price and the Per Cent Profit or Loss being given. Divide the selling price by 1 plus or minus the rate of profit or loss, as the case may require. (Prob. V, Per ct.) 45. A jockey sold two horses for $168.75 ^^^^^ ? ^^ ^^^ ^^® made i zY/c, on the other lost 1 2\% : what did each horse cost him ? Solution.— Cost of one=$i68.75 (sel. pr.)-f-i.i25 (i +rate)=$i50. Cost of other= $168.75 (sel. pr.)-4-.875 (i— rate)=$iq2.86. Ans. 46. By selling 568 bis. of beef at $15^ a barrel, a grocer lost 12^%: what was the cost? Ans. $10061.714. 47. Sold 5000 acres of land at $3-J an acre, and thereby gained 22^: what was the cost? 48. Sold a case of linens for £27, los., making a profit of 25^: what was the cost? 356. How find the cost, when the sslling price and the per cent profit are given ? 252 PROFIT AND LOSS. 49. If a newsboy sells papers at 4 cts. apiece, lie makes 33i% ' what do they cost him ? 357. To Mark Goods so that a given Per Cent may be deducted, and yet make a given Per Cent profit or loss. 50. Bought shoes, at $2.55 a pair: at what price must they be marked that 15^ maybe deducted, and jet be sold at 20^ profit ? Analysis.— The selling price is 120% of $2.55 (tlie cost); $2.55 x i.20=$3.o6, the price at which they are to be sold. But the marked price is 100% of itself; and ioo%— 15%=85%. The question now IS, $3.06 are 85^ of what sum V If 11^3.06 =-i^y%, 7^7; = $3.06 ^85, or ^.036 ; and |{}{l=$3.6o. Ans. (Art. 340.) Hence, the EuLE. — Find the selling price, and divide it dy 1 minus the given per cent to he deducted ; the quotient will he the marhed price, 51. Paid 56 cts. apiece for arithmetics: what must they be marked in order to abate 5;^, and yet make 25%? 52. If mantillas cost $24 apiece, at what price must they be marked that, deducting 8^, the merchant may realize 33j^ profit? 53. When apples cost I3.60 a barrel, what must be the asking price that, if an abatement of 1 2^% is made, there will still be a profit of i6|^^? 54. A merchant paid 87^ cts. a yard for a case of linen, which proved to be slightly damaged : how must he mark it that lie may fall 25^^, and yet sell at cost ? 55. A goldsmith bought a case of watches at $60: how must he mark them that, abating ^%, he may make 20^? 56. If cloths cost a tailor $4.50 a yard, at what price must he mark them, that deducting 10^ he will make 15 per cent profit? 357. When the selling price and the per cent profit or loss are given, how find the per cent profit or loss .nt any proposed price ? How find what to mark goods, that a given per cent m.xy be deducte J, and yet a given per cent profit or loss be -.flAli^ed "i PKOriT ANL LOSS. 253 QUESTIONS FOR REVIEW. 1. A grocer paid 23 cts. a lb. for a tub of butter weighing 65 lbs., and sold it at 18^ profit: hoAV much did he make? 2. Bought a house for I3865, and paid $1583.62 for re- pairs: how mucli must I sell it for to make i6f^? 3. Paid $2847 for ^ case of shawls, and $956 for a case of ginghams; sold the former at 22^^ advance, and the latter at 11^ loss : what was received for both ? 4. A jockey bought a horse for $125, which he traded for another, receiving $37 to boot; he sold the latter at 25^ less than it cost: what sum did he lose ? 5. What will 750 shares of bank stock cost, if the bro- kerage is 1%' and .the stock 2>% above par ? 6. Sold 3 tons of iron, at 15 1 cts. a pound, and charged 2,1% commission : what were the net j)roceeds ? 7. Bought wood at $4^ a cord, and sold it for %G\: re- quired the per cent profit ? 8. A shopkeeper buys thread at 4 cts. a spool, and sells at 6 J cts.: what per cent is his profit, and ho\? much would he make on 1 000 gross ? 9. Bought 1650 tons of ice, at $12 ; one half molted, and sold the rest at $1 a hund. : what per cent was the loss? 10. .The gross proceeds of a consignment of apples was $1863.75, and the agent deducted $96,911 for selling: re- quired the per cent commission and the net jiroceeds? 11. A lady sold her piano for J of its cost: what per cent was the loss ? 12. If I pay $2 for 3 lbs. of tea, and sell 2 lbs. for $3, what is the per cent profit ? 13. A man sold his house at 20^ above cost, and thereby made $1860: required the cost and the selling price? 14. A miller sells flour at 15;^ more than cost, and makes $1.05 a bar.: what is the cost and selling price ? 15. Lost 25 cts. a pound on indigo, which was 12^% of the cost : required the cost and the selling price. 254 PROFIT AN^t) LOSS. 1 6. A commission merchant received $260 for selling a quantity of provisions, which was 5^: required the amount of sales and the net proceeds. 17. A broker who charges i%, received |6o for selling a quantity of uncurrent money : how much did he sell ? 18. A grocer makes $1.25 a pound on nutmegs, which is 100^ profit : what does he pay for them ? 19. A merchant sold a bill of white goods for $7000, and made ssl% : required the cost and the sum gained. 20. Sold a hogshead of oil at 93J cts. a gallon, and made i8J^: required the cost and the profit. 21. A man, by selling flour at $i2| a barrel, makes 25^: what does the flour cost him ? 22. Made 12^^ on dry goods, and the amount of sales was $57725 : required the cost and the sum made. 23. Sold a quantity of metals, and retaining 3j^ com- mission, sent the consignor $15246: required the amount of the sale and the commission. 24. Sold 250 tons of coal at $6-|, and made 12^^: what per cent would liave been my profit had I sold it for $8 a ton ? 25. A merchant sold out for $18560, and made 15;^ on his goods : what per cent would he have gained or lost by selling for $15225 ? 26. Paid $40 apiece for stoves : what must I ask that I may take off 20^, and yet make 20;^ on the cost ? 27. If a bookseller marks his goods at 25^ above cost, and then abates 25^, what per cent does he make or lose ? 28. A grocer sells sugar at 2| cts. a pound more than cost, and makes 20% profit : required the selling price. 29. A man bought 2500 bu. wheat, at $if ; 3200 bu. corn, at 87 J cts. ; 4000 bu. oats, at 25 cts.; and paid $450 freight: he sold the wheat at 5;^ profit, the corn at 11% loss, and the oats at cost ; the commission on sales was $% • what was his per cent profit or loss ? INTEREST. 358. Interest is a compensation for tlie use of money. It is computed at a certain per cent, per annum, and em- braces five elements or parts : tlie Principal, tlie Bate, the Interest, the Time, and the Amount. 359. The I^rincij^al is the money lent. The Mate is the per cent per amimn. The Interest is the percentage. The Time is the period for which the principal draws interest. The Amount is the sum of the principal and interest. Notes. — i. The term per annum, from the Latin per and annus, signifies b^/ the year. 2. Interest differs from the preceding applications of Percentage only by introducing time as an element in connection with per cent. The terms rate and rate per cent always mean a certain number of hundredths yearly, and pro rata for longer or shorter periods. 360. Interest is distinguished as Simple and Compound. Simple Interest is that which arises from the principal only. Compound Interest is that which arises both from the principal and the interest itself, after it becomes due. 361. The Legal Mate of interest is the rate allowed by law. Rates higher than the legal rate, are called usury. Notes. — i. In Louisiana the legal rate is - - - 5%- In the N. E. States, N. C, Penn., Del., Md., Va., W. Va., Tenn., Ky., 0., Mo., Miss., Ark., Flor., la., 111., Ind., the Dist. of Columbia, and debts due the United StatttS, - ^%. In N. Y, N. J., S. C, Ga., Mich., Min., and Wis., - -]%. In Alabama and Texas, - - - - - - - 8%, In Col., Kan., Neb., Nov., Or., Cal. and Washington Ter., - lofo. 2. In some States the law allows higher rates by special agreement. 3. When no rate is specified, it is understood to be the legal rate. 358. What is interest? 359. The principal ? The rate? The int. ? The time? The amt. ? 360. What is simple interest ? Compound ? 356 INTEREST. THE SIX PER CENT METHOD. 362. Since the interest of $i at 6 per cent for 1 2 months or I year, is 6 cents, for i month it is i-hoelfth of 6 cents or ^ cent ; for 2 months it is 2 halves or i cent ; for 3 mos. I J cent, for 4 mos. 2 cents, etc. That is, The interest of $1, at 6 per cent, for any number of mo7iths, is half as many cents as months. 363. Since the interest of $1 at 6 per cent for 30 days or I mo., is 5 mills or | cent, for i day it is -j-q of 5 mills or J mill ; for 6 days it is 6 times J or i mill ; for 1 2 days, 2 mills ; for 15 days, 2^ mills, etc. That is. The interest of ^1, at 6 per cent, for any number of days. is i-sixth as many mills as days. Hence, 364. To find the Interest of $1, at 6 per cent, for any given time. Take half the number of months for cents, and one sixth of the days for mills. Their sum will be the interest required. Notes. — i. When the rate is greater or less than 6 per cent, the interest of $1 for the time is equal to the interest at 6 per cent increased or diminished by such a part of itself as the given rate exceedjS or falls short of 6 per cent. Thus, if the rate is ']%, add \ to the int. at 6;/ ; if the rate is 5^', subtract I, etc. 2. In finding i-sixth of the days, it is sufficient for ordinary pur- poses to carry the quotient to tenths or hundredths of a mill. 3. When entire accuracy is required, the remainder should be placed over the 6, and annexed to the multiplier. 1. Int. of 81, at "]% for 9 m. 18 d. ? A71S. $.056. 2. Int. of $1, at s% for 11 m. 21 d. ? Ans. I.04875. 3. What is the int. of Si, at 6% for 7 m. 3 d. ? 4. What is the int. of $1, at 6% for 11 m. 13 d. ? Note. Wherein does intereet differ from the preceding applications of per- centage ? 361. The legal rate ? What is ueuiy ? 362. What is the interest of $1 nt 6 per cent for months ? 363. For days ? 364. IIow find the int. of $1 for nny [riven time ? Note. When the rate is greater or lees than 6 per cent, to what id the intereet of f i equal? IKTEKEST. 267 Rem. — The relation between the principal, the interest, the rate, the time, and the amount, is such, that when any three of them are given, the others can "be found. The most important of these problems are the following : PROBLEM I. 365. To find the Interest, the Princ pal, the Rate, and Time being given. 1. What is the interest of $150.25 for i year 3 months and 18 days, at 6^? Analysis. — The interest of $1 for 15 m.=.o75 operation. 18 d. =.003 Prin. $150.25 I 7. 3 m. 18 d. =^78" -oyS Now as the int. of $1 for the given time and 120200 rate is $.078 or .078 times the principal, the int. 105 175 of $150.25 must be .078 times that sum ; and %ii.niq^ $150.25 X .o78 = $ii. 71950. Hence, the EuLE. — Multiply the ^^rmc'?};^^/ ly the interest of $1 for the time, expressed decimally. (Art. 336.) For the amount, add the interest to the priyicijxd. Rem. — I. The amount may also be found by multiplying tha principal by i plus the interest of %i for the time. (Art, 337.) 2. When the rate is greater or less than 6%, it is generally best to find the interest of the principal ai 6% for the given time ; tlien add to or subtract from it such a part of itself, as the given rate exceeds or falls short of 6 per cent. (Art. 364.) 3. In finding tlie time, first determine the number of entire calendar months ; then the number of days left. 4. In computing interest, if the mills are 5 or more, it is customary to add I to the cents ; if less than 5, they are disregarded. Only three decimals are retained in the following Answers, and each is found by the rule under which the Ex. is placed. 2. What is the amt. of 1 150. 60 for i y. 5 m. 15 d. at 6% / Analysis. — Int. of $1 for i y. 5 mos. on 7 mos., equals .085 ; 15 d. it equals .0025 ; and .085 + .0025 = .0875 the multiplier. Now I150.60 X.0875 — $13 1775, int. Finally, $150.60 + $13. I775=$i63. 7775, ^72.*. 365. How find the interest, when the principal, rate, and time are given? How find the amount, when the principal and interest are given ? liem. How else is the amount found ? 2^H II^TEREST. ,5. Find the interest of $31.75 for i yr. 4 mos. at 6%. 4. What is the int. of $49.30 for 6 mos. 24 d. at 6^ ? 5. What is the int. of $51.19 for 4 mos. 3 d. at y% ? 6. What is the int. of I142. 83 for 7 mos. 18 d. at 5^? 7. What is the int. of I741.13 for 11 mos. 21 d. at 6%? S. What is the int. of $968.84 for i yr. 10 mos. 26d. at 6% ? 9. What is the int. of I639 for 8 mos. 29 d. at j%? 10. What is the int. of $741.13 for 7 mos. 17 d. at 5^? ^i. What is the int. of |i 237.63 for 3 mos. 3d. at S%? 12. What is the int. of $2046 J for 13 mos. 25 d. at 4%? 13. What is the int. of $3256.07 for i m. and 3 d. at 6^? " 14. Find the amount of $630.37 J for 9 mos. 15 d. at 10^? 15. Find the amount of I75.45 for 13 mos. 19 d. at 7;^? J 6. Find the amount of $2831.20 for 2 mos. 3 d. at g%? 17. Find the amount of $356.81 for 3 m. 11 d. at si?c^ 18. Find the amount of $2700 for 4 mos. 3 d. at 6l-%? 19. Required the amount of $5000 for ^^ days at 7^. 20. Required the int. of $12720 for 2 mos. 17 d. at 4^%. 21. What is the amt. of $221.42 for 4 mos. 23 d. at 6%? 22. What is the int. of $563. 16 for 4 mos. at 2^ a month ? Suggestion. — At 2% a montli, the int. of $1 for 4 mos. is $.08. 23. What is the int. of $7216.31 for 3 mos. at i^a month? 24. Find the int. of $9864 for 2 mos. at 2^% a month ? 25. Find the amt. of $3540 for 17 mos. 10 d. at 7^%? \ 26. What is the interest on $650 from April 17th, 1870, to Feb. 8th, 187 1, at 6%? Analysis. — 1871 y. 2 m. 8 d. J Int. $1 for 9 m.=.o45 $650 1870 y. 4 m. 17 d. I " " 21 d. =.0035 .0485 Time, o y. 9 m. 21 d. j Multiplier, .0485 $31,525 27. What is the interest on $1145 from July 4th, 1867, to Oct. 3d, 1868, at 7%. 28. What is the interest on a note of $568.45 from May 2ist, 1861, to March 25th, 1862, at 5^? V29. Required the amount of $2576.81 from Jan, ?»ist, 1871, to Dec. i8th, 1 87 1, at 75^ INTEREST. 259 METHOD BY ALIQUOT PARTS. 366. To find the Interest by Aliquot Parts, the Principal, Rate, and Time being given. I. What is the interest of $137 for 3 y. i m. 6 d. at 5^? Analysis. — As tlie rate is 5 % , tlie interest for S 1 3 7 prin. I year is .05 (-[ nu) of the principal, - - - -0 5 I'^.te. Now$i37x.o5 = $6.85, I2)$6.85 int. I y. Again, the int. for 3 y. is 3 times as much as for i y ., 3 J- And $6.85 (int. i y.) x 3 = $20.55, - - - 20.55 int. 3 y. The int. for i m. is -jV of int. for i yr. ; and $6.85^i2 = .57, 5) -57 " ini. The int. for 6 d. is i of int. for i m.; and $.57-5 = . 114, .114 '' 6 d. Adding these partial interests together, we have $21,234, Ans. the answer required. Hence, the Rule. — For one year. — lluUipli/ the principal hy the rate, expressed decimally. For tivo or more years. — Multi2)ly the interest for 1 year ly the number of years. For months. — Take the aliquot part of i yearns interest. For days. — Tahe the aliquot part of i month's interest. That is, for i m., take -f^ of the int. for i y.; for e. m., l\ for 3 m., J, etc. For I d., take -}q of the int. for i m. ; for 2 d., -^^ ; for 6 d., I; for 10 d., ^, etc. Note. — This method has the advantage of directness for different rates; but in practice, the preceding method is generally shorter and more expeditious. 2. What is the int. of $143.21 for 2 y. 5 m. 8 d. at 7^? 3. What is the int. of $76.10 for i y. 3 m. 5 d. at 6^^? 366. How find the interest for i j'ear, at any given rate, by aliquot parts? How for 2 or more years ? For months ? For days ? What part for i month ? for3mo8.? For6mos.? For i day? For 5 d.? For 10 d.? For 15 d.? Foraod.? 4. 260 IKTEKEST. 4. What is the int. of $95.31 for 8 m. 20 d. at 7^? 5. What is the int. of $110.43 fer i yr. 6 m. 10 d. at 4%? 6. What is the int. of $258 for 3 yrs. 7 m at S%? 7. What is the interest of $205.38 for 5 yrs. at 6\%? 8. Find the interest of $361.17 for 11 months at 8^? 9. Find the interest of $416.84 for 19 days at 7^^ ? 10. At 'j^%, what is the int. of $385.20 for i yr. 13 d. ——II. At 5^^, what is the int. of $1000 for i y. i m. 3 d. ? 12. At 8^, what is the int. of I1525.75 for 3 months? 13. Eequired the interest of I12254 for 2 J years at 8%? 14. AVhat is the amount of $20165 ^^^ 5 ^^- ^7 ^- ^^ 7%^ METHOD BY DAYS. 367. To find the Interest by DaySf the Principal, Rate, and Time being given. I. What is the interest of $350 for 78 days at 6^? Analysis. — The interest of $1 at 6^ for i day $350 is Jy of a mill or ^ifou dollar. (Art. 363.) There- 78 fore for 78 days it must be 78 times BiAyo^t^ooo "2800 dol., or ^l.to times the principal. Again, since oAcn the int. of $1 for 78 days is ^JiTid times the princi- — ^ — pal, the interest of $350 for the same time and o|ooo)$27_^300 rate must be ^Jusd times that sum. In the opera- Ans. $4.55 tion, we multiply the principal by 78, the numera- tor, and divide by the denominator 6000. Hence, the Rule. — Multiply the principal hy the numler of days, and divide hy 6000. Hie quotient ivill he the i^iterest at 6%. For other rates, add to or suhfract from the interest at 6 per cent, such a part of itself as the required rate id greater or less than 6%. (Art. 364.) Note. — This rule, though not strictly accurate, is generally used by private bankers and money-dealers. It is based upon the suppo- sition that 360 days are a year, which is an error of five days or y^ of a year. Hence, the result is 7^, too larj^e. When entire accuracy is required, the result must be diminished by 7^3- part of itself. 367. Eow compute interest by days ? Note. Upon what is this method founded ? INTEREST. 2G1 2. A note for $720 was dated April 17th, 187 1 : what was the interest on it the i6th of the following July at 7^ ? Omitting the day of the first date. (Art. 339, n.) $720 prin. April has 30 days— 17 d.=r 13 d. 90 days. May " 31 d. 6!ooo)64|8oo June " 30 d. 6)io.8oint. 6^^. July " 16 d. 1.80 " ifc. Time " =90 d. ^/iS. $12.60 int. 7^. 3. What is the interest of $5 1 7 for ^^ days at 6% ? 4. What is the interest of $208.75 for 6;^ days at 6%? 5. What is the interest of I631.15 for 93 days at 'j%? 6. Find the interest of $1000 for 100 days at 5^? 7. Find the amount of 1 1260. 13 for 120 days at 6^? 8. Kequired the interest of $3568.17 for 20 days at 7^? 9. Eequired the amount of $4360.50 for 3 days at '/%? 10. Find the interest of $5000 from May 21st to the 5th of Oct. following at 6% ? 11. Find the interest of $6523 from Aug. 12th to the 5th of Jan. following at 7;^'? 12. Find the interest of $7510 from Jan. 5th to the loth of the following March, being leap year, at 6% ? PROBLEM II. 368. To find the Mate, the Principal, the Interest, and the Time being given. I. A man lent his neighbor I360 for 2 yrs. 3 mos., and received $48.60 interest: what was the rate of interest? Analysis. — The int. of $360 for $360 x .01 = $3.60 int. i y. I year at I % is $3. 60; and for 2 i^y. 3.60 X 2J =$8.10 " 2\'y. $8.10. Now, if $8.10 is 1% of the $8.10)148.60 principal, $48 60 must be as many Ans. 6 per ct. per cent as $8.10 are contained times in $48.60, which is 6. Therefore the rate was 6 % . Hence, the Rule. — Divide the given interest ly the interest of the principal for the time, at i per cent. 36S. How find the rate, when the principal, rate, and time are given f 262 Il^TEREST. Note. — Sometimes the amount is mentioned instead of the prin^ cipal, or the intercut. In either case, the principal and interest mav be said to be given. For, the amt.=the prin. + int. ; hence, amt.— int,=the prin. ; and amt.— prin. =:the int. (Arts. 365, loi, Def. 17.) 2. If $600 yield 1 10.50 interest in 3 months, what is the rate per cent ? 3. At what rate will $1500 pay me $52.50 interest semi-annually ? 4. At what rate will $1000 amount to $1200 in 3 y. 4 m. ? 5. A lad at the age of 14 received a legacy of I5000, which at 21 amounted to $7800 : what was the rate of int. ? 6. A man paid $9600 for a house, and rented it for $870 a year : what rate of interest did he receive for his money ? 7. At what rate of int. will $500 double itself in 1 2 years ? 8. At what rate will $1000 double itself in 20 y. ? In 10 y.? 9. At what rate will $1250 double itself in 14^ years? 10. At what rate must $3000 be put to double itself in 1 6f years? PROBLEM III. 369. To find the Time, the Principal, the Interest, and the Rate being given. I. Loaned a friend $250 at 6%, and received $45 interest : how long did he have the money ? Analysis.— The int. of $250 at $250 X .06 = $15 int. for 1 v. 6% for I yr. is $15. Now if $15 $15)845 int. int. require the given principal ^nsTT VearS. I yr. at 6%, to earn $45 int., the ' ^ '' same principal will be required as many years as $15 are contained times in $45; and |45-^$i5=3. He therefore had the money 3 years. Hence, the Rule. — Divide tlie given interest hy tlie interest of the principal for i year, at the give7i rate. Note. How find the principal, wlien the amonnt and interest are given ? How find the interest, when the amount and principal are given ? 369. How find tho time, when the principal, interest, and rate are given ? IKTEKEST. 363 Notes.— I. If the quotient contains decimals, reduce them to months and days. (Art. 294.) 2. If the amount is given instead of the principal or the internet, find the part omitted, and proceed as above. (Art. 368. n.) 2. In what time will $860 amount to $989 at 6% ? , ANALYSIS.-The amount $989-$86o=$i29.oo the int., and the int. of $860 for I year at 6^ is $51.60. Now $I29-^$5I.6o=2.5, or 2^ years. 3. In what time will $1250 yield $500, at 7^ ? 4. In what time will $2200 yield $100, at 6% ? 5. In what time will $10000 yield $200, at %% ? 6. In what time will $700 double itself, at 6% ? Solution.— The int. of $700 for i year at 6%, is $42 ; and $700^ $42 = i6| years. ^n«. 7. How long must |i 200 be loaned at 7^ to double itself? 8. In what time will 87500 amount to $15000, at 6% ? 9. In whd,t time will $10000 amount to $25000, at 8^? PROBLEM IV. 370. To find the Principal^ the Interest, the Rate, and tho Time being given. 1. What principal, at 6%, will produce $60 int. in 2| yrs. ? Analysis. — 2^ y.— 30 raos. ; there/ore the int. of $1 for the given time at 6% is 15 cents. 2^ y.=r3o m. Now as $.15 is the int. of $1 for the given j^^ of $ I = $11? time and rate, $60 must be the int. of as many ^^. dollars as $.15 are contained times in $60; ' ^l—. '- and $6o-r-$.i5 = $4oo, the principal required. ^^5. I400 prin. Hence, the Rule. — Divide the given interest hy the interest of%i for t\e given time and rate, expressed decimally. 2. At 6fc, what principal will yield $100 in i year? 3. At 7^, what principal will yield $105 in 6 months? 370. How find the principal, when the interest, rate, and time arc given S64 IKTEEEST. 4. What sum, at 5%, will gain $175 in i year 6 months? 5. What sum must be invested at 6^ to pay the ground rent of a house which is $150 per annum? 6. A gentleman wished to found a professorship with an annual income of I2800 : what sum at j% will produce it? 7. A man invested his money in 6% Government stocks, and received $300 semi-annually: what was the sum invested ? 8. A man bequeathed his wife $1500 a year: what sum must he invest in 6% stocks, to produce this annuity ? PROBLEM V. 371. To find the Principal, the Amount, the Rate, and the Time being giv€n. 1 I. What principal will amount to I508.20 in 3 y. at 6% : Analysis.— 3 y.=36 m. ; therefore the int. of $1 for the given time at 6% is 18 cts., 3 years = 36 m. and the anit.r=$i.i8. (Art. 365.) Now as jnt. $1= $.18 $1.18 is the amt. of $1 principal for the given ^.^ %i=i%i 18 time and rate, $508,20 must be the amount s» * s\ 274 COMPOUJs^D INTEREST. 4. What is the com. int. of $1000 for 2 y. 7 m. 9 d. at 6^? 5. What is the interest of $1360 for 2 years at 7^, compounded semi-annually ? 6. What is the amount of $2000 for 2 years at 4%, com- pounded quarterly ? COMPOUND INTEREST TABLE. Showing the amount of %i, at 3, 4, 5, 6, and 7;;^ com- pound interest, for any number of years from i to 25. Yrs. I. 3%. 4%. 5%. tfo. r/e. 1.030000 1.040000 1.050000 1.060000 1.07 000 2. 1.060900 I.081 600 1. 102 500 1. 123 600 1. 14 490 3- 1.092 727 1. 124 864 1. 157 625 1. 191 016 1.22 504 4 1. 125 509 1. 169 859 1.215 506 1.262477 I.3IO79 5- 1. 159 274 1. 216 653 1.276 282 1.338226 1.40255 6. 1. 194 052 1.265 319 1.340096 I.4I85I9 1.50073 7- 1.229 ^74 1-315932 1.407 100 1.503630 1.60578 8. 1.266 770 1.368569 1-477 455 1.593848 I.71818 9- 1-304773 1.423 312 1-551328 1.689479 1.83 845 10. 1-343916 1.480 244 1.628895 1.790848 1.96 715 II. 1.384234 1-539 451 1.710339 1.898 299 2.10485 12. 1.425 761 1.601 032 1-795 856 2.012 196 2.25 219 13- 1.468534 1.665 074 1.885 649 2.132 928 2.40 984 14. 1.512590 1.731676 1.979932 2.260 904 2-57853 15- 1.557967 1.800 944 2.078 928 2-396558 2.75 903 16. 1.604 706 1.872 981 2.182875 2.540352 2.95 216 17. 1.652848 1.947 900 2.292 018 2.692773 3.15 881 18. 1.702433 2.025 817 2.406 619 2-854339 3-37 293 19. 1-753506 2.106 849 2.526950 3.025*600 3.61652 20. 1.806 III 2.191 123 2.653 298 3-207 135 3.86 968 21. 1.860295 2.278768 2-785 963 3-399 564 4.14056 22. 1.916 103 2.369919 2.925 261 3-603537 4.43 040 23- 1-973587 2.464 716 3-071 524 3-819750 4-74052 24. 2.032 794 2.563304 3.225 100 4-048 935 5.07 236 25- 2.093 778 2.665 ^Z^ 3.386355 4.291 871 5-42 743 COMPOUND INTEREST. '^75 394. To find Compound Interest by the Table. ^ 7. Yfhat is the amount of $900 for 6 yrs. at 7;?, the int ^ compounded annually ? A¥hat is the compound interest ? Solution — Tabular amt. of $1 for 6 yrs. at 75^, $1.50073 The principal, 9°° Amt. for 6 yrs., = $1350.65700 The principal to be subtracted from amt., 900 Compoundlni.ioT 6 yv^., = $450,657 Hence, the Rule. — I. Multiply the tabular amount of%ifor the given time and rate hy the principal; the product tvill he the amount. II. From the amount subtract the principal, and ilie remainder will be the compound interest. Note. — If the given number of years exceed that in the Table, find the amount for any convenient period, as half the given years ; then on this amount for the remaining period. 8. What is the interest of $800 for 9 years at 6'^, com- pounded annually ? 9. What is the int. of $1100 for 12 years at 7;^, com- pounded annually ? 10. What is the int. of $1305 for 16 years at 5^, com- pounded annually? 11. What is the amount of $4500 for 15 years at 4^, compounded annually? 12. What is the amount of $6000 for 25 years at 7;^, compounded annually ? 13. What is the amount of $3800 for 30 years at 6^, compound interest ? 14. What is the compound interest of $4240 at 5% for 40 years ? 15. What is the amount of $1280 for 50 years at 7^ compound interest ? 16. What will $100 amount to in 60 years at 6% com- pound interest ? 276 DISCOUKT. DISCOUNT. 395. Commercial Discount is a deduction of a certain per cent from the price-list of goods, the face of bills, &c., without regard to time. 396. True Discount is the difference between a debt bearing no interest and \i^ present worth. 397. The l^resetit WortJi of a debt payable at a future time without interest, is the sum, which, put at legal interest for the given time, will amount to the debt. 397, a. To find the Net Proceeds of Commercial Discount. 1. Sold goods marked $1560^ a-t 20^ discount, on 4 m., then deducted $% for cash. Required the net proceeds? Analysis. — $1560 x .20 = $312.00, and $1560 — $312 = $1248. $1248 X .05 =$62.40, and $1248— $62.4o=$ii85.6o, net. Hence, the EuLE. — Deduct the commercial discount from the list price, and from the remainder tahe the cash discount. 2. What is the net value of a bill of $3500, at 15^ dis- count, and 5^ additional for cash? Ans. $2826.25. 3. What is the net value of a bill amounting to I5280, at 1 2 i^ discount ? A71S. I4620. 398. To find the Frese^it Worth of a debt, the Rate and Time being given. 1. What is the present worth of $250.51, payable in 8 months without interest, money being worth 6% ? Analysis. — The amount of $1 for 8 $i,o4==:amt. I2 mos., at 6%, is $1.04; therefore $1 is the $1.04)1250.5 1 debt present worth of $1.04, due in 8 mos., $240 871; Ans and$25o.5i^$i.04=$240.875. Hence, the Rule. — Divide the debt hy the amount of %i for the given time and rate ; the quotient will he the present loorth. 2. Find the present worth of $300, due in 10 m., Avhen interest is 7^. 395. What is discount? 396. Commercial? True? 397. Present worth? 397, a. How And net proceeds of commercial discount? 398. Present worth ? Discoui^T. 277 3. Find the present worth of $500, due in i year, when interest is 8^. 4. What is the present worth of a note for $1250, pay- able in 6 mos., interest being 6% ? 5. A man sold his farm for $2500 on i year without interest: what is the present worth of the debt, money being 7^? 6. What is the present worth of a legacy of $5000, payable in 2 years, when interest is 6% ? 399. To find the True Discount, the Rate and Time being given. 7. What is the true discount at 6^0 on a note of $474.03 due in 6 months and 3 days ? ANALYSis.-The amount of $1 for %i.o^ok)%a.a.ox 6 mos. and 3 days is $1.0305. (Art. 365.) ^ --^?~- Therefore, the present worth of the note ^40° is $474.03 -r- 1.0305, or $460. (Art. 398.) $474.03— 46o = $i4.03 But, by definition, the debt, minus the present worth, is the true discount ; and $474.03— $460 (present worth)=$i4,o3, the Ans. Hence, the EuLE. — First find the present worth; then suUrad it from the deht. 8. Find the discount on I2560, due in 7 months, at 6%. 9. Find the discount on $2819, due in 9 months, at <,%. 10. Bought $2375 worth of goods on 6 months: what is the present worth of the bill, at 8^ ? The discount ? 11. A.t 6% what is the present worth of a debt of $3860, half of which is due in 3 months and half in 6 months ? 12. What is the difference between the int. of $6000 for I y. at 6%, and the discount for i y. at 6% ? 13. Bought a house for $5560 on i-J year vithout interest : what would be the discount at 7^, if paid down ? 14. If money is worth 7^, which is preferable, $15000 cash, or $16000 payable in a year without interest? •^99, How find the true discount, when the rate and time are given ? [ ' (yx eAM^^ y \ ^ -^^ 278 BANKS AI^D BAKK DISCO U2hT. BANKS AND BANK DISCOUNT. 400. JSanJcs are incorporated iustitutions which deal in money. They are of four kinds: banks of Dejjosit, Discount^ Circulation, and Savings. 401. A J^anh of Deposit is one that receiyes money for safe keeping, subject to the order of the depositor. A Hanh of Discount is one that loans money, discounts notes, drafts, etc. A Danh of Circulation is one that makes and issues hills, which it promises to pay, on demand. A Savings Sank is one that receives small sums on deposit, and puts them at interest, for the benefit of depositors. 402. Dank Discount is simple interest paid in advance. Notes. — i. A note or draft is said to bo discounted wlien the interest for the given time and rate is deducted from tlie face of it, and the halance paid to the holder. 2. The part paid to the holder is called the proceeds or avails of the note; the part deducted, the discount. 3. The time from the date when a note is discounted to its ma- turity, is often called the Term of Disco ant, 403. To find the Bank Disco^mt, the Face of a Note, the Time, and the Rate being given. I. What is the bank discount on $450 for 4 m. at 6;^? Ana-lysts. — The int. of $1 for 4 m. 3 d. is $.0205 ; and $450 x .0205 —$9,225, the bank discount required. (Art. 380.) Hence, the Rule. — Compute the interest on the face of the note at the given rate, for 3 days more than the given time. To find the proceeds : — Suhtract the discount from the face of the note, 400. What is a bank? 401. A bank of deposit? Discount? Circulation? Bavingsbank? 402. Wiiat is bank discount ? 403. How find tlie bank discount, when the face of a note, the rate and time are given ? IIow find the proceeds ? BANKS AND BAKK DISCOUKT. 579 Note. — If a note is on interest, find its amount at maturity, and taking this as tlie/«ce of the note, cast the interest on it as above. 2. Find the term of discount ainl proceeds of a note for $500, on 90 d., dated June 5th, 1873, nnd discounted July 3d, 1873, at ']%. Ans. 65 d.; Pro., I493.68. 3. Find the proceeds of a note of $730 due in 3m. at 6%. 4. Find the proceeds of a draft for $1000 on 60 d. at 6%. 5. Find the maturity, term of discount, and proceeds of a note of $1740, on 6m., dated May i, '73, and dis. Aug. 21st, '73? at 5^. Ans. Nov. 4th, '73 ; Time, 75 d. ; Proceeds, — . 6. What is the difference between the true and bank discount on $5000 due in lyear at 6%? yr- 7. A jobber buying $7^00 worth of goods for cash, sold them on ^ mos. at 1 2-^% advance, and got the note dis- counted at y/o to pay the bill : how muoh did he make ? 404. To find the Face of a Wote, that the proceeds at Bank Discount shall be a specified sum, the R. and T. being given. 8. For what sum must a note be drawn on 6 mos. that at 6% bank discount, the proceeds shall be $500 ? Analysis. — The bank discount of $1 for 6 mos. 3 d. at 6% is $.0305 ; consequently the proceeds are $.9695. Nuw as §.9695 are the proceeds of Si, $500 must be the proceeds of as many dollars as $.9695 is contained times in $500; and $5oo^$.9695 = $5i5.729, the face of the note. Hence, the EuLE. — Dioide the rjivoi j^i^oceeds 1)]) the proceeds of %\ for the given time and rate. 9. What must be the face of a note on 4 mos. that when discounted at ']% the proceeds may be $750 ? 10. What was the face of a note on 60 days, the pro- ceeds of which being discounted at 5^, were I1565 ? 11. If the avails are $2165.45, the time 4 mos., and the rate of bank discount 8;:^, what must be the face of the note? 12. Bought a house for $7350 cash : how large a note on 4 mos. must I have discounted at bank 6% to pay this sum? 404. How And how large to make a note to raise a Bpecifled sura, when the rate ftfid time are ifhrcn ? 2B0 STOCK INVESTMEiJ^TS. STOCK INVESTMENTS. 405. Stocks are tlie funds or capital of incorporated companies. An Incorporated Cofnpany is an association authorized by law, to transact business. Note. — Stocks are divided into equal parts called shares, and tlie owners of tlie shares are called stockholders. These shares vary from $25 to $500 or $1000. They are commonly $100 each, and will be BO considered in the following exercises, unless otherwise stated. 406. Certificates of Stoclz are written statements, specifying the number of shares to which holders are entitled. They are often called scrip. 407. The Var value of stock is the sum named on the face of the scr'.p, and is thence called its nominal value. The Market value is the sum for whicli it sells. • Notes. — i. When shares sell for their nominal value, they are at •par ; when they sell for riiore, they are ahove par, or at a premium ; when they sell for less, they are helow par, or at a discount. 2. When stocks sell at par, they are often quoted at 100 ; when 8 % ahov:e par, at 108 ; when 8 % 'below par, at 92. The term par, Latin, signifies equal ; hence, to be at par, is to be on an equality. 408. The Gross Eai'nings of a Company are its entire receipts. The Net Earnings are the sums left after deducting all expenses. 409. Instalments are portions of the capital paid by the stockholders at diflPerent times. 410. Dividends are portions of the earnings distributed among the stockholders. They are usually made at stated periods ; as, annually ; semi-annually, etc. 411. A Hond is a writing under seal, by which a party binds himself to pay the holder a certain sum, at or before a specified time. 405. What are stocks? An incorporated company? Note. Into what are stocks divided ? What are stockholders ? 406. What are certificates of stock ? 407. What is the par value of stock? The market value? Note. When are stockn at par? Above par? Below par? The meaning of the term par? 40S. What are the gross earnings of a company ? The net earuiu^'s ? 409. Whal nre Instalments ? 410. Dividends? 411. What is a bond? STOCK IlirVESTMEKTS. 28:1 UNITED STATES BONDS. 412. United States Bonds are those issued by Government, and are divided into two classes : those pay- able at a give7i date, and those payable within the limits of two given dates, at the option of the Government. Note. — The former are designated by a combination of the numerals, which express the rate of interest they bear, and the year they become due ; as " 6s of '8i." The latter are designated by combining the numerals expressing the two dates between which they are to be paid ; as " 5-203." 413. A Coupon is a certificate of interest attached to a bond, which, on the payment of the interest, is cut oif and delivered to the payor. 414. The principal XJ. S. bonds are the following : 1. "6s of '81," bearing 6% interest, and payable in 1881. 2. " 5-20S," bearing dfo interest, and payable in not less than 5 or more tlian 20 years from their date, at the pleasure of the Govern- ment. Interest paid semi-annually in gold. 3. " 10-40S,' bearing 5 % interest, redeemable after 10 years from their date, interest semi-annually in gold. 4. " 5s of '81," bearing 5% interest, redeemable after 1881, interest paid quarterly in gold. 5. "4^s of '86," bearing ^\% interest, redeemable after 1886, the interest paid quarterly in gold. 6. "4s of 1901," bearing 4% interest, the principal payable after 1901, the interest paid quarterly in gold. 415. State, City, I^ailroad Bonds, etc., are payable at a specified time, and are designated by annexing the numeral denoting the rate of interest they bear, to the name of the State, etc., by which they are issued ; as. New York 6s ; Georgia 7s. 416. Computations in stocks and bonds are founded upon the principles of percentage; the par value being the base, the ^;er cent premium the rate, the premium, etc., the percentage, and the market value the amount. 412. What are U. S. bonds? How many classes? Note. How are the former designated ? The latter ? 413. What is a coupon ? 414. What is meant by TJ. S. ^ of '81 f By U. S. 5-203 ? By U. S. 10-40B ? By new U. S. 58 of '81 ? ^2 STOCK IKYESTMENTS Note.— Tlie comparative profit of inveBtments in U. S. Bonds de- pends upon their market value, the rate of interest tliey bear, and the premium on ^old. That of Railroad and other Stocks upon their market value, and the per cent of their dividends. 417. To find the Preiiiium, Discount, Tnstahnienf, or Dividend^ the Par Value and the Rate being given. 1. What is the premium on 20 shares of the New Orleans National Bank, at 8^ ? . Analysis. — 2os.=$2ooo; and $2000 x .o8 = $i6o, Ans. Hence, the Rule. — Multij^ily the par value hy the rate, expressed decimally. (Art. 336.) 2. What is the discount on 27 shares of Michigan Central Railroad at g% ? 3. What is the premium on three 81000 U. S. 6s of ^81, when they stand at 17 J;^ above par ? 4. The Maryland Coal Company called for an instalment of i$%\ what did a man pay who owned 35 shares? 5. What is the discount on $4000 Tennessee 6s, at 10^? 6. The Virginia Manufacturing Company declared a dividend of 17/^: to what sum were 28 shares entitled? 418. To find thf^ 3Iai'kef Value of Stocks and Bonds, the Rate and Nominal Value being given. 7. What is the market value of 45 shares of New York Central, at 4% premium ? Analysis.— 45s. = $4500; and $4500x1.04 (i + the rate)=$468o, the value required. Hence, the Rule. — Multip)ly the nominal value hy 1 plus or minus the rate. Or, multiply the market value of i share hy the number cf shares. Note. — In finding the net value of stocks, the hrokerage, postage stamps, and other expenses must be deducted from the market value, 8. What is the worth of 1 14000 of Kentucky 6s, at 92 ? 417. How find the premium, etc., when the par value and rate are given ? STOCK II^VESTMENTS. 283 9. What is the worth of an investment of $9500 in (J. S. 5-20S, at io|^;^ premium ? 10. What is the net value of 58 shares of New Jersey Central, at no; deducting the express and other charges $1.89, and brokerage at i% ? 11. What will be realized from $7500 Texas 7s, at 15^ discount, deducting the brokerage at ^%, and $1.39 for postage and other charges. 12. What will $10000 U. S. 5-20S cost at S^^o premium, adding brokerage at I'r, and other expenses $2.37^? 13. What is the value in currency of $15750 gold coin, the premium being 2 2j;7^? 14. A person has $10000 U. S. 5-20S: what will he ra- ceive annually in currency, when gold is 12^' premium? 419. To find the Mate, the Par Value, and the Dividend, Premium, or Discount being given. 15. The capital stock of a company is $100000; its gross earnings for the year are $34500, and its expenses $13500: deducting from its net earnings $1000 as surplus, what per cent dividend can the company make ? Analysis. — $13500 + $1000 = $14500, and $34500 — $14500 = $20000, the net earnings. Now 20000-7- $100000=. 20 or 20^, the rate required. Hence, the EuLE. — Divide the premium or discount, as the case may be, hy the par value. (Art. 339.) 16. Paid $750 premium for 50 shares of bank stock: what was the per cent ? 17. The discount on 100 shares of the Pacific Eailroad is $625 : what is the per cent below par? 18. The net earnings of a company with a capital of I480000 are $35000; reserving $3000 as surplus, what per cent dividend can they declare ? /.18. How find the market value when the rate and nominal value are given S 41). How find the rate, when the premium, discount, or dividend are given? 284 STOCK i:N^VESTME:NrTS. 420. To find how much Stock can be bought for a specified sum, the Rate of Premium or Discount being given. 19. How much of Kansas 6s, at 20^ discount, can be bought for $5200? Ai!^'ALYSis. — $5200-^.80 (val. of $1 stock)=$65oo, Ans. Hence, the Rule. — Divide the sum to be invested hy i plus or minus the rate, as the case may he. (Art. 349.) Or, divide the given sum hy the marhet value of%\ of stock, 20. How many $100 U. S. 10-40S, at ^% premium, can be bought for I4200 ? 2 1. What amount of Virginia 6s, at 90, can be purchased for 1 1 0800? 421. To find what Sum must be invested in Bonds to realize a given income, the Cost and Rate of Interest being given. 22. What sum must be invested in Missouri 6s at 90, to reahze an income of % 1 800 annually ? Analysis. — At 6% the income of $1 is 6 cts. ; and $1800-4- .06= $30000, the nominal value of the bonds. Again, at .90, $30000 of bonds will cost .90 times $30000= $27000, the sum required. Hence, the Rule. — I. Divide the given income hy the annual interest 9/ 1 1 of honds; the quotient ivill he the jiominal value of the honds. II. Multiply the nominal value hy the marhet value of$i of honds J the product ivill he the sum required. 23. What sum must be invested in IJ. S. 5-20S, at 106, to yield an annual income of ^2500 in gold ? 24. How much must one invest in Wisconsin 8s, at 95, to receive an annual income of $3000 ? Ans. $35625. 25. How much must be invested in Mississippi 6s, at 80, to yield an annual interest of $4200 ? 420. How find the quantity of stock that can be bought for a epecifled sum, when the rate is given ? EXCHANGE. 422. JExchange is a method of making payments between distant places by Bills of Exchange. 423. A Bill of Exchange is an order or draft directing one person to pay another a certain sum at a specified time. Note. — The person who signs the bill is called the drawer or maker ; the one to whom it is addressed, the drawee; the one to whom it is to be paid, the payee ; t*he one who sends it, the remitter. 424. Bills of Exchange are Domestic or Foreign. Domestic Bills are those payable in the country where they are drawn, and are commonly called Drafts. Foreign Bills are those drawn in one country and payable in another. 425. A Sight Bill is one payable on its presentation, A Time Bill is one payable at a specified time after its date, or presentation. 426. The Bar of Exchange is the standard by which the value of the currency of difierent countries is compared, and is either i^itrinsic or no?mnal. An Intrinsic Bar is a standard having a real and fixed value represented by gold or silver coin. A N^oininal Bar is a conventional standard, having any assumed value which convenience may suggest. 427. When the market price of bills is the same as the face, they are at par ; when it exceeds the face, they are above par, or at a premium; when it is less than the face, they are helow par, or at a discount. (Art. 407, n.) Notes.— I. The fluctuation in the price of bills from their par f*lue, is called the Course of Exchange. 42a. What is exchange T 423. A bill of exchange ? 424. What are Jomestic bflla T Foreign bills ? 425. What are eight bills ? Time bills ? 436. What ia the par of exchange ? An intrinsic par ? A nominal par ? 28G DOMESTIC EXCHANGE. 2. The rate of Exchange between two places or countries depen^'a upon the circumstances of trade. If the trade between New York and New Orleans is equal, exchange is at par. If the former owls the latter, the demand for drafts on New Orleans is greater than the supply; hence they are above par in New York. If the latter owes the former, the demand for drafts is less than the supply, con- sequently drafts on New Orleans are helou) par. 428. An accex>tance of a bill or draffc is an engage- ment to pay it according to its conditions. To show this, it is customary for the drawee to write the word accepted across the face of the bill, with the date and his name. Note. — Bills of Exchange are negotiable like promissory notes, and the laws respecting their indorsement, collection, protest, etc., are essentially the same. DOMESTIC OR INLAND EXCHANGE. 429. To find the Cost of a Draft, its Face and the Ra'.e of Exchange being given. 1. What is the cost of a sight draft on New York, for I4500, at 2\% premium ? Analysis.— At 2\% premium, a draft of |i will I4500 cost $1 + 2^ cts. = $i.o25, or 1.025 times the draft. I.025 Hence, the cost of $4500 draft will be 1.025 times a ^TtTs^co its face ; and $4500 X i.o25=$46i2. 50, ^?i«. '^ 2. "VYhat is the cost of a sight draft on St. Louis of $5740, at 4% discount ? Analysis. — At 4^ discount, a draft of $1 will $5740 cost $1—4 cents=$0 96, or .96 times the draft .q6 Therefore, the cost of $5740 draft will be .06 times i7T7r~rr~ ^ ^, „ ats face ; and $5740 x .96=15510.40, Hence, the ^^ EuLE. — Multiply the face of the draft ly the cost of $1 of draft, expressed decimally. 427. Whan are bills at par ? When ahove par ? When below ? Note. What is the fluctuation in the price of exchange called ? Upon what does the rate of excuange depend ? Explain? 428. The acceptance of a bill? 429. How find the 'JOBt of a draft, when the face and rate are given ? DOMESTIC EXCHA2^aE. 287 Notes. — i. When payable at sight, the worth of $i of draft is $i p?MS or minus the rate of exchange. 2. On time drafts, both the exchange and the bank discount are computed on their face. Dealers in excliange, however, make but. one computation ; the rate for time drafts being enough less than sight drafts to allow for the bank discount. 2. A merchant in Galveston bouglit a draft of $2000 on Philadelphia at 60 days sight : what was the draft worth, the premium being 3^^, and the bank discount Gfc 3. What cost a sight draft of $3560, at 2^ discount? ^ 4. Eequired the worth of a draft for 84250 on Chicago, at 90 days, sight drafts being 1% discount and interest "]%. ^ 5. The Bank of New York having declared a dividend of 4^, a stockholder living in Savannah drew on the bank for his dividend on 50 shares, and sold the draft at \\% premium : how much did he realize from the dividend ? 430. To find the Face of a Draft, the Cost and Rate of Exchange being given. 7. Bought a draft in Omaha on New York, payable in 90 days, for $3043.50, exchange being 2>% premium, and interest (i% : what w^as the face of the draft ? Analysis. — The cost of $r of |j. ^_ ^a/__^j_q^ draft is $1 plus the rate, minus pjg^ ^^ ^^ fo^, ^^^^^o oj^^ the bank discount for the time. .-. i r. a ■\dl is Now $i + 3%-|i.o3; the bank Cost of $i dft. = $l.oi45 discount on $1 for 93 d. is $0.0155, i-OM5)^3043-5o and $1.03 - $0.0155 = $10145. Face of dft.=: $3000. Now if I1.0145 will buy $1 of draft, $3043.50 will buy a draft of as many dollars as 1.0145 is con- tained times in $3043.50. $3043.50-^1.0145 — $3000, the draft re- quired. Hence, the KuLE. — Divide the cost of the draft ly the cost of %i of draft, expressed decimally. 8. What is the face of a sight draft purchased for $1250, the premium being 2^% ? 430. How find the face of a draft, the cost and rate being given ? 288 ¥OREIGKEXCHAKGE. 9. What is the face of a draft at 60 days sight, purchased for 1 1 500, when interest is 8^, and the premium 2%? ^A -^10. What is the face of a sight draft, purchased for $2500, the discount being 4^^ ? y/ — 1 1. What is the face of a draft on 4 m., bought for I3600, ^ the int. being 6^, and exchange 2% discount ? J"^ 12. A merchant of Natchez sold a draft on Boston at *^ r^% prem., for $3806.25 : what was the face of the draft ? FOREIGN EXCHANGE, 431. A Foreign Bill of Uxchanr/e is a Bill drawn in one country and payable in another. A Set of Exchange consists of three bills of the same date and tenor, distinguished as the First, Second, and Third of exchange. They are sent by different mails, in order to save time in ease of miscarriage. When one is paid, the others are void. 432. The Legal JPar of Exchange between Great Britain and the United States, is $4.8665 gold to the pound sterling.* 433. To find the Cost of a Bill on England, the Face and the Rate of Exchange being given^ I. What is the cost of the following bill on London, at $4.8665 to the £ sterhng ? ^354> 1 28- New Yobs, July 4th, 1874. At sight of this first of exchange (the second and third of the same date and tenor unpaid), pay to the order of Henry Crosby, three hundred and fifty-four pounds, twelve shillings sterling, value received, and charge the same to the account of 0. J. King & Co. To George Peabodt, Esq., London. 431. What is a foreign bill of exchange ? What is the legal par of exchang* with Great Britain ? * Act of Congress, March 3d, 1873— To t*^^ eflbct Jan. ist, 1874. FOREIGN EXCHAI^GE. 289 Analysis. — Reducing tlie given sliillings to opbbation. the decimal of a pound, £354, 12s. — £354.6. 4.8665 (Art. 295.) Now if £1 is worth $4.8665, £354-6 3_54'^ are worth 354.6 times as much; and $4.8665 x Ans, $1725.661 354.6=$i725.66i, the cost of the bill. 2. What is the value of a bill on England for ^£43 6, 5 s. 6d., at $4.85 J to the £ sterling ? Analysis. — £436, 5s. 6d.=£436.275, and the operation. market value of exchange is $4.8525 to the £. $4.8525 Now $4.8525 X436.275=$2ii7.02, the cost of the 436.275 bill. Hence, the A^g. "$2117.02 Rule. — Multiply the market value of £1 sterling ly the face of the bill ; the product will be its value in dollars and cents. Notes. — i. If there are shillings and pence in the given bill, they should be reduced to the decimal of a pound. (Art. 295.) 2. Bills on Great Britain are drawn in Sterling money. 3. The New Par of Sterling Exchange $4.8665, is the intrinsic value of the Sovereign or pound sterling, as estimated at the United States Mint, and is 9? % greater than the old par. 4. The Old Par, which assumed the value of the £ sterling to be $4.44^, is abolished by law, and all contracts based upon it after January ist, 1874, are nvM and void. 3. What is the cost of a bill on DubUn for ^£381, at $4.87!- to the ie sterling ? 4. What is the cost in currency of a bill on England for ^6750, exchange being at par, and gold 2>^l% premium? 5. B owes a merchant in Liverpool ^£1500; exchange is $4.93!; to transmit coin will cost 2^ insurance and freight; and when delivered its commercial value is $4.80 to a } sound: which is the cheaper, to buy a bill, or send the gold ? How much ? 6. A New Orleans merchant consigned 568 bales of cot- ^ ton, weighing 450 lbs. apiece, to his agent in Liverpool; the agent paid id. a pound freight, and sold it at i2d. a 433. How find the cost of a bill on England, when the face and rate are given ? 290 FOREIG^q" EXCHANGE. pound ; he charged 2 J^ commission, and £S, 6s. for stor- age: what did the merchant realize for his cotton, ex- change on the net proceeds being 1491^ to the £ ? 434. To find Ihe Face of a Bill on England, the Cost and the Rate of Exchange being given* 7. What is the face of a bill on England which cost $1725.72, exchange being at par, or I4.8665 to the £? ANALYsis.-Smce $4.8665 buys £1 of $4.866s)li72t;.72 bill, $1725.72 will buy as many pounds as — ^— — — — $4.8665 are times in $1725.72, or £354.612 *'354-oi2 =£354, I2S. 2|d. Hence, the ^^^5. £354, 12s. 2jd. KuLE. — Divide the cost of the hill ty the market value of £1 sterling ; the quotient will he the face of the hill. 8. What is the face of a bill on London which cost $2500, exchange being $4.88| to the £'^ 9. What amount of exchange on Dubhn can be obtained for $3750, at $4.84^ to the £ sterling? 10. A merchant in Charleston paid $5000 for a bill on London, at $4.87^ to the £-. what was the face of the bill? 11. Paid 1 7 500 for a bill on Manchester: what was the face of it, exchange being I4.86J- to the iB? 434, a. In quoting Exchange on Foreign Countries, the gen- eral rule is to quote the money of one country against the money of other countries. Thus, exchange is quoted On Austria, in cents to the Florin (silver) =$0476. On Frankfort, in cents to the Florin (gold) =$0.4165. On the German Empire, in cents to the Mark (gold)= $0.2382. On North Germany, in cents to the new Thaler (silver) =$0.7 14. On Russia, in cents to the Rouble (silver) =$0.7717. Note. — Bills of Exchange between the United States and foreign countries are generally drawn on some of the great commercial centers ; as London, Paris, Frankfort, Amsterdam, etc. 434. IIow find the face of a bill on Bngland, when the cost and rate are given ? FOREIGN" EXCHANGE. 291 435. Bills Oil France are drawn in French cur- rency, and are calculated at so manj francs and centimes to a dollar. Note. — Centimes are commonly written as decimals of a franc Tims 5 francs and 23 centimes are written 5.23 francs. (Art. 224, n.) 436. To find the Cost of a Bill on France, the Face ana Rate of Exchange being given. 12. What is the cost of a bill on Paris of 1500 francs, exchange being 5.25 f to a dollar? Analysis. — Since 5.25 francs will buy a 5.25 fr.) 1500.00 fr. bill of $1, 1500 francs will buy a bill of as Ans "^28=; 7^ 4- many dollars as 5.2', is contained times in 1500; and 1500-7-5. 25 = $285. 71, the cost required. Hence, the Rule. — Divide the face of the Mil ly the number of francs to %i exchange. 13. What is the cost of a bill of 3500 francs on Havre, exchange being 5.18 francs to a dollar? 437. To find the Face of a Bill on France, the Cost and the Rate of Exchange being given. 14. A traveler paid I300 for a bill on Paris; exchange being 5.16 francs to $1 : what was the face of the bill ? Analysis. — If $1 will buy 5.16 francs, $300 will buy 300 times as many ; and 5.16 fr. x 300=1548 francs, the face of the bill required. Hence, the Rule. — Multij^ly the number of francs to %i exchange bt, the cost of the bill. 15. Paid $2500 for a bill on Lyons, exchange being 5.22 francs to $1 : what was the face of the bill ? • 16. A merchant paid $3150 for a bill on Paris, exchange being 5.23 francs to |i : what was the face of the bill ? 435. How is exchangee on France calculated ? 436. How find the cost of a bill on France, when the face and rate are given? 437. How find the face, when the cost and rate are gireu ? 438. Insurance is indemnity for loss. It is dis- tinguished by different names, according to the cause of the loss, or the object insured. 439. Fif^e Insurance, is indemnity for loss by fire. Marine Insurance, for loss by sea. Life Insurance, for the loss of life. A.ccident Insurance, for personal casualties. Health Insurance, for personal sickness. Stocic Insurance, for the loss of cattle, horses, etc. Notes. — i. The party who undertakes the risk is called the Insurer or Underwriter. 2. The party protected by the insurance is called the Insured. 440. A JPolicy is a writing containing the evidence and terms of insurance. The Premiiini is the sum paid for insurance. Note. — The business of insurance is carried on chiefly by In- corporated Companies. Sometimes, however, it is undertaken by individuals, and is then called out-door-insurance. 441. Insurance Companies are of two kinds : Stock Companies, and Mutual Companies. A Stock Insurance Company is one which has a paid up capital, and divides the profit and loss among its stockholders. A Mutual Insurance OoJW^atii/ is one in which the lossea are shared by the parties insured. 442. Premiums are computed at a certain per cent of the sum insured, and the operations are similar to those in Percentage. Note. — Policies are renewed annually, or at stated periods, and the premium is paid in advance. In this respect insurance diflFera from commission, etc., which have no reference to time. 438. What is insurance? 439. Fire insurance? Marine? Life? Accident? Health? Stock? 440. What is a policy ? The premium? 441. How many kinds of insurance companies? A stuck insurance company? A mutual? 442. How are premiums computed ? II^SURAKCE. 293 FIRE AND MARINE INSURANCE. 443. To find the Preniui^n,f the Sum insured and the Rate for the period being given. 1. What premium must I pay per annum for insuring ^1750 on my house and furniture, at ^%'t Analysis. — $1750 x .005 (rate)=$8.75, Ans. Hence, tlie EuLE. — Multiply the sum insured ly the rate. (Art. 2,2)^.) Note. — The rate of insurance is sometimes stated at a certain number of cents on $100. In such cases tlie rate should be reduced io the decimal of $1 before multiplying. Thus, if the rate is 25 cents on $100, the multiplier is written .0025. (Art. 295.) 2. At 35 cents on $100, what is the insurance of $1900? 3. At i\%, what is the premium on $2560? 4. At 2^^, what is the premium on $3750? 5. At 25 cents on $100, what is the premium on $4280? 6. At 50 cents on $100, what is the premium on $5000? 7. At 3^^, what is the cost of insuring $6175 ? 8. What is the premium for insuring a ship and cargo valued at $35000, at 2Y/c ? 9. What is the cost of insuring a factory and its con- tents, valued at I48250, at 3 J;|, including %i\ for policy ? 444. To find what Sum must be insured to cover both the Property and Premium, the Rate being grven. 10. For what must a factory worth $20709 be insured, to cover the property and the premium of 2^% ? Analysis. — The sum to be insured includes the property plus the premium. But the property is 100% of itself, and the premium is 2.\fc of that sum ; therefore $20709=100%— 2^%, or .97^ times the sum to be insured. Now, if $20709 is .97^ times the required sum, once that sum is $20709 -=-.971, or $21240, Ans. (Art. 337.) Hence, the EuLE. — Divide the value of the property by 1 minus the rate. Note. — This and the preceding problem cover the ordinary cases of Fire and Marine Insurance. Should other problems occur, they may be solved like the corresponding problems in Percentage How find the premium, when the sum insured and ttie rate arc given I 294 I2^SURAKCE. 11. A merchant sent a cargo of goods worth $15275 to Canton: what sum must he get insured at 2>%, that he may suffer no loss, if the ship is wrecked ? 12. A house and furniture are worth $27250 : what sum must be insured at 2% to cover ihQ property and premium ? 13. What sum must be insured at <^% to coyer the premium, with a vessel and cargo worth I35250? LIFE INSURANCE. 445. Life Insurance Policies are of different hinds, and the premium varies according to the expectation of life. ist. Life Policies^ whicli are payable at the death of the party- named in the policy, the annual premium continuing through life. 2d. Lfife l^oUcieSf payable at the death of the insured, the annual premium ceasing at a given age, 3d. Term Policies , payable at the death of the insured, if he dies during a given term of years, the annual premium continuing till the policy expires. 4th. JEhidotvinent Policies, payable to the insured at a given age, or to his heirs if he dies before that age, the annual premium continuing till the policy expires. Note. — The expectation of life is the average duration of the life of individuals after any specified age. 1. What premium must a man, at the age of 25, pay annually for a life policy of $5000, at 4^% ? Analysis. — 4^% =,045, and $5000 x .045 = $22 5, Ans. (Art. 443.) 2. What is the annual premium for a life policy of $2500, at 5^? 3. A man at the age of 35 years effected a life insurance of I7500 for 10 yrs., at 3^^: what was the amt. of premium? 4. A man 65 years old negotiated a life insurance of $8000 for 5 years, at 1 2}^ : what was the amt. of premium ? 5. At the age of 30 years, a man got his life insured for ^75000, at 4% per annum, and lived to the age of 70 : which was the greater, the sum he paid, or the sum received? 444. How find what sum must be insured to cover the property and premium 1 TAXES. 446. A Tax is a sum assessed upon the person or property of a citizen, for public purposes. A Property Tax is one assessed upon jjroperty. A JPersonal Tax is one assessed upon the perso?i, and is often called a poll or capitatioji tax. Note. — The term poll is from tlie German polU, tlie liead ; capita- tion, from the Latin caput, the head. 447. I*roperty is of two kinds, real and personal. Ileal T*roperty is that which is fixed; as, lands, houses, etc. It is often called real estate. JPersonal Property is that which is movable; as, money, stocks, etc. 448. An Assessor is a person appointed to appraise property, for the purpose of taxation. 449. An Inventory is a list of taxable property, with its estimated value. 450. Pfoperty taxes are computed at a certain per cent on the valuation of the property to be assessed. Tiiat is. The valuation of the property is the base ; the sum to be raised, the percentage ; the per cent or tax on $i, the rate ; the sum collected minus the commission, the net proceeds. Poll taxes are specific sums upon those not exempt by law, without regard to property. 451. To assess a Property Tax, the Valuation and the Sum to be raised being given. I. A tax of $6250 was levied upon a corporation of 6 per- sons; A's property was appraised at $30000; B's, I37850; C's, $40150 ; D's, $50000 ; E's, $55000 ; Fs, $37000. What was the rate of the tax, and what each man's share ? 446. What is a tax ? A property tax ? A personal tax ? Note. What is a per- sonal tax called ? 447. Of how many kinds is property ? What is real property ? Personal ? 448. What is an assessor ? 449. An inventory ? 450. How are prop- erty taxea computed? How are poll taxes levied? 451. How is a property tax assessed, when the valuation and the sum to be raised are given ? 296 TAXES. AnsALYsis. — The valuation=$300oo + $37850+ $40150+ $500004- $55000 + $37000= $250000, The valuation $250000, is the hase ; and the tax $6250, the percentage. Therefore $625o-t-$25oooo=.025, or 2hfc, the rate required. Again, since A's valuation was $30000, and the rate 2^5^, his tax must have been 2i% of $30000; and $30000 x .025 =$750.00. The tax of the others may be found in like manner. Hence, the EuLE. — I. MaJce an inventory of all the taxable property. II. Divide the sum to le raised ly the amount of the inventory y and the quotient will he the rate. III. Multiply the valuation of each man^s property hy the rate, and the product will le his tax. Notes. — i. If a poll tax is included, the sum arising from the polls must be subtracted from the sum to be raised, before it is divided by the inventory. 2. If the tax is assessed on a large number of individuals, the operation will be shortened by first finding the tax on $1, $2, $3, etc., to $9 ; then on $10, $20, etc., to $90 ; then on $100, $200, etc., to $900, etc., arranging the results as in the following ASSESSORS' TABLE. $1 pays $.025 $10 pay $.25 $100 pay $2.50 2 a .050 20 .50 200 5.00 3 a .075 30 ii •75 300 ii 7-50 4 a .100 40 ii 1. 00 400 .10.00 5 a •125 50 ii 1-25 500 ii 12.50 6 a .150 60 ii 1.50 1000 ii 25.00 7 a •175 70 ii 1-75 2000 ii 50.00 8 a .200 80 ii 2.00 3000 ii 75.00 9 i( .225 90 ii 2.25 4000 a 100.00 2. B's yaluation=:$3785o=$3oooo + $7ooo + $8oo + $5o, By the table the tax on $30000 =$7 5 0.00 " " 7000= 175.00 " " 800=: 20.00 50= 1^ Therefore, we have B's tax, =$946.25 Note. If a poll tax is included, how proceed ? TAXES. 297 3. Required C, D, E, and F's taxes, both by the rule and the taUe. 4. A tax levied on a certain township was $16020; the valuation of its taxable property was $784750, and the number of polls assessed at $1.25 was 260. What was the rate of tax ; and what was A's tax, who paid for 3 polls, the valuation of his property being I7800 ? 5. The State levied a tax of $165945 upon a certain city which contained 1260 polls assessed 75 cents each, an in- ventory of I5427600 real, and $72400 personal property. What was G's tax, whose property was assessed at $15000 ? 6. What was H's tax, whose inventory was $10250, and 3 polls ? 7. A district school-house cost $2500, and the valuation of the property of the district is $50000 : what is the rate, and what A's tax, whose property is valued at $3400 ? 452. To find the Antoiuit to be assessed, to raise a net sum, and pay the Commission for collecting it. 8. A certain city required $47500 to pay expenses: what amount must be assessed in order to cover the expenses, and the commission of 5^ for collecting the tax ? Analysis. — At 5 % for collection, $1 assessment yields $.95 ; tlicrefore, to obtain $47500 net, requires as many dollars assessment as $.95 are contained times in $47500; and $47500-?-. 95 (1 — 5^) = $50000, the sum required to be assessed. (Art. 341.) Hence, the KuLE. — Divide the net sum ly i miiius the rate; the quotient will be the amount to be assessed. (Art. 348.) 9. What sum must be assessed to raise a net amount of $3500, and pay the commission for collecting, at 4% ? 10. What sum must be assessed to raise a net sum of $5260, and pay for the collection, at 4^% commission ? 11. What sum must be assessed to raise a net amount of $10500, and pay the commission for collecting, at 5^? 452. How find the amount to be assessed, to cover the sum to be raised and the commission for collection ? DUTIES. 453. Duties are sums paid on imported goods, and are often called customs. 454. A Custom House is a building where duties are re- ceived, ships entered, cleared, etc. 455. A Port of Entry is one where there is a Custom House. The Collector of a Fort is an officer who receives the duties, has the charge of the Custom House, etc. 456. A Tariff is a list of articles subject to duty, stating the rate, or the sum to be collected on each. 457. An Invoice is a list of merchandise, with the cost of the SGveral articles in the country from which they are imported. 458. Duties ^re of two kinds, specitic and ad yalorem. 459. A Specific Duty is a fixed sum imposed on each article, ton, yard, etc., without regard to its cost. An Ad Valorem Duty is a certain joer cent on ilie value of goods in the country from which they are imported. Note. — The term ad valorem is from the Latin ad and valorem, according to value. 460. Before calculating duties, certain allowances are made, called tare, tret or draft, leakage, and breakage. Tare is an allowance for the weight of the box, bag, cask, etc., containing the goods. Tret is an allowance in the weight or measure of goods for wa£te or refuse matter, and is often called draft. Leakage is an allowance on liquors in casks. JBreaJcage is an allowance on liquors in bottles. Notes. — i. Tare is calculated either at the rate specified in the invoice, or at rates established by Act of Congress. 2. Leakage is commonly determined by gauging the casks, and Breakage by counting. 3. In making these allowances, if the fraction is less than i it is rejected, if ^ or more, i is added. 45^. What are duties? 454. A custom house? 455. What is a port of entry? A collector? 456. A tariff? 457. What is an invoice? 45S. Of how many kinds are duties ? 459. A specific duty ? An ad valorem ? 460. What is tare ? Tret ? Leakage ? Breakavties, the quantity of goods, and the Sum levied on each article being given. 1. What is the specific duty on 12 casks of brandy, each containing 40 gal., at $1 J per gal., allowing 2% for leakage ? Analysis. — 12 casks of 40 gallons each, =480.0 gal. The leakage at 2fo on 480 gallons, == 9.6 gal. The remainder, or quantity taxed, =470.4 gal. Now $1.50 X 4704 = $705.6o, the duty required. Hence, the Rule. — Deduct the legal alloivance for tare, tret, etc., from the goods, and muUij^ly the remainder hy the sum levied on each article. 2. What is the duty, at $1.25 a yard, on 65 pieces of brocade silk, each containing 50 yards ? 3. What is the duty on 87 hhds. of molasses, at 20 cts. per gallon, the leakage being 3^ ? 4. What is the duty, at 6 cts. a pound, on 500 bags of coffee, each weighing 6d> lbs., the tare being 2% ? PROBLEM II. 462. To calculate Ad Valorcni Dufia^f the Cost of the goods and the Rate being given. 5. What is the ad valorem duty, at 2^%, on a quantity of silks invoiced at $3500 ? Analysis. — Since the duty is 25% ad valorem, $35°° it is .25 times the invoice; and $3500 x .25 = 1875. .25 Hence, the J^^s. $875.00 KuLE. — Multiply the cost of the goods hy the given rate, expressed decimally. 6. What is the duty, at 2,zi% ^^ valorem, on 1575 yds. of carpeting invoiced at $1.80 per yard? 7. What is the ad valorem duty, at 40^, on no chests of tea, each containing 67 lbs., and invoiced at 90 cts. a pound, the tare being 9 lbs. a chest ? 461. How calculate specific duties when the quantity of goods and the sum evieJ on eac h article are given ? 462. How ad valorem ? INTEE]S"AL EEYE]^UE. 463. Intei^nal Hevenue is the income of the Government from Excise Duties, Stamp Duties, LicenseSj Special Taxes, Income Taxes, etc. 464. Excise Duties are taxes upon certain home productions, and are computed at a given per cent on their value. 465. Stamp Duties are taxes upon written instruments ; as, notes, drafts, contracts, legal documents, patent medicines, etc. 466. ^ License Tax is the sum paid for permission to pursue certain avocations. 467. Special Taxes are fixed sums assessed upon certain articies of luxury ; as, carriages, billiard tables, gold watches, etc. Income Taxes are those levied upon annual incomes. Note. — In determining income taxes, certain deductions are made for house rent. National and State taxes, losses, etc. 468. To compute Income Taxes, the Rate being given. 1. What is a man's tax whose income is $5675 ; the rate being 5^, the deductions $1000 for house rent, $350 national tax, and $1 100 for losses ? Analysis. — Total deductions are $1000 +$350 + $1100= $2450; and $5675 — $2450= $3225 taxable income. Now 5% of $3225 = 3225 X .05 = $161. 2 5, the tax required. Hence, the EuLE. — From the ioicome subtract the total deductions, and multiply the remainder hy the rate. Note. — If there are special taxes on articles of luxury, as carriages, etc., they must be added to the tax on the income. 2. What was A's revenue tax for 1869, at 5^ ; his income being $4750; bis losses $1185, and exemptions $1200 ? 3. In 1870, A had $10500 income; 35 oz. taxable plate at 5 cts., I watch $2, and i carriage I2 : what was his tax at K%, allowing $2100 exemption ? 463. What is internal revenue? 464. Excise duties? 465. Stamp duties? 466. A license tax? 467. Special taxes? Income taxes? 468. How compute income taxes, when the rate is given ? EQUATION" OF PAYMENTS. 469. Equation of l*ayments is finding the at-em^^ time for payment of two or more sums due at different times. The average time sought is often called tiie mean, or equated time. 470. Equation of Payments embraces two classes of examples, ist. Those in which the items or bills have the same date, but different lengths of credit. 2d. Those in which they have different dates, and the same or different lengths of credit. PROBLEM I. 471. To find the Avei'age Tinier when the Items have the same date, but different lengths of credit. I. Bought Oct. 3d, 1870, goods amounting to the fol- lowing sums: $50 payable in 4 m., $70 in 6 m., and $80 in 8 m. : what is the average time at which the whole may be J)aid, without loss to either party? Analysis. — The int. on $50 for 4 m.=the int. ^i-q >^ 4=200 on $1 for 50 times 4 m, or 200 m. Again the int. mq ^ 6 = 420 on $70 for 6 m.=the int. on $1 for 70 times 6 m., 80 X 8 = 640 or 420 m. Finally, the int. on $80 for 8 m.=tlie — — int. of $1 for 80 times 8 m., or 640 m. Now 200 ^°° ) l2bo m. +420 m. + 640 m. = i26o m. ; therefore I am A.ns. 6-j^ m. entitled to the use of $1 for 1260 m. But the sum of the debts is $50 + $70 + $80= $200. Now as I am entitled to the use of $1 for 1260 m., I must be entitled to the use of $200 for 7^0 part of 1260 m., and 1260-5-200=6 1\ m., or 6 m. 9 d., the average time re- quired. Hence, the Rule. — Multiply each item ly its length of credit, and divide the sum of the products ly the sum of the items. The quotient will he the average time. Notes. — i. When the date of the payment is required, add the average time to the date of the transaction. Thus, in the preceding Ex., the date of payt. is Oct. 3d + 6 m. 9 d., or April 12th, 1871. 469. What is equation of payments ? Note. What is the average time called ? 470. How many classes of examples in Equation of Payments? 471. How find the avera^re time, when the items have the same date, but different credita 302 EQUATION OF PAYMENTS. 2. This rule is applicable to notes as well as accounts. It is founded upon tlie supposition that hank discount is the same as simple interest. Though not strictly accurate, it is in general use. 3. If one item is cash, it has no ^m6,-and no p7'oduct; but in find- ing the sum of items, this must be added with the others. 4. In the answer, a fraction less than | day, is rejected ; if | day or more, i day is added. 2. Bought a house June 20th, 1870, for $3000, and agreed to pay I down; | in 6 m., the balance in 12 m.: at what date may the whole be equitably paid ? 3. A owes B $700, payable in 4 mos. ; $500, in 6 mos. : $800, in 10 mos.; and $1000 in 12 mos.: in what time may the whole be justly paid? 4. Bought a bill of goods March loth, 1868, amounting to $2500; and agreed to pay $500 cash, $750 in 10 days, $600 in 20 days, $400 in 30 days, and $250 in 40 days. At what, date may I equitably pay the whole ? 5. A jobber sold me on the ist of March I12000 worth of goods, to be paid for as follows : J cash, | in 2 mos., J in 4 mos., and the remaining J in 6 mos. When may I pay the whole in equity ? PROBLEM II. 472. To find the Average Time, when the items have diiferent dates, and the same or different lengths of credit. 6. Bought the following bills of goods: March loth, 1870, $500 on 2 m.; April 4th, $800 on 4 m.; June 15th, $1000 on 3 m. What is the average time? ANALYSI8.-We first find May 10, 00 X $500= 0000 d. when the items are due by ^ ^^ g^ ^ ^800= 68800 d. adding the time of credit to g^ ^^ 128 X liooo^ 128000 d. the date of each, and for con- -^^ , — — — , venience place these dates ^2300 ) 196800(1. in a column. Taking the Average time, 85^1 d. earliest date on which either item matures as a standard, we find the number of days from this standard date to the maturity of the other items, and place them on the ri^ht, with the sign ( x ) and the items opposite. EQUATIOK OF PAYMENTS. 303 Multiplying each item and its number of days together, the sum of the products shows that the interest on the several items is equal to the interest of $i for 196800 days. Now if it takes 196800 days for $1 to gain a certain sum, it will take $2300, j-h-u of 196800 d. to gain this sum ; and 196800-^-2300= 85H, or 86 days from the assumed date. Now May 10 + 86 d.=Aug. 4th, the date when the amt. is equitably due. Hence, the Rule. — I. Find the date when the several items hecome due, and set them in a column. Take the earliest of these dates as a standard, and set the number of days from this date to the maturity of the other items in another column on the right, with the items opposite. II. Multiply each item ly its number of days, and divide the sum of the lyroducts by the sum of the items. The quotient will be the average time of credit. NoTES.-^i. Add the average time to the standard date, and the result will be the equitable date of payment. 2. The latest date on which either item falls due may also be taken as the standard ; and having found the average time, subtract it from this date ; the result will be the date of payment. 3. The date at which each item becomes due, is readily found by adding its time of credit to the date of the transaction. 7. A bought goods as follows: Apr. lotli, $310 on 6 m. ; May 2ist, I468 on 2 m. ; June ist, $520 on 4 m. ; July 8th, $750 on 3 m.: what is the average time, and at what date may the whole debt be equitably discharged ? Suggestion. — The second item matures earliest; this date is therefore the standard. Ans. Av. time 59 d. ; date Sept. i8tli. 8. Sold the following bills on 6 months credit: Jan. 15th, $210; Feb. nth, $167; March 7th, $320.25 ; April 2d, $500.10: when may the whole be paid at one time? 9. Bought June 5th, on 4 m., groceries for I125 ; June /ist, $230.45; July 12th, I267; Aug. 2d, $860.80: what is the amt. and time of a note to cover the whole ? 472. How find the average time, when the items have different dates, and the Bane or differeiit creaits? Note. How find the date of the payment? How find the date at which each item becomes due ? 104 AVERAGING ACCOUNTS. AVERAGING ACCOUNTS. 473. An Account is a record of the items of debii and credit in business transactions. Note. — The term debit is from tlie Latin debitus, owed. 474. A 3Ierchandise JBalauce is the difference between the debits and credits of an account. A Cash balance is the difference between the debits and credits, with the interest due on each. Notes. — i. Bills of goods sold on time are entitled to interest after they become due ; and payments made before they are due are also entitled to interest. 475. To find the Average Time for paying the balance of an Account which has both debits and credits. I. Find the cash Bal. of the following Acct., and when due. Dr. Wm. Gokdon in Acct. with John Raistdolph. Cr. 1870. Feb. 10. May II. July 26. ForMdse., 4 m. " " 2 " $450.00 500.00 360.00 1870. Mar. 20. iJuly 9. jSept. 15. By Sundries, 3 ni. " Draft, 60 d. " Cash, $325-00 150.00 400.00 Analysis. — Setting down the date when each item is due, take June loth, the earliest of these dates, as the standard, find the number of days from this standard to the maturity of each item on both sides, and place it on the right of its date with the sign ( X ), then the items, add- ing 3 days grace to the time of the draft. Debits. June 10, 00 d. X I450 = 00000 d. Aug. II, 62 d. X 500 =: 31000 d. Sept. 26, 108 d. X 360 =■ 38880 d. Amt. debits, $1 3 10, Int. 69880 d. Credits. June 20, 10 d. X $325 — 3250 d. Sept. 10, 92 d. X 150 = 13800 d. " 15, 97 d. X 400 = 38800 d. Amt. credits, $875, Int. 5 5 850 d. Cash bal. $435, " 14030 d. $435)14030 d. = 32 d.+. Ans. Bal. I435, due July 12, '70. 473. What is an account? Note. What is the meaning of the term debit! 474 What is a merchandise balance ? A cash balance ? AVERAGING ACCOUNTS. 305 Muldplying each item on both sides by its number of days, the int. on the debits is equal to the int. of $i for 69880 days, and the int. on the credits is equal to the int. of $1 for 55850 days. (Art. 474.) The balance of int. on the Dr. side is 69880 d. — 55850 d. = 14030 d.; that is to the int. of $1 for 14030 d. The balance of items on the Dr. side is $1310— $875 = 1435. Now if it takes 14030 days for $1 to gain a certain sum, it will take I435, 4^5 of 14030 d. and 14030^435=321?, or 32 d. But the assumed standard June ioth + 32 d.=July 12th. Therefore the balance due Randolph the creditor is $435, payable July 12th, 1870. If the greater sum of items and the greater sura of products were «n opposite sides, it would be necessary to subtract the average time from the assumed date. Hence, the Rule. — I. Set down the date luhen each item of deUt and credit is due; and assuming the earliest of these dates as a standard, ivrite the number of days from this standard date to the maturity of the respective ite7ns, on the right, with the sign ( x ) atid the items themselves ojjposite. II. Multiply each item hy its numher of days, and divide the difference between the sums of products by that between the sums of items; the quotient tvill be the average time. III. If the greater sum of items and the greater sum of products are both on the same side, add the average time to the assumed date; if on opposite sides, subtract it; and the result tvill be the date when the balance of the account is equitably due. Notes. — i. The average time may be such as to extend to a date either earlier or later than that of any of the items. (Ex. 2.) 2. In finding the maturity of notes and drafts, 3 days grace yhoulJ be added to the specified time of payment. 3. In finding the extension to which the balance of a debt ia entitled, when partial payments are made before it is due, Multiply each payment by the time from its date to the maturity of the debt, and divide the sum of the products X)y the halance remaining unpaid. 4. When no time of credit is mentioned, the transaction is under- stood to be for cash, and its payment due at once. 475. How find the average time for paying the balance, when there are both debits and credits? BOG AVERAGIl^Q ACCOUKTl 2. Find the balance of the following Acct., and when due. Dr. A. B. in account with C. D. Cr. i860. i860. Aug. II. For Mdse., $160.00 Sept. 2. By Sundries, $7500 Sept. 5. « « 240.00 Oct. 10. " your Note on 30 d.. 100.00 Oct. 20. " I Horse, 175.00 Nov. I. " Cash 110.00 Dr, OPKBATION. Or, Due. Bays, Items. Products. Due. Days. Items. Products. Aug. II. S160 0000 Sept. 2. 22 - $75 1650 Sept. 5. 25 240 6000 Nov. 9. 93 100 9300 Oct. 20. 70 175 12250 Nov. I. 82 no 9020 575 18250 285 19970 _2^1 1720-^290=6 d. (nearly), av. time. 18250 Bal=$290 Aug. 1 1 —6 d. = Aug. 5, bal. is due. 1720 Suggestion. — In this example, the greater sum of items and the greater sum of products are on opposite sides ; hence, the average time must be subtracted from the standard date. 3. A man bought a cottage for $2750, payable in i year; in 3 mos. he paid $500, and 3 mos. later I750 : to what extension is he entitled on the balance ? Solution. — The sum of the product is 9000 ; and the balance of the debt $1 500. Now 9000-^ 1 500=6. Ans. 6 mos. after the maturity of the debt. (Note 3.) 4. A merchant sold a bill of $4220 worth of goods on 8 mos.; 2 months after the customer paid him $720, one month later $850, and 2 months later liooo: how long should the balance in equity remain unpaid ? 5. What is the balance of the following account, and when is it due ? Br. HEifliBT Swift in Acct. with Homee Morgan. Cr. 1865. 1865. March 10. For Sundries, $250 April I. By Bal. of Aoct., $iiol April 15. " Flour on 60 d., 420 May 21. " Dft.onaod., 3001 June 20. " Mdse. on 30 d,, boo iJuly I. « Cash, 560 1 RATIO. 476. Matio is the relatmi which one number has to another with respect to magnitude. The Terms of a ratio are the numbers compared. They are often called a couplet. 477. Ratio is commonly expressed by a colon ( : ) placed between tlie two numbers compared. Thus, the ratio of 6 to 3 is written 6:3. 478. The jirst term is called the antecedent, the second t!.o consequent. The comparison is made by considering what multijple ot part the antecedent is of the consequent. Notes. — i. The dgn of ratio (:) is derived from tlio sign cf division (-f-), the horizontal line being dropped. 2. The terms are so called from the order of their position. They must be of the same kind or denomination; otherwise they cannot be compared. 479. Ratio is measured by 9>. fraction, the numerator of which is the antecedent, and the denominator the consequent ; or what is the same thing, by dividing the antecedent by the consequent. 480. The xialue of a ratio is the value of the fraction by which it is measured. Thus, comparing 6 with 2, we say the ratio of 6 : 2 is f , or 3. That is, the former has a magnitude which contains the latter 3 times ; therefore the value of the ratio 6:2 is 3. 481. A Simple Matio is one which has but two terms ; as, 8 : 4. A Compound Matio is the product of tivo or more simple ratios. Thus : 4 : 2 1^ are simple But 4x9:2x3 9:33 ratios. is a compound ratio. Note. — ^The nature of compound ratios is the same as that of simple ratios. They are so called to denote their origin, and are usually expressed by writing the corresponding terms of the simple ratios one under another, as above. 476. What is ratio? The numbera compared called? 477. How is ratio com- monly expressed ? The first terra called ? The second? Note. Why ^ 479. How ig ratio mcaijured ? Tho value of a ratio ? 481. Simple ratio? Cosapoundt SOS RATIO. 482. Ratio is also distinguished as direct and inverse or reciprocal A direct ratio is one which, arises from dividing the antecedent by the consequent. An inverse ratio is one which arises from dividing the consequent \>j the antecedent, and is the same as the ratio of the reciprocals of the two numbers compared. (Art. io6.) Thus, the direct ratio of 4 to i2=:-j4j, or i" ; the inverse ratio of 4 to I2=\% or 3. It is the same as the ratio of their reciprocals, \ to -^2. 483. 1'1^6 ratio between two fractions having a common denomi- nator is the same as the ratio of their numerators. Thus, the ratio of I : I is the same as 6 ; 3. In finding the ratio of two fractions which have different denomi- nators, they should be reduced to a common denominator ; then take the ratio of their numerators. (Art. 153.) 484. Ill finding the ratio between two compound numbers, they must be reduced to the same dcTiomination. Note. — Finding the ratio between two numbers is the same in principle as finding what part one is of the other, the number denoting the part being the antecedent. Thus, 2 is | of 4— | ; and the ratio of 2 to 4 is |=i. (Art. 173.) 485. Since ratios are measured by fractions whose numerators are the antecedents, and denominators the consequents, it follows that operations have the same effect upon the terms of a ratio as upon the terms of a fraction. (Art. 144.) That is, 1. Multiplyi7ig tlie antecedent multiplies the ratio; and dividing the antecedent divides the ratio. 2. Multiplying the consequent divides the ratio; and dividing the consequent midtiplies the ratio. 3. Midtiplying or dividing hoth the antecedent and con- sequent hy the same numher, does not alter the value of the ratio. 482. Direct? Inverse? 483. The ratio of two fractions having a common denominator ? 484. In finding the ratio of two compound numbers, Avhat must be done ? 485. What effect do operations on the terms of a ratio have ? Multi- plying the antecedent ? Dividing it ? Multiplying the consequent ? Dividing it ? ■\7hnt efTcct has multiplying or dividing both the antecedent and consequent by the tame number? PROPORTION. 309 What are the ratios of the following couplets: 1. 12 14. 4. 6 ; 24. 7. £5 : los. 6d. 2. 28:7. 5. 8 : 40. 8. 10 y. : 6 ft. 3 in. 3.36:12. 6.9:51. 9. 25 g. : 2 qt. I pt. 10. Eednce the ratio of 14 to 35 to the lowest terms ? Solution. — 14 : 35 equals ^ ; and i*=f, or 2 : 5. (Art. 146.) Reduce the following ratios to the lowest terms ? 11. 154:28. 13. 73:511. 15. 238:1428. 12. 39:165. 14. 113:1017. 16. 576:1728. 17. Reduce the ratio f :|- to the lowest integral terms? f =il. and f =if. Now ^f : 1^- is the same as 4 : 5. (Art. 483.) 18. Multiply the ratio of 2 1 : 7 by 4 : 8. Ans. S4: ^6, or i^ 19. What is the value of 5:8x4: 10x7:9? PROPORTION. 486. JPropoi^tion is an equality of ratios. 487. Every proportion must have at least four terms', for, the equality is between tivo or more ratios, and each ratio has two terms, an antecedent and a consequent. A proportion may, however, be formed from three num- hers; for, one of the numbers may be repeated, so as to form two terms. 488. Proportion is denoted in two ways ; by a double colon ( : : ), and by the sign of equality ( = ), placed between the ratios. Thus, each of the expressions 4 : 2 : : 6 : 3, and 4:2 = 6:3 indicates a proportion ; for, f =f. The former is read, " 4 is to 2 as 6 to 3," or " 4 is the same part of 2, that 6 is of 3." The latter is read, " the ratio of 4 to 2 equals the ratio of 6 to 3." Note. — The sign ( : : ) is derived from the sign (=), the points being the extremities of the parallel lines. 486. What is proportion ? 487. How many terms has every proportion ? Caa three mimbers form a proportion ? How ? 310 PKOPOTITIOK. 489. The four numbers which form a proportion, are called proportionals. The first and last are the extremes^ the other two the ineans. When a proportion has but three numbers, the second term is called a mean proportional between tlie other two. 490. If four numbers are proportional, the product of the extremes is equal to the product of the means. Hence, 491. The relation of the four terms of a proportion to each other is such, that if any three of them are giyen, the other or missing term may be found. 492. To find the 3Iissing Term of a Proportion, the other three Terms being given, 1. Let 6 be the first term of a proportion, 3 and 10 the two means; tlie other extreme equals 3xio-r-6=5; for, the product of the meansr=the product of the extremes; and the product of two factors divided by one of them, gives the other. (Art. 93.) 2. Let 3, 10 and 5 be the last three terms of a proportion, the first term equals 3 x 10-^-5 = 30-7-5 or 6. 3. Let 6 and 5 be the extremes of a proportion, and 3 one of the means ; the other mean equals 6 x 5-7-3=30-^3 or 10. 4. If 6 and 5 are the extremes, and 10 one of the means, the other mean equals 6 x 5-7-10=30^10 or 3. Hence, the EuLE, — I. If one of the extremes and the two means are given, divide the product of the means hy the given extreme, II. If one of the means and the tivo extremes are given, divide the product of the extremes hy the given mean. Find the missing term in the following proportions : 1. 52 : 13 1:62: — . 5. 4 rods : 11 ft. : : 18 men : — . 2. 15:90:: — : 72. 6. 24 yd. :3 yd. :: — :$i2. 3. 60 : — : : 100 : 33J. 7. 20 gal. : — : : $40 : $8. 4. _: 25 :: |: I;. 8. — : 40 lb.: : £2 : 8s. 4Sg. What are the four numbers forming a proportion called? 491. If three terms of a proportion are given, what is true of the fourth ? If the two means and one extreme are given, how find the other extreme ? If the two extremes and one mean are given, how find the other mean ? SIMPLE PROPOETIO]^. 493. Simple I^ropor^tion is an equality of two simple ratios. Simple JProportion is applied cliiefly to the solu- tion of problems having three terms given to find a, fourth, of which the third shall be the same multiple or part, as the Jlrst is of the second. 494. To solve Problems by Simple Proportion. 1. If 5 baskets of peaches cost $io, what will 3 baskets cost? Analysis. — The question assumes statement. that 3 baskets can be bought at the 5 bas. : 3 bas. :: ^10 : Ans. same rate as 5 baskets ; therefore 3 5 bas. has the same ratio to 3 bas. S)'!^© as the cost of 5 bas. has to the cost of ^y j 3 bas. That is, 5 bas. : 3 bas. : : $10 : cost of 3 bas. We have then the two means and one extreme of a proportion to find the fourth term, or other extreme. (Art. 492.) Now the product of the means $10 x 3 = $3o; and $30-7-5 (the other extreme) =$6, the cost of 3 baskets. Hence, the Rule. — I. Tahe that numler for the third term^ which is the same hind as the answer. II. When the answer is to he larger than the third term, place the larger of the other two numbers for the second term; but when less, place the smaller for the second tertn, and the other for the first. III. Multiply the second and third terms together, and divide the product by the first ; the quotient will be the fourth term or answer. (Arts. 490, 491.) Proof. — If the product of the first and fourth terms equals that of the second and tliird, the answer is right. Notes. — i The arrangement of the given terms in the form of a proportion is called " Stating the question." 2. After stating the question, the factors common to the first and §€cond, or to the first and third terms, should be cancelled. 493. Simple Proportion ? 494. How aolve problems by Simple Proportion ? C12 SIMPLE PKOPOIlTIOiT. 3. If tlie first and second terms contain different denominations^ tliey must be reduced to the same. If the tliird term is a compound number, it must be reduced to the lowest denomination it contains. 495. Reasons, — i. The reason for placing that number for the tliird term, which is the same kind as the answer, and the other two numbers for the first and second, is because money has a ratio to money, but not to the other two numbers ; and the other two numbers have a ratio to each other, but not to money. 2. Of the two like numbers, the smaller is taken for the second term, and the larger for the first, because 3 baskets being less than 5 baskets, will cost less ; consequently, the answer or fourth term must be less than the third, the cost of 5 baskets. 3. If it were required to find the cost of a quantity greater than that whose cost is given, the answer would be greater than the third term ; consequently the greater of the two similar numbers must then be taken for the second term, and the less for the first. 4. The reason for multiplying the second and third terms together and dividing the product by the first, is because the product of the means divided by one of the extremes, gives the other extreme ot ansioer. (Arts. 93, 492.) 2. If 9 yards of cloth cost $54, what will 23 yards cost? Statement.— g yd. : 23 yd. : : $54 : Ans. And ($54 X 23)-^9=$i38, the Ans. By Cancellation. — Since 9 yds. cost Operation. $54, I yd. will cost ^ of $54, or $Y ; ^V" ^ 2^ = A7is. and 23 yds. will cost $ V" x 23=ii-^i^ 6, 5^ x 2 3 ^ „ , 9 '- =^ = $138, ^ns. =$13S, Ans. ^ '^ Proof.— <) yd. : 23 yd. : : $54 : $138 ; for 9 x 138=23 x 54. 3. If 7 barrels of flour cost $56, what will 20 barrels cost? 4. What cost 75 bushels of wheat, if 15 bushels cost %;^;^? 5. What cost 150 sheep, if 17 sheep cost $51 ? 6. If 5 lb. 8 oz. of honey cost $1.65, what will 20 lb. cost? 7. Paid £1, 15s. 6d. for 6 pounds of tea: what must be paid for a chest containing 65 lb. 8 oz. ? 495. Why take the number which is of the same kind as the answer for the third term, and the other two for the first and second ? When place tlie larger of the other two numbers for the second? Why? When the smaller? Why ? Why does the product of the second and third terms divided by the first give the answer? What is the arrangement of the temis in the form of a propor- tion called ? If the first and second terms contain different denominations, hoT^ proceed ? If the third is a compound nnmber, how ? SIMPLE PROPORTIOJ^. 313 SIMPLE PROPORTION BY ANALYSIS. 496. The chief diflBculty experienced bj the pupil in Simple Proportion, lies in " stating the question." This difficulty arises from a want of familiarity with the relation of numbers. He will be assisted by analyzing the examples before attempting to state them. 8. If 7 hats cost $42, liow mucli will 12 hats cost? Analysis. — i hat is i seventh of 7 hats ; therefore i hat will cost I seventh as much as 7 hats ; and \ of $42 is $6. Again, 12 hats will cost 12 times as much as i hat, and 12 times $6 are $72. There- fore, 12 hats will cost $72. Or, 7 hats are f^ of 12 hats ; therefore the cost of 7 hats is -^^ the cost of 12 hats. But 7 hats cost $42 ; hence $42 are -,^2 the cost of 12 hats. Now, if -f^- of a number are $42, tV of that number is | of $42, which is $6 ; and || are 12 times $6, or $72. By Proportion.— 'J h. : 12 h. : : $42 : Ans. That is, 7 h. are the same part of 12 h. as $42 are of the cost of 12 hats. 497. Solve the following examples both by analysis and proportion, 9. If II men can cradle 33 acres of grain in i day, how many acres can 45 men cradle in the same time ? 10. When mackerel are $150 for 12 barrels, what must I pay for 75 barrels ? ^ ^ 11. How far will a railroad car go in 12 hours, if it goes at the rate of 15 miles in 40 minutes ? 12. A bankrupt owes $3500, his assets are $1800: how much will a creditor receive whose claim is $560 ? 13. At the rate of 18 barrels for %62„ what will 235 bar- rels of apples cost ? 14. If $250 earns I17J interest in i year, how much will 1 1 900 earn in the same time? 15. If a car wheel turns round 6 times in T^^t yards, how many times will it turn round in going 7 miles ? 16. If a clerk can lay up 1 1500 in i^ year, how long will It take him to lay up $5000 ? 17. If 6 men can hoe a field of corn m 20 hours, how long will it take 1 5 men to hoe it ? c 14 314 SIMPLE PROPOETIOK. 1 8. An engineer found it would take 75 men 220 days to build a fort; the general commanding required it to be built in 15 days: how many men must the engineer smploy to complete it in the required time ? 19. If 5 oz. of silk can be spun into a thread 100 rods long, what weight of silk is required to spin a thread that will reach the moon, 240000 miles distant ? 20. How many horses will it take to consume a scaffold of hay in 40 days, if 12 horses can consume it in 90 days ? 21. If I acre of land costs $15, what will 2^-} acres cost ? 22. If f of a ton of iron costs £f, what will -^^ of a ton cost ? 2^. If loj lb. sugar cost $1 J, what will 3of lb. cost ? 24. If I of a chest of tea costs $35.50, what will 15^ chests cost ? 25. What will 48! tons of hay cost, if 1 2^ tons cost |i 26 J ? 2 6. If y'2 of a ship is worth $ 1 6 2 5 of, what is t\ of it worth ? 27. What will it cost me for a saddle horse to go 100 miles if I pay at the rate of $7^ cts. for 3 miles ? 28. What must be the length of a slate that is 10 in. wide, to contain a square foot ? 29. How many yards of carpeting J yard wide will it take to cover a floor 15 ft. long and 12 feet wide ? 30. If a man's pulse beats 68 times a minute, how many times will it beat in 24 hours ? 31. If Halley's comet moves 2° 45' in 11 hours, bow far will it move in 30 days ? "32. If a pole 10 ft. high cast a shadow 7^ ft. long, how high is a flag-staff whose shadow is 60 ft. long? SS. At the rate of 3 oranges for 7 apples, how many oranges can be bought with 150 apples? '- 34. If an ocean steamer runs 1250 miles in 3 days 8 h., how far will she run in 8 days ? 35. If 12 men can harvest a field of wheat in 11 days, how many men are required to harvest it in 4 days ? 36. The length of a croquet-ground is 45 feet ; and lU width is to its length as 2 to 3 : what is its width ? Simple propoktioi^. ?15 37. A man's annual income from U. S. 6s is $1350 when gold is 112J : whafc was it when gold was 160 ? ^8. George has 10 minutes start in a foot-race; and runs 20 rods a minute: how long will it take Henry, who runs 28 rods a minute, to overtake him? 39. If the driving wheel of a locomotive makes 227 revolutions in going 206 rods 6 ft., how many revolutions will it make in running 18 miles 240 rods? 40. If 3 lbs. of coffee cost $1.20, and 10 lbs. of coffee are worth 6 lbs. of tea, what w^ill 60 lbs. of tea cost ? 41. A can chop a cord of wood in 4 hours, and B in 6 hours : how long will it take both to chop a cord ? 42. A reservoir has 3 hydrants; the first will empty it in 8 hours, the second in 10; the third in 12 hours: if all run together, how long will it take to empty it ? 43. A man and wife drank a keg of ale in 18 d. ; it would last the man 30 d. : how long would it last the woman ? = 44. A fox is 100 rods before a hound, but the hound runs 20 rods while the fox runs 18 rods: how far must the hound run before he catches the fox ? 45. A cistern holding 3600 gallons has a supply and a discharge pipe ; the former runs 45 gallons an hour, the latter ^3 gallons : how long will it take to fill the cistern, when both are running ? 46. A clerk who engaged to work for I500 a year, com- menced at 12 o'clock Jan. ist, 1869, and left at noon, the 2ist of May following: how much ought he to receive ? 47. A church clock is set at 12 o'clk. Saturday night; Tuesday noon it had gained 3 min. : what will be the true time, when it strikes 9 the following Sunday morning ? 48. A market-woman bought 109 eggs at 2 for a cent, and another 100 at 3 for a cent; if she sells them at the rate of 5 for 2 cents, what will she make or lose ? 49. Two persons being :^:^6 miles apart, start at the same time, and meet in 6 days, one traveling 6 miles a day faster than the other: how far did each travel ? OOMPOUl^D PEOPORTIOK * > : : 1 2 : 2, is a componnd proportion. 498. Compound JProportioii is an equality of a som2)Ound and a simjjle ratio. Thus, 4: 9 It is read, '* The ratio of 4 into 9 is to 2 into 3 as 12 to 2." I. If 4 men saw 20 cords of wood in 5 days, how many Cards can 1 2 men saw in 3 days ? Analysis. — The answer is to statement. be in cords; we therefore make 4 m. :12 m. ) . o > A^q 20 c. the third term. The other 5 d. : 3 d. [ * * p:iven numbers occur in pairs, two 20 C. X 12 X3 = 720C. of a kind ; as, 4 men and 12 men, 4x5 = 20 5 days and 3 days. We arrange 720 C.-j-20 = 36 C. Ans. these pairs in ratios, as we should in simple proportion, if the answer depended on each pair alone. That is, since 12 m. will saw more than 4 men, we take 12 for the second term and 4 for the first. Again, since 12 men will saw less in 3 days than in 5 days, we take 3 for the second term and 5 for the first. Finally, dividing the product of the second and third terms 20 c. X 12 X 3 = 720 c. by the product of the first terms 4 x 5 = 20, wo have 36 cords for the answer. Hence, the EuLE. — I. Mahe that number the third term which is of the same Icind as the ansiuer. II. TalcG the other numbers in pairs of the same Tcind, and arrange them as if the answer depended on each couplet, as in simple proportion. (Art. 494.) III. Multiply the second and third terms together, and divide the product hy the product of the first terms, cancelling the factors common to the first and second, or to the first and third terms. Tlie quotient will he the ansiver. Proof. — If the product of the first and fourth terms equals that of the second and third terms, the luorh is right, 498. What is Compound Proportion ? Explain the first example. The rule. Proof. Note. How proceed when tlie firpt and second terms contain different denominations ? When the third does ? How else may questions in Compound Proportion be solved ? COMPOUND PROPORTIO]^. 317 Notes. — i. The terms of eacli couplet in fhe compound ratio must be reduced to the same denomination, and the third term to the lowest denomination contained in it, as in Simple Proportion, 2. In Compound Proportion, all the terms are given in couplets or pairs of the same kind, except one. This is called the odd term, or demand, and is always the same kind as the answer. 3. It should be observed that it is not the ratio of 4 to 2, nor of 9 to 3 alone that equals the ratio of 12 to 2 ; for, 4-5-2 = 2 and 9-5-3=3, while i2-j-2=6. But it is the ratio compounded of 4 x g to 2x3, which equals the ratio of 12 to 2. Thus, (4X g)-=-(2 x 3)=i6 ; and 12-^2=6. (Art. 498.) 4. Compound Proportion was formerly called " Double Rule of Three." 499. Problems in Compound Proportion may also be solved by Analysis and Simple Proportion. Take the preceding example : By Analysis. — If 4 men saw 20 cords in 5 d., i m. will saw \ of 20 c, which is 5 c, and 12 m. will saw 12 times 5 c, or 60 cords, in the same time. Again, if 12 men saw 60 c. in 5 days, in i d. they will saw \ of 60 c , or 12 cords, and in 3 d. 3 times 12 c, or 36 cords, the answer required. By Simple Proportion. — 4 m. : 12 m. : : 20 c. : the cords 12 m. will saw in 5 d. ; and 20 c. x 12-5-4=60 c. in 5 days. Again, 5 d. : 3 d. : : 60 c. : the cords 12 men will saw in 3 d. And 60 c. x 3^5 = 36 cords, the same as before. 2. If 4 men earn $219 in 30 clays, working 10 hours a day, how muoli can 9 men earn in 40 days, working 8 hours a day ? sTATEMEiirT. By CanceUatkm. S^m. 3 ^Bd. B.h. 4 : : $219 : Anfi, 4 m.: 9 m. ) 4 m. 30 d. : 40 d. V : : $2 19 : A7is. %, 15, SH d. loh. :8h. ) \%h. {I219 X 9 X40 X 8)-^ (4 X 30 X 10) = $5251, ^Iws. 5 15219x4x3=1: $5251, J^^5. 3. If 6 men can mow 28 acres in 2 days, how long will it take 7 men to mow 42 acres ? 4. If 8 horses can plow 32 acres in 6 days, how many hcrses will it take to plow 24 acres in 4 days ? 5. If the board of a family of 8 persons amounts to $300 in 15 weeks, how long will $1000 board 12 persons? 318 COMPOUND PROPORTIO]!^. 6. What will be the cost of 28 boxes of candles contain - 'ng 20 pounds apiece, if 7 boxes containing 15 pounds ij)iece can be bought for $23.75 ? 7. If the interest of $300 for 10 months is I20, what will be the interest of $1000 for 15 months? 8. If a man walks 180 miles in 6 days, at 10 h. each, how many miles can he walk in 15 days, at 8 h. each ? 9. If it costs $160 to pave a sidewalk 4 ft. wide and 40 ft. long, what will it cost to pave one 6 ft. wide -md 125 ft. long? 10. If it requires 800 yards of cloth f yd. wide to supply 100 men, how many yards that is f wide will it require to clothe 1500 men? 11. If 75 men can build a wall 50 ft. long, 8 ft. high, and 3 ft. thick, in 10 days, how long will it take 100 men to build a wall 150 ft. long, 10 ft. high, and 4 ft. thick ? 12. If it costs $56 to transport 7 tons of goods no miles, how much will it cost to transport 40 tons 500 miles ? 13. If 30 lb. of cotton will make 3 pieces of muslin 42 yds. long and f yd. wide, how many pounds will it take to make 50 pieces, each containing 35 yards 1} yd. wide ? 14. If the interest of $600, at 7^, is $35* for 10 months, what Avill be the int. of $2500, at 6'/c, for 5 months ?. '/' i$. If 9 men, working 10 hours per day, can make 18 sofas in 30 days, how many sofas can 50 men make in 90 days, working 8 hours per day ? i>'' ^ 16. If it takes 9000 bricks 8 in. long and 4 in. wide to pave a court-yard 50 ft. long by 40 ft. wide, how many tiles 10 in. square will be required to lay a hall-floor 75 ft long by 8 ft. wide ? ." 17. If in 8 days 15 sugar maples, each running 12 quarts of sap per day, make 10 boxes of sugar, each weighing 6 lb., how many boxes weighing 10 lb. apiece, will a maple grove containing 300 trees, make in 36 days, each tree running 16 quarts per day? A71S. 720 boxes. PARTITIVE PROPORTION. 319 PARTITIVE PROPORTION. 500. Partitive Proportion is dividing a mimber into two or more parts having a given ratio to each other. 501. To divide a Number into two op more parts which shall be proportional to given numbers. 1. A and B found a purse of money containing $35, which they agree to divide between them in the ratio of 2 to 3 ; how many dollars will each have ? By Proportion.— The sum of the statement. proportional parts is to each separate 5 : 2 : : $35 : A's S. part as the number to be divided is to - . ^ . . |„- . j^jg g^ each man's share. That is, 5 (2 + 3) is to 2 as $35 to A's share. Again, 5 W5 X2)h-5=$I4Ass. is to 3 as $35 to B's share. Hence, the ($35X3)^5 = ^21 B's S. Exile. — I. Take the miynber to he divided for the third term; each proj^ortional part successively for the seco7id term; and their sum for the first. XL The product of the second and third terms of each proportion, divided by the first, will he the corresponding part required. By Analysis. — Since A had 2 parts and B 3, both had 2 + 3, or 5 parts. Hence, A will have | and B \ of the money. Now % of $35 =:$I4, and \ of $35 =$21. Hence, the EuLE. — Divide the given number by the sum of the pro- portional "^umbers, and multiply the quotient by each one's jjroportional part. 2. It is required to divide 78 into three parts Vfhich shall be to each other as 3, 4, and 6. Ans. 18, 24, 2,^. 3. A man having 200 sheep, wished to divide them into three flocks which should be to each other as 2, 3, and 5 : how many Avill each flock contain ? 4. A miller had 250 bushels of provender composed of oats, peas, and corn in the proportion of 3, 4, and 5 1 : how many bushels were there of each kind ? 500. How divide a number into parts having a given ratio ? 320 PARTNERSHIP. 5. A father divided 497 acres of land among his four sons in proportion to their ages, which were as 2, 3, 4, and 5 : how many acres did each receive ? 502. The principles of Partitive Proportion are appli- cable to those classes of problems commonly arranged under the heads of Partnership, BanTcruptcy, General Average, etc., in which a given number is to be divided into parts having a, given ratio to each other. ' PAETNEESHIP. 503. JPartnersJiip is the association of two or more persons in business for their common profit. It is of two kinds ; Simple and Compound. 504. SiTTiple I^artnership is that in which the capital of each partner is employed for the same time. 505. Compound JPartnersJiip is that in which the capital is employed for unequal times. Note. — The association is called a firniy Jiouse, or company ; and the persons associated are termed partners. 506. The Capital is the money or property employed in the business. The JProfits are the gains shared among the partners, and are called dividends. Notes. — i. The profits are divided as the partners may agree. Other things being equal, when the capital is employed for the same time, it is customary to divide the i)rofits according to the amount of capital each one furnishes. 2. When the capital is employed for unequal times, the profits are usually divided according to the amount of capital each furnisifaes, and the time it is employed. 502. To what are the principles of Partitive Proportion applicable ? 503. What is Partnership ? Of how many kinds ? 504. Simple Partnership ? 505. Com- pound ? Note. What is the association called ? 506. What is the capital ? The profltP? What called? PARTi^EESHIP. 321 PROBLEM I. 507. To find each Partner's Share of the Profit or Loss, when divided according to their capital. 1. A and B entered into partnership ; the former furnish- ing $648, the latter $1080, and agreed to divide the profit according to their capital. They made I432 : what was each one's share of the profit ? Analysis. — The capital equals $648 + §io8o=$i728. $1728 (capital) : $648 (A's cap.) : : $432 (profit) : A's share, or $162. $1728 " : $1080 (B's cap.) : : $432 " : B's share, or $270. Or thus: The profit $432-7-11728 (the cap.)=.25 ; that is, the profit is 25^ of the capital. Therefore each man's share of the profit is 25% of his capital. Now $648 x .25 = $162, A's share ; and $1080 X .25 = $270, B's share. Hence, the EuLE. — The wJiole capital is to each partner'' s capital, as the whole profit or loss to each partners share of the profit or loss. Or, Fhid ichat per cent the profit or loss is of the whole capital, and mtdtiply each mail's capital hy it. (Art. 339.) 2. A and B form a partnership, A furnishing $1200, and B $1500; they lose I500: what is each one's share of the loss ? 3. A and B buy a saw-mill, A advancing $3000, and B $4500; they rent it for I850 a year: what should each receive ? 4. The net gains of A, B, and C for a year are $12500; A furnishes $15000, B $12000, and $10000 : how should the profit be divided ? 5. Three persons entering into a speculation, made $15300, which they divided in the ratio of 2, 3, and 4: how much did each receive ? 6. A, B, and C hire a pasture for $320 a year; A put in 80, B 120, and 200 sheep: how much ought each man to pay ? 507. How find each partner's profit or loss, when their capital is employed the same time ? OA.2 PARTI^EKSHIP. PROBLEM II 508. To find each Partner's Share of the Profit or Loss, when divided according to capital and time. 7. A and B enter into partnership ; A furnishes $400 for 8 months, and B $600 for 4 months; they gain $350: what is each one's share of the profit ? Analysis. — In this case the profit of the partners depends on two conditions, viz. : the amount of capital each furnishes, and the time it is employed. But the use of $400 for 8 months equals that of 8 times $400, or $3200, for I m. ; and $600 for 4 m. equals 4 times $600, or $2400, for I m. The respective capitals, then, are equivalent to $2400 and $3200, each employed for i m. Now, as A furnished $3200, and B $2400, the whole capital equals $3200+ $2400= $5600. Therefore, $5600 : $3200 : : $350 (profit) : A's share, or $200. $5600 : $2400 : : $350 " : B's share, or $150. Or thus: The gain $350-7-15600 (the cap.) = .06^, or 6}%. Therefore, $32oox .o6i=$20o, A's share; and $240ox .o6l = $i50, B's share. Hence, the EuLE. — 3fuUiply each partner^s capital ly the time it is employed. Consider these products as their respective capi- tals, and proceed as in the last problem. Or, Fi7id ivhat per cent the profit or loss is of the ivhole capital, and multiply each man's capital hy it. (Art. 339.) Note. — The object of multiplying each partner's capital by the time it is employed is, to reduce their respective capitals to equivalents for the same time. 8. A, B, and C form a partnership ; A furnishing I500 for 9 m., B I700 for i year, and $400 for 15 months; they lose $600 : what is each man's share of the loss ? 9. Two men hire a pasture for $50 ; one put in 20 horses for 12 weeks, the other 25 horses for 10 weeks: how much should each pay ? 508. How find each one's share, when their capital is employed for unequal timet? ? Note. Why multiply each one's capital by the time it is employed? BAN-KRUPTCY. 323 10 A., B, C, and J) commenced business Jan. ist, 1870, when A Airnisiied $1000; March ist, B put in $1200; July ist, put in $1500; and Sept. ist, D put in $2000; during the year they made $1450 : how much should each receive ? 11. A store insured at the Howard Insurance Co. for $3000; in the Continental, $4500; in the American, $6000, was damaged by fire to the amount of $6750: what share of the loss should each company pay ? Remark. — The loss should be averaged in proportion to the risk assumed by each company. 12. A quartermaster paid $2500 for the transportation of provisions ; A carried 150 barrels 40 miles, B 170 barrels 60 miles, C 210 barrels 75 miles, and D 250 barrels 100 miles : how much did he pay each ? 13. A, B, and C, formed a partnership, and cleared I12000; A put in I8000 for 4 m., and then added $2000 for 6 m.; B put in $16000 for 3 m., and then withdrawing half his capital, continued the remainder 5 m. longer; C put in $13500 for 7 m. : how divide the profit. ba:nkeuptcy. 509. Bankruptcy is inability to pay indebtedness. Note. — A person unable to pay his debts is said to be insolvent, and is called a bankrupt. 510. The Assets of a bankrupt are the property in his possession. The Liabilities are his debts. 511. The ^et JProceeds are the assets less the ex- pense of settlement. They are divided among the creditors according to their claims. 509. "What is bankruptcy ? ^10. What are assets ? Liabilities? 511. Net pro- ceeds? 512. How find each creditor's dividend, when the liabilities and the net proceeds are given ? 324 ALLIGATION". 512. To find each Creditor's Dividend, the Liabilities and Net Proceeds being given. I. A merchant failed, owing B $1260, $1800, and D $1940; his assets were $1735, and the expenses of settling $435 : how much did each creditor receive ? Analysis. — The liabilities are $1260 + $1800 + $1940=15000 ; and the net proceeds $173.5— $435 =$1300. Now, $5000 : $1260 : : $1300 : B's dividend, or $327.60. $5000 : $1800 : ; $1300 : C's " or $468.00. $5000 : $1940 : :^i300 : D's " or $504.40. Or thus: The net proceeds $1300-7- $5000=. 26, or 26%, the rate he is able to pay. (Art. 339.) Now $1260 x .26 = $327.60 B's ; $1800 X .26= $468 C's ; $1940 X .26= $504.40 D's. Hence, the EuLE. — The wliole liabilities are to each creditor's claim, as the net j^roceeds to each creditors dividend. Or, Find what per cent the net proceeds are of the liabilities, and multiply each creditors claim by it. 2. A bankrupt owes A $6300, B I4500, and D $3200; his assets are $5250, and the expenses of settling $1500: how much will each creditor receive ? 3. A. B. & Co. went into bankruptcy, owing $48400, and having $13200 assets; the expense of settling was $1 100. What did D receive on $8240 ? ALLIGATIOlsr. 513. Alligation i^oiU^o'kmdi^, Medial d^ndi Alternate. Alligation Medial is the method of finding the mean value of mixtures. Alligation Alternate is the method of finding the proportional parts of mixtures having a given value. Notes. — i. Alligation, frran the Latin alligo, to tie, or hind together, is so called from the manner of connecting the ingredients by curve lines in some of the operations. 2. Alternate, Latin alternatus, by turns, refers to the manner of connecting the prices above the mean price with those helow. 3. The term medial is from the Latin mMius, middle or average. 513. What ie alligation medial ? Alternnt-e? ALLIGATION. 32^ ALLIGATION MEDIAL. 514. To find the 3Iean Value of a mixture, the Price and Quantity of each ingredient being given. 1. Mixed 50 lbs. of tea at 90 cts., 60 lbs. at $1.10, and 80 lbs. at $1.25 : what is the mean yalue of the mixture ? Analysis.— The total value of the first $.90 X 50= $45.00 kind .90x50 =$45. GO, the second $1.10x60 $1.10x60;= $66,00 =z$66.oo, the third $1.25 x 8o=$ioo.oo ; $1.25 X 8o = $ioo.oo therefore the total value of the mixture = f^ \ $2 1 1. 00 $211 .00, But tlie quantity mixed= 190 lbs. a ^^ ^^ Now, if 190 lbs. are worth $211, i lb. is worth r^o of $211, or $i.n + . Therefore the mean value of the mixture is $1.11 + per pound. Hence, the EuLE. — Divide the value of the whole mixture ly the sitm of the ingredients mixed. Note. — If an ingredient costs nothing, as water, sand, etc., its value is o ; but the quantity itself must bo added to the other ingredients. 2. If I mix 3 kinds of sugar worth 12, 15, and 20 cts. a pound, what is a pound of the mixture worth ? 3. A farmer mixed 30 bu. of corn, at $1.25, with 25 bu. of oats, at 60 cts., and 10 bu. of peas, at 95 cts. : what was the average yalue of the mixture ? 4. A grocer mixed 5 gal. molasses, worth 80 cts. a gal., and 107 gallons of water, with a hogshead of cider, at 20 cts. : what was the average worth of the mixture ? 5. A goldsmith mixed 12 oz. of gold 22 carats fine with 8 oz. 20 carats, and 7 oz. 18 carats fine: what was the average fineness of the composition ? 6. A milkman bought 40 gallons of new milk, at 4 cts, a quart, and 60 gallons of skimmed milk at 2 cts. a quart, which he mixed with 1 2 gallons of water, and sold tho whole at 6 cts. a quart : required his profit ? Note. Why is this rule called alligation ? Why alternate ? W^hat is the import of medial? 514. How find the mean value of a mixture, when the price and quantity are given ? 3-^ 3 lb. 11^ J I " 326 ALLIGATION ALTERNATE. ALLIGATION ALTERNATE. PROBLEM I. 515. To find the I*ro2)ortloiial Parts of a Mixture, the Mean Price and the Price of each ingredient being given. 7. A grocer desired to mix 4 kinds of tea worth 3s., 8s., IIS., and i2S. a pound, so that the mixture should be worth 9s. a pound: in what proportion must they be taken? Analysis. — To equalize tlie gain and loss, we opekation. compare the prices in pairs, one being aboce and the other helow the mean price, and, for conve- nience, connect them by curve lines. Taking the first and fourth; on i lb. at 3s. the gain is 6s. ; on I lb. at i2S. the loss is 3s., which we place op- posite the 12 and 3. Therefore, it takes i lb. at 3s. to balance the loss on 2 lb. at I2S., and the proportional parts of this couplet are as I to 2, or as 3 to 6. But 3 and 6 are the differences between the mean price and that of the teas compared, taken inversely. Again, i lb. at 8s. gains is., and i lb. at us. loses 2s., which we place opposite the 11 and 8. Therefore, it takes 2 lb. at 8s. to bal- ance the loss on i lb. at us., and the proportional parts of this couplet are as 2 to i. But 2 and i are the differences between the mean price and that of the teas compared, taken inversely. The parts are 3 lbs. at 3s., 2 lbs. at 8s., i lb. at us., 6 lbs. at 12s. If we compare i\\e first and third, the second and fourth, the pro- portional parts will be 2 lbs. at 3s., 3 lbs. at 8s., 6 lbs. at us., and I lb. at I2S. Hence, the EuLE. — I. Write the prices of the ingredients in a columUy with the mean price on the left, and talcing thenn in pairs, one less and the other greater than the mean price, connect them hy a curve line. II. Place the difference hetiveen the mean price and that of each ingredient opposite the price with which it is com- p>ared. The sum of the differences standing opposite each price is the proportional part of that ingredient. 515. How And the proportional parts of a mixture, wheii the mean price aud the price of each ingredient are given ? ALLIGATION" ALTERNATE. 327 Rem. — Since tlie results sliow tlie proportional parts to be taken, it follows if eacli is multijjlied or divided by the same number, an endless variety of answers may be obtained. 2D Method. — Since the mean price is gs. a pound, / I lb. at 3S. gains 6s. ; hence, to gain is. takes ^ lb.=^. lb. ) I lb. at 8s. gains is. ; hence, to gain is. takes i lb.=f lb. J I lb. at IIS. loses 2s. ; hence, to lose is. takes ^ lb.=| lb. ( I lb. at I2S. loses 3s. ; hence, to lose is. takes ^ lb. = | lb. Reducing these results to a common denominator, and using the numerators, the proportional parts are i lb. at 3s., 6 lbs. at 8s., 3 lbs. at IIS., and 2 lbs. at 12s. Hence, the Rule. — Take the given prices in pairs, one greater, the other less than the mean price, and find how much of each article is required to gain or lose a unit of the mean price, setting the result on the right of the corresponding price. Reduce the results to a co7nmo7i denominator, and the numerators will be the proportional parts required. Notes. — i. If there are three ingredients, compare the price of the one which is greater or less than the mean price with each of the others, and take the sum of the two numbers opposite this price. 2. The reason for considering the ingredients in pairs, one alove, and the other heloic the mean price, is that the loss on one may be counterbalanced by the gain on another. [For Canfield's Method, see Key to New Practical.] 8. A miller bought wheat at I1.60, $2.10, and I2.25 per bushel respectively, and made a mixture worth $2 a bushel: how much of each did he buy ? 9. A refiner wished to mix 4 parcels of gold 15, 18, 21, and 22 carats fine, so that the mixture might be 20 carats fine : what quail tity of each must he take ? 10. A grocer has three kinds of spices worth 32, 40, and 45 cts. a pound : in what proportion must they be mixed, that the mixture may be worth 38 cts. a pound? 11. A grocer mixed 4 kinds of butter worth 20 cts., 27 cts., 35 cts., and 40 cts. a pound respectively, and sold the mixture at 42 cts. a pound, whereby he made 10 cts. a pound : how much of a kind did he mix ? 12 — ^ 3 lb. 14- ^15" 21 — ^]6" 23- ^ 4" 328 ALLIGATIOi^ ALTERJS^ATE. PROBLEM II. 516. When one Ingredient, the Price of each, and the Mean Price of the Mixture are given, to find the other Ingredients. 12. How much sugar worth 12, 14, and 21 cts. a pound must be mixed with 12 lbs. at 23 cts. that the mixture may be worth 18 cts. a pound ? Analysis. — If neither ingredient were limited, tlie proportional parts would be 3 lbs. at 12 cts., 5 lbs. at 14 cts., 6 lbs. at 21 cts., and 4 lbs. at 23 cts. But the quantity at 23 cts. is limited to 12 lbs., which is 3 times its difference 4 lbs. Now the ratio of 12 lbs. to 4 lbs. is 3. Multiplying each of the proportional parts found by 3 the result will be 9 lbs., 15 lbs., 18 lbs., and 12 lbs. Hence, the EuLE. — Find the proportional parts as if ilie quantity of neither ingredient were limited. (Art. 515.) Multiply the parts thus found ly the ratio of the given ingredient to its proportional part, and the products will he the corresponding iyigredients required. Note. — When the quantities of tico or viore ingredients are given, find the average value of them, and considering their sum as one quantity, proceed as above, (Art. 515.) 13. How much barley at 40 cts., and corn at 80 cts., must be mixed with 10 bu. of oats at 30 cts. and 20 bu. of rye at 60 cts., that the mixture may be 55 cts. a bu. ? Suggestion. — The mean value of 10 bu. of oats at 30 cts. an(^ 20 bu. of rye at 60, is 50 cts. a bu. Ans. 30 bu. barley, and 24 bu. corn. 14. How much butter at 40, 45, and 50 cts. a pound respectively, must I mix with 30 lbs. at 65 cts. that the mixture may be worth 60 cts. a pound ? 15. How many quarts of milk, worth 4 and 6 cts. a quart respectively, must be mixed with 50 quarts of water s( that the mixture may be worth 5 cts. a quart ? 516. When one ingredient, the price of each and mean price are given, ho\» find the other ingredients ? ALLIGATIOIT ALTER 1^-ATE. 329 PROBLEM III. 517. To find the Ingredients, the Price of each, th® Quantity mixed, and the Mean Price being given. 1 6. How much water must be mixed with two kinds of Bourbon costing $4 and $6 a gal., to make a mixture of T50 gal. the mean price of which shall be I3 a gal. ? Analysis. — The price of the water is o. Disregarding the quantity to be mixed, and proceeding as in Problem I., the proportional part? are 4 g. water, 3 g. at $4, and 3 g. at $6, the sum of which is 4 g- ^ 3 g. + 3 g.=io gallons. But the whole mixture is to be 150 gallons. Now the ratio of 150 g. to 10 g. equals \^^^, or 15. Multiplying each of the parts previously obtained by 15, we have 4 gal. X 15=60 gal. water; 3 gal. x 15=45 gal. at $4; and 3 gal. x 15=45 gal. at $6. Hence, the EuLE. — Find the proportional parts without regard tc the quantity to he mixed, as i7i Problem I. Multiply each of the proportional parts thus found hj the ratio of the given mixture to the sum of these parts, and the several products will he the corresponding ingredients required, 17. A grocer mixed 100 lb. of lard worth 6, 8, and 12 cts. a pound, the mean value of the mixture being 10 cts. : how many pounds of each kind did he take ? 18. Having coffees worth 28, 30, 2>^, and 42 cts. a pound respectively, I wish to mix 200 lbs. in such proportions that the mean value of the mixture shall be 7,6 cts. a pound : how many pounds of each kind must I take ? 19. A grocer wished to mix 4 kinds of petroleum worth 40, 45, 50, and 60 cts. a gal. respectively: how much of each kind must he take to make a mixture of 300 gallons, worth 52 cts. a gallon? SI 7. How find the ingredientB, when the price of each, the quantity mised, and the mean price are given ? INVOLUTION. 518. Involution is finding a 'power of a number. A I^oivev is the product of a number multiplied intc itself Thus, 2x2 = 4; 3x3 = 9, etc., 4 and 9 are powers. 519. Vo-WQY^ 2a'Q (^lYidLQdiYnio different degrees ; as, first, second, tliird, fourth, etc. The name shows lioiv many times the number is taken as a factor to produce the power. 520. The First ^ower is the root or number itself The Second I^oiver is the product of a number taken twice as a factor, and is called a square. The Thi7*d JPower is the product of a number taken three tiynes as a factor, and is called a cube, etc. Notes. — i. The second power is called a square, because the pro- cess of raising a number to the second power is similar to that of finding the area of a square. (Art. 243.) 2. The third power in like manner is called a cube, because the process of raising a number to the third power is similar to that of finding the contents of a cube. (Art. 249.) 521. Powers are denoted by a small figure placed above the number on the right, called the index or exponent ; because it shows Iww many times the number is taken as a factor, to produce the power. Note. — The term index (plural indices), Latin indicere, to proclaim. Exponent is from the Latin exponere, to represent. Thus, 2' ==2, the first power, which is the number itself. 2^=2 X 2, the second power, or square. 2^=2 X 2 X 2, the third power, or cube. 2'*=2 X 2 X 2 X 2, i\\Q fourth power, etc, 522. The expression 2^ is read, "2 raised to the fourth power, or the fourth power of 2." 1. Read the following: 9^, 12', 25^, 245^ 38110, 465'% looo^*. 2. 6^ X 74, 2S^ X 48'^ 1408—753, 256^0 -r- 975. 518. What is involntion ? A power •' 519. How are powers divided ? 520. What is the first power ? The second ? Third ? Note. Why is the second power called a square? Why the third a cube ? 521. How are powers denoted? IJs^VOLUTIOIT. 331 3. Express tlie 4th power of 85. 5. The 7tli power of 340. 4. Express the 5th power of 348. 6. The 8th power of 561. 523. To raise a Number to any required Power. 7. What is the 4th po^v^r of 3 ? Analysts. — The fourth power is the product of a number into itself taken four times as a factor, and 3 x 3 x 3 x 3=81, the answer required. Hence, the KuLE. — Multiply the number into itself, till it is taken as many times as a factor as there are units in the index of the required poiver. Notes. — i. In raising a number to a power, it should be observed that the number of multiplications is always one less than the number of times it is t^^'^en as a factor; and tlierefore one less than tho number of the index. Thus, 4^=4 x 4 x 4, the 4 is taken three times as a factor, but there are only two multiplications. 2. A decimal fraction is raised to a power by multiplying it into itself, and pointing oJ0F as many decimals in each power as there are decimals in the factors employed. Thus, .i' = .oi, .22=.oo8, etc. 3. A common fraction is raised to a power by multiplying each term into itself. Thus, (f)^— ,Sg. 4. A mixed number should be reduced to an improper fraction, or the fractional part to a decimal; then proceed as above. Thus, (2|)* _(.^)2_iA. or 2^=2.5 and (2.5)-2=6.25. 5. All powers of i are i ; for i x i x i, etc. = i. Compare the square of the following integers and that of their corresponding decimals ; 8. 5> 6, 7, 8, 9, 10, 20, 30, 40, 50, 60, 70, 80, 90. 9. .5, .6, .7, .8, .9, .01, .02, .03, .04, .05, .06, .07, .08, .09. Eaise the following numbers to the powers indicated : 10. 53. 13- 4^. 16. 2.033. 19. r. II. 2^. 14. 8^ 17. 4.00033. 20. 13.- 12. 1323. 15- 25^ 18. 400.053. 21. 2-J4. 523. How raise a number to a power? Note. In raising a number to a power, how many multiplications are there ? How is a decimal raised to a power ? A common fraction? A mixed number? 524. How find the product of two or more powers of the same number ? 332 INVOLUTIOiq-. 524. To find the Product of two or more Powers of th© same Number. 2 2. What is the product of 4^ multiplied by 4^ ? Analysis.— 4=^=4 X 4 X 4, and 42=4x4; therefore in tlie product of 4^ x 4^, 4 is taken 3 + 2, or 5 times as a factor. But 3 and 2 are the given indices ; therefore 4 is taken as many times as a factor as there are units in the indices. Ans. 4^. Hence, the EuLE. — Add the indices^ and the sum ivill he the index of the product. 23. Mult. 2 3 by 22. 25. Mult. 4^ by 4^ 24. Mult. 3"^ by 3 3. 26. Mult. 5"^ by 5 2. FORMATION OF SQUARES. 525. To find the Square of a Number in the Terms of it* Parts. I. Find the square of 5 in the terms of the parts 3 and 2. Analysis. — Let the shaded part of the diagram represent the square of 3 ; — each f.ide being divided into 3 inches, its con- tonts are equal to 3 x 3, or 9 sq. in. The question now is, what additions must be made and how made, to preserve the form of this square, and make it equal to the square of 5. 1st. To preserve the form of the square it is plain equal additions must be made to two adjacent sides ; for, if made on one side, or on opposite sides, the figure will no longer be a square. 2d. Since 5 is 2 more than 3, it follows that two rows of 3 squares each must be added at the top, and 2 rows on one of the adjacent sides, to make its length and breadth each equal to 5. Now 2 into 3 plus 2 into 3 are 12 squares, or twice the product of the two parts 2 and 3. But the diagram wants 2 times 2 small squares, as represented by the dotted lines, to till the corner on the right, and 2 times 2 or 4 is the square of the second part. We have then 9 (the sq. of the ist })art), 12 (twice the prod, of the two parts 3 and 2), and 4 (the square of the 2d part). But 94-12 + 4=25, the square required. t)yoLUTiOK. 333 Again, if 5 is divided into 4 and r, the square of 4 is 16, twice the prod, of 4 into i is 8, and the square of i is i. But 16 + 8 + 1=25. 2. Eequired the square of 25 in the terms of 20 and 5. Analysis.— Multiplying 20 by 20 gives 2^— 20 + 5 400 (the square of the ist part); 20x5 25 20 + 5 plus 20 X 5 gives 200 (twice the prod, of J^ 400 + 100 the two parts) ; and 5 into 5 gives 25 (the "^ ^ , ^^^ , ^ square of the 2d part). Now 400 + 200 + 25 - — — - =625,or252. Hence, universally, 625=400 + 200 + 25 The square of any number expressed in the terms of its parts, is equal to tlie square of the first part, plus twice the product of the two parts, plus the square of the second part. 3. What is the square of 23 in the parts 20 and 3 ? 4. What is the square of 2^ or 2 +^ ? Ans. 6|. 5. What is the square of f or f + i ? Ans, 4, EVOLUTION. 526. Evolution is finding a root of a number. A Hoot is one of the equal factors of a number. 527. Boots, h'ke powers, are divided into degrees; as, the square, or second root ; the cube, or third root ; the fourth root, etc. 528. The Square Moot is one of the two equal factors of a number. Thus, 5 x 5 = 25 ; therefore, 5 is the square root of 25. 529. The Cube Moot is one of the throe equal factors of a number. Thus, 3x3x3^27; therefore, 3 is the cube root of 27, etc. 525. To what is the square of a number equal in the terms of its part»? 506. What is evolution ? A root ? 528. Square root ? 529. Cube root ? 334 EVOLUTION. 530. Roots are denoted in two loays: ist. By prefixing to the number the character ( |/), called the radical sigyi, with a figure over it ; as V4, \/8. 2d. By a fractional exponent placed above the number on the right. Thus, V9, or 9*, denotes the sq. root of 9. Notes. — i. The figure over the radical sign (|/) and the denomi- nator of the exponent respectively, denote the name of the root. 2. In expressing the square root, it is customary to use simply the radical sign (-i/), the 2 being understood. Thus, the expression 1/25 = 5, is read, " the square root of 25 = 5." 3. Tlie term radical is from the Latin radix, root. The sign (t ) is a corruption of the letter r, the initial of radix. 531. A Perfect JPower is a number whose exact root can be found. An Imperfect JPower is a number whose exact root can not be found. 532. A Surd is the root of an imperfect power. Thus, 5 is an imperfect power, and its square root 2.23 + is a surd. Note. — All roots as well n^ powers of i, are i. Eead the following expressions : 1. 1/40. 3. 119^ 5. i.5i 7. V256. 9. Vff. 2. V15. 4. 243^- 6. V29. 8. V45.7. 10. V-B|. 1 1. Express the cube root of 64 both ways ; the 4th root of 25 ; the 7th root of 81 ; the loth root of 100. 533. To find how many figures the Square of a Number contains. ist. Take i and 9, the least and greatest integer that can be expressed by one figure; also 10 and 99, the least and greatest that can be expressed by two integral figures, etc. Squaring these numbers, we hare for The Roots: i, • 9, 10, 99, 100, 999, etc. The Squares: i, 81, 100, 9801, loooo, 998001, etc. 530. How are roots denoted? 531. A perfect power? Imperfect? 532. h Burd ? 5»j3. How many figures has the square of a number ? SQUARE ROOT. 335 2d. Take . i and .9, the least and greatest decimals that can be expressed by one figure; also .01 and .99, the least and greatest that can be expressed by tivo decimal figures, etc. Squaring these, we have for • TheEoots: .1, .9, .01, .99, .001, .999, etc. The Squares: .01, .81, .0001, .9801, .000001, .998001, etc. By inspecting these roots and squares, we discover that Tlie square of the number contains twice as ma7i7j figures as its root, or ttuice as many less one. 534. To find how many figures the Square Root of a Number contains. Divide the number into periods of two figures each, placing a dot over units^ place^ another over hundreds, etc. The root will have as many figures as there are periods. Remark. — Since the square of a number consisting of tens and units, is equal to the square of the tens, etc., when a number has two periods, it follows that the left hand period must contain the squart of the tens 01 first figure of the root. (Art. 525.) EXTRACTION OF THE SQUARE ROOT. 535. To extract the Square Hoot of a Number. I. A man wishes to lay out a garden in the form of a square, which shall contain 625 sq. yards: what will be the length of one side ? Analysis. — Since 625 contains two periods, its root operation. will have two figures, and the left hand period con- r '( ^ tains the sgw^re of the tens' figure. (Art. 534, ^^m.) ^ But the greatest square of 6 is 4, and the root of 4 ^ is 2, which we place on the right for the tens' figure 45)2 25 of the root. Now the square of 2 tens or 20 is 400, 225 and 625—400=225. Hence, 225 is twice the product of the tens' figure of the roct into the units, plus the square of the units. (Art. 525.) 534. How many figures has the square root of a nnmLer ? 336 EXTRACTIOl?^ OF SQUARE ROOT. 20 yds. syds. m T! >. w^.:^.^..^ fO - ^ O ^^^:;=^"— ==tr}f-g yds. syds. But one of tlie factors of this product is 2 times 20 or 40 ; there* fore the other factor must be 225 divided by 40 ; and «25-f-40=5. the factor required. (Art. 93.) Taking 2 times 20 into 5=200 (i, e., twice the product of the tens into the units' figure of the root) from 225, leaves 25 for the square of the units' figure, the square root of which is 5 Hence 25 is the square root of 625, and is there- fore the length of one side of the garden. 2d Analysis. — Let the shaded part of the diagram be the square of 2 tens, the first figure of the root ; then 20 x 20, or 400 sq. yds., will be its contents. Sub- tracting the contents from the given area, we have 625—400=225 sq. yds. to be added to it. To preserve its form, the addition must be made equally to two adjacent sides. The question now is, what is the width of the addition. Since the length of the plot is 20 yds., adding a strip i yard wide to two sides will take 20 + 20 or 40 sq. yds. Now if 40 sq. yds. will add a strip i yard wide to the plot, 225 sq. yds. will add a strip as many yds. wide as 40 is contained times in 225 ; and 40 is contained in 225, 5 times and 25 over. That is, since the addition is to be made on two sides, we double the root or length of one side for a trial divisor, and find it is con- tained in 225, 5 times, which shows the width of the addition to be 5 yards. Now the length of each side addition being 20 yds., and the width 5 yds., the area of both equals 20 x 5 + 20 x 5, or 40 x 5 = 200 sq. yards. But there is a vacancy at the upper corner on the right, whose length and breadth are 5 yds. each ; hence its area =5 x 5, or 25 sq. yards; and 200 sq. yd. + 25 sq. yd. = 225 sq. yd. For the sake of finding the area of the two side additions and that of the corner at the same time, we place the quotient 5 on the right of the root already found, and also on the right of the trial divisor to complete it. Multiplying the divisor thus completed by 5, the figure last placed in the root, we have 45x5 = 225 sq. yds. Subtracting this product from the tfiividend, nothing remains. Therefore, etc. Hence, the 535. What is the first step in extracting the square root of a number ? The eecond? Third? Fourth? How proved? iVofe. If the trial divisor is not con- tained in the dividend, what must be done ? If there is a remainder after the root of the last period is found, what? How many decimals does the root of a decimal fraction have I EXTRACTION OF SQUARE ROOT. 337 EuLE. — I. Divide the number i7ito periods of two figures each, putting a dot over units, then over every second figure towards the left in whole numbers, and towards the right in decimals, II. Find the greatest square in the left hand period, and place its root on the right. Subtract this square from the period, and to the right of the remainder bring down the next period for a dividend, III. Double the part of the root thus found for a trial divisor; and finding how many times it is contained in the dividend, excepting the right hand figure,, amiex the quotient both to the root and to the divisor, IV. Multiply the divisor thus increased by this last figure placed in the root, subtract the product, and bring down the next period. V. Double the right hand figure of the last divisor, and proceed as before, till the root of all the periods is found. Proof. — Multiply the root into itself, (Art 528.) Notes. — i. If the trial divisor is not contained in the dividend, annex a cipher both to the root and to the divisor, and bring down the next period. 2. Since the product of the trial divisor into the quotient figuro cannot exceed the dividend, allowance must be made for carrying, if the product of this figure into itself exceeds 9. 3. It sometimes happens that the remainder is larger than the divisor ; but it does not necessarily follow from this that the figure in the root is too small, as in simple division. 4. If there is a remainder after the root of the last period is found, annex periods of ciphers, and proceed as before. The figures of the root thus obtained will be decimals. 5. The square root of a decimal fraction is found in the same way as that of a whole number ; and the root will have as many decimal figures as there are periods of decim^ils in the given number. 6. The left hand period in whole numbers may have but one figure ; but in decimals, each period must have two figures. (Art. 533.) Hence, if the number has but one decimal figure, or an odd number of decimals, a cipher must be annexed to complete the period. 536. Reasons. — i. Dividing the number into periods of two figures each, shows how many figures the root will contain, and 15 338 EXTRACTIOi^ OF SQUAllE KOOT. enables us to find its first figure. For, the left hand period containa the square of this figure, and from the square the root is easily found. (Art. 534, Rem.) '1. Subtracting the square of the first figure of the root from the left hand period, shows what is left for the other figures of the root. 3. The object of doubling the first figure of the root, and dividing the remainder by it as a trial divisor, is to find the next figure of tho root. The remainder contains twice the product of the teiis into the units; consequently, dividing this product by double the tens' factor, the quotient will be the other factor or units' figure of the root. 4. Or, referring to the diagram, it is doubled because the remainder must be added to two sides, to preserve the form of the square. 5. The right hand figure of the dividend is excepted, to counter- balance the omission of the cipher, which properly belongs on the right of the trial divisor. 6. The quotient figure is placed in the root; it is also annexed to the trial divisor to complete it. The divisor thus completed is multiplied l)y the second figure of the root to find the contents of the additions thus made. The reasons for the steps in obtaining other figures of the root mi^ be shown in a similar manner. 2. What is the square root of 381.0304? Ans. 19.52. Suggestion. — Since the number contains decimals, we begin at the units' place, and counting both ways, have four periods ; as, 381.0304. The root will therefore have 4 figures. But there are two periods of decimals; hence we point off two decimals in the root. 3. What is the square root of 1 01 2036 ? Ans. 1006. 4. What is the square root of 2 ? A7is. 1.4 142 1 +. Extract the square root of the following numbers : 5. 182329. II. .1681. 17.5. 23. 19.5364. 6. 516961. 12. .725. 18.7. 24. 3283.29. 7- 595984. 13. .1261. 19. 8. 25. 87.65. 8. 3.580. 14. 2.6752. 20. 10. 26. 123456789. 9. .4096. 15. 4826.75. 21. II. 27. 61723020.96. 10. .120409. 16. 452.634. 22. 12. 28. 9754.60423716. 536. Why divide the number into periods of two figures each ? Why subtract the square of the first figure from the period ? Why double the first figure of the root for a trial divisor ? Why omit the right hand figure of the dividend ? Why place the quotient figure on the right of the trial divisor? Why multiply il.e trial divisor thus completed by the figure last placed in the root? APPLICATION'S OF SQUARE ROOT. 339 637. To find the Square Root of a Common Fraction. I. When the numerator and denominator are both perfect squares, or can he reduced to such, extract the square root of each term separately. II. When they are imperfect squares, reduce them to decimals, and proceed as above. Note. — To find the square root of a mixed nuiriber, reduce it to an improper fraction, and proceed as befora 29. What is the square root of y^g^? Analysis.— 1%=^, and t/^=|, Arts. 30. What is the square root off? Ans..'j'j4S +' 31. What is the square root of if? Ans. 1.1726 +. rind the square root of the following fractions : 32. Hi' 35. 6i. 38. Iff. 41. i^V • 33' f 36. i3i- 39- i-U' 42. 27^^. 34. iV 37. lyf- 40. it-f 43- 5ifi- APPLICATIONS. 538. To find the Side of a Square equal in area to a given Surface. 1. Find the side of a square farm containing 40 acres. Analysis. — In i acre there are 160 1 J . ,. OPERATION. sq. rods, and m 40 acres, 40 times k ^ r 160, or 6400 sq. rods. The ^^6400= 40 A. X 160 = 6400 sq. r. 80 r. Therefore the side of the farm is 1/6400 = 80 r. Ans. 80 linear rods. Hence, the Rule. — I. Extract the square root of the given surface. Note. — The root is in linear units of the same name as the given surface. 2. What is the side of a square tract of land containing II 02 acres 80 sq. rods ? 537. How find the square root of a common fraction? Note. Of a mixed number? 538. How find the side of a square equal to a given surface ? 340 APPLICATIOif S or SQUARE ROOT. 3. How many rods of fencing does it require to inclose a square farm which contains 122 acres 30 sq. rods ? 4. A bought 141 6 1 fruit trees, which he planted so r:s to form a square : how many trees did he put in a row ? 5. A general has an army of 56644 men: how many must he place in rank and file to form them into a square ? 539. A Triangle is a figure "having three C ddes and three angles. A Hi (flit-angled Triangle is one that contains a right angle. (Art. 260.) The side opposite tlie right angle is called tlio hypothenuse ; the other two sides the hose and perpendicular. 640. The square described on the hypothenuse of a right- angled triangle is equal to the sum of the squares descriied on the other two sides.^ 541. The truth of this principle may be illustrated thus : Take any right-angled triangle ABC; let the hypothenuse h, be 5 in., the base h, 4 in., and the per- pendicular p, 3 in. it will be seen that the square of h contains 25 sq. in., the square of h 16 sq. in., and the square of ^ 9 sq. in. Now 25 = 16 + 9, which accords with the proposition. In like manner it may be shown that the principle is true of all right-angled triangles. Hence, 542. To find the IIyj)ot7ienuse, the Base and Perpendicular being given. To the square of the base add the square of the per- pendicular, and extract the square root of their sum. 539. What is a triangle? A right-angled triangle? Draw a right-angled triangle? The side opposite the right-angle called? The other two sides? 540. To what is the square of the hypothenuse equal? 541. Illustrate this prin- ciple by a figure? 542. How find the hypothenuse M-hen the base and per- pendicular are given ? \A ) n m 1=1 ^mm^ B - - * Thomson's Geometry, IV, 1 1 ; Euclid, I, 47. /- applicatio:n^s of squabe boot. 341 643. To find the Base, the Hypothenuse and Perpendicular being given. From the square of the hypothenuse take the square of the perpendicular, and extract the square root of the remainder. 544. To find the Perpendicular, the Hypothenuse and Base being given. From the square of the hypothenuse take the square of the hase, and extract the square root of the remainder. Note. — The pupil should draw figures corresponding with the conditions of the following problems, and indicate the parts given : 6. The perpendicular height of a flag-staff is ^6 ft.: what length of Hne is required to reach from its top to a point in a level surface 48 ft. from its base ? SOLUTlOlf.— The square of the base . . =48 x 48=2304 '* " perpendicu]ar=36 X 36 = 1296 The square root of their sum =1/3600 =60 ft. Ans. 7. The h}"pothenuse of a right-angled triangle is 135 yds., the perpendicular 81 yds.: what is the base? 8. One side of a rectangular field is 40 rods, and the distance between its opposite corners 50 rods: what is the length of the other side ? 9. Two vessels sail from the same point, one going due south 360 miles, the other due cast 250 miles: how far apart were they then ? 10. The height of a tree on the bank of a river is 100 ffc., and a line stretching from its top to the opposite side is 144 ft. : what is the width of the river ? 11. The side of a square room is 40 feet: what is the distance between its opposite corners on the floor ? 12. A tree was broken 35 feet from its root, and struck the ground 21 ft. from its base: what was the height of the tree ? 543. How find the base when the other two sides are given ? 544. How find the perpendicular when the other two sides are given ? 342 SIMILAR FIGURES. SIMILAR FIGURES. 545. Similar Figures are those which have the same form, and their like dimensions proportional. Notes. — i. All circles, of wliatever magnitude, are similar. 2. All squares, equilateral triangles, and regular polygons are similar. And, universally, All rectilinear figures are similar, wlien their several angles are equal each to each, and their like dimensions proportional. 3. The like dimensions of circles are their diameters, radii, and circumferences. 546. The areas of similar figures are to each other as the squares of their like diraensions. And, Conversely, the liJce dimensions of similar figures are to each other as the square roots of their areas. 13. If one side of a triangle is 12 rods, and its area 72 sq. rods, what is the area of a similar triangle, the corresponding side of which is 8 rods ? Solution. — (12)2 : (8)2 : : 72 : Ans., or 32 sq. rods. 14. If one side of a triangle containing s^ sq. rods is 8 rods, what is the length of a corresponding side of a similar triangle which contains 81 sq. rods? So:.UTiON. — v^36 : |/8i : : 8 : Ans., or 12 rods. 15. If a pipe 2 inches in diameter will fill a cistern in 42 min., in what time will a pipe 7 in. in diameter fill it ? 547. To find a Mean Proportional between two Numbers. 16. What is the mean proportional between 4 and 16 ? Analysis. — When three numbers are proportional, the product of the extremes is equal to the square of the mean. (Arts. 487, 490.) But 4 X 16=64 ; and 1/64=8. Ans. Hence, the EuLE. — Extract the square root of their product. Find the mean proportional between the following : 17. 4 and 36. 19. 56 and 72. 21. Jf and -^. t8. 36 and 81. 20. .49 and 6.25. 22. -^ and ^^V 545. What are similar figures ? 546. How are tkeir areas to each other V FORMATIOK OF CUBES. 543 FORMATION OF CUBES. 548. To find the Cube of a Number In the terms of its parts. I. FiaJ the cube of 32 in the terms of the parts 30 and 2. ANALYSl!^.— 32=30 4-2, and 325 = (30 + 2)x(30 4-2)x(30 + 2.) 96 : 64: :3^_+2 •3^^ + (30x2) : (30 X 2) 4- 2^ :302 + 2Xv30 1024 32 = 30 +2 2) + 22 3072 =30^ + 2 X (30'' X 2) +-(30x2*^) 2048 — (30- X 2) + 2 X v30 X 2'^) + 28 32768 = 30^ + 3 X (3o''^ X 2) + 3 X ,;30 X z^) + 2- 3 feet. 3X.3X3 Or thus : Let the diagram represent a cube whose side is 30 ft. ; its contents= tlie cube of 3 tens, or 2700 cu. ft. What additions must be made to this cube, and hoic made, to preserve its/orw, and make it equal to the cube of 32. ist. To preserve its cubical form, the additions must be equally made on tliree adjacent sides ; as the top, front, and right. 2d. Since 32 is 2 more than 30, it follows that this cube must be made 2 ft. longer, 2 ft. wider, and 2 ft. higher, that its lengt^, breadth, and thickness may each be 32 ft. But as the side of this cube is 30 ft., the contents of each of these additions must be equal to the square of the tens (so^) into 2, the units, and their sum must be 3 times (30* x 2)=3 x gcx) x 2=5400 cu. ft. But there are three vacancies along the edges of the cube adjacent to the additions. Each of these vacancies is 30 ft. long, 2 ft. wide, and 2 ft. tliick ; hence, the contents of each equals 30 x 2*, and the sum of their contents equals 3 times the tens into the square of the units = 3 times (30 X 2^)= 3 X 30 X 4=360 cu. ft. But there is still another vacancy at the junction of the corner additions, whose length, breadth, and thickness are each 2 ft., and whose contents are equal to 2^=2x2x2=8. The cube is now complete. Therefore 32^= 27000 (30^) + 5400 (3 times 30^x2) + 360 (3 times 30x2^)4-8 (2^) = 32768 cu. ft. In like manner it may be shown that. 548. To what IB the cube of a number consisting of tens and units equal 344 FORMATION OF CUBES. TJie cube of any number consisting of tens and units is equal to the cube of the tens^ plus 3 times the square of the tens into the miits^ plus 3 times the tefis into the square of the units, plus the cube of the units. 549. To find how many figures the Cube of a Number contains. ist. Take i and g, also 10 and 99, 100 and 999, etc., the least and greatest int^ers that can be expressed by one, two, three, etc., figures. 2d. In like manner talie .1 and .9, also .01 and .99, etc., the least and greatest decimals that can be expressed by one, two, etc., decimal figures. Cubing these, we have t I and 1^=1, - - - .1 and .l3=.I 9 " 9^-729, - - - -9 " .93 = . 729 10 " 10^=1000, - - - .01 " .01 3 =.000001 99 " 993=970299, - - -99 " .993=. 970299 100 " 1003=1000000, - - .001 " .ooi^ .000000001 999 " 999^=997002999, - .999 " .9993=. 997002999 By comparing these roots and their cubes, we discover that T/ie cube of a number canfiot have more than three times as many figures as its root, nor but tivo less. Hence, 550. To find how many figures the Cube Soot contains. Divide the number into periods of three figures each, putting a dot over units, then over every third figure towards the left in whole numbers, and towards the right in decimals. Remakes. — i. Since the cube of a number consisting of tens and units is equal to the cvhe of the tens, plus 3 times the square of the tens into the units, etc., when a number has two periods, it follows that the left hand period must contain the cube of the tenSy or first figure of the root. 2. The left hand period in whole numbers may be incomplete, having only wie or two figures; but in decimals each period must always have Viree figures. Hence, if the decimal figures in a given number are less than three, annex ciphers to complete the period. How many figures in the cube root of the following : 2. 340566. 4. 576.453. 6. 32.7561. 3. 1467. 5. 5.7321. 7. .456785. 549. How many figures has the cube of a number? EXTEACTIOI^ OF THE CUBE ROOT. 345 EXTRACTION OF THE CUBE ROOT. 551. To extract the Cube Root of a Number. I. A man having 32768 marble blocks, each being a cubic foot, wishes to arrange them into a single cube: what must be its side ? Analysis. — Since 32768 contains two periods, operation. we know its root will have two figures; also that ■^2768(-^2 the left hand period contains the cube of the tens 27 or first figure of the root. (Art. 550, Rem) The greatest cube in 32 is 27, and its root is 3, ' °° which we place on the right for the tens or first ^ figure of the root. Subtracting its cube from the 5768 5768 first period, and bringing down the next period 2884 to the right of the remainder, we have 5768, which by the formation of the cube is equal to 3 times the square of the tens' figure into the units, plus 3 times the tens into the square of the units, plus the cube of the units. We therefore place 3 times the square of the tens or 2700, on the left of the dividend for a trial divisor ; and dividing, place the quotient 2 on the right for the units' figure of the root. To complete the divisor, we add to it 3 times 30 the tens into 2 units=i8o; also 4, the square of the units, making it 2884. Multi- plying the divisor thus completed by the units' figure 2, we have 2884 X 2 = 5768, the same as the dividend. Ans. 32 ft. Or thus : Let the cube of 30, the tens of the root, be represented by the large cube in the set of cubical blocks.* The remainder 5768, is to be added equally to three adjacent sides of this cube. To ascertain the thickness of these side additions, we form a trial divisor by squaring 3, the first figure of the root, with a cipher annexed, for the area of one side of this cube, and multiply this square by 3 for the three side additions. Now 30^=30 x 30=900; and 900x3 = 2700, the trial divisor. Dividing 5768 by 2700, the quotient is 2, which shows that the side additions are to be 2 ft. thick, and is placed on the right for the units' figure of the root. 551. The first step in extracting the cube root f The second? Third? Fourth? Fifth? Note. If the trial divisor is not contained in the dividend, how proceed ? If there is a remainder after the root of the last period is found, how f * Every school in which the cube root is taught is presumed to be furnished with a set of Cubical Blocks. '' S46 EXTRA CTIOK OF THE CUBE EOOT. To represont these additions, place the corresponding layers on the top, front, and right of the large cube. But we discover three vacancies along the edges of the large cube, each of which is 30 ft. long, 2 ft. wide, and 2 ft. thick. Filling these vacancies with the corresponding rectangular blocks, we discover another vacancy at the junction of the corners just filled, whose length, breadth, and thickness are each 2 ft. This is filled by the small cube. To complete the trial divisor, wo add the area of one side of each of the corner additions, viz., 30 x 2 x 3, or 180 sq. ft., also the area of one side of the small cube=2 x 2, or 4 sq. ft. Now 2700+ 180 + 4= 2884. The divisor is now composed of the area of 3 sides of the large cube, plus the area of one side of each of the corner additions, plus the area of one side of the small cube, and is complete. Finally, to ascertain the contents of the several additions, we multiply the divisor thus completed by 2, the last figure of the root ; and 2884x2 = 5768. (Art. 249.) Subtracting the product from the dividend, nothing remains. Hence, the Rule. — I. Divide the number into periods of three figures each, putting a dot over units, then over every third figure towards the left in ivhole numbers, and towards the right in decimals. II. Find the greatest cube in the left hand period, and place its root on the right. Subtract its cube from the period, and to the right of the remainder bring down the next period for a dividend. III. Multiply the square of the root thus found ivith a cipher annexed, by three, for a trial divisor ; and finding how many times it is contained in the dividend, write the quotient for the second figure of the root. IV. To complete the trial divisor, add to it three times the product of the root 'previously found with a cipher annexed, into the second root figure, also the square of the second root figure. V. Multiply the divisor thus completed by the last figure placed in the root Subtract the product from the divi- dend ; and to the right of the remainder bring down the next period for a neio dividend. Find a neio trial divisor, as before, and th^ proceed till the root of the last period is found. EXTRACTIOi^ OF THE CUBE ROOT. 347 Notes. — i. If a trial divisor is not contained in tlie dividend, put a cipher in tlie root, tivo ciphers on the right of the divisor, and bring down the next period. 2. If the product of tho divisor completed into the figure last placed in the root exceeds the dividend, the root figure is too large. ►Sometimes the remainder is larger than the divisor completed ; but it does not necessarily follow that the root figure is too small. 3. When there are three or more periods in the given number, the first, second, and subsequent trial divisors are found in the same manner as when there are only two. That is, disregarding its true local value, we simply multiply the square of the root already found with a cipher annexed, by 3, etc, 4. If there ip. a remainder after the root of the last period is found, annex periods of ciphers, and pioceed as before. The root figures thus obtained will bo decimals. 5. The cube root of a decimal fraction is found in the same way as that of a whole number ; and will have as many decimal figures as there are periods of decimals in tho nutuber. (Art. 549.) 552. Reasons. — i. Dividing the number into periods shows how many figures the root contains, and enables us to find the firnt figure of the root. For, the left hand period contains the cube of the first figure of the root. (Art, 550.) 2. The object of the trial divisor is to find the next figure of the root, or the thickness of the side additions. The root is squared to find the area of one side of the cube whose root is found, the cipher being annexed because the first figure is tens. This square is multi- plied by 3, because the additions are to be made to three sides. 3. The root previously found is multiplied by this second figure 10 find the area of a side of one of the vacancies along the edges of the cube already found. This product is multiplied by 3, because there are three of these vacancies ; and the product is placed under the trial divisor as a correction. The object of squaring the second figure of the root is to find the area of one side of the cubical vacancy at the junction of the corner vacancies, and with the other correction this is added to the trial divisor to complete it. 4. The divisor thus completed is multiplied by the second figuro of the root to find the contents of the several additions now made. 552. Why divide the number into periods of three fl^irep ? "What is the ohject of a trial divisor ? Why square the root already found ? Why annex a cipher to it ? Why multiply this square by 3 ? Why is the root previously found multi- plied by the second figure of the root? Why multiply this product by 3? Why square tbe second fignre of the root ? Why multiply the divisor rrhen completed by the second flg:ure of the root * 348 EXTRACTION OF THE CUBE ROOT. 2. What is the cube root of 1 30241.3 ? Anaxysis. — Having completed the operatiok. period of decimals by annexing two i-:jo24I.':joo(c;o 6 + ciphers, we find the first figure of the j 2 c root as above. Bringing down the next period, the dividend is 5241. '^ The trial divisor 7500 is not contained " 'in the dividend; therefore placing a ^ 5241.300 4554216 cipher in the root and two on the 759*^3^ right of the divisor, we bring down 687084 Rem. the next period, and proceed as before. 3. Cube root of 6 141 25 ? 6. Cube root of 3 ? 4. Cube root of 84.604 ? 7. Cube root of 21.024576 ? 5. Cube root of 373248 ? 8. Cube root of 17 ? 9. What is the cube root of 705919947264? 10. What is the cube root of .253395799 ? 11. What is the side of a cube which contains 628568 cu. yards ? Note. — The root is in linear units of the same name as the given contents. 12. What is the side of a cube equal to a pile of wood 40 ft. long, 15 ft. wide, and 6 ft. high ? 13. What is the side of a cubical bin which will hold 1000 bu. of corn; allowing 2150.4 cu. in. to a bushel ? 553. To find the Cube Root of a Common Fraction, If the numerator and denominator are perfect cules, of can he reduced to such, extract the cube root of each. Or, reduce the fraction to a decimal, and proceed as before* Note. — ^Reduce mixed numbers io improper fractions, etc. 14. What is the cube root of /^ ? . Ans. %. 15. Cube root of f? ^?zs. .75+. 16. Cube root of ^^^^ ? 18. Cube root of ^\^i^ ? 1 7. Cube root of \Uil ? 1 9. Cube root of 8 1 1 ? 553 How find the cube root of a common fraction ? Note. Of a mixed number f SIMILAR SOLIDS. 349 APPLICATIONS. 554. Similar Solids are those which have the same form, and their like di7nensions 'projportional. Notes. — i. The like dimensions of spheres are their diameters, radii, and circumferences ; those of cubes are their sides. 2. The like dimensions of cylinders and cones are their altitudes, and the diameters or the circumferences of their bases. 3. Pyramids are similar, when their bases are similar polygons, and their altitudes proportional. 4. Polyhedrons (i. e., solids included by any number of plane faces) are similar, when they are contained by the same number ot similai polygons, and all their solid angles are equal each to each. 555. The C07itents of similar solids are to each other aa the cubes of their lihe dimensions; and, Conversely, the like dimensions of similar solids are a& the cube roots of their contents. 1. If the side of a cuhical cistern containing 1728 cu. in. is 12 in., what are the contents of a similar cistern whoso side is 2 ft. Ans. i ^ : 2^ : : 1728 cu. in. : con., or 13824 cu. in. 2. If the side of a certain mound containing 74088 cu. ft. is 84 ft., what is the side of a similar mound which con- tains 17576 cu. ft.? 3. If a globe 4 in. in diameter weighs 9 lbs , what is the weight of a globe 8 in. in diameter? A71S. 72 lbs 4. If 8 cubic piles of wood, the side of each being 8 ft.^ were consolidated into one cubic pile, what would be the length of its side .? 5. If a pyramid 60 ft. high contains 12500 cu. ft., what are the contents of a similar pyramid whose height is 20 ft. ? 6. If a conical stack of hay 15 ft. high contains 6 tons, what is the weight of a similar stack whose height is 1 2 ft. ? 554. What are similar solids ? Note. What are like dimen?ion8 of sphere? 1 Of cylinders and cones ? Of pyramids ? 555. What relation have similar solids ? AEITHMETICAL PROGEESSIOK 556. An AintJunetical I^rogression is a series "jf numbers which increase or decrease by a common dif- ference. Note. — The numbers forming the series are called the Terms, ^he first and last terras are the Extremes; the intermediate terms the Meav^. (xirts. 476, 489.) 557. The Cojnmon Difference is the difference between the successive terms. 558. ^n Ascendiufj Series is one in which the successive terms increase ; as, 2, 4, 6, 8, 10, etc., the common difference being 2. A Descetiding Series is one in which the successive terms decrease ; as, 15, 12, g, 6, etc., the common difference being 3. 559. In arithmetical progression there are five parts or elements to be considered, viz.: the first term, the last term, the number of t:rms, the common difference, and the sum of all the terms. Theeo parts are so related to each other, that if any three of them are given, the other two may be found. 560. To find the Last Term, the First Term, the Number of Terms, and the Common Difference being given. I. Eequired the last term of the ascending series having 7 terms, its first term being 3, and its common difference 2. Analysis. — From the definition, each succeeding term is found by adding the common difference to the preceding. The series is : 3, 3 + 2, 3 + (2 + 2), 3 + (2 4- 2 + 2), 3 + (2 + 2 + 2 + 2), etc. Or 3» 3 + 2, 3 + (2 X 2), 3 + (2 X 3), 3 + (2 X 4), etc. That is, 561. The last term is equal to the first term, increased by the product of the common difference into the number of terms less I. Hence, the Rule. — Multiply the common difference hy the number of terms less i, and add the product to the first term. 556. What is an arithmetical progression? Note. The first and last terms called? Tiie intervening;? e,^j. The common difference? 558. An ascendinsf peries? Descending? 560. tlow find the last term, when the first term, tlie number of terms, and the common difference are given? ARITHMETICAL P R O G R E S S I O iq". 351 ;N'OTES. — I. In a descending series the product must be subtracUd from tlie first term ; for, in this case each succeeding term is found by subtracting the common difference from the preceding terms, 2. Any term in a series may be found by the i)receding rule. For, the series may be supposed to stop at any term, and that may be considered, for the time, as the last. 3. If the last term is given, and the first term required, invert the order of the terms, and proceed as above. 2. If the first term of an ascending series is 5, the common difference 3, and the number of terms 1 1, what is the last term ? Ans. 2>S- 3. The first term of a descending series is 35, the com- mon difference 3, and the number of terms 10 : what is the last ? 4. The last term of an ascending series is 77, the number of terms 19, and the common difference 3: what is the first term ? Ans. 2t^. 5. What is the amount of $150, at 7^ simple interest, for 20 years ? 562. To find the Number of Tenns, the Extremes, and the Common Difference being given. 6. The extremes of an arithmetical series are 4 and 37, and the common difference 3 : what is the number of terms ? Analysis. — The last term of a series is equal to the first term increased or diminished by the product of the common difference into the number of terms less 1. (Art. 561.) Therefore 37—4, or 33, is the product of the common difference 3, into the number of terms less i. Consequently 33-^3 or 11, must be the number of terras less I ; and 11 + i,or 12, is the answer required. (Art. 93.) Hence, the Rule. — Divide the difference of the extremes ly the com- mon difference, and add i to the quotient. 7. The youngest child of a family is i year, the oldest. 21, and the common difference of their ages 2 y. : how many children in the family ? 562. How find the number of terms, when the extremes and common difiercnce are given ? 352 AKITHMETICAL PKOGRESSIOK. 563. To find the Coinnion Difference^ the Extremes and the Number of Terms being given. 8. The extremes of a series are 3 and 21, and the number of terms is 10 : what is the common difference ? Analysis. — The difference of the extremes 21—3=^18, is the pro- duct of the number of terms less i into the common difference, and 10— I, or g, is the number of terms less i ; therefore 18-5-9, or 2, is tlie common difiference required. (Art. 93.) Hence, the Rule. — Divide the difference of the extremes ly the number of terms less i. 9. The ages of 7 sons form an arithmetical series, the youngest being 2, and the eldest 20 years: what is the difference of their ages ? 564. To find the Sum of all the Terms, the Extremes and the Number of Terms being given. 10. Required the sum of \hQ series having 7 terms, tlie extremes being 3 and 15. Analysis — (i.) The series is 3, 5, 7, g, 11, 13, 15. (2.) Inverting the same, 15, 13, 11, 9, 7, 5, 3. (3.) Adding (i.) and (2.), 18 4- 18 + 18 + 18 + 18 + 18 f i8=twice the sum. (4.) Dividing (3.) by 2, 9 + 9 + 9 + 9 + g + 9 + 9=63, the sum. By inspecting these series, we discover that half the sum of the extremes is equal to the average value of the terms. Hence, the Rule. — Multiply half the sum of the extremes by the number of terms. Remark. — From the preceding illustration we also see that, The sum of the extremes is equal to the sum of any two terms equidistant from them ; or, to twice the sum of the middle terni , if the number of terms he odd. 11. How many strokes does a common clock strike in 12 hours? 563. How find the common difference when the extremes and number of terma are given ? 564. How find the sum of all the terUiS, when the extremes twid number of terms are given ? GEOMETEIOAL PEOGRESSIOK 565. A Geometrical Progression is a series of numbers which increase or decrease by a common ratio. Note. — The series is called Ascending or Descending, according as the terms increase or decrease. (Art. 558.) 566. In Geometrical Progression there are also five parts or elements to be considered, viz.: the first term, the last term, the number of terms, the ratio, and the sum of all the terms. 567. To find the Last Term, the First Term, the Ratio, and the Number of Terms being given. 1. Required the last term of an ascending series having 6 terms, tlie first term being 3, and the ratio 2. Analysis. — From the definition, the series is 3* 3x2, 3 X (2 X 2), 3 X (2 X 2 X 2), 3 X (2 X 2 X 2 X 2), etc. Or 3, 3x2, 3 X 2'^, 3x2*^, 3 X 2^, etc. That is. Each successive term is equal to the first term multiplied by the ratio raised to a power whoso index is one less than the number of the term. Hence, the Rule. — Multiphj the first term ly that imiver of the ratio whose index is i less than the number of terms. Notes. — i. Any term in a scries may be found by the preceding rule. For, the series may be supposed to stop at that term. 2. If the last term is given and the first required, invert the order of the terms, and proceed as above. 3. The preceding rule is applicable to Compound Interest; the principal being the first term of the series ; the amount of $1 for 1 year the ratio ; and the number of years plus i, the number of terms. 2. A father promised his son 2 cts. for the first example he solved, 4 cts. for the second, 8 cts. for the third, etc. : what would the son receive for the tenth example ? 3. What is the amount of $1500 for 5 years, at 6% com- pound interest ? Of I2000 for 6 years, at 7^ ? 565. What is a geometrical progression ? 567. How find the last term. wheA the first term, the ratio, and number of terms are given ? 354 GEOMETEICAL PROGRESSION". 568. To find the Sum of all the Terms, the Extremes and Ratio being given. 4. Eequired the sum of the series whose first and last terms are 2 and 162, and the ratio 3. Analysis. — Since eacli succeeding term is found by multiplying tlie preceding term by the ratio, tlie series is 2, 6, 18, 54, 162. (i.) The sum of the series, =2-1-6+18 + 54+ 162. (2.) Multiplying by 3, = 6 + 18 + 54 + 162 + 486. (3.) Subt. (i.) from (2.), 486—2=484, or twice the sum. Therefore 484-5-2=242, the sum required. But 486, the last term of the second series, is the product of 162 (the last term of the given series) into the ratio 3 ; the difference between this product and the first term is 486—2 or 484, and the divisor 2 is the ratio 3 — 1. Hence, the EuLE. — Multiply the last term hj the ratio, and divide the difference 'between this product and the first term hy tlie ratio less i. 5. The first term is 4, the ratio 3, and the last term 972 : what is the sum of the terms ? 6. What sum can be paid by 1 2 instalments ; the first being $1, the second $2, etc., in a geometrical series ? 569. To find the Siifti of a Descending Infinite Series, the First Term and Ratio being given. Bemakk. — In a descending infinite series the last term being infinitely small, is regarded as o. Hence, the EuLE. — Divide the first term hy the difference betwee7i the ratio and i, and the quotient will be the sum required. 7. What is the sum of the series |, J, J, A, continued to infinity, the ratio oeing ^? Ans. i}. Note.— The preceding problems in Arithmetical and Geometrical Progressions embrace tlieir ordinary applications. The others involve principles with which the pupil is not supposed to be acquainted. 56R. How And the sum of the terms, when the extremee and ratio are given ? %^ ft. long and 12 feet wide to contain 27 cords ? 23. Sold 60 bales of cotton, averaging 425 lb., at 22 J cts., on 9 m., at 7^ int. : what shall I receive for the cotton ? 24. When it is noon at San Francisco, it is 3 h. i m. 39 S3C. past noon at Washington : what is the difference in the longitude ? 25. A man bought a city lot 104 by 31 J ft., at the rate of $2 2 J for 9 sq. ft. : what did the lot cost him ? 26. If ^ of a ton of hay cost £3^^, what will JJ ton cost ? 27. What sum must be insured on a vessel worth 1 165 00, to recover its value if wrecked, and the premium at 2% ? 28. How many persons can stand in a park 20 rods long and 8 rods wide ; allowing each to occupy 3 sq. ft. ? 29. A builder erected 4 houses, at the cost of $4 2 84 J each, and sold them so as to make 1 6% by the operation : what did he get for all ? 30. A man planted a vineyard containing 16 acres, the vines being 8 ft. apart : what did it cost him, allowing he paid 6 J cts. for each vine ? 31. If a perpendicular pole 10 ft. high casts a shadow of 7 ft., what is the height of a tree whose shadow is 54 ft. ? IG 3&2 MISCELLAISTEOUS EXAMPLES. 32. A miller sold a cargc of flour at 20^ profit, by which he made $2500: what did he pay for the flour? ^S. Paid $1.25 each for geographies: at what must I mark them to abate 6^, and yet make 20^ ? 34. Sold goods amounting to 1 1500, i on 4 m., the other on 8 m. ; and got the note discounted at 7^: what were the net proceeds ? 35. A line drawn from the top of a pole 36 ft. high to the opposite side of a river, is 60 ft. long: what is the width of the river ? 36. A school-room is 48 ft. long, ^6 ft. wide, and 11 ft. high : what is the length of a line drawn from one corner of the floor to the opposite diagonal corner of the ceiling ? 37. The debt of a certain city is 8212624.70: allowing 6^ for collection, what amount must be raised to cover the debt and commission ? 38. A publisher sells a book for 62} cents, and makes 20 fo : what per cent would he make if he sold it at 75 cents ? 39. What will a bill of exchange for £534, los. cost in dollars and cents, at 14.87! to the £ sterling ? 40. Three men took a prize worth $27000, and divided it in the ratio of 2, 3, and 5 : what was the share of each ? 41. An agent charges s% for selling goods, and receives $^35-5o commission : what are the net proceeds? 42. A, B, and C agreed to harvest a field of corn for $230; A furnished 5 men 4 days, B 6 men 5 days, and 7 men 6 days : what did each contractor receive ? 43. A and B have the same salary ; A saves J of his, but B spending $40 a year more than A, in 5 years was $50 in debt: what was their income, and what did each spend a year ? 44. If a pipe 6 inches in diameter drain a reservoir in 80 hrs., in what time would one 2 ft. in diameter drain it ? 45. A young man starting in life without money, saved $1 the first year, $3 the second, $9 the third, and so on, for 12 years: how much was he then worth ? ANSWERS ADDITION. Ex, Ans. rage 25. 2. T3839 3. 18250 4. 20000 5. 20438 6. 212269 Page 26, 1. 2806 2. $1941 3. 25285 lbs. 4. 14756 yds. 5- 98937 r. 6. 2051834 ft. 7. 2460 A. 8. 23459 9. 185462 10. 76876 II- 33367 12. T79589 13. 273070 Ex. Asa. 14. 2616263 15- 9539381 16. $4668 rage 27. 17. 1376 yds. 18. $6332 19. 1695 lbs. 20. 2668 g. 21. 10438 22. 8636 23. 10672 24. 3874 25. 15246 26. 100980 27. 1207053 28. $9193 29. 3998 bn. 30. $107601 31. $38058 32. 2844 Ex. Ans. Ex. SS. 13800 rage 28. 34. 159755 848756 182404 1039708 1 1 485 9929 13720 1328464 8237027 25148 IIIIIIIO 22226420 $1460 1925 y- $8190 6987 lbs. 93 yrs. $22338 35- ?>^- 37- 39- 40. 41. 42. 43- 44. 45- 46. 47. 48. 49. 50- 51- 52. Page 29. 53. $i6829,D's; $33658, alL 156 str. 7213 bu. 366 d. $5296 50529 3674 A. $437-44 ^571-54 $376.02 63. $476.19 64. $501.31 65. $475-89 Page SO. 66. $1704.28 67. $16988.71 68. $16580.34 69. $179403.71 70. $157011.73 54- 55- 56. 57- 58. 59- 60. 61. 62. SUBTRACTION Page 35. 2. 346 3- 147 4. 3106 5. 2603 6. 509 Page 36. I- 53637 2. 305 r. 3. 67 lbs. 4. 3779 y- 5. 1719A. 6. 11574 7- 22359 8. 27179 9. 267642 10. 235009 11. 5009009 12. 5542809 13. 2738729 14. 51989 lbs. 15- 16. 17. 18. 19. 20. 21. 22. 23- 24. 25- 309617 T. 209354 A. 34943 1235993 3633805 33230076 349629696 $18990 $1915 $415026 $200005 Page 37. 26. $279979 27. iiiiiii 1111111112 289753017- 746. 270305844- 28516. 226637999^ 876130. 364 ANSWERS. Ex. Ans. 33' 1990005 34. 995500 35- ^4564 Fage 38. 1. $1200 2. $1107 3. 2365 sold, 1 1 95 left. 4. ^4331 5- 4471 6. 1279 Ex. Ans. 2,6. 999001000 39. 2235 A. 40. 26530000 Ex. Ans. 41. 1732 yrs. 42. 84 yrs. 43. 1642 y. Ey. Aks. 44. 413000000 45. I3II5027 46, 8253204 QUESTIONS FOR REVIEW. 7- 3771 S. 803559 9. $14292 10. $963 11. $12523 12. 1700 13- 4434 14. 112122 15. 127680 rage 39. 16. 108888 17. 126950 18. 38022 19. 43898 20. 33885 21. 762 22. 1680 MULTIPLICATION. 20. 21. r((ge 46. 10224 19705 64896 761824 rrtge 48. 15- 3563875009 16. 2923420500 17. 1572150300 18. 450029050401 19. 1924105179680 80625 lbs. ^55350 22. 22360 bu. 23. $222984 24. 78475 m. 25. 1351680 ft. 26. $48645 27. $33626 28. $2367500 29- 54075 yds. 30. $1786005 31. $242284 32. 432 m. Page 47. 2. 242735 3. 1230710 4. 3627525 5. 20136672 rage 50. 3. 65100 4. 290496 5- 509733 6. 3263112 7. 145152011.111. 8. $295008 9. $197500 Page 51. 21. 1491000 22. 3328000 23. 166092000 24. 740000 lbs. 25. $184000 26. 2600000 cts. 27. $1050000 28. 604800000 t. 29. $7350000 Page 48. 6. 332671482 7. 941756556 8. 1524183620 9. 2751320848 23. 2198 24. $4730 25. $522. ?6. $1802 27. $13970 B's $27440 C's 28. 5673 29. $737 I 30. $18775 10. 131247355 11. 351410400 12. 5321254S0 13. 1651148750 14. 1583318550 30. 4022635 1915 148000 31. 64003360044100000 32. 22812553589100000 Page 52. 2,2,- 12615335010000000 34. 312159700930000000 35. 263200000756000000 2^. 20776000000000 37. 42918958404000 38. 370475105000000 39- 652303302651 40. 400800840440000 42. 17784 43. 29484 44. 45975 45. 107872 46. 454842 47- 1585873 ANSWERS. 365 SHORT DIVISION Ex. Ans. Ex. Ans. Ex. Ans. Pafje 59, 13. 8243 h. 2 2, I280604-I- 2. 218392 14. 211 s. 23. IOOI162 3. 186782 15. 8978I r. 24. 746367A 4. I72258I 16. $10671 25. I2OO381 5. 149647! 17. 122 yds. 26. 2346842 6. 662107! 18. 14140^ lbs 27. 3562695I- 7. 6865861- 19. $8121 28. 5848142! 8. 923808 29. 8447232^ 9. 922969! 10. 5762314 Page 60. 30. 59363694A 31. 64519169-A II. 60663768^2^ 20. 1067102-J 32. 37941 w. 12. 680021033^^2 21. 933539 ?>?>' $6412 Ex. Ans. 34. 9065I t. 35. 52481-bar. 36. 3438t'2 y^- 37. 64692- yds. 38. I9405 39. 5812 sq.yd. 40. 2500 hrs. 41. 70440 b. 42. I14531 43- Z2>S'^^2> A- 44. i392of bar. 45. 8090 cows LONG DIVISION. Page 6.1, 3- 2312A 4. 4091 5- 20761^ 6. 7. 8. 9- 10. II. 12. 13- 14. 15- 16. 17- 18. 2 1 06 14 10778/^ io774jf 9759ff 1025 2 /y 5087 2|i 6i294fi ioi77iff io8o88f| 8871211 9497off 243f|- 454 am. Page 64. 20. 297963^ I99i6i|-f- 236851-ff 21. 22. 23- 24. 450 sh. 25. 4iiA\s. ft. 26. I2 294f|j 27. 28. 29. 30- 31- 2^' 33- 34. 35- 37- 3^' ). $94 54of|t 0. $6o2if| i44Hitcu.ft. SOyft fiy lbs. 8787ifll iii7iffff I2030y2^T^ TTr.<^T 6276 3o878Mflfi I7767lfff^f I948II3 4 4I332 3-9~5-3T2-S" l*a T3 (vg-j T^ug" 4- 5- 6. M I98fi 2640 2970 2.16 /• T^-^0^ "5 9 4'0? 5940? T970 Q 14080 18480 9 2 4 3.80 8 ^* 2 46^¥0? '24Z^iUy ^^5^^} 2T^^0 n 1 575 00 25 2000 2 1262 5 9- T83 5 0> 2¥33ir(5"J 2^3T07r TO I39I04 172368 10 3 488 1°' "5^0 81)3" 2 J 3irSl)3^J ^^O'SITJ^ rage 112. TO 330 140 231 ^2. T^s", totj Toy T, 74 3 6 9048 176 85 8 ^3* TT44:? TT44> TTi:^? TTi:? T/l JL44 I29I5 1400 ^4' 2^2 0? "2^X0"? 2y2^U" T C 1463 304 154 ^5* 2^^> 2^^> 2^5" 16. 20. 21. 22. 23- 24. 25' 594 2024 4 8 I I 13 1428 40 -8-4-? -&4-7 -g^ 324 756 T-44F? T^^J 27 40 7-2? 7U 384 T4^ ih 66 60 36 60 T3 5 30 -y-s-j To T05 98 , 3 5 750 476 loooj Tinnr^ looo o. 56 63 60 ■81F> "8^? "81^ III ^J 4? 4" 18 3 2 5 43"? ^T? 4T 54 28 567 T2^? T2^? TT6" 14 15 2 '2 0? 2 0' "T0~ 120 23 1 1760 ^^"^J "2-^4' -TS'T" 2 5 8 7 175 2 0' "215^? TU} To' ,W II. 13^ 14. 4 4 5 103 ^TT? ^xy? '2Uf Tir 90 140 105 ^T0> TTTT^ 2TTr> 168 180 2T0J 2T0 12 S"^J 87 2400 I_04 5 7 2 > ^Ty> Ti7 T r- 465 680 44 16. -f^, ■^, -^ tn 23 6 2690 1441? tQ 3 2 5 3 40 ^^' TUj T(J) tu Ai^SWERS. 369 ADDITION OF FRACTIONS. Ex. Ans. Ex. Ans. Ex. Ans. Ex. Ans. Ex, Ans. Page lid. 12. 82^V 17. 32oi 22. I3t'^ 26. 5 1 1- y. 13. 2T¥ff 18. 75ifl 23. $7i 27. $489! 8, 9. Given 14. If^ 19. 554H-S 24. $7 If 29. 9t¥t 10. 8yV 15. i35t2 20. io|| 25. 2Mlbs. 30. 6Jf II. 22 J 16. io3fJ 21. 5itl SUBTRACTION OF FRACTIONS. Paf/c 118. 10. II. 12. 13. 14. 31 m 1 1 TO 15. 2tW? 16. 3 m 17. 3:^ 18. 3tV(7 19. 2^^ 20. 32I; lbs. 21. 87 2^^ A. Page 119. 22. Given 23" 24. 25. 26. 38i 37i 43 f 67f 27. lo2TVy 28. J _ 29. Given 30. i 31. 32. 33. si 47 57 TT 53 34. 35- 36. iiIt\ 37. 4o|gal 38. $291 40. i^ 41. I 174IJ « 13 2-S MULTIPLICATION OF FRACTIONS. Page 121. 18. $15924 13. 643lf II. ? 2. $4 19. I17374 14. 1256J 12. A 3- $1 3. 2j 20. $45374 15. 6i2i 13. A 4. 1 49 A 4. 7A 21. $5418 16. 1 009 J 14. AS 5- 3714 5. i89i 17. 21721^^ 15. 3il 6. 600 6. 3574 7. 583* Page 123. 18. 700031 16. 17. 78i 9 cts. 7. 8. 3986|-f i2377t'5 8. 36 9. 277i 3- 4. 24| 26A Page 124=, 1 8. 19. I36 203il 9. 10. 1731711 'o. 48Hf 5. 336 3.1 20. 703-i II. ii4J Li.. 2566 J 6. 406 4. i 21. 1 9531- 12. %of 12. 4094M 7. 542^ 5. 6f 22. 15352H 13. 4334 eta 13. 275 8. $26f 6. fi 24. i 14. $181^ 14. $6of 9. $250f 7.4 25. 4 15. 750 tin, 15. $23lJ 10. $6429! 8. f 3000 cop. 16. $I2l8J II. 549fl 9. 44! Page 125. 16. $1918^ 17. iB8o4f 12. 1407 10. 14 I. H 17. 44839l§ 870 ANSWERS, DIVISION OF FRACTIONS, Ex. Ans. Ex. Ans. Ex. Ans. Ex. Ans. Ex. Ans. Page P26. 28. 6d>ii Page 130. 3- $i| II. ih A 11 29- 315x1 30- ii5toV 2. 2f| 3.ii .1 T I 4. 17H 5. 15 times 12. 13- 50/2 bu. 245iTy- 4- T35- 7- A ^- T2T 9- TT g- lo. xMIt II- tWt 12. ff-ill 3i.^8i« 6- Hi 14. 86o^^2iii' 32. $I26J 33. ^IH 4- I^V 5. 41-1 7- M 8. lA 15- 8103!- 9- 'Ul 10. 50 J lbs. 11. 3o| A. 12. 10 lots 13- 42T¥y y- T 4 16 Page 133. Page 128. 2. I26f 3- 672 4-350 5-375 7. 2| 9- A lo. 4j 16. 41 ^^--L 120 27 '7 s. '120 13. ^Sh 6. 1479 12.4 14. J-^ 17. « 14. Tli?-T 7- 447f 13- ^zUHh 15- 273 18. A 15. ^V bar. 8. 1645! '5- ifJ 400 19- 4i 16. ItJct 9. 21630 1 6- 5il Page 132, 20. 80 diws 17. ^/. 10. 56727TT 17- 3A I. I43I err. 21. $4ii " 18. l7f 12. 135 1 8- 7il 2. 1619214:0. 22. 23M 13- 477 19. 6 $i9A-er. 23- illJ Page 127. 14. 266 20. i4ofJ r. 3- 50I 24. iS^ 19- 6 A 15. 804 21. lojfjs.r. 4- 32|- A. 25. 50 vests 20. 5iM 17- 8A 23. f^ 5- M 26. 44f?c. 21. if 18.8 24. ij 6. 3iJ 27. 12x^2 hr. 22. Ill 19. I2lf 25- 2t¥? 7. H sold; 28. i6f 23. loJJI 20. Ilf^ 27.tV^V fjown; 29. 60 bu. 24- 122V0- 22. 9 $24783 30- 244f yd. 25. i5tM 23- 3tt l^a^re 131. 8. 86|J A. 31- 2 26. 15HM 24. 3-i-? I. 5 J mo. 9. AV 32. /A-V 27- 20^Jj^ 25- 4A 2. 8} lbs. 10. li 33- 3t¥5 FRACTIONAL RELATIONS OF NU MBERS. P<^ff/e J54. 5- J 8.^ II. Y^bii. Page 135, 3. h A 6.^ 9- \i 12. fjton 15 . $253 4- A 7- i 10. f wk. 13-1 \t . f59i ANSWERS. 371 Ex. Ans. 17. $132 19. 20. 21. 22. i ITS- ^00" 3 3 24. $5-- 25- 27- 28. 29. 30. 32. >$2^ \o29 3 2. 3 9_6. 5 7 7_ ~4 200 i4 33^ Ex. Ass. 33- if 34- A 35- fM Page 136, 37. i 3^- I 39- I 40. 1 41. tV 42. 1 44. A Ex. Aus. 45- in: 46. f I lb. 47. 2 ft. 48. tV lb. 49- T? 5^- 2i50 52. ^'^ 3. 84 4. 90I Ex. Ans. 5- 2I2i 6. 138 7. 1017-J 8. 729I- 9. 2283I 10. 2283! 11. 270 12. $14720 13- 33250 16. I ° 17. I 19. 62I I Ex. Ans. 20. 83J J*«gre 15<5?. 22. 288 24. 9ii ycis. 25- I3 each 27. iijt. 28. 100 cts. 8 cigars 30- 1 30 J 32- 7 times 34. 3i REDUCTION 2Xf/e ^4^- rage 144:, 8. 4.7 9. 21.06 10. 84.45 11. 93.009 12. 7.045 13. 10.00508 14. 46.0007 15. 80.000364 17. .06 ; .063; ,0109 18. .305;. 00021; .000095 19. .004; .0108; .46; .000065; .0001045 20. 69.004; 10.0075 ; 160.000006 3. t'A 4. m 5- 1 6. 863 7- iV 8. 100 9- 1250 10. ^Ww II. TODFO" 12. tttJ^ T-7 _ _9 I ^3* TSiiUTT 14* 20 T e -JO 17 1 ^5* iSOOOO" 16. I2S00O0 TT I 9 8> I 17- T5 t8 15 5 7 9 I^- 6250000 OF DECIM Page 145, 1. Given 2. .25 3- -4 4- .75 5. .8 6. .625 7. .25 8. .875 9. .8 10. .95 11. .6 12. .0875 13. .02 14. .000375 15. .0078125 16. .00875 17. .01125 ALS. Page 146, 19. Given 20. .3333-^ 21. .8ss3-h 22. .2857 4- 23. .4444 + 24. .2727 4- 25. .64284- 26. .6041664- 27. .5466 + 28. 75.6 29. 136.875 30. 261.68 31. 346.8133 4-. 32. 465.0025 33- 523-o'^39o625 34. 740.01375 35. 956.0078125 ADDITION OF DECIMALS Page 147. I 5. 14.38916 9- 92.00537 | T, 2. Given 6. 118.792 10. 37.417 ^3. 881.6217 I 7. 892.688 II. 2.3948 4. 139.26168 ; 8. 2.76231 1 12. 23.25553 ; Page 148, 13- 575-729105 14. 53.1 bii. IS- 75-97 1^ 372 A 3S^ S W E R S . SUBTRACTION OF DECIMALS. Ex. Ans. ! Ex. Ans. Ex. Ans. 1 Ex. / NS. Bage 14=8. Page Hi). 12. .8969755 19. 4K/955 I. Given 6. 7.831 13. .5496933 20. .000098 2. 7.831 7. 6.60249 14. .876543211 21. $443-825 3. 6.60249 8. 17.3675 15. .01235679 22. 139.83 A. 4. 17-3675 9. 77.94794 16. .099 23. 99.063 5. 17.94794 10. 78.569966 17. .00999 24. 7-.333 ni. II. 2.896216 1 18. 99.999 107-. 40703 MULTIPLICATION OF DECIMALS. JPage 150. 15. 431.25 lbs. 26. .oooo<^<:>252 3- -1453 16. $222.9375 27. .OOI 4. .000151473 17. 469.0625 bu. 28. I [0000 1 5. .0000016872 18. $10639.75 29. .OOOOOOOOOO' 6. 21800.6 19. .0126 31. 3205.05 7. .012041505 20. $686.71875 32. 8003.56 8. 20.08591442 2>Z' 2.43 9. 318.0424 Page 151. 34. 5^8 10. 721.36 21. .0025 35. 5 II. .00004368 22. .000004 36. $50 12. 1.50175036 23. .00000049 37. $600 13. .000721236 24. .000000603 38. 27.625 bu. 14. .020007 25. .0000003 39. 65.625 m. DIVISION OF DECIMALS. rage 153. 8. .13 + 9. .7115 + 16. 10 17. .01 24. 27 stoves 26. 4-3753 2. .007 10. .9768 18. .000002 27. .063845 3- 4 II. .00675 19. .0005 28. .0000253 4. 600 12. .0000576 20. 50 29. .0000005 5- -4154 + 13. -015 21. .00000027. 30. S.0005 6. 30.153 + 14. 625.5 2 2. 46 coats 31. $.0475 7. 2.4142 + 15. 10 ^23. 44.409 +r. ADDITION OF U. S. MONEY. Page 158. 3. $780.25 4. $200.09 7. $17.28 8. $69.1925 Page 169. 10. $54.50 I. Given 5. $224.53 9. $13131.72 II. $120.48 2. $48,625 6. $ 761.4785 12. $76.82 X AKSWERSS. 373 SUBTRACTION OF U. S. MONEY. Ex. Ans. Ex. Ans. Ex. Ans. Ex. Ans, Page 159. 5. $585.25 7. $171.9625 10. $7.4275 2. $533-105 '^ 8. $411,075 II. $494,945 3. $604,625 Page 160. 9- $.955 12. $28.75 : 4. $524.50 6. $948.33 1 MULTIPLICATION OF U. S. MONEY. rage 100, Page 161. ! 12. $.00056 18. $45 2. $747-50 7. $100.9125 13- $5 19. $80.4375 3. $1569.24 8. $14.124375 14. $.001011 20. $1890 4. $290,625 9. $67.8375 15. $6.5625 21. $548,625 5. $60,165 10. $310.596255 16. $58,905 22. $178.75 6. $459-25 II. $4630.70025 17. $56.25 23. $14208 DIVISION OF U. S. MONEY. rage 162, 4. $2.75 5. $2,921 9. .01 10. .1 14. $2,965 + 15- 925405 + I. Given 6. 100 II. 100 16. 293.039 + 2. $1,964 7. 361.455 + 12. loooo 17. $.0625 3. $.05 8. 4000 13. 600000 18. $.0547 + COUNTING-ROOM EXERCISES.. rage 163. Pfige 165 . 4. $3183.07- r 6. $5201.70 I. $13958.38 bal. 2. $206.83 Page 166. 7- $717-77 8. $395-37 2. $159857. i6bal.i 3. $4367.12 5 5- $71.15 ANALYSIS. rage 167. 13. $ii4.o6J 20. 6132 A. 10. $31^ 4. $789.25 14. $47.82 am t. 21. $478,125 1. II. $291!: 5. $4312.50 15. $3007750 2. $66J 12. $35 6. $43l 16. 946.1+ lb. 3. $65 13- $27f 7- $8924 17. 78971b. s'd; 4. $22 14. $107^ 8. $o.33j 149 lb, av. 5. $409j 15. $ii5f 9. $5,264 18. 36.45 bu. 6. $I2i 10. $31.50 7. $150 Page 169. ir. 102 bibles Page 168. 8. $2o8| 17. $210 12. 148^ vests 19 I247 1 9. $5811 18. $547i 374 AN"SWERS. Ex. Ans. Ex. Ans. Ex. Ans. Ex. Ans. 19. $576 20. $168 21. 8240 23. 13.66 yds. 24. II 25 lbs. 25. 1800 coc's 26. 2690 pine's Pfige 170, 2. $78,177 3. $240,625 4. $789,625 5. $1169.83! 6. $1203.93! 7. $40.4875 8. $22.3I-J: 9. $100.2472 10. $4.51 11. $i4.i8J 12. $15,975 3- 4- 6. 7- 8. 9- (o. II. 12. 13- 14. 15. 16. 17. 18. 19. 20. 21. 22. 23- 24. 26. 28. 29. rage 100, 614 far. 39618 far. £S, I OS. 7d. i2S. 9d. 2 far. £41, 5s. 2d. 2 far. 1 66 1 yds. £225 4440 pwts. rage 191, 7865 pwts. 80 lbs. 7 oz. 7 lbs. I oz. i8pwt. 18 grs. i49f rings $27 2653 oz. 6721944 oz. 2 t. 725 lb. II t. 1 141 lb. I2 0Z $700.55 $280 profit 24048 drams 7lb.6oz. I dr. 2SC. 1928495 ft. 7 ra. 79 r. 1^ ft. 443440 rods REDUCTIO N Page 192, 30. 31. 32. 3Z' 34. ZS- 36. 37. 38. 39- 40. 41. 42. 43- 44. 45. 46. 47. 48. 49. 50- 51- 52. 53. 54. 495782 111. 126720 in. $1480 $1556.10 456 eighths 2608 sixteenths 144^ yds. i23|yds. 22 vests $16.25 Pi'o^it 160 r. 256 sq. ft. 5 A. 40 s. r. 20 s. y. 697641 4 J sq. ft. 10240000 sq. r. 10 A. 108 sq. r. $2 2875 profit 3983040 cu. in. 32000 cu. ft. 10 c. ft. 985 c.in. 64 C. 86 cu. ft. 5259 qts. 24051 pts. Page 193, 6641 bu. 254 bu. 2 p. 3 q. $412.08 55. ^12.75 56. 790 gi. 57. 603 qts. 58. 5713 qts. 59. 34616 gi. 60. 10834 gal. I pt. 61. 168 bottles 62. $126.00 6^. 2612530 sec. 64. 176010 min. 65- 31556929-7 sec. 66. 9W. 6d. 2oh.i5m, 67. 639 y.225d.5 h. 6^. 19 y. 164 d. 6 h. 20 min. 69. $212,625 70. 13° 29' 21" 71. 36 s. 7° 17' 72. 855631'' 73. 1296000" 74. 163I: dozen 75- 1500 eggs 76. 694-^ gross 77. 9360 pens 78. 67 lbs. 79. 1800 sheets 80. 41 6f quires APPLICATIONS OF WEIGFITS AND Page 194, 1. Given 2. 7-i A. 3. 165 ft. Page 195, 4. 72 rods 5. $46500 pr. 6. 41 16 s. ft. 2 sq. ft. 1200 bu. 9. Given MEASURES. ' 10. 192 bulbs 11. 3630 gr.v. $190,575 3. 12. 2\ A. ANSWERS. 3T, Ex. Aks. Ex. Ans. Ex. Ans. Ex. Ans. Page 196. 23. 147 b. ft. 24. $278.10 35. $112 2,6. %2>Z1'^^ 47- i795ff g. 14. $324 37. 86751 br. Page 201. 15. $8.88 Page 198. 38. $3600 48. 56^Vin. 16. $402.27! 28. $14.85 29. $23.40 30- ^4f 49- 57-857+ b. 17. ^22.50 Page 200. 50. 42 1 ^\ eft 18. I45.i3f 40. 56 yds. 51. 9.955|in.. 41. 44f yds._ 52. 3750 lbs. Fafje 197. Page 199. 42. i2f yds. 43. 320 sods 55- ^7x1^ 56. 9tVV lbs. 21. i8|b. ft.; 32. 2|if cords 44. 48 yds. 57. 871 rings $1.40 val. 2,:^. i6i tons 45. 640 tiles 58. $72H 22. 12 J b. ft. 34. $133.20 46. 13 J rolls 59. $8o5|f DENOMINATE FRACTIONS Page 202. 2. f qt. 3. If d. 4. 2^ hr. 5. If oz. 6. y^^- s. in. Page 203. 8. $ 9. £-J T5"0~(J TTJ?" TTT2 oz. II- tIo gal. 12. m. 3- 16000 t- 16. 13s. 4d. 17. 3 pk. 6 qt. 18. 1250 lbs. 19. 10 oz. 10 p. 20. 3 fur. 13 r. I yd. 2 ft. 6 in. 21. 146 s.r. 2os.y. I s. ft. 72 s. in. 22. 112 cu. ft. 24. n gal. -,- T 229 IV, 26. 11 bu. 27. 2Vx) ton 28. Iff sq. yd. 29- tW^ 30- 2 3_3. ^J.. 504 c. 32. M 34. 2S. 6d.o.4 + far. 10 oz. 18 p. 22.56 grs. 2 furlongs 32 rods. 2 qt. I pt. 38. 39' 36" 39. I d. i6h.29 m. 16.8 s. 40. 5051b. 1.92 oz. 31. i 35- 37 41. 17.28 grs. 42. £5, i2s. 6d. 0.4032 far. Page 206. 44. .5423+ lb. .005 ton .45539 + ni. .04 lb. .5bl. .409+ r. •25 .48 bl. 1.44 r. .166 + wk. .28182 + .41666 + 0. 45 46 47. 48. 49. 50- 51- 52. 53- 54. 55. METRIC NOTATION AND Page 212. 5370.9845 dekameters; 537.09845 hektometers ; 53.709845 kilometers; 537098.45 decimeters. NUMERATION. 2. 450.5108 dekagrams; 450510.8 centigrams; 45.05108 hektograms; 4.505108 kilograms. 376 AN"SWERS, REDUCTION OF METRIC WEIGHTS AND MEASURES. Ex AN8. Ex Ans. Ex. Ans. I. 2. 3. 4- rage 213. Given 437500 sq. m. 867000 grams 26442 liters 5- 6. 7. 8. 9- 256100 sq. m. 865 2000 cu. dm. 4256250 grams Given. 65.2254 hektars. 10. II. 12. 0.087 kilos. 1.48235 kg. 39.2675 kl. APPLICATIONS OF METRIC WEIGHTS AND MEASURES rage 215. 3. 39.14631 m. 4. 1 9.8 13 1 J gals. 5. 15.89 bu. 6. 4.2324 oz. 7. 303.68365 lbs. 8. Given 9. 148.87775 A. 10. 4237.92 cu. ft. 12. 58.2934- m. 13. 6236.959+ kg. 14. 236.585 + liters. 15. 72.492+ kl. 16. 143.223+ kg. 17. 6000.06+ s. m. 18. 16.378+ hect. 19. 410.748+ c. m. 20. 27958.715 +c.m f. Page 217. 3. £11, los. od. 2 f. 4. 26T.3cwt.83lb. 3 oz. 45 bu. o pk. 2 qt. 74J yds. 1093 lb. 5 oz. 55 gal. 2 qt. COMPOUND ADDITION. 9. 196 bu. 2 pk. 7 q. 10. 6 C. 80 cu. ft. 11. 98 bu. 3 pk. 2 q. rage 218, 12. Given 13. 23 wk. I d. 17 h. 58 m. 14. 64A. 7s,r. io|s.y 17. i5Cwt.5olb. 40Z 18. 2 pk. 4 qt. 19. 9 oz. ipwt. logr. 20. 3 s. 6 d. ,^.2 far. 21. 2 A. 52 sq r. 22. 2 0. 81 en ft* 1209! cu. in COMPOUND SUBTRACTION. Tage 219, 2. I fur. 39 r. I yd. 2^ ft. 3. 7 lb.4oz. 17 p. 9grs. 4. 7 T. 8 cwt. 26 lb. 5. 8 gal. I qt. ipt. 2gi. Tage 220. 6. 159 A. 12 sq. r. 222^ sq. ft. 7. 46 cu. ft. 1689 cu. in. 8. 1 45 A. 2 1 sq. r. 9. 46J yd. 10. 3 m. 5 fur. 38 r. 5 yd. o ft. I in. 2. 9 s. 9 d. 3. 2pk.5qt. i.2p, 4. 4 J pt. 5. I lb. 2 oz. 2 p. 6. 1 14.4 sq. r. 7. 31.25 lbs. rage 221-2, 3. 67 y. 9 m. 22d. 2. 113 days 3. 74 days 4. 150 days 5. 224 days 6. loi days Page 223. 10. 86° 19' 24" 11. 7° 24' 7" 12. 18° 2' 13. 5° 6' 46" 14. 15° 4' 16" 15. 19^" 35' 16. 56° II' 17. 70° 29' 18. 10° 19' 38" ANSWERS. 377 COMPOUND MULTIPLICATION. Ex. Ans. rage 224=. 2. 98T.i7CWt.28lb. 3. £151, 15s. 9jd. 4. 33 oz. 15 pw. log. Page 225. 7. 331 gal. 2 qt. Ex. Ans. 8. 22 C. 91 cu. ft. 9. £23, 15s. 3|d. 10. 562 m.4fu. 241'. 11. 1937 bu. I pk. 12. 22 0. 57 cu. ft. 13. 61 T. 844 lbs. 14. 1307 r. 8 qr. 8 s. Ex. Ans. 15. 161° 37' 30'' 16. 431 h. 15 m. 17- ^ZZ sq. r. 21 sq. yd. f sq. ft. 18. 73 T. 1492 lbs. 19. 15 7 1 bu. 2 p. 4q. — 1946 gal. 3 q. I p. 20 COMPOUND DIVISION. rage 227. 3. 4 fur. 8 r. 2 yd. 2^ ft. 4. 6 gal. 3 qt. o pt. 3j gi. 5. 25 bu. o pk. I qt. f pt. 6. 15 A. 106 sq. r. 5 sq.yd. 2^ sq. ft. 7. 26f spoons 8. 8800 rails 9. 1049/y times 10. 12 books 11. 2 A. 64 sq. r. 12. 6 bu. I pk. I qt. COMPARISON OF TIME AND LONGITUDE. rage 228. 1. Giyen. 2. 43 m. 32.13+ s. 3. 1 1 o'c. 1 2 m. 4 s. 4. i2o'c.37m. 8.4 s. 5. 49 m. 20 s. 6. I o'c. 30 m. E. ; 10. 6° 9' 10 o^c. 30 m. W. II. 6° 45' 5" 7- 54 m. 19.8 sec. 12. 45° 24' 45" 8. 13111.44.86+8. 13- 56 m. 49+ sec> rage 229. 14. 12 o'clk. 34 HL 9- Given 47 sec. PERCENTAGE. Fage 232. 13- -50; -25; -75; -20; .40; -60; .80 14. .10; .70; .90; .05; .35; .12; .06 15. .661; .i6|; .125; .625; .875; .58J; .91^ rage 234. 3. $24.21 4. 10.8 bu. 5. 22.6 yds. 6. 8.64 oxen 7. 8.25 yds. 8. $11,584 9. 100.8 lbs. 10. 63.14 T. 11. 52.5 men 12. 145.656 m. 13. $857,785 14. £.106 15. 1.25 1. 16. 1.38 k. 17. 840 lbs. 18. $1000 19. 78.75 bu. 20. .35 lbs. 21. 9.765 gals. 22. 840 men 24. $215 25. i57.'2 lbs. 26. 32.25 m. 27. 116 1. 28. 580 sheep 29. 156 bu. 30. $925 378 AKSWEKS. Ex. Ans. Ex. An3. Ex. Ans. Ex. Ans. ,51. 576 CU. ft. 14. $10200 -PaWo 7. £450 5. 809 S80 s. D. 3- ^S% 8. 600 6. 1820 4. $2300.91 4. zzWo 9- 360 fr. 7. Given 5. $2850 5- ^S% 10. 8000 8. f 6. 504 bales 6. gi% II. $200 9. T% B's 7. 2018.75 gal. 7. 2>\% 12. $i6666f 10. 5175 m. 8. 13 1 2 m. 8. 10^ 13- $17600 II. $3648 9. $47040 9. 80^ 14. 54.4 yds. 12. 6875 p. 10. 129 t. 10. 10^ 15. $50 " 13' $3350 II. $1410.50 - II. 2M% 16. 100 14. 228 sheep 12. $228 12. 42f^ 17- $5000 15. 400 pupils 13. $4823.33^ 13. 3o|i^ . 18. $1920 16. 5800 m. COMMISSION AND BROKERAGE. Page 242. Page 244. 3. $31-14315 12. $9000 4. $366,225 com. ; 13. $4212 sales; $10902.225 paid $4001.40 net p. 14. $1500 col.; Page 243. $1432.50 paid 5. $4310.145 15. $9000 sale ; 6. $238.66! $8865 rec'd 8. i% 17. $5100 9- 5% 18. $6834.872 sales; 10. i% $170,872 com. PI lOFIT AND LC Page 247-9. II. $453.60 3. $22.20 12. $2942.50 4. $22 13. $9520 5. $.0875 14. 1 1^ cts. 6. $16.65 15. $1 a pair- 7. $1.40 16. $2.77^ eacli 10. $2925 A.; 17. $4.90 a yd. $2075 B. 18. $4331.25 Page 245, 6. 19. $15488.89 + 20. $26635.294 + 22. $3010.75 23. $2419.234- 24. $4926 1.083+ in.; $738,916+ com. 25. $13687.50 g. a.; $360.75 ch.; ^13326.75 net p. 26. $12127.83 net p. 19. 21. 22. 23. 25. 26. 27. 28. $12810 28f^ 60% 100% Ao% ANSWERS. 379 Ex AN9. Ex. Ans. Ex. Ans. rage 250, rage 252, 14. $7 cost; 29. so% 51- $.736 marked p. $8.05 selFg pr. 30- Given 52. I34.78+ m. pr. 15- $2 cost; 31- So% 53. $4.80 m. pr. $1.75 sell'g pr. 32. 50^ 54. $1.1 6f m. pr. 33- 33i% 55. $75 each Page 254, 34. $12.60 pr.; iW/o 56. ^5-75 per yd. 16. $5200 sales; $4940 net 36. $9750 Page 253. 17. $24000 sales I. $2,691 prof. 18, $1.25 cost Page 251, 2. $6356.72^ 19. $5250 cost; 39- $1 cost per lb. 3. ^3487-575 sh.; $1750 gain 40. $22500 cost; $850.84 ging.; 20. .7894^1 per gal.; $30000 selFg pr. • $4338.415 both . 1 48 1 gam per g.; 41. 1 1 7500 spend; 4. $22 loss $49.73 wh. cost; $2 1 000 amt. sales 5- $77812.50 cost $9.33 wh. gain 42. $12000 A's in v.; 6. $897.45 net pr. 21. $10 a barrel $9375 B's inv. 7. 38f^ 22. $51314 cost; 43- $250000 cost; 8. 5 6 J^ prof.; $6413! gain $275000 amt. s. $3240 prof. 23- $15798.963 sale; 44. $.02 cost; 9- i6|-;?^ loss $552,963 com. $.025 selling pr. 10. •052 or 5^-^c.; 24. 38A^ 45* Given $1766.835 net p. 25. S^% loss 46. $10061.71^ II. 25^ loss 26. $60 m. p. 47- $13319.672 12. 125^ 27. 6\% loss 48. £22 cost 13. $9300 cost ; 28. 15 cts. 49- 3 cents $11160 sell'gpr. 29. iot?o^1oss INTEREST. Page 258. 3- ^2.54 4. $1,676 5. $1,224 6. $4,523 7- $43-356 8. $110.77 9- $33-42 10. $23,364 11. $25,577 12. $94.35 13. $17.91 14. $680.28 15. $81.44 + 16. $2875.792 17. $362,315 18. $2759.962 19. $5032.083 20. $122.40 21. $226.69 22. $45,053 23. $216,489 24. $493-2o 25. $3923.47 27. $99.96 28. $24.00 29. $163,842 ; $2740.652 Page 259. 2. $24.44 3- $6,252 Page 260. 4. $4.82 5. $6.75 6. $73.96 7. $64.18 8. $26.49 9. $1.65 10. $29,933 11. $60.04 12. $30.52 13. $2450.80 14. $20819.80 Page 261, 3- $2.8435 80 ANSWERS, Ex. Ans. Ex. Ans. Ex. Ans. Ex. Ans. 4. $2,192 4. 6^ 7. 14-f-yrS. . 2. $235.85 5- Iii-4i3 5. 8^ 8. i6f yrs. 3- ^327-36 6. $13.89 6. 9t\% 9. 18 J yrs. 4. $8928.57 7. ^1285.33 7. 8i^ 5. $892.86 ^8. $13,876 8. 5^; 10^ 2. $i666| ^ 6. $5582.142 9. $4363.044 9. 7^ 3. I3000 10. $ii4.i6f 10. 6^ Page 268,9, II. $185.18 Page 264, I. Given 12. $81,358 Page 263, 4. I2333-J 2. $1519.71 3- 5f y-> or 5 5. $2500 3. % 38.63 Page 262, y. 8 m. 1 7 d. 6. $40000 5. $270.19 2. {% 4. 9 m. 2 d. 7. $10000 6. $388.23 3- 1% 5. J y., or 3 m. 8. $25000 7. $516.32 COMPOUND INTEREST Page 273, 4, 1. Given 2. $112.52 3. $161.63 4. $164.61 5. $200.63 6. $2165.713 7. Given 9- 10. II. 12. %5i-58 $1377.41 11543-65 $8104.25 $32564.58 13. $21825.26 14. $29849.56 15- ^37704-95 16. $3298.77 DISCOUNT. Page 277. 1. Given 2. $283.47 + 3. $462.96 4. $1213.59 5. $2336.45 6. $4464.28 Given $86.57 $101,892 $91.35 dis; $2283.65 p. 11. '^3775-264 12. $20,377 7- 8. 9- 10. 13. $528.33 14. $46.73 for. Page 279. 3. $718,685 4. $989.50 5. $1721.875 6. $17.25 7- ^735-70 8. Given 9. $768.44 10. $1578.81 11. $2226.23 12. $7503.83 Page 282. 2. $243 3. $525 4- 1^525 5. $400 6. $476 STOCKS 8. $12880 AND BONDS Page 283. 9. $10497.50 10. $6^6^.61 11. $6336.11 12. $10852.37^ 13. $19293.75 14. $672 cur. 16. 15^ 17. H% 18. 6i% Page 284. 20. 40 bonds. 21. $12000 23. $44i66| 24. $35625 25. $56000 A N S W E K S . m EXCHANGE. Ex. Ans Ex. Ans. Ex. Ans. Ex. Ans. B^({/e '4H7. 10. $2617.801 rage 290, II. J2l542,IIS. 2. $2049 II. $3751.95 6. $56125.124- 6d. 3. $3488.80 12. $3750 7. Given. 4. I4 1 30.647 8.iE5ii, 15 s. JPagre J?<>i. 5. $203 Tage 289, 4 d. 3.2 far. 12. Given 8. $1219.51 I, 2. Given. 9. £774, 7 s. 13. $675.68 3. $1858.80! lod. 1.28 far. 14. Given Taqe 288, 4. $4866.^0 10. i8io26, 8 s. 15. 13050.00 f. 0. $1491.053 5. $56.62ig. 7.2 d. 16. 16474.50 f. INSURANCE. rage 293, 1. Given 2. $6.65 3. $38.40 4. $84,375 5. $10.70 6. $25.00 7. $197.60 8. $875 9. $1569.625 10. Given 11. $15747.423 12. $27806.122 + ^3- $37105-263 + 1. Given 2. $125.00 3. $2437.50 4. $5000 5. $45000 gr. rage 297. $1003.75, C'stax $1250, D's " $1375, E's " $925, Fs « Page 299, 2. $4062.50 3. $1063.314 TAXES. 2% rate; $159.75, A's tax Z% I'ate ; $450, G's tax 6. $309.75 H's tax j II. $11052.63 + 7. 5^ rate; $170, A's tax 9. $3645-831- 10. $5507.853 + DUTIES 4. $1999.20 6. $945 7. $2296.80 Page 300. 2. $118.25 3- ^425.75 EQUATION OF PAYMENTS. Page 302. 2. 4 m. from J. 20, or Oct. 20 3. 8 m. 18 d. 4. Moll. 10+17 d.=:Mch. 27 5. June I, or in 3 m. AVERAGING Page 306. 1-3. Given 4. 7 m. extension ; Bal., $1650 Page 303, 6. Given 8. July 15+51 d.=Sept. 4 9. $1483.25 ami; 42.9 d. av. time; due Nov. 17 ACCOUNTS. 5. $300 bal. debts ; 22320 bal. prod. ; 74 d. av. time Due May 23. 38^ AFSWEES, SIMPLE PROPORTION. Ex. Ans. JPage S12, 3. $160 4. I165 5. $450 6. $6 7. £i9,7s.6id. 9- 135 A. 10. $93 7 J 11. 270 miles 12. $288 13. $822! Ex. Ans. 14- ^T-33 15. 2240 times 16. 5 years 17. 8 hours rage 314. 18. iioomen 19. 240000 lb. 20. 27 horses 21. $612 22. £ij- 23. I5.06J 24. I880.40 25. $490.52^ Ex. Ans. 26. $5223A\ 27. $12.50 28. 14.4 in. 29. 26f yds. 30. 97920 t. 31. 180° 32. 80 ft. T,^. 64! orang. 34. 3000 mi. 35. 33 men 36. 30 ft. Page 315, 37' ^1920 Ex. Ans. 38. 25 min. 39. 6600 rev. 40. $40 41. 2 1 lirs. 42. 3x7-1^^8. 43. -45 days 44. 1000 rods 45. 300 hrs. 46. $191.78 47. 8 m. 51 s. before 9 48. 3^ cts. loss 49. 150 m. less; 186 m. gr. COMPOUND PROPORTION Page 317. 3. 2f days 4. 9 horses 5. 33 J weeks JPage 3 IS. 6. |i26.66f 7. $100 8. 360 miles. 9. I750 10. 10285! yd. 11. 374 days 12. $i454A 13. 750 lbs. 14. $62.50 15. 240 sofas 16. 864 tiles PARTITIVE PROPORTION. rage 319. 40 s.: 60 s. ; 100 s. 4. 60 bn. oats ; 80 b. p. ; no b. c. 5. 71; io6i; 142; i77iA. PARTNERSHIP.— Paj7e 321-3. 2. $2 2 2f, A's loss; $277j, B's " 3. $340, A's share ; $510, B's " 4. l5o67|f,A'sg'n; $4054^. B's " ^ZZim, C's « 5. $3400; I5100; and I6800 6. $64, A's share ; rage 324. 2. $1687.50, A's sh.; $96B's;$i6oC's 8. $142.85!, A'ssh. $266.66f, B's " $190.47-11, C's" 9. $24.48^1, one ; $25.51^^, other 10. $424^^, A'ssh.; $3i8Jf, C's " $282ff, D's " BANKRUPTCY. $i2o5.35f,B'ssh.; $857.i4f,D's « 11. $1500, HU; $2250, Cont; $3000, Am. 12. $263.38ifi|,A's; $447.76tVAB's; ^69i.39AVV.C's; $io97.45A%V,D. 13. $402iffi, A's; $3846^11, B's; ^4I3It¥j. C's 3. $2060, D rcc'd ; $12100 net proc. ANSWERS. 383 ALLIGATION Ex. Ans. rage 325. 2. 151 cts. 3- 95 A f'-ts. 4. 9JI cts. 5. 2o|f carats 6. $15.68 Page 327- 8. 35 bu. ist; 40 bu. 2d ; 40 bu. 3d Ex. Ans. 9. 2 p. at 15; 1 p. at 18; 2 p. at 2 1 ; 5 p. at 22 10. 9 lb. at 3 2 c; 6 lb. at 40 ; 6 lb. at 45 11. 8 lb. at 20 c; 3 lb. at 27; 5 lb. at 35-; 1 2 lb. at 40 Ex. Ans. Page 328. 14. 3 J lbs. ea. 15. 5oq.at4C.; 300 q. at 6 Page 329. 17. 20 lb. at 6 c.; 20 lb. at 8 ; 60 lb. at 12 Ex. Anb. 18. 54yyb. 28 c; 18^ lb. 30; 54Alb. 38; 72Alb. 42 For other ans. see Key, i9-53Tg-4oc.; 53i g. 45 ; 53ig- 50; 140 g. 60 INVOLUTION —Page 331, 8. 25; 36; 49; 64; 81; 100; 400; 900; 1600; 2500; 3600, 4900; 6400; 8100. 9. .25; .36; .49; .64; .81; .0001; .0004; .0009; .0016; .0025; .0036; .0049; .0064; .0081. 10. 125 11. 64 12. 2299968 13. 1024 14. 4096 15- 15625 16. 8.365427 17. 64.014401080027 18. 64024003.000125 20. Iff 21. ^2¥^ EXTRACTION OF THE SQUARE ROOT. Page 338. 5.427 6. 719 7. 772 8. 1.892 + 9. .64 10. .347 11. .41 12. .8514+ 13. •355 + 14. 1.635 + 15. 69.47 +- 16. 21.275 + 17. 2.236 + 18. 2.64 + 19. 2.828 + 4 Page 339-4 1. 2. 420 rods. 3- 559.28 r. 4. 119 trees 5. 238 men 7. 108 yds. 8. 30 rods 20. 3.16 + 21. 3.316 + 22. 3.46 + 23. 4.42 24. 57-3 25- 9-36 + 26. mil. II + 27. 7856.4 APPLICATIONS. 9. 438.29+ m. 10. 103.61 + ft. 11. 56.56+ ft. 12. 75.81+ ft. Page 342. 15. St min. 28. 98.7654 Page 339 32. H 33' -745 + 34. .866 + 35. 3^' 2.529 + 3-^3 + 16 24 30 37. 4.1683 38' 39' 40. .8545 + 41. i.oi8 + 42.^ 43. ¥ 16. 17. 18. 19. 20. 21. 22. Given 12 54 63.49 + 1-75 8 ^^ 72 TTCF ANSWERS. EXTRACTION OF THE CUBE ROOT Ex. Ans. Ex. Ans. Ex. Ans. Ex. Ans. l*nge 348. 7. 2.76 13. 129.07+ in Page 34=9. 1-2. Given 8. 2.57 + 14. Given. I. Given 3. 85 9. 8904 16. .746 + 2. 52 ft. 4. 4.38 -f 10. .632 + 17. f* 4. 16 ft. 5- 72 II. 85.6+ yds. 18. A 5. 462.96 + c f. 6. 1.44 + 12. 15.3+ ft. 19- 4.3 + 6. 3.072 tons ARITHMETICAL 5. I360 ^ II children 15 8 i 7. PROGRESSION. - 9. 3 yrs. 10. Given II. 78 strokes GEOMETRICAL PROGRESSION. Page 353, 1. 96 2. I10.24 3. .^2007.3383664; $3001.460703698 4. 242 Page S54, 5- 1456 6. 1^4095 MENSURATION .—Page 350-9, 1. Given 2. 3600 s. y. 3. 144008.!'. 4. 80 rods 7. $90.90^ 9. 204.20335 r. 10. 47.746-mMf ft. 11. 3i.83i3fl«^r. 12. 4417.8609375 s. ft. 13. 3183.1 s. r. 15- 3817-03185 cu. ft. 16. 52 2f CU. ft. 18. 756 sq. ft. 20. 14684558.20796 5. m. 22. 26065 1 609333 J cii.m. M I S C E L L 1. $1083! 2. I? 1 20 cost; $45 loss 3. 692 less; 1434 great. 4. 22 ft. 5. 240 days 6. 143^ miles 7. $20, A's; $40, B's ; ^140, C's 8. 79-59¥bu. 9. 5.26^% C. 10. |;3 11. |i 64.06 J c. AN ECUS EXAMPLES.— JPrtflre ^45-93j pr. 21. 150 rods Z^' 12. 12 o'cl. 31 22. 8 ft. 37.^ m. 20^ s. 23. I6038.72 38. 13. $120 cost; 24. 45° 24' 45" 39- 830 profit 25. ^8125 40. 14. 35 m. 8 s. 26. £3#f past 7 A.M. 27. $16836.734 15. I24 28. 14520 per. 41. 16. 3 farmer's; 29. $19878.92 42. 24 neigh Vs 30. $680,625 17. 540 sheets 31. 77ift. 18. $40. A's; 32. $12500 43- $60, B's Z2>' ^1-59 19. 1980 pick. 34. $1446.625 44. 20. 15^ 35- 48 ft. 45- 360-2, 61 ft. I226196.489 44^ $2607.291 $5400, 1 St.; $8100, 2d.; ^13500? 3^^ $2574.50 $50, A; ^75, B; $105, $120 each $130, B 5 hrs. $265720 ^4 VB 35846 M249549 QMOX EDUC. THE UNIVERSITY OF CALIFORNIA LIBRARY