THE LIBRARY OF THE UNIVERSITY OF CALIFORNIA BEQUEST OF Alice R. Hilgard ECLECTIC EDUCATIONAL SERTES. SAY'S ARITHMETIC, THIRD BOOK. PRACTICAL ARITHMETIC. BY INDUCTION AND ANALYSIS. BY JOSEPH [RAY, M.D. LATE PROFESSOR OF MATHEMATICS IN WOODWARD COLLEGE. ONE THOUSANDTH EDITION IMPBOVED. CINCINNATI j WILSON, H INKLE & CO. PHIL'A: CLAXTON, REM8EN & HAFFELFINGEB. NEW YORK: CLARK & MAYNARD. THE BEST AND CHEAPEST MATHEMATICAL WORKS. BY PROFESSOR JOSEPH RAY. TYPE ENLARGED NEW STEREOTYPE PLATES. Each BOOK of Ray s Arithmetical Course, also of the Algebraic* is a complete work in itself, and is sold separately. FTttST BOOK. PRIMARY LESSONS: simple and progressive Mental Lessons and Tables for little learners. SECOND BOOK. OTELLECTUAL ARITHMETIC, by Induction and Analysis; the most complete and interesting intellectual arithmetic ex- tant. " THIRD BOOK. PRACTICAL ARITHMETIC, by Induction and Analysis ; a sim pie and thorough work for schools and private students. KEY TO RAY'S ARITHMETIC, THIRD BOOK. HIGHER ARITHMETIC. Principles of Arithmetic analyzed an practically applied. For advanced students. ELEMENTABY ALGEBRA. RAY'S ALGEBRA, FIRST BOOK, for Common Schools and Acad- emies ; a simple and thorough elementary treatise. HIGHER ALGEBEA. RAY'S ALGEBRA. SECOND BOOK, for advanced students in Acad- emies, and for Colleges ; a lucid and comprehensive work. KEY TO RAY'S ALGEBRA, FIRST AND SECOND BOOKS; complete in one volume, 12mo. Entered according to Act of Congress, in the year Eighteen Hundred and Fifty- Seven, by WIXTHROP B. SMITH, in the Clerk's Office of the District Court of tho United States, for the Southern District of Ohio. ELECTEOTYPED IT FRANKLIN TYPE FOUNDEBY, CINCINNATI, 0. PREFACE. * FEW, it any works on Arithmetic, have received the approbation tvhich has been bestowed upon "Ray's Arithmetic, Part Third;" and the constantly increasing demand for it having rendered another renewal of the stereotype plates necessary, it has been made the occasion for remodeling and greatly improving the volume in all its parts. The Inductive and Analytic methods here adopted, lead the learner to an understanding of the principles from ffhich the rules are derived, and teach him to regard rules as results rather than reasons : he will understand the " why and wherefore " of every operation performed, and gain a thorough knowledge of the principles of Arithmetic with its applications, while the faculties of the mind are strengthened antt disciplined. The leading features of the work are : 1st. Every principle is clearly explained by an analysis or solution 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. For this reason, those processes of reduction that require the use of fractions, are introduced after fractions. 3d. The subject of Fractions, a thorough understanding of which is almost a knowledge of Arithmetic, has received that attention which its use and importance demand. 4th. The subject of Proportion is introduced immediately after Decimals: this enables the instructor to treat Percentage and its various applications, either by proportion, or by analysis, as he may prefer. plOOrTlC'ei MoobUoo () 4 PREFACE. 6th. Particular attention has been given to render the practical; (he weights and -measures are referred to, and con* form to the legal standards; while pounds, shillings, and pence t *>eing no longer used in actual business, are only introduced under Exchange. < While Federal' Money may be considered in connection witi decimals, yet it is truly a species of compound numbers, and is so regarded in all the ordinary computations of business. Hence, the propriety of assigning it the place which it occupies in this work. While cancellation is introduced, it is not made a hobby, or an arithmetical machine, by which results can be obtained merely in a meclunical manner. The objec: throughout has been to combine practical utility witfr. scientific accuracy; to present a work embracing the best methods^ with all real improvements. How far this object has been secured, is submitted to those engaged in the laborious and responsible work of edu3ation. J8&" An APPENDIX has been added to this edition, presenting the Metrical System of Weights and Measures. This has been com- piled chiefly from modern French works on this subject ; and was submitted, in manuscript, to the critical examination of PROF. H. A. NETYTOX, of Tale College, to whom we are under great obli- gations for such corrections, additions, and emendations, as his intimate knowledge of the subject suggested. A NEW BOOK. RAY'S TEST EXAMPLES. Three Thousand Test Examples in Arithmetic; Prac- tical Problems for the Slate or Blackboard, for Drill exercises and Review. Designed to save the teacher much time and labor by furnishing, rco^y to his hand, a large number and variety of Drill Exer-- cises for frequent and thorough review. Two Editions : With Answers, Without Answers, CONTENTS. SIMPLE NUMBERS. Notation and Numeration, 9 Addition, 19 Subtraction, 26 Multiplication, 33 Division, 43 Kevicw of Principles, 61 Promiscuous Examples, 63 General Principles of Division, 65 Cancellation, 67 COMPOUND NUMBERS. Definitions, 70 United Siates or Federal Money, 71 Notation and Numeration of li. S. Money, 72 Reduction of U.S. Money, 73 Addition of U. S. Money, 75 Subtraction of U. S. Money, 76 Multiplication of U. S. Money, 77 Division of U. S. Money,. '. 78 Promiscuous Examples in U. S. Money, 81 Eeduction of Compound Numbers, 83 Dry Measure, 83 Troy or Mint Weight, 87 Apothecaries Weight, 83 Avoirdupois Weight, '. 89 Long or Linear Measure, 90 Land or Square Measure, 90 To find the Area of a Eectangle, .... 92 Solid or Cubic Measure, -. 94 Cloth Measure, . 95 Wine or Liquid Measure,... s... i)6 Ale or Beer Measure, 97 Time Measure, 97 Circular Measure, 99 Miscellaneous Table,. 99 Promiscuous Examples in Reduction, 100 Addition of Compound Numbers, , 103 (Subtraction of Compound Numbers, 107 To find the period of Timo between any two Dates, 110 Multiplication of Compound Numbers, ,.... Ill Division of Compound Numbers, 114 fiongitude and Tiine 4 ... ,...,., .......,..,.... v ..... 117 6 CON'lENTS. FACTORING. PAGE, Definitions, 120 Six Propositions, 122 Examples for Practice, 125 GREATEST COMMON DIVISOR. First Method, by Factoring, 127 Second Method, by Division, 128 LEAST COMMON MULTIPLE. First Method, by Factoring, 131 Second Method, by Division, 132 COMMON FRACTIONS. Origin and Nature of Fractions, 134, Definitions, 138 General Principles, 139 Reduction of Fractions, , .; 142 To reduce a fraction to its lowest terms, 142 To reduce an improper fraction to a whole cr mixed number, 143 To reduce a whole or mixed number to an im;>n per frnctic.n, 144 To reduce compound to simple fractions, 146 To reduce compound to simple fractions by Cuncellatit n, 147 To reduce fractions to a common denominator. 148 To reduce fractions to the least common dcnomlriuior, 150 Addition of Fractions, 152 Subtraction of Fractions, 154 Multiplication of Fractions, 156 Division of Fractions, 162 Fractional Exercises in Compound Niunb er.-- 1G9 Reduction of Fractional Compound Numb ;;.- v 171 Addition and Subtraction of Fractional Compound Number.-. 17-t Promiscuous Examples, 175 DECIMAL FRACTIONS. Origin and nature of Decimals, 177 Decimal Numeration and Notation, 180 Addition of Decimals, ; 183 Subtraction of Decimals, 181 Multiplication of Decimals, 18") Division of Decimals, 187 Reduction of Decimals, 183 Promiscuous Examples, 10 > RATIO. The nature of Ratio, 195 Method of expressing Ratio, 190 PROPORTION. The nature of Proportion,... 197 Simple Proportion,...,..,,... ... , . ,,.. 199 CONTENTS. 7 PROPORTION CONTINUED. PAGE. Eule of Cause and Effect, '. . 205 Compound Proportion, 206 ALIQUOTS, OR PRACTICE 209 PERCENTAGE AND ITS APPLICATIONS. To find any per cent, of a Number, ' 212 To find what per cent, one Number is of another, '.... 214 Commission, , 215 Insurance,... 217 Stocks, 218 Brokerage, 219 INTEREST. Simple Interest, 220 General Eule for Interest, 225 Another method for Interest, 227 Partial Payments, 229 Problems in Interest, 233 Compound Interest, 230 Discount, 239 Profit and Loss, 245 Assessment of Taxes, 250 American Duties, i 252 PARTNERSHIP, 253 EQUATION OP PAYMENTS,; . 25* ALLIGATION MEDIAL, 260 ANALYSIS, 201 EXCHANGE OP CURRENCIES, 269 DUODECIMALS, ... 273 INVOLUTION, 275 EVOLUTION, , 277 Extraction of the Square Eoot, 279 Applications of the Square Eoot, 284 Extraction of the Cube Eoot, 286 ARITHMETICAL PROGRESSION, 293 GEOMETRICAL PROGRESSION, 296 PERMUTATION, ,.., 299 MENSURATION. Definitions, 300 Measurement of Surfaces, 802 Plasterers', Painters', and Carpenters' Work, ,*. 303 Measurement of Bodies or Solids, 806 Masons' and Bricklayers' Work, ,... 308 Gauging, 811 PROMISCUOUS QUESTIONS, , 814 OBSERVATIONS TO TEACHERS. INTELLECTUAL ARITHMETIC should be thoroughly studied by all, and especially by the young, before commencing PRACTICAL. For this purpose, attention is called to "Ray's Arithmetic, Second Book," which has been carefully prepared, and is now published With important improvements. When admissible, pupils studying Arithmetic should be taught in classes; the presence of the class being a stimulus to both teacher and pupil. This arrangement also economizes time, since the same oral illustrations, necessary for the instruction of a single pupil, serve for a class. The time occupied at each recitation ought not to be less than thirty minutes, nor more than one hour. The class should not be too large; and, if possible, the attainments of its members equal. Every school should have a blackboard, on which pupils can solve the questions, and explain the method of solution. A prime object in recitations is to secure attention; to do this, the exercises must be interesting, and all 'must be kept employed. Let as many be called out as can obtain positions at the black- board, and let all solve the same question at once. When the solutions are completed, let some one be called on to explain the process, giving the reason for each step of the opera- tion. Exercises thus conducted animate the class; and by re- quiring the learner to explain every process, and assign a reason for every step, he learns to rely on his own reasoning powers. In assisting pupils to overcome difficulties, it is preferable to do it indirectly, by making such suggestions, or asking such, questions, as will enable the learner to accomplish the object. Frequent Reviews will be found of great benefit. The pupil should be rendered familiar with the answers to the questions in the REVIEW at the foot of the page. This review is intended to aid the teacher, but not to prevent his asking other questions, or presenting different illustrations. ARITHMETIC, 1. A Unit, or one, is a single thing of any kind; a^ one apple, one dollar, one pound. 2. Number is a term signifying one or more units ; as, one ; five, seven cents, nine men. 3. Numbers are expressed by ten characters, called Figures; as, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0. 4. Arithmetic treats of numbers, and is the art of computing by them. The fundamental rules are five ; Notation and Numeration, Addition, Subtraction, Multi- plication, and Division. I. NOTATION AND NUMERATION. ART. 1. A single thing is a Unit, or one; written, 1 One unit and one more, are two; .. 2 Two units and one more, are three; .. 3 Three units and one more, are four; .. 4 Four units and one more, are five ; .. 5 Five units and one more, are six; .. 6 Six units and one more, are seven; .. 7 Seven units and one more, are eight; .. 8 Eight units and one more, are nine; .. 9 ART. 2. These nine characters are called digits, or significant figures, because they denote something. The character, 0, called cipher, naught, or zero, is em- ployed to denote nothing: thus, to show that there are no cents in a purse, write, the number of cents is 0. REMARK. The cipher is sometimes termed an auxiliary digit, because it helps the other digits in expressing numbers. REVIEW. ART. 1. What is a single thing? What are one unit and one inoro of the same kind f Two units and one more, &c. ? 2. What are these nine characters called ? Why ? What does naughty or zero denote ? KE#. What is the cipher termed ? Wh^ f 10 RAY'S PRACTICAL ARITHMETIC. ORDER OF TENS, ART. 3. Nine units and one more are called Ten: it forms a unit of a second or higher order called Tens. Ten is represented by the same figure, 1, as a single thing, or unit of the first order; but, To distinguish ten, the 1 is written in the second order, the second place from the right hand : The first order, or right hand place, is filled with cipher. The cipher, in the first place, denotes that there are no units of the first order. The number ten is written thus ; ... 1 One ten and one unit, are . eleven ; 11 One ten and two units, . . twelve; 12 One ten and three units, . . thirteen; 13 One ten and four units, . . fourteen; 14 One ten and five units, . . fifteen; 15 One ten and six units, . . sixteen; 16 One ten and seven units, . . seventeen; 17 One ten and eight units, . . eighteen; 18 One ten and nine units, . . nineteen; 19 Two tens, twenty ; 20 Two tens and one unit, . . . twenty-one; 21 Two tens and five units, . . twenty-five; 25 Three tens, thirty; 30 Four tens, forty; 40 Five tens, fifty; 50 Six tens, sixty; 60 Seven tens, .... . seventy ; 70 Eight tens, eighty ; 80 Nine tens, ninety; 90 REVIEW. 3. What are nine units and one more called ? Ten forms a unit of what order ? How is it written ? When two figures are written together, what is the place on the right called ? Ans. The first place, or first order, or unit's place. What the place on the left? Ans. The second place, or second order, or tens' place. 3. Li writing ten, why is a naught put in unit's place ? What are one ten and one unit ? How written ? What are two tens ? How written ? NOTATION AND NUMERATION. 11 ART. 4. The numbers between 20 and 30, 30 and 40, and twenty or one thousand, no hun- dreds, two tens, and no units. 41. Twenty-five thousand and six or two tens of thou- sands, five thousands, no hundreds, no tens, and pi^ units. 3d Bk, 2 42. Three hundred and forty- five or thre^ hundreds, four tens, and five units. 43. Seven hundred and sixty or seven hundreds, six tens, and no units. 44. Three thousand four hun- dred and six or three thousands, four hundreds, no tens, and six units. 45. Forty-two thousand and thirty or four tens of thousands, two thousands, no hundreds, three tens, and no units. 46. Thirty thousand or three tens of thousands, no hun- dreds, no tens, no units. 47. One hundred and sixty- three thousand or one hundred thousand, six tens of thousands, three thou- sands, no hundreds, no tens, and no units. 48. Three, hundred and forty- one thousand, five hundred and sixty-three or three hundred thousands, four tens of thousands, one thousand, five hundreds, six tens, and three units. 49. Four tens of millions, five millions, no hundreds of thousands, eight tens of thousands, three thousands, no hundreds, two tens, and six units. 50. Eight hundreds of million^ seven tens of millions, no millions, seven hundreds of thousands, four tens of thousands,three thousands, five hundreds, seven tens, and nine units. 18 RAY'S PRACTICAL ARITHMETIC. 51. Two thousand eight hun- dred and four. 52. Four thousand and twenty- nine. 53. Six thousand and six. 54. Twenty-two thousand seven hundred and sixty-five. 55. Eighty thousand, two hun- dred and one. 56. Ninety thousand and one. 57. Thirty thousand and thirty. 58. Four hundred and ten thousand, two hundred and five. 59. Eight hundred thousand, six hundred and sixty-nine. '^60. Nine hundred thousand and one. 61. Five hundred thousand and fifty. 62. One hundred thousand and ten. 63. Nine hundred and nine million and ninety thou- sand. 64. One hundred million, ten thousand and one. 65. Ninety-one million, seven thousand and sixty. 66. Seventy million and four. yen hundred millions, ten thousand and one. 68. One billion, ono million and forty. 69. Forty billions, two hundred thousand and five. 70. Seven hundred and twenty- six billions, fifty millions, one thousand, two hundred and forty-three. , EOMAN NOTATION. ART. 16. The common method of representing num- bers, by figures, is termed the Arabic. Another method, by means of letters, is termed the Roman. The letter I stands for one; Y^five; X, ten; L, fifty \ C, one, hundred ; D, five hundred ; and M, one thousand. Numbers are represented on the following principles : 1. Every time a letter is repeated, its value is repeated : thus, II denotes two, XX denotes twenty. 2. When a letter of less value is placed before one of greater value, the less is take a from the greater ; if placed after it, the value of the greater is increased : thus, IV denotes four, while VI denotes six. REVIEW. 15. "What is Notation? What is tho'Eulc? 16. What is the common method of Notation termed ? What other method ? What number is represented by the letter I ? By V ? .By X ? By L ? By C ? By D ? By M ? What is the effect cf repeating a 1 16. What the effect of placing a letter of less value before one of greater value ? After one of greater value ? "What of placing a lino over a letter ? ADDITION OF SIMPLE NUMBERS. 19 3. A lino over a letter increases its value a thousand times. Thus, V denotes 5000, JVf denotes one million. TABLE OF BOMAN NOTATION. I II . III . IV . V . VI . IX . X . XI . XIV . XV . XVI . XVII XVIII XIX. XX . ART. 17. The preceding illustrations show the three methods of expressing numbers : IST. By words, or ordinary language. 2D. T$j fyurcs, termed the Arabic method. 3D. By letters, termed the Roman method. One. XXI . . Twenty-one. Two. XXX . . Thirty. Three. XL. . . Forty. Four. L . . . Fifty. Five. LX. . . Sixty. Six. XC. . . Ninety. Nine. C . . . One hundred. Ten. COCO. . Four hundred. Eleven. D . . Five hundred. Fourteen. DC. . . Six hundred. Fifteen. DCC . . Seven hundred. Sixteen DCCC . Eight hundred. Seventeen. DCCCC . Nine hundred. Eighteen. M . One thousand. Nineteen. MM . . Two thousand. Twenty. MDCCCLVI . 1856. II. ADDITION. 1. If you have 2 cents and find 3 cents, how many will you then have ? Ans. 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. 3. John gave 6 cents for an orange, 7 cents for pencils, and 9 cents for a ball : how many cents did they all cost ? Ans. 22 cents. ART. 18. The process of uniting two or more numbers info one number, is termed Addition. The number obtained by addition, is the Sum or Amount. 20 RAY'S PRACTICAL ARITHMETIC. REMARK. When the numbers to be added are of the same de- nomination, that is, all cents, or all yards, c., the operation ia called Simple Addition. ART. 19. OF THE SIGNS. The Sign +, called plus, means more. "When between two numbers, it shows that they are to be added ; thus, 4 + 2 means that 4 and 2 are to be added together. The sign of equality, =, denotes that the quantities between which it stands equal each other. The expression 4+2=6, means that the sum of 4 and 2 is 6 ; read, 4 cciid 2 are 6, or 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 REVIEW. 17. What are tho three methods of expressing numbers? IS. What is Addition? What the sum or amount? EEM. What is Simple Addition? 19. What does the sign plus mean? What does it show ? What does the sign of equality denote ? Give an example. ADDITION OF SIMPLE NUMBERS. 21 ART. 20. 1. James had 63 cents, and his father gave him 35 cents : how many cents had he then ? Place the units and the tens of one number under the units and the tens of the other, that figures of the same unit value may be more easily added. SOLUTION. Write the numbers as in the 3 margin ; then say 5 units and 3 units are 8 *J * units, which write in units' place; 3 tens and ;? ~ c 1 s * 6 tens are 9 tens, which write in tens' place. _ cents> The sum is 9 tens and 8 units, or 98 cents. An*. 9 8 cents. In this example, units are added to units, and tens to tens, since only numbers of the same kind, that is, having the same unit value, can be added. Thus, 3 units and 2 tens make neither 5 units nor 5 tens ; as, 3 apples and 2 plums are neither 5 apples nor 5 plums. 2. I own 3 tracts of land : the first contains 240 acres ; the second, 132 acres; the third, 25 acres: how many acres in all ? Since units of different orders can not be added to- gether, place units under units, tens under tens, &c., that figures to be added may be in the most convenient position. SOLUTION. Begin at the right, and say 5 units 240 acres, and 2 units are 7 units, which write in units' 132 acres, place; 2 tens and 3 tens and 4 tens are 9 tens, 25 acres, which write in tens' place : Lastly, 1 hundred and 2 hundreds are 3 him- 3 9 < a dreds, which write in hundreds' place, and the work is complete. Questions to be solved and explained as above. The sign $ stands for the word dollars, and when used is placed before the figures. 3. There are 43 sheep in one pasture, 21 in another, . and 14 in another : how many sheep in all ? Ans. 78. REVIEW. 20. Why are units placed under units, and tens under tens? Onn mini VMS of different unit values bo added ? Why not? Givo au ex- iuiii)le. What sign is used for the word dollars? 22 RAY'S PRACTICAL ARITHMETIC. 4. I owe one man $210, another S142, and another 35 : what is the amount of the debts ? Ans. 387. 5. Find the sum of 4321, 1254, 3120. Ans. 8605. 6. The sum of 50230, 3105, 423. Ans. 53758. ART. 21. Where the sum of the figures in a column docs not exceed 9, it is written under the column added. When the sum of the figures exceeds 9, two or more figures are required to express it. To explain the method, TAKE THIS EXAMPLE. 1. James bought a reader for 74 cents, an atlas for 37 cents, a slate for 25 cents : how much did all cost ? IST SOLUTION. By adding the figures in * the first column, the sum is 16, which is 1 Reader * * cents. ten and G units. Write the G units in tho Atlas * cents ' order of units, and the 1 ten in the order of Slate _ tens. I Q The sum of the figures in the tens' column 1 2 is 12 tens, which is 1 hundred and 2 tens. , Write the 2 tens in the order of tens, and Ans ' * ^6 cents ' the 1 hundred ir the order of hundreds. Lastly, unite the figures in the column of tens. The sum is 1 hundred, 3 tens, and G units, or 136 cents. 2o SOLUTION. The preceding example is usually per- 7 4 formed thus : 5 units 7 units and 4 units are 16 units, 3 7 which is 1 ten and 6 units. Write the 6 units in units' 2 5 place, and carry the 1 ten to tens' place. Then, 1 ten 2 tens 3 tens and 7 tens are 13 tens, which is 1 hundred and 3 tens; write the 3 tens in tens' place, the 1 hundred in hundreds' place, and the work is completed. The 1 ten derived from the sum of the figures in tho 1st column, and added to the 2d, is said to be carried. EEYIETV. 21. When the sum of a column docs not exceed 9, -where is it written? If greater than 9 ? What is understood by carrying the tens? In what does it consist? Why docs the addition begin with tho units' column ? 22. What is the rule for addition ? The proof? ADDITION OF SIMPLE NUMBERS. 23 When the sum of the figures in a column exceeds 9, reserving the tens, or left hand figure, and adding it to the figures in the next column, is called carrying the tens. 2. Add the numbers, 3415, 503, 1870, and 922. SOLUTION. Write units of the same order under 3415 each other. Then say, 2 and 3 arc 5, and 6 are 10 503 units } which is no (0) units, written in units' place, 1870 and 1 ten, carried to the tens ; 1 and 2 are 3, and 7 922 are 10, and 1 arc 11 tens, which is 1 ten, written in tens' place, and 1 hundred, carried to tho hundreds; ^7 *" 1 and 9 are 10, and 8 are 18, and 5 are 23, and 4 are 27 hundreds, which is 7 hundreds, written in hundreds' place, and 2 thousands, carried to the thousands ; 2 and 1 are 3, and 3 are G thousands, written in thousands' place.) Carrying the tens is simply adding tens to fens, hundreds to hundreds, &c., on the principle (Art. 20), that only numbers of the same unit^alue can be added. The addition begins at the right hand column, with tho unit of the lowest order, so that, If the sum of the units in any column exceeds 9, the tens can be carried to the sum of the next higher order. AKT. 22. BULE For Addition. 1. Write the numbers to be added, so that figures of the same order may stand under each other, units under units, tens under tens, &c. 2. Begin at the right hand, and add each column sepa- rately. Place the units obtained Inj adding each column under it, and carry the tens to the next higher order. At the last column write the whole amount. PROOF. Separate tho numbers into two or more divisions; find the sum of each; then add the amounts together. If their sum equal that found by the rule, tho work is correct REMARKS. 1. Another method of proof consists in adding the columns downward, taking the figures in a different order from that in which they were taken before. RAY'S PRACTICAL ARITHMETIC. 2. For the method of proof by casting out the 9's, (too difficult for a work of this grade,) see " Ray's Higher Arithmetic" 3. Add the numbers 853, 516, 29, 6, 34, 580. SOLUTION. Writing the numbers and add- ing them, their sum is 2018. PROOF. If the numbers be separated by a line between 29 and 6, the sum of the num- bers in the first division will be 1398, and in the second, 620. Adding these numbers together, tkeir sum is 2018, as before. 853 516 -1398 34 620 ; 580 5018 2018 4. Find the sum of 3745, 2831, 5983, 7665. In adding long columns of figures, is necessary to retain the numbers carrii This may be done by placing them small figures under their proper colum: as s, 2, 1, in the margin. it 3745 3d. 2831 in 5983 as, 7665 20224 321 EXAMPLES FOR PRACTICE. (5) (6) (?) (3) (9) (10) 184 204 103 495 384 1065 216 302 405 207 438 6317 135 401 764 185 348 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 (11) (12) (13) (14) (15) 3725 5943 82703 987462 6840325 5834 6427 102 478345 7314268 4261 8204 6005 610628 3751954 7203 7336 759 423158 6287539 ADDITION OF SIMPLE NUMBERS. 25 16. 23+41+74+83+16= how many? Ans. 237. 17. 45+19+32+74+55= liow many? Ans. 225. 18. 51+48+76+85+4 = how many? Ans. 264. 19. 263+104+321+155 = how many? Ans. 843. 20. 94753+2847+93688+9386+258+3456 are how many ? Ans. 204388. 21. January has 31 days, February 28, March 31, April 30, and May 31 : how many days are there in these five months? Ans. 151. 22. June has 30 days, July 31, August 31, September 30, October 31 : how many days in all? Ans. 153. 23. 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 ? Ans. 365. 24. 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 ? Ans. 246 yds. 25. I owe one man $245, another $325, a third $187, a fourth $96 : how much do I owe ? Ans. $853. 26. General Washington was born A. D. 1732, and lived 67 years : in what year did he die? Ans. 1799. 27. From the creation of the world to the flood was 1656 years ; thence to the siege of Troy, 1164 years ; thence to the building of Solomon's Temple, 180 years ; thence to the birth of Christ, 1004 years : in what year of the world did the Christian era begin ? Ans. 4004. 28. A has 4 flocks of sheep ; in the 1st are 65 sheep and 43 lambs; in the 2d, 187 sheep and 105 lambs; in the 3d, 370 sheep and 243 lam^s; in the 4th, 416 sheep and 95 lambs : how many sheep and lambs has he? Ans. 1038 sheep, and 486 lambs. 29. 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 dol- lars did he pay? Ans. 105 bar!., and $1072. 26 RAY'S PRACTICAL ARITHMETIC. 30. The first of four numbers is 287 ; the second, 596 ; the third, 841 ; and the fourth, as much as the first three : what is their sum? Ans. 3448. 31. The Pyramids of Egypt were built 337 years before the founding of Carthage ; Carthcige was founded 49 years before the destruction of Troy ; Troy was destroyed 431 years before Rome was founded; Carthage was de- stroyed 607 years after the founding of Rome, and 146 before the Christian era. How many years before Christ were the Pyramids built? Ans. 1570. 32. Add three thousand and five ; forty-two thousand, six hundred and twenty-seven ; 105 ; three hundred and seven thousand and four ; 800,791 ; three hundred and twenty thousand, six hundred. Ans. 1474132. 33. Add 275,432 ; four hundred and two thousand and thirty ; three hundred thousand and five ; 872,026 ; four million, two thousand, three hundred and forty- seven. Ans. 5851840. 34. Add eight hundred and eighty millions, eight hundred and eighty-nine ; 2,002,002 ; seventy-seven million, four hundred and thirty-six thousand ; two hun- dred and six million, five thousand, two hundred and seven ; 49,003 ; nine hundred and ninety million, nine- teen thousand, nine hundred and nineteen. Ans. 2155513020. III. SUBTRACTION. ART. 23. 1. If you have 9 apples, and give 4 away, how many are left? Ans. 5. Why? Because 4 and 5 are 9. 2. Frank had 15 cent;; after spending 7, how many \vereleft? Ans. 8. Why? 3. If you take 8 from 13, how many are left? Ans. 5. The operation in the preceding examples is termed Subtraction. Hence, Subtraction is the process of finding the difference between two numbers. SUBTRACTION OF SIMPLE NUMBERS. 27 The larger number is called the Minuend; the less, the Subtrahend; and the number left after subtraction, the Difference or Remainder. REMARKS. 1. The word Minuend means, to be diminished; Subtrahend, to be subtracted. 2. In Addition (See Art. 20), numbers of the same kind are added; in Subtraction they are taken from each other; therefore, Subtraction is the reverse of Addition. 3. When the given numbers are of the same denomination, the operation is called Simple Subtraction. ART. 24. The Sign , is called minus, meaning less. Placed between two numbers, it denotes that the one on the right is to be taken from that on the left. Thus, 8 5 = 3, shows that 5 is to be taken from 8 ; it is read, 8 minus 5 equals 3. Here, 8 is the minuend, 5 the subtrahend, and 3 the remainder. 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 64 = 2 7 5 = 2 5 2 = 3 6 3 = 3 7 4 = 3 S_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 = 7 114 = 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 = 1 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 [0 6 = 4 11 7 = 4 12 8 = 4 13 9 = 4 U_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 K3_9 = 7 14 6 = 8 15 7 = 8 16 8=^8 n-.g^s 15 6 = 9 16 7 = 9 17 8 = 9 18 9 = 9 28 BAY'S PRACTICAL ARITHMETIC. ART. 25. When numbers are small, the difference be- tween them may be ascertained in the mind ; when large, the operation is most easily performed by writing them. EXAMPLES. 1. A man having $135, spent $112 : what sum had he left? Since only things of the same unit value can be added (Art 20), the difference between things of the same unit value only can be found ; hence, Place units under units, tens under tens, &c., that the figures between which the subtraction is to be made, may be in the most convenient position. SOLUTION. After arranging the numbers, s Bay 2 (units) from 5 (units) leave 3 (units), J: % * which put in units' place ; then 1 (ten) from 1 3 5 minuend. 3 (tens) leaves 2 (tens), which put in tens' 2 subtrahend, place ; then 1 (hundred) from 1 (hundred) 2 3 remainder, leaves 0, and there being no figures on the left of this, the place is vacant ; hence the number of dollars left is 23; that is, 135112=23. 2. A farmer having 245 sheep, sold 123- now many had he left? Am. 122. 3. A man bought a farm for $751, ancT sold it for $875 : how much did he gain ? Ans. $124. What is the difference 4. Between 734 and 531 ? ... An*. 203. 5. Between 8752 and 3421? . . . Ans. 5331. 6. Between 529 and 8? ... Ans. 521. 7. Between 79484 and 25163? . . Ans. 54321. REVIEW. 23. What is Subtraction? Give an example. What is the minuend ? The subtrahend ? Kemainder ? EEII. What docs minuend mean ? What subtrahend ? 23. Why is subtraction the reverse of addition? 24. What does the sign minus mean ? When placed between two numbers what does it denote ? SUBTRACTION OF SIMPLE NUMBERS. 29 ART. 26. When the lower figure in each order is not greater than the upper, the less is subtracted from the greater, and the difference written beneath ; but, When the lower figure in any order is greater than tho upper, a difficulty arises, which we will now explain. James had 13 cents ; after spending 5, how many had he left? 13 5 can not be subtracted from 3, but can be from 5 13 ; 5 from 13 leaves 8; hence he had 8 left. 1. From 73 subtract 45. SOLUTION. Here, 5 units can not be taken from T - * 8 units. Take 1 (ten) from the 7 (tens), and add 7 3 this 1 (ten) or 10 units to the 3 units, which will 45 make 13 units in the units' place; then, subtract _.. 9 g the 5 units, and there will remain 8 units, to bo put in units' place. Since 1 ten is taken from tho tens, units. 7 tens, there remain 6 tens in th.' tens' place. Sub- Q 13 tract 4 tens from 6 tens and put the remainder in 4 5 tens' place. The difference of the two numbers is - thus found to be 2 tens and 8 units, or 28. Instead of actually taking 1 ten from the 7 tens, and adding it to the 3 units, as is done in the margin, the operation is performed in the mind: thus, 5 from 13 leaves 8, and 4 from 6 leaves 2. In such cases, the value of the upper number i not changed, since the 1 ten which is taken from the order of tens is added to the number in the order of units. EXPLANATIONS. 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. After increasing the units by 10, instead of considering the next figure of the upper number as diminished by 1, 25. When the numbers are small, how is their difference easily found ? When they are large ? In writing number! for subtrao- tkm, why place units under units, tens under tens? 30 RAY'S PRACTICAL ARITHMETIC. the result trill be the same, if the next figure of the lowef 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 ; 5 from 7 leaves 2, the same as 4 from 6. Hence, when a figure in the lower number is greater than that above it, add 10 to the upper figure, then sub- tract the lower figure from the sum ; and, To compensate for the 10 added to the upper figure, increase the next lower figure by 1. This process depends on the principle, that if any two numbers be equally increased, their difference will remain the same. The 10, added to the upper number, is equal to 1 of the next higher order added to the lower number ; ten units of any order being always equal to 1 of the ordei next higher. 2. Find the difference between 805 and 637. SOLUTION, (!ST METHOD.) Writing the less 805 number under the greater, with units of the same 637 order under each other, it is required to subtract the 7 units from 5 units, which is impossible. ^"* The five can not be increased by borrowing from the next figure, because it is 0; therefore, borrow 1 hun- dred 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. Subtracting 7 units from 15 units, 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 EEYIEW. 26. When the lower figure- is less than the upper, how is the subtraction performed? What do you understand by borrowing ten? Illustrate the process, by subtracting 45 from 73. After borrowing ten, what may be done in order to avoid diminishing the next upper figure by 1 ? On what principle does this process depend ? SUBTRACTION OF SIMPLE NUHBEES. 31 hundreds from 7 hundreds, there remains 1 hundred, to be written in hundreds' place. The difference is 168. Or, (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, say 7 from 8 leaves 1, and the whole remainder is 168, as before. QUESTIONS TO BE SOLVED BY BOTH METHODS. 3. From 73 take 48 Ans. 25. 4. From 340 take 150 Ans. 190. 5. From 508 take 325. .... Ans. 183. 6. From 4603 take 3612. .... Ans. 991. 7. From 8765 take 7766 Ans. 999. REMARKS. 1, The SECOND method is generally used ; it is more convenient, and less liable to error, especially when the upper number contains ciphers. 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. . When each figure in the lower number is less than the one above it, the subtraction may commence at the left hand. ART. 27. If 5 subtracted from 8 leaves 3, then 3 added to 5 must produce 8 ; that is, If the difference of two numbers be added to the less, the sum will be equal to the greater. EULE For Subtraction. 1. Write the less number under the greater, placing units under units, tens under Uns, &c. 2. Beginning at the right hand, subtract each figure from the one directly over it, and write the remainder beneath. 3. If the lower 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 tho remainder to the subtrahend ; if the sum is equal to the minuend, the work is correct. For proof by casting out the 9's, see " Rays Iligher Arithmetic? 52 RAY'S PRACTICAL ARITHMETIC. (8) (9) (10) (11) Minuends, 7640 860012 4500120 3860000 Subtrahends, 1234 430021 2910221 120901 Remainders, 6406 429991 1589899 373909S Proof . . . 7640 860012 4500120 3860000 12. Take ' 1234567 from 4444444. Ans. 3209877. 13. Take 15161718 from 91516171. Ans. 76354453. 14. Take 34992884 from 63046571. Ans. 28053687. 15. 153425178 53845248= Ans. 99579930. 16. 10000000010001001= Ans. 89998999. 17. Take 17 cents from 63 cents. Ans. 46 cents. 18. A carriage cost 137, arid a horse $65 : how much more than the horse did the carriage cost? Ans. 72. 19. A tree 75 feet high was broken : the part that fell was 37 feet long : how high was the stump ? Ans. 38 ft. 20. America was discovered by Columbus in 1492 : how many years had elapsed in 1837 ? Ans. 345. 21. I deposited in the bank $1840, and drew out 475 : how many dollars had I left? Ans. 81365. 22. A man has property worth $10104, and owes debts to the amount of 87426 : when his debts are paid, how much will be left? Ans. $2678. 23. A man having 8100000, gave away 11 : how riiany had he left? Ans. 99989. 24. Subtract 19019 from 20010. Ans. 991. 25. Required the excess of nine hundred and twelve thousand and ten, above 50082. Ans. 861928. 26. Take 4004 from four million. Ans. 3995996. 27. Subtract 1009006, from two million, twenty thou- sand, nine hundred and thirty. Ans. 10li924. REVIEW. 26. EEM. After borrowing ten, which of the two methods is generally used? Why? Why begin at the right hand to 27, Giyo tho rule for subtraction. Method of proof. IV. MULTIPLICATION. ART. 28. 1. If 1 orange cost 2 cents, what will 3 cost? ANALYSIS. Three oranges will cost 3 times as much as one. That 25, 2 cents taken 3 times: 2+2+2=6. Ans. 2. If 1 lemon cost 3 cents, what will 4 lemons cost? 3+3+3+3=12. Ans. 3. In an orchard there are 4 rows of trees, in each row 21 trees : how many trees in the orchard? SOLUTION. IST. By writing 21 four times, \^ roWj 21 trees, as in the margin, and adding, the whole 2d row, 21 trcet. number of trees is 84. 3J row> 21 trees. 2n. Instead of writing 21 four times, write 4ihrow 21 trees, it once, place the number 4 under, it being the number of times 21 is to be taken, and 84 say, 4 times 1 (unit) are 4 (units), which put in units' place: 21 then, 4 limes 2 (tens) are 8 (tens), to be put in tens' place; 4 the result is 84, the same as found by addition. The latter method is termed Multiplication. DEFINITIONS. Multiplication is a short method of Addition, when tie numbers to be added are equal. Multiplication is also taking one number as many times as there are units in another. The number to be taken is the multiplicand. The number denoting how many times the multiplicand io taken, is the mnltiptitr. The result is the product. Thus, 4 times 5 are 20 ; 5 is the multiplicand, 4 the multiplied and 20 the product. The multiplicand and multiplier are together called J\u'tfii'*, because they m;ike or produce the product. It::viE\v. 28. \Vluit is multiplication? What is tha multiplicand T Vv'vu ilo-ja the multiplier denote? What is iho pr 'luer ? \V!i:it arc the mulr'niPc'ind nnd multiplier tog-jiher called? Why? Eitat. When ia iiiu:tipii<"!?i(.n termed sitnpl'j? ad JJk. 3 33 34 RAFS PRACTICAL AttlTHMETIC, REMABK. When the multiplicand is of one denomination, the operation is called Simple Multiplication. ART. 29. The Sign X, read multiplied by, or times, denotes that the numbers between which it is placed are to be multiplied together. Thus, 4X312, shows that 4 multiplied by 3, or 4 times 3, are 12, or equal 12. In the table, the sign X , may be read times : thus, 2 times 2 are 4 ; 2 times 3 arc G ; and so on. MULTIPLICATION TABLE. ix o= o 2X 0= 3X 0= 4X 0= 1X1=1 2X 1= 2 3X 1= 3 4X 1=4 1X2=2 2X 2= 4 3X 2= 6 4X 2= 8 IX 3= 3 2X 3= 6 3X 3= 9 4X 3 = 12 1X4=4 2X 4= 8 3X 4 = 12 4X 4 = 16 IX 5= 5 2X 5 = 10 3x 5 = 15 4X 5 = 20 1X6=6 2X 6 = 12 3X 6 = 18 4X 6 = 24 1X7=7 2X 7-14 3x 7 = 21 4X 7 = 23 1X8=8 2X 8 = 16 3X 8 = 21 4X 8 = 32 IX 0= 9 2X 9 = 18 3X 9 = 27 4X 9=36 1 X 10 = 10 2X 10 = 20 3X 10 = 30 4 X 10 = 40 1 X 11 = 11 2X 11=2J 3 X 1 1 = 33 4X 11=41 IX 12 = 12 2X 12 = 21 3X12 = 36 4X 12=43 5 X 0= 6 X 0= 7X 0= 8X 0= 5X 1= 5 6X 1= 6 7X1=7 8>( 1= 8 5X 2 = 10 6x 2 = 12 7x 2 = 14 8X 2 = 16 5X 3 = 13 6X 3 = 18 7X 3 = 21 8X 3 = 2fc 5X 4 = 20 6X 4 = 21: 7X 4 = 28 8X 4 = 82 5 X 5 = 25 6X 5 = 30 7x5 = 35 8X 5 = 40 5 X 6 = 30 6x 6 = 36 7X 6=42 8x 6 = 43 f> x 7 = 35 6X 7=42 7X 7 = 49 8X 7 = 56 5X 8 = 40 6X 8 = 43 7X 8 = 56 8X 8 = 64 5X 9 = 45 6 X 9 = 5 t 7X 9 = 63 8X 9 = 72 5X10 = 50 6X 10 = 60 7 X 10 = 70 8X 10 = 80 5X 11 = 55 6X11=66 7 X 11 = 77 8X 11=88 5X12=60 6X 12 = 72 7X 12 = 81 8X12 = 96 MULTIPLICATION OF SIMPLE NUMBERS. 35 9X 0= 10X 0= 11 X 0= 12X 0= 9X 1= 9 10X 1= 10 11X 1= 11 12X 1= 12 9X 2= 18 10X 2= 20 11 X 2= 22 12 X 2= 24 9X 3= 27 10 X 3= 30 11 X 3= 33 12 X 3= 36 9X 4= 36 10 X 4= 40 11 X 4= 44 12 X 4= 48 9X 5= 45 10X 5= 50 11 X 5= 55 12X 5= 60 9X 6= 54 10 X 6= GO 11 X 6= 66 12 X 6= 72 9X 1= C3 10X 7= 70 11 X 7= 77 12 X 7= 84 9X 8= 72 10 X 8= 80 11 X 8= 88 12 X 8= 96 9X 9= 81 10 X 9= 90 11 X 9= 99 12X 9=108 9X10= 90 10X10=100 11X10=110 12X10=120 9X11= 99 10X11=110 11X11=121 12X11=132 9X12=108 10X12=120 11X12=132 12X12=144 ART. 30. Here are several rows of stars. ^ ^ ^ ^ Counting upward, there are 5 rows of 4 stars ^ ^ ^ ^ each ; 5 rows contain 5 times as many as 1 row; ^ ^ ^ ^ 5 times 4 stars are 20 stars. ^ ^ ^ ^ Counting across, there are 4 rows of 5 stars ^ ^ ^ ^ each ; 4 rows contain 4 times as many as one row ; 4 times 5 stars are 20 stars, the same a: before. Hence, the product of two numbers is not altered by changing the order of the factors. What is the difference between 6X5 and 5X6? 9X8 and 6X12? 10X12 and 12X10? REMARK 1. The product is always of the same kind or denomi- nation as the multiplicand. PRINCIPLES AND EXAMPLES. 1. What is the product of 5 multiplied by 3 ; that is, what is the amount of 5* taken 3 times? Ar,s. 15. 3 times 5 are equal to 5+5+5=15. Here, the multiplicand, 5, is an Abstract number; that is, it denotes no particular thing^ and the product, 15, is also abstract. KEVIEW. 29. What does the sign ( X ) of multiplication denote ? Give an example. GO. Docs it alter the product to make either of the factors the multiplier? Illustrate this, 80 KAY'S PRACTICAL ARITHMETIC. 2. "What will 3 yards of muslin cost, at 5 cents a yard? ANALYSIS. Three yards will cost three times the price of one yard ; that is, 5 cents taken 3 times, which is 5 cts.+5 cts.+5 cts. = 15 cts. Ans. Here, the multiplicand is a Concrete number ; that is, it denotes some particular thing, as Cents; the product, 15 cents, denoting the same thing, is also concrete. Hence, in general, if the multiplicand is money, the product will he money of the same name ; if it is pounds, the product will be pounds, c. EEM. 2. The multiplier shows the 'number of time* the multiplicand 13 to bo taken ; henco, it must always be considered an abstract number. In Example 2, 5 cents are not multiplied by 3 yards, but are taken three times; as three yards will cost tkr^e times the price of 1 yard. To speak of multiplying 3 dollars by 2 yards, or 25 cents by 25 cents, is as absurd aj to propose to multiply 3 apples by 2 potatoes. ART. 31. When the Multiplier does not exceed 12. EXAMPLES. 1. How many yards of cloth are there in 3 pieces, each containing 123 } T ards? SOLUTTOX. Having placed fhe multi- OPERATION. plier under (he multiplicand, as in Iho .123 multiplicand, margin, say, 3 times 3 units are 9 un't*, 3 multiplier, which write in units' place; 3 limes 2 tens ~ " are G tens, which write in tens' place; 3 3b9 product, times 1 hundred are 3 hundreds, which write in 1m r. deeds' p^nc?. 2. What will 2 houses cost at $231 each? AUK. S4G2. 3. What wnl 3 horses cost at 8132 each ? Ans. 8300. 4. What is the product of 201 X I? Aits. 804. 5. What is the product of 2301 X3? An*. 6303. "REvinw. 00. "HEM. 1. What is the denomination of tin product ? Sli vr that wh>n the multiplicand is ;in abstract number, the pn duet is also nb:tmct. Sliow th:it whoa the multiplicand is a concrete 'number, the p r -duct inu^t be concrete. iteM, 2. What ducd the multiplier show ? What must it be considered ? MULTIPLICATION OF SIMPLE NUMBERS. 37 6. At 13 an acre, what will 5 acres of land cost? SOLUTION'. Say, 5 times 3 (units) are 15 OPERATION. (units); write the 5 (units) in units' place, and 43 carry the 1 (ten); then, 5 times 4 (tens) are 20 5 (tens;, and 1 (ten) carried make 21 (tens), which r write as in the margin. 7. What is the product of 347 X 12 ? 347 12 The product of each of the figures by 12 is known from the Mul. Table; multi- 41 04 ply then by 12, as by a single figure. PROOF. Since 11 times 347 and 1 time 347 347 are c a barrel? An*. S75G. 51 What wiU 823 barrels of pork cost, at 12 a barrel ? An*. Sflc7J. 55. What will G75 pounds of cheese cost, at 13 cents a pound? Aii*. b775 cents. 53. What will 4DG bushels of potatoes co>t, at 24 cents a bushel? -Us. 11 904 cents. 57. If a man travel 23 miles a day, bow many milci will he travel in 152 days? *1/^. 425G miles. 53. There are 17GO yards in one mile; bo*,v manyyarda are there in 209 miles/ Ans. 3G7b40 yd . 59. There are 2t hours in a day, and 3(55 days in a year: if a ship sail 8 miles aa hou;, ho'.v far will bhc sail in a year? Ans. 700^-0 miles. GO. Sound moves 1133 feet in a second : how far will it move in 103 seconds? Ans. 123170 feet. 61. Multiply two thousand and twenty-nine by one thousand and seven. Ans. 2043203. G2. Multiply eighty thousand four hundred and one by ixty thousand and seven. Arts. 4824622807. C3. Multiply one hundred and one thousand and thirty- two by 20001. Ans. 2020741032. CONTRACTIONS II? MULTIPLICATION, /J^ CASE I. ^ ART. 33. When the Multiplier is a Composite Nm scr. Any number produced by multiplying together yro or more numbers, is termed a composite number; an*/ The numbers which, multiplied together, produce a number, are its component parts^ or factors, MULTIPLICATION OF SIMPLE NUMBERS. 41 Thus, 21 is a composite number : the factors are 7 and 3, because 7 multiplied by 3 produces 21. So, also, 12 is a coui posite number; GX-=i-J or, 3X412; or, 2X-X3=12. 1. "What will 15 oranges cost, at 8 cents each? Since 5X3 15, it follows that 15 is a composite number tf which the factors are 5 and 3. ANALYsis.-Since 15 are 3 Cost of l orangej g cents< times 5, 1-3 oranges will cost r 3 times as much as 5 oranges. Thureiore, instead of muiti- Cost of 5 oranges. 40 cents, plying 8 by 15, first find the 3 cost of o oranges, by multiply- r ' f * % Cost of lo oranges, 120 cents, ing 8 cents by 5, then take 3 times that product lor the cost of 15 oranges. PROOF. 8x15^120. Rule for Case I. Separate the multiplier into two or more factors. 2. Multiply the multiplicand by one of ike factors, and this product by another factor ', till every factor is used; the last product will be the one required. REM. Do not confound the factors of a number with the parts into which it may be separated. Thus, the factors of 15 are 5 and 3, while the parts into which 15 mny be separated, are any numbers \vhose sum equals 15; as, 14 and 1; or, 2, 9, and 4. EXAMPLES FOR PRACTICE. 2. vYhat \rill 24 acres of land cost, at 124 an acre ? AHS. 82976. 3. How far will a ship sail in 56 weeks, at the rate of 2395 miles per week ? Ans. 134120 miles. 4. How many pounds of iron arc there in 54 loads, each weighing 2873 pounds? A us. 155142 pounds. 5. How many gallons of wine in 63 vats, each contain- ing 1673 gallons? Ans. 105399 galls. 6. Multiply 2374 by 72. Ans. 206928. 7. Multiply 8074 by 108. Aw. 871992, 42 RAY'S PRACTICAL ARITHMETIC. CASE II. ART. 34. Wlien (he Multiplier is 1 with ciphers annexed to it; as, 10, 100, 1000, &c. Placing one cipher on the right of a number, (Art. 8), changes the units into tens, the tens into hundreds, and so on, and, therefore, multiplies the number by 10. Annexing two ciphers, changes units into hundreds, tens into thousands, &c., and multiplies the number by 100. Annex one cipher to the right of 25, and it becomes 250, the product arising from multiplying it by 10. Aniaex two ciphers to 25, and it becomes 2500, the pro- duct of 25 X 100. Hence the Hul3 for Caso II. Annex as many ciphers to the Multi- plicand as there are ciphers in the multiplier, and the number thus formed will be the product required. 1. Multiply 245 by 100. Ans. 24500. 2. Multiply 138 by 1000. Ans. 138000. 3. Multiply 428 by 10000. Ans. 4280000. 4. Multiply 872 by 100000. Ans. 87200000. CASE III. ART 35. When there are ciphers at the right hand of one or Loth of the factors. Find the product of 2300X170. ANALYSIS. The number 2300 may be regarded 2300 as a composite number, of which the factors are 23 170 and 100; and 170 as a composite number, of which 7 ~~ (he fictora are 17 and 10. By Art. 33, the product of 2300 by 170 will be fmnd by multiplying 23 by 17. and this product by 391000 100, and the- resulting product by 10; that is, by finding the product of 23 multiplied by 17, and then annexing tc the product 3 ciphers, as there are 3 cipher? at U*e right pf botb factors. llense the DIVISION OF SIMPLE NUMBERS. 43 Rule for Case III. Multiply without regarding the ciphers on the right of the factors ; then annex to the product as many t'.phers as are at the right of both factors. 1. Multiply 2350 by 60. 2. Multiply 80300 by 450. 3. Multiply 10240 by 3200. 4. Multiply 9600 by 2400. 5. Multiply 18001 by 26000. 6. Multiply 8602 by 1030. 7. Multiply 3007 by 9100. 8. Multiply 80600 by 7002. 9. Multiply 70302 by 80300. 10. Multiply 004000 by 10200. 1 For additional problems, see Ray's An*. 141000. An*. 36135000. Am. 32768000. Ans.. 23040000. An*. 468026000. Ans. 8860060. -4ns. 27363700. An*. 564361200. An*. 5645250600. Ans. 9220800000. Test Examples. B. V. DIVISION. ART. 36. 1. If you divide 6 apples equally between 2 buys, how many will each have ? ANALYSIS. It will require two apples to give each boy 1 ; hence, each boy will have as many apples as there are times '2, apples in 6 apple*, that is 3. How many times 2 in 6? Ans. 3. Why? Because 3 timts 2 arc 6. 2. If you divide 8 peaches equally between 2 boys, how many will each have? Ans. 4. Why? 3. How many times 2 in 10? Why? REYIETT. 33. What is a composite number? What are its component parts or factor? ? Give an example. II w multiply by a composite num- ber, Rule ? Illustrate this method. REJJ. In what respect do the/actors of a number differ from its party / Give an example. 84. If one cipher is placed on the right of a number, how are the orders |f two cipher ? llow multiply by 10 } 100, 1000, &c, ? 44 BAY'S PRACTICAL ARITHMETIC. DEFINITIONS. The process by which the preceding cs amples are solved, is called Division ; hence, Division is the process of finding how many tunes oni number is contained in anot-wr. The number by which to divide, is the divisor. The number to be divided, is the dividend. The number denoting koto many times the divisor is contained in the dividend, is the quotient. ART. 37. How many times 3 in 12? An*. 4 times. Here, 3 is the divisor, IJ the dividend, anJ 4 the quotient. Since 3 is contained in 12 four tiinas, 4 times 3 ;uvi TJ; that is, the divisor and quotient multiplied, produce the dividend. Hence, since 3 and 4 are factors of the product 1'2, the divisor and quotient correspond to the factors in Multiplication; the dividend, to the product. Factors. Product. BY: MULTIPLICATION, ... 3X4 =12 Dividend. Divisor. Quotient. By DIVISION, 12 divided by 3 = 4 Or .... 12 divided by 4 = 3 Hence, Division is the prows of finding one of the fac- turs of a product, when the other factor is known. ART. 33. If 7 cents be divided equally 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 remainder. REMARKS. 1. Since the remainder is a part of the dividend, it must bo of the saui3 denomination. If the dividend bs dollars, the remainder will be dollars: if pounds, the remainder will be prumls. 2. The remainder is always Ictat than the divisor ; for, if it were equal to, or greater than it, the divisor would be contained at least once more in the dividend. 3. If the dividend denotes things of one denomination only, the opera- tion is called Simple Division. REVIEW. 36. What is Division? What is the number by which to divide? What the number to be divided? What the number Denoting how many times the divisor is contained in the dividend ? DIVISION OF SIMPLE NUMBERS. 45 ART. 39. A boy has 8 cents : how many lemons can he buy at 2 cents each ? ANALYSIS. lie can buy 4, because 4 lemons at 2 cents each, will cost 8 cents. If he did not know that 4 times 2 are 8, the operation would be thus : The boy would give 2 cents for 1 lemon, and then have G cents left. After giving 2 cents for the 2d leinon, he would have 4 cents left ; Then giving 2 cents for the third, he would have 2 cents left; Lastly, after giving two cents for the fourth, he would have nothing left: having taken 2 cents 4 times from 8 cents, and each time received one lemon. The natural method of performing this operation is by Subtraction ; but, When it is known 7iow many times 2 can be subtracted from 8, instead of subtracting 2 four times, say, 2 in 8 four times, and 4 times 2 arc 8 ; which, subtracted from 8 once, nothing is left. The last method is by Division, and it differs from the first in this: that the subtractions, instead of being per- formed separately, arc all made at once. Hence, Division may be termed a short method of mak- ing many subtraction* of the same number. The divisor is the number subtracted ; the dividend 1st lemon 8 cents. 2 cents. Left, . . . 2d lemon . G cents. 2 cents. Left, . . . 3d lemon . 4 cents. 2 cents. Left, . . . 4th lemon . 2 cents. 2 cents. Left, . . . . cents. KKVIETT. 37. Thrors in 12, 4 tim^ ; what is 3 called? 12? 4? To what is the product rf the d' visor an;l quotient equal ? To what do they correspond ? To what docs the dividend correspond ? What is divisicn ? rs. When there H fi number left after dividing, what is it called? UKM Of what denomination is tho remainder? Why? Why is the re- in tinder always tests than tho d visor? When the dividend denotes things of ono don' minution only, what is the operation called? f9. To what natural nr'fhrd do the operations in division belong? 11- lutrnto by an cx'iinnle. What in;iy division bo termed? What U the divider f Dividend? ( 46 RAY'S PRACTICAL ARITHMETIC. the number from which the subtractions are made ; th& quotient shows how many subtractions have been made. ART. 40. DIVISION is DENOTED BY THREE SIGNS : 1st. 3)12 means that 12 is to be divided by 3. 2d. - 1 ! means that 12 is to be divided by 3. 3d. 12-H-3 means that 12 is to be divided by 3. Use the 1st sign when 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 : place the quotient on the right of this. The sign, -, in the Table, is read divided by. 8 2)8 I 1 15)45(3 45 15 "5"" 213=7. DIVISION TABLE. 1 -1= 1 2- -2= 1 3- -3= 1 4- -4= 1 2- -1= 2 4- -2= 2 6- -3= 2 8- -4= 2 3- -1= 3 G- -2= 3 9- -3= 3 12- -4= 3 4- -1= 4 8- -2= 4 12- -3= 4 1G- -4= 4 5- -1= 5 10- -2= 5 15- -3= 5 20- -4= 5 G- -1= 6 12- -2= G 18- -3= G 24- -4= 6 7- -1= 7 14- -2= 7 21- -3= 7 28- -4= 7 8- -1= 8 16- -2= 8 24- -3= 3 32- -4= 8 9- -1= 9 18- -2= 9 27- -3= 9 3G- -4= 9 10- -1 = 10 20- -2 = 10 30- -3=10 40- -4 = 10 5- -5= 1 6- -G= 1 7_ -7= 1 8- -8= 1 10- -5= 2 12- -G= 2 14- -7= 2 16- -8= 2 15- -5= 3 18- -G= 3 21- -7= 3 24- -8= 3 20- -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= G 3G- -G= G 42- -7= 6 4S- -8= 6 35- -5= 7 42- -G= 7 49- -7= 7 56- -8= 7 40- -5= 8 48- -G= 8 56--7= 8 64- -8= 8 45- -5= 9 54- -6= 9 63-^-7= 9 72-f-8= 9 60- -5 = 10 G0--G = 10 70-^-7=10 80 --8==10 DIVISION OP SIMPLE NUMBERS, 9- -9= 1 10- -10= 1 ll-r-ll= 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- -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 PRINCIPLES AND EXAMPLES. ART 41. 1. I wish to put 15 hats into boxes, each con- taining 3 hats : how many boxes do I need ? IST SOLUTION. I need as many boxes as 3 hats are contained times in 15 liats ; that is, 5 boxes. Hats. Hats. 3)15(5, boxes. 15 2. Having 15 hats, I wish to separate them into 5 equal lots : how many hats will there be in each lot ? 2i> SOLUTION. Puth'ngonehat into each lot will require 5 hats; hence, there will be as many hats in each lot, as there are times 5 hats in 15. lints. Hats. 5) 15(3, hats in each. 15 The first solution shows, that by Division a number can be separated into parts containing a certain number of units, and the number of parts found. The second solution shows, that by Division a given number can be separated into a certain number of equal parts, and the number of units in each part found. REVIEW. 40. How many signs aro used to denote division ? What is the first? Second? Third? Illustrate their meaning. 41. What docs the first solution show? What the second? RFM. TIow does it appear that the divisor and dividend are both of tho saim donoini- ]j nation? Is tho. quotient an abstract or a concrete number? What does it show ? What may it represent ? 43 BAY'S PRACTICAL ARITHMETIC. MENTAL EXERCISES. Solvo tho examples on the left liks the first, and those on the righi, liko the second of the preceding SOLUTIONS. 3. Ho\v many lemons at 3 cents each, can you buy for 12 cents? 5. How many barrels of flour, at 4 dollars a barrel, can you purchase for $20? 7. At six dollars a yard, how many yards of cloth can you purchase for $30? 9. How many lead pencils, at o cents apiece, can you buy for 3-3 cents? 11. A man has C3 pounds of butter, and wishes to put it into boxes, each containing 7 pounds: how many boxes will be required ? 13. Into how many parts, of 3 each, can 24 be separated ? 4. If you pay 12 cents for 4 lemons, how much will each cost? 6. If you pay $20 for 5 barrels of flour, how many dollars will a barrel cost? 8. If you pay $30 for 5 yards of cloth, how many dol- lars will a yard cost? 10. If you pay 3J cents for 7 lead pencils, how much will that be for each? 12. A man has G3 pounds of butter, to put into 9 boxes: how many pounds must he put into each box? 14. If 24 is separated into 8 equal parts, how many will there be in each part? REM. 1. The divisor and dividend are both of the same denomination. This follow * from that view of division, which shows it to be a short method of making several subtraction* of tho saon number. 8;nae it is only numbers of the sum? denomination whoc difference can bo found, thosj on!y of tho same denomination can be divided. 2. Tho quotient is an abstract number, and show.' how mn,iy i/'mcs tho divisor is contained in the divided. But, it may represent the number of units in aomo concrete number, 03 in cx.implos 1 and 2, Art. 4.1. ART. 42. EXAMPLES. 1. Two in 6 how many times? Ans. 3 times. 2. How many times is 2 contained in G3? ANALYSIS. Two in G units of the first order, 3 2)00 tinjps: in units of ihn second order, (10 limes as - lursre,) it is contained 10 times as m.-iny times. 3. How many times? is 2 contained in GOO? DIVISION OF SIMPLE NUMBERS. 49 ANALYSIS. Two is contained 3 times in 6 units 2)600 tf the first order ; but, in G units of the third order, which are 100 times as large, it is contained 100 300 times as of Lea ; that is, 300 times. 4. Three feet mako one yard: how many yards in GO feet? Ans. 20. 5. Two pints mako ono quart: how many quarts in 400 pints? Ans. 200. 6 How many times 3 in GOOO? AM. 2000. 7. How maay times 4 in 80000 ? Ans. 20000. ART. 43. 1. How many times is 2 contained in 468? Here, the dividend is composed of 3 numbers; 4 hundreds, G tens, aud 8 units; that is, of 400, CO, and 8. Divisor. Dividend. Quotient. Now, 2 in 400 is contained 200 times. 2 2 in in 60 8 30 4 times, times. Hence, 2 in 468 is contained 234 times The same result can be obtained without separating the dividend into parts : Thus, 2 in 4 (hundreds) 2. How many times 3 3. How many times 4 4. How many times 2 5o How many times 4 6. How many times 3 7. How many times 2 RE\'TE\v.-43. Explain how many times 2 is contained in 468. 3d Bk. 4 un I red) times, Dividend. ce; then, 2 in Divisor 2)468 w; : te in tens' (units) times, Quotient 234 in 693? Ans. 231. in 848? Ans. 212. in 46S2? Ans. 2341. in 8408? Ans. 2102. in 3G936? Ans. 12312. in 884G8? Ans. 44234. 50 BAFS PRACTICAL ARITHMETIC. ART. 44. SHORT DIVISION. 1. How many times is 3 contained in 129 ? SOLUTION. Here, 3 is not contained in 1; but 3)129 consider the 1 and 2 as forming 12 tens, then 3 is . in 12 tens, 4 (tens) times, which write in tens' 43 $>lace; 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 7 (hundreds), 3^735 2 (hundred) times, and 1 (hundred) over; the 1 him- ^ dred united with the 3 tens, makes 13 tens, in which 245 '8 is contained 4 (tens) tunes and 1 (ten) left; this 1 ten united with the 5 units, makes 13 units, in which 3 is contained 5 times. OF REMAINDERS. 3. ITow many times is 3 contained in 743? After dividing, there is 2 left, which 3)743 ought to be divided by the divisor 3: 2 4 7. ..2 Hem. But the method of doing this will not be explained until the pupil has studied Fractions. The division is merely indicated by placing the divisor under the remainder, thus, f . The quotient is written thus, 217*; read, 247, and two divided ly three; or, 247, with a remainder, ttco. Instead of performing all the operation mentally, the work may be written, as in the foliowing solution: SOLUTION. In this operation, sny 3 in 7 opr.B ATTOV. (hundreds), 2 (hundred) times; then niuHi- 3)743(247. ply 3 by 2 (hundreds), and subtract the (j product, C (hundreds), from 7 (hundreds), ^~~ which leaves a remainder, 1 (hundred); to ^ ^ !ns * this remainder unite the 4 (tens), making 14 (tens), which contains 3, 4 (tens) times, 93 units, with a remainder 2 (tens). 9 ^ To this rem. unite the 3 units, and 3 in 23 Cunits), 7 times, with a remainder 2. DIVISION OF SIMPLE NUMBERS. 51 DEFINITIONS. When the division is performed men- tally, and merely the result written, it is termed Short Division; when the entire work is written, Long Divi- tion. Short Division is used when the divisor does not exceed 12. 4. How many times 3 in 4G2? . . . Ant. 154. 5 How many times 5 in 1170? . . . Am. 234. 6. How many times 4 in 948 ? ... An*. 237. ART. 45. RULE For Short Division. 1. Write the divisor at the left of the dividend, with a curved line between them. Begin at the left hand, divide successively each figure of the dividend by the divisor, and write the result 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 doe* 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. EXPLANATION. To Prefix, means to place before, or at the left hand. To Annex, is to placo after, or at the right hand. PROOF. Multiply the quotient hy the divisor, add the re- mainder, if any, to the product: if the work is correct, the gum will bo equal to the dividend. Rfiir. This method of proof depends on the principle, Art. 37, that A dividond is a product, of which the divisor and quotient are factors. If the rcinaindar he subtracted from the dividend, and the result divided by the quotient, tho quotient thus obtained will bo tho divisor. REVIEW. 4. How many tim?s 3 in 129 ? Explain i*-. II .w is a re- mainder to be written ? When is the operation term:? 1 short division f When long division ? When is short division used ? When long ? What is the difference between long ani short division ? 4.5. In dividing, how are tho numbers written, Rule ? Where do yoP begin to divide I Where is each quotient figure placed ? 52 RAY'S PRACTICAL ARITHMETIC. 7. Divide 65 3 cents by 3. Dividend. Divisor . 3)653 Quotient. .217 Rem. 2. 217 PROOF. 3 (8) G)454212 7)874293 An*. 75702 6 124893 7 Proof. 454212 874293 651= cents divided, 2 = remainder. 653 = the dividend. (10) (11) 9)2645402 8)3756031 293933s 469503] 9 8_ 2645402 3756031 OF PARTS OF NUMBERS. NOTE. When any number is divided into two equal parts, *)no of the parts is called one-half of that number. If divided into three equal parts, ono of tho parts is called ons-third. If into four equal parts, one fourth. If into five equal parts, one-ffth; and so on. Hence, To find one-half of a number, divide by 2; to find one-third, divide by 3; one fourth, divido by 4; onejijih, by 5, 24. If oranges cost 3 cents each, bow many can be bought for 894 cents ? Ans. 298 oranges. 25. If 4 bushels of apples cost 140 cents, how much is that a bushel ? An*. 35 cts. 26. If flour cost $4 a barrel, how many barrels can be bought for $812 ? Ans. 203 bad. 27. A carpenter receives $423 for 9 months' work : how much is that per month ? Ans. $47. 28. There are 12 months in 1 year : how many years are there in 540 months? Ans. 45 yrs. 29. There are 4 quarts in 1 gallon : how many gallons are there in 321276 quarts ? Ans. 80319 gals. 30. At $8 a barrel, how many barrels of flour can bo bought for $1736 ? Ans. 217 barl. 31. There are 7 days in one week : how many weeks ara there in 734566 days? Ans. 104938 wks. 32. A number has been multiplied by 11, and the pro- duct is 495 : what is the number? Ans. 45. 33. The product of two numbers is 3582 : one of tK* numbers is 9 : what is the other? Ans. 398. 34. Find one-half of 56. . . Ans. 23. 35. Find one-half of 3725. . . Ans. 1862-J. 36. Find one-third of 147. . . Ans. 49. 37. Find one-fourth of 500. . . Ans. 125. 38. Find one-fifth of 1945. . . Ans. 389. 39. Find one-sixth of 4476. . . Ans. 746. 40. Find one-seventh of 2513. . . Ans. 359. 41. Find one-eighth of 5992. . . Ans. 749. 42. Find one-ninth of 8703. . . Ans. 977. 43. Find one-tenth of 1090. . Ans. 109. 64 RAY'S PRACTICAL ARITHMETIC. 44. Find one-eleventh of 4125. . . . Ans. 375, 45. Find one-twelfth of 5556. . . . Ans. 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? Am. 12 47. James found 195 cents, and gave to Daniel one- fifth of them : Daniel gave one-third of his share to hig sister : how many cents did she receive ? Ans. 13. ART. 46. LONG DIVISION. 1. Divide 3465 dollars equally among 15 men. SOLUTION. When the divisor exceeds 12, it becomes necessary to write the process, as in "OPERATION," Example 3, page 50. Fifteen is not contained in 3 (thousands), therefore, there will be no thousands in the quotient. Take 34 (hundreds) as a partial dividend; 15 is contained in it 2 (hundred) times; that is, 15 men have each $200, which requires in all ^ - ^ . 15 X 2 (hundred) = 30 hundreds. Jill j J g Subtract 30 (hundreds) from 34 (him- 15)3465(231 dreds) and 4 (hundreds) remain, to which 3 hund. bring down the 6 (tens), and you have 46 t 46 (tens) for a second partial dividend. 46 contains 15, 3 (tens) times, giving each man 3 ten (30) dollars more, requiring * * units, for all, 15X3 (tens)45 tens of dollars. Subtract 45 and bring down the 5 (units), you have 15 (units) for a 3d partial dividend, in which the divisor, 15, is contained once, giving to each man 1 (unit) dollar. Hence, each man receives 2 hundred dollars, 3 ten dol- lars, and 1 dollar ; that is, 231 dollars. By this process, the dividend is sep- Divisor. Part*. Quotients, arated into parts, each containing the 15 3000200 divisor a certain number of times. 450 30 The first part, 30 (hundreds), eon- 1 5 tains th divisor 2 (hundred) times ; ^ . ^. ^"i the second part, 45 (tens), contains it S (tn) times; the third part, 15 (units), contains it 1 (unit) time. DIVISION OF SIMPLE NUMBERS. 55 The several parts together, equal the given dividend: the several partial quotients make up the entire quotient. 2. In 147093 days, how many years, each 365 days? SOLUTION. Taking 147 365)147095(403 year*. (thousand) for the first par- 1460 tial dividend, we find it will - ~ ~ _ .not contain the divisor; hence 1 OQ^ use four figures. Again, after multiplying and subtracting, as in the preceding ex- ample, and bringingdown the 9 (tens), the partial dividend, 109 (tens), will not contain the divisor; hence, write a (no tens) in the quotient, and bring down the 5 (units): the last partial dividend is 1095 (units), which contains the divisor 3 (units) times. 3. Divide 4056 by 13. ...... Ans. 312. RULE For Long Division.!. Place 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 theproduct under that part of the dividend from which it was obtained. 4 Subtract this product from the figures above it; to the re- mainder bring down the next figure of the dividend, and divide a* before, until all the figures of the dividend are brought down. 5. If, at any time, after bringing down a figure- the number thus formed is too small 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, ft REVIEW. i>>. Repeat tho Rule for Long Division. TThere is the divi- sor placed ? Whero the quotient ? How commence dividing? 4$. After obtaining tho first quotient figure, how proceed ? After bring- ing down a figure, if the partial dividend thus form 3d is too small to contain the (Jivisor, what is to ba dono ? What the method of proof ? 56 RAY'S PRACTICAL ARITHMETIC. NOTES. 1. The product must never be greater than the partial dividend from which it is to be subtracted. When so, the quotient figure is too large, and must be diminished. 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 small, and must be increased. 3. Example 1 shows that the order of each quotient figure is the same as the lowest order in the partial dividend from which it was obtained. It will be a useful exercise to name the order of each quotient figure immediately after obtaining it. $S?* For casting out the 9's, see "Ray's Higher Arithmetic." 4. Divide 78994 by 319. 319)78994(247 638 1519 1276 PROOF. 247 Quotient 319 Divisor. 2434 2233 201 Rem. 5. Divide 11577 by 14. 6. Divide 48690 by 15. 7. Divide 1110900 by 23. 8. Divide 122878 by 67. 9. Divide 12412 by 53. 10. Divide 146304 by 72. 11. Divide 47100 by 54. 12. Divide 71104 by 88. 13. Divide 43956 by 66. 14. Divide 121900 by 93. 15. DivLle 25312 by 112. 16. Divide 381600 by 123. 17. Divide 105672 by 20 L 18. Divide 600000 by 1234. 78793 Add 201 Remainder. 7 89 94 = the Dividend. . . . Ans. 826{|. . . . Ans. 3246. . . . Ans. 4S302.lt. . . . Ans. 1834. 23410. 2032. Ans. Ans. Ans. Ans. Ans. Ans. 12313 Ans. 22 Ans. 3102fV A ns. 8721?. 808. 666. 518. Ans. DIVISION OF SIMPLE NUMBERS. 57 19. Divide 47.263488 by 4674. An*. 10112. 20. Divide 26497935 by 2034. AM. 13027gJ|. 21. Divide 48903952 by 9876. Am. 4952. 22. Divide 40191GO by 12345. Am. 328. 23. Divide 552160000 by 973. Am. 567482^- 24. At $15 an acre, liow many acres of land can be bought for 3465 ? Ans. 231 acres. 25. If a man travel 26 miles a day, in how many days will he travel 364 miles ? Ans. 14 days. 26. If 31083 be divided equally among 19 men, how many dollars will each have? Ans. 157. 27. A man raised 9523 bushels of corn on 1 07 acres : how much was that on one acre ? Ans. 89 bush. 28. In 1 hogshead there are 63 gallons : how many hogsheads in 14868 gallons? Ans. 236 hds. 29. A President receives $25000 a year (365 days) : how much is that a day ? Ans. $68 a day, and $180 over. 30. The yearly income from a railroad is $37960 : how much is that per day? (365 days=l year.) Ans. $104. 31. The product of two numbers is 6571435 ; one of the factors is 1235 : what is the other? Ans. 5321. 32. Divide one million two hundred and forty-seven thousand and four hundred by 405. Ans. 3080. 33. Divide 10 million four hundred and one thousand by one thousand and six. Ans. 10338 CONTRACTIONS IN DIVISION. CASE I. ART. 4.7. When the Divisor is a Composite Number. 1. A man paid $255 for 15 acres of land : how much was that per acre ? ANALYSIS By taking D ^ ars 5255, y 3u ob- 3 j 2 5 5 = the Talue Qf 15 acres tain the value or onp,-tlurd } (o acres.) of t lie land; divi- 5)85 = the value of 5 acres. dine this quotient (85) by 6, gfves U,e\alu. of 1 acre. 1 7 - * Talue f l acre ' 58 BAY'S PRACTICAL ARITHMETIC. The preceding analysis shows that instead of dividing by the composite number 15, whoso factors are 3 and 5, wo may first divide by one factr, then divide the quotient thus obtained by the other factor. 2. Find the quotient of 37, divided by 14. SOLUTION. Dividing by 2, the quo- 2^)37 tient is 18 twot and. 1 unit remaining. Dividing by 7, the quotient is 2, with 7)18 and 1 over. a remainder of 4 twos ; the whole re- 7> i A . . ., . . A A and 4 twos left. mainder then, is 4 twos plus 1, or U. Rule for Case I. 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 ftrst divisor; to the product add the first remainder; the amount will be the true remainder. NOTE. When the divisor can be resolved into more than two factors, you may divide by them successively. The true remainder will bo found by multiplying each remainder by all the preceding divisors, except that whiuih produced it. To their sum add the remainder from first divisor. 3. Divide 2583 by 63. (03 = 7X0) Ans. 41. 4. Divide 6976 by 32. (32 = 4x8) Ans. 218. 5. Divide 2744 by 23. (23 = 7x4) Ans. 98. 6. Divide 6145 by 42. (42 = 0X7) Ans. UGJ-J. 7. Divide 19008 by 132 Ans. 144. 8. Divide 7840 by 64 Ans. 1223*. 9. Divide 14771 by 72 Ans. 205 ?J. 10. Divide 10206 by 81 Ans. 126. CASE II. ART. 48. To divide by 1 with ciphers annexed; as 10, 100, 1000, REVIEW. 46. How do you divide by 10, 100, 1000, 4. Whai is found by multiplication? G've an example. When rhe divisor and quotient are given, how is the dividend found ? 55. What ia found by division ? What does it enable ns to do ? G're exnmnlos If the dividend and quotient are given, how is the divisor found ? Hiving the product of three numbers, and two of them given, hew is tho other found ? 64 RAY'S PRACTICAL ARITHMETIC. 4. In an army of 57068 men, 9503 arc killed; 586 join the enemy ; 4794 arc prisoners ; 1234 die of wounds ; 850 are drowned: how many return? Ans. 40101 5. On the first of the year a man is worth 8123078; during the year he gains 8706 ; in January he spends 237, in February $301 ; in each of the remaining ten months, he spends as much as in the first two : how much had he at the end of the ye<*r ? Ana. 125866. 6 In a building there are 72 rooms ; in each room 4 windows, and in each window 24 lights : how many lights are there in the house? Ans. 6912. 7. A merchant has 9 pieces of cloth, of 73 yards each : and 12 pieces, of 88 yards each : how many yards in all ? Ans. 1713 yards. 8. I spend 99 cents a day : how many cents will I spend in 49 years, of 3o5 days each? Ans. 1770615. 9. An Encyclopedia consists of 3D volumes ; each vol- ume has 774 pages of two columns each ; each column 67 lines; each line 10 words ; and every 10 words 47 letters: how many pages, lines, words, and letters, in the work ? A ( 301 8J pages, 4044924 lines, S ' (40449240 words, 190111428 letters, 10. The Bible has 31173 verses : in how many days can I read it, reading 86 verses a day? Ans. 362|^ days. 11. I bought 28 horses for 81400; 3 died: for how much each must I sell the rest, to incur no loss ? Ans. 856. 12. How many times can I fill a 15 gallon cask, from 5 hogsheads of 63 gallons each? Ans. 21 times. 13. A certain dividend is 73900 ; the quotient 214; the remainder 70 : what is the divisor? An*. 315. 14. Multiply the sum of 148 and 56 by their difference ; divide the product by 23. Atts. 816. 15. How much cloth, at 6 a yard, will pay for 8 horses at 860 each, and 14 cows at 815 each ? Ans. 115 yards. 16 A cistern of 360 gallons, has 2 pipes; one will fill it in 15 hours, and the other empty it in 20 hours. If both pipes are left open, how many hours will the cistern be in filling? Ans. CO hours, REVIEW OF PRINCIPLES. 65 17. Two men paid $6000 for a farm ; one man took 70 acres at $30 an acre, the other the remainder, at $25 an acre : how many acres in all ? Ans. 226 acres. SUGGESTION. In the four following examples, obtain the re- quired number by reversing the operations. 18. What is the number, from which, if 125 be sub- tracted, the remainder will be 222 ? Ans. 347. 19. What is *lie number, to which, if 135 be added, the sum will be 500 ? . Ans. 365.. 20. Find a number, from which, if 65 be subtracted, and the remainder divided by 15, the quotient will be 45 * Ans. 740. 21 What is the number, to which if 15 be added, the sum multiplied by 9, and 11 taken from the product, the remainder will be 340 ? Ans. 24. 22. If 98 be subtracted from the difference of two numbers, 27 will remain ; 246 is the less number : what is the greater ? Ans. 371 G s, GENERAL PRINCIPLES OJ DIVISION, ART. 57. The value of the quotient depends on the relative values of divisor and dividend. These may be changed by Multiplication and Division, thus : 1st. The Dividend may be multiplied, or the Divisor divided. 2cl. The Dividend may be divided, or the Divisor multiplied. 3d. Both Dividend and Divisor may be multiplied, or both divided, at the same time. ILLUSTRATIONS. Let 24 be a dividend, and 6 the divi- sor; the quotient is 4: (24 -7-6 = 4). If the dividend (24), be multiplied by 2, the quotient will be multiplied by 2: for, 24X2 48; and 48 -j- 6 = 8. which is the former quotient (4), multiplied by 2. Now, if the divisor (6), be divided by 2, the quotient will ue multiplied by 2; for, 6-r-2 = 3; and 24 -H 3 = 8, which is the former quotient (4), multiplied by 2. Hence, PRINCIPLE 1. If the dividend be multiplied, or the divisor be divided, the quotient will be multiplied. 3d Bk. 5 66 RAY'S PRACTICAL ARITHMETIC. ART. 58. Take the same Example, 24 6 = 4. If the dividend (24), be divided by 2, the quotient will be divided by 2: for, 24-7-2 = 12; and 12-7-6 = 2, which is the former quotient (4), divided by 2. And, if the divisor (6), be multiplied by 2, the quotient will be divided by 2; for, 6X2 = 12; and 24 -j- 12 = 2, which is the former quotient (4), divided by 2. Hence, PRINCIPLE II. If the dividend be divided, or the divisor be multiplied, the quotient will be divided. ART. 59. Take the same Example, 24-7-6 = 4. If the dividend (24), and divisor (6), be multiplied by 2, the quotient will not be changed; for, 24X2=48: and 6X2 = 12; 48-7-12=4, the former quotient (4), unchanged. And if the dividend (24), and divisor (6), be divided by 2, the quotient will not be changed ; for, 24 -7- 2 = 12 ; and 6-7-2 = 3; 12 ~3 = 4; the former quotient (4), unchanged. Hence, PRINCIPLE III. If both dividend and divisor be multiplied or divided by the same number, the quotient will not be changed. ART. 60 If a number be multiplied, and the product divided by the same number, the quotient will be the original number. For, 24X2=48; and 48 2 = 24, the original number, on the principle, that if the product be divided by the multi- plier, the quotient will be the multiplicand. Also, If a number be divided, and the quotient multiplied by the same number, the product will be the original number. For, 24 -7- 2 = 12 ; and 12 X 2 = 24, the original number, on the principle, that if the quotient be multiplied' by the divisor, the product will be the dividend. Hence, the operations of multiplication and division by tho game number, destroy (cancel) each other. REVIEW. 57. On what does the value of the quotient depend ? How may the divisor and dividend be changed ? If the dividend be multiplied, what effect on the quotient ? If the divisor be divided ? 58. If the dividend be divided, what effect on the quotient? If the divisor be multiplied ? 59. If both divisor and dividend be multiplied by the same number, what effect on the quotient ? If both be divided by the same number ? A VI. CANCELLATION. Most TEACHERS defer this subject until after Factoring. 1. I bought 3 oranges at 10 cents each, and paid for them with pears at 3 cts. each : how many pears did I give ? SOLUTION. Ten cents multiplied by 3, OPERATION. give 30 cents, -the cost of the oranges. 1 Then it will take as many pears, as 3 cents 3 are contained times in 30 cents ; that is, ^ 30-^3 = 10, the number of pears. Here, 10 is multiplied by 3 and the pro- ^ns. 1 pears, duct divided by 3; but a number is not changed by multiplying it, and then dividing the product by the same number) (Art. 60) ; hence, multiplying by 3, and then dividing by 3, may be omitted, and 10 taken as the result ; hence, ART. 61 a . When a number is to be mul- OPERATION. tiplied and then divided by the same num.- 1 X $ ber, both operations may be omitted, and " =10. aline drawn across the common multiplier r and divisor, as in the margin. EEM. In the above example, 10 and 3 form the dividend, and 3 the divisor. In arranging the numbers, place the dividend above a horizontal line, and the divisor below it. 2. How many barrels of molasses at $13 a barrel, will pay for 13 barrels of flour at $4 a barrel ? Ans. 4. 3. If I buy 41 cows at $11 each, and pay in horses at $41 each, how many horses are required? Ans. 11. 4. If I buy 10 lemons at 3 cents each, and pay ID oranges at 5 cts. each, how many oranges will I give ? SOLUTION. Ten times 3 cents are 30 cents, OPERATION. the cost of the lemons : 30 cts. divided by 5 cents, equal 6, the number of oranges. 1 But, as 10 is a composite number, whose factors RNOA are 5 and 2, (5X2 = 10), indicate. the operation as in the margin on the left. Ans. 6. g X 2 X 3 Since 5, 2, and 3 are to be multiplied together, and their product divided by 5, omit 5 both as multiplier and diviso? (Art. 60), draw a line across it, and only multiply 2 by 3. 68 RA5TS PRACTICAL ARITHMETIC. 5. Multiply 17 by 18, and divide the product by 6. SOLUTION. Instead of mul- OPERATION. tiplying 17 by 18, and dividing Dividend, 17X0X3 the product by 6, separate -- =51. 18 into the factors 6 and 3. Divisor, (6X3 = 18), then cross the factor 6, which is common to both mul- tiplier and divisor: after which, multiply 17 by 3/ 6. In 15 times 8, how many times 4? Ans. 30. 7. In 24 times 4, how many times 8 ? Ans. 12. 8. In 37 times 15, how many times 5? Ans. 111. 9. Multiply 36 by 40, and divide the product by 30 multiplied by 8. OPERATION. Dividend, 36X40_0X3X 2 Divisor, 30X8 SOLUTION. Indicate the operation to be performed, then resolve (separate) the numbers into factors. Now the quotient will not be changed (Art. 59) by dividing both divisor and dividend by the same number, which is done by erasing the same factors in both. SECOND SOL. Indicate the ope- OPERATION. ration as in the margin. 6 5 As 8 is a factor of 40, divide 40 Dividend, 00X40 by 8, and write the quotient 5 over == * ) . 40; cross out (cancel) 40 and 8. Divisor, $ X g ' .Then, as 36 and 30 have a com- p mon factor 6, divide each by 6, and write the quotients 6 and 5 as in the operation : cancel 36 and 30. Next, canceling the common factor 5 in both dividend and divisor, the result is 6, as before. 10. In 36 times 5, how many times 15? Ans. 12. 11. In 14 times 9, how many times 6? Ans. 21. ART. 61 . The process of shortening the operations of arithmetic, by omitting equal factors from the dividend EEVIEW. 60. If a number be multiplied, and the product divided by the same number, what will be the quotient ? On what principle ? If a number be divided and the quotient multiplied by the same number, what will the product be ? On what principle ? CANCELLATION. 69 and divisor, is termed Cancellation. It depends on the principle explained in Arts. 58 and 59. NOTE. To cancel is to suppress or erase. When the same factor is omitted in both dividend and divisor, it is said to be canceled. '/ RULE FOR CANCELLATION. When there are common factors in a dividend and its divisor, shorten the operation by canceling all the factors common to both: proceed with the remaining factors as the question may require. REM. 1. Canceling is merely dividing both dividend and divisor by the same number, which (Art. 59) does not alter the quotient. 2. The pupil should observe that one factor in the dividend will cancel only one equal factor in the divisor. 3. Some prefer to place the numbers forming the dividend on the right of a vertical line, and those forming the divisor on the left. 1. Multiply 42, 25, and 18, together, and divide the product by 21 X 15. Ans. 60. 2. I sold 23 sheep at $10 each, and was paid in hogs at $5 each : how many did I receive ? Ans. 46. 3. How many yards of flannel at 35 cents a yard, will pay for 15 yards of calico at 14 cts. ? Ans. 6 yards. 4. What is the quotient of 21X11X6X26, divided by 13X3X14X2? Ans. 33. 5. The factors of a dividend are 21, 15, 33, 8, 14, and v 17; the divisors, 20, 34, 22, and 27: required the quotient. Ans. 49. 6. 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 ? Ans. 38. NOTE. Other applications of Cancellation will be found in Fractions, Proportion, &c. The pupil will apply it more readily, when acquainted with. Factoring. REVIEW. ()\ a . REM. How are numbers arranged for cancellation? > 61 0. What is cancellation? Upon what principle docs it depend? What does cancel mean ? . When is a factor canceled ? What is the rule ? -- REM. What is canceling ? 70 BAY'S PRACTICAL ARITHMETIC, VII. COMPOUND NUMBERS. To TEACHERS. While placing Fractions immediately after Simple W*iole Numbers is philosophical, and appropriate in a Higher Arithmetic for Advanced pupils, the experience of the author convinces him that, in a r 'ook for Young learners, Compound Numbers should be introduced here, ustead of after Fractions, as is done by some authors. His reasons are, 1st. The operations of Addition, Subtraction, Multiplication, and Divi- ?ion of Compound Numbers, are analogous to the same operations in ^imft Numbers, and serve to illustrate the principles of the fundamental rule The principle of Notation is the same in each. 2d. The subject of Fractions is important and difficult. Before studying it, most pupils require more mental discipline than is demanded in the ele* mentary 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. Ijgir TEACHERS who prefer it, can direct their pupils to defer Compound Numbers until they have studied Fractions to page 169. DEFINITIONS. ART. 62. When two numbers have the same unit, they are of the same kind or denomination : thus, 3 dollars, 5 dollars, are of the same denomination ; both dollars. When they have different units, they are of different denominations : thus, 3 dollars, and 5 cents, are of different denominations ; dollars and cents. ART. 63. A simple number denotes things of the same unit value: thus, 3 yards, 2 dollars, 5 pints, are each simple numbers. All abstract numbers are simple. ART. 64. A compound number is two or more numbers of different unit values used to express one quantity : thus, 3 dollars 5 cents, 2 feet 3 inches, are each compound. REX. 1. In Compound Numbers, denomination or order, denotes the REVIEW. 62. When are two numbers of the same denomination 1 Give an example. When of different denominations? Give an example. 63. What does a Simple Number ienote ? Give an example. 64. What is a Compound Number ? Give an example. COMPOUND NUMBERS. U. S. MONEY. 71 name of the unit considered. Thus, dollar and cent are denominations of money ; foot and inch, of length ; pound and ounce, of weight. 2. Compound Numbers are analogous to Simple Numbers in this particu- lar ; a certain number of units of each order is collected into a group, and forms a unit of a higher order or denomination. But, They differ in this : that in compound numbers 10 units of one order do not uniformly make one of the next higher. 3. The simplest class of Compound Numbers is Federal money, because we pass from one denomination to another according to the scale of tens. FEDERAL OR UNITED STATES MONEY, ART. 65, Is the currency of the United States, estab- lished by the Federal Congress, in 1786. While U. S. money may be treated decimally, it is a species of Compound Numbers, being so regarded in ordinary business transactions. Its denominations, or the names of its different orders, are mill, cent, dime, dollar, eagle. Ten units of each denomination make one unit of the next higher denomination. TABLE. 10 mills, marked m., make 1 cent, marked ct. 10 cents :....! dime, d. 10 dimes .1 dollar, $. 10 dollars . . . -* 1 eagle, E. Also, 5 cents . . * make one-half dime. . 25 cents one-quarter of a dollar. 50 cents one-half of a dollar. 75 cents three-quarters of a dollar. 100 cents one dollar. The coins of the United States are of copper, silver, and gold. Their denominations are, 1st. Copper : cent^half cent. (3 cent piece, silver and copper. ) 2d. Silver: dollar, half dollar, quarter dollar, dime, half dime 3d. Gold : $20 piece, eagle, half eagle, quarter eagle, three dollar piece, dollar. The mill is not coined. EEVIEW. 64. KEM. 1. What does denomination or order denote ? 2. In what are Simple and Compound Numbers analogous ? In what do they differ ? 3. What is the simplest class of Compound Numbers ? * 72 RAY'S. PRACTICAL ARITHMETIC. NOTATION AND NUMERATION. ART. 66. Accounts are kept in dollars, cents, and mills ; or in dollars, cents, and parts of a cent. Eagles and dollars are called dollars / dimes and cents, cents. NUMERATION TABLE. E 'S .043 read 4 cents and 3 mills, or 43 mills. .214 read 21 cents and 4 mills, or 214 mills. 3.0 4 5 read 3 dollars 4 cents and 5 mills. 76.250 read 76 dollars and 25 cents. 6 8 1.3 4 5 read 681 dollars 34 cents and 5 mills. The second line may also be read, 2 dimes, 1 cent, 4 mills ; the fourth line, 7 eagles, 6 dollars, 5 cents. This method of reading is not customary. The third line may also be read, 304 cents and 5 mills, or 3045 mills; the fourth line, 7625 cents, or 76250 mills; the lower line, 68134 cents and 5 mills, or 681345 mills. A period (.), is used as a separating point, to separate the cents and dollars. Some use the comma. Thus, 2 dollars, 2 dimes, 2 cents, or 2 dollars and 22 cents, are written, $2.22 ART. 67. The Table shows that cents occupy the first two places to the right of dollars, and mills the place to the right of cents, the third from dollars. Hence the Hule for Numeration. Bead the number to the left of the period as dollars, and the first two figures on the right of the period as cents ; and if there be a third figure, as mills. REVIEW. 65. What are the denominations of United States money? How many units of either denomination make a unit of the next higher ? Repeat the table. How many cents in a half dime ? In a quarter dollar ? In a half dollar ? In a dollar ? Of what are the coins of the United States ? Which are copper ? Which silver ? Which gold ? COMPOUND NUMBERS. U. S. MONEY. 73 EXAMPLES TO BE COPIED AND THEN READ. $18.62 5 $ 70.01 5 $6.12 $ 29.00 $20.32 4 $100.28 3 $3.06 $100.03 $79.05 $150.00 2 $4.31 $ 20.05 $46.00 3 $100.00 3 $5.43 $ 40.00 7 ART. 68. RULE FOR NOTATION. Write the dollars as in whole numbers ; place a period on the right of dollars, next to this write the cents, then the mills. If the cents are less than ten, place a cipher next to the dollars; if there are no cents, put two ciphers in the place of cents. EXAMPLES TO BE WRITTEN. 1. Twelve dollars, seventeen cents, eight mills. $12.17 8 2. Six dollars, six cents, six mills $ 6.06 6 3. Seven dollars, seven mills $ 7.00 7 4. Forty dollars, fifty-three cents, four mills. . $40.53 4 5. Two dollars, three cents $ 2 03 6. Twenty dollars, two cents, two mills. . . $20.02 2 7. One hundred dollars, ten cents $100.10 8. Two hundred dollars, two cents $200.02 9. Four hundred dollars, one cent, eight mills. $400.01 8 REDUCTION OF TT. S. MONEY. ART. 69. Reduction consists in changing the denomi- nations or orders, without altering the value. 1st. Reduction Descending is changing numbers from a higher to a lower denomination ; as, from dollars to cents. 2d. Reduction Ascending is changing numbers from a lower to a higher denomination ; as, from cents to dollars. REVIEW. 66. In what denominations are accounts kept? What are eagles and dollars together called? What dimes and cents? How are dollars and cents separated? 67. What places do cents occupy? What place mills ? What is the Rule for Numeration ? 68. How write any sum of U. S. money, Rule ? 69. In what does Reduc- tion consist ? What is Reduction descending ? What ascending ? 74 RAY'S PRACTICAL ARITHMETIC. ART. 70. 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, multiply by 10; that is, annex one cipher. Hence, conversely, to reduce mills to cents, divide by 10; that is, cut off one figure on the right. As there are 100 cents in 1 dollar, in any number of dollars there are 100 times as many cents as dollars. Therefore, to reduce dollars to cents, multiply by 100 ; that is, annex two ciphers. Hence, conversely, to reduce cents to dollars, divide by 100 ; that is, cut off two figures on the right As there are 1000 mills in a dollar, in any number of dollars there are 1000 times as many mills as dollars. Therefore, to reduce dollars to mills, multiply by 1000 ; that is, annex three ciphers. Hence, conversely, to reduce mills to dollars, divide by 1000; that is, cut off three figures on the right. ART. 71. As the operations of reduction consist in - multiplying or dividing by 10, 100, or 1000, the work can be shortened by simply moving the point. ILLUSTRATIONS. In Multiplying, move the point as many places to the right, as there are ciphers in the multiplier. Thus, $2.50 = 250 cents; $2.50 5=2505 mills. In Dividing, move the point as many places to the left, as there are ciphers in the divisor. Thus, 275 cents = $2 . 75 ; 4285 mills = $4.28 5 1. Reduce 17 cts. to mills. Ans. 170 m. 2. Reduce 28 cts. to mills. Ans. 280 m. REVIEW. 70. How are cents reduced to mills ? Why ? Mills to cents ? Dollars to cents? Why? Cents to dollars? Dollars to mills? Why? Mills to dollars ? 71. In what do the operations of reduction consist ? How can the work be shortened ? Illustrate. COMPOUND NUMBERS. U. S. MONEY. 75 3. Reduce 43 cts. and 6 m. to mills. Ans. 436 m. 4. Reduce 70 cts. and 6 in. to mills. Ans. 706 m. 5. Reduce 106 mTto cents. Ans. 10 cts. 6m. 6. Reduce 490 mills to cents. Ans. 49 cts. 7. Reduce 9 dollars to cents. Ans. 900 cts. 8. Reduce 14 dollars to cents. Ans. 1400 cts. 9. Reduce 104 dollars to cents. Ans. 10400 cts. 10. Reduce $60 and 13 cts: to cents. - Ans. 6013 cts. 11. Reduce $40 and 5 cts. to cents. Ans. 4005 cts. 12. Reduce 375 cts. to dollars. Ans. $3.75 13. Reduce 9004 cts. to dollars. Ans. $90.04 14. Reduce 4 dollars to mills. Ans. 4000 m. 15. Reduce $14 and 2 cts. to mills. Ans. 14020 m. 16. Reduce 2465 mills to dollars. Ans. $2.46 5 17. Reduce 3007 mills to dollars. Ans. $3.00 7 18. Reduce 3187 cents to dollars. . Ans. $31.87 ADDITION OF TI. S. MONEY. ART. 72. 1. Add together 4 dollars, 12 cents, 5 mills; 7 dollars, 6 cents, 2 mills; 20 dollars 43 cents; 10 dol- lars, 5 mills; and 16 dollars, 87 cents, 5 mills. RULE 1. Write the numbers to be added, units of the same denominations under each other; dollars under dollars, cents under cents, mills under mills, because only num- bers of the same denomination can be added. 2. Add as in Addition of Simple Num- bers, and -place the separating point directly under the separating points above. OPERATION. $ cts. m. 12 5 06 43 00 87 4 7 20 10 16 2 5 5 $58.49 7 PROOF. The same as in Addition of Simple Numbers. REVIEW. 72. How write numbers in addition of U. S. money? Why? ow is the addition performed ? Why ? Where place the point? 76 ftAY'S PRACTICAL ARITHMETIC. 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? Ans. $68.45 . 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. Ans. 222.19 9 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? Ans. $1077.58 5. Bought a gig for $200 ; a watch for 43.87 5; a suit of clothes for $56.93 7 ; hat for $8.50 ; and a whip for $2.31 3 ; what was the amount? Ans. $311.62 5 6. A person has due him, five hundred and four dollars, six cents, three mills ; $420, 19 cents, 7 mills ; one hundred and five dollars, fifty cents ; $304 and 5 mills ; $888, forty-five cents, five mills : how much is due to him? Ans. $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 --ent, one mill; $1000.10; forty dollars, four mills; and $64.58 7. Ans. 2000. SUBTRACTION OF TJ, S. MONE* ART. 73. From one hundred dollars, five cents, three mills, take $80, 20 cents, 7 mills. Rule. Place the less number under the greater, dollars under dollars, cents inn n^ 3 under cents, &c. Subtract as in Simple Qn ' 90 7 Numbers, placing the separating point _ ! _ tinder the j^oints above. AUK. S 1 9 . 8 4 6 PROOF. As in Subtraction of Simple Numbers. 2. From $29.34 2 take $17.26 5 Ans. $12.07 7 3. From $46.28 take 817.75 Ans. 28.53 4. From $20-05 take $ 5.50 Ans. $14.55 COMPOUND NUMBERS. U. S. MONEY. 77 5. From $3, take 3 cts. . . . Ans. $2,97 6. From $10, take 1 mill. . . . Ans. $9.99 \, 7. From $50, take 50 cts., 5 mills. Ans. $49.49 5 8. From one thousand dollars, take one dollar, onfl cent, and one mill. Ans. $998.98 9 9. B owes 1000 dollars, 43 cents, 5 mills; if he pay , : nine hundred dollars, sixty-eight cents, seven mills, how } much will he still owe? ' Ans. $99.74 8 #4 MULTIPLICATION OF U. S. MONEY. ART. 74. 1. What will 13 cows cost, at 17 dollars, 12 cents, 5 mills each ? SOLUTION. Consider the cost, $17.12 5, as OPERATION. reduced to its lowest denomination, viz.: 17125 $17.12 5 mills. Then, since 13 cows will cost 13 times as 1 3 much as 1, multiply the cost of one, by 13, which - - gives for the cost of 13 cows, 222625 mills, the 51375 product being of the same denomination as the multiplicand, Art. 30. Finally, reduce the mills $222 62 5 to dollars, Art. 70. Hence, the e. Multiply as in Simple Numbers ; the product wiu oe the answer in the lowest denomination of the multiplicand, which may then be reduced to dollars by pointing. PROOF. As in Multiplication of Simple Numbers. 2. Multiply $7, 83 cts. by 8. Ans. $62.64 3. Multiply $12, 9 cts., 3m. by 9. Am. $108 83 < 4. Multiply $23, let., 8m. by 16. Ans. $368.28 8 5. Multiply $35, 14 cts. by 53. Ans. $1862.42 6. Multiply $125, 2 cts. by 62. Ans. $7751.24 7. Multiply $40, 4 cts. by 102. Ans. $4084.08 8. Multiply 12 cts., 5m. by 17. Ans. $2.13 5 9. Multiply $3, 28 cts. by 38. Ans. $124.64 REVIEW. 73. How are numbers written in subtraction ? Why ? How is the subtraction performed? Where is the separating point placed? 74. How is multiplication performed ? How is the product pointed ? 78 RAY'S PRACTICAL ARITHMETIC. 10. What cost 338 barrels of cider, at 1 dollar, 6 cents a barrel? Am. $358.28 11. Sold 38 cords of wood, at 5 dollars, 75 cts. a cord: to what did it amount? Ans. 218.50 12. At 7 cts. a pound, what cost 465 pounds of sugar? . Ans. 832.55 // NOTE. Instead of multiplying 7 cents by 465, multiply 465 by 7, which gives the same product, Art. 30. But, in fixing the denomination of the product, remember that 7 cents is the true multiplicand. 13. What cost 89 yards of sheetrng, at 34 cts. a yard ? Ans. 830.26 14. What will 24 yards of cloth cost, at 5 dollars, 67 3cnts a yard ? Ans. 136.08 15. I have 169 sheep, valued at 2.69 each: what is the value of the whole? Ans. 454.61 16. If I sell 691 bushels of wheat at 1 dollar, 25 cts. a bushel, what will it amount to ? Ans. 863.75 17. I oold 73 hogsheads of molasses, of 63 gallons each, at 55 cts. a gallon : what is the surn? Ans. 2529.45 18. What cost 4 barrels of sugar, of 281 pounds each, at 6 cents, 5 mills a pound? Ans. 73.06 19. Bought 35 bolts of tape, of 10 yards each, at 1 cent a yard : what did it cost? Ans. 3.50 20. If I earn 13 cts. an hour, and work 11 hours a day, how much will I earn in 312 days? Ans. 446.16 21. I sold 18 bags of wheat, of 3 bushels each, at 1.25 a bushel : what is the amount? Ans. 67.50 22. What cost 150 acres of land, at 10 dollars, 1 mill per acre? Ans. 1500.15 23. What cost 17 bags of coffee, of 51 pounds each, at 14 cents, 7 mills per pound? Ans. 127.44 9 ART. 75. DIVISION OF u, s. MONEY. The object in Division of United States money is, 1st. To find how many times one sum of money is contained in another of the same order or denomination. Or, 2d. To divide a sum of money into a given number of equal par is ^ Art. 41. COMPOUND NUMBERS. U. S. MONEY. 79 1. How much cloth at 7 cts. a yard, will SI. 75 buy,? ANAL. As 1 yard costs 7 cts., there will be as many yards as 1 cts. are contained, times in 175 cts. 1757=25. Here, the divisor and dividend are the same denomination, cents ; and the quotient, how many yards, is an abstract number. 2. Divide 65 dollars equally among 8 persons. SOLUTION. In this case it is required to OPERATION. divide $G5 into 8 equal parts, that is, to find $ cts. m. one-eighth of it. One-eighth of $65 is $8, with a 8)65.00 remainder $1 = 100 cts. One-eighth of 100 cts. is 12 cts., with a remainder of 4 cts. which equals o . 1 2 5 Mills. ^O mills. One-eighth of 40 mills is 5 mills; hence, 8)65000 one-eighth of $65 is $8.12 5 ~ The operation may be performed by reducing the dollars to mills, then dividing by 8, and after this, $8.12 5 reducing the quotient to dollars. 3. A farmer received 29.61 cents, for 23 bi^hels of wheat : how much was that per bushel ? SOLUTION. To divide $29.61 OPERATIC*. into 23 equal parts, annex a 23)29610(1287 mills. cipher, which reduces it to 29610 23 - n ft pk g )252qt> j pt> this remainder is 1 pint. To reduce quarts to pecks, di- Pk. in a bu. 4)31 pk. 4 qt. vide by 8, and 4 quarts are left. ~ , o u To reduce pecks to bushels, di- vide by 4, and 3 pecks are left. Ans. 7 bu. 3 pk. 4 qt. 1 pt. The remainder is always of the same denomination as the dividend, Art. 38. Hence, ART. 81. GENERAL RULES. To REDUCE FROM A HIGHER TO A LOWER ORDER, Bllle. Multiply the highest denomination given, by that num- ber of the next lower, which makes a unit of the higher ; add to the product the number, if any, of the lower denomination. Proceed in like manner with the result thus obtained, till the whole is reduced to the required denomination. To KEDUCE FROM A LOWER TO A HIGHER ORDER, Rule. Divide the given quantity by that number of its own denomination which makes a unit of the next higher. Proceed in like manner with the quotient thus obtained, till the whole is reduced to the required denomination. The last quotient, with the several remainders, if any, annexed, will be the answer. PROOF. Reverse the operation : that is, reduce the answer back to the denomination from which it was derived. If this result is the same as the quantity given, the work is correct. 11. Reduce 2 bu. to pints Ans. 128 pt. 12. 12 pk. to pints Ans. 192 pt. 13. 8 bu. to quarts. .... Ans. 256 qt. 14. 1 bu. 1 pk. to pints. . . . Ans. 80 pt. 15. 2 bu. 2 qt. to pints. . . . Ans. 132 pt. EEVIEW. 81. What is the general rule for reducing from a higher to a lower order ? From a lower to a higher ? What the method of proof ? REDUCTION OF COMPOUND NUMBERS. 87 16. Reduce 4 bu. 2 pk. 1 qt. to pints. Ans. 290 pt. 17. 7 bu. 3 pk. 7 qt. 1 pt. to pt. Ans. 511 pt. 18. 3 bu. 1 pt. to pints. ... Ans. 193 pt. 19. 384 pt. to bushels. . . . Ans. 6bu 20. 47 pt. to pecks. Ans. 2 pk. 7 qt. 1 pt. 21. 95 pt. to bu. Ans. 1 bu. 1 pk. 7 qt. 1 pt. 22. 508 pt. to bu. Ans. 7 bu. 3 pk. 6 qt. v>4, ^ ART. 82. TROY OR MINT WEIGHT Is used in weighing gold, silver, jewels, liquors, &o. TABLE. 24 grains (gr.) make 1 pennyweight, marked pwt 20 pennyweights ... 1 ounce, ...... ...... oz. 12 ounces ........ 1 pound, .......... o Ib NOTE. The standard unit of weight in the United States, is the pound, containing 5760 grains. For interesting historical and other information with respect to coins, see " Ray's HigJier Arithmetic" TEACHERS should ask questions on each Table, as on Dry Measure. 1. Reduce 13 Ib. 11 oz. 16 pwt. 14 gr. to grains. SUGGESTION. "When the de- 167 oz. (Brought up.) nominations to be added are 20 small, add while multiplying; when large, beginners should oo4U pwt. add after multiplying. * P wt - to be added - OPERATION. 3356 P wt Ib. oz. pwt. gr. _ _ 13 11 16 14 13424 12 6712 156 oz. 80544 gr. 11 oz - to be added - 1 4 gr. to be added, 1 6 7 oz. ( Carried up.) 80558 gr. KEVIEW. 82. For what is Troy Weight used? Eepeat the Tabla, NOTE. What is the standard unit of weight in the United States ? 88 RAY'S PRACTICAL ARITHMETIC. 2. Reduce 4 Ib. to grains. . . . Am. 23040 gr. 3. 5 Ib. 4 oz. to ounces. . . . Ans. 64 oz. 4. 9 Ib. 3 oz. 5 pwt. to pwt. . . Ans. 2225 pwt. 5. 14 Ib. 11 oz. 19 pwt. 23 gr. to gr. Ans. 86399 gr. 6. 8 Ib. 9 oz. 13 pwt. 17 gr. to gr. Ans. 50729 gr. 7. 171 gr. to pennyweights. Ans. 7 pwt. 3 gr. 8. 505 gr. to ounces. Ans. 1 oz. 1 pwt. 1 gr. 9. 12530 gr. to pounds. Ans. 2 Ib. 2 oz. 2 pwt. 2 gr, 10. 805 pwt. to pounds. Ans. 3 Ib. 4 oz. 5 pwt. 11. 25591 gr. to pounds. Ans. 4 Ib. 5 oz. 6 pwt. 7 gr. ART. 83. APOTHECARIES WEIGHT Is used by Apothecaries in compounding medicines. TABLE. 20 grains (gr.) make 1 scruple, marked 9. 3 scruples 1 dram, 3* 8 drams 1 ounce, g. 12 ounces 1 pound, Ib. The grain, ounce, and pound, in Apothecaries and Troy weight, are tin same : but the ounce is differently divided. feUF Questions should be asked on the Table, as before. 1. Reduce 3R> to grains. . . . Ans^ 17280 gr. 2. 4 Ib 5 5 2 gr. to grains. . . . Am. 23342 gr. 3. 7 Ib 2 9 to scruples. . . . Am. 2018 9. 4. 7K> 2 g 1 9 to grains. . . . Ans. 41300 gr. 5. 67 g to pounds Ans. 5 Ib 7 g. 6. 431 3 to pounds Ans. 4ft5g73. 7. 9759 to pounds Ans. 31b 4g 55. 8. 6321 gr. to pounds. Ans. Ilbl15l9lgr. 9. 30941 gr. to pounds. Ans. 5R>4g33291gr. 10. 29239 gr. to pounds. . . Ans. 51b 7 5 19 gr. EEVIEW. 83. For what is Apothecaries "Weight used? Eepeat the Table. What is said of the grain, ounce, and pound, of Apothecaries and Troy weight ? What is differently divided ? ^. REDUCTION OF COMPOUND NUMBERS. 8fc ART. 84. AVOIRDUPOIS WEIGHT Is used in weighing heavy articles ; as, groceries, coarse metals and medicines at wholesale. TABLE. 16 drains (dr.) . . . make 1 ounce,. . marked oz. 16 ounces 1 pound, Ib. 25 pounds . 1 quarter, . . . . qr. 4 quarters or 100 Ib. . . 1 hundred weight, . cwt. 20 hundred weight ... 1 tun, . . . . . . T. NOTES. 1. The standard Avoirdupois pound of the United States is determined from the Troy pound, and contains 7000 grains Troy. 2. Formerly, 28 pounds were allowed for a quarter, 112 pounds for ... hundred weight, 2240 pounds for a tun ; these, called the long hundred and long tun, are chiefly used at the Custom House. 1. Reduce 2 cwt. to pounds. . . . Ans. 200 Ib. 2. 3 cwt. 3 qr. to pounds. ... Ans. 375 Ib. 3. 1 T. 2 cwt. to pounds. . . . Ans. 2200 Ib. 4. 3 T. 3 qr. to pounds. . . . Ans. 6075 Ib. 5. 4 cwt. 1 qr. 19 Ib. to pounds. . Ans. 444 Ib. 6. 5T. 3qr. 15 Ib. to pounds. . . Ans. 10090 Ib. 7. 2 cwt. 3 qr. 2 Ib. 12 oz* to ounces. Ans. 4444 oz. 8. 2 cwt. 17 Ib. 3 dr. to drams. . . Ans. 55555 dr. 9. IT. 6 cwt. 41b. 2oz. 10 dr. to dr. Ans. 666666 dr. 10. 4803 Ib. to cwt Ans. 48 cwt. 3 Ib. 11. 22400 Ib. to tuns Ans. 11 T. 4 cwt. 12. 2048000 dr. to tuns. . . . Ans. 4 T. 13. 64546 dr. to cwt. Ans. 2 cwt. 2 qr. 2 Ib. 2 oz. 2 dr. 14. 97203 oz. to tuns. Ans. 3T. 3qr. 3oz. 15. 544272 dr. to T. Ans. IT. Icwt. Iqr. 1 Ib. loz. 16. In 52 parcels of sugar, each containing 18 Ib., how many hundred weight? Ans. 9 cwt. 1 qr. 11 Ib. REVIEW. 84. For what is Avoirdupois Weight used? Eepeat the Table. NOTE 1. From what is the standard Avoirdupois pound deter- mined ? 2. What is the long hundred and long tun ? Where used ? 90 RAY'S PRACTICAL ARITHMETIC. ART. 85. LONG OR LINEAR MEASURE: Used in measuring lengths, breadths, thickness, &c. TABLE. 12 inches (in.). . make 1 foot, . . marked ft. 3 feet ....... 1 yard, ...... yd. 5 J yards or 1 6 J feet . . 1 rod, perch, or pole, rd. 40 rods or 220 yards . 1 furlong, ..... fur. 8 furlongs or 1760 yd. = 5280ft. ) -, ., j ' ' l mile ' ALSO, 60 geographic, or 69J statute miles, make 1 degree. 360 degrees make a great circle, or circumference of the earth. 3 miles make 1 league, used in measuring distances at sea. 4 inches make 1 hand, used in measuring the height of horses 6 feet make 1 fathom, used in measuring the depth of water. NOTE. The standard unit of length is the yard. 1. Eeduce 2yd. 2ft. Tin. to inches. Ans. 103 in. 2. 7yd. 11 in. to inches ..... Ans. 263 in. 3. 12 mi. to rods ..... . ... ute. 3840rd. 4. 7 mi. 6 fur. to rods ..... Am. 2480 rd. 5. 9 mi. 31 rd. to rods ..... Am. 2911 rd. 6. 133 in. to yards. . . . Ans. 3 yd. 2 ft. 1 in. 7. 181 in. to yards ..... , Ans. 5 yd. 1 in. 8. 2240 rd. to miles ....... Am. 7 mi. 9. 2200 rd. to miles ..... Ans. 6 mi. 7 fur. 8 For examples involving Fractions, see page 170. ART. 86. LAND OR SQUARE MEASURE Is used in measuring land, or any thing in which both length and breadth are considered. 1 inch. ART. 87. A figure having four equal sides, and four right angles (corners), is a square. Hence, A square inch is a square, each side of which is a linear inch; that is, 1 inch in length. REVIEW. 85. For what is Long Measure used? Repeat the Table. NOTE. What is the standard unit of length ? REDUCTION OF COMPOUND NUMBERS. 91 3 feet. 3 square in. A square foot is a square, each side of which is a linear foot. A square yard is a square, each side of which is a linear yard (3 feet). The figure shows that 1 square yard, that is, 3 feet square, contains 9 square feet. The number of small squares in any large square, is equal to the number of units in one side of the large square multiplied by itself. Thus, In a square figure, each side of which is 8 inches, there are 64 square inches : in 1 foot square, or 12 inches on each side, there are 12X 12 144 square in. 3 in. square. ART. 88. By 3 inches square, we mean a square figure, each side 3 inches : but, 3 square inches are 3 small squares, each an inch long, and an inch wide ; and, 3 inches square contain 9 square inches. The difference between 4 in. square, and 4 square in., is 12 square inches; between 6 miles square, and 6 square miles, is 30 sq. miles. TABLE. 144 square inches make 1 square foot, marked sq. ft. square feet , . . . 1 square yard, . , . sq. yd. SQl square yards . . . 1 sq. rod or perch, . . P. 40 perches 1 rood, ...... R. 4 roods 1 acre, A. 640 acres ...... 1 square mile, e . . sq. mi. ALSO, 7 T 9 er of times it contains its unit of measure. The unit of measure for surfaces, is a square whos? dde is a linear unit, as a square inch, a square foot, &c. 1. How many square inches in a board 4 incnes long and 3 inches wide ? SOL. Dividing each of the longer sides into 4 equal parts, the shorter sides into 3 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 the board is equal to the pro- duct obtained by multiplying together the numbers representing the length and breadth; that is, 4 X 3 = 12. Hence the RULE FOR FINDING THE AREA OF A RECTANGLE. Multiply the length by the breadth: the product will be the superficial contents, REVIEW. 88. What is meant by 3 inches square ? By 3 sq. inches ? What is the difference between 4 inches square, and 4 square inches ? 89. What is a Rectangle ? What is the Area of a figure ? What is the unit of measure for surfaces ? REDUCTION OF COMPOUND NUMBERS. 93 NOTE. Both the length and bread tn, if not in units of the same denomination, must be made so, before multiplying. 2. In a floor 16 feet long and 12 feet wide, how many square feet ? Ay*. 192 sq. ft. 3. How many square yards of carpeting will cover a room 5 yds. long and 4 yds. wide ? Ans. 20 sq. yd. 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 ? Ans. 59 sq, yd. 5. How many square yards in a ceiling 18 feet long and 14 feet wide ? Ans. 28 sq, yd. 6. In a field 35 rods long and 32 rods wide, how many acres ? Ans. 7 A. ART, 90. The Area of a Rectangle being equal to the product of the length by the breadth, and as the product of two numbers, divided by either of them, gives the other (Art. 37) ; therefore, Rule. If the area of a rectangle be divided by either side, the quotient will be the oilier side. ILLUSTRATION. In Example 1, Art. 89, 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. NOTE. Dividing the area of a rectangle by one of its sides ; is really dividing the number of squares in the rectangle by the number of squares on one of its sides. Thus, In dividing 12 by 4, the latter is not 4 linear inches, but the number of inches in a rectangle 4 in. long and 1 in. wide. See figure, Art. 89. 1. A floor containing 132 square feet, is 11 feet wide : what is its length ? Ans. 12 ft. 2. A floor is 18 feet long, and contains 30 square yards : what is its width? Ans. 15 ft. 3. A field containing 9 acres, is 45 rods in length : what is its width ? Ans. 32 rd. 4. A field 35 rods wide, contains 21 acres : what is its length ? Ans. 96 rd. REVIEW. 89. What is the Rule for finding the Area of a Rectangle ? 90. The area and one side being given, how may the other side be found ? 94 BAY'S PRACTICAL ARITHMETIC. ART. 91. SOLID OR CUBIC MEASURE: Used in measuring th^gs having length, breadth^ and thickness; as, timber, stone, earth. ART. 92. A Cube is a SOLID, hav- ing 6 equal faces, which are squares. 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 in the figure, it is a cubic or solid yard. The base of a cube, being 1 square yard, contains 3X3 = 9 square feet; and 1 foot high on this base, contains 9 solid feet ; 2 feet high contain 9 X 2 = 18 solid feet ; 3 feet high contain 9X3 =27 solid feet. Also, it maybe shown that 1 solid or cubic foot contains 12X12X12 = 1728 solid or cubic inches. Hence, the number of small cubes in any large cube, is equal to the length, breadth, and thickness, multiplied together. ART. 93. Any solid, whose corners resemble a cube, is a rectangular solid : boxes and cellars are of this form. 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. cu.yd. tn. 27 cubic feet . < .1 cubic yard, . 40 feet of round or ) 50 feet of hewn timber j * ' ' J an > 128 cubic feet 8X4X4 = 8 ft. ), , n long, 4 ft. wide, and 4ft. high, j L ra ' NOTE. 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. 1. Reduce 2 cu.yd. to cubic inches. Ans. 9331 2 eu. in. 2. 28 cords of wood to cu.ft. Ans. 3584 cu. ft. REVIEW. 91. For what is Cubic Measure used ? 92. What is a cube ? A cubic inch? Cubic yard? How many sq, ft. in one side of a cubic yard ? How many cubic feet in a solid 3 ft, long, 3 ft. wide, and 1 ft. thick? If 2 ft. thick? If 3 ft. thick? REDUCTION OF COMPOUND NUMBERS. 95 3. Reduce 34 cord of wood to C. ft. Ans. 272 C. ft. 4. 1 cord of wood to cu, in. Am. 221184 cu. in. 5. 63936 cu. in. to cu. yd. Ana. 1 cu. yd. 10 cu. ft 6. 492480 cu., in. to tn. round timber. Ans. 7 tn. 5 cu. ft. 7. How many cubic feet in a rectangular solid, 8 ft long, 5 ft. wide, 4 ft. thick ? See Art. 93. Ans. 160 cu. ft. 8. How many cubic yards of excavation in a cellar 8 yd. long, 5 yd. wide, 2 yd. deep? Ans. 80 cu. yd. 9. How many cubic yards in a cellar, 18 feet long, 15 feet wide, 7 feet deep ? Ans. 70 cu. yd. 10. In a pile of wood 40 feet long, 12 feet wide, and 8 feet high, how many cords ? Ans. 30 C. ART. 94. CLOTH MEASURE lb viood in measuring cloth, muslin, ribbons, tape, &c. TABLE. 2| inches (in.) make 1 nail, . . . marked na. 4 nails (or 9 in.) . . 1 quarter of a yard, qr. 4 quarters (36 in.) .1 yard, yd. 3 quarters 1 Ell Flemish, . . . . E. Fl. 5 quarters 1 Ell English, . . . . E. En. 6 quarters 1 Ell French, . . . . E. Fr. This is a species of long measure, the yard being 36 inches in both. 1. Reduce 19yd. to nails Ans. 304 na. 2. 14E.F1. to nails. . ..... Ans. 168 na. 3. 5 yd. 2 qr. 3 na. to nails. . . . Ans. 91 na. 4. 13 E. En. 1 qr. to nails. . . . Ans. 264 na. 5. 23 E. Fr. 3 qr. 2 na. to nails. . . Ans. 566 na. 6. 159 na. to yards. . . . Ans. 9 yd. 3 qr. 3 na. 7. 287 na. to E. Fr. . . Am. 11 E. Fr. 5 qr. 3 na. 8. 6 yd. to Ells Flemish Ans. 8 E. Fl. Reduce the yards to quarters ; then the quarters to Ells Flem. ./ *' EEYIEW. 92. To what is the number of small cubes in any large cube equal? 03, What is a rectangular solid? How is the solid contents found ? Kepeat the Table. 94. For what is'Cloth Measure used? Eepeat the Table. Of what is Cloth Measure a species ? 96 RAY'S PRACTICAL ARITHMETIC. 9. Keduce 9yd. 3qr. to Ells Flemish. Am. 13E.FL 10. 12 yd. 1 qr. to E. Fl. . . Ans. 16 E. Fl. 1 qr. 11. 37 E. Fl. to yards Ans. 27 yd. 3 qr. 12. 36 E. En. to Ells French. . . . Ans. 30 E. Fr. 13. 22 E. En. 4 qr. to Ells Fr. . . Ans. 19 E. Fr. 14. 47 E. En. to yards. . . .' Ans. 58 yd. 3 qr. t ART. $5. WINE OE LIQUID MEASURE: For measuring all liquids except beer, milk, and ale TABLE. 4 gills (gi.) make 1 pint, marked pt. 2 pints 1 quart, . . . . qfc. 4 quarts 1 gallon, .... gaL 63 gallons 1 hogshead, . . . hhd. 4 hogsheads ... 1 tun, T. Also, 3 1-J gallons ..... 1 barrel, . . . . bL 42 gallons 1 tierce, . . . . tr. 84 gallons 1 puncheon, . . pn. 126 gallons 1 pipe, p. The Standard Unit of Liquid Measure is a gallon of 231 cubic inches. 1. Eeduce 17 gal. to pints. . . . Ans. 136 pt. 2. 13 gal. to gills Ans. 416 gi. 3. 2 hhd. to pints Ans. 1008 pt. 4. 5 T. to gills Ans. 40320 gi. 5. 3T. 3 hhd. to gallons Ans. 945 gaL 6. 1 hhd. 60 gal. 1 pt. to pints. . . Ans. 985 pt. 7. 2 hhd. 17 gal. 3qt. to gills. . . Ans. 4600 gi. 8. 2T. 62 gal. 1 pt. to gills. . . Ans. 18116 gi 9. 96 gi. to gallons Ans. 3 gal. 10. 6048 gi. to hogsheads Ans. 3 hhd. 11. 32256 gi. to tuns Ans. 4T. 12. 4050 gi. to hogsheads. Ans. 2 hhd. 2 qt, 2 gi. BE VIEW. 95. For what is Liquid Measure used ? Eepeat the Table. NOTE. What is the standard unit of Liquid Measure ? REDUCTION OF COMPOUND NUMBEBS. 07 13. Reduce 30339 gi. to T. Ans. 3T. 3 hhd. 3 gal. 3gi. 14. 10125 gi. to T. Ans.lT. Ihhd. Igal. Iqt. Ipt. Igi. 15. 3 puncheons to gills ..... Ans. 8064 gi. 16. 5 pipes to quarts ...... Ans. 2520 qt. 17. 5712 pt. to tierces ....... Ans. 17 tr. ART. 96. ALE OB BEER MEASURE Is generally used in measuring ale, beer, and milk. TABLE. 2 pints (pt.) make 1 quart, marked qt 4 quarts ..... 1 gallon, .... gal. 36 gallons ..... 1 barrel, .... bl. 54 gallons ..... 1 hogshead, . . hhd. NOTE. The beer gallon contains 282 cubic inches. In most placet milk is now sold by Wine Measure. 1. Reduce 4 hhd. to pints ..... Ans. 1728pt. 2. 7 hhd. 3 qt. to pints ..... Ans. 3030pt. 3. 1000 pt. to barrels ..... Ans. 3bl. 17gal. 4. 443pt. to hhd. . Ans. Ihhd. Igal. Iqt.-lpt. ART. 97. TIME MEASURE. Time is a measured portion of duration. The parts into which time is divided are shown in this TABLE. 00 seconds (sec.) make 1 minute, marked min. GO minutes ...... 1 hour, .... hr. 24 hours ....... 1 day, ..... da. 305 days 6 hours, (365 J da ) 1 solar year, . . . yr, 100 years ....... 1 century, .... cea Also, 7 days - ...... 1 week, ..... wk. 4 weeks . . . . ... 1 month, (nearly) mon. 1*2 calendar months ... 1 year, ..... yr. 105 days ....... .1 common year, . yr. 30i> days ........ 1 leap year, . . . yr. Sec Notes on page 98. REVIEW. 96. For what is Beer Measure used? Repeat the Table. )7. What, is Time? Repeat the Table. 3d Bk. 7 08 RAY'S PRACTICAL ARITHMETIC, NOTE 1. The EXACT length of the SOLAR year is 365 days, 5 hours, 48 minutes, 48 seconds: but, is usually considered to be 865 days, 6 hours. Hence, One year being regarded as 305 days, 6 hours, the odd G hours of each year, make, in 4 years, 24 hours, an additional day. This gives 360 days to every fourth year, called LEAP YEAR : and, All leap years may be exactly divided by 4; thus, 1848, 1852, 1856, are leap years; while 1847, 1850, 1855, are not so: and, The additional day in leap years is added to February, making this month 29 days, instead of 28, as in common years. NOTE 2. The difference between the SOLAR year and 365 days and 6 hours, is about three-fourths of a day in 100 years, or 3 days in 400 years. Hence, For greater accuracy, it is agreed that of the centennial (hundredth) years, only those which are exactly divisible by 400 shall' be leap years: thus, 1900 will be a common 3 r ear, and 2000 a leap year. Teachers and advanced students will find interesting and instructive historic information on this subject in " Itay's Higher Arithmetic" NOTE 3. The year, as used by civilized nations, is divided into 12 calendar months, and numbered in their order as follows : January, lstmonth,cl days. July, 7th month, 31 day& February, 2d .. 28 March, 3d .. 31 April, 4th .. 30 May, 5th .. 31 June, 6th .. 30 August, 8th .. 31 September, 9th .. 30 October, 10th .. 31 November,! 1th .. 30 December, 12th .. 31 Thirty days hath September, April, June, and November : All the rest have thirty-one, save February, which alone Hath twenty-eight ; and one day more we add to it, one year in four. 1. Reduce 2hr. to seconds. . . . Ans. 7200 sec. 2. 7 da to minutes. . . . . . Ans. 10080 min. 3. Ida. 3hr. 44 min. 3sec. to sec. Ans. 99843scc. 4. 9wk. 6 da. 10 hr 40 min. to min. Ans. 100000 min. REVIEW. 97. NOTE 1. What is the exact length of the solar year? What are leap years? February has how many days in lenp years? 2 Which of the centennial years are leap years ? Why ? 3. How is the year divided ? Give tho number of days in each month. REDUCTION OF COMPOUND NUMBERS. 99 5. Reduce Imon. 3 da. 4min. to min. Ans. 44644mm. 6. -Reduce 1 yr. 20 da. 19 hr. 15 min. 33 sec. to seconds, allowing 365 days to a year. Ans. 33333333 sec. 7. How many seconds in an exact solar year ? (See Note 1, page 98.) Ans. 31556928 sec. 8. Reduce 10800 sec. to hours. . . . Ans. 3hr. 9. 432000 sec. to days ..... Am. 5 da. 10. 7322 sec. to hours. . Ans. 2hr. 2 min. 2 sec. 11. 4323 min. to days. . . . Ans. 3 da. 3 min. 12. 20280mm. to weeks. . . Ans. 2wk. 2hr. 13. 41761 min. to mon. Ans. Imon. Ida. ^ ART. 98. CIRCULAR MEASURE Is used in estimating latitude and longitude, and in measuring the motions of the heavenly bodies. A circle is divided into 360 equal parts, called degrees; cc^h degree into 60 parts, called minutes; each minute into 60 parts, called seconds. TABLE. 60 seconds ( /x ) make 1 minute, marked '. 60 minutes ... . .1 degree, . . . . . 30 degrees ..... 1 sign, ..... s. 12 signs or 360 ... 1 circle, . . . . c. 1. Reduce 5 3' to minutes. . . . Ans. 303'. 2. 8 41' 45" to seconds. . . . Ans. 31305". 3. 3 9 25' to minutes ..... Ans. 5425'. 4. 1" to seconds. .... Ans. 1296000". 5. 214" to minutes. ..... Ans. 4' 4". 6. 915' to degrees. . . . . Ans. 15 15'. 7. 1861' to signs ..... Ans. 1 s 1 1'. ART. 99. MISCELLANEOUS TABLE. 24 sheets of paper .... make 1 quire. 20 quires . ....... .... 1 ream. 2 reams .......... < . 1 bundle. REVIEW, 98. For what is Circular Measure used? How is every Circle divided ? Eepcat the Table. 100 RAY'S PRACTICAL ARITHMETIC. OP BOOKS. A sheet folded in 2 leaves is called a folio. A sheet folded in 4 a quarto, or 4to. A sheet folded in 8 . . an octavo, or 8vo. A sheet folded in 12 . a duodecimo, or 12mo. A sheet folded in 18 an 18mo. OF THINGS. 12 things make 1 dozen. 12 dozen 1 gross. 12 gross, or 144 dozen 1 great gross. 20 things 1 score. 19(5 pounds of flour 1 barrel. 200 pounds of beef or pork 1 barrel. 100 pounds of fish 1 quintal. 18 inches 1 cubit. 22 inches (nearly,) 1 sacred cubit ART. 100. PROMISCUOUS EXAMPLES. 1. What cost 2bu. plums, at 5 cts. a pt. ? Arts. 86.40 2. 3bu. 2pk. peaches, at 50 cts. a pk. ? Ans. $7-' 3. 3pk. 3qt. barley, at Sets, a pt. ? Ans. 81.62 4. At 15 cents a peck, how many bushels of apples "can be bought for 3 ? Ans. 5 bu. 5. If salt cost 2 cents a pint, how much can be bought with SI . 66 ? Ans. I bu. 1 pk. 1 qt, 1 pt. 6. I put 91 bushels of wheat into bags of 3bu. 2pk. each : how many bags were required ? Ans. 26. Reduce both quantities to pecks, and then divide. 7. What is the value of 1 Ib. 3pwt. of gold ore, at 3 cents a grain? Ans. 174.96 8. How many spoons, each weighing 2oz. 5pwt., can be made from 21b. 5oz. 5pwt. of silver? Ans. 13. 9. How many rings, weighing 5 pwt. 7 gr. each, can be made from 1 Ib. 8 oz. 18pwt. Igr. of gold? Ans. 79. 10. If lib 45 of calomel be divided into doses of 15 grains each, and sold at 12 cts. 5 in. a dose, what will it amount to ? Ans. 50. REDUCTION OF COMPOUND NUMBERS. 101 WHAT WILL BE THE COST OF 11. lib Ig 13 opium, at4cts. a 9? Am. 812.60 12. 6cwt. Iqr. raisins, at Bets, a Ib. ? Ans. $18.75 13. IT. Icwt. rice, at $2.25 a qr. ? Ans. $189. 14. 71b. 8oz. copper, at 5cts. an oz.? Ans. $6. 15. A physician put up 316 doses of rhubarb, of 20 gr oach : how much did he use ? Ans. 1 Ib 1 1 3 1 9- 16. How many nails, weighing 4 drams each, are in a parcel weighing 15 Ib. 9oz. 12 dr. ? Ans. 999. 17. I bought 44cwt. 2qr. 21b. of cheese; each weighed 91b. 15 oz. : how many cheese did I buy? Am. 448. 18. How many kegs, of 84 Ib. each, can be filled from a hhd. of sugar of 14 cwt. Iqr. 31b. ? Am. 17. 19. How many boxes, containing 12 Ib. each, can be filled from 7 cwt. 2qr. 61b. of tobacco ? Ans. 63. 20. If a family use 31b. 13 oz. of sugar in a week, how long will 6 cwt. 10 Ib. last them? Ans. 160 wk. 21. What will 2 A. 3R. 5 P. of land cost, at 20 cts. a perch? Ans. $89. 22. What cost 2sq. yd. 2sq ft. of ground, at 5 cents a square inch ? Ans. $144. 23. How many leaves, 3 in. long and 2 in. wide, can be cut from Isq. yd. of paper? Am. 216. 24. A farmer has a field of 16 A. 1R. 13 P., to divide into lots of 1A. IE. IP. each: how many lots will it make? Ans. 13. ^25. How many cu. in. in a block of marble 2ft. long, 2ft. high, 2ft. wide? Ans. 13824cu.in., orScu. ft. 26. One cubic foot of water weighs 1000 oz. Avoirdu- pois : what do 5 cu. ft. weigh ? Ans. 312 Ib. 8oz. 27. What is the weight of a quantity of water occupy- ing the space of 1 cord of wood, each cu. ft. of water weigh' ing 1000 oz. Avoirdupois? Am.. 4T. 28. A cubic foot of oak weighs 950 oz. Avoirdupois : what do 2 cords of oak weigh? Am. 7T. 12 cwt. 29. How many pieces, of 13yd. 3qr. 2na. each, are there in 666 yards ? Ans. 48. 30. How many suits of clothes, each containing 5yd. Iqr., can be made from 147 yards? Ans. 28, 102 RAY'S PRACTICAL ARITHMETIC. 31. How many coat patterns of 2yd. Iqr. Ina. each t can be cut from 37yd. of cloth? Ans. 16. 32. How many suits of clothes of 3yd. 2qr. each, can be cut from 70 Ells Flemish of cloth ? Ans. 15. 33. Fina the cost of 1 hogshead of wine, at 5 cents a gill. Ans. $100.80 34. Find the cost of 5 bl. of molasses, each containing 31 gal. 2qt., at 10 cents a quart. Ans. 63. 35. At 5 cents a pint, what quantity of molasses can be bought for $2 ? Ans. 5 gal. 36. At let. 5m. a gill, how much cider can be bought for 12? Ans. 25 gal. 37. How many dozen bottles, each bottle holding 3qt. Ipt., can be filled from Ihhd. of cider? Ans. 6 doz. 38. How many kegs of 4 gal. 3qt. Ipt. each, can be filled from 1 wine hhd. ? Ans. 12 ; and 4 gal. 2qt. left. 39. How many bottles, each holding Igal. Iqt. Ipt., can be filled from 165 wine gal.? Ans. 120. 40. What will 1 hogshead of beer cost, at 3 cents a quart? Ans. 6.48 41. The human heart beats 70 times a minute : how many times will it beat in a day? Ans. 100800. 42. How many seconds in the month of February, 1840 ? (See Note l v page 98.) Ans. 2505600 sec. 43. How many hours longer is January than February, A. D. 1839? (See Note 1, page 98.) Ans. 72 hr. 44. What are the leap years between 1837 and 1850? (See Note 1, page 98.) Ans. 1840, 1844, 1848. ^45. The exact length of the solar year is 365 da. 5 hr. 48min. 48 sec. : how many such years are contained in 1609403328 seconds? Ans. 51 years. 46. How much time will a man gain in 60 yr., of 3G5 da. each, by rising 30 min. earlier each day? Ans. 456 da. Ghr. 47. If a ship sail 8 miles an hour, how many miles will she sail in 3wk. 2 da. 3hr. ? Ans. 4440 mi. 48. If a planet move through 2 in a day, how long will it require to move tjirough 2 9 4? 4*. 32 da, ADDITION OF COMPOUND NUMBERS. 103 49. In what time will one of the heavenly bodies move through a quadrant, (90), at the rate of 43' 12" per minute? Ans. 2hr. 5min. WHAT WILL BE THE COST 50. Of 2 reams paper, at 20 cts. a quire? Ans. $8. 51. 3 quires paper, at 2 cts. a sheet? Ans. $1.44 52. 3 dozen apples, at 2 cts. each ? Ans. 72 cts. 53. 4 doz. doz. oranges, at 3 cts. each? Ans. $17.28 54. 5 gross buttons, at 5 cts. a doz. ? Ans. $3. 55. 1 bl. of flour, at 4 cts. a pound? Ans. $7.84 5G. 1 bl. pork, at 12 cts. 5 m. a pound? Ans. $25, 57. A farmer started for market with G dozen dozen ~ ~ 7 the 1 (quart) to the column of quarts. Then, adding the numbers in the column of quarts, the sum >s 10 quarts, which, being reduced, make 1 peck and 2 quarts; write the 2 quarts in the column of quarts, and carry the 1 (peck) to t 1 -" column of pec^s* T.he number iu the column of pecks added. 104 RAY'S PRACTICAL ARITHMETIC. make 5 pk., which reduces to 1 bushel and 1 pk. ; write the 1 pk. in the column of pecks, and carry the 1 bu. to the column of bushels : the sum of the numbers in this column is 72, which is written in the same manner as the sum of the numbers in the left hand column in Simple Addition. bu. pk. qt pt. bu. pk. qt pt. (2)3201 (3.) 7371 4061 G 2 1371 9241 9 2 G 1 24 4 Rule for Addition,!. Write the numbers to be added, placing units of the same denomination under each other. 2. Begin with the lowest order, add the numbers, and divide their sum by the number of units of this denomination, which make a unit of the next higher. Write the remainder under the column added, and carry the quotient to the next column. 3. Proceed in the same manner with all the columns to the last f under which write its entire sum. PROOF. The same as in Addition of Simple Numbers. NOTE. In writing Compound Numbers, if any intermediate denomination is wanting, supply its place with a cipher. R EM . In adding Simple Numbers, we carry one for every ten, as 18, 47 ll' 44 . , 47 11' 44 Yi-V ^^ // - 106 ' RAY'S PRACTICAL ARITHMETIC. 21 Five loads of wheat measured thus: 21 bu. 3pk. 2qt. Ipt.; 14 bu. 5qt. ; 23 bu. 2pk. Ipt. ; 18 bu. Ipk. Ipt.; 22 bu. 7qt. Ipt. : how much in all? Ans. lOObu. 22. A farmer raised of oats 200 bu. 3pk. Ipt.; barley, 143 bu. 2qt, Ipt,; corn, 400 bu. 3pk. ; wheat, 255 bu. Ipk. 5qt. : how much in all? Ans. 1000 bu. 23. 1 bought a silver dish weighing 21b. lOoz. 15pwt. 21 gr. ; a bowl weighing lib. loz. 16pwt. 14gr. ; a tank- ard, 21b. 8oz. 5pwt. 12gr. : what' did all weigh? Ans. 61b. 8oz. 17pwt. 23 gr. 24. A druggist mixed together four articles : the first weighed 3343 19; the second, 43 3 3 29; the third, 43 18 gr. ; and the fourth, 63 53 29 18 gr. : what was the weight of all ? Ans. 1 Ib 3 5 2 3 IGgr. 25. A grocer sold 5hhd. of sugar : the first weighed 8cwt. Iqr. lllb.; the 2d, 4cwt. 2qr. 141b.; the 3d, 5cwt 191b.; the 4th, 7cwt. 3qr. ; the 5th, 7cwt, 3qr. 91b. : what did all weigh? Ans. 33cwt. 3qr. 31b. 26. Add together, 131b. 11 oz. 15dr. ; 171b. 13oz. lldr. ; 14 Ib. 14oz.; 16 Ib. 10 dr. ; 191b. 7oz. 12dr.; and 17 Ib. 9oz. 9 dr. Ans. 99 Ib. 9oz. 9 dr. 27. Two men depart from the same place : one travels 104 mi. Ifur. lOrd. due east; the other, 95 mi. 6 fur. 30 rd. due west : how far are they apart? Ans. 200 mi. 28. A man has 3 farms: in the 1st are 186 A. 3R 14 P.; in the 2d, 286 A. 17 P.; in the 3d, 113 A. 2H. 9 P. : how much in all? 'Ans. 586 A. 2ft. 29. Add together, 17sq.yd. 3sq ft. HDsq.in. ; 18 sq.yd. 141sq.in. ; 23sq.yd, 7 sq.ft. ; 29sq.yd. 5 sq.ft. 116sq.in. Ans. 88 sq.yd. 8sq.it, SSsq.in. 30. A has 4 piles of wood : in the first 7 C. 78 cu. ft. ; the 2d, 16C. 24cu.ft. ; the 3d, 35C. 127eu.ft,; the4th,29C. lOcu.ft. : how much in all? Ans. 88 C. lllcu.ft. 31. Bought 5 pieces of cloth ; the first contained 17yd. 3qr. 2na. f the 2d, 13yd. 2qr, Ina.; the 3d, 23yd. 2na.; the 4th, 27yd. Iqr. 2na. ; the 5th, 29yd. Iqr. 2na.: find the ainpunt, 4ns ? lllyd, Iqr. SUBTRACTION OP COMPOUND NUMBERS. 107 32. I sold to A, 73hhd. 43 gal. 3qt. Ipt. of wine ; to B, 27hhd. 3 gal.; to C, 15hh 8 13 14gr. 17. I bought 46 Ib. 9oz. of rice: after selling 191b. 4 dr., how much remained ? Ans. 27 Ib. 8oz. 12 dr. 18. A wagon loaded with hay weighs 32cwt. 2qr. 161b. ; the wagon alone weighs 8cwt. 2qr. 171b. : what is the weight of the hay ? Ans. 23cwt. 3qr. 24 Ib. 19. It is about 25000 miles round the earth: after a man has traveled 100 mi. Ifur. Ird., what distance will remain? Ans. 24899 mi. 6 fur. 39 rd. 20. I had 146 A. 2R. of land. I gave my son 86 A. 2 K. 14 P. : how much was left ? Ans. 59 A. 3 R. 26 P. 21. From 80. 50cu.ft. of wood, there is taken 3C. 75cu.ft. : how much is left? Ans. 40. 103cu.ft. 22. From 25E.FL, take 14E.F1. Iqr. 3na. Ans. 10E.F1. Iqr. Ina, 23. From 11 yards of cloth, there is cut 3yd. 2qr. 2na. : what remains ? Ans. 7yd. Iqr. 2na. 24. A hhd. of wino leaked; only 51 gal. 1 qt. 2gi. re-, mained : how much was lost ? Ans. 11 gal. 2 qt. 1 pt. 2gi. 25. From 5da. lOhr. 27min. 15sec., take 2da, 4hr. 13 min. 29 sec. Ans. 3 da. 6hr. 13min. 46 sec. liEviETv'. 102. Why placo units of the same denomination under each other? At which column begin to subtract? Why? How is the sub- traation performed ? Repeat the Rule. What is the proof ? 102. REM. In subtraction of Compound Numbers when the lower num- ber of any order id greater than the upper, what is to be don ? 110 HAY'S PRACTICAL AHITHMETIC. ART. 103. To find the Time befween any Two Dates. Subtract the first date from the last, numbering the months according to their order. See note 3, page 98. NOTE. In finding the time between two dates, and also in Interest, consider oC days 1 month, and 12 months 1 year. 26. A note, dated April 14th, 1835, was paid Feb- ruary 12th, 1837 : find the time between these dates. SOLUTION. In writing the dates, observe OPERATION. that February is the 2d month, and April, yr. mon. da. the 4th. In subtracting, 14 days can not 1&37 2 12 be taken from 12 ; theref re add 30 to the 12 1835 4 14 days, subtract, and carry 1 to the 4 months. " ^ ^ "^ As 5 months can not be taken from 2 months, add 12 to the latter, subtract, and carry 1 to the years. 27. The Independence of the United States was declared July 4th, 1776 : what length of time had elapsed on the 5th of March, 1857? Ans. 80 yr. 8 mon. Ida, 28. A certain man was born June 24th, 1822 : what was his age Aug. 1st, 1848? Ans. 26 yr. Imon. 7 da. 29. A man was born Nov. 25th. 1807 ; his son was born June 28th, 1832 : what is the difference of their ages? Ans. 24 yr. 7 mon. 3 da. 30. The latitude of the Cape of Good Hope is 33 55' 15" south ; that of Cape Horn, 55 58' 30" south : find the difference of latitude. Ans. 22 3' 15". 31 The latitude of Gibraltar is 36 6' 30" north; that of North Cape, in Lapland, 71 10' north: what their difference of latitude ? Ans. 35 3' 30". 32. A ship departs from latitude 10 25' 48" north, and sails north to latitude 50 : through how many de- grees of latitude has she sailed? Ans. 39 34' 12". 33. Take 5 quires 1 1 sheets from a ream of paper, and tow much will remain? Ans. 14 quires 13 sheets. 34. A man started to market with 6 doz. doz. eggs : he broite half a doz. doz. ; how many were left? A.is. 792, COMPOUND NUMBERS. - 111 MULTIPLICATION OF COMPOUND NUMBERS, ART. 104. The process of taking a number consisting of different denominations, a certain number of times, is termed Multiplication of Compound Numbers. 1. A farmer takes to mill 5 bags of wheat, each con- taining 2 bu. 3 pk. 4 qt. : how much in all ? SOLUTION. Begin at the OPERATION. lowest. denomination: 5 time9 > b u p^ qt. 4 quarts are 20 quarts, which 234 amt. iu 1 bag. reduced, make 2 pecks and 4 5 number of bags quarts; write the 4 (quarts) in the column of quarts, and 14 1 4 amt. in 5 bags. carry the 2 pecks. Next, multiply the 3 pk. by 5, making 15 pk., to which add the 2 pk. carried, making 17 pk., which, reduced, give4bu. and 1 pk. ; write the 1 (pk.) in the column of pk., and carry the 4 bu. Then, multiply the 2 bu. by 5, add to the product the 4 bu. curried, and we have 14 bu. to be written in the column of bu. In Compound, as in Simple Multiplication, multiply the lowest denomination first, so as to carry from a lower to a higher order. 2. Multiply 2 bu. 5 qt, 1 pt. by 6. Ant. 13 bu. 1 qt. 3. Multiply 2 bu. 2 pk. 2 qt. by 9. Am. 23 bu. 2 qt. Rule for Multiplication.!. Write the multiplier under the lowest denomination of the multiplicand. 2. Multiply the lowest denomination first, and divide the product ly the number of units of this denomination, which 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. REVIEW. 103. How find the time between any two dates ? 104. Wha: is Multiplication of Compound Numbers? Why multiply the lowest der nomination first ? Repeat the KuU. 112 RAY'S PRACTICAL ARITHMETIC. 3. Proceed in like manner with all the denominations, writing the entire product at the last. PROOF. The same as in Sinnlt) Multiplication. REM. 1. In Simple Multiplication, we carry one for every ten, because ten units of a lower order make one unit of the next higher. In Compound Multiplication, we carry one for the number of units in each lower order which make a unit of the next higher. 2. The multiplier is always an abstract number, and shows how many times the multiplicand is to be taken. 4. If 4bu. 3pk. 3qt. Ipt. of wheat make Ibl. of flour, how much will make 12 bl. ? Am. 58 bu. Ipk. 2qt. 5. What is the weight of 6 silver spoons, each weigh- ing 2oz. llpwt. 6gr. ? Jin. lib. 3oz. 7pwt. 12 gr. 6. What is the weight of 10 bars of silver, each 10 oz. lOpwt, lOgr.? Am. 81b. 9oz. 4pwt, 4gr. 1, I put up 8 packages of medicine, of 4 5 23 15 gr. each : what did all weigh ? Ans. 2 Ib 8 7 3 19. 8. Find the weight of 9hhd. of sugar, of 8cwt. 2qr. 12 Ib. each. Ans. 3T. 17 cwt, 2qr. 81b. 9. How much hay in 7 loads, each weighing 10 cwt. 3qr. 141b.? Ans. 3T. 16 cwt. 23 Ib. 10. If a ship sail 208 mi. 4fur. 16 rd. a day, how far will she sail in 15 Jays? Ans. 3128 mi. 2 fur. 11. If a man travel 30 mi. 4fur. 10 rd. a day, how far will he travel in 12 da. ? Ans. 366 mi. 3 fur. 12. Multiply 130 \. 3E. 30 P. by 4. Ans. 523 A. 3E. 13. Multiply 23 cu. yd. Dcu.ft. 228cu.in. by 12. Ans. 280cu.yd. leu. ft. 1008 cu. in. ^ 14. How many yards in 6 pieces of muslin, of 26yd. 2qr. 2na. each? Ans. 159yd. 3qr. 15. Multiply 62 gal. Iqt. Ipt. by 8. Ans. 499 gal. 16. How many gallons in 5 casks, each containing 123 gal. 2qt. Ipt.? Ans. CISgal. Ipt. REVIEW. 104. REM. 1. In Compound Multiplication, for what number do you carry one? 2. Is the multiplier concrete or abstract? MULTIPLICATION OF COMPOUND NUMBERS. 113 17. In a solar year are 365 da. 5hr. 48min. 48 sec. how long has B lived, who is 12 years old? Ans. 4382 da. 21 hr. 45min. 36 sec. 18. Multiply 4 11' 15" by 8. Am. 1 s 3 C 30'. 19. How many buttons in 3 great gross? Ans. 5184. 20. How much water in 48 casks, each contain ir* 62 gal. 1 qt. 1 pt. 1 gi. ? Here, 48 = 6 X 8. (See Art. 47.) gal. qt. pt. gi. 62 1 1 1 85 6 o 374 1 1 2 = water in 6 casks. 8 2995 2 = water in 48 casks. 21. Multiply 2bu. 3pk. 5qt. by 24. u4rcs.69bu. 3pfc 22. 3 mi. 5 fur. 16rd. X 60. Ans. 220mi. 4 fur. 23. 6 A. 3R. SOP. X 56. Ans. 388 A. 2H. 24. 8cwt. 2qr. 141b. 12oz. 13dr. X 22. Ans. 9T. lOcwt. Iqr. 9oz. 14 dr. 25. 3gal. 2qt. Ipt. Igi. X 112. (112 = 8X2X7.) Ans. 6hhd; 31 gal. 2qt. NOTE. When the multiplier exceeds 12, and is not a composite number, it is most convenient to multiply each denomination, and reduce it separately, and write only the result. 26. Multiply 16cwt. 2qr. 24 Ib. by 119. Ans. 99 T. 12cwt. 61b. 27. 37yd. 3qr. 2na.X89. Ans. 3370yd. 3qr. 2na. 28. 47 gal. 3 qt. lpt.X59. fc Ans. 44hhd. 52 gal. 2qt. Ipt 29. A travels 27 mi. 3 fur. 35 rd. in 1 day : how far will be travel in liuon. of 31 days? Ans. 852 mi. 5rd. 30. In 17 piles, each containing 70. 98cu. ft. : what quantity of wood? An*. 132 C. 2cu.ft. 3d Bk. 8 114 RAY'S PRACTICAL AKITHMETIC, DIVISION OF COMPOUND NTJL BEES. ART. 105. The process of dividing numbers consisting of different denominations, is termed Division of Com- pound Numbers. , The Divisor may be cither a Simple, or Compound Number. This gives rise to two cases : FIRST CASE. To find how often one Compound Number is contained in another Compound Number. SECOND CASE. -To divide- a compound number into a given number of equal parts. (See example 1, below.) NOTE. Examples of the First Case are solved by reducing both divisor and dividend to the same denomination, and then dividing. They are treated of under Reduction. See examples 6, 8, page 100. Examples of the Second Case, are usually considered under the head of Compound Division. ART. 106. 1. Divide 14 bu. Ipk. 7qt. Ipt. of wheat equally among 3 persons. SOLUTION. Divide the highest denom- OPERATION. ination first, as in Simple Numbers, that bu. pk. qt, pt. if there be a remainder, it maybe reduced 3)14 1 7 1 to the next lower order, and added to it. . T o ^ T First, Sin 14 is contained 4 times, and 2 bushels left; write the 4 in the order of bushels, and reduce the remaining 2 bushels to pecks, to which add the 1 in the order of pecks, and the sum is 9 pecks, which, divided by 3, gives a quotient of 3 pecks, to be written in the order of pecks. Next, divide 7 quarts by 3, and the quotient is 2 quarts, with 4 quart remainder; write the 2 quarts in the order of quarts; reduce the 1 quart to pints, and add it to the 1 in the order of pints; the sum is 3 pints, and this, divided by 3, gives a quotient of 1 pint, which write in the order of pints. (2.) bu. pk. qt. (3.) da. hr. min. sec, 7)33 2 6 5)17 12 56 15 Ans. 4 3 2 Am. 3 12 11 15 DIVISION OP COMPOUND NUMBERS. 115 Rule for Division, 1. Write the quantity to be divided in the order of its denominations, beginning with the highest; write the divisor on the left. 2. Begin with the highest denomination, divide each number separately, and icrite 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. HEM. In Simple Numbers when a remainder occurs in dividing any order except the lowest, it is prefixed to the figure *n the next lower order, which is equivalent to multiplying it by 10, and add- ing to tlie product the figure in the next lower order. Hence, in both Simple and Compound Division, each remainder is multiplied by that number of units of the next lower order which make a unit of the same order as the remainder. Bach partial quotient is of the same denomination as that part of the dividend from which it is derived. 4. Divide 67 bu. 3pk. 4qt. Ipt. by 5. Ans. IS^u. 2pk. 2qt. Ipt. 5. Eight silver tankards of the same size, weigh 14 Ib. 8oz. IGpwt. 16 gr. : what is the weight of each? Ans. I Ib. 10 oz. 2pwt. 2gr. 6. What will 1 dollar weigh ; the weight of 10 dollars being 8oz. 12pwt. 12 gr.? Ans. 17pwt. 6gr. 7. Eleven bl. of sugar weigh 35cwt. Iqr. 17 Ib. 3oz. 7 dr. : find the weight of one. Ans. 3ewt. 22 Ib. 5 dr. 8. I traveled 39nri. 7fur. 8rd. in. 7 hours: at what rate per hour did I travel? Ans. 5 mi. 5 fur. 24 rd. 9. Divide 62yd. 3na. by 5. Ans. 12yd. Iqr. 3na. REVIEW. 105. What is Division of Compound Numbers? "What may the divisor ba ? What is the first Case ? What the second ? NOTE. How are examples of the first Case solved ? 103. In dividing a Compound Number, why divide the highest denomi- nation first ? When a remainder occurs, how proceed, Rule ? 116 RAY'S PRACTICAL ARITHMETIC. NOTE. If the divisor exceeds 12, and is a composite number, A 3. The diff. of Ion. between two places is 4 *! 4': what the diff. of time? Ans. 4hr. 44 min. 16 sec. REVIEW. 103. At what rate docs *ho pun appear to move in a day? In an hour ? In a minute ? In a r acona ? What do degrees, minutes, and seconds of longitude divided by 15 give ? Eepeat the Rules. DIVISION OF COMPOUND NUMBERS. H9 4. The diff. of Ion. between New York and Cincinnati is 10 35.' : what the diff, of tijne ? Ans. 42min. 20sec. 5. The diff. of time between Cincinnati and Philadelphia is 37min. 20 sec. : what the diff. of Ion. ? Ans. 9 20'. ypo y^ y 6: The diff. of time between New York and St. Lcuis islhr.4min. 56 sec. : what the diff. of Ion. ? Ans. 16 14'. , 3 ' I S" 1. The din. of time between London and Washington is 5hr. 8min. 4sec. : what the diff. of Ion.? Ans.W 1'. DIFFERENCE IN TIME. ART. 109. It is noon (12 o'clock), at any place when the sun is on the meridian of that place ; and, As the sun appears to travel from the east toward the west, when it is noon at any place, it is after noon cast of that place, and before noon west of that place : Hence, a place has later or earlier time than another, according as it is east or west of it. Therefore, When the time at one place is given, the time at another, EAST of this, is found by ADDING their difference of time: Or, if WEST, by SUBTRACTING their difference of time. 8. When it is noon at Cincinnati, what is the time at Philadelphia? Ans. 37niin. 20 sec. past noon. 9. When it is 11 o'clock A. M. at New York, what is the time in Ion. 3Ct east of New York ? Ans. 1 P. M. 44. 10. When 12 o clock (noon) at Philadelphia, what is the time at Cincinnati? Ans. llhr. 22min. 40sec. A.M. 11. When it is 11 o'clock A. M. at New York, what is the time at St. Loirls? Ans. 9hr. 55min. 4sec. A. M. 12. Wheeling, Va., is in Ion. 80 42' west: the mouth of the Columbia river in Ion. 124 west : when it is 1 o'clock, P. M., at Wheeling, what is the time at the mouth of Columbia river ? Ans. 10 hr. 6min. 18 sec. A. M. , _ _ - . - REVIEW.-109. When is it noon at anyplace? What the time cast or \tost of that place ? Why is the time later fast f Why earlier wst ? Having the timo at one place, how find the time at another ? 120 . BAT'S PRACTICAL ARITHMETIC. ' VIII. FACTORING. ART. 110. DEFINITIONS. 1. An integer is a whole number ; as, 1, 2, 3, &c. DEF. 2. Whole numbers are divided into two classes ; prime numbers, and composite numbers. DEF. 3. A prime number can be exactly divided only by itself and unity, (1). Thus, 1, 2, 3, 5, 7, 11, &c., are prime. DEF. 4. A composite number (Art. 33) can be exactly divided by some other number besides itself and unity. Thus, 4, 6, 8, 9, 10, &c., are composite. DEF. 5. Two numbers are prime to each other when unity (1), is the only number that will exactly divide both. Thus, 4 and 5 are prime to each other. REM. Two prime numbers are always necessarily prime to each other. Also, two composite numbers are sometimes prime to each other : thus, 4 and 9 are prime to each other. DEF. 6. An even number can be divided by 2 without a remainder. Thus, 2, 4, G, 8, &c., are even. DEF. 7. An odd number can not be divided by 2 with- out a remainder. Thus, 1, 3, 5, 7, &c., are odd. REM. All even numbers except 2, are composite numbers, while odd numbers are partly prime and partly composite. DEF. 8. A divisor of a number will exactly divide it; x that is, without a remainder: thus, 2 is a divisor of 4; 6 of 10, &c. A divisor of a number is a measure f that number. DEF. 0. One number is divisible by another, when the former contains the latter without a remainder. Thus, 6 is dimtible by 2. REVIEW. 110. What is an integer? How are the whole numbers .livirted? What is a prime number? Give examples. What a com- posite? Give examples. When are two numbers prime to each otfcerf U-ive examples. What is an even number ? An odd ? Give examples. FACTORING. 121 DEF. 10. A multiple (dividend), of a number is the product arising from taking it a certain number of times : thus, G is a multiple of 2, because it is equal to 2 taken 3 times. Hence, A multiple of a Dumber can be divided by it without a re- mainder. Therefore, every multiple is a composite number. DEF. 11. A factor of a number is a number that will exactly divide it : thus, 4 is a factor of 8, 12, 16, &e. REM. The terms, factor, divisor and measure, all mean the same thing. Every composite number being the product of two or more factors, each factor must exactly divide it, (Art. 37). Hence, every factor of a number, is a divisor of that number. DEF. 12. A prime factor of a number is a prime number that will exactly divide it : thus, 3 is a prime factor of 12 ; while 4 is a factor of 12, but not a prime factor. Therefore, all the prime factors of a number, are all the prime numbers that will exactly divide it: thus, 1, 3, and 5, are alltho prime factors of 15. Every composite number is equal to the product of all its prime factors : thus, all the prime factors of 10 are 1, 2, and 5; 1X2X5-10. DEF. 13. An ALIQUOT part of a number, is a number that will exactly divide it: thus, 1, 2, 3, 4, and 0, are aliquot parts of 12. RESOLVING NUMBERS INTO PRIME FACTORS. ART. 111. The smaller composite numbers may be resolved into their prime factors by inspection; thus, 6=2X3; 8=2X2X2; 9=3X3; 10=2X5. In the case of larpre numbers, their factors are found by trial , that is, by dividing by each of the prime num- RKvniw. 110. R^M. Arc the even numbers primo cr composite? Aro tho td-l number* ? What is a divisor of a number ? Give examples. When is ono number divisible by another ? Give examples. 110. What is a multiple ? Give examples. A factor? Give examples. REM. What terms besides divisor are used in the same sense ? Why is every factor a divisor ? What is a prime factor ? Give an example. 122 RAY'S PRACTICAL ARITHMETIC. bers 2, 3, 5, 7, &c. ; the prime factors of any number, being all the prime numbers that will exactly divide it. In determining either the factors, or the prime factors, of a number, observe the following principles. PRINCIPLE 1. A factor of a number is a factor of any multiple of thai number. Thus, 3 is a factor of 6, and of any number of times 6; for, 6 is 2 threes, and any number of times 6 will be twice as many times 3. PRINCIPLE 2. A factor of any two numbers is also a factor of their sum. Since each number contains the factor a certain number of times, their oum must contain it as many times as both numbers. Thus, 2 being a factor ot 1 6 and 8, it is a factor of their sum ; for, 6 id 3 twos, and 8 is 4 twos, and their sum is 3 twos-\-& twos t =l twos. ART. 112. From these two Principles, are derived SIX PROPOSITIONS. PROP. 1. Every number ending with 0, 2, 4, G, or 8. is divisible by 2. ILLUSTRATION. Every number ending with 0, is either 10 or some number of tens; and, since 10 is divisible by 2, any num- ber of tens will be divisible by 2. Priu. 1. Again: any number ending with 2, 4, G, or 8, maybe considered a certain number of tens, plus the figure in units' place : And, as each of the two parts of the number is divisible by 2, therefore, Prin. 2, the number itself is divisible by 2. Conversely : No number is divisible by 2, unless it ends with a 0, 12, 4, 6, or 8. PROP. II. Every number is divisible by 4, when the number denoted by its first two digits is divisible by 4. iEW. 110. "What is an aliquot part of a number? Give examples. Hi. U:>w may the smaller composite numbers be resolved into prime fac- tors ? What aro tho primo factors of G ? Of 8? Of 9 ? Of 10? 111. Li determining the factors of a number, what two principles ar u?ed ? Explain the first principle. Tho second, FACTORING. 123 ILLUSTRATION. Since 100 is divisible by 4, any number of hun- dreds is divisible by 4 ; and any number of more than two places of figures, may be regarded as a certain number of hundreds, plus the number denoted by the first two digits. Then, since both parts of the number are divisible by 4, Prin. 2, the number itself is divisible by 4. Conversely : No number is divisible by 4, unless the number denoted by its first two digits is divisible by 4. PROP. III. Every number is divisible by 5, when its right hand digit is or 5. ILLUSTRATION. Ten being divisible by 5, and every number consisting of two or more places of figures/ being composed of tens, plus the figure in the units' place: Therefore, if this is 5, both parts of the number are divisible "by 5 ; hence, Prin. 2, the number itself is divisible by 5. Conversely : No number is divisible by 5, unless its right hand digit is or 5. PROP. IV. Every number whose first digits are 0, 00, &c., is divisible by 10, 100, nbe: -which will exactly divide the given number and- one of its parts, will also exactly divide the other. In this case, either of the parts is the difference between the given number and the other part. 4TII PRIN. The greatest common divisor of tiro numbers, is a divisor of their remainder after division. ee Solu. 1. Find the G. C. D. of 16 and 44. SOLUTION. As 1G is .a divisor of OPERATION. itself, if it be .a divisor of 44, it will 16)44(2 be the G. C. D. required, since no number can have a divisor greater I 9> )16fl than itself. But 10 is contained twice in 44, with a remainder 12; lience, 16 is not the G. C. D. Since 4)12(3 the G. C. D. is a divisor of 16 and 44, by first Prin., it will be a GREATEST COMMON DIVISOR. 129 divisor of 16 X 2 = 32 ; hence, by 3d Prin., it must be a divisor O f 4432 = 12; that is, the G. C. D. of two number* is also a divisor of their remainder after division. Hence, the G. C. D. of 10 and 44 is also a com. divisor of 12 and 1G, and it can not exceed 12. Since 12 is a divisor of itself, if it. be a divisor of 16, it must be the G. C. D. sought. By dividing 12 into 10, the remainder is 4 ; hence, 12 is not the G. C. !>.; but, by Prin. 4th, the G. C. D. of 12 and 1C is a divisor Df i, their remainder after division; hence, the G. C. D. of 10 and 44 can not exceed 4, and must be a divisor of 4 and 12. By dividing 12 by 4, there is no remainder; hence, 4 is a divisor of 12, and therefore, of 12 X l-f-4 = lt>, Prin. 2d. And, Since 4 is a divior of 12 and 16, it must be a divisor of 16 X 2+12 = 44 ; and since the G. C. D. can not exceed 4, and 4 is a divisor of 16 and 44, therefore, 4 is the G. C. D. sought. 2. What is the G. C. D. of 14 and 35? Ans. 7. 3. What is the G. C. D. of 9 and 24? Ans. 3. Etlle II. Divide the greater number by the less, and that divisor by the remainder, and so on, always dividing the last divisor by the last remainder, till nothing remains-, the last divisor will be the greatest common divisor. NOTE. To find the G. C. D. of more than two numbers, first find the G. C. D. of two of them, then of that com. divisor and one of the remaining numbers, and so on -for all the numbers; the last com. divisor will be the G. C. D. of all the numbers. FIND THE GREATEST COMMON DIVISOR OP 4. 42 and 495. Ans. 3. 8. 37 and 759. Ans. 1. 5. 47 and 323. Ans. 19. 9. TS73 and 69087 . Ans. < i) 285 and 465. Ans. 15. 10. 1814 and 1776. A 118. ', . 532 and 1274. Ans. 14. 11. 693 and 1815. Ans. 33. Ivnvinv. 11*. On what principles does the second mothod flf finding tho G. C. D. depend ? Explain the third principle. Find the G. C. D* of 3 ani 24, and explain the fourth principle. What is Rule IT? 117. NOTB. Hyw LJ tho G. C. D. of more than two numbers found T 3d Bk. 130 RAY'S PRACTICAL ARITHMETIC. 'l2. Find the G. C. D. of 2145 and 3471. . . Ans. 39. 13. Of 840 and 17017 Ans. 7. 14. Of 66284 and 153452 Ans. 908. 15. Of 40, 55, and 105 Ans. 5. 16. Of 70, 154, and 819 Ans. 7. 17. Of 120, 168, and 1768 Ans. 8. 1 For additional problems, see Ray's Test Examples. X. LEAST COMMON MULTIPLE. ART; 118. A multiple (dividend), of a number, is a number that can be divided by it without a remainder. Thus, 12 is a multiple of 3, because 3 is contained in 12 an exact number of times, 4. Art. 110, Def. 10. A common multiple (dividend), of two or more num- bers, is a number that can be divided by each, without a remainder. Thus, 24 is a common multiple of 3 and 4. The least common multiple of two or more numbers, is the least number that can be divided by each without a remainder. Thus, 12 is the least common multiple of 3 and 4. REM. 1. As the Com. Mul. of two or more numbers contains each of them as a facfor, it is a composite number. 2. As the continued product of two or more numbers is divisible by each of them, a Com. Mul. of two or more given numbers may always be found by taking their continued product; and, Since any multiple of this product will be divisible by each of the given numbers, (Art. Ill, 1st Prin.), an unlimited number of Com. Multiples may be found for any given numbers. REVIEW. 118. What is a multiple of a number? Give an example. What is a- cc.mmcm, multiple cf two or more numbers ? What the haft common multiple ? EEM. 1. Is a common multiple of two or more numbers, a prime or composite number? Why ? 2. How mny a Com. Mul. always bo found ? How many Com. Multiples may numbers hare ? Why ? LEAST COMMON MULTIPLE. 131 ART, 119. To find the least common multiple of two or more numbers. FIRST METHOD. One number is divisible by another, when it contains all the prime factors of that number. Thus, 30 is divisible by 6, because 30 r= 2X3X5, and 6 = 2X3; the prime factors of 6, which are 2 and 3, being also factors of 30. One number is not divisible by another, unless it con- tains all the prime factors of that other. Thus, 10 is not divisible by 6, because 3, one of the prime factors of 6, is not a factor of 10. Hence, a common multiple of two or more numbers must contain all the prime factors in those numbers; and, To be the least common multiple, (L. C. M.), it must not con- tain any prime factor not found in some one of the numbers. 8@^L. C. M. should be read, least common multiple. 1. What is the L. C. M. of 6 and 10? SOLUTION. By factoring. 6 = 2X3. riPTTT? A TTHV and 10=2 X 5. A number composed 6 = 2X3 of the factors 2, 3, and 5, will contain 1 4 > y 5 all the factors in each of the numbers 6 and 10, and will contain no other 2X3X5 = 30 Ans. factor; therefore, cross out (cancel) the factor 2, in one of the numbers; the product of the remaining factors, 2X3 X 5 = 30, will be the L. C. M. of 6 and 10. 2. What is the L. C. M. of 6, 8, and 12? SOLUTION. By factoring the numbers, tne prime factor 2 oc- 6=^ X3** curs once in 6, three times in 8, 8 2 y 9 V ? and twice in 12 ; hence it must 1 2 = 2 X i X occur three times, and only three times, in the L. C. M. ; therefore, 3X2X2X2 = 24 Ans. after reserving it as a factor , three times, cancel it, in the other numbers. The prime factor 3 occurs once in 6 and once in 12; hence, it must occur once, and only once, in the L. C. M. After reserving it once as a factor, cancel the other factor 3. The L. C. M. is then found by multiplying together the figures not canceled. RAY'S PRACTICAL ARITHMETIC. 3. Find the L. C. M. of 12 and 30. . . Am. 60. 4, Of G, 10, and 18 Ans. 90. Httle I. Separate the numbers into prime factors; then multiply together ONLY such of those factors, as are necessary to form a product that will contain all the prime factors in each number, using no factor oftener than it occurs in any one number. NOTE. The solution of Ex. 2, shows that the same factor must be taken (he greatest number of times it occurs in either number. After factoring, cancel (cross out) the needless factors. FIND THE LEAST COMMON MULTIPLE OP 5. 6, 8, 9. Ans. 72. 6. 6, 15, 35. Ans. 210. 7. 10, 12, 15. Ans. 60. 8. 9, 15, 18, 24. Ans. 360. 9. 8, 15, 12, 30. Ans. 120. 10. 14, 21, 30, 35. Ans. 210. SECOND METHOD. ART. 120. The L. C. M. of two or more numbers, contains all the prime factors of each of the numbers once, and no other factors. For, if it did not contain all the prime factors of any num- ber, it would not be divisible by that number; and, if it con- tained any prime factor not found in either of the numbers, it would not bo the least common multiple. Thus, the L. C. M. of 4 (2X2), and G (2X3), must con- tain the factors 2, 2, 3, and no others. 1. Find the^L. C. M. of C, 9, and 12. SOLUTION. Arranging the num- OPERATION". bers as in the margin, we find 2)G 9 12 that 2 is a prime factor common ~7T I to two of them. 5 lf _ I Hence, 2 must be a factor of 132 the L. C. M.; therefore, place 'it 9 , 9 on the left, and cancel it in the 2 X o numbers of which it is a factor, by dividing by it. Next, observe that 3 is a factor eotnmon to the quotients and the remaining number, and hence, (Art. Ill,) id a factor of the given numbers, and must be a factor of the L. C. M.; LEAST COMMON MULTIPLE. 133 therefore, place it on the left, and cancel it in each of the numbers i;i the 2d line, by dividing by it. As the numbers 3 and 2, in the / 8J line, have no common factors to cancel, we do not divide them. Thus \vo find, that 2, 3, 3, and 2, i.re all the prime factors in tho given numbers; hence, their product, 2X^X^X2 = 36, is theL. C..M. of 6, 9, and 12. In this operation, let the learner notice, 1st. The number 3G is a common multiple, because it ecu- tains all the prime factors in each of the numbers ; it is the least C. M., because all the needless factors were canceled by dividing. 2d. To cancel needless factors, divide by & prime number. By dividing by a composite number, in some cases, all the needless factors are not canceled ; thus, in the preceding ex* ample, 6 will exactly divide two of the numbers ; but, In dividing by 6, a factor, 3, is left uncanceled in the mul- tiple 9, and thus the L. C, M. is not obtained. 2. Find the L. C. M. of 6 and 10. ... Ans. 30. 3. Of 15, 21, 35 ........ Ans. 105. Rule II. 1. Place the numbers in a line, divide by any prime number that will divide two or more of them without a remainder, and place the quotients and undivided numbers in a line beneath. 2. Divide this line as before : continue to divide till no number greater than 1 will exactly divide two or more of the numbers. 3. Multiply together the divisors and the numbers in the lowest line, and their product will be the least common multiple. HEM. If the given numbers contain no common factor, their product will be the L. C. M. Thus, the L. C..M. of 4, 5, and 9, is REVIEW. 119. When is one number divisible by another? Give an example. When not divisible ? Give an example. What factors iniiot the Cum. Mul. contain ? 119. What prime factors must the L. G. M. not contain? Find tha ti. C. M. of 6, 10, and 18, and explain the operation. What is Erie I? 119. Nom How ofien iuu4 the saiuo focwr be 1'o.uQd in the I<. C, M- 7 134 RAY'S PRACTICAL ARITHMETIC. FIND THE LEAST COMMON MULTIPLE OP 4. 9 ,12. ^4?is. 36. 8. 6, 10, 15, 18 Ans. 90. 5. 14 ,21. Ans. 42. 9. 7, 11, 13, 3. Ans. 3003. 6. 6 J 9 ,15. Ans. 90. 10. 63, 12, 84, 7. Ans. 252. 7. 4, 14 ,35. Ans. ] 40. 11. 54, 81, G3. Ans. 1134. 12. Of 8, 12, 20, 24, 25. An s. 600. 13. 10, 24, 25, 32, 45. . . An s. 7200. 14. 9 8', 72, 64, 21, 18. . . . . A 71 s. 28224. 15. 2, 3, 4 ; 5> 6, 7, ,, 9. An s. 2520. XL COMMON FRACTIONS. ART. 121. A single thing, (Art. 1), is called a unit, or orie, which may be divided into equal parts. Thus, suppose 3 apples are to be equally divided behveen 2 boys: after giving one to each, there would remain one to be divided into two equal parts, to complete the division. The equal parts into which a unit is divided are fractions. ART. 122. When a unit, or single thing, is divided into two equal parts, one of the parts is one-half. If it is divided into three equal parts, one of the parts is one-third; two of the parts, two-thirds. If divided into four equal parts, one of the parts is one-fourth; two of the parts, two-fourths; and three of the parts, three-fourths. If divided into five equal parts, the parts are fifths; if into six equal parts, sixths, and so on. Hence, When a unit is divided into equal parts, the parts are named from the number of parts into which the unit is divided. REVIEW. 120. Ex. 1. Why divide by 2 ? By 3 ? Why multiply together the numbers 2, 3, 3, and 2 ? Why is 30 a Com. Mul. of B, 9, and 12 ? Why the least ? To cancel needless factors, why net divide by a composite number ? What i^ Rule II ? 120. REM. If the numbers contain no common factor, how is their L. C. M. found? 121. How do you divide 3 apples equally between 2 boys? When a uoit ia divided into equal parts, what are Uie parts called ? COMMON FRACTIONS. 135 ART. 123. The value of one of the parts depends on the number of parts into which the unit is divided. Thus, if 3 apples of equal size be divided, one into 2, another into 3, and another into 4 equal parts, the thirds will be less than the halves, the fourths less than the thirds. ART. 124. Fractions are divided into two classes, Common and Decimal. Common Fractions are expressed by two numbers, one above the other, with a horizontal line between them. Thus, one-half is expressed by I ; two-thirds by \ . The number below the line is the denominator : it denominates, or gives name to the fraction. It shows the number of parts into which the unit is divided. 1 The number above the line is the numerator : it numbers tho parts, showing how many parts are taken. Thus, in the fraction J, the denominator, 5, shows that the unit is divided into Jive equal parts, and the numerator, 3, shows that the fraction contains 3 of those parts. The numerator and denominator together, are called the terms of the fraction. Thus, the terms of f, are 3 and 5. ART. 125. ANOTHER METHOD. In the definition of numerator and denominator, refer- ence is had to a unit only. This is the simplest method of considering a fraction ; but, there is another mode : ILLUSTRATION. To divide 2 apples equally among 3 boys, divide each apple into three equal parts, making 6 parts in all; then give to each boy 2 of the parts, expressed by |. REVIEW. 122. When a unit is divided into two equal parts, what is one part called ? When divided into three equal parts, what is ono part called ? What two parts ? When divided into four equal parts, what is ono part called ? Two parts ? Throe parts ? When a unit is divided into equal parts, from what are the parts named ? 123. On what does the value of one of the parts depend? Which is greater, 1-half or 1-third ? 1-third or 1-fourth ? 1-fourth or 1-fifth ? 12i. Into what two classes are fractions divided? How are common fractions expressed ? What is the number below the line ? Why ? What} the number above the }ine ? Why ? What are the. terms ? 136 RAY'S PRACTICAL ARITHMETIC. In selecting one boy's share, take the 2 parts from one applo, or 1 part from each of the two apples; hence, | expresses either 2 thirds of 1 thing, or 1 third of 2 things. Also, | expresses 3 fifths of 1 thing, or 1 fifth of 3 things. Therefore, the numerator of a fract'on expresses the number of units to be divided ; and the denominator the divisor , or what part in taken from each. Hence, A fraction is expressed in the form of an unexecuted division, In which, The DIVIDEND is the NUMERATOR; The DIVISOR is the DENOMINATOR; The QUOTIENT is the FRACTION itself. J is the quotient of 1 (numerator) 3 (denom.); is the quotient of 4 (numerator) 5 (denom.); J is the quotient of 7 (numerator) ~ 6 (denom.). ART. 126. Since fractions arise from division, one of the siynis of division (Art. 40,) is used in expressing them ; the numerator being written above, and the de- nominator below, a horizontal line. Thus, three-fifths is written |; two-sixths is written . TO READ COMMON FRACTIONS. Read the number of parts taken, as expressed by ihe numerator] then the size of the parts, as expressed by the denominator. TO WRITE COMMON FRACTIONS. Write the numerator, place a horizontal line below it, under which write the denominator. REM. In reading, | moans two-thirds of crie. There are two other methods, <' Art. 125) : thus, | mny bo resu\,fne-t/iitd (/ two, or two divided by three ; but these methods are rarely used. FRACTIONS TO BE READ. * i. i I?- I*, -If 2 5 r- A- U- 18- ift. T 4 - s'&- T3 8 n- REVIEW. 125. What do two-thirds express ? What does the numerator of a fraction express? The denominator? In what form is a fractu-n expressed? What is the dividend ? The divisor? The quotient? Giro examples. 12ti. What sign is used in fractions ? Why ? How .s a Woa fraction read ? Give examples. Uow written ? COMMON FRACTIONS. 137 FRACTIONS TO BE WRITTEN. Three-serenes : Five-ninths : Six-tenths : One-twelfth: Two-thirteenths : Xine-sixteenths: Eleven-twentieths : Fourteen-twenty-ninths : Thirty-one-ninety-thirds : Twenty - three - one - hund* red-and-fourths. If an orange be cut into 8 equal parts, what fraction express 3 of the parts ? If 3 apples be divided equally among 4 boys, what part of an apple will be given to each boy? What part of 1 apple, is a third part of 2 apples ? What expresses the quotient of 5, divided by 8 ? ART. 127. A whole number may be expressed in the form of a fraction, by writing 1 below it for a denominator. Thus, 2 may be written f and is read two-ones. 3 may be written f and is read three-ones. 4 may bo written J and is read four-ones. But, 2 ones are 2; 3 ones, 3, &c., hence, the value of the number is not changed. ART. 123. If 2 apples be divided, each into four equal parts, there will be 8 parts in all. Three of the parts (fourths) arc expressed by | ; 4 parts by | ; 5 parts by f. When the number of parts taken is less than the number into which the unit is divided, the value expressed is less than one, or the whole thing; When the number of parts taken is equal to the number into which the unit is divided, the value, as J, is equal to 1 ; When greater than the number into which the unit is divided, the value, |, is greater than 1. Hence, LiiviEW. 12fi. Of tho fractions to be read, which expresses p.irts of the largest size? Which tho smallest? Which tho least number? Which the greatest ? Which tho same ? Which parts of tho same size? 127. IIcw rany a \vholo number be expressed in fractional form? Docs this chrin^o its value ? Why n<^t ? 328. When is the value of a fraction less than 1 ? Whea equal ? Y/hen greater ? Illustrate by examples. 138 RAY'S PRACTICAL ARITHMETIC. 1st When the numerator is less than the denominator , th& value of the fraction is less than 1. 2d. When the numerator is equal to the denominator, the value of the fraction is equal to 1. 3d. When the numerator is greater than the denominator, the value of the fraction is greater than 1. DEFINITIONS. ART. 129. 1. A Fraction is an expression of one or more of the equal parts of a unit, or one thing. 2. A Proper Fraction has a numerator less than the de- nominator; as, i, f, and f. 3. An Improper Fraction has a numerator equal to, or greater than the denominator; as, |, and |. HEM. A proper fraction is so termed, because it expresses a value less than one. An improper fraction is not properly a fraction of a unit, the value expressed being equal to, or greater than one. 4. A Simple Fraction is a single fraction, either proper or improper; as, l, f, and |. 5. A Compound Fraction is a fraction of a fraction, or several fractions joined by of; as, \ of J, f of f of f. 6. A Mixed Number is a fraction joined to a whole number; as, 2i, 31, and 5|. 7. A Complex Fraction has a fraction in one or in both of 3^ 1 -I 21 its terms ; as, > > > PARTS OP FRACTIONS. ' ART. 130. If a line, H - 1 - -- 5 - H as A B, be divided into two equal parts, one of the parts, as A C, is termed one- half (4) : that is, one-half of 1, or the whole thing. REVIEW. 129. What is a fraction ? A proper fraction ? An im- proper fraction ? HEM. Why is a proper fraction so termed ? An im- proper fracrtion ? What is a simple fraction ? A compound fraction ? A mixed number ? A complex fraction ? Give examples of each. 130. Wha| fc the haif of one-half? Why? The &ir< 2 = 1; and 3 X 3 = 9_ ' 5X2 10 5X3 15 ILLUSTRATION. Multiplying both terms of the fraction | by 2, gives, y 6 fj, in which the parts are twice as many, but only one-half the size. Multiplying both terms of | by 3, gives -$~; three times as many parts, each part one-third the size. Hence, PROP. V. Multiplying both terms by- the same number^ changes its form, but does not alter its value. ART. 136. If the numerator of a fraction be divided by any number, its value (PROP, n,) will be divided by that number; if the denominator be divided, the value (PROP, iv,) will be multiplied by that number. Hence, if both terms are divided by the same number, the decrease from dividing the numerator, equals the increase from dividing the denominator: and the value is not changed. hus,=; and ^ = 12 2 6 12-T-3 4' REVIEW. 132. What is the effect of dividing the numerator of a frac- tion, without changing the denominator? 133. What of multiplying the doBcminnior without changing the numsrator? 134. What of dividing the denominator, without changing the nume* rator ? J 35. What of multiplying both terms by tho same number ? 142 RAY'S PRACTICAL ARITHMETIC. ILLUSTRATION. Dividing both terms of the fraction -^ by 2, it gives |; in which there are one-half as many parts, but each part is twice the size. Dividing both terms of -^ by 3, gives |, one- third as many parts, each part three times the size. Hence, PROP. VI. Dividing both terms by the same number, changes its form, but does not alter its value. To TEACHERS. By considering the numerator a dividend, the denomi- nator a divisor, and the value of the fraction the quotient (Art. 125), the preceding propositions may be regarded as inferences from Art. 57, 58, 59. This short method is not best adapted to young pupils. f ART. 137. REDUCTION OF FRACTIONS Is changing their form without altering their value. CASE I. ART. 138. To reduce a fraction to its lowest terms. A fraction is in its lowest terms, when the numerator and denominator are prime to each other. Art. 110, Def. 5. Thus, | is in its lowest terms, while f is not. 1. Reduce |$ to its lowest terms. SOLUTION. Since the value of a FIRST OPERATION. fraction is not altered by dividing 24 ^)l2 4 both terms by the same number, 2)q-7r= lTjL~K **' (Art. 136), and, as two is a common factor, divide both terms by it; the fraction then becomes ||. Again, since 3 is a factor of 12 and 15, divide them both by it; the result, 4, can not be reduced lower. SECOND OPERATION. Instead of dividing by 2, and then by 3, 24__4 . divide at once by 6, the greatest com. divisor "^"SO 5 of the two terms, and the result is the same. Solve the two following Examples by loth methods. NOTE. All subsequent Examples having a star, &, are intended to illustrate the principles on which the next succeeding rule is founded. The pupil should solve them and explain the operation, referring, at the conclusion of the exercise, to the rule which follows. COMMON FRACTIONS. 143 *2. Reduce J- to its lowest terms. . . . Ans. f. *3. Reduce |g to its lowest terms. . . . Ans. f . Rule for Case I. Divide the numerator and denominator by any common factor; divide the resulting fraction in the same manner, and so on till no number greater than 1 will exactly divide both terms. Or, Divide the numerator and denominator by their greatest common divisor; the resulting fraction will be in its lowest terms. REM. When the terms of a fraction are small, the first method is most convenient ; when large, the second method. REDUCE TO THEIR LOWEST TERMS, . Ans. 2 9 5 . EXPRESS IN ITS SIMPLEST FORM, 16. The quotient of 391 divided by 667. Ans. |. 17. The quotient of 585 divided by 1287,', /' Ans. T 5 T . 18. The quotient of 796 divided by 14129. Ans. j r CASE II. ART. 139. To reduce an improper fraction to a whole or mixed number. In 4 halves (*) of an apple, how many apples? in 6 thirds () ? in 8 fourths (f) ? in f ? in ^ ? In 8 pecks, that is, in f of a bushel, how many bushels ? in | ? in L ? in J J ? in 7 - 3 ? REVIEW. 136. What is the effect of dividing both terms of a fraction by the same number ? 137. What is Reduction of Fractions ? 38. When is a fraction in its lowest terms ? Give an example. How la a fraction reduced to its lowest terms, ~ " " 4. fl- Ans. f. 10. iff. 5. If Ans. |. 11. Ill- 6. - 6 - Ans. . 12. Ill- 7. 4- Ans. |. 13. 873 70S?' 8. 96 772' Ans. . 14. _777_ 9. T25- Ans. 1 1. 15. , 9 & 9 g. 144 RAY'S PRACTICAL ARITHMETIC. 3. In 9 fourths (|) of a dollar, how many dollars ? SOLUTION. Since 4 fourths make one dollar, OPERATION. there are as many dollars as there are times 4 4)9 fourths in 9 fourths; that is, 21 dollars. Ant. S2 1. 4. Reduce * 5 7 to a mixed number. OPERATION. SOLUTION. Since 5 fifths make 1 (unit), KMT there will be as many ones as there are times 5 '_ _ in 17; that is, 3|. *5. In f g of a dojlar, how many dollars ? An*. 2 T 3 5 . *6. Reduce 2 3 5 to a mixed number. Ans. 8^. Rule for Case II. Divide the numerator by the denomi- nator : the quotient will be the whole or mixed number. 7. In 6 3 3 of a dollar, how many dollars? Ans. 13]. 8. In Y of a yard, how many yards ? Ans. IS j. 9. In *| 5 of a mile, how many miles? Ans. 15J. 10. In V 4 of a day, how many days? Ans. 25^ j. REDUCE TO AVIIOLE OR MIXED NUMBERS, 11. ig. Am. 1. 75' 12. v^. Ans. 31. 13. \V Ans. 14^7. 14. 5 yV. Ans. 46 T : V. ii ii IIP; 0137 /f H< . 91 1 79 iO ' 598 * - 1 ^gg* 16. ifff. Ans. GO T 3 7 V n37 8 1 J } jc 1 00 i TJ ** '* " ll/7 -, CASE III. ART. 140. To reduce a whole or mix^d number to an in^rojjer fraction. 1. In 2 apples, how many halves ? In 3? In 4? 2. In 2 apples, how many thirds ? In 3? In 4? 3. In 2 apples, Low many fourths? In 3? In 4? 4. In 2-\ apples, how many halves? In 3.^? In 4^? 5. In 2] apples, how many thirds ? In 2 j ? In 3.1 ? REVIEW. 1C8. Why is tho vnluo of a fraction not nlfrrvl by hcing f?duccd to ltd lowest terms ? 1^9. How id an improj)f fraction reduced w a wholo or mixed number, Rule ? COMMON FRACTIONS. 145 6. In 5 1 dollars, how many fourths? Or, reduce 5^ to an improper fraction. SOLUTION. Since there 5| OPERATION. ^re 4 fourths in $1, in 5 dollars 4 there are 5 times as many ; 5 times 4 fourths are 20 fourths, 2 j> = fourths in 5 d 11ar8 ' and 20 fourths + 3 fourths,= _3== fourths in fraction. 23 fourths. Ans. ^. 23 = fourths in 5 f. *7. In 8J apples, how many fourths? Ans. 3 ? 5 . *S. Keduce 121 to an improper fradtioa. Ans. . Hllle for Case III. Multiply the whole number by the de- nominator of the fraction ; to the product add the numerator, and write the sum over the denominator. REM. The analysis of question G, shows that the whole number is really the multiplier, and the denominator the multiplicand; but the result will be the same (Art. 30), if the denominator be taken as the multiplier. 9. In 5 T 3 dollars, how many tenths? Ans. J. 10. In 15 J yards, how many sixths? -4ns. 9 g 3 . 11. la 26Af days, how many 24ths? Ans. 6 2 3 7 . REDUCE TO IMPROPER FRACTIONS, 12. Si. Ans. 13. 5]. Am. U. 3U. Ans. i . . 17. lyMfo. Ans. 18. 14/ T . Ans. 19. 10 r { T . Ans. 15. 46} Ans. 3 | 3 . ART. 141. To reduce a whole number to a fraction hai'ii'ff a given denominator. 1. Reiucc 3 to a fraction whose denominator is 4. SOLUTION. bi nee there are 4 OPERATION. fourths in 1, in 3 there will be 3 4 = fourths in L times 4 fourths = 12 fourths; and hence, 3 = ^ 2 Ans. 12 = fourths in 3. 3dBk. 10 146 RAY'S PRACTICAL ARITHMETIC, Rule. Multiply together the whole number and the denomt* nator; beneath the product write the denominator. 2. Reduce 4 to a Frac. \vhose denom'r is 7. Ans. 2 7 8 % 3. Reduce 8 to ninths Ans. 7 ^. 4. 19 to nineteenths Ans. yy. 5. 37 to a fraction whose denom. is 23. Ans. S^J. j> CASE IV. ART. 142. To reduce compound to simple fractions. 1. Reduce f of to a simple fraction. ANALYSIS. | of | is 2 times as OPERATION. much as 1 of |, and i of | is 4 times J2 f 4 = 2 X 4_ 8 as much as 1 of ; but J ; of i T V 3 5 ~~ 3 X 5 ""1 5 (Art. 130); and hence, | ofi = 4 times J 3 = -* 6 (Art. 131), and f of f = 2 times T % = y \. In this operation, the numerators are multiplied together, as also are the denominators. *2. Reduce | of to a simple fraction. Ans. ^J. ^3. Reduce of | to a simple fraction. Hule for Case IV. Multiply the numerators together for a new numerator, and the denominators together for a new denominator. If mixed numbers occur, reduce them to improper fractions. 4. Reduce of | of 2| to a simple fraction. SOL. 2| = U, and I of f of y = \ \ 5. Reduce T 7 T of | to a simple fraction. Ans. -*|. 6. | of | to a simple fraction. Ans. -^4- 7. off of 1^ to a simple fraction. Ans. j|g. REVIEW. 140. How is a mixed number reduced to an improper frao-' tion, Rule ? 141. How is a whole number reduced to a fraction having * given denominator, Rule ? COMMON FRACTIONS, 14? 8. Reduce f of f of f to a simple fraction. SOLUTION. After indicating the OPERATION. operation, the numerator of the result will be 2X3X4; the denominator, 2 jf ji__ 3X4X5. ~Jj* X 5~~ The value of a fraction not, being i altered by dividing both terms by the same number (Art. 136), Cancel the factors (3 and 4,) common to both terms. As 3 = 3X1, and 4 = 4X1, the factors 1 and 1 will remain after canceling 3 and 4. Hence, the products of the remaining factors are 2x1X1, and 1X1X5, which give the terms of the required fraction in its simplest form. * 9. Reduce J of | of | to a simple fraction. Ans. 2 5 ? . *10. Reduce f of | of %\o a simple fraction. Ans. |. ART. 143. Hence, to reduce compound to simple fractions by Cancellation, Indicate the operation ; cancel all the factors common to both terms, and multiply together the factors remaining in each. REM. As all the factors common to both terms are canceled by the operation, the result will be in its simplest form. 11. Reduce f of | of / 3 of || to a simple fraction. SOLUTION. First, cancel the OPERATION. factors 3 and 4 in the numera- 2 tor, nnd 12 in the denominator, C| A V T8 2 Since 9 is a factor of 18, can- cel the factor 9 in both terms, and write the remaining factor, 2, above 18; ns 7 is a factor of 35, can- cel the fnctor 7 in both terms, and write the remaining fac- tor, 5, below 35. Then multiply the remaining factors as before. REVIEW. 142. How are compound reduced to simple fractions, Rule ? U3. How reduced by Cancellation ? Why is tho value of the fraction nofc iltwed ? KJSM. Why Is the result in ita lowest terms ? 148 RAY'S PRACTICAL ARITHMETIC. REDUCE TO SIMPLE FRACTIONS, 12. i of | of f Ans. i. 13. I of | of 1J. 4ns. 1. 14. f of f of lij. Ans. 1. 15. * off of If. Ans. 2. 16. -9. of T 7 g of 1. Ans. I 17. 1 of | of Jof 5. Ans, \. 18. I off of | of 1 of I Of 4 Of f 9 7t. 4/iS. -S 1 :,. For method of reducing complex to simple fractions, see page 167. CASE V. /ART. 144. To reduce fractions of different denominators to equivalent fractions having a common denominator. 1. Reduce I, |, and | to a common denominator. SOLUTION. The value of a OPERATION. fraction not being altered by 1X3X4 _ 1 2 new numer. multiplying both terms by the 2X 3X 4~~24 new denom. same number (Ait. 135), multiply the numerator and denominator *|**=4f ^ nUmer * of etch by the denominators of 3 X 2X4 24 new denom. the other fractions ; this will ren- 3 X 2 X 3__1J3 new nuraer. der the new denominator of each 4x2X3 24 new denom. the same,- since, in each case, it will consist of the product of the same numbers, that is, of all the denominators. *2. Reduce and ? to a com. denom. Ans. ^g, | *3. i, |, and | to a com. denom. Ans. |, |g, | Rule for Case V. Multiply loth terms of each fraction by the product of all the denominators except its own. NOTE. First reduce compound to simple fractions, and whole or mixed numbers to improper fractions. 4. Heduce J, |, and Z to a common denominator. SUGGESTION. Since the denomi- OPERATION. nator of each new fraction consists 1X3X5 = 15 1st num. of the product of the same num- 4X2X5 40 2d num. bers, (all the denominators of the 7X2X3=42 3d num. given fractions,) we multiply them 2X3X5=30 together but once. COMMON FRACTIONS. 149 Observe, that, in each case, the result is obtained by multiplying the numerator and denominator by the same number. 5. |,|, and |. Ant. |, |, |J. 6. J. }. and J-. ^.ns. 4?, 77%, A. i. ' J * 4 ^v 4 / *.4.' * 4 7. -|, f, and f . Ans. f , 4|, f. 8. *,$,*r*nd4. ^n. JA, A, A%, AV REDUCE TO COM. DENOMINATORS, i, *. an <* f. >4 20 Ans. 4 , i, -J, and |. jins. ||, I f> and f . ^.ns. |^, ^ i I and f. JLns. j^(j) "i2(j) i 1, ?, and {{. ^ffg> lift -ft, f and T 5 . ^ s - T 6 oW> : A of |, 2J, and 3. An*. o 9 4, -| of g, and f of | of|ofl. 4** 9. ?, , ?, and g. ^ns. |JS, |JS, |j-g, |f | 10. 7 f , f and T 5 . -4w. T V \, T V^ T , T Vo 6 r . 11. I of -|, 2^ and 3. .4/is. ^, ||, ||. 12. I of g, and i of | of | of If. ^1/is. ||, ^ ^ART. 145. When the given fractions are expressed in small numbers, and the denominator of either fraction is a multiple of the denominators of the others, reduce them to a common denominator ; thus, Multiply both terms of each fraction by such a number as will render its denominator the same as the largest denomi- nator; obtain this number by dividing the largest denominator by the denominator of the fraction to be reduced. 1. Reduce i and i to a com. denom. OPERATION. j. X 3 o . SOLUTION. Since the largest denom., 6, is a 2X3 6" multiple of 2, multiplying both terms of -j^ by 3, reduces it to . Am. % and k A == 'i - o 6 By similar process, Reduce to Com. Denominators, EXAMPLES. ANSWERS. EXAMPLES ANSWERS. o *J . - 1 , and -| = 4' 3 4' G. i 5 G' A = - 4 ii. T 7 2- 3. ! ' y - 4 I, 1 7. i ii' 4 7' -h 1 ~ 1 4' T 8 4, T 9 ?- 4. i ,',_ ~ !b 3 f 6* 8. |, 81 M = fl, \l 1 i 1 IT = 7% T 4 a. T S 0- 9. o f, n = 1 8 3' A. Ji UEVIEW. 144. How are two or more fractions reduced to a common denominator ? Why is the value of each fraction not altered ? Why does this operation render the new denominator of each the same ? 150 RAY'S PRACTICAL ARITHMETIC. CASE VI. ART. 146. To reduce fractions of different denominator^ 1(5 40' 40' 40' ' f > T> 2?P M- ( See Note !) -^"S. 2 % ig, J, |J, For additional problems, see Ray's Test Examples. ART. 147. ADDITION OF FRACTIONS Is the process of uniting two or more fractional numbers. 1. What is the sum of J and f and f ? SOLUTION. Since the denominntors are the same, the numerators express parts of the same size: therefore, add 1 fifth, ] f 1 cent, 2 fifths, > as you would add < 2 cents, Z_ffths, } (j^ cents, The sum is 6 fifths (|) in one case ; 6 cen ts in the other. Hence, to add fractions having a Com. Denominator, find the sum of the numerators; write the result over the denominator. EXAMPLES FOR MENTAL SOLUTION. 2. Add 4' 2 4' and _3 4' 5. Add f' i. 5 6^ 3. i, 2 |, 5' 6. A' nr ' i%) 7%- 4. '' 2 7' 5 I- 7. A r-fe -i 8 ^ 5 TI. T?' REVIEW. 140. NOTE 1. Bpfore coramoncing the operation, what is require! ? 2. If there are compound fraction 5 " or mixed numbers ? NoTBo. What miy 0-5 omittcil ? Why ? 4. What is the objcetin reducing fractions to a crmmon denominator? 147. What is Addition of Fractions ? How add fractions having a common denominator ? Why ? COMMON FRACTIONS. 153 ART. 148. 1. What is the sum of i and -J ? SOLUTION. Since the denominators are dif- OPERATION. ferent, the numerators do not express things ^ =: | of the fame unit value, and they can not be JL 4 added together. The sum of 1 half and 1 third, 3 i 2 5 A is neither 2 halves nor 2 thirds. But, reduced to a common denominator, (Art. 144), 1 half =3 sixths, and 1 third = 2 sixths ; their sum is 5 sixths (|). *2. Add * and ^ Ans. fo *3. Add and Ans. U=1 T V Rule for Addition, Reduce the fractions to a common denominator ; add their numerators together, and place the sum over the, common denominator. NOTES. 1. Reduce compound to simple fractions, and each frac- tion to its lowest terms, before commencing the operation. 2. Mixed numbers may be reduced to improper fractions, and then added ; or the fractious and whole numbers may be added separately, then united. 3. After adding, reduce the result to its lowest terms. Art. 133. 4. What is the sum of f, |, and ? OPERATION. 2)3 4 6 12-^-3 = 4, and 4X2= 8 3)3 2 3 12-f-4 = 3, and3x3= 9 121 12-r-6 = 2, and2x5=10- A n m REVIEW. 148. Why can not 1-half and 1-third be added, without re- ducing them to the same denominator ? What the rule for addition ? New 154 HAY'S PRACTICAL ARITHMETIC. 8. Add | and. Am. 1 T V|11. Add J, f, {J. ^ins. 2^. / 9. Add |, f ^ns. if. 12. Add I ', 7 \. ^In*. $$i. 10. Add * . 14. Add -,V, T ' 5 , T \, and ^ ..... 15. Add 2 and 3 ....... 4n. 5J. 16. Add |,~7i, and 8| ...... 4n*. 17^- 17. Add 16|, 12}, 8|, and 2]. . . . An*. 40^. . Add | of |, and \ of 4 of &. . . Am. f . 19. Add i of 1A, ^ of 21, y of i, |. . Ans. 2]|g. 20. Add |, T V/o> 7 | n , and ^^. . . Ans. 1. I* For additional problems, see Ray's Test Examples. ART. 149. SUBTRACTION OF FRACTIONS Is the process of finding tlie difference between two fractional numbers. 1. What is the difference between and f ? SOLUTION. Since the denominators are the same, the numerators express parts of the same size: therefore, subtract *2 sevenths from 5 sevenths as you would 2 cents from 5 cents. Thus, 5 sevenths, 5 cents, 2 sevenths, 2 cents, Difference 3 sevenths (|) in one case; 3 cents in the other. Hence, to find the difference between two fractions having a common denominator, Find the difference between their numerators, and write the result, over the common denominator. ** QUESTIONS FOR MENTAL SOLUTION, 2. What is the difference between | and f ? | and .' f and*-? -ft and A? V *nd { ? V and | ? IEW. 1:8. NOTE. If there are Compound Fractions, what i? re- quired ? What if each fraction is not ia its lowest ternu ? How are mixed Bujubors added ? COMMON FRACTIONS. 155 ART. 150. 1. Find the difference between J and f . SOLUTION. Since the denominators are OPERATION. different, the numerators do not express i~i things of the same unit value: hence, one can 2~6 not be subtracted from the other. Art. 25. Thus, the difference between 1 half and 2 g t~i Ans. third* is neither 1 half nor 1 third; but, reduce them to a common denominator (Art. 144), and 2 thirds = 4 sixths, and 1 half =3 sixths; their difference being 1 sixth (J). *2. The difference between J and |. Ans. y 1 ^. *3. Between A and i Ans. -f d . Kule for Subtraction. Reduce the fractions to a common denominator, find the difference of their numerators, and place it over the common denominator. NOTE. Reduce compound to simple fractions, and each fraction to its lowest terms, before commencing the operation. After sub- tracting, reduce the result to its lowest terms. 4. What is the difference between | and -^ ff ? OPERATION. 6 = ^X3 30-f- 6 = 5, and 5X5 = 25) New 10 = 2X5 30-r-10 = 3, and 3x3= 9 j Nume'rs. 3X2X5 = 30, Least Com. Mult. 5. 8 _ 3 i = Ans. i- 10. i i_ = Ans. TyV 6. 1 -7- --- Ans. i 11. 5 3 Ans. ij. 1 tf 1 2* G 8 ii4 7. 1 1 = Ans. r'V 12. -k = ^Ins. T ? 5 . 8. I 1 = Ans. TjV 13. 4 1 T5 TO = J.AIS. . 9. 4 1 = Ans. *ff- 14. fl-A = ^s. J, PV-EM. In finding the difference between mixed numbers, either reduce them to improper fractions, and to a common denominator, and then make the subtraction; or, find the difference between the whole numbers and the fractions separately. REVIEW. 149. What is subtraction of fractions ? IIow find the dif- 'crcnso between two fractions having a com. denominator? Why? 156 BAY'S PRACTICAL ARITHMETIC. 15. From 3 subtract 1. SUGGESTION. As 4 sixths can OPERATION. not be taken from 3 sixths, 3^ =.7 = ^1 Or, 3-1=33 borrow 1 from 3, reduce it to -10 J 10 .. ^ -, ^ sixths, add them to the 3 sixths, 3 ~ a ~ jL__ A 3 _ making 9 sixths; then subtract. =: -*-o- *<>' 16. From 5 subtract |. OPERATION. SUGGESTION. The subtraction may ^ j 5 ~ ^ be made by reducing 5 to thirds; or, by * o borrowing 1 from 5, and reducing it to -J. 3 thirds, as in the above example. Ans. - 3 =4^. 4^. 17. 4f 2] = 2J. ^8. 8J-3= 4^. 19. 5^ 4-J= li. 20. 7-4|= 20 21. 5 1A = 3i. 22 V 8 = 7. '23. 1145 = 24. 8 3 T 3 5 = 25. foflO of6 = 26. ioff -Jofg = 27. 14]-Jofl9= 28. of72 of21 = For additional problems, see Ray's Test Examples. '/ART. 151. MULTIPLICATION OF FEACTIONS. CASE I. To multiply a fraction by a whole number. This operation consists in taking the fraction as many times as there are units in the multiplier. 1. If 1 apple cost I of a cent, what cost 3 apples? SOLUTION. Three apples cost 3 times as much as one; that is, J taken 3 times: !+-[- = -|X3^f. (Art. 131.) 2. If 1 lemon cost | of a cent, what cost 4 lemons ? Ans. I taken 4 times; |+|+| + | = |X4= 1 ^=2|. REVIEW. 1"0. Why can not one-half be taken from two-thirds without reducing to the samo denominator ? What the rule for subtraction ? 150. NOTE. What is required if there are compound fractions? What if each fraction is not in its lowest terms ? REM. HOW find the difference between two mixed numlers ? Between a whole number and a fraction ? COMMON FRACTIONS. 157 ANOTHER SOLUTION. Since 3 fifths express three parts of the same size, we may multiply 3 fifths, as we would multiply 3 cents. Thus, 3 fifths And, 3 cents multiplied by 4 ' multiplied by 4 the product is 12 fifths ( l f). the product is ]2 cents. Hence, To multiply a fraction by a whole number^ multiply its numerator. EXAMPLES FOR MENTAL SOLUTION. 7. s X 6 = Am. 2;}, 8. f X 8 = Ans. 5J-. 9. X 6 = Ans. 4. 3. \ X 3 = 4. -J X 4= Ans. I 1 .. 5. | X 6 == Ans. 4. 6. | X 10 = Ans. 6. 10. f X 10 = Ans. 6]. 11. What is the product of -J multiplied by 4 ? SOLUTION. Since a fraction is multiplied by multiplying its numerator, (Art, 131,) multi- I><4=\ 8 = 2=3J- Ans. ply 7 (eighths) by 4, and the or, product is 28 (eighths); which, ?X4= , 2 = ^ = 3^. reduced, equals 3-1. Or, since a fraction is multiplied by dividing its denominator, (Art. 134,) if the denominator, 8, is divided by 4, the result is 2 3^, the same as by the first method. *12. g X 3 = Ans. 2|, | *13. ^ X 4 *= Ans. 2J. Rule for Case I. IST. Multiply the numerator of the fraction by the whole number, and under the product icrite the denominator. Or, 2o. Dinde the denominator by the whole number, ichen it can be done without a remainder; over the quotient write the numerator. 14. JJX12 = Ans. 7}|. | 15. 2 2 T X 7 = Ans. |. NOTE. When the denominator of the fraction, and the whole number contain common factors, employ Cancellation. 158 RAY'S PRACTICAL AHITHMETIC. 1C. J^X 9= Ans. 1}. 17. gf X 8= Ans. 3*. 18. -X21= -4ns. 13i. 19. j ; J ; XlO= Ans. 7*. 20. f X 9= Ans. 5. 21. 3' X 4= (Seesug.) Cue. In multiplying a inixel by a OPERATION. 3J whole number, multiply the fractional 4 part and the whole number separately, Ans 1 3 l i hen unite their products; or, icJuca A o ^ the mixed number to an improper fr^c- ' j == a ' tion, then multiply. V> X 4= 4o = 131 :** 22. 18J x 8 = ^ns. 150. 23. 16* X 3 = Ans. 50. 24. 103X7= .47*s. 75|. 25. lOji X 9 = Ans. 974. * o CASE II. ART. 152. To multiply a ivhole number by a fraction. Multiplying by a whole number, is taking the multiplicand as many times as there are units in the multiplier. Multiplying by a fraction, or part of a unit, is taking a part of the multiplicand. Therefore, Multiplying by i, is taking 1 half the multiplicand. Multiplying by , is taking 1 third of the multiplicand. Multiplying by |, is taking 2 thirds of the multiplicand, &c. Hence, To multiply ly a fraction, is to take such a part of the multiplicand, as the multiplier is part of a unit. 1. At 12 cts. a yard, what cost -J of a yard of ribbon ? SOLUTION. If 1 yard cost 12 cents, 1 third of a yard will cost 1 third as much, that is, (Art. 54,) 12 X^ = 4 cents. Ans. 2. At 12 cts. a yard, what cost f of a yard of ribbon ? SOLUTION. TVo-tkirda -will cost twice as much as one-lh\\^] but, 1 third costs 4 cents, hence, 2 thirds cost 4X2 = 8 cents, that is, (Art. 54,) 12X = 8 cents. Ans. KSVIETV. 151. In what docs the multiplication of a fraction Y>y r, whole number consist? "What is tho first, method? The second nr^hod V 131. NOTE. When tho denominator and whole number have common fac- tors, how shorten the process ? How multiply a mixed by a whoie number ? COMMON FRACTIONS. 8. What is the product of 20 multiplied by f ? SOLUTION. Three- fourths are 3 times 1 fourth; 1 fourtb Df 20 is 5, and 3 times 5 are 15: Ans. Or, \ of 20 is 2 ^> ; and 3 times 2J> are 2J> x 3 = 6 ^ = 15. EXAMPLES FOR MENTAL SOLUTION. 4. 8X1= Ans. 6. 5. 12 X I = Ans. 8. 6. 10 X f = Ans. 4. 7. 14 X ? = Ans. 10. 8. 16 X f = Ans. 14, 9. 5 X | = Ans. 3f . 10. 7 X f = -4ns. 4|. 11. 8X|= -4ns. 4? . Rule for Case II. IST. Divide the whole number ly tht denominator of the fraction : multiply the quotient ly the numerator. Or, 2a Multiply the whole number ly the numerator of the fraction, and divide the product by the denominator. REM. 1. The 2d rule is best, when the denominator of the fraction is not a factor of the whole number. 2. Since the product, of two numbers is the same, -whichever is the multiplier, (Art. 30,) the examples in this and the preceding case may be performed by the same rule. 3. When the multiplier is a whole number greater than 1, the multiplicand is increased ; when it is a proper fraction, the mul- tiplicand is decreased, the product being the same part of the multiplicand, that the multiplier is o'f unity. 4. Multiplying by a fraction always involves division. Thus, multiplying by one-third, is the same as dividing by three. 12. 28 X 4 = Ans. 16. 15. 31 X 2 3 "~ Ans. 20 2. 13. 36 X 9 = Ans. 28. 16. 29 X ! = Ans. 21 f. 14. 50 X -,'0 = Ans. 45. 17. 37 X 1 = Ans. 29|. REVIEW. 152. In what does multiplication by a whole number consist ? In what the multiplication by a fraction, or part of a unit ? What U aaulti plying by one-half ? By one-third? Two-thirds? 152. In multiplying fey a fraction, what part of the multiplicand is takes f What is Bole for Case II, 1st ? 2d Rule ? BXM. 1. Which ia boat ? 160 RAY'S PRACTICAL ARITHMETIC. 18, Multiply 8 by 3~. OPERATION. o Q Q2 = 1_1 In multiplying a whole by a ^ ? a 3 mixed number, multiply by the 8X V = S -j 8 29^. integer and the fraction sepa- 24 rately, then unite their pro- 5'rriSX? iucts; or, reduce the mixed number to an improper frac- 29| Ans. tion, then multiply 21. 55X93=^5.518;*. 22. 64X82= 4ns. 568. 19. 25X83= ^Ins. 215. 20. *a CARE III. ART. 153 a . To multiply one fraction ly another. 1. \V T hat is the product of * by f ? SOLUTION. To multiply % by f, is to OPERATION. take f of f (>*U 152), and this is equal 4 X = if- ANALYSIS. Now, | of $ is 3 times J of $' and J of 4 is 4 times i of i; but i of ^ 3^ ( Art - 130); hence, 4 of ^ is 4 times 3*5, and | of I ^s 3 times ^5 ||- This is multiplying the numerators together anJ the denominators together. *2. What is the product of | by \ ? . . Ans. 2 s a . p ^3. The product of -| by f ? .... ^ls. i|. Ullle for Case III. Multiply together the numerators for a new numerator, and the denominators for a new denominator. HEM. Multiplying a fraction by a fraction, is the same ai reducing a compound to a simple fraction. 4. -^ X ? - 5- -?i> x B "^ ^ ns - T 5 o 5 d- . 152. UEM. 2. "Why m.iy the examples in this and the preceding caso bo solved by the snmo rule ? 152. KEM. 3 \Vbon is the multiplicand increased by multiplying? "When (Increased t 4. "What operation id involved in multiplying by a fraction? 11. 2' X 21. ^ns. 61. COMMON FRACTIONS. 161 NOTE 1. Reduce mixed numbers to improper fractions. 8. Multiply 21 by 3. SOL. 21=|, and 3^ = 7. then | x 7 = y _- 77 9. 8|X|. Ans. 3|. 10. T %X17 T 3 T . Ans. 15 T 6 T . 12. 16iXl6i. Ans. 272|. NOTE 2. After indicating the operation, employ CANCELLATION, when possible, as shown in Art. 61. 13. Multiply 1 of 4 by J of 5. . . Ans. 1. 14. Multiply f of 41 by 1 of 3^. . . Ans. If 15. Multiply | of 91 by of 17. . . Ans. 78|. 16. Multiply J^ of 71 by ji of 86 T 3 T . Ans. 417^|. 17. Multiply 3, 2|, f, and f of 5 together. Ans. 17|. 19. IX f X|X|X|X| of 6 = Ans. I. J20. fX|Xl|X^ of|X|X|of 20 = ^4n. f. 21. 2iX6|-X3iX T 7 3 of 2Xf = Ans. 24. For additional problems, see Ray's Test Examples. 153 . MISCELLANEOUS EXAMPLES. What will be the cost 1. Of 21 Ib. of meat, at 11 cts. a Ib. ? Ans. 2| ctsi 2. Of 3 yd. linen, at $f a yd. ? of 5 yd. ? of 7 yd. ? of 6^ yd. ? 5| yd. ? ^Lrcs. to last, $3|. 3. Of 31 Ib. of rice, at 4| cts. a Ib. ? ^ircs. 16 cts. 4. Of 3 i tuns of iron, at $18f per T. ? ^1/is. $60. 5. Of If yd. of muslin, at $^ per yd. ? Ans. $J . 6. Of 2i Ib. of tea, at $| per Ib. ? Ans. $2. 7. Of 5| cords of wood, at $l-i per C.? Ans. $6f . REVIEW. 152. How may a whole be multiplied by a mixed number? 3 a . How multiply one fraction by another, Rule ? NOTE 1. What is re- quired when mixed numbers occur? 2. When employ Cancellation? 3d Bk. 1 1 162 RAY'S PRACTICAL ARITHMETIC. 8. At the rate of 5^ miles an hour, how far will a man travel in 7| hours ? A ns. 42 f mi. 9. I own | of a steamboat, and sell | of my share : what part of the boat do I sell ? . Ans. f . 10. At $6| per yard, what cost f of a piece of cloth containing 5^ yards? Ans. $8^. 11. | off of 161, X | of | of 15,=what? Ans. 34j. 4&" For additional problems, see Ray's Test Examples. ART. 154. DIVISION OF FRACTIONS. CASE I. To divide a fraction by a whole number. The object in dividing a fraction by a whole number, is to separate the fraction into a given number of equal parts, and find the value of one of the parts ; or, to find what part a traction is of a whole number. 1. If 3 yards of ribbon cost 9% , what cost 1 yard ? ANALYsis.-If 3 yards cost $ of OPERATION. a dollar, 1 yard, being l of 3 yards, fi ON fi will cost i of 9=3 of a dollar. | -f- 3 = J| = f Ans. Or, since a fraction is divided by multiplying its denominator . Or, (Art. 133), multiply 7 by 3, and the result, , reduced, is , as by the | -4- 3 = ^ =^=2. first method. Here, ^ is divided into 3 equal parts, and the wzfoe of each part is |; thus, =f -ff + ?; the number of parts corresponds to the aivisor, and the vafoe of each, to the quotient. 2. At 2 dollars a yard, what part of a yard of cloth can be bought for | a dollar ? ANALYSIS. Had it been required to OPERATION. find how many yards, at $2 a yard, could 111 be bought for $6, the $6 should be divided 2 "^ 2 X 2 == 4 by $2 ; and, to find the part of a yard that $| will pay for, divide $ by $2 : to divide Ans. \ yd. by 2, multiply the denominator (Art. 133) ; the quotient is one-fourth, COMMON FRACTIONS. 163 *3. If 4 yards of muslin cost f of a dollar, what will 1 yard cost ? Ans. $|. *4. If 1 orange cost 3 cents, what part of an orange can be purchased for ^ a cent? Ans. . Rule for Case I. Divide the numerator by the whole num- ber, when it can be done without a remainder ; write the quotient octr the denominator. Otherwise, Multiply the denominator by the whole number, and write the product under the numerator. 5. If 4 yards of cloth cost || of a dollar, what will 1 yard cost? Ans. $ T 3 g. 6. If a man travel T 9 T of a mile in 3 hours, how far does he travel in 1 hour ? Ans. j\ mi. 7. If 5 yards of tape cost ^ of a dollar, what will 1 yard cost? Ans. T ^. 8. If 7 pounds of coffee cost <|f of a dollar, what will 1 pound cost ? Ans. $ 2 2 3 . 9. At 4 dollars a yard for cloth, what part of a yard will if of a dollar buy? Ans. 2 \yd. 10. At 5 dollars a tun, what part of a tun of hay will ^ a dollar purchase ? Ans. y 1 ^ T. /"ll. At 6 dollars a barrel, what part of a barrel of flour will $2| pay for ? *-p , . , . OPERATION. Keduce a mixed number to an improper fraction, and it may be divided by a whole 5 = 1> number, the same as a proper fraction. ^ -*-6 = f Ans. 12. If 5 bushels of wheat cost 3| dollars, what cost / 1 bushel ? Ans. | of a dollar. 13. If 7 ounces of opium cost 8| dollars, what COSY, 1 ounce? * Ans. |=$1^. 14. If || be divided into 9 equal parts, what will each f part be ? Ans. T \. REVIEW. 154. What is the object in dividing a fraction by a whole number ? In Ex. 1, what does the divisor show ? The quotient ? 154. In Ex. 2, what does the quotient show ? How divide a fraction by a whole number, Rule for Case I ? 164 RAY'S PRACTICAL ARITHMETIC. 15. 44 -5- 8 = Am. 16. | -*_ 5 = Ans. 17. 12-f-ll = Ans. 18. 3 19. 47f -=-15=^*5. 20. 130f-f-18==.4n. (;ASE II. ''''ART. 155. ^b divide a whole number by a fraction. Dividing a whole number by a fraction, is finding how many times the fraction is contained in the whole number. 1. At of a cent for 1 lemon, how many can be bought OPSKATION. SOLUTION. In 4 cents there are 12 thirds of 1 cent (Art. 141). If 1 lemon costs 2 thirds of a cent, there will be as TMrds many lemons as 2 thirds are contained times in 12 thirds; that is, 6. Ans. 6 lemons. Ans. 6 Or thus : | is contained in 4 as many times as thgre are thirds in 4, that is, 12 times; and 2 thirds in 4, one-half as many times as 1 third; 12-=-2=6 times. In this operation, the whole number is reduced to the same name the same parts of a unit as the divisor, that the divisor and dividend may be of the same de- nomination. Art. 41, Rem. The whole number is multiplied by the denominator of the fraction, and the product divided by the numerator. *2. At A a cent each, how many apples can be bought for 3 cents"? Ans. 6 apples. *3. At | of a dollar per yard, how many yards of cloth can you buy for 6 dollars? Ans. 8 yards. Rule for Case II. Multiply the whole number by the denom- inator of the fraction, and divide the product by the numerator. REVIEW. 155. What is dividing a whole number by a fraction ? How many times is one-third contained in 1 ? In 2 ? In 3 ? In 4 ? How many times two-thirds in 1 ? In 2 ? In 8 ? In 4 ? COMMON FRACTIONS. 165 4. 4-4- = Ans. 10. 5. 16-*-! = Ans. 21' . 6. 8 -r- T 2 ^ = Ans. 60. 7. 6-j- = J.W8. 14. 8. 13 -f- | = .4ns. 21~. 9. 21-r- T 7 T = J.n. 33. 1 1 '10. In one rod there are 5| yards : how many rods are there in 22 yards ? Reduce a mixed number to OPERATION . 51 = 1 1 an improper fraction, and a whole number may be divided ~ ^.ssL -= ^ 4 rods. 4ns. by it, as by a proper fraction. f \\. At 2| dollars for 1 yard of cloth, how many ya v ds lean be bought for $6 ? 4ws. 24 yd. 12. At 3| cents a lb., how many pounds of rice can ^ be bought for 30 cts. ? Ans. 8 lb. ( 13. How many times 4| in 50? Ans. 11^ } . V 14. Divide 56 by 5|. Ans. lOf. CASE III. ART. 156. To divide a fraction by a fraction. The object in dividing a fraction by a fraction, is to find how many times the divisor is contained in the divi- dend ; or, what part the divisor is of the dividend. 1. At T 2 ^ of a dollar per yard, how many yards of muslin can be bought for $ T 9 ^ ? SUGGESTION. Find how often 2 tenths OPERATION. are contained in 9 tenths, as you would find Tenths 2)9 tenths, how often 2 cents are contained in 9 cents ; . ~4.T~ that is, by dividing 9 by 2 : 9-^-2=4-^. Hence, when two fractions have a common denominator, obtain their quotient by dividing the numerator of the dividend by the numerator of the divisor. 2. How many times f in | ? OPERATION. 2X4 3X3 . We can not find how often 2 3 X 4 = T ^ ' 4 X 3 J ^ ' inches are contained in 3 feet, without expressing the divisor T 9 5 -j- T ^j = 9 -r- 8 = 1 g Ans. and dividend in the same de- nomination, inches; so, also, to find how often 2 thirds ar 166 RAY'S PRACTICAL ARITHMETIC. contained in 3 fourths, reduce them to the same denomination, twelfths : | are 8 twelfths, and | are 9 twelfths ; and 8 twelfths are contained in 9 twelfths, 9-i-8 = lJ. No use is made of the common denominator, the numerator of the divisor being multiplied by the denominator of the dividend, and the numerator of the dividend by the denominator of the divisor. This is easily performed by inverting the terms of the divisor, then proceeding as in Multiplication of Fractions, Art. 153. TVnc 3 2 3V3 9 11 A. -LHUSj - -T 3 4 A ^- Q J-g Jfts. ANOTHER SOLUTION. If | be divided by 2, the quotient, o (Art. 133), is j ; but, since 2 is 3 times |, the divisor used 4^ * is 3 times the given divisor; hence, multiply this quotient by 3 to obtain the true quotient: this gives | X 4 = lg Ans. *3. At | of a dollar each, how many knives can you buy for | of a dollar ? Ans. 2. *4. At i of a dollar per yard, how many yards of ribbon can be purchased for | of a dollar? Ans. 3|. Rule for Case III. Invert the divisor ; multiply the nu- merators together for a new numerator, and the denominators for a new denominator. NOTE. Reduce compound to simple fractions, and mixed num- bers to improper fractions, before commencing the operation. FIND THE QUOTIENT OF = Ans. 3 1 25. 5. 3 . 1 = Ans. 3. 10. 21 - J ff = 6. V~*-\ = Ans. 2. 11.. 41-:- I* = 7. i-t-3 = Ans. ^. 12. 4| -f- 5| = 8. 5 '2 = Ans. 11. 13. 2|--7^ = 9. -i -5-J = Ans. 11. 14. 2-*" A = 15. Divide of | by ^ of | 16. Divide I of 51 by | of 171. EEVIEW. 155. How do you divide a whole number by a fraction? 156. What is the object in dividing a fraction bv a fraction? COMMON FRACTIONS. 167 ART. 157. The rules in the three preceding Articles may be embraced in this General Rule for Division of Fractions, Express the divisor and dividend in the form of a fraction ; invert the di- visor ; cancel all the factors common to both terms ; then multiply together the numbers remaining in the numerator for a new numerator, those in the denominator for a new denominator. 1. Divide of f by f of f . OPERATION. In this operation, cancel the factors 2 2 . and 3 on both sides above and below the f X o X p X 4= 12 line then multiply together the factors remaining on each side. 2. il --21 = Ans. ft. 5. 14 -*-8f = Ans. If. 3. 8|-7-35 = Ans. ft. . 6. f^~^~|f = Ans. j 9 ^. 4. 8 -5- jf = Ans. 8. 7. . 12f-^-4| = Ans. 3. 8. Divide T 5 g of f of 12 T 3 by 1 of 81. . . Ans. f . REDUCTION OF COMPLEX TO SIMPLE FRACTIONS. ^.RT. 158 a . A complex fraction is the expression of an unexecuted division (Art. 125), in which the divisor cr dividend, or both, are fractions. U . Thus, si indicates that 1| is to be divided by 2|. Hence, to reduce a complex to a simple fraction, Regard the numerator as a dividend, the denominator as a divisor, and proceed as in Division of Fractions, Art. 157. 1. Reduce of to a simple fraction. iVssi'i OPERATION. 4 4 C 5 _i_J __. | >< 3 = || AnS. In this operation, after reducing the mixed numbers to i n- proper fractions, the numerator of the result is the product of tiio REVIEW. 156. When two fractions have a common denominator, how obtain their quotient ? How find how often 2 inches are contained in 3 feet ? How often two-thirds are contained in three-fourths ? How divide a frac- tion by a fraction ? 157. What is the General Rule 2 168 RAY'S PRACTICAL ARITHMETIC. extreme terms, 5 and 3 ; the denominator, the product of the mean terms, 4 and 7. Hence, To REDUCE A COMPLEX TO A SIMPLE FRACTION, Reduce mixed numbers to improper fractions, then multiply together the extreme terms for a numerator, and the mean terms for a denominator. Reduce these complex to simple fractions : 2. JL = V 3 - 4= Ans. 4. _ =^77 6 ^^ ' _ 3| TT' 3 ! . _* AM* 2 5|~ Complex fractions may be multiplied together, or, one divided by another, by first reducing each to a simple fraction. Indicate the operation, and cancel. SlVTnlfiT^l^r ~ htr Ans. ^j. ^3 ^2 ( 9. Multiply Q 5 of o 2 "^y " I 2 5 1 10 "HlVirlA 3 \*TT _ ^ . . 158^. MISCELLANEOUS EXAMPLES, 1. At ^ a dollar per yd., how many yards of silk can b* bought for $31 ? Am. 6 yd. 2. At | of a dollar per pound, how many pounds of tea can be purchased for $2 T 3 ? Ans. 3|lb. 3. Find the quotient of divided by 2 ; by ; by | ; by i ; b y 4 ; b y ruW ^ ns - to last ? 50 - 4. At 3| dollars per yard for cloth, how many yards can be purchased with $42^ ? Ans. 11^ yd. 5. By what must | be multiplied, that the product may be 10? Ans. 26f . REVIEW. 158 a . What is a complex fraction? How reduced to a simple fraction ? COMMON FRACTIONS. 169 6. Divide 3| by f of Ij ....... Ans. 5J. x 7. Divide T 4 T of 27 by ^ of 21]. . . 4w. Iff. 11 2^- / 8. Divide - by f ART. 159. EXAMPLES IN u. s. MONEY. 1. Add $16. 06]; $9.12^; $5.43f; 82.81]- Ans. $33. 43 1 2. I paid for books $9.121; paper $4.43f ; a slate $0.37|; quills $1.621: what did I pay? Ans. $15.5(i] 3. Having $50.25, I paid a bill of $27.18f : how much had I left? Ans. $23.06] 4. From $32.31] take $15.12^ Ans. $17.18| 5. From $5.81] take $1.18| Am. $4.62^ Find the cost of 6. 9yd. of muslin at 12 lets, a yd. Ans. 81.12 7. 21 Ib. of sugar at 6] cts. a Ib. Ans. $1 .31] 8. 15yd. of cloth at $3.18| per yd. Ans. $47.81] 9. 51 yd. of linen at $0.62i per yd. Ans. $3.43| 10. 12^- yd. of ribbonat 18f cts. per yd. Ans. $2.34| 11. 13^ yd. of calico at 16f cts. per yd. Ans. $2.25 12. 10] yd. of cloth at $3.37-i a yd. Ans. 834.59?- 13. 17| dozen books at $3.75 per doz. Ans. $66.25 14. At 18|cts. per yd., how many yards of muslin can be purchased for $2.25 ? Ans. 12yd. ^15. At 37i cents per bu., how many bushels of barley can you buy~for $5.81] ? Am. 15^bu. 16. If five yards of cloth cost $11.56{, what cost one yard? Ans. $2.31] 17. Seven men share $31.06] equally: what is the share of each man ? Ans. 84 . 43 170 RAY'S PRACTICAL ARITHMETIC. EXAMPLES IN LONG MEASURE. 18. Reduce 5 mi. to inches. . . Ans. 316800 in. 19. 2 mi. 2 rd. 2 ft, to feet. . . Ans. 10595ft. 20. 20 yd. to rods. OPERATION. 5i yards = 11 halves, 20 yards = 40 halves. 11)40 ~~3 rd. 7 half yd. left, = 3| yd. Ans. 3 rd. 31 yd. SUGGESTION. In reducing numbers from a lower to a higher denomination, when the divisor is a fractional number (Art. 155), reduce both divisor and dividend to like parts of a unit. The remainder being of the same denomination as the dividend, (Art. 38), will be a fraction, which reduce to a whole or mixed number. Here, the remainder, 7 half yd., reduced, makes 3^ yd. ' 21. Reduce 15875 ft. to miles. Ans. 3 mi. 2 rd. 2ft. ' 22. 142634 in. to miles. Ans. 2 mi. 2 fur. 2 yd. 2 in. 23. How many steps, of 2 ft. 8 in. each, will a man take in walking 2 miles? Ans. 3960. 24. How many revolutions will a wheel, of 9 ft. 2 in. circumference, make, in running 65 mi. ? Ans. 37440. ^ EXAMPLES IN SQUARE MEASURE. 25. Reduce 1 A. 3R. 16 P. 25 sq. yd. to square yards. Ans. 8979 sq. yd. 26. 7506 sq. yd. to A. Ans. 1 A. 2 R. 8 P. 4 sq. yd. 27. 5 chains 15 links, to in. Ans. 4078| in. / 28. How many acres in a field 404 rd. Ions:, and 32 rd. wide? Ans. 8 A. 16 P. EXAMPLES IN TIME MEASURE. In these examples, the year is supposed to be 365^ days. '29. Reduce 4 years to hours. . . . Ans. 35064 hr. > 30. 914092 hr. to cen. Ans. 1 cen. 4yr. 101 da. 4hr. 31. In what time will a body move from the earth to the moon, at the rate of 31 miles per day, the distance being 238545 miles ? Ans. 21 yr. 24| da. COMMON FRACTIONS. 171 BEDUCTION OF FRACTIONAL COMPOUND NTTMBEBS, CASE I. ART. 160. To reduce a fraction of a higher denomina- tion^ to a fraction of a lower. 1. Reduce ^ of a peck to the fraction of a pint. SOLUTION. To reduce pecks to pints, OPERATION. we multiply by 8 -to reduce them to pk. qt. qt. pt. quarts, then by 2 to reduce them to i v 1 1 v 9 2. pints. In like manner, multiply the fraction or, of a peck by 8, to reduce it to the frac- i \/ Q \/ o o j. A . , -. , T>-T /\ O A, < = -^ pt. ^a/lS, tton of a quart, then by 2, to reduce - it to the fraction of a pint. Hence, fractional numbers may be reduced from a higher to a lower denomination, (Art. 81), by multiplying by that number of the next lower order which makes a unit of the higher. *2. Reduce ^ of a bu. to the fraction of a qt. Ans. |. Rule for Case I, Multiply as in Reduction of Whole Numbers, Art. 81, according to the rules for the multiplication of fractions. REM. The work in Cases I and II may often be shortened by Cancellation. ^3. Reduce - Ib. Av. to the fraction of an oz. Ans. . 4. T'e of a Ib. Troy, to the fraction of an oz. Ans. I- 5. & of a yd. to the fraction of a na. Ans. 4. 6. 7 T2HO of an A. to the fraction of a P. Ans. s- 7. sfo of a dollar to the fraction of a ct. Ans. (5 7' 8. T5H4 of a da. to the fraction of a min. Ans. JQ 1 1 ' 9. ife of a bu. to the fraction of a pt. Ans. 1- REVIEW-. 100. How reduce pecks to pints? How the fraction of _ peck to the fraction of a pint? How are fractional numbers reduced from a higher to a lower denomination ? RAY'S PRACTICAL ARITHMETIC. CASE II. ART. 161. To reduce a fraction of a lower denomina- tion, to a fraction of a higher. ^ 1. Reduce of a pint to the fraction of a peck. SOLUTION. To reduce pints to pecks, OPERATION. we first divide by 2 to reduce them p t. qt. qt. pk. to quarts, and then by 8 to reduce %^_2\ ; . 1 - i-8=- 1 - : them to pecks. In like manner, divide the fraction by 2, to reduce it to the fraction of a 1 * 2* g=2 1 ? P k - Ans. quart, and then by 8, to reduce it to the fraction of a peck. Hence, fractional numbers may be reduced from a lower to a higher denomination (Art. 81), by dividing the given fraction by that number of its own denomination which makes a unit of the next higher. * 2. Reduce f of a qt. to the fraction of a bu. Ans. J . Rule for Case II. Divide, as in Reduction of Whole Numbers, Art. 81, according to the rules for the division of fractions. 3. Reduce |- of a na. to the fraction of a yd. Ans. ^. 4. I of a gr. T. to the fraction of a Ib. Ans. 5. | of a 9 to the fraction of a Ib. Ans. 6. | of a pt. to the fraction of a bu. Ans. 7. ^ of an oz. to the fraction of a cwt. Ans. 8. of an in. to the fraction of an E.En. Ans. - 9. | of a min. to the fraction of a da. Ans. T ^ 10. '! of a dr. to the fraction of a qr. -4ns. . 160. What is Case 1? What is the rule for Case 1? 161. How reduce pints to pecks? How the fraction of a pint to the frac- tion of a peck ? How are fractional numbers reduced from a lower to a higher denomination ? What is Case 2 ? The Rule for Case 2 ? COMMON FRACTIONS, 173 CASE III. ART. 162. To find the value of a fraction in integer* (whole numbers) of a lower denomination. 1. Find the value of | of a day in integers. SOLUTION. J of a da., are | of 24 OPERATION. hr.; and | of 24 hr. are found by 24 hr. 60 min. ^nultiplying by 2 and dividing by 5 (Art. 152). This gives 9| hr. 5)43 5)180 Again. | of an hour are the same as | of GO minutes, which are 36 min. 9f hr. 3 6 min. Hence, | of a day =9 hr. 36 min. Ans. 9 hr. 36 min. *2. Value of i of a ini. ui integers. -4ns. 6fur. 16rd, Rule for Case III. Take the number of units of the next, lower denomination which forms a unit of the denomination of the fraction; multiply U by the numerator, and divide the product by the denominator. If this division produce a fraction, find its value in the same manner, and so on : the several quotients will be the answer. 3. What is the value of f of a dollar? Ans. 60cts. 4. Of | of a mile? Ans. 3 fur. 8rd. 5. Of i of a Ib. Troy ? . . . . Ans. 9oz! 12pwt. 6. Of 4 of a Ib. Av. ? .... Ans. 9 oz. 2* dr. 7. Off of an acre? .... Ans. 2R. 20 P. 8. Of of a T. of wine ? - Ans. 3hhd. 31 gal. 2qt. CASE IV. ART. 163. To reduce a quantity having one or more denominations, to flie fraction of another quantity composed of one or more denominations. 1. Reduce 2 feet 3 inches to the fraction of a yard : that ife, 2rf. oin. is what part of 1 yard? SoiUTioy.~2ft. 3 in. = 27 inches: 1 yd. = 36 inches; and since I mob. >& ^ of 36 inches, 27 inches are f J = j. A- f yd. 174 RAY'S PRACTICAL ARITHMETIC. 2. Find what part 2ft. 6 in. is oj SOLUTION. Reducing both quantities 2. Find what part 2ft. Gin. is of 6ft. Sin. OPERATION. ft. 6 in. = 30 in. to the same denomination, the first is 30, and the second 80 in. Since 1 in. is ^ of 80 in., 30 in. will be f g = |. Ans. Jjj- = f. *3. Reduce 2pk. 4 qt. to the fraction of a bu. Ans. |. *4. What part is 2yd. Iqr. of 8yd. 3qr.? Ans. ^. Rule for Case IV. Reduce both quantities to the lowest denomination in either ; the less will be the numerator, and the greater the denominator of the required fraction, which reduce to its lowest terms. 5. Reduce 13 hr. 30 min. to the frac. of a da. Ans. . 6. 3 fur. 25 rd. to the fraction of a mile. Ans. ||. 7. 2ft. Sin. to the fraction of a yard. Ans. f . 8. What part is 96 pages of 432 pages ? An^ f . 9. 15 mi. 3 fur. 3rd. of 35 mi. 7 fur. 7rd.? Ans. f 10. A man has a farm of 168 A. 28 P. : if he sell 37 A ^ 2R. 14 P., what part will he sell ? . Ans. $L. NOTE. If one or both quantities contain a fraction, reduce then/ to a common, denominator, and compare their numerators. 11. What part is 7oz. Igdr. of 1 Ib. Av.? Ans. |. 12. 2 qt. 1 1 pt. of 1 bu. 1 qt. If pt. ? Ans. ^ . 13. lyd. 1ft. l T 9 T in. of 3yd. 2ft. 8fin.? Ans. if if. ADDITION AND SUBTRACTION OF FRACTIONAL COMPOUND NUMBERS. ART. 164. 1. Add | of a yard, to | of a foot. SOLUTION. Find the value of each of the OPERATION. quantities in integers (Art. 162), and then 3 , ^ o' obtain their sum by the rule for addition in i '" Art. 101. If their difference be requi.-ed, &** = subtract according to the Rule in Art. 10 Ans. 3 1 COMMON FRACTIONS. 175 Rule for Addition or Subtraction, First, find the value of each fraction in integers (Art. 162) : Then add or subtract, as may be required, according to the rules for the addition and subtraction of compound numbers. NOTE. If fractions occur in the lowest denomination, add or subtract by the rules for addition and subtraction of fractions. 2. Add | of a da. to f of an hr. Ans. 16hr. 45min. 3. Add together |wk. |da. |hr. Ans. 2 da. 15min. 4. |wk. | da. f hr. f min. Ans. 5 da. 6hr. 40 sec. X 5. j gal. of wine, J 3 hhd. Ans. 6 gal. 1 pt. 1| gi. 6. How much land in 2f B., | A., and 3R. 28$ P.? Ans. 2A. 1R. 9'iP. 7. | of a da., less T ! s of an hr., equals what? Ans. 18 hr. 36min. 40 sec. 8. f of $1, less | of a dime? Ans. 55cts. 9. | of an oz., less | of a pwt. ? Ans. llpwt. 3gr. TO. | mi., less T 7 T fur.? Ans. Ifur. 5rd. 10ft. lOin. 11. fda., less'fhr. ? Ans. 19hr. 54min. IT^sec. 12. f E.En., less f yard? Ans. 3qr. jfna. ART. 165. PROMISCUOUS EXAMPLES. 1. Reduce f f-ff f to its lowest terms. . . Ans. {^. 2. Find the sum of | and f ; of f and f ; of T 3 ^ and J4 j of 2J, 3, T \, and / T . Ans. to last, 6' f . ' 3. What is the difference between | and | ? -? and T 4 T ? 34 and If? 3| and | of 3 ? ^ws. to last, 2^J. 4. Add f of T 7 a to | of T 7 3 ...... . 162. Two-fifths of a day, are two-fifths of what? How find two-fifths of 24 hours ? What is Case 3 ? What is the Eule ? 163. What is Case 4? Whafr the Rule ? NOTE. If one or both quan- tities contain a fraction, how proceed ? 164. What is the rule for tha addition or subtraction of fractional compound numbers ? 176 RAY'S PRACTICAL ARITHMETIC. 5. To the quotient of lf-i-2i, add the quotient of5i-f-3|. Ans. 2fj. 6. What number divided by f will give 10 for a quo- tient? Ans. 6. 7. What number multiplied by f will give 10 for a product? Ans. 16|. 8. What number is that, from which if you take f of itself, the remainder will be 16 ? Ans. 28. 9. What number is that, to which if you add f of itself, the sum will be 20 ? Ans. 14. 10. A boat is worth $900 ; a merchant owns f of it, and sells | of his share : what part has he left, and what is it worth ? Ans. -f% left, worth $375. 11. I own -j 7 ^ of a ship, and sell | of my share for $1944| : what is the whole ship worth? Ans. $10000. 12. What part of 3 cents, is f of 2 cents ? Ans. |. 13. What part of 368, is 176 ? Ans. j . 15. What number,+ T 3 IJ of -^ of 4 T 9 5 ,=1 ? Ans. T 3 g . 16. From the quotient of f -s-f, subtract the quotient n f n 1 Avto 5 9 >I g yy Ji.nS. 24 3 . 17. If I walk 2044 rods in T 7 ^ of an hour, at that rate how far will I walk in Ijihr. ? Ans. 8468 rd. 18. What part of 1| feet, is 3^ inches? Ans. f. 19. Two men bought a bl. of flour ; one paid $3, and the other $3= : what part of it should each have? Ans. One y 4 ^, the other -^V 20. A has $2400 ; | of his money,+$500, is f of B's : what sum has B ? Ans. $1600. 21. John Jones divided his estate among 2 sons and 3 daughters, the latter sharing equally with each other. The younger son received $2200, which was T 5 ^ of the share of the elder, his share being |f of the whole estate : find the share of each daughter. Ans. $1356|. XII. DECIMAL FRACTIONS. ART. 166. A common fraction (Art. 124), is one whose denominator may be any number ; as, 2, 3, 4, &c. A decimal fraction is one whose denominator is 10, or a number of 10's multiplied together ; as, 100, 1000, &c. EXPLANATION. A common fraction arises from dividing a unit into any number of equal parts. A decimal fraction arises from dividing a unit by 10, or some multiple of 10 by itself; and its denominator is not usually expressed in figures. AKT. 167. If a unit be divided into 10 equal parts, each part will be 1 tenth; thus, T \y of 1 = r ^. If each tenth be divided into 10 equal parts, the unit will be divided into 100 equal parts, each part being 1 hundredth; thus, y 1 ^ of T 1 = T ^. If each hundredth be divided into 10 equal parts, the unit will be divided into 1000 parts, each part being 1 thousandth of the whole ; thus, y 1 ^ of T ^ = J-QQ-Q. A comparison of the fractions y 1 ^, T ^Q, y^ 1 ^, &c., shows that the value of each is one-tenth of that which precedes it ; that is, they decrease in value tenfold. ART. 168. The orders of WHOLE numbers decrease from left to right tenfold ; thus, 1 hundred is one-tenth of 1 thousand ; 1 ten is one-tenth of 1 hundred ; and 1 unit is one-tenth of 1 ten : in like manner, The orders may be continued from the place of units toward the right, by the same law of decrease ; and, The first order on the right of units will then express tenths of units, that is tenths; the next order, tenths of tenths, or hundredths ; the next, tenths of hundredths, or thousandths ; and so on : Hence, if a point (.) be placed to separate units and tenths, J0 may be written in the order of tenths ; thus, .1, as 1 (unit) is written in the order of units : and, T 2 u> 7%> T 4 o> & c -> ma J be written, .2, .3, .4, &c. REVIEW. 166. What is a Com. Fraction ? A Decimal ? 167. What ia 1 tenth? 1 hundredth ? 1 thousandth ? What the value of each ? 3d Bk. 12 1T7 178 RAY'S PRACTICAL ARITHMETIC. Also, yfo, T g u , T {fo, c., may be written in the order of hwidredths, thus: .01, .02, .03, &c. : here, There being no tenths, place a cipher in the order of tenths. As the orders decrease from left to right, in the same manner as in whole numbers, Decimal Fractions may be expressed in figures, without writing their denominators. NOTE. ^ j^ j^ and. 3, .24, .205 are alifa decimal fractions. ART. 169. A figure in the first decimal place or order, expresses a fraction whose denominator is a unit with one cipher annexed ; thus, .2 = T 2 ^ : and, A figure in the second decimal place, expresses a fraction whose denominator is a unit with two ciphers annexed ; Thus, .02 = T ^ : also, A figure in the third place expresses a fraction whose denominator is a unit with three ciphers annexed ; Thus, .002 = j-fc-Q ; and so on : hence, The denominator of a Decimal Fraction is 1, with as many ciphers annexed as there are decimal places in the numerator. ART. 170. TABLE OF DECIMAL PLACES OR ORDERS. . 6 is read - 6 tenths. .25 25 hundredths. .013 13 thousandths. .0305 305 ten-thousandths. .00108 108 hundred-thousandths. .378659 378659 millionths. 5.0000124 5 and 124 ten-millionths. 26.00000005 26 and 5 hundred-millionths. REM. The orders called ten-thousandths, hundred-thousandths, are, really, tenths of thousandths, hundredths of thousandths, ART. 190. PROMISCUOUS EXAMPLES. 1. What cost 9 yards of muslin, at 80. 4 per yd., and 12 yards at $0.1875 per yd.? Ans. $5.85 2. What cost 2.3 yards of ribbon, at $0.45 per yd., and 1.5 yards at $0.375 per yd.? Am. $1.5975 3. At 82.0875 per yd., what cost 16 yd. of cloth What 16; yd.? Ans. to last, $43.671875 4. At ?0,75 per bushel, how much wheat can be bought for 8:^5.25? Ans. 47 bu. liKviKw. 188. What is Case 4? What the Eule for Case 4 1S, Wha* is Case 6 ? What the Eule for Caao 5 ? 3d 15k, 13 194 BAFS PRACTICAL ABITHMETIC. 5. At $2.5625 per -yard, how much cloth can you buy for $98.4? An*. 38 . 4 yd. 6. What cost 6 jwt. 2qr. of hops, at $3.25 per cwt? SOLU. Reduce 2qr. to the decimal of a cwt.; then, Gcwt. 2qr. =6.5cwt.; $3.25X6.5=^7W. $21.125 - 7. What will be the cost of 7hhd. 23 gal. of *t $49 per hhd. ? Am. $360 . 8888+ 8. Of 343yd. 3qr. linen, at $.16 per yd. ? Am. 855. ^9. What cost 14 bu. 3pk. 4qt. of corn, at 80.625 per bushel? Am. $9.296875 10. What will 13 A. 2E. 35 P. of land cost, at $17.28 per acre? Am. $237.06 <*\\. At 81.24 per yard, how much cloth can be bought for $19.065? Arts. 15.375yd. = 15yd. 1 qr. 2na. 12. At $0.3125 per bu., how much corn can be bought for $9.296875? Ans. 29.75bu. = 29bu. 3pk. 13. At $4.32 per A., how much land can you buy for $59.265? Am. 13.71875 A. = 13A. 2R. 35P. 14. Add .34yd. .325qr. .4na. Am. Iqr. 3. Una. To add or subtract decimals of different denominations, first re- duce them to the same denomination. In this example, reduce the dec. of a yd. to qr., then add the dec. of a qr.; next reduce this result to na., and add the dec. of a na. 15. From 1 . 53 yards take 1 . 32 qr. Ans. 1 yd 3 . 2 na. ^16. .05 of a year, (365.25 day?,) take .5 of an hour. Ans. 18 da. 5hr. 48min. , 17. .41 of a da. take .16 of an hr. Ans. 9hr. 40min. 48 sec. 18. In .4T .3hhd .8gal. how many pt.? Am. 964pt. 19. Find the value of .3 of a year, (,.r,5.25 day?.) in integers. An*. 109 da. 13 hr. 48 mirk 20. In .005 of a year how many sec.? Ans. 157783. 21. What decimal of a C. is 1 cu. in? Ans. . 000004+ RATIO OF NUMBERS, 195 22. At $690.35 per mile, what cost a road 17 mi. 15 rd. long ? Ans. $12027 . 19140625 In practice, only 3 or 4 places of decimals are generally used. t^j* For additional problems, see Ray's Test Examples. XIII. RATIO. ART. 191. Katio is the quotient arising from dividing one quantity by another of the same denomination ; Thus, the ratio of 2 to 6 is 3 ; as, G -~ 2 gives the quotient 3. The ratio of 2 to 8 is 4 ; of 2 yd. to 10yds., 5. ILLUSTRATIONS. 1. Two quantities to be compared, or to hava a ratio to each other, must be of the same kind, and in the sam* denomination, that the one may be some part of the other. Thus, 2 yards has a ratio to G yards. But, 2 yards has no ratio to G dollars, the one being no part of the other. 2. Since ratio is the relation of two numbers expressed by their quotient; and since the quotient o'f 2 and G may be G divided by 2, or 2 divided by G, either may be used to express their ratio. Thus, in comparing two lines, one of which is 2, the other G inches long, if the first is taken as the unit (1) or standard of comparison, the second is three, that is, it is 3 times the first. If G is taken as the unit of comparison, 2 is one-third. In finding the ratio between two numbers, the French take the first as the divisor, the English the last. The French method being regarded the most simple, is now generally used. 3. Finding the ratio of two numbers, is finding what part, or what multiple one is of the other. The following arc equivalent expressions : What is the ratio of 2 to G ? What part of 2 is 6 ? What multiple of 2 is G ? 4. The ratio of $2 to $10 is 5; of $2000 to $10000 is nl?o5: hence, ratio shows only the relative magnitude of tvo quantities. REVIEW. 190. What is ratio? Give examples. ILLUS. 1. CV.n quantities, not nf tho Frnn kind, have a ratio ? Why not? 2. What is the ratio of 2 to 6 ? 3. What arc equivalent expressions ? ILLUS. 4. What does ratio show ? In finding tho ratio between two quaiv titiea of tho same kind, bub of ditferect denominations, what ie required J 196 RAY'S PRACTICAL ARITHMETIC. 1. What is the ratio of 3 to 6? of 3 to 9? 2. Of 2 to 12? of 3 to 15? of 7 to 21? 3. Of 6 to 18? of 6 to 30? of 5 to 30? 4. Of 2 to 3? Ans. -3 = H. of 2 to 5? 5. Of 3 to 4? of5~to 8? of 4 to 6? 6. Of 12 inches to 36 inches ? of 2 feet to 9 feet? 7. Of 3 inches to 1 foot 9 inches ? . . Ans. 7. When the quantities are of the same kind, but of different denominations, reduce them to the same denomination. 8. What is the ratio of 3 in. to 2 ft.? of 4 in. to 3 yd.? 9. Of 15 to 25? Ans. If. of 25 to 15? Ans. |. 10. Of 4 to 10? of 10. to 4? of 6 to 16? of 16 to 6? ART. 192. A ratio is formed by two numbers, each of which is called a term, and both together, a couplet. Thus, 2 and 6 together form a couplet of which 2 is the first term, and 6 the second. The first term of a ratio is called the antecedent; -the second, the consequent. ART. 193. RATIO is EXPRESSED IN TWO WAYS : 1st. In the form of a fraction, of which the antecedent is the denominator, and the consequent the numerator. The ratio of 2 to G is expressed by f ; of 3 to 12, by ^. 2d. By a colon (:) between the terms of the ratio. Thus, the ratio of 2 to G is written 2 : G ; of 3 to 8, 3 : 8. ART. 194. Since the ratio of two numbers is expressed by & fraction, of which the antecedent is the denominator, and the consequent the numerator, whatever is true of a fraction, is true of a ratio. Hence, REVIEW. 192. By what is a ratio formed? What \* each number called ? What both together ? What is the first terra called ? The 2n ? 193. In how many ways is ratio expressed? What is the first method ? The second ? Give examples of each. PROPORTION. 197 1st. To multiply the consequent, or divide the antecedent, multiplies the ratio. Arts. 131 and 133. Thus, The ratio of 4 to 12 is 3; of 4 to 12x5, is 3x5; and The ratio of 4-f-2 to 12, is 6, which is equal to 3X2. 2d. To divide the consequent or multiply the antecedent, di- vides the ratio. Arts. 132 and 134. Thus, The ratio of 3 to 24 is 8; of 3 to 24-^2, is 4, =8-=-2; and The ratio of 3 x2 to 24, is 4, which is equal to 8-f-2. 3d. To multiply or divide both consequent and antecedent by the same number, does not aiter the ratio. Arts. 134, 135. Thus, The ratio of 6 to 18, is 3; of 6X2 to 18X2, is 3; and The ratio of 6 2 to 18-5-2, is 3. ART. 195. A single ratio, as 2 to 6, is a simple ratio. A compound ratio is the product of two or more, simple ratios. Thus, The ratio u> multiplied by the ratio f, is ^X? = ?2=4. In this case, 3 multiplied by *:, is said to have to 10X6, the ratio compounded of the ratios of 3 to 10 and 5 to 6. ART. V 196. Ratios may be compared with each other, by reducing to a common denominator the fractions by which they are expressed : thus, To find the greater of the two ratios, 2 to 5, and 3 to 8, we have | and f, which, reduced to a common denominator, are ^ and ^ ; and, as ^ is less than l /, the ratio of 2 to 5, is less than the ratio of 3 to 8. ^ XIV. PROPORTION. ART. 197. Proportion is an equality of ratios. Four numbers are proportional, when the first has the same ratio to the second that the third has to the fourth. REVIEW. 191. How is a ratio affected by multiplying the conse- quent, or dividing the antecedent? By dividing the consequent, or multiplying the antecedent? By multiplying or dividing both consequent and antecedent by the same number ? Why ? Illustrate each,. 198 RAY'S PRACTICAL ARITHMETIC. Thus, the two ratios, 2 : 4 and 3 : 6, form a proportion, since | |, each being equal to 2. ART. 198. PROPORTION is WRITTEN IN TWO WATS: 1st. By placing a double colon between the ratios. Thus, 2 : 4 : : 3 : 6. 2d. By placing the sign of equality between them. Thus, 2:4 = 3:6. The first is read, 2 is to 4 as 3 is to 6 ; or, 2 has the same ratio to 4, that 3 has to 6. The second is read, the ratio of 2 to 4 equals the ratio of 3 to G. REM. 1. The least number of terms that can form a proportion ia/owr, ?ince there are two ratios each containing two terms. But, one of the terms in each ratio may be the same; thu*, 2: 4:: 4: 8. The number repeated is called a MEAN proportional between the other two terms. 2. The terms ratio and proportion are often confounded with each other. Two quantities having the same ratio as 3 to 4, are impro- perly said to be in the proportion of 3 to 4. A ratio subsists between two quantities; a proportion only between four. ART. 199. The first and last terms of a proportion are called the extremes ; the second and third, the means. Thus, in the proportion 2 : 3 : : 4 : 6, 2 and 6 are the extremes, and 3 and 4 the means. ART. 200. In every proportion, the product of the means is equal to the product of the extremes. ILLUSTRATIONS. If we have 3 : 4 : : 6 : .8, the ratios of each couplet being equal ) 4-^8 (Art. 206), we must have J 3~~6 Reducing these fractions to a com- ) 4_X 6 __ 3X8 mon denominator (Art. 155), gives . ) 18 18 REVIEW. 195. What is a simple ratio? A compound ratio? Givo examples of each. 198. How compare ratios with each other ? 197. What is proportion ? When are four numbers proportional ? Give oxnmples. 198. How is proportion written ? How is the first read ? The second ? BEM. 1. What the least number of terms that can form a proportion? Ar Wi*it of the terms rutio and proportion f SIMPLE PROPORTION. 199 The denominators being the same, and the values of the fractions being equal, the numerators must be equal ; that is, 4X6, the product of the means, must = 3x8, the product of the extremes. REM. The preceding shows that four numbers are not in pro- portion, when the product of the extremes is unequal to the pro- duct of the means. Thus, 2, 3, 5, and 8, are not in proportion, for 3X5 is not equal to 2X8. Proportions have numerous properties, the full discussion of which belongs to Algebra. See "Ray's Algebra, First Book." PROPORTION is divided into Simple and Compound. f ART. 201. SIMPLE PROPORTION. Simple Proportion contains only simple ratios, Art. 195. It is sometimes called the HULE OF THREE, as three terms are given to find a fourth. REM. Some authors divide Proportion into direct and inverse / & dis- tinction of no utility, and always embarrassing to the learner. ART. 202. Since (he product of the means equals the product of the extremes, and the product of two factors divided by either of them, gives the other, Art. 37, Therefore, If the product of the means be divided by one oftht extremes, the quotient will be the other extreme. Or, If the product of the extremes be divided by one of the means, the quotient will be the other mean. Thus, in the proportion 2 : 3 : : 4 : 6, 3X4 2 6, one of the extremes. 3X46 2, the other extreme. 2X6 3 = 4, one of the means. 2 X 6 -r- 4 = 3, the other mean. REVIEW. 199. What arc the first and last terms called? The second ttnd third ? 200. If four numbers are proportional, to what is the product of the means cquiil ? REM. When are four numbers not in proportion ? How is proportion divided ? 20 1 , What does Simple Proportion contain ? What is U called ? Why f 102, When 3 terms of a proportion axe given, how find th fourth \ 200 RAY'S PRACTICAL ARITHMETIC. Hence, If any three terms of a proportion are given, the fourth may be found by multiplying together the terms of ike sam* name, and dividing their product by the other given term. 1. The first three terms of a proportion are 2, 8, and 6 : what is the fourth term ? SOLUTION. The preceding shows OPERATION. that the 4th term will be found by 2 : 8 : : 6 taking the product of the 2d and 3d 8X6 terms, and dividing by the 1st. = *-*. Ans. Or, by the nature of proportion, Art. 197, the ratio of the 3d term to the Or ' 4th = the ratio of the 1st to the 2d. f = 4, and 6 X 4 = 24. Hence, If the third term be multiplied by the ratio of the first to the second, the product will be the fourth term. EXAMPLES TO BE SOLVED BY EITHER METHOD. 2. The first three terms of a proportion are 5, 7, and 10 : what is the fourth term ? Ans. 14. 3. The last three terms are 8, 6, and 16: -what is the first ? Ans. 3. 4. The first, third, and fourth terms, are 5, 6, and 12: what is the second? Ans. 10. 5. The first, second, and fourth terms, are 3, 7, and 14: what is the third ? Ans. 6. 6. Seven is to 14, as 9 is to what number ? Ans. 18. '''ART. 203. 1. If 2 Ib. of tea cost 4, at the same rate, what will be the cost of 6 Ib.? SOLUTION. Two pounds have the same ratio OPERATION. to Gib., that the cost of 21b. (4), has to the Ib. Ib. $ cost of 6 Ib. Therefore, the first three terms are 2 : 6 : : 4 given, to find the fourth (Art, 202). 6 To find the result, multiply the 3d term by the 9 2d, and divide by the 1st; or, multiply the 3d "'2 term by the ratio (3) of the first to the 2d. Ans. $12. ' In stating this question, (arranging the terms,) we REVIEW. 202. If the third term of a proportion be multiplied by the ratio of the first to the Second, what will be the product ? SIMPLE PROPORTION. 201 may, with equal propriety, place the $4 as the first term. Thus, $4 : 2d term :: 21b. : Gib. The operation of finding the 2d term in this arrange- ment, is like that of finding the 4th in the preceding. It is more convenient to arrange the terms so that the required term shall be the fourth; then, Since a ratio can subsist only between quantities of the same denomination, Art. 1Q1, lllus. 1, The third term of a proportion must be of the same denomi nation as that in which the answer is required. 2. If 3 men can dig a cellar in 10 days, in how many days will 5 men dig it? SOLUTION. Since the answer is to be days, OPERATION. write 10 days for the SD terra. The result will m - ni - da. be found by multiplying the 3o term by the 2o, 5 : 3 : : 10 and dividing by the IST (Art. 201); and, Since it will require 5 men a less number of 5^30 days than 3 men, place the less number of men for the 2o term, and the greater for the IST. Am. C. 3. If 3 men can dig a cellar in 10 days, how many men will dig it in 6 days? SOLUTION. Since the answer is to be men, STATEMENT. write 3 men for the third term. Since it will da. da. m. require & greater number of men in G days than 6 : 10 : : 3 in 10 days, place the greater number of days for . . the second term, and the less for the first. *4. If 3 yd. cloth cost $8, what cost 6 yd.? Ans. $16. *5. If 5 bl. flour cost $30, what cost 3 bl. ? Ans. $18. GEITE2AL EULE FOR CIIIPLE PROPORTION. 1. Write for the third term that number which is of the denomination required in the answer. 2. If, from the nature of the question, the answer should be greater than the third term, place the greater of the other two numbers for the second term^ and the less for the first; but, SOS RAY'S PRACTICAL ARITHMETIC. If the answer should be less than the third term, place tht less number for the second term, and the greater for the first 3. Multiply the third term ~by the second^ and divide the product by the first ; the quotient will be the answer in the de- nomination of the third term. NOTES. 1. If the first and second terms contain different, denom- inations, reduce them to the same; if the third term consists of more than one denomination, reduce it to the lowest given. 2. Multiplying the 2d and 3d terms together, and dividing by the 1st, is really multiplying the 3d by the ratio of the 1st to the 2d. It is most convenient, especially -when the ratio i? a whole num- ber, to express it in its simplest form before multiplying. 3. After dividing, if there be a remainder, reduce it to the next lower denomination, and divide again, and so on. 6. If 31b. 12oz. tea cost 83.50, what cost 11 Ib. 4oz.? SOLUTION State the question ; and, OPERATION. to express the 1st and 2d terms in the ^3 1 9 1 i 4 3 ^rt same denomination, reduce both to oz. Then multiply the third term ($3.50) J? Z A . * cfrt . . q KA by 180, and divide the product, by 60. Y ' But. it is shorter to multiply at once _ by 3, the ratio of 60 to 180. Ans. $10.50 ' 7. If 21b. 8oz. of tea cost $2, what quantity can you buy for 5? Am. 61b. 4oz. Reduce the 3d term to ounces ; the answer will then be in ounces. 8. If 4 hats cost $14, what cost 10 hats? Ans. $35. 9. If 3 caps cost 69 cents, what cost 11 ? Ans. $2.53 '^10. If 4yd. cloth cost $7, what cost 9? Ans. 15.75 11. If 8yd. cloth cost 832, what cost 12 ? Ans. $48. REVIEW. 203. In stating a question in Fimplo proportion, what num- ber is put for the 3d term, Rule ? How are the other two numbers arranged ? How is th* answer obtained ? Of what denomination h the answer ? 203. NOTE 1. If the first and second terms contain different denomina- tion", what is required ? What if the third term contains more than on denomination ? What other explanations in ^otes 2 aaU 3 ? SIMPLE PROPORTION. 203 12. If 12 yd. cloth cost $48, what cost 8 yd.? Ans. $32. 13. If $32 purchase 8 yards of cloth, how many yards will $48 buy? Ans. 12yd. 14. If $48 purchase 12yd. of cloth, how many yards can be bought for $32 ? Ans. 8yd. 15. A man receives $152 for 19 months' work : how much should he have for 4 mon. work ? -SUGGESTION. Since the product of the second and third terms is to be divided by "To* ."^i .1^9 the first, therefore, (Art. 71,) when the first R term contains one or more factors common i r ON/ 4 $ to either of the other terms, shorten the '- = 32 Ans. Operation by canceling the common factors. JL ft ^iG. If 8 men perform a piece of work in 24 days, in what time can 12 men perform it? Ans. 16 days. 17. If 60 men perform a piece of work in 8 days, how many men will perform it in 2 days? Ans. 240. 18. If 15 oz. of pepper cost 25cts., what cost 6 Ib. ? SUGGESTION. Instead of OPERATION. !. i ., / 11 i /> . oz. Ib. cts. multiplying the o Ib. by lo, to -i * , n .. 05 reduce it to ounces, indicate the IF- ' / Jr'i / OK .... . . lOlU/\iOi:^tJ multiplication, and cancel the factors common to the divisor $ V 1 6X 2 ^ and dividend. In this manner, ' =160cts. An$. operations may be shortened. 1 5 rf P 19. If 6gal. of molasses cost 65cts., what cost 2hhd. ? Ans. $13.65 20. If 5cwt. 3qr. 10 Ib. of sugar cost $21.06, what will 35cwt.-lqr. cost? Ans. $126.90 ^21. If lyd. 2qr. of cloth cost $2.50, what will be the :-ost of 1 qr. 2 na. ? Ans. $0 . 62 5 22. If 90 bu. of oats supply 40 horses 6 days, how long v.ill 45.0 bu. supply them ? Ans. 30da. 2:>. If 6 men build a wall in 15 days ? how mnny men uild it in 5 dap ? 4fW- 18 m ea f ^ 204 RAY'S PRACTICAL ARITHMETIC. 24. If 15 bu. of corn pay for 60 bu. of potatoes, how much corn can be had for 140 bu. potatoes? Ans. 35 bu. 25. If 3cwt. Iqr. of sugar cost 22.60, what will be the cost of 16 cwt. 1 qr.? A us. 8113. 26. If a perpendicular staff, 3ft. long, cast a shadow 4ft. 6 in., what ij the height of a steeple, whose shadow measures 180 ft. ? Ans. 120 ft. 27. If a man perform a journey in 60 days, traveling 9 hours each day, in how many days can he perform it by traveling 12 hours a day? A us. 45 da. 28. A merchant failing, paid 60cts. on each dollar of his debts. He owed A 82200, and B $1800: what did each receive ? Ans. A 1320. B 1080. X 29. A merchant having failed, owes A 8800.30; B $250; C 8375.10; D 8500; F 8115. His property, worth 8612.12, goes to his creditors : how much will this pay on the dollar? Ans. 30 cts. * 30. If the 4 cent loaf weigh 9 oz. when flour is 88 a bl., what will it weigh when flour i.s 6 a bl.? Ans. 12 oz. 31. I borrowed 250 for 6 mon. : how long should I lend 8300, to compensate the favor ? Ans. 5 mon. 32. A starts on a journey, and travels 27 miles a day; 7 days after, B starts, and travels the same road, 36 mi. a day: in how many da. will B overtake A? Ans. 21 da, 33. If William's services are worth 815 J a mon., when he labors 9 hr. a da., what ought he to receive for 4? mon., when he labors 12 hr. a da. ? Ans. 891 .91^ If 5 Ib. of butter cost 8, what cost fib.? Ans. 9s- 35. If 6 yd. cloth cost 85 1, what cost 73 yd.? Ans. $6jjg. 33. If-] bu. wheat cost 8j, what cost A bu.? Ans. 9 -fa. 37. If I.; yd. cloth cost 0^, what cost 2 yd.? Ans. $J, 33. If 82D| buy 59.\ yards of cloth, how much will 831] buy? Ans. 62 i yd. 3D. If .85 of a gallon of wine cost 81.36, what will be the cost of .25 of a gallon? Ans. 80.40 40. If 61.31b. of tea cost 844.9942, what will ] cost of 1.07 lb.? 4w*. O.T SIMPLE PROPORTION 205 41. If 4 of a yard of cloth cost 8f , what will T 9 T of an Ell English cost? Ans. 8J. 42. If | of a yd. of velvet cost $4?, what cost 17| yd. ? Ans. $178.38 43. A wheel has 35 cogs ; a smaller wheel working in it, 26 cogs : in how many revolutions of the larger wheel, will the smaller gain 10 revolutions? Ans. 28|. 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? Ans. 96| gal. 45. If the velocity of sound be 1142 feet per sec., and the number of pulsations in a person 70 per min., what the distance of a cloud, if 20 pulsations are counted be- tween the time of seeing a flash of lightning and hearing the thunder? Am. 3 mi. 5 fur. 145 yd. 2| ft. 46. The length of a wall, by a measuring line, was 643 ft, 8 in., but, the line was found to be 25ft. 5. 25 in. long, instead of 25 feet, its supposed length : what the true length of the wall? Ans. 654ft. 11.17 in. ^NOTE. A bushel of -wheat is 60 lb.; of rye, 56 lb.; corn, 56 lb.; oats, 32 lb.; barley, 48 lb.; timothy seed, 42 lb.; clover seed, 60 lb.; flax seed, 56 lb. See Dry Measure, Note 2, page 83. The simplest method for the following examples, is by Simple Proportion. The first term will be the number of pounds to the bushel; second, the' weight; third, the price. What will be the cost of 47. 2136 lb. wheat, at 75 cts. per bu.? Ans. $26.70 48. 1225 lb. wheat, at 81 cts. per bu.? Ans. $16.53+ 49. 1000 lb. rye, at 63 cts. per bu.? Ans. 911. 25 50. 3000 lb. oats, at 24 cts. per bu. ? Ans. $22 . 5^ ART. 204. OF CAUSE AND EFFECT. The rule for Simple Proportion may be presented under another form, based on the obvious truth, that, When a cause produces a certain Effect, if the Cause be in- creased or diminished, the Effect will be increased or fauiidshtd in the same ratio. 206 ^AY ? S PRACTICAL ARITHMETIC. Thus, if the cause be doubled, the effect will be doubled, If trebled, the effect will be trebled Hence this PRINCIPLE. Any CAUSE is to another similar CAUSE, as the EFFECT ofthejirst CAUSE is to the EFFECT of the second CAUSE. ILLUSTRATION. If 3 men make 6 rods of road in a day, how many rods can 4 men make? 3 men. : 4 men. : : G rods. : 8 rods. Ans. 1st cause. 2J cause. 1st effect. 2d effect. The effect of the 2d cause, being the 4th term of a proportion, is found by multiplying the 3d term by the 2d, and dividing the product by the 1st. Art. 202. NOTE. In reviewing, the questions may be stated by this principle ; or it may be used at first instead of the Rule, Art. 203. ART. 205. COMPOUND PSOPOBTI01T. A Compound Proportion contains one or more com- pound ratios. Art. 195. 9 . g i Thus, 7 Q f : * 7 : 42, is a compound proportion. Or, 2X4 : 6X3 : : 7 : 42. . This, sometimes called the DOUBLE RULE OF THREE, is applied to the solution of questions requiring more than one statement in Simple Proportion. 1. If 2 men earn $20 in 5 days, what sum can 6 men earn in 10 days? Here, the sum earned depends on two things: the number of men, and the number of days; that is, on the ratio of 2 men to G men, and on the ratio of 5 days to 10 days. First find the sum earned by G men in the same time with 2 men. men. men. $ $ 2 : G : : 20 : GO sum earned by G men in 5 days. Knowing the sum earned in 5 days, the sum earned in 10 days can be found thus: days, clay 3. $ $ 5 : 10 : : 60 : 120 = sum earned by G men in 10 days. REVIEW. 204. On what is Cause anil Effect based? What is the Principle ? Hwry find the effect uf the 2d eauso ? COMPOUND PROPORTION. 207 By examining these proportions, it is seen that the ratio of the third term to the fourth, depends on two ratios 1st, the ratio of 2 men to 6 men ; 2d, the ratio of 5 da. to 10 da. The first ratio is 3, and the second 2; their product, 3X2=0, Is the ratio of the third term to the required term. Hence, The ratio, which the 3d terra has to the 4th, is compounded of the ratios of 2 to 6 and of 5 to 10: write the two simple proportions Thus, 2 men : G men } . * L . r .. , 5 days : 10 days j * ' 20 : fourth term. Or, 2 X 5 : 6 X 10 : : 20 : fourth term. 20X0X10 = 1200; 2X5 = 10; and 1200 10 = $120. Ans. *2. If a man travel 24 mi. in 2 da., by walking 4 lir. a day; at the same rate, how far will he travel in 10 da., walking 8 hr. a day? Ans. 240 mi. * Rule for Compound Proportion.! . Write for the Zd term, that number which is of the denomination required in the answer. 2. Take any two numbers of the same kind, and arrange them as in Simple Proportion. Art. 203, Kule 2, page 01. 3. Arrange any other two numbers of the same kind, in like manner, till all are used. 4. Multiply the third term by the continued product of the tecond terms ; divide the result by the continued product of the first terms : the quotient ivill be the fourth term, or answer. NOTES. 1. If the terms in any couplet are of different denomi- nations, reduce them \o the same. If the third term consists of more than one denomination, reduce it to the lowest named. 2. The examples may be solved by two or more statements in Simple Proportion, or by Analysis, Art. 263. Also, by Cause and Effect, Art. 204, see statement, page 203. W. 20"). What does Compound Proportion contain? To what is it applied? Li stating a question, what number is put for the third term, Rule ? How are the other numbers arranged ? 203. How is the fourth term found? NOTE 1. If the terms In any coupbt are of different denominations, what is required? What if the third term contains more than one denomination 7 208 RAY'S PRACTICAL ARITHMETIC. 3. If 6 men, in 10 days, build a wall 20ft. long, 3ft. high, and 2ft. thick: in how many days could 15 mon build a wall 80ft. long, 4ft. high, and 3ft. thick? STATEMENT BY CAUSE AND EFFECT. 1st cause. 2d cause. 1st effect. 2d effect. 6 : 15 : : 20 : 80 10 X 3 4 2 3 The terms of the 2d cause and 1st effect form the divisors ; those of the 1st cause and 2d effect, the dividends. The x shows the place of the required term. OPERATION BY CANCELING. EXPLANATION*. When the terms forming the divisor and the dividend, contain one or more common factors, shorten the operation by Cancellation. In such cases, arrange the divisors on the left of a vertical line, amd the multipliers {dividends) , on the right. Men * Length . Eight . Thickness Days . . H 4X4X2=32 da. Ans. 4. If 16 men build 18 rods of fence in 12 days, how many men can build 72 rd. in 8 da. ? Ans. 96 men. 5. If 6 men spend 150 in 8mon., how much will 15 men spend in 20rnon. ? Ans. $937.50 6. I travel 2 17 mi. in 7 days of 6 hr. each, how far can I travel in 9 da. of llhr. each? Ans. 511^ mi. 7. If $100 gain 86 in 12 months, what sum will 875 gain in 9nion. ? Ans. $3.375 8. If 100 Ib. be carriecV20mi. for 20cts., how far mry J.01001b. be carried for $60.60 ? Ans. 60mi. 0. To carry 12cwt. 3qr. 400mi., costs 857.12: what will it cost to carry 10 tuns 75 mi. ? Ans. $168. 10. If 18 men, in 15 da., build a wall 40 rd. long, 5ft. high, 4ft. thick ; in what time could 20 men build a wall 87 rd. long, 8ft, high, and 5ft. thi^ ? Ans. 58|gda. PRACTICE. 209 11. If 180 men, in 6 da. of 10 hr. each, dig a trench 200 yd. long, 3 yd. wide, 2 yd. deep ; in how many days can 100 men, working 8hr. a day, dig a trench 180yd. long, 4 yd. wide, and 3 yd. deep ? Ans. 24.3 da. 1 For additional problems, sco Ray's Test Examples. '"'XV. ALIQUOTS, OR PRACTICE. ART. 206. One number is an aliquot part of another, when it will exactly divide it (Art. 110). Thus, 5 cents, 10 cts., 20 cts., c., are aliquot parts of $1. TABLE OF ALIQUOT PARTS OF A DOLLAR. 100 cents - . .a dollar. 50 cents = J of a dollar. 25 cents = of a dollar. 12^ cents = ^ ; of a dollar. 6| cents =-Jj of a dollar. 20 cents = of a dollar. 1 cents = y 1 ^ of a dollar. 5 cents = 7^0 of a dollar. 33i cents = i of a dollar. ICi cents = 1 of a dollar. CAGE I. ART. 207. To find (he cost, of articles, when (he price is an aliquot part, or aliquot parts of a dollar. 1. What cost 24 yards of muslin, at 25 cts. a yd.? SOLUTION. If the price was $1 a yard, the cost would be $24; and, since 25 cents is of a dollar, the cost at 25 cts. a yard will be 1 the cost at $1; und | of 24 =$0. Ans. 2. What cost 1G yards of calico, at 374 cts. a yard? SOLUTION. If the price was $1 a yard, the cost would be $16. At 25ct.s. a yard, the cost would be i the cost at $1; i of$16=$4. Ajrain ; since 12^cts. is ^ of 25cts., the cost at 12cts. a yard, will be -^ the cost at 25cts. ; J of $4 =$2. But. the cost at 37cts. a yard, is equal to the sum of the costs at 2ycts. and at V2\ cts.; $4-f-S2 = $6. Ans. EKVIEW. 206. When is one number an aliquot part ? Give examples. 210 RAY'S PRACTICAL ARITHMETIC. *3. What cost 24yd. silk at 62A cts. a yd.? Ans. $15. Rule for Case I. Take such aliquot parts of the cost at $1, as may be necessary to find the cost at the given price- 4. What cost 48 yd. of linen, at 68| cts. per yd.? OPEEATION. $48 = cost of 48 yards, at $1. 60 cts. = i 12^ cts. = \ 6} cts. = 24= " " " at 50 cts. 6 = " " " at 12 cts. 3 = " " " at 6 j cts. Ans. $33 = " " " at C8| cts. What will be the cost 5. Of 173 bu. corn, at 25 cts. a bu.? Ans. $43.25 6. 45 Ib. cheese, at 31] cts. a lb.? Ans. $14.06} 7. 54yd. calico, at 43 J cts. a yd.? Ans. $23. 62^ 8. 32 bu. rye, at 93| cts. a bu.? Ans. $30.00 9. What cost 20yd. cloth, at $3.12A per yd.? The cost at $1 a OPERATION. $20 = cost of 20 yd. at $1 yard is multiplied by the number o ? dollars ( 3 ) ; and for the 12^ cts., an 60 = costat$3.00 r A - .-> . 2. 50= cost at .12 1 , aliquot part, (J) of Ans. $62.50= cost at $3.12^ the cost at $1 is taken j the results are added. Find the cost of ANSWERS. 10. 80 gal. of wine, at $2. 374 a gal. $190.00 ^11. 36 bl. of flour, at $8.8?i a bl. $319.50 12. 77 gal. of wine, at $l.G2i a gal. $125.12.} 13. 175 A. of land, at $14. 374 an A. 2515. 62| 14. 224 bl. of flour, at $3.43| a bl. $770.00 15. 462yd. of cloth, at $1.06] a yd. $490. 87A 115. 185yd. of cloth, at $1.331 a yd. $246. 66 17. 150yd. of s;itin,at $3.66|ayd. $550.00 18, 24yd. of silk, at $1.16jayd. $28.00 PRACTICE. 21] ART. 208. CASE II. To find the cost of a quantity con- sisting of several denominations. 1. What will be the cost of 4 A. 1R. 20 P. of land, at $16. 40 per A:? OPERATIC*. $16. 40 = cost of 1 A. 4 65. 60 = cost of 4 A. 4. 10 = cost of 1 R. 2. 05 = cost of 20 P. Ans. $71. 75 = cost of 4 A. 1 R. 20 P. for Case II. Multiply the price by the number of the denomination at which the price is rated, and find the cost of the lower denominations, by taking aliquot parts. Add the different costs ; the sum will be the cost of the whole. Find the cost of ANSWERS. 2. 14A. 2 R., at 310. 00 per A. $145.00 3. 28 A. 3R., at S12.50 per A. $359.37^ 4. 5 A. 1 R. 10 P., at $12 per A. $63.75 5. 12 A. 1R. 10 P., at $18 per A. $221.62' 6. 14 A. 3R. 25 P., at $12. 50 per A. $186.3i 7. 3yd.2qr., at $1.75 per yd. $6.12| 8. 4 yd. 3 qr., at $1.50 per yd. $7.12| _J). 56yd.3qr., at $17.25 per yd. $978. 93| To. 83 bu. 3 pk. 2 qt., at $6 a bu. $502.87^ 11. 24 bu. 3 pk. 7 qt., at $4 a bu. $99.87* 40bu.3pk. 7qt. 1 pt., at $3.20 a bu. $131.15 17bu. Ipk. Iqt. Ipt., at $2. 56 a bu. $44.28 5 Ib. 11 oz. butter, at 24 cts. per Ib. $1.3G* 31b. 13 oz. 15 dr. spice, at $2. 56 alb. $9.9l 17 A. 3 R. 39 P., at $3 . 20 per A. $57 . 58 J6 For additional problems, see Ray's Test Examples. KE VIEW. 207. What is Case 1 ? What tho .Rule for Case 1 7 203. What is Caso 2 ? What the Bale for Cwo 2 1 212 RAY'S PRACTICAL ARITHMETIC. XVI. PERCENTAGE. ART. 209. Percentage embraces calculations in which reference is made to a hundred as the unit ; and, The per cent, or percentage of any number or quantity, \s so many liundredths of it. Thus, 1 per cent, is T J ; 2 per cent. T | ; 3 per cent. yg^. Since 7 J = .01, T go = .02, and so on, (Art. 176), &arefcre, per cent, may be expressed decimally ; thus, 4 per cent, is .04 5 per cent, is .05 9 per cent, is .09 10 per cent, is .10 25 per cent, is .25 75 per cent, is .75 100 per cent, (fgg, or the whole,) is 1.00 125 per cent, (jgg, or -f^ more than the whole,) is 1.25 per cent, is a half of 1 per cent., of yj^, is .005 per cent, is J of 1 per cent., J of yj^, is .00125 3 per cent, is .03^,^.035 is 035 EXAMPLE. "Write 7 per cent, decimally. 4j per ct. 5| per ct, 101 per ct. 37i per ct. 120 per ct. THE SIGN J&, is used in business, instead of the words " per cent. ;" thus, 5 per cent, is written 5 %. ART. 210. CASE I. To find any per cent, of a number. 1. What is 2 % (2 per cent.) of 125 ? ,. SOL. To take 2 per cent, is to take 2 Imn- drcdths. Hence, multiply by 2 and divide by 100; or^mltiply by -g^ expressed decimally. J, iS 2 5G~ *2. TOiat is 7 % of 175 dollars ? Ans. 812.25 REVIEW. 209. What does Percentage embrace ? What is the per ent. of *ny number ? Give examples. PERCENTAGE. 213 Rule for Percentage. Multiply by the rate per cent, ex- pressed decimally- the product will be the per cent, required. Or, by PROPORTION (Art. 203). As 100 is to the rate per ceni so is the given number or quantity to the required per cent. 3. What is 5' % of $150? SUGGESTION. When the rate per cent, is a common fraction, or mixed number, multiply as explained in Art. 152 ; and in pointing the product, count only two decimals in the multiplier. OPERATION. $150 05j 7.50 = 5 .50= J Ans. 8.00 = 5 percent. per cent. percent. ANSWERS. 4. What is 6 < of $250? $15.00 5. What is 7 i of $162? $11.34 6. What is 5 < Yo of $118? $5.90 7. What is 8 ; of $11? $0.88 8. What is 1 t of $278? $2.78 "*9. What is 2]< ?o of $68? $1.53 To7 What is 4^ i & of $220.50? $9.922+ 11. What is '$1 of $115.42? $8.656+ 12. What is 5| c / of $243.16? $13.981 + ^3. What is 3J5 of $1250? $40.625 14. What is 25 ^ g of $25? $6.25 15. What is 10U 5 of $2002? $2032.03 16. What is 20S~ 9 & of $650? $1352.00" 17. What is 1000 5 of $24.75? $247.50 18. What is TO 9 of $400? $0.40 19. What is 1 9 of $464? $1.74 20. is pr 9 of $1950? S1.62J REVIEW. 209. How may per cent, be expressed ? Give examples. What is 100 per cent. ? What is 125 per cent. ? A per cent. ? 3^ per cent. ? What sign is used for per cent, f Give examples. 214 RAY'S PRACTICAL ARITHMETIC. * 21. If 8i % of 72 be taken from it, how much will remain? Am. $65.88 22. I had $800 in bank, and drew out 36 % of it : how much had I left? Ans. $512. 23. What is the difference between 13'- % of $56, and 14| % of $51? Ans. 13cts. > 24. A merchant in expending $1764, paid 23 % for cloth ; 31 % for calicoes ; 9 % for linens ; 3| % for silks ; the remainder for muslin : how much did he pay for each ? Ans. to last, $595.35 25. A grocer bought 4 bags of coffee, of 75 Ib. each : 12^ % was lost by waste : what was the rest worth, at 14cts. per Ib. ? Ans. $36.82 26. A flock of 160 sheep increased 35 % in 1 year : how large was it then? Ans. 216. 27. I had 320 sheep ; 5 % were killed : after selling 25 % of the rest, how many were left ? Ans. 228. 28. A's salary is $800 a year : he spends 18 % for rent ; 15 - 84 2$ per cent. *2. What % of 15, is 03? A*s. 20. REVIEW. 210. How is the perccntago of a number found, Rule ? How, when the rate per cent, is a mixed number or a fraction ? 211. "What is Case 2 ? What the Rule for Case 2 ? COMMISSION. 215 Rule for Case II. Divide the number which is considered the percentage, by \ per cent of the other number; the quotient will be the required rate per cent. Or, by PROPORTION. As the given number is to the percentage, so^is 100 to the required rate per cent. 3. What % of $50, is $6? Ans. 12. 4. What % of $75, is $4.50? Ans. 6, 5. What % of $9, is $3? Ans. 33j. X 6. What % of $25, is $0.25? Ans. 1. 7. What % of $142.60, is $7.13? Ans. 5. 8. What % of $9, is $9 ? Ans. 100. 9. What % of $9, is $13.50 ? 4. 150. 10. What % of $243, is $8.505? Ans. 3.5=3. ll. What % of $2, is 2 mills? ^TZS. T \j. 12. What % of $3532, is $13.245? AM. f. 13. A man had $300, and spent $25 : what % wa* that of the whole? -4ns. 8*. 14. A man owed $500, and paid $75 of it: what % of | the whole debt did he pay ? Ans. 15. i 15. What % of any number is f of it? Ans. 60. 1G. A miller takes for toll, 6qt. from every 5bu. of grain ground : what % does he get ? Ans. 3|. ,5^** For additional problems, see Rays Test Examples. ^ APPLICATIONS OF PERCENTAGE. ART. 212. The principal applications of percentage are 18 * QC = Int - for * 7*- of this is $1.50 Ans. $1 . 5 0=Int. for 1 mon, *2. The Int. of $240 for 2 mon., at 8 %. Ans. $3.20 On this principle, the Int. for 3 mon. is -*%=$ the Int. for a yr.: for 4 mon. it is ~ 4 2 = ; for 6 mon. _ 5 2 , &c. 3. Find the interest of $288 for 1 day, at 5 %. SOLUTION.-AS 1 da. is JL of **** OPERATION. (') ) a mon., the Int. for 1 da. is ^ of the Int. for 1 mon. The Int. 12)14. 40=Int, for 1 yeav. for 1 mon. is $1.20, and J n of this is $.04 Ans. 30)1.20=lnt.forlmon t $.04=Int, forl day. *4. The Int. of $360 for 2 days, at 6 % . Ans. , 12 On the same principle, the interest for 3 da. is ^^j -J^of the Int. for 1 mon.; for 4 dn. it is JL T 2 .; for 5 da., ^ ; for G da.j **is J^i; for 7 da, fo * Rule for Case II. Find the interest cf the given sum for one year, (Art. 210). To find the Int. for any number cf months, take such a part of this, as the given number cf months is part of a year. To find the Int. for any number cf days, tale such apart of the Int. for 1 mon. } as the days are part cfa month. 224 RAY'S PRACTICAL ARITHMETIC. 5. Find the interest of $50 for 5 mon., at 6 SUGGESTION. After find- ing the interest for 1 year, find the Int. for months, by taking aliquot parts (Art. 208); or, Multiply by the number of mon., and divide by 12, (Art. 152). OPERATION. $50 .06 3.00=Int. lyr. 4mon.= ^] 1 .00 = Int. for 4 mon. 1 mon.=| ,25 = Int. for 1 mon. $1 .25 = Int. for 5 mon. To find the Int. for da., take aliquot parts of the Int. for X mon. or, multiply by the number of da., and divide by 30. Jbma tne interest 01 ANSWERS. 6. $86.00 for 3 mon., at 6 %. . . $1.29 7. $50.00 for 4 mon., at 8 %. . . $1.33] 8. $150. 25 for 6 mon., at 8%. . . $6.01 9. $360.00 for 7 mon., at 5 %. . . $10.50 10. $204.00 for 11 mon., at 7 %. . . $13.09 .Jl. $726.00 for 10 days, at 6 %. . . $1.21 12. $1200. 00 for 15 days, at 6 %. . . $3.00 13. $180. 00 for 19 days, at 8 %. . . $0.76 14. $240. 00 for 27 days, at 7 %. . . $1.26 .15. $100. 80 for 28 days, at 5 %, . $0.392 Find the Amount of ANSWERS. 16. $228. 00 for 9 mon., at 6 %. . $238.26 17. $137. 50 for 8 mon., at 6 %. . 143.00 18. $150. 00 for 18 days, at 5 %. . 150.375 19. 360.00 for 11 days, at 6 %. . 360.66 20. 264.00 for 9 days, at 6 %. . 264.396 1* For additional problems, see Ray's Test Examples. REVIEW. 221. How find the interest of any sum for any number months, Eule for Case II ? For any number of days ? SIMPLE INTEREST. ART. 222. GENERAL RULE FOR INTEREST. 229 1. For one year. Multiply the Principal by the rate per cent., expressed decimally. 2.^ For more years than one. Multiply the Interest for 1 year by the number of years. 3. For months. Take such part of the Interest for 1 year, as the number of months is part of one year. 4. For days. Take such part of tJie Int. for 1 month, as the number of days is part of one month. 5. For years, months, and days, or for any two of these periods. Find the Int. for each period, and add the results. 6. To find the amount. Add the Principal to the Int. Or, by PROPORTION. As 100 is to the rate per cent.j so is the Principal to the Int. for 1 yr. Then, As 1 yr. is to the given time, so is the Int. for 1 yr. to the Int. for the given time. NOTE In computing Int., regard 30 days 1 month, and 12 months, 1 year. Custom has made this lawful. 1. Find the interest of $360 for 2yr., 7mon. 25 da., at 8 %, per annum. OPERATION. OPERATION BY PROPORTION. $360 2yr. 7 mon. 25da. = C55Ja. .08 100 : 8 : : $360 28.80 = Int. 1 yr. 100 8 4 36X4 98 80 2 5 36p 5 57.6!) = Tnt, 2 yr. 360 : 955 : : $28.80 6 mon. = - 14.40 = rnt. mon. $60 9^ 101 1 mon.== ' 2.40 = Tnt. 1 mon. fl 28.^0 .40 15 da, = 1 10 da. =1 1.20 = Int.. IT, da. .80 = Int. 10 da. 191 X .40 = 876.40 Ans. d Tho operation mny be further $76.40 = Tnt. for shortened by writing the proper- 2yr. Tinon. '2oda. tions together, and canceling. What is the Interest of ANSWERS. 2. $350.00 for 7yr. 3 mon., at 4 % ? $101.50 3. $150.00 for 4yr, 2 mon., at 6 % 1 $ 37.50 SdBk, 15 226 tvAYS PRACTICAL ABITHMETXC. ft Find the 4. $375 f 5. 92 . $500 7. 8560 , 8. $750 9. 456 10. 8216 11. 380 Find the 12. $300 ,13. $250 ,14. $205 . 15. $150 16. 8210 117. 8 57 Interest of .40 for lyr. .75 for Syr. .00 for lyr. . 00 for 2 yr. , 00 for 4 yr. ,00 for 3yr. 00 for Syr. ,03 for 3yr. Amount of ,00 for 3yr. . 00 for 1 yr. . 25 for 2 yr. .62 for 3yr. .25 for 2yr. .85 for 2yr. 8 mon., at 6 %. 5 mon., at 6 j. 1 mon. 18 da., at 4mon. 15 da., at 3 mon. 6 da.-, at 5 mon. 18 da., at 7 mon. 27 da., at '. 8% ANSWERS. $ 37.54 $ 19.01+ $ 34.00 $106.40 $192.00 $ 79.04 8122.22 9 mon. 9 da., at 15 % . $215 . 175 Sraon., at G %. 7 mon., at 6 %. 8 mon. 15 da., at 6 % 5 mon. 12 da., at 5 % 7 mon. 20 da., at 7 % 3 mon. 23 da., at 5 % ANSWERS. 8366.00 8273.75 8238.603+ $176.601 + 249.087+ $ 64.542+ 18. Find the interest of 8150, from January 9, 1847, to April 19, 1840, at 6 %. Ans. $20.50 NOTE. To find tlio Time between two dates, see Art. 103. 19. The interest of 8240, from February 15, 1848, to April 27, 1849, at 8 %. Ans. 823.04 ^ 20. The interest of 8180, from May 14, 1843, to August 28, 1845, at 7%. Ans. $28.84 21. The interest of 8137.50, from July 3, to Novem- ber 27, at 9 %. Am. 84.C5 22. The amount of $125.40, from March 1, to August 28, at 8}^. ( Ans. 8130.64+ f 23. The amount of 8234.60, from August 2, 1847. to March 9, 1348, at 5| %. An* 8242.024+ 24. The amount of $153.80, from Oct. 25 1846, to July 24, ib47, at 5 %. Ans. $159.546 + REVIEW. 222. How find the interest cf any sum at any rate for 1 year, Rule? For two or more years? For months? For dnys? For years, months, and day*? JIow find tho iaterest by proportion? SIMPLE INTEREST. 227 ANOTHER METHOD FOR INTEREST. ART. 223. To find the interest o/$l, a* 6 % , for any time. At 6 #, the Int. of $1 for 1 year, (12 mon.) is Gets. $ .06 For 1 mon. it will be T 6 2 ct., or 5 mills 005 For 2inon., X2=1 cent 01 For 3 mon., ^X3 = H cents 015 For 4 mon. , \ X 4 = 2 cents. (And so on. ) . . .02 Again: since 30 days = l mon. For 1 da. the Int. is fo of ct. = fa ct = J m. . . $ .000 For 6 da., as & = , it is of ct. = J n ct. = 1 m. n 001 P Hence, to find the Interest of $1, at 6 J&, for any time, Take as many cents as equal half the even number of months; take 5 mills for each odd month, 1 mill for each six days, and one sixth ofle ; thqt neither interest nor shall draw interest, 230 RAY'S PRACTICAL ARITHMETIC. 1. $350. Boston, July 1st, 1845. For value received, I promise to pay to Edward Sargent, or order, on demand, Three hundred and fifty dollars, vvith interest at 6 %. LOWELL MASON. On this note the following payments were indorsed: March 1st, 1846, rec'd $44. Jan. 1st, 1847, rec'd $26. Oct. 1st, 1846, rec'd $10. Dec. 1st, 1847, rec'd $15. What was due, on settlement, March 16th, 1848? OPERATION. Principal, $350.00 Interest to 1st payment (3 mon.), 14.00 Amount due March 1st, 1S4G, 364.00 1st payment (greater than Int.) deducted, .... 44.00 Balance due March 1st, 1846, $320.00 Int. on bal. to Oct. 1st, 1846, (7 mon.) . . $11 .20 2d payment, (less than Int. due,) .... 10.00 Surplus Int. unpaid, Oct. 1st, 1846, ... 1.20 Int. continued on bal. from Oct. 1st, 184G, to ) Jan. 1st, 1847, (3 mon.) {4.80 6.00 Amount due Jan. 1st, 1847, 326.00 Deduct 3d payment, (greater than Int. due,) ... 26.00 Balance due Jan. 1st, 184 7, 8300.00 Int. on bal. to Dec. 1st, 1847, (11 raon.) . . 616.50 4th payment, (less than Int. due.) .... 15.00 Surplus Int. unpaid Dec. 1st, 1347, ... 1.50 Int. continued on bal. from Dec. 1st, 1347, to ) March 16ih, 1848, (3 mon. 15 da. N . . . J 5.25 6.75 Balance due, on settlement, March IGth, 1848, . . 306.75 SUGGESTION*. When any payment is less than the Interest on tl Indorsed, May 1, 18-10, $ 18. Sept, 4, 1840, $ 20. Dec. 10, 1840, $ 15. April 10, 1841, $ 21. July 13, 1841. $118. Dec. 23, 1841, $324. was due, Ctet, Jst, J843? , 4*, $663,89 232 RAY'S PRACTICAL ARITHMETIC. ART. 227. In Computing Interest, when PARTIAL PAYMENTS have been made, the following rule is applied where settlement is made in a year, or in less than a year, from the commencement of Int. Find the amount of the principal for the whole time; then find the amount of each payment from the time it was made to the time of settlement. Add tor/ether the amounts of the several payments, and subtract their sum from the amount of the prin~ cipal; the remainder will be the sum due on settlement. 1. A note of 320 is dated Jan. 1st, 1846. Int. 6 %. Indorsed, May 1, 1846, $50. Nov. 1G, 1840. $100. What was due Jan. 1st, 1847? An&. 186.45 2. A note of 8540 is dated March 1st, 1847, Int. 8 %. Indorsed, May 1, 1847, $.0. July 1, 1847, $100. Aug. 1, 1847, $150. Oct. 11, 15547, $180. What was due Jan. 1st, 1848? Ans. 39. ART. 223. CONNECTICUT RULE. "Compute the interest to the time of the first pnyment; if that be 1 year or more from the time the Int. commenced, add it to the principal, and dvluct the payment from the sum total. "If there he afier payments made, compute the Int. on the bal- ance di? to the next payment, and deduct the payment as above; and, in like manner, from one payment to another, till all the pay- ments are absorbed, provided the time between one payment and another he 1 year or more. But, "If any payment? be made before 1 yr.'s Int. hns nccrned, then compute the Int. on the principal sum due on I he Obligation, for 1 jr., add it to the Prin., and compute the Int. on the sum paid, from the time it was paid to the end of the yr., add it to the sum paid, and deduct that sum from the Prin. and Int. added as ahove.* "If any payments be made of a less sum than the Int. arisen at *If a year dies not exten 1 boyml the time of payment; but, if it docs, thin find the amount of the principal, remair..ng unpaid, up to tho tima of settlement, likewise the amount of the payment or payments from the timo they were paid to the time of settlement, and deduct the sum of these several amounts from the amount of ttie principal, SIMPLE INTEREST. 233 the time of such payment, no Int. is to be computed, but only on the principal sum for any period." KIRBY'S REPORTS. 1. A note of $875 is dated Jan. 10th, 1831. Int. 6 %. Indorsed, Aug. 10, 1834, $2(50. Dec. 16, 1835, $300. ' Mar. 1, 1836, $ 50. July 1, 1837, $150. What was due Sept. 1st, 1838? Ans. $446.983+ PROBLEMS IN INTEREST. ART. 229. There are four parts or quantities connected with each operation in Interest : these are, the Principal: Rate per cent. : Time: Interest or Amount. If any three of them arc given, the oilier may be found. PROBLEM I. To find the INTEREST or AMOUNT, the Principal, Rate per cent, and Time being given. This, the most important problem in Interest, is illustrated in Art 220 to 223. ART. 230. PROBLEM II. To find the TIME, the Prin- cipal, Hate per cent, and Interest being given. 1. I loaned $200 at G %, and received 36 for interest: how long was the money loaned ? SOLUTION. The interest of $200 (Art. 220), OPERATION. at G per cent., for 1 year, is $12; then, since tin 200 given principal in 1 year produces $12 Int., it 06 will require the same principal ns many times 1 year to produce $30 Int., as $12 arc contained times in $30; that is, 3. Ans. 3 yr. 83G-r-12=3. *2. In what time will SCO, at 5 %, gain 12 interest Ans. 4yr RULE FOR PROBLEM II. Divide tlie given interest by the Int. of the principal at the given rate per cent, for one year ; the quotient will be the time. Or, BY PROPORTION. As the Int. of the principal for one year is to the given Int., so is one year to the required time. REVIEW. 229. What quantities arc connected with each operation in interest ? How ninny must be given, that the remaining may bo found? What is the most important problem in interest ? 230. What is Problem II ? WUat the ruio ? JIow, by proportion ? 234 RAY'S PRACTICAL ARITHMETIC. NOTE. When the quotient contains a common fraction, or & decimal, find its value in months and days. Art. 1G2 and 188. 3. A man loaned $375, at 8%, and received $90 interest: how long was it loaned? Ans. Syr. 4. How long will it take $225, at 4%, to gain 8G6 interest? Ans. 7* yr. = 7yr. 4mon. 5. How long will it take $250, at 6%, to gain 34.50 interest? Ans. 2.3yr,= 2yr. 3mon. 18da. ^6. How long will it take $60, at 6$g, to gain $13.77 interest? Ans. 3.825 yr. = Syr. 9mon. 27 da. 7. $500, at 10 Jg, to amount to $800? Ans. 6yr. SUGGESTION. By subtracting the principal, $500, from the amount, $800, the remainder, $300, is the interest of the principal for the required time: then proceed as before. 8. In what time will $GOO, at 9jg, amount to $798? Ans. Syr. 8mon. 9. In what time will $200, at 6 %, amount to 8400; or, in what time will any sum of money double itself, at 6 % ? Ans. 16 yr. 8mon. SUGGESTION. To find in what time any principal will double itself at any given rate per cent, Take any sum, as $100, for a principal, and find the time in which, at the given rate per cent., the Int. produced would equal the principal. IN WHAT TIME WILL ANY PRINCIPAL DOUBLE ITSELF, 10. At 4 c / per annum? Ans. 25 yr. 11. At 5 % ? Ans. 20 yr. ! 14. At 10 % ? Ans. 10 yr. 12. At 7 % ? Ans. 14;- yr. L 15. At 12 % ? Ans. 8 ; ^ yr. 13. At 8 % ? Ans. 12 h yr. 1C. At 16 % ? Ans. Q\ yr. 17. In what time, at 15 %, will 4.65 + 7. $200 for 2yr., at 6 %, payable semi-annually. 8. Find the amount of $500 for 2yrs., at 20 % com- pound interest, payable quarterly. Ans. 738.727 + 9. What is the compound interest of $300 for 2 yr. 6 num., at 6 %1 Ans. $47.192+ 10. What the compound interest of 1000 for 2 yr. 8 mon. 15 da., at 6 % ? Ans. $171. 3C3 ^AuT. 234. Since the amount of 2, for any given time, will be twice the amount of SI j the amount of 3, tluee times as much, &c. ; Therefore, If a table be formed containing the amounts of $1 for 1, 2, 3, &c.. years, any of these amounts multiplied by a given Prin- cipal will give its amount at Compound Interest for the same time and rate. Kce Tablo, page 238. 238 BAt'S PRACTICAL ARITHMETIC. TABLE Showing the amount of $1, at 3, 4, 5, G, 7 and 8 per cent., Com. pound Interest, for any number of years, from 1 to 25. Yr. 3 per cent. 4 per cent. 5 per cent. C per cent. 7 per cent. 8 per cent 1 1.03 1.04 1.05 1.C3 1.C7 1.03 2 1.060 i:osiG 1.1023 1.1233 1.1449 1.1664 3 1.092727 1.124804 1.157625 1.191010 1.225043 1.2597M 4 1.125509 1.169859 1.215506 1. 262477 1.310786 1.360483 5 1.159274 1.21C60J 1.276282 1.338226 1.402551 1.4G9328 6 1.194052 1.265319 1.340093 1.418519 1.500730 1.586874 7 1.229874 1.315932 1.407100 1.50363'J 1.G05781 1.713824 8 1.266770 1.338569 1.477455 1.593843 1.718186 1.850930 9 1.304773 1.423312 1.551328 1.689479 1.838459 1.99D004 10 1.343916 1.480244 1.628895 1.790843 1.967151 2.158924 11 1.384234 1.539454 1.710339 1.898299 2.104851 2.331638 12 1.425701 1.601032 1.795856 2.012196 2.252191 2.518170 13 1.468534 1.665074 1.885640 2.132923 2.409845 2.719623 14 1.512590 1.731676 1.979932 2.260904 2.578534 2.937193 15 1.557967 1.800944 2.078923 2.396553 2.759031 3.172169 16 1. 604706 1.872981 2.182875 2.540352 2.952163 3.425942 17 1.652848 1.947900 2.292013 2.692773 3.158815 3.700018 18 1.702433 2.025817 2.406619 2.854333 S.379932 3.996019' 19 1.753506 2.106849 2.526950 3.025603 3.616527 4.315701 20 1.806111 2.191123 2.653293 3.207135 3.8CD684 4.CC0957 21 1.860295 2.278768 2.785903 3.399564 4.140562 5.033833 22 1.916103 2:369919 2.925261 3.603537 4.430401 5.436540 23 1.973587 2.464716 3.071524 3.819750 4.740529 5.871463 24 2.032794 2.563304 3.225103 4.048935 5.072366 C.341180 25 2.093778 2.665836 3.386355 4.291871 5.427432 6.848475 11. What, by the Table, will be the Amount of S70 for 9yr., at 5 % compound interest? SOLUTION. By the Table, the nmount of $1 for 9yrs., at 5 pr. ct, is $1.551328; and $1.55132SX70=$108.5929G 12. Of 345 for 10 yr., at G % ? Am. 8617.844- 13. 200 for 41 yr., at 6 % ? EXPLANATION. Take any two periods whose sum is 41 years, thus, 20yr.-|-21 yr.=41 yr., and find the amount for the 1st period: then regard this a new principal, and find its amount for the 2d period : the last amount will be the Am. Ths Tabular number for 20 yr. is 3.207135; this, multiplied by 200, gives $041.427, amount for 20 yr. : Tabular number for 21 yr. 183.3993G4, which X 041.427 = 2180.572-{- Ana. DISCOUNT. 239 What, by the Table, will be the Interest 14. Of $890 for 30 yr., at 6 % ? Ann. $4221.70+ 15. $200 for 70 yr., at 5 % ? Ans. $5885. 28+ ART. 235. DISCOUNT Is a deduction made for the payment of money before it is due. For example, If a debt of $106, due one year hence without interest, be paid at the present time, the sum paid, with one years interest added, should make $106. And, If the rate per cent, is G, this sum would be 100 ; for, the amount of 8100 at interest for 1 year, at 6 %. is $106. Art. 220. ^ The PRESENT WORTH of a debt payable at a future time without interest, is that sum which, at a specified rate % for the same time, would amount to the debt. The DISCOUNT is the sum deducted for present payment. ART. 236. 1. Find the Present Worth of $224, due 2 yr. hence, without interest, money being worth 6 A, 25 33 1 per cent.= : |, i 3 10 1 2) = JL , * W 3 ' 20 S' 50 ~ " is ' 1 iD> 5> * REVIEW. 244. What is the Rule for Case 2 ? 245. What part of any thing is 5 per cent.? 8 and 1 third? 10? 12 and a half? 16 and 2 thirds? 20? 25? 33 and 1 third? 50 per cent? PROFIT AND LOSS. 247 To mark goods for different rates of profit, Add to the cost price such part of itself as the rate per cent, is part of 100. 6. To make 10 ^ profit, -what must calico be marked, that cost lOcts. per yard? 15cts.? 20 cts. ? 30 cts. ? 40 cts. ? 50 cts. ? 60 cts. ? Arts, to last, 66 cts. 7. To make 12i % profit, how mark muslin that cost 8 cts. per yard. ? 12 cts. ? . 16 cts. ? 20 cts. ? Ans. to last, 22| cts. 8. To make 20 % profit, how mark ribbons that cost 10 cts. per yd. ? 15 cts. ? 25 cts. ? Ans. to last, 30 cts. 9. To make 25 $ profit, how mark cloth that cost $1 per yard? 1.20? $1.50? $2? $3? $4? $6? Ans. to last, $7.5^) 10. To make 33| % profit, how mark ginghams that cost 25 cts. per yd.? 50 cts. ? Ans. to last, 66* cts. 11. To make 50 % profit, how mark shawls that cost $2 ? $3 ? $4 ? $5 ? 97 ? Ans. to lust, $10.50 j ART. 246. CASE in. To find the RATE PER CENT, of profit or loss, when the cost and selling price are given. 1. I sold cloth at 5 a yard, that cost $4 a yard: what was the gain % ? SOLUTION. Take the difference between the cost and selling price, the gain is $1 on what cost $4; then find what per cent. $1 is of $4; this, (Art. 211), is 1 = .'25 = 25 per cent. Ans. Elllo for Case ILL Take the difference between the cost price and selling price } and jind what per cent, this is of the cost price. 2. A man paid $75 for a horse, and sold him for $105 : what % did he gain? Ans. 40 %. 3. I bought a piece of cloth for $30, and sold it for $40 : what was the % profit? Ans. 33 J %. REVIEW. 245. How mark goods to sell at different rates of profit? 246. What is Case 3 ? What the Eule ? 248 RAY'S PRACTICAL ARITHMETIC. 4. Cloth cost 25 cts. a yard, and sold for 30 cts. a yd. : what the gain % ? Ans. 20 % . 5. Muslin that cost 20 cts. a yard, is sold at 21 cts. a yard: what the % profit? Ans. 5 %. 6. Bought cloth at $8 a yard, and sold it at SO a yard : what % profit did I make? Ans. 12i %. 7. Muslin that cost 30 cts. a yard, is sold at 24 cts. a yard : what % is lost ? Ans. 20 % . 8. A bought 40 bales of cotton, at $40 each, and sold it at a profit of $704 : what did he make ? Ans. 44 % . ART. 247. CASE iv. To find the COST PRICE, when the selling price and rate per cent, of profit and loss are given. 1. Cloth sold at $5 a yard, pays 25 % profit: required the cost price per yard. SOLUTION. When the cost is $1, and 25 per cent, is gained, the selling price is $1.25 (Art. 244); hence, as many times as $1.25 is contained in the selling price, so many times is $1 con- tained in the cost: the cost price is as many dollars as $1.25 is contained times in $5. $5 -f- $1.25 = 4. Ans. $4. PROOF. 25 % of $4 is $1; and $4 + $1 $5. 2. If, by selling cloth at $9 a yard, 10 33 T 2 T cts. a yd. Gain, 33| %. 7. Sold a bushel of rye for 1, and gained 25 % ; purchased a bu. of wheat with the 1, and sold it at a loss of 25 % : what did I lose ? Ans. 5 cts. #8. A merchant bought 14 pieces of cloth at 9.60 each; sold 5 pieces at $14= . 40 each, and 4 pieces at 12 each: at how much a piece must he sell the remainder, to gain 20 % on the whole? Ans. 8.256 9. Sold wine at $1.29 a gal., and lost 14 % : at what price per gal. must it sell, to gain 14 _ ? ^ . 10. 5764801? 2401. 20. T%y ii- 11. 43046721? 6561. 21. 30|? 5.1. 12. 987656329? 31427. 22. 10? 3.162277 + 13. 289442169? 17013. 23. 2? 1.41421+ 14. 234.09? 15.3 24. |? .81649+ 15. 145.2025? 12.05 25. 6|? 2.5298+ 284 RAY'S PRACTICAL ARITHMETIC. 26. How many rods on each side of a square field of 6241 sq. rd.? Ans. 79 rds. THE SQUARE ROOT BY FACTORING. ART. 288. Since any square is the product of two equal factors, if a perfect square be separated into its prime fac- tors (Art. 113), its square root will be composed of half the equal factors. Thus, 441=3X3X7X7; hence ^441 =3X7=21. Therefore, to obtain the square root of any perfect square, resolve the number into its prime factors, and take the product of one of each two equal factors. By Factoring, find the Square Root 1. Of 16. Ans. 4. 5. Of 16X25. Ans. 20. 2. 36. Ans. 6. 6. 36X49. Ans. 42. 3. 100. Ans. 10. 7. 64x81. AnS. 72. 4. 225. Ans. 15. 8. 121X25. Ans. 55. ART. 289. APPLICATIONS OF THE SQUARE ROOT. DEFINITIONS. A triangle is a figure bounded by three straight lines. When one side is perpendicular to another, the angle between them is a right angle^ and the triangle is a right-angled triangle. The side opposite the right angle is called the hypotenuse; the other two sides, the base and perpendicular. Thijg, A B C is a right-angled triangle; A B being the base, B C the perpendicular, and A C the hypotenuse. ART. 290. It is a known principle, that the square of the hypotenuse of a right-angled, triangle is equal to the sum of the squares of the other two sides. REVIEW. 287. What is the rule for square root? NOTES. How proceed when any trial divisor is not contained in the corresponding dividend? 287. When there is a remainder after bringing down the last period, how proceed ? How extract the square root of a common fraction ? How, if both terms are not perfect squares ? How, if a mixed number ? SQUARE ROOT. 285 Thus, in the triangle ABC, the side A B is 3 feet; B C, 4 feet; and A C, 5 feet; the sum of the squares of A B and B C,.9+16, is equal to 25, the square of the side A C. RULE FOR FINDING THE HYPOTENUSE V7HEN THE BASE AND PERPENDICULAR. ARE KNOWN. To the square of the base, add the square of the perpendicular ; the square root of the sum will give the hypotenuse. RULE FOR FINDING EITHER SIDE WHEN THE HYPOTENUSE AND THE OTHER SIDE ARE KNOWN. From the square of the hypotenuse, subtract the square of the other given side; the square root of the remainder will give the required side. 1. The base and perpendicular of a right-angled tri- angle are 30 and 40: what the hypotenuse? Ans. 50. 2. The base and perpendicular of a right-angled tri- angle are 81 and 108: what the hypotenuse? Ans. 135. 3. The hypotenuse of a right-angled triangle is 100; the base, 60: what the perpendicular? Ans. 8.0. 4. A castle 45 yd. high, is surrounded by a ditch 60 yd. wide : what length of rope will reach from the outside of the ditch to the top of the castle ? Ans. 75 yd. 5. It is 36 yd. from the top of a fort, standing by the edge of the water, to the opposite side of a stream 24 yd. wide: how high- is the fort? Ans. 26. 83+ yd. 6. A ladder 60 ft. "long, reaches a window 37 ft. from the ground on one side of the street, and without moving it at the foot, will reach one 23 ft. high on the other side: find the width of the street. Ans. 102.64+ ft. ^ _ REVIEW. 288. How extract tho square root of a perfect square by factoring? 289. What is a triangle? A right angle? Aright-angled triangle ? W hat the hypotenuse ? What tho other two sides ? 290. What is true of every right-angled triangle ? Give an example. How find the hypotenuse when the base and perpendicular are known ? How find either side when the hypotenuse and tho other side are known ? 280 KAY'S PRACTICAL ARITHMETIC. 7. A tree, 140 ft. high, is in the center of a circular island 100ft. in diameter; a line GOO feet 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 ? An*. 533 . 43 -f ft. 8. A room is 20ft. long, 16ft. wide, and 12ft. high: what is the distance from one of the lower corners to the opposite upper corner? Ans. 28. 28 -f- ft. ART. 291. Since the area or superficial contents of a square equals the square of one of its sides, (Art. 87), hence, the RULE FOR FINDING THE SIDE OP A SQUARE EQUAL IN AREA TO ANY GIVEN SURFACE. Extract the square root of the given area; Hie root will be the side of the required square. 1. The superficial contents of a circle are 4096 : what the side of a square of equal area? Ans. 64. 2. A square field measures 4 rd. on each side : what the length of one side of a square field having 9 times as many sq. rd. ? Ans. 12 rd. 3.. There are 43560 sq. ft. in 1 A. : what is each side of a square, containing 1 A, i A, \ A ? Ans. 208. 71 + ft.; 147.58+ft.; and 104.35+ft. 4. A man has 2 fields; 10 A. and 12^ A. : find the side of a sq. field equal in area to both. Ans. 60 rd. EXTRACTION OF THE CUBJ! ROOT. ART. 292. To extract the cube root of a number, is to resolve it into THREE equal factors ; or, to find a number which, when multiplied by itself twice, will pro duce the given number. Thus, 4 is the cube root of 64, because 4X4X4 04. Roots. 1, 2, 3, 4, 5, 6, 7, 8, 9. Cubes. 1, 8, 27, 64, 125, 216, 343, 512, 720, KEVIEW. 291. How find the side of a square equal in area to an? given surface ? 292. What is it to extract the cube root of a number 1 CUBE ROOT. 287 ART. 293. From Art. 278, it follows that the cube root of a number expresses the side of a cube whose solid contents are equal to the given number. Hence, extracting the cube root, is finding the side of n. cube when its solid contents are known; or, arranging a givo number of cubes, so as to form the largest cube possible. ART. 294. From Art. 280, it follows that, The cubo root of 1 is 1 ; The cubo root of 1000 is 10; The cube root of 1000000 is 100; and so on : hence, The cube root of a number between 1 and 1000, consists of one figure ; between 1000 and 1000000, of two figures; between 1000000 and 1000000000, of three, &c. : hence, RULE FOR POINTING. If a dot (.) be placed over every 3d figure of any given number, beginning with units, the number of dots will denote the number of figures in the cube root. ART. 295. 1. Extract the cube root of 13824; or, suppose 13824 cubic blocks, each 1 in. long, 1 in. wide, and 1 in. thick, are to be arranged in the form of a cube. SOLU. First separate OPERATION. 13824(20+4 the given number into 8000 =*?4 periods, by placing dota over the 4 and 3. 20 X 20 X 3 = 1200 The root will consist of 20 X 4X3= 240 two figures. 4X 4- = 16 We next find that the largest cube contained 1456 5824 5824 Root, in 13 (thousand) is 8 (thousand), the cube root of which is 2 (tens), which place on the right, as in extracting the square root. Subtract the cube of 2 (tens), which is 8 (thousand), from the iven number, and 5824 remain. While solving this example by figures, attend to arranging the cubic blocks. After finding that the cube root of the given number REVIEW. 202. Of what numbers are the nine digits tho cube roots ? 293. "What doos tho cubo root of a number express ? 2S4. What tho cubo root of a number between 1 and 1000 ? Why ? Of a numbor between 1000 and 1000000? Why? What tho rule for pointing? 288 RAY'S PRACTICAL ARITHMETIC. will contain two places of figures, (tens and units,) and that the figure in the tens' place is 2, form a cube, A, 20 ('2 tens) inches long, 20 in. wide, and 20 in. high; this cube will contain (Art, 92; 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. It is obvious that to increase the figure FTG. 1. A, and at the same time preserve it a cube, the length, breadth, and bight, must each receive an equal addition. Then, since each side is 20 in. long, \.-.^ ////''// square 20, which gives 20 X 20 = 400, for .J the number of sq. in. in each face of the pi'jj^X cube; and since an addition is to be made I' J' 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 mean's of it, the thickness of the additions may be determined. By examining Fig. 2 (or the blocks, see Note 6), 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 hight are each the same as the additional thickness; and also a cube, D, whose length, breadth, and hight, 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 remaining cubes (5824). Now find the thickness of the additions. It will always be something less than the number of times the trial devisor (1200) is FIG. 2. contained in the dividend (5824). By trial we find 1200 is contained 4 times in 5824. Place the 4 in the quo- tient, and proceed to find the contents of the diiferent solids: these added together, make the number to be sub- tracted, called the subtrahend. The solid contents of the first three additions, B, B. B. are found, (Art. 93) by multiplying the number ot sq. in. 12 the face by the thickness; ______ i CUBE ROOT. 289 now there are 400 sq. in. in the face of each, and 400 X 3=1200 sq. in. in one face of the three; then multiplying by 4, (the thickness,) gives 4800 cu. in. for their contents. The solid contents of the three oblong solids, C, c, c, are found (Art. 93) 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 its length (4) by its breadth (4), and that product by the thickness (4); this gives 4 X 4 X 4 = 64 cu. in. ADDITIONS. If the solid contents of the several B B B =4800cu. in. additions be added together, their sum, Q c c == ggQ 6824 cu. in., will be the number of D = 64 small cubes remaining after forming the first cube, A. Sum 5824 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. In finding the solid contents of the additions, in each case the last multiplier is the thickness. To produce the same result more conveniently, find the area of one face of each of the additional solids, then the sum of the areas, (ns in the numerical operation,) and multiply it by the common thickness. The sum of the areas of ono face of each of the additional solids, is termed the COMPLETE DIVISOR. Thus, In the preceding operation, 14G is the complete divisor. NOTE. As the 1st figure of the root is always in the tens' place with regard to the 2d, annex to it a cipher before it is squared; or, omit the cipher, and multiply the square by 300 instead of 3. REVIEW. 295. How obtain the 1st figure of tho root? Why square it? Why multiply by 3? What is tho product called? Why? 295. IL w obtain the 2d figure of tho root ? Why multiply tho 1st figuro by tho 2d ? Why multiply their product by 3 ? Why square tho 2J figure of the root ? How find the subtrahend ? What is tho complete divisor ? 295. NOTE. Why is a cipher annexed to tho 1st figure cf tho root before squaring? If the cipher is omitted, what must be done? What if the cipher is omitted in multiplying the 1st figure of the root by the 2d? 3dBk. 19 ' 290 RATS PRACTICAL ARITHMETIC. For the same reasoii, annex a cipher to the figure first obtained, before multiplying it by the 2d (the thickness) : Or, omit the cipher, and multiply by 30 instead of 3. Thus, 20 X 20 X 3 = 1200 2 X 2 X 300 = 1200 20 X 4X3= 240 Or, 2 X 4 X 30= 240 4X 4 = 16 4X4 = 16 1456. 1456. *2. What is the cube root of 1728? Arts. 12. 3. Find the cube root of 413493625. OPERATION. 413493625(745 Ans. 343 7X7X300=14700 7X4X 30= 840 4X4 = 16 15556 70493 = dividend. 62224 = subtrahend. 74 X 74 X 300=1642800, 8269625 = dividend. 74 X 5X 30= 11100! 5X 5 = 25! j 1653925; 8269625= subtrahend. EXPLANATION. By the rule for pointing (Art. 294), the root will contain 3 figures. Find 'he 1st and 2d figures of the root ae in the preceding examples, 'i'ten consider 74 as so many tens, and find the 3d figure in tae JO-me manner as the 2d was obtained. 4. Find the cube ~oot of 515.849608 !: 515.849608(8.02 Ans. 512 19200 3849 80 x. 80 X 300 = 1920000 SOX 2X 30^ 4800 2X 2 = 4 1924804 3849608 3849608 EXP. After obtaining the 1st figure, and bringing down the 2d period, we find the trial divisor is not contained in the dividend ; CUBE ROOT. 291 therefore, place a in the root, and bring down another period. Hence, the cube root of a decimal is found in the same manner as that of a whole number, the periods being reckoned both ways from the decimal point. ART. 296. TO EXTRACT THE CUBE ROOT, Kule, 1. Separate the given number into periods of '3 places each, by placing a dot over the units, a dot over the thousands, and so on. (The left period often has only one or two figures.) 2. Find the greatest cube in the left period, and place its root on the right, as in division. Subtract the cube of the root from the left 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 this divisor is contained in the dividend, and write the result in the root. Multiply the last figure of the root by the rest, and by 30; square the last figure of the root, and add these two products to the trial divisor; the sum will be the COMPLETE divisor. 4. Multiply the complete divisor by the last figure of the root, and subtract the product from the dividend; to the remainder bring down the next period for a new dividend, and so proceed until all the periods are brought down. NOTES. 1. When the product of the complete divisor by the last figure of the root is larger than the dividend, the figure of the root must be diminished. 2. After bringing down all the periods, if there be a remainder, the operation may be continued by annexing periods of ciphers. 3. If the divisor is not contained in the dividend, write a cipher in the root, and bring down another period for a new dividend. 4. When there are decimals in the given number, separate them into periods by placing dots over the tenths, ten-thousandths, and so on. The reasons for this are similar to those in Art. 286, Ex. 4. 5. To extract the cube root of a common fraction, reduce it to its lowest terms; then, if both terms are perfect cubes, extract the cube root of each; but, if either term is an imperfect cube, reduce the fraction to a decimal, and then extract the root. VERY TEACHER should have a set of cubical blocki. 292 RATS PRACTICAL ARITHMETIC. To TEACHERS. Instead of finding the subtrahend by the rule, it may be obtained by finding separately the contents of each solid, then adding the whole together. This method, in connection with the blocks, is best adapted to give a clear idea of the nature of the operation. What is the Cube Root of, ANSWERS. ANSWERS. 5. 91125? 45. 15. 53.157376? 3.76 6. 195112? 58. 16. .199176704? .584 7. 912$73? 97. 17. !J j? f. 8. 1225043? 107. Is. 5.744 7 H- 9. 13312053? 237. 19. rW/A ? IS- 10. 102503232? 468. 20. 5J3|? if. - 11. 529475129? 809. 21. 2?~ 1.25992+ 12. 958585256 ? 986. 22. 9? 2.08008+ v 13. 14760213677 ? 2453. 23. 200? 5.84803+ 14. 128100283921 ? 5041. 24. 9i? 2.092+ 25. The contents of a cubical cellar are 1953.125 ctt. ft. i find the length of one side. Ans. 12.5ft. 26. In 1 cu. ft., how many 3 in cubes ? J.?is.-64. 27. How many cubical blocks, each side of which is one-quarter of an inch, will fill a cubical box, each side of which is 2 inches ? Ans. 512. 28. Find the difference between half a solid foot, and a solid half foot. Ans. 648 cu. in. 29. Find the side of a cubical mound equal to one 288ft. long, 216ft. broad, 48ft. high. Ans. 144ft. 30. The sido of a cubical vessel is 1 foot : find the side of another cubical vessel that shall contain 3 times as much. Ans. 17.306+in. EEVIEW. 296. What the rule for extracting the cube root? NOTES. When the subtrahend is larger than the dividend, what is required ? 298. When there is a remainder, how continue the operation ? How proceed when there are decimals in the given number ? How extract the cube root of a common fraction ? ARITHMETICAL PROGRESSION. 293 ART. 297. It is a known principle, that spheres are to each other as the cubes of their diameters j and that All similar solids are to each other as the cubes of theh corresponding sides. Hence, the solid contents, or weight of two similar solids, hav*. to each other the same ratio as the cubes of their like parts. 31. A metal ball Gin. in diameter weighs 32 Ib. : what is the weight of one of the same metal, whose diameter is 3 in - ? Ans. 4 Ib. 32. If the diameter of Jupiter is 11 times that of the earth, how many times larger is it? Ans. 1331 % ART. 298. THE CUBE BOOT BY FACTORING. The cube root of any perfect cube may be extracted ty resolving the given number into its prime factors, and multiplying together one of each three equal factors. 1. Find the cube root of 216. . . Ans. 3x2=6. 2. Find the cube root of 27X64. . . . Ans. 12. 3. Find the cube root of 125x343. . . Ans. 35. X XXV. AEITHMETICAL PROGRESSION. ART. 299. An Arithmetical Progression, or Series, is a series of numbers which increase or decrease, by a common difference. If the series increase, it is called an increasing series ; if it decrease, a decreasing series. Thus, 1, 3, 5, 7, 9, 11, c., is an increasing series. 20, 17, 14, 11, 8, 5, &c., is a decreasing series. The numbers forming the series are called terms ; the first and last terms are the extremes ; the other terms, the mean*. BEVIEW. 297. What ratfo havo the solid contents of two similar bodies ? 298. How extract the cube root of a perfect cube by factoring ? 299. What is an arithmetical progression ? When is the series increas- ing? Decreasing? Give examples. What are the extremes ? Theme*o*t 294 RAY'S PRACTICAL ARITHMETIC. ART. 300. In every arithmetical scries, five things are considered: 1st, the first term ; 2d, the last term ; 3d, the common difference; 4th, the number of terms; 5th, the sum of all the terms. CASE I. ART. 301. To find the LAST TERM, when the first term, the common difference, and the number of terms are given. 1. I boiight 10yd. of muslin, at Sets, for the 1st yd., 7cts. for the 2d, llcts. for the 3d, and so on, with a com. difference of 4cts. : what did the last yd. cost? SOLUTION. To find the cost of the second yard, add 4cts. once to the cost of the first; to find the cost of the third, add 4 cts. twice to the cost of the first ; to find the cost of the fourth, add 4 cts. three times to the cost of the first, and so on. Hence, To find the cost of the tenth yard, add 4 cts. nine times to the cost of the first; but 9 times 4 cts. are 36 cts., and 3cts.-}-36 cts.=39cts., the cost of the last yard, or last term of the progression. 2. The first term of a decreasing series is 39 ; the com. diff. 4; the number of terms 10 : find the last term. SOLUTION. In this case, 4 must be subtracted 9 times from 39, Which will give three for the last term. Hence, the Rule for Case I. Multiply the common difference by the number of terms less one ; if an increasing series, add the product to the 1st term: if a decreasing series, subtract the product from the 1st term : the result will be the required term. 3. Find the last term of an increasing progression : the first term 2; the common difference 3; and the number of terms 50. Ans. 149. 4. I bought 100yd. muslin, at 9 cts. for the 1st yard., 14 cts. for the 2d, and so on, increasing by the com. dif- ference 5 cts. : find the cost of the last yd. Ans. $5 . 04 5. What is the 54th term of a decreasing series, the 1st term 140, and com. din 7 . 2? Am. 34. 6. A lends 3200 at simple interest, at 8 % per annum : REVIEW. 800. What rive things are considered in every series ? ARITHMETICAL PROGRESSION. 295 at the end of the 1st year $216 will be due; at the end of the 2d year, 8232, and BO on : what sum will be due at the end of 20 years? Ans. $520. 7. What is the 99th term of a decreasing series, the 1st term 329, and com diff. g? Ans. 243|. CASE II. ART. 302. To find the COMMON DIFFERENCE, when the extremes 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 the com. diff. ? SOLUTION. The difference of the first and last terms is always equal to the com. diff. multiplied by the number of terms less one (Art. 301 ) ; therefore, If the difference of the extremes be divided by the number of terms less one, the quotient will be the com. diff. Hence, the Hule for Case II. Divide the difference of the extremes by the number of terms less one; the quotient will be the com. diff. 2. The extremes are 3 and 300; the number of terms 10: find the com. diff. Ans. 33. 3. A travels from Boston to Bangor in 10 da. ; he goes 5 mi. the first day, and increases the distance traveled each day "by the same number of miles ; and on the last day he goes 50 mi. : find the daily increase. Ans. 5 mi. ART. 303. It is obvious that if the difference of the extremes be divided by the com. ^diff., the quotient, increased by unity (1), will be the number of terms. 1. The extremes are 5 and 49; the com. diff. 4: finA number of terms. Ans. 12. CASE III. ART. 304. To find tlie SUM of all the terms of the series, wlien the extremes and number of terms are given. "REVIEW. 301. What is Case 1? What is tho Eulo for Cwe 1? 802* What is Case 2 ? What is the Rule for Case 2 ? 296 RAY'S PRACTICAL ARITHMETIC. 1. Find the sum of G terms of the series whose first term is 1, and last term 11. SOLUTION. The series is . . 1, 3, 5, 7, 9, 11, The order inverted is ... 11, 9, 7, 5, 3, 1, The sum 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 extremes), as there are terms. Hence, the Rule for Case III. Multiply the sum of the extremes by the number of terms ; half the product will be the sum of the series. 2. The extremes are 2 and 50 ; the number of terms, 24 : find the sum of the series. Ans. 024. 3. How many strokes does the hammer of a clock strike in 12 hours ? Ans. 78. 4. Find the sum of the first ten thousand numbers in the series, 1, 2, 3, 4, 5, &c - Ans ' $' 8. Find the sum of an infinite series, greatest term 100; ratio 1.04 Ans. 2600. 9. The sum of the infinite series J, -J, 5 V> &c. Ans. ^. 10. The sum of the infinite series J, J, J, &c. Ans. 1. XXVII. PEKMUTATION. ART. 309. Permutation teaches the method of finding in how many different positions any given number of things may be placed. Thus, the two letters a and 6 can be placed in two positions, ab and ba; but if we add a third letter c, three positions can be made with each of the two preceding: thus, Cab, acb, abc, and c6a, 6ca, 5ac, making 2X3=6 positions. By taking a fourth letter d, four positions can be made out of each of the six positions, making 6X4=24 in all. 300 RATS PRACTICAL ARITHMETIC. Rllle for Permutation, Multiply together the num- bers, 1, 2, 3, &c.,from 1 to the given number; the last product will be the required result. 1. In how many different ways may the digits 1, 2, 3, 4, and 5 be placed? Ans. 120. 2. What number of changes may be rung on 12 bells ? Ans. 479001600. 3. What time will 8 persons require to seat themselves dif- ferently every day at dinner, allowing 365 days to the year ? Ans. HOyr. 170 da. 4. Of how many variations do the 26 letters of the alpha- bet admit ? Ans. 403291461 126605635584000000. XXVIII. MENSURATION. ' To TEACHERS. As this short article on Mensuration is intended for pupils who may not have an opportunity of studying a more extensive course, only the more useful parts are presented. The definitions and illustrations are given in plain and familiar terms, not with a view to mathematical precision. ART. 310. DEFINITIONS. 1. An ANGLE is the inclination of two straight lines meeting in a point, which is called the VERTEX. It is the degree of the opening of the lines. 2. When one straight line stands on another so that it makes with it two equal angles, each of Right angle. Right these angles is a RIGHT ANGLE ; and the straight* angle. line which stands on the other is said to be PER* PENDICULAR to it, or at KIGHT ANGLES to it. obtnso /Acte 3. An OBTUSE ANGLE is greater than ** angle. / angle, right angle : and an ACUTE ANGLE is less than NOTE. An angle is named by 3 letters, the middle one beinjj placed at the vertex, and the other two on the lines which form REVIEW. 308. What is Case 2? What the Eule ? NOTE. What is the last term of a decreasing series of which the number of terms it infinite ? 809. What is Permutation ? What the Eule ? MENSURATION. 801 the angle. In the diagram the obtuse angle is called the angle A C B, and the acute angle, the angle A C D. 4. PARALLEL STRAIGHT LINES are everywhere equally distant from each other. 5. A SURFACE has length and breadth, without thickness. 6. A PLANE is a surface, in which, if any two points be taken, the straight line joining them will be wholly in the plane. 7. A FIGURE is a portion of surface inclosed by one or more boundaries. 8. If a figure has equal sides, it is E-QUI-LAT'-ER-AL ; if it has equal angles, E-QUI-AN X -GU-LAR. 9. A TRIANGLE is a figure bounded by 3 straight lines. The side on which the triangle stands, is the BASE. The perpendicular height is the shortest distance from the base to the op- posite angle. Thus, A B C is a triangle ; A C is the base, and B D the perpendicular hight. 10. A QUAD-RI-LAT'-ER-AL is bounded by 4 straight lines. 11. A POLYGON is bounded by more than 4 straight lines. 12. A PAR-AL-LEL'-O-GRAM is a quadrilateral M , whose opposite sides are parallel. Thus, M N O P is a parallelogram. 13. A RECT'-AN-GLE is a quadrilateral whose opposite sides are parallel; its angles right angles. Thus, R E C T is a rectangle. 14. A SQUARE is a quadrilateral whose sides are equal to each other ; its angles right angles. Thus, 8 Q U A is a square. _^ 15. A RHOM'-BUS is a quadrilateral whose sides are equal to each other ; its angles, not right angles. See Fig. Dei. 12. 16. A TRAP'-E-ZOID is a quadrilateral having only two sides parallel. Thus, Z O I D is a trapezoid; tLe sides Z D and 1 being parallel 17. A TRA-PE'-ZIUM is a quadrilateral having no two sides parallel to each other. Thus, T B A P is a trapezium. 302 RAY'S PRACTICAL ARITHMETIC. 18. A DI-AG'-O-NAL is a line joining two angles of a figure not next to each other. Thus, S U (Fig. Def. 14) is a diagonal of the square. 19. A CIRCLE is a figure bounded by a curve line, called the circumference, every part of which is equally distant from a point within, called the center. A DIAMETER is a straight line passing through the center and terminated both ways by the circumference. A RADIUS is T a straight line drawn from the center to the circumference; it is half the diameter. Thus, A D B E is the circumference ; A B the diameter ; A F or B F the radius. 20. A TANGENT is a straight line which touches the circum- ference only in one point, called the point of contact. Thus, T P is a tangent. 21. An ARC of a circle is any part of the circumference, as A F. A CHORD is a straight line joining the extremities of an arc. 22. MENSURATION is the art of finding the surface, and also the solid contents of bodies. 23. The AREA of a figure is the surface which it contains. The quantity of this surface 13 denoted by the number of times it contains a given surface called the measuring unit. The MEASURING UNIT for surfaces is a square surface, whose side is some one of the common measures of length, such as a square inch, a square foot, &c. See Arts. 87, 88, and 89. MEASUREMENT OF SURFACES. ART. 311. To find the superficial contents or AREA of a Parallelogram, Kectanglc, Square, or Khombus, Rule, Multiply tlie length by the perpendicular breadth, the product will be the area. NOTE. The learner must recollect (Art, 27G) that feet in length multiplied by feet in breadth, produce square feet; and the same of the other denominations of lineal measure. 1. Row many square feet in a floor 17 feet long and 15 feet wide? Ans. 255 sq.ft. MENSURATION. 303 2. Find the sq. 'ft in a board 2 ft. 3 in. wide and 1 2 ft 6 in. long. Ans. 28. 1 25 = 28 sq. ft 3. The sq. ft. in a board 15 in. wide and 16ft. long. Ans. 20 sq. ft. 4. How many sq. ft in a board 1 ft 2 in. in mean breadth, 12ft. Gin. long? Ans. 14sq. ft. 84sq. in., or 14 T 7 2 sq. ft 5. At $1.50 per sq. ft, what cost a marble slab; the length 5ft. 7 in.; breadth 1ft 10 in. ? Ans. $15.354+ 6. How many acres of land in a parallelogram ; the length. 120 rd.; the breadth, 84 rd.? Ans. 03 A. 7. How many acres in a square field, each side of which is 65 rd.? Ans. 26 A. 1 R. 25 P. 8. How many acres in a field in the form of a rhombus ; each side measures 35 rd. ; the perpendicular distance across it, 16 rd.? Ans. 3 A. 2R. 9. Each side of the base of a pyramid is 693 ft. long : how many acres does it cover? Ans. 11 A. 4 P. 10. Find the difference between a floor 30ft sq., and two others each 15ft sq. Ans. 450 sq. ft 11. If a room is 10ft long, how wide must it be to contain 80 sq. ft ? (See Art. 90.) Ans. 8 ft 12. A board is 10 inches wide: what must be its length to contain lOsq. ft? Ans. 12ft 13. How many yd. of carpet lyd. wide, will cover a floor 6yd. long, 5yd. wide? Ans. 20yd. 14. How many yd. of carpet llyd. wide, will cover a floor 21ft. Sin. long, 13ft 6in. wide? Ans. 25^ yd. 15. How many yd. of flannel f yd. wide, will line 3 yd. of cloth, IJyd. wide? Ans. 6yd. Plasterers', Pavers', Painters', and Carpenters' Work. ART. 312. Several kinds of artificers' work are measured by the preceding rule. Plasterers', Pavers', and Painters' work, is computed in sq. yards: Glaziers' work, by the sq. ft, or by the pane. Carpenters' and Joiners' work, some parts by the sq. yard ; other parts by the SQUARE, which contains lOOsq. ft. 1. How many square yards in a ceiling 25ft 9 in long, and 21 ft 3 in. wide? Ana. 60 sq. yd. 7 sq. ft+ 304 RAY'S PRACTICAL ARITHMETIC. 2. At 20cts. a sq. yd., what will it cost to piaster a ceil- ing 22 ft 7 in. long, 13 ft. 11 in. wide ? Am. $6.9844- 3. A room is 20ft 6 in. long, 1(5 ft. 3 in. broad, 10 ft 4 in. high : how manv yd. of plastering in it, deducting a fireplace 6ft. 3 in. by 4ft 2 in. ; a door 7ft by 4ft 2 in., and two windows, each 6 ft by 3 ft. 3 in. ? Ans. 1 10 sq. yd. 8j% sq. ft 4. A room is 20ft. long, 14ft. 6 in. broad, and 10ft 4 in. high : what will the coloring of the walls cost, at 27 cts. per. sq. yd., deducting a fireplace 4ft by 4ft. 4 in., and two win- dows, each 6ft by 3 ft 2 in. ? Ans. $19.73 5. At 18 cts. per sq. yd., find the cost of paving a walk 35 ft 4 in. long, 8 ft 3 in. broad. Ans. $5.83 6. What will it cost to pave a rectangular yard, 21 yd. long, and l5yd. broad, in which a footpath, 5ft 3 in. wide, runs the whole length of the yard ; the path paved with flags, at 36 cts. per sq. yd., and the rest with bricks, at 24 cts. per sq.yd.? Ans. $80.01 7. At 10 cts. a sq. yd., what the cost to paint the walls of a room 75 ft 6 in. in compass, 12ft 6 in. high ? Ans. $10.486+- 8. A house has 3 tiers of windows, 7 in a tier: the height of the first tier is 6ft 11 in. ; of the 2d, 5ft 4 in. ; the 3d, 4 ft 3 in. ; each window is 3 ft. 6 in. wide : what cost the glazing, at 16 cts. per sq. ft ? Ans. $64.68 9. A floor is 36 ft. 3 in. long, 16 ft. 6 in. wide : what will it cost to lay it, at $3 a square ? Ans. $17.943+- 10., A room is 35ft long, and 30ft wide: what will the flooring cost, at $5 per square, deducting a fireplace 6ft by 4 ft. 6 in., and a stairway, 8 ft. by 10 ft 6 in. ? Ans. $46.95 11. At $3.50 per square, what cost a roof 40ft long, the rafters on each side 18ft 6 in. long? Ans. $51.80 & ART. 313. To FIND THE AREA OF A TRIANGLE. Rule. Multiply the base by the perpendicular hight, and take half the product for the area. Or, when the sides are given, the following RULE : 1st Add the three sides together, and take half the sum. 2d. From the half sum take the 3 sides severally. 3d. Multiply the half sum and the 3 remainders together, and extract the square root of the product^ which gives the area. MENSURATION. 305 1. Find the area of the triangle, E F G H, the base, F H, is 15 feet; the perpendicular height, G E, 1 2 feet. Ans. 90 sq. ft. 2. The contents of a triangular space, the base 44 rd. , perpendicular height 1 8 rd. , Ans. 2 A. 1 11. 36 P. 3. How many acres in a triangular field ; the base 80 rd. ; perpendicular height, 67 rd. ? Ans. 16 A. 3 R. NOTE. The area of any field or piece of land may be found by dividing it into triangles, and measuring the base and perpendic- ular height of each triangle thus formed. 4. "What cost the glazing of a triangular skylight, at 12 cts. per sq. ft., the base, 12 ft. 6 in., the perpendicular height, 16 ft 9 in. ? Ans. $12.56} 5. Find the area of a triangle, the sides being 13, 14, and 15ft. Ans. 84 sq.ft. 6. The area of a triangle, the sides 2, 3, and 4 feet respect- ively. Ans. 2.9047375+sq. ft ART. 314. To FIND THE AREA OF A TRAPEZOID. Multiply the sum of the parallel sides by the perpen- dicular breadth, take half the product. 1. The parallel sides of a trapezoid, F C G D, are 35 and 26 inches; its breadth 1 1 in. ; required the area Ans. 335^ sq. in. c 2. A field is the form of a trapezoid; one of the parallel sides is 25 rd., the other 19 rd. ; the width 32 rd. : how many ucres in it? Ans. 4 A. 1 R. 24 P. ART. 315. To FIND THE CIRCUMFERENCE OF A CIRCLE, WHEN THE DIAMETER IS GIVEN. Rule. Multiply the diameter by 3. 14 16, the product will the circumference. 1. The diameter A B of the circle A D B E is 48 feet : what is the circumference ? Ans. 150.7908 a 2. The diameter of a wheel is 4 feet :^ the circumference. Ans. 12 ft. 6.7968 in. 3d Bk. 20 308 RAT'S PRACTICAL ARITHMETIC. 3. What is the circumference of the earth, the mean di' ameter being 7912.4 mi. ? Ans. 24857.59584 ml ART. 316. To FIND THE DIAMETER OF A CIRCLE, WHEN THE CIRCUMFERENCE IS GIVEN. Rule. Divide the circumference by 3.1416, the quotient will be the diameter. 1. The circumference of a circle is 15 feet: what is the diameter ? Ans. 4 ft. 9.295+in. 2. If the girt of a tree is 12 feet 5 inches, what its thick- ness or diameter ? Ans. 3 ft. 11.428-|- in. ART. 317. To FIND THE AREA OF A CIRCLE. Rule. Multiply the diameter by the circumference, and take one-fourth of the product. Or, Multiply the square of the diameter by .7854; or, for greater accuracy, by .785398 Or, Multiply the square of the radius by 3.1416 1. Find the area of a circle, the diameter being 42 feet. Ans. 1385.4456 sq.ft. 2. Find the area of a space on which a horse may graze, when confined by a cord 7^ rods long, one of its ends being fixed at a certain point. Ans. I A. 16.715P. ART. 318. To FIND THE DIAMETER OF A CIRCLE, WHEN THE AREA is GIVEN. Rule. Divide the area by .7854 ; the square root of the quotient will be the diameter. 1. The area of a circle is 962.115: what its diameter and circumference? Ans. diam. 35 : circum. 109.956 2. What length of halter will fasten a horse to a post in the center of an acre of grass, so that he can graze upon the 1 A and 10 more ? Ans. 7.1364-f-rd., or 117 ft. 9-f-in. ART. 319. MEASUREMENT OF BODIES OH SOLIDS. DEFINITIONS. 1. A BODY or SOLID, has length, breadth, and thickness or depth. p 2. A PRISM is a solid whose ends, or bases, are parallel; its sides, parallelo- gram*. Such a body is termecf a RIGHT MENSURATION. 307 PRISM when each of its bases is perpendicular to its other side* ; and it is TRIANGULAR, QUADRANGULAR, &c., according as its base is a triangle, quadrangle, &c. Thus, P is a triangular prism. 3. A PAR-AL-LEL-O-PI'-PED is a prism whoso bases and also its other sides are parallelo- grams. Thus, B is a parallelepiped. 4. A parallelepiped is RIGHT when each of its faces is a rectangle. A common chest, a bar of iron, brick, &c., are instances of right parallelepipeds. When each face of a right parallelepiped, as A, is a square, it is termed a cube. A cube has 6 equal square faces. J 5. A CYLINDER is a round prism, having circles for its ends. Thus, C is a cylinder, of which the line E F passing through the centers of both ends, is called the axis. 6. A PYRAMID is a solid having any plane figure for a base, and its sides triangles, whose vertices meet in a point at the top, called the VERTEX of the pyramid. A pyra- mid is TRIANGULAR, QUADRANGULAR, &c., according as its base is a triangle, quad- rangle, $900. 320 RAY'S PRACTICAL ARITHMETIC. 89. Two Globes, each 5 in. in diameter, and 2 cubes, each 5 in. in length, were melted into one cube : how long was the side of this cube ? Ans. 7.24-j-in. 90. Sold a cow for $25, losing 16f % ; bought another and sold it at a gain of 16 % ; I neither gained nor lost on the two : what the cost of each? Ans. 1st, $30: 2d, $31.25 91. After a battle, in which an army of 24000 men were engaged, it was found that the number of slain was \ of those who survived ; the number wounded was equal to \ of the slain: find the number wounded. Ans. 1500. 92. I cut 95yd. of cloth into 3 pieces, so that the first was three times the second, and the third one-fourth the first: what was the length of each? Ans. 1st, 60; 2d, 20; 3d, 15yd. 93. A merchant purchased 60yd. of cloth, at $4 a yard, on a credit of 6 mon. ; he sold it immediately for $250 : what did he gain, money being worth 6 % ? Ans. $ 16.99 -f- 94. A drover expended $1500 in horses, cows and sheep: the horses cost $120, the cows $30, and the sheep $5 each: there were six times as many sheep as cows, and i as many horses as sheep: find the number of each. Ans. 10 horses; 5 cows; 30 sheep. 95. The tax in a certain town was 1 ct. 6m. on the dollar, and polls $1 each ; A, who paid for 4 polls, was charged $328 tax: what the value of his estate ? Ans. 20250. 96. A, B, and C do a piece of work in a certain time : A and B together do | of it ; B and C together do f : what part can B do alone? Ans. |. 97. A and B together received $1000: if B had received $100 less, his share would then have been one-half of A's: how much did each receive? Ans. A, $600; B, $400. 98. A and B have just $500 : f of A's money is $50 less than f of B's : what sum has each ? Ans. A, $200; B ; $300. 99. If | of the time past noon is equal to | of the time to midnight, what is the hour? Ans. 3 P. M. 100. A lady spent in one store J of all her money and $1 more ; in another, ^ of the remainder and $^ more ; in another, ^ of the remainder and $1 more; and in another, of the re- mainder and