I VANANlWE«RliAltth&CO ^ 1 M c J N r4 \T ^ILW ^'ORK. ^v^! '',^&i^/-iA -i ^ {jiA-^X {\y\J- '/A/ly-cNxr' 1/ ^ ..^X/^ 5-V V , ^-.^V'/ ^/zU^' ■y '^ / 1 >7 ' c ,<. ( .^ .6-^' 0\. yu- -'t-'t...^ / \^4-^ ' ^r^^i - uh sMji^ We H>A,^ 'J CLEC'TIC EDUCATIONAL HEEJJiS. ^£/^ CRIES. i .^'"Yl"^^^^^'^''^ cC NEW PRACTICAL ARITHMETIC A Revised Edition of the Practical Arithmetic JO^^'E'PH" KA'Y/K/R-,: * Ijjie Professor in Woodward College. YAN ANTWERP, BRAGG & CO. 137WALNUT Street, CINCINNATI. 28 Bond Street NEW YORK. RAY'S MATHEMATICAL SERIES, ^f^^f^ ^_ ^e/^V, Ray'^s Neiv Primary Arithmelic. Ray^s New Intellectual Arithinetico Rafs New Practical Aritlimetic. Ray^s New Elementary Arithmetic. Rafs Higher Arithmetic, Ray's Test Examples. Ray^s Neiv Elemeittary Algeh'a. Rays New Higher Algebra. Ray^s Plane and Solid Geometry. Ray^s Geojnetry and Trigonouietry. Rafs Analytic Geometry>. Ray^s Elements of Astronomy. Ray's Surveying and Navigatian. Ray's D>j^'er2ntial i0^d\l\ii^gral\ Calculus. Copyright 1877 BY Van Antwerp, Bragg & Co. EDUCATION OEfY* ECLECTIC PRESS, VAN ANTWERP, BRAGG & CO. CINCINNATI. yti tM{ PREFACE. Changes in the methods of instruction in our schools and in the modes of transacting business have made it necessary to revise Ray's Practical Arithmetic. No other work on Arithmetic ever had so extensive use or wide- spread popularity. Teachers every-where, throughout the length and breadth of the land, are familiar with its pages, and millions of pupils have gained their arithmetical knowledge from the study of its principles. More than ten thousand editions of it have gone forth from the press. In view of these facts, it has been the constant aim in making this revision to preserve carefully those distinctive features of the former editions, which constituted the peculiar philosophical method of its learned author, viz.: 1st. Every principle is clearly explained by an analysis or solu- tion of simple examples, from which a Rule is derived. This is followed by graduated exercises designed to render the pupil familiar with its application. 2d. The arrangement is strictly philosophical; no principle is anticipated; the pupil is never required to perform any operation until the principle on which it is founded has first been explained. The changes made fall naturally under two heads: (1) those which adapt the book better to the advanced methods of instruction; (2) those which exhibit present methods of computation in business. In the first place, special attention is invited to the beauty and elegance of the typogranhy^^^ The^^ffiyant matter of the volume, 961b;>y (iii) IV PKEFACE. the definition, the solution, or the rule, is at once clearly indicated by a difference of type. A running series of articles, with numbered paragraphs, enhances the convenience of the text-book for recitation and for reference. The analytic solutions and written operations have been carefully separated. All obsolete Tables of Weights and Measures, such as Beer Measure and Cloth Measure, and all obsolete denominations, such as drams, roods, etc., are discarded. The Metric System of Weights and Measures is presented in accordance with its now widely extended usage, and is assigned its proper place immediately after Decimals. A few subjects, such as Factoring and the principles of Frac- tions, have been entirely rewritten, and in many instances the definitions and rules have been simplified. The subject of Percent- age has been much expanded, and an endeavor has been made to systematize its numerous applications; many novel and interesting features, both of subject-matter and classification, will here be met with for the first time. The subjects of Interest and Discount have received that careful attention which their importance demands. The publishers desire to express their thanks to the many teachers whose suggestions and corrections are embodied in the present edition. Especial mention is due Prof, M. W. Smith and Mr. A. P. Morgan for many valuable features of this revision. In conclusion, the publishers wish to reiterate that the object throughout has been to combine practical utility with scientific accuracy; to present a work embracing the best methods, with all real improvements. How far this object has been secured is again submitted to those engaged in the laborious and responsible work of education. Cincinnati, August, 1877. TABLE OF CONTENTS. PAGE The Arabic System of Notation 9 The Koman System oe Notation 20 Addition 22 Subtraction 31 Multiplication 39 Contractions in Multiplication 47 Division 50 Short Division . . , . 54 Long Division 59 Contractions in Division . 64 General Principles of Division 67 Compound Numbers 71 United States Money 72 Merchants' Bills 83 Reduction of Compound Numbers 84 Dry Measure 84 Rules for Reduction . . 87 Liquid Measure 88 Avoirdupois Weight 89 Long Measure 90 Square Measure 90 Solid or Cubic Measure 94 Time Measure 96 (V) Vi CONTENTS. PAGM Miscellaneous Tables 97 Addition of Compound Numbers . . , . . 102 Subtraction of Compound Numbers .... 106 Multiplication of Compound Numbers .... Ill Division of Compound Numbers 113 Longitude and Time 115 Factoring 118 To Find the Greatest Common Divisor .... 123 To Find the Least Common Multiple .... 125 Cancellation 127 Fractions 131 Principles 135 Reduction of Fractions 137 Addition of Fractions 144 Subtraction of Fractions 147 Multiplication of Fractions . . . . . . 149 . Compound Fractions 152 Division of Fractions 154 Complex Fractions 157 Fractional Compound Numbers 159 Practice 165 Decimal Fractions 168 Reduction of Decimals 175 Addition of Decimals 178 Subtraction of Decimals 179 Multiplication of Decimals 180 Division of Decimals . . . . . . . 183 Decimal Compound Numbers . . . . . . 186 The Metric System 189 Measures of Length 190 Land or Square Measure 192 Measures of Capacity 192 Measures of Weight 193 Table of Values ........ 194 CONTENTS. vii PAGK Percentage 197 Formulas for the four cases of Percentage. . . . 203 Applications of Percentage 205 Mercantile Transactions 206 Commission 206 Trade Discount 208 Profit and Loss 210 Stock Transactions 213 Brokerage 214 Assessments and Dividends . . . . . .215 Stock Values 216 Stock Investments 217 Interest 219 Simple Interest 221 The Twelve Per Cent Method 229 Formulas for the five cases of Interest . . . . 237 Compound Interest 237 Annual Interest 239 Partial Payments 241 Discount 247 Bank Discount 247 True Discount . .256 Exchange 260 Domestic Exchange 261 Foreign Exchange 262 English Money 262 French Money ' . . 263 German Money 263 Canadian Money 263 Insurance 265 Fire and Marine Insurance 265 Life Insurance . . 267 viii CONTENTS. PAGK Taxes 269 State and Local Taxes 269 United States Revenue 273 Internal Revenue . . 274 Duties or Customs .... ... 274 Ratio .276 Principles 280 Proportion 282 Simple Proportion 285 Compound Proportion . . . . . . . 289 Partnership 291 Bankruptcy 293 General Average " . . . 293 Partnership with Time 294 Equation of Payments 295 Average 297 Involution , 298 Evolution . . . 300 Square Root 302 Cube Root 309 Mensuration 316 Measurement of Surfaces 316 Measurement of Solids 323 Applications of Mensuration 328 Progressions 331 Arithmetical Progression 331 Geometrical Progression 334 Article 1.' 1. A Unit is a single thing of any kind; as, one, one apple, one dollar, one pound. 2. A Number consists of one or more units ; as, one, five, seven cents, nine men. 3. Arithmetic treats of nimibers, and is the art of computing by them. 4. Numbers are expressed in two ways ; first, by words; second, by characters. 5. A System of Notation is a method of expressing numbers by characters. 6. Two systems of Notation are in use, the Arabic and. the Roman. The Arabic system is used in all our arithmetical calculations. THE ARABIC SYSTEM OF NOTATION. 2, 1. To express numbers, the Arabic Notation em- ploys ten characters, called figures; namely, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0. Eemark 1. — The Arabic System of N'otation is so called because its characters appear to have been introduced into Europe bj' the (9) 10 RAY'S NEW PRACTICAL ARITHMETIC. Arabians; tut i'c is now o-cnv^r.illy acknowledged that they originated in India. Rem. 2^; — ^T'ae 'A^f'J^hia Not'ition is also called the Decimal System and the ^ClnrvrTio'ii '%,s^m. ' ' 2. The Order of a figure is the place it occupies in a number. UNITS OF THE FIRST ORDER, OR UNITS. 3. 1. A unit or single thing is owe, One unit and one more are two^ Two units and one more are three^ Three units and one more are four^ Four units and one more are five. Pive units and one more are six^ Six units and one more are seven. Seven units and one more are eight, Eight units and one more are nine, written 1. 2. 3. 4. 5. 6. 7. 8. 9. 2. These nine characters are called significant figures, because they denote something. 3. The character 0, called naught, stands for nothing; its use is to fill vacant orders. The is also called cipher and zero. 4. When a figure stands alone or in the first place at the right of a number, it represents one or more units of the first order. 5. Units of the first order are called simply units; and the place they occupy is called the units' place. UNITS OF THE SECOND ORDER, OR TENS. 4. 1. Nine units and one more are called ten; it Iso is represented by the figure 1 ; but the one is NOTATION. 11 made to occupy the second place from the right by writing a in the units' phice. 2. One ten is written thus . . Two tens are twenty^ written Three tens are thirty, '^ Four tens are forty, " Five tens are fifty, '' Six tens are sixty, " Seven tens are seventy, '' Eight tens are eighty, " Nine tens are ninety, ^' 10. 20. 30. 40. 50. 60. 70. 80. 90. 3. When a figure in a number stands in the second place from the right, it represents one or more units of the second order. 4. Units of the second order are called tens; and the place they occupy is called the tens' place. TENS AND UNITS. , 5. 1. The numbers between 10 and 20, 20 and 30, etc., are expressed by representing the tens and units of which they are composed. 2. One ten and one unit are eleven, written 11. and two units are twelve, " 12. and three units are thirteen, ^' 13. and four units are fourteen, " 14. and five units are fifteen, " 15. and six units are sixteen, " 16. and seven units are seventeeti, " 17. and eight units are eighteen, " 18. and nine units are nineteen, '• 19. unit are twenty -one, '' 21. units are twenty -two, '' 22. One ten One ten One ten One ten One ten One ten One ten One ten Two tens and one Two tens and two 12 KAY'S NEW PK ACTIO AL AKITHMETIC. ]S^UMBERS TO BE WRITTEN. 1. Twenty-three; twenty-four; twenty-five; twenty- six; twenty-seven; twenty -eight ; twenty- nine. 2. Thirty-seven ; forty-tw^o ; fifty-six ; sixty-nine ; sev- enty-three ; eighty-seven ; ninety-four. 3. Eighty-three; forty-five; ninety-nine; fifty-one; thir- ty-six: seventy-eight; sixty -two. 4. Fifty-five ; ninety -three ; eighty-one ; sixty-seven ; forty-nine ; seventy-four ; thirty-eight. 5. Seventy-six ; forty-four ; eighty -two ; fifty-seven ; thirty-five ; ninety-one ; sixty-three. NUMBERS TO BE READ. 1. 71; 32 53 84 65 46; 2. 58; 34 79 66 41 85; 3. 75; 43 88 61 59 33; 4. 39; 72 54 86 47 98; 5. 6S; 77 31 89 52 96; 97. 92. 95. 64. 48. UNITS OF THE THIRD ORDER, OR HUNDREDS. 6. 1. Ten tens are one hundred; it is represented by the figure 1 written in the third order, the orders of tens and units being each filled with a cipher. One hundred is wa^itten thus, 100 Two hundred " " 200. Three hundred *v '' 300 Four hundred '' " 400 Five hundred '' " 500 Six hundred '' " 600. Seven hundred '' " 700. Eight hundred " " 800. Nine hundred '' " . 900. NOTATION. 13 2. Units of the third order are called hundreds; and the place they occuj^y is called the hundreds' place. HUNDREDS, TENS, AND UNITS. 7. 1. The numbers between 100 and 200, 200 and 300, etc., are expressed by representing the hundreds, tens, and units of which they are composed. 2. One hundred and one unit are one hundred and one, written 101. One hundred and one ten are one hundred and ten, written 110. One hundred and one ten and one unit are one hun- dred and eleven, written 111. One hundred and two tens are one hundred and twenty, written 120. One hundred, two tens, and five units are one hundred and twenty -five, written 125. NUMBERS TO BE WRITTEN. 1. One hundred and thirty; one hundred and forty; one hundred and fifty ; one hundred and sixty ; one hun- dred and seventy; one hundred and eighty. 2. One hundred and twenty-three ; four hundred and fifty-six ; seven hundred and eighty-nine ; one hundred and forty-seven ; two hundred and fifty-eight ; three hundred and sixty -nine. 3. One liundred and two; three hundred and forty- five ; six hundred and seventy-eight ; two hundred and thirty-four; five hundred and sixty-seven; eight hundred and ninety. 4. Four hundred and fifty -three ; seven hundred and eighty-six; nine hundred and twelve; two hundred and thirty ; four hundred and fifty ; six hundred and seventy. 14 RAY'S NEW PRACTICAL ARITHMETIC. 5. One hundred and fifty-three; four hundred .and eighty-six; seven hundred and twenty-nine; one hun- dred and three; four hundred and six; seven hundred and nine. NUMBERS TO BE READ. 1. 210 320 430 540; 650 760. 2. 213 546 879 417; 528 639. 3. 201 435 768 324; 657 980. 4. 543 876 192 329; 548 765. 5. 513 846 279 301; 604 907. UNITS OF HIGHER ORDERS. 8. 1. Ten hundreds are one thousand; it is repre- sented by 1 in the fourth order; thus, 1000. 2. Ten thousands form a unit of the fifth order ; thus, 10000 ; one hundred thousands, a unit of the sixth order ; thus, 100000, etc. 3. Invariably, ten units of any order make a unit of the next higher order. 4. The names of the first nine orders may be learned from the followino- Table of Orders. 9th. 8th. 7th. 6th. 5th. 4th. 3d. 2d. Ist 'T3 t o T3 O 1^ W H C o ^ a o NOTATION. 15 DEFINITIONS AND PRINCIPLES. 9. 1. The first nine numbers are represented by the nine figures, — 1, 2, 3, 4, 5, 6, 7, 8, 9. 2. All other numbers are represented by combinations of two or more of the ten figures, — 1, 2, 3, 4, 5, 6, 7, 8, J, 0. 3. The numbers that end with 2, 4, 6, 8, or are called even numbers. 4. The numbers that end with 1, 3, 5, 7, or 9 are called odd numbers. 5. The value of a figure is the number of units it ex- presses. 6. The value of a figure is always local; that is, it depends upon the place it occuj)ies in a number. Rem. — The principle of local value is Avhat peculiarly distinguishes the Arabic System of Notation from all other systems that have existed. 7. The number a figure expresses w^hen it stands in units' place is called its simple value. 8. The value of a figure is increased tenfold by remov- ing it one place to the left. 9. The value of a figure is decreased tenfold by remov- ing it one place to the right. GROUPING OF ORDERS INTO PERIODS. 10. 1. For convenience in writing and reading num- bers, the diflPerent orders are grouped into periods of three orders each. Rem. — A number is pointed off into periods of three figures each by commas. 2. The first three orderp — units, tens, hundreds — con- stitute the first, or unit period. 16 RAY'S NEW PRACTICAL ARITHMETIC. 3. The second group of three orders — thousands^ ten thousands, hundred thousands — constitutes the second, or thousand period. 4. The third group of three orders constitutes the third, or million period. 5. The periods from the first to the twelfth inclusive may be learned from the following Table of Periods. No. Name. No. Name. First Unit. Seventh Quintillion. Second Thousand. Eighth Sextillion. Third Million. Xinth Septillion. • Fourth Billion. Tenth Octillion. Fifth Trillion. Eleventh Nonillion. Sixth Quadrillion. Twelfth Decillion. 6. The grouping of the orders into periods is shown in the following Table. 9 o en 4. Billion. o a? '-^ ^ IS o S H pq ^ S B 3. Million o mII 2. 1. Thousand. Unit. TU fl 't:! o ^ O W H H 3 3 G H H P NOTATION. 17 7. It is plain that each period is composed of unitSj lens J and hundreds of that iwriod. To Write Xumhers in the Arabic System, 11. 1. Write six hundred and fifty -four trillion three hundred and twenty -one billion nine hundred and eighty- seven million six hundred and fiftj-four thousand three hundred and twenty-one. -d g ^- G c3 o o 6 5 4, 3 2 1, 1) 8 7, 6 5 4, 3 g a -« 0= 5 ^ OJ ^ WhP Wh^ ^ G s •: c p MhP WhP WhP Rule. — Begin at the left, and write each period as a number composed of himdreds, tens, and units — filling the vacant orders with ciphers. Rem. — In the left hand period, however, when the hundreds or the hundreds and tens are wanting, the vacant orders are not filled with ciphers. NUMBERS TO BE WRITTEN. 2. Two thousand; thirty thousand, four hundred thou- sand. 3. Five million ; sixty million ; seven hundred million. 4. Eight billion; ninety billion; one hundred billion. 18 KAY'S NEW PRACTICAL ARITHMETIC. 5. One thousand two hundred ; two thousand one hun- dred. G. Three thousand four hundred and fifty ; six thou- sand seven hundred and eighty-nine. 7. Tw^elve thousand three hundred and forty-five. 8. Six hundred and seventy-eight thousand nine hun- dred and twelve. 9. One milHon three hundred and fil'ty-seven thou- sand nine hundred and twenty-four. 10. vSixty-eight million one hundred and forty-three thousand seven hundred and ninety-two. 11. One thousand and one; one thousand and ten; one thousand one hundred. 12. One thousand one hundred and one ; one thousand one hundred and ten ; one thousand one hundred and eleven. 13. Two thousand and three; four thousand and fifty. 14. Forty -five thousand and tw^enty-six. 15. Eighty thousand two hundred and one. 16. Ninety thousand and one. 17. Four hundred and ten thousand two hundred and five. 18. One hundred thousand and ten. 19. Three million seventy thousand five hundred and nine. 20. Forty -five million eighty -three thousand and twenty-six. 21. Nine hundred and nine million ninety thousand. 22. Seven hundred million ten thousand and tw^o. 23. Forty billion two hundred thousand and five. 24. Seven hundred and twenty-six billion fifty million une thousand two hundred and forty-three. 25. Eighty billion seven hundred and three million five hundred and four. NOTATIOJ^. 19 12. Numeration is the reading of numbers when ex- pressed according to a system of notation. To Read JYutnbers in the Arabic Sr/stein. 1. Eead 654321987654321. I § HP 1^ H 6 5 4, 3 2 1, 9 8 7, 6 5 4, 3 2 1. a, V. Of .-§ ^ fl 1— ' O H HP OD +J W^p W^P W^P H^P wIp Rule. — 1, Begin at the right, and point off the number info periods of three figures each. 2. Begin at the left, and read each jjeriod as a number com- posed of hundreds, tens, and units, giving the name of the period. Rem. 1. — The left hand period will sometimes contain hut one oi two figures. Rem. 2. — It is customarj^ to omit the name of the unit period. NUMBERS TO BE READ. 2. 41582; 763491; 2519834; 375486921; 4923176358. 3. 37584216974; 432685729145; 6253971438267. 4. 1300; 2540; 6070; 8009; 13200; 1005. 5. 682300; 8600050; 3040; 50004; 704208. 6. 7085; 62001; 400009; 2102102; 9001003. 20 RAY'S NEW PRACTICAL ARITHMETIC. 7. 130670921; 6900702003; 23004090701; 9420163070. 8. 570000010326049; 200103478511992485. 9. 45763000020108000507. 10. 800820020802008. THE ROMAN SYSTEM OF NOTATION. DEFINITIONS. 13. 1. To express numbers, the Roman Notation em- ploys seven letters; namely, I, V, X, L, C, D, M. Rem. — The Roman System of Notation is so called because it was the method of expressing numbers used by the ancient Romans. It is now used to mark the chapters of books, the dial plates of clocks, etc. 2. In the Koman Notation, numbers arc expressed in four ways. 1st. Each of the seven letters expresses a numher, as fol- lows: I, one; V, five; X, ten; L, fifty; C, one hundred ; D, five hundred ; M, one thousand. 2d. Seven numbers are expressed by repetitions of the let- ters 7, JT, and C. Thus, II represent two; Hi, three; XX, twenty ; XXX, thirty; CC, two hundred; CCC, three hundred ; CCCC, four hundred. 3d. Four numbers are expressed by a subtractive combi- nation, as follows: IV, four; IX, nine; XL, forty; XC, ninety. 4th. All other numbers are formed by additive combina- tions of two or more of the preceding eighteen numbers, the smaller being always situated on the right of the larger number. For example, YI is six; XYII, seventeen; LXXYIII, seventy -eight ; CLXXXIX, one hundred and eighty-nine \ MDCCCLXXYII, eighteen hundred, and seventy -seven. NOTATION. 21 Write in the Koman J^otation, 1. The numbers from one to twenty. 2. The numbers from twenty to thirty. 3. 30 40 50 60; 70 80; 90. 4. 57 29, 61 38; 46 72; 93. 5. 100 101; 106 117; 129 168. 6. 199 246, 309 482; 527 693. 7. 734 859 975 1001; 1010. 8. 1048 1119 1285 1326. 9. 1492, 1776; 1861 1900. THE FUNDAMENTAL RULES. DEFINITIONS. 14. 1. An integer is a whole number. 2. Numbers are either abstract or concrete. 3. An abstract number is a number simply, as 5, 12, 20. 4. A concrete number is a number applied to one or more objects; as 1 apple, 5 pounds, 12 men. 5. The name of the object of a concrete number is its denomination. Thus, in 5 pounds, the denomination is pounds. 6. Numbers are either simple or compound. 7. A simple number is a single number, either abstract or concrete; as 3, 7 dollars, 1 pint. 8. A compound number is made up of two or more concrete numbers of different denominations ; as 3 pecks 7 quarts 1 pint. 9. There are four primary operations of Arithmetic; namely. Addition^ Subtraction, Multiplicatio7i, and Division; — these are called the Fundamental Rules. 15. 1. If you have 2 cents and find 3 cents, how many will you then have? Ans. 5 cents. Why? Because 2 cents and 3 cents are 5 cents. 2. I spent 12 cents for a slate, and 5 cents for a copy- book : how many cents did I spend ? Ans. 17 cents. Why? 3. John gave 6 cents for an orange, 7 cents for pen- cils, and 9 cents for a ball : how many cents did all cost? A71S. 22 cents. Why? 4. Joseph gave 5 cents for a daily paper, 10 cents for a weekly paper, 25 cents for a monthly magazine, 30 cents for a book of poems, and 40 cents for a novel : how much did he spend? 110 cents. 16. 1. The operation in these examples is termed Ad- dition; hence, Addition is the process of uniting two or more numbers into one number. 2. The number obtained by addition is the Sum or Amount. 3. When the numbers to be added are simple, the operation is called Addition of Simple Numbers. 4. The sign of Addition (-|-), called j)lus, means more; when placed between two numbers, it shows that they are to be added ; thus, 4 + 2 means that 4 and 2 are to be added together. (22) ADDITION OF SIMPLE NUMBERS. 23 5. The 8ign of equality (==:) denotes that the quantities between which it sstands are equal; thus, the expression 4 + 2 = 6 means that the sum of 4 and 2 is 6 : it is read, 4 plus 2 equals 6. Addition Table. 2 + 0= 2 3 + 0= 3 4 + 0= 4 5 + 0= 5 2 + 1= 3 3 + 1= 4 4+1= 5 5 + 1= 6 2 + 2= 4 3 + 2= 5 4+2= 6 5 + 2= 7 2 + 3= 5 3 + 3= 6 4 + 3= 7 5 + 3= 8 2 + 4= 6 3 + 4= 7 4 + 4= 8 5 + 4= 9 2 + 5= 7 3 + 5= 8 4 + 5= 9 5 + 5 = 10 2 + 6= 8 3 + 6= 9 4 + 6 = 10 5 + 6 = 11 2 + 7= 9 3 + 7 = 10 4 + 7 = 11 5 + 7 = 12 2 + 8 = 10 3 + 8 = 11 4 + 8 = 12 5 + 8 = 13 2 + 9 = 11 3 + 9 = 12 4 + 9 = 13 5 + 9 = 14 6 + 0= 6 7 + 0= 7 8 + 0= 8 9 + 0= 9 6 + 1= 7 7 + 1= 8 8 + 1= 9 9 + 1 = 10 6 + 2= 8 7 + 2= 9 8 + 2 = 10 9 + 2 = 11 6 + 3= 9 7 + 3 = 10 8 + 3=11 9 + 3 = 12 6 + 4 = 10 7 + 4 = 11 8 + 4 = 12 9_[_4:^13 6 + 5 = 11 7 + 5 = 12 8 + 5 = 13 9 + 5 = 14 6 + 6 = 12 7 + 6 = 13 8 + 6 = 14 9 + 6 = 15 6 + 7 = 13 7 + 7 = 14 8 + 7 = 15 9 + 7 = 16 6 + 8 = 14 7 + 8 = 15 8 + 8 = 16 9 + 8 = 17 6 + 9 = 15 7 + 9 = 16 8 + 9 = 17 9 + 9 = 18 17. When the sum of the figures in a eokimn does not exceed 9, it is written under the column added. 24 KAY'S NEW PRACTICAL AKITHMKTIC. Examples. 1. I own 3 tractH of land : the first contains 240 acres ; the second, 132 acres; the third, 25 acres: how many acres in all? SoLUTiOiV. — Since units oi different orders can not be added together, write units of the same order in the same column, so that the figures to be added may be in the most convenient ])ositio7i. Begin at the right, and say 5 and 2 are 7 units, 2 4 acres, which write in units' place; 2 and 3 are 5, and 4 132 acres, are 9 tens, which write in tens' place; 1 and 2 are 3 2 5 acres, hundreds, which write in hundreds' place. 39 7 acres. 2. I owe one man $210, another S142, and another $35: what is the sum of my debts? $387. 3. Find the sum of 4321, 1254, 3120. 8695. 4. Find the sum of 50230, 3105, 423. 53758. 18. When the sum of the figures in a column ex- ceeds 9, two or more figures are required to exi)re8S it. Example. 1. Add the numbers 3415, 503, 1870, and 922. 3415 503 Solution. — Write units of the same order in the same column. Then say 2 and 3 are 5, and 5 are 10 units, which are no ( ) units, written in the units' place, and 1 18 70 ten, carried to the tens; 1 and 2 are 3, and 7 are 10, and 1 922 are 11 tens, which are 1 ten, written in the tens' place, and 6 710 1 hundred, carried to the hundreds; 1 and 9 are 10, and 8 are 18, and 5 are 23, and 4 are 27 hundreds, which are 7 hundreds, written in the hundreds' place, and 2 thousands, carried to the thou- sands; 2 and 1 are 3, and 3 are 6 thousands, written in the thou- sands' place. 4 ADDITION OF SIMPLE NUMBERS. 25 CaiTying the tens is simply adding tens to tens, hun- dreds to hundreds, etc., on the principle that only units of the same order can be added. For convenience, the addition begins at the right hand column, with the units of the lowest order, so that, if the sum of the figures in any column exceeds 9, the tens can be carried to the sum of the next higher order. Rem. — To illustnitc; the greater convenience of adding the units' column first, take the above example. Solution. — Commencing the addition with the thou- 3415 sands' column, the sum is 4; next adding the hundreds, ^^^ the sum is 26 hundreds, which equal 2 thousands and 6 f.<^o hundreds; next adding the tens, the sum is 10 tens, equal to 1 hundred; and finally adding the units, the sum is 10 units, equal to 1 ten. As these sums have also to be 10 added, this much extra work must be done in order to __1_9 complete the solution. 6710 19. Rule. — 1. Write the mnnhers to he added, so that figures of the same order may stand in the same coluinn. 2. Begin at the right hand, and add each column sepa- rately. Place the units obtained by adding each cohwm under it, and carry the tens to the next higher order. Write down the entire sum of the last column. Proof. — Add the columns downward, commencing with the column of units. 1. Find the sum of 3745, 2831, 5983, and 7665. In adding long cokimns of figures, it is necessary 3 745 to retain the numbers carried. This may be done by 2831 5 9 83 placing them in smaller figures under their proper 7«ak columns, as 3, 2, 1, in the margin. 9 9 2 4 321 26 liAY'tS NEW PRACTICAL AKITHMETIC. Examples. (2) (3) (4) (5) (6) (7) 184 204 103 495 384 1065 216 302 405 207 438 6317 135 401 764 .^-85 348 5183 320 311 573 825 843 7102 413 109 127 403 483 3251 101 43 205 325 834 6044 1369 1370 2177 2440 3330 28962 (8) (9) (10) (H) (12) 3725 5943 82703 987462 6840325 5834 6427 102 478345 7314268 4261 8204 6005 610628 3751954 7203 7336 759 423158 6287539 13. 11 + 22 + 33 -(- 44 + 55 = how man}^ ? 165. 14. 23 + 41 -I 74 + 83 + 16 = how many ? 237. 15. 45 + 19 + 32 + 74 + 55== how many? 225. 16. 51 + 48 + 76 + 85 + 4 = how many ? 264. 17. 263 + 104 + 321 + 155 = how many ? 843. 18. 94753 + 2847 + 93688 + 9386 + 258 + 3456 are how many ? 204388. 19. January has 31 days ; February, 28 ; March, 31 ; April, 30 ; and May, 31 : how many days are there in these five months? 151. 20. June has 30 days; July, 31; August, 31; September, 30; October, 3l : how many days in all? 153. 21. The first 5 months have 151 days, the next 5 have 153 days, November has 30, and December, 31 : how many days in the whole year? 365. ADDITION OF SIMPLE NUMBEES. 27 22. I bought 4 pieces of muslin : the first contained 50 yards, the second, 65, the third, 42, the fourth, 89 : how many yards in all? 246 yd. 23. I owe one man S245, another $325, a third $187, a fourth $96 : how much do I owe ? $853. 24. General Washington was born A. D. 1732, and lived 67 years: in what year did he die? 1799. 25. Alfred the Great died A. D. 901 ; thence to the signing of Magna Charta was 314 years ; thence to the American Eevolution, 560 years: in what year did the American Eevolution begin? 1775. 26. A has 4 flocks of sheep ; in the first are 65 sheep and 43 lambs ; in the second, 187 sheep and 105 lambs ; in the third, 370 sheep and 243 lambs ; in the fourth, 416 sheep and 95 lambs: how many sheep and lambs has he? 1038 sheep, and 486 lambs. 27. A man bought 30 barrels of pork for $285, 18 barrels for $144, 23 barrels for $235, and 34 barrels for $408 : how many barrels did he buy, and how many dollars did he pay? 105 bbl., and $1072. 28. The first of four numbers is 287 ; "^^^he second, 596 ; the third, 841 ; and the fourth, as much as the first three : what is their sum? 3448. 29. The Pyramids of Egypt were built 1700 years before the founding of Carthage ; Carthage was founded 47 years before and was destroyed 607 years after the founding of Eome, or 146 years before the Christian era. How many years before Christ were the Pyramids built? 2500. 30. Add three thousand and five; fort^^-two thousand six hundred and twenty-seven ; 105 ; three hundred and seven thousand and four; 80079; three hundred and twenty thousand six hundred. 753420. 31. Add 275432 ; four hundred and two thousand and 28 KAY'S NEW PRACTICAL ARITHMETIC. thirty ; three hundred thouBand and five ; 872026 ; four million two thousand three hundred and forty -seven. 5851840. 32. Add eight hundred and eighty million eight hun- dred and eighty-nine ; 2002002 ; seventy-seven million four hundred and thirty-six thousand; two hundred and six million five thousand two hundred and seven ; 49003; nine hundred and ninety million nineteen thousand nine hundred and nineteen. 2155513020. 33. North America has an area of 8955752 square miles; South America, 6917246 square miles; and the West Indies, 94523 square miles: what is the area of the entire continent? 15967521 sq. mi. 34. A man pays $600 for a lot, $1325 for building materials, $30 for digging the cellar, $120 for stone- work, $250 for brick-work, $140 for carpenter- work, $120 for plastering, and $115 for painting: how much did his house and lot cost him? $2700. 35. A man bequeaths $7850 to his wife, $3275 to each of his two sons, and $2650 to each of his three daughters : what is the amount of his bequest? $22350. 36. A merchant spent $8785 for dress goods, and $12789 for sheetings. He sold the dress goods at a profit of $878, and the sheetings at a profit of $1250: for how much did he sell the whole? $23702. 37. A merchant began business with $7000 cash, goods w^orth $12875, bank stock worth $5600, and other stocks worth $4785. In one year he gained $3500 : what w^as he worth at its close? $33760. 38. A house has two parlors, each requiring 30 yards of carpet; four bed-rooms, each requiring 25 yards; a dining-room and sitting-room, each requiring 20 yards: how many yards are required to carpet the entire house? 200 yd. ADDITION OF SIMPLE NUMBERS. 29 20. An excellent practice, in order to secure readiness and accuracy, is to add two columns at once. The fol- lowing example illustrates the method : (!)■ Beginning with 47, add the 3 tens above, which equal 77 ; then the 4 units, making 81 ; then the 6 tens above, 141; and the 5 units, 146; then the 7 tens above, 216; and the 9 units, 225^ then the 9 tens above, 315, and finally the 2 units, 317. Put down the 17, and carry the 3 hundreds to the hundreds' column. Then 93 and 3 to carry are 96, and 60 are 156, and 2 are 158, and 40 are 198, and 8 are 206, and 60 are 266, and 7 are 273, and 70 are 343, and are 351, wnk'h write in its proper place. 7892 6 7 79 4865 6234 9347 35117 Examples. (2) (3) w (5) (H) 3686 9898 4356 893742 234567 4724 8989 6342 743698 765432 6583 4545 7989 437821 987654 5798 5454 4878 643567 456789 6953 6363 6749 892742 778899 27744 35249 30314 3611570 3223341 (7) (8) (9) (10) 5493275 4819 18356 849627 6182463 9263 49276 532472 9538719 2752 94678 293784 2645834 8375 36525 468135 8256386 6498 42983 926547 32116677 31707 241818 3070565 30 RAY'S NEW PRACTICAL ARITHMETIC. (11) (12) (13) (14) (15) 7421 6873 4729 237285 884261 6322 2196 6234 64371 724353 798 583 5781 2143 416213 4352 79 3143 842 598624 547 684 7182 55 784344 674 4348 6989 789 627517 2315 7896 7222 4621 843641 7218 233 6643 15115 47821 1847 594 7859 647890 52348 5721 6483 6742 77442 2932 6848 7542 8982 84931 4751 4722 3967 3451 894623 896 5976 29 8692 446217 722 6843 478 7341 134162 823344 1234 1717 6822 192317 874132 62833 43702 97812 2802803 6685899 21. 1, If you have 9 apples, and give 4 away, how many will you have left? Ans. 5 apples. Why? Because 4 apples from 9 apples are 5 apj^les. 2. Frank had 15 cents ; after spending 7, how many were left? Aiis. 8 cents. Why? 3 If you take 8 from 13, how many are left? Ans. 5. 4. If I have 25 cents, and spend 10 of them for a lead- l^encil, how much will I have left ? Ans. 15 cents. 5- Twelve from twenty leaves how many? A71S. 8. 22. 1. The operation in the preceding examples is termed Subtraction ; hence. Subtraction is the process of finding the difference between two numbers. 2. The larger number is called the Minuend; the less, the Subtrahend ; and the number left after subtraction, the .Difference or Remainder. 3. When the given numbers are simple, the operation is called Subtraction of Simple Numbers. 23. The sign of Subtraction ( — ; is called minus, meaning less; when placed between two numbers, it de- notes that the number on the right is to be taken from the one on the left; thus, 8 — 5 = 3 means that 5 is to be taken from 8, and is read, 8 minus 5 equcds 3. (31) 32 RAY'S NEW PRACTICAL ARITHMETIC. Subtraction Table. 2 — 2 = 3 — 3 = 4 — 4 = 5-5 = 3-2 = 1 4 — 3 = 1 5-4 = 1 6-5 = 1 4-2 = 2 5 — 3 = 2 6 — 4 = 2 7 — 5 = 2 5 — 2 = 3 6 — 3 = 3 7 — 4 = 3 8 — 5 = 3 6 — 2 = 4 7-3 = 4 8-4 = 4 9-5 = 4 7 — 2 = 5 8 — 3 = 5 9 — 4 = 5 10 — 5 = 5 8 — 2 = 6 9 — 3 = 6 10 — 4 = 6 11 — 5 = 6 9 — 2 = 7 10-3 = 7 11 — 4 = 7 12 — 5 = 7 10 — 2 = 8 11 — 3 = 8 12 — 4 = 8 13 — 5 = 8 11 — 2 = 9 12 — 3 = 9 13 — 4 = 9 14 — 5 = 9 6 — 6 = 7 — 7 = 8 — 8 = 9 — 9 = 7--6 = l 8 — 7 = 1 9 — 8 = 1 10 — 9 = 1 8 — 6 = 2 9-7 = 2 10 — 8 = 2 11 — 9 = 2 9 — 6 = 3 10 — 7 = 3 11 — 8 = 3 12 — 9 = 3 10 — 6 = 4 11 — 7 = 4 12 — 8 = 4 13 — 9 = 4 11 — 6 = 5 12 — 7 = 5 13 — 8 = 5 14 — 9 = 5 12 — 6 = 6 13 — 7 = 6 14 — 8 = 6 15-9 = 6 13 — 6 = 7 14 - 7 = 7 15 — 8-= 7 16 — 9 = 7 14 — 6 = 8 15 — 7 = 8 16 — 8 = 8 17-9 = 8 15 — 6 = 9 16 — 7 = 9 17 — 8 = 9 18 — 9 = 9 24. When each figure of the subtrahend is not greater than the corresponding figure of the minuend. Examples. 1. A man having $135, spent $112: how much had he left? SUBTRACTION OF SIMPLE NUMBERS. 33 Solution. — Since the difference between units of the same order only can be found, write units of the same order in the same column, so that the figures between which the subtraction is to be made may be in the most convenient position. Begin at the right, and say 2 from 5 leaves 3, which put in units' place; I from 3 leaves ^^^ minuend. 2, which put in tens' place; 1 from 1 leaves il^ subtrahend. 0, and, there being no figures on the left of ^^ remamder. this, the place is vacant. 2. A farmer having 245 sheep, sold 123: how many sheep had he left? 122. 3. A man bought a farm for $751, and sold it for $875: how much did he gain? $124. What is the difference 4. Between 734 and 531? 203. 5. Between 8752 and 3421? 5331. 6. Between 79484 and 25163? 54321. 7. Between 49528 and 16415? * 33113. 25. When the lower figure in any order is greater than the upper, a difficulty arises, which we will now explain. Examples. 1. James had 13 cents; after spending 5, how manv cents had he left? 1 3 5 can not be subtracted from 3, but it can be from 13; ^ 6 from 13 leaves 8. "qT 2. From 73 subtract 45. Prac. 3. 34 KAY'S NEW PRACTICAL ARITHMETIC. Solution. — 5 units can not be taken from 3 units. Take 1 (ten) from the 7 (tens), and add this 1 (ten) or 7 3 10 units to the 3 units, which makes 13 units; then, 4 5 subtract the 5 units, and there will remain 8 units, to be 2 8 put in units' place. Since 1 ten is taken from the 7 tens, there remain but 6 tens. Subtract 4 tens from 6 tens and pui the remainder, 2 tens, in tens' place. The difference is 28. Rem. 1. — Instead of actually taking 1 ten from the 7 tens, and adding it to the 3 units, the operation is perfi.rmed mentally; thus, 6 from 13 leaves 8, and 4 from 6 leaves 2. Rem. 2. — In such cases, the value of the upper number is not changed, since the 1 ten which is taken from the order of tens is added to the number in the order of units. Rem. 3. — Taking a unit of a higher order and adding it to the units of the next lower, so that the figure beneath may be subtracted from the sum, is called borrowing ten. Rem. 4. — After increasing the units by 10, instead of considering the next figure of the up}>er number as diminished by 1, the result will be the same, if the next figure of the lower number be increased by 1; thus, in the previous example, instead of diminishing the 7 tens by 1, add 1 to the 4 tens, which makes 5; thus, 5 from 13 leaves 8, and 5 from 7 leaves 2. Rem. 5. — This process depends upon the fact that having borrowed 1 from the 7 tens, we have to subtract from it lx)th 1 ten and 4 tens, or their sum, 5 tens. 3. Find the difference between 805 and 637. Solution — 1st Method. — Writing the less number 8 05 under the greater, with units of the same order in the 6 3 7 same column, it is required to subtract the 7 units from 16 8 5 units. The five can not be increased by borrowing from the next figure, because it is 0; therefore, borrow 1 hundred from the 8 hundreds, which leaves 7 hundreds in hundreds' place; this 1 hundred makes 10 tens; then, borrowing 1 ten from the 10 tens, and adding it to the 5 units, 9 tens will be in the tens' place, and 15 units in the units' place. SUBTRACTION OP SIMrLE NUMBERS. 35 Subtracting 7 from 15, 8 units are left, to be written in units* place; next, subtracting 3 tens from 9 tens, there are left 6 tens, to be written in tens' place; lastly, subtracting 6 hundreds from 7 hun- dreds, there remains 1 hundred, to be written in hundreds' place. 2d Method. — If the 5 units be increased by 10, say 7 from 15 leaves 8; then, increasing the 3 by 1, say 4 from can not be taken, but 4 from 10 leaves 6; then, increasing 6 by 1, sa^- 7 from 8 leaves 1. Rem. 1. — The second method is generally used; it is more con- venient, and less liable to error, especially when the upper number contains ciphers. Rem. 2. —Begin at the right to subtract, so that if any lower figure is greater than the upper, 1 may be borrowed from a higher order. Rem. 3. — If the difference of two numbers be added to the less number, the sum will be equal to the greater. Thus, if 5 subtracted from 8 leave 3, then 3 added to 5 will equal 8. 26, Rule. — 1. Write the less number under the greater, placing figures of the same order in the same column. 2. Beginning at the right hand, subtract each figure from the one directly over it, and write the remainder beneath. 3. If the loiver figure exceeds the upper, add ten to the upper figure, subtract the lower from it, and carry one to the next lower figure, or take one from the next upper figure. Proof. — Add the remainder to the subtrahend ; if the sum is equal to the minuend, the work is correct. Minuends, Subtrahends, (1) 7640 1234 6406 7640 Examples. (2) 860012 430021 (3) 4500120 2910221 (4) 3860000 120901 Remainders, 429991 1589899 3739099 Proof, 860012 4500120 3860000 36 RAY'S NEW PRACTICAL ARITHMETIC. 5. Take 1234567 from 4444444. 8209877. 6. Take 15161718 from 91516171. 76354453. 7. Take 34992884 from 63046571. 28053687. 8. 153425178 — 53845248==? 99579930. 9. 100000000 — 10001001==? 89998999. 10. Take 17 cents from 63 cents. 46 cents. 11. A carriage cost $137, and a horse S65 : how much more than the horse did the carriage cost? $72. 12. A tree 75 feet high was broken ; the part that fell was 37 feet long: how high was the stump? 38 ft. 13. America was discovered by Columbus in 1492 : how many years had elapsed in 1837? 345. 14. I deposited in the bank $1840, and drew out $475 : how many dollars had 1 left? $1365. 15. A man has property worth $10104, and owes debts to the amount of $7426 : when his debts are paid, how much will be left? $2678. 16. A man having $100000, gave away $11 : how many had he left? $99989. 17. Subtract 19019 from 20010. 991. 18. Eequired the excess of nine hundred and twelve thousand and ten, above 50082. 861928. 19. Take 4004 from four million. 3995996. 20. Subtract 1009006 from two million twenty thou- sand nine hundred and thirty. 1011924. 21. Subtract four hundred and five thousand and twenty-two from 2000687. 1595665. 22. What is the difference between thirteen million two hundred and one and 17102102? 4101901. 23. A man invested in business $30,000; at the end of the first year he found that all his assets amounted to only $26,967; how much had he lost? $3,033. 24. Take 9238715 from 18126402. 8887687. 25. Take 9909090009 from 19900900900. 9991810891. ADDITION AND SUBTRAQTION. 37 Examples in Addition and Subtraction. 1. 275 + 381 + 625—1098==? 183. 2. 6723 — 479 — 347— 228 = ? 5669. 3. In January, 1876, a merchant bought goods to the amount of $2675; in February, S4375; and in March, $1897 ; after making one payment of $3000, and another of $4947, how much did he still owe? $1000. 4. I owe three notes, whose sum is $1300 — one note being for $250, and another for $650: what is the amount of the third note? $400. 5. Mr. Jones deposited $450 in bank on Monday; on Tuesday, $725; oji Wednesday, $1235; on Thursday, $4675; and on Friday, $1727. On Saturday morning he drew out $5935, and Saturday afternoon, ^877 : how much money had he left in bank? $2000. 6. At the end of one year I found I had spent $2300. Of this amount, $350 were paid for board, $125 for cloth- ing, $375 for books, $150 for incidentals, and the remain- der for two acres of ground : how much did the two acres cost? $1300. 7. A speculator bought three houses. For the first he gave $4875 ; for the second, $2250 more than for the first ; and for the third he gave $3725. He afterward sold them all for $20838: how much did he gain? $5113. 8. A man owns j^roperty valued at $49570, of which $16785 are in personal property, and $24937 in real estate ; the remainder was deposited in bank : how much has he in bank? $7848. 9. A merchant bought a bill of goods for $7895, and paid $175 for freight, and $3 for dray age. He sold the goods for $10093: how much did he gain? $2020. 10. A farmer invested $10000, as follows: in land, $5750; in horses, $925; in cattle, $1575; in hogs, $675; 38 RAY'S NEW PRACTICAL ARITHMETIC. and the remainder in im^jlements and tools : how much did he invest in implements and tools? 81075. 11. A speculator on Monday gained $4625 ; on Tues- day, $3785 ; on Wednesday he lost $6955 ; on Thursday he lost $895; on Friday he gained $985, and on Satur- day he lost $1375: how much did he gain during the entire week? $170. 12. The following 18 Mr. Brown's private account for two weeks: First week, received $50 for salary, and spent $25 for clothing, $7 for board, $2 for washing, and $5 for sundries. Second week, received $50 for salary, loaned $35 to Tom Jones, paid $7 for board, $2 for washing, and $8 for sundries. How much did Mr. Brown have at the end of the two weeks? $9. MULTIPLICATION OP SIMPLE NUMBERS. 43 EXAiMPLES. (J) (8) (9) (10) Multiplicand, 5142 4184 3172 41834 Multiplier, 5 2 6 5 7 Product, 25710 15104 15860 292838 11. Multiply 49 by 3. 147. 12. Multiply 57 by 4. 228. 13. Multiply 128 by 5. 640. 14. Multiply 367 by 6. 2202. 15. Multiply 1427 by 7. 9989. 16. Multiply 19645 by 8. 157160. 17. Multiply 44386 by 9. 399474. 18. Multiply 708324 by 7. 4958268. 19. Multiply 96432 by 10. 964320. 20. Multiply 46782 by 11. 514602. 21. Multiply 86458 by 12. 1037496. When the Multiplier Exceeds 12. 22. What is the product of 43 X 25 ? Analysis. — Since 25 is equal to 2 tens and 4 3 5 units — that is, 20 -f 5, — multiply by 5 and 2 5 write the product, 215; then multiply by the 21 5 = 4 8 X ^ 2 tens, and set the product, 8 hundreds and 6 8G =43X20 tens, under the 2 hundreds and 1 ten. 1075 = 43X25 Multiplying by 5 units gives 5 times 43, and multiplying by 2 tens gives 20 times 43; add them, because 5 times 43 and 20 times 43 equal 25 times 43. Hence, multiply by the units' figure of the multiplier, and write the product so that the right-hand figure will fall in units' place; then multiply by the tens' figure, and write the right-hand figure of the product in the tens' place. 44 KAY'S NEW PRACTICAL ARITHMETIC. Therefore, in multiplying by a figure of any order, write the last figure of the product in the same order as the multiplier. Note. — The products of the multiplicand by the separate figures of the multiplier are called pariial producU. General Rule. — 1. Write the multiplier under the miilti' plicand, placing figures of the same order in a column. 2. Multiply the midtiplicand by each figure of the multi- plier in succession, beginning with units, always setting the right hand figure of each product under that figure of the multiplier which produces it. 3. Add the partial products together : their sum will be the product sought. Proof. — Multiply the multiplier by the multiplicand: the product thus obtained should be the same as the first product. 23. Multiply 2345 by 123. SOLUTION. PROOr. 123 multiplier. 2 345 multiplicand. 234 6 multiplicand. 12 3 multiplier. 615 = 1 2 3 X 5 7035=^2345X 3 492 =123X 40 469 ^2345X 20 869 =123X 300 2345__ =2 345X1 00 246 =123X2000 288435 = 2345X123 288435 = 123X2345 24. Multiply 327 by 203. Remark. — "When there is a cipher in the multiplier, 82 7 leave it, and multiply by the other figures, being careful 20 3 to place the right-hand figure of each partial product 981 under the multiplying figure. 6 5 4 66381 MULTIPLICATION OF SIMPLE NUMBERS. 45 Examples. 25. 235 26. 34() 27. 425 28. 518 29. 279 30. 869 31. 294 32. 429 33. 485 X13 = X19 = X29 = X34. X37 = X49. X57. X62: X76: : 3055. : 6574. : 12325. : 17612. : 10323. : 42581. : 16758. = 26598. = 36860. 34. 624 35. 976 36. 342 37. 376 38. 476 39. 2187 40. 3489 41. 1646 42. 8432 X 85: X 97 = X364: X526 = X536. X215: X276 = X365: X635: 43. Multiply 6874 by 829. 44. Multiply 2873 by 1823. 45. Multiply 4786 by 3497. 46. Multiply 87603 by 9865. 47. Multiply 83457 by 6835. 48. Multiply 31624 by 7138. : 53040. 94672. : 124488. : 197776. : 255136. : 470205. : 962964. : 600790. : 5354320. 5698546. 5237479. 16736642. 864203595. 570428595. 225732112. 49. What will 126 barrels of flour cost, at $6 a bar- rel? $756. 50. What will 823 barrels of pork cost, at $12 a bar- rel? $9876. 51. What will 675 pounds of cheese cost, at 13 cents a pound? 8775 cents. 52. What will 496 bushels of oats cost, at 24 cents a bushel? 11904 cents. 53. If a man travel 28 miles a day, how many miles will he travel in 152 days? 4256 miles. 54. There are 1760 yards in one mile : how many yards are there in 209 miles? 367840 yards. 55. There are 24 hours in a day, and 365 days in a year: if a ship sail 8 miles an hour, how far will she sail in a year? 70080 miles. 46 RAY'S NEW PRACTICAL ARITHMETIC. 56. Multiply two thousand and twenty-nine by one thousand and seven. 2048203. 57. Multiply eighty thousand four hundred and one by sixty thousand and seven. 4824622807. 58. Multiply one hundred and one thousand and thirty- two by 20001. 2020741032. 59. A grocer bought 2 barrels of sugar, each weighing 215 pounds, for 8 cents a pound : how much did he pay for the sugar? 3440 cents. 60. A grocer bought a barrel of molasses, containing 36 gallons, for 45 cents a gallon ; and • sold it for 55 cents a gallon: how much did he gain? 360 cents. 61. A commission merchant sold 2650 bushels of wheat for a farmer, at 95 cents a bushel, and charged him 2 cents a bushel for selling: how much money Was duo the farmer? 246450 cents. 62. A farmer bought 6 horses of one man for 75 dol- lars each, and 5 horses of another for 125 dollars each, and sold them all for 150 dollars each : how many dol- lars did he gain? $575. 63. A merchant bought one box of goods for 250 dollars, two more for 325 dollars each, and three more for 175 dollars each; he sold them all so as to gain 356 dollars: for how much did he sell them? $1781. 64. A farmer bought 24 sheep, at 5 dollars a head; 36 hogs, at 14 dollars a head; and 9 cows, at 45 dollars a head : when he sold them all, he lost 275 dollars : for how much did he sell them? $754. 65. To 75 X 37 add 85 X 54, and subtract 5284. 2081. 66. To 69 X 53 add 48 X 27, and subtract 4279. 674. 67. I bought 50 bags of coffee, averaging 63 poupds in a bag, paying 34 cents a pound : how much did it cost? 10719© cents. MULTIPLICATION OF SIMPLE NUMBERS. 47 CONTRACTIONS IN MULTIPLICATION. CASE I. 32. When the multiplier can be separated into factors. 1. What will 15 oranges cost, at 8 cents each? Analysis. — Since 15 is 3 Cost of 1 orange, 8 ct. times 5, 15 oranges will cost 3 5 times as much as 5 oranges. Cost of 5 oranges, 4 ct. Therefore, instead of multiply- 3 ing 8 by 15, first find the cost of Cost of 15 oranges, 120 ct. 6 oranges, by multiplying 8 cents by 5; then take 3 times that product for the cost of 15 oranges. Rule. — 1. Separate the multiplier into tico or more factors. 2. Multiply the multiplicand by one of the factors, and this product by another factor, till every factor is used; the last product will be the one required. Kem. — Do not confound the factors of a number with the ;?rtrts into which it may be separated. Thus, the factors of 15 are 5 and 3, while the parts into which 15 may be separated are any numbers whose sum equals 15: as, 7 and 8; or, 2, 9, and 4. Examples. , 2. What will 24 acres of land cost, at $124 an acre? $2976. 3. How far will a ship sail in 56 weeks^ at the rate of 1512 miles per week? 84672 miles. 4. How many pounds of iron are there in 54 loads, each weighing 2873 pounds? 155142 pounds. 5. Multiply 2874 by 72. 206928. 6. Multiply 8074 by 108. 871992. 48 KAY'S NEW PRACTICAL ARITHMETIC. CASE IT. 33. When the multiplier is 1 with ciphers annexed; as 10, 100, 1000, etc. 1. Placing one cipher on the right of a number (8, 3) changes the units into tens, the tens into hundreds, and so on, and, therefore, multiplies the number by ten; thus, annex one cipher to 25, and it becomes 250. 2. Annexing two ciphers changes units into hundreds, tens into thousands, etc., and multiplies the number by one hundred; thus, annex two ciphers to 25, and it becomes 2500. Rule. — Annex as many ciphers to the multiplicand, as there are ciphers in the 7nultiplier, and the nwnber thus formed will be the product required. 1. Multiply 245 by 100. 2. Multiply 138 by 1000. 3. Multiply 428 by 10000. 4. Multiply 872 by 100000. 5. Multiply 9642 by 1000000. 6. Multiply 10045 by 1000000. 24500. 138000. 4280000. 87200000. 9642000000. 10045000000. ^CASE III. 34. When there are ciphers at the right of one or both of the factors. 1. Find the product of 625 by 500. Analysis. — The multiplier may be considered as composed of two factors: 5 and 100, Multiplying by 6^5 •5, the product is 8125; and the product of this number 500 by 100 is 812500, which is the same as annexing two 312 5 00 <'iphcrs to the first product. MULTIPLICATION OF SIMPLE NUMBERS. 49 2. Find the product of 2300 X 170. 2300 Analysis. — The number 2300 may be regarded as 17 composed of the two factors 23 and 100; and 170, of 161 the two factors 17 and 10. 2 3 391000 The product of 2300 by 170 will be found by multi- plying 23 by 17, and this product by 100, and the resulting product by 10 (33); that is, by finding the product of 23 by 17, and then annexing 3 ciphers to the product, as there tire 3 ciphers at the right of both factors. Rule. — Multiply without regarding the ciphers on the right of the factors ; then annex to the product as many ciphers as are at the right of both factors. 3. Multiply 2350 by 60. 141000. 4. Multiply 80300 by 450. 36135000. 5. Multiply 10240 by 3200. 32768000, 6. Multiply 9600 by 2400. 23040000. 7. Multiply 18001 by 26000. 468026000, 8. Multiply 8602 by 1030. 8860060. 9. Multiply 3007 by 9100. 27363700, 10. Multiply 80600 by 7002. 564361200. 11. Multiply 70302 by 80300. 5645250600. 12. Multiply 904000 by 10200. 9220800000, m ..gmm^ ^i^ 35. 1. If you divide 6 apples equally between 2 boys, how many will each boy have? Analysis. — It will require 2 apples to give each boy 1. Hence, each boy will have as many apples as 2 is contained times in (i, which are 3. How many times 2 in 6? Ans. 3. Why? Because 3 times 2 are 6. 2. If you divide 8 peaches equally between 2 boys, how many will each have? Ans. 4 peaches. Why? 3. How many times 2 in 10? Ans. 5. Why? The process by which the preceding examples arc solved is called Division. DEFINITIONS. 36. 1. Division is the process of finding how many times one number is contained in another. 2. The divisor is the number by which to divide ; the dividend is the number to be divided ; the quotient is the number denoting how many times the divisor is contained in the dividend. DIVISION OF SIMPLE NUMBERS. 51 Thus, 3 is contained in 12, 4 times; here, 3 is the divisor, 12 the dividend, and 4 the quotie7it. 3. Since 3 is contained in 12 four times, 4 times 3 are 12; that is, the divisor and quotient multiplied pro- duce the dividend. 4. Since 3 and 4 are factors of the product 12, the divisor and quotient correspond to the factors in multi- plication ; the dividend, to the product. Therefore, Bi- insion is the process of finding one of the factors of a product, when the other factor is known. 37. A boy has 8 cents : how many lemons can he buy, at 2 cents each? Analysis. — He can buy 4, because 4 lem- 8 cents, ons, at 2 cents each, will cost 8 cents. 1st lemon, 2 cents. The boy would give 2 cents for 1 lemon, Left, G cents, and then have 6 cents left. 2d lemon, 2 cents. After giving 2 cents for the 2d lemon, he Left, 4 cents, would have 4 cents left. 3d lemon, 2 cents. Then, giving 2 cents for the 3d, he would Left, 2 cents, have 2 cents left. 4th lemon, 2 cents. Lastly, after giving 2 cents for the 4th, he Left, cents, would have nothing left. The natural method of performing this operation is by subtraction; but, when it is known how many times 2 can be subtracted from 8, instead of subtracting 2 four times, say 2 in 8 four times, and 4 times 2 are 8. Therefore, Division may be termed a short method of making many subtractions of the same number. The divisor is the number subtracted ; the dividend, the number from which the subtraction has been made ; the quotient shows how many subtractions have been made. 52 KAY'S NEW PKACTICAL ARITHMETIC. 38. 1. Division is indicated in three ways: 1st. 3)12, wliieb means that 12 is to be divided by 3. 12 2d. ^ which means that 12 is to be divided by 3. D 3d. 12-^3, which means that 12 is to be divided by 3. 2. In using the first sign wiien the divisor does not exceed 12, draw a line under the dividend, and write the quotient beneath; if the divisor exceeds 12, draw a curved line on the right of the dividend, and place the quotient on the right of this. 3. The sign (-^) is read divided by. Examples. 2)8 4 15)45(3 45 ¥-« 21-^3=7. Division Table. 1-5-1= 1 2-^2= 1 8--3^ 1 4-4= 1 2--l= 2 4h^2= 2 6--8= 2 8-j-4= 2 3--l= 8 6 --2= 8 9--3= 3 12^4=^ 3 4--l== 4 8-^2= 4 12-:~8= 4 16-^4= 4 5-r-lr^ 5 10-1-2= 5 15 H- 3= b 20^4= 6 0^1^ G 12^2= 6 l8--8=: 6 24-^4= 6 1 7-^-1= 7 14 -r- 2= 7 21 - 3 = 7 28-4-4= 7 8--l=r 8 16-2= 8 24 -f- 3= 8 32 4-4= 8 9 H- 1 — 9 18-^2= 9 27-4-8= 9 86^4= 9 10--1==10 20 --2 = 10 80 -^ 3 = 10 40-4-4 = 10 11--1==11 22 H- 2 = 1 1 83 ---3 = 11 44-4-4 = 11 12 --1=12 24 -=-2 = 12 36-3 = 12 48 -r- 4 = 12 DIVISION OF SIMPLE NUMBERS. 53 1 5--5=r 1 6 — 6= 1 7 — 7= 1 8- -8= 1 10 --5= 2 12 — 6= 2 14-^7=, 2 16- -8= 2 15^5= 3 18 — 6= 3 21 — 7= 3 24- -8= 3 20 -f- 5== 4 24 — 6= 4 28 — 7= 4 32- -8= 4 25^-5= 5 30-^6= 5 35 — 7= 5 40- -8= 5 30 --5= 6 3G^6= 6 42^7= 6 48- -8= 6 S5-^6= 7 42 — 6= 7 49 — 7= 7 56- -8=7 40-v-5=:r 8 48 — 6= 8 56 — 7= 8 64- -8= 8 45--5=? 9 54 — 6= 9 63 — 7= 9 72- -8= 9 50 --5 = 10 60 — 6 = 10 70 -- 7 = 10 80- -8 = 10 55-5 = 11 66 — 6 = 11 77 — 7 = 11 88- -8 = 11 60 --5 = 12 72 — 6 = 12 84 — 7 = 12 96- -8 = 12 9--9= 1 10 — 10= 1 11—11= 1 12- -12= 1 18-9= 2 20- -10= 2 22 — 11= 2 24- -12= 2 27 — 9= 3 30- -10= 3 33_^11^ 3 36- -12= 3 36 — 9= 4 40- -10= 4 44_^11^ 4 48- -12= 4 45 _^ 9=: 5 50- -10= 5 55 — 11= 5 60- -12= 5 54_j_.9= 6 60- -10= 6 66 — 11= 6 72- -12= 6 63 — 9= 7 70- -10= 7 77 — 11= 7 84- -12= 7 72 — 9= 8 80- -10= 8 88 — 11= 8 96- -12= 8 81 — 9 = 9 90- -10= 9 99 — 11= 9 108- -12= 9 90-9 = 10 100- -10 = 10 110 — 11=10 120- -12 = 10 99 — 9 = 11 110- -10 = 11 121 — 11=11 132- -12=11 108 — 9 = 12 120- -10 = 12 132 — 11=12 144- -12 = 12 39. If 7 cents be divided as equally as possible among 3 boys, each boy would receive 2 cents, and there would be 1 cent left, or remaining undivided. The number left after dividing, is called the re- raainder. Rem. — 1. Since the remainder is a part of the dividend, it must be of the same denomination. If the dividend he dollars, the remainder will he dollars; if pounds, the remainder will he pounds. 54 RAY'IS NEW PRACTICAL ARITHMETIC. Rem. 2.— The remainder is always lens than the divisor; for, if it were equal to it, or greatei* the divisor would be contained at least once more in the dividend. Rem. 3. — If the dividend and divisor are simple numbers, the operation is called Division of Simple Numbers. Short Division. 40. When the division is performed mentallj^, and merely the result written, it is termed Short Division. Short Division is used when the divisor does not exceed 12. 1. How many times is 2 contained in 468? Here, the dividend is composed of three numbers; 4 hundreds, 6 tens, and 8 units; that is. of 400, 60, and 8. Divisor. Dividend. Quotient. Now, 2 in 400 is contained 200 times. 2 in 60 '' '' 30 times. 2 in 8 ^' " 4 times. Hence, 2 in 468 is contained 234 times. The same result can be obtained without actually separating the dividend into parts : Thus, 2 in 4 (hundreds), 2 times, which write Dividend, in hundreds' place; then, 2 in 6 (tens), 3 times. Divisor, 2)468 which write in tens' place; then, 2 in 8 (units), 4 Quotient, 2 34 times, which write in units' place. 2. How many times 3 in 693? 231. 3. How many times 4 in 848? 212. 4. How many times 2 in 4682? 2341. 5. How many times 4 in 8408? 2102. DIVISION OF SIMPLE NUMBERS. 55 6. How many times 3 in 3693G? 12312. 7. How many times 2 in 88468? 44234. 41. 1. How many times is 3 contained in 129? Solution. — Here, 3 is not contained in 1 ; but 3 is 3)129 contained in 12 (tens), 4 times, which write in tens' 4 3 place; 3 in 9 (units), 3 times, which write in units' place. 2. How many times is 3 contained in 735? Solution. — Here, 3 is contained in seven (hundreds), 2 times, and 1 hundred over; the 1 hundred, united 3)735 with the 3 tens, makes 13 tens, in which 3 is contained 245 4 times and 1 ten left; this 1 ten, united with the 5 units, makes 15 units, in which 3 is contained 5 times. 3. How many times is 3 contained in 618? SoLUTiON.-r-Here, 3 is contained in 6 (hundreds), 2 times; as the 1 in ten's place will not contain 3, a cipher 3)618 IS placed in ten's place; the 1 ten is then added to the 8 206 units, making 18 units, and the quotient figure 6 is placed in units' place. 4. How many times is 3 contained in 609? Here, the solution is the same as in the above ex- 3)609 ample; there being no tens, their order is indicated by 0. 20 3 5. How many times is 3 contained in 743? After dividing, there is 2 left, the division of which is merely indicated by placing the divisor under the 3)743 remainder; thus, f. The quotient is written thus, 24 7§ 247f; read, 247, and two divided by three; or, 247, with a re7nainder, two. 56 KAY'S NEW PRACTICAL ARITHMETIC. 6. How many times 3 in 462? 154. 7. How many times 5 in 1170? 234. 8. How many times 4 in 948? 237. Rule. — 1. Wiite the divisor at the left of the dividend, tvith a curved line between them^ and draw a line beneath the dividend. Begin at the left hand, divide successively each figure of the dividend by the divisor, and write the re- sult in the same order in the quotient. 2. If there is a remainder after dividing any figure, 'prefix it to the figure in the next lower order, and divide as before. 3. If the number in any order does not contain the divisor, place a cipher in the same order in the quotient, prefix the number to the figure in the next lower order, and divide as before. 4. If there is a remainder after dividing the last figure, place the divisor under it, and annex it to the quotient. Proof. — Multiply the quotient by the divisor, and add the remainder, if any, to the product: if the work is correct, the sum will be equal to the dividend. Rem. — This method of proof depends on the principle (36, 4] that a dividend is a product, of which the divisor and quotient are factors. 9. Divide 653 cents by 3. SOLUTION. TROOF. 217 Dividend. 3 Divisor, 3)653 (151= cents divided. Quotient, 2 1 7 § 2 = remainder. ¥53 = dividend. DIVISION OF SIMPLE NUMBEKS. 57 (10) 6)454212 (11) 7)874293 (12) 8)3756031 Ans. 75702 6 124899 7 469503J 8 Proof, 454212 874293 3756031 PARTS OF NUMBERS. Note. — When any number is divided into two equal parts, one of the parts is called one-half of that number. If divided into three equal parts, one of the parts is called 07ie- third; if into four equal parts, one-fourth; if into live equal parts, one-fifth; and so on. Hence, to find one-half of a number, divide by 2; to find one-third^ divide by 3; one-fourth, divide by 4; one-ffth, by 5, etc. 4326. 13541 If. 1687601 . 196855. 4311 7^. 1234753f 754065. 1003634. 1830023-V 54841. 3472834. 24. If oranges cost 3 cents each, how many can be bought for 894 cents? 298.. 25. If 4 bushels of apples cost 140 cents, how much is that a bushel? 35 ct. 26. If flour cost 84 a barrel, how many barrels can be bought for $812? 203. 13. Divide 8652 by 2. 14. -IHvide 406235 by 3. 15. Divide 675043 by 4. 16. Divide 984275 by 5. 17. Divide 258703 by 6. 18. Divide 8643275 by 7. 19. Divide 6032520 by 8. 20. Divide 9032706 by 9. 21. Divide 1830024 by 10. 22. Divide 603251 by 11. 23. Divide 41674008 by 12. 58 liAY'S NEW FKAGTICAL ARITHMETIC. 27. A carpenter receives $423 for 9 months' work : liow much is that a month? $47. 28. There are 12 months in 1 year: how many years are there in 540 months? 45. 29. There are 4 quarts in 1 gallon : how many gallons are there in 321276 quarts? 80319. 30. At $S a barrel, how many barrels of flour can be bought for $1736? 217. 31. There are 7 days in one wxek : how many weeks are there in 734566 days? " . 104938. 32. A number has been multij^lied by 11, and the i)ro- duct is 495 : what is the number ? 45. 33. The product of two numbers is 3582 : one of the numbers is 9 : what is the other ? 398. 34. Find one-half of 56. 28. 35. Find one-half of 3725. ' 1862^. 36. Find one-third of 147. 49. 37. Find one-fourth of 500. 125. 38. Find one-fifth of 1945. 389. 39. Find one-sixth of 4476. 746. 40. Find one-seventh of 2513. .^ 359. 41. Find one-eighth of 5992. * 749. 42. Find one-ninth of 8793. 977. 43. Find one-tenth of 1090. r 109. 44. Find one-eleventh of 4125. 375. 45. Find one-twelfth of 5556. 463. 46. I divided 144 apples equally among 4 boys; the eldest boy gave one-third of his share to his sister : what number did the sister receive? 12. 47. James found 195 cents, and gave to Daniel one- fifth of them: Daniel gave one-third of his share to his sister: how many cents did she receive? 13. 48. One-eleventh of 275 is how much greater than one-eighth of 192? 1. DIVISION OF SIMPLE NUMBERS. 59 Long Division. 42, When the entire work of the division is written down, it is termed Long Division. Long Division is commonly used when the divisor exceeds 12. 1. Divide 3465 dollars equally among 15 men. Solution. — Fifteen is not contained in 3 (thousands) ; therefore, there will be no thou- 15)3465(231 sands in the quotient. Take 34 (himdreds) as 3 hund. Si partial dividend; 15 is contained in 34, 2 4 6 tens, times; that ig, 15 men have 200 dollars each, 4 5 which requires in all 15 X 2 = 30 hundreds of 15 units, dollars. 1 5 Subtract 30 hundreds from 34 hundreds, and 4 hundreds remain; to which bring down the 6 tens, and you have 46 (tens) for a second partial dividend. 46 contains 15, 3 times; that is, each man has 30 dollars more, and all require 15 X ^ =^ 45 tens of dollars. Subtract 45, and bring down the 5 units, which gives 15 (units) for a third partial dividend; in this the divisor is contained once, giving to each man 1 dollar more. Hence, each man receives 2 hundred dollars, 3 ten dollars, and 1 dollar; that is, 231 dollars. By this process, the dividend is sep- arated into parts, each part contain- Divisor. Parts. Quotients ing the divisor a certain number of 15 3000 200 times. 4 5 3 The first part, 30 hundreds, con- 15 1 tains the divisor 2 times; the second 3 4 65 231 part, 45 tens, contains it 3 times; the third part, 15 units, contains it 1 time. The several parts together equal the given dividend, and the several partial quotients make up the entire quotient. 60 RAY'S NEW PRACTICAL ARITHMETIC. 2. In 147095 days, how many years, each of 365 days? Solution.— Taking 147 (thou- 865 ) 1 4 7005 ( 403 years, sands) for the first partial dividend, 14 60 we find it will not contain the di- 109 5 visor; hence use four figures. 109 5 Again, after multiplying and sub- tracting, as in the preceding example, and bringing down the 9 tens, the partial dividend, 109 (tens), will not contain the divisor; hence, write a cipher (no tens) in the quotient, and bring down the 5 units; the last partial dividend is 1095 (units), which contains the divisor three times. 3. Divide 4056 by 13. 312. Rule. — 1. Flace the divisor on the left of the dividend^ draw a curved line between them, and another on the right of the dividend. 2. Find, how many times the divisor is contained in the fewest left hand figures of the dividend that will contain the divisor^ and place this number in the quotient at the right. 3. Multiply the divisor by this quotient figure; place the ^product under that part of the dividend from which it was obtained. 4. Subtract this product from the figures above it ; to the remainder bring down the next figure of the dividend, and divide as before, until all the figures of the dividend are brought down. ' 5. If at any time, after bringing down a figure, the num- ber thus formed is too smcdl to contain the divisor, place a cipher in the quotient, and bring down another figure, after which divide as before. Proof. — Same as in Short Division. DIVISION OF SIMPLE NUMBERS. 61 Rem. — 1. The product must never be gr eater than the partial dividend from which it is to be subtracted; when so, the quotient figure is too large, and must be diminished. Rem. — 2. After subtracting, the remainder must always be less than the divisor; when the remainder is not less than the divisor, the last quotient figure is too S7nall, and must be increased. Rem. — 3. The order of each quotient figure is the same as the lowest order in the partial dividend from which it was obtained. 4. Divide 78994 by 319. SOLUTION PROOF. 3 1 9 ) 7 8 9 9 4 ( 247Hi 247 Quotient 638 319 Divisor. 1519 2223 12 76 247 2434 741 2233 78793 201 Rem. Add 201 78994 Remainder. = the Dividend. 5. Divide 11577 by 14. -.«26|f. 6. Divide 48690 by 15. 3246. 7. Divide 1110960 by 23. 48302^1. 8. Divide 122878 by 67. 1834. 9. Divide 12412 by 53. 234if. 10. Divide 146304 by 72. 2032. 11. Divide 47100 by 54. 872i|. 12. Divide 71104 by 88. 808. 13. Divide 43956 by 66. 666. 14. Divide 121900 by 99. 1231ff 15. Divide 25312 by 112. 226. 16. Divide 381600 by 123. 3102A\. 17. Divide 105672 by 204. 518. 18. Divide 600000 by 1234. 486^3V 19. Divide 1234567 by 4321. 285itff 20. Divide 50964242 by 7819. 6518. 62 KAY'S NEW PRACTICAL ARITHMETIC. 21. Divide 48905952 by 9876. 4952. 22. Divide 4049160 by 12345. 328. 23. Divide 552160000 by 973. 567482^VV 24. At $15 an acre, how many acres of land can be bought for $3465? . 231 acres. 25. If a man travel 26 miles a day, in how many days will he travel 364 miles? 14 days. 26. If $1083 be divided equally among 19 men, how many dollars will each have? $57. 27. A man raised 9523 bushels of corn on 107 acres: how much was that on one acre? 89 bu. 28. In 1 hogshead there are 63 gallons : how many hogsheads in 14868 gallons? 236. 29. The President receives $50000 a year (365 days): how much is that a day? $136 and $360 over. 30. The yearly income from a railroad is $379600 : how much is that a day? (365 da. =:= 1 yr.) $1040. 31. The product of two numbers is 6571435 ; one of the factors is 1235: what is the other? 5321. 32. Divide one million two hundred and forty-seven thousand four hundred by 405. 3080. 33. Divide 10 million four hundred and one thousand by one thousand and six. 10338yyg?_ 34. A colony of 684 men bought a tract of land, con- taining 109440 acres : if equally divided, to how many acres was each man entitled? 160 acres. 35. A farmer raised 8288 bushels of corn, averaging 56 bushels to the acre: how many acres did ho plant? 148 acres. 36. The capital of a joint-stock company is $262275, and is divided into 269 shares : what is the value of each share? $975. 37. The earth, at the equator, is about 24899 miles* in DIVISION OF SIMPLE NUMBERS. 63 circumference, and turns on its axis once in 24 hours : how many miles an hour does it turn? 1037^^. 38. A railroad 238 miles long, cost $3731840: what was the cost per mile? $15680. 39. A fort is 27048 feet distant from a city; the flash of a cannon w^as seen 24 seconds before the sound was heard : how many feet a second did the sound travel ? 1127 feet. 40. Light travels at the rate of 11520000 miles a min- ute : how many minutes does it require for the light of the sun to reach the earth, the sun being 92160000 miles distant? 8 minutes. Examples for TIkvtew. 41. Subtract 86247 from 94231 and divide the re- mainder by 16. 499. 42. Divide the sum of 46712 and 6848 by 104. 515. 43. Divide the product of 497 X 583 by 71. 4081. 44. To the difference between 2832 and 987 add 678, and divide the sum by 87. 29. 45. Multipl}^ the diflPerence between 4896 and 2384 by 49, and divide the product by 112. 1099. 46. Multiply the sum of 228 + 786 by 95, and divide the product by 114. 845. 47. Multiply the sum of 478 and 296 by their differ- ence, and divide the product by 387. 364. 48. A horse-dealer received S7560 for horses ; he sold a part of them for S3885 ; if he sold the rest for $175 apiece, how many horses did he sell the second time? 21 horses. 49. A farmer expended at one time $7350 for land, and at another, $4655, paying $49 an acre each time : how many acres did he buy in both purchases? 245 acr^. 64 KAY'S NEW PRACTICAL ARITHMETIC. 50. A refiner bought 58 hogsheads of sugar, at $77 a hogshead, and afterward sold them for $5742: how much did he gain on each hogshead? $22. 51. A man bought 240 acres of land, at $26 an acre, giving in payment a house valued at $2820, and horses valued at $180 apiece: how many horses did he give? 19 horses. 52. A speculator bought 25 acres of land for $10625, and after dividing it into 125 village lots, sold each lot for $250: how^ much did he gain on the whole? On each acre? On each lot? $20625. $825. $165. CONTRACTIONS IN DIVISION. CASE I. 43. When the divisor can be separated into factors. 1. A man paid $255 for 15 acres of land: how much was that per acre? Solution. — 15 acres are 3 Dollars, times 5 acres; dividing $255 by 3)255 = the value of 15 acres. 3 gives $85, the value of 5 6)85 — the value of 6 acres, acres; dividing $85 by 6 gives 1 7 = the value of 1 acre. $17, the value of 1 acre. The solution shows that instead of dividing by the number 15, whose factors are 3 and 5, we may first divide by one factor, then divide the quotient thus obtained by the other factor. 2, Find the quotient of 37, divided by 14. Solution. — Dividing by 2, the quotient is 18 twos and 1 unit remaining. Divid- 2)37 ing by 7, the quotient is 2, with a re- 7)18 and 1 over, mainder of 4 tioos; the whole remainder ^^ and 4 twos left. then is 4 ttcos plus 1, or 9. DIVISION OF simplp: numbers. 65 Rule. — 1. Divide the dividend by one of the factors of the divisor; then divide the quotient thus obtained by the other factor. 2. Multiply the last remainder by the first divisor ; to the product add the first remainder ; the amount ivill be the true remainder. Rem. — When the divisor can be resolved into more than two factors, you may divide by them successively. The true remainder will be found by multiplying each remainder by all the preceding divisors, except that which produced it. To their sum add the re- mainder from first divisor. 3. Divide 2583 by 63. 4. Divide 6976 by 32. 5. Divide 2744 by 28. 6. Divide 6145 by 42. 7. Divide 19008 by 132. 8. Divide 7840 by 64. 9. Divide 14771 by 72. 10. Divide 10206 by 81. 11. Divide 81344 by 121. 12. Divide 98272 by 108. CASE II. 44. To divide bj^ 1 with ciphers annexed; as 10, 100, iOOO, etc. To multiply 6 by 10, annex one cipher, thus, 60. On the principle that division is the reverse of multiplica- tion, to divide 60 by 10, cut off a cipher. Had the dividend been 65, the 5 might have been separated in like manner as the cipher; 6 being the quotient, 5 the remainder. The sam-; will apply when the divisor is 100, 1000, etc. Prac- 5 (t)3 = = 7X9) 41. (32 = = 4X8) 218. (28 = = 7X4) 98. (42 = = 6X7) 146if. 144. 122ff. 205||. 126. 672/A. 66 KAY'S NEW PRACTICAL ARITHMETIC. Rule.— Cut off as viany figures from the right of the dividend as there are ciphers in the divisor ; the figures cut off will be the remainder, the other figures, the quotient. 1. Divide 34872 by 100. 1|00 Ol'KR )348| ATIO.V. 72 Quo. 72 Rem 348 2^ 3. 4. 5. 6. Divide Divide Divide Divide Divide 2682 by 10. 4700 by 100. 37201 by 100. 46250 by 100. 18003 by 1000. 268ft. 47. 372t^. 462AV CASE ITT. 45. To divide when there are ciphers on the right of the divisor, or on the right of the divisor and dividend. 1. Divide 4072 by 800. Solution. — Regard 800 as com- posed of the factors 100 and 8, and divide as in the margin. In dividing by 800, separate the two right hand figures for the re- mainder, then divide by 8. 2. Divide 77939 by 2400. OPERATION. I|00)40i72 8^ 40 5 Quo... 72 Rem. 8100 )40 172 5 Quo... 72 Rem. Solution. — Since 2400 equals 24 X 100, cut off the two right hand figures, the same as dividing by 100; then divide by 24. Dividing by 100, the remainder is 39; dividing by 24, the remainder is 11. To find the true remainder, multiply 11 by 100, and add 39 to the product (Art. 43, Rule); this is the same as annexing the figures cut oflf, to the last remainder. operation. 24100)7 79139(32 1^^1 72 48 DIVISION OF SIMPLE NUMBEKS. 67 3. Divide 62700 by 2500. Solution. — The example above. same as for the OPERATION. 25iOO)627|00(25//o°^ 50 T2T 125 2 Rule. — 1. Cut off the ciphers at the light of the divisor^ and as many figures from the right of the dividend. 2. Divide the remaining figures in the dividend by the remaining figures in the divisor. 3. Annex the figures cut off to the remainder^ which gives the triie remainder. 4. Divide 73005 by 4000. 5. Divide 36001 by 9000. 6. Divide 1078000 by 11000. 7. Divide 40167 by 180. 8. Divide 907237 by 2100. 9. Divide 364006 by 6400. 10. Divide ' 76546037 by 250000. 11. Divide 43563754 by 63400. 1844>4>A 98. 223jV^. 432^H^. 306/A^V\\- 687^Wo. GENERAL PRINCIPLES OF DIVISION. 46. The value of the quotient depends on the rela- tive values of divisor and dividend. These may be changed by Multiplication and by Division, thus : 1st. The dividend may be multiplied, or the divisor divided. 2d. The dividend may be divided, or the divisor multi- plied. 3d. Both dividend and divisor may be multiplied, or both divided, at the same time. 68 RAY'S NEAV PRACTICAL ARITHMETIC. Illustrations. Let 24 be a dividend, and 6 the divisor; the quotient is 4. 24-^6 = 4. If the dividend, 24, be multiplied by 2, the quotient will be multi- plied by 2; for, 24X2 = 48; and 48-v-6=:8, which is the former quotient, 4, multiplied by 2. Now, if the divisor, 6, be divided by 2, the quotient will be multi- plied by 2; for, 6 --2 = 3; and 24-f-3 = 8, which is the former quotient, 4, multiplied by 2. Principle I. — If the dividend be multiplied, or the divisor be divided, the quotient will be multiplied. 47. Take the same example, 24 -f- 6 = 4. If the dividend, 24, be divided by 2, the quotient will be divided by 2; for 24 -=-2 = 12; and 12-h6 = 2, which is the former quo- tient, 4, divided by 2. And, if the divisor, 6, be nudtiplied by 2, the quotient will be divided by 2; for, 6 X 2 = 12; and 24 h- 12 = 2, which is the former quotient, 4, divided by 2. Prin. II. — If the dividend be divided, or the divisor be mxdtiplied,, the quotient will be divided. 48. Take the same example, 24 -f- 6 = 4. If the dividend, 24, and divisor, 6, be multiplied by 2, the quotient will not be changed; for, 24 X 2 = 48 ; and 6X2 = 12; 48^-12 = 4; the former quotient, 4, unchanged. And if the dividend, 24, and divisor, 6, be divided, by 2, the quo- tient will not be changed; for, 24 -f- 2 = 12; and 6-f-2 = 3; 12^3 -=4; the former quotient, 4, unchanged. Prin. III.^ — If both dividend and divisor be jmdtiplied or divided by the same number, the quotient will not be changed. DIVISION OF SIMPLE NUMBERS. 69 Promiscuous Examples. 49. 1. In 4 bags are $500; in the first, $96; in the second, $120 ; in the third, $55 : what sum in the 4th bag? $229. 2. Four men paid $1265 for land : the first paid $243 ; the second $61 more than the first; the third $79 less than the second: how much did the fourth man pay? $493. 3. 1 have five apple trees: the first bears 157 apples; the second, 264; the third, 305; the fourth, 97; the fifth, 123 : I sell 428, and 186 are stolen : how many apples are left? 332. 4. In an army of 57068 men, 9503 are killed; 586 join the enemy; 4794 are prisoners; 1234 die of wounds; 850 are drowned: how man}- return? 40101. 5. On the first of the year a speculator is worth $12307 : during the year he gains $8706 ; in January he sj^ends $237 ; in February, $301 ; in each of the remain- ing ten months he spends $538: how much had he at the end of the year? $15095. 6. The Bible has 31173 verses: in how many days can I read it, by reading 86 verses a day? 362|^. 7. I bought 28 horses for $1400: 3 died; for how much each must I sell the rest to incur no loss? $56. 8. How many times can I fill a 15 gallon cask, from 5 hogsheads of 63 gallons each? 21 times. 9. A certain dividend is 73900; the quotient 214; the remainder, 70: what is the divisor? 345. 10. Multiply the sum of 148 and 56 by their difier- ence ; divide the product by 23. 816. 11. How much woolen cloth, at $6 a yard, will it take to pay for 8 horses at $60 each, and 14 cows at $45 each? 185 yd. 70 RAY'S NEW PRACTICAL ARITHMETIC. 12. Two men paid $6000 for a farm: one man took 70 acres at S30 an acre, tlie other the remainder at $25 an acre : how many acres in all ? 226. 13. My income is $1800 a year. If I spend $360 a year for provisions, $300 for rent, $150 for clothing, $100 for books, and $90 for incidentals, in how many years can I save $10400? 13. 14. A man bought 40 acres of ground at $15 an acre, and 80 acres at $25 an acre. He sold 90 acres for $4500, and the remainder at $60 an acre : for how much did he sell the whole land? How much did he gain? $6300. $3700. 15. A merchant bought 275 3^ards of cloth at $4 a yard; he sold 250 yards at $5 a yard, and the re- mainder at $6 a yard : how much did he gain? $300. 16. A broker buys 125 shares of stock for $85 a share, and 75 shares at $115 a share. He invests it all in other stock at $175 a share : how many shares does he get by the last purchase? 110. 17. A farmer sends to a dealer 20 horses and 15 mules to be sold. The dealer sells the horses for $125 each, and the mules for $150 each, charging $95 for selling. The farmer then buys 50 head of cattle at $45 each, with part of the money, and deposits the remainder in bank: how much does he deposit in bank? $2405. To Teachers. — While placing Fractions immediately after Simple Whole Numbers is philosophical, and appropriate in a higher arithmetic for advanced pupils, the experience of the author convinces him that, in a book for young learners, Compound Num- bers should be introduced before, instead of after, Fractions — for the following reasons: 1st. The operations of Addition, Subtraction, Multiplication, and Division of compound numbers are analogous to the same operations in simple numbe/s, and serve to illustrate the principles of the Fun- damental Rules. 2d. The subject of Fractions is important and difficult. Before studying it, most pupils require more mental discipline than is furnished by the elementary rules; this is acquired by the study of Compound Numbers. 3d. The general principles involved in their study, do not require a knowledge of Fractions. The examples involving fractions are few, and are introduced, as they should be, with other exercises in that subject. Teachers, who prefer it, can direct their pupils to defer Compound Numbers until they have studied Fractions as far as page 159. DEFINITIONS. 50. A compound number is made up of two or more concrete numbers of different denominations ; as, 3 pecks 7 quarts 1 pint. Rem. 1. — The different denominations of a compound number must belong to the same table; thus, in the example given, the (71) 72 RAY\S NEW PRACTICAL ARITHMETIC. pecks may be reduced to quarts or pints, and the pints and quarts are j)arU of u peck. 3 pecks 7 dollars would not be a compound number. Rem. 2. — Compound numbers resemble simple numbers in tho following particulars: the denominations of compound numbers cor- respond to the orders of simple numbers, and a certain number of units of a lower denomination make one unit of the next higher denomination. Rem. -5. — Most compound numbers differ from simple numbers in this; ten units of each lower denomination do not uniformly make one unit of the next higher denomination. Rem. 4. — In United States Money and the Metric System of Weights and Measures, however, ten units of a lower denomination do make one imit of the next higher denomination. 51. 1. The operations with compound numbers are Bediictlon, Addition, Subtraction, Multiplication, and Di- vision. 2. Reduction is the process of changing the denom- ination of a number without aUering its value. Thus, 5 yards may be changed to feet; for, in 1 yard there are 3 feet; then, in 5 yards there are 5 times 3 feet, which are 15 feet. 3. Eeduction takes place in two ways: 1st. From a higher denomination to a lower. 2d. From a lower de- nomination to a higher. UNITED STATES MONEY. 5"2. United States money is the money of the United States of America. Table. 10 mills, m., make 1 cent, marked ct. 10 cents " 1 dime, " d. 10 dimes ^' 1 dollar, " %. 10 dollars " 1 eagle, " E. UNITED STATES MONEY. 73 Rem. 1. — United States money was established, by act of Con- gress, in 1786. The first money coined, by the authority of the United States, was in 1793. The coins first made were copper cents. In 1794 silver dollars were made. Gold eagles were made in 1795; gold dollars, in 1849. Gold and silver are now both legally standard. The trade dollar was minted for Asiatic commerce. Rem. 2. — The coins of the United States are classed as bronze, nickel, silver, and gold. The name, value, composition, and weight of each coin are shown in the following Table. r 1 COIN. VALUE. COMPOSITION. WEIGHT. 48 grains Troy. BRONZK. 1 One cent. ! 1 cent. 95 parts copper, 5 parts tin & zinc. NICKEL. 3-cent piece. 5-cent piece. 3 cents. 5 cents. 75 parts copper, 25 parts nickel. 75 " *' 25 •' 30 grains Troy. 73.1G " SILVER. Dime. Quarter dollar. Half dollar. Dollar. 10 cents. 25 cents. 50 cents. 100 cents. 90 parts silver, 10 parts copper. 90 " " 10 " 90 " " 10 " 90 '* " 10 *' 2.5 grams. 6.25 " 12.5 4123^ grains Troy. GOLD. Dollar. Quarter eagle. Three dollar. Half eagle. Eagle. Double eagle. 100 cents. 23^ dollars. 3 dollars. 5 dollars. 10 dollars. 20 dollars. 90 parts gold. 10 parts copper. 90 " " 10 " 90 " " 10 " 90 •' " 10 " 90 " " 10 " 90 " " 10 " 25.8 grains Troy. 64.5 77.4 129 258 516 Rem. 3. — a deviation in weight of | a grain to each piece, is allowed by law in the coinage of Double Eagles and Eagles; of \ of a grain in Half Eagles and the other gold pieces: of 1^ grains in all 74 KAY'S NEW PRACTICAL ARITHMETIC. silver pieces; of 3 grains in the iivc-cent piece; and of 2 grains in the three-cent piece and one cent. Rem. 4. — The mill is not coined. It is used only in calculations. 53. 1. A sum of money is expressed as dollars and cents, and, when written in figures, is always preceded by the dollar sign ($). Rem. — Calculations are sometimes carried out to mills, but, in bus- iness transactions, the final'result is always taken to the nearest cent. 2. A period (.), called the decimal point, is used to separate the dollars and cents. 3. Eagles are read as tens of dollars, and dimes are read as tens of cents. Thus, $24.56 is read 24 dollars 56 cents; not 2 eagles 4 dollars 5 dimes 6 cents. $16,375 is read 16 dollars 37 cents 5 mills. 4. Hence, the figures to the left of the decimal point ex- press a, number of dollars; the two figures to the right of the decimal point, a number of cents; and the third, figure to the right, mills. Rem. — If the number of cents is less than 10, a cipher must be put in the tens' place. EXAMPT^ES TO BE ^VRITTEX. 1. Twelve dollars seventeen cents eight mills. 2. Six dollars six cents six mills. ^ 3. Seven dollars seven , mills. 4. Forty dollars fifty -three cents five mills. 5. Two dollars three cents. 6. Twenty dollars two cents two mills. 7. One hundred dollars ten cents. 8. Two hundred dollars two cents. 9. Four hundred dollars one cent eicfht mills. UNITED STATES MONEY. 75 EXAMPLES TO BE READ. $18,625 $ 70.015 ^6.12 $ 29.00 320.324 $100.28 $3.06 $100.03 $79.05 $150.05 $4.31 $ 20.05 $46.00 $100.00 $5.43 $ 40.125 REDUCTION OF IT. S. MONEY. 54. 1. As there are 10 mills in 1 cent, in any num- ber of cents there are 10 times as many mills as cents Therefore to reduce cents to mills — Rule. — Multiply the number of cents by ten ; that is^ annex one cipher. 2. Conversely, to reduce mills to cents — Rule. — Divide the iiumber of mills by ten ; thcit is^ cut off one figure from the right. 3. As there are 10 cents in 1 dime and 10 dimes in 1 dollar, there are 10 X 10 = 100 cents in 1 dollar; then, in any number of dollars there are 100 times as many cents as dollars. Therefore, to reduce dollars to cents — Rule. — Multiply the number of dollars by one hundred; that is, annex two ciphers. 4. Conversely, to reduce cents to dollars-^ Rule. — Divide the number of cents by 07ie hundred ; that is, cut off tiro figures from the right. 5. As there are 10 mills in 1 cent and 100 cents in 1 dollar, there are 100 X lOirrrlOOO mills in 1 dollar; then, in any number of dollars there are 1000 times as many mills as dollars. Therefore, to reduce dollars to mitls — 76 RAY'S NEW PRACTICAL ARITHMETIC. Rule. — Multiply the number of dollars by one thousand; that is, annex three ciphers. 6. Conversely, to reduce mills to dollars. Rule. — Divide the number of mills by one thousand : that is, cut off three figures from the right. 55. The reduction of mills or cents to dollars may be made simply with the decimal point. Thus, 1st. If the sum is mills. Rule. — Put the decimal point between the third and fourth figures from the right. 2d. If the sum is cents. Rule. — Put the decimal point betiveen the second and third figures from the right. 1. Eeduce 17 ct. to mills. 170 m 2. Ileduce 28 ct. to mills. 280 m 3. Keduce 43 ct. and 6 m. to mills. 436 m 4. Reduce 70 ct. and 6 m. to mills. 706 m. 5. Reduce 106 m. to cents. 10 ct. 6 m. 6. Reduce 490 mills to cents. 49 ct. 7. Reduce 9 dollars to cents. 900 ct. 8. Reduce 14 dollars to cents. 1400 ct. 9. Reduce 104 dollars to cents. 10400 ct. 10. Reduce $60 and 13 ct. to cents. 6013 ct. 11. Reduce $40 and 5 ct. to cents. 4005 ct. 12. Reduce 375 ct. to dollars. 83.75 13. Reduce 9004 ct. to dollars. S90.04 14. Reduce 4 dollars to mills. 4000 m. 15. Reduce S14 and'*2 ct. to mills. 14020 m. 16. Reduce 2465 mills to dollars. S2.46 5. 17. Reduce 3007 mills to dollars. $3.00 7. 18. Reduce 3187 cents to dollars. $31.87. 19. Reduce 10375 mills to dollars. $10,375. UNITED STATES MONEY. 77 ADDITION OF U. S. MONEY. 56. 1. Add together 4 dollars 12 cents 5 mills; 7 dollars 6 cents 2 mills; 20 dollars 43 cents; 10 dollars 5 mills; 16 dollars 87 cents 5 mills. Rule. — 1. Write the numbers and add as orERATiox. in simple numbers. $• ct. m. 2. Place the decimal point in the sum under 7 * o fi 9 the decimal points above. 2 ! 4 3 10.005 Proof. — The same as in Addition of Sim- 16.875 pie Numbers. $58749T 2. What is the sum of 17 dollars 15 cents ; 23 dollars 43 cents; 7 dollars 19 cents; 8 dollars 37 cents; and 12 dollars 31 cents? S68.45. 3. Add 18 dollars 4 cents 1 mill; 16 dollars 31 cents 7 mills; 100 dollars 50 cents 3 mills; and 87 dollars 33 cents 8 mills. S222.199. 4. William had the following bills for collection: $43.75; $29.18; $17.63; $268.95; and $718.07: how much was to be collected? $1077.58. 5. Bought a gig for $200 ; a watch for $43.87 ; a suit of clothes for $56.93 ; a hat for $8.50 ; and a whip for $2.31: w^hat was the amount? $311.61. 6. A person has due him, five hundred and four dol- lars six cents; $420.19; one hundred and ^yq dollars fifty cents ; $304 ; $888.47 : what is the whole amount due him? $2222.22. 7. Add five dollars seven cents; thirty dollars twenty cents three mills ; one hundred dollars five mills ; sixty dollars two cents; seven hundred dollars one cent one mill; $1000.10; forty dollars four mills; and $64.58 7. $2000. 78 HAY'S NEW PRACTICAL ARITHMETIC. SUBTRACTION OF U. S. MONEY. 57. 1 From one hundred dollars five cents threo mills, tuke eighty dollars twenty cents and seven milla Rule. — 1. Write the numbers and subtract as in Simple Numbers. operatiox. 2. Flace the decimal point in the remainder $. et. m. under the decimal points above. 100.053 80.207 Proof. — The same as in Subtraction of $19,846 Simple Numbers. 2. From $29,342 take 817.265. $12,077. 3. From $46.28 take $17.75. $28.53. 4. From $20.05 take $5.50. $14.55. 5. From $3, take 3 et. $2.97. 6. From $10, take 1 mill. $9,999. 7. From $50, take 50 ct. 5 mills. $49,495. 8. From one thousand dollars, take one dollar one cent and one mill. $998,989. 9. B owes 1000 dollars 43 cents; if he pay nine hun- dred dollars sixty-eight cents, how much =^^%e still owe? "" $99.75. MULTIPLICATION OF U. S. MONEY. 58. 1. What will 13 cows cost, at 47 dollars 12 cents 5 mills each? Rule. — 1. Multiply as in Simple Numbers. operation. 2. Put the decimal point in the same place $47,125 in the product^ as it is in the multiplicand. 1_3 141375 47125 Proof. — The same as in Multiplication of Simple Numbers. $612,6 2 5 UNITED STATES MONEY. 79 2. Multiply S7.835 by 8. $62.68 3. Multiply $12, 9 et. 3 m. by 9. $108,837, 4. Multiply $23, 1 ct. 8 m. by 16. $368,288 5. Multiply $35, 14 ct. by 53. $1862.42, 6. Multiply $125, 2 ct. by 62. $7751.24. 7. Multiply $40, 4 ct. by 102. $4084.08, 8. Multiply 12 ct. 5 m. by 17. $2,125, 9. Multiply $3.28 by 38. $124.64 10. What cost 338 barrels of cider, at 1 dollar 6 cents a barrel? $358.2B. 11. Sold 38 cords of wood, at 5 dollars 75 cents a cord: to what did it amount? $218.50. 12. At 7 ct. a pound, what cost 465 pounds of sugar ^ Note. — Instead of multiplying 7 cents l)y 465, mul- operation tiply 465 by 7, which gives the same product, Art. 30. 4 6 1 But, to place the decimal point, remember that 7 cents .0 ? is the true multiplicand. $ 'S^2.6 1 13. What cost 89 yards of sheeting, at 34 ct. a yard? $30.26. 14. What will 24 yards of cloth cost, at $5.67 a yard? $136.08. 15. I have 169 sheep, valued at $2.69 each : what is the value of the whole? $454.61. 16. If I sell 691 bushels of wheat, at $1.25 a bushel, what will it amount to? $863.75. 17. I sold 73 hogsheads of molasses, of 63 gallons each, at 55 ct. a gallon: what is the sum? $2529.45. 18. What cost 4 barrels of sugar, of 281 pounds each, at 6 cents 5 mills a pound? $73.06. 19. Bought 35 bolts of tape, of 10 yards each, at 1 cent a yard : what did it cost ? $3.50. 20. If I earn 13 ct. an hour, and work 11 hours a day, how much will I earn in 312 days? $446.16. 80 KAY'S NEW PEACTICAL AKITHMETIC. 21. I sold 18 bags of wheat, of 3 bushels each, at $1.25 a bushel: what is the amount? $67.50. 22. What cost 150 acres of land, at 10 dollars 1 mill per acre? 81500.15. 23. What cost 17 bags of coffee, of 51 pounds each, at 24 cents 7 mills per pound? $214.14 9. DIVISION OF U. S. MONEY. 59. Case I. — To find how many times one sum of money is contained in another. 1. How much cloth, at 7 cents a yard, will $1.75 buy? OPEllATIQN. Solution.— Aij many yards as 7 cents is contained 7)175 times in 175 cents, which are 25. 2 5 Rule. — 1. Beduce both sums of money to the same denom- ination. 2. Divide as in Simple Numbers. 2. How much rice, at 9 cents a pound, can be bought for 72 cents? 8 lb. 3. How many towels, at 37 cents and 5 mills apiece, can be bought for $6? 16. 4. How many yards of calico, at 8 cents a 3^ard, can be bought for $2.80? 35 yd. 5. How many yards of ribbon, at 25 cents a yard, can be purchased for $3? 12 yd. 6. At $8.05 a barrel, how many barrels of flour will $161 purchase? 20 bl. 7. At 7 cents 5 mills each, how many oranges can be bought for $1.20? " 16. 8. At $1,125 per bushel, how many bushels of wheat can be purchased for $234? 208 bu. UNITED STATES MONEY. 81 Case II. — To divide a sum of money into a given number of equal parts. 1, A man worked 3 days for $3.75, what were his daily wages? OPERATION Solution. — His daily wages were $3.75 -^ 3 r=: $1.25. 3 ) 3.7 5 $T25 2. A farmer sold 6 bushels of wheat for $d : how much a bushel did he get? Solution. — He got for eadi bushel $9—6. $9 di- operation. vided by 6 gives a quotient $1, with a remainder 6)9.00 $3 = 300 cents. 300 cents divided by 6 gives a quo- $1.50 tient 50 cents. Rule. — 1. Divide as in Simple Nvmbers, 2. Put the decimal point in the same j^lace in the quotient as it is in, the dividend, Kem. 1.— If the dividend is dollars, and the division not exact, annex two ciphers after the decimal point for cents; and, if nec- essary, a third cipher for mills. Rem. 2. — Should there be a remainder after obtaining tlie mills, it may be indicated by the sign -f placed after the quotient. 3. Divide 65 dollars equally among 8 persons. S8.125. 4. A farmer received $29.61 for 23 bushels of wheat: how much was that per bushel? $1,287 H-. 5. If 4 acres of land cost $92.25, how much is that an acre? $23,062+. 6. Make an equal division of $57.50 among 8 persons. $7,187+. 7. A man received $25.76 for 16 days' work: how much was that a day? $1,61. 82 RAY'S NEW PRACTICAL ARITHMETIC. 8. I bought 755 bushels of apples for $328,425: what did they cost a bushel? $0,435. 9. My salary is $800 a year : how much is that a day, there being 313 working days in the year? $2.555-|-. 10. Divide ten thousand dollars equally among 133 men: what is each man's share? $75.187 -f. 11. A man purchased a farm of 154 acres, for two thousand seven hundred and five dollars and 1 cent: what did it cost per acre? $17,565. 12. I sold 15 kegs of butter, of 25 pounds each, for $60: how much Avas that a pound? 16 ct. 13. I bought 8 barrels of sugar, of 235 pounds each, for $122.20: what did 1 pound cost? $0,065. Promiscuous Examples. 60. 1. I oAve A $47.50; B, $38.45; C, $15.47; D, $19.43: what sum do I owe? $120.85. 2. A owes $35.25 ; B, $23.75 ; C, as much as A and B, and $1 more: what is the amount? $119. il A paid me $18.38; B, $81.62; C, twice as much as A and B: how much did I receive? $300. 4. I went to market w4th $5 ; I spent for butter 75 cents, for eggs 35 cents, for vegetables 50 cents, for flour $1.50: how much money was left? $1.90. 5. A lady had $20; she bought a dress for $8.10, shoes for $5.65, eight yards of delaine at 25 cents a yard, and a shawl for $4: what sum was left? 25 ct. 6. I get $50 a month, and spend $30.50 of it: how much will I have left in 6 months? $117. 7. A farmer sold his marketing for $21.75 : he paid for sugar $3.85, for tea $1.25, for coffee $2.50, for spices SI .50: how much had he left? $12.65. UNITED STATES MONEY. 83 8. T owe A S37.06; B, $200.85; C, S400; D, $236.75, and E $124.34; my property is worth $889.25 : how much do I owe more than I am worth? $109.75. 9. Bought 143 pounds of coffee, at 23 cents a pound : after paying $12.60, what was due? $20.29. 10. A owed me $400: he paid me 435 bushels of corn, at 45 cents a bushel: what sum is due? $204.25. 11. If B spend 65 cents a day, how much will he save in 365 days, his income being $400? $162.75. 12. Bought 21 barrels of apples, of 3 bushels each, at 35 cents a bushel: what did they cost? $22.05. 13. What cost four pieces of cambric, each containing 19 yards, at 23 cents a yard? $17.48. 14. If 25 men perform a piece of Avork for $2000, and spend, while doing it, $163.75, what will be each man's share of the profits? $73.45. 15. If 16 men receive $516 for 43 days' work, how much does each man earn a day? 75 ct. 16. C earned $90 in 40 days, working 10 hours a day: how much did he earn an hour? 22 ct. 5 m. 17. A merchant failing, has goods worth $1000, and $500 in cash, to be equally divided among 22 creditors: how much will each receive? $68.18-|-. MERCHANTS' BILL8. A Bill or Account, is a written statement of articles bought or sold, with their prices, and entire cost. 18. Bought 9 pounds Coffee, at $0.32 per lb. $ 4 pounds Tea, " 1.25 do. 45 pounds Sugar " .09 do. 17 pounds Cheese " .20 do. "What is the amount of my bill? $15.33 84 RAY'S NEW PRACTICAL AKITHMETIC. 19. Bought 22 yards Silk, at $1.75 per yd. S 18 yards Muslin, '' .15 do. 25 yards Linen, '- .65 do. 6 yards Gingham, " .18 do. What is the whole amount? S58.53 20. Bought— 4 pounds Prunes, at SO. 18 per lb. $ . 8 pounds Peaches, '' .23 do. 7 pounds Rice, *' .11 do. 6 pounds Oat-meal, " .09 do. 13 pounds Java Coffee, " .35 do. 26 pounds Sugar, " .12 do. What is the whole amount? $11.54 21. Bought 43 yards Muslin, at $0.13 per yd. $ 28>^5iiaTd8 Calico, '- .09 do. 23 yards Alpaca, '' .23 do. What is the whole amount? $13.40 REDUCTION OF COMPOUND NUMBERS. DRY MEASURE. 61. Dry Measure is used in measuring grain, vege- tables, fruit, coal, etc. Table. 2 pints (pt.) make 1 quart, marked qt. 8 quarts " 1 peck, '' pk. 4 pecks '' 1 bushel, '• bu. Rem. 1. — The standa7'd iinii of Dry Measure is the bushel; it is a cylindrical measure 18^ inches in diameter, 8 inches deep, and con- tains 21501 cubic inches. REDUCTION OF COMPOUND NUMBERS. 85 Rem. 2. —When articles usually measured by the above table are sold by weight, the bushel is taken as the unit. The following table gives the legal weight of a bushel of various articles in avoirdupois pounds: ARTICLES. LB. - -■-- EXCEPTIONS. Beans. 60 Blue Grass Seed. 14 Clover Seed. 60 N. J., 64. Coal (mineral). 80 Ind., 70. Corn (shelled). Flax Seed. 56 56 Cal., 111. Mo., 52. 111., N. J., N. Y., 55. Hemp Seed. Oats. 44 32 rConn., 28; Me., N. H., Mass., N. J., 30; (Ky., 33^; Oregon, 34; la., 32 ; Mo, 35. Potatoes. 60 Rye. 56 La., 52; Cal., 111., 54. Salt. 50 N. Y., 56. Timothy Seed. 45 N. Y., 44; Wis., 46. Wheat. 60 Conn., 56. To Teachers. — Numerous questions should be asked on each table similar to the following: 1. How many pints in 2 quarts? In 4? In 6? In 8? In 10? 2. How many quarts in 3 pk.? In 5? In 7? In 9? 3. How many pecks in 9 bii.? In 11? In 13? In 15? In 17? In 19? 4. How many quarts in 10 bu.? In 12? In 14? In 18? In 25? In 56? 5. How many pecks in 16 qt.? In 24? In 32? In 40? In 48? In 64? 6. How many bushels in 32 qt.? In 64? In 96? 7. How many pints in 1 bu. ? In 2 ? In 5 ? St) KAY S NEW PRACTICAL ARITHMETIC. 62. The preceding examples show that — To reduce quarts to pints, multiply the number of quarts by the number of pints in a quart. To reduce pecks to quarts, or bushels to pocks, multi- ply in the same manner. Hence, to reduce from a higher to a lower denomina- tion, multiply by the number of units that make one unit of the required denomination. They also show that — To reduce pints to quarts, divide the number of pints by the number of pints in a quart. To reduce quarts to pecks, or pecks to bushels, divide in the same manner. Hence, to reduce from a lower to a higher denomina- tion, divide by the number of units that make one unit of the required higher denomination. 1. Eeduce 3 bushels to pints. OPERATION. Solution. — To reduce bu. to pk. multiply by 4, ^ bu. because there are 4 pk. in 1 bu., or 4 times as many pk. as bu. To reduce pk, to qt. multiply by 8, because there are 8 qt. in 1 pk. To reduce qt. ^ . to pt. multiply by 2, because there are 2 pt. in 2 * 1 qt. foTpt. 2. Eeduce 192 pints to bushels. OPERATION. Solution. — To reduce pt. to qt. divide by 2, be- 2)192 pt. cause there are 2 pt. in 1 qt. To reduce qt. to pk. 8)96 qt. divide by 8, because there are 8 qt. in 1 pk. To 4)12 pk. reduce pk. to bu. divide by 4, because there are 4 3 bu. pk. in 1 bu. The two preceding examples show that reduction from a higher to a lower denomination^ and from a loicer to a higher denomination, prove each other. f2pk. REDUCTION OF COMPOUND NUMBERS. 87 3. Eeduco 7 bu. 3 j)k. 6 qt. 1 pt. to joints. OPERATION. bu. pk. qt. pt. Solution. — Multiply the bu. by 4, 7 3 6 1 making 28 pk., and add the 3 pk. Then _4 multiply the 31 pk. by 8 and add the 6 3 1 pk. in 7 bu. 3 pk qt.; multiply the 254 qt. by 2 and add the z. 1 pt.; the result is 509 pt. 2 5 4 qt. in 31 pk. 6 qt. 2 5 9 pt. in the whole. 4. Eediice 509 pt. to bushels. Solution. — To reduce pt. to qt. divide by operation. 2, and there is 1 left; as the dividend is pt. 2)509 the remainder must be pt. To reduce qt. to 8)254 qt. 1 pt. pk. divide by 8, and 6 qt. are left. To reduce 4)31 pk. 6 qt. pk. to bu. divide by 4, and 3 pk. are left. The 7 bu. 3 pk. answer is, therefore, 7 bu. 3 pk. 6 qt. 1 pt. 03, RULES FOR REDUCTION. I. FROM A HIGHER TO A LOWER DENOMINATION. 1. Multiply the highest denomination given, by that num her of the next lower which makes a unit of the higher. 2. Add to the product the number, if any, of the lower denomination. 3. Proceed in like manner icith the result thus obtained, tdl the u'hole is reduced to the required denomination. II. FROM A LOWER TO A HIGHER DENOMINATION. 1. Divide the given quantity by that number of its own denomination which makes a unit of the next higher. 2. Proceed in like manner with the quotient thus obtained, till the whole is reduced to the required denomination. 88 KAY'S NEW PKACTlCAl^ AKITHMETIC. 3. The last quotient, with the several remainders^ if any, annexed, will be the answer. Proof. — Eevcrsc I he operation: that is, reduce the answer to the denomination from which it was derived. Jf this result is the same as the quantity given, the work is correct. 5. lieduce 4 bu. 2 pk. 1 qt. to pints. 290 pt. 6. Eeduce 7 bu. 3 pk. 7 qt. 1 pint to pints. 511 pt. 7. Reduce 3 bu. 1 pt. to pints. 193 pt. 8. Reduce 384 pt. to bushels. (> bu. 9. Reduce 47 pt. to pecks. 2 pk. 7 qt. 1 pt. 10. Reduce 95 pt. to bushels. 1 bu. 1 pk. 7 qt. 1 pt. 11. Reduce 508 pt. to bushels. 7 bu. 3 pk. G qt. LIQUID MEASURE. 64. Liquid Measure is used for measuring all liquids. Table. 4 gills (gi.) make 1 pint, marked pt. 2 pints ^* 1 quart, '• qt. 4 quarts " 1 gallon, " gal. Kem. — The standard unit of liquid measure is the gallon, which contains 231 cubic inches. 1. Reduce 17 gal. to pints. 136 pt 2. Reduce 13 gal. to gills. 416 gi. 3. Reduce 126 gal. to pints. 1008 pt 4. Reduce 1260 gal. to gills. 40320 gi. 5. Reduce 1120 gi. to gallons. 35 gal. 6. How many gallons in 1848 cubic inches? 8 gal. 7. How many gallons in a vessel containing 138138 cubic inches? 598 gal. REDUCTION OF COMPOUND NUMBEKS. 89 AVOIRDUPOIS WEIGHT. 65. Avoirdupois Weight is used for weighing all ordinary articles. Table. 16 ounces (oz.) make 1 pound, '' lb. 100 pounds " 1 hundred-weight/' cwt. 20 cwt., or 2000 lb.," 1 ton, ^ " T. Rem. 1. — The standard avoirdupois pound of the United States is determined from the Troy pound, and contains 7000 grains Troy. Rem. 2. — At the Custom House (and in some trades) 2240 pounds are considered a ton. 1. Reduce 2 cwt. to pounds. 200. 2. Eeduce 3 cwt. 75 lb. to pounds. 375. 3. Keduce 1 T. 2 cwt. to pounds. 2200. 4. Eeduce 3 T. 75 lb. to pounds. 6075. 5. Eeduce 4 cwt. 44 lb. to pounds. 444. 6. Eeduce 5 T. 90 lb. to pounds. 10090. 7. Eeduce 2 cwt. 77 lb. 12 oz. to ounces. 4444. 8. Eeduce 2 cwt. 17 lb. 3 oz. to ounces. 3475. 9. Eeduce 1 T. 6 cwt. 4 lb. 2 oz. to ounces. 41666. 10. Eeduce 4803 lb. to cwt. 48 cwt. 3 lb. 11. Eeduce 22400 lb. to tons. 11 T. 4 cwt. 12. Eeduce 2048000 oz. to tons. 64 T. 13. Eeduce 64546 oz. to cwt. 40 cwt. 34 lb. 2 oz. 14. Eeduce 97203 oz. to tons. 3 T. 75 lb. 3 oz. 15. Eeduce 544272 oz. to tons. 17 T. 17 lb. 16. What is the total weight of 52 parcels, each con- taining 18 lb.? 9 cwt. 36 lb. 17. What is the weight of 180 iron castings, each weiffhin^ 75 lb.? 6 T. 15 cwt. 90 KAY'S NEW PRACTICAL A KIT II MET I C. LONG MEASURE. 66. Long Measure is used in measuring distances, or lengtli, in any direction. Table. 12 inches (in.) make 1 foot, marked ft. 3 feet '• 1 yard, '- yd. 5^ yards, or 16^ feet, '* 1 rod, '' rd. 320 rods " 1 mile, " mi. Rem. — The standard unit of length is the yard. The standard yard for the United States is preserved at Washington. A <•(>))> of this standard is kept at each state capital. 1. Reduce 2 yd. 2 ft. 7 in. to inches. 103 in. 2. Reduce 7 yd. 11 in. to inches, 263 in. 3. Reduce 12 mi. to rods. 3840 rd. 4. Reduce 7 mi. 240 rd. to rods. 2480 rd. 5. Reduce 9 mi. 31 rd. to rods. 2911 rd. 6. Reduce 133 in. to yards. 3 yd. 2 ft. 1 in. 7. Reduce 181 in. to yards. 5 yd. 1 in. 8 Reduce 2240 rd. to miles. 7 mi. 9. Reduce 2200 rd. to miles. 6 mi. 280 rd. 10. Reduce 1 mi. to yards. 1760 y(\. 11. Reduce 1 mi. to feet. 5280 fL SQUARE MEASURE. 67. Square Measure is used in measuring any thing which has both length and breadth ; that is, two dimen- A figure having 4 equal sides and 4 right angles is a square. REDUCTION OF COMPOUND NUMBERS. 91 A square inch is a square, each side of which is 1 inch in length. A square foot is a square, each side of which is 1 foot. A square yard is a square, each side of which is 1 yard (3 feet). One Square Foot. Suppose the figure to represent a square yard. It will then be 3 feet each way, and contain 9 square feet. Each foot will he 12 inches each way, and contain 144 square inches. The number of small squares in any large square is, therefore, equal to the number of units in one side multiplied by itself. Rem. — By 8 feet square is meant a square figure, each side of which is 3 feet, or 9 square feet; but by 3 square feet is meant 3 squares each one foot long and one foot wide; therefore, the difference in area between a figure Zfeet square and one containing 3 square feet, is 6 square feet. Table. 144 square inches make 1 square foot, marked sq. ft Q finnar.^ fonf »i 1 Qmior^o a7«it1 " en ^', 9 square feet 30J square yards '^" square rods acres 160 640 1 square yard, 1 square rod, 1 acre, 1 square mile, 1. Eeduce 8 sq. yd. to square inches. 2. Reduce 4 A. to square rods. 3. Eeduce 1 sq. mi. to square rods. 4. Reduce 2 sq. yd. 3 sq. ft. to sq. in. 5. Eeduce 5 A. 100 sq. rd. to sq. rd. 6. Eeduce 960 sq. rd. to acres. sq. yd. sq. rd. A. '• sq. mi 10368 sq. in. 640 sq. rd. 102400 sq. rd. 3024 sq. in. 900 sq. rd. 6 A. 92 KAY'S NEW PRACTICAL ARITHMETIC. 7. Eeduce 3888 sq. in. to square yards. 3 sq. yd. 8. Eeduce 20000 sq. rd. to acres. 125 A. 9. Reduce 515280 sq. rd. to square miles. 5 sq. mi. 20 A. 80 sq. rd. 10. Eeduce 4176 sq. in. to sq. yd. 3 sq. yd. 2 sq. ft. 68. A Rectangle is a figure having four sides and four right angles. See the figure below. The unit of measure for surfaces, is a square whose side is a linear unit; as a square inch, a square foot, etc. The Area or Superficial contents of a figure, is tlie number of times it contains its unit of measure. 1. How many square inches in a board 4 inches long and 3 inches wide? Explanation.— Dividing each of the longer sides into 4 equal parts, the shorter sides into o equal parts, and joining the opposite divisions by straight lines, the surface is divided into squares. In each of the longer rows there are 4 squares, that is, as many as there are inches in the longer side; and there are as many such rows as there are inches in the shorter side. Hence, The whole number of squares in tbe board is equal to the product obtained by multiplying together the numbers representing the length and breadth; that is, 4 X 3 = 12. Rule for Finding the Area of a Beetangle. — Multiply the length by the breadth ; the product uill be the area. Kem. — Both the length and breadth, if not in units of the same denomination, should, be made so before multiplying. 2. In a floor 16 feet long and 12 feet wide, how many square feet? 192. REDUCTION OF COMPOUND NUMBERS. 93 3. How many square yards of carpeting will cover a room 5 yards long and 4 yards wide? 20. 4. How many square yards of carpeting will cover two rooms, one 18 feet long and 12 feet wide, the other 21 feet long and 15 feet, wide ? 59. 5. How many square yards in a ceiling 18 feet long and 14 feet wide? 28. 6. In a field 35 rods long and 32 rods wide, how^ many acres? 7. 7. How much will it cost to carpet two rooms, each 18 feet long and 15 feet wide, if the carpet costs $1.25 per square yard? $75. 8. What will it cost to plaster a ceiling 21 feet long and 18 feet wide, at 17 cents per square yard? S7.14. 69. The Area of a Eectangle being equal to the prod- uct of the length by the breadth, and as the product of two numbers, divided by either of them, gives the other (36, 4); therefore, Rule. — If the area of a rectangle he divided by either side, the quotient will he the other side. Illustration. In Example 1, 68, if the area 12 be divided by 4, the quotient 3 is the width; or, divide 12 by 3, the quotient 4 is the length. Rkm. — Dividing the area of a rectangle by one of its sides, is really dividing the number of squares in the rectangle "by the num- ber of squares on one of its sides. In dividing 12 by 4, the latter is not 4 linear inches, but the num- ber of square inches in a rectangle 4 in. long and 1 in. wide. See figure, Art. 68. 94 RAY'S NEW PRACTICAL ARITHMETIC. 1. A floor containing 132 square feet, is 11 feet wide: what is its length? 12 il. 2 A. floor is 18 feet long, and contains 30 square yards: what is its width? 15 ft. 3. A field containing 9 acres, is 45 rods in length: what is its width? 32 rd. 4. A field 35 rods wide, contains 21 acres : what is its length? 96 rd. SOLID OR CUBIC MEASURE. 70. Solid or Cubic Measure is used in measuring things having length, breadth, and thickness; that is, three dimensions. A Cube is a solid, having equal faces, which are squares. Rem. — If each side of a cube is 1 inch long, it is called a cubic inch; if each side is 3 feet (1 yard) long, as represented in the figure,. it is a cubic or solid yard. The base of a cube, being 1 square yard, contains 3 X '^ = 9 square feet; and 1 foot high on this base, contains 9 solid feet; 2 feet high contains 9 X ^ = 18 solid feet; 3 feet high contains 9X8 = 27 solid feet. Also, it may be shown that 1 solid or cubic foot contains 12 X 12 X 12 = 1728 solid or cubic inches. ^. ^ ^^ -,^ ■r Hence, the number of small cubes in any large cube, is equal to the length, breadth, and thickness, multiplied together. Rem. — Any solid, whose corners resemble a cube, is a rectangular solid; boxes and cellars are generally of this form. KEBUCTION or COMPOUND NUMBERS. 95 The solid contents of a rectangular solid are found, as in the cube, by multiplying together the length, breadth, and thickness. Table. 1728 cubic inches (cu. in.) make 1 cubic foot, marked cu. ft. 27 cubic feet " 1 cubic j^ard, ♦' cu. yd. 128 cubic feet = 8X4X4 = 8 ft. long,")^ , ^^ ^ ^^1 cord, " C. 4 ft. wide, and 4 ft. high, make) Hem. 1. — A cord foot is 1 foot in length of the pile which makes a cord. It is 4 feet wide, 4 feet high, and 1 foot long; hence, it contains 16 cubic feet, and 8 cord feet make 1 cord. Rem. 2. — A perch of stone is a mass 16J ft. long, 1^ ft. wide, and 1 ft. high, and contains 24J cu. ft. 1. Reduce 2 cu. yd. to cubic inches. 93312 cu. in. 2. Reduce 28 cords of wood to cu. ft. 3584 cu. ft. 3. Reduce 34 cords of wood to cu. in. 7520256 cu. in. 4. Reduce 1 cord of wood to cu. in. 221184 cu. in. 5. Reduce 63936 cu. in. to cu. yd. 1 cu. yd. 10 cu. ft. 6. How many cubic feet in a rectangular solid, 8 ft. ' long, 5 ft. wide, 4 ft. thick? 160 cu. ft. 7. How many cubic yards of excavation in a cellar 8 yd. long, 5 3^d. wide, 2 yd. deep? 80 cu. yd. 8. How many cubic yards in a cellar, 18 feet long, 15 feet wide, 7 feet deep? 70 cu. 3'd. 9. In a pile of wood 40 feet long, 12 feet wnde, and 8 feet high, how many cords? 30 C. 10. What will be the cost of a pile of wood 80 feet long, 8 feet high, and 4 feet thick, at S5.50 per cord? $110. 11. What will be the cost of excavating a cellar 24 ft. long, 15 ft. wide, -and 6 ft. deep, at $1.25 per cubic yard or load? $100. 96 RAY'S NEW PRACTICAL ARITHMETIC. TIME MEASURE. 71, Time Measure is used in measuring time. Table. 60 seconds (sec.) make 1 minute, marked min. 60 minutes 24 hours 365 days, 6 hours 100 years Also, 7 days 4 weeks 12 calendar months 365 days 366 days '' 1 hour, '' 1 day, '^ 1 year, ^' 1 century make 1 week, '' hr. '' da. yr. '' cen. marked wk. 1 month (nearly), '^ mon. 1 year, '' yr. 1 common year. 1 leap year. Rem. 1. — The exact length of the moan solar, or tropical year, is 365 days, 6 hours, 48 minutes, 46 seconds. To correct the error of considering 365 days as the length of the year, the following rule has been adopted: Every year whose number is not divisible by 4 consists of 365 days. Every year whose number is divisible by 100, but not by 400, consists of 365 days. Every year, except the even centuries, whose number is divisible by 4, and the even centuries divisible by 400 consist of 366 days. The year containing 366 days is called Leap year, and the extra day is added to February, giving it 29, instead of 28 days. Rem. 2. — Among nearly all civilized nations the year is divided into 12 calendar months, and numbered, in their order, as follows: January, February, March, April, May, June, 1st month, 81 days. 2d " 28 " 8d " 31 " 4th " 30 '« 5th «' 31 '' 6th " 30 " July, 7th month, 31 days. August, 8th ** 31 " September, 9th " 30 " October, lOth " 31 " November, 11th " 30 " December, 12th " 31 " REDUCTION OF COMPOUND NUMBERS. 97 1. Eeduce 2 hr. to seconds. 7200 sec. 2. Eeduce 7 da. to minutes. 10080 min. 3. Reduce 1 da. 3 hr. 44 min. 3 sec. to seconds. 99843 sec. 4. Reduce 9 wk. 6 da. 10 hr. 40 min. to minutes. 100000 min. 5. Reduce 1 mon. 3 da. 4 min. to minutes. 44644 min. 6. Reduce 10800 seconds to hours. 3 hr. 7. Reduce 432000 seconds to days. 5 da. 8. Reduce 7322 seconds to hours. 2 hr. 2 min. 2 sec. 9. Reduce 4323 minutes to days. 3 da. 3 min. 10. Reduce 20280 minutes to weeks. 2 wk. 2 hr. 11. Reduce 41761 min. to months. 1 mo. 1 da. 1 min. MISCELLANEOUS TABLES. I. MEASURES OF WEIGHT. 72. Troy Weight is used in weighing gold, silver, and jewels. 24 grains (gr.) make 1 pennyweight, marked pwt. 20 pennyweights " 1 ounce, '^ oz. 12 ounces " 1 pound, '^ lb. The Standard Unit of all weight in the United States is the Troy pomid, containing 5760 grains. Apothecaries "Weight is used only in compounding medicines. 20 grains (gr.) make 1 scruple, marked 9. 3 scruples '' 1 dram, " 3. 8 drams '^ 1 ounce, '' 5. 12 ounces " 1 pound, " lb. 98 RAY'S NEW PKACTICAL AKITHMETIC. The following are also used by apothecaries : 60 minims (or drops) rT\^. make 1 fluid drachm, marked f. 3. 8 fluid drachms " 1 fluid ounce, " f. §. 10 fluid ounces '* 1 pt. (octarius) " O. 8 pints '' 1 gal. (congius) " cong. II. MEASURES OF LENGTH. The following measures are often mentioned and most of them are still used in special professions: 12 lines =1 inch. 3 barleycorns = 1 inch 4 inches = 1 hand. 9 inches = 1 span. 3 feet = 1 pace. 6 feet =1 fathom. 3 miles = 1 league. ' 69^ miles (nearly) = 1 degree. Surveyors use a chain four rods long, divided into links of Ty^^ inches each. Engineers divide the foot into tenths and hundredths. The yard is also divided similarly in estimating duties at the custom houses. A degree is divided into 60 nautical or geographic miles. A nautical mile or knot is, therefore, nearly 1^ com- mon miles. Circular Measure is used in measuring circles. 60 seconds C) make 1 minute, marked '. 60 minutes ^' 1 degree, " °. 360 degrees " 1 circle. Rem. — The circumference is also divided into quadrants of 90° each, and into siqjis of 80° each. KEDUCTION OF COMPOUND NUMBERS. 99 III. MISCELLANEOUS TABLE. 12 things make 1 dozen, marked doz. 12 dozen " 1 gross, " gr. 12 gross " 1 great gross. 20 things " 1 score. 100 pounds of nails, make 1 keg. 196 pounds of flour " 1 barrel. 200 pounds of pork or beef make 1 barrel. 240 pounds of lime " 1 cask. 24 sheets of paper make 1 quire. 20 quires " 1 ream. 2 reams " 1 bundle. A sheet folded in 2 leaves is called a folio. 4 '• '' " a quarto, or 4to. 8 " '' '' an octavo, or 8vo. 12 '' " " a duodecimo, or 12mo. 16 " " '^ a 16mo. Examples in Miscellaneous Tables. 73. 1. Reduce 5 lb. 4 oz. Troy to ounces. 64. 2. Eeduce 9 lb. 3 oz. 5 pwt. to pwt. 2225. 3. Eeduce 8 lb. 9 oz. 13 pwt. 17 gr. to gr. 50729. 4. Eeduce 805 pwt. to pounds. 3 lb. 4 oz. 5 pwt. 5. Eeduce 12530 gr. to pounds. 2 lb. 2 oz. 2 pwt. 2 gr. 6. Eeduce 4 lb. 5 g 2 gr. to grains. 25442. 7. Eeduce 7 lb. 2 § 1 9 to grains. 41300. 8. Eeduce 431 3 to pounds. 4 lb. 5 5 7 5. 9. Eeduce 975 9 to pounds. • 3 lb. 4 § 5 5. 10. Eeduce 6321 gr. to pounds. 1 lb. 1 g 1 3 1 9 1 gr. 11. Eeduce 4 cong. 7 f g to fluid drams. 4152. 100 KAY'S NEW PRACTICAL AKITHMETIC. 12. Eediice 5 O. 6 f. § 3 f. 3 to minims. 41460. 13. Keduce !H69 f. 5 to gallons. 2 cong. 3 O. 4 f. 3 5 f. 3. 14. Reduce 3 yd. to barleycorns. 324. 15. How many lines in 1 foot 6 inches? 216. 16. What is the height of a horse of 16^ hands? 5 ft. 6 in. 17. A field measures 24 chains in length and 15 chains in breadth: how many acres in it? 36. 18. A cistern contains 267 cubic feet 624 cubic inches : how many gallons does it hold? (Art. 64, Eem.). 2000. 19. Reduce 8° 41' 45" to seconds. 31305. 20. Reduce 61° 59' 28" to seconds. 223168. 21. Reduce 915' to degrees. 15° 15'. 22. Reduce 3661" to degrees. 1° 1' 1". 23. What cost 6 gross of screws at 5 cents a dozen? $3.60. 24. A man is 4 score and 10: how old is he? 90 yr. 25. At 18 certs a quire, what will 3 bundles of paper cost? $21.60. 26. How many sheets of paper will be required for a a 12mo. book of 336 pages? 14. 27. An octavo work in 5 volumes has 512 pages in Vol. 1, 528 in Yol. 2, 528 in Vol. 3, 512 in Vol. 4, and 496 in Yol. 5: how much paper was used for one copy of the whole work? 6 quires 17 sheets. Promiscuous Examples. 74. 1. What cost 2 bu. of plums, at 5 ct. a pint? $6.40. 2. What cost 3 bu. 2 pk. of peaches, at 50 ct. a l>eck? $7. KEDUCTION OF COMPOUND NUMBERS. 101 3. What cost 3 pk. 3 qt. of baVl^y, 'ai 3* ct. a pint? 4. At 15 ct. a peck, how many bashels of apples can be bought for 83? 5 bii. 5. If salt cost 2 ct. a pint, how much can l)e bought with SI. 66? 1 bu. 1 pk. 1 qt. 1 pt. 6. I put 91 bu. of wheat into bags containing 3 bu. 2 pk. each : how many bags were required ? 26. Rem. — Reduce both quantities to pecks, and then divide. 7. How many spikes, weighing 4 oz. each, are in a parcel weighing 15 lb. 12 oz. ? 63. 8. I bought 44 cwt. 52 lb. of cheese ; each cheese weighed 9 lb. 15 oz. : how many cheeses did I buy? 448. 9. How many kegs, of 84 lb. each, can be filled from a hogshead of sugar weighing 14 cwt. 28 lb.? 17. 10. How many boxes, containing 12 lb. each, can be filled from 7 cwt. 56 lb. of tobacco? 63. 11. If a family use 3 lb. 13 oz. of sugar a week, how long will 6 cwt. 10 lb. last them? 160 wk. 12. What will 2 acres 125 square rods of land cost, at 20 cents a square rod? $89. 13. A farmer has a field of 16 A. 53 sq. rd. to divide into lots of 1 A. 41 sq. rd. each : how many lots will it make? 13. 14. How many cu. in. in a block of marble 2 ft. long, 2 ft. high, 2 ft. wide? 13824. 15. One cu. ft. of water weighs 1000 oz. avoirdupois: what do 5 cu. ft. weigh? 312 lb. 8 oz. 16. What is the weight of a quantity of water occupy- ing the space of 1 cord of wood, each cubic foot of water weighing 1000 ounces avoirdupois? 4 T. 102 KAY'S NEW PRACTICAL ARITHMETIC. 17' jV cufei6' fo(>t I af! ' oak weighs 950 oz. avoirdupois: Ay.bat >l5c>"3 ;Coi\lw,*of; oak'\\:eigh? 7 T. 1:^ cwt. *' 18* \^ihd' the *eo»t^ of' 63' gallons of wine, at 20 cents a pint. $100.80. 19. Find the cost of 5 barrels of molasses, each con- taining 31 gal. 2 qt., at 10 cents a quart. $63. 20. At 5 cents a pint, what quantity of molasses can be bought for $2? 5 gal. 21. How many dozen bottles, each bottle holding 3 qt. 1 pt., can be filled from 63 gal. of cider? 6 doz. 22. How many kegs, of 4 gal. 3 qt. 1 pt. each, can be filled from 58 gal. 2 qt.? 12. 23. If a human heart beat 70 times a minute, how many times will it beat in a da}^? 100800. 24. How many seconds in the month of February, 1876? 2505600 sec. 25. If a ship sail 8 miles an hour, how many miles will it sail in 3 wk. 2 da. 3 hr. ? 4440 mi. 26. A horse is fed 1 peck of oats daily. If oats cost 44 cents a bushel, how much will it cost to feed him a year of 365 days? $40.15. 27. A flour dealer bought 40 barrels of flour for 3 ct. a pound, and sold it for 5 ct. a pound: how much did he gain? $156.80. ADDITION OP COMPOUND NUMBERS. 75. When the numbers to be added are compound, the operation is called Addition of Compound Numbers. 1. A farmer sold three lots of wheat: the first lot contained 25 bu. 3 pk. ; the second, 14 bu. 2 pk. ; the third, 32 bu. 1 pk. : how much did he sell? ADDITION OP COMPOUND NUMBERS. 103 Solution. — Place units of the same denomination in the same column (Art. 17). Beginning with pecks, and adding, the sum is 6, which is reduced to bushels by dividing by 4, the number of pecks in a bushel, and there being 2 pecks left, write the 2 under the column of pecks, and carry the 1 bushel to the column of bushels; adding this to the bushels, the sum is 72, which write under the column of bushels. OPERATION. bu. pk. 25 3 14 2 32 1 72 (2) (3) bu. pk. qt. pt. bu. pk. qt. pt, 3 2 1 7 3 7 1 4 6 1 6 2 1 3 7 1 9 2 4 1 "9 2 6 1 24 4 Rule. — 1. Write the numbers to be added, placing units of the same denomination in the same column. 2. Begin with the lowest denomination, add the numbers, and divide their stim by the number of units of this denom- ination ivhich make a unit of the next higher. 3. Write the remainder under the column added, and carry the quotient to the next column. 3. Proceed, in the same manner with all the colurnns to the last, under which write its entire sum. Proof. — The same as in Addition of Simple Numbers. Rem. 1. — In writing compound numbers, if any intermediate de- nomination is wanting, supply its place with a cipher. Rem. 2. — In adding simple numbers we carr}^ one for every ten, because ten units of a lower order always make one of the next higher; but, in compound numbers, the scale varies, and we carry one for the number of the lower order, which makes one of the next higher. 104 RAY'S NEW PKACTICAL AKITHMETIC. Examples DRY MEASURE. (4) (5) bu. pk. qt. bu. pk. qt. pt. 4 3 7 8 1 7 1 5 2 2 7 3 2 1 7 1 6 9 2 7 1 LIQUID MEASURE. (6) (7) qt. pt. gi- gal. qt. pt. gi- 7 1 3 40 3 1 3 6 2 16 1 2 9 1 3 71 2 1 2 AVOIRDUPOIS WEIGHT. (8) (9) T. cwt. lb. oz. cwt. lb. oz. 45 3 53 10 16 85 14 14 14 75 15 15 90 13 19 17 18 13 18 74 12 LONG MEASURE. (10) (11) mi. rd. yd. ft. in. 28 129 4 2 11 64 280 3 1 9 17 275 5 1 8 ADDITION OF COMPOUND NUMBERS. 105 SQUARE MEASURE. (12) A. sq. rd. 41 51 sq. yd. 15 (13) sq. ft. 8 sq. in. 115 64 104 20 7 109 193 155 14 5 137 CUBIC MEASURE. (14) (15) C. cu. ft. cu. in. cu. yd. cu. ft. cu. in. 13 28 390 50 18 900 15 90 874 45 17 828 20 67 983 46 20 990 TIME MEASURE. (16) (17) da. Ill*, min. sec. mo. wk. da. lir. min. sec. 16 18 28 47 3 23 51 40 13 15 49 59 12 4 19 30 37 19 16 53 42 3 1 5 13 27 18 18. Five loads of wheat measured thus : 21 bu. 3 pk. ; 14 bu. 1 pk. ; 23 bu. 2 pk. ; 18 bu. 1 pk. ; 22 bu. 1 pk. : how many bushels in all? 100 bu. 19. A farmer raised of oats 200 bu. 3 pk. ; barley, 143 bu. 1 pk. ; corn, 400 bu. 3 pk. ; wheat, 255 bu. 1 pk. : how much in all? 1000 bu. 20. A grocer sold 5 hogsheads of sugar : the first weighed 8 cwt. 36 lb. ; the second, 4 cwt. 64 lb. ; the third, 5 cwt. 19 lb. ; the fourth, 7 cwt. 75 lb. ; the fifth, 7 cwt. 84 lb. : what did all weigh? 33 cwt. 78 lb. 21. Add 13 lb. 11 oz.; 17 lb. 13 oz. ; 14 lb. 14 oz. ; 16 lb. ; 19 lb. 7 oz. ; and 17 lb. 9 oz. 99 lb. 6 oz. 106 RAY'8 NEW PRACTICAL ARITHMETIC. 22. Two men depart from the same place : one travels 104 mi. 50 rd. due east; the other, 95 mi. 270 rd. due west: how far are they apart? 200 mi. 23. A man has 3 farms: in the first are 186 A. 134 sq. rd. ; in the second, 286 A. 17 sq. rd. ; in the third, 113 A. 89 sq. rd. : how much in all? 586 A. 80 sq. rd. 24. Add 17 sq. yd. 3 sq. ft. 119 sq. in.; 18 sq. yd. 141 sq. in. ; 23 sq. yd. 7 sq. ft. ; 29 sq. yd. 5 sq. ft. 116 sq. in. 88 sq. yd. 8 sq. ft. 88 sq. in. 25. A has 4 piles of wood : in the first, 7 C. 78 cu. ft. ; the second, 16 C. 24 cu. ft.; the third, 35 C. 127 cu. ft.; the fourth, 29 C. 10 cu. ft.: how much in all? 88 C. Ill cu. ft. 26. I sold 4642 gal. 3 qt. 1 pt. of wine to A ; 945 gal. to B; 1707 gal. 1 pt. to C; 10206 gal. 1 qt. to D: how many hogsheads of 63 gal. each did I sell ? 277 hogsheads 50 gal. 1 qt. SUBTRACTION OF COMPOUND NUMBERS. 76. When two given numbers are compound,, the operation of finding their difference is called Subtraction of Compound Nximhers. 1. I have 67 bu. 2 pk. of wheat: how much will re- main after selling 34 bu. 3 pk. ? Solution. — Write the less number under the greater, placing units of the same denomination in the same column. 3 pk. can not be taken from 2 pk., but 1 bu. being taken from 67 bu. reduced to pk., and added to the 2 pk., gives 6 pk. 3 pk. from 6 pk. leaves 3 pk.; 34 bu. from 66 bu. leaves 32 bu. The "32 3~ difference is, therefore, 32 bu. 3 pk. ^_ PERJl bu. lTION. pk. 67 2 34 3 SUBTRACTION OF COMPOUND NUMBERS. 107 Rem. — Instead of diminishing the 67 bu. by 1, the result will be the same to increase the lower number 34 bu. by 1, as is done in subtraction of simple numbers. (2) (3) bu. pk. qt. pt. bu. pk. qt. pt. From 12 1 5 Take J 2^ 1 1 0_ 1^ 3 1 7 1 3 3 7 1 Rule. — 1. Write the less number under the greater, placing units of the same denomination in the same column. 2. Begin with the lowest denomination, and, if possible. take the lower number from the one above it. 3. But, if the lower number of any denomination be greater than the upper, increase the upper number by as many units of that denomination as make one of the next higher ; sub- tract as before, and carry one to the lower number of the next higher denomination. 4. Proceed in the same manner with each denomination. Proof. — The same as in Subtraction of Simj^le Num- bers. Rem. — The resemblance between subtraction of simple, and ^^f compound numbers, is the same as in Addition 75, Rem. 2. Examples, liquid measure. gal. qt. pt. gal. qt. pt. gi. From 17 2 1 43 1 1 2 Take Hi 3 0, 23 3 1 3 108 RAY'S NEW PRACTICAL ARITHMETIC. AVOIRDUPOIS WEIGHT. (6) T. cwt. lb. From 14 12 50 Take 10 13 75 T. cwt. lb. oz. 16 7 18 14 5 6 75 15 LONG MEASURE. (8) mi. rd. From 18 198 Take 11 236 (9) yd. ft. in. 4 1 10 2 1 11 SQUARE MEASURE. ( 10) (11) A. 8q. rd. sq. yd. sq. ft. sq. in. From 327 148 19 6 72 Take 77 155 CUBIC 16 MEASURE. 6 112 ( 12) (13) C. cu. ft. cu. yd. cu. ft. cu. in. From 28 116 18 7 927 Take 19 119 TIME (14) 9 MEASURE. 15 (15) 928 hr. min. sec. da. hr. min, . sec. From 18 43 27 245 17 40 37 Take 17 51 45 190 11 44 42 SUBTKACTION OF COMPOUND NUMBEKS. 109 16. If 2 bu. 1 pk. 1 qt. be taken from a bag containing 4 bushels of hickory nuts, what quantity will remain ? 1 bu. 2 pk. 7 qt. 17. From 100 bu. take 24 bu. 1 pt. 75 bu. 3 pk. 7 qt. 1 pt. 18. I bought 46 lb. 4 oz. of rice : after selling 19 lb. 8 oz., how much remained? 26 lb. 12 oz. 19. A wagon loaded with hay weighs 32 cwt. 66 lb. ; the wagon alone weighs 8 cwt. 67 lb. : what is the weight of the hay? 23 cwt. 99 lb. 20. It is 24899 miles round the earth : after a man has traveled 100 mi. 41 rd. what distance will remain? 24798 mi. 279 rd. 21. I had a farm containing 146 A. 80 sq. rd. of land. I gave my son 86 A. 94 sq. rd. : how much was left ? 59 A. 146 sq. rd. 22. From 8 C. 50 cu. ft. of wood, 3 C. 75 cu. ft. are taken: how much is left? 4 C. 103 cu. ft. 23. A cask of wine containing 63 gal. le^J^ed; only 51 gal. 1 qt. 2 gi. remained : how much was lost ? 11 gal. 2 qt. 1 pt. 2 gi. 24. From 5 da. 10 hr. 27 min. 15 sec. take 2 da. 4 hr. 13 min. 29 sec. 3 da. 6 hr. 13 min. 46 sec. 77. In finding the time between any two dates, con- sider 30 days 1 month, and 12 months 1 year. 1. A note, dated April 14, 1875, was paid February 12, 1877 : find the time between these dates. Solution. — In writing the dates, observe operation. that February is the 2d month of the year yr. mon. da. and April the 4th; then, from 1877 yr. 2 mo. 187 7 2 12 12 da. subtract 1875 yr. 4 mo. 14 da. The re- 18 7 5 4 14 mainder is 1 vr. 9 mo. 28 da. 19 2 8 110 KAY'S NEW PKACTICAL ARITHMETIC. 2. The Independence of the United States was declared July 4, 1776 : what length of time had elapsed on the Ist of September, 1876? 100 yr. 1 mo. 27 da. 3. The first crusade ended July 15, 1099; the third crusade, July 12, 1191 : find the difference of time be- tween these dates. 91 yr. 11 mo. 27 da. 4. Magna Charta was signed June 15, 1215 ; Mary, Queen of Scots, was beheaded February 8, 1587 : find the difference of time between these dates. 371 yr. 7 mo. 23 da. 5. The battle of Hastings was fought Oct. 14, 1066; William, Prince of Orange, landed at Tor Bay Nov. 5, 1688: what w^as the difference of time between the two events? 622 yr. 21 da. 6. The battle of Austerlitz was fought December 2, 1805; the battle of Waterloo, June 18, 1815: find the difference of time. 9 yr. 6 mo. 16 da. 78. To fiijd the time between two dates in days. 1. Find the number of days from May 10 to Oct. 21. Solution. — Of May, there remains 31 — 10 == 21 days; there are 30 days in June, 31 in July, 31 in August, 30 in September, and 21 in Octo- ber; then the number of days from May 10 to October 21, is 21 -f 30 + 31 + 31 + 30 + 21 == 164. 2. Find the number of days from March 17 to Septem- ber 12. 179. OPERATION. 31 May, 10 21 June, 30 July, Aug., Sept., Oct., 31 31 30 21 164 MULTIPLICATION OF COMPOUND NUMBEKS. HI 3. A note dated April 18, 1877, is due June 20, 1877: how many days does it run? 63. 4. A note dated Sept. 5, 1877, is due Dec. 7, 1877 : how many days does it run? 93. 5. Find the number of days from Oct. 12, 1877, to May 25, 1878. 225. 6. Find the number of days from Aug. 20, 1875, to March 8, 1876. 201. MULTIPLICATION OF COMPOUND NUMBERS. 79. When the multiplicand is a compound number, the operation is called Multiplication of Compound Numbers. 1. A farmer takes to mill 5 bags of wheat, each con- taining 2 bu. 3 pk. : how much had he in all ? Solution. — Begin at the lowest denomination for convenience. Multiply the 3 pk. by 5, making 15 pk., bu. pk. which, reduced, gives 3 bu. and 3 pk.; write the 3 pk. 2 3 under the pecks, and carry the 3 bu. Then, multiply 5 the 2 bu. by 5, add to the product the 3 bu., and write 13 3 the 13 bu. under the bushels. Rule. — 1. Write the multiplier under the lowest denom- ination of the multiplicand. 2. Multiply the loivest denomination first, and divide the product by the number of units of this denomination ivhich make a unit of the next higher, write the remainder under the denomination multiplied, and carry the quotient to the product of the next higher denomination. 3. Proceed in like manner with all the denominations, writing the entire product at the last. Proof. — The same as in Simple Multiplication. 112 llA^'S NEW PRACTICAL ARITHMETIC. Rem. — There are two differences between multiplication of simple and of compound numbers: 1. In simple numbers it is more con- venient to use one figure of the multiplier at a time; in compound numbers it is better to use the eritire tnultiplier each time. 2. In simple numbers the scale is miiforvi; in compound numbers it varies with the table. Examples. 2. Multiply 2 bii. 1 pk. 1 qt. 1 pt. by 6. 13 bu. 3 pk. 1 qt. 3. Multiply 2 bu. 2 pk. 2 qt. by 9. 23 bu. 2 qt. 4. If 4 bu. 3 pk. 3 qt. 1 pt. of wheat make 1 bl. of iiour, how much will make 12 bl.? 58 bu. 1 pk. 2 qt. 5. Find the weight of 9 hogsheads of sugar, each weighing 8 cwt. 62 lb. 3 T. 17 cwt. 58 lb. 6. How much hay in 7 loads, each weighing 10 cwt. 89 lb.? 3 T. 16 cwt. 23 lb. 7. If a ship sail 208 mi. 176 rd. a day, how far will it sail in 15 days? 3128 mi. 80 rd. 8. Multiply 23 cu. yd. 9 cu. ft. 228 cu. in. by 12. 280 cu. yd. 1 cu. ft. 1008 cu. in. 9. Multiply 16 cwt. 74 lb. by 119. 99 T. 12 cwt. 6 lb. 10. Multiply 47 gal. 3 qt. 1 pt. by 59. 2824 gal. 2 qt. 1 pt. 11. A travels 27 mi. 155 rd. in 1 day: how far will he travel in one month of 31 days? 852 mi. 5 rd. 12. In 17 piles of wood, each pile containing 7 C. 98 cu. ft. : what is the quantity of wood? 182 C. 2 cu. ft. 13. Multiply 2 wk. 4 da. 13 hr. 48 min. 39 sec. by 75. 49 mo. 3 wk. 3 hr. 48 min. 45 sec. 14. A planter sold 75 hogsheads of sugar, each weigh- ing 10 cwt. 84 lb., to a refiner, for 6 ct. a pound. The refiner sold the sugar for 8 ct. a pound : how much did he gain? $1626. DIVISION OF COMPOUND NUMBEES. 113 15. A cotton-factor sold 425 bales of cotton, each weighing 4 cwt. 85 lb., for 13 ct. a pound. He paid $24735 for the cotton : how much did he gain ? $2061.25. DIVISION OF COMPOUND NUMBERS. 80. When the dividend is a compound number, the operation is called Division of Compound Numbers. The divisor mvij be either a Simple or a Compound Number. This gives rise to two cases: First. — To find how often one Compound Number is contained in another Compound Number. This is done by reducing both divisor and dividend to the .^ame denomination before dividing (Examples 6 and 8, Art. 74). Second. — To divide a Compound Number into a given number of equal parts. This is properly Compound Di- vision. 1. Divide 14 bu. 2 pk. 1 qt. by 3. Solution. — Divide the highest denomination operation. first, so that, if there be a remainder, it may be re- bu. pk. qt. duced to the next lower denomination, and added 8)14 2 1 to it. 3 in 14 is contained 4 times, and 2 bu. are 4 3 3 left; write the 4 under the bushels, and reduce the remaining 2 bu. to pk., to which add the 2 pk., making 10 pk. This, divided by 3, gives a quotient of 3 pk., with 1 pk. remaining; which, reduced to qt., and 1 qt. added, gives 9 qt. This, divided by 3, gives a quotient 3, wdiich is written under the quarts. (2) (3) bu. pk. qt. da. hr. min. sec. 7)33 2 6 5 )17 12 56 15 4 3 2 3 12 11 15 PPAC. 8. 114 KAY'S NEW riiACTICAL ARITHMETIC. Rule. — 1. Write the quantity to be divided in the order of its denominations, beginning with the highest; place the divisor on the left, 2. Begin ivith the highest denomination, divide each num- ber separately, and write the quotient beneath. 3. If a remainder occurs after any division, reduce it to the next lower denomination, and, before dividing, add to it the number of its denomination. Proof. — The same as in Simple Division. Rem.— Each partial quotient is of the same denomination as that part of the dividend from which it is derived. 4. Divide 67 bu. 3 pk. 4 qt. 1 pt. by 5. 13 bu. 2 pk. 2 qt. 1 pt. 5. Eleven casks of sugar weigh 35 ewt. 44 lb. 12 oz. : what is the average weight of each? 3 cwt. 22 lb. 4 oz. t). I traveled 39 mi. 288 rd. in 7 hr. : at what rate per hour did I travel? 5 mi. 224 rd. 7. Divide 69 A. 64 sq. rd. by 16. 4 A. 54 sq. rd. 8. 490 bu. 2 pk. 4 qt. -^ 100. 4 bu. 3 pk. 5 qt. 9. 265 lb. 10 oz. -^50. 5 lb. 5 oz. 10. 45 T. 18 cwt.-^17. 2 T. 14 cwt. 11. 114 da. 22 hr. 45 min. 18 sec. -1-54. 2 d«a. 3 hr. 5 min. 17 sec. 12. 10 cwt. 27 lb. 13 oz. -f-23. 44 lb. 11 oz. 13. 309 bu. 2 pk. 2 qt.-^78. 3 bu. 3 pk. 7 qt. 14. 127 gal. 3 qt. 1 pt. 3 gi. — 63. 2 gal. 1 gi. 15. 788 mi. 169 rd. -^319." 2 mi. 151 rd. 16 A farmer has two farms, one of 104 A. 117 sq. rd.; the other, 87 A. 78 sq. rd. He reserves 40 A. 40 sq. rd., and divides the remainder equally among his 3 sons : what is the share of each son? 50 A, 105 sq. rd. LONGITUDE AND TIME. 115 17. A farmer's crop consisted of 5000 bu. 3 pk. of corn one year, and 7245 bu. 2 pk. the year following. He sold B022 bu. 1 pk. and placed the remainder in 8 cribs, each crib containing an equal amount : how many bushels in each crib? 528 bu. 18. A speculator bought 6 adjoining pieces of land, each containing 4 A. 80 sq. rd. He divided the whole into 54 lots, and sold them at $5 a sq. rd. : how much did he get for each lot? S400. 19. Add 35 lb. 9 oz.. 75 lb. 14 oz., 85 lb. 15 oz.; from the sum take 186 lb. 14 oz. ; multiply the remainder by 8; divide the product by 64: what is the result? 1 lb. 5 oz. LONGITUDE AND TIME. 81. Difference of longitude and time between different 1^ laces. The circumference of the earth, like other circles, is divided into 360 equal parts, called degrees of longitude. The sun appears to pass entirely round the earth, 360°, once in 24 hours, one day ; and in 1 hour it passes over 15°. (360° -f- 24 .:= 15°). As 15° equal 900', and 1 hour equals 60 minutes of time, therefore, the sun in 1 minute of time passes over l^' of Si degree, (900'~60=r=: 15'). As 15' equal 900", and 1 minute of time equals 60 seconds of time, therefore, in 1 second of time the sun passes over 15" of a degree. (900"^ 60 = 15"). Table for Comparing Longitttde and Time. 15° of longitude -=.- 1 hour of time. 15' of longitude ^^^ 1 min. of time. 15" of longitude = 1 sec. of time. 116 KAY'S NEW PRACTICAL ARITHMETIC. 1. How many hr. min. and sec. of time correspond to 18° 25' :W of longitude? 1 hr. 13 min. 42 sec. Analysis. — By inspection of the table, it is evident that, Degrees ( ° ) of longitude, divided by 15, give hours of time. Minutes ( ^ ) of longitude, divided by 15, give minutes of time. Seconds (^^) of longitude, divided by 15, give seconds of time. Hence, if 18° 25^ 30^^ of Ion. be divided by 15, the quotient will be the time in hr. min. and sec. corresponding to that longitude. To find the time corresponding to any difference of longitude : Rule. — Divide the longitude by 15, according to the rule for Division of Compound Numbers , and mark the quotient hr. min. sec, instead of ° ' ". Conversely: To find the longitude corresponding to any difference of time. Rule. — Multiply the time by 15, according to the rule for Multiplication of Compound Numbers^ and mark the product ® ' " instead of hr. rain. sec. 2. The difference of longitude between two places is 30*^ : what is their difference of time? 2 hr. 8. The difference of longitude betw^een two places is 71° 4': what is the difference of time? 4 hr. 44 min. 16 sec. 4. The difference of longitude between New York and Cincinnati is 10° 35': what is the difference of time? 42 min. 20 sec. 5. The difPerence of time between Cincinnati and Phil- adelphia is 37 min. 20 sec. : what is the difference of longitude? 9° 20'. LONGITUDE AlNl) TIME. 117 6. The difference of time between !N^ew York and St. Louis is 1 hr. 4 min. 56 sec. : what is the difterence of longitude? 16° 14^ 7. The difference of time between London and Wash- ington is 5 hr. 8 min. 4 sec. : what is the difference of longitude? 77° I'o DlB^FERENCE IN TiME. 82. It is noon (12 o'clock), at any i)l^ce when the sun is on the meridian of that place. As the sun appears to travel from the east toward the west, when it is noon at any place, it is after noon east of that place, and before noon icest of that place. Hence, a place has later or earlier tirne than another, according as it is east or west of it. Therefore, When the time at one place is given, the time at another, if EAST of this, is found by adding their difference of time; if WEST, by SUBTRACTING their difference of time. 8. When it is noon at Cincinnati, what is the time at Philadelphia? 37 min. 20 sec. past noon. 9. When it is 11 o'clock A. M. at New York, what is the time in longitude 30° east of New York? 1 P. M. 10. When 12 o'clock (noon) at Philadelphia, what is the time at Cincinnati? 11 hr. 22 min. 40 sec. A. M. 11. When it is 11 o'clock A. M. at New York, what is the time at St. Louis? 9 hr. 55 min. 4 sec. A. M. 12. Wheeling, W. Ya., is in longitude 80° 42' west: the mouth of the Columbia river, in longitude 124° west: when it is 1 o'clock P. M. at Wheeling, what is the time at the mouth of Columbia river? 10 hr. 6 min. 48 sec. A. M. DEFINITIONS. 83. 1. Factors of a number are two or more num- bers, the product of which equals the given number (Art. 28, 2). Thus, 2 and 3 are factors of 6, because 2X3 = 6; 2, 3, and 6 are factors of 30, because 2 X 3 X ^ = 30- Kem. 1. — One and the number itself are not considered factors of a number. Kem. 2. — A number may be the product of more than one set of factors. Thus, 2 X 6== 12, 3 X 4 = 12, and 2X2X^ = 12. 2. A multiple of a number is a product of which the number is a factor. Thus, 6 is a multiple of 3; 30 is a multiple of 5. 3. INTumbers are divided into two classes, prime and composiie. 4. A prime number has no factors. Thus, 5, 11, 17 are prime numbers. 5. A composite number has two or more factors. Thus, 6, 12, 30 are composite numbers. 6. A prime factor is a factor which is a prime p umber. Thus, 3 is a prime factor of 12. (118) FACTORING. 119 7. A factor is common to two or more numbers when it is a factor of each of them. Thus, 3 is a common factor of 12 and 15. Rem. — Sometimes the smallest of two or more numbers may be the common factor. Thus, 6 is a common factor of 6, 12, and 18. 8. Two or more numbers are prime to each other ^ w^hen they have no common factor. Thus, 9 and 10 are prime to each other. 9. A common divisor (C. D.) of two or more num- bers is any common factor. Thus, 2, 3, and 6 are each a common divisor of 12 and 18. 10. The greatest common divisor (G. C. D.) of two or more numbers is the greatest common factor. Thus, 6 is the greatest common divisor of 12 and 18. 11. A common multiple (C. M.) of tw^o or more num- bers is any multiple of all of them. Thus, 6, 12, 18, etc., are common multiples of 2 and 3. 12. The least common multiple (L. C. M.) of two or more numbers is the least multiple of all of them. Thus, 6 is the least common multiple of 2 and 3. 13. Factoring is the process of resolving composite numbers into their factors. To Find the Prime Jfiniihers. 84. All the prime numbers except 2 are odd num- bers. 120 RAYS NEW PKACTICAL ARITHMETIC. Rule. — 1. Write the odd tuanUrs in a series 1, ;5, 5, 7, 1), etc. 2. After 3 eixtse every 'M number; after 5 erase every bth number ; after 7 erase every 7th number; after 11 erase every llth number y etc. 3. Then 2 and the numbers that remain are the prime numbers. KxERCiSE. — Find the prime numbers from 1 to 100. 85. The operations of Factoring depend upon the following PRINTIPI.ES. 1. A factor of a number exactly divides it. Tlius, 5 is a factor of 30 and is contained in it G times. 2. A multiple of a number exactly contains it. Thus, 30 is a multiple of 5 and contains it G times. 3. A factor of a number is a factor of any midtiple of that number. Thus, 3 being a factor of 6 is a factor of 12, 18, 24, etc. 4. A composite number is equal to the product of all its prime factors. Thus, the prime factors of 30 are 2, 3, and 5; 2 X ^ X ^ = 30. 86. In resolving numbers into their prime factors it will be found convenient to remember the following facts in reference to the prime numbers 2, 3, and 5. FACTORING. 121 1. Two is a factor of every even number. Thus, 2 is a factor of 4, 6, 8, 10, etc. 2. Three is a factor of a number when the sum of its digits is 3 or some multiple of 3. Thus, 3 is a factor of 2457; for 2 -[- 4 + 5 + 7 = 18, which is 6 times 3. 3. Five is a factor of every number ichose unit figure is or 5. Thus, 5 is a factor of 10, 15, 20, 25, etc. Rem. — Whether the prime numbers 7, 11, 13, etc., are factors of a number or not is best ascertained by trial. To Resolve a J^umher into its Prime Factors. 87. 1. Resolve 30 into its prime factors. Solution.— 2 is a factor of 30 (Art. 86, 1). Di- viding 30 by 2, the quotient is 15. 3 being a factor of operation. 15 (Art. 86, 2) is also a factor of 30 (Art. 85, Prin. 2)30 3). Dividing 15 by 3 the quotient is 5, a prime num- 3)15 ber. Then, 2, 3 and 5 are the prime factors of 30. 5 Rule. — 1. Divide the given number by any prime number that unll exactly divide it. 2. Divide the quotient in the same manner ; and so con- tinue to divide, until a quotient is obtained which is a prime number. 3. The several divisors and the last quotient will be the prime factors of the given number. Rem. — It will be most convenient to divide each time by the smallest prime number. 122 RAY'S NEW PRACTICAL ARITHMETIC. Resolve the following into their prime factors: 2. 4. ^ 2 ^1 -J. 23. 39. 3, 13 3. 8. 2, 2, 2. 24. 40. 2, 2, 2, 5 4. 9. 3, 3. 25. 42. 2, 3, 7 5. 10. 2, 5. 26. 44. 2,2,11 6. 12. 2, 2, 3. 27. 45. 3, 3, 5 7. 14. 2, 7. 28. 46. 2, 23 8. 15. 3, 5. 29. 48. 2, 2, 2, 2, 3 9. 16. 9 ■^5 2, 2, 2. 30. 49. 7, 7 10. 18. 2, 3, 3. 31. 50. 2, 5, 5 11. 20. 2, 2, 5. 32. 70. 2, 5, 7 12. 22. 2, 11. 33. 77. 7, 11 13. 24. 9 2, 2, 3. 34. 91. 7, 13 14. 25. 5, 5. 35. 105. 3, 5, 7 15. 26. 2, 13. 36. 119. 7, 17 16. 27. 3, 3, 3. 37. 133. 7, 19 17. 28. 2, 2, 7. 38. 154. 2, 7, 11 18. 32. 2 2 2^2 39. 210. 2, 3, 5, 7 19. 34. 2/17; 40. 231. 3, 7, 11 20. 35. 5, 7. 41. 330. 2, 3, 5, 11 21. 36. 2 2, 3, 3. 42. 462. 2, 3, 7, 11 22. 38. 2, 19. 43. 2310. 2, 3, 5, 7, 11 88. To find the prime factors common to two or more numbers. 1. What prime factors are common to 30 and 42? Solution. — Write the numbers in n line. 2 is a prime factor of both 30 and 42 (Art. 86, 1). Di- operation. viding by 2, the quotients are 15 and 21. 3 is a 2)30 4 2 prime factor of both 15 and 21 (Art. 86, 2); and 3)15 Tl consequently of both 30 and 42 (Art. 86, Prin. 3). 5 7 Dividing by 3, the quotients 5 and 7 are prime to each other (Art. 83, 8). Then 2 and 3 are the common factors. FACTORING. 123 Rule.— 1. Wiite the given mimbers in a line. 2. Divide by any prime number that will exactly divide all oj them; divide the quotients in the same manner ; and so continue to divide until two or more of the quotients are prime to each other. 3. Then the several divisors will be the common factors. What prime factors arc common to 2, 3, 5. 9 9 9 — , -J, -J. 2, 2, 3. 2, 3, 3. 3, 3, 3. 2, 2, 5, 2, 3, 7. 2, 2. 2, 3. 3, 3. 2, 5. 3, 5. 5, 5. 2, 7. 11. 13. 17. 19. 23. 89. Finding the G. C. D. of two or more numbers depends upon the following Principle. — The G. C. D. of two or more numbers con- tains all the prime factors common to the numbers, and no other factor. 2. 60 and 90? 3. 56 and 88? 4. 72 and 84? 5. 54 and 90? 6. 81 and 108? 7. 80 and 100? 8. 84 and 126? 9. 52, 68 and 76? 10. 66, 78 and 102? 11. 63, 99 and 117? 12. 50, 70 and 110? 13. 45, 75 and 105? 14. 75, 125 and 175? 15. 42, 70 and 98? 16. 33, 55, 77 and 121? 17. 39, 65, 91 and 104? 18. 34, 51, 85 and 102? 19. 38, 57, 95 and 114? 20. 46, 69, 92 and 115? 124 RAY'S NEW PRACTICAL ARITHMETIC. Thus, the G. C. D. of 12 and 18 is 6; it contains the common fac- tors 2 and 3; it must contain both of them, else it would not be the greatest C. D.; it can contain no other factor, else it would not divide both 12 and 18. 1. Find the G. C. D. of 80 and 42. First Method. OPERATION. Solution. — The prime factors common to 30 and 2)30 42 42 are 2 and 3 (Art. 88); their product is 6; then 3)15 21 the G. C. D. of 30 and 42 is 6 ( Prin.). 5 7 Rule. — 1. F'uid the prime factors common to the given numbers. 2. Multiply them together. 3. The product will be the greatest common divisor. Second Method. Solution. — Dividing 42 by 30, the re-^ mainder is 12; dividing 30 by 12, the re- operation. mainder is 6; dividing 12 by 6, the re- 30)42(1 mainder is 0. Then 6 is the G. C. D. of 3 30 and 42. For, 30 r= 6 X 5 a"d 42 = 6 T^) 3 ( 2 X 7; then, because 5 and 7 are prime to 2 4 each other, 6 must contain all the prime 0)12(2 factors common to 30 and 42; it- is, there- 12 fore, their G. C. D. (Prin.). Rule. — 1. Divide the greater number by the less, the divisor by the remainder, and so on, always dividing the last divisor by the last remainder, until nothing remains. 2. The last divisor nill be the greatest common divisor. Rem. — To find the G. C. D. of more than two numbers, first find the G. C. D. of two of them, then of that common divisor and one of the remaining numbers, and so on for all the numbers; the last common divisor will be the G. C. D. of all the numbers. FACTORING. 125 Find the greatest common divisor of the following numbers : 12. 18. 20. 27. 30. 16. 24. 36. 31. 26. 23. 19. 17. 39. .227. 12. 5. 8. 90. Finding the L. C. M. of two or more innnbers depends upon the following * Principle.— TAe L. C. M. of two or more numbers con- tains all the prime factor's of each number and no other factor. Thus, the L. C. M. of 12 and 18 is J^6; its prime factors arc % % 3, and 8; it must contain all these factors, else it would not contain both the numbers; it must contain no other factor, else it would not he the least CM. 2. 16, 24 and 40. 3. 24, 36 and 60. 4. 36, 54 and 90. 5. 40, 60 and 100. 6. 54, 81 and 108. 7. 60, 90 and 120. 8. 32, 48, 80 and 112. 9. 48, 72, 96 and 120. 10. 72, 108, 144 and 180. 11. 62 and 93. 12. 78 and 130. 13. 161 and 253. 14. 247 and 323. 15. 391 and 697. 16. 2145 and 3471. 17. 16571 and 38363. 18. 72, 120 and 132. 19. 75, 125 and 165. 20. 64, 96, 112 and 136. 126 EAY'S NEW PRACTICAL ARITHMETIC. 1. Find the L. C. M. of 4, 6, 9 and 12. Solution. — The prime factors of 4 are 2 and 2; those of 6 are 2 and 3; of 9, 3 and 3; and of 12, 2, 2, and 3. Then, the prime factors of the L. C. M. are 2, 2, 3, o, and no other factor ( Prin.). Hence, 36 is the L. C, M. OPERATION. 4 = 2X2 6 = 2X3 9 = 3X3 12 = 2X2X3 2X2X3X3 = 36 The process of factoring and selecting the prime factors for the L. C. M. is very much simplified hy the operation in the form of Short Division, as shown. OPERATION. 2)4 6 9 12 2 12 3 9 6 3)3 9 3 2 X 2 X 3 X 3 = 36. Rule. — 1. Wii'te the given numbers m a line, 2. Divide by any prime munher that will exactly divide two or more of them. 3. Write the quotients and undivided numbers in a line beneath, 4. Divide these numbers in the same manner^ and so con- tinue the operation until a line is reached in which the numbers are all prime to each other. 5. Then the product of the divisors and the numbers in the last line will be the least , common multiple. Rkm. — "When the quotient is 1 it need not be written. Find the least common multiple of 2. 4, 6 and 8. 3. 6, 9 and 12. 4. 4, 8 and 10. 5. 6, 10 and 15. 6. 6, 8, 9 and 12. 24. 36. 40. 30. 72. CANCELLATION. 127 7. 10, 12, 15 and 20. 60. 8. 9, 15, 18 and 30. 90. 9. 12, 18, 27 and 36. 108. 10. 15, 25, 30 and 50. 150. 11. 14, 21, 30 and 35. 210. 12. 15, 20, 21 and 28. 420. 13. 20, 24, 28 and 30. 840. 14. 45, 30, 35 and 42. 630. 15. 36, 40, 45 and 50. 1800. 16. 42, 56 and 63. 504. 17. 78, 104 and 117. 936. 18. 125, 150 and 200. 3000. 19. 10, 24, 25, 32 and 45. 7200. 20. 2, 3, 4, 5, 6, 7, 8 and 9. 2520. 21. 16, 27, 42, and 108. 3024. 22. 13, 29, 52, and 87. 4524. 23. 120, 360, 144, 720, and 72. 720. CANCELLATION. 91. 1. I bought 3 oranges at 5 cents each, and paid for them with pears at 3 cents each : how many pears did it take? OPERATION, Solution I.— 5 cents multiplied by 3 are 15 5 cents, the price of the oranges. 15 divided by 3 3 is 5, the number of pears. 3)15 5 From a consideration of this example and its solution we have the following Principle.— A mimber is not changed by multiplying it and then dividing the product by the mtdtipUer. 128 RAY'S NEW PRACTICAL ARITHMETIC. For the example, then, we may offer the following solution and operation : Solution II. — Indicate the multiplica- tion and division; then, erase or cancel the multiplier 3 and the divisor 3 hy drawing a line across them; and write the result, equal to 5. OPERATION. Rem. — The product 5 X ^ forms a dividend of which 3 is the divisor. 2. If 1 buy 10 pears at 3 cents each, and pay for them with oranges at 5 cents each : how many oranges will it take ? Solution. — 5 is a factor of 10, for 10 = 5X2; then, cancel the divisor 5 and also the factor 5 'in 10 by canceling 10 and writing the remaining factor 2 above it. The product of the remaining factors is 6. OPERATION. 2 —==6 3. Divide 15 X 21 by 14 x 10. Solution. — 5 is a common factor of 15 and 10; then, cancel 15, writing 3 above it, and 10, writing 2 below it. 7 is a common factor of 14 and 21; then, cancel 14, writing 2 below it, and 21, writing 3 above it. The product of the factors re- maining in the dividend is 9, and of those remaining in the divisor is 4; the quotient of 9 divided by 4 is 2\. Therefore, OPERATION. 3 3 2 2 --f=-2i Cancellation is a process of abbreviation by omitting the common factors of the dividend and divisor. CANCELLATION. 129 Rule. — 1 . Cancel the factors common to both the dividend and divisor. 2. Divide the product of the factors remaining in the dividend by the product of the factors reynainijig in the divisor. 3. The result will be the quotient required. 4. How many barrels of molasses, at S13 a barrel, will pay for 13 barrels of flour, at $4 a barrel ? 4. 5. Multiply 17 by 18, and divide by 6. 51. 6. In 15 times 8, how many times 4? 30. 7. In 24 times 4, how many times 8? 12. 8. In 37 times 15, how many times 5? 111. 9. Multiply 36 by 40, and divide the product by 30 multiplied by 8. 6. 10. In 36 times 5, how man}^ times 15? 12. 11. Multiply 42, 25, and 18 together, and divide the product by 21 X 15. 60. ^ 12. I sold 23 sheep, at $10 each, and was paid in hogs, at $5 each : how many did I receive ? 46. y 13. How many yards of flannel, at 35 cents a yard, will pay for 15 yards of calico, at 14 cents? 6 yd. V 14. What is the quotient of 21 X H X 6 X 26, divided by 13X3X14X2? 33. ^ 15. The factors of a dividend are 21, 15, 33, 8, 14, and 17 ; the divisors, 20, 34, 22 and 27 : required the quotient. 49. V 16. I bought 21 kegs of nails of 95 pounds each, at 6 cents a pound ; paid for them with pieces of muslin of 35 yards each, at 9 cents a yard : how many pieces of muslin did I give? 38. 17. What is the quotient of 35 X 39 X -10 divided by 26 X 30 X 42 ? If. Prac. 9. 130 * KAY'S NEW PKACTICAL ARITHMETIC. 18. What is the quotient of 26 X ^3 X 35 divided by 4X9X25? - 33^'. 19. What is the quotient of 6 X 9 X 15 >< 21 divided by 4X6X10X14? S^^. 20. What is the quotient of 21 X 24 X 28 X 35 divided by 14 X 18X20X25? 3||. <33?^»Jt'/^]f^ FRACTIONS. ' "%!'<* 92. A unit may be divided into equal parts ; thus, 1st. An apple may be divided equally between two boys, by cut- ting it into two equal parts. 2d. An apple may be divided equally among three boys, by cut- ting it into three equal jpartfi. 3d. In like manner, an apple may be divided into/o?/r,^ve, six, or any number of equal parts. These equal parts into which a unit may be divided are called fractions. DEFINITIONS. 93. 1. A fraction is one or more equal parts of a unit. 2. To express fractions by words and figures. When a unit is divided into two equal parts, Each part is called one-half, written i. Both parts are called iifo -halves, " f. When a unit is divided into three equal parts. Each part is called one-third, written ^. Two parts are called two-thirds, " f. All the parts are called three-thirds, " f. When a unit is divided into four equal parts. Each part is called one-fourth, written \. Two parts are called two-fourths, " f. Three parts are called three-fourths, " f . All the parts are called four -fourths, " |. (131) 132 KAY'S NEW PKACTICAL AK1THMP:T1C. When a unit is divided into five equal parts, Each part is called one-fifth, written \, Two parts are called two-fifths, " J. Three parts are called threefifths, " |. Four parts are called /owr-//f As, " ^. All the parts are called five-fifths, " |. When a unit is divided into six, seven, eight, etc., equal parts, each part is called one-sixth, ^, one-seventh, |, one-eighth, \, etc. 94. 1. A fraction is expressed in words by two num- bers; the first numbers the parts, the second names them; the first number is called the numerator, the second is called the denominator. 2. A fraction is expressed in figures, by writing the numerator above the denominator with a line between them. 3. The numerator and denominator are styled the terms of the fraction. 4. The denominator shows into how many equal parts the unit is divided, and the numerator, how many of the parts are taken. 95. When a unit is divided into equal parts, the size of each part depends upon the number of the parts. Thus, if apples of equal size be divided, one into two equal parts, another into three equal parts, a third mio four equal parts, etc., a half will be larger than a third, a third larger than a fourth, etc. Hence, 1st. The less the number of parts into which a unit is divided, the greater the size of each part. 2d. The greater the number of parts into which a unit is divided, the less the size of each part. COMMON FKACTIONS. I33 96. 1. A fraction may also be regarded as a part of one or more units. Thus: 1st. Two apples may be divided equally among three boys. Each boy will receive, either one-third of each of the two apples, or two-thirds of one of the apples; therefore, ^ of 2 is §. Hence, § may be considered either as two-thirds or as one-third of two. 2d. Two apples may be divided equally between two boys. Each boy will receive, either one-half of each of the two apples, or one of the two apples; therefore, ^ of 2 is |, or 1. Hence, | may be considered either as two halves or as one-half of two. 3d. Three apples may be divided equally between two boys. Each boy will receive, either one-half of each of the three apples, or one apple and one-half of another; therefore, J of 3 is |, or 1^. Hence, | may be considered either as three halves or as one-half of three. 2. A fraction is a part of one or more units. 3. The numerator expresses the ntimber of units. 4. The denominator expresses the part of each to be taken. 97. 1. A fraction may also be regarded as an ex- pression of division, in which the numerator is the dw- idend and the denominator the divisor. Thus: 1st. f is 2 divided by 3; here, the division can only be indicated, 2d. I is 4 divided by 2; in this case, the division can be performed exactly, giving a quotient 2. 3d. f is 5 divided by 2; in this case, the division can not be per- formed exactly, the quotient being 2^. 2. A fraction is an indicated division. The numerator is the dividend and the denominator is the divisor. 134 KAY'S NEW PRACTICAL ARITHMETIC. 3. A whole number may be expressed in tli(' form of a fraction, by writing the number for the umiierator and 1 for the denominator. Thus, 2 may be written |; for 2 divided by 1 is 2; 3 may be writ- ten |; 4 may be written f, etc. 98. The value of a fraction is its relation to a unit. 1. When the numerator is less than the denominatoi-, the value of the fraction is less than 1. Thus, ^, J, §, etc., arc less tlian 1. 2. When the numerator is equal to the denominator, the value of the fraction is equal to 1. Thus, f, |, f, etc., equal 1. 3. When the numerator is greater than the denomina- tor, the value of the fraction is givater than 1. Thus, |, |, f, etc., are greater than 1. 4. A proper fraction is one whose value is less th'an 1. 5. An improper fraction is one whose value is equal to or greater than 1. 6. A mixed number is a whole number and a fraction. 99. 1. A fraction may be divided into equal ])ai'ts. Thys, after an apple has been divided into two equal parts, each half may be divided into two equal parts; the whole apple will then be divided into four equal parts; therefore, ^ of ^ is \. Such expressions as ^ of ^, ^ of ^, etc., are termed compoimd fractions. 2. A compound fraction is a fraction of a fraction. COMMON FRACTIONS. 135 100. 1. Fractions sometimes occur in whicli the numerator, the denominator or botli are fractional. Thus, -r> oT> ^» are such expressions; they are called complex fractions. They are read 3^ divided hy 4, etc. 2. A simple fraction is one in which both terms are entire. 3. A complex fraction is one in whi^h one or both of the terms are fractional. 101. The operations with fractions depend upon the following ' Principles. 1. A fraction is multiplied by multiplying the numerator. Thus, if the numerator of | be multiplied bj' 3, the result will be f ; in ^ the parts are of the same size as in |, but there are three times as many. 2. A fraction is divm^d hy dividing the numerator. Thus, if the numerator of | be divided hy 3. the result will be ^; in ^ the parts are of the same size as in f , but there are only one- third as many. I 3. A fraction is divided hy multiplying the denominator. Thus, if the denominator of | be multiplied by 3, the result will be |; in I there are the same number of parts as in |, but the parts are only one-third as large. 4. A fraction is multiplied hy dividing the denominator. Thus, if the denominator of | be divided by 3, the result will be |; in I there are the same number of parts as in |, but the parts are three times as larsre. 136 RAY'S NEW PRACTICAL ARITHMETIC. 5. Multiplying both terms of a fraction by the same num- ber does not change its value. Thus, if both terms of | be multiplied by 2, the result is y«^; in ^ there are twice as many parts as in J, but they are only one-half as large. G. Dividing both terms of a fraction by the same number does not change its value. Thus, if both terms of /^ be divided by 2, the result will be |; in I there are only one-half as many parts as in 3^5, but they are twice as large. These six j)rinciple8 may be stated more briefly, as follows : ^ I. A fraction is multiplied, 1st. By multiplying the numerator. 2(1. By dividing the denominator. II. A fraction is divided, 1st. By dividing the numerator. 2d. By 7nultiplying the denominator. III. The value of a fraction is not changed, 1st. By multiplying both terms by the same number. 2d. By dividing both terms by the same number. The operations with fractions are Beduction, Addition, Subtraction, Multiplication and Division. COMMON FRACTIONS. 137 REDUCTION OF FRACTIONS. 102. Reduction of Fractions is changing their form witiiout altering; their value. There are six cases. CASE I. 103. To reduce an integer to an improper fraction, having a given denominator. 1. In 3 apples, how many halves? OPERATION. Solution. — In 1 apple there are 2 halves; then, in | X 3 = f 3 apples there are 3X2 halves = 6 halves. Rule.—l. Multiply the integer by the given denominator ; under the product write the denominator. 2. In 4 apples, how many halves? 3. In 2 apples, how many thirds? 4. In 3 apples, how many fourths? 5. In 4 apples, how many fifths? ^-f- 6. In 6 inches, how many tenths ? fj 7. In 8 feet, how many twelfths? ff 8. Eeduce 4 to sevenths. t^ 9. Eeduce 8 to ninths. V' 10. Eeduce 19 to thirteenths. W 11. Eeduce 25 to twentieths. ^ 12. Eeduce 37 to twenty-thirds. W CASE II. 104r. To reduce a mixed number to an improper fraction. 138 RAY'S NEW PRACTICAL ARITHMETIC. 1. In 31 apples, how many halves? OPERATION. Solution. — In 1 apple there are 2 halves; then, in |X3=| 3 apples there are 3 X ^ halves = 6 halves. 6 halves and 1 half are 7 halves. 1 + 1 = 1 Rule. — 1. Multiply the integer by the denominator of the fraction; to the product add the numerator, and under the sum write the denominator. 2. In 4^ apples, how man}^ halves? f 3. In 2\ apples, how many thirds? J 4. In 2| apples, how many thirds? | 5. In 5 1 dollars, how many fourths? ^^- 6. Reduce 8J to an improper fraction. ^- 7. Reduce 12f to an improper fraction. -^ 8. Reduce 15f to an improper fraction. ^ 9. Reduce 26^ to an improper fraction. ^^- It). Reduce 3^ to an improper fraction. ^~^- 11. Reduce 46| to an improper fraction. ^-p 12. Reduce 21^i| to an improper fraction. ^"Ml^ 13. Reduce lyVA ^^ ^'^ improper fraction. |^^| 14. Reduce 14^^ to an improper fraction. "Ti^ 15. Reduce lOy^ to an improper fraction. tVt CASE ITT. 105. To reduce an improper fraction to an integer or mixed number. 1. In I of an apple, how many apples? OPERATION. Solution.— There are 2 halves in 1 apple; then, in 2)6 6 halves, there are 6^2=3 apples. ~3 COMMON FRACTIONS. 139 2. In f of a dollar, how many dollars? OPERATION. Solution. — There are 4 fourths in 1 dollar; then, 4 ) 9 in 9 fourths, there are 9 -=- 4 =; 2^ dollars. 2^ Rule. — 1. Divide the numerator by the denominator ; the quotient will be the integer or the mixed number. 3. In f of an apple, how many apples? 2. 4. In -1^ of an apple, how many apples? 3. 5. In ^-^ of a dollar, how many dollars? $3|. 6. In y^ of- a dollar, how many dollars? S3f. 7. In ^ of a bushel, how many bushels? 2^ bu. 8. In ^ of a dollar, how many dollars? ^^y^ij^. 9. In -2^ of an ounce, how many ounces? 8^ oz. 10. In -5^ of a dollar, how many dollars? $131 11. Eeduce ^^ to a mixed number. 18|. 12. Reduce ^\^ to a mixed number. 15|. 13. Reduce ^^ to a mixed number. 25|^i. 14. Reduce ^^^ to an integer. 40. 15. Reduce ^^U^ to an integer. 31. 16. Reduce ^j^- to a mixed number. l^y^T* 17. Reduce ^^ to a mixed number. 46-j^. 18. Reduce ^-^-^ to a mixed number. 2im. 19. Reduce ^-f^ to a mixed number. 6^T%- 20. Reduce ^^^- to an integer. 199. 21. Reduce -^^ to a mixed number. I^tot* CASE IV. 106. To reduce a fraction to higher terms. A fraction is reduced to higher terms by multiplying both terms by the same number. This does not change its value (Art. 101, Prin. 5). 140 RAYS NEW PRACTICAL ARITHMETIC. 1. Eeduee | to thirtieths. OPERATION. Solution. — 30 divided by 5 is 6. Multiplying 3 0-t-5=: 6 both terms of | by 6, the result is ff 6X4 = 24 Kule. — 1. Divide the required denominator by the denom- inator of the given fraction. 2. Multiply both terms of the fraction by the quotient; the result will be the required fraction. 2. Eeduce i to fourths. f 3. Reduce 1 to sixths. *• 4. Reduce 1 to twelfths. T%- 5. Reduce 5 to twenty -fourths. If- 6. Reduce f to twenty-eighths. If 7. Reduce A to eighty-fourths. if- 8. Reduce i to seventy -seconds. n- 9. Reduce f to sixtieths. n- 10. Reduce A to hundredths. - 1%- 11. Reduce 9 2'U to a fniction whose denominator is 720. T5ir- 12. Reduce H to a fraction whose denominator is 2016. im- 13. Reduce H to a fraction whose denominator is 1935. 99 Q 1935- 14. Reduce If to a fraction whose denominator is 8118. nn- 15. Reduce If to a fraction whose denominator is 5134. 4832 5134* 16. Reduce H to a fraction whose denominator is 2332^ I am- 17. Reduce « to a fraction whose denominator is 2541. mi- COMMON FRACTIONS. ^ 141 CASE V. 107. To reduce a fraction to its lowest terms. 1. A fraction is reduced to lower terms by dividing both terms by the same number. This does not change its value. (Art. 101, Prin. 6). 2. A fraction is in its lowest terms when the numer- ator and denominator are prime to each other. (Art. 83, 8). 1. Eeduce |^ to its lowest terms. First Method. Solution. — 2 is a common factor of 24 and 30 operation. (86, 1 ). Dividing both terms of |^ by 2, tbe result 2 4 _ 1 2 is -^f. 3 is a common factor of 12 and 15 (86, ^~30~T5' 2). Dividing both terms of ^| by 3, the result is o\ 12 4 f . 4 and 5 are prime to each other. 1 5 ~~ 5 Rule. — 1. Divide' both terms of the given fraction by any common factor. 2. Divide the resulting fraction in the same manner. 3. So continue to divide until a fraction is obtained whose terms are prime to each other. Second Method. OPERATION. 24)30(1 Solution. — The greatest common divisor of 2 4 24 and 30 is 6. Dividing both terms of |J by ~6~) 2 4(4 6, the result is -|-. 2 4 '30 5 142 RAY'S NEW PRACTICAL ARITHMETIC. Rule. — 1. Divide both terms of the given fraction by their greatest common divisor. 2. The resulting fraction will be in its lowest terms. 2. Eeduce u to its lowest terms. i 3. Eeduce n to its lowest terms. I 4. Reduce H to its lowest terms. |. 5. Reduce n to its lowest terms. I 6. Reduce T% to its lowest terms. I 7. Reduce n to its lowest terms. I 8. Reduce A-V to its lowest terms. f- 9. Reduce 1% to its lowest terms. if. 10. Reduce m to its lowest terms. i^T- 11. Reduce m to its lowest terms. if- 12. Reduce m to its lowest terms. If 13. Reduce iV^ to its lowest terms. A. 14. Reduce i¥A to its lowest terms. 1^^. 15. Reduce ^m to its lowest terms. .^T 16. Reduce m to its lowest terms. u- 17. Reduce 585 1287 to its lowest terms. fV- 18. Reduce Tiffs to its lowest terms. A- 19. Reduce mi to its lowest terms. i¥r. 20. Reduce mi to its lowest terms. CASE VI. m- 108. To reduce two or more fractions to their least common denominator. 1. Two or more fractions have a common denominator when they have the same denominator. 2. A common denominator of two or more fractions is a commx)n multiple of their denominators (83, 11). COMMON FRACTIONS. 143 3. The least common denominator of two or more fractions is the least common multiple of their denom- inators (83, 12). 1. Eeduce nominator. , I, and to their least common de- OPERATION. 2)4 6 9 12 Solution. — The least common multiple of the denominators 4, 6, 9, and 12 is 36 (90). Each fraction, then, must be reduced to thirty-sixths (106). i = U* 5 3 8 3 2 nnc\ H 3 3 -6 — 3^' ¥ — ^S' ^^^^ T2 — 3T- 2)2 3 9 3)3 9 2X2X3 X 3 = 36. 36^4r= 9 36 - 6 = .6 9X3 = 27 6 X 5 = 30 4- 27 ? 3^ i=U 86 -=-9= 4 36 --12= 3 4^X 8 = 32 3X11 = 33 l-=M ih = U Rule. — 1. Find the L. C. M. of the denominators of the fractions for their least common denominator. 2. Reduce each fraction to another having this denomin- ator. Kem. 1. — Integers must be reduced to the common denominator by Art. 103, Rule. Rem. 2. — Before commencing the operation, mixed numbers must be reduced to improper fractions (104). Rem. 3. — Each fraction must be in its lowest terms (107 ). Rem. 4.— Two or more fractions may be reduced to any common denominator in the same way. Eeduce to their least common denominator: 2. 3. 4. h f- 1- ■l, *. 1- i. f, *• A, T%, A- ii if, II- M- TO"? M- 144 RAY'S NEW PKACTICAL AKITHMETIC. 5. .3 I fV- 6. i h J- 7. 3 5 5 T' ^' ¥• 8. h 3 i- 9. I 1. A- 10. h f. H- 11. h i i *• 12. I I f. f- 43. h A> A. a- 14. I 1. I H- 15. 2 i, s A- 16. 2i f- 4, 5f. H. M, e- if, fl, M- M, M, n- 1. t, h A, «, A- H, H, iJ- n, M- M, M- m, llf, fflf. m- A. if, if. ii- #i M, IS, M- H, If, il fj- faj i!, -¥/, ■w- ft. M, li. n- 17. 2i, 3i, 4i, 5. 18- A, ii, ii, if. If TTTfJ TT¥> Trf > TTT' ITT* !"• T' TTT' TT> 85» A> Tff- TtVtT' 126"I5"' 12 6 0' TTslj'' TJB^TF' T^VlT- 2^- fj iV' A' ii' if' ff- ^T)T7' 71)^17' ¥llir' ^17' ^TTTTj luiT' ADDITION OF FRACTIONS. 109. Addition of Fractions is the process of finding the sum of two or more fractional numbers. There are two cases. CASE I. 110. When the fractions have a common denominator. 1. Add i I and f. OPERATION. Solution.— The sum of 1 fifth, 2 fifths, and 3 i _[- 2 _^ 3 r= 6 fifths, is 6 fifths, f are equal to 1^ (Art. 106). l = H COMMON FRACTIONS. 145 Explanation. — Since the denominators are the same, the numer- ators express parts of the same size; therefore, add 1 fifth, 2 fifths, and 3 fifths, as you would add 1 cent, 2 cents, and 3 cents; the sum, in one case, being 6 fifths, in Ihe other, 6 cents. Rule.— 1. Add the numerators; under the sum write the common denominator. Rem. 1 — The result, if an improper fraction, must be reduced to an integer, or a mixed number (Art. 105). Rem. 2. — The result must be reduced to its lowest terms (Art. 107). 3. Add 4. Add I, I 3^ f If 5. Add 4, 4, i, I. 22. h I> i- h h f, f h h h f- i, s h 1- A, tV> A, H- 5 T3? A. A- H- tV. iV H, if- 9 20? TO? ¥1 H- fi Add -3_ _7_ _8_ iil. 9 6 7. Add -i-V, T%, A- H- 2t^ 8. Add tV„ a. H, if- 2f QAHH-JL1113 17 91 V. ^UU 2 0? "20"? 2U? "2"0^- ^"^^ 10. Add if, il if, ||. 2f CASE II. 111. When the fractions have not a common denom- inator. 1. Add I, I, and |i. Solution. — Reducing the fractions operation. to a common denominator (Art. 108), | ^^ff f = f f ih = II I = f f^ f -= fi and H = !l; then, the ff + f | + H = || sum of f^, ff, and |f is |f . || are U^^U equal to 2 f|. Explanation. — Since the denominators are different, the numer- ators do not express parts of the same size; therefore, the fractions can not be added till they are reduced to a common denominator. Prac. 10. 146 RAYS NEW PRACTICAL ARITHMETIC. Bule. — 1. Reduce the fractions to a common denominator. 2. Add the numerators^ and under the sum write the com- mon denominator. , Rem. 1. — Integers and fractions may bo added separately and their sums then united. Rem. 2. — The integral and the fractional parts of mixed numbers may be added separately and their sums then united. 2. Add \ and i- 3. Add \ and i- 4.. Add \ and f. 5. Add 1 and *• 6. Add 1 and 1- 7. Add i and +*. 8. Add 21 and 31. A- Solution. — The sum of \ and | is |; JrrrlJ; operation. write the \ under the column of fractions and carry 2\ the 1 to the column of integers. The sum of 1, 3, 3^ and 2 is 6. • 6^ A7y.s. 9. Add 10. Add 11. Add 12. Add 13. Add 14. Add 15. Add 16. Add 17. Add 18. Add 19. Add 20. Add 21. Add ^' ¥' h h W 1 1 2 ¥' J' IT- i n. 8f- tV ' tV Vt' iV- H . A- ii, M- tV , 2f, 3|, 3f 16 1, 12f, 8f, 2i. h i i.^i, i- I iV' "JIT' i4«> TirV^ tV ' 16' T2"' IfV, 2H- f. 2i, ^, 6i, 8i. H , 4f 2i, 2A- 2\. 2A\. 10f|. 1. 21H- 9f|. COMMON I'KACTIONS. U7 SUBTRACTION OF FRACTIONS. 112. Subtraction of Fractions is the process of finding the difference between two fractional numbers. There are two cases. CASE I. 113. When the fractions have a common denomin- ator. 1. From f subtract ^. OPERATION. Solution. — 2 'sevenths from 5 sevenths leaves 3 ^ — 7 = 7 sevenths. Explanation. — Since the denominators are the same, the num. erators express parts of the same size; therefore, subtract 2 sevenths from 5 sevenths as you would subtract 2 cents from 5 cents; the re- mainder, in one case, being 3 sevenths, in the other 3 cents. Kule. — 1. From the greater numerator subtract the less; under the remainder write the common denominator. 2. From | subtract \. \. 3. From | subtract |. \. 4. From | subtract |. \. 5. From -^-^ subtract f\. \. 6. From 3^ subtract If. operation. Solution. — f can not be taken from J; so borrow 3^ 1 from 3. 1 equals f ; f and | are f ; | from f leaves lf_ I; | — J. 2 from 8 leaves 1. l^ Ans. 7. From 4^ subtract 2|. H- 8. From 8-I subtract 3f 4|. 9. From 23^ subtract 17^^ ^. UB KAY'S NEW PRACTICAL ARITHMETIC. CASK II, 114. When the fractions have not a common denom- i nator. 1. From yV subtract |. Solution.— Ri^diicing the fractions to a common denominator (Art. 108), f = ^J and T% = H; then, f^ from |-J leaves ^% OPKRATION. Explanation. — Since the denominators are different, the nu- merators do not express parts of the satne size; therefore, one fraction can not he subtracted from the other till they are reduced to a common denominator. Bule. — 1. Beduce the fractions to a common denomin- ator. 2. Fro7n the greater numerator subtract the less, and under the remainder icrite the common denominator. 2. From ^ subtract i- 3. From \ subtract i- 4. From \ subtract 1- 5. From 1 subtract h 6. From 1 subtracc A- 7. From ^ subtract 1- 8. From 1^ subtract h 9. From j-4^ subtract tV 10. From ^^ subtract T^- 11. From 3^ subtract If- .8 15* fi- t's- tt- Solution. — \ equals |, and | equals |. | can Tiot OPHUATloN^. be taken from f ; so borrow 1 from the 3. 1 equals 3| f ; § fi^nd I arc |; | from | leaves §. 2 from 3 leaves 1, If 12. From 5 subtract |. 13. From H subtract 4|. 14. From "^1 subtract 4f. 15. From 141 subtract 12| 16. From 5A subtract 2i^. COMMON FKAOTIo/s. 149 If 2H- 17. From 4^i^ subtract 3^. 4| 18. From 56^ subtract 421 14_i__. 19. From 60| subtract 41^^. 19f 20. From 97| subtract 48|. 48|. MULTIPLICATION OF FRACTIONS. 115. Multiplication of Fractions is the process of finding the product of two or more fractional numbers. 1. If 1 apple cost I of a cent, what will 3 apples cost? OPERATION. Solution. — They will cost 3 times -| of a cent | X f = ¥' = -L2_ of a cent (Art. 101, Prin. 1). -i/ equals y- = 2f Explanation. — 3 apples will cost | -j- | -f | = ^^z. of a cent; hence, 3 times ^-=y,^-. 2. At 12 ct. a yard, what will | of a yard of ribbon cost? Solution. — \ of a yard will cost \ of 12;=rij2 ^t. operation. then, I of a yard will cost 2 times -V— ^ ^^- ( ^^^- ^ )' V" X f = "¥- ¥ = 44. ^ ^ .2^4^44 3. What will ^ of a yard of cloth cost, at | of a dollar pe** yard? dollar; then, ^ of a yard will cost 4 times ^3^ = i| of |X t= 35 a dollar. 15a RAY'S NEA^ PHACTICAL ARITHMETIC. Explanation. — f of I of a dollar is -^^ of a dollar (Art. 99); then, } of I of a dollar is 3 times -^^ = /^ of a dollar (Ex. 1). 4. MuUiply f by f Solution. — J is the same as ^ of 4 (Art. 96). f operation. multiplied by 4 is f (Art. 101, Prin. 1); then, f multi- f n^ |_ ^8^ plied hy ^ of 4 is J of 1 = ^5 (Ex. 3, Explanation). Rule. — 1. Multiply together the numerators of the given fractions for the numerator of the product. 2. Multiply together the denominators of the given frac- tions for the denominator of the product. Rem. 1. — Express integers in the form of fractions (Art. 97, 3). Rem. 2. — Reduce mixed numbers to improper fractions (Art. 104). Sometimes it may be more convenient to multiply by the integral and fractional parts separately. Rem. 3. — Indicate the operation and apply the Rule for Cancella- tion wherever it is practicable (Art. 91, Rule). 5. Multiply f by 3. 2\ 6. Multiply 8 by |. ^ 7. Multiply I by f. ^, 8. Multiply I by 4. 2|, 9. Multiply 5 by f . 3| 10. Multiply f by |. operation. Solution. — Indicating the operation and ap- ^ plying the Rule for Cancellation (Art. 91), the | X f — " f result is §. ^ 11. Multiply I by 6. 4. 12. Multiply 20 by f. 15. 13. Multiply ^\ by if H- 14. Multiply I by 10. 6. COMMON FRACTIONS. 151 15. Multiply 12 by |. 16. Multiply j\ by f. 17. Multiply f by 6. 18. Multiply 7 by |. 19. Multiply 21 by 3i. 2JL 91- To. Solution. — Reducing 2 J and SJ to im- proper fractions (Art. 104), they are | and |. Multiplying together | and |, the result is 6-3 ^:^ 71^ OPERATION. --n 20. Multiply 18f by 8. OPERATION. 144 + 6 = 150. 144 150 21. Multiply 8 by 3|. 22. Multiply 2^ by 2|. 23. Multiply 10| by 7. 24. Multiply 25 by 8|. 25. Multiply -^% by 17^. 26. Multii:)ly lOf by 9., 27. Multiply 64 by 8f. 28. Multiply 8f by f 29i. H- 7^. 215. 15^. 97|. 568. 3f. Multiply together : 29. 30. 31. 32. 33. 34. 35. 2A- TJ' 16' 2-1- -A- li ■^16' IT' *9- 21. If 6|, 2|, 2h 3|, 4f, 2i, 2A, 3i,.lT^. i _3_ 8. 5 ^ 6 8' 10? 9? ^' 3' T* 1 9 4 7 .5 2 fi 4' T' -^J 9^' 4' 3' "• 4091. 49if 22. 152 KAY'S NEW PRACTICAL AKITHMETlC. 36. f f, If, h h h h 20. 37. 2i, 6|, ^, ^3, 2, f. 24. 116. Fractioiuil multiplication. parts of integers are obtained by 1. What is I of 2? Solution.— J of 2 is f (Art. 96); then, f of 2 is 2 times ^ = i. i^^l OPERATION. i=n 3f. 2f 8. 10. 12f 18f Hi. 117. Compound fractions (Art. 99) are reduced to simple fractions by multiplication. 1. Reduce | of ^ to a simple fraction. Solution. — Multiplying | by f (Art. 115, Rule), operation. the result is j%. 2y^* = j\ 2. What is f of 5? 3. What is ! of 7? 4. What is 4 of 10? 5. What is 1 of 12? 6. What is 1 of 15? 7. What is 1 of 21 ? 8. What is -J^ of 25? 9. What is ^ of 27? 0. What is ^ of 28? 2. Reduce ^ of f to a simple fraction. 3. Reduce f of |^ to a simple fraction. 4. Reduce i of | of 2f to a simple fraction. 5. Reduce ^^ of f to a simple fraction. 6. Reduce f of f to a simple fraction. I of 4 of 1|^ to a simple fraction. 7. Reduce 3 3 14 3^ 15 ■3T' COMMON FRACTIONS. 153 8. Eeduce f of f of i to a simple fraction. 9. Reduce ^ of | of f to a simple fraction. 2T 10. Bediice f of |- of |^ to a simple fraction. 11. Reduce f of | of y7_ of |f to a simple fraction. 12. Reduce ^ of | of | to a simple fraction. 13. Reduce ^ of | of 1^ to a simple fraction. •^, 14. Reduce f of ^ of l\l to an integer. 1 15. Reduce f of 2f of If to an integer. 2, 16. Reduce ^3 of -/^ of 1|^ to a simple fraction. 17. Reduce ^ of 4 of | of 5 to a simple fraction. \, 18. Reduce ^ of | of f of | of f of | of j\ to a sim- ple fraction. iV* Miscellaneous Examples. 118. What will be the cost 1. Of 2^ lb. of meat, at l^ ct. a lb.? 30f ct. 2. Of 3 yd. linen, at $| a yd. ? Of 5 yd.? Of 7 yd.? Of 61 yd.? 5fyd.? ' S3f 3. Of 3^ lb. of rice, at 4|- ct. a lb. ? 16 ct. 4. Of 3| tons of iron, at $18f per T.? $60. 5. Of If yd. of muslin, at $^\ per yd.? ${. 6. Of 21 lb. of tea, at $f per lb.? $2. 7. Of 5| cords of wood, at $1| per C? $6|. ' 8. At the rate of b^ miles an hour, how far will a man travel in 7f hours ? 42| mi. 9. I own I of a steamboat, and sell f of my share : what part of the boat do I sell ? f . 10. At $6| per 3^ard, what cost | of a piece of cloth containing 5^ yards? $8^. 11. f of I of 161- X I of I- of 15 =: what? 34f. 12. What is the sum of f + J and | X I"? l^' 154 RAY'S NEW PRACTICAL ARITHMETIC. DIVISION OF FRACTIONS. 119. Division of Fractions is the process of finding the quotient of two fractional numbers. 1. If 3 yards of ribbon cost f of a dollar, what will 1 yard cost? OPERATION. Solution. — 1 yard will cost J of f =f of a dollar ^ (Art. 117). ?Xi-f Explanation. — ^ is to be divided into 3 equal parts. Each part will be 2 (Art. 101, Prin. 2); for f :== f -f- ^-f ^. 2. At 2 dollars a yard, what part of a yard of cloth can be bought for f of a dollar? Solution. — For 1 dollar \ a yard can be bought, operation. and for ^ of a dollar \ of \^^^^ of a yard (Art. |X| = A 117); then, for | of a dollar 3 times -^^^-^^ of a yard can be bought. Explanation. — Were it required to find how many yards, at $2 a yard, could be bought for $6, then 6 would be divided by 2; hence, to find the part of a yard that $| will pay for, \ must be divided by 2. To divide \ by 2, multiply the denominator (Art. 101, Prin. 3). 3. At I of a cent for 1 apple, how many can be bought for 4 cents? Solution. — For J of a cent \ an apple can be operation. bought, and for |, or 1 cent, 3 times | = | of an ^ • apple; then, for 4 cents, there can be bought 4 times f X J = 6 I ^= 6 apples. 4. At f of a cent for 1 apple, how many apples can be bought for f of a cent? COMMON FRACTIONS. 155 of a cent J an apple can be operation. bought, and for |, or 1 cent, 3 times i = | of an | X 1 = f apple; then, for \ of a cent J of f t=r | of an apple 1 = 1} can be bought (Art. 117), and for | of a cent 3 times f = 1-1- apples. 5. Divide | by 4. Solution. — f is the same as ^ of 4 (Art. 96). operation. I divided by 4 is j\ (Art. 101, Prin. 3); then, f sy^^ = \^ divided by i of 4Js 5 times j\ = i^ (Art. 115, Ex. 1). Rule. — Mulflply the dividend by the divisor with its terms inverted. Rem. 1, — Express integers in the form of fractions (Art. 97, 3). Rem. 2. — Reduce mixed numbers to improper fractions (Art. 104). Rem. 3. — Indicate the operation and apply the Rule for Cancel- lation whenever it is practicable (Art, 91, Rule). 6. If 4 yards of muslin cost f of a dollar, what will 1 yard cost? Sf 7. At i a cent each, how many apples can be bought for 3 cents? 6. 8. At ^ of a dollar per 3^ard, how many yards of mus- lin can be bought for Sy% ? ^' 9. If 1 orange cost 3 cents, what part of an orange could be purchased for ^ a cent? \. 10. At f of a dollar per yard, how many yards of cloth can you buy for 6 dollars? 8. 11. At \ of a dollar per yard, how many yards of ribbon can be purchased for f of a dollar? 3f. 12. If 7 pounds of rice cost i|. of a dollar, what will 1 pound cost? $^. 156 KAY'S NEW PRACTICAL ARITHMETIC. 13. Divide 4^- by If Solution. — Keducing 4^ and IJ to improper frac- tions (Art. 104), we have | and |. Dividing | by J, the result is 31. 14. Divide 2| by 6. 15. Divide 22 by 5J. 16. Divide ^ by tV. 17. Divide 4f by 8. 18. Divide 6 by 2f 19. Divide 4f by 51. 20. Divide 124 by 11. 21. Divide 30 by 3f . 22. Divide 2\ by 7f 23. Divide 3| by 7. 24. Divide 50 by ^. 25. Divide Iby A. 26. Divide 47| by 15. 27. Divide 56 by 5f 28. Divide H V 21. 29. Divide 130| by 18. 30. Divide i of 1 by 1 off OPERATION. li=J Y = 3| f- 4. 40. f- If- If 8. ^- 1 I HA- 25. lOf A- Explanation. — Invert the terms of both operation. I and ^ as in the case of the divisor being a jXfXIXi" simple fraction. Of 17^. 31. Divide f of f by f of 32. Divide i of 5^ by f 33. Divide j\ of | of 12f^ by ^ of 8|. 34. Divide f of | by f of ^ of 5. 35. Divide j\ of f of 12^3^ by i of 4^^ of 20. 5 6* i- 120. What part one number is of another is found by division. COMMON FRACTIONS. 1. 1 is what part of 2? Solution. — 1 is i of 2; for | of 2 is |, or 1 (Art. 98, 2d). 2. 2 is what part of 3? Solution. — 1 is ^ of 3; then, 2 is 2 times 1=1 of 3. 3. ^ is what part of 3? Solution. — 1 is i of 3; then, i is J of -J- 4. I is what part of f ? Solution. — J is i of f, and |, or 1, is 4 times =n:| of f; then, i is i of | = | of |, and |- is 2 imoQ 4 8 /-»f 3 157 i of 3. t-l of f; then, times J = I of f . 5. 3 is what part of 4 ? 6. f is what part of 5? 7. ^ is what part of i? 8. I is what part of |? 9. 3| is what part of 5 ? 10. I is what part of f ? 11. 8| is what part of 11? 12. fi is what part of |f? OPERATION. :Xi OPERATION. 2 V 1 ^- 1 /\ 3 3 OPERATION. OPERATION. 2 \/ 4 8 3 A 3 — ¥ 2- 4 5^- 3 1 5 16- I. 9- _9 121. Complex fractions, (Art. 100) are reduced to simple fractions by division. 1. Eeduce -| to a simple fraction. Solution.— Reducing IJ and 2i to improper frac- tions (Art. 104), we have { and |. Dividing | by J (Art. 119), the result is if. OPERATION. 91 — ^ ^3 — -i 158 RAY'S NEW PRACTICAL ARITHMETIC. 2. Eeduce -^ to a simple fraction. ^. 2 3. Keduce -^ to a simple fraction. y2^. 2 4. Keduce — to a simple fraction. ^. 31 5. Reduce j| to a simple fraction. ^J|. 21 6. Eeduce -^ to a simple fraction. J|. 7. Eeduce — to a simple fraction. |. 97 8. Eeduce -r^ to a mixed number. 44. 8J 9. Eeduce ^^ to a mixed number. If. ^^ 75 10. Eeduce ^ to a simple fraction. |^. Miscellaneous Examples. 122. 1. At ^ a dollar per yard, how many yards of silk can be bought for $3^? 6^. 2. At f of a dollar per pound, how many pounds of tea can be purchased for ^2^? 3|. 3. At 3| dollars per yard for cloth, how many yards can be purchased with $42^? 11^. 4. By what must | be multiplied that the product may be 10? 26|. 5. Divide 3f by f of If 5f 6. Divide ^ of 271 by ^ of 21f I29. 7. Multiply li by i. A- COMMON FRACTIONS. 159 8. Multiply JA of 5^ by ^. |». 'T2 ^T^ 6 li 2i 9. Divide ^2 by -f . 2 10 Divide — bv — ^. 32- 11. FRACTIONAL COMPOUND NUMBERS. 128. 1. Add $16j\', $9-1; $53-V; $2j|. SSB^-V 2. I paid for books $9|-; for paper, ^4^7^; for a slate, $|; for pens, $lf; what amount did I expend? S15^. 3. Having $50^, I paid a bill of $27-^^: how much had I left? S23Jg. 4. From $32.31^ take $15.12i. $17.18f. 5. From $5.81^ take $1.18f. ^4.621 Find the cost of 6. 9 yd. of muslin, at 121 ct. a yd. S1.12f 7. 21 lb. of sugar, at 6^ ct. a lb. $1.31f 8. 15 yd. of cloth, at $3.18f per yd. $47,811 9. 51 yd. of linen, at $0,621 per yd. $3.43|. 10. 121 yd. of ribbon, at 18| ct. per yd. $2.34|. 11. 131 yd. of calico, at 16| ct. per yd. $2.25. 12. 101 yd. of cloth, at $3,371 a yard. $34.59f. 13. 17| dozen books, at $3.75 per dozen. $66.25. 14. At 18| ct. per yard, how many yards of muslin can be purchased for $2.25? , 12 yd. 15. At 371 ct. per bushel, how many bushels of barley can you buy for $5.81|? 15^ bu. 16. If 5 yards of cloth cost $11.56^, what cost one yard? $2.31^. 17. Seven men share $31.06^ equally: what is the share of each man? $4.43f. IHO KAY'S NEW PKACTR AL AKITHMETIC. 18. Eeduce 5 mi. to inches. 316800 in. 19. Keduee 2 mi. 2 rd. 2 ft. to feet. 10595 ft. 20. Reduce 20 yd. to rods. ^ 3 rd. 3| yd. 21. Ecduce 15875 ft. to miles. 3 mi. 2 rd. 2 ft. 22. Reduce U2634 in. to miles. 2 mi. 80 rd. 2 yd. 2 in. 23. How many steps, of 2 ft. 8 in. each, will a man take in walking 2 miles? 3900. 24. How many revolutions will a wheel, of 9 ft. 2 in. circumference, make in running 65 miles? 37440. 25. Reduce 1 A. 136 sq. rd. 25 sq. yd. to square yards. 8979 sq. yd. 26. Reduce 7506 sq. yd. to A. 1 A. 88 sq. rd. 4 sq. yd. 27. Reduce 5 chains 15 links to in. 4078| in. 28. How many acres in a field 40^ rd. long and 32 rd. wide? 8 A. 16 sq. rd. 29. Reduce 4 years to hours. 35064 hr. 30. Reduce 914092 hr. to cen. 1 cen. 4 yr. 101 da. 4 hr, 31. In what time will a body move fi'om the earth to the moon, at the rate of 31 miles per day, the distance being 238545 miles? 21 yr. 24| da. 124. A fraction is reduced to a lower denomination by multiplication (Art. 63, Rule I). 1. Reduce J^ of a peck to the fraction of a i)int. Solution. — To reduce .}^ of a jx'ck to the opkration. fraction of u pint, multiply by 8 unci by 2. The ^ X I X r =- I result is | of a pint. ^ 2. Reduce -^^ bu. to the fraction of a quart. |. 3. Reduce -^^ lb. to the fraction of an ounce. ^. 4. Reduce yL. lb. Troy to the fraction of an ounce. |. 5. Reduce ^ rd. to the fraction of a foot. |^. 6. Reduce y^Vo" "^' ^^ ^^^^ fraction of a square rod. |. COMMON FK ACTIONS. 161 7. Reduce $3!^ to the fraction of a cent. f. 8. Eeduce y^^g^ da. to the fraction of a minute. |^. 9. Reduce -^^ bu. to the fraction of a pint. |. 125. In reducing a fraction to a lower denomination, when the result is a mixed number, proceed only with the reduction of the fractional part. This is called find- ing the value of a fraction in integers. 1. Find the value of f of a day in integers. Solution. — To reduce | of a day to hours, mul- tiply by 24; the result is 9| hr. To reduce | of an hour to minutes, multiply by 60; the result is 30 min. I of a day, then, is 9 hr. 36 min. 2. Find the value of 4 mi. in integers. 3. Find the value of $|- in integers. 4. Find the value of f mi. in integers. 5. Find the value of ^ lb. Troy in integers. 9 oz. 12 pwt. 6. Find the value of -^^ T. in integers. 8 cwt. 75 lb. 7. Find the value of | A. in integers. 100 sq. rd. 8. Find the value of -J of 63 gallons of wine in in- tegers. 55 gal. 1 pt. 12G. A fraction is reduced to a higher denomination by division (Art. 63, Rule IT). 1. Reduce | of a pint to the fraction of a peck. Solution.— To reduce | of a pint to the frac- operation. tion of a peck, divide by 2 and by 8. The result | X ^ X i = 2? is 2V of a peck. operation. -I X -\^- -^ 9f |X¥ = 36 256 rd. 60 ct. 128 rd. 2. Reduce 4 qt. to the fraction of a bushel. ^. luce 4 Prac. 11. 3. Reduce 4 ft. to the fraction of a rod. y 162 RAY'S NEW PKACTICAL ARITHMETIC. 4. Eeduce ^\ oz. to the fraction of a pound. rAir- 5. Keduce ^ lb. to the fraction of a ton. t^Vtt- 6. Eeduce f pt. to the fraction of a bushel. -^^. 7. Eeduce ^ oz. to the fraction of a hundred-weight. 2800- 8. Eeduce f in. to the fraction of a rod. ^J^ 9. Eeduce | min. to the fraction of a day. rwru- 10. Eeduce yf^ ^^- to the fraction of a hundred-weight. 127. To find what part one compound number is of another, reduce them to the same denomination and pro- ceed as in Art. 120. 1. 2 ft. 3 in. is what part of a yard? OPERATION. Solution. — 2 ft. 3 in. equals 27 in. I yd. 2 ft. 3 in. = 27 in. equals 36 in. 27 in. are |J of 36 in. fj 1 yd.=r=36in. equals j. 2 ft. 3 in., then, is } of a yard. ii = i 2. 2 ft. 6 in. is what part of 6 ft. 8 in.? | 3. 2 pk. 4 qt. is what part of a bushel ? | 4. What part is 2 yd. 9 in. of 8 yd. 2 ft. 3 in. ? ^ 5. What part of a day is 13 hr. 30 min. ? ^ 6. What part of a mile is 145 rd.? fj 7. What part of a yard is 2 ft. 8 in.? « 8. 15 mi. 123 rd. is what part of 35 mi. 287 rd.? f. 9. A man has a farm of 168 A. 28 sq. rd. ; if he sell 37 A. 94 sq. rd., what part of his farm will he dispose of? A\- 10. What part of a pound is 7^ oz. ? ^. 11. 2 qt. lipt. is Avhat part of 1 bu. 1 qt. If pt. ? 1 6 12. 1 yd. 1 ft. 1^ in. is what part of 3 yd. 2 ft. 8f in ? 1 91 « 1^- • 5 4T3' COMMON FKACTIONS. 163 128. To add and subtract fractional compound num- bers, find the value of the fractions in integers and then proceed as in Addition and Subtraction of Compound Numbers. 1. Add I yd. and f ft. Solution. — -| yd. equals 2 ft. 3 in.; | ft. equals 10 in.; the sum of 2 ft. 3 in. and 10 in. is 3 ft. 1 in. (Art. 75). OPERATION. yd. :rrr2 ft. 3 in. ft. = 10 in. 3 ft. 1 in. 2. From | da. subtract | hr. Solution. — | da. equals 5 hr. 20 min.; I hr. equals 50 min.; 50 min. subtracted from 5 hr. 20 min. leaves 4 hr. 30 min. (Art. 76). OPERATION. I da. ==5 hr. 20 min. I hr. = 50 min. 4 hr. 30 min. 3. Add I da. and f hr. 16 hr. 45 min 4. Add \ wk. i da. and \ hr. 2 da. 15 min 5. Add I wk. I da. | hr. and | min. 5 da. 6 hr. 40 sec 6. Add \^ gal. and ^2 ^t. 7. From ^ da. subtract Jg hr. 8. From H subtract SA. 3 qt. 1 pt. 2 gi. 18 hr. 36 min. 40 sec. 55 ct. 9. From | lb. subtract ^ oz. ^oz. 10. From | da. subtract ^ hr. 2 hr. 34 min. 17| sec. 129. Promiscuous Examples. 1. Reduce y g | j to its lowest terms. 2 Arid -^ 8 91 ^2 3. From 34^ subtract 1^. 4. From 3| subtract ^ of 3^. 5. Add f of j\ and | of -j^. a- 16^ RAY'S NEW PRACTICAL xVRITIlMETIC. G. Add 1| -^ 2.1- and ^ -^ 3i 2f|. 7. What niiinber divided by f will give 10 for a quo- tient? 6. 8. What number multiplied by | will give 10 for a product? 16|. 9. What number is that, from which if you take ^ of itself, the remainder will be 16? 28. 10. What number is that, to which if you add j^ of itself, the sum will be 20? 14. 11. A boat is worth S900 ; a merchant owns | of it, and sells ^ of his share : what part has he left, and what is it worth? * 3^ left, worth $375. 12. I own j^ of a ship, and sell ^ of my share for $1944|: what is the whole ship worth? 810000. 13. What part of 3 cents is | of 2 cents? |. 14. What part of 368 is 170? i|. 15. From |^ subtract the sum of ^, y^y, and Jj^y. 1 007 16. From 1 subtract -^^ of ^^ of 4yV y\. 17. From | ^ f subtract | -^ |f • 2%- 18. If I ride 2044 rods in y^^ of an hour, at that rate how far will I ride in 1|4 hr. ? 8468 rd. 19. What part of 1^ feet are 3^ inches? |. 20. Two men bought a barrel of flour ; one paid S3^, and the other $3| : what part of it should each have ? One ^^^, the other y^^^. 21. A has $2400 ; | of his money, + $500, is | of B's : what sum has B? $1600. 22. John Jones divided his estate among 2 sons and 3 daughters, the latter sharing equally Avith each other. The younger son received $2200, which was -^ of the share of the elder, w^hose share was ^f of the whole estate : find the share of each daughter. $1356^. 130. An aliquot part is an exact divisor of a number. Aliquot Parts of 100. 5 =iV 124 =i 25 =i 6i=TV ie| = * 33J = i " =tV 20 =1- 50 =1 The following multiples of aliquot parts of 100 are often used ; 184=^3^, 37i=a, 40=f, 60=f, 624=t, 75=f, 87*=^ 1. What will 24 yd. of muslin cost at 25 ct. a yd. ? OPERATION. Solution. — Since 25 ct. is \ of a dollar, the cost will 4)24 be ^ as many dollars as there are yards. \ of $24 is $6. $ 6 2. I spent $1,121 for muslin at 12| ct. a yd. : how many yd. did I bii} ? OPERATION. Solution. — Since V2h ct. is I of a dollar, there will 1"^ he 8 times as many yards as there are dollars. 8 times 8 H = 9yd. 9" yd. 8. What cost 12^- yd. of ribbon at ]8| ct. a yd? S2.34g. 166 KAY'S NEW PKACTICAL ARITHMETIC. 4. Paid $2.25 for muslin at 18J ct. a yd. how many yd. did I buy? 12 yd. 5. What will 5^ yd. of linen cost at S0.62i a yd.? $3.43f. 6. Paid $66.25 for books at $3.75 a dozen : how many doz. books did I buy? 17| doz. 7. What will 80 gal. of wine cost at $2.37^ a gal.? $190. 8. A number of men divide $39 so that each one re- ceives $4.87^: how many men are there? 8. 9. What will 36 barrels of flour cost at $8.33^ a bar- rel ? $300. 10. How many yd. of cloth at $1.33^ a yd. can be bought for $246. 66|? 185 yd. 11. What will 4 A. 60 sq. rd. of land cost at $16.50 an acre? Solution. — Since 1 A. costs $1G.50, operation. 4 A. cost $16.50X4:= $66. Since 160 $16.50 sq. rd. = 1 A., 40 sq. rd. = \ A. The ^ cost of 40 sq. rd. will be J of $16.50= 66.00 $4,121. The cost of 20 sq. rd. will be J J of $16.50 = 4.12^ of the cost of 40 sq. rd., or $2.06^. The J of 4.12^ = 2mI total cost is $66 -f $4.121 -f $2.06}= $72.18J $72.18|. 12. At $18.33^ per acre, how much land can be bought for $229,162? mA. 13. What will 11 A. 120 sq. rd. of land cost, at $125.60 per acre? $1475.80. 14. At $250 a lot, containing 50 X 150 ft., how much land can be bought for $10000? 6 A. 141 sq. rd. 28 sq. jd. 108 sq. in. 15. What will 83 bu. 3 pk. 2 qt. of grass seed cost, at $6.20 a bu.? $519.63|. PKACTICE. 167 16. At $0.75 a bushel, how raany bushels can be bought for S167.50? 223 bu. 1 pk. 2 qt. li pt. 17. What will 3| yd. cost, at $1.75 a yard? $6.12f 18. At SI. 50 a yard, how much cloth can be bought for S7.12I-? 4f yd. 19. What will 45 lb. 12 oz. of butter cost, at $0,371 per pound? $17.15|. 20. At $0,121 per pound, how much sugar can be bought for $2.93f ? 23-i lb. 21. What is the cost of 2 T. 9 cwt. of wool at 37^ ct. a pound? $1837.50. 22. What is the cost of 100 readers at $3.90 a dozen? $32.50. 23. What is the cost of 3f dozen knives at $5.40 a dozen? $20.25. 24. A farmer sold 6^ doz. chickens, at $0,331 apiece, and 37^ lb. butter, at $0.37^ per pound : he received $36 in money, and the remainder in sugar, at $0.12^ per pound: how many pounds of sugar did he get? 32i lb. 131. The orders of integers decrease from left to right in a tenfold ratio. Thus, in the mmibcr 1111, 1 thousand is 10 times 1 hundred,! hundred is 10 times 1 ten. and 1 ten is 10 times 1 unit. ORDERS OP DECIMALS. 132. 1. The orders may be continued from the order units toward the right by the same law of decrease. 2. Let tlie order units be separated from the order that follows by a point (.). 3. Then, in the number 1.111, Ist. Since the 1 to the left of the point is 1 unit, the 1 to the riu;ht of the point is 1 tenth; for 1 unit is 10 times J^. 2d. Since the first order from the unit is 1 tenth, the second order from the unit is 1 hundredth; for ^^ is 10 times yi^. 3d. Since the second order from the unit is 1 hundredth, the third order from the unit is 1 thousandth; for yi^ is 10 times yoVo- 4th, In like manner it may be shown that 1 in the fourth ^^ decimals. 170 1. RAY'S NEW PRACTICAL ARITHMETIC. yi^ is written .1 j\ are written .2 4 TIT 5 TIT .4 .5 A are written .6 A u -' .7 8 u u .8 T% u u .9 Hence, i^*/ien the denominator is 10, fAere is one decimal order. 2. Yw^ ^^ wi-itten .01; there being no tenths, a cipher is written in the vacant order. yf^ are written .02 j^^ are written .06 yfir " " -03 T^ " " -07 T^T " " •O't Ths " " -08 t4i7 " '• -05 T^tj " " -09 .Hence, when the denominator is 100, there are tiro deci- mal orders. 3. yo^oTj is written .001 ; there being no tenths and no hundredths, ciphers are written in the vacant orders. lOOlF 3 5 Tirxro .003 .004 .005 Att ^i*e written .006 T7T0¥ 8 9 TOGO .007 .008 .009 Hence, when the denominator is 1000, there are three decimal orders. 4. In like manner; 1 OUTFO is written TTr¥77"00^ .0001 .00001 .000001 DECIMAL FRACTIONS. 171 Hence, the number of orders in the decimal is always the same as the number of ciphers in the denominator of the common fractio7i. 5. j\ and j^-^ are jW written .11 1 1 1 1 f\V(^ 1111 '^ 1111 Hence, tenths and hundredths are read as hundredths; tenths, hundredths, and thousandths are read as thou- sandths ; tenths, hundredths, thousandths, and ten-thou- sandths are read as ten-thousandths, etc. 6. The numerator of a decimal is the number it ex- presses disregarding the decimal point. 7. If there are vacant orders before the numerator, ciphers are written in them. 8. The name of the right hand order is the name of the decimal. To Wj^ite Decimals. 135. 1. Write two hundred and sixty-five thousandths. Number Written. .265. Explanation. — First, write the numerator, 265, as an integer. The figure 5 must stand in the order thousandths (134, 8); then, 6 must be hundredths and 2 must be tenths; the decimal point, therefore, is phiced before the figure 2 (133, 2). 2. Write two hundred and sixty-five millionths. Number Written. .000265. Explanation. — Write the numerator, 265, as an integer. The figure 5 must stand in the order millionths (134, 8); then, 6 must be hundred-thousandths, 2 must be ten-thousandths, and ciphers must be written in the orders thousandths, hundredths, and tenths (134, 7 ); the decimal point is placed before tenths ( 133, 2 ). 172 KAY'8 NEW PRACTICAL ARITHMETIC. 3. Write two huiidrod and Hixty-fivc hundredths. NuMJJKR Written. 2.65. Explanation. — Write the numerator, 2G5, as an integer. The figure 5 must stand in the order hundredths; then, must be tenths; the decimal point, therefore, is placed between the figures 2 and 6. 4. Write four huiidred and ninety-eight and two hun- dred and sixty-five miUlonths. NuMBKR Written. 498.000265. Explanation. — First write the decimal as in Ex. 2; then write the integer, placing it at the left of the decimal point. Rule. — 1. Write the numerator as an integer. 2. Place the decimal point so that the name of the right hand order shall be the same as the name of the decimal. Note. — Pupils should be rendered familiar with the decimal orders so as to name them readily, in supcession, both from loft to right, and from right to left. Rem. 1. — When the decimal is a proper fraction it is sometimes necessary to prefix ciphers to the numerator ( Ex. 2 ). Rem. 2. — When the decimal is an improper fraction, the decimal point is placed between two of the figures of the numerator ( Ex. 3). Rem. 3. — In a mixed number, the decimal point is placed after the units order of the integer ( Ex. 4 ). Write the following decimal numbers: 5. Twenty-six hundredths. G. Thirty-five himdredths. 7. Eighty-seven hundredths. 8. Four hundred and nineteen hundredths. 9. Five thousandths. 10. Fifty-four thousandths. 11. Three hundred and four thousandths. DECIMAL FK ACTIONS. 173 12. Seven thousand two hundred and ninety -three thousandths. 13. Twenty-five and forty -seven thousandths. 14. Tw^o hundred and five ten-thousandths. 15. Four thousand one hundred and twenty -five ten- thousandths. 16. Mne hundred -thousandths. 17. Nine hundred thousandths. 18. Six hundred and five hundred-thousandths. 19. Twenty thousand three hundred and four hundred- thousandths. 20. Seven millionths. 21. Two hundred and three millionths. 22. Three hundred thousand and four millionths. 23. Twenty-four ten-millionths. 24. Eighty thousand and six ten-millionths. 25. Two hundred millionths. 26. Two hundred-millionths. 27. Nine hundred and seven hundred-millionths. 28. Twenty million twenty thousand and three hun- dred-millionths. 29. One million ten thousand and one hundred mill- ionths. 30. One million ten thousand and one hundred-mill- ionths. 31. One hundred and six and thirty-seven thousandths. 32. One thousand and one thousandth. 33. Two hundred and twenty-five thousandths. 34. Two hundred units and twenty-five thousandths. 35. Two thousand nine hundred and twenty-nine millionths. 36. Two thousand nine hundred units and twenty- nine millionths. 37. One million and five hillionths. 174 HAY'S NEW PRACTICAL ARITHMETIC. 38. Two hundred and two ten-biUionths. 39. Two hundred units and two ten-hillionths. 40. Sixty-five and six thousand and five inilUonths. Change the following common fractions to decimals: 4.1 3 7 9 17 2 3 4 1 5 3 '^^^ tW' tVh' iWtt' tWttj AVtf' tVV^- A*:{ 3 10 1 5 3 5 3 To Read Decimals. 136. 1. Read .2G5. Number Read. — Two hundred and sixty-five thousandths. Explanation. — Disregarding the decimal point, the number is two hundred and sixty-five; this is the numerator of the decimal (134, 6). The right hand order of the decimal is thousandths; this is the name of the decimal (134, 8). 2. Read .000265. Number Read. — Two hundred and sixty-five millionths. Explanation. — Disregarding the decimal point, the number is two hundred and sixty-five; this is the numerator of the decimal. The right hand order is millionths; this is the name of the decimal. 3. Read 2.65. Number Read. — Two and sixty-five hundredths, or two hundred and sixty five hundredths. Rule. — 1. Disregarding the decimal pointy read the num- ber as an integer. 2. Give the name of the right hand order. DECIMAL FRACTIONS. 175 Note. — Before commencing to read the decimal, the name of the right hand order should be ascertained (135, Note, under Rule). Rem. — A mixed number may be read either as an integer and a fraction, or as an improper fraction (Ex. 3). Eead the following decimal numbers : 4. .028; .341; 2.327; 50.005; 184.173. 5. .0003; .0625; .2374; .2006; .0104. 6. 3.0205; 810.2406; 10720.0905. 7. .00004; .00137; .02376; .01007. 8. .001768; .040035; 70.360004. 9. .1010101; .00040005; .00100304. 10. .31456; .000133; 60.04; 45.1003. 11. 357.75; .4928; 5.945; 681.0002. 12. 70.1200764; 954.203; 38.027. 13. 1007.3154; 7496.35491768. 14. .00715; 3.00005; 28.10065701. 15. 13.0008241094710947. Change the following decimals to common fractions, 16. .9; .13; .19; .29; .37; .73. 17. .91; .347; .513; .691; .851; .917. 18. .007; .0207; .00079; .001007. 19. 1.36; .3421; .03401; .0900. 20. .001; .5302; 8.01; .000053. 137, The operations with decimals are Reduction^ Ad- dition, Subtraction, Multiplication and Division. REDUCTION OF DECIMALS. 138. Keduction of Decimals is changing their form without altering their value. There are four cases. 176 KAY'S NEW PRACTICAL ARITHMETIC. CASE I. 139. 1. Annexing decimal ciphers to an integer does not change its vdlue. Thus, 7.00 is the sume as 7; for 7.00 is 7 and no hundredths (Art. 136, Rule). 2. Conversely : Omitting decimal ciphers from the right of an integer does not change its value. Number 1 of this case evidently corresponds to Case I, Art. 103, and 2 to Case III, Art. 105. CASE II. 140. 1. Annexing ciphers to a. decimal does not change its value. Thus, .70 is the same as .7; for y^ — y^o^. 2. Conversely: Omitting ciphers from the right of a decimal does not change its value. Number 1 of this case evidently corresponds to case IV, Art. 106, and 2 to Case V, Art. 107. CASE III. 141. To reduce a decimal to a common fraction. 1. Eeduce .75 to a common fraction. Solution. — 75 hundredths written as a common operatiox. fraction is j^j^^. -^^^ reduced to its lowest terms (Art. .75 = -^^-^ 107),isf. i%=l Bule. — 1. Write the decimal as a common fraction. 2. Reduce the fraction to its lowest terms. T6"- JL9_ 12 8- ^5U- DECIMAL FKACTIONS. 177 2. Reduce .6 to a common fraction. |. 8. Reduce .25 to a common fraction. \. 4. Reduce .375 to a common fraction. f. 5. Reduce .035 to a common fraction. 2-^^. 6. Reduce .5625 to a common fraction. 7. Reduce .34375 to a common fraction. \ 8. Reduce .1484375 to a common fraction. 9. Express 4.02 as an integer and common fraction 4 10. Express 8.415 as an integer and common fraction CASE IV. 142. To reduce a common fraction to a decimal. 1. Reduce f to a decimal. Solution. — Annexing a decimal cipher to 3, it is operation. 3.0; 30 tenths divided by 4 is 7 tenths, and 2 tenths 4 ) 3.00 remaining. Annexing a cipher to .2 it is .20; 20 .7 5 hundredths divided by 4 is 5 hundredths. The result is .75. Explanation. — J is 3 divided by 4 (Art. 97 ). Annexing a decimal cipher to 3 does not change its value (Art. 139). Annexing a cipher to .2 does not change its value (Art. 140). B.ule. — 1. Annex decimal ciphers to the numerator. 2. Divide by the denominator. 3. Point off as many decimal orders iyi the quotient as there are decimal ciphers annexed to the numerator. 2. Reduce I to a decimal. .8 3. Reduce f to a decimal. .625 4. Reduce 2V ^^ ^ decimal. .28 Prac. 12. 178 KAY'S NEW PRACTICAL AKITHMETIC. 5. Eeduce -^jj to a decimal. .075 6. Eeduce ^| to a decimal. .9375 7. Eeduce yrsir *^ *^ decimal. .0008 8. Eeduce -^^j^ to a decimal. .0225 9. Eeduce ^l^ to a decimal. .00390625 10. Eeduce |^ to a decimal. .83 -\- 11. Eeduce ^ to a decimal. .09 + 12. Eeduce ^\ to a decimal. .12 -|- ADDITION OF DECIMALS. 143. Addition of Decimals is the process of finding the sum of two or more decimal numbers. 1. Add 375.83; 49.627; 5842.1963; 813.9762. Solution. — Write the numbers so that the operation. four decimal points may be in a column, the 3 7 5.83 units 5, 9, 2, 3 in the first cohimn to the left, the 4 9.62 7 tenths 8, 6, 1, 9 in the first column to the right, 5 842.196 3 etc.; then, adding as in simple numbers, place the 813.9762 decimal point in the sum between 1 and 6 under 7081.6295 the column of decimal points. Rule. — 1. Write the ninnhers so that the decimal points and figures of the same order may stand in the same cohunn. 2. Add as in simple numbers. 3. Place the decimal point in the sum under the column of decimal points. 2. Add 37.1065; 432.07; 4.20733; 11.706. 485.08983 3. Add 4 and 4 ten-thousandths; 28 and 35 thou- sandths; 8 and 7 hundredths; and 9404 hundred-thou- sandths. 40.19944 DECIMAL FEACTIOISIS. 179 4. Find the sum of 3 units and 25 hundredths; 6 units and 4 tenths; and 35 hundredths. 10. 5. Add 21.611; 6888.32; 3.4167. 6913.3477 6. Add 6.61; 636.1; 6516.14; 67.1234; and 5.1233. 7231.0967 7. Add 4 and 8 tenths ; 43 and 31 hundredths ; 74 and 19 thousandths; 11 and 204 thousandths. 133.333 8. Add 45 and 19 thousandths; 7 and 71 hundred- thousandths ; 93 and 4327 ten-thousandths ; 6 and 401 ten-tiiousandths. 151.49251 9. Add 432 and 432 thousandths; 61 and 793 ten- thousandths; 100 and 7794 hundred-thousandths; 6.009; 1000 and 1001 ten-thousandths. 1599.69834 10. Add 16 and 41 thousandths; 9 and 94 millionths ; 33 and 27 hundredths; 8 and 969 thousandths; 32 and 719906 milUonths. 100. 11. Add 204 and 9 ten-thousandths; 103 and 9 hun- dred-millionths; 42 and 9099 millionths; 430 and 99 hundredths ; 220.0000009. 999.99999999 12. Add 35 ten-thousandths; .00035; 35 millionths, and 35 ten-milHonths. .0038885 SUBTRACTION OF DECIMALS. 144. Subtraction of Decimals is the process of find ing the difference between two decimal numbers. 1. From 729.835 subtract 461.5738. Solution. — Write the numbers so that the two decimal points may be in a column, the units 9 operation. and 1 in the first column to the left, the tenths 8 7 2 9.835 and 5 in the first column to the right, etc.; then, 4 61.5 7 38 subtracting as in simple numbers, place the deci- 2 6 8.26 12 mal point in the remainder between 8 and 2 under the column of decimal points. 180 RAY\S NEW PRACTICAL ARITHMETIC. Rem. — The ten-thousandth phice in the minuend may be regarded as occupied by a cipher (Art. 140). Rule. — 1. Write the numbers so that the decimal points and figures of the same order may stand in the same column. 2. Subtract as in simple numbers. 3. Place the decimal point in the remainder under the column of decimal points. 2. From 97.5168 subtract 38.25942. 59.25738 3. From 20.014 subtract 7.0021. 13.0119 4. From 5.03 subtract 2.115. 2.915 5. From 24.0042 subtract 13.7013. 10.3029 6. From 170.0035 subtract 68.00181. 102.00169 7. From .0142 subtract .005. .0092 8. From .05 subtract .0024. .0476 9. From 13.5 subtract 8.037. 5.463 10. From 3 subtract .00003. 2.99997 11. From 29.0029 subtract 19.003. 9.9999 12. From 5 subtract .125. 4.875 13. From 1 thousand subtract 1 ten-thousandth. 999.9999 14. From 1 subtract 1 millionth. .999999 15. From 25 thousandths take 25 millionths. .024975 MULTIPLICATION OF DECIMALS. 145. Multiplication of Decimals is the process of finding the product of numbers involving decimals. 146. Placing the decimal point in the product de- pends upon the following DECIMAL FRACTIONS. 181 Principle. The nvmber of decimal orders in the product is equal to the number of decimal orders in both the factors. Thus, let the factors be .2 and .03; then, the number of decimal orders in the product will be three. For, .2=z^q and .0o = y3_. then, the product of .2 by .03 will be the same as the product of ^^ by y^^ But, rX Too — To^o> and T^oo = :.006. Therefore, .2X-03- .006, in which there are three decimal orders. Examples. 147. 1. Multiply 2.149 by 6.34. Solution. — Multiply as in simple numbers, 2140 by 634. There are three decimal orders in 2.140, and two decimal orders in 6.34; hence, there must be five decimal orders in the product (Art. 146). There- fore, the product is 13.62466. OPERATION. 2.1 4 6.3 4 "85 6 6447 1 2804 ] 3.6 2 4 6 6 2. Multiply .0276 by .035. Solution. — Multiply the numerator (Art. 134, 6) 276 by the numerator 35; the result is 0660. There are four decimal orders in .0276, and three decimal orders in .035; hence, there must be seven decimal orders in the product (Art. 146); three ciphers, then, must be prefixed to 0660. Therefore, omitting the .0000660 cipher on the right (Art. 140, 2) the product is .000066. 3. Multiply 2.075 by 100. Solution. — Write 2075 and place the decimal operation. point between 7 and 5, two places farther to the right 2 7.5 than it is in 2.075. 182 RAY'S NEW PRACTICAL ARITHMETIC. Rem. — To multiply 207.5 by 100, annex a cipher and mcve the decimal point two places to the right. Kule, — 1. Multiply together the numerators of the deci- mals as in Simple Numbers. 2. Point off as many decimal orders in the jiroduct as there are decimal orders in both factors. Rem. 1. — When the number of figures in the product of the numerators is less than the number of decimal orders required, prefix ciphers. (Ex. 2.) Rem. 2. — After placing the decimal point, omit ciphers at the right of the decimal part of the product. (Ex. 2.) Rem. 3. — To multiply a decimal by 10, 100, 1000, etc., remove the decimal point as many places to the right as there are ciphers in the multiplier. If there be not enough figures annex ciphers. 4. Multiply 5. Multiply 6. Multiply 7. Multiply Multiply Multiply Multiply 11. Multiply 12. Multiply 13. Multiply 14. Multiply Multiply Multiply Multiply Multiply Multiply Multiply 21. Multiply 22. Multiply 8. 0, 10 15 IG, 17, 18. 19. 20. 33.21 by 4.41. 32.16 by 22.5. .125 by 9. .35 by 7. .2 by .8. .02 by .4. .15 by .7. 125.015 by .001. .135 by .005. 1.035 by 17. 19 by .125. 4.5 by 4. .625 by 64. 61.76 by .0071. 1.325 by .0716. 4.87 by 10. 5.3 by 100. 17.62 by 100. 1.01 bv 10. 146.4561 723.6 1.125 2.45 .16 .008 .105 .125015 .000675 17.595 2.375 18. 40. .438496 .09487 48.7 530. 1762. 10.1 DECIMAL FKACTIONS. 183 23. Multiply .0001 by 100. .01 24. Multiply 1 tenth by 1 hundredth. .001 25. Multiply 1 hundred by 1 ten-thousandth. .01 26. Multiply 43 thousandths by 21 ten-thousandths. .000090a 27. Multiply 40000 by 1 millionth. .04 28. Multiply .09375 by 1.064. .09975 DIVISION OF DECIMALS. 148. Division of Decimals is the process of finding the quotient of two numbers involving decimals. 149. Placing the decimal point in the quotient de- pends upon the following Principle. The number of decimal orders in the quotient is equal to the number of decimal orders in the dividend^ less the num- ber in the divisor. Thus, let .006 be divided by .03; then, the number of decimal orders in the quotient will be one. For .006=:y^%^ and .0?)=:j|}o; then, the quotient of .006 by .03 will be the same as the quotient of To%o divided by ^f^. But, \o%o - if o = I'o ; ^^^^^l -^,= :2. There- fore, .006 -^- .03 = .2, in which there is one decimal order. Examples. 150. 1. Divide 2.125 by .5. Solution. — Divide as in simple numbers 2125 by 5. operation. There are three decimal orders in 2.125, and one .5)2.12 5 decimal order in .5; hence, there must be two decimal 4.2 5 orders in the quotient (Art. 149). Therefore, the quotient is 4.25. 184 KAY'S NEW PRACTICAL ARITHMETIC. 2. Divide .048 by .006. Solution. — Divide the numerator (Art. 134, 6) 48 operation. by the numerator 6. There are three decimal orders .006).0 4 8 in .048, and three decimal orders in .006; hence, 8 there will be no decimal orders in the quotient (Art. 149). Therefore the quotient is 8. 3. Divide .3 by .004. OPERATION. Solution. — Annex two ciphers to .3; then solve .004 ).800 as in Ex. 2. 75" 4. Divide 83.1 by 4. Solution. — Annex two ciphers to the decimal operation. (Art. 140, 1) in order that the division may be per- 4 ) 8 3.1 00 formed exactly; then solve as in Ex. 1. 20.7 7 5 5. Divide 2.11 by 3. Solution. — Annex one or more ciphers to the dec- operation. imal (Art. 140, 1) in order to carry the division as 3)2.1 10 far as is wanted; then solve as in Ex. 1. ,7 03-|- 6. Divide 475.(1 by 100. Solution. — Write 4756 and place the decimal operation. point between 4 and 7, two places farther to the left 4.7 5 6 than it is in 475.6. Rem. — To divide 4.756 by 100 prefix a cipher; thus, .04756. Rule. — 1. Divide the numerator of the dividend by the numerator of the divisor as in simple numbers. 2. Point off as many decimal orders in the quotient as the number of orders in the dividend exceeds the number in the divisor. DECIMAL FKACTIONS. 185 Eem. 1. — When the number of decimal orders in the dividend is the same as the number in the divisor, the quotient is an integer (Ex. 2). Rem. 2. — When the number of decimal orders in the dividend is less than the number in the divisor, for convenience in pointing off, make them the same by annexing ciphers to the dividend ( Ex. 3). Rem. 8. — When the division is not exact, it may be continued to any required number of decimal places (Ex. 5). Rem. 4. — To divide a decimal by 10, 100, 1000, etc., remove the decimal point as many places to the left as there are ciphers in the divisor. If there be not enough figures, prefix ciphers (Ex. 6, Rem). 7. Divide 1.125 by .03. 37.5 8. Divide 86.075 by 27.5. 3.13 9. Divide 24.73704 by 3.44. 7.191 10. Divide 206.166492 by 4.123. 50.004 11. Divide .96 by .24. 4. 12. Divide .0425 by .0085. 5. 13. Divide 21 by .5. 42. 14. Divide 2 by .008. 250. 15. Divide 37.2 by 5. 7.44 16. Divide 100.8788 by 454. .2222 17. Divide .000343 by 3.43. .0001 18. Divide 9811.0047 by .108649. 90300. 19. Divide .21318 by .19. 1.122 20. Divide 102048 by .3189. 320000. 21. Divide .102048 by 3189. .000032 22. Divide 9.9 by .0225. 440. 23. Divide 872.6 by 100. 8.726 24. Divide 4.5 by 1000. .0045 25. Divide 400 by 10000. .04 26. Divide 1 tenth by 10. .01 27. Divide 1 by 1 tenth. 10. 28. Divide 10 by 1 hundredth. 1000. 29. Divide 1.7 by 64. .0265625 186 liAY'S NEW PRACTICAL ARITHMETIC. 30. Divide .08 by 80. .001 31. Divide 1.5 by 7. .2142857 + 32. Divide 11.1 by 32.76. .3388278 + 33. Divide .0123 by 3.21. .00383177 + DECIMAL COMPOUND NUMBERS. 151. A decimal is reduced to a lower denomination by multiplication (Art. 63, Rule I). 1. Reduce .05 gal. to the decimal of a pint. orKRATION. .06 Solution. — To reduce .05 gal. to the decimal of a 4 pint, multiply by 4 and by 2. The result is .4 pint. .20 2_ .4 2. Reduce .035 pk. to the decimal of a pint. .56 pt. 3. Reduce .0075 bu. to the decimal of a quart. .24 qt. 4. Reduce .005 yd. to the decimal of an inch. .18 in. 5. Reduce .00546875 A. to the decimal of a square rod. .875 sq. rd. 152. To find the value of a decimal in integers (Art. 125). 1. Find the value of .3125 bu. in integers. ^ OPERATION. Solution. — To reduce .3125 bu. to pecks, multiply .8 12 5 by 4; the result is 1.25 pk. To reduce .25 pk. to 4 quarts, multiply by 8; the result is 2 qt. Therefore, 1.2 5 00 .3125 bu. equals 1 pk. 2 qt. 8 2:00 2. Find the value of .75 yd. in integers. 2 fl. 3 in. 3. Find the value of .3375 A. in integers. 54 sq. rd. DECIMAL FRACTIONS. 187 4. Find the value of .7 lb. Troy in integers. 8 oz. 8 pwt. 5. Find the value of .8125 bii. in integers. 3 pk. 2 qt. 6. Find the value of .44 mi. in integers. 140 rd. 4 yd. 1 ft. 2.4 in. 7. Find the value of .33625 cwt. in integers. 33 lb. 10 oz. 153. A decimal is reduced to a higher denomination by division (Art. 63. Rule II). 1. Reduce .64 pt. to the decimal of a gallon. OPERATION. Solution. — To reduce .64 pt. to the decimal of a 2).6 4 gallon, divide by 2 and by 4. The result is .08 gal. 4 ).3 2 708 2. Reduce .72 qt. to the decimal of a bushel. .0225 bu. 3. Reduce .77 yd. to the decimal of a mile. .0004375 mi. 4. Reduce .25 pt. to the decimal of a gallon. .03125 gal. 5. Reduce .6 pt. to the decimal of a bushel. .009375 bu. 6. Reduce .7 rd. to the decimal of a mile. .0021875 mi. Promiscuous Examples. 154. 1. What is the cost of 9 yd. flannel, at $0.40 per yard, and 12 yd., at S0.75 per yard? S12.60. 2. What is the cost of 2.3 yd. of ribbon, at $0.45 per yard, and 1.5 yd., at $0,375 per yard? $1.5975. 188 KAYS NEW PKACTICAL ARITHMETIC. 3. What is the cost of 16.25 yd. of cloth, tit $2.6875 per yard? $43.671875. 4. At $0.75 per bushel, how much wheat can be boui^ht for $35.25? 47 bii. 5. At $2.5625 per yard, how much cloth can be bought for $98.40 ? 38.4 yd. 6. What will 6 cwt. 50 lb. of hops cost at $3.25 per hundred-weight? $21,125. 7. What will 14 bu. 3 pk. 4 qt. of corn cost, at $0,625 per bushel? $9.296875. 8. What will 13 A. 115 sq. rd. of land cost, at $17.28 per acre? $237.06. 9. At $0.3125 per bushel, how much corn can be bought for $9.296875? 29 bu. 3 pk. 10. At $4.32 per acre, how much land can be bought for $59,265? 13 A. 115 sq. rd. 11. If 63 gal. of wine cost $49, what will 464 gal cost? $360.88 + 12. Add .34 3^d., 1.07 ft. and 8.92 in. 2 ft. 10 in. 13. Add .625 gal. and .75 qt. 3 qt. .5 pt. 14. From 1.53 yd. subtract 2 ft. 3.08 in. 2 ft. 4 in. 15. From .05 yr. subtract .5 hr. 18 da. 5 hr. 48 min. 16. From .41 da. subtract .16 hr. 9 hr. 40 min. 48 sec. 17. Find the value of .3 yr. in integers. 109 da. 13 hr. 48 min. 18. What is the cost of 343 3^d. 2 ft. 3 in. of tubing, at $0.16 per yard? $55. 19. At $690.35 per mile, what is the cost of a road 17 mi. 135 rd. long? $12027.19140625. THEjMETRreiSYSTEM. DEFINITIONS. 155. 1. The Metric System is so called from the meter, the unit upon which the system is based. Eem. — The French originated this system of weights and measures at the close of the last centurj^, and its use in France became obligatory in 1841. The metric system is now legal in nearly all civilized countries, and, in several, it is making its way rapidly into general use. In 1866, its use was legalized, in the United States, by act of Congress. It is in general use by scientific men throughout the world. 2. All the units of the other measures are derived in a simple manner from the meter. Thus, 1st. The Meter is the unit of Length. It is the base of the Metric System, and is very nearly one ten-mill- ionth (.0000001) part of the quadrant extending through Paris from the equator to the pole. 2d. The Ar is the unit of Land Measure. It is a square whose side is 10 meters. 3d. The Liter is the unit of capacity. It is a vessel wdiose contents are equivalent to a cube the edge of which is .1 meter. 4th. The Gram is the unit of Weight. It is the weight of a cube of pure water whose edge is .01 meter. (189) 190 KAY'S NEW rilACTlCAL ARITHMETIC. 3. The name of each denomination indicates at once its relation to the unit oi' the measure. 27 Thus: 1st. The luinies uf the louver dononiiiiutions are formed by prefixing to the natiie of the unit the Latin numerals mdli (.001), ctntl (.01), and dcci (.1). For example, a miUimeter is one thousandth of a meter; a centigi-a^n is one hundredth of a gram; and a deciliter is one tenth of a liter. 2d. The names of the liiffhrr drnoiiiinatidiis arc formed by prefix! Hi;- to tin- unii tin- a ml; miiiuTal.- (Iil;ii (10), heldo (100), /.v/o i l(»n(» . mimI inur'in i KMHXh. For example, a dekameter i- I'^n nirt.T.-; a hrl;ii_,r,frr i> ..nc hundred liters; a kilocjrain is one tliousand grams; and a >ni/ria7neter is ten thousand meters. 4. Since in the Metric System 10, 100. 1000, etc., units of a lower denomination make a unit of a higher denomination, it follows thai. Ist. A number is reduced in g- 10 dekagrams " 1 hektogram, " Hg. 10 hektograms *' 1 kilogram, " Kg. 10 kilograms " 1 myriagram, " Mg. 10 myriagrams " 1 quintal, " Q 10 quintals, or 1000 kilograms, " 1 metric ton, " M.T. Rem. — The weights commonly used are the gram, kilogram, and metric ton. The gram is used in mixing medicines, in weighing the precious metals, and in all cases where great exactness is required. The kilogram — or, as it is commonly called, the "kilo"— is the usual weight for groceries and coarse articles generally; it is very nearly 2i pounds Av. The metric ton is used for weighing hay and other heavy articles; it is about 204 lb. more than our ton. Prac. 13. 194 KAY'S NEW PRACTICAL ARITHMETIC. 1. Eediice 1428.06 g. to kilograms. 1.42806 Kg. 2. Eeduco .28 Kg. to grams. 280 g. 3. Reduce 1713.5 Kg. to metric tons. 1.7135 M.T. 4. lieduce .00654 Hg. to centigrams. 65.4 eg. 5. Reduce 192.7 dg. to dekagrams. 1.927 Dg, IGO. The legal and approximate values of those de- nominations of the Metric System which are in common use are presented in the following Table : DENOMINATION. LEGAL VALUE. APP. VALUE. Meter. 39.37 inches. 3 ft. 3| inches. Kilometer. .62137 mile. f mile. Square Meter. 1.196 sq. yards. 10| sq. feet. Ar. 119.6 sq. yards. 4 sq. rods. Hectar. 2.471 acres. 21 acres. Cubic Meter. 1.308 cu. yards. 35J cu. feet. Ster. .2759 cord. \ cord. Liter. 1.0567 quarts. 1 quart. Hektoliter. 2.8375 bushels. 2 bu. 3J pecks. Gram. 15.432 gr. T. 15^ grains. Kilogram. 2.2046 lb. Av. 2^ pounds. Tonneau. 2204.6 lb. Av. 1 T. 204 lb. Note. — The legal value is used in solving the following examples. 1. How many yards, feet, etc., in 4 m. ? Solution. — In 4 meters there are 4 times 39.37 in. which are 157.48 in., 157.48 in. reduced to integers of higher denomina- tions are 4 yd. 1 ft. 1.48 in. OPERATION. 39.37 4 12)157.48 3)13 ft. 1.4 8 in 4 yd. 1 ft. THE METRIC SYSTEM. I95 2. What is the value of 36 lb. in kilograms? OPERATION. Solution.— In 36 pounds there 2.2 4 G ) 3 6.0 ( 1 G.3 2 9 + are as many kilograms as 2.2046 2 2046 are contained times in 36 which 139540 are 16.329 + . 1322 7 6 72640 66138 65020 44092 2 9 2 8 198414 3. What is the value of 20 Km. ? 12.4274 mi. 4. How many hektars in 160 acres? 64.75-|- Ha. 5. What is the value of 49 m.? 9 rd. 4 yd. 3.13 in. 6. What is the value of 15 g.? 9 pwt. 15.48 gr. 7. How many hektoliters in 42 bu.? 14.8+ HI. 8. How many cords in 500 sters? 137.95 C. 9. How many square yards in a roll of paper 9 m. long and .5 m. wide? 5.382 sq. yd. 10. 32 1. are how many gallons? 8.4536 gal. Miscellaneous Examples. 161. 1. What is the sum of 127 dl., 4.87 1., 1563 cl, and 234.5 dl.? 56.65 1. 2. What will be the cost of 45 Ha. of land, at $3.32 an ar? S14940. 3. A merchant paid $457.92 for cloth, at S3 a meter: how many meters did he buy? 152.64 m. 4. A block of marble .72 m. long, .48 m. wide, and .5 m. thick cost $.864 : what is the cost of the marble per cubic meter? $5.' 196 RAY'S NEW PRACTICAL ARITHMETIC. 5. A manufacturer bought 380 stcrs of wood for $454.10: how much was that a 8tcr? SI. 195 6. How many hektoliters of oats in 4685 sacks, each containing 1.6 HI.? 7496 HI. 7. I bought 346.75 Kg. of coffee for $194.18 : what did I pay per kilogram? $0.56 8. The nickel 5-cent coin weighs 5 g. and is 2 cm. in diameter: what would be the weight of enough of these coins hiid in a row, to make a meter in length ? 250 g. 9. How much lining 1.85 m. wide will it take for a garment made of 6.5 m. of cloth 1.25 m. wide? 4.39+ m. 10. How many kilometers from Cincinnati to Dayton, the distance being 60 miles. 96.56-|- Km. 11. A map is 29 mm. long and 22.4 mm. wide : what space does it cover? 649.6 mm^. 12. The distance between two. towns is 13.24037 Km.: how many steps of .715 m. each, must I take to walk that distance ? 18518 steps. Note. — To illustrate the difference between the metric system and our common system of measures, a similar example may be given, substituting 8 mi. 72 rd. 4 yd. 1.7 in. for the distance, and 28.15 in. for the length of one step. 162. 1. Any per cent of a number is so msiuy hun- dredths of it. Thus, 1 per cent of a number is yi^ of it, 2 per cent is y^^, etc. Rem. — Per cent i^ from the Latin per centum, by the hundred. 2. The sign of per cent is %, read per cent. Thus, 5 % is read five per cent. 3. In all operations with per cent, it may be expressed in two ways: 1st. As a common fraction; 2d. As a decimal. Thus the following expressions are equivalent: One per cent. 1/., is T^t> or .01 Two per cent, 2%, is TOO ~ A' or .02 Three per cent. 3f., is Too or .03 Four per cent. 4/„ is TOO = TJ' or .04 Five per cent, 5^„ is Too — 2^' or .05 Six per cent, 6fc, is t! = A' or .06 Rem. 1. — Per cent, which is expressed as a mixed number, may be reduced to equivalent expressions by Arts. 121 and 142. Thus, 4i 9 4^ ^ = Tqq» which may be reduced to ^qqI also, i\ ^ = .045. no7) 198 RAY'S NEW PRACTICAL ARITHMETIC. Express the following as common fractions and as decimals : 1. 2. 8. 4. 5. 6. 10% 20% 30<^ 70 2i% yV and .10 ^-^ and .15 i and .20 f^ and .30 -1- and .50 i^ and .025 7. ^>i% tV a^''^ .0625 8. 12i% -J- and .125 9. 18|% A and .1875 10. 37i% f and .375 11. 56i% A and .5625 12. 874% 1 and .875 Rem. 2.--C()inmon fractions may be reduced to hundredths by Art. 108, snul tluMi read as per cent. Thus, ^ = .165 or 16§ ol How many per cent are equivalent to the following fractions ? 1. h- 8% 6. \. 25% 2. A- 12% 7- f 40% 3. ^V 16% 8. f. 75% 4. A- 3i% 9. ^. 33^% 5. tV- 8J% DEFINI 10. Jj. TIONS. 43|% 163. 1. Percentage embraces the various operations with per cent. 2. In Percentage three quantities are considered. (1) the Base^ (2) the Rate^ and (3) the Percentage. 3. The Base is the number upon which the per cent is estimated. 4. The Rate is the per cent when expressed as a common fraction or as a decimal. 5. The Percentage is the resujt of taking the per cent of the base. 6. Any two of these quantities being given, the third may be found. There are four cases. PEECENTAGE. 199 CASE I. 164. Given the base and the rate, to find the per- centage. 1. What is 25% of 32? OPERATION. SoLUTiox.— 25^0 is i (Art. 162). i of 32 is 8. 2b^r- centage. Examples. 1. A broker bought for me 75 shares New York Cen- tral and Hudson Eiver stock : required the brokerage at J%. $18.75. 2. The brokerage for buying 50 shares of Chicago and Eock Island stock was $6.25 : what was the per cent? 1^. 3. At \% brokerage a broker received $10 for making an investment in bank stock : how many shares did he buy? 40. ASSESSMENTS AND DIVIDENDS. 215 4. A broker buys 17 shares Milwaukee and St. Paul preferred stock : what is his brokerage, at J^ ? $4.25. 5. The brokerage on 95 shares of Vermont Central stock is $11. 87^: what is the per cent? -J^. 6. A broker received $9.50, or a brokerage of ^%, for buying Union Pacific stock : how many shares did he purchase ? 38. ASSESSMENTS AND DIVIDENDS. 178. 1. An assessment is a sum of money paid by the stockholders. Kem. — In the formation of a company for the transaction of any business, the stock subscribed is not usually all paid for at once; but assessments are made from time to time as the needs of the business require. The stock is then said to be paid for in installments. 2. A dividend is a sum of money paid to the stock- holders. Rem. — The gross earnings of a company are its total receipts in the transaction of the business; the net earnings are what is left of the receipts after deducting all expenses. The dividends are paid out of the net earnings. Examples. 1. I own 35 shares of bank stock; if the bank de- clare a dividend of 4^, what will I receive? $140. 2. A man pays an assessment of 7^%, or $300, on his insurance stock: how many shares does he own? 40. 3. A mining company declares a dividend of 15^ : what does Mr. Jones receive who owns 80 shares of stock? $1200. 4. A man owns 60 shares of railroad stock : if the company declare a dividend of 5^ payable in stock, how much stock will he then own? 03 shares. 216 KAY'S NEW PRACTICAL ARITHMETIC. 5. A gas company has a capital stock of $lt)0000; its gross earnings are $15700, and its expenses $4500 annu- ally : what per cent does it pay the stockholders? 1%- STOCK VALUES. 179, 1. The market value of stocks, bunds, and gold is the price at which they sell. Rem. — Stock is above pnr, or at a premium, when it sells for more than the par value; stock is below jxir, or at a disconni, when it sells for less than the par value. 2. The market value of stocks, bonds, and gold is estimated at a certain per cent of the ])ar value. Thus, "gold, 106^," means that the gold dollar is worth 106J^ of the currency dollar, or is at a premium of 6| ^. " New York Cen- tral and Hudson River, 91 J," means that the stock of this railroad sells for 91^ ol of the par value, or is at a discount of Sh 4). 3. The par value is the base; the premium or discount is the percentage; the market value, the amount or difference. Examples. 1. What will be the cost of 150 shares ($50 each) of Harlem, at 139|, brokerage \% ? $10500. 2. Bought $8000 in gold at 110, brokerage ^^ : what did I pay for the gold in currency? $8810 3. My broker sells 50 shares of Chicago and North western, brokerage ^% ; he remits me $2475 : at what per cent did the stock sell? 49f^. 4. What will be the cost of 25 1000-dollar 5-20 U. S. Bonds of 1867, at 114J, brokerage i% ? $28593.75. 5. I paid $1560 for Milwaukee & St. Paul, at 19i brokerage \% : how many shares did I buy? 80. STOCK INVESTMENTS. 217 6. When gold is at 105, what is the value in gold of a dollar in currency? ^^2T ^^' 7. When gold was at 112J, what was the value of a dollar in currency? 88|- ct. 8. In 1864, the "greenback" doilar was worth only 35f ct. in gold : what wa^ the price of gold ? 280. 9. A merchant paid $8946.25 for gold, at 105, broker- age \^c • l^<)w much gold did he buy? $8500. 10. My broker sells a certain amount of gold, and remits me $25734.37^? His brokerage, at xV%' ^^® $15,621: what was the price of the gold? 103. STOCK INVESTMENTS. 180. 1. The income is the annual profit from the investment. Eem;— The income from most of the United States bonds is in coin or its equivalent. 2. The cost of the investment is the base; the income is the 'percentage. Examples. 1. If I invest $39900 in 6^ bonds, at par, what will be my income? $^394. 2. If I invest $39900 in 6% bonds, at 105, what will be my income? $2280. 3. if I invest $39900 in 6% bonds, at 95, what will be my income? $2520. 4. What is a man's income who owns 20 1000-dollar IT. S. 6^ bonds, when gold is 107? $1284. 5. What income in currency would a man receive by investing $5220 in U. 8. 5-20 6^ bonds, at 116, when gold is ^105? ^283.50. 218 RAY'S NEW PRACTICAL ARITHMETIC. 6. What per cent of income do U. S. 4 J per cents, at 108, yield when gold is 105? 4g%. 7. If I receive an annual dividend of G^ on Michigan Central stock, which cost me but 37^, what per cent of income do I receive on my investment? 16^. 8. What sum invested in U. S. S's of 1881, at 118, yielded an annual income of $1921 in currency, when gold was at 113? $40120. 9. How many shares of stock bought at 95J, and sold at 105, brokerage \% on each transaction, will yield a profit of $925? "^ 100. 10. What must be paid for 6^ bonds to realize an income of 8^^? 75^. 11. When U. S. 4% bonds are quoted at iOG, what yearly income will be received in gold from the bonds that can be bought for $4982? $188. 12. If I pay 87 J for railroad bonds that yield an an- nual income of 7^, w^hat per cent do I get on my in- vestment? %%. 13. What could I afford to pay for bonds yielding an annual income of 7^ to invest my money so as to realize 6%? 116f. DEFINITIONS. 181. 1. Interest is money paid for the use of money. Rem. — The interest is paid by the borrower to the lender. 2. The principal is the money for the use of which interest is paid. 3. The amount is the sum of the principal and in- terest. 4. A promissory note is a written promise to pay a certain sum of money at a specified time. Rem. — The borrower always gives the lender his note for the money. The following is a common form: $500.00. Dayton, O., June IG, 1877. One year after date I promise to pay Clu^*les Thomas, or order, five hundred dollars, with interest at 8^,, for value received. James Q. Dean. Rem. — "When a note is made to draw interest from date, the words "from date" are frequently inserted after the word "interest." 5. The face of the note is the principal. 6. Legal interest is interest at a per cent that is allowed b}^ law. (219) 220 KAY'S NEW PRACTICAL ARITHMETIC. Rem. — The per cent of interest that is legal in the different States and Territories, is exhibited in the following TABLE. NAME OF STATE. Alabama Arizona Arkansas California Colorado Connecticut Dakota Delaware District Columbia. Florida Georgia Idaho Illinois Indiana Iowa Kansas Kentucky Louisiana Maine -^ Maryland Massachusetts Michigan Minnesota Mississippi 8^« 6% 6^ 7f* the principal, for 5 yr. it is j\ of (j cL = ^-^ the principal, and the amount is \^ of the ,3yXf = A principal. Then, }| of the principal are $650, }g -f fV^if y^^ of the principal is $50, and the principal is 50 $500. IMx^$='jOO Rule. — Multiply the rate by the time, and divide the amount by 1 -\- the j)roduct. 2. What principal in 9 yr., at 5%, will amount to $435? $300. COMPOUND INTP]REST. 237 3. The amount for 4 yr., at 5%, is $571.20: what is the interest? S95.20. 4. The amount for 6 yr., at 7%, is $532.50: what is the interest? $157.50. 5. The amount for 2 yr. 9 mo., at 8%, is $285.48: what is the j)rincipal? $234. 6. The amount for 2 yr. 6 mo., at 6%, is $690: what is the interest? $90. ' 7. The amount for 3 yr. 4 mo. 24 da., at 7%, is $643.76: what is the principal? $520. 8. The amount for 4 yr. 3 mo. 27 da., at 4%, is $914.94: what is the interest? $134.94. 189, Formulas for the five cases of Interest. Let h represent the principal, t the time, r the rate, and i the interest. Then, Case I. b Xr Xt=^i. Case II. I -f- (6 X — ^• Case TIT. (i ^ t) ~ b = r. Case lY. i -^{rXt) = b, * ' Case Y. ^--^"^-^^^b. 1 + (^- X COMPOUND INTEREST. 190. In Compound Interest the principal is increased yearly b}^ the addition of the interest. Rem. 1. — Sometimes tlie interest is added semi-amiually, or quarterly. Rem. 2. — The way in which interest is legally compounded is, at the end of each year, to take up the old note and give a new one with a face equal to both the principal and interest of the former note. 238 KAY'S NEW PRACTICAL ARITHMETIC. 1. Find the compound interest of $300 for 3 3^r., at 6%. Solution. — The interest of $300 for 1 yr., at 6^, is operation. $18; the amount is $18+ $3 $3 3 7.0 8 $300 r^ $318. The interest 6 .0 6 of $318 for 1 yr., at 6^^, 18.0 2 0.2 2 4 8 is $19.08; th' amount is 3 00 3 3 7.0 8 $10.08 + $318 = $337.08. -3X8. 357^3048 The interest of $337.08 for .0 6 3 18.0 300 3 18. .0 6 1 9.0 8 318 8 3 7.08 1 yr., at 6^^, is $20.2248; 19.08 $57.3048 the amount is $20.2248 + $337.08 = $357.3048. Then, the compound interest is $357.3048 — $300 = $57.30. Kule. — 1. Find the amount of the given principal for the first yearj and make it the principal for the second year, 2. Find the amount of this principal for the second year^ make it the principal for the third year, and so on for the given number of yearns. 3. From the last amount subtract the given principal ; the remainder will be the compound interest. Rem. 1. — When the interest is payable half-yearly, or quarterly, find the interest for a half, or a quarter year, and proceed in other respects as when the interest is payable 3'^early. Rem. 2. — When the time is years, months, and days, find the amount for the years, then compute the interest on this for the months and days, and add it to the last amount. Find the amount, at 6%, compound interest, 2. Of $500, for 3 years. S595.51. 3. Of $800, for 4 3^ears. $1009.98. ANNUAL INTEREST. 239 Find the compound interest 4. Of $250, for 3 yr., at 6%. $47.75. 5. Of $300, for 4 yr., at 5%. $64.65. 6. Of $200, for 2 yr., at 6%, pajable semi-annually. $25.10. 7. Find the amount of $500, for 2 yr., at 20% com- pound interest, payable quarterly. $738.73. 8. What is the compound interest of $300, for 2 yr. 6 mo., at 6%? $47.19. 9. What is the compound interest of $1000, for 2 yr. 8 mo. 15 da., at 6%? $171.35. 10. What is the amount of $620 at compound interest semi-annually, for 3 yr. 6 mo., at 6%? $762.52. 11. What is the diiference between simple interest and compound interest on $500, for 4 yr. 8 mo., at 6% ? $16.49. ANNUAL INTEREST. 191. Annual Interest is interest on the principal, and on each annual interest after it is due. Rem. 1. — This interest is sometimes semi-annual or quarterly. Rem. 2. — Annual interest may be collected when the note or bond, reads " with interest payable annually." Rem. 3. — The annual interest is sometimes represented by interest notes; these are given at the same time as the note for the principal, and draw interest if not paid when due. Rem. 4. — The annual interest on bonds is sometimes represented by interest notes, called coujwns; these are detached from the bond and presented for payment when the interest is due. 1. ]^o interest having been paid, find the amount due in 4 yr. 8 mo. 24 da., on a note for $400, with interest at 6%, payable annually. OPERATION. 2 ):5 6 8 400 .2 8 4 .0 6 400 2 4.00 1 1 3.6 2)1.072 yr. mo. da. .5 3 6 3 8 24 24 2 8 24 2144 1 8 24 1072 8 24 1 !\8 6 4 8 11 1 1 3.6 400. $5 2 6.4 6 4 240 KAY'S NEW PKACTICAL ARITHMETIC. Solution. — The interest of $400 for 4 yr. 8 mo. 24 da., at 6^, is $113.60. One annual in- terest of $400, at 6^c» is $24. The first annual interest re- mains unpaid 3 yr. 8 mo. 24 da.; the second, 2 yr. 8 mo. 24 da.; the third, 1 yr. 8 mo. 24 da., and the fourth, 8 mo. 24 da.; hence, interest must be reckoned on $24 for 8 yr. 11 mo. 6 da.; this is $12,864. The amount, then, is $12,864 -f $113.60 -I- $100 = $526.46. Bule. — 1. Find the interest of the principal for the time during xchich no annual interest is paid. 2. Find the interest of one annual interest for the sum of the times each annual interest remains unpaid. 3. The sum of the two interests will be the interest due, and this, added to the principal, will be the amount due. 2. ^o interest having been paid, find the amount due in 3 yr. on a note for S800, with Interest at 8%, payable annually. S1007.36. 3. The interest having been paid for 2 yr., find the amount due in 5 yr. on a note for $750, with interest at 10%, payable annually. 8997.50. 4. No interest having been paid for 4 yr., find the interest due in 6 yr. on a bond for $10000, with interest at 5%, payable annually. $2150. 5. No interest having been paid, find the amount due Sept. 1, 1877, on a note for $500, dated June 1, 1875, with interest at 6%, payable semi-annually. $571.10. PARTIAL PAYMENTS. 241 6. [S1200.] Milwaukee, Wis., May 12, 1873. For value received, on demand, I promise to pay John G. Morgan, or order, twelve hundred dollars, with interest at 6%, payable annually. H. W. Slocum. No interest having been paid, what was the amount due on this note, Sei:)tember 20, 1877? $1545.66. 7. [$1500.] New Orleans, La., October 10, 1872. On the first day of May, 1877, for value received, I promise to pay Andrew^ Jackson, or order, fifteen hun- dred dollars, with interest, payable annually, at 5%. George Quitman. No interest having been paid, what amount was due at maturity? $1872.75. 8. What is the difierence between simple and annual interest on $1000 for 5 yr., at 6% ? $36. 9. What will be due on six 500-dollar city bonds run- ning 3 yr., with interest at 6%, payable semi-annually, if the interest should not be paid? $3580.50. 10. The interest on U. S. 4 per cent bonds is payable quarterly in gold ; granting that the income from them might be immediately invested, at 6%, w^hat would the income on 20 1000-dollar bonds amount to in 5 yr., with gold at 105? $4798.50. PARTIAL PAYMENTS. 192. 1. A partial payment is a sum of money, less than the face, paid on a note. 2. The receipt of a partial payment is acknowledged by indorsing it on the back of the note. Prac. 16. 242 KAY'S NEW TKACTICAL ARITHMETIC. 3. The indorsement consists of the date and amount of the pajMuent. 4. The rule of the Supreme Court of the United States, in reference to Partial Payments, is as follows : '^ When partial payments have been made, apply the payment, in the first place, to the discharge of the interest then due. ^' If the payment exceeds the interest, the surplus goes toward discharging the principal, and the subsequent in- terest is to be computed on the balance of principal remuin- ing due. " If the payment be less than the interest, the surplus of interest must not be taken to augment the principal, but interest continues on the former principal, tintil the period when the payments, taken together, exceed the interest due, and then the surplus is to be applied toward discharging the principal ; and interest is to be computed on the balance, as aforesaid.'' — Kent, C. J. Rem. — This rule is founded on the principle that neither interest nor payment shall draw interest. 1. [SIOOO.] Boston, Mass., May 1, 1875. For value received, on demand, I promise to pa}^ to Alonzo Warren, or order, one thousand dollars, with interest at 6%. William Murdock. On this note partial payments were indorsed as fol- lows : November 25, 1875, $134; March 7, 1876, $315.30; August 13, 1876, $15.60; June 1, 1877, $25; April 25, 1878, $236.20. What w^as the amount due on settlement, September 10, 1878? FAUTIAL PAYMENTS. 243 Solution. —The time from May 1, 1875, to November 25, 1875, is 6 mo. 24 da.; the interest of $1000 for this time is $34; the payment, $134, exceeds the interest; the amount is $1034; $1034 — $134 = $900, the sec- ond iirincijiol. The time from No- vember 25, 1875, to March 7, 1876, is 3 mo. 12 da.; the interest of $900 for this time is $15.30; the payment, $315.30, exceeds the in- terest; the amount is $915.30; $915.30—1315.30 =r$600, the third jirin- cipal. The time from March 7, 1876, to August 13, 1876, is 5 mo. 6 da.; the interest of $600 for this time is $15.60; the pay- ment, $15.60, equals the interest; the amount is $615.60; $615.60— $15.60 ^^ $600, the fourth prin- cipal. The time from August 13, 1876, to June 1, 1877, is 9 mo. 18 da.; the in- terest of $600 for this time is $28.80; the pay- ment, $25, is less than the interest; continue to princij^al. OPERATION. 1875 11 25 $134 1000 1875 5 1 .0 3 4 6 24 34 2 ).0 6 8 .0 3 4 1000 1034 134 1876 3 7 $ 3 1 5.3 9 00 1875 1 1 2 5 .017 3 12 15.3 2 ).0 3 4 .017 900 9 1 5.3 3 1 5.3 1876 187 6 8 13 3 7 5 6 $15.6 600 .0 2 6 15.6 600 2).0 5 2 6 1 5.6 .02 6 15.6 1877 6 1 $25 600 .0 4 8 .2 8.8 1876 8 13 9 18 2).0 9 6 .0 4 8 600 1878 1877 4 2 5 6 1 10 2 4 $ 2 3 6.2 2 6 1.2 .0 5 4 3 2.4 2 8.8 6 1.2 2).108 6 00. .0 5 4 6 6 1.2 2 6 1.2 1878 9 10 400 1878 4 25 .0225 4 15 9 2).0 4 5 .0225 400 409 find the interest on the fourth 244 KAY'S NEW PRACTICxVL AKITIIMETIC. The time from June 1, 1877, to April 25, 1878, is 10 mo. 24 da.; the interest of $000 for this time is $32.40; the sum of the payments, $261.20, exceeds the sum of the interests, $01.20; the amount is $661.20; $661.20 — $261. 20 ==$400, the ffth principal. The time from April 25, 1878, to September 10, 1878, is 4 mo. 15 <;la.; the interest of $400 for this time is $9; the amount due on set- tlement is $409. RULE. I. When each payment equals or exceeds the interest. 1. Find the time from the date of the note to the date of the first payment 2. Find the amount of the given j^rincipal for this time. 3. From this amount subtract the payment ; the remainder is the second principal. 4. Find the time from the date of the first payment to the date of the second payment. 5. Then proceed with the second principal as with the first, anil so on to the date of settlement. II. When one or more payments are less than the interest. 1. Continue to find the interest on the same principal until a date is reached, when the sum of the payments equals or exceeds the sum of the interests. 2. Then subtract the sum of the payments from the amount; the remainder is th?. next principal. Rem.— Sometimes it may be determined, by inspection, that the payment is less than the interest; when this can be done, it is not necessary to find the intermediate time and interest, but interest may at once be found to the date when it is apparent that the sum of the payments exceeds the interest. PARTIAL PAYMENTS. 245 2. [S350.] Boston, Mass., July 1, 1875. For value received, I promise to pay Edward Sargent, or order, on demand, three hundred and fifty doHars, with interest at 6%. James Gordon. Indorsements: March 1, 1876, ^44; October 1, 1876, SIO ; January 1, 1877, S26 ; December 1, 1877, $15. What w^as the amount due on settlement, March 16, 1878? $306.75. 3. A note of $200 is dated January 1, 1875. Indorse- ment: January 1, 1876, $70. What was the amount due January 1, 1877, interest at 6% ? $150.52. 4. A note of $300 is dated July 1, 1873. Indorsements: January 1, 1874, $109; July 1, 1874, $100. What was the amount due January 1, 1875, interest at 6%? $109.18. 5. A note of $150 is dated May 10, 1870. Indorse- ments: September 10, 1871, $32; September 10, 1872, $6.80. What was the amount due November 10, 1872, interest at 6% ? $132.30. 6. A note of $200 is dated March 5, 1871. Indorse- ments: June 5, 1872, $20; December 5, 1872, $50.50. What was the amount due June 5, 1874, interest at 10% ? $189.18. 7. A note of $250 is dated January 1, 1875. Indorse- ments: June 1, 1875, $6; January 1, 1876, $21.50 What was the amount due July 1, 1876, interest at 7%' $248.40, 8. A note of $180 is dated August 1, 1874. Indorse ments: February 1, 1875, $25.40; August 1, 1875, $4.30 January 1, 1876, $30. What was the amount due July 1, 1876, interest at 6% ? $138.54. 9. A note of $400 is dated March 1, 1875. Indorse ments: September 1, 1875, $10; January 1, 1876, $30 24G RAY'S NEW PRACTICAL ARITHMETIC. July 1, 1876, $11; September 1, 1876, $80. What was the amount due March 1, 1877,. iuterest at 6%? 8313.33. 10. A note of $450 is dated April 16, 1876. Indorse- ments: January 1, 1877, $20; April 1, 1877, $14; July 16, 1877, $31; December 25; 1877, $10; July4, 1878, $18. What was the amount due June 1, 1879, interest at 8% ? $466.50. 11. A note of $1000 is dated January 1, 1870. In- dorsements: May 1, 1870, $18; September 4, 1870, $20; December 16, 1870, $15; April 10, 1871, $21; July 13, 1871, $118; December 23, 1871, $324. What was the amount due October 1, 1873, interest at 6^? $663.80. 193. When partial paj'ments are made on notes and accounts running a 3'ear or less, the amount due is commonly found by the MERCANTILE RULE. 1. Find the amount of the principal from the date of the note to the date of settlement. 2. Find the amount of each payment from its date to the date of settlement. 3. From the amount of the principal subtract the sum of the amounts of the payments. 1. A note of $320 is dated Jan. 1, 1876. Indorsements: May 1, 1876, $50; Nov. 16, 1876, $100. What was the amount due Jan. 1, 1877, interest at 6^ ? $186.45. 2. An account of $540 was due March 1, 1877. Credits: May 1, 1877, $90; July 1, 1877, $100; Aug. 1, 1877, $150; Oct. 11, 1877, $180. W^hat was the amount due on settlement Jan. 1, 1878, interest at S%? $39. DEFINITIONS. 194. 1. Discount is interest paid in advance. 2. There are two kinds of discount, Bank Discount and True Discount. BANK DISCOUNT. 195. 1. Banks lend money on two sorts of notes. (1) accommodation notes, and (2) business notes. Rem. — These notes are frequently termed accommodation paper, and business paper. 2. An accommodation note is made payable to the bank which lends the money. TvEM. — The following is a common form of an accommodation note: ^500. Chicago, III., October 20, 1877. Ninety days after date, we, or either of us, promise to pay to the Second National Bank of Chicago, 111., ^ve hundred dollars, for value received. O. S. West. W. B. Sharp. 18 / Due January /oi , 1878. •^ ^^^ (247) 248 RAY'S NEW PRACTICAL ARITHMETIC. 3. A business note is payable to an individual. 4. A business note may be nefjotiahle or not negotiable. 5. A negotiable note is one that can be bought and sold. Rem. — The followinu:; aro common forms of business notes: 1st. JV'ot ne^otiahlc. $200. Buffalo, N. Y., March 21, 1877. On demand, I promise to pay Charles II. Peek, two hundred dollars, ibr value received. G. W. Clinton. This note is payahle only to Charles H. Peck; it is due at once, and bears interest from datt?. 2d. JS^egotidble, $1000. St. Louis, Mo., May 1, 1877. One year after date, I promise to pay to David King, or order, one thousand dollars, for value received. Elmer B. Archer. The words "or order" make this note negotiable. If David King transfers it, he must indorse it — that is, write his name across the back of it. This note bears no interest till after it is due. 3d. JYeiotiable. $150. Washington, D. C, August 10, 1877. On or before the first day of May, 1878. I promise to pay Amos Durand, or bearer, one hundred and fifty dollars, with interest at 10^ from date, for value received. John Sherwood. The words " or bearer " make this note negotiable without indorse- ment. This note bears interest from date, it being so specified. PANK DISCOUNT. 249 A 6. A note is payable, or nominally due, at the end of the time specified in the note. 7. A note matures, or is legally due, three days after the specified time. 8. The three days after the specified time are called days of grace. Rem. 1. — Banks lend money only on short time; rarely beyond 3 months. To find ivhen a note matures: 1st. When the time is expressed in days : Rule. — Count the days from the date of the note and add three days. 2d. When the time is expressed in months: Rule. — Count the months from the date and add three days. Rem. 2. — In Delaware, Maryland, Pennsylvania, Missouri, and the District of Columbia, the day of discou7it is the first day of the time. Rem. 8. — When a note in bank is not paid at maturity, it goes to protest — that is, a written notice of this fact, made out in legal form, by a notary public, is served on the indorsers, or securit^^ 9. The bank discount is simple interest taken in advance. 10. The proceeds is the money received on the note. 11. In Bank Discount four quantities are considered: (1) The /ace of the note, (2) the per cent, (3) the time^ and (4) the discount. 12. Any three of these quantities being given, the fourth may be found. We will consider two cases. 250 KAY'S NEW PRACTICAL ARITHMETIC. CASE I. 196. Given the face of the note, the per cent, and the time to find the discount and the proceeds. 1st. When the note does not hear interest. 1. Find the date when due, bank discount, and pro- ceeds of the following accommodation note, discounted at 6%: $700. Mobile, Ala., June 25, 1877. Sixty days after date we, or either of us, promise to pay to the First National Bank, of Mobile, Ala., seven hundred dollars for value received. Charles Walker. Walter Smith. OPERATION. Solution. — The note is due August /oy , nTTTF / ^ ' .0 1 o 1877 (Art. 78). The interest of $1 for 63 days, 7 00 at 6^, is $0.0105, and the interest of $700 is 73 5 $0.0105 X 700 = $7.35; this is the discount ; then, 7 ~ $700 — $7.35 = $692.65, the proceeds. 7*3 ^ ir92ir5 Rule. — 1. Find the interest on the face of the note for the given time; this is the hank discount. 2. From the face of the note subtract the discount ; the remainder is the proceeds. Find the date when due, bank discount, and pro- ceeds of 2. A note of $100, dated June 20, payable in 60 days, and discounted at 6%. August ^ /22' SI. 05, $98.95. BANK DISCOUNT. 251 3. A note of $120, dated October 12, payable in 30 (lays, and discounted at 8^. November ^7^4, S0.88, $119.12. 4. A note of $140, dated January 15, payable in 4 months, and discounted at 6^. May ^^jig, $2.87, $137.13. 5. A note of $180, dated April 10, payable in 6 months, and discounted at 4%. October ^^/^3, $3.66, $176.34. 6. A note of $250, dated December 1, payable in 5 months, and discounted at 8^. May y^, $8.50, $241.50. 7. A note of $375, dated August 4, payable in 30 days, and discounted at 6%. September ^g , $2.06, $372.94. 8. A note of $600, dated February 12, 1876, paj^able in 60 days, and discounted at 9%. April ^^^^5 , $9.45, $590.55. 9. A note of $1200, dated February 20, 1877, payable in 90 days, and discounted at 10^. May ^^24' ^^1' ^^^^^• 10. A note of $1780, dated January 11, 1872, payable in 90 days, and discounted at 6^. April ^^/i3, $27.59, $1752.41. Find the date when due, time of discount, bank dis- count, and proceeds of the following business notes: 252 KAY'S NEW PRACTICAL ARITHMETIC. 11. [$600.] San Francisco, Cal., Sept. 15, 1876. One 3^ear after date, I promise to pa}^ to the order of Abel E. Worth, at the First National Bank of San Francisco, Cal., six hundred dollars, for value received. George M. Burgess. Discounted May 21, 1877, at 10%. Sept. ^^/-^g, 1877, 120 days, S20, $580. 12. [SIOOO.] Nashville, Tenn., May 8, 1877. Ninety days after date, I promise to pay Albert E. Kirk, or order, one thousand dollars, for value received. Jacob Simmons. Discounted June 22, 1877, at 6%. August /g, 48 days, $8, $992. 13. [S1500.] Pittsburgh, Pa., July 10, 1877. Six months after date, I promise to pa}'' Alex. M. Guthrie, or bearer, fifteen hundred dollars, for value received. Orlando Watson. Discounted October 25, 1877, at 6^. January ^^3, 1878, 81 days, $20.25, $1479.75. 2d. When the note hears interest. 1. Find the dale when due, time of discount, bank discount, and proceeds of the following business note: $800. Dayton, O., January 5, 1877. Six months after date, I promise to pay to the order of Charles Stuart, at the Dayton National Bank, of Day- ton, O., eight hundred dollars, with interest at 6^, for value received. Francis Murphy. BANK DISCOUNT. 253 Discounted April 15, 1877, at 8^. OPERATION. 2).0 6 1 800 .0 30 5 .0305 2 4.4 00 3)84 800 3).0 2 8 824T0 .0 9^ .01 8f 2 5 4 9 6 .0181 659520 82440 82 4.4 1 5.3 8 8 8 1 5.3 9 $8 9.0 1 Solution. — The note is due July ^/g , 1877. The time of dis- count, from April 15 to July 8, is 84 days. The amount of $800 for 6 mo. 3 da., at ^o]^, is $824.40. The bank discount of $824.40 for 84 days, at '^c/^, is $15.39. The j^roceeds are $809.01. Rule. — 1. Find the amount of the note for the given time. 2. Find the bank discount and proceeds of this amount. Eem. — In the following examples, remember that in leap years February has 29 days. Find the date when due, time of discount, bank dis- count, and proceeds of 2. A note of $150, dated May 20, 1875, payable in 6 months, with interest at 6%, and discounted September 9, 1875, at S%. November ^^/23 , 1875, 75 days, $2.58, $152. 3. A note of $300, dated August 5, 1876, payable in 1 year, with interest at S%, and discounted April 16, 1877, at 6%. August ^g, 1877, 114 days, $6.16, $318.04. 4. A note of $450, dated March 4, 1877, due January 1, 1878, with interest at 6%, and discounted August 13, 1877, at 10%. January ^^4, 1878, 144 days, $18.90, $453.60. 254 KAY'S NEW Pll ACTIO AL ARITHMETIC. 5. A note of $650, dated May 16, 1876, due Sept. 1, 1878, with interest at 9%, and discounted April 25, 1878, at 6%. Sept. Y4, 1878, 132 days, $17.26. $767.29. 6. A note of $840, dated September 1, 1875, payable in 6 months, with interest at 10^, and discounted De- cember 20, 1875, at 8%. March Y4, 1876, 75 days, $14.71, $867.99. 7. A note of $1400, dated July 19, 1875, due May 1, 1876, with interest at 6^, and discounted Jan. 17, 1876, at 10%. May y4, 1876, 108 days, $44, $1422.50. 8. A note of $2400, dated Oct. 16, 1876, due Jan. 1, 1878, with interest at 8^, and discounted July 26, 1877, at 10%. January '^/ji^, 1878, 162 days, $118.51, $2515.09. 9. [$3500.] Macon, Ala., October, 15, 1877. One year after date, I promise to pay Adam Moore, or order, thirty-five hundred dollars, with interest at 6%, for value received. Joseph Stephens. Discounted May 15, 1878, at 9^. October ^^j^g, 1878, 156 days, $144.76, $3566.99. 10. [$6000.] Frankfort, Ky., 3Iay 10, 1875. One year after date, 1 promise to pay Henry Warren, or order, six thousand dollars, with interest at S%. for value received. Amos E. Burton. Discounted November 21, 1875, at 10^. May ^^L, 1876, 174 days, $313.39, $6170.61. BANK DISCOUNT. 255 CASE II. 197e Given the per cent, the time, and the proceeds, to find the face of the note. 1. For what sum due 90 days hence, must I give a note to a bank, that, when discounted at 6%, the pra ceeds will be S177.21? OPERATION. 3)93 1.0 00 Solution.— The bank discount ^ )>0 3 1 -0155 of $1 for 93 days, at 6^^, is $0.0155, .0155 .9845 and the proceeds $1 — $0.0155 — $0.9845. Then, $177.21 is the pro- .0845)177.21(180 ceeds of 177.21 -- .9845 = $180. 9845 787CO 78760 Rule.— 1. Find the proceeds of %l for the given time at the given per cent. 2. By this divide the given proceeds. 2. The proceeds of a note discounted at a bank for GO days, at 6^, were $197.90: what was the face of the note? $200. 3. For what sum must a note be made, so that when discounted at a bank, for 90 days, at 6%, the proceeds will be $393.80? $400. 4. What must be the face of a note, that when dis- counted at a bank for 5 months^ at 8^, the proceeds may be $217.35? $225. 5. The proceeds of a note are $352.62, the time 4 months, and the discount at (3^ : what is the face? S360. 256 KAY'S NEW PRACTICAL ARITHMETIC. 6. 1 wish to borrow $400 from a bank for 30 days: what must be the face of my note, that, when discounted at 6%, I may receive this amount? $402.21. 7. I wish to obtain from a bank $500 for 60 days: for what sum must I give my note, at 8^ discount? $507.10. 8. I wish to use $1500 for 6 months; if I can obtain money from a bank, at a discount of 10%, for what sum must I give my note to realize this amount? $1580.33. 9. A note dated February 19, 1876, payable January 1, 1877, and bearing 8% interest, was discounted Octo- ber 12, 1876, at 6fc ; the proceeds were $1055.02: what was the face of the note? $1000. TRUE DISCOUNT. 198. 1. The present worth of a note is a sum of money, which, being on interest for the given time at a given i)er cent, will amount to the same as the note. 2. The true discount is the difference between the present worth and the amount of the note. Rem. 1. — Notes, d^^bts, and running accounts are discounted by True Discount. Rem. 2. — Banks sometimes discount by the method of True Discount. 199. Given the face of the note, the time, and the per cent, to find the present worth and discount. 1. Find the present worth and discount, at 6^, of a note of $430.50, due in 2 yr. 5 mo. 18 da. TRUE DISCOUNT. 257 Solution. — The amount of $1 for 2 yr. 5 mo. 18 da., at 6^o, is $1,148. Then, the present worth of $430.50 is 430.50 --1.148 = $375; and the dis- count is $430.50 — $375 = $55.50. OPEKATIOK. 2 ).2 9 6 .14 8 1.0 1.14 8 .1 4 8 ) 4 3 0.5 ( 3 7 5 3444 8610 8036 5740 5 740 4 3 0.5 375 $5 5.5 2. Find the present worth and discount, at 8^, of a note of $500, due in 3 yr., and bearing interest at 6%. Solution. — The amount of $500 for 3 yr., at 6^^, is $590. The amount of $1 for 3 yr., at 8^^, is $1.24. Then, the present worth of $590 is 590 ^ 1.24 = $475.81; and the discount is $590 — $475.81 = $114.19. Prac. 17. OPERATION. 500 .0 8 .0 6 3 3 0.0 :24 3 1.0 ¥0 1.2 4 500 590 1.2 4 ) 5 9 ( 4 7 5. 81 496 590 4 7 5.8 1 1 1 4.1 9 940 868 720 620 1000 9 92 80 258 HAY'S NEW PKACTICAL ARITHMETIC. Kule. — 1. Find the amount 0/ SI for the given time at the given per cent. 2. By this divide the amount of the note; this is the present worth. 3. From the amount of the note subtract the present worth; this is the discount. Hem. — When the note does not bear interest, of course the amount is the same as the face of the note. 3. Find the present worth and discount, at G^, of a note of $224, due in 2 yr. 8200, $24. 4. Find the present worth and discount, at 6^, of a note of $300, due in 2 yr., and bearing interest at 8^. $310.71, $37.29. 5. Find the present worth and discount, at 6^, of a debt of $675, due in 5 yr. 10 mo. $500, $175. 6. Find the present worth and discount for 5 mo., at 10%, of an account of $368.75. $354, $14.75. 7. A note of $800, dated September 10, 1876, due January 1, 1878, and bearing interest at 6%, was dis- posed of for the present worth, at 10%, July 19, 1877: what was the present worth at this date and the dis- count? $825.65, $37.15. 8. A merchant bought a bill of goods amounting to $775, on 4 months' credit: if money is worth 10% to him, what might he pay for the goods in cash? $750. 9. Bought a bill of goods, amounting to $260, on 8 •months' credit: if money is worth 6%, what sum will pay the debt in cash? $250. 10. A merchant buys a bill of goods amounting to $2480: he can have 4 months' credit, or 5% off, for cash: if money is worth only 10% to him, what will he gain by paying cash? $45.47. TRUE DISCOUNT, 259 11. Find the present worth, at 5^, of a debt of S956.34, one-third to be paid in 1 yr., one-third in 2 yr., and one third in 3 yr. 8870.60. 12. Omitting the three days of grace, what is the difference between the true discount and the bank dis- count of $535, for 1 yr., at 7%? S2.45. 13. A man was offered $1122 for a house, in cash, or $1221, payable in 10 mo., without interest. He chose the latter : how much did he lose, if money is worth 12% to him? $12. 14. A man offers to sell his farm for $8000 in cash, or for $10296, payable in three equal installments at the end of 1, 2, and 3 years, without interest: considering money to be w^orth 10^, what will be the gain to the buyer by paying cash? $620. 15. A note of $2000, dated July 4, 1876, due May 1, 1878, and bearing interest at 8^, was cancelled October 25, 1877, by payment of the present worth at 6^ : what was the present worth, at this date, and the discount? $2223.08, $68.92. Xe HANG EI. 200. 1. A draft, or bill of exchange, is a written order, from one person to another, for a certain amount of money. Rem. 1. — The person upon whom the bill is drawn is called the drawee; the person in whose favor it is drawn is called the payee. Rem. 2. — When the draft is to be paid upon presentation, it is called a sight draft; when it is to be paid at the end of a certain time, it is called a time draft. 2. Exchange is the method of making a payment by means of a draft, or bill of exchange. 3. There are two sorts of exchange : domestic or Inland, and foreign. 4. Domestic exchange takes place between localities in the same country. Rem. — The following is a common form of an inland bill of ex- change, which is commonly termed a rfraft or check: $500. Cincinnati, O., May 1, 1877. At sight, pay to John Jones, or order, five hundred dollars, for value received, and charge to account of Silas Thompson. To Charles Smith & Co., Xew York. (260) EXCHANGE. 261 5. Foreign exchange takes place between localities in different countries. Rem. — The following is a common form of a foreign bill of exchange: £500. Cincinnati, O., May 1, 1877. At sight of this first of exchange (second and third of the same tenor and date unpaid), pay to Amos Car- roll, or order, five hundred pounds sterling, for value received, and charge to account of Stanley Bingham. To James Smith & Co., London. A foreign bill of exchange is usually drawn in dupli- cate or triplicate, called a set of exchange; the different copies, termed respectively the firsts second, and third of exchange, are then sent by different mails, that miscar- riage or delay may be avoided. When one is paid, the others are void. 6. The acceptance of a bill of exchange is the agree- ment b}' the drawee to pay it when due. KEM.-Abill is accepted by the drawee's writing the word "ac- cepted," with his name, across the face of the bill; the bill is then an acceptance. 201, To find the cost or face of a domestic bill of exchange (Art. 170, Eule). 1. What is the cost of a sight draft on New York for $1400, at ^% premium? ^ $1407. 2. What is the cost of a sight draft on Boston, for S2580, at \% discount? $2567.10. 262 RAY'S NEW PRACTICAL ARITHMETIC. 3. What is the face of a sight draft on Wheeling, which cost $375.87, at J% premium? $375.40. 4. What is the cost of a sight draft on Chicago, for $2785, at 1% discount? 82778.04. 5. What is the face of a sight draft, which cost $1852.55, at \\% discount? $1876. G. What is the cost of a draft on New Orleans for $5680, payable in 60 days, exchange being at ^% premium, and interest 6% ? $5649.08. 7. What is the cost of a draft on IS'ew York for $1575, payable in 30 days, exchange being at ^^% premium, and interest 6%? $1578.13. 8. The face of a draft, payable in 60 days, is $2625; exchange being at \\% premium, and interest 6%, what is the cost of the draft? $2636.69. FOREIGN EXCHANGE. 202, Foreign bills of exchange are drawn in the money of the country in which they are to be paid. Rem. — The foreign exchange of the United States is chiefly with Great Britain, France, Germany, and Canada. ENGLISH MONEY. The unit of English money is the pound sterling. 4 farthings make 1 penny, marked d. 12 pence " 1 shilling, '' s. 20 shillings '' 1 pound, " £. Rem. — The usual coins are: gold, sovereign = 1 £, and half sov- ereign; silver, crown = 5 s., half crown, florin ==2 s., shilling, six- penny, and three-penny; copper, the penny, half penn}^ and farthing. EXCHANGE. 263 FRENCH MONEY. The unit of French money is the franc, marked ft\ 10 centimes make 1 decime. 10 decimes ^' 1 franc. Rem. — The usual coins are: gold pieces for 100, 40, 20, 10, and 5 francs; silce?- pieces for 5, 2, 1, h, and \ francs; bronze pieces for 10, 5, 2, and 1 centimes. GERMAN MONEY. The unit of German money is the mark, which is divided into 100 pennies (pfennige). Rem.- — The usual coins are: gold pieces for 20, 10, and 5 marks; silver pieces for 2, 1, and ^ marks; nickel pieces for 10, 5, and 1 pennies. Canadian money is in dollars and cents, corresponding with United States currency. The par of exchange is the comparative value of the standard coins of two countries. Rem. — The commercial value of foreign exchange may be above or below the par value. Quotations are always in gold. The par value of the pound is $4.8665. Its quoted commercial value varies from $4.83 to $4.90 gold. The par value of the franc is $0,193. It is usually quoted at about 5 fr. 14f centimes, equal to one dollar gold. The par value of the mark is $0,238. The commer- cial quotations, always for four marks, vary from $0.95 to $0.98. To find the cost or face of a foreign bill of exchange : 264 RAY'S NEW PRACTICAL ARITHMETIC. 1. What will a sight bill on London, for £500 IOh., cost in 'New York, exchange being at $4.87? OPERATION. 10 S.r=:£.5 Solution. — Since 20 s. = £1, 10 s. r.= £.5. If £1 is worth $4.87, £500.5 are worth $4.87 X 500.5 =z $2437.44. 5 0.5 4.8 7 3~5T35 40040 20020 $2437.435 2. How large a bill on London can be bought for $1808.04, exchange being at $4.88? Solution.— Since £1 is worth $4.88, as many pounds can be bought for $1808.04 as $4.88 is contained times in $1808.01. It is contained 870 times, with a remainder. Reduce the remainder to shillings by multiplying by 20. 4.88 is contained in the prod- uct 10 times. The bill will be for £870 10s. OPERATION. 4.8 8 ) 1 8 8.0 4 ( 3 7 1464 3440 341o 244 20 4.88) 4 8 80( 10 s. 488 3. What will a bill on London for £890 8s. .cost, ex- change being at $4.86? $4327.34. 4. How large a bill on London can be bought for $2130.12, exchange being at $4.88? £436 lOs. 5. What will a bill on Paris cost for 1290 francs, ex- change being 5 fr. 15 centimes to $1? $250.49. 6. How large a bill on Paris can be. bought for $1657.60, exchange being at 5 fr. 16 centimes? 8553 fr.22. 7. What will a bill on Berlin cost for 12680 reichsmarks, exchange being $.97 per 4 reichsmarks? $3074.90. 8. How large a bill on Frankfort can be bought for $1470, exchange being at .98? 6000 m. DEFINITIONS. 203. 1. Insurance Companies agree, for specified sums of money, to pay a certain amount to the person insured on the occurrence of a certain event. 2. The policy is the written contract given by the company. Eem. — The persons insured are called the policy holders. The companies are sometimes stj^led the underwriters. 3. The premium is the sum paid to the company for insurance. 4. Pire Insurance is indemnity for a certain amount in case of loss by fire. 5 Marine Insurance is indemnity for a certain amount in case of loss'by the dangers of navigation. 6. Life Insurance is an agreement to pay a specified sum at the death, or at a certain time in the life, of the insured. FIRE AND MARINE INSURANCE. 204-. The premium in fire and marine insurance is a certain percentage of the amount insured (Art. 170, Eule). (265) 260 RAY'S NEW PRACTICAL ARITHMETIC. Rkm. — Insurance companies will seldom insure property at its full value. The insurance is commonly upon § or ^ of the value. 1. What is the cost of insuring a liouse worth $3375, at f of its value, the premium being H^ and the policy costing $1 ? Solution. — ^ of the value of the house is $2250. The premium is U^;^ of $2250, which is $33.75; adding $1, the cost of the policy, the sum is $34.75; the cost of insurance. OrERATION ) 3 3 7 5 2250 1125 .ou 2 2250 2250 1125 3 3.7 5 1.0 3 4.7 5 2. What is the cost of insuring a house worth $5000, at f of its value, the premium being ^^, and the policy costing S1.50? $20.25. 3. A store is valued at $12600, and the goods at S14400; § of the value of the store is insured at f ^ and ^ the value of the goods at 2% ; the cost of the two pol- icies is $1.25 apiece: what was the total cost of in- surance? $209.50. 4. A man owns a manufactory valued at $21000, and a dwelling-house worth $7200: what will it cost. to in- sure the manufactory, at ^ of its valuiB, at IJ^, and the house, at its full value, at f^, the two policies costing $1.25 each? $23G.50. 5. A man's dwelling, valued at $5600, was burned ; it had been insured, in a certain company, 20 years, for f of its value, at l^% : how much did he receive from the company more than the sum total of the annual premiums? $2940. 6. A man secures a policy of insurance, on his house, for $3600, furniture for $1600, and library $800; the INSURANCE. 267 premium is ^^, mid cost of policy $1.25: what is the cost of the insurance? $53.75. 7. A hotel is insured, for | of its value, at 1J% ; the policy costs $1.25 and the total cost of insurance is $151.25 : at what sum is the hotel valued? $15000. 8. The cost of insuring a house worth $4500, for ^ of its value, was $32.75 ; the cost of the policy was $1.25 : what was the per cent of insurance? ^^. 9. A farmer, with an insurance of $1000 on his house, and $1500 on his barn, in the Yermont Mutual, pays an annual assessment of $3.50 : what is the per cent of the premium? -J-^%. LIFE INSURANCE. 205. 1. Life Insurance policies are of two principal kinds (1) life policies, (2) endowment policies. 2. A life policy is payable at the death of the person insured. 3. An endowment policy is payable at a specified time, or at death if it occurs within this time. Rem. — In life insurance the premium is commonly a regular annual payment, dependent, in amount, upon the age of the in- dividual when he effects his insurance. The tables of a company show the annual premium, at any age, for $1000 of insurance, 1. A man at the age of 40 insures his life for $5000 ; the company's annual premium on $1000, for a life policy at this age, is $31.30; if he dies at the age of 70, how much money will he have paid the company? OPERATION. Solution. — Since the annual premium on $1000 $31.30 is $31.30, on $5000 it is $31.30 X ^ = $1^6.50; then, 5 the amount paid, in 30 yr., will be $156.50 X 30 15 6.50 --=$4695. 8 $r6¥5.W 268 RAY'S NEW PRACTICAL ARITHMETIC. 2. Mr. Harris, aged 35, takes out an endowment policy in a life insurance company for $10000, payable in 10 years ; the cost of the annual premium on $1000, at bis age, is $105.58 : if he lives to receive the endowment, what will be the cost of the paid-up policy, without interest? $10553. 3. At the age of 50, the cost of a life policy, payable annually, is $47.18 on $1000; the cost of an endowment policy, payable in 20 years, is $60.45 on $1000 ; at the end of 20 years, how much more will have been paid on a policy of $8000 by the endowment plan than by the life plan? $2123.20. 4. At the age of 44, a man insures his life to the amount of $12000 in favor of his wife; the company's annual premium at this age, for a life policy, is $36.46 on $1000 : if the man dies after the payment of 5 pre- miums, how much more than he paid out, will his widow receive ? $9812.40 5. At the age of 21, a young man takes out a life policy for $5000, upon which the annual premium is $19.89 on $1000 : if he lives to the age of 75, how much will it cost him to keep up his insurance? $5370.30. 6. At the age of 30, to secure an endowment policy for $1000, payable in 10 3 ears, costs an annual premium of $104.58 ; what will be the amount of the ten pay- ments at the end of the time, allowing interest at 6% ? $1390.91. 7. At the age of 38, a gentleman took out a policy for $6000, on the life plan, paying annually $29.15 on $1000. After keeping up his premiums for 15 years, he suffered his policy to lapse : how much money had he paid out, allowing interest at 6% ? $3882.78. DEFINITIONS. 206. 1. A tax is money paid by the citizens of a coimtry for the support of government or for other pub- lic purposes. 2. A tax is either direct or indirect. 3. A direct tax is one which is levied upon the per- son or property of the citizens. 4. A tax upon the person is called a poll tax; upon property, a property tax. 5. An indirect tax is one which, in some way, is levied upon the business of the citizens. 6. The taxes of the United States, considered in refer- ence to their nature and purpose, are of two classes, (1) State and Local Taxes; (2) United States Revenue. STATE AND LOCAL TAXES. 207. 1. The money for State and local purposes arises chiefly from direct taxation. Kem. — Some revenue accrues to the State from the rent of school lands, from licenses, fines, etc. (269) 270 KAY'S NEW PKACTICAL ARITHMETIC. 2. For the purposes of taxation, property is ([classed as Real Estate and Personal Property. 3. Real Estate is property which is fixed, as lands, houses, etc. 4. Personal Property is that which is movable, as furniture, merchandise, etc. 5. The valuation is the estimated worth of the prop- erty. # Rem.^ — The vnluation is generally the basis upon which to estimate the tax. In some states, however, the specific tax upon the polls must first be subtracted; in Massachusetts, a sixth part of the tax is assessed upon the polls, provided it does not exceed $2 for each indi- vidual; in Vermont, the basis is what is called the Grand List, which is ascertained by dividing the valuation by 100 and adding $2 for each poll. 6. The valuation is made by an officer called an assessor. Rem. — This official makes out a list called an assessment roll; it contains the names of the persons to be taxed, along with the valua- tion of their property. 208. To find the rate of taxation. The rate of taxation is expressed as so many mills on each dollar of taxable property, or as such a per cent of it. 1. The property of a certain town is valued at $1049905 ; there are 483 persons subject to poll-tax. In a certain year the total taxes of the tow^n are $13323.36 ; the poll-tax being $1.50 for each person, w^hat is the rate of taxation upon the property? TAXES. 271 Solution. — The poll-tax is $1.50x483r=$724.50; then, the property tax is $13323.36— ^ ^^ ^ r n $724.50 = $12598.86. Then, ' '"^ ' since the tax on $1049905 is 1 '^ ^ 2 3.3 6 $12598.86, the tax on $1 is 7 2 4.5 $12598.86 -- 1049905 = $0,012, 1 4 9 9 5 ) 1 2 5 9 8.8 6 (.0 1 2 12 mills, or 1 1 ^c- 10 4 9 9 5 2 099810 2099810 Rule. — 1. Multiply the tax on each poll by the number of polls ; the product is the poll-tax. 2. From the total amount of tax subtract the poll-tax ; the remainder is the property tax. 3. Divide the property tax by the valuation ; the quotient is the rate of taxation. Rem. — Of course, where there is no specific poll-tax, the total amount of the tax is to be divided immediately by the valuation. 2. A tax of $2500 is assessed upon a certain district to build a school-house. The property of the district is valued at $618000, and there are 28 persons subject to poll-tax: if the poll-tax is $1, what will be the rate of taxation? 4 mills on $1, or 1%. 3. Upon a valuation of $2876475 the tax is $18409.44: there being no poll-tax, what is the rate? 6.4 mills on $1. 4. The total valuation of property in the State of Wisconsin, for j.874, was $421285359; the tax levied upon this valuation was $656491.61: what was the rate to the hundredth of a mill? . 1.56 mills on $1. ITl KAY'S NEW PRACTICAL AR1THMETI(\ 209. To apportion the tax among the tax-])ayer8. I. A tax of $1373.64 is aBsessed upon a village, the property of which is valued at $748500 ; 57 persons pay a poll-tax of $1.25 each; find the rate of taxation, and construct a tax table to $9000. TAX TABLE. Rate, 1.74 mills on $1. PROP. TAX. PROP. TAX. PROP. TAX. PROP. TAX. $1 $0,002 . $10 $0,017 $100 $0,174 $1000 $ 1.74 2 .003 20 .035 200 .348 2000 3.48 3 .005 30 .052 300 .522 3000 5.22 4 .007 40 .070 400 .696 4000 6.96 5 .009 50 .087 500 .870 5000 8.70 6 .010 GO .104 600 1.044 6000 10.44 7 .012 70 .122 700 1.218 7000 12.18 8 .014 80 .139 800 1.392 8000 13.92 9 .016 90 .157 900 1.566 9000 15.66 Rem. — In order to facilitate the calculation of each person's tax, it is customary to construct such a table. It is not necessary to carry it out in any column farther than the nearest mill. 1. James Turner's property is valued at $7851, and he pays poll-tax for 2 persons: what is his tax? OPERATION. Solution. — By the table, the tax on $7000 is $12.18; on $800, $1,392; on $50, $0,087; and on $1, $0,002; then, the tax on $7851 is $12.18 + $1,392 -f- $0,087 + $0,002 =r $13.66; this is his prop- erty tax. The poll-tax is $1.25 X 2 = $2.50. Then, James Turner's tax is $13.66 + $2.50 at $16.16. $ 7851 12.18 1.3 9 2 .0 8 7 .0 2 13.6 6 2.5 16.16 UNITED STATES REVENUE. 273 Explanation. — It is evident that the operation is equivalent to multiplying $7851 by the rate, 1.74, and adding the poll-tax. 2. John Brown's property is valued at $2576, and he pays poll-tax for 1 person : what is his tax ? S5.73. 3. Henry Adams' property is valued at $9265, and he pays poll-tax for 3 persons: what is his tax? $19.87. 4. Amos Clarke's property is valued at $4759, and he pays poll-tax for 1 person: what is his tax? $9.53. 5. Emily Wood's property is valued at $8367 : what is her tax? $14.56. II. The tax to be raised in a city is $64375 ; its tax- able property is valued at $16869758 ; find the rate of taxation to thousandths of a mill, and construct a tax table to $90000. Eate 3.816 mills on $1. 1. William Mill's property is valued at $56875: what is his tax? $217.04. 2. Samuel Young's property is valued at $27543 : what is his tax? $105.10. 3. Charles O'Neil's property is valued at $83612 : what is his tax? $319.06. 4. Adolph Meyer's property is valued at $72968 : what is his tax? $278.45. 5. Louis Ganot's property is valued at $69547 : what is his tax? $265.39. UNITED STATES REVENUE. 210. 1. The United States Revenue arises wholly from indirect taxation ; it consists of Internal Revenue and the revenue from Duties or Customs. 2. The Internal Revenue arises from the sale of pub- lic lands, from a tax upon certain manufactures, from the sale of postage stamps, etc. 274 KAY'S NEW rilACTICAL ARITHMETIC. 3. Duties or Customs are taxes on goods imported from foreign countries. INTERNAL REVENUE. 211. 1. The public lands are disposed of at SI. 25 per acre: what will the government receive for a town- ship containing 36 sq. miles? $28800. 2. Letter postage is 3 ct. for each half-ounce, or frac- tion thereof: what is the postage on a letter weighing IJ oz.? 9 ct. 3. The postage on books is 1 ct. for each 2 oz., or fraction thereof: what is the postage on a book weigh- ing 1 lb. 5 oz.? 11 ct. 4. The tax on ])roof spirits is 70 ct. per gallon : what is the tax on a barrel of 40 gallons? $28.00. 5. The tax on cigars per 1000 is $5 : how much does this enhance the price of a single cigar? ^ ct. 6. The tax on beer is $1 per barrel of 31 gal. Each wholesale dealer in malt liquors pays a special tax of $50, and each retail dealer a special tax of $20 ; in a certain city there are 12 wholesale dealers, 250 retail dealers, and the annual manufacture of beer is 30000 bbl. : what is the revenue to government? $35600. DUTIES OR CUSTOMS. 212. 1. Duties are of two kinds, specific and ad valorem. 2. A specific duty is levied upon the quantity of the goods. Rem. — In levying specific duties, allowance is made ( 1 ) for waste called draft, (2) for the weight of the box, cask, etc.^ containing the goods, called tare. The waste of liquors, imported in casks or barrels, is called leakage) that of liquors imported in bottles, break- age. Gross weight is the weight before deducting dry ft and tare; net weight is the weight after deducting draft and tare. DUTIES. ' 275 3. An ad valorem duty is levied upon the cost of the goods. Rem. — The cost of the goods is shown by the foreign invoice, or it is determined by appraisement at the custom-house. 4. Duties must be paid in coin. Kem. — The duty is computed on the net weight and on the total cost of the article in the foreign country. The dutiable value upon which the duty is estimated, is always the nearest exact number of dollars, pounds, etc. 1. The gross weight of a hogshead of imjiorted sugar is 1760 lb.; allowing 12^% tare, what is the duty at If ct. per pound? S26.95. 2. A manufacturer imported from Spain 40 bales of wool, of 400 lb. each, tare ^% ; the cost was 45 ct. per pound: what was the duty, at 9 ct. per pound and 10^ ad valorem? $2052. 3. A merchant imported a case of glassware ; the cost of the ware in France was 365.15 francs, the case and charges were 57.15 francs, and the commission 5^ : what was the duty at 40^ in U. S. money, reckoning the franc at 19-\ ct.? $34.40. Rem. — The total cost being $85.58, the dutiable value is $86.00. 4. A book-seller imports a case of books ; their cost in Germany was 1317.04 marks, case and charges 34.36 marks, and commission 6^ : what was the duty at 25^ in U. S. money, the mark being estimated at 23.8 ct. ? $85.25. 5. A merchant imports six cases of woolen cloth, net weight 1500 lb. ; the cost in England was £500, cases and charges £8 48. 6d., commission 2^^ : what was the duty, at 50 ct. per To. and 35% ad valorem in U. S. money, estimating the pound at $4.8665? $1637.25. DEFINITIONS. 213. 1. Ratio is the relation of two numbers ex- pressed by their quotient. Thus, the ratio of to 2 is 6 — 2^3; that is, 6 is 3 times 2. Rem. — The established custom in several departments of math- ematics makes it advisable to change the treatment of ratio as given in former editions of Ray's Arithmetics. 2. The ratio of two numbers is indicated by writing the sign ( : ) between them. Thus, 2 : 6 is read the ratio of 2 to 6. • 3. The two numbers are styled the terms of the ratio. 4. The first term is called the antecedent, and the second term the consequent. 5. 6 : 2 is 3, a ratio between two abstract numbers. $6 : S2 is 3, a ratio between two concrete numbers of the same denomination. To find 2 yd. : 2 ft., reduce the 2 yd. to ft; 6 ft. : 2 ft. is 3. A ratio can not exist between 2 ft. and (276) because KATIO. 277 they can not be reduced to the same denomination. Hence, 1st. The terms of the ratio may be either abstract or concrete. 2d. Whe7i the terms are concrete, both must be of the same denominatioji. 3d. The ratio is always an abstract number. 6. Eatios are either simple or compound. 7. A simple ratio is a single ratio. Thus, 2 : G is a simple ratio. 8. A compound ratio consists of two or more simple ratios. 2 • 6 ^ Thus, o ! Q MS a compound ratio. 9. In Eatio three quantities are considered: (1) the antecedent J (2) the co7isequent, and (3) the ratio. Any two of these being given, the third may be found. 214. Given the terms, to find the ratio. 1. What is the ratio of 6 to 3? OPERATION. Solution. -The ratio of to 3 is 6 divided by 3, G : 3 equal to 2. 6 -- 3 = 2 2. What is the ratio of § to ^? OPERATION. Solution. — The ratio of f to f^is | divided by |, I = f or I multiplied by f , equal to |. I "^ f Rule. — Divide the antecedent by the consequent. Rem. — When the terms are of different denominations, they must be reduced to the same denomination. 278 KAYS NEW PKACTICAL AKITIIMETIC. What is the ratio of 3. 12 to 3? 4, 4. 30 to 5? 6, 5. 35 to 7? 5 6. 56 to 8? 7 7. 5 to 10? ^. 8. 7 to 21? J. 9. 12 to 18? |. 10. 15 to 20? f 11. 15 to 25? f 12. 25 to 15? l| 13. 36 to 28? 14. 49 to 35? 15. ^ to §? 16. 4 to 1? 17. i to 4? 18. I to i? 19. n to 1? 20. H to 2J? 21. H to 2i? 22. 6A to 4f ? If I- 2. If ^■ What is the ratio of 23. S18 to $6? 24. 54 days to 9 days? 25. 96 men to 12 men? 26. 221 bu. to 17 bu.? 27. 1 fl. 9 in. to 3 in.? 28. 5 yd. 1 a. to 5 fl. 4 in.? 215. Given the ratio and the consequent, to find the antecedent. 3. 6. 8. 13. 7. 3. 1. 7 is the ratio of what number to 4? OPERATION. Solution. — The number is 4 multiplied by 7, 4 X " =^28 equal to 28. Rule. — Multiply the consequent by the ratio. 4 is the ratio of what number to 13? |- is the ratio of what number to 27 ? -j^ is the ratio of what number to 52? 2|- is the ratio of what number to 24 ? 45^ is the ratio of what number to If? 52. 15. 28. 63. 7^ EATIO. 279 7. 3 is the ratio of what to 75 ct. ? $2.25. 8. ^ is the ratio of what to 4 lb. 8 oz. ? 3 lb. 15 oz. 9. 2.6 is the ratio of what to $4. $10.40. 216. Given the ratio and the antecedent, to find the consequent. 1. 5 is the ratio of 45 to what number? OPERATION. Solution. — The number is 45 divided by 5, 45 -=-5 = 9 equal to 9. Rule. — Divide the antecedent by the ratio. 2. 4 is the ratio of 56 to what number? 14. 3. ^ is the ratio of 42 to what number? 60. 4. 2f is the ratio of 23f to what number ? 8i 5. 7| is the ratio of $27.20 to what ? $3.60. 217. To find the value of a compound ratio. 6 : 2) 1 Find the vahie of the compound ratio q • 3 f OPERATION. Solution.— The product of the antecedents 6 6X^ = 5 4 and 9 is 54, the product of the consequents 2 and 2X3= 6 3 is 6; then, the value of the compound ratio is 5 4-^6= 9 54 divided by 6, equal to 9. Rule. — Divide the 'product of the antecedents by the prod- uct of the consequents. Rem. — Multiplying the antecedents together and the consequents together, evidently reduces the compound ratio to a simple one; 6 2 ^ . thus, in the above example the compound ratio g " o >is equivalent to the simple ratio 54 : 6. 280 RAY'S NEW PRACTICAL ARITHMETIC. Find the value 2. Of the compound ratio q .' ^ [ 3, 3. Of the compound ratio o^f ! of [ 8. 4. Of the compound ratio f ! if [ f- - ^n .1 1 *• 8 men : 2 men. ) ,, 5. Oi the compound ratio it> j • 24 d 6. Of the compound ratio ^^u^ ; f^^^^ I 6. 5 : 2) 7. Of the compound ratio 7:3^ 10^. 9:5) 218. The terms of a ratio correspond to the terms of a fraction, the antecedent to the numerator, the conse- quent to the denominator. Thus, ill 2 : 3 the ratio is §, in which the antecedent 2 is the numerator and the consequent 3 the denominator. Hence (Art. 101) we have the following Principles. I. A ratio is multiplied 1st. By multiplying the antecedent. 2d. By dividing the consequent. II. A ratio is divided Ist. By dividing the antecedent. 2d. By multiplying the consequent. III. A ratio is not changed 1st. By midtiplying both terms by the same number. 2d. By dividing both terms by the same number. KATIO. 281 219. To reduce a ratio to its lowest terms. 1. Eeduce 16 : 24 to its lowest terms. OPERATION. Solution.— The G. C. D. of 16 and 24 is 8; divid- 8)16:24 ing both terms of 16 : 24 by 8, it becomes 2 : 3 (Art. 2~: 3 218, III, 2d). Rule. — Divide both terms of the ratio by their greatest common divisor. 2. Eeduce 20 3. Eeduce 10 4. Eeduce 34 5. Eeduce 95 6. Eeduce 75 7. Eeduce 217 25 to its lowest terms. 4 : 5. 30 to its lowest terms. 1 : 3. 51 to its lowest terms. 2 : 3. 133 to its lowest terms. 5 : 7. 125 to its lowest terms. 3:5. 279 to its lowest terms. 7 : 9. 220. To clear a ratio of fractions. 1. Clear 1^ 2\ of fractions. Solution. — The L. C. M. of the denominators 2 and 3 is 6; multiplying both terms of K] : 2\ by 6, it be- comes 9 : 14 (Art. 218, III, 1st). operation. U:2^ 6 ~^T~1"4 Rule. — Mvltiply both terms of the ratio by the least com- mon ymdtiple of the denominators of the fractions. 2. Clear 3| 4f of fractions. 75 88 3. Clear n lOf of fractions. 45 64 4. Clear 1 1 of fractions. 15 14. 5. Clear 61% 9^2^. of fractions. 189 284. -zz:'. z^*;— -''-O, DEFINITIONS. 221. 1. Proportion is an expression for the equality of two ratios. Thus, 2 : 4 and 3 : 6 may form a proportion, for the ratio of each is \. 2. The proportion is indicated by writing : : between the ratios. Thus, 2 : 4 : : 3 : 6 is read 2 is to 4 as 3 is to 6. 3. A proportion is either simple or compound. 4. In a simple proj^ortion both the ratios are simple. Thus, 2 : 4 : : 3 : G is a simple proportion. 5. In a compound proportion one or both the ratios are compound. Thus, o ! 4 [ • • r ." Q [ 's a compound proportion. G. Every proportion consists of four terms. 7. The first and fourth terms of a proportion are called the extremes. (282) PROPORTION. 283 8. The second and third terms of a proportion are called the means. 9. The last term is said to be a fourth proportional to the other three taken in order. Thus, in the proportion 2 : 4 : : 3 : G, the extremes are 2 and 6; the means are 4 and 3; and 6 is a fourth proportional to 2, 4, and 3. 10. When three numbers form a proportion, the second number is said to be a mean proportional between the other two. Thus, in the proportion 2 ; 4 : : 4 : 8, 4 is a mean proportional between 2 and 8. 222. The operations of proportion depend upon the following Principle. — In every proportion the product of the ex- tremes is equal to the product of the means. Thus, in the proportion 2 : 4 : : 3 : 6, 2 X 6 = 4 )< 3; "i the proportion 3 ! 4} : = 5 ! ^ | 2 X 3 X ^ X 8 == 3 X 4 X 4 X ^i -^nd the same may be shown for any other proportion. Hence (36, 4), 1st. Tf the product of the means be divided by one of the extremes, the c/uotlent Kill be the other extreme. 2d. If the product of the extremes be divided by one of the means, the quotient will be the other mean. 223. Given three terms of a proportion, to find the fourth. 1. What : G : : 4 : 8? 284 RAYS NEW PRACTICAL ARITHMETIC. Solution. — The product of the means G and 4, operation. is 24; then, 24 divided by 8, one of the extremes, GX4 = 24 equals 8, the other extreme (222, 1st). : 5|, : 12] • 2. 4 : what : : '^ ' ^ I ? 10 Solution. — The product of tlie ex- tremes, 4 X S X 12, divided by 3 X 10, one of the means equals 8, the other, mean (222, 2d). 24--8. operation. 2 4 =-8 Rule. — Divide the 'product of the terms of the same name by the other given term. Rem. — Indicate the operation and cancel whenever it is prac- ticable (91). 3. 2:8:: 6 : what? 4. 5 : 7 : : 10 : what? 5. What : 8 : : G : 1()? 6. 5 : what : : G : 12? 7. 3:7:: what : 14? 8. 7 : 14 : : : what? 9. ^;Jj::what:45? 10. 5 : 8) .3 :4 |, 4 : 10 ) * 7 : what j " 11. 10 : what) . . 22 : 33 | 14 : 21 j 2G : 39 1 12. 13. 14. f : f : : 1 : what? 1 : what : : A : li? What : 4§ : : 71 : 10^? 15. 4 : G : : G : Avhat? 24. 14. 3. 10. G. 18. 20. 21. 15. A- I- 3^. 9. PKOrOKTlON. 285 224. Proportion, when applied to the sohition of con- crete problems, has been styled " The Rule of Three,'' because three terms are given to find the fourth. The use of Proportion was formerly so extensive that it was often called " The Golden Bide.'' The solution of a problem by proportion consists of two parts: 1st. The statement; that is, the proper arrangement of the numbers into a proportion. 2d. The operation of finding the required term. Rem. — In arranging the numbers in a proportion, it is cus- tomary, though not necessary, to make the number or quantity re- quired a fourth pi'oportional to the other three; then, the first three terms of the proportion always are given to find the fourth. I. SIMPLE PROPORTION. 1. If 2 yd. of cloth cost $4, what will 6 yd. cost? OPERATION. Solution. — Since the number re- 2 : 6 : : 4 : what? quired, or fourth term of the propor- 3 tion, is dollars, the third term is $4. ^ X 4 Since the cost of 6 yd, will be greater " ^ than the cost of 2 yd., 6 yd. is the second term of the proportion, and 2 yd. the first term. Dividing the product of 6 and 4 by 2 (Art. 223, Rule), the required term is $12. Rem.— In this example, the number of dollars is in a direct ratio to the number of yards; that is, the greaier the number of yards, the greater the number of dollars they will cost. 286 KAY'S NEW PliACTICAL ARITHMETIC. 2. If 3 men can dig a cellar in 10 days, in how many days can 5 men dig it? OPERATION. Solution. — Since the number re- 5 : 3 : : 1 : what? quired, or fourth term of the propor- 2 tion, is days, the third term is 10 da. 3 X /l^ /, Since 5 men will dig the cellar in a 5 less number of days than 3 men, 3 men is the second term of the proportion and 5 n)en the first term. Dividing the product of 3 and 10 by 5 (Art. 223, liule), the required term is 6 da. Kem. — In this example, the number of days is in an inverse ratio to the number of men; that is, the yreaicr the number of men, the less the number of days in which they will dig the cellar. Rule. — 1. For the third term, write that number which is of the same denomination as the mimher required. 2. For the second term, tcrite the gueat'er of the two remaining numbers, when the fourth term is to be greater than the third; and the less, when the fourth term is to be less than the third. 3. Divide the j^^oduct of the second and third terms by the first; the quotient will be the fourth tenn, or number re- quired. 3. If 3 men can dig a cellar in 12 days, how many men will dig it in 6 days? (i 4. If 3 yd. cloth cost $8, what cost 6 yd.? 816. 5. If 5 bl. flour cost $30, what cost 3 bl.? $18. 6. If 3 lb. 12 oz. tea cost $3.50, what cost 11 lb. 4 oz.? $10.50. 7. If 2 lb. 8 oz. of tea cost $2, w^hat quantity can you buy for $5 ? 6 lb. 4 oz. 8. If 4 hats cost $14, what cost 10 hats? $35. PROPORTION. 287 9. If 3 caps cost 69 cents, what cost 11 caps? $2.53. 10. If 4 yd. cloth cost $7, what cost 9 yd. ? $15.75. 11. If 8 yd. cloth cost $32, what cost 12 yd.? $48. 12. If 12 yd. cloth cost $48, what cost 8 yd. ? $32. 13. If $32 purchase 8 yd. of cloth, how many yards will $48 buy? 12. 14. If $48 purchase 12 yd. of cloth, how many yards can be bought for $32? 8. 15.' A man receives $152 for 19 months' work: how much should he have for 4 months' work? $32. 16. If 8 men perform a piece of work in 24 days, in what time can 12 men perform it? 16 days. 17. If 60 men perform a piece of work in 8 da., how many men will perform it in 2 days? 240. 18. If 15 oz. of pepper co^ 25 ct., what cost 6 lb.? $1.60. 19. If 6 gal. of molasses cost $2.70, what cost 26 gal.? $11.70. 20. If 5 cwt. 85 lb. of sugar cost $42.12, what will 35 cwt. 25 lb. cost? $253.80. 21. If 11 yd. of cloth cost $2.50, what will be the cost of 11 yd.?"' $1,871 22. If 90 bu. of oats supply 40 horses 6 da., how long will 450 bu. supply them? 30 da. 23. If 6 men build a wall in 15 da., how many men can build it in 5 da.? 18. 24. If 15 bu. of corn pay for 30 bu. of potatoes, how much corn can be had for 140 bu. potatoes? 70 bu. 25. If 3 cwt. 25 lb. of sugar cost $22.60, what will be the cost of 16 cwt. 25 lb.? " $113. 26. If a perpendicular staff, 3 ft. long, cast a shadow 4 ft. 6 in., what is the height of a steeple whose shadow measures 180 ft.? 120 ft. 27. If a man perform a journey in 60 da., traveling 9 288 KAY'S NEW PRACTICAL ARITHMETIC. lir. each day, in how many days can he perform it by traveling 12 hr. a day? 45. 28. A merchant, failing, paid 60 ct. on each dollar of his debts. He owed A S22()0, and B $1800: what did each receive? A S1320. B $1080. 29. A merchant, having failed, owes A $800.30; B $250; C $375.10; D $500; F $115; his property, worth $G12.12, goes to his creditors: how much will this pay on the dollar? 30 ct. 30. If the 4-ccnt loaf weigh 9 oz. when flour is $8 a bl., what will it weigh when flour is $6 a bl.? 12 oz. 31. I borrowed $250 for G mo. : how long should I lend $300 to compensate the favor? 5 mo. 32. A starts on a journey, and travels 27 mi. a day ; 7 da. after, B starts and travels the same road 3G mi. a day: in how many days will B overtake A? 21. 33. If William's services are worth $15§ a mo., when he labors 9 hr. a day, what ought he to receive for 4|^ mo., when he labors 12 hr. a day? $91.91^. 34. If 5 lb. of butter cost $|, what cost f lb. ? $3%. 35. If 6 yd. cloth cost $5f , what cost 7f yd.? $6|f 3G. If J bu. wheat cost $f , what cost | bu. ? $^. 37. If If yd. cloth cost $2^, what cost 2 yd. ? $|. 38. If $29f buy 59^ yd. of cloth, how much will $31J buy? G2iyd. 39. If .85 of a gallon of wine cost $1.36, what will be the cost of .25 of a gallon? $0.40. 40. If 61.3 lb. of tea cost $44.9942, what will be the cost of 1.08 lb. ? $0.79. 41. If ^ of a yard of cloth cost $|, what will ^V ^^ ^ yard cost? ^^. 42. If f of a yard of velvet cost $4|, what cost 17f yd.? $178.38J. 43. A wheel has 35 cogs ; a smaller wheel working in PilOFOKTlON. 289 it, 26 cogs : in how many revolutions of the larger wheel will the smaller gain 10 revolutions? 28f. 44. If a grocer, instead of a true gallon, use a measure deficient by 1 gill, what will be the true measure of 100 of these false gallons? 96J gal. 45. If the velocity of sound be 1142 feet per sec, and the number of pulsations in a person 70 per min., what is the distance of a cloud, if 20 pulsations are counted between the time of seeing a flash of lightning and hearing the thunder? 3 mi. 22G rd. 2 yd. 2\ ft. 46. The length of a w^all, by a measuring line, \v^s 643 ft. 8 in., but the line was found to be 25 ft. 5. 1 in. long, instead of 25 feet, its sup2)osed length : wliat was the true length of the wall? 654 ft. 11.17 in. II. COMPOUND PROPORTION. 225. 1. If 2 men earn $20 in 5 da., what sum can 6 men earn in 10 da.? \.i}-- OPERATIOIS^. 2 : what? Solution.— Since the num- 2 ber required or fourth term of 5 the proportion is dollars, the 3 2 third term is $20. Since 6 0X|0X2O men can earn a greater num- ber of dollars than 2 men, 6 ?X^ 120 men is in the second term of the proportion and 2 men in the first term; and since in 10 da. a greater number of dollars can be earned than in 5 da., 10 da. is in the second term of the proportion and 5 da. in the first term. Dividing the product of 6, 10, and 20 by the product of 2 and 5 (Art. 223, Rule), the required term is $120. 2. If 6 men, in 10 da., build a wall 20 ft. long, 3 ft. high, and 2 ft. thick, in how many days could 15 men build a wall 80 ft. long, 2 ft. high, and 3 ft. thick? I'rao. 19. 290 RAY'S NEW PRACTICAL ARITHMETIC. OPERATION. 10 : what? Solution. — Since the 15 rumber required, or fourth 2 term of the proportion, is 3 days, the third term is 10 2 days. Since 15 men can 94 o build a wall in a less nun,- ^x^^X^ X ^XX P , « ber of days than 6 men ^ ^ l^ 6 men is in the second ,^y/^f^y^jj/\ ^ term of the proportion, and ^ 15 men in the first term; since to build a wall 80 Tt. long will take a greater number of days than to build a wall 20 ft. long, 80 ft. is in t second term of the proportion and 20 ft. in the first term; since to build a wall 2 ft. high will take a less number of days than to build a wall 3 ft. high, 2 ft. is in the second term of the proportion and 3 ft. in the first term; and since to build a wall 3 ft. thick will 1 .. e a greater number of days than to build a wall 2 ft. thick, 3 ft. is in the second term of the proportion and 2 ft. in the first term. Dividing the product of 6, 80, 2, 3, and 10 by the product of 15, 20, 3, -nd 2 (Art. 223, Rule), the required term is 16 da. Rule. — 1. For the third term, \trite that number which is 0^ the same denomination as the number required. 2. Arrange each pair of the numbers forming the com- pound ratio as if with the third term, they formed a simple proportion. 3. Divide the product of the numbers in the second and tnird terms by the product of the numbers in the first term ; the quotient will be the fourth term or number required. 3. If a man travel 24 mi. in 2 da., by walking 4 hr. a day: at the same rate, bow far will be travel in 10 da., walking 8 br. a day? 240 mi. 4. If 16 men build 18 rods of fence in 12 days, bow many men can build 72 rd. in 8 da. ? 96. 5. If 6 men spend $150 in 8 mo., bow mucb will 15 men spend in 20 mo.? 8^37.50. PARTNERSHIP. 291 6. I travel 217 mi. in 7 days of 6 hr. each : how far can I travel in 9 days of 11 hr. each? 511 J mi. 7. If $100 gain $6 in 12 mo., what Hum will $75 gain in 9 mo.? $3.37f 8. If 100 lb. be carried 20 mi. for 20 ct., how far will 10100 lb. be carried for $60.60? 60 mi 9. To carry 12 cwt. 75 lb. 400 mi., costs $57.12: what will it cost to carry 10 tons 75 mi.? $168. 10. If 18 men, in 15 da., build a wall 40 rd. long, 5 ft. high, 4 ft. thick, in what time could 20 men build a wall 87 rd. long, 8 ft. high, and 5 ft. thick? 58|f da. 11. If 180 men, in 6 days, of 10 hr. each, dig a trench 200 yd. long, 3 yd. wide, 2 yd. deep, in how many days can 100 men, working 8 hr. a day, dig a trench 180 yd. long, 4 yd. wide, and 3 yd. deep? 24.3 PARTNERSHIP. 226. 1. A Partnership is an association of persons for the transaction of business. Such an association is called a firm, or house, and each member, a partner. 2. The capital, or stock, is the amount of money or property contributed by the firm. 3. The assets are the amounts due a firm, together with the property of all kinds belonging to it. 4. The liabilities of a firm are its debts. 5. The net capital is the difference between the assets and liabilities. 1. A and B engaged in trade; A's capital was $200; B's, $300; they gained $100: find each partner's share. Solution— The whole capital is $200 + $300 =r $500; of this A 292 K/iY'S NEW PRACTICAL ARITHMETIC. owns |ggr=|, and B owns f gg — | of the capital; hence, A's gain will be f of $100:= $40, and B's gain will be ^ of $100=: $60. Or, Solution.— The whole capital is $200 -f $300 = $500; then, $500 : $200 :: $100 : $40, A's share; $500 : $300 :: $100 : $00, B's share. Rule. — Take such part of the whole gain or loss, as each jmrtnefs stock is part of the whole stock. ' Or, Rule. — As the whole stock is to each partnefs stocky so is the whole gain or loss to each partner's gain or loss. Rem. —This rule is applicable when required to divide a sum into parts having a given ratio to each other; as in Bankruptcy, General Average, 6tc. 2. A and B form a partnership, with a capital of ^800 : A's part is S300 ; B's, $500 ; they gain $232 : what is the share of each ? A's, $87 ; B's, $145. 3. A's stock was $70; B's, $150; C's, $80; they gained $120: what was each man's share of it? A's, $28 ; B's, $60 ; C's, $32. 4. A, B, and C traded together: A put in $200; B, $400; C, $600: they gained $427.26: find each man's share. A's, $71.21; B's, $142.42; C's, $213.63. 5. Divide $90 among 3 persons, so that the parts shall be to each other as 1, 3, and 5. $10, $30, and $50. 6. Divide $735.93 among 4 men, in the ratio of 2, 3, 5, and 7. $86.58; $129.87; $216.45; $303.03. 7. A person left an estate of $22361 to be divided among 6 children, in the ratio of their ages, which are 3, 6, 9, 11, 13, and 17 yr. : what are the shares? $1137; $2274; $3411; $4169; $4927; $6443. 8. Divide $692.23 into 3 parts, that shall be to each other as ^, f, and f $127.60; $229.68; $334.95. BANKRUPTCY. 293 BANKRUPTCY. 227. A Bankrupt is one who has failed to pay his debts when due. Rem. — The assets of a bankrupt are usually placed in the hands of an assignee, whose duty it is to convert them into cash, and divide the net proceeds among the creditors. 1. A man, failing, owes A $175; B, $500; C, $600; D, $210; E, $42.50; F, $20; G, $10; his property is worth $934.50: what will be each creditor's share? A's, $105; C's, $360; E's, $25.50; B's, $300; D's, $126; F's, $12.00; G's, $6. 2. A man owes A $234; B, $175; C, $326: his prop- erty is worth $492.45 : what can he pay on $1 ; and what will each creditor get? 67 ct. on $1; A, $156.78; B, $117.25; C, $218.42. 3. Mr. Smith failed in business, owing $37000. His assignee sold the stock for $25000, and charged $4650 for expenses: how much did he pay on the dollar? 55%. GENERAL AVERAGE. 228. General Average is the method of apportioning among the owners of a ship and cargo, losses occasioned by casualties at sea. 1. A, B, and C freighted a ship with 108 tuns of wine. A owned 48, B 36, and C 24 tuns ; they Were obliged to cast 45 tuns overboard : how much of the loss must each sustain? A, 20; B, 15; C, 10 tuns. 294 RAY'S NEW PRACTICAL ARITHMETIC. 2. From a ship valued at SI 0000, with a cargo valued at $15000, there was thrown overboard goods valued at $1125: what % was the general average, and what was the loss of A, whose goods were valued at $2150? General average, 4J%; A's loss, $9G.75. PARTNERSHIP WITH TIME. 229. 1. A and B built a wall for $82; A had 4 men at work 5 days, and B 3 men 7 days : how should they divide the money? Solution. — The work of 4 men 5 da. equals the work of 4 X S» or 20 men 1 da.; and the work of 3 men 7 da., equals the work of 3X7, or 21 men 1 da.; it is then required to divide $82 into two parts, having the same ratio to each other as 20 to 21; hence, A's part is If of $82 = $40; B's part is f| of $82 :=r $42. 2. A put in trade $50 for 4 mo.; B, $60 for 5 mo.; they gained $24: what w^as each man's share? Solution.— $50 for 4 mo. equals $50X4 = $200 for 1 mo.; and $60 for 5 mo. equals $60X5 = $300 for 1 mo. Hence, divide $24 into two parts having the same ratio as 200 to 300, or 2 to 3. This gives A I of $24 = $9.60, and B | of $24 = $14.40. Rule. — Multiply each partnefs stock by the time it was employed; then take such part of the gain or loss as each partner's product is part of the sum of all the products. 3. A and B hire a pasture for $54 : A pastures 23 horses 27 da.; B, 21 horses 39 da.: Avhat will each pay? A, $23.28|; B, $30.71i. 4. A put in $300 for 5 mo. ; B, $400 for 8 mo. ; C, $500 for 3 mo.: they lost $100; find each one's loss. A's, $24.19i|; B's, $51.61^; C's, $24.19^. EQUATION OF PAYMENTS. 295 5. A, B, and C hire a pasture for $18.12: A pastures 6 cows 30 da. ; B, 5 cows 40 da. ; C, 8 cows 28 da. : what shall each pa}^? A, S5.40; B, $6; C, S6.72. 6. Two men formed a partnership for 16 mo. : A put in, at first, $300, and, at the end of 8 mo., $100 more ; B put in, at first, $600, but, at the end of 10 mo., drew out $300; they gained $442.20: find each man's share. A's, $184.80; B's, $257.40. 7. A and B are partners : A put in $800 for 12 mo., and B, $500. What sum must B put in at the end of 7 mo. to entitle him to half the year's profits? $720. EQUATION OF PAYMENTS. 230. Equation of payments is the method of finding the mean or average time of making two or more pay- ments, due at different times. 1. A owes B $2, due in 3 mo., and $4, due in 6 mo. : at wlvdt period can both ' sums be paid so that neither party will be the loser? Solution. — The interest on $2 for 3 mo. equals operation. the interest on $1 for 3 X 2 = 6 mo.; the interest 2 X ^ == ^ on $4 for 6 mo. equals the interest on $1 for 6 X 4 4 X 6 == 2 4 == 24 mo.; then, the interest on $2 + $4 = $6 6 y^O equals the interest on $1 for 6 mo. -j- 24 mo. = 30 5 mo.; hence, $6 must be on interest 30 ^- 6 = 5 mo. Rule. — 1. Multiply each payment by the time to elapse till it becomes due. 2. Divide the sum of the products by the sum of the pay- ments ; the quotient will be the equated time. Rem. — When one of the payments is due on the day from which the equated time is reckoned, its product is 0; but, in finding the sum of the payment, this must be added with the others. 2m KAY'S NEW PRACTICAL ARITHMETIC. 2. A owes B $2, due in 4 mo., and $6, due in 8 mo.; find the average time of paying both sums. 7 mo. 3. A owes B ^8, due in 5 mo., and $4, due in 8 mo. : find, the mean time of payment. 6 mo. 4. A buys $1500 worth of goods ; S250 are to be paid in 2 mo., $500 in 5 mo., $750 in 8 mo.: find the mean time of payment. 6 mo. 5. A owes B $300; 1 third due in 6 mo.; 1 fourth in 8 mo.; the remainder in 12 mo.: what is the average time of payment? 9 mo. 6. I buy $200 worth of goods ; 1 fifth to be paid now ; 2 fifths in 5 mo. ; the rest in 10 mo. : what is the average time of paying all? 6 mo. 231. In finding the Average or Mean time for the pay- ment of several sums due at difixirent times, any date may be tal^en from which to reckon the time. 1. A merchant buys goods as follows, on 60 days credit: May 1st, 1848, $100; June 15th, $200: what is the average time of payment? Ju\y 30th. Solution. — Counting from May 1, it is operation. 60 days to the time of the first payment, $100 X ^>0— 6000 and 105 days to that of the second; then, $200 X 105=^21000 the equated time is 90 days from May 1st, $300 ) 27000 that is, July 30th. ^90 2. I bought goods on 90 days credit, as follows: April '2d, 1853, $200; June Ist, $300: what is the average time of payment? Aug. 6th. 3. A merchant bought goods as follows: April 6, 1876, on 3 mo., $1250; May 17, 1876, on 4 mo., $4280; June 21, 1876, on 6 mo., $675: what is the average time of payment? Sept. 12, 1876. AVERAGE. 297 AVERAGE. 232. Average is the method of finding the mean or average price of a mixture, when the ingredients com- posing it, and their jDrices, are known. 1. I mix 4 pounds of tea, worth 40 ct. a lb., with 6 lb., worth 50 ct. a lb. : what is 1 lb. of the mixture worth ? OPERATION. Solution. — 4 lb. at 40 ct. per lb. are worth 4 X -40 = 1.60 $1.60, and 6 lb. at 50 ct. are worth $3.00; 6 X -50 rrr 3.00 then, 4 -f- 6 =:^ 10 lb. are worth $4.60; hence, 10 ) 4.60 "746 Rule. — Divide the whole cost by the whole number of ingredients ; the quotient will be the average or mean price. 2. Mix 6 lb. of sugar, at 3 ct. a lb., with 4 lb., at 8 ct. a lb., what will 1 lb. of the mixture be worth? 5 ct. 3. Mix 25 lb. sugar, at 12 ct. a lb., 25 lb., at 18 ct., and 40 lb., at 25 ct. : what is 1 lb. of the mixture worth? 19| ct 4. A mixes 3 gal. water, w^ith 12 gal. wine, at 50 ct. a gal.: what is 1 gal. of the mixture worth? 40 ct. 5. I have 30 sheep : 10 are worth $3 each ; 12, $4 each ; the rest, $9 each : find the average value. S5. 6. On a certain day the mercury in the thermometer stood as follows: from 6 till 10 A. M., at 03°^ from 10 A. M. till 1 P. M., 70° ; from 1 till 3 P. M., 75° ; from 3 till 7 P. M., 73° ; from 7 P. M. till 6 A. M. of the next da}^, 55° : what was the mean temperature of the da}^, from sunrise to sunrise? 62X°. DEFINITIONS. 233. 1. Involution is the multiplication of a number into itself one or more times. 2. A power is the product obtained by involution. 3. The first power is the number itself 4. The second power, or square, is the product ob- tained by taking the number twice as a factor. Thus, 2 X 2 == 4, is the second power or square of 2. 5. The third power, or cube, is the product obtained by taking the number three times as a factor. Thus, 2 X 2 X 2 = 8 is the third power or cube of 2. Rem. — The second power is called the square, because the area of a square is the product of two equal factors (Art. 68). The third power is called the cube, because the solid contents of a cube is the product of three equal factors (Art. 70). 6. The higher powers of a number are denominated respectively the fourth power^ fifth power ^ sixth power ^ etc. Thus, 2 X 2 X 2 X 2 — 16, is the fourth power of 2; . 2 X 2 X 2 X 2 X 2 == 32, is the fifth power of 2; 2X2X2X2X2X2 =6 4, is the sixth power of 2, etc. (298) INVOLUTION. 299 7. The exponent is a number denoting the power to which the given number is to be raised. Thus, in 3 2, read 3 square, the 2 denotes the square of 3; hence, 32=9. In 53, reads cube, the 3 denotes the cube of 5; hence, 6^ z= 125. 7* is read 1 fourth power j 9^, % Jifth power, etc. 234. To raise a number to any power. 1. Find the cube of 75. Solution. — 75 multiplied by 75 is 5625; this is the square of 75. 5625 multiplied by 75 is 421875; this is the cube of 75. OPERATION. 75 75 375 525 5625 5625 7_5 28125 39375 421875 Rule. — Obtain a product in which the number is taken as a factor as many times as there are units in the exponent of the power. 2. Find the square of 65. 4225. 3. Find the cube of 25. 15625. 4. Find the fourth power of 12. 20736. 5. Find the fifth power of 10. 100000. 6. Find the sixth power of 9. 531441. 7. Find the eighth power of 2. 256. 8. Find the square of f . |. 9. Find the cube of |. If- 10. Find the fourth power of f . m- 11. Find the fifth powder of |. 2%- 12. Find the square of 16i. 2721 13. Find the cube of 12f 19531-. 14. Find the fourth power of .25 .00390625 15. 143 =what? 2744. 16. 194= what? 130321. 17. (21)5 zz^ what? 69A%- ^ EVOLUTION DEFINITIONS. 235. 1. Evolution is the process of resolving a num- ber into two or more equal factors. 2. A root of a number is one of the two or more equal factors. 3. The square root of a number is one of two equal factors. Thus, 3 is the square root of 9; for 9 = 3 X 3. 4. The cube root of a number is one of three equal factors. Tlius, 3 is the cube root of 27; for 27 =r 3 X 3 X 3. 5. The higher roots of a number are denominated respectively the fourth root, fifth root, etc. Thus, 3 is the fourth root of 81; for 81 = 3 X 3 X ^ X 3. 3 is the fifth root of 243; for 243 = 3 X 3 X 3 X 3 X 3. 6. The radical sign |/ placed before a number shows that its root is to be extracted. 7. The index is a number placed above the radical sign to show the number of the root. (300) EVOLUTION. 301 Rem. — It is customary, however, to omit 2, the index of the square root. Thus, VH^ is read the square root of 25; hence, l/^=r5. ^"^ is read the cube root of 27; hence, f~2T—Z. ^T6 is read the fourth root of 16; hence, Vl6:=2. 8. A perfect power is one whose root can be ob- tained exactly. Thus, 25 and if are perfect squares; 27 and ^V are perfect cubes; 16 and Jg are perfect fourth powers. 9. The squares and cubes of the first ten numbers are exhibited in the following TABLE. Numbers. 1 2 3 4 5 6 7 8 9 10 Squares. 1 4 9 16 25 36 49 64 81 100 Cubes. 1 8 27 64 125 216 343 512 729 1000 Rem. — The numbers in the first line are the square roots of the corresponding numbers in the second line, and the cube roots of those in the third line. 10. An imperfect power is one whose root can be obtained only approximately. Thus, ]/Tz=: 1.41421 +. 302 RAY'S NEW PRACTICAL ARITHMETIC. SQUARE ROOT. 236. To find the number of figures in the square root. 1. The square root of 1 is 1, and the square root of 100 is 10 (Art. 235, 9, Table); between 1 and 100 are all numbers consisting of one or two figures, and between 1 and 10 are all numbers consisting of one figure ; there- fore, When a number consists of one or two figures^ its square root consists of one figure. 2. The square root of 100 is 10, and the square root of 10000 is 100; between 100 and 10000 are all num- bers consisting of three or four figures, and between 10 and 100 are all numbers consisting of two figures ; there- fore, When a number consists of three or four figures^ its square root consists of tico figures. 3. In like manner it may be shown that, When a number consists of five or six figures, its square root consists of three figures. And so on ; therefore, 1st. If a number be pointed off into periods of tivo figures each, the number of periods will be the same as the number of figures in the square root. 2d. The square of the units will be found in the first period, the square of the tens in the second period, the square of the hundreds in the third period, etc. SQUAKE ROOT. 303 237. To point off a number into j^eriods of two figures each. 1. Point off 368425. 368425. 2. Point off 6.843256. 6.843256 3. Point off 83751.42963. 83751.429630 Rule. — Place a point over the order units, and then over every second order from units to the left and to the right. Rem. 1. — The first period on the left of the integral part of the number will often contain but a single figure. Rem. 2. — When the first period on the right of the decimal part contains but a single figure, a cipher must be annexed to complete the period. 4. Point off 864326 ; 4.758462 ; 7584.3769. 5. Point off 97285.46138 ; 75300 ; .046827 ; .0625 ; .625. 238. To extract the square root of a number. 1. Extract the square root of 256. OPERATION. Solution. — Point off 256 into periods of two 2 5 6(16 figures each by placing a point over 6 and 2 1 (Art. 237, Rule). 26)156 156 The largest square in 2 (Art. 235, 9, Table) is 1; its root is 1; place the root 1 on the right and subtract the square 1 from 2; the remainder is 1, to which bring down the next period 56. Double the root 1 and place the result 2 on the left of 156 for a trial divisor. Find how many times 2 is contained in 15 (making allowance for subsequent increase of the trial divisor); the result is 6; place 6 in the root on the right of 1 and also on the right of 2, the trial divisor; then 26 is the complete divisor. Multiply 26 by 6 and subtract the product 156 from 156; the remainder is 0. There- fore, 256 is a perfect square, and its square root is 16. 304 HAYS NEW PRACTICAL ARITHMETIC. GEOMETRICAL EXPLANATION. 1 1 1 II D - 6X6=36 1 1 1 1 1 1 1 1 I— B 10X6=60 1 C - 10X6=60 A. 10X10=100 After finding that the sq. root of the given number will con- tain two places of figures (tens and units), and that the figure in tens' place is 1 (ten), form a square figure (A) 10 in. on each side, which contains (Art. 67) 100 sq. in.; taking this sum from the whole number of squares, 156 sq. in. remain, which correspond to the num- ber, 156, left after subtracting above. It is obvious that to increase the figure A, and at the same time preserve it a square, both length and breadth must be increased equally; and, since each side is 10 in. long, it will take twice 10, that is, 20 in., to encompass two sides of the square A. For this reason, 10 is doubled in the numerical operation. Now determine the breadth of the addition to be made to each side of the square A. After increasing each side equally, it will require a small square (D) of the sa.vie breadth as each of the fig- ures B and C, to complete the entire square; hence, the superficial contents of B, C, and D, must be equal to the remainder, 156. Now their contents are obtained by multiplying their length by their breadth. Then the figure in the units' place — that is, the breadth of B and C — must be found by trial, and it will be somewhat less than the number of times the length of B and C (20) is contained in the re- mainder (156). 20 is contained in 156 more than 7 times; let us try 7: 7 added to 20 makes 27 for the whole length of B, C, and D, and this, multiplied by 7, gives 189 for their superficial contents; this being more than 156, the breadth (7). was taken too great. Next, try 6 for the length and breadth of D; adding 6 to 20 gives 26 for the length of B, C, and D; multiplying 26 by the breadth (6) gives 156 for the superficial contents of B, C, and D. Hence, the square root of 256 is 16; or, when 256 sq, in. are arranged in the form of a square, each side is 16 inches. SQUAKE KOOT. 305 2. Extract the 8qiiare root of 758.436. OPERATION. Solution. — Point off 758.43G ^ro^o^. A/o-rro , . T ,» ^ ' ^ "•'* <^0U( Z I .O O -4- into periods of two ngures each . by placiiiijj a point over 8 and then over 7 to the left, and 3 4 / j .i o » and to the right (Art. 237, ^ '^ ^ Rule), Then find the figures of 5 45)2943 the root as in Ex. 1. The last re- 2 7 2 5 mainder is 5351. Therefore, 758.43G 5503)21860 is an imperfect square, and its 16 5 9 square root is 27.53 -f. 5 3 5 1 Rem. — By bringing down one or more periods of decimal ciphers, the operation might be continued to any required number of decimal places in the root. 3. Extract the square root of ||f . Solution. — The square root of the numerator 256 is 16, and the square root of the denominator 625 is 25 (Ex. 1); then, the square root of ||f is if. 4. Extract the square root of f . Solution. — | reduced to a decimal is .375, The square root of .375, to five decimal places, is .61237 (Ex. 2); then, the square root of f is .61237 -f . Rule. — 1. Point off the given number into periods of two figures each. 2, Find the greatest square in the first period on the left ; place its root on the rights like a quotient in division; sub- tract the square from, the period, and to the remainder bring down the next period for a dividend. 3. Double the root found, and place it on the left of the dividend for a trial divisor. Find how many times the trial divisor is contained in the dividend, exclusive of the right hand, figure; place the quotient in the root^ and also on the right of the trial divisor. 306 RAY'S NEW PRACTICAL ARITHMETIC. 4. Multiply the complete divisor by the last figure of the root; subtract the product from the dividend, and to the remainder bring down the next period for a new dividend. 5. Double the ivhole root found, for a new trial divisor, and continue the operation in the same manner until all the periods are brought down. Rem. 1. — When the luiniher is an imperfect square, the operation may be continued to any required number of decimal places in the root by bringing down periods of decimal ciphers (Ex. 2). Rem. 2. — To extract the square root of a common fraction: (1) when both terms are perfect squares, extract the square root of the numerator and then of the denominator (Ex. 3); (2) wh(;n both terms are not perfect squares, reduce the fraction to u decimal and extract the square root of the decimal ( Ex. 4 ). Extract the square root of 5. 529. 23. 17. 915.0625. 30.25. 6. 625. 25. 18. .0196. .14 7, 6561. 81. 19. 1.008016. 1.004 g. 56644. 238. 20. .00822649. .0907 9. 390625. 625. 21. TT^- 2^. 10. 1679616. 1296. 22. t¥A- H. 11. 5764801. 2401. 23. 30i ^. 12. 43046721. 6561. 24. 10. 3.16227 -f 13. 987656329. 31427. 25. 2. 1.41421-f 14. 289442169. 17013. 26. 3- .81649 + 15. 234.09. 15.3. 27. 6f. 2.52982 -f 16. 145.2025. 12.05. 28. 384f 19.61049 + 239. To extract the square root of a perfect square by factoring. 1. Exti-Mct llie square root of 441. SQUAKE KOOT. 307 Solution. -441 =3X 3X 7X 7; hence, v^ 44l = 3 X 7 = 21. Rule. — Besolve the number into its prime factors, and find the product of one of each two equal factors. Extract the square root of 2. 16. 4. 6. 400. 20. 3. 36. 6. 7. 1764. 42. 4. 100. 10. 8. 5184. 72. 5. 225. 15. 9. 3025. 55. 240, Given two of the sides of a right-angled tri- ani^C^e to find the third side. ] . A triangle is a plane figure bounded by three straight lines, called its sides. Thus, D E F is a triangle; its sides are D E, E F, and D F. 2. When one of the sides is perpendicular to another, they form a right-angle, and the triangle is called a right-angled triangle. A Thus, in the triangle A B G, the side A C being perpendicular to the side B C, they form a right-angle at C; hence, ABC is a right- angled triangle. 3. The side opposite the right-angle is called the hypotenuse ; the other two sides, the base and the per- pendicular. Thus, in A B C, A B is the hypotenuse, B C the base, and A C the perpendicular. 308 KAY'S NEW PRACTICAL ARITHMETIC. 4. Proposition. — The square described on the hypotenuse of a right-angled triangle is equal to the sum of the square.^ described on the other two sides. Draw a right-angled triangle, ABC, with the side B C 4 in., and the side A C 3 in.; then, the side A B will be 5 in. Describe a square on each side of the triangle, and divide each square into smaller squares of 1 in. to the side. Then, the square described on A B will contain 25 square inches, and the two squares described on B C and A C will contain la-j- 9 =;= 25 square inches. I I c 5. From this proposition we deduce the following Rules. — Ist. To find the hypotenuse; To the square of the base add the square of the perpendicular, and extract the square root of the sum. 2d. To find the base or the perpendicular; From the square of the hypotenuse subtract the square -of the other given side, and extract the square root of the difference. 1. The base and perpendicular of a right-angled tri- angle are 30 and 40: what is the hj'potenuse? 50. 2. The hypotenuse of a right-angled triangle is 100, and the base 60: what is the perpendicular? 80. 3. A castle 45 yd. high is surrounded by a ditch 60 yd. wide : what length of rope will reach from the out- side of the ditch to the top of the castle? 75 yd. 4. A ladder 60 ft. long reaches a window 37 ft. from the ground on one side of the street, and, without mov- ing it at the foot, will reach one 23 ft. high on the other side: find the width of the street. 102.64+ ft. CUBE HOOT. 309 5. A tree 140 ft. high is in the center of a circular island 100 ft. in diameter; a line 600 ft. long reaches from the top of the tree to the further shore : what is the breadth of the stream, the land on each side being of the same level ? 533.43 -f- ft. 6. A room is 20 ft. long, 16 ft. wide, and 12 ft. high : what is the distance from one of the lower corners to the opposite upper corner? 28.28 -f ft. 241. Given the area of a square to find its side (Art. 67). Biile. — Extract the square root of the area. 1. The area of a square field is 6241 sq. rd. : what is the length of one side? 79 rd. 2. The surface of a square table contains 8 sq. ft. 4 sq. in.: what is the length of one side? 2 ft. 10 in. 3. The area of a circle is 4096 sq. yd. : what is the side of a square of equal area? 64 yd. 4. A square field measures 4 rd. on each side : what is the length of the side of a square neld which contains 9 times as many square rods? 12 rd. 5. What is the length of one side of a square lot con- taining 1 acre? 208.71+ ft. CUBE ROOT. 242. To find the number of figures in the cube root. 1. The cube root of 1 is 1, and the cube root of 1000 is 10 (Art. 235, 9, Table); between 1 and 1000 are all numbers consisting of one, two, or three figures, and between 1 and 10 are all numbers consisting of one figure ; therefore, 310 RAY'S NEW PRACTICAL ARITHMETIC. When a number consists of one, two, or three figures, its cube root consists of one figure. 2. The cube root of 1000 is 10, and the cube root of 1000000 is 100; between 1000 and 1000000 are all num- bers consisting of four, five, or six figures, and between 10 and 100 are all numbers consisting of two figures; therefore, When a number consists of four, five, or six figures, its cube root consists of two figures. 3. In like manner it may be shown that. When a number consists of seven, eight, or nine figures^ its cube root consists of three figures. And so on ; therefore, 1st. If a number be pointed off into periods of three figures each, the number of periods will be the same as the number of figures in the cube root. 2d. The cube of the units will he found in the first period, the cube of the tens in the second period, the cube of the hundreds in the third period, etc. 243. To point off a number into periods of three figures each. 1. Point off 876453921. 87^453921. 2. Point off 7.356849227. 7.356849227 3. Point off 37683.5624. 37683.562400 Rule. — Place a point over the order units, and then over every third order from units to the left and to the right. Rem. 1. — The first period on the left of the integral part of the number will often contain but one or two figures. Rem. 2. — When the first period on the right of the decimal part contains but one or two figures, ciphers must be annexed to complete the period. 4. Point off^ 138975462; 3.561325482; 684536.256403. CUBE KOOT. 311 5. Point off 2756.56843 ; 98451.3276; .856375; .0064. 244. To extract the cube root of a number. 1. Extract the cube root of 13824. OPERATION. 13824(24 2X2X300=:120() 2X4X 30= 240 4X4 ■= 10 1456 i824 5824 Solution. — Point off 13824 into periods of three figures each by placing a point over 4 unci 3 (Art. 243, Rule). The largest cube in 13 (Art. 235, 9, Table) is 8; its root is 2; place the root 2 on the right, and subtract the cube 8 from 13; the remainder is 5, to which bring down the next period 824. Square the root 2 and multiply it by 300; the result, 1200, is the trial divisor. Find how many times 1200 is contained in 5824; the result is 4; place 4 in the root on the right of 2. Multiply 2 b\^ 4 and by 30, and square 4; add the products 240 and IG to 1200; the sum 1456 is the complete divisor. Multiply 1456 by 4, and subtract the product 5824 from 5824; the remainder is 0. Therefore, 13824 is a perfect cube, and its cube root is 24. GEOMETRICAL EXPLANATION. After finding that the cube root of the given number will contain two places of figures (tens and units), and that the figure in the tens' place is 2, form a cube, A, Fig. 1, 20 (2 tens) inches long, 20 in. wide, and 20 in. high; this cube will contain, (Art. 70,) 20X20X20 = 8000 cu. in.; take this sum from the whole number of cubes, and 5824 cu. in. are left, which correspond to the number 5824 in the numerical operation. 312 RAY'S NEW PRACTICAL ARITHMETIC. It is obvious that to increase the figure A, and at the sarr.e timo preserve it a cube, the length, breadth, and height must each receive an equal addition. Then, since each side is 20 in. long, square 20, which gives 20 X 20 = 400, for the number of square inches in each face of the cube; and since an addition is to be made to three sides, multiply the 400 by 3, which gives 1200 for the number of square inches in the 3 sides. This 1200 is called the trial divisor; because, by means of it, the thickness of the additions is determined. By examining Fig. 2 it will be seen that, after increasing each of the three sides equally, there will be required 3 oblong solids, C, C, C, of the same length as each- of the sides, and whose thickness and height are each the same as the additional thickness; and also a cube, D, whose length, breadth, and height are each the same as the additional thickness. Hence, the solid contents of the first three rectangular solids, the three oblong solids, and the small cube, must together be equal to the remainder (5824). Now find the thickness of the additions. It will always be . something less than the number of times the trial divisor (1200) is contained in the dividend (5824). By trial, we find 1200 is contained 4 times in 5824; proceed to find the contents of the different solids. The solid contents of the first three addi- tions, B, B, B, are found by mul- tiplying the number of sq. in. in the face by the thickness (Art. 70); there are 400 sq. in. in the face of each, and 400 X S — 1200 sq. in. in one face of the three; then, multiplying by 4 (the thick- ness) gives 4800 cu. in. for their contents. The solid contents of the three oblong solids, C, C, C, are found by multiplying the number of sq. in. in the face by the thickness; now there are 20 X 4 = 80 sq. in. in one face of each, and 80 X 3 = 240 sq. in. in one face of the three; then multiplying by 4 (the thickness), gives 960 cu. in. for their contents. Lastly, find the contents of the small cube, D, by multiplying together its length, breadth, and thickness; this gives 4X4X4=- 04 cu. in. Fig. 2. CUBE ROOT. 313 If the solid contents of the several additions be added together, as in the margin, their sum, 5824 cu. in. will be the number of small cubes remaining after forming the first cube, A. Hence, when 13824 cu. in. are arranged in the form of a cube, each side is 24 in.; that is, the cube root of 13824 is 24. ADDITIONS. B B Br=4800 cu. in. C C C=: 960 " " D= 64 " " Sum, 5 8 2 4 Kem. — It is obvious that the additions in the margin may readily be arranged in the same way as in the operation of the example. 2. Extract the cube root of 413.5147. OPERATION. 413.5 14 700(7.4 5 + 343 7X 7X300r^ 14700 7X 4X 30= 840 4X4 = 16 15556 74X74X300 = 1642800 74X SX 3 0= 11100 5X 5 25 1653925 70514 62224 8290700 8269625 2 10 7 5 Solution.— Point off 413.5147 into periods of three figures each by placing a point over 3, and then over 4 and to the right (Art. 243, Rule). Then find the figures of the root as in Ex. 1. The last nv mainder is 21075. Therefore, 413.5147 is an imperfect cube, and its cube root is 7.45 -|-. Kem. — By bringing down one or more periods of decimal ciphers the operation might be continued to any required number of decimal places in the root. 3. Extract the cube root of yVg¥?- 314 KAY'S NEW PRACTICAL AKITIIMETIC. Solution. — The cube root of the numerator 2197 is 13 and the cube root of the denominator 13824 is 24; (Kx. 1); then, the cube root of tVbV? i« \l- 4. Extract the cube root of ^. Solution. — i reduced to a decimal is .8. The cube root of .8 to three decimal places is .928; (Ex. 2); then, the cube root of ^ is .928 +. Bule. — 1. Point off the given number into periods of three figures each. 2. Find the greatest cube in the first period on the left; place its root on the right, like a quotient in division ; sub- tract the cube frojn the period, and to the remainder bring down the next period for a dividend. 3. Square the root found, and multiply it by 300 for a trial divisor. Find how many times the trial divisor is con- tained, in the dividend, and place the quotient in the root. 4. Midfiply the preceding figure, or figures, of the root by the last and by 30, and square the last figure; add the products to the trial divisor ; the sum is the complete divisor. 5. Multiply the complete divisor by the last figure of the root; subtract the product from the dividend, and to the remainder bring down the next period for a new dividend. 6. Find a new trial divisor as before, and continue the operation in the same manner until all the periods are brought down. Rem. 1. — When the number is an imperfect cube, the operation may be continued to any required number of decimal places in the root by bringing down periods of decimal ciphers. (Ex. 2). Rem. 2. — To extract the cube root of a common fraction: (1) when both terms are perfect cubes, extract the cube root of the numerator and then of the denominator; (Ex. 3); when both terms are not per- fect cubes, reduce the fraction to a decimal and extract the cubo root of the decimal. (Ex. 4). CUBE ROOT. 315 Extract the cube root of 5. 91125. 45. 15. 53.1573^ 16. 3.76 6. 195112. 58. 16. .199176704. .584 7. 8. 9. 10. 912673. 1225043. 13312053, 102503232. 97. 107. 237. 468. 17. 18. 19. 20. tWAV if If. If- 11. 529475129. 809. 21. 2. 1.259+ 12. 958585256. 986. 22. 9. 2.080+ 13. 14760213677. 2453. 23. 200. 5.848+ 14. 128100283921. 5041. 24. 91 2.092+ 245. Given the solid contents of a cube to find its side (Art. 70). Rule. — Extract the cube root of the solid contents. 1. The contents of a cubical cellar are 1953.125 cu. ft.: find the length of one side. 12.5 ft. 2. Sixtj^-four 3-incli cubes are piled in the form of a cube: what is the length of each side? 1 ft. 3. A cubical box contains 512 half-inch cubes : what are the dimensions of the box inside? 4 in. 4. A cubical excavation contains 450 cu. yd. 17 cu. ft. : w^hat are its dimensions? 23 ft. 5. Find the side of a cube equal to a mass 288 ft. long, 216 ft. broad, and 48 ft. high. 144 ft. 6- The side of a cubical vessel is 1 foot: find the side of another cubical vessel that shall contain 3 times as much. 17.306+ in. X MENSURATION.)! I. MEASUREMENT OF SURFACES. DEFINITIONS. 246. 1. A line has length without breadth or thick- ness. 2. Lines are either straight or curved. ^^^^ ~^^- 3. When two lines meet, the}'^ form an angle. Rem. — The point at which the lines meet is called the vortex of the angle. 4. Angles are either acute., obtuse^ or right angles. 5. When two straight lines are perpen- dictular to each other, they form a right angle. 6. An acute angle is less than a right angle. 7. An obtuse angle is greater than a right angle. 8. When tw^o straight lines are ever^^- where equally distant they are parallel. 9. A surface has length and breadth without thickness. (316) MENSURATION. 317 10. Surfaces arc either plane or curved. Thus, the surface of a table or floor is plane; that of a ball or g^lobe is curved. 11. A plane figure is a portion of a plxme surface bounded by one or more lines. 12. A polygon is a plane figure bounded by straight lines. Rem. — The straight lines are called the sides of the polygon; the perimeter of a polygon is the sum of all its sides. 13. A triangle is a plane figure /1\ bounded by three straight lines. / I \^ Rem. — If one side be taken for the base, the perpendicular let fall upon the base from the opposite angle is called the altitude of the triangle. 14. A quadrilateral is a plane figure bounded by four straight lines. 15. There are three kinds of quadrilaterals : the tra- pezium^ the trapezoid^ and the parallelo- gram. 16. A trapezium is a quadrilateral v;ith no two sides parallel. 17. A trapezoid is a quadrilateral / I with only two sides parallel. / 18. A parallelogram is a quadri- lateral with its opposite sides equal / / and parallel. / / Rem.— If one side be taken as the base, the perpendicular let fall upon the base from the opposite side is called the altitude of the parallelogram. 318 KAYS NEW PKACTICAL ARITHMETIC. 19. A rhombus is a parallelogram with all its sides equal, and its angles not right angles. 20. A rectangle is a parallelogram with all its angles right angles. 21. A square is a rectangle with all its sides equal. 22. A polygon of five sides is called a pentagon ; of six, a hexagon ; of eight, an octagon, etc. 23. A diagonal is a line joining two angles not adjacent. 24. A circle is a. plane figure hounded hy a curved line, every point of which is equally distant from a point within called the center. 25. The circumference of a circle is the curved line which bounds the figure. 26. The diameter of a circle is a straight line pass- ing through the center, and terminated, both Avays, by the circumference. 27. The radius of a circle is a straight line drawn from the center to the circumference; it is half the diameter. 247. To find the area of a parallelogram (Art. 246, 18, 19, 20, 21). Rule. — Multiply the base by the altitude. Explanation. — The area of a paral- lelogram is equal to the area of a rect- angle, having an equal base and the same altitude; but the area of the rectangle is equal to its length multiplied by its breadth; (Art. 68); hence, the area of a parallelogram is equal to its base multiplied by its altitude. MENSURATION. 319 1. How many square feet in a floor 17 ft. long and 15 ft. Avide? 255 sq. ft. 2. How many acres of land in a parallelogram, the length of which is 120 rd., and the perpendicular breadth 84 rd. ? 63 A. 3. How many acres in a square field, each side of which is 65 rd. ? 26 A. 65 sq. rd. 4. How many acres in a field in the form of a rhom- bus, each side measuring 35 rd., and the perpendicular distance between two sides being 16 rd. ? 3 A. 80 sq. rd. 5. Find the difference in area between a floor 30 ft. square, and two others each 15 ft. square. 50 sq. yd- 6. A table is 3 ft. 4 in. long, and 2 ft. 10 in. wide: how many sq. ft. in its surface? Solution.— 3 ft. 4 in.r=3J or -i/- ft.; 2 ft. 10 in. = 2| or -i/ ft.; then, the surface of the table is }^oy^lJ_ — ^ gq, ft. Qr Solution. — 3 ft. 4 in.:r=40 in.; 2 ft. 10 in. = 34 in.; then the sur- face of the table is 40X34^=1360 sq. in.; 1360 -- 144=^9 sq. ft. 64 sq. in., or 9| sq. ft. 7. How many square feet in a marble slab 5 ft. 6 in, long and 1 ft. 8 in. wide? dj- sq. ft. 8. How many square yards in a ceiling 25 ft. 9 in, long, and 21 ft. 3 in. wide? 60 sq. yd. 7 sq. ft. 27 sq. in. 9. A room is 10 ft. long: how wide must it be to contain 80 sq. ft. ? 8 ft. 10. How many yards of carpet, 1^ yd. wide, will cover a floor 18 ft. long and 15 ft. wide? 20 yd. 11. How many yards of flannel, | yd. wide, will it take to line 3 yd. of cloth, li^ yd. wide? 6 yd. 320 KAY'S NEW PRACTICAL ARITHMETIC. 12. How mtiny yards of carpet, IJ yd. wide, will it lake to cover a floor 21 ft. 8 in. long and 13 ft. G in. wide? 25iyd. 13. A rectangular field is 15 rd. long: what must be its width to contain 1 A. ? lOf rd. 248. To find the area of a trapezoid (Art. 246, 17). Rule — Multiply half the sum of the jxiralld sides by the altitude. Explanation. — The base of a prirallclo- grarn having the same altitude and equal area is one-half the sum of the parallel sides of the trapezoid. 1. The parallel sides of a trapezoid are 2 ft. 2 in. and 2 ft. 11 in.; its altitude is 11 in.: what is its area? 2 sq. ft. 47^ sq. in. 2. A field is in the form of a trapezoid ; one of the parallel sides is 25 rd., and the other 19 rd. ; the width is 32 rd. : how many acres in the field? 4 A. 64 sq. rd. 3. How many square yards in a piece of roof 10 ft. 8 in. wide on the lower side, and 6 ft. 2 in. wide on the upper side, the length being 12 ft.? 11 sq. yd. 2 sq. ft. 249. To find the area of a triangle. Ist. When the base and altitude are given. Rule. — Multiply the base by the altitude^ and take half the product. Explanation.— The area of a triangle /\ 7 is one-half the area of a parallelogram / \^ / having the same base and altitude. / I _\^/ MENSUKATION. 321 2d. When the three sides are given. Rule. — 1. From half the sum of the three sides take each side separately, 2. Midtiply the half-sum and the three remainders to- gether ^ and extract the square root of the product. 1. The base of a triangle is 15 ft. and its altitude 12 ft. : what is its area ? 90 sq. ft. 2. One side of a triangular lot is 44 rd., and the per- pendicular distance from the angle opposite to this side is 18 rd. : how many acres in the lot? 2 A. 76 sq. rd. 3. What is the area of a triangle, of which the base is 12 ft. 6 in. and the altitude 16 ft. 9 in.? 11 sq. yd. 5 sq. ft, 99 sq. in. 4. Find the area of a triangle whose sides are 13, 14, and 15 ft. 84 sq. ft. 5. The sides of a triangle are 30, 40, and 50 ft. : what is the area? QQ sq. yd. 6 sq. ft. 250. To find the area of a trapezium (Art. 246, 16) or. other irregular figure. Rule. — 1. Divide the figure into triangles by diagonals. 2. Find the areas of the triangles, and add. them together. 1. Find the area of a field in the form of a trapezium, of which a diagonal is 50 rd. and the perpendiculars to the diagonal from the opposite angles 30 rd. and 20 rd. 7 A. 130 sq. rd. 251. 1. To find the circumference of a circle when the diameter is given. Rule. — Multiply the diameter by 3.1416. Prac. 21, 322 KAY'S NEW PRACTICAL ARITHMETIC. 2. Conversely : to find the diameter of a circle when the circumference is given. Rule. — Divide the circumference by 3.141G. 1. The diameter of a circle is 48 fl. : what is the cir- cumference? 150 fl. 9.56 in. 2. The circumference of a circle is 15 fl. : what is the diameter? 4 fl. 9.3 in. 3. The diameter of a wheel is 4 fl. : what is its cir- cumference? 12 fl. 6.8 in. 4. If the girth of a tree is 12 fl. 5 in., what is its diameter? 3 fl. 11.43 in. 5. What is the circumference of the earth, the diam- eter being 7912 mi.? 24856+ mi. 252. 1. To find the area of a circle, when the radius is given. Rule. — Multiply the square of the radius by 3.1416. 2. Conversely : to find the radius of a (drcle when the area is given. Rule. — Divide the area by 3.1416, and extract the square root of the quotient. 1. Find the area of a circle whose radius is 21 fl. 153 sq. yd. 8 sq. ft. 64 sq. in. 2. The area of a circle is 6 sq. fl. 98.115 sq. in. : what are its diameter and circumference? 2 ft. 11 in.; 9 fl. 1.9+ in. 3. How long a rope will it take to fasten a horse to a post so that he may graze over 1 A. of grass, and no more ? 7 rd. 2 ft. 3 in. MEASUREMENT OF SOLIDS. 323 4. Two circles, 10 and 16 ft. in diameter, have the same center : what is the area of the ring between their circumferences? 122 sq. ft. 75 sq. in. 5. The area of a circle is 1 square foot, what is its diameter? 13.54 in. II. MEASUREMENT OF SOLIDS. DEFINITIONS. 253. 1. A solid, or body, has length, breadth, and thickness. 2. A prism is a solid with two parallel bases, which are ^^olygons, and with its faces parallel- ograms. Rem. — A prism is triangular, quadrangular, etc., ac- cording to the shape of the base. 3. A right prism has its faces rectangles. 4. The altitude of a prism is the perpendicular let fall from one base upon the other. 5. The convex surface of a prism is the sum of the areas of its faces. 6. A parallelopipedon is a prism with its bases paral- lelograms. 7. A right parallelopipedon is a solid with six rectangular faces (Art. 70). 8. A cube is a solid with six equal square faces. 9. A pyramid is a solid with one base, which is a polygon, and with its faces triangles. 324 RAY'S NEW PRACTICAL ARITHMETIC. 10. A right pyramid has all its faces equal. 11. The slant height of a right pyramid is the distance from the vertex to the middle of each side of the base. 12. The three round bodies are the cylinder^ the cone^ and the sphere. 13. A cylinder is a solid with two paral- lel bases, which are circles, and with a curved surface. 14. The axis oi^ a cylinder is a line join- ing the centers of the two bases. 15. The convex surface of a cylinder is the area of its curved surface. 16. A cone is a solid with one base, which is a circle, and with a curved surface terminating in an apex. 17. A sphere is a solid with a curved surface, every point of which is equally distant from a point within called the center. 18. The volume of a body is its solid contents. 254. 1. To find the convex surface of a right prism. Rule. — Multiply the perimeter of the base by the altitude, 2. To find the convex surface of a cjdinder. Rule. — Multiply the circumference of the base by the altitude. 3. To find the entire surface of a prism, or of a cylinder. MEASUREMENT OF SOLIDS. 325 Rule. — To the convex surface add the areas of the two bases. 1. Find the surface of a cube, each side being 37 in. 6 sq. yd. 3 sq. ft. 6 sq. in. 2. Find the surface of a right prism, with a trian- gular base, each side of whicii is 4 ft. ; the altitude of the prism is 5 ft. 73.85 -f- sq. ft. 3. Find the surface of a box which is 3 ft. 6 in. long, 2 ft 9 in. wide, and 1 ft. 10 in. high.. 42^ sq. ft. 4. Find the surface of a cylinder, its altitude being 5 . ft. and the radius of the base, 2 ft. 87.96 + sq. ft. 255. To find the volume of a prism or of a cylinder. Rule. — Multiply the area of the base by the altitude. Rem. — The rule for finding the volume of a right parallelopipedon is given in Art. 70. 1. Find the volume of a right parallelopipedon, of which the length is 12 ft., the width 3 ft. 3 in., and the 4 ft. 4 in. 169 cu. ft. SoLTJTiON.— 3 ft. 3 in. = 3|: or -i^3 ft.; 4 ft. 4 m. = ^ or -i/ ft.; then, the volume of the parallelopipedon is 12 X -V" XV'^ -^^^ cu. ft. 2. How many cubic yards in a room 24 ft. long, 18 ft. 6 in. wide, and 10 ft. 7 in. high? 174 cu. yd. 1 cu. ft. 3. Each side of the base of a triangular prism is 2 ft.; its altitude is 14 ft.: what is the volume of the prism? 24J cu. ft. nearly. 4. Find the volume of a cylinder whose altitude is 12 ft. and the radius of the base 2 ft. 150.8 cu. ft. nearly. 326 RAY'S NEW PRACTICAL ARITHMETIC. 5. How many cubic inches in a peck measure, the diameter of the bottom being 9J in. and the depth 8 in. ? 537.6 -f cu. in. 256. 1. To find the convex surface of a right pyramid. Rule. — Multiply the perimeter of the base by the slant height^ and take half the product. 2. To find the convex surface of a cone. Rule. — Multiply the circumference of the base by the slant height^ and take half the jnoduct. 3. To find the entire surface of a p3'ramid or of a cjone. Rule. — To the convex surface add the area of the base. 1. Find the entire surface of a right pyramid, with a triangular base, each side of which is 5 ft. 4 in. ; the slant height of the pyramid is 7 fl. 6 in. 72.3 -f sq. ft. 2. What is the convex surface of a cone of which the slant height is 25 ft. and the diameter of the base 8 ft. 6 in.? 333.8 sq. ft. nearly. 3. Find the entire surface of a cone, of which the slant height is 4 ft. 7 in. and the diameter of the base 2 ft. 11 in. 27.6 + sq. ft. 257. To find the volume of a pyramid or of a cone. Rule. — Multiply the area of the base by the altitude^ and take one-third of the product. 1. Find the volume of a square pyramid, of which each side of the base is 5 ft. and the altitude 21 ft. 175 cu. ft. MEASUREMENT OF SOLIDS. 327 2. Find the volume of a cone, of which the altitude is 15 ft. and the radius of the base 5 ft. 392.7 cu. ft. 3.- A square pyramid is 477 ft. high ; each side of its base is 720 ft. : how many cubic yards in the pyramid ? 3052800 cu. yd. 4. The diameter of the base of a conical, glass house, is 37 ft. 8 in., and its altitude 79 ft. 9 in. : what is the space inclosed? 29622 + cu. ft. 258. To find the surface of a sphere. Rule. — Multiply the square of the diameter by 3.1416. 1. What is the surface of a sphere, of which the diam- eter is 1ft.? 3.14 + sq. ft. 2. What is the surface of a sphere, of which the diam- eter is 4 ft. 6 in. ? 63.6 + sq. ft. 3. What is the area of the earth's surface, on the supposition that it be a perfect sphere 7912 miles in diameter? 196663355.75 + sq. mi. 259. To find the volume o-f a sphere. Bule. — Multiply the cube of the diameter by one-sixth of 3.1416, or .5236. 1. Find the volume of a sphere 13 ft. in diameter. 1150.3 + CU. ft. 2. Find the volume of a sphere 2 ft. 6 in. in diam- eter. 8.2 cu. ft. nearly. 3. The volume of a sphere is 1 cu. ft. : what is its diameter? 14.9 in. nearly. 328 UAY\S NEW PRACTICAL AKITHMETIC. III. APPLICATIONS OF MENSURATION. 260. 1. Plastering, house-painting, paving, paper- hanging, etc., are meaBured by the square foot or square yard. 2. Glazing is measured by the square foot or by the pane. ' 3. Stone cutting is measured by the square foot. 4. Flooring, roofing, etc., are measured by the square yard or by the square of 100 sq. ft. 1. A room is 20 ft. 6 in. long, 16 fl. 3 in. broad, 10 ft. 1 in. high : how many yards of plastering in it, de- ducting a fire-place 6 ft. 3 in. by 4 ft. 2 in. ; a door 7 ft. by 4 ft. 2 in., and two windows, each 6 ft. b}' 3. ft. 3 in. ? 108 sq. yd. 8 sq. ft. 6 sq. in. 2. A room is 20 ft. long, 14 ft. 6 in. broad, and 10 ft. 4 in. high: what will the papering of the walls cost, at 27 ct. per square 3'ard, deducting a fire-place 4 ft. by 4 ft. 4 in., and two windows, each 6 ft. by 3 ft. 2 in.? SI 9.73. 3. What will it cost to pave a rectangular court, 21 yd. long and^VlS j&. broad, in which a foot-path, 5 ft. wide, runs the whole length: the path paved with flags, at 36 ct. per square 3'ard, and the rest with bricks, at 24 ct. per square yard ? ^79.80. 4. At 10 ct. a square yard, what will it cost to paint both sides of a partition 15 ft. 6 in. long, 12 ft. 6 in. high? S4.31. 5. A house has three tiers of window^s, seven in a tier: the height of the first tier is 6 ft. 11 in.; of the second, 5 ft. 4 in. ; of the third, 4 ft. 3 in. ; each win- dow is 3 ft. 6 in. wide: what will the glazing cost, at 16 ct. per square foot? $64.68. APPLICATIONS OF MENSUKATION. 329 6. A floor is 36 ft. 3 in. long, 16 ft. 6 in. wide : what will it cost to lay it, at $3 a square? $17.94. 7. At S3.50 jDer square, what will be the cost of a roof 40 ft. long, the rafters on each " side 18 ft. 6 in. long? $51.80. BOARD MEASURE. 261, 1. Board Measure is used in measuring all lumber which is sawed into boards, planks, etc. 2. A foot, board measure, is 1 foot long, 1 foot wide, and 1 inch thick. 3. Hence, to find the number of feet in a board, we have the following Rule. — 1. Find the surface of the hoard in square feet. 2. Multiply the surface by the thickness in inches. 1. How many feet in an inch board 16 ft. long and 1 ft. 3 in. wide? 20 ft. 2. How many feet in a two-inch plank 12 ft. 6 in. long and 2 ft. 3 in. wide? 56 J ft. 3. How many feet in a piece of scantling 15 ft. long, 4 in. wide, and 3 in. thick? 15 ft. 4. How many feet of inch boards will a stick of tim- ber 12 ft. long and 2 ft. square make? 576 ft. 5. How many feet in an inch board, 12 ft. 6 in. long 1 ft. 3 in. wide, at one end, and 11 in. wide at the other end? 13i| fl. MASONS' AND BRICKLAYERS' WORK. 262. 1. Stone masonry is usually measured by the perch, which is 24f or 24.75 cu. ft. (Art. 70). 2. Bricklaying is commonly measured by the 1000 bricks. 330 RAY'S NEW PRACTICAL ARITHMETIC. 1. How many perches in a stone wall 97 ft. 5 in. long, 18 ft. 3 in. high, 2 ft. 3 in. thick? 161.6 2. What is the cost of a stone wall 53 ft. G in. long, 12 ft. 6 in. high, 2 ft. thick, at ^2.25 a perch? $121.59. 3. How many bricks in a wall 48 ft. 4 in. long, 16 ft. 6 in. high, 1 ft. 6 in. thick, allowing 20 bricks to the cubic ft.? 23925. 4. How many bricks, each 8 in. long, 4 in. wide, 2.25 in. thick, will be required for a wall 120 ft. long, 8 ft. high, and 1 ft. 6 in. thick? 34560. 5. Find the cost of building a wall 240 ft. long, 6 ft. high, 3 ft. thick, at $3.25 per 1000, each brick being 9 in. long, 4 in. wide, and 2 in. thick. S336.96. MEASUREMENT BY BUSHELS OR GALLONS. 263. 1. To find the number of bushels (Art. 61). Rule. — Find the volume in cubic inches , and divide by 2150.4. 2. To find the number of gallons (Art. 64). Rule. — Find the volume in cubic iiiches, and divide by 231. 1. How many bushels in a bin 15 ft. long, 5 ft. wide, and 4 ft. deep? 241 +. 2. How many gallons in a trough 10 ft. long, 5 ft. wide, and 4 ft. deep ? 1496 +. 3. How many bushels in a cylindincal tub 6 ft. in diameter and 8 ft. deep? 181.76+. 4. How many barrels, of 31 J gal. each, in a cistern, in the form of a cylinder, of which the diameter is 4 ft. and the depth 6 ft.? 17.9+ bl. I. ARITHMETICAL PROGRESSION, 264. 1. An Arithmetical Progression is a series of numbers which increase or decrease by a common difference. 2. If the series increase, it is called an increasing series; if it decrease, a decreasing series. Thus, 1, 8, 5, 7, 20, 17, 14, 11, 11, etc., is an increasing series. 5, etc., is a decreasing series. 3. The numbers forming the series are called terms; the first and last terms are the extremes; the other terms, the means. 4. In every arithmetical series, five things are consid- ered : (1) i\\Q first term, (2) the last term, (3) the com- mon difference^ (4) the number of terms, and (5) the sum of the terms. CASE I. 1 265. To find the last term, when the first term, the common difference, and the number of terms are given. 1. I bought 10 yd. of muslin, at 3 ct. for the first yard, 7 ct. for the second, 11 ct. for the third, and so on : what did the last A^ard cost ? (331) 332 KAY'S NEW PRACTICAL ARITHMETIC. Solution. — To fihd the cost of the second yard, operation. add 4 ct. once to the cost of the first; to find the . 4 X 9 = 36 cost of the third, add 4 ct. twice to the cost of the 3 -f 3 6 = 3 9 first; to find the cost of the jourth, add 4 ct. three times to the cost of the first, and so on; hence, to find the cost of the tenth yard, add 4 ct. nine times to the cost of the first; but 9 times 4 ct. are 36 ct., and 3 ct. -f 36 ct. = 39 ct., the cost of the last yard, or last term of the progression. 2. The first term of a decreasing scries is 39, the com- mon difference 4, and the number of terms 10 : find the last term. OPERATION. Solution. — In this case, 4 must he subtracted 4 X 9 ==: 3 6 9 times from 39, which will give 3 for the last 3 9 — 36= 8 term. Rule. — 1. Multiply the common difference by the number of terms less one. 2. If an increasing series, add the product to the first term; if a decreasing series, subtract the product from the first term. 3. Find the last term of an increasing series in which the first term is 2, the common difference 3, and the number of terms 50. 149. 4. What is the 54th term of a decreasing series in which the first term is 140, and common difference 2? 34. 5. What is the 99th term of a decreasing series in which the first term is 329, and common difference ^? 243J. CASE II. 266. To find the common difference, when the ex- tremes and the number of terms are given. 1. The first term of a series is 2, the last 20, and the number of terms 7: what is the common difference? AKITHMETICAL FROGKESSION. 333 Solution. — The difference of the extremes 20 operation. and 2 is 18; 18 divided by 6, the number of 20 — 2 = 18 terms less 1, is 3, the common difference. 18^6= 3 Rule. — Divide the difference of the extremes by the num- ber of terms less one. 2. The extremes are 3 and 300 ; the number of terms 10 : find the common diiferenco. 33. 3. A travels from Boston to Bangor in 10 da. ; he goes 5 mi. the first day, and increases the distance trav- eled each day by the same number of miles; on the last day he goes 50 mi. : find the daily increase. 5 mi. CASE III. 267. To find the sum of all the terms of the series when the extremes and the number of terms are given. 1. Find the sum of 6 terms of the series whose first term is 1, and last term 11. Solution. — The series is . In inverted order it is The sum of the two is . . 12^ 12^ 12^ 12^ 12; 12. Since the two series are the same, their sum is twice the first series; but their sum is obviously as many times 12, the sum of the ex- tremes, as there are terms; hence, the sum of the series is 6 times 12 = 72 divided by 2 = 36. Rule. — Multiply the sum of the extremes by the number of terms; and take half the product. 2. The extremes are 2 and 50 ; the number of terms, 24 : find the sum of the series. 624. 3. How many strokes does the hammer of a clock strike in 12 hours? 78. 1, 3, 5, 7, 9, 11. 11, 9, 7, 5, 3, 1. 334 RAY'S NEW PRACTICAL ARITHMETIC. 4. Place 100 apples in a right line, 3 yd. fi-om each other, the first, 3 yd. from a basket: what distance will a boy travel who gathers them singly and places them in the basket? 17 mi. 69 rd. ^ yd. 5. A body falling by its own weight, if not resisted by the air, would descend in the first second a space of 16 ft. 1 in.; the next second, 3 times that space; the third, 5 times that space ; the fourth, 7 times, etc. : at that rate, through what space would it fall in 1 minute? 57900 ft. II. OEOMETRICAL PROGRESSION. 268. 1. A Geometrical Progression, is a series of numbers increasing by a common multiplier^ or decreasing by a common divisor. Thus, 1, 3, 9, 27, 81, is an increasing geometric series. 48, 24, 12, 6, 3, is a decreasing geometric series. 2. The common multiplier or common divisor, is called the ratio. Thus, in the first of the above series, the ratio is 3; in the second, 2. 3. The numbers forming the series are the terms; the first and last terms are extremes; the others, means. 4. In every geometric series, five things are considered : (1) the first term ; (2) the last term ; (3) the number of terms ; (4) the ratio ; (5) the sum of the terms. CASE I. 269, To find the last term, when the first term, the ratio, and the number of terms are giv^en. GEOMETRICAL PROGRESSION. 335 1. The first term of an increasing geometric series, is 2 ; the ratio, 3 : what is the fifth term ? Solution. — The first term is 2; the second, 2X3; the third, 2X^X3; the fourth, 2X8X3X3; and the fifth, 2X3X3X3X 3. Each term after the first, consists of the first term multiplied by the ratio as many times less one, as is denoted by the number of the term; then, the fifth term consists of 2 multiplied by 3 taken four times as a factor; but 3, taken 4 times as a factor, is the 4tt\\ power of 3. Hence, the fifth term is 2 X 3^ =. 162. 2. The first term of a decreasing geometric series is 192; the ratio, 2: what is the fourth term? Solution. — The first term is 192; the second term is 192-7-2; the third is 192 divided by 2X2; the fourth is 192 divided by 2 X 2X2; that is, 192 -f- 2=* = 24. Rule. — 1. Raise the ratio to a power whose exponent is one less than the number of terms. 2. If the series be increasing, multiply the first term by this power; if decreasing, divide the first t^m^by the power. 3. The first term of an increasing series is 2 ; the ratio, 2; the number of terms, 13: find the last term. 8192. 4. The first term of a decreasing series is 262144 ; the ratio, 4 ; number of terms, 9 : find the last term. 4. 5. The first term of an increasing series is 10; the ratio, 3: what is the tenth term? 196830. CASE II. 270. To find the sum of all the terms of a geometric series. 1. Find the sum of 5 terms of the geometric series, whose first term is 4, and ratio 3. 336 PvAY'S NEW PRACTICAL ARITHMETIC. Solution. — Write the terms of the series as below; then multiply each term by the ratio, and remove the product one term toward the right, thus : 4 + 12-f36-flO 8 + 324 =:=^ sum of the series. 12 + 36 + 108 + 324 + 97 2 =r^ sum X 3. Since the upper line is once the sum of the series, and the lower three times the sum, their difference is twice the sum; hence, if the upper line be subtracted from the lower, and the remainder divided by 2, the quotient will be the sum of the series. Performing this operation, we have 972 — 4=^968 divided by 2; the quotient is 484, the sum of the series. In this process, 972 is the product of the greatest term of the given series by the ratio, 4 is the least term, and the divisor 2 is equal to the ratio less one. Rule. — Multiply the greatest term by the ratio; from the product subtract the least term; divide the remainder by the ratio less 1. Rem. — When a series is decreasing, and the number of terms in- finite, the last term is 0. 2. The first term is 10 ; the ratio, 3 ; the number of terms, 7: what is the sum of the series? 10930. 3. A father gave his daughter on T^ew Year's day $1; he doubled it the first day of every month for a year: what sum did she receive? 84095. 4. I sold 1 lb. of gold at 1 ct. for the first oz. ; 4 ct. for the second, 16 ct. for the third, etc. : what sum did I get? $55924.05. 5. Find the sum of an infinite series, of which the greatest term is .3 and the ratio, 10; that is, of yu ~r toj^ + T0V0' ^tc. ^ ^. 6. Find the sum of the infinite series i, i, 2V? ^^c. J. 7. Find the sum of the infinite series i, ^, J, etc. 1. i^rJ J^ / 7J^ ^ . ^ 14 DA1 RETURN TO DESK FROM WHICH BORROWED EDUCATION - PSYCHOLOGY LIBRARY This book is due on the last date stamped below, or on the date to which renewed. Renewed books are subject to immediate recall. 7 DAY USE DURING JAN 1 6 ^^^^ REC^D LP M 7-66 -8 / .M JAN10RECn--^PW SUMMER SESSIONS VB 35633 /^ Z0^', 961659 U'^ ' ..-^ /S7 7 THE UNIVERSITY OF CALIFORNIA UBRARY 4 Joo^Uy"^.... I jq!^j0Sbm- y f / .< i ' % (^•i. ,' J^^^-u*^- '.^«^.,^ mmuv fm. McGuffey's Speller, McGuffey's Readers, Harvey's Speller, Harvey's Readers, Ray's Arithmetics, Ray's Algebras, Harvey's Grpmmars, Eclectic Geographies, Eclectic Penmanship, Norton's Physics, Brown's Physiology, Venable's U.S. History, Thalheimer's Histories. ^; X I Van Antwerp B'v