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Being the product of long years in the class room, it is thoroughly practical. The Globe is a working model of the Earth in its relation with the Sun. The horizon and other parts are 80 constructed, that in every illustration the Globe represents the real position and relation of the Eanli to the Sun. The Tellurian Globe combines in the most con- venient and substantial shape, and at much less expense, the best form of Globe, and the best form of Tellurian. It illustrates also in a more simple and philosophic manner than can be done by the use of both Globe and Tellurian, all problems relating to the phenomena of day and night, change of season, twilight, rising and setting of the sun, the apparent daily motion of the sun across the horizon, the motion of the earth in its orbit. The Tellurian Globe is made in [two sizes. Eight inch and Twelve inch. Bt dr. MALCOLM MACVICAR, Ph.D., LL.D., Principal of the State Normal School^ Potsdamj N.Y.j Handbook of the Mac Vicar Tellurian Jllobe, for the use of Teachers, Schools and Families ; oon- taimng a complete course of illustrations and problems in Geography and Astronomy. School Booics Published by Daicson Bros, By professor ANDREW, 0/ the University of McGill College, The New Dramatic Beader ; Comprising a Selec- tion of Pieces for Practice in Elocution, with intro- ductory hints to Readers. Price, 75 cents. By J. D. MORELL, LL.D., n. M. Inspector of Schools, England, A Complete Manual of Spelling on the Prin- ciples of Contrast and Comparison ; with numerous- Exercises. Price, 30 centta. By F. C. EMBERSON, M.A., Inspector of Schools in the Province of Quebec, The Art of Teaching: A Manual for the use of Teachers and School Commissioners. Price 50 cents* Canadian Elementary School Atlas; For the use of Junior Classes, containing 16 Maps. Price, 25 cents. A cheap and yet very complete elementary Atlas. The Maps, which are similar to those in Keith Johnston's ele- mentary Atlases, are clear and attractive. Lennie's English Grammar Carpenter's Spelling The Spencerian Series of Copy Books; A System of Penmanship very extenbively used in the United States and the Dominion of Canada. The System is comprised in twelve Numbers, and divided into four distinct Series, viz : — Nos« If 3* 3* 4 dc 6, Common School Sorles. No«. 6 & 7* Bunlneso Series. No«. 8 dc 9, • L.»dle»* Series. N«s. 10, 11 4c 12 Exerctoo Sorlea. 1^ National Library Bibliotheque nationale of Canada du Canada 0( D] ^^ A 'J// /^ / COMPLETE ARITHMETIC, ORAL AND WRITTEN. Designed for the Use of Common and High Schools AND Collegiate Institutes. REV. D. H. MACVICAR, LL.D., PRLVCIPAL, PrbSBTTBRIAH COLLBOl, MOHTRIAL. ■ « » ■ MONTREAL: DAWSON BROTHERS, PUBLISHERS 1879. OA /03 1^31 Entered according to Act of Parliament of Canada, in the year 1879, byl Dawson Brothers, in the Office of the Minister of Agriculture. PREFACE. THIS work is based upon the Complete Arith- metic prepared two years ago by M. JMacVicar, Mi.D., LL.D., Principal of the State Normal School, 'otsdam, N. Y., and the undersigned. In the present vohime the order is changed throughout, and the work is specially adapted to the wants of the Com- mon and High Schools, and Collegiate Institutes of 'unada. Attention is invited to the Properties of Numbers, rreatest Common Divisor, Fractions, Decimals, Com- [)ound Numbers, The Metric System, Business Arith- metic, Ratio and Proportion, Alligation, and Square md Cube Root, with the belief that the treatment will be found new and an improvement upon former uethods. In every subject the pupil is first required to baster elements or preparatory steps and propositions In order to fit him for the more advanced and com- )lex operations. A systematic drill is provided on Oral and Written Exercises and Review and Test Questions, by which [he principles and processes of numbers, with their iv F liEFAC E. applications to j)racticul busiiiesri, will be permanently fixed in the mind. In all cases in which it is possible, each process is presented objectively, so that the truth is exhibited to the eye and thus clearly defined in the mind. The entire drill and discussions are believed to be so arranged, and so thorough and complete, that by passing through them the pupil cannot fail to acijuire such a knowledge of principles and facts, and to receive such mental discipline, as will prepare him properly for the study of the higher mathematics. Invaluable aid in methods of presentation and the management of class work will be found in the Teacher's Edition of Dr. M. MacVicar's Complete Arithmetic, published by Taintor Brothers, Merrill &Co., N. Y. D. H. MacVICAE. Montreal, July, 1879. CONTENTS. Page Notation and Numeua- TION 1 Roman Notation 9 Revikw and Test Ques- tions 10 Addition 11 Canadian Money 18 Rkview and Test Ques- tions 20 Subtraction 21 Review and Test Ques- tions 27 Multiplication 28 Review and Test Ques- tions 40 Division 42 Division by Factors ... 55 Review and Test Ques- tions 60 Properties op Numbers 66 Exact Division 67 Prime Numbers 70 Factoring 72 Cajjcellation 74 Greatest Com. Divisor. 77 Least Com. Multiple. . 85 Pagk Review and Test Ques- tions 89 Fractions 90 Reduction 93 Addition 102 Subtraction 105 Multiplication 107 Division 114 Complex Fractions 119 Review and Test Ques- tions 126 Decimal Fractions 128 Notation and Numera- tion 129 Reduction 134 Addition 142 Subtraction 143 mui.tiplication 144 Division 140 Review and Test Ques- tions 152 Denominate Numbers.. 154 Units op Weight 155 Comparative Units op Weight 156 Units of Length 168 VI CONTENTS, Page Units op Surfaoe 171 Units of Volume. . . . 174 Units of Capacity. . . . 180 Comparative Units of Capacity 182 Units of Time 183 Units of Money 186 Metuic System 188 Decimal Related Units 188 Comparative Table of Decimal Units 191 Duodecimals 192 Longitude and Time . . 193 Review and Test Ques- tions 195 Business Arithmetic... 196 Aliquot Parts 199 Business Problems 202 Applications 217 Profit and Loss 218 Commission 220 Insurance 222 Stocks 224 Duties or Customs 228 Review and Test Ques- tions 5)29 Interest 230 Method by Aliquot Parts 232 Method by Six Per Cent 234 Method by Decimals . . 236 Page Exact Interest. 237 Compound Interest 242 Interest Tables 244 Annual Interest 246 Partial Payments 247 Discount 251 253 256 257 261 , Bank Discount '] Exchange Inland Exchange Foreign Exchange. . . Equation of Payments. 266 Review and Test Ques- tions 275 Ratio 276 Proportion 281 SiMPiJE Proportion 285 Compound Proportion. 289 Partnership 292 Alligation Medial. 296 i Alligation Alternate. 297 Involution 302 j Evolution 304 Progressions 320 Arithmetical Progres- sion 321 Geometrical Progres- sion 323 Annuities 325 Mensuration 328 Review and Test Ex- amples 342 Answers 358 ARITHMETIC. NOTATION AND NUMERATION. NUMBERS PROM 1 TO 1000. Art, 1. Numbers are expressed by means of ten figiirei, r.».o«,. 1 2 3 J,. 5 6 7 8 (j Snmea, Naught, Que, Two, Three, Four, Five, Six, Seven, Eight, Nine. Observe regarding the ten figures : 1. The uaugld is also called cipher or zero, and when written alone stands for no number. 2. The other nine figures are called digits or significant figures, and each stands for the number written under it. :3. Any number of objects not greater than nine is expressed [by one figure. Thus, 4 boys, 2 girls, 6 pens, 9 desks, 9 windows. One or luiitg is the foundation of all numbers, and they are jut liundreds. Each succeeding period is read in the same manner. EXAMPLES. 5 6. A large number can be read aa easilj^ as a number of three places, when the followiii<^ names of the fii-st order on the right of the successive periods are fixed in the memory ; PEKIODS. NAMES. PERIODS. NAMES. 1st. Units. 7th. Quintiliions 2d. Thousands. 8th. Sextillions. 3d. Millions. 9th. Septillions. 4th. Billions. 10th. Octillions. 5th. Trillions. nth. Nonil lions. Gth. Quadrillions. 12th. Decillions. From these illustrations, we obtain for reading numbers the follotving RULE. I. Begin at the right and separate the 7iumber, by inserting commasy into periods of three figures each. II. Begin at the left and read the hundredx, tens, and ones of each period, giving the name of the ones in each case except in the last peHod. EXAMPLES FOR PRACTICE. 5. Point off and read the following numbers : ■ '■ 1. 400. 704. 393. 3348. 5592. 9347. 6043. 2. 74085. 93061. 452034. 290620. 48207604. 3. 1401. 4033. 6306. 8300. 3080. 5906. 310S. 4. 65003. 64004. 99040. 30307. 406205. 340042: 5.507009. 85004. 230060. 903560. 100001. 6. 2060. 50040. 3040000. 2406007. 50:]0063. 7. 3000000. 40006003. 30304090. 400006000. 300000804. 8. 900800800800. 4005008004. 307000060080. 9. 804042. 85064. 9002005. 100100100101. 10. 3000050030. 8300400706005. 9000100130004. 11. 97304306590724059034. 3000600000596034006670. 6 NOTATION AND NUMERATION. WHITING LABGE NUMBEBS. G. Nurnhers are written one 'period at a time and in the order in which the periods are read. Observe regarding this proposition . 1. Each period in a number except the one at the left must contain three figures. Hence the places for which significant figures are not given must be filled with ciphers. Thus, three hundred seven million, four thousand, eighty- two, is written 307,004,082. Observe in this number a significant figure is given only for the hundreds and ones in the million's period, hence the ten's place is filled with a cipJier. For a like reason the ten's and hundred's place in the thousand's period and the hundred's place in the unit's period are filled with ciphers. 2. When a number is read, a period in which all the orders are wanting is not named. Care must therefore be taken to notice such periods and fill their places in each case with three ciphers. For example, in the number seven million three hundred four, the thousands period is not named, but when the num- ber is expressed in figures its place is tilled with three ciphers ; thus, 7,000,304. RULE. Begin at tJie left and write the figures expressing the hun- dreds, tens, and ones of each period in their proper order, filling with cipJiers all x)eriods or places where no signifixiant figures are given. EXAMPLES FOR PRACTICE. 7. Express in figures the following numbers : 1. Three hundred four. Five hundred sixty. Eight hun- dred ninety. Three hundred seventy-seven. 2. Tliree hundred five. Elight thousand thirty. DEFINITIONS, 7 3. Twelve hundred. Twenty-seven hundred. 4. Ten tens. One hundred tens. Ten tens and five. Two hundr(;d tens and sixteen. Eight hundred six tens. 5. Ten thousand. Ten thousand four. Twenty thousand. Twenty tliousand fifty-nine. 6. Eleven thousand eleven. One million five. 7. Eighty million seventy thousand ninety. 8. Three thousand three hundred six million. 9. Eight hundred thirty-two tens. Six hundred seven tens. Two thousand sixty-five tens. 10. Fifty-nine thousand million. Seven thousand and six million. Forty-four billion seven. 11. 77 million 1 thousand 5. 409 trillion 4 million 6. 12. 12 billion 205 thousand 49. 1 trillion 1 million 1. DEFINITIONS. 8. A Unit is a single thing, or group of single things, regarded as one ; as, one ox, one yard, one ten, one hundred. 9. Units are of two kinds — Mathematical and Com- mon. A matTiematical unit is a single thing which has a fixed value ; as, one yard, one quart, one Tumr, one ten. A common unit is a single thing which has no fixed value ; as, one house, one tree, one garden, one farm. 10. A Number is a unit, or collection of units; a.B, one man, three houses, four, six hundred. Observe, the number is " the how many" and is represented by whatever answers the question. How many? Thus, in the ^expression seven yards, seven represents the number. 1 1. The Unit of a Number is one of the things num- bered. Thus, the unit of eight bushels is one bushel, of five boys is one boy, of nine is one. 1 2. A Concrete Number is a number which is applied ! to objects that are named ; as, four chairs, ten beUs, 8 NOTATION AND NUMERATION. ill, 1J$. An Abstract Number is a number which is not applied to any named objects ; as, nine^ Jlue, thirteen. 14:. LiUce Numbers are such as have the same unit. Thus, four windows and eleven windows are like numbers, eight aud ten, three hundred and seven hundred, 15. Unlike Numbers are such as have different units. I Thus, twelve yards and five days are unlike numbers, also six j cents and nine minutes. 16. Fi ff tires are characters used to express numbers. 17. The Value of a figure is the number which itj represents. 18. The Simitle or Absolute Value of a figure is the | number it represents when standing alone, as 8. 19. The Local or Representative Value of a figure isj the number it represents in consequence of the place it] occupies. Thus, in 66 the 6 in the second place from the rigln | represents a number ten times as great as the 6 in the first | place. 20. Notation is the method of writing numbers by] means of figures or letters- 21. Numeration is the method of reading numbers] which are expressed by figures or letters. 22. A Seafe in Arithmetic is a succession of mathematical! units which increase or decrease in value according to a fixed] order. 23t A Deeiinal Scale is one in which the fixed order of| increase or decrease is uniformly ten. This is the scale used in expressing numbers by figures. 24. Arithmetic is the Science of Numbers and the Artf of Computation. B M A ^' N OTATIO X, 9 ROMAN NOTATION. 25. Cluiracters Used, — The liomau Notation expresses numbers by seven letters and a dasL. Letters,— I, V, X, L, C, D. M. r«//ie«.-One, Five, Ten, Fifty, u^«^,^ „ Fiv«^^ ^^^O,. ^^^^ 2C5. Laws of llonian Notation ,— The above sevoii letters and the dash are used in accordance with the following laws: 1. Repeating a letter repeats its value. Thus, I denotes one; II, two; III, three; X, ten; XX, two tens, or twenty. 2. ^Mle>l a letter ia placed at the left of one of greater value, the difference (f their values is tJie 7iumber expreaseil. Thus, IV denotes four ; IX, nine: XL, forty. 3. W/ien a letter is placed at the right of one of greater value, the sum of their values is the number expressed. Thus, VI denotes six ; XI, eleven ; LX, sixty. 4. A dash placed over a letter multiplies its value hy one thousand. Thus, XI denotes eleven thousand ; V, five thousand ; VI, six thousand. EXERCISE FOR PRACTICE. 27. Express the following numbers by Roman Notation : 1. Six. 2. Four. 3. Three. 4. Two. 5. Nine. 6. Sixteen. 7. Thirteen. 8. Seventeen. 13. Thirty -eight. 14. Thirty- nine. 15. Forty-six. 16. Forty-seven. 9. Nineteen. 10. Fourteen, 11. Twenty. 13. Seventy-five 17. One hundred twenty-seven. Seven hundred four. 18. Nine hundred forty-nine. Ninety-five. 19. One thousand. Nine thousand. Fifty thousand. 2 10 NOTATION AND NUMERATION, 20. Four thousand. One hundred thousand. Eight hundred thi)usand. Ninety thousand. 21. 2800. 1875. 0053. 7939. 4854. 10365. ^5042. 22. Read the following: MIXj MDL XIV ; X; D; MM;| MD ; DVII ; MDCCCLXXVI ; ML ; DLX. REVIEW AND TEST QUESTIONS. ii8. Study carefully and answer each of the following! questions : 1. Define a scale. A decimal scale. 2. How many figures are required to express numbers in tliej decimal scale, and why ? 3. Explain the use of the cipher, and illustrate by exampKsJ 4. State reasons why a scale is necessary in expressing! numbers. 5. Explain the use of each of the three elements— ^gr?/r^^' place, and comma — in expressing numbers. 6. Wliat is meant by the simple or absolute value of figures !| What by the local or representative value ? 7. How is the local value of a figure affected by changing itj from the first to the third place in a number ? 8. How by changing a figure from the second to the fourth ! From the fourth to the ninth ? 9. Explain how the names of numbers from twelve to twentj ar(i formed. From twenty to nine hundred ninety. 10. What is meant by a period of figures ? 11. Explain how the name for each order in any period isj foi-med. 12. State the name of the right-hand order in each of tli^ first six periods, commencing with units. 13. State the two things mentioned in (6) which must observed when writing large numbers. 14. Give a rule for reading numbers ; also for writing numbers. ADDITION 20. Tlie Addition Table consists of the sums of the lumbers from 1 to inclusive, taken two at a time. These 11118 must at first be found by counting ; but when found, |i«y Hhould l>e fixed in the memory so that they can be given Hight of the figures. 30. To find the mim of two or more numbers^ each expressed one figure, 1/y using the Addition Table, 1. Find the sum of 7, 9, and 8. SOLUTION,— (1) We know at once from the memorized remits of the Idiiion table, that the pum of 7 and 9 is 16 or 1 ^en and 6 units. \{'l) We add the 8 unitti to the 6 uuitB of the last result and know in kc sami! manner that the sum of the 8 and 6 le 14, or 1 (en and 4 unit*. jniting tliia ten with the ten found by adding the 7 and 9, we have 2 tens |i(l 4 units., or %\. Hence the sum of 7, 9, and 8 is 84. 2. The process in finding the sum of any column of figures )iisists in noting the tens which the column makes. Thus, suppose the figures in a column to be 9, 6, 8, 5, and 7. )mmencing with 9 we note that 9 and 6 make 1 ten and 5. ^' add the 8 to the 5 and we have another ten and 3, making tons and 3. We add the 5 to the 3, making 2 tens and 8. ^e now add the 7 to the 8 and we have another ten and 5, iking in all 3 tens and 5 units, or 35. |3. Be careful to observe that in practice each new number added to the excess of the tens mentally^ and nothing Lined but results jFor example, in finding the sum of a column consisting the figures 9, 2, 8, 5, 7 and 4, commencing with 9 the Hilts should be named, thus, ni?ie, eleven, nineteen, twenty- nir, thirty-one, thirty five. IS ADDITION. 4. The numbers to bo added are callt-d Addends. The result found is called the tyuni or Amount, and the process <'J riniHii;r the sum is called Addition. The tii'fjn +, read i)Ius, placed between numbers ; thus, G + 3 + 10, hIjovvs that these numbers are to be added. The sign — , read ('(juals, denotes that wluit is writt(?n l)ef()re it is equal to what is written after it ; thus, 8 + () =^ 14, is read 8 plus eijuals 14, 5. To become expert and accurate in adding you muht practice on columns first of three figures, then four, then fui , until you can giv(; the sums of such columns at sight, iou rmst al. first example, tlion oppoHite li and copy three nuinluMrt for the Hccond cxanipl". and no on to the bottom of tlie column. The fir.st hIx examplch copied from column A in this way are , (1) } 1 S (8.) 3 5 (8.) 3 5 (4.) (5.) (0.) 5 4 6 A 6 8 9 8 7 8. Copy examples with three numbers from each column in tlie name way, and i)ractice on finding the sums as directed for I memorizing the addition table. 3. Copy in the same manner examples with four numbers, fice numbers, and so on up to ten numbers. Continue to practice in this way until you can add rapidly land accurately. ILLUSTRATION OF PROCESS. »*{2. Prob. 1. — To find the sum of two or more numbers, )ach containing only one order of units, and all the same )rder. Find the sum of (1.) (2.) (3.) 60 600 90 900 ^ JOO 230 2300 Explanation.— 1. The Biim of 8, 9, and 6 is found by forming groups of ten. Thut», 8 and 9 make 1 ten and 7 ; 7 and 6 make 1 ten and 3 ; hence, 8, 9, and 6 make 2 tens and 3, or 33. 2. The snm of 8, 9, and 6 is the same whether these flt^ures express units, tens, or hundreds, etc. Hence, when their sum is found, if they express units, as in the lifit example, the sum is units ; if they express tens, as in the second ex- ample, the sum is tens ; [{hundreds, hundreds, etc. 6 9 _8 23 14 ADDITION, SIGHT EXERCISES. * • Find the sum of 1. 70 + 9 + 5. 5. 3500 + 80 + 2. 9. 500 + 300. 3. 900 + 50 + 8. 6. 7 006 + 800. 10. 8000 + 6000 3. 900 + 60 + 7. 7. 90 + 80. 11. 6000 + 5000 4. 3000 + 50 + 3. 8. 70 + 80. 12. 200 + 400. (13.) (14.) (15.) (16.) (17.) (18.) 30 200 9000 40000 800 9000 60 400 3000 60000 600 5000 80 700 7000 70000 700 3000 (19.) (30.) (31.) (22.) (33.) (34.) 40 900 5000 .50000 700 8000 50 300 3000 80000 300 3000 70 800 7000 60000 400 6000 33. Prob. 2.- -To find the sum of any two or more numbers. Find the sum of 985, 854, and 698. (1.) ANALYSIS. (3.) 985 = 900+80+5 985) 854 -^ 800 +50+4 854 ■ Addends 698 = 600+90+8 698 1 17^ 2537 Sum. 320 [ = 2300 + 220 + 17 2300) 2537 Explanation.— 1. The orders of units in the ntunbers to be added i indeiwndent of each other, and may ])e separated as ehown in the analymX 2. The Hum of each order u found by finding the eum of the figures e^ pressing that order (32). EXERCISES FOR PRACTICE, 15 00+300. 000 + 6000. .OOO + 5O00J 100 + 400. (18.) 9000 5000 3000 (24.) 8000 3000 6000 or mord Addends. Sam. 3. The Bums of the separate orders may be united into one sum, a9 phown in the aualyf is ; or, 4. By commencing with the units' order, the number of tens found can at once be added to the tens' order; so with the hundreds found by ailding the tens' order, etc., and thus the sum may be found in one operation, as shown in (2). From these illustrations we obtain the following RULE. ' iJ4:. / Write t?vc numbers to he added in such a manvfir that units of the same order will stand in the same column. IF. Add each column separately, commencing uith the units. III. When the sum of any column U expressed by two or more figure*, place the right-hnnd figure under the column, and add the nuinber expressed by the remaining figures to the next column. IV. Write under the last column its entire sum. Proof. — Add the numbers by commencing at the top of the columns. If tlce results agree, the work is probably correct. EXERCISES FOR PRACTICE. IJo. For practice with abstract numbers, copy from Table No. 1, page 12, examples as follows : Three Numbers of Ttiree Plaeest 1. Use any three consecutive columns, as A, B, C. Com- mence opposite 1 and copy three numbers for ilie first example, then opposite 2 and copy three more for the second example, and so on to the bottom of the table. The first six examples copied in this way are as follows : 1 ■ (1.) (2.) (3.) (4.) (5.) (6.) be added oil 1234 138 395 557 487 68G the awa/yfl^l |l38 395 557 487 686 848 he figures CM |395 557 487 686 848 797 16 ADDITION, 3. Copy in the same manner examples with three numbers from columns b, c, d ; c, D, e ; D, E, F ; E, F, G ; P, G, H ; G, H, I ; and H, i, J. Four Numbers of Four Flacea* 36. 1. Copy as before the numbers from any four consec- utive columns, as c, d, e, f. Commence in each case oppo- site 1 for the first number of the first example, opposite 2 for the first number of the second example, and so on to the bottom of the table. 2. Copy in the same manner examples from A, B, c, D ; B, c, D, E ; D, E, F, G ; E, P, G, H ; F, G, H, I ; and G, H, I, J. ■i Numbers of Five Fiaees. 37. Continue the practice by copying numbers of five places, as already directed. Commence with examples of five numbers, then six, then seven, and so on. ORAL EXAMPLES. 38. 1. A farmer sold 60 bushels of wheat to one man, 40 to another, and 20 to another ; how many bushels did he sell ? Solution.— He BOld as many bushels as the Bum of 60, 40, and 20, which is 120. Hence he sold 120 bushels. 2. Mr. Amaron owns 30 acres of land, Mr. Cruchet owns 50, and Mr. Easty 70 ; how many acres do they all own ? 3. R. \Vliillans sold a cow for $40 and fifteen sheep for $65 ; how much did he receive for the cow and sheep ? 4. A lady paid ,f 34 for a shawl, $45 for a dress, and $7 for a scarf ; how much did she pay for all ? 5. A boy bought 4 bails and paid 40 cents for each ball. How much did he pay for the three ? 6. W. D. Russel sold a tub of butter for $27, a cheese for $24, and some beans for $16 ; how much money did he receive 1 EXAMPLES, 17 7. A tailor sold a coat for $25, a vest for $6, and a hat for $5 ; how many dollars did he get for all ? 8. A lady gave $72 for a watch, ^32 for a chain, $2 for a key, and ij^8 for a case ; what did she give for all ? WRITTEN EXAMPLES. 39. 1. A newsboy sold 244 papers in January, 301 in Feb- ruary, 278 in March, and 390 in April ; how many papers did he sell in the four months? Ans. 1213. 2. A grocer paid $375 for coffee, $280 for tea, $564 for |suf]:ar, !^108 for dried apples, and $198 for spices; what was I the amount of the purchases ? Ans. $1525. 3. In a city containing 4 wards, there are 340 voters in the [first ward, 533 in the second, 311 in the third, and 425 in the I fourth ; how many voters in the city? 4 Norman D. Warren has a house worth $850, and five more 3ach worth $975 ; what is the value of the six "'' 5. In 1870 the population of Albany was G9452, Utiea >879S, Syracuse 43081, Rochester 63424, Buffalo 117778; ,'hat was the united population of these cities ? 6. What is the distance from the Gulf of St. Lawrence to jake Michigan, passing up the River St. Lawrence 750 miles, jake Ontario 180 miles, Niagara River 34 miles. Lake Erie ioO miles, Detroit River 23 miles, Lake and River St. Clair 15 miles, and Lake Huron 260 miles ? Ans. 1542 miles. 7. A man bought a house for $3420 ; he paid $320 to have jt painted, and $40 to have it shingled ; for what amount mst he sell it in order to gain $250? u4/i«. $4030. 8. Bought a horse for $275 and a carriage for $342 ; sold |he horse at an advance of $113 and the carriage at an advance ^f !{;65 ; how much did I get for both ? Ans. $795. 9. A grain dealer paid $1420 for a lot of flour, and $680 for lot of meal : he gained $342 on the flour and $175 on the leal ; how much did he receive for both lots ? 18 ADDITION, 10. Bought 3 house-lots ; the first cost $325, the second $15 more than the first, and the third as much as both the others ; what was the cost of the whole ? An%. $1330. CANADIAN MONEY. 40. Tlie sign $ stands for the word dollars* Thus, $13 is read 13 doUara. 41. The letters ct. stand for cents. Thus, 57 ct. is read fifty-seven cents. 43. When dollars and cents are written together, the cents are separated from the dollars by a ( , ). Thus, $42 and 58 ct. are written $42.58. 43. When the number of cents is less than 10, i cipher must occupy the first place at the right of the period. Thus, $8 and 4 ct. are written $8.04. 44. In arranging the numbers for adding, do. lars must be placed under dollars and cents under cents, in such a manner that the periods in the numbers stand over each other, thus : (1.) (3.) (3.) $370.84 $3497.03 $53.70 43.09 69.50 786. 706.40 240.84 9.08 ^A/■RITTEN EXAMPLES. 45. Read, arrange, and add the following : 1. $4.75 + $3083.09 + $72.50 + $9.32 + $384. 2. $93.48 + $406.30 + $8.07 + $5709.80. 3. $500 + $93.05 + $364.80 + $47.09. Express in figures the following : 4. Nine hundred six dollars and seventy-five cents. 5. Seventy-five dollars and thirty -eight cents. 6. Three hundred twelve dollars and nine cents. 7. Eighty -four cents ; seven cents ; three cents. DEFINITIONS, 1& 8. Find the sum of $206.08, $5.54, and $396.03. 9. A farmer sold a quantity of wheat for $97.75, of barley for .$42.06, of oats for $39.50. How much did he receive for the whole? '^^^' $179.31. 10. A man bought a horse for $345.50, a carriage for $182.90, and sold them so as to gain on both $85.50. How much were they sold for? Arts. $613.90. 11. Bought a house for $4268.90, furniture for $790.07, car- peting $380.60, and made repairs on the house amounting to $307.05. How much did the whole cost? Ans. $5746.62. 12. A man is in debt to one man $773.60, to another $600.50, to another $73.08, to another $305.04 ; how much does he owe in all ? Ans. $1751.22. 13. A furniture dealer sold a bedroom set for $125.86, a bookcase for $85.09, and 3 rocking-chairs for $5.75 each. How much did he receive for the whole ? Ans. $228.20. 14. D. N. Mac Vicar bought a saw mill for $8394.75, and sold it so as to gain $590.85 ; for how much did he sell it ? 15. A lady after paying $23.85 for a shawl, $25.50 for a dress, $2.40 for gloves, and $4.08 for ribbon, finds she has $14.28 left ; how much had she at first ? Am. $70.11. DEFINITIONS. 46. Ailditiofi is the process of uniting two or more num- bers into one number. 47. Adilenils are the numbers added. 48. The Sam or Amount is the number found by addi- tion. 40. The Process of Afldition consists in forming units of the same order into groups of ten, so as to express their amount in terms of a higher order. 60. The Sign of Addition is 4-, and is read plus. When placed between two or more numbers, thus, 8 + 3 + 6 + 2 + 9, it means that they are to be added. 20 ADDITION, ■ii 51. The Sign of Equdlity is =, and is read equals, or equal to ; thus, 9 + 4 = 13 is read, nine plus four equals thirteen. r>2. Principles. — /. Only numbers of the same denomind' Hon and units of the same order can be added. IT. TJie sum is of the same denomination as the addends. HI. The whole is equal to the sum of all the parts. REVIEW AND TEST QUESTIONS, 53. 1. Define Addition, Addends, and Sum or Amount, 3. Name each step in the process of Addition. 3. Why place the numbers, preparatory to adding, units under units, tens under tens, etc. ? 4. Why commence adding with the units' column ? 5. What objections to adding the columns in an irregular order? Illustrate by an example. 6. Construct, and explain the use of the addition table. 7. How many combinations in the table, and how found ? 8. Explain carrying in addition. What objection to the use of the word? 9. Define counting, and illustrate by an example. 10. Write five examples illustrating the general problem of addition, "Given all the parts to find the whols." 11. State the difference between the addition of objects and the addition of numbers. 12. Show how addition is performed by using the addition table. 13. What is meant by the denomination of a number? Wliat by units of the same order ? 14. Show by analysis that in adding numbers of two or more places, the orders are treated as independent of each other. SUBTRACTION. 54, The difference between two numbers is the amount that one number i.s greater than the other. ThuH, 7 is 2 greater than 5 ; hence 2 is the difference between 7 and 5. ILLUSTRATION OF PROCESS. *>5. Prob. I. — To find the difference between two numbers, each containing only one order of units and both the same order. Find the difference between (1.) 8 3 5 m (3.) 800 300 500 Explanation.— 1. The differ- ence between 8 and 3 is found by making 8 into two parts, one of which is 3, the other 5, the differ- ence. 2. The difference between 8 and 3 is the Bamo, whether these fi, which is $10758 for three of them, lie iiui>-t have paid for the fourtli the diflcrencc between $15960 aud $10758 which is $5202. . _ t- 8. A man's salary is $1300 a year, and he has money at interest which brings him $125 more ; if his expenses art.' (875, how much can he save ? Ans. $5j0. 9. A has $6185, B has $15181, C has $858 less than A and B together, and D has as much as all the rest ; how much has D? Ans. $41874. 10. Warren Xewhall deposited $302 in the Montreal Bank on Monday, $760 on Tuesday, and $882 on Thursday ; on Wednesday he drew out $380, on Friday $350, and on Satur- day $200 ; how much remained on deposit at the end of the week? Ans. $1074. 11. My property is valued at $7090, and I owe a debt of EXAMPLES' 25 |600, another of $1247, and another of |420 ; wlint am I worth V Ans. !*^4820. 12. A merchant paid $4570 for goods ; he sold a part of them for .t;3480, and the rust for $2724; how much did he gain by the transaction ? Ana. J? 1034. i:j. A man deposits $1110 in the bank at one time, and $1004 at another ; he then draws out $786 at one tinu*, $Go4 at another, $489 at another; how much still remained in the birnk? Am. $245. 14. J. Locke bought a farm for $4750, and built a house and barn upon it at a cost of $4475, and then sold tlie whole for $8090 ; how much did he lose ? Am. $1135. 15. A grain dealer bought 9710 bushels of grain ; he then sold 3348 bushels at one time and 5303 bushels at another ; how many bushels had he left ? An8. 1005 bushels. 10. Find the difference between $527.03 and $204.39. Explanation.— Write the Bnbtrabend under the min- uend, bo that dollars are under dollars and centi^ under cents. Snbtract as if the numbers were abstract, and place a period in the result between the second and third figures from the right. The figures on the left of the $527.03 204.39 $202.04 period express dollars and tho^e on the ri-^'ht cents 17. I received $352.07, and paid out of this sum to one man $73.12, to another $112.57 ; how much have I left of the money ? Ai,f>. $100.38. 18. A lady had $23.37, and paid out of this $7.19 for flour, $3.07 for sugar, $2.05 for butter; how much had she left? 19. A farmer sold $153 worth of wheat, $54.75 of barley, and $29.0") of oats. He paid out of the money receivetl to one man $:i2.13, to another $109.55; how much had he left? 20. Three men are to pay a debt of $6809. The first man pays $3905.38, the second $2001.70 ; how much has the third to pay? An». $901.92. 21. A merchant sold in one day $782.17 of goods. He re- ceived in cash $459.58; how much did he sell on credit? 22. A man owns five farms containing in all 3256 acres, and jsellstwoof them containing together 876 acres. How many I acres has he left? Ana. 2380. 3 96 aUBTRA CTION. DEFINITIONS. 62. Subtraction is tho process of finding the difference between two numbers. an. The 3luiU€lid is the greater of two numbers whose difference is to be found. <14:. Tho Siibtrahenil is the smaller of two numbers whose difference is to be found. 05. The Difference or Remainder is tlie result ob- tained by subtraction. GO. Tho Process of Subtraction consists in comparing two numbers, and resolving? the greater into two parts, one of which is equal to tho less and the other to the difference of the numbers. 67. The Sign of Subtraction is — , and is called minus. When placed between 'two numbers, it indicates that their dif- ference is to be found ; thus, 14 — 6 is read, 14 minus 0, and means that thes 2 are 5 times 2 are 11 times 2 are 3 times 2 are 9 times 2 are 8 times 2 are 12 timers 2 are 4 times 2 are G times 2 are G times 2 are 10 times 2 are 2. Find, by adding, the product of each example and write it after the w^rd " are." 3- Read very carefully the two sets several times, then erase the products and writ(^ them again from memory ns you read the example. Continue to erase and write the products in this way until they are firmly fixed in your memory. 4. Write on your slate a series of ticoa, and write under them in irregular order the numbers from 2 to 12 inclusive ; thus, 222 2 22 22 2 2 2 i ^ ^ '1 Q. 'Ill I U I M Write the product under each example as you repeat men- tally the number of twos. Continue to erase and write again and EXAMPLES. 29 erase read zts in I them 18, a^n, until each product ia called up to your mind just as soon as you look at the two numbers. 5. Pursue the same course in memorizing the products of 3's, 4'8, 5's, 6's, 7'8, 8'8, and 9's. MULTIPLIER ONE FIGURE. PREPARATORY STEPS. 74. Step I. — Find by using the Multiplication Table the product of eiich of tlie following : Thus, 5 X 7 = 35, 5 tens x 7 = 35 tens, 500 x 7 = 3500. Find the product of 1. 8 X 6 ; 8 tens x 6 ; 8 hundred x 6 ; 8000 x 6. 2. 9 X 7 ; 90 X 7 ; 900 X 7 ; 9000 x 7. 3. 3 X 5 ; 30 X 5 ; 300 X 5 ; 3000 x 5. 4. 7000 X 3 ; 500 X 6 ; 8000 x 4 ; 4000 x 4. 5. 60000 X 9 ; 900000 x 7; 5000000 x 5. 75. Step XL — The orders in a number are independent of each other ; hence, to find any number of times a given nuniber, we muUiply each order separately, thus : To find 6 times 748, we regard the 748 = 700 + 40 + 8. We know from memonzed results that G times 8 are 48, that 6 times 40 are 240, and that 6 times 700 are 4200. Having taken each of the three parts of 748 G times, the sum of these products must be 6 times 748. Hence, 48 + 240 + 4200 = 4488 1= 6 times 748. EXERCISE FOR PRACTICE. Multiply and explain, as shown in Step H, each of the 2 1 foUowing : * ' ^ w 1 1. 242 X 4. 6. 735 X 8. 9. 637 X 4. 1 2. 432 X 3. 6. 507 X 6. 10. 482 X 8. men- I 3. 321 X 2. 7. 389 X 5. 11. 795 X 9. nand ■ 4. 612 X 7. 8. 837 X 6. 12. 359 X 7. 1 *l j'V^r. \' m M 30 MULTIPLICATION. 76, The method of finding the mm of two or more times a given number by using memorized results is called Multipli- cation. The number taken is called the Multiplicand, and the number wliich denotes how many times the multiplicand is taken is called the Multiplier. ILLUSTRATION OF PROCESS. 77. Prob. I. — To multiply any number by numbers less than lo. How many are 4 times 369 ? (1.) ANALYSIS. i 9x4 = 369 X 4 = ] GO X 4 = ( 300 X 4 = 36 240 1200 1476 (2.) 869 4 1476 Explanation.— 1. The 869 is equal to the three parts, 9, 60, and ^K). 2. By taking each of these parts four timeB, the 369 is taken four times. Hence, to fiud 4 times ;i69, the 9 is taken 4 times ; then the 60 ; then the 300, as shown in the analysis. 3. Uniting the 36, the 240, and the 1200 in one number, we have 4 times 969. Ilence, 1476 is 4 times 309. 4. In practice, no analysis is made of the number. We commence with the units and multiply thus : (1.) 4 times 9 units are 36 units or 3 tens and 6 units. We write the 6 units in the units' place and reserve the 4 tens to add to the product of the tens. (2.) 4 times 6 tens are 24 tens, and the 3 tens reserved are 27 tens or 2 hundred and 7 tens. We write the 7 tens in the tens' place, aud reserve the 2 hundred to add to the product of the hundreds. (3.) We proceed in the same manner with hundreds, thousands, etc. From these illustrations, we obtain the following RULE. 78. Begin at the right hand and multiply each order of the multiplicand by the multiplier. Write in the product, in each case, the units of the result, and add the tens to the next higher result. EXAMPLES. 31 EXAMPLES FOR PRACTICE. Perform the multiplication in the following : 1. 837x3. 7. 986 x 2. 18. 579x9. 2. 5709x5. 8. 7093x5. 14. 90703 X 7. 3. 83095x6. 9. 50739x8. 15. 29073x8. 4. 39706x5. 10. 79060 X 6. 16. 40309 X 7. 5. 95083 K 4. 11. 79350x3. 17. 73290x8. 6. 70G39>8. 12. 60790x5. 18. 30940x6. i 7U. Continue the practice with abstract numbers by taking examples from Arithmetical Table No. 1, page 12, in the fol- lowing order: Thi*ee Figures in the Multiplicand. 1. Use three columns and copy for multiplicands each num- ber in Ihe colunms, commencing at the top of the Table. 2. Take as multii)lier the figure immediately under the right- hand figure of the multiplicand. T]ie first six examples taken in this way from columns A, B, C, are (1.) (2.) (3.) (4.) (5.) (6.) 234 138 395 557 487 686 8 5 7 7 6 8 3. Let examples be copied in this way from columns A, b, c ; B, C, D ; C, D, E ; D, E, F ; E, F, G ; F, G, n ; G, H, I ; and n, I, J. Four Figures in the Multijylicand, 1. Use four columns, and copy the multiplicands and multi- pliers in the same way as \nth three figures, taking the multi- pliers from the first column on the right. 2. Copy from columns A, b, c, D ; then B, C, D, E ; C, D, E, F ; D, E, F, G -, E, F, G, H ; F, G, H, I. 32 MULTIPLICA TION, Six Figures in the Multiplicand. 1. Copy, as already directed, examples from columns A, B, C, D, E, F ; then B, C, D, E, F, G ; C, D, E, F, G, H ; D, E, F, G, H, I ; and E, F, G, u, I, J. Take the multipliers from the right-hand column used. . 2. Let the examples from each of these sets be worked at your seat between recitations or out of school. ORAL EXAMPLES. 80. 1. Bought 4 barrels of tlour, at ^12 a barrel, and a barrel of crackers for $G ; how much did the whole cost ? Solution.— The whole cost four times $12, plus $G, whidi is $54, 2. If it requires 5 yards of cloth to make a coat, and 1 yard to make a vest, how many yards will make 9 of each ? 12 of each ? 7 of each ? ' 3. Bought 12 chairs at |3 each, a sofa at |47, and 8 taHes at $9 each ; how much did the whole cost ? 4. Gave $7 each to 5 men, paid for 10 yards of cloth at ^ a yard, and for a coat .$17 ; how much money have I spent ? 5. At 7 dollars a cord, what will 6 cords of v>^ood cost? 8 cords? 11 cords? 9 cords? 13 cords? ,, ,, WRITTEN EXAMPLES. 81. 6. How much will 7 acres of land cost, at $285 an acre? Ans. $1995. Solution.— 7 acres will copt 7 timcp $285. 7 time? $285=7 times $5 + 7 times $80 + 7 times $200 = $19!>5. Hence, 7 acres cost $1095. 7. What will be the cost of building 213 yards of iron fence, at 3 dollars a yard? Aits. Go9 dollars. 8. What will 647 cords of wood cost at $G a cord ? 9. There are 5280 f"5et in a mile; how many foot in 12 miles? Ans. 63360 feet. PREPARATOnr STEPS, 33 10. I sold 852 yds. of cloth at 3 dollars a yard ; bow much money did I receive ? Ana. $2556. 11. There are 4 fartliings in one penny ; how many farthings in 379 pennies ? Ans. 1516 farthings. 12. William Robb went to market with $485 ; he paid for 20 barrels of flour at $8 a barrel ; 16 boxes of soap at $3 a box ; and 3 tubs of butter at $12 a tub ; how much money did he have left ? Ans. $241. 13. Sold 89 bushels of beans at $2 a bushel, and 7 loads of hay at $19 a load ; how much did I receive for both ? 14. A merchant bought 12 hogsheads of molasses at $50 a hogshead, and sold the whole for $524 ; how much did he gain by the transaction ? Ans. $76. MULTIPLIERS 10 AND ABOVE. PREPARATORY STEPS. 83. Step I. — To multiply any number by 10, 100, 1000, and so on. 1. A figure is multiplied by 10 by moving it one place to the left, by 100 by moving it two places, etc. Thus, 4 expresses four, 40 expresses 10 fours, 400 expresses 100 fours, etc. 2. A cipher placed at the right of a number moves each significant figure in it one place to the left ; hence, multiplies it by 10. Thus, in 372 the 2 is in the first place, the 7 in the second, and the 3 in the third ; but in 3720 the 2 is in the second, the 7 in the third, and the 3 in the fourth place; hence, annexing the cipher has removed each figure one place to the left, and consequently multiplied each order in the nimiber by 10. 3. In like manner annexing two ciphers, three ciphers, etc., multiplies a number by 100, 1000. etc., respectively. 34 MULTIPLICA TION 83. Step ll.^rTo multiply by using the parts of the multiplier. 1. The multiplier may be made into any desired parts, and the multiplicand taken separately the number of times ex- pressed by each part. The sum of the products thus found is the required product. Thus, to find 9 times 12 we may take 4 times 12 which are 48, then 5 times 12 which are 60. 4 times 12 plus 5 times 12 are 9 times 12 ; hence, 48 plus 60, or 108, are 9 times 12. 2. WV 1 we multiply by one of the equal parts of the muiiiplier, we find one of the equal parts of the required product. Hence, by multiplying the part thus found by the number of such parts, we find the required product. For eram^^l , to find 12 times 64 we may proceed thus : ' ' ^ AVA1,T8I8. (2.) ■:4 X 4 - ■:'^r,[ = 04 y i :.. ... " ' 64 = 3 times 256. 4 256 64 X 12 ■-= ;Go 3 768 (1.) Observe, that 12 = 4 + 4 + 4 ; hence, 4 is one of the 3 equal parts of 12. - (2.) That 64 is taken 12 times by taking it 4 times + 4 times + 4 times, as shown in the analysis. (3.) That 4 times 64, or 256, is one of the 3 equal parts of 12 times 64. Hence, multiplying 256 by 3 gives 12 times 64, or 768. 3. In multiplying by 20, 30, and so on up to 90, we invari- ably multiply by 10 one of the equal parts of these numbers, and then by the number of such parts. For example, to multiply 43 by 30, we take 10 times 43, or 430, and multiply this product by 3 ; 430 x 3 = 1290, which is 30 times 43. ILLUSTRATION OF PROCESS, 35 We multiply in the same manner by 200, 300, etc., 2000, :3000, etc. ; multiplying first by 100, 1000, etc., then the product thus found by the number of lOO's, lOOO's, etc. ILLUSTRATION OF PROCESS. 84. Prob. II. — To multiply by a number containing only one order of units. 1. Multiply 347 by 500. (1.) ANALTSIB. (2.) First step, 347 X 100 = 34700 347 Secoud step, 34700 X 5 = 173500 500 173500 Explanation.— 500 is equal to 5 times 100 ; hence, by taking 347, as \xx first step^ 100 times, 5 times this result, or 5 times 34700, as shown in second step, will make 500 times 347. Hence 173500 is 500 times 347. 2. In practice we multiply first bv the significant figure, and annex to the product as many ciphers as there are ciphers in the multiplier, as shown in (2) ; hence the following RULE. 85. Multiply by the significant figure and annex as many ciphers to the result as there are ciphers in the multi2)lier. 86. EXAMPLES FOR PRACTICE. or ch (1.) (2.) (3.) (4.) Multiply 34 256 573 968 By 50 70 90 60 (5.) (6.) (7.) (8.) Multiply 3465 8437 2769 4763 By 600 300 800 200 36 MULTIPLICATION, (9.) (10.) (11.) (12.) Multiply 70 850 7300 8300 By 40 80 600 900 (13.) (14.) (15.) (16.) Multiply 326 184 972 262 By 80 700 500 20 87. Prob. III.— To multiply by a number containing^ two or more orders of units. 1. Multiply 539 by 374. (1.) ANALYSIS. r539 X 539 X 374 = ^539 X (539 X 374 4 = 2156 70 = 37730 300 = 161700 201586 (3.) 539 Multiplicand. Multiplier. 1st partial product. 2d partial product. 3d partial product. Whole product. Explanation.— 1. The multiplier, 374, is analyzed into the parts 4, 70, and 300, according to (83). 2. The multiplicand, 53{>, is taken flrt 17. A merchant bought 473 yards of cloth at $1.25 a yard ; 147 were damaged and had to be sold at $.07 a yard. He 8f)ld the remainder at $1.58 a yard ; did he gain or lose on the transaction, and how much? Ana. $21.99 gain. 18. A mechanic employed on a building 78 days received $2.75 a day. His family expenses during the same time were $1.80 a day ; how much did he save ? Ans. $09.42. 19. A merchant purchased 10 pieces of cloth, each containing 48 yards, at $2.75 a yard. He sold the entire lot at an advance of $.45 per yard. How much did he pay for the cloth, and what was his entire gain ? 20. Bought 107 bushels of wheat at $1.65 a bushel, and 287 bushels of oats at $.37 a bushel. I sold the wheat at a loss of 4 cents on a bushel, and 34 bushels of oats at a gain of 18 cents a bushel, the remainder at a- gain of 13 cents. What did I gain on the transaction ? a DEFINITIONS. 91. Multiplication is the process of ttiking one number as many times as there are units in another. 92. The Multiplicand is the number taken, or multi- plied. . • 93. The Multiplier is the number which denotes how many times the multiplicand is taken. 94. The Product is the result obtained by multipli- cation. 95. A Partial Product is the result obtained by multiplying by one order of units in the multiplier, or by any part of the multiplier. 96. The Total or Tlliole Product is the sum of all the partial products. 97. The Process of Multiplication consists, first. 40 MUL TIP Lie A TION. in finding partial products by usinf? the memorized results of the Multiplication Table ; second, iu uniting these partial products by addition into a total product. 08. A Ffictor is one of the cqnnl pttrta of a number. Thus, 12 is composed of six 2*8, four ^'s, three 4's, or two O's ; hence, 2, 3, 4, and are factors of 12. The multiplicand and mnltiplior are factorH of the product. Thus, 87 X 2.5 = 925. The product Wi5 \h conlpo^'ed of (weniy-Jlve 37'h, or t/iirly- seven 25'f . Ilcuce, both 37 and 2.5 are equal parts or factors of 926. 90. The Sign of Mult nd't cation is x , and is read times, or multiplied by. When placed between two number?, it denotes that either is to be mul- tiplied by the other. Thus, 8x6 shows that 8 is to be talien 6 times, or that 6 in to be taken 8 times ; hence it may be read either 8 times ti or 6 times 8. lOO. Principles. — / The midtiplieand may he either an abstract or concrete number. 11. The multiplier is alirays an abstract number. in. The 2)roduct is of the same denomination as the mxdtipli cand. KEVIEW AND TEST QUESTIONS. 101. 1. Define Multiplication, Multiplicand, Multiplier, and Product. 2. What is meant by Partial Product? Illustrate by an example. 3. Define Factor, and illustrate by examples. 4. What are the factors of G ? 14? 15? 9? 20? 24? 25? 27? 32? 10? 30? 50? and 70? 5. Show that the multiplicand and multiplier are factors of the product. 6. What must the denomination of the product always be, and whv ? REVIEW, 41 an of 7. Explain tlio procoaa in each of the following cases, and illustrate by examples : I. To multiply hy numbers less than 10. II. To multiply by 10, 100, 1000, and so on. III. To multiply l)y one order of units. IV. To inulti])ly by two or more order of units (Hti). V. To multij)ly l)y the factors of a number (83 — 2), 8. Give a rule for the third, fourth, and fifth cases. 9. (Jive a rule for the shortest method of working examples where both the multiplicand and multiplier have one or more ciphers on the right ? 10. Show how multiplication may be performed by addition. 11. Explain the construction of the Multiplication Table, and illustrate its use in multiplying. 12. Why may the cijdiers be omitted at the right of partial products? 13. Why commence multiplying the units' order in the multiplicand first, then the tens', and p<^ on ? Illustrate your answer by an example. 14. Multiply 8795 by G29, multiplying first by the tens, then by the hundreds, and last V)y the units. 15. Multiply 3572 by 483, commencing with the thousands of the multiplicand and hundreds f the multiplier. 10. Show that hnndrcds multiplied by hundreds will give ten thoymnds in the product. 17. Multiplying thousands by thousands, what order will the product be? 18. Name at sight the loirest order which each of the follow- ing examples will give in the product : (1.) 8000 X 3000 ; 2000000 x 3000 ; 5000000000 x 7000. (2.) 40000 X 20000 ; 7000000 x 4000000. 19. What orders in 3928 can be multiplied by each order in 473, and not have any order in the product less than thousands? ^ ■ DIVISION. < i 102. To apply the Multiplication Table in finding at (rigJit how many times a nuriiber expressed fry one figure is contained in any nutnber not greater than 9 times the given number. Pursue the following course : I. Write on your islute in irregular order the products of the Multiplieaiion Tablo, commencing with the products of 2. Write immediately before, the number whose products you La> e taken ; thus, ^JJO .i)l 2)11,, 2)6 2)12 2)10 /^ 2. Write under the line from memory the number of 2's in 10, in 4, in 14, etc. When this is done, erase each of these results, and rewrite and erase again and again, until you can give the quotients at sight of the other two numbers. 3. Look at the numbers and question yourself. Thus, you say mentally, t^cos in ten, and you follow with the answer, five ; tiros in four, tico ; twos m fourteen, seven. 4. Omit the questions entirely, and pass your eye along the exani])les and name the results ; tlius,^cf, two, seven, etc. lOJJ, Practice as above directed on each of the following : 1. 3)0 3)12 3)15 3)9 3)24 3)18 3)21 2. 4)12 4)24 4)32 ^L^ 4)16 4)28 4)20 3. 5)15 5)25 5)10 6)30 5)20 5)45 5)35 4. 0) 12 0)24 6)30 6)18 6)^42 6)54 6)30 5. 7)14 7 ) 35 7)49 7)21 7)42 7 ) 56 7)28 6. 8) 16 8)40 8)24 8)56 8)32 8)64 8)48 7. 9)27 9)45 9)18 9)5-1 9)72 9)36 9)03 PREPARATORY STEPS. 43 DIVISORS PROM 2 TO 12. PREPARATORY STEPS. 104. Step I. — To dicide when the quotient is expressed by two or mare peaces, but contains only one order of iinitn. 1. Regard the dividend as made iuto equal parts, divide one of these equal parts by the given divisor aud multiply the quotient by the number of eqm\] parts ; thus, Take for exampU' GO divided by 3. VVe know 6 is one of the 10 equal parts of GO. We know also that there are 2 threes in G, and that each G in the GO must contain 2 threes. Then as 60 contains 10 times G. it must contain 10 times 2 threes, or 20 threes. Hence the quotient of GO divided by 3 is 20. 2. The equal parts of the dividend which wo divide may be expresst^l by two or more fii^ures. Take, for example, 3500 divided by 7. Here we divide first the 35 by 7. We know that 35 is one of the 100 e. Step II. — Examples tcith one order in the quotient, where the quotient figure must be found by trial. In examples of this kind, we proceed thus : Divide 1709 by 287. PTRST TRIAL. 287 ) 1709 ( 8 2296 SKCONS TRIAL. 287 ) 1769 ( 7 2009 1. We divide as before by 2, the left-hand figure of the divisor, and find the quotient 8. This course will always give the largest pos- ftiUe quotient figure. Multiplying the divisor 287 by 8, we observe at once that the product 2296 is greater than the dividend 1769. Hence 287 is not contained 8 times in 1769. 2. We erase the 8 and 2296 and try 7 as the quotient figure. Multiplying 287 by 7, we observe agjiin that the product 2009 is greater than the dividend 1769. Hence 287 is not contained 7 times in 1769. EXAMPLES, 61 THIRD TRIAL. 287 ) 1769 ( 6 1782 47 3. We erase the 7 and 2009, and try 6 as the quo- tient figure. Multiplying 287 by 6, we obpcrve that the product 1722 in less than the dividend 1769. Subtracting 1722 from 1769, we have 47 remaining, a number lef b than the diviKor 287. Hence 287 is contained 6 timcn in 1769 and 47 remains. EXAMPLES FOR PRACTICE. 117. Find the quoticixts r.nd rcmaiuders in each of the fol- lowing : 1. 1194-27. 8. 4275 ^4')8. 15. 215400-5-356. 2. 236-^40. 9. o93(>-5-C43. 10. 430900 -T- 588. 3. lOO-f-39. 10. 9758 -^r.82. 17. 028400^898. 4. 410-J-G8. 11. 3657^739. 18. 80(5700 ^903. 6. 248-^38. 12. 7890-1-490 19. 190000^379. 6. 845-^07. 13. 4705 -^ (58. 20. 587500^-825. 7. 605^84. 14. 9850-^ 39. 21. 477400-V-493. ( , 22. 784- n32. 043(5-4-27. 7357 ■^857 : .i 23. 02G- 1-82. 8708^40. 327G8 ■4-760. 118. divisor. ILLUSTRATION OF PROCESS. Prob. II. — To divide any number by any given 1. Divide 21524 by 59. 59 ) 21524 ( 364 177 882 354 284 236 48 Explanation.— 1. We find how many times the (livif'or is contained in the few- est of the left-hand flgurcH of the dividend which will contain it. 59 is contained 3 times in 215, with a remainder 38; hence, according to (104-1), it ii* contained ^iOO limes in 81500, with a remainder :i800. 2. We annex the figure in the next lower order of the dividend to the remain- der of the previous division, and divide the number thus found by the divisor. 2 tens annexed to 380 tens make .382 tens. 59 is contained 6 times in 383, with a remainder 28; hence, according to (104-1), It is contained 60 times in 3820, with a remainder 280. 3. We annex the next lower figure and proceed as before. 52 DIVISION, juuci each fitcp. step. 69 X 300 ^ 17700 69 X GO = 3540 59 X 4 = 236 59 X 3G4 = 21476 4 units annexed to 380 nnlts make 284 nnits. 59 is contained 4 times in 384, Willi a remainder of 48, a number smaller tlmn tlic divinor, hence tlie division is completed, and we have found that 50 is contained 364 times in 31624, with a remainder 48. Observe carefully the following analysis of the process in the preceding example ; Multiplying the divisor by Part of dividend Part of divided dividend the part of the quotient divided each subtracted from the part undivided. 21524 17700 3834 3540 284 236 •f It From these illustrations we obtain the following RULE. no, /. Find lioic many times t7ie divisor is contained in the least number of orders at the left of the dividend that tnll contain it, and vyrite the result for the first figure of the quotient. IT. Multiply the divisor by this quotient figure, and subtra4it the result from the part of the dividend that was used ; to the remainder annex tJie next lower order of the dividend for a new partial dividend and divide as before. Proceed in this manner loith each order of the dividend. III. If there be at last a remainder, place it after tJie quotient, vAth tJie divisor underneath. Proof. — Multiply the divisor by tJie quotient and add the remainder, if any, to the product. This result will be equal to the dividend, when the division has been performed correctly. EXAMPLES, 63 120. EXAMPLES FOR PRACTICE. 1. 9225 -r- 45. 2. 18450 -f- 90. 3. 20840 4- 61. 4. 280135 -?- 89. 5. 17472 -5- 21. 6. 255708 -T- 81. 7. 72144 -i- 72. 8. 9590 -5- 70. 9. 137505 -T- 309. 10. 59644 -{- 62. 11. 467775 -^ 105. 13. 264375 -^ 705. 13. 1292928 -^ 312. 14. 289520 -f- 517. 15. 2750283 -i- 603. 16. 1143723 -4- 509. 17. 1& 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 13824 35904 142692 1678306 31809868 04109742 5332114 19014604 10205721 7977489 203812983 31907835 61142488 119836687 406070736 330445150 -i- 128. -J- 204. H- 517. H- 313. H-4004. -^ 706. -*- 4321. -f- 406. -f- 3243. -f- 923. -^5049. -^4005. -f- 4136. -^ 3041. -5- 8056. -H 3145. 121. Additional examples for practice should be taken from Arithmetical Table No. 2, page 47, as follows : . Dividend four ftgureSf Biviaor two. 1. Take the dividends in order from columns A, B, C, D ; B, C, D, B ; C, D, E, P ; D, E, F, G ; E, F, G, H ; F, G, H, I ; G, H, I, J. 2. Take as divisors in each set the figures immediately above the dividend, in the two right-hand columns of those used. Dividend six flgnres. Divisor three, 1. Take the dividends in order from columns A, B, c, D, E, F ; B, C, D, E, F, G ; C, D, E, P, G, H ; D, E, P, G, H, I ; E, F, O, H, I, J. 2. Take the divisors as before from the three right-hand columns of those used for dividend. 54 Diviaioir, WRITTEN EXAMPLES. 122t 1. A hogshead of molausoH containB G3 ^llons ; how many hogsheads in 1GU02 gallons ? Solution.— As ouo hogehcad contains 63 gallons, 10008 gallons will make a» many bogHhuadu an (Vi i» cuutalncd timcH in 10002. 16002 + G:i = 2&1. Ilcnce thuru arc 2&4 bogsticadb in HMyif. gallons. 2. An anny contractor furnished horses, at $72 each, to the amount of !j;llol204; how many did he furnish ? Aji8. 15712. 3. A man winlies to carry to market 2C2JJ bushels of jM)ta- toes ; if he carries 61 bushels at a load, how many loads will they makeV Ans. 43 loads. 4. A man paid |1548 for a farm at the rate of $43 an acre ; how manv acres did the farm contain ? Ans. 30 acres. 5. A certain township contains 192000 acres ; how many square miles in the township, there being 640 acres in a square mile? Ans. 300 miles. 6. How many acres of land at $200 an acre, can be bought for $53400 ? Ans. 267 acres. 7. A certain product is 43964 and one of the factors is 58 ; what is the other factor ? Ans. 758. 8. At what yearly salary will a man earn 40800 dollars in 34 years? Ans. $1200. 9. If light travels 192000 miles in a second, in how many seconds will it travel 691200000 miles? Ans. 3000. 10. Henry Morgan divided $47400 into 3 equal parts, one of wliich he gave to his wife ; the rest, after paying a debt of $3280, he divided equally among 4 children ; what did each child receive ? Ans. ^70S0. 11. A piano maker expended in one year for material $20041, and for labor $4925, paying each week the same amount ; whnt was his weekly expense ? 12. A farmer in Ontario raised in one year 13475 ' -wl**^' wheat ; the average yield was 49 bushels per acre ; » mui acres did he have sown ? I P h' /; /• . I A' . 1 r O A' 1' S T KP s. 65 in nc of cli DIVISION BY FACTORS. PREPARATORY STEPS. l!2fl. Step I. — Any munher may he e.rpreased in terms of one of its fdctors by takimj anuthev facto v as the Unit, (11.) Thus, 12 = 4 + 4 + 4 ; honco, 12 raay be expreeetxl as 2 fours, the four being tho tcnit of the number 3. Write the following numbers : 1. Express 12 as 2*8 ; as 'S'h ; as 4's ; as C's. 2. Express 80 as 3'8 ; as Oh ; as IBs ; as 12's ; as 6'b. 8. Ex])reHS 4.") as 5*8 ; as 3*8 ; as 9*8 ; as IS's. 4. Express 42 and 24 each as O'h. 5. Express 45 and 225 each as U's ; as o's ; as 3'8. 1124, Stkp II. — When a number is made into three or more factors, any tico or 7nore of them may be regarded as the unit of the 71 Umber expressed by the remaining factors. For example, 24 = 3 x 4 x 2. This may be expressed thus, 24 = 3 (4 twos). Hero the 3 expresses the number of 4 twos ; hence, (4 twos) is regarded as the unit of the number 3. Write the following : 1. Express 12 as (3 twos) ; as (2 twos) ; as (2 threes). 2. Express 30 as (3 twos) ; as {2 Jives) ; as (3 fees). . 3. Express 42 and 126 each as (2 snrns) ; as (7 threes) 4. Exi)res8 75, 225, and 375 each as (5 threes). 5. Express 6t), 198, and 264 each as (11 threes) and as (2 elevens). ' 125. Step III. — When the same factor is made the unit of both the dividend and divisor, the division is performed as if the numbers were concrete. Thus, 60 -4- 12 may be expressed, 20 threes h- 4 threes, and the division performed in the same manner "- in 6 feet -4- 3 feet. 4 threes are contained 5 times in 20 threes ; hence, 12 is contained 5 times in GO. 56 DIVISION, The division may be perfonned in this way when the factors are connected by the sign of multiplication ; thns, CO -5- 12 = (20 X 3) -5- (4 X 3). We can regard as before the 3 rs the unit of both dividend and divisor, and hence say, 4 threes are con- tained 5 times in 20 threes. Perform the division in eacli of the following eicamples, without performing the multiplication indicated : 1. 25 threes -*- 5 threes = ? 5. (64x 9)-*.( 8x 9) = ? 2. 42 eights -^ C eights = ? 6. (49xl3)-i-( 7x13) = ? 3. 88 twos -J- 11 twos - ? 7. (96 x 7)-T.(12x 7) = ? 4. 108 fives ^ 9 fives = ? 8. (78 X 11) -f- (20x11) = ? ILLUSTRATION OF PROCESS. 126. Pros. III.— To divide by using the factors of the divisor. Ex. 1. Divide 375 by 35 5 )J15 7 jives ) 03 fives Explanation.—*. The divisor .35 = 1 Jives, 2. Dividing the 315 by 5. we fiud that 315 := " 3. The OS Jives contain 9 timee TJlres ; hence 815 contains 9 times 1 Jives or 9 times 35. Ex.2. Divide 350 by 24. 2 [359 3 twos I 179 twos and 1 remaining = 1 4 (3 twos) I 59 (3 twos) and 2 twos remaining = 4 Quotient, 14 and 3 (3 twos) remaining = 18 True rcmaiuder, 23 Explanation.— 1. The divieor 34 = 4x3^2 — 4f3 fttot). 8. Dividing 359 by 2, wo find that 3.59 = 179 ftco" and 1 uuit rpmaining. 3. Dividing 179 twos by 3 twos, we find that 179 twos - r>9 {}i twos) and 3 ttvos rcma!.iing. 4. Dividing 59 (3 twos) by 4 (3 twos), we find that 50 !3 tivos) contain 4 (3 ttvos) 14 times :tnd 3 (3 twos) remaining. Hence 859, which is equal to 59 (3 tivos) and 2 ttr + 1, contains 4 (3 twos)^ or 94, }4 times, and 3 (3 twos) + S twos + 1, or S«, remaining. t€ ni Wi in PREPARATORY STEPS. 57 From these illustrations we obtain the following RULE. 127. /. Resolve the divisor into convenient factors ; divide the dividend by one of these factors, the quotient thus obtained by another, and so on until all the factors have been used. The last quotient will be the true quotient. II. The true remainder isfovnd by multiplying each remain- der, after the first, by all the divisors preceding its own, and finding the sum of these products and the first remainder. EXERCISE FOR PRACTICE. 128. Examples for practice in dividinpf by the factors of the divisor: t .376-^ 100. 10. 19437 -^ 40. iL 8975 -4- 100. 11. 13658 -f- 42. 3. 76423 -5- 1000. 13. 27780 -r- 60. 4. 92708 -j- 1000. 18. 7169 -J- 90. 5. 774-^ 18. 14. 4947 -T- 108. 6. 873 H- 24. 15. S0683 -5- 400. 7. 4829 ^ 28. 16. 75947 -J- 900. 8. 15836 -^ 30. 17. 8460 -H 180. 0. 7859-5- 84 18. 14025 H- 165. land 8). • ONE ORDER IN DIVISOR. PREPARATORY STEPS. 129. Step l.—To divide by 10, 100, 1000, etc. 1. Observe, the figure in the second place in a number denotes tens, and this figure, with those to the left of it, express the number of tens. Hence, to find how many tens in a numl)er, we cut off the right-hand figure. Thus, in 7309 tiie 6 denotes tens, and 736 the number of tens in 7369 ; hence 7369 -5- 10 = 786, and 9 remaining. 4 58 DIVISION, 2. In like maimer the figure in the third place denotes hundreds, the figure in the f mirth place thousands, otc. Hence by cutting off tiDO figures at the right, we divide by 100 ; by cutting off three, we divide by 1000, etc. The figures cut off are the remainder. Qlve the quotient and remainder of the following at sight : 587 -^ 10. 463 -5- 100. 8973 -5- 100. 50380 -*- 100. 73265 ^ 1000. 58307 -^ 10000. 130. Step II. — A number c-onsisting of only one order of nnits, contains two factors which can be given at sight. Thus, 20 = 2 X 10. 400 = 4 X 100. 7000 = 7 x 1000. Observe that the sJi-fuificant flf^urs of the number, in each case, is one factor and that the other factor i;* 1 with as many ciphers annexed as there are ciphers at the ri^ht of the eigniflcaut fl<{ure. ILLUSTRATION OF PROCESS. liil, Prob. IV.— To divide when the divisor consists of only one order of units. ' * 1. Divide 8T36 by 500. 5)87|36 17 and 230 remaining. Explanation.— 1. We divide first by the factor 100. This is done by cutting: off 36, the nnlts and tens at the right of thPdividon(?. 2. We divide the quotient, 87 hundreds, by the factor 5. which sivos a quotient of 17 and 2 hundred remaining, which added to 36 gives 236, the true remainder. EXAMPLES FOR PRACTICE. 132. Divide and explain each of the following examples : 13. 03nf*0 -4- 800. 1 1. 79:565 -- 3000. 15. 57842 -^ 5000. 10. 90000 ^ 900. 17. 40034 ^ 600. 18. 20306 -h 700. 1. 752 -*- 200. 7. 8365 -J- 1000. o /W. 593 -f- 30. 8. 5973 H- 400. 3. 80> -f- 50. 9. 62850 -J- 4000. 4. 938 -h 600. 10. 06462 -i- 6000. 5. 452 -i- 300. 11. 86352 -i- 900. 6. 983 H- 700. 12. 49730-*- 800. DE FIN IT IONS* 09 lenotes Hence 00 ; by cut off orht : . 1000. - 10000. ^rdcr of m. >e, is one 1 as Ihcre consists vide firpt donf by I'lis at the rh clvoH a js 236, the nplos : ^ 800. -4- :^000. -H r)000. ^ 000. -J- 000. -*- 700. DEFINITIONS. 133. Division is tho iirocess of finding how maDj times one number is contained in another. 134. The Dividend is the number divided. 135. The Divisor is the number by which the dividend is divided. 1 30. The Quotient is the result obtained by division. 137. The Remainder is the part of the dividend left after the division is performed. 138. A l*artial Dividend is any part of the dividend which is divided in one operation. 139. A Partial Quotient is any part of the quotient which expresses the number of times the divisor is contained in a partial dividend. 140. The I*roeess of Division consists, J^r.<(^ in finding the partial quotients by means of memorized results; second, lin nniltiplying the di\i8or by the partial quotients to find the partial dividends ; third, in subtracting the partial dividends jfrom the part of the di\idend that remains undivided, to find tlu' i)art yet to Ije divided. 141. Short Division is that form of division in which lo step of the process is written. 142. TjOnff Division is tlint form of division in which [he third step of the i)roce88 is written. 143. The Sign of Division is ■^-. ami is read divided by. \W\ou placed between two numbers, it dfuoti-s that the number •f »n' it is to be divide»or, [ith a uhort horizontal line between them ; Ihue, Y i"^ read, ;» ilividod by 5. 60 Dl VISION, 144. Principles.—/. The dividend and divia&r muttt be numbers of the same denomination. II. The denomination of t?ie quotient is determined by the nnture of the probk in soloed. III. The remainder is of the same denomination as the dividend. KEVIEW AND TEST QUESTIONS. 14f'>. 1. Dofiue Division, and illustrate each step in the process by exaraploa 2. Explain and illustnito by examples Partial Dividend, Partial Quotient, and Uemainder. 8. Pre|)arc two examples illustratinjir each of the following problems : I. Given all the parts, to find the whole. II. Given t\\o, whole and one of the parts, to find the other part. III. Given one of the equal parts and the number of parts, to find the whole. IV. Given the whole and the size of one of the parts, to find the number of parts. V. Given the? whole and the number of equal parts, to find tlu> Bi'/e of one of the parts. 4. Show that 45 can be f^xpressed as nines, as fives, as threes. 5. What is meant by true remainder, and how found? 6. Explain division by factors. Illustrate by an example. 7. ^Vhy cut off as many figures at the light of the dividend as there are ciphers at the right of the divisor ? Illustrate by an example. 8. Give a rule for dividing by a number with one or more cii)hers at the right. Illustrate the steps in the process by an exam])le. 9. Explain the difference between Long and Short Division, and show that the process in both cases is performed mentally. AP P LI C A TlOyS. Gl nu8t be by the aa the in the ivideiul, jUowing find the mber of le parts, parts, to IS threes. iple. Idivideud itratu by or more ss by an >i vision, jrtbrmed 10. Illustrate each of the following problems by three ex- amples : VI. Uiven tho final quotient of a continued division, the true remainder, and the several divisors, to find the dividend. VII. (jJiven the i)r()duct of a continued multiplication and the; several multipliers, to find the multi- plicand. VIII. Oiven tlw sum and the difference of two numl)er8, to find the numbers. APPLICATIONS. 140. Prob. I. — To find the cost when the number of units and the price of one unit is given. Ex. 1. What is the cost of 42 yards of silk, at |2.36 a yard ? Solution.— If one yard cost $2M, H yards must coet 42 times 12.36. Ileuce, *-J.3« x 42 = $mA'i, is the cost of 42 yards. Find the cost of the following : 2. 118 stoves, at jflS for each stove. Aus. |;3384. 3. 25J) yards of broadcloth, at ^2.84 per yard. 4. 436 bushels wh.at, at $1.7() a bushel. Atts. J|;767.3(>. 5. 2 farms, each containing 139 acres, at $73.75 jxt acre. «;. 84 tons of coal, at 17.84 per ton. Ans. ^058.56. 7. 218 barrels of apples, at $2.90 per barrel. 8. 432 yards cloth, at !f;1.75 i)er yard. 0. 34C bushels of wheat, at !?1.73 a bushel. 10. 897 pounds butter, at $.37 per pound. 147. Prob. II.— To find the price per unit when the cost and number of units are given. Ex. 1. Bought 25 cows for $1175 ; how much did each cost? SoLmoN.— since 85 cows coPt $1175, cnch cow cont an many dollars as 26 is contained times iu 1173. Hence, 1175+2.5 = 47, the number of dollara each cow cost. €2 AP PLICA TlOyS, Find the price of the following : 2. If 42 tons of hay cost $546, what is the price per ton ? 'S. Bought 268 yards cloth for $804 ; how much did I pay per yard? Ann. $y. 4. Paid $1029 for 147 barrels flour; what did I pay per barrel? A us. $7. 5. Sold 190 acres laud for $10102; how much did I receive an acre V Am. $52. 0. The total cost for conducting a certain school for 14 years was $252000 ; what was the yearly exi)ense ? Ans. $18000. 7. Received $980 for 28 weeks' work ; what was my wages per week? Ann. $35. 8. A merchant pays his clerks for half a year $1872, How much is this per week ? Ans. $72. 148. PiiOB. III.— To find the cost when the number of units and the price of two or more units are giyen. 1. At .$15 for 8 cords of wood, what is the cost of 39 cords ? Solution 1.— Since 3 cords C08t $15, 3f.i cords must cost as mauy tiiuet $15 SB .3 Is contained times in 39. Hence, Jirtit step, 39+3 = 13 ; second step, $15 X 13 = ♦1!>.5, tlic cost of 39 cords. Solution 2.— Since 3 cords cost $15, eacli cord coat as many dollars as 3 is contained times in 15 ; bcncc cacli cord cost |5, and 39 cords cost 39 times $5, or 1195. Wcncc, first step, 15+3 = 5; second step, $5x29 = $195, the cost of :39 cords. Observe careftilly the difference between these two solutions. Let both be used in practice. Tulve for each example the one by which the divisioa can be most readiiy performed. Solve and explain the following : 2. If stoves cost $135, what is the cost, at the same rate, of 84 stoves ? 8. Paid $28 for 4 barrels of flour ; how much, at the same rate, will I pay for 164 barrels ? Ana. $1148. 4. A farmer paid for 64 sheep $256 ; what is the cost, at the same rate, of 793? An9. $3173. A PP LIGATIONS, oa 5. A man travelled by railroad 5184 miles in days ; how many miles, at the uame rate, will he travel in 54 days ? n. A book-keeper receives for his service at the rate of $024 for 13 weeks ; what is his yearly salary? Ans. $2496. 7. A merchant bought 150 yards of cotton for $18; how much will he pay, at the same rate, for 1350 yards? 14t>. PiiOB. IV.— '."o find the number of units when the cost and the price per unit are given. 1. At $7 a ton, how many tons of coal can be bought for $6r)8? Solution.— since 1 ton can be bought for 17, there can be as many tons boui^'lit fur $038 a.s ^1 is contuined titncti in |ti58. Hence, $6G8-«-|7 c M, the uumber of tona that can be bought for |668. Solve and explain the following : 2. For $935, how many barrels of pears can be bought at $11 a barrel ? Ans. 85. 3. How many horses can be bought for $9928, at $130 per hor sc Ans. 73. 4. A mechanic received at one time from his enii)loyer $357. He was paid at the rate of $21 a week : how many weeks had \w worked? Ann. 17. 5. A umn paid for a farm $6134, at $38 per fvcre; how many acres does the farm contain ? 0. Wm. Henry paid for walnut lumber $27795, at $85 a thousand feet ; how many thousand feet did he buy ? 150. Prob. V. — To find the number of units that can be purchased for a given sum when the cost of two or more units is given. 1. Wlien 8 bushels of wheat can be bought for $12, how many bushels can be lM)ught for $(5348 ? PoLUTioN.— Since 8 bashcl!) can be bouirht for $12, there can be an many times 8 buHhelt* bought for ♦(i:M8 an $12 is contained time» in $;G:}-18. Ileuce, flrM step, $«»48-»-|t2 = 52{); nermd step, 529x8 = 4232, the uumber of buHhele that can be bought for $6348. :| 64 A PI' LI ('A rroNS, Solve and explain the following : 2. If 30 pounds of sugar cost $4, how many pounds can be bouglit fur $375 ? Ati^. 3375 |)ouuds. 3. The cost of 4 boxes of ortinges is $12. How many boxes, at the same rate, can be bought for $552 ? 4. A farmer sold butter at $35 a hundred jwunds, and received $1715 ; how many pounds did he sell ? 5. Wlien peaches are sold at $0 for 8 baskets, how many baskets must a man sell to receive !3!582 ? Aus. 770 baskets. 0. A carpenter was paid at the rate of $42 for 12 days, and received $588 ; how many days was lie employed ? 7. At $09 for 12 cords of wood, how many cords can be bought for $900 ? Ans. 108 cords. REVIEW EXAMPLES. 1/51, 1. I sold 75 pounds of butter at 20 cents a pound, and laid out the proceeds in coffee at 00 cents a pound ; how many jwunds of coffee did I buy ? A)is. 25 pounds. 2. Bought a quantity of wood for S3959, and sold it for $0095, thus gaining $3 on each cord sold ; how much wood did I buy? An8. 712 cords. 3. I sold a farm of 850 acres at $45 an acre, and another farm of 175 acres at $75 an acre ; how much more did the first farm bring than the second ? *4/<«. $2025. 4. I paid $8900 for 8 city lots, and sold them at a loss of $12 on each lot ; how much did I receive for 3 of them? 5. The expenses of a young lady at school were $75 for tuition, $20 for books, $08 for clothes, $17 for railroad fare, $5 a week for board for 42 weeks, and $30 for other exi^enses ; what was the total expense ? Ans. $420. 0. I bought 27 acres of land at $41 an acre, and 20 acres at $27 an acre, and sold the whole at $43 an acre ; how much did I gain or lose ? Ans. $470 gain. 7. What is the total cost of 45 acres of land at $17 an APPLICATIONS, G5 acre, two horses at $132 each, a yoke of oxen for $130, a horee- rake for .t«5, and a plough for 117 ? Aiis. $1241. 8. A grocer bought 28 barrels of apples for $84: how much will he pay at the same rate for 168 barrels ? Ans. $504. 9. The sum of two numbers is 73, and their difference 47 ; what are the numbers ? Solution.— The pum 73 is equal to the greater unmber plun the lesB, and the Ichs number plus the diflcrunce 47, are equal to the greater ; hence, if 47 be added to the Hum 73, wc have twice the greater number. Ilence, jtrst Htep, 73 + 47 = 130 ; stoond taid $22541 for two farms, and the difference in the cost of the farms was $3471. The ])rice of the farm for which he paid the smaller sum was $64 an acre, and of the other $87 an nrre. How many acres in each farm ? 17. A fami-r sold 62 bushels of wheat for $50, also 14 cords of wood at §5 a cord. 4 tons of hay at <^15 a ton, and 2 co\r9 at !530 apiece ; he took in payment *145 in money, a coat worth !j:50, a horse-rake worth $21, and the balance in clover- seed at $4 a bushel ; how many bushels of seed did he receive ? Ans. 6 bushels. PROPERTIES OF NUMBERS. DEFINITIONS. 1512. An Integer is a uumber that expresses bow many there are in a collection of whole things. Thus, 8 yards, 12 houses, 32 dollars, 10 tables, 18 windows, 25 horses, etc 153. An Exact Divisor is a number that will divide another number without a remainder. Thus, 3 or 5 is an exact divisor of 15 ; 4 or 6 is an exact divi- sor of 24 and 36, etc. All numbers with reference to exact divisors are either prime or composite. 154. A Prime Number is a number that has no exact divisor besides 1 and itself. Thus, 1, 3. 5, 7, 11. 13, 17, 19, 23, 29, 31, 37, etc., are prime numbers. 155. A Composite Number is a number that has other exact divisors l)e8ides 1 and itself. Thus, 6 is divisible by either 2 or 3 ; 12 is divisible by either 8 or 4 ; hence and 12 are composite numbers. 150. A Prime Divisor is a prime number used as a divisor. Thus, in 35 h- 7, 7 is a prime divisor. 157. A Composite Divisor is a composite number used as a divisor. Thus, in 18 + 6, is a comix)site divison EXACT DIVISION. «T EXACT DIVISION. 158. The following teats of exact division should be care- fully studied and fixed in the memory for future uae. Puup. I. — A didsor of any number is a dimorofany number oftimea that number. Thus, 12 = 3 fours. Hence, 13 x 6 = 3 fours x 6 = 18 fours. Hut 18 fours are divisible by 4 Hence, 12 x G, or 72, is divisi- ble by 4. Prop. II. — A divisor of each of tico or more numbers is a divisor of their sum. Thus, 5 is a divisor of 10 and 30 ; that is, 10 = 2 fives, and 30 = « fives. Hence, 10 + 30 = 2 fives + fives = 8 fives. But 8 fives are divisible by 5. Hence, 5 is a divisor of the sum of 10 and 30. Prop. III. — A divisor of each of two nmnbcrs is a divisor of thiir difference. Thus, 3 is a divisor of 27 and 15 ; that is, 27 = 9 threes and 15 = 5 throes. Hence, 27 — 15 = 1) threes — 5 threes = 4 times. But 4 threes are divisible by 3. Hence 3 is a divisor of the difference between 27 and 15. Prop. IV.—Ani/ number ending with a cipJier is divisible by the dicisors of 10, viz., 2 and 5. Thus, 370 = 37 times 10. Hence is divisible by 2 and 5, the divisors of 10, according to Prop. I. Prop. V. — Any nnmber is divisible by either of the divisoi's of 10, ichcn its right-hand figure is divisible by the same. Thus, 498 = 490 + 8. Each of these parts is divisible by 2, Hence the number 498 is divisible by 2, according to Prop. II. In the same way it may be shown that 495 is divisible by 5. 68 FROrERTlEU OF M U M U E It S . Puop. VI. — Any nnmber ending with two ciphers is divisible by the dmnuTH of lUO, via., 2, 4, 5, 10, 20, 25, arul 50. 'I'huH, 8!HK) = bU times 100. Hence is divisible by any of the divi.soiH ol" 100, according to Prop. 1. Piior. V'll. — Any number is divisible by any one of the dicisors of 100, irhen the number npnsacd by its two r'ujht-hund Jiy tires it! diinnible by the same. Thus, 4075 = 4000 + 75. Any divisor of 100 is a divisor of 4000 (Prop. VI). Hence, any divisor of 100 which will divide 75 is a divisor of 4075 (Prop. II). Pkop. VIII. — Any number ending irith three cipfiers is divin- ble by the divisors of 1000, viz., 2, 4, 5, 8, 10, 20, 25, 40, 50, 100, 125, 200, 250, and 500. Thus, 83000 = 83 times 1000. Hence is divisible by any of the divisors of 1000, according to Proj). I. Prop. IX. — Any nnmber is divitiible by any one of the dirisors of 1000, when the number expressed by its three right-hind figures is dirisihle by the same. Thus, 02025 = 02000 + 025. Any divisor of 1000 is a divisor of 02000 (Prop. VIII). Hence, any divisor of 1000 which will divide 025 is a divisor of 02025 (Prop. II). Prop. X. — Any number is divisible by 0, if the sum of its digits is divisible by 0. This proposition may be shown thus : (1.) 486 = 400 + 80 + 6. (2.) 100 = 00 + 1 •- 11 nines + 1. Hence, 400 = 44 nines + 4, and is divisible by with a remainder 4. (3.) 10 = + 1 = 1 nine + 1. Hence, 80 = 8 nines + 8, and is divisible by with a remainder 8. (4.) From the foregoing it follows that 400 + 80 + 6, or 486, is divisible by with a remainder 4 + 8 + 6, the sum of the digits. Hence, if the sum of the digits is divisible by 0, the number 486 is divisible by (Prop. II). ^XACT Divisloy, G9 divUible y of the : divisors I Jiyurea ivLsor of II divide is divisi- , 50, 100, by any of ic di'risors ight-Juind n divisor hicli will ' its digits Vnov.Xl.^Any number is divisible by 3, if the sum of its digiti* ut dicisible by 3. Thirt proposition i.s shown In the same manner as Prop. X ; as 3 divides 10, 100, 1000, etc., with a remainder 1 in each cu»e. Puoi*. XII.— .l/. Find exact divisors of each of the following numbers by jijiplylng the foregoing tests : 1. 470. 12. 9375. 23. 5478. o 975. 13. 15264. 24. 3825. 44 nines M 3. 4. 23ai. 4500. 14. 15. 37128. 28475. 25. 26. 8094. 3270. 5. 8712. 16. 47000. 27. 3003. + 8, and M 6. 9736. 17. 69392. 28. 8004. 1. 5725. 18. 34605. 29. 7007. or 486, is M 8. 8375. 19. 38745. 80. 1005. le digits. ■ 9. 6000. 20. 53658. 31. 9009. 1 number ■ 10. a500. 21. 25839. 32. 3072. 11. 3025. 22. 21762 33. 8008. n 70 PROPERTIES OF NUMBERS, PRIME :^:UMBERS. PREPARATORY PFxOPOSITIONS. lOO. Prop. I.— All even numbert are ditmble by 2 and consequently all even numbers, ejc^pt .', are composite. Ilenco, in finding the prime numbere, we cancel as composite all even numbers rxcept 2. Thus. 3, 4y 5, 0, 7, $f 0, 10, 11, a, and so on. Prop. II. — AV/cA unniber in the stries of odd numbers is 2 greater than the munber immediately precediny it. Thus, the numljers left after cancelling the even numbers are. 8 5 T • 11 13. and so on. 8 8 + 3 5 + 3 7 + 2 9 + 2 li . 2 Prop. III. — Tn the series of odd numbers, erery third num- ber from i{ is dir/nihle by H, erery FiFTn it umber from 5 is dirinihlc by 5, and so on irith each numbtr in the series. This proposition may l)e shown thus : According to Proj;. II, the serit-B of odd numbers incnmso by 2's. Hence the third number from U is found by adding 2 three times, thus: t 5 7 I 8 + 2 3 + 2 + 3 8 + 3-^2 + 2 From this it will be seen that 9, the third numlM»r from 3, is composed of Jl, p'us 5{ twos, and in divisible by U (Prop. II) ; and HO with the third numlM^r from 9. and so on. By the same courst^ of reasoning, onch fifth number in the series, counting from 5, may U; shown to be divisible l)y 5 ; and 80 with any other number in the wries ; hence the follow- ing method of finding the prime nambcrs. PRIME NUMBERS, 71 ILLUSTRATION OF PROCESS. 1CI1. Pnon— To find all the Prime Numbers from x to any given number. Find all the prime numbers from 1 to 03. 1 S 5 7 9 11 13 15 a 6 17 19 21 8 7 23 25 t 27 3 » 29 ai 33 3 11 35 6 7 37 39 3 13 41 48 45 3 S !• 15 47 49 7 51 3 17 53 55 S 11 01 9 ]• 50 ei 63 3 7 » SI Expi-akatios.— 1. Arrange the pcrlos of odd nnmb<>n> «n line?, at con- vcnii-ut dii'tuiicc)' from each other, an Bbown iu illustration. •i. Write 3 under every third number from 3, 5 under wcryjlffh nnrabcr from o. 7 under every stventh number from 7. nnd ho on with eaeh of tho Ml her i.umbe.'r. 3 The terms under which the nnmbere nre written are compof>ite, and ibc numberi" written umlerare their fuotorc, accordini; to Prop. III. All tlic remniniuf; number>* arc prime. Henre all tlu^ prinin nunilK»r8 from 1 to ftii rel,2, 3, 5, 7, 11, l:], 17. 10, 23, 20, 31, 37, 41, 43, 47, 53, 59, 01. EXAMPLES FOR PRACTICE. \iVZ, 1. Find all tin' prime niini1)er8 from 1 to 87. 2. Find all tin* prinv nunilM'rs froDi 32 to 100. •1 Find nil tin* primr ntiiiilxrs from 84 to 157. 4. Find all the prime num1)ers from 1 to 200. •°>. Find all tin- prim<' n: ^hers from 200 to 400. 0. Show by nn examine thnt eN'ery seventh number from sevi-n, hi the eorics of odd numbers, is divimibk' by seven. 72 PROPERTIES OF X r JI Ji E R S , i| FACTORI^^G. PREPARATORY STEPS. 10S5. Step I. — Find by irtupection all the exact diciaors of euch of the following numbers, and xcrite them in order on your date, thus 6 = 3x2, 10 = 5 x 2. 1. G 10 14 15 21 22 26 2. 33 34 35 38 39 46 51 3. 55 57 58 62 65 69 74 4. 77 79 82 85 86 87 01 5. 3 11 115 119 123 129 141 Refer to the results on your slate and obsi'rve (1.) Euch prime exact divisor is culled n prime factor of the number of whicli it is a divisor. (2.) Each number is equal to the prtxluct of its pnwjt' factors. Step II. — The same prime factor may enter into a mtmn . two or more times. Thus, 18 = 2 x 3 x 3. Hence the prime factor, 3, enters twice into 18. Resolve the following numbers into their prime factors, and name how many times each factor enters into a number. 1. 4 8 16 32 64 9 27 2. 18 20 28 40 44 45 50 3. 54 50 75 80 98 100 108 4. 71 25 49 125 121 213 343 one are DEFINITIONS. 164. A Factor U one of the equal parts of a number, or e of its exact divisors. Thus, 15 is romposed of djices, or 5 threes; hence, 5 and 3 e factors o»' 15. . FA CTORiy G. 73 165. A Pritne Factor is a prime number which is a factor of a given number. Thus, 5 is a prime factor of 30. IGG. A Composite factor is a composite number which lb a factor of a givi n nunil>er. Thus, (J is a com|x)site factor of 24. 107. Pactorhtff is the process of resolving a coini»o8ite number into its factors. H58. An K' poHcnt is a small figure placed at the ri^ht of a number and a littlo above, to show how many times tlu; nuuib»T is used as a factor. Tims, 8' = 3 X a X 3 X 3 X 3. The 5 at the right of 3 denotes that the 3 is used 5 times as a factor. KM), A Common Factor is a number that is a factor of each of two or more numbers. Thus, 3 is a factor of 0, J), 13, and 15; hence is a common fiictor. 170. The Greatest Common Factor \h the greatest number that is a factor of each of two or mon* numbers. Thus, 4 is the greatest number that is a factor of 8 and also of 13. lleu<.e 4 is the greatest common factor of b and 13. ILLUSTRATION OF PROCESS. 171. Find the prime factors of 402. Explanation— 1. We observe that the number 4*i3 is dlvicibU' by i, the Hinallei^t prlino nuinber. Ilonci? \v«« di- vide by a. 4. We obeerve that th« flrft quotient, SWt. in divinlble by 3, wliich lt« a prime number. Ilfut-e we divide by .3. 3. We observe tliaf the neennd (juoilent, 77. In dlvicible by 7. whieh \x a prime number. Ilenee we diviiU* )>y 7. 4. The third (luotient, 11,ir*a prime uumlM-r, IIcuco the prime tacturs )f 4ti2 are 8, 3, 7, and 11 ; that b, -«» = ^I x 8 < 7 x 11. 3 ) 4({3 3)23) 11 74 PROPERTIES OF NUMBERS, Any composite number may be factored in the same manner Hence the following RULE. 1712. Dioide tJi^ giren nvinher hf any prime number that is an exact divisoi', and the reuniting quotient by amtther, and so continve the dimion uvtil the quotient is a prime number. The several didsors and the bust quotient are the rt quirt d prims factms. EXAMPI-ES FOR PRACTICE. 17:J. Find the prime factors of the folio v k ing 1 lumbers: 1. «30. 12. 19175. 23. 9100. 2. 210. 13. 10028. 24. 5184. 3. 1380. 14. 1250. 25. 8030. 4. 402. 15. 10323. 20. 410.5. 5. 8130. 10. 2240. 27. 02500. 0. 1470. 17. 0400. 28. 81000. 7. 4301. 18. 4515. 29. 04000. 8. 3234. 19. 1000. 80. 45500. 9. 11025. 20. 2310. 31. 10875 10. 30030. 21. 7854. 32. 18590. 11. 14000. 22. 54.50. 33. 10380. CANCELLATION. PREPARATORY PROPOSITIONS. 174. Study carefully the following propositions : Prop. I. — Rejecting a factor from a number divides the num- ber by that factor. Thus, 72 = 24 X 3. Hence, njrcting the factor 3 fron. 72, we have 34, the (piotient of 72 diviiled by 3. P«OP. II. — Dividing both dividend and divisor by the same number docs not change (hv quotient. Thus, CC -^ 12 -- 20 thm-.i + 4 threes = 5. CA y CELL A Tioy, 75 Obporve that the unit three, in 20 threes -4- 4 threes, does not in nny way nft'ect tho pIzo of the (luotient ; therefore, it may l)e rejected RntI the quotient will not be clianped. Hence, dividing both the dividend (>0 and the divisor 12 by 3 does not change tho quotient. ILLUSTRATION OF PROCESS. 1 75. Ex. 1. Divide 402 by 42. Explanation.— We divide I'oth the dlvipor niid dividend l>y ♦>. Accoidiii;,' to Prop. 1!, the quotient Ic uot changed. Hence, T7+7 = 4anie manner, M and .'iS, i:} and 39, S-l and 3 are divided. The remainlnf,' factors, H inid 11. in the dividend are prime to each of tho rcinalninjj factors in tlie divis^or. Hence, no further dlvlHlon can l)e jjcr- formed. 2. We divide the product of 8 and 11, the remaining faotort* in tho divi- dend, by the product of .S and 7, the remaininu factors in the divii««)r, and find as a (jiiotlent IV. 1 ^^bich, accordin;? to ^114— II), b equal lu the quo- tient of 65 - 21 X .15 divided by .39 ^ 15 x .35. u .Ml similar cases may be treated in the same manner ; hence, the following RULE. 17(5. /. Caned nil the fdetorn that arc common to the (lin- den d dud dirisor. IF. Din'dr the prod net of the remninlnrj fartorK of (he dtiidend hjf the product of the remaining factors of the ditiitor. 77ie re- sult irill hi the qui>ti>'nt rcf fired. 76 P R P E RTI E S F X LMUERS, \A^RITTEN EXAMPLES. Ans. 16f. Am. \\. Ann. 20. Ans. 15 J 1. Ans. \m\. Ann. 5/, 4t.' 177. 1. Divide 847.') by 52"). 2. Divido OOOD by GOOO. 3. Divido 8 X 15 X 40 by 10 X 24. 4. Divide ;];528 by 210. 5. Divide 12.">00 by 75, 0. Divide 4!) x 25 x 12 by 10 x 30 x 5. 7. Divide 12 > lU x 27 by 42 x 14. Ann. 27. 8. Divido 04 x 81 X 25 by 24 x 27. Ana. 200. 9. Multiply 8 times 00 by 5 times 18 and divide the prcMluct by 33 times 72. Ana. 20. 10. What is the (luotient of 10 times 5 times 4 divided by 8 times 20? Ann. 2. 11. How many barrels of Hour, at 12 dollars a barrel, are worth ns much a»s \{\ rords of wood, at 3 dollars a cord '.' 12. If 10, 12, HI, and 42 are the factors of the dividend, and 12, 5, 24, and 7 aro the factors of the diviwjr, what is the quotient? Ann. 42. 13. When a laborer can buy 30 bu.shels of potatoes, at 4 shillings a buHhi^l, with the earnings of 24 days, how many shillinps does he earn a pounds of tea, at 72 cents a pound, would i>ay for 3 hogsheads t>f sugar, each weighing 1404 pounds, at 15 cents a pound ? Ana. 015 jwunds. 17. A man exchanged 75 bu.shels of onions, at 00 cents a bushel, for a number of boxes of tea, containing 25 pounds each, at 54 cents a ix)uud \ how many boxes did he receive ? GREATEST COMMON DIVISOR, 77 GREATEST COMMO::^ DIVISOR. PREPARATORY STEPS. 1 7H. Step I. — Find hy inspection an exact ditisor for each of the fulioiting setn of n u inherit : 1. 3, U, ir», and 12. 4. 18, 45, 27. and 72. 2. 7. 14, 21, and :m. 5. 80, 84, 108, and 60. X 8, 12, :{«. and 28. 6. 42, 70, 28, and 112. Stki' U.—Find hy inspection the grcdtrat nunilnr that is an 41 art dirisor of each of the f dinting jmirn of niimberti : 1. r,. 25. 3. «, 120. 5. 25, 750. 7 O 4. la, laoo. 0. 45, 9000. Find in the nam** manner the greatest exact divisor of the following : 7. 14, :J5. 0. 3(5, 96. 11. 84. 1^2. H. 25, 45. 10. 72, 108. 12. 8H. 121. Stkp III. — ErprcftM the vfitnbcrit in eacJi of the foregoing cvamjilen in terms of thtir grintcnt exact dirimr. ThiiH, the jifreatest exact divisor of 16 and 40 is 8, hence 16 may l>e expresaed as 2 eights, and 40 as 5 eights. DEFINITIONS. 1 79. A Common Divisor is a number that is an exact (livisor of each of two or more niiml)ers. Thus, 5 is a divisor of 10, 15, and 20. 18. 117. 4. 78. i:51, :50(J. (J. 11-2. 19(J. 272. 7. 187, 221. ;}23. H. 40."), r>(iT. 324. 0. 225. 525, 300. 10. «H. 102. 238 11. (i(). l:i2, 231 12. 105. 245, 315 13. 138. 181. 322 14. 105, 2j?0. ::4r) 15. 147. 483 10. 228. 2Ti;, 348 17. 8-10. 312. 408 18. 300. 315. 4U5. bJTHOD BY DIVISION. PREPARATORY PROPOSITIONS. 18(>. Lot tlio two followin*^ piojMwitions be carefully studied niul illustrated by other examples, bofon* attempting to find tlu' greatest common divisor by this ineth(xl. Phop I. — 77(/' greatest rommon din'mr of two mtmher^ U the (jrentcd common divisor of the smaller number and their differ- enee. Thus, 3 is the greatest common divisor of 15 and 27. Hence, 15 = 5 threes and 27 = 9 tlirees ; and threes — 5 threes — 4 threes. But 9 and 5 are prime to each otlier ; hence, 4 and 5 must be prime to each other, for if not, their common divi.sor will divide their sum, according to (158— II), an greatest common divis4)r of two numb«'rs is tho greatest common divisor of the smaller numlx^r and their ditlVrence. .«■■ It I- m •3t 80 PROPERTIES OF NUMBERS, 22-0= 10 10 - = 10 10-0= 4 Pnor. \\. — T?w f/rratest common divisor of two numbers U the grentvHt ronimon dirimrr of (he nmiiUcr number and the remainder after the division of the greater by the Uhh. TliJH proposition may Ix- illuHtruted tlius: 1. Sul)tnut (I from 22. thru from the dif- fcr(;iK*<>, 1(1, etc., until u remainder le»s thau in o1)taincd. 3. Obscrvt; tlial the number of tinu'H lias ))e«'n Hubtracted is tin- (juotientof 22 divided by 0, and honre that the remainder, 4, is the remainder after the divirtion of 22 by (I. W. According to Prop. I, the jjreatest common dlvJKor of 22 and is the grcatcHt ('onimon divisor of their difTerence, 10, and 0. It is also, a<'<'()rdin/j; to the same Proposition, the great- ent connnon divisor of 10 and 0. and of 4 and 0. But 4 is the n-mainder after division and the snuiUer number. Hence tlie ,'reatert common divisor of 28 and 17(1 h the name M tlic prratoHf common divisor of 28 and 8, ftl«o of 8 and 4. Bnt 4 In the eroatoHt common divii^or of 8 and t. Flence 4 1h the greatont common divisor of 28 and 17C. Note. — If the giv«>n numbers have not a eommon factor, they cannot have a common divisor ^'n*ater tlian unity, and are either prime numbers or prime to each other. GREATEST COMMON DIVISOR, 81 The greatoRt common divisor of any two numlKjre is found in the same manner ; hence the following 188. Iti'LE. — Divide the greater number by the lens, tJun the ItHS nuinbirht/ the veiriditKhr, then thi fast dirisor hi/ the UtH remniinh r, (tud ho on until nothing remains. The taut dicinor i* the grin tent common dicutur sought. To find tlip pn'fttoHt common divisor of throe or more num- Im-Mh l)y tills method we liuve tlio following 180. RiiiE. — Find the greatest common divisor of tiro of the iiunilwri*, then of the common dirimr thus found and a third number, and so on icith a fourth, ffth, etc., number. ARITHMETICAL DRILL TABLE NO. 3. ItM). Taljle forOral Exercises iiiCJreatest Common Divisor, jilid for Oral and Written Exercises in Least Common Multiple. 1. 4. 8. !>. M>. 11. 12. A. 18 21 20 20 18 21 ■'iO IS ')0 B. 10 9 27 32 35 ^2 42 10 ^5 70 55 72 48 c. u 15 S 36 40 54. 03 32 72 00 no 84 i>. G 6 16 15 45 12 18 F. O. 16 8 "0 fl 40 56 27 60 77 60 2 12 48 28 72 63 80 22 96 24 12 10 24 14 56 48 SI 44 88 36 12 28 30 36 35 24 64 36 66 24 108 1 ■ \'L •^, ■-h- ■*m ^ A/. IMAGE EVALUATION TEST TARGET (MT-3) 1.0 ^« II |M III 2.5 IM 2.2 1.8 1.25 1.4 1.6 •4 6" — ► V] <^ /a e. /, 4 > /A '♦v^ 7 Photographic Sciences Corporation 23 WEST MAIN STREET WEBSTER, NY. MS80 (716) 872-4503 \ ^^ ^ :\ \ ^^ ^■^^ %^ 9) o\ I I* ^ ^ ■:< M% ids ^ ' ^ \# V i t4 .^ 82 PROPERTIES OF NUMBERS, ARITHMETICAL DRILL TABLE NO. 4. 101. Table for Written Exercises in Greatest Common Divisor and Least Common Multiple. A. B. c. D. E. F. 1. 30 36 154 176 88 198 a. 48 210 72 54 84 126 n. 252 396 264 480 220 792 4. 60 120 40 420 175 195 6. 132 26j^ 396 462 594 528 6. UO 105 420 156 315 585 7. 96 280 112 192 336 840 8. 198 315 297 693 567 594 9. 210 350 240 300 720 630 10. 132 220 264 308 660 528 11. 168 480 504 420 252 540 13. 156 312 130 364 273 351 EXAMPLES FOR PRACTICE. 10i2. Find the greatest common divisor of the following: 1. 357 and 483. 6. 385 and 1085. 3. 195 and 465. 7. 356 and 808. 3. 418 and 330. 8. 195 and 483. 4. 803, 546,- and 124. 9. 546, 4641, and 364. 5. 455 and 1085. 10. 465, 1365, and 215. 19«$. Continue the practice in finding the greatest common divisor of abstract numbers by taking examples from the above Arithmetical Tables. Let all the examples taken e BEATS ST VOilMOy D I VI SOB. 83 I. 4. Common F. 198 12G 792 195 52S 58') 8JfO 50^ I GclO 528 5J^0 351 owing : common I from the los token from Table No. 3 be worked orally and in sets in the same manner as directed for written exercises. JExutnples with Two Auinbcrs. 194. FiEST Set. — Take examples for written exercises from the first line of Table No. 4, thus : (1.) 30 36 (2.) 36 154 (3.) 154 170 (4.) 176 88 (^.) 88 196 Observe that in each new example, the first number taken in the last example is omitted and a new number added. Take in the same manner examples from each line in the table. 19,1. Second Set. — Take examples from the first and second lines thus : (1.) 30 48 (3.) 154 72 (5.) 88 84 (2.) 30 210 (4.) 176 54 (6.) 198 126 To present other examples, omit the first line and use the second and thii'd, then the third and fourth, and so on to the bottom of the table. 190. TiiiKD Set. — Take the numbers from the first and thh'd line, then from the second andfoyrth, then from the tJiird ^nd fifth, etc., to the bottom of the table. Examples with Uiree Nntnbers. 197. First Set. — Take examples from each line, thus: (1.) 30 30 154 (2.) 36 154 176 (3.) 154 170 88 (4.) 170 88 198 198. Second Set. — Take the numbers from the first, second, and third lines, then from the second, third, and fourth, and so on to the bottom of the table. >?''>■§ m ■*' ■ - t; !l 84 PROPERTIES OF NUMBERS, WRITTEN EXAMPLES. 199. 1. I have rooms 12 feet, 15 feet, and 24 feet wide; what is the width of the widest carpeting that will fit any room in ray house? A us. 8 feet. 2. Divide the greatest common divisor of 48, 72, UO, and 120 by the greatest common divisor of 21, 30, 39, and 84, 3. In the city of Montreal, some of the sidewalks are 48 inches wide, some CO inches, and others 72 inches ; what is the widest flagging that can be used in each of these side- walks without cutting ? Anfi. 12 inches. 4. D. White owns in Hamilton 3 lots of equal depth, the first having a front of 72 feet, the second 144 feet, and the third 108 feet, which he wishes to divide into as many lots as pos- sible having equal fronts ; how many feet will each front con- tain ? Ans. 30 feet. 5. A teamster agrees to cart 132 barrels of flour for a mer- chant on Monday, 84 barrels on Wednesday, and 108 barrels on Friday; what is the largest number he can carry at a load, and yet have the same number in each ? Afis. 12 barrels. 6. I have a lot whose sides measure, respectively, 42 feet, 84 feet, 112 feet, and 126 feet ; I wish to enclose it with boards having the greatest possible unifonn length ; Avhat will be the length of each board ? Ano. 14 feet. 7. A merchant has three pieces of cloth containing respec- tively 42, 98, and 84 yards, which he proposes to sell in dress patterns of uniform size. What is the largest number of yards the dress i>atterns can contain so that there may be nothing left of either piece ? 8. If two farms containing each nn exact number of acres were purchased for $8132 and $0270 respectively, what is the highest uniform price per acre th.at could have been paid, and in this case how many acres in each farm ? LEAST COJfJlOX MULTIPLE, 85 LEAST OOMMOjS" MULTIPLE. PREPARATORY PROPOSITIONS. 200. Study carefully each of the following propositions : Prop. I. — A multiple of a number contaiTis as a factar each prime factor of the number as muiiy times as it enters into the number. Thusi, 60, which Is a multiple of 12, contains 5 times 12, or 5 times 2x2x3, the prime factors of 12. Hence, each of the prime factors of 12 enters as a factor into 60 as many times as it enters into 12. Prop. ll.— T7ie least common multiple of two or more given numbers must contain, as a factor, each prime factor in those numbers the greatest number of times that it enters into any one of tliem. Thus, 12 = 2 X 2 X 3, and 9 = 3x3. The prime factors in 12 and 9 are 2 and 3. A multiple of 12, according to Prop. I, must contain 2 as a factor twice and 3 once. A multiple of 9, according to the same propopition, must contain 3 as a factor twice. Hence a number which is a multiple of both 12 and 9 must contain 2 as a factor twice and 3 twice, which is equal to 2 X 2 X 3 X 3 = 36. Hence 36 is the least common multiple of 12 and 9. '•, ?! '4 ll 11'" '^i>\ DEFINITIONS. 201. A Multiple of a number is a number that is exactly divisible by the given number. Thus, 24 is divisible by 8 ; hence, 24 is a multiple of 8. 202. A Conunon Multifile of two or more numbers is a number that is exactly divisible by each of them. Thus, 36 Is divisible by each of the numbers 4, 9, and 12 ; hence, 36 is a common multiple of 4, 9, and 12. 2015. The Least Common Multiple of two or more numbers is the least number that is exactly divisible by each of them. Thus, 24 is the least number that is divisible by each of the numbers 6 and 8 ; hence, 24 is the least common multiple of 6 and 8. m 86 PROPERTIES OF NUMBERS, METHOD BY FACTORING. ILLUSTRATION OF PROCESS. 204. Prob. I. — To find, by factoring, the least com- mon multiple of two or more numbers. Find the least common multiple of 18, 24, 15, and 35. Explanation.— 1. We observe that 3 is a factor of 18, a4, and 13. Dividing these numbers by 3, we write the quo- tients with 35, in the second line. 2. Observing that 2 is a factorof 6 and 8, we divide as before, and find the third line of numbers. Dividing by 5, we find the fourth line of numbers, which are priiLe to each other; hence cannot be further divided. 3. Observe tlie divisors 8, 2, and 5 are all the factors that are common to any two or more of the given numbers, and th quotients 3. 4, and 7 are the factors that belong each only to one number. Therefore the divisors and quotients together contain each of the prime factors of 18, 24, 15, and 35 as many times as it enters into any one of these numbers. Thus, the divisors Sand 2, with the quotient 3, are the prime factors of 18; and so with the other numbers. Hence, according to (200—11), the continued product of the divisors 3, 2, and 5, and the quotients 3, 4, and 7, which is equal to 2520, is the least common multiple of 18, 24, 15, and 85. From this illustration we have the following 3 18 24 15 35 2 6 8 5 35 5 3 4 5 35 3 4 % RULE. 205. L Write the numbers in a line, and divide by any prime fci^tor that is contained in any two or more of them, placing the quotients and the undivided numbers in the line below. II. Oj)erate upon the second line ofnumbem in the mme man- ner, and so on until a line of numbers that are prime to each other is found. III. Find the continued product of the divisors used and the numbers in the last line ; this icill give tlie least common multiple of th^ given numbers. LEAST rOJfJfO.Y MULTIPLE, 87 Ml WRITTEN EXAMPLES. 200. 1. What is the least number of cents that can be exactly expended in oranges, whether they cost 4, o, or 6 cents apiece ? 2. What is the least common multiple of the nine digits ? 3. What is the smallest quantity of milk that will exactly fill either six-quart, nine-quart, or twelve-quart cans? 4. ^Vhat is the smallest sum of money that I can exactly lay out in calves at 14 dollars each, cowa at 38 dollars each, or oxen at 57 dollars each ? Ans. 798 dollars. 5. A can lay 42 rows of shingles on my house in a day, and B can lay 50 rows ; what is the least number of rows that will give a number of full days' work to either A or B ? 6. What is the width of the narrowest street across which stepping-stones either 3, 4, or 9 feet long will exactly reach ? 7. Three separate parties are measuring the distance from the city hall, Kingston, to the University, Toronto ; one party uses a chain 33 feet long, another a chain 66 feet long, and the third a chain 50 feet long, marking each chain's length with a stake ; at what intervals of space will three stakes he driven at the same place ? Ans. Every 1650 feet. M m^ METHOD BY GREATEST COMMON DIVISOR. ILLUSTRATION OF PROCESS. 207. Prob. II.— To find, by using the greatest com- mon divisor, the least common multiple of two or more numbers. Find the least common multi])le of 195 and 255. Explanation.—!. We find the greatest common divitor of 195 and 255, which is 15. 2. The greatest common divipor, 15, according to (182), contains all the prime factors that are common to 1'15 and 255. Dividing each of these numbers by 15, we find the factors that are not common, namely, 13 and 17. l-f -m Mm 1..^. 88 PROPERTIES OF N U Jif B E R S . 3. The common divisor 15 and the qnoticnt 13 contain all the prime factors of 195, and the common divisor 15 and the quotient 17 contain all the prime factors of 255. Hence, accor(!'ng to (^200—11), the continued product of the common divisor 15 and the quotients 13 and 17, which is 3315, is the least common multiple of 195 and 255. The least common . uiltiple of any two numbers is found in the same manner ; hence the following RULE. 208. /. Find the grefftent common dicisar of the two given numbers, and divide each of the numbers by this divisor. II, Find the continued product of the greatest common divisor and the qvotients ; this will give the least common midtiple of the two given numbers. To find the least common multiple of three or more numbers by this method, we have the foUowinpf RULE. 200* Find the least common mtdtiple of two of them; then find the least common multiple of the multiple thus found and the third number , and so on with four or more numbers. , i:XAMPLES FOR PRACTICE. Ill m [!* 210. Find the least common multiple of 1. 110 and 165. 2. 91 and 182. 3. 78 and 195. 4. 143 and 165. 5. 385 and 455. 6. 154 and 231. 7. 462 and 546. 8. 364 and 637. 211. For further practice take examples with two numbers from Table No. 4, page 82, as directed in (194), (195), and (196) ; and examples with three numbers from Table No. 3, as directed in (197) and (198). Continue to practice with abstract numbers until you can find the least common multiple of two or more numbers accurately and rapidly. LEAST COMMOX MULTIPLE, 89 BEVIEW AND TEST QUESTIONS. 212. 1. Define Prime Number, Composite Number, and Exact Divisor, and illustrate each by an exami)]e. 2. What is meant by an Odd Number? An Even Number? 3. Show that if an even number is divisible by an odd num- ber, the quotient must be even. 4. Name the prime numbers from 1 to 40. 5. Why are all even numbers except 2 composite ? 6. State how you would show, in the series of odd numbers, that every fifth number from 5 is divisible by 5. 7. What is a Factor ? A Prime Factor ? 8. What are the prime factors of 81 ? Of 64 ? Of 125 ? 9. Show that rejecting the same factor from the divisor and dividend does not change the quotient. 10. Explain Cancellation, and illustrate by an example. 11. Give reasons for calling an exact divisor a measure. 12. What is a Common Measure ? The Greatest Common Measure ? Illustrate each answer by an example. 13. Show that the greatest common divisor of 42 and 114 is the greatest common divisor of 42 and the remainder after the division of 114 by 42. 14. Explain the rule for finding the greatest common divisor by factoring ; by division. 15. Why must we finally get a common divisor if the greater of two numbers be divided by the less, and the divisor by the remainder, and so on ? 16. What is a Multiple ? The Least Common Multiple ? 17. Explain how the Least Common Multiple of two or more numbers is found by using their greatest common divisor. 18. Prove that a number is divisible by 9 when the sum of its digits is divisible by 9. 19. Prove that a number is divisible by 11 when the differ- ence of the sums of the digits in the odd and even places is zero. 5' '^ -» 1 FRACTIONS. PREPARATORY PROPOSITIONS. 213. Prop. I. — Any thing regarded as a whole can be divided Into unequal or equal parts ; thus, WHOLE. PARTS. Hi' ■iH' (1.) (2.) 1. Equal parts of a whole are called Fractions. 2. Into what kind of parts can a pear be divided ? A bushel of wheat? A slate? A garden ? Anything? 3. Make $12 into unequal parts in six ways, and into equal parts in five ways ? 4. In how many ways can 15 be made into equal parts? Into unequal parts ? Prop. II. — The same whole can be divided into equal parts of different s-izea ; thus, . • • WHOLE. EQUAL PARTS. Halves. Thirds. Fourths. Fifths. 1. Observe, the equal parts are named by using the ordinal corresponding with the number of parts. Thus, when the whole is made into three parts, one part is called a third, when PREPARATORY PROPOSITIONS, 91 into four parts, one part is called a fourth, and so on to any mimber of parts. 2. When the whole is made into ten equal parts, what is one part called ? Into sixteen equal parts ? Into twenty-four ? Into forty-three ? 3. What are the largest equal parts that can be made of a whole ? The next largest ? The next largest ? 4. What ib meant by one-half of an apple? One-third? One-fifth ? 5. What is meant by two-thirds of a line? Of an hour? Of a day? 0. How would you find the fourth of anything? The seventh? The tenth ? 7. Find the third of 6. Of 12. Of 15. Of 24. Of 48. 8. If a whole is made into twelve equal parts, how would you name three parts? Seven parts? Five parts? Nine parts ? 9. How many halves make a whole? How many thirds f How many sevenths? How many tenths? How many fif- teenths ? Prop. III. — Equal parts of a whole, or Fractions, are expressed by two numbers written one over the other, with a line between them ; thus, Numerator, lL Shows the number of equal parts In the fraction. Dividing Line, ^ Shows that 4 and 5 express a fraction. Denominator, ^ Shows the number of equal parts in the whole. Read, Four -fifths. ,' t 1. Read the following : 3 7 6 9 "5> F» ^» TTF? I If. * 2. What does f signify ? f? If? 23 9 :Vi) • 3. Express in numbers three-fifths • nine- ■thirteenths ; eleven- titteenths. 4. Read the following : fi M-> \l Uh ih ni 5. Write in numbers eight-twentieths ; twelve-sixteenths; j fifteen -seventieths ; nine- ■fortieths. ■ 92 FHA t'TIONS. C. What does Numerator mean ? Denominator f Dividing lino ? Tirniit of a fraction ? 7. How is a fraction oxpreflsod by numbers? 8. Name tlie tenns of I Viy. ifl. n- Vh ill 9. Express in numbers seven-nintlis ? Nineteen forty-fifths. Prop. IV. — The me or value of the same kind of equal parts depends upon the size or value of the whale of which they are parts; tlius, f WHOLE. EQUAL PARTS. Hnlvpft, IlalvvH, 1. Tlio equal parts in the illustration, although halves in both cases, are unequal in size, because the wholes are unequal in size. 2. Which is the larger, the half of $4 or the half of |6 ? 8. Which is the smaller, the fourth of 13 inches, or the fourth of 20 inches, and why? 4. If oft'ored the half of either of two farms, which would you take, and why ? Prop. V. — The size or value of the equal parts of a whole diminish as the number of parts increase^ or increase as the number of parts diminish ; thus. WHOLE. EQUAL PABT8. Tttirda. Fourths* Fifths. 1. Which is the greater, one-half or one-third? One-fourth or one-fifth ? One-sixth or one-ninth, and why ? 2. How much is J of $48 smaller than J of it? 3. Upon what two things does the value of one-half, one- third, one-fourth, one-fifth, etc., depend? Illustrate your answer by two examples. PHE1*AKAT0RY PROPOSITIONS, 93 DEPIinTIONS. 214« A Ft*actional Unit is one of the equal parts of fiuything regarded as a whole. iil5. A Fraction is one or more of the equal parts of anything regarded as a whole. 21($. The Unit of n Fraction is the unit or whole which is considered as divided into equal parts. 217. The Nnincrator is the number above the dividing line in the expression of a fraction, and indicates how many equal parts are in the fraction. 218. The Denominator is the number below the dividing line in the expression of a fraction, and indicates how many equal jmrts are in the whole. 210, The Terms of a fraction are the numerator and denominator. 220. Taken together, the tei^ms of a fraction are called a Fraction, or Fractional Namher, 221. Hence, the word Fraction, means one or more of the equal parts of anything, or the expression that denotes one or more of the equal parts of anything. REDUCTION". PREPARATORY STEPS. 222. Step I. — A fraction is represented by lines thus : • 3 ^^iM ^■HH Part taken. 3 KMiM^-^a^M Ttliole. Observe carefully the following : 1. In f , the denominator 3 expresses the whole, or 3 thirds, and the numerator 2 expresses two parts of the same size. 'ii.ll ft . if 'f „■'■■!? * 94 FEA C TIONS. Hence, 3 equal lines for the denominator and 2 equal lines for the numerator, of the same length as those in the denom- inator, represent v orrectly the whole, the parts taken, and the relation of the parts to each other, as expressed by the frac- tion 8 5- 2. Represent by lines f ; f ; y? ; iV J « ; i^'a- Why can the numerator and denominator of a fraction be represented by equal lines ? Step II. — Show hy representing t?ie fraction with lima that one-half 18 equal to two-fourths ; XhxxB, . - %.-\ 2 -■ Part takeiit Whole, 1. By observing the illustration, it will be seen that the value of the numerator and denominator is not changed by making ea^h part in each into two equal parts. It will also be seen that when this is done the numerator contains 2 parts and the denominator 4. Hence ^ = f . 2. Show in the same manner that one half is equal to three- sixths, four-eighths, five-tenths, and so on. Step III. — Any fractional unit can, without changing its value, he divided into any desired number of equal parts. Study carefully and explain the following illustrations : 'tgeu u t 2. 1 8 ^ ^^ 2 6 1 ' ■ sum 1 1 4 8 18 — — • ■ Be ■ by tb RED UCTIoy. 96 :m ORAL EXAMPLES. 223. 1. How many tenths in I of an orange ? How many tifteenths ? How many twentieths ? etc., and why ? 2. How can ^ of a yard be made into sixteenths of a yard ? 3. How many twelfths in^? In^? In|? In^? 4. Make | into twenty-firsts, and explain the process. 5. Show by lines that J = j*^ ; that | = tj^ ; that i = ^. 6. Change |, without altering its value, into a fraction con- taining 7 equal part 3 ; 10 equal parts ; 25 equal parts. PBINCIPIiES OP BEDUCTION. 224. Let each of the following principles be illustrated by the pupil with a number of examples : Prin. I. — The numerator and denominatoi' of a fraction represent, each, parts of the same size ; thus. 7 4 5 (1.) (3.) Observe in illustration (1) the denominator 7 represents the whole or 7 sevenths, and the numerator 3 represents 3 sevenths ; in illustration (2), the denominator represents b fifths^ and the numerator A fifths. Hence the numerator and denominator of a fraction represent parts of the same size. Prin. H. — Multiplying both the terms of a fraction by the same number does not change the value of the fraction ; thus, 2 3 2x4 8x4 8 12 Be particular to observe in the illustration that the amount expressed by the 2 iu the numerator ur the 3 in the denominator of | is not «: 96 FRA CTIO N8, changed by making each part into 4 equal parts; therefore, J and t\ express, each, the same amoant of the same whole. Hence, multiplying the numerator and denominator by the same number means, so far as the real fraction is concerned, dividing the equal partf in ea£h into as many equal parts as there are units in the number bji which they are multiplied. Prin. III. — Dividing both terms of a fraction by the same number does not change the value of the fraction ; thus, , 12-5-3 8 4 The amount expressed by the 9 in the numerator or the 12 in the denom- inator of t'?! Is not changed by putting every 3 parts into one, as will be seen from the illustration. Hence, x"i and | express each the same amount of the same whole, and dividing: the numerator and denominator by the same number means jmt- ting as many parts in each into one as there are units in the number by which they are divided. /»!■ hi DEFINITIONS. *2*2i5. The Value of a fraction is the amount which it represents. , .^ 226. Reduction is the process of changing the terms of a fraction without altering its value. 227. A fraction is reduced to Higher Terms when its numerator and denominator are expressed by larger numbers. Thus, i = tV- 228. A fraction is reduced to Lower Terms when its numerator and denominator are expressed by smaller numbers. Thus, j% = I 229. A fraction is expressed in its Lowest Terms when its numerator and denominator are prime to each other. u ILLUSTRATION OF PROCESS, 97 Thus, in f , the numerator and denominator 4 and 9 are prime to each other; hence the fraction is expressed in its lowest terms. 230. A Common Denominator is a denominator that belongs to two or more fractions. 231. The Least Cotnmon Denominator of two or more fractions is the least denominator to which they can all be red: V wed. 232. A Proper Fraction is one whose numerator is less than the denominator ; as f , f . 233. An Impro2)er Fraction is one whose numerator is equal to, or greater than, the denominator ; as f , |. 2*34. A 3Iixed Nwinher is a number composed of an integer and a fraction ; as 4f , 18f . ILLUSTRATION OF PROCESS. 235. Prob. I.— To reduce a whole or mixed number to an improper fraction. 1. Reduce 3| equal lines to fifths. WHOLES. FIFTHS. = IS fifths = 18 Explanation.— Each whole line is equal to 5 J^fths, as ehown in tiie illustration ; 3 lines must therefore bo equal to 15 Jlfths. \hjifths + Z fifths = 18 fifths. Hence in 3J lines there are V of a line. From this illustration we have the following : ^H^4 RULE. 23C$. Multiply the whole luimher by tlie given denominator, and to the product add the numerator of the given fraction, if any, and icrite the result over the given denominator. 98 FRA CTIONS, EXAMPLES FOR PRACTICE. 237. Reduce orally the following : 1. In 5 pounds of sugar how many fourt?i8 of a pound ? Solution.— In 1 pound of sugar there are \ fourths ; hence, in 5 pounds there are 6 times i/ourths, which Is \^ of a pound. 2. In 7 tons of coal how many ninths of a ton ? d. Bow msLUj tenths in $Qd1 In 42 yards? In 17 pounds? 4. Express 20 slb fourths. As sevenths. As hundredths. 5. In $9f how many sevenths of a dollar ? Solution.— In $1 there are 7 sevenths. In $9 there must therefore be 9 times 7 sevenths or 63 sevenths. 63 sevenths + 3 sevenths are equal to 66 sevenths. Hence, in $9f there are V of a dollar. • ' .s. 6. In 12^ acres how many twelfths of an acre? 7. How many eigJitlis in 9| ? In 11 1 ? In 7f ? In 5| ? Reduce the following to improper fractions : 8. 83i. 12. 340«. 16. nn 9. 45|. 13. 462g. 17. 3ff. 10. 76f. 14. 1875. 18. 4t^ 11. 13H. 15. 463^\. 19. 3tU- 238. Prob. II.— To reduce an improper fraction to an integer or a mixed number. 1. Reduce 9 fourths of a line to whole lines. » =r 9^4 = 2i \il Explanation.— A wholo line is compoeed of Af&iiriht<. Hence, to make the ^fourths of a line into whole lines, we put every /ot/r parts into on^, as shown iu the illuntration, or divide the 9 by 4, which gives 2 wholes and 1 of ihefourtlis remaining. Hence the following RULE. 239. Divide tlie numerator hy the denominator. EXAMPLES. 99 EXAMPLES FOR PRACTICE. 7B 1 240^ Reduce and explain orally the following: 1. How many bushels are -^/ of a bushel? ^■'- V y? y^ 2. In $-"/, how many dollars ? In V o^ a yard, how many- yards V In -^^ of a foot, how many feet ? 3. How many miles in ^^ of a mile ? 7 • Reduce to whole or mixed numbers the following ; 4. H.I 8 2S • 5. 508 Br* 6. w. 7. w. 8. ni 9. ^fF. 10. ^tW. 11. ntl^ 12. W^ 13. -'bW- 14. 15. 16. 17. 18. 50705 9 0067 nsoi 80«6 3F~' 241. Prob. III. — To reduce a fraction to higher terms. 1. Reduce f of a line to twelfths. I 2x4_ 8_ I * 3x4~ 12 Explanation.— 1. To make a whole, which is already in thirds, into 12 equal part8, each third must be made miofcmr equal parts. 2. The numerator of the given fraction expresses 2 thirds, and the denom- inator 3 thirds; making each third in both mXo four equal i)arts (224—11), as shown in the illustration, the new numerator and denominator will each contain 4 times as many parts as in the given fraction. Hence, \ of a line is reduced to tice^ths by multiplying both numerator and denominator by 4. Hence the following rule for reducing a fraction to higher terms : %l RULE. _.. 242. Divide the required denominator by the denominator of the given fraction, and multiply the terms of the given frac- tion hy the quotient. I 100 FEA CTIONS, EXAMPLES FOR PRACTICE. 243. Reduce and explain orally the following : 1. How would you make liahea of an apple into fourths? luU) .sixths? Into tenths? Into siosteentTis ? 2. How many twelft/ts in f of a rx)rd o' wood ? 8. Explain how i^, ^, and f can be reduced to twentieths. 4. Show by the use of lines that f = * = ^ = -^^ = |§. 5. Reduce |, f , f, ^, and ^f each to forty -seconds. 6. In f how many ninety-eighths ? 7. Change f , f^, ^, f , and ^^ each to 360ths. 8. Reduce f , ^, ^, ^f , and ft to 165ths. 244. Pros. IV.— To reduce a fraction to lower terms. Reduce ^^j of a given line to fourths. 12^3 "■ f 4 ExPLAKATiON.— 1. To make into 4 equal parts or fourths a whole which is already in 12 equal parte, or twelfths, every 3 of the 12 parts must be put into one. 2. The numerator of the given fraction expresses 9 twelfths, and the de- nominator 12 twelfths ; putting every 3 twelfths into one, in both (224— III), as shown in the illustration, the new numerator and denominator will each contain one-third as many parts as in the given fraction. Hence tt of a line is reduced to fourths by dividing both numerator and denominator by 3. Hence the following rule for reducing a fraction to its lowest terms : BULB. 245. Reject from tJie terms of the giten fraction aU their common factors. Or, Divide the terms of the given fraction hy their greatest common divisor. EXAMPLES. 101 EXAMPLES FOR PRACTICE. 24G. Reduce and explain orally the following : 1. In ^ of a bushel, how many thirds of a bushel? 2. How can twelfths of a bushel be made into fourths of bushel? \u\jo tJiirdaf Into halcesf 3. Reduce ^^ of a dollar to Jijths of a dollar. 4. Show by the use of lines that ^^ = ig = yV = I = i« 5. Reduce ^\ to its lowest terms. ■^^. |g. ||. C. Express if in parts 8 times as great in value. Reduce the following to their lowest terms : 7. VV\. 10. Uh 13. fM. 16. 8. ^Vtr- 11. Uh 14. ,%¥.. 17. To* 9. Ut 12. 847 88 15. fro* U^ U* 324 18. 61 8 4 347. Prob. V. — To change fractions to equivalent ones having a common denominator. 1. Reduce f and f of a line to fractions having a common denominator. *" i 3x4 ~ 12 (1.) ' '* fa m ■A' "•1 I- ill 8 4 8x3 4x3 9^ 13 (3.) Explanation.— 1. We And the least common multiple of the denomina- tors 3 and 4, which is 12. 2. We reduce each of the fi^ctions to twelfths (141), as shown in illua- trations (1) and (2). Hence the following i'';i BUTiE. 248. Find the least common multiple of all the denominators for a common denominator; divide this by each denominator ^I'f: 102 FRA CTIONS. separately, and multiply the corresponding numerator by the quotient, and lorite the product over the common denominator. EXAMPLES FOR PRACTICE. 2249. Reduce and explain orally the following : 1. Reduce | and j^ to sixths, f and f to twelfths. 3. Change f and f to fractions having the same denominator, and explain each step in the process. 3. Express f , /^ and f as fortieths. 4. What is the least common denominator of |, f , and j ? Observe, fractions have a least common denominator when their denom- inators are alike and there is no factor common to all the nnmerators and the common denominator. Reduce the following to their least common denominator : 5. f, 1,1, and |. 9. i, |, f , A, /f, and H- 6. I j%, and \l 10. H. M» H. aiid ^. 7. |,f,f, and,^. 11. f, I, uV. FT. is4^. and VW- 8. ^y, Af,andf 13. h j^, \h ^, and ^. ADDITION. PREPARATORY PROPOSITIONS. 250. Prop. I. — Fractional units of the same kind, th/xt are fractions of the sams whole, a/re added in the same manner as integral units. Thus, f of a yard can be added to | of a yard, because they are each fifths of one yard. But | of a yard cannot be added to I of a day. Solve orally the following : 1. f + i + f . 2. f + f + f . 3. A + A + H- Y + T + f- 4. 5. {1 + -h + ^z- 6. A + if + yV EXAMPLES. 103 the ator, lenom- jrs and Lor V^- manner ^se tliey added 7. In I + i^ + V of a yard, how many yards 7 8. How many are %{^ + ^^V + $t^ + $ii + lA ^ 9. Find the sum of f J + i | + sV + ^V + ?t miles. 10. Why cannot f of a bushel and | of a peck be added as now expressed ? Prop. II. — Fractions expressed in different fractional units must he changed to equivalent fractions having the same frac- tional unit, before tJiey can he added. For example, f and f of a foot cannot be added until both fractions are expressed in the same fractional unit. Thus, f of a foot is equal -^^^ of a foot, and f of a foot is equal -^^^ of a foot ; A + 1 ? ®^ ^ ^^* — \h o' 1 A f®®** Hence the sum of | + 5 of a foot — 1^*^^ feet. Find orally the sum of the following : 1. l^^ii- 4. tV + M- 7. f + f + A. 2. f + f- 5. % + ^Z' 8. 1 + A + |. 3. h + 1- 6. ^ + ^' 9. 1 + f + ii- 251. Prob. I.— To find the sum of any two or more given fractions. 1. Find the sum of | + f + |. ErPLANATioN. — 1. We reduce the fractions to the same fractional unit, by reducing them to their least common denominator, which is 72 (247). 2. We find the sum of the numera- tors, 155, and write it over the common denominator, 72, and reduce Y>^ to 2i^. 4 32 y -72 5 60 6 ~72 7 63 8 "72 155 72 = 2M. The sum of any number of fractions may be found in the same manner ; hence the following ii m '"i ^i BULE. /. Change the fractions to equivalent ones having the host common denominator, then add the numerators, write the 104 FRA CriONS. remit over the common denominator^ and reduce, when poss^le, to lower terms or to a whole or mixed number. II. When there are mixed nnmbers or integers, add the frac- tions and integers separately, then add the results. TATRITTEN EXAMPLES. !1 •I - I .'! i 253. Find the sum of each of the following : 1. 2. 3. 4. 5. 6. 7. 8. i,f,i?,andH. h h I. h and \, ■I, f , J, \, and \. I, \, and ,V », I, and \. 8 1 a S> ¥» ?> 1 n 4) 6« and §. and f . 8 H 15 anA 6 3 9> Iff' Tff> »"" 'Si' 10. 11. 13. 13. 14. 15. 16. 17. 18. a* i, 7], and 8 I, T«^, ^i and ^. If. 2J, and 4i. 8^, 2§, 3§, and 4f. 4^, 2}, ^, and ^. H,4J,T\,andH. 4|, lOJ, and 83-JI. 8J, 25^, 19, and 68A- 68|, 28i. 32?, 7^\, and 6Bi 9. 19. John Munro has lOJ acres of land in one field, 10^ in another, and llj^ in a third ; how many acres has he in the three fields ? Ans. 32 y^ acres. 20. There are three tubs of butter, weighing, respectively, 44J pounds, 56^ pounds, and 78| pounds ; how much butter in the three tubs ? Ans. Yl^ ^^ \iowjiAQ. 21. I have a board 7f feet long, another 11 1 feet long, and a third 9^ feet long; what is their united length ? 22. How many yards in three remnants of silk, containing, respectively, 2} yards, IJ yards, and 2| yards? 23. William earned 3f dollars, his father gave him 5^^ dol- lars, and his brother gave him lyV dollars more than his father ; how much money did he have in all ? Ans. 14^ dollars. 24. Three pieces of cotton contain, respectively, 43f, 54f, and 87f yards ; how many yards in all ? 25. H. Weston travelled 42 y^^ miles on Monday, 30f miles on Tuesday, 48J^ miles on Wednesday, and 25 J miles on Thursday ; how far did he travel during the four days ? EXAMPLES, 105 ^ie. afi- in the iCres. itively, .tier in nds. and a gaining, ij^^dol- I father; lUars. ^4f , and lileB on lorsday ; SUBTRACTION, 254. Prop. I. — Fractional units of the same kind that are fractions of the same whole are subtracted iu the same man- ner as integral units. Thus, 7 ninths — 5 ninths = 2 ninths, or J — j == !• Perfonn orally the subtraction in the following: j» Toflf Prop. II. — Fractions expressed in different fractional units must be reduced to the same fractional unit hefm'e subtracting. 1. f-f. 3. if - tV. 5. Jl - if. 7. iiJ 3. if - ^. 4. il - H- ft io« 40 "• TO 8 10 5- 8. 87 TOO Thus, in I — -ji'y we reduce the | to sixteentJis ; | if - Tir = T^ir ; hence, | - A = VV- Perform orally the subtraction in the following : i|, and 1. f-i^. 3. ii~M 4 2 B • -ff Tff* 6. 7 ' — _ 7 T5 ~ H* 8. 9. B 4 Tff' 18 11 Si Bii' s- *1^^, Prob. I.— To find the difference of any two given fractions. 1. Find the difference between | and ^^. 7 5 21 10 11 Explanation. — 1. We reduce Q 12^^24 24^^24 *^® given fractions to their least common denominator, wliich is ^. 2. We find the difference of the numerator?, 21 and 10, and write it over the common denominator, giving Hi the required difference. 2. Find the difference between 35f and 16|. 35f = 35^ Explanation.— 1. We reduce the 1 and \ to their Jl Qs __ 1 g » least common denominator. * — ^ 2. i\ cannot be taljen from A ; hence, we increase 18ii the t's by \\ or 1, taken from the .35. We now sub- tract xV from ?§, leaving H- 3. We subtract 16 from the remaining 34, leaving 18, which united with \\ gives 18|}, the required difference. 8 m mm if m 106 FRA VTIONS, The difference between any two fractions or mixed numbers may be fuuud in the same manner ; hence the following RULE. 250. /. Reduce the given fractions to equivalent ones having tfie least common denominator ; then Jind t/ie difference of f/ie numerators and vyrite it over the common denominator. II. When there are mixed numbers, subtract the fraction JirM, then the integer. If the fraction in the minuend is smaller than that in the sub- trahend, increase it by one from the integral part of the minu- end ; then subtract. WRITTEN EXAMPLES. \i 257. Perform the following subtractions : 1. H-f 5. 37j\-33/y. 9. 73| - 29H. 2. f-§. 6. 63-4^. 10. 84^ - 37f . 3. 7i-4J. 7. 13 6^-9jV 11. 511i - 34H 4. 9f-6|. 8. 50ii^-47yV 13. 65/^ - 59i§ 13. From a cask of vinegar containing 31^ gallons, 16g gal- lons were drawn ; how many remained? Ans. 15 J gallons. 14. If flour be bought for $9tV a barrel, and sold for $12^, what is the gain per barrel ? Ans. d^^ dollars. 15. If a grocer buy, 4| and 6^ barrels of flour, and then sells 1^ and 4^ barrels, how many does he still have? 16. The sum of two numbers is 59§, and the greater is 30|f ; what is the other number ? Ans. 2S\^. 17. P. Jones is to build 45| miles of railroad, and has com- pleted 25 1 miles ; how many miles has he to build? 18. James found $2f , earned $1|^, and had |2} given him ; how much more money had he then than George, who earned $6| and spent $4^? 19. I bought two tubs of butter, the tubs and butter together weighing lllf pounds, and the tubs alone weighing 7f and 8 pounds respectively ; what was the weight of the butter ? PKUPA R ATORr P It O P O S TTI X S . 107 MULTIPLICATION. PREPARATORY PROPOSITIONS. 258. Tlie following propopitiona must be mastered por- fectly, to undcrstaiul luul explain the process in multiplication and division of fractions. Prop. I. — Multiplying the numerator of a fraction, while the denominator remains unchanged, multiplies the fraction ; thus, 2x4 t J - 5 gal- |ns. il2|, Lrs. sells I30II ; Hi com- liim ; larned retlier land 8 Observe that since the denominator is not changed, the "izo of the jtarts remain the same. Hence the fraction ? is multiplied by 4, as shown in the illustration, by multiplying the numerator by 4, Prop. II. — Dividing the denominator of a fraction while the numerator remains unchanged multiplies the fraction ; thus, 2^ 2 12-1-4 • 3 (1.) (2.) Observe that In (1) the whole is made into 12 equal parts. By putting every 4 of these parts into one, or dividing the denominator by 4, the svhole, as shown in (2), is made into 3 equal parts, and each of the 2 parts in the numerator is 4 times 1 ttvdffh. Hence, dividing the denominator of i\ by 4, the number of parts in the numerator remaining the same, multiplies the fraction by 4. Prop. III. — Dividing the numerator of a fraction ichile the denominator remains uncJianged divides the fraction ; thus, 6-^3 _ 2 9 " 9 (1.) ^of (2.) '••• 1- iiiniMiMMiiM 108 FRA CTIONS. .'.Ill In (1) the numerator 6 expresses the parts taken, and one-third of these 6 partii, as shown by comparing (1) and (2), the denominator remaining the same, is one-third of the value of the fraction. Hence, the fraction % is divided by 3 by dividing the numerator by 3. Prop. IV. — Multiplying the denominator of a fraction ichile the numerator remains unchanged divides the fraction ; thus, 3 5x3 3^ 10 (1.) (2.) In (1) the whole is made into 5 equal parts ; multiplying the denominator by 2, or making each of these 5 parts into 2 equal parts, as shown in (2), the whole is made into 10 equal parts, and the 3 parts in the numerator are one-half the size they were before. Hence, multiplying the denominator of | by 2, the numerator remaining the same, divides the fraction by 2. EXERCISES. 259. Show by the use of lines or objects that ^ll 1. 2. 3. 10. 4x 3 12 7 7 5 5 18- f-6" '3 14- h3 7 17 ~ "17 4 4 4. 5x 8 3" '15' 7. 5. 13- 3 4-4 x4 = 1. 8. 6. 12 = 1. 9. ^ is how many times ^ ^ 7 ^ 7x3 , and why ? = 2}. 9^ 20-5-5 15-4-3 _^ 19 ~ 19' 3x5 5 = 3. 3 3x5 11. Why is STT ^ = KT? ^ Explain by lines. 20-5-5 20 7 7 12. -T-a is how many times it^ — 5 , and why ? lo lo-5-b 8 3 13. Why is ^ greater than ^ — r ? Explain by lines. 1^1 EXAMPLES. 109 ILLUSTRATION OF PROCESS. 260. Prob. L — To multiply a fraction by an integer. 1. Multiply I by 7. Solution. —1. According to (258—1), multiplying the numerator, the denominator remaining the same, multiplies the fraction. Hence, 7 times :i8equalto^'*'' = ^ = 3S. - 2. According to (258—11), a fraction is also multiplied by dividing the denominator. Hence the following RULE. 201. Multiply the numerator of the fraction by th^ given integer^ or divide the denominator. . = ov = 3. EXAMPLES FOR PRACTICE. 202. Multiply orally the following, reduce the results to their lowest terms, and explain as above. 1. f X 3. 4. I X 9. 7. 2. I X 12. 5. A X 6. 8. ^ X 5. 3. I X 12. f X 4. I X 9. 6. f X 8. 9. \\ 14. Multiply the following and reduce. Cancel when possible. 10. yVn ^ 8. 13. Hf X 48. 16. T'xmr x 90. 11. 12. II X 9. T^ X 10. 14. tWt7 X 50. 15. ^"A X 100. 17. AViT X 100. 18. Ml X 75. 203. Prob. II — To find any given part of an integer. 1. Find ^ of $395. Solution.— 1. We find the \ of $395 by dividing it by 5. Hence the first step, $393 -«- 5 = $79. 2. Since $79 is 1 fifth of $395, four times $79 will be 4 fifths. Hence the second step, $79 x 4 = $316. To avoid fractions until the final result, wc multiply by the numerator first, then divide by the denominator ; hence the following RULE. 204. Divide by the denominator and multiply by the niimer- T¥ff 7 Tff* If 4. 5. 6. 89 V "4 5 X SOO 60 TTT* n. I! X H. 7. 8. 9. 3 7 4^ 1 00 lOOff 8477' TOO' Find the continued product of 10. ff . M, H, irW and li. 11- i»|, i 6» I. f. and^. 12. t. 2|, 3A, 5A, and Cxb- 13. 14. 15. \\h /rV. and f f • ^\\, \l, and /,V 31. 4L and 7^ '5» ^tf» 25. |;6| a cord, and 16. A lady bought 15 yards of silk at $2| a yard, and 7| yards of lace at $3| a yard ; what was the cost of both ? 17. What is the cost of 12 cords of wood at 8 tons of coal at $11| a ton ? Am. $171.50. 18. A province has an area of 37680 square miles, and the average population to a square mile, in 1870, was 35^*5 J ^^at is its population ? 19. What is the cost of 45^^ tons of iron, at $27 J per ton ? 20. At $3 1 a yard, what is the cost of 15g ? Of 32| ? 21. A merchant sold 12| yards of cloth at $2i a yard ; 28| at $1| ; and 52 J at |3| ; what did he receive for the whole ? 22. Find the product of f of 25 J, and y\ of l^A- 23. Find the product of 4| of f of 12, and 7f of 15. 24. Bought 19 pounds of butter at 23^ cents a pound, giving in return 27f pounds of lard at 15 cents a pound, and the rest in cash ; what did I give in cash ? ^n«. 20 cents. 25. C. Smart has two fields containing respectively 11 1 acres and 21^ acres ; how much hay will he take from both fields, at the rate of If tons an acre for the first, and 2f tons an acre for the second ? Ans. 69yV tons. 26. Find the valae of ($37f - $13f ) x (f of 8 - 2 J). 27. Find the value of (| of 8) — (* of 9 - 2^). 28. Find the value of 2| + 3| 84 4.1 ^ M i m 'V 114 FEA C TIO N S, DIVISION". ILLUSTRATION OF PROCESS. 276. Prob. I.— To divide a fraction by an integer. 1. Divide f by 4 ExPLAKATioN. — 1. According: to (258—111), a fraction is divid- ed by dividing the numerator. Hence we divide {J by 4, as shown in (1), by dividing the numerator 8 by 4. 2. According to (358— IV), a fraction is divided by multiplying the denominator. Hence we divide % by 4, as shown in (2), by multiplying the denominator by 4, and reducing the result to its lowest terms. Hence the following EXILE. ^ 277. Divide the numerator, or multiply the denominator^ by tlie given integer. (1.) 8 • 9 ' •4 8-4- ~9 4_ 2 9 8 8 8 ?, (2.) 9 • 4- ~9 X 4^ 36 ~ 9 EXAMPLES FOR PRACTICE. 278. Divide orally and explain the following. Dividing the numerator in every case where it can be done, in preference to multiplying the denominator. 1. If -^ 4. 3. H -^ 3. 5. A -- 4. 2. tf -f- 7. 4. f -*- 8. 6. T^if -*- 6. Perform the division in the following : 57. 9. tf H- 25. 7. 885 TS'S 11. 12. 1^^ - 50. 8. Ht -^ 32. 10. 11^ ^ 75. 13. If 7 yards of calico cost $f , what will 1 yard cost? 14. At $f for 4 boxes of figs, what will 1 box cost ? 15. Show that multiplying the denominator of f by 4, divides the fraction. Explain by lines. 16. The product of two numbers 1^ 149 1, and one of them is S3 ; what is the other ? EX A MP L E S, 115 17. If a compositor earns $45|^ in 18 days, how much does he earn in 1 day ? In 9 days V In 5 days 1 In 27 days ? Find the value 18. Of (f of f — xV) -5- 8- 20. Of (I of ^ - 2f ) -5- 12. 19. Of (i^ + I) -*- 32. 21. Of (f of 10| + if) -f- 32. 270. Prob. II. — To divide by a fraction. 1. How many times is f of a given line contained in twice the same line ? 2 lines ^^ \ z= FIRST STEP. 10 . ,. -p of a Ime. 5 If ^■B^H SECOND STEP. 3 I 10 5 3 - ^i ^J ,1^ — ^i ides Explanation.— 1. We can And how many times one number is con- tained in another, only when both are of the same denomination (144). Hence we first reduce, as shown in First Step, the 2 lines to 10 J{fths of a line ; the s&me fractianal denomination as the divisor, Sflffhs. 2. The 3 fifths in the divisor, as shown In Second Step, are contained in the 10 fifths in the dividend 3 times, and 1 part remaining, which makes \ of a time. Hence 2 equal lines contain I of one of them 3i times. Observe the following regarding this solution : (1.) The dividend is reduced to the same fractional denomination as the divisor by multiplying it by the denominator of the divisor ; and when reduced, the division is performed by dividing the numerator of the divi- dend by the numerator of the divisor. (2.) By inverting the terms of the divisor these two operations are expressed by the sign of multiplication. Thus, 2-t-3 = 2x », which means that 2 is to be multiplied by .5, and the product divided by 3. < m ii m I mmm 116 FJRA CTI0N8, 2. How many times is ) of a given line contained in f of it ? FIB8T STEP. 2 3 4 6 — 1 2 8 " 6 ■ ■ ■ -t ' ■- ~ 4 3 ~ 4 6 SXCOMD 8TSP. 3 -6 = Explanation.— 1. We reduce, as shown in First Step, the dividend | and the divisor | both to sixths (144—1). 3. Wo divide the J by | by dividing the numerator of the dividend by the numerator of the divisor. The f is contained in |, as shown in Second Step, 1| times. Hence \ is contained 1| times in §. 280. When dividing by a fraction we abbreviate the work by inverting the divisor, as follows : 1. In reducing the dividend and divisor to the same fractional unit, the product of the denominators is taken as the common denominator, and each numerator is multiplied by the denom- inator of the other fraction ; thus, 5 2 5x3 8x7 15 14 15 Numerator of dividend. 7* 3~7x3' 3x7~21 *31~14 Numerator of divisor. 2. By inverting the divisor, thus, f -«- f = f x f = ||, the numerators 15 and 14 are found at once, without going through the operation of finding the common denominator. Hence the following KULE. 281. Invert the terms of the divisor and proceed as in mid- iiplication. EXAMPLES, ii; lend, tor. the 3nce lUlr EXAMPLES FOR PRACTICE. 282. Solve orally the following and explain as above : 4 2. f + f 3. 8-*-#. 4. 13 -^ f. 5. t-*-f. 6. 9 -I- 1. 7. 90 -I- H- 8. H + f. 9. 200 4- iM. 10. 1 is how many times i ? i? ^? ^? \t \t 11. At $J a bushel, how many bushels of com c&n be bought for |9 ? Solution.— As many bushels as %{ is contained times In |9. 19 are equal to $Vi and $| is contained in $V, 10? times. Hence, etc. 12. At $5 a yard, how many yards of serge can be bought for|3? For $10? For $15? For $7? For $25? For $9? 13. For $12 how many poimds of tea can be bought at $f per pound? At$|? At$f? At$f? At$|? At$|? 6 9 3 9 7 «; ? 10 ' IT 14. 5 are how many times f ? 15. If f of a ton of coal cost $3, what will 1 ton cost ? Solution.— Since J of a ton cost $3, \ will cost \ of $3, or f J, and 1 ton, or I, will cost 9 times $|, or $V, equal to $6|. Hence, etc. Or, 1 ton will cost as many times $3 as | of a ton is contained times in 1 ton, 1 ton + 5 = s = gi. Hence, 1 ton will cost 2i times $3, or $6*. 16. At $1 for I of a pound of tea, what is the cost of 1 pound ? Of 7 pounds ? Of y»j of a pound ? Of | pounds ? 17. If |- of a cord of wood cost $4, what will 1 cord cost ? 4 cords ? 11 cords ? | of a cord ? ^^ of a cord 1 18. How many bushels of wheat can be bought for $8, if f of a bushel cost $| ? If f of a bushel cost $^5 ? 19. At $1 a yard, how much cloth can be bought for %f^ ? Solution.— As many yards as %\ is contained times in $r"u. %l equals $^, and %^ is contained 1^ times in %^a- Hence, etc. 20. At $yV 8- bushel, how many bushels of potatoes can be bought for %l ? For $|| ? For $f ? For $^ ? For $| ? 21. How many pounds of sugar at $|^ can be bought for $|? For$|? For$H? For$||? For$|? 5 VH ii 118 PlfA CTIO NS. "?«; 22. If 3'^ of a yard of cloth can be bought for $ i\, how much will 1 yard cost ? 5 yards ? 7| yardw ? 23. Geo. Graham expended f of $480 in purchasing tea at $f per pound, and the balance in purchasing coffee at %l i^er pound. How many pounds did he buy of each ? Perform the division in the follo^ving. Invert the divisor and cancel common factors. (175.) 24. U + 11- 29. N6 .85 ^iiT — if' 34. KH + m- 25. .»!• -*- M- 30. 573 -5- T^jf. 35. 11 -*- iM. 26. If + M- 31. 862 -f- f 4. 36. 1000 ^ m- 27. m + n. 32. 100 . 46 37. 3000 ^ f «t. 28. 324 -J- f . 33. 573 -f- Ii 38. m -*- If. 283. Prob. III. — To divide when the divisor or divi- dend is a mixed number, or both. 1. Divide 48 by 4f . ' t^' (1.) 48 -f- 4f - 48 -^ V- ExPLANATioN.-l. We re- duce the divisor 4^ as shown in (1), to the improper ft-ac- (2.) 48-4- V = 48 X A = 10? 2. We invert the divisor, as shown in (2), according to (280), and mul- tiply the 48 by /„ giving 10? as the quotient of 48 divided by 4?. 2. Divide 8? by 3|. (1.) 8? -*- ^ = \^ (2.) ¥ -^ V = s. '^ = i^ = 2f . Explanation.— 1. We reduce the dividend and divisor, as shown in (1), to improper fractions, giving V + V. 2. We invert the divisor, Vi as shown in (2), according to (280), and cancel 31 in the numerator 62 and denominator 81 (1 76), giving V, or ^. Hence, 8? -•- 3| = 2?. From these illustrations we obtain the following BULE. 284. Reduce mixed numbers to improper fractions ; then invert the divisor and proceed as in multiplication, cancelling any factors that are common to any numerator and a denom- inator. EXAMPLES, 119 mul- -l.We Indand in(l), ;tlons. )), and WRITTEN EXAMPLES. 285* Perform and explain the division in the following : 1. 7«-f-2t. 8. 732-i-14f. 15. 5,«„»a-^2THff. 2. 2A-*-4J. , 9. 36^-^8^. 16. 873-*-^?. 3. 9i'ir-*-5|. 10. 85,\-J-23. 17. 302-f-/„V 4 89^-7|. 11. 37t\j-6tV 18. 5. 8624-421. 12. lOOOA-f-T^B-. 19. H of 15? 4-5. 6. 43TVa-^Ti?in7. 13. '^Tun'^^TTnjTr' 20. f,of5^-^^ 7. 100iV-*-5^. 14. 936+5x§T,. 21. § of 28 -f-^ oft. 22. At $f for ij^^ of an acre of land, what is the cost of 1 acre ? Of ViT of an acre ? Of ^ of an acre ? Of 29^ of an acre ? 23. If a bushel of wheat cost $1§, how much can bo bought for$12|? For$28«? For$273|? 24. Jas. Johnston expended $597| in buying cloth at $2f a yard. He afterwards sold the whole of it at $3f a yard ? How much did he gain by the transaction ? COMPLEX FKACTIOFS. 28G. Certain results are obtained by dividing the numera- tor and denominator of a fraction by a number that is not an exact divisor of each, which are fractional in form, but are not fractions according to the definition of a fraction. These frac- tional forms are called Complex Fractions. The following examples, which illustrate the three classes of complex fractions, should be carefully studied : 2- Ex. 1. Show that y\ of a line is equal to -j of the same line. then 1 eUing 1 tnom- 1 8 4-3 12-^3 2| ~ 4 1 HHH MMH HBB HI ■ ■ ^^ mmm w^mm m^ m ti ■i i 120 COMPLEX FRACTIONS, Explanation.— 1. Dividing tho uumorator and deiiotuiiintor of ,% by 8 makcH every 3 parts in each into 1 part, as nbuwu in tlio illuutratlou, but doc8 not change tho value of the fraction (22-1-III). 3. The denominator or whole contiiinfei 4 of tbeHu partn, and the uumcra- tor 2 of them and \ of one of them, ati will be seen by the illuBtration. 23 Benco, {^ of a line itj equal to ~ of the same line. 4 Ex. 2. Show that /j of a line is expressed by — . 13 5 5 2| Explanation.— 1. Dividing the numerator and denominator of A ^7 5 makes every 5 parts in each into 1 part, as shown in the illustration. 2. The denominator or whole contains two of these parts uud } of one of them, and the numerator contains 1 part, as shown in the illustration. Hence, t'i of a line is represented by the fractional expression — . Ex. 3. To show that \% of a line is expressed by 8f lO-i-4 13 -i- 4 1$ , Explanation.— 1. Dividing the numerator and denominator of \% by 4 makes every U parts in each into Ipart^ as shown in the illustration. 2. The denominator or whole contains 3 of these parts and \ of one of them, and the numerator contains 2 of them and \ or { of one of them. 2i Hence, \% of a line is represented by the fractional expression 3i- From these illustrations we have the following definitions: 287. A Comj)l€X Fraction is an expression in the form of a fraction, having a fraction in its numerator or denominator, ^ 6f r 4 or in both ; thus, ^, ^, • Of COMPLEX FRACTIONS, 121 288. A Slmjde Fraction is a fraction having a whole number lor its numerator and for its denominator. PROBLEMS IN COMPLEX FRACTIONS. 289. Pros. I.— To reduce a complex fraction to a sim- ple fraction. 4ff Reduce =v to a simple fraction. 7f 4^ 4| X 12 56 Explanation.— 1. Wo find the least com- 7? ~ 7? X 13 ~ 93 ™°° multiple of the denominatoru of the partial fractions | and J, which i» 12. 3. Multiplying both terms of the complex fraction by 12 (23'i— II), which is divisible by the deuominators of the partial fractions, % and J, reduces each term to a whole number. 4] x 12 = 56 ; 7i x 12 = 93. Thcrc- 41 fore jl| Is equal to the simple ft-action 1%. Hence the following \% by 4 one of them. ms: form Lnator, EULE. 290. Multiply both terms of the complex fraction by the least common multiple qfaU the denominators of the partial fractions. 291. The three classes of complex fraction.s are forms of expressing three cases of division ; thus, (1.) ^ = 5|-*-7. (2.) ^ = 32-j-9f A mixed number divided by an integer. An Integer divided by a mixed number. 8f- 2| (^•) of ~ 8f "*"2|. A mixed number divided by a mixed number. Hence, when we reduce a complex fraction to a simple frac- tion, as directed (290), we in fact reduce the dividend and divisor to a common denominator, and reject the denominator by indicating the division of the numerator of the dividend by the numerator of the divisor ; thus, 9 I !'>{ It;-- X-f. w iW 122 FRACTIONS. !■ .'{ ^ \ t 1 it , .; ! ; %-^ r\. (1.) ^ (2.) 5f s Of 5fxl2 69 ,. ^ /o«ON = sl .. io = oT, . according to (289). 2| X 12 32 -21 = Y- ^ S. and -Y- - f = f I .13 f I, the game result as obtained by tbe method of muUiplying by the least common multiple of the denominators of the partial fractions. EXAMPLES FOR PRACTICE. C , 1292. Reduce to simple fractions, and explain as above : 1. 2. 13| 16f 13^ 23tVi7 5 1?I?I, 32^ 6. When the numerator or denominator contains two or more terms connected by a sign, perform the operation indicated by the sign first, then reduce to a simple fraction. -» Reduce the following to simple fractions : 3^ ^ ^ 8. 9. «l + 5f 4J -2i_ 6? -n (SI— ^) X 2 5 (O X I) + (I of I) 10. 11. 12. (^ of 9) + it of 2) fof5 (? o^ f ) - A (22 of 2) - ^j, 1000 293. Prob. II. — To reduce a fraction to any given de- nominator. 1. Examples where the denominator of the required fraction is a factor of the denominator of the give7i fraction. Reduce |f to a fraction whose denominator ii 8. • 17_ 24 ~ fraction. 17 24 Explanation.— We observe that 8, the denominator of the required fraction, is a factor of 24, the denominator of the given Hence, dividing both terms of J{ by 3, the other factor of 24, the 3_5| 3~ 8" 53 fraction is reduced (224—111) to -~, a fraction whose denominator is 8. o 3. 4. 5. COMPLEX FRACTIONS, 123 2. Examples wTicre the denominator of the nqni red fraction is not a factor of the denominator of the given fraction. Keduce j^j to a fraction whose denominator is 10. 8 _ 8 X 10 _ 80 ^ ^^ 13 ~ 13 X 10 ~ 130 80 -T- 13 6fs (2) ^^- ^ '' 130 130 13 10 Explanation. — 1. Wc intro- tliicc the given denominator 10 as a factor into the denominator of Vj by multii^ying, a» t^hown in (1), both terms of the fraction by 10(224-11). 2. The denominator 130 now contains the factors 13 and 10. Hence, dividing both terms of the fraction ^''s'i, by 13 (224— III), as t^hown in (2), 6 * the result is ", a fraction whoso denominator is 10. From these examples we obtain the following RULE. jnde- Hion is 8, the m, is a [e given 24, the Is 8. 294. Mvltiply both terms of the fraction by the given denom- inator , and then divide them by the denominator of the fraction. Observe that when the given denominator is a factor or multiple of the denominator of the fraction, it is not necessary to multiply by it, as will be seen in the first example above. EXAMPLES FOR PRACTICE. 295. 1. Reduce ^, ^\, -j?, ||, !|^, and fg each to thirdt. 2. How many sevenths in f[ ? In ^ ? In ^ ? 3. In If how many twentieths? How many ninths? 4. 5. Reduce |, i;, Reduce \, J, 10' 4 »' and \ each to sevenths. and A to hundredths. 6. Express as hundredtlis 7. How many tenths in % ? In 8. How many thousondths in ^? 9. How manv hundredths in 5 8 7 8 7 •1.V and m. 13 ? In 17? In t'i In 10 Tn '> '1 9 in ^^ ,f T In-,^? In In AS? In In 4^? s 1^.' H? \V^ In m In7f? In9f? 10. Reduce to hundredths ? 43 . UITT' 11. Reduce to hundredths 2|. 7' 5^ 4 I J Q4 t 8" 40A 3* 05 i^ iii \- ■ m 124 FRA C T I NS, REVIEW EXAMPLES. P! 290. 1. Reduce f , ^, -r^, ^^, and || each to twenty-eighths. 2. How many thirtieths in f , and why ? In | ? 3. Reduce to a common denominator — , ^, and tV. 4. State the reason why 5x4 (258). 9-5-4 9 5. Redux i to a fraction whose numerator is 12 ; is 20 ; is 2 ; is 3; is 7 (224). 6. Find the sum of |, y^, J, J, and | J. 7. Reduce to a common numerator f and f (241). 8. Find the value of (5 of A - iV) ^ (| + 3t)- 9. If ^ of an estate is worth $3460, what is i of it worth ? . 10. $4 is what part of $8 ? Of $12 ? Of $32 ? Of $48 ? Write the f?olution of this example, with reason for each step. 11. If a man can travel a certain distance in 150 days, what part of it can he travel in 5 days ? In 15 days ? In 25 days ? In 7^ days ? In ^ days? In 12^ days ? 12. A's farm contains 120 acres and B's 280 ; what part of B's farm is A's ? Ans. |. 13. 42 is 7^ of what number ? Write the solution of this example, with reason for each step. 14. $897 is I of how many dollars? , r 15. 5 of 76 tons of coal is y^^ of how many tons ? 16. A piece of silk containing 73 yards is I of another piece. How many yards in the latter ? 17. 84 is {^ of 8 times what number ? Write the solution of this example, with reason for each step. 18. Bought a carriage for $286, and sold it for ^ of what it cost ; how much did I lose ? 1^! EX A MP L ES. 125 trt Ice. it 19. A lias $G94 in a bank, which is i of 3 times the amount B has in the same bank ; what is B's money ? 20. Two men are 86J miles apart ; when they meet, one has travelled SJ miles more than the other ; how far has each travelled V 21. If -/^ of a farm is valued at $4T32|, what is the value of the whole farm ? 22. The less of two numbers is 432|, and their diflFerence 123 iV . Find the f^reater number. 23. A man owning J of a lot, sold f of his share for $2800 ; what was the value of the lot? 24. What number diminished by f and | of itself leaves a remainder of 32 ? A us. 504. 25. I sent j of my money to Quebec and sent ^ of what I had left to Gait, and had still remaining $400. How much had I at first? Afi8. $1800. 26. Sold 342 bushels of wheat at $lf a bushel, and expended the amount received in buying wood at $4J a cord. How many cords of wood did I purchase? A718. 1232 cords. 27. If 5 be added to both terms of the fraction f?, how much will its value be changed, and why ? 28. If I of 4 pounds of tea cost $2i, how many pounds of tea can be bought for $7i ? For $12f ? " For $f ^ ? 29. I exchanged 47J bushels of corn, at $5 per bushel, for 24| bushels of wheat ; how much did the wheat cost a bushel ? 30. Bought f of S^ acres of land for ^ of $3584; ; \vhat was the price per acre ? 31. A can do a piece of work in 5 days, B can do the same work in 7 days ; in what time can both together do it ? 32. A fisherman lost f of his line ; he then added 8 feet, which was f of what he lost ; what was the length of the line at first? Anf). 15 feet. 33. C. Poison bought cloth to the value of $2849?, and sold it for y'^ of what it cost him, tliereby losing $5 a yard ? How many yards did he purchase, and at what price per yard ? I hi iiii .'*i *H !*ta I ili K.n 126 FRA CTIO NS. - 84. A tailor having 276| yards of cloth, sold f of it at one time and f at another ; what is the value of the remainder at $3 a yard ? 35. A man sold -^^ of his farm at one time, f at another, and the remainder for $180 at $45 an acre ; how many acres were there in the farm ? 36, A merchant owning i| of a ship, sells ^ of his share to B, and f of the remainder to C for $000^ ; what is the value of the ship? BEVIEW AND TEST QUESTIONS. 297. 1. Define Fractional Unit, Numerator, Denominator, Improper Fraction, Reduction, Lowest Terms, Simple Fraction^ Common Denominator, and Complex Fraction. 2. What is meant by the unit of a fraction ? Elustrate by an example. 3. When may \ be greater than \ ? \ than \1 4. State the three principles of Reduction of Fractions, and illustrate each by lines. 5. Illustrate with lines or objects each of the following propositions: . : v' I. To diminish the numerator, the denominator remaining the same, diminishes the value of the fraction. II, To increase the denominator, the numerator remaining the same, diminishes the value of the fraction. III. To increase the numerator, the denominator remaining the same, increases the value of the fraction. IV. To diminish the denominator, the numerator remaining the same, increases the value of the fraction. 6. What is meant by the Least Common Denominator ? 7. When the denoi linators of the given fraction are prime to each other, how is the I^east Common Denominator found, and why ? RE VIEW. 127 8. State the five problems in reduction of fractions, and illus- trate each by the use of lines or objects. 9. Show that multii)lying the denominator of a fraction by any number divides the fraction by that numljer (258). 10. Show by the use of lines or objects tlie truth of the following: ^ing ing to id. (1.) ^ of 2 equals f of 1. (3.) I of 1 equals ^ of 3. (3.) I of 5 equals | of 1. (4.) ^ of 9 equals 4 times \ of 9. 11. To give to another person f of 14 silver dollars, how many of the dollar-pieces must you change, and what is the largest denomination of change you can use ? 13. Show by the use of objects that the quotient of 1 divided by a fraction is the given fraction inverted. 13. Why is it impossible to perform the operation in | + f , or in ^ + f , without reducing the fractions to a common denominator? 14. Why do we invert the divisor when dividing by a frac- tion ? Illustrate your answer by an example. 5 15. What objection to calling — a fraction (21*>)? 16. State, and illustrate with lines or objects, each of the three classes of so-caUed Complex Fractions. 17. Which is the greater fraction, 5 or ^}l, and how much ? 18. To compare the value of two or more fractions, what must be done with them, and why ? 34 4 2i 3 5^ 2? 19. Compare -^ and ^ ; -j^ and ^h 5 '?rl ^^^ 51 » »^d show in 7 o y 10 I* oi each case which is the greater fraction, and how much. 30. State the rule for working each of the following examples: (1.) 3| + 4f + 8|. (3.) (7| + 5^)-(8-3i). (3.) 5 X I of I of 27. (4.)8?x5| (5.) V ^ ?• (^•) I'^a"'"^' Explain by objects. Explain by objects. 21. Illustrate by an example the application of Cancellation in multiplication and division of fractions. 1 t I' i-.y DECIMAL FRACTIONS. DEFINITIONS. 298. A unit is separated into dec'unal parts when it is divide.' ij> ' mthd; thus, f <(i UNIT. DECIMAL FARTS. 399. A T}(hnal FractionaZ Unit is one of the decirnalpa.'hsui ^.'-^^Jning. 300. By makiLg ?, whole or unit into decimal parts, and one Gi these jare into ' 'loal parts, and so on, we obtain a series ot dist'/iict ct Ir .^ v/-. • ■•• or .537 may be read 5 tent/is 3 hundredths and 7 thou^ sandths. Analyze the following ; thus, .0709 = y^ + yi^f ciy. 1. .046. 2. .909. ^' 3. .0027. 4. .01207. 5. .04063. 6. .05095. 7. .0300702. 8. .0003092. 9. .0060409. s 't !' at 15 lO. Prop. V. — A decimal is read correctly by reading it as if it were an integer and giving the name of the right-hand order. 900 4. 70 I 6 TSITQ- Hence is read, nine hun- Thus, .975 — xwQ^ "^ ToiTTy dred seventy-five thousandths. 1. Observe that when there are ciphers at the left of the decimal, according to (305), they are not regarded in reading the number ; thus, .002 is read sixty-two thousandt?is. 2. The name of the lowest order is found, according to (303), by prefixing 1 to as many ciphers as there are figures in the decimal. For example, in .00209 there are five figures ; hence the denominator is 1 with five ciphers ; thus, 10CK)00, read hundred-thousandths. From these illustrations we obtain the following RULE. 311. Read the decimal as a ichole number ; then pronounce the name of the lowest or right-hand order. 132 DECIMAL FRACTIONS. Read the following: 1. .004. 2. .00902. 3. .4097. 4. .0C419. 5. .02006. 6. .30007. 7. .5307. 8. .01007. 9. .0058. 10. .000904. 11. .040972. 12. .000252. 13. .020304. 14. .0090409. 15. .00030503. 16. .00200059. 17. .0000007. 18. .00034657. 19. What is the denominator of .000407 ? How is it found ? What is the numerator, and how read ? 20. What effect have the ciphers in .0085 ? 21. Write a rule for expressing a decimal by writing its denominator. 312. The relation of the orders of units in an integer and decimal will be seen from the following table : ♦ DECIHAL NUMERATION TABLE. 09 o 5 CD a o S P 5 5 a a m o I « (X a s H 5 s S3 O I a 5 o h 5 -a a o K 5 n -o i • J3 a ai 6 T! • J3 • a o a o 5 e e es tt. S o .a •^^ c c .a *^ o 'a m MM .2 S a ;^ P H HH H H EC Eh 5 • 5 5 5 5 5 5 5 CO g 'F^ (1> 3 a e 5 5 ORDERS OP INTEGERS. Y ' ORDERS or DECUfAL FRACTIONS. Observe carefully the following : {-£ 1. The ZTnit is the standard in both cases. The integral orders are multiples of one iiiiit^ and the decimal orders are decimal fractions of one unit. 2. Figures that are equally distant from the units' place on the right or left, have corresponding names ; thus, tenths cor- respond to tens, hundredths to hundreds, and so on. IS. ral ire EXAMPLES, 133 3. In reading an integer and decimal together, " and " should not be lifted anywhere but between the integer and fraction. Thus, 0582.643 should be read, nine thousand five hundred eighty-two and six hundred forty-three thousandths. 4. Dimes, cents, and mUls being respectively tenths, hun- dredths, and thousandths of a dollar, are written as a decimal. Thus, $.347 is 3 dimes, 4 cents, and 7 mills. In reading dimes, cents, and mills, the dimes are read as cents. Thus, $62,538 is read, 62 dollars, 53 cents, 8 mills. EXAMPLES FOR PRACTICE. 313. Read the following : 1. $285.56. r 2. $920,905. 3. $203.06. 4. $70,007. 5. $300.02. 6. $9,807. 19. 20. 21. 5 ruo- I^TiHnsTr* 7. 20040.20104. m 9.00006. 8. 90309.00703. 14. 10.1. 9. 3001.0201. 15. 100.0003. 10. 50400.000205. 16. 35.00045. 11. 2070.00301. 17. 9.30005. 12. $9005.009. 18. 10.000001. g without writing the denominator : 23. 407Tiy^*Wij. 25. 10201,^^71. 23. 703^«^%^. 26. 4030,|UiT- 24. 9Y7y5^J^. 27. lOOjjjijj^^^ny' 28. Write with figures : Eighty-two thousandths ; four hun- dred five millionths ; eight ten- thousandths. 29. Three thousand six hundred-millionths ; ninety-one millionths ; six hundred four thousand one billionths. 30. Eighty.four and seven ten-thousandths; nine thousand six and five hundred nine ten-millionths ; six and five mil- lionths. 31. Nine thousand thirty-seven and three hundred seven billionths ; one million one and one thousand one ten-millionths. ■'V ■ 'I V '1 r t Ml '. -i.t m 134 DECUfAL FRACTIONS* EEDUCTIOJSr. PREPARATORY PROPOSITIONS. The following preparatory proposi 'ions Bliould be very eare- fvlly studied. 314. Prop. \.— Annexing a cipher or multiplying a num- ber by 10 introduces into the number the two prime factors 2 and 5. Thus, 10 being equal 2 x 5, 7 x 10 or 70 = 7 x (2 x 5). Hence a number must contain 2 and 5 as a factor at least as many times as there are ciphers annexed. 315. ^ROP. II. — A fraction in its lowest terms, whose de- nominat&r contains no other pnme factors than 2 or 5, can be reduced to a simple decimal. Observe that every cipher annexed to the numerator and denominator makes each divisible once by 2 and 5 (314). Hence, if the denominator of the given fraction contains no other factors except 2 and 5, by annexing ciphers the numera- tor can be made divisible by the denominator, and the fraction reduced to a decimal. Thus, | = U^^ (224—11). Dividing both tci-ms of the fraction by 8 (224— III), we have l^^ = y«^Vj = .875. r 1 1. 2. 3. Reduce to decimals and explain as above : ¥• ^ TU' 7. f. 4. 1^. 7. 5. H. 8. 6. A. 9. ^V r¥ir. 10. Ih 13 11. A. 14 12. #2. 15 49 m- 1 s 1 6. How many ciphers must be annexed to the numerator and denominator of f to reduce it to a decimal ? 17. Reduce -l to a decimal, and explain why the decimal must contain three places. 18. If reduced to a decimal, how many decimal places will | make ? Will ^'V make ? Will -^^ make, and why ? '> the Iff* UtOT lal HI PRKPARATORY P R O P S ITI XS . 135 511 0, Prop. III.— ^ frnrtion in its lowest terms, whose de* nominator contdina any other prime factors ttiau 4 or 5 can be reduced only to a complex decimal. Observe that in this case annexing ciphers to the numerator and denominator, wliich (;J14) introduces only the factors 2 and 5, carmot make the numerator divisible by the giv»'n de- nominator, which contains other prime factors than 2 or 5. Hence, a fraction will remain in the numerator, after divid- ing the numerator and denominator by the denominator of the given fraction, however far the division may be carried. Thus, \\ = iUSg (224—11). Di\'iding both numerator and denominator by 21, we have ^[^ = ^^ = .523H, a com- plex decimal. Reduce and explain the following : 1 . How many tenths in f V In f ? In f ? In | ? In {'^ ? 2. Reduce to hundredths f ; 5 ; ^\; ^7^ ; ^1^ ; IS- 3. How many thousandths in ^ ? In | ? Tn ^ ? In j% ? J51 7. Pkop. IV. — The same set of figures must recur indeji- niteJy in the same order in a co'mpi r decimal ichich cannot he reduced to a simple decimal. Thus ^~ - ^^^^ - ^'^^^'^ - G363 V ihus, ^^ _ j^^^^ _ ^^^^ - .WbdfV- Observe carefully the following : 1. In any division, the number of different remainders that can occur is 1 less than the number of units in the divisor. Thus, if 5 is the divisor, 4 must be the greatest remainder we can have, and 4, 3, 2, and 1 are the only possible different remainders ; hence, if the division is continued, any one of these remainders may recur. 2. Since in dividing the numerator by the denominator of the given fraction, each partial dividend is formed by annex- ing a cipher to the remainder of the previous division, when a S'-» w I If ; ^ ' 136 DECIMAL FsRACTIOXS, remainder recurs the partial dividend must again be the same as was used when this remainder occurred before ; hence the same remainders and quotient figures must recur in the same order as at first. 3. If we stop the division at any point where the given numerator recurs as a remainder, we have the same fraction remaining in the numerator of the decimal as the fraction from which the decimal is derived. Thus, or ZTT = 7 11 11 ' 70000 700 63t'i 1100 1000 " = .63T'r ; 110000 = 10000 = •^^^^'^' ^'^^ ^ ^"- Jl'18. Prop. V. — The 'value of a fraction which can only he reduced to a complex decimal is expressed, nearly, as a simple decimal, by rejecting the fraction from the numerator. 3 27 '- Thus, — = ~~ (316). Rejecting the ^ from the numer- ator, we have ^^^, a simple fraction, which is only y\ of ^J^ 27A smaller than the given fraction ^ or -jk^^ Observe the following : 1. By taking a suflBcient number of places in the decimal, the true value of a complex decimal can be expressed so nearly that what is rejected is of no consequence. Thus, - = -^yoi^QQQ^; rejecting the -^j from the numer- ator, we have xVinHrffffVff' o' .27272727, a simple decimal, which is only ^ of 1 hundred-millionths smaller than the given fraction. 3. The approximate value of a complex decimal which is expressed by rejecting the given fraction from its numerator is c illed a Circultding Decimal, because the same figure or set of figures constantly recur. Ihe lat DEFINITIONS. 137 319. Prop. VI. — Diminishing the numerator and denom- inator by the same fractional part of each does not change the value of a fraction. Be particular to .master tLe followingf, as the reduction of circulating decimals to common fractions depends upon this proposition. 1. The truth of the proposition may be shown thus : 9^_^_tof9_9^-3_6_3 12 ~ 13 — i of 12 ~ 12-4"" 8 ~ 4* Observe that to diminish the numerator and denominator each by | of itself is the same as multiplying each by f . But to multiply each by f , we multiply each by 2 (t224— II), and then divide each by 3 (224 — III), which does not change the value of the fraction ; hence the truth of the proi>osition. 2. From this proposition it follows that the value of a frac- tion is not changred by subtracting 1 from the denominator and the fraction itself from the numerator. Thus, ^ = I ! = ^^ Observe 5 — 1 4 that 1 is the 1 of the denominator 5, and | is | of the numerator 3 ; h«nce, the numerator and denominator being each diminished by the same fractional part, the value of the fraction is not changed. |i a. le lis is of DEFINITIONS. 320. A SimjUe Decimal is a decimal whose numerator is a whole number ; thus, ^Yff or .93. Simple decimals are alno called Finite Decimals. 321. A Complex Decimal is a decimal whose numer- 26* ator is a mixed number: as -— or .263. 100 ^ 10 138 DECIMAL FRACTIONS. There are two classes of complex decimals : 1. Those whose valae can be expressed as a simple decimal (315), as .aaj = .235 ; .32: = .8275. 2. Those whose value cannot be expressed as a simple decimal (316), as .53J = .58333 and bo on, leaving, however flar we may carry the decimal places, i of 1 of the lowest order unexpressed. See (3 1 7). 322. A Circulating Decimal is en approximate value for a complex decimal which cannot be reduced to a simple decimal. Thus, .666 is an approximate value for .6662 (318). 323. A Repetend is the figure or set of figures that are repeated in a circulating decimal. 324. A Clrculafiuff Dechnal is exjtressed hy writ- ing the repetend once. When the repetend consists of one figure, a point is placed over it ; when of more than one figure, points are placed over the first and last figures ; thus, .333 and so on, and .592592+ are written .3 and .592. 325. A Pure Circulating Decimal is one which commences with a repetend, as .8 or .394. 326. A Mixed Circulating Decimal is one in which the repetend is preceded hy one or more decimal places, called the finite part of the decimal, as .73 or .004725, in which .7 or .004 is the finite part. ILLUSTRATION OF PROCESS. 327. PuoB. I.— To reduce a common fraction to a decimal. Reduce I to a decimal. 3 8 3000 800() 375 1000 = .375. Explanation. — 1. We annex the same number of ciphers to both terms of the fraction (224— Prin. II), and divide the resulting terms by 8, the eigniflcant figure In ..^-.3;.T,>»W':-,VWr*iJ- EXE R CISE, 139 the denominator which must give a decimal denominator. Hence, » expressed decimally if .375.' 2. In case annexing ciphers does not make the numerator divisible (316) by the tiisnificant flguree in the denominator, the number of places in the decimal can be extended indefinitely. In practice, we abbreviate the work by annexing the ciphers to the numerator only, and dividing by the denominator of the given fraction, pointing off as many decimal places in the result as there were ciphers annexed. Hence the following Lich lied or a RULE. 328. / Annex ciphers to the numerator and divide by the denominator. II. Point off as many places in th^ result as there are ciphers annexed. EXERCISE FOR PRACTICE. 329. Reoixce to simple decimals; 813 ml' S7 1. If. 8. /j- 5. II. 7. 2 7 4 27 ft «» ft •'*'"* Reduce to a complex decimal of four decimal places: ft B 11 23 1.^ !« 15 10 11 19 13 14 ,** 16 Find the repetend or approximate value of the following : 17. if. 20. II. 23. 11. 2C. 8?. 18. 2. 21. U. 24. 1%. 27. ^{^j^. 19. i|. 22. y. 25. II. 28. Ui^. 330. Prob. II.— To reduce a simple decimal to a com- mon fraction. Reduce .35 to a common fraction. 86 7 ExPLANATKN.—Wc wHte thc decimal with the dcnominiiK r, and reduce the fraction (244) to ite lo\vei*t tt-im** ; hence the following .35 = 100 20 1 lex to 14- In RULE. 331. Express the decimal by irHting the denominaior, then reduce the fraction to its lowest terms. r r f,*- : H 140 DECIMAL FB ACTIONS. mt EXAMPLES FOR PRACTICE. 332. Reduce to common fractions in their lowest terms: 1. .840. 2. .215. 3. .750. 4. .0125. 5. .0054. 6. .0064. 7. .008025. 8. .00096. 9. .00075. 10. .00512. 11. .0625. 12. .00832. 333. Pros. III.— To find the true value of a pure cir- culating decimal. Find the true value of .72. 72 _ _ 7^ _ _ 72 _ 8 100 ~ 100 - 1 ~" 99 ~ 11 7'2 — .1^ — ExPLAKATioN.— In taking ,72 as the approximate value of a given fraction, we have subtracted the given fraction from its own numerator, as* shown in (318— V). Hence, to find the true vahie of t'u\, we must, according to (319— VI, 2), subtract 1 IVom the denominator 100, which makes the denominator as many 9'8 as there are places in the repetend ; hence the following RULE. 334r. Write the figures in the repetend for the numerator of the fraction, and as many 9's as there are places in the repe- tend for the denominator, and reduce the fraction to its lowest terms. EXAMPLES FOR PRACTICE. 335* Find the true value of 1. 78. 4. 856. 7. 324. 10. 2718. 2. 36. 5. 372. 8. i89. 11. 5368. 3. 54. 6. 135. 9. 836. 12. 8163. Find tlie true value as improper fractions of 13. 37.8i. 16. 53.324. 19. 29.i88i. 14. 9.i08. 17. 89.54. 20. 63.2745. 15. 3.504. 18. 23.758. 21. 6.636. EXAMPLES. 141 (1) .318 r= .3J?j 336. PRO" IV.— To find the true value of a mixed cir- culating decimal. Find the true value of SiS. _ 3J2 _ 315 ^ 7 »0«j - 10 - 990 22 Explanation.— 1. We find, according to (333) the true value of the repetend .Ols, which Is .OiJ. Annexing this to the .3, the finite part, we have .31si the true value of .Sis in the form of a complex decimal. gin 2. We reduce the complex decimal .3JJ, or '^* to a eimple fraction hy multiplying, according to (289), both terms of the fraction hy 99, giving ^"* - {J5 = ,^,. Hence the true value of .3*18 is ^V 3'* (2) .318 Given decimal. 3 Finite part. Abbreviated Solution.— Obsenr'e 3iS 315 m ■h' that in simplifying *^*, wc multiplied both terms by 9ft. Instead of multi- plying the 3 by 9ft, wo may multiply by 100 and subtract 8 from the product. Hence we add the 18 to .300, and subtract 3 from the resiUt, which gives us the true numerator. Hence the following RULE. 337. I. Find the true value of the repetend, annex it to the finite imH, and redurs the complex decimal thus formed to a simple fraction. To abbreviate the work : //. From the given decimal subtract the finite part for a numerator, and for a dcnominatoi' icrite as many O's as there are figures in the repetend, with as many ciphers annexed as there are figures in the finite 2)art. \\ ;:» ■r EXAMPLES FOR PRACTICE. 338. Find the true value of 1. .959. 4. .00641. 7. .008302685. 2. .7i2. 5. .04328. 8. .000035739. 3. .486. 6. .03287. 9. .020734827. 143 D E CI MA L Fit ACTIO AS, [:' '• Find the true value, in tlie form of an improper fraction, of 10. 9.753. 12. 7.86. 14. 5.39. 11 5.328. 13. 2.43. 15. 12.227. ADDITIOjN^. PREPARATORY PROPOSITION. J^JJl). Any two or inore decimals can he reduced to a common denomiiuitor hy annexing ciphers. Thus, .7 = 3^. a"^> according to (224—11), ^'^ = ^^^ = ^'^ooj = I'oOflVff. and so on ; therefore, .7 = .70 = .700 = .7000. Hence any two or more decimals can be changed at once to the same decimal denominator by annexing ciphers. ILLUSTRATION OF PROCESS. ;54:0. Find the sum of 34.8, 6.037, and 27.62. Explanation.— 1. We arran^ the num- bers BO that units of the same order stand In the same column. 2. We reduce the decimals to a common denominator, as shown in (1), by annexing ciphers. .3. We add as in integers, placing the decimal point before the tenths in the t^um. In practice, the ciphers are omitted, as shown in (2), but the decimals are re^rded as reduced to a common denominator. Thus the 3 hundredths in the second number and the 2 hundredths in the third, when added are written, as shown in (2>, as 50 thousandths ; in the same manner, the 8 tetiths and 6 tentfis make 1400 thousandths, or 1 unit and 400 thousandths. The 1 unites added to the units and the 4 written in the tenths' place as 400 thoii^atidfhs. From this it will be seen that the addition of decimals is subject to the same laws (250 — I and II) and rule (252) as other fractions. (1.) (2.) 34800 348 6.037 6.03T 27.620 27.62 68.457 68.457 I 1 S UBTRA CTION, 143 EXAMPLES FOR PRACTICE. 341. Find the sum of the following, and explain as above 1. 9.07, 36.000, 84.9, 5.0036, 23.608, and .375. 2. 38.9. 7.05, 59.82, 365.007, 93.096,' and 8.504. 3. $42.08, $9.70, $89.57, ^396.02, and $.89. 4. 395.3, 4.0701, 9.96, and 83.0897. 5. .039, 73.5, .0407, 2.602, and 29.8. 6. 8.0093, .805, .03409, 7.69, and .0839. 7. .80003, 3.09. 13.36. 97.005, and .9999. 8. $.87, $32.05. $9. $75.09, $.67, and $3.43. SUBTRACTIO]^. 343. Find the difference between 83.7 and 46.392. 83.700 45.392 38.308 Explanation.— 1. We arrun<,'e the numbers so that units of the same order stand in the same column. 2. We reduce the decimals?, or regard them as reduced to a common denominator, and then subtract as in whole numbers. The reason of this course is the same as given in addition. The ciphers are also usually omitted. EXAMPLES FOR PRACTICE. 343. Subtract and explain tlie following : 1. 39.073 - 7.0285. 6. 54.5 - 37.00397. 2. 834.9 - 52.47. 7. 379.000001 - 4.0396. 3. $67.09 - $29.83. 8. 4. 83.003 - 45.879. 9. 5. $95.02 - $78.37. 10. 96.03 - 89.09005. .7 - .099909. .09 - .0005903. t 144 DECIMAL FRACTIONS. 11. A man paid out of $3432.95 the following sums : $342.06, $593,738, $729,089, $1363.43, $296,085, $37,507. How much has he left? ^7W. $73,091. 13. In a mass of metal there are 183.741 pounds ; ^ of it is iron, 35.305 pounds are copper, and 3.0009 pounds are silver, and the balance lead. How much lead is there in the mass ? 13. A druggist sold 74.53 pounds of a costly drug. He sold in March 10| pounds, in April 25.135, in May 31f, and the balance in June. How many pounds did he sell in June ? Find the decimal value of 14. {^ - 3|) + (7/j - If) - (9.23 - 8.302). 15. {$85J - $37|) + (f of $184.20 - 16. $859,085 - ($138| + %^) + >S4 I/, MULTIPLICATIOIT. 344. Multiply 3.37 by 8.3. (1.) 3.27 X 8.3 = ^^^x«-^. (2.) 327 83 100 "^ 10 27141 1000 = 37.141. Explanation.— 1. Observe that 3.27 and 8.3 are mixed nnmberei; hence, according to (271), they are reduced before being multiplied to improper ^'actions, as shown in (1). 2. According to (265), \U x ??, as shown in (2), equals 27.141. Hence, 27.141 is the product of 3.27 and 8.3. The work is abbreviated thus : (3.) 3 37 ^® obeerve, as shown in (2), that the product of 3.27 and 8.3 must contain as many decimal places as there are deci- mal places in both numbers. Hence we multiply the num- bers as if integers, as shown in (8), and point off in the product as many decimal places as there are decimal places in both numbers. Hence the following a3 981 2616 27.141 tl he t0( ya the ] 2 3 3 EXAMPLES. 146 RIJIiE. 345. Multiply as in integers, and from the right of th£ pro- duet point off as many fgttres for decimals as there are decimal places in the multiplicand and multiplier. EXAMPLES FOR PRACTICE. 346. Multiply and explain the following : 1. 13.4 X. 37. 3. 7.3x4.9. 3. 35.08x6.2. 4. 183.65 X. 7. 6. 73. 406 X. 903. 7. 340007x8.43. 8. . 4903 X. 06. 9. 5.04 X .072. 11. 12. 13. .0007 X .036. .009 X. 008. .0405 X .09. 5. $97.03x42. 10. . 935 x. 008. 14. . 307 X. 005. 15. .00101 X .001. Multiply and express the product decimally ; 16. 3Jby6|. m 12 J by 3| hundredths. 17. $35 1 by 9|. 20. 7f thousandths by f . 18. |.05f by 18 J. 21. 9| tenths by .00031. 22. What is the value of 325.17 pounds of iron at $.023 per pound? ^n«. 7.47891 dollars. 23. A merchant sold 86.43 tons of coal at |9.23 a ton, thereby gaining $112.12 ; what was the cost of the coal ? 24. What would 12.34 acres of land cost at $43.21 per acre ? 25. A French gramme is equal to 15.432 English grains; how many grains are 14| grammes equal to ? 26. A merchant uses a yardstick which is .00538 of a yard too short ; how many yards will he thus gain in selling 438 yards measured by this yardstick ? 27. A metre is equal to 39.3708 inches ; how many inches are there in 1.325 metres ? Find the value of the following : . 28. $240.09 X (2 .3^ -f of If ). 29. (I of 12? - .9031 + 1.001) x 375. 30. ($375| - $87,093) x (f of 36 - | of ^y. '■'ill I 146 DECIMAL FRACTIONS, 81. A dealer in wood and hay boup^ht 2005 tons of hay at $14.75 a ton, and 2387^ cords of wood at ^4.50 a cord ; how much did he pay for all ? Ana. $46070. 32. Bought 18 books at |1.37| each, and sold them at a gain of .50 J cents each ; what did I receive for the whole? 33. A boy went to a grocery with a $10 bill, and bought 3J pounds of tea at $.90 a pound, 7 pounds of flour at 5^.07 a pound, and 4 pounds of butter at $.35 a pound ; how much change did he return to his father ? Ans. $4.96. 34. What would 15280 feet of lumber cost, at $2.37| for each 100 feet ? Ans. $302.90. Divisioisr. .1 PREPARATORY PROPOSITIONS. 347. Prop. I. — Wlien the divisor is greater than the divi- dend, the quotient expi'esses the part the dividend is of the divisor. Thus, 4 -*- 6 = f = f. The quotient § expresses the part the 4 is of 6. 1. Observe that the process in examples of this kind consists in reducing the fraction formed by placing the divisor over the dividend to its lowest terms. Thus, 33 -f- 56 = ^j, wldch reduced to its lowest terms gives 4. 2. In caso the result is to be expressed decimally, the j)roces8 then consists in reducing to a decimal, according to (JJii7), the fraction formed by placing the dividend over the divisor. Thus, 6 -*- 8 = ^, reduced to a decimal equals .625. Divide the following, and express the quotient decimally. Explain the process in each case as above. 1. 7H-20. 4. 154-32. 7. 8^-11. 10. 3-*-20 2. 3-4-4 5. 13-J-40. 8. 5-5-7. 11. 4-5-13. 8. 5-*-8. 6. 9-T-80. 9. 5-5-6. 12. 7-5-88 ILLUSTRATION OF PROCESS, 147 348. Prop. II. — TJie fraction remaining after the division of one integer by another expresses the part the remainder is of tJie divisor. Thus, 43 -f- 11 = 3i\. The divisor 11 is contained 3 times in 42 and 9 left, which is 9 parts or ^^ of the divisor 11. Hence we say that the divisor 11 is contained 3^"^ times in 42. Wo express the -^^ decimally by reducing it according to (3127). Hence, SV'r = 3.8i. Divide the following and express the remainder decimally, carrying the decimal to four places : 1. 473 -f- 23. 4. 65 -h 17. 7. 3000 -i- 547. 2. 324 -}- 7. 5. 89 h- 103. 8. 5374 -h 183. 3. 783 -^ 97. 6. 37 -^ 43. 9. 1000 h- 101. 349. Prop. III. — Division is possible only when the divi- dend and divisor are both of the same denomination (144 — I). For example, fV -J- t^tj. or .3 -f- .07 is impossible until the dividend and divisor are reduced to the same fractional denom- ination ; thus, .3 -r- .07 = .30 -t- .07 = 4f = 4.285714. , ILLUSTRATION OF PROCESS. 360. Ex. 1. Divide .6 by .64. (1.) (2.) .6 -i- .64 =r .60 -*- .64 60 60 -4- 64 = ^ = .9375 04 Explanation.— 1. We reduce, as shown in (1), the dividend and divisor to the same decimal unit or denomination (2T9). 2. We divide, according to (279), as Bhown in (2), the numerator 60 by the numerator M, which gives U- Reducing U to a decimal (32T), we have .6 -•- .&4 = .9375. Ex. 2. Divide .63 by .0022. .63 -^ .0022 = .6300 -4- .0022 6300 -*- 22 = 286x\ = 386.36 (1.) (2.) m Ji«i 148 DECIMAL FRACTIONS, \^ m m EXPLANATION.—I. We rcducc, as f«hown in (1), the dividend nnd divisor to the Humc decimal unit by auuezin^ ciptiurK to the dividend (.3311). 2. We divide, according to (270;, a« fliowu iu (.2), the numerator 6300 by the numerator 22, ^'iving as a quotient 28«,V. 3. We reduce, according to (3'it7), tho A in the quotient to a decimal, giving tlie repetend .30. Hence, .03 + .0023 = 880.^. Ex. 3. Divide 10.831 by 3.7. (1.) (3.) (8.) 10.831 10.831 3.7 = 10.831 -T- 3.700 3.700 = 10831 3700 1000 1000 37100 ) 1G8|31 ( 0.33 103 " 63 54_ 81 81 ExpLANATioN.~l. We reduce, as shown in (1), the dividend and divlflor to the same decimal unit by annexing ciphers to the divisor (330). 2. The dividend and divisor each express thousandths, as shown in (2). Hence we reject the denominators and divide as in integers (5170). 3. Since there are ciphers at the right of the divisor, they may be cut off by cutting off the same number of figures at the right of tlie dividend (131). Dividing by 87, we find that it is contained times iu 168, with 6 remaining. 4. The 6 remaining, with the two figures cut off, make a remainder of 621 or ,Voo. This is reduced to a decimal by dividing both terms by 27. Hence, as shown in (3), we continue dividing by 27 by taking down the two figures cut off. The work is abbreviated thus : We reduce the dividend and divisor to the same decimal tmit by cutting off from the right of the dividend the flgurcH that express lower decimal units than the divisor. We then divide as shown in (3), prefixing the remainder to the figures cut off and reducing the result to a decimal. From these illustrations we obtain the following RULE. 351. Reduce the dividend and divisor to the same 4mal unit ; divide as in integers and reduce the fractional remaindc in the quotient, if any, to a decimal. I I EXAMPLES, 14» \al I EXAMPLES FOR PRACTICE. 852. In the following examples carry the answer In each case to four decimal places : 1. Divide 27^ by 4.03 ; by .72 ; by 2.3J. 2. Divide 53.28 by 3.12 ; by 7.3 ; by 9.034. 8. Divide |.93 by $.847 ; $73.09^ by |.75J ; |.37J by $.74. 4. Divide $726.42 by $.37; by $3.08 ; by $.953. Wliat is the value of 5. $75.83 -*- $100. 8. $10000 -^ $.07. 6. {\ of .73) -*- .09. 9. 8.345 -*- 2.0007. 7. 734| H- 4.5|. 10. (8J + 12.07) + (15.03 - |). 11. (fTlKJ X 64) -^ (I of f of 12|). ' 12. ($354.07- 5 of $10.84) -i- I of $7.08. 13. (§ of $324.18 - $1) -i- $2.0005. 14. ($3.052 -*-?)- (I of $1.08 -f- ,«). 15. At $2.32, how many yards of cloth can be bought for $373.84? 16. The product of two numbers is 375.04, and one of them is 73.009 ; what is the other ? Ans. 5.1369 + . 17. How much tea can be bought for $134.84, if 23g pounds cost $17.70? ^«.?. 179.786G+ pounds. 18. A farmer sold 132f bushels wheat at $1.35 per bushel, and 184 bushels corn at $.78^ per bushel. He bought coal with the amount received, at $9.54 a ton. How many tons did he buy? 19. A merchant received $173.25. $32.19, and $89.13. He expended the whole in buying silk at $1.37^ per yard. How many yards of silk did he buy? 20. What decimal part of a house worth $3965 can be bought for $1498.77? Ana. 21 What is the value of 27f acres of land when .S acre worth $48 ? m I mi 150 D E CIMA L FRACTIO XS, 22. Geo. Bain lost .47 of his capital, and had to use .13 more for family expenses, and had still remaining $5380. What was his original capital ? Ana. $13450. 23. Henry Barber owns | of a cotton mill and sellc .8 of his share for $1650 ; what decimal part of the mill does he still own, and what was the mill worth ? REVIEW EXAMPLES. 353. Answers involving decimals, unless otherwise stated, are carried to four decimal places. What is the cost 1. Of 4.5 acres of land, if 100 acres cost $7385 ? 2. Of .7 J of a pound of tea, if 7 pounds cost $6.95? 8. Of 9 J cords of wood, at $12.60 for 2.8 cords ? 4. Of 13.25 yards of cloth, if 3.75 yards cost $9 93 J? 5. Of 5384 feet lumber, at $5.75 per 100 feet? 6. Of 31400 bricks, at $8.95 per 1000 bricks ? 7. Of 158i. pounds butter, if 9.54 pounds ccst 5^:3.239? Reduce each of the following examples to decimals : 8. 9. 10. 11. n H. 5 12. 13. 14. 15. 5|\— 5|. ? of 1|. (3g-Hi)x| 8 f of .3 8^ 4.3 16. 17. 18. 19. I of ^ of If 25 90. Seven car-loads of coal, each containing^ 13.75 tons, were sold at $8.53 per ton. How much was received for the whole ? 21. Four loads of hay weighed respectively 2583.07, 3007f, 35675, and 3074^ pounds; what wa.^ tht total weight? 22. At $1.75 per 100, wjiat ia the cost of 5384 oranges? REVIEW EXAMPLES, 151 A- 4 i) ^ere 23. What is the cost of carrying 893850 pounds of com from Chicago to Montreal, at $.35| per 100 pounds? 24. If freight from Sarnia to Halifax is $.39^ per 100 pounds, what is the cobt of transporting 3 boxes of goods, weighing respectively 783 », 325 g, and 286; pounds? 25. A piece of broadcloth cost $195.38^, at $3.27 per yard. How many yards does it contain ? 26. Expended *460.80 in purchasing silk, .3 of it at $2.25 y>eT yard, J of it at $1.80 per yard, and tlie balance at $o.45 per yard. How many yards did I buy of each quality of silk V 27. A person Laving $1142.49f wishes to buy an equal num- ber of bushels of wheat, corn, and outs ; the wheat at $1.37, the corn at $.87.^, and the oats at $.35J. How many bushels of each Ciin he buy ? 28. What is the value of (1^^1—3) .48 29. A produce dealer exchan^jed 48? bushels oats at 39f cts. por bushel, and 13| barrels of apples at $3.85 a barrel, for butter at 371 cts. a pound ; how many pounds of butter did he receive ? 30. A fruit merchant expended $523.60 in purchasing apples at $3.85 ;i barrel, which he afterwards sold at an advance of $1.07 i>er barrel ; what was his p^ain on the sale ? 31. A grain merchant bought 1830 bushels of wheat at $1.25 a bushel, 570 bushels corn at 731 cts. a bushel, and 468 bushels oats at 35^ cts. a bushel. He sold the wheat at an advance of 17i cts. a bushel, the com at an advance of 9^ cts. a bushel, and the oats at a loss f)f 3 cts. a bushel. How much did he pay for the entire quantity, and what was his gain on the transaction ? 32. The cost of constructing a certain road was $5050.50. There were 35 men employed upon it 78 days, and each man recoived the same amount per day ; how much was the daily wages? ,^ w 152 DECIMAL FRACTIONS. REVIEW AND TEST QUESTIONS. 354. 1. Define Decimal Unit, Decimal Fraction, Repetend, Circulating Decimal, Mixed Circulating Decimal, Finite Deci- mal, and Complex Decimal. 2. In how many ways may \ be expressed as a decimal frac- tion, and why V 3. What effect have ciphers written at the left of an integer? At the left of a decimal, and why in each case (i505) ? 4. Show that each figure in the numerator of a decimal represents a distinct order of decimal units (iiOO). 5. How are integral orders and decimal orders each related to the units (313)? Illustrate your answer by lines or objects. 6. Why in reading a decimal is the lowest order the only one named? Illustrate by examples (310). 7. Give reasons for not regarding the ciphers at the left in reading the numerator of the decimal .000403. 8. Reduce | to a decimal, and give a reason for each step in the process. 9. When expressed decimally, how many places must j^/j give, and why ? IIow many must ^^ give, and why ? 10. Illustrate by an example the reason why \\ cannot be expresiised as a simple decimal (31(1). 11. State what fractions can and what fractions cannot be expressed as simple decimals (315 and 316). Illustrate by examples. 12. In reducing f to a complex decimal, why must the numer- ator 5 recur as a remainder (317 — 1 and 2) ? 13. Show that, according to (234—11 and III), the value oi\\ will not be changed if we diminish the numerator and denominator each by | of itself. 14. Show that multiplying 9 by 1| increases the 9 by | of itself. b RE VIE W . 153 ;r- 15. Multiplying the numerator and denominator of \l each by If produces what change in the fraction, and why V 16. Show that in diminishing the numerator of I by I and the denominator by 1 we diminish each by the same part of itself. IT. In taking .3 as the value of J, what fraction has bo<>u rejected from the numerator? What must be rejected from the denominator to make .3 = J, and why? 18. Show that the true value of .81 is g^. Give a reason for each step. 19. Explain the process of reducing a mixed circulating decimal to a fraction. Give a reason for each step. 20. How much is .33333 less than |, and why? 21. How much is .571438 less than ^, and why? 22. Find the sum of .73, .0040, .089, 6.58, and 9.08703, and explain each step in the process (350 — I and II). 23. If tentJis are multiplied by hundredths, how many deci- mal places will there be in the product, and why (IJ44) ? 24. Show that a number is multiplied by 10 by moving the decimal point one place to the right ; by 100 by moving it two places ; by 1000 three places, and so on. 25. State a rule for pointing off the decimal ])lnces in the product of two decimals. Illustrate by an exauiple, and give reasons for your rule. 26. Multiply 385.23 by .742, multiplying fird by the 4 Jiun- d/redtJiS, then by the 7 tenths, and luat by the 2 thoumndtha. 27. Why is the quotient of an integer divided by a pri>per fraction greater than the dividend ? 28. Show that a number is divided by 10 by moving the decimal point one place to the left ; by 100 l)y moving it two places; by 1000, three places; by 10000, four places. and so on. 29. Divide 4.9 by 1.305, and give a rea.son for each step in the process. Carry the decimal to three i)lace8. 30. Give a rule for division of decimals. 11 '1 ill m • ill ''' 'l 1 II DENOMINATE NUMBERS. DEFINITIONS. 355. A Relafed Unit is a unit which has an invariable relation to one or more other units. Thus, 1 foot = 12 inches, or J of a yard ; hence, 1 foot has an invariable relation to the units inch and yard, and is therefore a related unit. 356. A Detioniinate Nutnher is a concrete number (1 li) whose unit ( 1 1 ) is a related vnit. Thus, 17 yards is a denominate number, because its unit, yard, has an invariable relation to the units foot and inch, 1 yard making always 3 feet or 36 laches. 357. A DcHominate Fraction is a fraction of a related unit. Thus, 5 of a yard is a denominate fraction, 358. The Orders of related units are called Denom^ inntions. Thus, yards, feet, and inches are denominations of length ; dollars and cents are denominations of money. 351>. A Cowpoitnd Number consists of several num- bers expressing related denominations, written together in the order of the relation of their units, and read as one number. Thus, 23 yd. 3 ft. 9 in. is a »mix)und number. 360. A Standard Unit is a unit established by law or custom, from which other units of the same kind are derived. TABLES, 156 iable 08 an efore mber unit, inch, [)f a um- the law are I Thus, tho standard unit of nioasurcs of extension is the yard. By dividing the yanl into 3 equal parts, we obtain the wxAXfoot ; into 36 equal parts, we obtain the unit inch ; mul- tiplyinjf it by 5|, we obtain the unit rod, and so on. 301. Related units mav be classified into dx kinds : 1. Extension. 2. Capacity. 8. Weight. 4. Time. 5. Angles or Arcs. 6. Money or Value. .*M>2. lieditction of Denominate JVttmbers is the process of changing their denomination without altering their value. UNITS OF WEIGHT. <163. The Troy pound is, according to law, the Stanih avil Unit of weight. TKOY WEIGHT. TABI.K OP TNTTS. )i\ gr. = 1 dwt. 20 dwt. = 1 oz. 12 oz. - 1 lb. 3.2 gr. = 1 carat 1. Denotni nations. — Grains (J '*^l 15G D K y QMINATE .V V M li KRS. AVOIRDUPOIS WEIGHT. 1. DriiomluotioiiH. — DramH (dr.X Ounces (oz.), PoundH (lb.>, QuartcTt» (qr.j, Iluudredwcij^htH (cwt.), Tons (T.). 2. KquivnlvHtH.—l ton = 20 cwt. = 8000 lb. = 32000 oz. 3. l/Jvf .— Used in weighing ^rocerlot*. all coarse and heavy articles, and drugs at wholesale. The term aiuitdiijjoiii is derived from aroirii (goods or chattels) and jxddd (weight). 4. In wholesale trausactions in coal and iron and in the Custom IIoubc, 1 qnarter = 28 lb., 1 cwt. ^ 112 lb., 1 T. = 2^0 lb. This is usually called till' iMug Ton table. 5. In general, 1 stone (I st.) = 14 lb. Avoirdupois, but for butchers' meat or fish, 1 stone = 8 lb., 1 firkin of butter = 56 lb., 1 fodder of lead^lSJJ cwt., 1 great pound of silk = 24 ounces, 1 pack of wooT = 240 lb. TABLK or UNITS. 10 drams =r uz. lU OZ. = lb. 25 lb. = qr. 4 cir. = cwt. 20 cvrt. ^^ T. APOTHECARIES' WEIGHT. TABLE OF UNITS. 20 Jfl". = 1 sc. or ^. 3 3=1 dr. or 3. 8 3 =1 oz. or z 12 oz. = 1 lb. 1. Itvnotnitmtiotta. — Oraiiis (gr.). Scruples (d), Drams (3), Ounces (5), rounds (lb.). 2. KquivnU'Ut.s.—Wi. \ - 512= 3 06 = D 288 - gr. 5700. 3. r/.<»e.— Used in medical prescrip- tions. 4. Medical proscriptions are usually written in Roman notation. The nnmbor is written after the symbol, and the final i is always written j. Thus, 5 vij is 7 ounces. .Conijtftfative Table of Units of Weight, By Act of the British Parliament in 1820, the brass weight of 07iC pound Troy of the year 1758, kept by the Clerk of the House of Commons, was made the unit or stundanl measure of weight, from which all other weights are derived and computed. This hrriAs weight havinj^ been lost or destroyed by fire in 1834, the Imperial standard pound is determined from the weight of a cubic inch of distilled water, as given in 3GtS — 5. TBOT. AVOIRDUPOIS. APOTHECARIKS. 1 pound = 5760 grains = 7000 grains = 5760 grains. 1 ounce = 480 " = 437.5 " =480 " t: X A M J' L KS. 157 PROBLEMS ON RELATED UNITS. ;{(M:. Pkob. I. — To reduce a denominate or a com- pound number to a lower denomination. 7 oz. 9 dwt. to pennyweights. Reduce 23 lb 23 lb. 7 oz. 12 283 oz. 20 5009 (Iwt. o uwt. Solution.— 1. Since 12 oz. make 1 lb., in any number of poumlH iben- are 12 timer- ao many ounccH an pounds. Ilenrc we multi- ply the 2;i lb. by 12, and add the 7 oz.,K'iving sasoz. 2. A^ln, since 20 dwt. make 1 oz., in any number of ouncee there are 2() times n» many peunyweij^hti* as ouncet*. IIcucc we multiply tlie 283 oz. by 20, and add the 9 dwt., giving 5669 dwt. RULE. JJO.". /. Multiply the numher of the highest denomination fficen, hytJw number of units of the next lower lie nomination that make I (f the higher, and to the product add the number given of thr loiter denomination. IF. Proceed in this me manner itith the fractional part of each successive product. III. The integral parts of tlie several products, with the frac- tion, if any, in the last product, arranged in proper order, is tlie required result. .ii- ■■'5F lift ■K' lit IGO D ENO .V T XA T E X UMBE R S. EXAMPLES FOR PRACTICE. 37t2. Find the value in lower denominations : 1. Of ,^0 of a dram. 2. Of 5 of a ton. Of ,\ of a pound Troy. Of .0 (»f a ixiund Avoir. 3. 4. 5. Of 5 of a pound Apotli. 0. Of .85 of a ton. 7. Of .73 of an ounce Troy. 8. Of .94 of a dram. 9. Of ^ of a quarter. 10. Of ^\ of a hundredwei|?ht 11. Of tJ pounds Troy. 12. Of 3.7 hundredweights. 13. Of 13? tons. 14. Of 5.1)4 pounds Apoth. 15. Of .730 1 of a i)ound Troy, 10. Of .9350 of a ton. 17. In I of a i)Ound Avoir, how much Troy weight? IS. Keduco .84 of a hundredweight to Troy weight. 19. How much will ij of a cwt. make expressed in Troy weiglit V Expressed in Apothecary weight ? J57IJ. Prob. IV. — To reduce a denominate fraction or decimal of a lower to a fraction or decimal of a higher denomination. , . Reduce g of a dram to a fraction of a pound. (1.) f dr. = Joz. X ■!l — 8 — Iff oz. (2.) :^ijOZ. = jljlb. X /^ = 3 J, lb. SoLrTioN.— 1. Since 8 drams = 1 ounce, 1 dram ie equal J of an oz., and { of a drain is equ.il | of J oz. Ilcncc, as shown in (1), J dr. = /„ oz. 2. Since 12 ounces = 1 pound, 1 ounce is equal ,», of a pound, and, as t*liown in (2\ ,", of an ounce is equal ,*„ of ^^ 11>m or yJa lb. Hence, | dr. = 1 lb. ilencc the following I RULE. ;J74. /. Find the part which a unit of the given denomi- nation is of a unit of the next higher denomination, and multi- ply this fraction hy t?i£ given fraction or decimal. II. Proceed in the same manner icith the result and each surccmve result, until reduced to the denomination required. Reduce t?ie result to its lowest terms or to a decimal. EXAMPLES, 161 EXAMPLES FOR PRACTICE, 375. Reduce and explain oruUy : 1. f dr. to a fructiou ul'u [xmud. 2. ^ Bc. to a t'ractiou ot'u i>ouud. 3. .7 oz. to a fraction of a pound Troy. 4. .8 lb. to a fraction of a ton. 5. .0 dwt. to a fraction of a pound. 6. ,''o H). to a fraction of a hundredweight. 7. lleducc I dwt., ^^ gr., \l oz., .JJ2 oz., ,74 dwt., and .04 gr. each to the fraction of a pound Troy. 8. Ueduce ^ of a scruple to a fraction of a |M)iind. J ExrLANATioN. — Since we 4 V J- ^' 1 V L 1 IJv divide the j,''veu fraction by the f X 'ft A t X 1*2 — TSTT "'• uiiinberH in the iiHceii,'»»calo •* KUcceH-ively (IITH) between the {riven and tljc required dCDomination, we may arrange them as shown iD tlie margin, and cancel. 0. Reduce .3 oz., .84 lb., and i cwt. eacli to the fraction of a ton. 10. What fraction of a pound is fj of a dram ? .8 of a sc. ? 11. Reduce to a fraction of a pound Troy .42 gr. ; .9() dwt. 12. Reduce to the fraction of a ton g cwt. ; .J) cwt. ; ^ lb. ; .34 cwt.; .861b.; .10 oz. ; J oz. 13. 4 of an ounce ia what fraction of 1 lb. ? of 1 cwt. ? 370. Prob. V. — To reduce a compound number to a fraction of a higher denomination. RtKiuce 34 30 D2toa fraction of a pound. (1.) 14 36 92=3116: lb. 1 = 3288. (2.) Wh = n ; lience, 5 4 3 6 32 = lb. 'il. Solution.— 1. Two numbcrB can be comi)ared only when they are the same denomination. Hence we reduce, as ehown In (1), the ^4 36 ^2 and the lb. 1 lo Hcruplei<, the lowest denomination mentioned in cither number. m '.^ 162 DENOMINATE NUMBERS, 8. M I« ^aiM'liiKoqHHl olio, luid lb. 1 lu'ln« niual oaHH, 54 3 Dai* the Huiiu' part ol'lb. 1 ait jllU iti uf j'^ vvUicli in ^1|, or ||. Uoucu J 4 3 6 02 lb. Jt.or.OOUm. RULE. JJ77, Reduce the given number t<) itn lowest denomination for the ninmrato^r of the required fraction, and a unit of the required denoniinntioii to the .same denomination for the denani' inator, and reduce the fraction to its loircat terms or to a decimal. 'i EXAMPLES FOR PRACTICE. Si78. 1. VVhttt fractiou of u pound Troy uro 7 oz. 12 dwt. 8gr.? 3. Reduce 5 cwt. 4 lb. to a fraction of one ton. 8. Wbut fractiou of u hundredweight are 04 lb. 12 oz. t Reduce to the fractiou of a pound : 4. 10 oz. 8 dwt. 10 gr. 7. 5 oz. 4 dr. 3 sc. 20 gr. 5. 9 oz. 5 dr. 3 8C. 8. 4 oz. 18 dwt. 20 gr. 6. 6 dr. 1 8C. 18 gr. 9. 11 oz. 19 dwt. 23 gr. 10. What part of lb. Troy are 3 lb. 8 oz. 10 dwt. 1 11. Reduce to the fractiou of a ton 8 cwt. 04 lb. 12. Wliat part of 4 cwt. are 1 cwt. 35 lb. V 2 cwt. 50 lb. V 13. Roduct^ 8 cwt. 00 lb. to the decimal of 1 ton ; of 8 tons. 14. Reduce 8 oz. 10 dwt. to the decimal of pounds. 15. Reduce 8 cwt. 3 qr. 10 lb. to the decimal of a ton. Abbreviated Solution.— Since the 16 poands arc reduced to a decimal of a quarter by redno ing i! to a decimal, wc auiiex two ciphers to the 16, ae ehown in the margin, and divide by 35, giviuu .04 qr. To this result wo prefix the 3 quarters, giving 3 64 a64 qr., which is equivalent to -' - hundredweights ; hence wc divide by 25 ) 10.00 lb. 4 )^.U qr. 20 ) 8.91 cwt. .4455 T. 4, as shown in the margin, giving .91 cwt. To the resalt wc again prefix the 8 cwt., giving 8.91, which is eqnlva* lent to ^;JJJ of a ton, equal .4455 T. Hence, 8 cwt. 3 qr. 16 lb. = .4456 T. E X AM PL E S . 10» 10. IUhIuco U oz. 1G (Iwt. 20 ^r. to the decimal of a pound. 17. Whut decimal of 31 lb. Troy is 2 II). H oz. 10 dwt. ? IH. OZ. 10 dwt. 12 ^r. iin* wimt dccimul of ii pound Y lU. Uiulitce 12 cwt. 2 qr. IH lb. to tbo dcciniul of u ton. 20. What (b'cinml of a pound arc 5 J) 3 5 .j2 t'r. IHV 21. Kt'ducc 8 oz. dr. 2 hc. to llir decinial of a pound. 22. Hfduco 7 lb, 5 oz. Avoir, to a dccintal of 12 lb. 5 oz. :{ dwt. 'i'roy. 23. 1 lb. oz. y dwt. Im what pail of 3 lb. Apotli. Witi^ht? )un(ls jrs to ie by Iving Ide by t*i71>. Piu»n. V\. — To find the sum of two or more de- nominate or compound numbers, or of two or more denom- inate fractions. 1. Find tbo Rum of 7 cwt. 84 lb. 14 oz., 5 cwt. 97 lb. 8 oz., and 2 cwt. U lb. 15 oz. Solution.—!. Wo writ*; iiuinborH of the Hame clonomiiiiitiuii utulfr. r> oz. We write the 5 oz. under the uuQceH aud a(l(i tlic 2 Ih. to the pounds. We proceed in the Hame manner with each denomination until the entire Bum, 15 cwt. 'J3 lb. 5 uz., in luuud. cwt. lb. oz 7 84 14 6 97 8 2 9 15 15 92 6 2. Find the sum of J lb., f dr., and J sc. Solution.— 1. According to (250)» only rructiunal unitn of the Hame kind and of the nanie whole ran he added ; hence we reduce I lb., t dr., and ] hc. to intej?erg of lower z. ot'])Utter during the spring, 1 T. 12 cwt. IH lb. G oz. during the summer, 2 T. 7 c\vt. 10 lb. 6 oz. during the autumn, and 3 T. 2 cwt. 98 lb. 15 oz. during the winter ; how much did he buy during the entire yeai ? 12. What is the sum of 8.7 lb., 3.34 oz., and 411 dwt. ? 13. Find the sum of .8 cwt. and .5 <«. '- 14. A grocer sold 4 lots of tea containing respectively 7 cwt. 391b. 13 oz., 5 cwt. 84 1b. 15 oz., 13 cwt. 93 lb. 7 oz., and 7 cwt. 74 1b. 11 oz.; what was the entire weight of the tea t*old V EXAMPLES. 165 3qT. 5 cwt. . 9 oz. AiDing id the irt. )3 of a ,11). 10 ).7 rO It lis? -01. silver, ). 8 «)/. (I (Iwt. lb. -^ f^7.. 1111; the bU 3 T. le buy i7 cwt. and llie tea lb. oz. dwt '.»7 .^ 4 15 13 9 18 JJ81. PuoB. VII. — To find the difference between any two denominate or compound numbers, or denominate fractions. Find the difference between 27 lb. 7 oz. M dwt. and IIJ lb. 9 02. 18 dwt. HoLUTioN.— 1. Wt' write inimbcri* of tho samt' (lenomiuntion uiulcr each other. a. VVc eiubtruct ae in r>iinpl(' luunbiT!*. Wlu'ii the number of auy deuomiiiatiou of the fiil)tra- 13 9 17 hcud cHunot be taken from the nnniber «>!' the same denomiuatiou iu tlie minuend, \vc add art in ftimple numbers (.57) one from tlie next hiuh«'r denomination. Thun, IH dwt. cannot l>e tiiken from 15 dwt. ; we add 1 of the 1 oz. to tlie \'i dwt., niakin<; :io dwt. 18 dwt. from ;^5 dwt. leaves 17 dwt., which we write under tl.e pennywei;rht!'. We tiroceed in the f»ame manner witli each denomination until tlie entire difTerencf. 1.3 lb. *.» oz. 17 dwt., Is found. To Hubtract denominate fractiouti, wc reduce as directed in addition, and then subtract. EXAMPLES FOR PRACTICE. JJ812. Find the differei)ce between 1. 29 lb. 1 oz. l;j dwt. and 17 lb. s oz. 19 dwt. 12 pr. 2. lb. 13 =7 :r> v)l 15 ^^r. and lb. 7 : i) :G v)2 12 gr. 3. 25 T. 10 cwt. 2 qr. 19 lb. uiid 13 T. 18 cwt. ^2 lb. 13 oz. 4. 5; lb. and 2:; dwt. 8. 8.36 T. and 19.75 cwt. 5. IJ T. and 2;' cwt. 9. 9.7 oz. and 5.3 dwt. 6. V, lb. and I dr. 10. 3« cwt. and 7^ lb. 7. 45 lb. and % oz. 11. 7.5 lb. and 4.75 sc 12. A druppst had 13 1b. 4 oz. 5 dr. of a certain niodirine, and sold at one time 3 lb. 7 oz. (5 dr. 2 sc. at another 5 lb. 9 oz. 4 dr. 1 sc. 10 pr. How much huH he left? 13. Out of a stnek of ha;' "ontaininp 10 T. 9 cwt 1 (jr. 12 lb. three loads were sold contninlnp. respectively, 3T. A cwt.. 2 T. 19 cwt. 2 i\r. 9 ib., and 3 T. 13 cwt. 14 lb. Uuw much hay is left in the btackV * I I 166 DEXOMINATE NUMBERS, 383* Pros. VIII.~To multiply a denominate or com- pound number by an abstract number. Multiply 18 cwt. 74 lb. 9 oz. by 6. 18cwt. 74 lb. 9 oz. 6 5T. 12 cwt. 471b. 6 oz. Solution.— We multiply as in simple numbiTH. commcncinp: with the loweHt (Icuoiniiintion. Thus, 6 times 9 uz. equals* M oz. We reduce tlie .54 oz. to pounds (36T), equal 3 lb. 6 oz. We write the fi oz. under the ounces, and add the .3 lb. to the product of the poands. We proceed In this manner witli each denomination until the entire product, 6 T. 12 cwt. 47 lb. 6 oz. is found. EXAMPLES FOR PRACTICE. 384. 1. Multiply 7 cwt. 2 qr. 18 lb. 5 oz. by 9 ; by 12; by 63. 2. Multiply 3 lb. 9 oz. 12 dwt. 17 gr. by 4 ; by 7. 3. Each of 8 loads of hay contained 2 T. 7 cwt. 19 lb. ; what is the weight of the whole ? 4. A grocer sold 12 firkins of Imtter. oach containing 03 lb. 13J oz. How much did they all contain ? 6. A druggist bought 25 boxes of a certain medicine, each l)ox containing 2 lb. 5 oz. 7 dr. 1 sc. 19 gr. ; what was tlie weight of the whole? , Multiply and reduce to a compound number : 6. 8» lb. by 14. 7. 21 lb. by 9. 8. 9.50 cwt. by 7. 9. 10.95 lb. by 5. 10. 13 lb. 7f oz. by 8. 11. 6.84 T. by .9. 12. 2.13 dr. by .4. 13. 7.03 cwt. by .34 14. S T. by l 15. I lb. by .7. 16. ^ cwt. by ^. 17. .9 dwt. by .9. 18. Tf a load of coal by the long ton weigh 2 T. 8 cwt. 3 qr. 11 lb., what will Ik* the weight of 32 loads? 19. A drayman delivered on board of a boat 12 loads of coal, each containing ^ T. 7 cwt. 16 lb. How much coal did he put on board? I EXAMPLES, 167 r com- ply as in mcnclng ulimilon. ilt* Muz. > pounds 1 add the he entire (185. Prob. IX.— To divide a denominate or compound number by any abstract number. Divide 29 lb. 7 ok. 3 dr. by 7. dr. 7)2U oz. 7 dr. o Solution.— 1. The object of the division is to find \ of the compound number. ThiA is dune by fliidiiiK the \ of each deuomiuatiou 4 2 6 separately. Hence tlie procet«8 is the same as in finding one of the equal parts of a concrete numbet*. ThiiH, the ) of 2*) lb. U 1 lb. and 1 lb. rcmaininj,'. We write the 4 lb. in the quotient, and reduce the 1 lb. to ounces, which added to 7 oz. make 1\« oz. We now And the \ of the 19 oz., and proceed as before until the entire quotient, 4 lb. 2 oz. (3 dr., io found. i by 12; b. ; what he oaib. line, ( was ach the |wt. 3 qr. of coal, he put I EXAMPLES FOR PRACTICE. 1. If 20 lb. 7 oz. IG dwt. are made into 7 equal parts, how much will there be in each i>art? 2. Divide 9 T. 15 cwt. 3 qr. 18 lb. by 2 ; by 5 ; by 8 ; bj i2. 3. A drugjjfist made 12 powders of ?! ' 5 of a certain med- icino ; what was th»; weight of each powder? 4. Divide lb. 3 =7 3 4 32 ])y 4 ; by ; by 12 : by 32. 5. The ig^re^nte weip^ht of 43 equal sacks of coffee is 2 T. 7 CNNl. 2 ([I 12 lb. ; what is the weight of each sack ? C. Divide 5 T. U cwt. 2 qr. 8 lb. by 3 cwt. 1 (jr. 12 lb. Reduce both the dividend and divisor to the fianic denomination, and divide UK in bimple munbers. 7. Divide lb. 6 5 9 3 7 32 by 3 7 !) 2 gr. 10. 8. How many boxes, each coiitaininj, Chain (ch.), Mile (ml.). 2. Equivalonts.—i ml.=80ch.-320rd. = 8001)1. .1 /'Af .— Us'-'d In nitasnrlng roads and boundiiries i.f l.'iinl. 4. The t'ltit of measure lu the ( unfer'n Chain, v\hlch CDUiaius 10() links, equal 4 rods or 60 Teet. 2.1 1. = 1 rd. 4 rd. - 1 rli. 80 cli. = 1 mi. CLOTH MEASURE, 169 urface, n.)»Fect ml.)- a. = 5281) ft aiul «co. 5 feet = 1 Roman nace. 12 m Pi tl < >i n 170 D£yo Miy A T E y um u ers. f EXAMPLES FOR PRACTICE, 380. Reduce and explain the following: 1. 38465 yd. to miles. 4. 84 rods to linke. 2. 8 inileH to yards. 6. ^ of a rod to inches. <3. 7 rods to inches. 6. j^ of a ch. to links. 7. 15 degrees to statute miles. 8. 8.76 geographical miles to statute miles. 9. 12 rd. 4 yd. 2 ft. to inches. 10. 210 geographical miles to statute miles. 11. 2 mi. 5 ch. 8 rd. to links. ' 12. .78 of a mile to a compound number. 18. .85 of a yard to a decimal of a mile. 14. 7 yd. 2 ft. to a decimal of 8 rd. . 15. Find the difference between 3 mi. 5 ch. 2 rd. 13 1., and -^^ of (1 mi. 7 ch. 8 rd. 18 1.) 16. Find the sum of ,; of a mi., .85 of a ch., and 3 ch. 2 rd. 17. The f(Mir sides of a tract of land measure respectively 3 mi. 5 ch. 2 rd., 2 mi. 7 ch. 3 rd. 13 1., 3 mi. 17 1., and 2 mi. 2 rd. ; what is the distance round it? 18. On a railroad 182 mi. 234 rd. 4 yd. 2 ft. long, tliere are 18 stiitioiiH at equal distances from each other. How far are the Htations ai)art, there being a station at each end of the road ? 19. A '^iiip moving r'ue north sailed 15.7 degrees. Ilow far did she :;iil in statute miles? 20. A ship sailing on the equator moved 45 leagues. How many degrees is she from the place of storting, and what is the distance in statute mile.s V 21. ij of a rod is what part of 3 chains? 22. 1 link is what decimal of 1 foot? 28. In 125 geog. miles how many statute mllea? 34. 3 ft. are what decimal of 8 hmIs ? 25. 82 fathoms are what decimal of a mile? D EFi y I T I o ys. 171 u:n^its of surface. ;J90. A square yard is tho Sfandat'd Unit of mirface measure. ;5U1. A Surface lias two i\\nuinsums-~/cnf;t?i and hronlth. ;iU2. A Squni'f is a ;>;^//if mrface bounded by four equal lines, and having four right angles. lV,y*\, A Jtf'ftattyle is any plane surface having four sides and four right angles. «$!>4. The Unit of Measure for surfaces is usually a sciuare, each side of which is one unit of u known length. Thut^, in 14 ^'q. ft., the unit of measure in a square foot. iJJ)r». The Area of a rectangle is the S'/rfare included within its boun- ^/ ^^^^ '""-• darie.s, and isexpres.sed by the number of times it contains a given unit of meanure. Q Pli. 11 '.»^ r. ■lit m ■i%\ '■' j? m ■*f IT'f V 1 (I , 172 DENOMINATE NUMBERS. SQUARE MEASUHE. TABUS OP UNITS. 144 sq. iu. =: 1 sq. ft. 9 sq. ft. = 1 sq. yd. 30 J »!• yd. = 1 sq. rd., or P. IGO Ki. ixl. = 1 A. GIO A. = 1 sq. mi. 1. Denominations. — Square lucb (»q. iu.), Squaro Yard (ttq. yd.), Bquurc Uutl (sq. nl.), Acru (A.), Square Mile (fq. ml.). 2. IJtiuivalenta.—l eq. mi. = WO A.-10-.MOO bq. rd. = 3UU7tiU> sq. yd. = 2787*100 eq. ft. = 40Ut8»iOO eq. iu. 8. A square ig a four-f>idcd figure whot^c sides and angles are equal. Tbii* table in constructed Troui tbo table uf liuoar mcauore by multiplying each dimension by itt, Towu- Bhlp (Tp.). 2. i:qnicalentH.—\ Tp. = :« pq. ml. r= 23040 A. = 2;J0400 eq. ch. = 3086400 P. = 2304000000 h^q. 1. 8. f '«e.— Used in computing the area of land. 4. The Unit of land meacnre is tlic acre. The measurement of a tract of land is usually recorded in square miles, acres, and hundrtdtha of an acre. 80 u\. mi. = 1 Tp. EXAMPLES FOR PRACTICE. 308. Reduce and explain the foUowinjj : 1. .83 of an A. to sq. yards. 4. .08 of an A. to sq. links. 2. 5 sq. mi. to sq. yards. 8. 8 acres to sq. ft. 5. .007 mi. to sq. links. 6. 3 sq. mi. to sq. chains. E X A MP L E S. 173 luting the area 7. 35 pq. yd. to a dpcimal of an acre. 8. 14 P. to t^ decimal of a sq. mi. D. J of a sq. mi. to a cnm]>oiind uumber. 10. .0005 of an A. to sq. feet. 11. 5 of a Tp. to a compound number. 12. .0008 of a Bq. mi. to a compound number. Find the sum of 13. f of an A., § of a sq. rd., and 3 A. 158 sq. rd. 25 sq. yd. 14. I of (1 Tp. 18 sq. mi. 584 A.), and ^\ of (378 A. 9 sq. ch. 12 P.) 15. Find tlie diiferonce between (^ of G (»i mi. + § of an A.), and (I of an A. -^ i{ of a i)ole). 16. Subtract 1 -q. 1. from 1 aero ; from 1 township. 17. A tract of land containing' 084 A. 7 8<|. ch. 13 P. was divided into 7 ctjual farms ; what was tlie size of each farm ? What is tlio area of rectangles of tlie following dimensions : 31. 7.5ch. by 3ch. 81.? 23. 4 yd. 2 ft. 4 in. by 3| yd. ? 33. 3.4 yd. by 0^ yd. ? 18. 15 yd. by 12 yd. ? 10. 10^ yards square ? 20. 93 yd. by 18| yd. ? 34. How many yards of carpeting. 3 ft. 3 in. wide, will be required for 3 rooms 18 ft. by 34 ft., and 4 rooms 12 ft. by 10 ft. C in. ? .1/^^. 309J yd. 85. How many boards 13 ft. long and 4 in. wide required to tloor a room which is 48 ft. by 33 ft.? Ans. 384 boards. 36. How many square feet of luinb<>r required for the floors of a house containing 3 rooms 15 ft. by 19 ft., 5 rooms 14 ft. by IG ft., and 3 rooms 13 ft. by 15 fr. ? Ans. 3230 sq. ft. 27. Find the cost of carptting a house containing rooms as follows : 4 rooms 15 ft. by 10 ft. 6 in., carpet f yd. wide at $1.26 iwr yard ; 3 rooms 18 It. by 25 ft., carpet I yd. wide at $3.45 per yard ; and 5 rooms 13 ft. 8 in. by 16 ft., carprt 1 yd. wide at $1.08. Ans. $620. m Ml 174 DENO MIX A TE i\ UM R E h' S » 28. Find tlie cost of glazing 10 windows, each 9 ft. 10 in. by 5 ft. 8 in., ftt ^.1)4 a square fcK)t. 21). How many tiles 10 inches square will lay a floor 32 ft. 6 in. l)y 28 ft. in. ? Ans. 1090.08. 30. Tlie ridge of the roof of a building is 44 ft. lonff, and the diHtance from each cave to the ridge is 19 ft. 'i in. How many shingles 4 in. wide, laid 6J in. to the weather, will be required to roof the building, the first row Ining doul)le? 31. Find the cost of lathing and plastering a house at ^.52 per square yard, containing the following rooms, no allowance being made for doors, windows, and baseboard ; 3 rooms 14 ft. by 18 ft., and 2 roreadtfi, and thicknesH. 400, A lU'ctnngulftr Solid is a body bounded by six rectdiiglcs called faces. 4-01. A Cube is a rectangular soUO, l)ounded by six eijual squares. 4012. The Unit of Measure is a cube whoso edge is a unit of some known length. 403. The Volume, or Solid Contents of a body, is CUBIC 3tt:ASUIiL\ 175 expressed by the number of times it (•ontninn a given nnii of vuuAurc. For cxumplf, tin; contents of u cubic yurd is ex- proHstul us '^7 cuUicfeet. Tlui!', bIucc each face of a m.^ jml'j^ N. cuhic yard ccu^aIiih U »q. ft., if a HCi-.tlou 1 ft. thick 18 takcu it iiiiiKt COD tain 3 timcB H cii. ft.,orUcu. ft., at) Hhown iu thu diuKram. And hIir-u tho cubic yard is 8 feet thick, it tniiMt cou- tain 3 HcctiuuH, oach cun- tainiuK *J cu. ft., which it) 87 en. ft. lloucu, the volume or con- ten*n of A cubic yard uzprcHHcd in cubic fei't itt found by taking; tlio product of iIju nuinborrt denoting itu 3 dlmencioni* in font. Thu contents of any rectangular Holid in found in the eame manner ; hence the following RULE. 404. Mnd the product of the numbers denoting the three dimensions expressed in the lowed denomination named. This result is the voiuniCf and can be reduced to any required de- nomination. Mi >■« To find a required dimension : RULE. 405. Divide the volume by the product of the numbers de- noting the other tico dimensions. The volume, before division, must bo rcdnrod to a cubic unit correa- pouding with the Bquare unit of the product of tlio two diinuut-iou.>«. V;'\ CUBIC MEASURE. 1. />*'n«i»*/»if»fi*oij.«i.— Cubic Inch (cu. in). Cubic Foot (cu. ft.), Cubic Yurd (cu. yd.). a. l^iiiirtiffHtM — 1 cu. yd. = 87 cu. ft. = 4»MW<)cu. in. 8. r'#«».— Used in computing tho volumo or conteut» of aollds. TABLE OP UNITS. 1728 cu. in. = I cu. ft. 27 cu. ft. = 1 cu. vd. r i . ■iil IMAGE EVALUATION TEST TARGET (MT-S) 1.0 I.I 1.25 m MM 1^ '^ IM 122. !'' IM '""^ .'■: IM 12.0 1.8 U 111.6 *^ &.. y <^ / o^ 'h Photographic Sciences Corporation ^ # 4t>^ ,\ iV :\ \ ^ -> ^'^x €> <> 23 WEST MAIN STREET WEBSTER, NY. 14580 (716) 872-4503 '^<^ ^ .<\ '<>', «y <\ i ft i 176 DENOMINATE NUMBERS. V I EXAMPLES FOR PRACTICE. 406. Reduce and explain the following : 1. 97 cu. ft. to cu. in. 3. 4 cu. yd. 394 cu. ft. to en. izu 2. .09 of a cu. yd. to cu. ft. 4. .0007 of a cu. yd. to cu. ia. 5. Find the sum of f of a cu. yd. and .625 of a cu. ft. 6. .8 of a cu. ft. to a decimal of a cu. yd. 7. f of a cu. ft. to a decimal of a cu. yd. 8. Find the difference between | of a cu. yd. and .75 of a cu. ft. Find the contents of rectangular solids of the following dimensions : 9. A solid 7 ft. 9 in. long by 3 ft. 4 in. by 4 ft. 6 in. 10. A cube whose edge.is 3 yd. 2 ft. 8 in. 11. A solid 34 ft. long by 1 ft. 6 in. by 2 ft. 9 in. 12. A solid 12 yd. 1 ft. 9 iu. long by 2 yd. 2 ft. by 2 ft. 8 in. 13. How many cubic feet in a stick of timber 38 ft. long by 2 ft. 3 in. by 1 ft. 9 in.? 14. A cistern 9 ft. sq. contains 1092 cu. ft. ; what is its depth? 15. A stick of square timber contains 189 cu. ft. ; 2 of its dimensions are 1 ft. 9 in. and 2 ft. 3 in. ; what is the other? 16. How many cubic yards of earth in an embankment 283 ft. by 42 ft. 8 in. by 18 ft. 6 m. ? 17. How many cubic feet of air in a room 74 ft. 9 in. long, 52 ft. 10 in. wide, and 23 ft. 6 in. high? 18. A bin contains d2Q\ cu. ft. ; 2 of its dimensions are 9 ft. 8 in. and 7 ft. 6 in. ; what is the other? 19. A vat is 7 ft. 2 in. by 4 ft. 9 in. by 3 ft. 4 in. How many cubic feet does it contain? 20. In digging a cellar 48 ft. 6 in. by 39 ft. 8 in., and 8 ft. 4 in. deep, how many cubic yards of earth must be removed? EXAMPLES. 177 21. What will be the cost of the following bill of square timber, at f .33J per cubic foot : (1.) 3 pieces 13 ft. by 9 in. by 7 in. ? (2.) 8 pieces 15 ft. 6 in. by 10 in. by 8 in. ? (3.) 4 pieces 33 ft. by 8 in. by 9 in. ? (4.) 6 pieces 3G ft. by 1 ft. 6 in. by 1 ft. ? (5.) 9 pieces 18 ft. 9 in. by 1 ft. 3 in. by 9 in. ? (6.) 12 pieces 15 ft. by 7^ in. by 9^ in. ? 22. How many perches in a wall 37 ft. long, 23 ft. 6 in. high, and 2 ft. G in. thick ? Table of Units for Measuring Wood and' Stone* 16 cu. ft. = 1 Cord Foot (cd. ft.) \ Used for measuring 8cd.ft.or ) ^ icord(cd.) [ „,'»"' 128 cii. ft. ^ ; wood and stone. 2i}l cu. ft. = 1 perch (pch.) of stone or masonry. 1 cu. yd. of earth is called a load. 1. The materials for masonry are npually estimated by the cord or perch^ the work by the perch and cuMc foot, also by the square foot and square yard. 2. In estimating the mason work in a building, each wall is measured on the outside, and no allowance is ordinarily made for doors, windows, and cornices, unless specified in contract. In estimating the material, the doors, windows, and cornices are deducted. 3. Brickwork is usually estimated by the thousand bricks, which are of various sizes. 4. Excavations and embankments are estimated by the cubic yard. How many EXAMPLES FOR PRACTICE. 407. Reduce and explain the following : 1. 42 cords to cd. feet. 3. 36 cords to cu. feet. 2. 64 pch, to cu. feet. 4. .84 pch. to cu. feet. 178 DENOMINATE NUMBERS. n I. 6. 7 of a cd. to cu. feet. 6. .73 of a cu. ft. to a decimal of a cd. 7. f of a cord to a decimal of a cu. yd. 8. ^ of a cu. ft. to a decimal of a pch. 9. .85 of a cord to a decimal of 3 cu. yd. 10. I of 8 cd. to a decimal of 13 cd. 11. Find the sura of f pch., | cd., and 11 cd. ft. 38 cu. ft. 12. A pile of wood containing 84 cd. 7 cd. ft. 12 cu. ft. was made into 5 equal piles ; what was the size of each ? 13. How many cords in a pile of wood 196 ft. long, 7 ft. 6 in. high, and 8 ft. wide? A Cord is a pile of wood, stone, etc., 8 ft. long, 4 ft. wide* and 4 ft. high. A Cord Foot is 1 ft. long, 4 ft. wide, and 4 ft. high, or \ of a cord, aa shown in the cut. 14. What is the cost of a pile of ^tone 28 ft. long, 9 ft. wide, and 7 ft. high, at |3.85 per cord ? 15. A load of wood containing 1 cord is 3 ft. 9 in. high and 4 ft. wide ; what is its length ? 16. How many perches of masonry will 18 cd. 5 cd. ft. of stone make, allowing 22 cu. ft. of stone for 1 perch of wall ? 17. How many cords of stone will be required to enclose with a wall built without mortar a lot 28 rods long and 17 rods wide, the wall being 5 ft. high and 2 ft. 9 m. thick t \n EXAMPLES. m BOARD MEASURE. TABLE OF TTNITS. 12 B. in. = 1 B. ft. 12 B. ft. = 1 cu- ft. 408. A Bofird Foot is 1 ft. long, 1 ft. wide, and 1 in. thick. Hence, 12 hoard feet equals 1 cu. ft. 409. A Board Inch is 1 ft. long, 1 in. wide, and 1 in. thick, or ^V of a hoard foot. Hence, 12 hoa/rd inches equals 1 hoard foot. Observe carefully the following : (1.) 4 feet long. •a « Square foot. 1 1 1. Diagram (1) represents a l/Ourd where botli dimensions are feet. Hence tlie product of the two dimensionB gives the square feet in surface (396), or the number of board feet when the lumber is not more than 1 hich thick. 2. Diagram (2) represents a board where one dimension is feet and the other inches. It is evident (408) that a board 1 foot long, 1 inch thick, and any number of inches wide, contains as many board incites as there are inches in the width. Hence the number of square feet or board feet in a board 1 inch thick is equal to the length in feet multiplied by the width in inches divided by 12, the number oi board inches in a board foot 3. In case the lumber is more than 1 inch thick, the number of board feet is equal to the number of square feet in the surface multiplied by the thickness- ID 4 X 2 = 8 sq. ft. or 8 B. ft. (2.) 4 feet long. 1ft. by 9 in. « 4x9=36 B. In.; 36 B. in.-i-12=3B. ft. '•">. EXAMPLES FOR PRACTICE. 410. Find the contents of boards measuring 1. 24 ft. by 13 In. 4. 9 ft. by 32 in. 7. 5 ft. by 18 in. 2. 28 ft. by 15 in. 5. 13 ft. by 26 in. 8. 34 ft. by 15 in. 3. 18 ft. by 16 in. 6. 17 ft. by 30 in. 9. 25 ft. by 14 in. fi-W IP! 180 DENOMINATE NUMBERS, f'/i Find the contents of boards measuring 10. 15 ft. by 1 ft. 3 in. 12. 19 ft. by 2 ft. 4 in. 11. 27 ft. by 1 ft. 6 in. 13. 23 ft. by 1 ft. 5 in. 14. Find the contents of a board 18 ft. long and 9 in. wide. 15. How many board feet in a stick of square timber 48 ft. long, 9 inches by 14 inches. 16. Find the length of a stick of timber 8 in. by 10 in., which will contain 20 cu. ft. Opbbation.— (1728 x 20) +(8 x 10) = 432 ; 432+12 = 3G ft., the length. 17. A piece of timber is 10 in. by 13 in. What length of it will contain 26 cubic feet ? Find the cost of the following : 18. Of 234 boards 14 ft. long 8 in. wide, at $3.25 per hundred. 19. Of 5 sticks of timber 27 ft. long, ^ in. by 14 in., at $1.75 per hundred feet board measure. 20. Of 84 plank 20 ft. long, 11 in. wide, 3 in. thick, at $1.84 per hundred feet board measure. UNITS OF CAPACITY. 411. The Standard Units of capacity are the Gallon for Liquid, pjid the Bushel for Dry Measure. II LIQUID MEASTTBE. 1. Deuoiuinations.—Gil\a (gi.), Pints (pt.), Quarts (qt.), Gallons (gal.), Barrels (bbl.). 2. Equivalents.— 1 gal.=:4 qt.=8 pt. = 32 gi. 3. Use.— Used in measuring liquids. 4. The capacity of cisterns, vats, etc., is usaally estimated by considering a barrel 31J gal. ; but barrels are made of various sizes, from 30 to 56 gallons. The hogshead, butt, tierce,' pipe, and tun are names of casks, and have usually their capacity in gallons marked upon them. TABLE OF UNTTS. 4 gi. = 1 pt. 2 pt. =1 qt. 4 qt. = 1 gal. 81| gal. = 1 bbl. DJiT 3IEASURE, 181 AFOTHECABIES' FLUID MEASURE. in. in. in. wide. mber 48 ft. Oin.,wliicli } length, ength of it TABLE OF UNITS. R 60 = f 3 1 f 3 8 = f 5 1 f § 16 = O. 1 8 = Cong. 1 O. 1. Denominations. — Minima or drops (iTl), Fluid Drachm (f 3), Fluid Ounce (f 5), Pint (O., for octarim, the Latin for one-eighth or pint), Gallon (CoDg., for congius, the Latin for gallon). 2. JiJqtiivalents, — Cong. 1 = O. 8 = f 5 128 = f 3 lOiM = ta 61'M0. 8. Use.— JJ^ed in prescribing and compounding liquid medicine. 4. The symbols precede the numbers, as in Apothecaries* Weight, as shown in the table of units. DRY MEASURE. ler hundred, in., at $1.75 ick, at $1.84 TABLE OP irniTS. tlie Oallon 2pt. = 8qt. = 4pk. = Iqt. Ipk. Ibu. 1. Denotninations.~-riatB (pt.), Quarts (qt.)t Pecks (pk.), Bushels (bn.). 2. Equivalents*— Ihn. = 4pk. = 32 qt. = 64 pt. 8. Use.— Used, in measuring grain, roots, fhiits, salt, etc. 4. Heaped rieamre., in which the bushel is heaped in the form of a cone, is used in measuring potatoes, com in the ear, coarse vegetables, large fruits, etc. Stricken measure is used in measuring grains, seeds, and small fruits. 5. A bushel of oats = 34 lb. ; of buckwheat, barley, timothy = 48 lb, ; of flaxseed = 50 lb. ; of rye and Indian corn = 56 lb. ; of wheat, potatoes, peas, beans, onions, or red clover seed = 60 lb. bfl (gl.), P^nt' [(gal.). Barrels :4 qt.=8 pt. = ig liquids. [b, vatB, etc., is els are made of h^ce,'iApe, and lallons marked EXAMPLES FOR PRACTICE. 412. Solve and explain orally the following : 1. How many gills in 4 qt. ? In ? gal. ? In 7 qt. ? In 8 qt. 1 pt. ? In 3 gal. 3 qt. ? 2. How many pints in 3 bu. ? In 3 pk. 5 qt. ? In 1 bu. 2 pk. 7qt.? 3. What is the sum of O. 5 f 3 12 f 3 7 and f § 8 f 3 3 ITilS? 4. Multiply 3 pk, 5 qt. 1 pt. by 3 ; by 5 ; by 10 ; by 7 ; by 12. Reduce 5. 93584 pt. to barrels. 8. 93654 pt. to bushels. 6. 28649 pt. to bushels. 9. 57364 gi. to barrels. 7. TTl 8405 to gallons. 10. f 3 7649 to gallons. m t,: * . 1 I'll . 1 * ! 1 182 DENOMINATE NUMBERS, 11. 3 qt. 1 pt. to a decimal of a gallon. # 12. f of o qt. 1 pt. to a decimal of 2 bushels. 13. f 3 7 TT[ 15 to a decimal of Cong. 3. 14. A merchant bought 5860 bushels wheat in Toronto at $1.25, and sold the whole in Halifax at the same price. How much did he gain on the transaction ? 15. A grocer bought 12 firkins of butter, each containing 73 lb. 13 oz., at 36 cts. a pound ; 7 bu. 3 pk. clover seed, at $1.15 a peck ; and 5 loads of potatoes, each load containing 43 bu. 8 pk.', at $.32 a bushel. How much was the cost ? > Comparative Table of Units of Capacity, CUBIC INCHES CXJBfC rNCHES CUBIC INCHEQ IN ONE GAIJ.ON. IN ONE QUAKT. IN ONK PINT. Imperial, 277.274 Liquid Measure U. S. 231 57f m Dry Measure (| pk.) 268| 671 m 1. The Imperial Bushel of Great Britain contains 2218.192 cu. in. and the Standard Hushel of the United States contains 2150.42 en. in. 2. An English Quarter contains 8 imp. bu. or 8J U. S. bu. A quarter of 8 U. S. bu., or 480 lb., is used in shipping grain from New York. 3. A Register Ten is 100 cu. ft.; used in measuring the internal capacity or t^ nnage of a vessel. A Shipping Ton Is 40 cu. ft. 4. A cubic foot of pure water weighs 1000 oz. or 62i lb. Avoir. \\ if' EXAMPLES FOR PRACTICE. 4: 13* 1. How many U. S. bushels in a bin of wheat 6 ft. long, 5 ft. 6 in. wide, and 4 ft, 9 in. deep ? How many cubic feet in a space that holds 2. 1000 U. S. bushels? 5. 240 English quarters? 3. 1000 imp. bushels? 6. 18 T. 16 cwt. of pure water? j 4. 120 bbl. water? 7. 804 bu. 3 pk. U. S. bu.? 8. A cistern containing 5300 gal. of water is 10 ft. square. | How deep is it? ^r<«. 7.085 + . 9. How many ounces in gold are equal in weight to 9 pounds] 14 ounces of iron ? |:t;-^ TIME MEASURE, 183 Toronto at price. How ;li containing over seed, at ad containing ,lie cost ? ipacity* CUBIC INCHES IM ONK FINT. 281 331 918.192 cu. in. and )0.42 cu. in. bu. A quarter of V York, iring the internal cu. ft. Avoir. of wheat 6 ft. q[uarters 1 . . of pure water? U. S. bu.? IS 10 ft. square. I Ans. 7.085 + . Iglit to 9 pounds TABLE OP UNITS. 60 sec. = 1 min. GO min. = Ibr. 24 hr. = Ida. 7 da. = 1 wk. 365 da. = 1 common yr. 366 da. = 1 leap yr. 100 yr. = 1 cen. UNITS OF TIME. 414. The mean solar day is the StiuuUiril Unit of time. 1. DenoniinationH. — Seconds (sec), Minutes (min.), Hours (hr.), Days (da.), Weeks (wk.), Months (mo.), Years (yr.), Centuries (cen.). 2. There are 12 Caleudur Months In a year; of these, April, June, September, and November, have 80 da. each. All the other months except February have 31 da. each. February, in common years, has 28 da., in leap years it has 29 da. .3. In computing interest, 30 days are usually considered one month. For bnsinci's purposes the day begins and ends at 12 o'clock midnight. 415. The reason for common and leap years wWl be seen from the following : The true year is the time the earth takes to go once arofund the sun, which is 365 days, 5 hours, 48 minutes and 49.7 seconds. Taking 365 days as a commcm. year, the time lost in the calendar in 4 years will lack only 44 minutes and 41.2 seconds of 1 day. Hence we add 1 day to February every fourth year, making the year 366 days, or l^enp Year. This correc- tion is 44 rain. 41.2 sec. more than should be added, amounting in 100 years to 18 hr. 37 min. 10 sec. ; hence, at the end of 100 years we omit adding a day, thus losing again 5 hr. 22 min. 50 sec, which we again correct by add- ing a day at the end of 400 years; >iLence the following rule for finding leap year: BULE. 416. Enery year, except centennial years, exajctly divisible by 4, is a lenp year. Every centennial year exactly divisible by 400 is also a leap year. This will render the calendar correct to within one day for 4000 years. 417. Prob. X. — To find the interval of time between two dates. How many yr., mo., da. and hr. from 6 o'clock P. M., July 19, 1862, to 6 o'clock A. M., April 9, 1876. 184 DENOMINATE NUMBERS. i f ■■■ t yr. mo. da. hr. 1876 4 9 7 1862 7 19 18 13 8 19 13 Solution.— 1. Since the latter date denutuH tin; jjrcuifr period of time, it is the minuend, aud the earlier date, the subtrahend. 2. Since each year commonce? with January, aud each day with 12 o'clock midnight, 7 o'clock a. m., April 9, 1876, is the 7th hour of the 9th day of the fourth month of 1876 ; and o'clock p. m., July 19, 18«nom{nae{on«.— Mills (m.), Cents (ct.), Dimes (d.), Dollars ($), Eagles (B.). 2. The United States coin, as fixed by the ''New Coinage Act" of 1878, is as follows: Gold, the double-eagle, eagle, half-eagle, quar- ter-eagle, three-dollar, and one-dollar ; Silver 1 1 the trade-dollar, half-dollar, quarter-dollar, and | ten-cent ; NicJcel, the five-cent and three-cent ; Bronze^ one-cent. 3. Cotnposition of Coina.—Gfold coins of Britain consist of 22 parts I 2nire gold and 2 parts of copper. Silver coins consist of 87 parts pure silver and 8 parts of copper. Gold coin of the United States contains .9 pure gold and .1 silver and copper. Silver coin contains .9 pure silver and .1 pure cop- per. Nickel coin contains .25 nickel and .75 copper. Bronze coin contains | .95 copper and .06 zinc and tin. 4. The 7Va«fe-dWtor weighs 420 grains and Is designed for commercial j purposes solely. Observe, the Canadian bronze cent is one inch in diameter, and one hundred cents weigh one pound Avoirdupois. The mill is not coined but is used in computation. Copper coinage is not a legal tender for any sum above 30 cents, nor silverj coinage for more than $10. TABLE OP UKITS. 10 m. ^^ let. 10 Ct. =z Id. 10 d. =: $1. $10 = IE. OEUMAN MONEY, 187 ENGLISH MONEY. 424. "Wxeponnd sterling is the Standanl Unit of English luney. It is equal to $4.8660 Canadian money. TABUS OF UNITB. 4 far. 12 d. 20 e. 2 8. 5 s. = Id. = Is. 1 Sov. or £1. = Ifl. = 1 cr. = 1 1. Denomiiuitiotui. — FartliinK*) (f^r.), PcunleB (d.), ShiUiugH («.), Soverei^'u (fov.), Pound (£), Florin (fl.), Crown (cr.). 2. The Coins in general uhc in Great Brltnln are as followt> : Gold, sovereign uml liult- Bovert'ign ; Silver, crown, half-crown, florin, BhilllnK, Blx-penny, and three-penny; Cop- per, penny, half -penny, and farthing. United States ;d for commercial PEBNCH MONEY. 425. The silver franc is the Standard Unit of French loney. It is equal to $.193 Canadian money. TABLK OP UOTTS. 10 ra. r= 1 ct. 10 ct. = 1 dc. 10 dc. = 1 fr. 1. DenonUuatt4yHa. — Millimes (m.), Cen- times (ct.), Decimes (dc), France (fr.). 2. Equivalents , — \ tr. = 10 dc. = 100 ct. = 1000 m. 8. The Coin of France is as followH : Gold, 100, 40, 20, 10, and 5 francs ; /Silver, 5, 2, and AW, ■«/, «V, iU, i»UU u uuuvn, ifttvi-if c, •, at franc, and 50 and 25 centimes ; Jtronxe, 10, 5, 2, and 1 centime pieces. GERMAN MONEY. 1 426. The mark is the Standard Unit of the New Oer- in Empire. It is equal to 23.85 cents Canadian money, and divided into 100 equal parts, one of which is called a \fennig» 11. The C;'( hi u '4 :( b :4: ■r ^\ it'^ I UNITS OP CAPACITY. 438. The Litre is the principal unit both of Liqnid and Measnre. tt is equal to a vessel whose volume is equal to a cube Whc edge is one-tenth of a metre. TABLE OF UNITS. 10 Millilitres, ml. = 1 Centilitre = .6102 cu. in. :=: .338 fl. oz 10 Centilitres, el. = 1 Decilitre = 6.1022 " " = .845 gill. 10 Decilitres, dl. = 1 Zitre = .908 qt. = 1.0567 qt. 10 LlTBBS, L. = 1 Dekalitre = 9.06 " = S.6417 gal. 10 Dekalitres, Dl. = 1 Hectolitre = 2.8372+ bu. = 26.417 " 10 Hectolitres, HI. = 1 Kilolitre = 28.372+ " = 2ft4.17 " 10 Kilolitres, Kl. = 1 Myrialitre =283.75+ " =2&41.7 " The Hectolitre is used in measuring large quantities in both liquid aij dry measure. UNITS OF WEIGHT. 430. The Oranirne is the principal unit of weight, and is equal | the weight of a cube of distilled water whose edge is one centimetre. TABLE OF UHITS. 10 Milligrammes, mg. = 1 Centigramme = .15432 + oz. Tro 10 Centigrammes, eg. =3 1 Decigramme = 1.54S34+ ii bh 10 Decigrammes, dg. = 1 Granune = 15.43248 + (i (I 10 Grammes, G. = 1 Decagramme = .8527 + oz. Avo 10 Decagrammes, Dg. =S 1 Hectogramme = 3..527.S9+ l( u 10 Hectogrammes, Hg. = 1 Kilogramme or Kilo. = 2.20462 + lb. 10 Kilogrammes, Kg. = 1 Myriagramme = 28.04681 + »« 10 Myriagrammes, Mg. = 1 Quintal = 820.46219+ 11 10 Quintals, = 1 Touneau or Ton. =2204.6212 + Ik l:l'i i^' The Kilogramme or Kilo., which Is little more than 8) lb. Ayoir.J the conwKni weight in trade. Heavy articles are weighed by the To neaUf which is 804 lb. tnore than a common ton. #! Ii ER8, EXAMPLES. 191 wl unit formeasari Comparative Table of Units, 1 Inch = .0254 metre. 1 Ca. Foot = .2832 Hectolitre. IFoot = .3(M3 " 1 Cu. Yard -- .7646 Steres. 6.316+ cu. ft. 3.079+ cu. yd. lYai-d = .9144 1 Cord = 8.625 Steres. IMIle = 1.0()i)3 Kilometres. 1 Fl. Ounce = .()29-i8 Litre. 1 Sq. Inch = .0006452 sq. metre. 1 Gallon = 8.T82500. 7. 75000-i-1500. 2. 3900-fr-180, 5. 89600-J-800. «. 32000^400. 3. 8400-4-30a 6. 3420-H5400. i9. 9999-*-63. SHORT METHODS. 199 )-i-360. r-4-600. f anotlicr* t denomina- ng comsaou XirV xt¥j^^- -•J- X 11 xi. X j|^ X y. r Another. r common to VI. as directed : 10 ^..0027. 008. )y another. d and divisor reciO ted: )00-h1500. )00^400. ALIQUOT FABTS. 464. An Aliquot Part of a number is any number, integral or mixed, which will exactly divide it. Thus, 3, 2|, ^, are aliquot parts of 10 4:05. The aliquot parts of any number are found by divid- ing; by 3, 8, 4, 5, and so on, up to 1 less than the given number. Thus, 100 ^2 = 50; 100-!-3 = 83J; 100-+-4=25. Each of the quotients 50, 33}, and 25, is an aliquot part of 100. 460. The character @ is followed by the prioe of a unit or one article. Thus, 7 cords of wood @ $4.50 means 7 cords of wood at $450 a cord. 467. Commit to memory the following aliquot parts of 100, 1000, and $1. Table of Aliquot Parts, no = \ 25 = \ 20 = \ 14f = \ 12| = i 10 =tV of 100. 500 333i 250 200 166| = jt > of 1000. 142? 125 lllj 100 = i 50 ct. 33} ct. 25 ct. 20 ct. 16|ct 14| ct. 12} ct. lljct. 10 ct. = Oof$L 408. Prob. IX. — To multiply by using aliquot parts. t Multiply 459 hy33|. 3 ) 45900 ExHiAjrArnow.— We multiply by 100 by annexing two cipherB to the multiplicand, or by moving the decimal 15300 point two placee to the right. Bat 100 being eqoal to 8 times the multiplier 33}, the product 46B00 is 8 timeg m large as the required product ; hence wc divide by 3. f 1 v' ;■' '" J ::•'•■'' i f^^4 . 11 * u 1 r 1 1 i! 1 1 t 1 1 (■' ' {>, 1:^-; •'■1 ';i i M. ! I !! I! •■ iU .j i^^ 200 BUSINESS ARITHMETIC. Perform the following multiplications by aliquot parts : 2. 805x125. 5. 234x333|. 8. 58.9x250- 6. 809 X Hi. 9. 7.63x142? 7. 73x111^. 10. 4.88x81. 8. 85.8 xl6«. 4. 974x50. Solve the following examples orally, by aliquot parts. 11. What cost 48 lb. butter @ 25 ct. ? @ 50 ct. ? © 33^ ct. ? Solution.— At $1 a pound, 48 would cost $48. Hence, at 38J cts. a pound, which Ib i of |1, 48 pounds would coat I of |48, which in $16. 12. What cost 96 lb. sugar @ 12^ ct. ? @ 14? ct. ? @ 1G| ct. ? 18. What is the cost of 24 bushels wheat @ $1.33^? Solution.— At $1 a bai>hel, 34 bushels cost $24 ; at 33^ ct, which Is \ of $1 a bushel, 34 bushels cost $8. Hence, at $1.83| a bushel, 34 bushels cost the sum of $24 and $S, which is 132. 14. What cost 42 yards cloth @ $1.16|? @ $2.14f ? 15. What cost 72 cords of wood @ $4.12J ? @ $3.25 ? Find the cost of the following, using aliquot parts for the cents in the price. 16. 2940 bu. oats @ 33 ct. ; @ 50 ct. ; @ 25 ct. 17. 100 tons coal @ $4.25; @ $5.50; @ $6.12^ ; @ $5.33 J. 18. 280 yd. cloth @ $2.14f ; @ $1.12^ ; @ $3.25 ; @ $2.50. 19. 150 bbl. apples @ $420 ; @ $4.50 ; @ $4.33J. 20. 834 bu. wheat @ $1.33^ ; @ $1.50; @ $1.25 ; @ $1.16|. 21. 896 lb. sugar @ 12^ ; @ 14f ; @ 16|. 22. What is the cost of 2960 yd. cloth at 37^ ct. a yard? 35 = I of 100, hence 4)2960 12J=iof 25, hence 2) 740 370 374 ¥ Explanation.— At |1 a yard, 2960 yd. will cost $2960. But 25 ct. is i of $1, hence \ of ^2960, which is $740, is the cost at 35 ct. a yd. 2. Again, 13} ct. is the ^ of 25 ct., hence $740, the cost at 25 cts., divided by 2, gives the cost at 12j ct., which is $370. But 25 ct. + 12i ct. = 37J; hence, $740 + $370, or $1110, is the cost at 37 J ct. $1110 SHORT METHODS, 201 t parts : 6b.9 X 250. 7.63 X 142?. 4.38 x8i. parts. 1 (itiSSict.? :e, at 38 i cts. a ich is $16. ;.? @lG|ct.? .331? ct, which iB \ of a4 boBhelB cost 14?? $3.25? t parts for tbe ; @ $5.33i. ; @ |2.50. J5 ; @ $1.16|. a yard? loN.-At»layard, Icoet $2960. But I, hence ^ of $2960, \ is the cost at Yt\ ct. iB the \ of le cost at Vi\ ct, I $370, or$U10,ts 23. 495 bu. barley @ 75 ct. ; @ 03| ct. ; @ 87J ct. 24. 870 lb. tea (t^ GO ct. ; % Q%\ ct. ; @ 80 ct. ; @ 87J ct. 25. 4384 yd. cloth @ 12^ ct. ; @ 16 ct. ; @ 30 ct. ; ^ 86 ct. Obsurve^ that 10 ct. = i>o of 100 ct, and 5 ct. = ) of 10 ct. 20. 680 lb. coffee @ 87^ ct. ; @ 76 ct. ; @ 60 ct. 460* PnoB. X.— To divide by using aliquot parts. 1. Divide 7258 by 83J. 72.58 8 217.74 ExPLAKATiON.— 1. We divide by 100 by moving tbe deci- mal point two places to the left. 8. Since 100 is 3 times 331, the given divisor, the quo- tient 72.68 iB only i of the required quotient : huuco we multiply the 73.58 by 8, giving 317.74, the required quotient. 5. 894.8+125. 8. 460.854-250. 6. 98.54-4-50. 9. 90.638-J-25. 7. 879.0+33i. 10. 73090 -^333i. Perform by aliquot parts the division in tbe following : 2. 8.375-*-16|. 3. 9764-5-5. 4. 8730-S-8J. Solve the following examples orally, using aliquot parts : 11. At 33 J ct., how many yards of cloth can be bought )r $4? Solution.— Since $1, or 100 ct., is 8 times 33i ct., we can buy 3 yards for . Hence for $4 we can buy 4 times 3 yd., wtiich is 12 yd. Observe^ that In this solution we divide by 100 and mnltiply by 3, the imber of time^ 83i, the given price, is contained in lOOL Thus, $4=400 ct., -<- 100 = 4, and 4x3= 13. In the solution, the redaction of the $4 to Ints is omitted, as we recognize at sight tliat 100 ct., or $1, is contained pmes in $4. 13. How many yards of cloth can be bought for $8 @ 12J ct. ? 14? ct.? @33ict.? @16|ct.? @25ct. ? @10ct.? @o0ct.? j 8 ct. ? @ 5 ct. ? @ 4 ct. ? |l3. How much sugar can be bought at 12^ ct. per pound for For $8? For|12? For $30? For $120? 14. How many pounds of butter @ 83J ct. can be bought for '? For $10? For $40? u 1 ! M. :^l Jl ^■- m ,. /' ^1 ^^H ' " ( i m m In 4 ' ■ \ .. i r f i i ! i 202 BVSTNESS ARITHMETIC. Solve the following, performing the division by aliquot parts : 15. How many acres of land can be bought for $8954 at ^25 per acre ? At $50 ? At $33^ ? At $125? At $16.-; ? At $250 V 16. How many bushels of wheat can be bought for S6354 at $1.25 per bushel ? At $2.50 ? Obsen-e, $1.23 = | of $10 and |2.50 = V of J'lO. Hence by movino: the decimal point one place to the left, which will give the number of bu. at $10, and multiplying by 8, will give the number of bu, at $1.25. Multiply- ing by 4 will give the number at $2.50. 17. How many yards of cloth can be bought for $2642 at m\i ct. per yard? At 14f ct.? At 25 ct.? At $3.33.^? At $2.50? At$l.lH? At$1.42f? 18. What is the cost of 138 tons of hay at $12i ? At $14f ? At $16| ? At $25 ? At $13.50 ? At $15. 33^ ? At $17.25 ? BUSIE'ESS PEOBL^MS. DEFINITIONS. 70. Quantity is the amount of anything considered ii)| a business transaction. 47 1 . Pricey or Rate, is the value in money allowed for p| given unit, a given number of units, or a giv€?t part of quantity. Thus, in 74 bu. of wheat at |2 per buehel, the oHce is the value of a uritl of the quantity ; in 8735 feet of boardb at 45 ct. p* r 100 fe'^t, the price is thej value of 100 units. 472. When the rate is the value of a given number of] units, it may be expressed as a fraction or decimal. ThuB, cloth at $3 for 4 yards may be expressed as $2 per yard ; 7 foij every 100 in a given number may be expressed thti or .07. Hence, f of 1 means 5 for every 8 in 64 or 5 per S of 64, and .08 means 8 per 100. 'j^u.J'S'itKm BUSINi:SS PROBLEMS, 203 •r consWered in 473. Cost is the value in money allowed for aL entire quantity. Thne, in 5 barrels of apples at $4 per barrel, $4 is the price, and $4 <5or |20, tlie entire value of the 5 barrel:?, in the cost. 474. /*er Cent meana l^cr I hundred. Thus, 8 per ceut of $G00 means $8 out of every $100, which is $48. Hence a given per cent is the price or rale per 100. 475. The Sign of Per Cent is %. Thus, 8% is read, ii per cent. since per cent means per hundred, any given ptr cent may bo expressed with the sign jt or in the form of a decimal or common fraction ; thus, 1 per cent is written 1% or .01 or ,i_. •07 - T^a. 1.00 " IS". 7 per cent " 8t or some other fractionai pai are used. BU/SINESS PROBLEMS, 205 pea ? 7 per five I already given ai id second, to gWJ that are of con\ Lroblem should Bon clearly unde^ Ix, X, and XI, a lead of Percent 1, as the Bolution sr fractional pai FBOBLEMS. 483. Prob. I.— To find the cost when the number of inits and the price of one unit are given. 1. What is the cost of 35 lb. tea @ $^ ? Solution.— Since 1 lb. cost |f , 35 lb. vrili cost 86 times $f , which is (260) $25. Find the cost and explain the following orally : 2. 74 bu. apples @ $|. 6. 19 boxes oranges @ |4|. 3. 34 yd. cloth @ |2|. 4. 6f yd. cloth @ ||. 5. 44i lb. butter @ |f . 5|- 7. 17 tons coal («) 8. 98 cords wood @ $4y^^. 9. 9| yd. cloth @ f^V- Find the cost of the following, and express the answer in iollars and cents and fractions of a cent : 10. 52 yd. cloth @ |3J. 11. 18 bblsi apples @ |4f. 12. 84 bu. oats @ $f. 13. 83 lb. coffee @ $f 14. 32\\ lb. sugar @ $gV 15. 63 iV lb. butter Qi ^t^. 16. 169 acr. land @ |27i. 17. 25f cords wood @ $6f 18. How much will a man earn in 19| days at $2| per day ? 19. Sold Wra. Henry 36 J lb. butter @ 28| ct., 17y<^ lb. ^offee @ $.33|, and 39^1 lb. sugar @ $.14|. How much was lis bill? 20. A builder has 17 carpenters employed @ $2.25 per day. [ow much does their wages amount to for 24| days 1 484. Prob. II. — To find the price per unit when the ^ost and number of units are given. 1. If 9 yards cost $10.80, what is the price per yard ? Solution.— Since 9 yards cost 110.80, 1 yard will cost I of it, or |10.80+ I = $1.20. Hence, 1 yard cost $1.30. Solve and explain the following orally : 2. If 7 lb. sugar cost f 1.08, what is the price per pound ? 3. At $4.80 for 12 yards of cloth, what is the price per yard? !^ HI -^■• ,^r J Mili! ;li III 'ill: WA 1," ; 1 ; 1 206 BUSINESS ARITHMETIC, 4. If 12 lb. of butter cost $3.84, how much is it a pound ? 5. Paid $3.42 for 9 lb. of coffee. How much did I pay per pound ? Solve and explain the following : tf. A farm containing 282 acres of land was sold for $22184 What was the rate per acre ? Ans. $78.66 -j- . 7. A piece of cloth containing 348 yd. was bought for $515.91. What did it cost per yard? Ans. $1.4825. 8. Bought 236 bu. oats for $90.80. What did I pay a bu. ? 0. If 85 cords of stone cost $371,875, what is the price per cord? Ana. $4.?75. 10. A farmer sold 70000 lb. of hay for $542.50. How much did he receive per ton ? Ana. $15.50. 11. A merchant bought 42 firkins of butter, each containing 63| lb., for $735.67. What did he pay per pound ? 12. There were 25 mechanics employed on a building, each receiving the same wages ; at the end of 28 days they were paid in the aggregate $1925. What was their daily wages? 485. Prob. ni. — To find the cost when the number of units and the price of any multiple or part of one unit is given. 1. What is the cost of 21 ^b. sugar at 15 ct. for ^ lb. ? SoLTTTioTir.— Since I lb. cost 15 ct., 21 lb. must cost as many times.l5 ci. ae I lb. isi contained times in it. Hence, Fivi^l utep, 21-;- J = 27 ; Second step, $.15 X 27 = $4.05. Find the cost of the following : 2. 124 acres of land at $144 for 2f acres ; for 1| A. 3. 486 bu. wheat at §11 for 8 bushels ; at $4.74 for 3 bushels ; at $.72 for f of a bushel. 4. 265 cords of wood nt $21.05 for 5 cords. 5. 135 yd. broadcloth at $8.97 for 2i^ yd.; at $12.65 for 3| yd. B USINJSSS PROBLEMS, 307 74 for 3 bushels ; .; at $13.65 for 6. Wliat is the cost of 987 lb. coal, at 35 ct. i er 100 lb. ? Solution.— Ab the price is per 100 lb., we find the number of hundreds in 087 by moving the decimal point two places to the left. The pn'oe innl- tipliod by this result will give the required cost. Hence, $.35 x 9.87=$3.4545. the cost of 987 lb. at 35 ct. per 100 lb. Find the cost of the following bill of lumber 7. 2345 ft. at 11.35 per 100 (C) feet; 3«28 ft. at $.98 per C. ; 1843 ft. at $1.90 per C. ft. ; 8364 ft. ai $2.84 per C. ; 4384 ft. at !^27.o0 per 1000 (M) ft. ; 19364 ft. at $45.75 per M. 8. What is the cost of 84690 lb. of coal at $6.45 per ton \2000 1b.)? Observe, that pounds are changed to tone by moviii^ the decimal point ^ places to the left and dividing by 2. 9. What is the cost of 96847 lb. coal at $7.84 per ton ? 486. Prob. IV. — To find the number of units when the cost and price of one unit are given. 1. How many yards of cloth can be bought tor $28 @ $f ? Solution.— Since 1 yard can be bought for $^, as many yards can be bonuht for $28 as ^ is contained times in it. Hence, $28-+-$f - 49 yd. Find the price and explain the following orally : 2. How many pounds of coffee can be bought for $60 at $^ per ixjund ? At $f ? At $j% V Ar, $|? ? At $.33^ ? At $.4 'if 3. For $40 how many bushels of corn can be bought at |^ per bu. ? At $| ? At ^j\ V At $f;; ? At .$.8 ? At $« ? 4. How many tons of coal can be bought for $56 at .$4 a ton c At$7V At $8? At $14? At$6V At$9V At $5? Solve the foUowinff : 5. The cost of digging a drain at $3§ per rod is $187 ; what \s the length of the drain ? Ans. 51 rd. G. How many bur»hols of wheat at $li) can be purchased for fl840 V At $lf- ? At $li ? At $1 1 ? At $lf ? 7. The cost of a piece of cloth is $480, and the price per yard ^I'l ; how many yards does it contain ? 1 ' ■ ! i I it I!! f ■ i If f:^ i *^T -.V - ■n>^' ■-••.I' .|., .; :#' J, '} *• |i , ■'!■ V ' ■ ' !■ 208 BUSINESS ARITHMETIC. 8. A grain dealer purchased a quantity of wheat at $1.2( per bushel, and sold it at an advance of 9^"^ cents per bushel receiving for the whole $616.896 ; how many bushels did h( purchase? 9. A grocer purchased $101.65 worth of butter, at 35f cents a pound ; how many pounds did he purchase V Ans. 285 lb. 10. How many yards of cloth can be bought, at $2.75 a yard, for $1086.25? Ana. 395. 11. A farmer paid $14198 for his farm, at $65f per acre how many acres does the farm contain? Ans. 217 A. 487. Pros. Y. — To find the number of units that cai be purchased for a given sum when the cost of a multiply or part of one unit is given. 1. At 19 ct. for f of a yard, how many yards can be bouglit for $8.55? SoLUTTON— 1. Since § yd. cost 19 ct., J must cost \ of 19 ct., f . 9} ct., an( I, or 1 yard, mast cost 3 times 9| ct., or 28| ct. 2. Since 1 yard cost 28J ct., as many yards can be bought for $8.55 a 38} ct. are contained times in it. Hence, $8.55 -»- $.285 = 80, the number o yards that can be bought for $8.55, at 19 ct. for % yd. 2. A town lot was sold for $1728, at $3 per 8 sq. ft. front of the lot is 48 ft. What is its depth ? Am. 96 ft. 3. How many bushels of com can be bought for $28, 32 ct. for I of a bu. ? At 28 ct. for | bu. ? 4. How many tons of coal can be bought for $277.50, at for I of a ton ? At $8 for f of a ton? Ana. 37 T. 5. A piece of doth was sold for $34.50, at 14 yards per $j How many yards did the piece contain? Ana. 483 yd. 6. A cellar was excavated for $408.24, at $4.41 for evel 7 cu. yd. The cellar was 54 ft. by 36 ft. How deep was it ? 7. A pile of wood was bought for $275.60, at $1.95 for 3 cor feet. How many cords iw the pile ? Ana. 53 cd. 8. A drove of cattle was sold for $3738, at $294 for eve^ 7 head. How many head of cattle in the drove ? Ana. 89. \ ITIC, BUSINESS PROBLEMS. JiOD of wheat at |1.20l 488. Prob. VI.— To find the cost when the quantity % cents per bushelMs a compound number and the price of a unit of one my bushels did hewenomination is given. butter, at 35f centE leY Ans. 2851b. fht, at $2.75 a yardj An8. 395. ., at $65f per acre] Ans. 217 A. of units that caii cost of a multiply rards can be bouglii iofl9ct.,f .9ict., and be bonght for $8.55 a] = 30, the number < 3 per 8 sq. ft. ? Ans. 96 ft. jbought for $28, for $277.50. at Ans. 37 T. I at 14 yards per $| Ans. 483 yd. I at $4.41 for evei [ow deep was it ? at $1.95 for 3 coi Ans. 53 cd. at $294 for evei )ve? Ana. 89. 1. What is the cost of 8 bu. 3 pk. 2 qt. of wheat, at $1.44 per t)ushel? Solution. — ;. Since $1.44 is the price per bushel, $1.44x8, or $11.53, is the cost of 8 bushels. 2. Since 2 pk. = J bu., $1.44 -*• 2, or 72 cts., is the cost of 2 pk., and the \ of 72 ct., or 36 ct. . is the cost of 1 pk. 3. Since there are 8 qt. in 1 pk., 2 qt. = J pk. Hence the cost of 1 pk., 36 ct -•- 4, or 9 ct., Is the cost of 2 qt. 4. The sum of the cost of the parts lust equal the cost of the whole quantity. Heuce, $12.69 is the cost of I bu. 3 pk. 2 qt., at $1.44 per bu. 2 ) $1.44 8 11.52 2) 72 4) 36 9 Cost of 8 bu. " ♦* 2 pk. *' " 1 pk. " " 2 qt. $12.69, Ans. Find the cost of the following orally : 2. 7 lb. 8 oz. sugar, at 12 ct. per pound ; @ 14 ct. ; @ 26 ct. 3. 7f yd. ribbon @ 15 ct. ; @ 40 ct. ; @ 25 ct. 4. 19 bu. 3 pk. 6 qt. of apples, @ $1 per bushel. 5. 13 lb. 12 oz. butter, at 34 ct. per pound ; at 40 ct. Solve the following : 6. What will 5 T. 15 cwt. 50 lb. sugar cost, at $240 per ton? 7. Find the cost of 48 lb. 9 oz. 10 dwt. of block silver, at $12 |er pound. Ans. $585.50. 8. Sold 48 T. 15 cwt. 75 lb. of hay at $15 per ton, and 32 bu. pk. 6 qt. timothy seed at $3.50 per bushel. How much did I sceive for the whole ? Ar}S. $847.9 + . 9. How much will a man receive for 2 yr. 9 mo. 25 da. service, \i $1800 per year? Am. $5075. 10. Find the cost of excavatir.g 240 cu. yd. 13^ cu. ft. of |arth, at 50 cts. per cubic yard. 11. How much will it cost to grade 8 mi. 230 rd. of a road, It $4640 per mile ? Ans. $40455. ! ^^H ) J 1 ^^H ^ 1 . i 1 ij. ; i i M III li^H; 1!H; ill 210 BUSINESS MUTHMETIU, 489. Pkob. Vll. — To find what part one number is of another. 1. What i>art of 12 is 4? Solution.— 1 is ,'4 ol !2 and 4 being 4 timcB 1, is 4 timt!^ ,", of 12, whicli is 1 9 = 3 ; lienoe, 4 is J of 13. Observe, that to atstertuin what part one number iw ofauoiher, we may at once write the lormer a» the numenitor and the latter an thu denominator of a fraction, and reduce the fraction to its lowest terms (5i45). 3. What part is 15 of 18 V Of 25 ? Of 24 ? Of 45 ? 8. What part is 3G of 48 ? Of 38? Of 42 V Of 72 ? 4. I is what part of f V Solution.— 1. Only unltv of the same integral and fractional denomina- tion can be compared (144); hence we reduce % and ? to \\ and J?, and place the numerator 14 over t.'te numerator 18, giving \% — l\ hence J is 2 off. 2. We may express the relatioii of the fractions in the form of a complex fraction, and reduce the result to a simple fraction (289). Thus, 3 K = I. Hence, ? is J off. 5. 7^ is how many times | ? 6. 5y inches is what part of 2,', yards? (See 370.) 7. 29 /tj rods is what part of 1 mile ? 8. f is what part of 11 ? | is what part of 2 J ? 9. 11^ is how many times 22 ? 10. What part of a year is 24 weeks ? 8 weeks 10 days ? 11. A man's yearly wages is $950, and his whole yearly ex- penses $590.80. What part of his wages does he save each year ? 12. 3|% Is what part of 9% ? 7] % is what part of 8|% Y 13. A man owning a farm of 210^ acres, sold 117^ acres. What part of hi^^ whole farm has he still left ? 14. 4% is what part, of 12 ;^ ? 8 % is what part of 14% "? 15. Out of |750 I paid |240. What part of my money have I still left ? ■ Alls. 11, or .68. I ! i BUSINESS PROBLEMS, 211 ne number is of iiuur I't of 12, which partof 8|%r 3old 117^ acres. 16. Illustrate in full the process in the 14th and 15th ^amples. 400. Prob. VIII. — To find a given fractional part of a Iven number. 1. Find? of 238. I SoLtTTioN.— We find | of 2!« by dividing it by 7 ; hence 2;« -*- 7 = 34, the )f 238. But I Is 3 times \ ; hence »4 x 3 = 102, the ? of 238. 1 2. Find ^ of 34() ; ^^^ of 972 ; Vtt of 560. k Find I of 48 : of 96 ; of 376 ; of 1035. k Find \ of $75 ; I of $824.60 ; ^^ of $3.25. \Obiierve, that | of $75 meauB Buch n number of dollars as will contain $4 every $5 in $75 ; hence, to find the ] oi" |T5, we divide by 5 and multiply quotient by 4. |5. Find 7% of 828. SoLcnoN.— 1. 1% means tJ^. We find ^Jo by moving the decimal ^nt two places to the left (459). Hence 1% or j §7; of 328 is equal to X 7 = 22.96. 2. We usually multiply by the rate first, then poini off two decimal places jlhe product, which divides it by 100. J. How much is -? of 157 acres ? -^ of 84 bu wheat ? 5 7 r. What is 8% of $736 ? 4% of 395 lb. butter 'f Find 8. ^% of 278 lb. 9. 5% of 300 men. 10. 7% of 28 yd. Find 11. 4ff^ of 284 mi. 12. 12i%of732. 13. f%of$860. 14. Find the amount of $832 + 1% of itself. 15. Find the amount of $325 + 7% of itself. 16. A piece of cloth contained 142 yd. ; 15% was sold : hoi;^ [uy yards yet remained unsold ? 17. A firkin of butter contained 72g lb. ; | of it was sold : many pounds are there left ? '■i ' ■ III 212 BUSINESS ARITHMETIC, •'V f Jl ■: ■'.;( ,1 1 18. J. Henderson's fann contained 284 acres, and H. Lee's 8J% less. How many acres in H. Lee's farm? 10. If tea cost 96 ct. per pound and is sold at a loss of 12^ fc, what is the selling price ? 20. A merchant bought 276 yards cloth at $3.40 per yard. He sold it at 25% profit. How much did he realize, and what was his selling price ? 491. Prob. IX.—To find a number when a fractional part is given. 1. FiDd the number of which 84 is }. Solution .—Since 84 Is J of the number, } of 84 mnet be J ; hence 84 -♦- 7 = 13 is the i of the required number. But 9 times I is equal to the whole ; hence, 12 x 9 = 108, the required number. 8. Find the number of yards of cloth of which 135 yd. is /^ . 8. $36 is I of how many dollars ? $49 is I of how many dollars ? 4. John has $756, which is | of Norman's money ; how many dollars has Norman ? Ans. $1323. 5. The profits of a grocery for one year are $3537, which is 1^ of the capital invested. How much is the capital ? 6. Find the number of dollars of which $296 are $8,'^ , or .08. First Solution.— Since $296 are tSt? of the number, ^ of $290, or $37, are rio ; hence Igg, or the whole, is 100 times $37, or $37 x 100 = $3700. Second Solution.— Since $296 are twt of the number, J of $296 is rh>^ and i of too times $296 is fgg, or the required number. Hence, $296 x 100 = $29600. and i of $29600 = $3700, the required number. From these solutions we obtain the following rule for finding a number when a decimal part of it is given : I BIJIiE. 492. Move the decimal point as many places to the right as ihere are pku^es in the given decimal, annexing ciphers if neces- sary, and divide the result by the number expressed by the sig- nificant figures in the given decimal. JUSTNESS PROBLEMS. 2ia A. |8%,or.08. j9G,or|37,are $3700. $296 is \hr,y jce, $296 X 100 Find what number 7. 16 is 8% of. 8. 34 is 6% of. 9. 84 is 7% of. 10. $73 arc 9% of. 11. 120 yd. are 5% of. 13. 50 bu. are 8/o of. find what number 18. I is 4% of. 14. $§ is 7^ of. 15. \ pk. is 8% of. 16. .7 ft. is 5% of. 17. .09 is 4% of. 18. .48 is 13% of. Find what number 19. 8} is 9% of. 30. $3.10 is 6% of. 31. 7|is9% of. 23. 27ii8 5% of 23. ^yd. is 8% of. 34. .96 is 12% of. 25. A grocer purchased 186 lb. butter on Saturday, which is 6% of the entire quantity purchased during the week. What was the week's purchase ? 26. A merchant sells a piece of cloth at a profit of 80 ct. a yard, which is 30% of what it cost him. "What was the buying price per yard ? 27. A man's profits for one year amount to $2840, which is 8% of the amount he has invested in business. What is his investment ? 38. A mechanic pays $13 a month for house rent, which is 16% of his wages. What does he receive per month ? 29. 12% of f is 9% of what number? 30. If in a certain town $3093.75 was raised from a |% tax, what was the value of property in the town ? 81. An attorney receives $1.75 for collecting a bill, which is 2^ per cent of the bill. What is the amount of the bill ? 32. A man having failed in business is allowed to cancel his debts by paying 30% . What does he owe a man who receives $270? Ans. $1350. 33. A man sold his house for $1000, which was 13% of the sum he received for his farm. What was the price of the farm? Ans. $8333.33^. 34. How many acres in a farm 14% of which contains 43 acres? 35. J. Simpson has 35 % 6f his property invested in a house,^ 10% in a farm, 5% in a bam, and the rest in a grove worth $4800. What is the amount of his property ? 'ft H j ^14 BUSJ JV A'S a A H J TH M E Tl V . ml m I 49c}. PiiOB. X. — To express the part one number is of another in any given fractional unit. 1. How many fftha of 3 is 8 ? Solution.— Since } is J of 8, there must be as xaanyffths of 8 In 8 as ^ is contained times lu it. 8 + J 13S. Uence, 8 Ib 1^ of three. 5 Solve the folio wiug orally : 2. How many fourths of is 7 ? Is 5 ? Is 13 V Is 20 V 3. How many hundredths of 36 is 9 ? Is 4 ? In 18? Is 13 V 4. $13 are how many tenths of $5 V Of !|8 V Of $15 ? 5. 42 yards are how many sixthB of 2 yd. ? Of 7 yd. ? Of 3 yd.? C. ^^hai per cent of $11 arc $3, or $8 are how many huu dredthsof $11? First Solution.— Since ^V is pjjj of $11, there must be ab many hun- dredths of $11 in $3 as ^o\j is contained times in $8. |3 + tVo = 8 x W"- = Vt" = 27t\. Hence, $3 are ?^I^, or 27A^ of $11. Second Solution.— Since (489) |8 are y*, of |11, we tmve only to reduce rV to hundredths to find wliat per cent $3 are of $11. A — iVJo = Zhl = 27t\?S. Hence, $3 are 27A5« of $11. From these solutions we obtain the following rule for finding what per cent or what decimal part one nmubcr is of another : BULB. 494. Express the former number as a fraction of the lattei (489), and redu^ce this fraction to hundredths or to the required decimal (327). Find what per cent 7. $36 are of $180. 8. 13 is of 73. 9. 16 is of 64. 10. $46 are of $414. 11. 13 oz. are of 5 lb. 12. 7 feet are of 8 yards. 13. If is of $|; of $5;of|2|. 14. 2 bu. 3 pk. are of 28 bu. 15. 284 acres are of 1 sq. mi. 16. 31b. 13 oz. are of 9 lb. 17. 48 min. are of 3 hr. 18. f of a cu. ft. is of 1 cu. yd. H, BUSINESS i' n O B L E M 8, nb nber is a In 8 aB ^ uany liuu have only to » — •"-I'- ll. IT - It"" le for finding of another : of the Uttti or to the ^5 ; of |2|. )f 28 bu. \ eq. mi. )f91b. lir. lof 1 cu. yd. 19. A morcliant invoetc^l ^$3485 in gomls which ho had to aell for ij^JOTS. Wliat per <'eDi of lii« investment did he lose? 20. Find what i>er cent 3 bu. 3 i>k. are of 8 bu. 3 pk. 5 iit. 21. A man paid $24 for the i;se of |3()0 for one year. What rate per cent did he pay ? 22. Find what por wnt | of a t^q. yd. is of I of a sq. yd. 23. A dmiorgist paid 84 ct. an ounce for a certain medicine, and sold it at $1.36 an ounce. What jxr cent profit did lie make? SoLUTioN.-tl.86-I.S4= i.^«; M = JiJ2.! = ^^ -= 61Jfj<. 24. J. RoHS deposited *2500 ju a bank, and again deposited enough to make the whole amount to |2750. What per ceut of the first deposit wad the last? Anti. 10. 25. When a yard of silk is bought lor |1.20 and sold for $1.60, what per cent is the profit of the buying price? 26. A man owed me $350, but fearing he would not [)ay it I a.j^reed to take .$306.25 ; what per cent did I allow him ? 27. A farmer owning 386 acres sold 148 acres. What pur cent of his original farm does he still own ? 38. Gave away 77i^ bushels of potatoes, and my whole crop was 500 bushels ; what ^ of the crop did I give away ? 29. A man pays $215.34 per acre for 4i acres of land, and lets it a year for $33,916 ; what % of the cost is tlie rent ? 495. Prob. XI. — To find a number which is a given fraction of itself greater or less than a given number. 1. Find a number which is § of itself less than 28. SoLTTTiON.— 1. Since the required number is I of itself, and in | of iteelf less than 28, hence 28 is j + i or j of it. 2. Since 28 is I of the number, | of 28, or 4, is {. Hence |, or the whole of the required number, is 5 times 4 or 20. Solve the following orally : 2. What number is | of itself less than 15 ? Less than 40 ? Less than 75 ? Less than 26 ? Less than 32 ? » 216 BUSINESS ARITH3IETIC, m 3. What number increased f of itself is equal 100 ? la equal 80 ? Is equal 120 ? Is equal 13 ? Is equal 7 V 4. Find a number which diminished by § of itself is equal 56. Is equal 70. Is equal 15. Is equal 5. Solve and explain the following : 6. What number increased by 7% or il^ of itself is equal 642? Solution.— 1. Since a number increar-cd by 7^ or , l„ of it?elf iB \%% + iJo = \ll of itself, 642 is \Zl or 107^ of the rt-quired number. 2. Since 642 is \%l of the required number, for every 107 in 642 there must be 100 in the required number. Hence, 642-«-107-t;. and 6 x 100=600, the required number. Observe^ that 642 -»- 1.07 is the same as dividing by 107 and mnltiplyinj," by 100 (349); hence the following rule, when unnmbcr has been increased or diminished by a given pf t cent or any decimal of itself: RULE. 496. Divide the given number, according an it is more or less than the required number^ by 1 increased or diiujiinhed by the given decimal. 6. A regiment after losing 8% of its number contained 736 men ; what was its original number? Ans. 800. 7. A certain number increased by 80^ of itseli is 331.2 ; what is that number? Aiis. 184. 8. By running 155^ faster than usual, a locomotive runs 644 miles a day ; what was the usual distance per day ? 9. What number diminished by 25% of itself i.s G54? 10. What number increased by 15% of itself is equal 248.40? 11. A man who has had his salary increased h',( now receives $1050 a year ; what was his former salary ? Ans. ^1000. 12. A tailor wlls a coat for $8, thereby gaining 25% ; whai did the coat cost him ? Ans. $6.40. 13. A teacher lays up 12|% of his salary, which leaves him $1750 to spend ; what is his salary ? .1 ns. $2000. 14. T. Laidlaw sold his farm for $3960, wlilrh was 10;^ lesj than he gave for it, and he gave 10% more than it was worth ; what was its actual value ? Ans. $4000. RULES FOR PERCENTAGE. 217 Is eqtial IB equal : is equal riBigg + iSo n 642 there 6^100=600, APPLIOATIOISrS. 497. Profit and Loss, Commission, Insurance, Stocks, Taxes, and Duties, are applications of Business Problems VIII, IX, X, XI. The rate in these subjects is usually a per cent. Hence, for convenience in expressing rules, we denote the quantities by letters as follows : 1. /{ repreecnts the Base^ or naraber on which the percontagc is reckoned. 2. li represents the Rate per cent ezprci^sed decimally. 3. I* represents the I*ereentage, or the part of the Base which is de- noted by the Jiate. 4. A repreeentH the Amonnt, or sum of the Base and Percentage. 6. JO represents the Hifferenee, or Base less the Percentage. FormtUcB, or Rules for Percentage, 498. PaoB. VIIL P = JB X JJ. Read, j 499. PROB.IX 500. PROB.X. 501. Pbob. XL B = B = 1 + R 1—R' Read Read, I Read, \ Read I'Ae percentage is equal to the base multiplied by the rate. ( The base is equal to the per- ' I centage divided by the rate. The rate is equal to the per- centage divided by (he base. The base is equal to the amount divided by 1 plus the rate. The base is equal to the d\ff"'nce divided by 1 minus the rate. •I 502. Refer to the problems on pages 211 to 214 inclusive, and answer the following questions regarding these formula; : 1. Wliat is meant hj B x R, and why is P = B x II? Illus- trate your answer by an example, giving a reason for each step. 2. Why is P-t-i? equal Z?? Give reasons in full for your answer. 3. If B is 135%, which is the greater, P or B, and why? 15 y-.i. 218 BUSINESS ARITHMETIC, f>- ) 4. Why is R equal to P -5- J5, and how must the quotient of P -*- JB be expressed to represent B correctly ? 5. If R is 248%, how would you express R without the sign %? 6. What is meant by J. ? How many times JR in P (491) ? How many times 1 in ^ ? How many times \ + R must there be in A and why ? 7. Why is B equal to ^ -r- (1 + i2) ? 2) is equal to B minus how many times B (490) ? 8. Why is B equal to 2) -5- (1 - i2)? Give reasons in full for your answer. ¥' T PEOPIT AND LOSS. 503* The quantities considered in Profit and Loss corres- pond with those in Percentage ; thus, 1. The Costt or Capital invested, is the Base. 2. Tlie Per cent of Profit or Loss is the Rate. 3. The Profit or Loss is the Percentage. 4. The Sellinfj Price when equal the Coat plus the Profit is the Amount ; when equal the Cost minus the Loss is the Difference, EXAMPLES FOR PRACTICE. 504. 1. A firkin of butter was bought for $19 and sold atj a profit of 16 % . What was the gain ? Formula, P = B x R, Read, ProJU or Loea — Cost x Rate %, Find the profit on the sale 2. Of 84 cd. wood bought @ |4.43^, sold at a gain of 20^. 3. Of 320 yd. cloth bought @ $1.50, sold at a gain of 17%. 4. Of 873 bu. wheat bought @ $1.25, sold at a gain of \^%.\ Find the loss on the sale 5. Of 180 T. coal bought @ $7.85, sold at a loss of 8J%. PROFIT AND LOSS, 219 tient of out tlie xst there gaud sold at] ;ainof20f.. rain of 17 fc. gain of mP )8 6. Of 134 A. land bought @ $84.50, sold at a loss of 21^ %. 7. If a farm was bought lor $4800 and sold for $729 more than the coat, what was the gain per cent? Formula, R = P+B. Read, Hats % Gain = Profit -t- Cost. 8. If ^ of a cord of wood is sold for f of the cost of 1 cord, what is the gain per cent ? 9. A piece of cloth is bought at $3.85 per yard and sold at $2. 10 per yard. What is the loss per cent V 10. Find the seUi/ig price of a house bought at $5385.90, and sold at a gain of 18 ^ . Formula, A = B x (1 + R). Read, 8eUing Price = Coat x (1 + Rate % Gain). 11. Corn that cost C5 ct. a bushel was sold at 20% gain. What was the selling price ? Ans. 78 ct. a bu. 12. A grocer bought 43 bu. clover seed @ $4.50, and sold it ill small quantities at a gain of 40 /t. What was the selling price per bu. and total gain ? 13. Bought 184 barrels of flour for $1650, and sold the whole at a loss of 8 % . What was the selling price per barrel ? Formula, D - B (1-R). Read, Selling Price = Cost y.{\— Hate % Loss). 14. C. Baldwin bought coal at $6.25 per ton, and sold it at a loss of 18%. What was the selling price ? 15. Flour was bought at $8.40 a barrel, and sold so as to lose 15%. What was the selling price? 16. Sold a house at a loss of $879, which was 15% of the cost. What was the cost ? Formula, B = P-^R. Read, CokI = Profit or Loss + Rate %. 17. A grain merchant sold 284 barrels of flour at a loss of !fG74.50, which was 25% of the cost. What was the buying [and selling price per barrel? 18. A drover wished to realize on the sale of a Hock of |23f5 sheep $531, which is 30% of the cost. At what price per |liead must he sell the flock ? 19. Two men engaged in business, each having $4380. A ^ 220 BUSINESS ARITHMETIC. r(... If I fh ! a ^, and the remainder at a gain of 24 ;6, what is my gain? 23. A man bought 24 acres of land at $75 an acre, and sold it at a profit of 8.V % . What was his total gain ? 23. Fisk and Gould sold stock for $3300 at a profit of 33^%. What was the cost of it ? 24. A merchant sold tsloth for $3.84 a yard, and thus made 20 */€ . What was the cost price ? 25. Bought wood at $3.25 a cord, and sold it at an average gain of 30 '/r . What did it bring per cord ? 26. Bought a barrel of syrup for $20 ; what must I charge a gallon in order to gain 20% on the whole ? 27. If land when sold at a loss of 12^% brings $11.20 per acre, what would be the gain per cent if sold for $15.36 ? COMMISSION. 505. A Coinmissioit Merchant or Agent is a person who transacts business for anotlier for a percentage. 50f>. A Broker is a person who buys or sells stocks, bills of exchange, etc., for a percentage. 507. Coniinission is the amount paid a commission mer- chiint or agent for the transaction of business. 508. lirohemge is the amount paid a broker for the transaction of business. 509. The JVf t Proceeds of any transaction is the sum of money that is left after all expenses of commission, etc., are paid. i more ind, for 53|% of )f 24^6, d sold it iu8 made n average L cbarge a $11.20 per L36? is a person 81001^9,131118 isslon ineT- iter for the is the sum Mission, etc., COMMISSION, 221 510, The quantities considered in Commission correspond with those in Percentage ; thus, 1. The amount of money invested or collected is the Itaxe. 2. The per anf allowed for eervicen in the Ratv. 3. The Commission or Brokerage in the I'erci'tttnffe, 4. The sum invested or collected, plus thccommiusion^ is the Amount; minus the commission is the Uifferenve. EXAMPLES FOR PRACTICE. 511, Let the pupil write out the formulae for each kind of examples in Commission in the same manner as they are given in Profit and Loss. What is the commission or brokerage on the following : 1. The collection of $3462.84, commission 2^ % ? 2. The sale of 484 yds. cloth (.. ^ :^ 2f>. In commercial language, the two classes of U. S. bonds are distinguished from each other thus : (1.) U. S, 6*M, bonds payable at a fixed time. (2.) U. s. 6*'« 5'ito, boHdf payable, at the option of the Government, at any time from 5 to 20 years from their date. 530. The liability on bank stocks in Canada is limited to double the amount of capital subscribed. On other stocks the liability of shareholders is limited to the amount of subscribed capital. EXAMPLES FOR PRACTICE. 53 .1 . Let the pupil write out the formula for each class of examples, as shown in Profit and Loss : 1. Find the cost of 120 shares of the Toronto Bank stock, the market value of which is 108, brokerage 1%. Solution.— Since 1 phare cost 108% + J^, or lOSi^ of $100 = 108.1, the coet of 1-20 shares wUl be |108i x 120 = $13020. 2. What is the market value of 86 shares in the Freehold Loan Company, at 3^% premium, brokerage -^ '/c ? 3. Find the cost of 95 shares bank stock, at G% premium, brokerage * ^ . 4. How many shares of the Dominion Telegraph stock at S^ip discount can be bought for f 7030, brokerage ^ % ? Solution.— Since 1 share cost 100;«-8;«+ Jj<, or 92.\;« of $100 = $92.50, as many shares can be bought as $92.50 are contained times in $7020, which is 76. How many shares of stock can be bought 5. For $10092, at a premium of 5;^, brokerage iii I 226 B UiSINESS ARITHMETIC, A 6. For $13428, at a discount oil%, brokerage \'/c ? 7. For $108rive, the stock yielding 7% ? 29. 0. E. Bonney sold $0000 Pacific Railroad G's at 107, and with a part of the proceeds bought St. Lawrence County b(jnds at 90, yielding 6% dividends sufficient to give an annual income of 8180 ; how much has ho left ? 30. What rate of income can be derived from money invested in the stock of a company paying a semi-annual dividend of 5%, purchased at 84V %, brokerajje i% ? 31. What must I pay for bonds yielding 4J % annually, that my investment may ])ay 6^ ? 32. What must be paid for stocks paying 5 per cent, that the investment may return 8^ ? 33. A man bought a farm, giving a note for $3400, payable in gold in 5 years ; at the expiration of the time gold waa 175% : what did his farm cost in currency ? I'V I .«fjs':: 228 BUSI^'ESS ARITHMETIC, DUTIES OR CUSTOMS. 5!$2. Duties or Customs are taxes levied by the govern- ment upon imported goods. 533. A Sjiecific Duty is a certain sum imposed upon an article without regard to its value. 534. An Ad Valorem Duty is a per cent assessed upon the value of an article in the country irom which it is brought. 535. A Tariff is a schedule giving the rates of duties fixed by law. 536. The following deductions or allowances are made before computing specific duties : 1. Tare.— An allowance for the box, cask, bag, etc., containing the merchaudiBe. 2. T,eakaffe.—An allowance for waste of liquors imported in cas^lts or barrels. 8. Breakage.— An allowance for loss of liquors imported in bottles. r" I! EXAMPLES FOR PRACTICE. 537. 1. What is the duty on 420 l)oxos of raisins, each containing 40 pounds, bought for 8 ceyits a pound, at 20 per cent ad valorem ? 2. Imported 21 barrels of wine, each containinpr 31 gfallons ; 2% being allowed for leakage, what is the duty at 40 cents per gallon ? 3. A merchant imported from Havana 100 boxes oranges @ $2.25 per box ; 75 hogsheads of molasses, each containing 63 gal., @ 23 cents i^er gal. ; 50 hogsheads of sugar, each containing 340 lb., @ 6 cents per lb. The duty on the molasses was 25%, on the sugar 30%, and on the oranges 20%. What was the duty on the whole ? UK VIEW, 220 4. What i.H the duty on 320 yards of cloth, invoiced at |1.15 |)er yard, at 20 ;c ad valorem? ^overn- ,pon an aBsoBScd ich it is )f duties re made talnlng tbe in cat^kB or bottles. isins, each iO X)er cent |l gallons ; cents per oranges (a ling 63 gal., [containing was 25%, lat was the 5. At VZ'/o ad valoroiu, what is the duty on 100 barrels of kerosene, invoiced at $.18 a gallon, 2% leakage? BEVIEW AND TEST QUESTIONS. 5J58. 1. When a fraction is to be divided by a fraction, why can the factors that are common to the denominators of the dividend and divisor bo cancelled ? 2. How does moving the decimal point one or more places to tlio left or right affect a number, and why? 3. Sliow that multiplying by 1000 and Hubtracting three times the multiplicand from the product is the same as multi- plying by 997 . 4. Define Base, Percentage, Amount, and D Terence. 5. When the amount and rate per cent is given to find the base, why add the rate expressed decimally to 1 and divide by tlip result . 6. Represent the quantities by lettt s and ^^ rite a formula for solving each of the following problems (41>7) : I. Given, the Cost and the Profit, to find the rate per cent profit. II. Given, the rate per cent profit and the selling price, to find the buying price. III. Given, the amount of money sent to an agent to pur- chase goods and the rate per cent commissio.i, to find the amount of the purchase. IV. Given, the rate at which stocks can be purchased, to find how much can be secured for a given sum. V. Given, the rate at which stocks can be purchased and the rate per cent of dividend, to find the rate per cent of income on the investment. VI. Given, the premium on gold, to find how much can be purchased for a given sum in currency. h- i\ If, INTEREST. i»:?;i DEFINITIONS. 539. Interest is a sum paid for the ?/«« of motiey Thus, I owe Wm. Henry $200, which he allows me to nee for one year after it in due. At the end of the year I pay hira Ihe $200 and $14 for its use. The $14 is called the Interest and the $200 the Principal. 540. Principal is a sum of money for the use of which interest is paid. 541. Rate of Interest is the numlierof units of any denomination of money paid for the use of 100 units of the same denomination for one year or some given interval of time. 542. The Amount i? the sum of the principal and interest. 543. Simple Interest is interest which falls due when the principal is paid, or when a partial payment is made. 544. Legal Interest is interest reckoned at the rate per cent fixed by law, 545. Usury is interest reckoned at a higher rate than is allowed by law. 540. The legal rate of interest for Canada is six per cent ; for England, five per cent ; and lor Ireland, six per rent. The following tahle ^ivea the lc<;al rates of interest in the different States of the U.S. Where two rates are given, any rate between these limitK in allowed, if apecifled in wiiting. When no rate is named in a paper involving interest, the UgcU or lowest rate is alwaye nndorstood. SIMPLE INTEREST, 231 • one yew iU for itB jf whicb ts of any its of the I of time. Id interest. due wben [ade. le rate pet ite than is per cent ; 'ent. the different u allowet^ if Iving Interest. STATES. RATE %. STATES. RATB %. Ala.. ..| Aik....| 8i I (i Any, 10 Any' I Cal .10 Any 1 ■ iCuiiii... ' 1 i 1 c.i.... 10 Anv 1 Dnkota. 7 Anyl 1 Do! ... . 6 1 D.C.... 6 10 ■ Flor. .. 8 Any ■ Goo... 7 10 1 Idabc* . . 10, lU lud... Iowa . . Kan... Ken. . . La.. .. Maine. Md. . . . Ma^s . . Mich.. Minn.. Mibs. . 6 (i 6 7 6 5 6 6 6 7 7 6 10 : 10 10 12 i 10 , 8 ' Any! I Any ; 10 12 10 ! STATES. RATE %. \ 1 STATES. RATEiC Mo 6 10 S.C... 1 7 Any Montana 10 , Tenn. I ^ 10 N.H.... 6 1 Texas. . ^ 8 12 N. J.... 7 Utah... no Any N. Y.... 7 Vt 6 X. C... 6 8 Va 1 6 12 Neb 10 15 1 i W.Va.. 6 1 Nevada . 10 Any| W. T... 10 Any ;01iio....i 6 8 Wis.... 7 10 Ore^'on ., 10 , 12 , \Vy.. .. 12 i Penii....| 6 i 7 1 |Ri 1 e Any 54:7. PiioB. I. — To find the simple interest of any given sum for one or more years. 1. Find the interest on $384 for 5 years, at 7%. Solution.— 1. Since the intcreet of $100 for one year is $7, the Interest of $1 for one year is $.07. Hence the interest of $1 for 5 years is |,07 x 5 ^ $.35. 2. Since the interest of $1 for h yr. i? $.35, the interest of 13^4 for the same time must ho 384 times $.35, or J134.40. Hence the following RUIiE. 548. /. Fetid the interest of $1 at the given rate for the given time, and multiply this resnU by the number of dollars in (he glccn principal. II. To find the amount, add the interest and principal. EXAMPLES FOR PRACTICE. 540. Find the interest on the tollowing oraDy : 1. 11200 for 3 years at d%. 2. $800 for 2 years at 4ji. :{. $200 for 5 years at 6^. 4. $90 for 2 years at 7%. ■». $600 for 4 years at 5%. 0. ?;T0 for4 youwatS^. 7. $100 for 12 years at 9%. 8. $400 for 8 years at 5%. 9. $1000 for 5 years at 8%. 10. .*;G00for lyeursat 10%. 1 1 . $500 for 5 yj'ars at o % . 12. $20 for 3 years at 9%. "i -, 232 BUSINESS ARITHMETIC, i.'.'i ■ Find the interest on the following : 13. $784.25 for 9 years at 4$^ . 20. !s:293.50 for 6 years at 45 % 14. $245.36 for 3 years at 7/o. 21. .$375.84 for 3 years at 9J%. 15. $836.95 for 2 years at ^%. 22. $600.80 for 9 years at 8^%. IG. $705.86 for 7 years at « % . 23. $899.00 for 12 years at 7| % . 17. $28.95 for 1 A yr. at 4? % . 24. $50.84 for 5 years at 1? % . 18. $896.84 for 3^ yr. at 2^ % . 25. $262.62 for 6 years at 6,\ % . 19. $414.14 for 4 years at |%. 26. $95.60 for j year at 71%. m METHOD BY ALIQUOT PABTS. 550. Prob. II. — To find the interest on any sum at any rate for years, months, and days by aliquot parts. 1. In business transactions involving interest, 30 days are usually considered one month, and 12 months one year. Hence the interest for days and months may be found according to (488), by regarding the time as a compound number ; thus. Find the interest and amount of $840 for 2 yr. 7 mo. 20 da., at 7%. $840 Principal. .07 Rate of Interest. 6 mo. =1 of 1 yr., hence 2) 58.80 2 Interest for 1 yr. 117.60 Int. for 2 yr. 1 mo. = J of 6 mo., hence 6) 29.40 " 6 mo. 15 da. =1 of 1 mo., hence 2) 4.90 " 1 mo. 6 da. =1 of 15 da., hence 3) 2.45 " 15 da. .811 •• 5 da. $155.16f 2 yr. 7 mo. 20 da 840.00 Principal. 5.16| Amt. for 2 yr. 7 mo. 20 da. IS51. The interest, by the method of aliquot parts, is usually found by finding first the interest of $1 for the given time, and SIMPLE INTEREST . 233 i 4^, % ,11%. It 6^%. sum at rts. days are .. Hence lording to r ; tiius, 10. 20 da., multiplying the given principal by the decimal expressing the interest of $1 ; thus, Find the interest of $G80 for 4 yr. 9 mo. 15 da. at 8^ . 1. We first find the iutereet of $1 for the given time ; thus, 8 ct. =• lut. of $1 for 1 yr., 8 ct. X 4 = Int. for 4 yr. =32 ct. 6mo. = J of 1 yr., hence, Jof8ct= " 6mo. = 4ct. 8 mo. = J of 6 mo., " io/'4ct. = " 3mo. = 2 ct. 15 da. =iof3mo., " iof2ct. = " 15 da. = .03^ m. Hence the interest on $1 for 4 yr. 9 mo. 15 da. = $.3835. 2. The decimal .383J expresses the part of $1 which is the intercBt of |1 for the given time at the given rate. Hence, $680 x .3831 = $260,663, la the interest of $680 for 4 yr. U mo. 15 da., at 8^ ; hence the following RULE. 552. /. Find by aliquot parts tlie interest of S 1 f 07' t?ie given rate and time. II. Multiply the principal hy the decimal expressing the inter- est jcr $1, and the product will be the required interest. II '". To find the amount, add t/ie interest to the principal. It. lino. 20 da. ' mo. 20 da. L is usually In time, and EXAMPLES FOR PRACTICE. 553. Find the interest 1. Of $560.40 for 2 yr. 10 mo. 18 da. at 7% ; at 9%. 2. Of 1284 for 3 yr. 8 mo. 12 da. at 6% ; at 8J%. 3. Of $296.85 for 4 yr. 11 mo. 24 da. at 8% ; at 5%. 4. Of $2940.75 for 3 yr. 11 mo. 17 da. at 7% ; at %\%, 5. Of $860 for 1 yr. 7 mo. 27 da. at ^% ; at 7^%. G. Find the amount of $250.70 for 2 yr. 28 da. at 8%. 7. Find the amount of $38.90 for 3 yr. 13 da. at 9%. 8. Paid a debt of $384.60, which was upon interest for 11 mo. 16 da. at 7%. Wliat was the amount of the payment? 9. A man invested $795 at 8% for 4 yr. 8 mo. 13 da. How much was the amount of principal and interest ? 10. Find the amount of $1000 for 9 yr. 11 mo. 29 da. at 7%. 16 234 BUSINESS ARITHMETIC, W: METHOD BY SIX PER CENT. PREPARA.TORY STEPS. 554. Step I.— To find the interest for any number of months at 6%. 1. Since the intereet of $1 for 12 months, or 1 yr., at 6%, is 6 cents, the interest for two months, which is J of 12 mouths, must be 1 cent, or y^ part of the principal. 2. Since the interest for 2 months is -^l-^ of the principal, the interest for any number of months will be as many times ^^^ of the principal as 2 is contained times in the given number of months. Hence the following RULE. 555. /. Move the decimal point in the principal two places to tJie left (459), prefixing ciphers, if necessary. 11. Multiply this result by one-half the number of months. Or, Multiply y^^^ of the principal by the number of months and divide the result by 2. M is r t EXAMPLES FOR PRACTICE. 5.">0. Find the interest at 6% 1. Of $973.50 for 10 mo. 2. Of !?896 for 8 mo. 3. Of $486.80 for 18 mo. 4. Of .$432.90 for 13 mo. 5. Of $304.40 for 7 mo. 6. Of $398 for 1 yr. 6 mo. =18 mo. 7. Of $750 for 2 yr. 8 mo. 8. Of $268 for 2 yr. 6 mo. 9. Of $186 for 4 yr. 2 mo. 10. Of $873 for 1 yr. 11 mo. 557. Step II.— To find the interest for any number of days at 0%. 1. Since the interest of $1 for 2 months at 6% is 1 cent, the interest for 1 month, or 30 days, must be ^ cent or 5 mills. And since days are I of 30 days, the interest for 6 days must be 1 of 5 mills, or 1 mill, which is y^'^^ of the principal. SIMPLE INTEREST, 235 Ur of 0, IS aaouths, ipal, the imes T^TJ imber of 2. Since the inte.ost for 6 days is y^^^y of the principal, the interest for any number of days will be as many times j^Vtt of the principal as 6 is contained times in the given number of days. Hence the following RDTiE. *>58. /. Move the decimal point in the principal three 2*1(1 cos to the left (459), prefixing ciphers, if necessary. II. Multiply this result by one-sixth the number of days. Or, Midtiply jtjVtf ^f ^^^ principal by the number of days and divide the result by 0. EXAMPLES FOR PRACTICE. ipal ttvo ry. onths. imths and 559. Find the interest at 6 % 1. Of $384 for 24 da. 2. Of $790 for 12 da. 3. Of $850 for 15 da. 4. Of ^mo for 16 da. 5. Of $935 for 27 da. 6. Of .$584 for 19 da. 7. Of $809 for 28 da. 8. Of $730 for 22 da. 9. Of $840 for 14 da. 10. Of ^396 for 17 da 10. =18 mo. 10. 10. 10. |mo. iberofdays 1 cent, the [or 5 mills- , days must cipah 560. Prob. III. — To find the interest on any sum at any rate for years, months, and days, by the six per cent method. Find the interest of $542 for 4 years 9 months 17 days at 8 per cent. RoLTTTicN — 1. The interest of $542 for 4 years at 6%, according to (547), is 1542 X .06 X 4 = $130.08. 2. The interest for 9 months, according to (554), is tJo of $542, or $5.42 mnltiplied by 9. and this product divided by 2 = $24.39. 3. The interest for 17 days, according to (567), is tj^oo of $542, or $.512 multiplied by 17, and this product divided by 6 = $1.535 + . Hence $130.08 + $24.39 + $1.54 = $156.01, the interest of $542 for 4 years fl months and 17 days. 4. Having found the interest of $542 at 65«, to find the interest at 8;^ we have 8^ = 6r; + 2t, and 2%Ul of 6^. Hence, $156.01 + i of $156.01 = $208,013. the interest of $542 at 6% for 4 yr. 9 mo. 17 da. ■BHI" 236 BUSI^'^ESS ARITHMETIC, I't: EXAMPLES FOR PRACTICE. 561. Find the interest by the 6% method f . Of $890.70 for 4 yr. 10 mo. 15 da. at 7% ; at 10% : at 4% 2. Of .$384.96 for 2 yr. 8 mo. 12 da. at 6fc ; at 9% ; at 8%. 3. Of $280.60 for 11 mo. 27 da. at 8;^ ; at 4% ; at 7%. 4. Of $890 for 9 mo. 13 da. at 6i% ; at 8^% ; at 9^%. 5. Of $480 for 2 yr. 7 mo. 15 da. at 9% ; at 12% ; at 4|%. a METHOD BY DECIMALS. 5G12. In tliis method the time is regarded as a compound number, and the months and days expressed as a decimal of a year. When the principal is a small sum, suflBcient accuracy will be secured by carrying the riocimal to three places ; but when a large sum, a greater number of decimal places should be taken. ^ 50>{. Prob. IV. — To find the interest on any sum at any rate for years, months, and days, by decimals. What is the amount of $450 for 5 yr. 7 mo. 16 da., at 6% ? Explanation. — 1. We express, according to (378—15), the days and months as a decimal of a year, ap shown in (1). 2. We find the interest on $450, the given principal, for 1 year, which l8 $27, an shown in (2). 8. Since |97 is the interest on $450 for 1 year, the Interest for 5.627 years is 5.627 limes $27, which is $151,929, ae shown in (1.) 4. The amount Is equal to the principal pins the interest (542); hence, $151.93+$450 = |()01.93iBtb& amount. Hence the following (1.) (2.) 30 ) 16 da. $450 12) 7.533 mo. .06 5.627 yr. $37.00 27 39 389 113 54 $151,929 Interest ■ 450 $601.93 Amount. SIMPLE INTEREST, 237 Lt4^ RULE. 564. /. To find the interest, mnltiph/ the principal by the rate, and this product by the time, expressed in years and deci- mals of a year. II. To find the amount, add the interest to the principal. f k'. ■ ^% '0* mpound aial of a racy ^^^ put wlien hould be sum at Lis. /e express, Jhe days and a year, a? Ut on $450, |year,wWch Lrest on |450 Ir 5.627 years 1 is $151,921), Iqual to tlic lre«t (542); 1601.93 19 the lowing EXAMPLES FOR PRACTICE. 505. Find the interest by the decimal method 1. Of $374.05 for 2 yr. 9 mo. 15 da. at 6% : at 9% ; at 4%. 2. Of $200 for I yr. 8 mo. 12 da. at 5/«, ; at 8% ; at 7^. 3. Of $790.80 for 5 yr. 3 mo. 7 da. at 7% ; at 11 % ; at 3%. 4. Of $700 for 11 mo. 27 da. at ^yo ; at 7|% ; at 2f %. 5. Of $460.90 for 3 yr. 5 mo. 13 da. at 6J % ; at 8] % ; at d^fo. 6. Of .^890 for 7 yr. 19 da. nt iiyi ; at 8% ; at 5%. 7. Of $580.40 for 17 da. at &^'/o ; at 9A% ; at 5^%. EXACT INTEREST. 560. In the foregoing methods of reckoning interest, the year is regarded as 300 days, wliich is 5 days less than a c^ym- man year, and days less than a leap year ; hence, the interest when found for a part of a year is incorrect. Thus, if tho interept of $100 is |7 for a common year or 865 days, tlie interet0. Prob. v.— To find the principal when the inter- est, time, and rate are given. Observe, that the interent of any principal for a given time at a given rate, is the interest of $1 taken (547) as many times as there are dollars in the principal ; hence the following RULE. 570. Divide the given interest by the interest of $1 for the given tim£ at the given rate. one ither for a jr of It 12%. )f $89& ind the )ni Jan- is given [icli will li 13 to Ic inter- SIM r LE I y T E li E S T, EXAMPLES FOR PRACTICE. 239 671. 1. What sum of money will gain $110.25 in 3 yr. 9 mo. at 7% ? Solution.— The interest of $1 for 3 yr. 9 mo. at 1% is 1.3625. No\v since $.20-25 \9 the interest of $1 for the given time at the given rate, $110.25 is the interest of as many dollars for the same time and rate as $.2()25 is contained times in 1110.25. Hence $110.36 -»- .SinaSt = $420, the required principal. What principal or sum of money 2. Will gain $63,488 in 2 yr. 9 mo. 16 da. at 8% ? 3. Will gain $95,456 in 3 yr. 8 mo. 25 da. at 7% ? 4. Will gain $106,611 in 3 yr. 6 mo. 18 da. at 6i % ? 5. Will gain $235,609 in 4 yr. 7 mo. 24 da. at 9% ? 6. Will gain $74,221 in 2 yr. 3 mo. 9 da. at 1\% ? 7. WIU gain $30,636 in 1 yr. 9 mo. 18 da. at ^% ? 672. Prob. VI. —To find the principal when the amount, time, and rate are given. Observe, that the amount is the principal plus the interest, and tliat the interest contains the interest (547) of $1 as many times as* tliere are dollars in the principal ; consequently the amount must contain (495) $1 plus the interest of $1 for the given time at the given rate aa many times as there are dollars in the principal ; hence the following BIJLE. 673. Divide the amount by the amount of SI for the given time at the given rate. I! i.^ It a given [dollars in for the EXAMPLES FOR PRACTICE. 674. 1. What sum of money will amount to $290.50 in 2yr. 8mo. 12 da. at 6% ? Solution.— The amount of $1 for 2 yr. 8 mo. 12 da. at 6;? is $1,162. Now since $1,163 is the amount of $1 for the given time at the given rate, $290.50 is the amount of as many dollars a5> $1.16-2 is contained times in it. Hence, $290.50 ■*■ $1,162 = $250, is the required priucipoL 240 BUSINESS ARITH3IETIC, 2. What is the interest for 1 yr. 7 mo. 13 da. on a sum of money which in this time amouuts to $487.65, at 7^ ? 3. What principal will amount to $310.60 in 3 yr. 6 mo. 9 da. at 5% V 4. At 8% a certain principal in 2 yr. 9 mo. 6 da. amounted to $09982. Find the principal and the interest. 5. What Bimi of money at 10^ will amount to $436.02 in 4yr. 8 mo. 23 da.? 575. Prob. VII. — To find the rate when the principal, interest and time are given. Observe^ that the given interest mupt be as many times 1% of the given principal for the given time as there arc onito in the rate ; hence the following BULB. 57C$. Divide the given interest hy the interest of the given principal for the given time at 1 per cent. EXAMPLES FOR PRACTICE. 577. 1. At what rate will $260 gain $45.50 in 2 yr. 6 mo. ? Solution.— The interest of $260 for 2 yr. 6 mo. at \% ie $6.60. Now since $6.50 is \% of $250 fur the given time, $45.50 is as many per cent as $6.50 is contained times in $45.50 ; hence, $45.50 -«- $6.50 = 7, is the required rate. At what rate per cent 2. Will $524 gain $206.63 in 5 yr. 7 mo. 18 da.? 8. Will $732 gain $99,674 in 2 yr. 3 mo. 7 da.? 4. Will $395.80 gain $53,873 in 2 yr. 8 mo. 20 da.? 5. Will $873 gain $132.S9 in 1 yr. 10 mo. 25 da.? 6. Will $908.50 gain $325,422 in 4 yr. 2 mo. 17 da. ? 7. A man purchased a house for $3186, which rents for $418 32. What rate per cent does he make on the invest- ment? 8. Which is the better investment and what rate per cent •• ♦ am of 5 mo. ntedto i6.02 in incipal, the given hence the tU given r. 6 mo.? .50. No^ Iny per cent 1= 7, Is the SIMPLE INTEREST, 241 per annum, i?4:3G0 which yields in 5 years $1635, or fa860 which yields in 9 years $2692.45 ? 9. At wliat. rate jht cent per annum will a sum of money donl)l(; itself in 7 years ? Solution.— Since iu 100 years at 1% any Bum doubles itself, to double Ittx'lf iu T yoais the rate per cent must be a» many times lj{ as 7 is contained times in 100, which is 14j^. Ilence, etc. 10. At what rate per cent per annum will i\ny sum double itself in 4, 8, 9, 12, and 25 years n^spcclivcly ? 11. Invested }*^3648 in a business that yields $1659.84 in 5 years. What per cent annual interest did I receive on my investment ? 12. At what rate per cent ])er annum will any sum triple or quadruple itself in 6, 9, 14, and 18 years respectively ? r>78. Prob. VIII.— To find the time when the princi- pal, interest, and rate are given. Observe, that the interest it? found (563) by multiplying the intercfit of the given i)rincipal for 1 year at the given rote by the time expropscd in years ; hence the following BULE. 579. /. Divide the given interest hy the interest of the given jtrincipal fur I year at the given rate. II. Rcdi'cc {"AVI), when called for, fractions of a year to months and days. •It t !'.h 1( rents foT le invest- per cent EXAMPLES FOR PRACTICE. 580. 1. In what time will $350 gain $63 at 8% ? Solution.— The interest of $350 for 1 yr. at 8% is $28. Now since $28 is tlie interest of ^S.jO at B:i for 1 year, it will take as many years to gain $63 M $28 is contained times in $63 ; hence $63 -t- $28 = 2} yr., or 2 yr. 3 mo., tlic required time. In what time will 2. $460 gain $80.50 at 5% ? 3. $80 jfaiu $UGat7i%? 4. $260 gain $96.80 at Bfo ? 5. $690 gain $301,392 at 7f % ? 6. $477 f^aia $152.64 at 12% ? 7. $385 gain $214.72 at 8f % ? i 1 : . ! 1 i 1 ml hi- ' ■.)!. ■ 242 B USINESS ARITHMETIC, f'l 8. My total gain on an investment of $8G0 at 1% per annam, is $455.70. How long has the investment been made ? 9. How long will it take any sum of money to double itself at 7% per annum? SoLUTioN.~At 100;( any enm will double itself In 1 year ; hence to double itHuirat 1% it will require as many years as 1% is coutuluud tiuicb iu l(>j;>, which in 14jf. Hence, etc. Obaervey that to find how long it will take to triple, quadruple, etc., any earn, we muat take 200^, 300^, etc. 10. At 7% the interest of $480 is equal to 5 times the prin- cipal. How long has the money been on interest V 11. How long will it take any sum of moneyut5%,8%,6^%, or 9% per annum to double itself? To triple itself, etc. ? R '1 I i: COMPOUND INTEREST. 581. Cow pound Interest is interest upon principal and interest united, at given intervals of time. Observe^ that the interest may be made a part of the principal, or com- pounded at any interval of time agreed ujton ; as, annually, semi-annually, quarterly, etc. 5822. PROB. IX. — To find the compound interest on any sum for any given time. Find the compound interest of $850 for 2 yr. 6 mo. at 6%. $850 Prin. for lat yr. 1.00 $901 Prin. for 2tl yr. 1.06 ^ii|! $955.06 Prin. for 6 mo. 1.03 $983.71 Total amount. $850 Given Prin. $133.71 Compound Int. Explanation.— Since at 6^ the anupmf is 1.06 of the principal, we multiply $850, the principal for the first year, by 1.06, iri v lug $901, the amount at the end of tlie first year, which forms the principal for " second year. In the same manner u'«' $955.06, the amount at the ) second year which forms the ojpui . the 6 mouths. 2. Since Q% for one year is 3K to 6 months, we multiply $955.06, the prin* tl for the 6 months, by 1.03, which gives the total amount at the end of the 2 yr. 6 mc. anum, ) itself ) double iu laj;.', etc., any le prin- principal or com- l-aunually. t on any at 6%. tho anuynf Itiply ?850, byl.O(i,i:iv- pal for •' \\» 9% to^ ti Ihe prin< A Ich gives the layr. 6mc\ COMPOUND lyTEREST, 243 8. From the total amount wo sabtract $850, the fjiven principal, which prives im.Tl, the compound Intcrctft of |»fiU for 2 years 6 mouths at 6,i. Hence the following RULE. 58;{. /. Find the amount of the principal for the first intercal of time (it the end of which interest is due, and make it the principal for the second internal. II. Find the amount of this principal for the second interval of time, and so continue for each successive interval and fraction of an interval, if any. III. Subtract the given principal from t?ie last amount and the remainder will be the compound interest. EXAMPLES FOR PRACTICE. 584. 1. Find the compound interest of $380.80 for 1 year at S^/c , interest payable quarterly. 2. Find the amount of $870 for 2 years at 6^ compound interest. 3 What is the compound interest of $650 for 3 years, at 7^ , payable annually ? 4. What is the amount of $1500 for 2 years 9 months at 8% compound interest, payable annually ? 5. What is the difference in the simple interest and compound interest of $480 tor 4 yr. and 6 mo. at 7 % ? 6. What is the amount of $600 for 1 year 9 months at 5 % compound interest, payable quarterly? 7. What is the annual income from an investment of $2860 t 7^0 compound interest, payable quarterly ? 8. A man invests $3750 for 3 years at 7^ compound intercc.;, ^livable semi-annually, and the same amount for the same time at 7|% simple interest. Which will yield the greater amount of interest at the end of the time, and how much? 9. Wh will be the compound interest at the end of 2 yr. 5 mo- or note for $600 at 7^ , payable semi-annually ? ! 1 r 1 .1 ; 244 BUSINESS ARITHMETIC. K.I INTEREST TABLES. 585. Interest, both simple and compound, is now almost invariably reckoned by means of tables, which give the interest or amount of f 1 nt different rates for years, months, and days. The following illustrate the nature and use of such tables : Table showing the simple interest of $1 nt 0, 7, and Sfe ,for years, months, ami days. w. 756. 8J<. 6<. T%. 8jt. Tears. 1 Yenra. 4 M .83 .06 .07 .08 .38 2 .12 .14 .16 5 .30 .35 .40 3 .18 .21 .24 6 .36 .43 .48 3fontha. n ^ * .' HI out ha. ■_ 1 .005 .00583 .00006 7 .035 .04088 .01666 3 .01 .01166 .01388 8 .01 .04666 .05333 3 .015 .01750 .02000 9 .045 .06350 .06000 4 .03 .02333 .02666 10 .06 .06833 .0r>») 5 .025 .02916 .a3333 11 .065 Jt»l\6 .07333 6 Jiaya. 1 m .03500 .04000 Dftya. 16 .00016 .00019 .00022 .ooaw .00311 .00355 3 .00033 .00088 .00044 17 .002S8 .0»«*l .wm 3 .00050 .00058 .00066 18 .00300 .(nti'4) owoo 4 .00066 .00077 .00088 19 .00316 .003f;9 .0(K«J 5 .00083 .00097 .00111 SO .00333 .00388 .00444 6 .00100 .00116 .(X)I33 21 .00350 .0040S .00166 7 .00116 .00130 .00156 22 .003(ki .mizi .0048>^ 8 .00138 .00163 .00177 33 .0038:) .00417 .00511 9 .00150 .00175 .00200 21 .00100 .00-166 .00533 10 .001 f;6 .OOIM .m-tn 36 .00416 .00180 .00555 11 .00183 .0021 H .00^.14 36 .00433 00505 .00577 13 mim .00233 .0020(5 1 37 .OWflO .00525 .0J«00 i 18 .00316 .00253 .00288 j 38 .0046(( .00541 .00622 1 14 .002:» .(KI2T2 .00311 39 .00483 .0066:] OOGM , 15 .00250 .00291 .0G3{8 1 1 INTEREST TABLES, 245 almost intcreBt d days. Metliotl of using the Simple Interest Table, 586. Find the inierest of $250 for 5 yr. 9 mo. 18 da. at 7%. 1. Wejlnd the interest 0/ %l/or the given time < .;^6 interest In table for [ = -j .l>525 *' " " •' ' I .0035 " " '* " .;^6 interest in table for 5 yr. 9 mo. 18 da. Interest of |1 for 5 yr. 9 mo. 18 da. id .4Uti of $1. 2. Since the interest of #1 for 5 yr, ft mo. 18 da. is .406 of $1, the interest of $^50 for the same time \^ .406 of $250. Hence, $230 x .406 = $101.50, tlie required interest. M. .» .40 .48 - — » 01666 p6' iO' .05333 .PCOOO o' .Of.666 i6 .07333 — 11 .00355 *» 1 .no:n7 r4> owoo (19 .fl01"<2 SB .00144 OH ' .0(M66 \ i7 .00468 17 .00511 bG .00533 80 .00&55 05 .oosn •a .00600 i 4-1 .0*522 1 (i; J OOOM EXAMPLES FOR PRACTICH. 587. Find by using the table the interest, at 7%, of 1. $880 for 3 yr. 7 mo. 23 da. 4. $325.80 for 5 yr. 13 da. 2. $438 for 5 yr. 11 nio. It) da. 5. .'?:00.50 for 11 mo. 28 da. 3. $283 for 6 yr. 8 mo. 27 da. 6. $395.75 for 3 yr. 7 mo. Table shomug the amount of ^i at 0, 7, and S'/c compound interest from 1 to t? years. YUS. . 1 2 3 4 5 6 w. 1%. 9fi. ; TBS. i ^%' 1%. 8?. 1.713824 l.a50930 1.9,"». A Partial Paffinent is a payment in part of a note, bond, or other oblif^ation. o90. An Itulorsement is a written acknowletlgment of a i>?trtial payment, placed on the back of a note, bond, etc., ("luting tliu time and amount of the same. \\ r* 248 BUSINESS A Ji ITU ME TIC. 'H' 1. MERCANTILE KULE. 597. The method of reckoning partial payments known as the Mercantile Rule is very commonly used in computing interest on notes and accounts running for a year or less. The role is as follows : KULE. />1>8, / Find the amoitnt of the note or debt frmn the time it begins to bear interest^ and of each payment until the date of settlement. II. Subtract the sum of the aniountu of jyayinents from, the amount of the note or debt ; the rtinuindcr ucUl be tJte hulauce due. Obsen'e, that an accnrate application of the rule requiree that the exact interest fhoultl be found according to (5GG). m 4-V EXAMPLES FOR PRACTICE. 599. 1. $900. Woodstock, Sept. 1st, 1870. On demand I promise to pay li. M. Mac Vicar, or order, nine hundred dollars wUh interest, calue receiced. Wakren Mann. Indorsed as follows: Oct. 18th, 1876, !i;l~)0; Dec. 23, 1876, $200 ; March 15th, 1877, $300. What is dne on the note July 19th, 1877 ? 2. An accouut amounting to $485 wns duo Se[)t. 3, 1875, and was not settled until Aug. 15, 1876. The pnynients made upon it wore : $125, Dec. 4, 1875 ; $84, Jan. 17, 1870 ; .s95, June 23, 1870. What was duo at the time of settlement, allowing inter- est at 7% ? 3. A note for $000 bearing (uterest at Q'/c from Jtdy Ist, 1S74, wus paid May 16th, ls75. The indorsements wore: July 12th: 1874. $185; Sept. 15, 1874, $70; Jan. 13.1875, c^230 ; i.l.I Marc]\ 2, 1875, $115. What was due on the note at the time of payment ? PARTIAL PAYMENTS, 240 own a& aputing 8. The the time i date of from tJie aiice due. t the exact t, 1870. \rder, nine Mann. 22, 187G, the note 1875, and iiatle upon June 23, ring iater- l8t, lb74, [julylStli; .230; 11*' tlie time of 4 $250. Hamilton, ^fnrc?^ 25, 1876. Niuety-eijfbt days after date I promise to pay E. D. Brooks, or order, two hundred fifty dollars with interest, value received. Silas Jones. Indorsements : $87, April 13, 1876 ; $48, May 9. What is to pay when the note is due ? UNITED STATES RULE. OOO. The United States courts have adopted the following rule for reckoning the interest on partial payments. It is also very frequently adopted in Canada. BULE. 601. / Find the amount of the given principal to the time of the first pat/ment ; if the payment equals or exceeds the inter, est then due, subtract it from the amount obtained and regard the remainder as the new principal. II. If the payment is less than the interest due, find the amount of the given principal to a time when the sum of the pay- ments equals or exceeds the interest then due, and subtract the sum of the payments from this amount, and regard the remain- der as the new principal. m. Proceed with this new principal and teUh each succeeding principal in the same manner. 602. The method of applying the above rule will be seen firom tbe following example : 1. A note for $900, dated Montreal, Jan. 5th, 1876, and paid Dec. 20th, 1876, had endorsed upon it the following pay- ments : Feb. 23d, 1876, $40 ; April 26th, $6; July 19th, 1870. $70. How much was the payment Dec. 20th, 1876, interest at7%? 17 I Hi T ' »u ■ BB If ■ }i HHR {• • - '' RJ^f ^ W' ' ■ m'l ^if' ■ |F^' . 250 BUSINESS ARITHMETIC, fiOliUTTON. FiVHt Step. 1. The flrRt principal in the face of the note . 3. We find the interest from the datn of the note to the first pay meiit, Feb. ij, 1B7() (49 da.), at 7< Amount 3. The first payment, $40, being greater than the interest then due, iii subtracted fl-om the amount /Second pnncipal Seeotul Step, 1. The second principal la the remainder alter subtracting the flrot pajrmcnt from the amount at that date .... 2. The interest on $868.43, from Feb. 38 to April 36, 187(1 (m da.), is $10,463 8. Thi» interest being greater than the second pay- ment (fB), we find the interest on $8«8.48 from April 26 to July 1!>, 1876 (84 da.), which is? $900 8.4;j $908.43 40.00 $8(i8.43 $868.43 Interest from first to third payment Amount. .... 4. The oum of the second and third pjijmentH bein^ i;."«* the interest due, we (subtract it from the amount Third pnncipal 13.961 $24,414 "- than !24.414 $8«.)2.814 76 $816,844 ,1 Tfilrd Step, We find the interest on 1816.844, from July 19 to Dec. 20, 1876 (154 da.), which is . . ... . 24.a')T Payment due Dec. 50, 1876 I840.M0 In the above example, the interest has been reckomd according to iiHUi) ; in the following, 360 days have been regarded as a year. EXAMPLES FOR PRACTICE. IP 2. A note for $16l}0 at 8% interest was dated March 18, 1872, and was paid Au^r. 13, 1875. Tlie following sums were endorsed upon it: $160, Fib. 12, 1873 ; $48, March 7, 1874; and $.ir>0, Aug. 25, 1874. How much was paid Aug. 13 1 ^ISC O UXT. 51 14 24.414 $8«.»2.8}4 \n . 7« • 1816.814 251 '^"- "l-n i, Fob. 9, 1870, iut,. J,' afj/^* ' ^"" '"»<"' '^'« - follows : A,.ri, 3. ',870%ir T'f "^"^ ^«' '««■ '-"'o^^' '•^« ; 1580, May 7, ,874 andTn'/r ''' ""' ^ *'2''. A"fc'. s! - 0- upon it Sept. ,,;Z. ^Z^::, t/;7' «ow n.tcU DISCOITN-T. «>04. The Present frovtl, ->f "• '"""f '•''•««l at interest nf a ,!» " 'T'' '' ^'''^ ««""' ^•ve„ ,„„ „h,„ ., ^^^^^^^ ^^a legal rate, w,ll amount to the "'"■'e^'^aJXafa'rur'im^a'Jr" '*'"'"'' "^ »"«' <" •imc and itB present worth - "• """• ^' -^° «- fe present won. of .„, ^' 15 252 BUSINESS A RITIIM ET/C. I- .t'i EXAMPLES FOR PRACTICE. 007. What is ihe present worth 1. Of $360 at 7^ , due in 2 yr. ? At 5% , due in 8 mo. ? 2. Of $800 at J^ , due in 6 mo. ? At 8% , duo in 9 mo. ? 3. Of $490 at 8;^, due in 42 da.? At 7 ;^, due in 128 da. « What is the true discount 4. Of $580 at 7%, due in 90 da.? At 8i%,due in 4yr. 17 da. ? 5. Of $860 at 7%, due in 93 da. ? At 12 '/c, due in Syr. 19 da.? 6. Of $260 at 6J%, due in 120 da.? At 9%, due in 2 jr. 25 da. ? 7. Sold my farm for $3800 cash and a mortgage for $6500 running for 3 years without interest. The use of money being worth 7% per annum, what is the cash value of the farm? 8. What is the true discount at 8% on a debt of $3200, due in 2 yr. 5 mo. and 24 da. ? 9. What is the difference between the interest and irve dis- count at 7;^ of $460. due 8 months hence? 10. A man is offered a house for $4800 cash, or for $5250 payable in 2 yr. 6 mo. without interest. If he accepts the former, how much will he lose when money is worth 8^ ? 11. A merchant buys $2645.50 worth of goods on 3 mo. credit, but is offered 3% discount for cash. Which is the better bargain, and how much, when money is at 7% per annum ? 12. Which is more profitablo, and how much, to buy wood at $4.50 a cord cash, or at $4.66 payable in 9 months without interest, money being worth 85^ ? 18. A ^rain dealer sold 2400 bu. of wheat for $3600, for which he took a note at 4 mo. without interest. W^hat was the cash price per bushel, when money is at 6^ ? B A y K DISCO V y T. 253 .1 iO.t in 4yr. . in 3 yr. le in 2 yr. for $6500 of money uo of the 13200, due true dis- for $5250 |ccept8 the on 3 mo. tich is the It 7fo per buy woo^ |i8 without 13600, for xat was the BANK DlSCOUin'. 008. BaHh lyiscounf is the interest on the Xe.ce of a note ftir tlie time it has to run, including three days grace. 1. Thirt dcdnctiou iu made by a bank for advancing the amount of the note before It it« due. S. A note to be dincounti'd at a bank must Ui>nally be made payable to the order of some person who must endorse it. 3. When a note bcarx interest, tlie discount ia computed on its face plus the interest for the time it lias to run. « What are the bank discount and proceeds of a note 1. Of 1790 for 154 da at 6% ? For 2 nio. 12 da. at 7% ? a. Of $380 for 3 mo. 15 da. at 7j^c ? For nio. 9 da. at S% V 8. Of $1000 for 80 da. at 7^^ ? For 140 du. at 8i '/c t 4. What is the difference between the bank and true discount on a note of $1000 at 7;^ , payable in 90 days ? 5. Valuing my horse at $212, 1 sold him and took a note for $2.'l5 payable in 60 days, which I discounted at the bank. How much did I gain on the transaction? 6. A man bought 130 acres of land at $16 per acre. He paid for the land by discounting a note at the bank for $2140.87 for 90 du. at 0%. How much cash has he left? Find the date of maturity, the time, and the proceeds of the following notes : (7.) $480 90. Chatham, Mar. 16, 187»». Seventy days after date I promise to pay to the order of D. MacVicar, ftnir hundred eighty -^^^ dollars, for value received. Discounted Mar. 29. N. L. Sage. (8.) $590. Sarnia, May 13, 1876. Three mouths after date I promise to pay to the order of Wm. Flint, five hundred ninety dollars, for value received. Discounted June 2. Peter M.vcKenztk. (9.) $1600. BELLEvnxE, Jan. 19, 1876. Seven months after date we jointly and Beverally agree to pay James Richards, or order, one thousano- mestic Exchange^ and between different countries Fotdgn Exchange. «20. A Draft or Bill of Exc/tanf/e is a written order for the payment of money at a specified time, drawn in one place and payable in another. 1. The Drawer of a bill or draft is the person who signs it ; the Drawee^ the person directed to pay it ; the Payee, the person to whom the money in directed to be paid ; the Indoreer, the person who trutitifers his right to a bill or draft by indorsing it ; and the Holder, the person who has legal pos- sesHion of it. 2. A Sight Draft or BiU is one which requires payment to be made when presented to the payor. 3. A Time Drqft or Bill is one which requires payment to be made at a ppecifled time after date, or after Hght or \}e\ng preiented to the payor. Three dnys of praco are usually allowed on bills of exchange. 4. The Acceptance of a bill or draft is the agreement of the party on whom it i?i drawn to pay it at mnturity. This is indicated by writing the wonl "Accepted" across the face of the bill and nigning it. When a bill is protested for non-acceptance, the drawer is bound to pny it immediately. 5. Foreign bills of exchange are usually drawn in duplicate or triplicate, and sent by different conveyances, to provide against miscarriage, each copy being valid until the bill is paid. These are distinguished from one another by being called the Jlnt, eeoond^ and third of exchange. / A' L A y I) E X (' H A NG E, 257 • otlief ^ In New trouble, khich ho : bnnk. \d or 1)0- gt. •n order i in ono DrawMy Imoney i«* ight to a jegal po8- Iti when kade at a [party on king the |d to pny ipUcate, 3, each |he Jlrtt, 021. The Par of Exchange is the relative value of the coins of two coimtries. 1. ThiH, the par of exchange between Canada and Britain Is the uumlMjr of (lollnrH, the ptandard unit nf Canadiun money, which is equal to the pound Htcrling, the etaudard unit of Eu^Hi'li money. *2. The itnl rate qf exchange depends ou the balance of trade aud is calU'd tho course offxchange. 3. The value of the pound t>tcrling wac, by Act of Purliamcnt, fixed at | IJ. ItH intrinsic value Ih now fixed at $-1,861. But raten of exchange are still quoted In commercial papers at a certain per cent on the old jnir of trchautje. Hence when exchange Is at a premium of ft; per cent, it is at par between Great Britain and Canada, because $4; + U} per cent = $4,862. i INLAND EXCHANGE. Ot22. Inland Exchange is a method of paying debts or other obligations at distant places in the same country, without transmitting thu money. Form of Sight atid Time Draft. £2700. Montreal, July 25. 1878. At fifteen days sight, pay to the order of Taintor Brothers d by the cashiir of the bank. ObBcrvo cnrefuUy the followinj^ : 1. Ill rnno ortrannactionH hi'iwi'cn dlHtiint rontitric«, the tninoiiortatiou ofHpeclu from iiit.* oik; to the otlit r would he MtteiHlcd with (rxp<'Ui»e, rt«k, aud lotiH of InfpnsHt. To uvold tlicnc ill(•(lIlv<'Ilit•^<•«•^, Hill- of Rxchantto ar- uwd (610) In payliij; dehtH rcriprocnily ditc In "iirli rountrii -. A Himila- n)ethu('. Ih fruqueutly fullowud with rctipeci tu truueuctiouH \, Wu the miqiu cuiintr* , thuH : 8. BankH hcU draftH upon others in which tlicy have i\i\toAX» in money or equivalent Hecurity. ITt^nce Y)nnk^4 throu;.'hoiit tlir country, in order to give them this fhclllly, hrtvo ;4urh dt-pofit-' at ••(•utre» of trad«', »uch an K utreal, Toronto, Hamilton, etc. 8. A hank Draft will u -nally lie purchased by hanks in any jmrt of the country, in rABo the piTHon oflcrinn it it* fttlly tdeniifl'-^l a» the party to whom the draft Ih jjayahhr Ilencf, a debt or other liability may !>e dl-- charfi^od at any place by u dral'l on a bank at any centre of trade. 4. A draft may be made payable to the percon to whom It li« iieot. or f ■ the person buylnj,' it. In the latter ease the piMt, it niu-il write on the Imck " Pay to the order of" (name uf party to whom it \* sent), aud Hl(:n his o-imUu- the t-aiuo mntry, In ade, «uch It t of the I»Hrty t'» l>f dl«- jul. or t ' lii-l wrilf Lot ). and money limited. Lloyed lu TniTiD Method. — 77*/' pmty (Imring to transmit mimiy^ maki'S a draft or ti^rder for thf amount upon a party oiriny him, at the pltir,- irJure the muMy is to he sent, and remits thin an prnioudy dirertrd. 1. By tli?rt inofhoon Ih nuid to draw upon nnotlior. Such (Irafls Htiniild \w i)r«?«'iii«*d for pnynu-nt an noon an received, and if not puid or acciptctl xliould ho. protcHic*! for non-payment ininit>(llat<-ly. 2. Lnrjfi! I)^l*ln«•^^^ flrniH havr deponitH in Imnkn nt buHincKH centrcn, nnd credit with otJ^T bii»«inof»« flrniH; h, tludr drafts nro U!*eonH«>qu«Mitly, lianki* at niitrnio nuj-t actually >cnd mon<-y to Now York by Kxpro.-* or oilit'r « onveyanc*'. He cc, for th<' (•xp«'n><«; thun incurred and other troiMtle In handling thu m . icy, a mihuII premium \* charged at Bullalo on New York draftn. EXAMPLES FOR PRACTICE. 025. 1. What Ih the cost of a m^hi draft for *2400. at \ % prenuuin ? SoLUTlOK.— Cowt r $2400 li of ♦2100 $2.1ir,. 2. What ie the cost of a dnift for i|;a*JO(). at J% premium? Solution.— Co«t - |u-floo . i«of$:BO() #.'1201. Find the cost of sight draft ;{. Fo" $h:14. pn-mium 2%. «. For %Vm, disrount 1%. 4. For Y'>'JO(>, pnmiiun \'/c. 7. For *aH 1.5(1, dineount |%. T). Kor |i:W.HO, pn'mium J%. H. For $*21)."».:{0. dlHrount li}%. U. Thf «M»Ht of u i»ij.'ht di .'t piirchaH'(l at V/,< pn luium Ib $U)3 'iU ; what i8 the face of the draft ? HoLi'TioN — At lj!S prpaiium, fl of ihe face of tlir (haft cont (l.Ol.'S. Ilenci- tlif face of Ihi- draft ia ai« nctuy doilure au tl.Ul5 in cuutuinud limcH in $4m.«>, which Ic |MW. ^p 260 D U SI NESS A li 1 TUM E TIC. Find the face of a draft which cost h'i'V ■ % 1 t|,1 h liil ( ■! 10. $575.40, premium 2;%. 11. $731.70, pr-jmium \\%, 13. $483.20, premium \^o. 18. $810.88, discount 1%. 14. $273,847, (liHcount %%. 15. $31.>.«5. discount \\'/o. 10. Wliat is the c» face. The man rr([uireH, therefore, to purclmKC a draft whoM» face pin** %% of it (MinnlH $4W.4H. Hence, accor(llni» to (495—5), the amount paid, or face of the draft, ia fl'.m.lH -»- l.OOS ~ (4%. 22. Excliange being nt 98? (\\'/( discount), what iu the cost of a draft, time 4 mo., interest at 5% ? 23. The face of n draft which was ])urcha8ed at li% premium is $2500, the time 40 da., rate of interest allowed \'/c ; what was its cost ? 24. My .agent in Halifax »oJd a consignment of goods for $8200, commission on the sale 21 [■', . lie reniitted the proceeds by draft on Montreal, at a premium of i^'/c. What is the amount remitted ? ^t^'if cost lium lit NVllB Is for ireeds Is the FOREIGN EXCIIAXOE. 2G1 FOREIGN EXCHANGE. 026. Foreign Kxi'hanffv, is a metliml of paying debts or other obligatious iu foreign couutrieti without truuumittiug the money. Ob^ert'e, Xhfii foreign exchange is bamHl upon the fact that diflcrent coun- tries exchange proiiucts, Hecurltlc«, etc., with each other. ThiiH, Cauada t^e'ln v\ heat, etc., to England, and Kut;land in return hcUs niuiiufactured ^uode, etc., to Canada, lienct;, partieH iu eacli country Ifcconu indebted to parties in the other. For thin reanon. a nuTcliant in Canada can pay for goodii purchased in England by buying' an order upon a firm in England which id indebted to a firm in Cauada. Form of a liill or Set of Exchange, £400. Ottaw.\, Ju/y 13, t870. At Hif/ht of this First of Exchange (second mtd third of the mme date and tenor unpaid), pay to the ordtr of E, J), lilakvslee Four Hundred Pounds Sterling, foi' value ''erdced, and charge the wiine to the account of WiLM.vMs, BnowN & Co. To Martin, Williams & Co., London. The per-« used luHtead of "Sbconu," utul the j(an?ntlicr.is n-adts, '' MrHt and Third of the Hame date and tenor unpaid." A siuiilar cliange is made in th.' Third Bill. Oii7. Exrhnnge with Europe is conducted chiefly through prominent flnancinl centres, as London, Paris, Berlin, Antwerp, Amsterdam, etc. iV2H, ijuotatioits are the puhli.shod rates at which bills of exchange, stocks, bonds, etc.. are bouglit and sold in the money market from day to day. ThcHc (juot.itlons give tin- ni.irket ;.'old value in stcrllnjj money of one or more unltH of the foreiirn coin Thus, quotJitlons on London give the value of £1 nterling in dollars ; on Paris. Antwerp, and (jeueva, the value of ♦! \i\ francii : on IIainbur<;. B«'r- lln, Bremen, and Frankfort, the value uf 4 inarhi Iu cetUs ; on Amsterdam, Uie value of a yuildtr in centit. >., I 2G2 nirsixESS arithmetic. If*. ,1 ■1T u m Ot20f The following table gives the par of exchange, or gold value of foreign monetary units : Table op Par ok Exchange. COUNTRIK8. Austria I Bt'lgiuin I Bolivia 1 Brn/JI , Bci^^ota Unified States Cunt. ill Ainurica.. . Chill Denmark Ecuador Egypt Franco Great Britain Greece German Empire. .. Japan ' India I Italy Liberia ' Mexico Ni'therlands Norway ' Pern i Portugal RuHHla Sandwich InlandH. . Spain Sweden Swltzi-rlond . Trl|.oll Tuni*" Turkey , U. 8. ofCoIomb'a. MONET ART UNIT. STANDARD. Florin Franc Dollar MilrelBoflOOOreis ... PoHO Dollar , Dollar Peso Crown Dollar Pound of 100 piantrcB . . Franc Pound sterling , Drachma Mark Yen Rupee of 16 annas Lira Dollar Dollar Florin Crown Dollar Mllrcla of 1000 reiH . . . . Rouble of 100 copcckn. DolKir Peseta of 100 ccntlmcH Crown. Franc Matibiib of 20 plantrc .-( Pla(»lre of iO caroiibs . , Piastre Pt'HO Silver Gold and silver (Jold and silver 0(.ld Gold Gold Silver Gold Gold Silver Gold Gold and silver Gold Gold and silver G(dd Gold Silver Gold and silver Gold Sliver (iold and silver Gold SIher Gold Silver ... Gold Gohl and sliver Gold Gohl and silver SIIVIT Silver Gold Silver VAUIK IN CANAniAN MCNEY. .4ft, 8 .19,8 .%, 5 .%, 6 11.00 .'.»1, 8 .91,2 .26,8 .91,8 4.97,4 .19, .3 4.86. 66 .19, 3 .28,8 .90,7 .48,6 .19,8 1.00 .99,8 .38,6 .26, 8 .91, 8 1.08 .73,4 1.00 .19, 3 .26, 8 .19. .\ .82, St .11. H .04, 8 1 ■"'• 8 FOREiaX EXC H ANQ E, 2C3 ■•'•i METHODS OP DIRECT EXCHANGE. i%*My. Dh'i'vt Ejrjhfthf/e is a mcthotl of making pay- lutMitH in a torcigD ooimtry at the quoted rate ot exchange with tliat country. FinsT MKTnoD. — The person deftiring to transmit the money purrhani^a (t Set of tlrrhange for the amount on the eouutry to ichich the money is to he sent, and forirordu the three hUls by different ukuIs or rmttes to their destination. Skcoxd Metthod. — The person desirimj to transmit the money instrnrts his ci'editor in tJie foreign eountry to tliunv uponhim, that is, to sf'fl a set of exchange npon him, irhich he pays in his own country when presented. ! r 'fc i' I'* t EXAMPLES FOR PRACTICE. 031 . 1. Wliat is the cost in currency of a bill of excbanffe on Liverpool for £285 Os. 6d., (>xchange bein^ quoted at $4.88, and gnUi at 1.12, brokerage \'/l ? £285 9s. fid. = £285.475 $4.88 X 285.475 - $1393. 118 $1.1225 X I.^OJJ.llS .: 1503.77 4- SoLUTioN.— 1. Wi* fiMtlure the 9t>. (WI. to n decimsil of ^l. Ilonro t'2K'S"N.rKl. €2Hr..J7.'S. 2. Since tl ♦•» SK, t:*<>.475 inust be<'(iiii»l KRS v 285.475 ;=: $i;W3.118, the i;oUl value of tin* hill with- (»ut hrokora;jc. ;{. Hiiicc $1 iroid i« equal 91.12 cnrroncy. and tho hrolcor/ipo is J .thoront <>r$l ^(dd in currency 1h $1.12%. Hunce tho hill co.xt lo curreucy tl.l225 x 1W3.118 - ILVil.T? +• . What is the cost of a bill on 2. Loivion for .€430 Sj^. 3d., sterling at 4.84|, brokerage 1% ? 8. Paris for 4500 francs at .198, brokerage ^'/y t 4. Geneva, Switzerlaud. for 80!H> francs at .189? 5. Anfirrrp for 4000 rriincs at .175, in currency, golil at 1.09 Y 6. Amsterdam for 84^K) g lilders at 41 {, brokerage \'/i) V 7. FrnJcfort for 2500 marks, quoted at .97|? 264 BUSINESS ARITHMETIC, p.i^ i^.* 8. A merchant in Kingston instructed his agent at Berlin to draw on him fur a bill of goods of 43000 marks, exchange at 24?,. gold being at 1.08^, brokerage ] % ; what did the merchant pay in currency for the goodB t METHODS OF INDIRECT EXCHANGE. (iii*2. Indirect Ka'chunye is a method of making payments in a foreign country by taking advuntuge of the rate of exchange between that country and one or more other countries. Observe carefully the following : 1. The advanta);c of indirect over direct exchange under certain finan- cial conditions which gomctimei*, owing to varlouo cauttcH, exitst between different countrieu, may be ithown bh fullowa : Suppose exchau|;e in Moutruul to be at par on London, but on Parin at ITcontH fur 1 franc, and at PariB on Ltmdon at 24 francH for jb'l. Witli thcMO conditions, a bill on London for £10U will cobt in Moutiual #-186.(>5 ; but a bill on London fur £100 will co»t in Parltt 24 iVancH x 100 = 3400 francH, and u bill on Pariu for 3400 francH will cost in Montreal 17 cents x MM)^ $408. Ilencc €100 can be sent from Montreal to London by direct exchange for f48«.f computing indirect exchange 1^^ called ArbitraiUm Qf ExfhaiKje. Wli' there ih onlv one intermediate place, it is called simple Arldf ration ; v ,ere there are two or more iutermediate places it is called VotnjtoKnd Arbitration. Eitiier of the following methods may be pursued : FiiisT Method. — The person deHiring to transmit the money may buy n bUl of exchange for the ammint on an ivtermeduitc pi^icc, tehkh he nenda to Jus agmt at that plare itith instructwus to buy a bill itith the yroieetla on the pLiee to trhich the money i* to he aeni, and toforicard it to thf prund numbers and business. Probs. Vili, IX, X. und XI. EXAMPLES FOR PRACTICE. 0«i3. 1. Exchange in Hamilton on London is 4.8o, and on Paris in London is 24^ ; what is the cost of transmitting 631)04 francs to Paris through London ? Solution.— 1. We find the copt of a b!U of oxclmnsro In London for639ftl francs Since i\\ franco r= £1, 63994 -4-3 remit 4800 guihlere from Mon- treal to Amsterdam through Paris aneini>; (putted as follows : Montreal on Paris, 18/. ; at Paris <^n Lon- don, 241 francs to a £ ; and at London on Amsterdam, 18 26G BUSI^'£:ss arithmetic. H%, i: J 1- 1 > ". t . ']^-i 12 J guilders to the £. How much more would it cost by direct exchaugo at 391 cents for 1 guilder? 4. A Canadian residing in Berlin wishing to obtain $0000 from Canada, directs his agent in I'aris to draw on Montreal and remit the proceeds by draft to Berlin. Exchange on Montreul at Paris being .18, aud on Berlin at Paris 1 mark for 1.2 francs, the agent's commission being \% hoi\\ for drawing and remit- ting, how much would he gain by drawing directly on Canada at 34J cents i)er mark ? EQUATION OF PAYMENTS, ($34. An Acvomit is a written statement of the debit and credit transactions between two persons, with their dates. The debit or left-hand side of an account (marked Dr.) shows the Hams duo to the Creditor, or person keeping the account ; the credit or right- hand side (marked C'r.) showB the Hum« paid by the Debtor, or person agaiutit whuui the account is made. 035. The lialdiice of an account is the difference between the sum of the items on the debit and credit sides. G'KI. Equation of Paymeuts is the process of finding a date at which a debtor may pay a creditor in one payment several sums of money due at different times, without loss of interest to either party. y one payment. (5;>8. The Matuflfy of any obligation is the date at which it b(^Ci)nieH due or draws interest. «.'59. The Tevm of Credit is the interval of time from till' date a debt is contracted until its maturity. 040. The Averaf/e Term of Credit is the interval of time from the viaturity of the first item in an account to the Eqiuitcd Time. ■P=T? Ei^UATIO y OF PA I'M hW T:i, 2G7 lirect 10000 i\ and lit real rancB, remit- 'auada eUt and the pttTtiB or rigbt- >r person jctween finding taynieni loss ol" Beveral [t whicli le from >rval of it to tliu PREPARATORY PROPOSITIONS. 041. The method of ppttling accounts hy tqitdtiini of pdy. ments depends upon the following propositions ; hence they shouM he carefully studied : Prop. 1. — When, by agreement, no intenst u< to be paid on a debt from u ttpecijied time, if any part of the amount i-i paid by the debtor, he is entitled to iitUrcut untU the cjrplrat, m of the specified time. Thne, A owoh B $100, payable In 12 months without Intorct. whirh means' that A is entitled by agreement to the iice of $100 of B's mniH-y l«nyt« any part of it before the expiration of the 13 months, he lis entitled iutcrest. Obsertt, that when credit is given without charging interest, the i)rofltB or advantage of the transaction are such as to give the creditor an equiva- lent for the loss of the interest of his money. Prop. II. — After a debt is due, or tht time expires for which by agreement no interest is charged, the creditor is entitled to interest on the avMunt until it is paid. Thus, A owes B $300, duo in 10 days. When the 10 days expire, the f :X)0 should be paid by A to B. If not paid, B loses the use of the moucy, aiid is hence entitled to interest until it is paid. Prop. III.—- W7ien a term of credit is allowed upon any of the items of an account, the date at irhicJi such items are due or commence to draw interest is found by adding its term of credit to the date of each item. Thns, goods purchased Marrh 10 on 10 days' credit would be due or draw interest March 10 + 40 da., or April 11). 042. Prob. I. — To settle equitably an account con- taining only debit items. R. Bates bought merchandipe of IT. P. Emerson as follows r May IT, 1875, on 3 montlis' credit, $205; July 11, on 25 days, $4G0 ; Sept. 15, on 05 days, $G50. 208 B US ly K a a a ii i tum e tic. m 1 Find the equated time niul tli<> ninouut that will equitably settle the account at the dut«' when the last Item is due, 1% intcnst being allowed on each item from maturity. N . f h ' 1 ^^'1 p - % ' .; , » 1 ,;>( * ¥^ JV /'. i,^' ■ A 'nraf m BOLUTIOM BT INTKHB8T MBTUOD. 1. Wo flnd tbo datu of maturity of each item thus : $3(i5 on 3 mo. ix due May 17+8 mo. = Aug. 17. f t6() «>ii S.'i da. irt duo July 11 +35 da. - Au)?. 5. $tifiU on U5 da. Ih duu Sept. 13 r 65 da. = Nov. 19. i. An the HnmH of the debt are duo nt thono dnt<'«. It is evident that when thoy nil romain unpaid until the latest maturity, II. V. Kuxthod is entitled to Ic^'al iuteroBt On laea from An;;. 17 to Nov. 1» ^ 94 do. On |460 from Au^. 5 to Nov. 19 ::= lUii du. The t050 being due Nov. 19 bearti no intervnt before thiN date. 3. On Nov. 19, II. P. Ememon U entitled to receive |i;n5, the Hum of the IttMii* of till! dfl)t and tlie iutcn-hi on |3(j5 fur 94 da. pluB the interest ou $m\ for l()(i (!a. at "7%, which Ih $14.13. Ilcncc thi' account may bo eciuifnbly nettled on Nov. 19 by It. Baton pny- ing II. 1*. KmerHon |1375 + |14.19 - $1389.12. 4. Slnco n. P. EmorHon Ik entitled to rocelvo Nov. 19, tvm ♦• |14.1'i lntol•(•^t, it la evident that if he Ih paid ♦l})7.'i a Kuftlelont tinx- Ix'forc Nov. li» to yii'M $1 1.13 Interent ut thin (inte, the del)t will b«' eiinitably nettled. But |l.'i7.*>, accordintf to (."STO), will yield $14.13 In 53 + a fraction of a day. Hence the equated lime of hottlomcnt in Sept. 80, which in W day* pre- ▼iouH to Nov. 19, the aHBumed date of settlement. SOLUTION BT PBOnncT MSTIIOD. 1. Wo find in the same manner as in the Interest mtthod the date* of maturity and the number of days each item bear:' iuterobt. 3. A»-nn>in(f Nov. 19, the latent maturity, an the date of settlement. It In evhiont that II. P. KmerMon nhouid be paid nt tliis date flliVS, the «um or the item" of the account and the Interest on |t3(S5 for tM d.iyH pluH the inter- 681 on ^-ttiO for llHi dayx. 3. Since the interest on $3«'»5 for 94 dayn at any fjlven rate in equal to the Interent on ♦3Mrv€, tliat R. BatuH may dihcbargu uquital>ly thu iuUebtedueHM iu OQO of throe way» : (1.) liy paying Xov. 10, the InUst maturity, $1875, the mm of the ifetru qf the acemmt, and th€ Intermt (/ $73070/"/ 1 day. In thii case the payment Ih $1:175 + $14.1-.J intcroHt -- $10K!).i:2. (9.) By paying $1075, tA$ $um qf the itetn* in ca*h, on ^Sejtt. '41, tht KqiTATEO TIME. (3.) Hy ffiHng Mji note for 81.175, the mim qf the itemn qf the account^ hi-itring intere/it from Stpt. ai, fhf equated time. ObHorve, tblb Ih equivalent to puyiuK thu $1075 iu cai^ \\ 9) o^ <> I i tl 270 B US INE SS A li I Til ME TIC. ''Kb. ■ V IV. Multiply each item by the number of days from its ma^ turity to the latest maturity in t/ie accoujit, and dicide the sum of these products by the sum (f the items ; the quotient is the ma/iber of days which must be counted back from the latest ma- turity to give the equated time. V. The first mMurity subtracted from tJie equated time gives the average term of credit. if r ^;f! ! • j EXAMPLES FOR PRACTICE. 044. 1. Henry Ross purchases Jan. 1, 1876, $1600 worth of goods from James Mann, payable as follows : A])ril 1, 1876, $700; June 1, 1876, $400; and Dec. 1, 1876, $500. At what date can he equitably settle the bill in one payment ? When the interval between the maturity of each item and the date of settlement is months, as in this example, the months should not be reduced todays; thus, Solution.— 1. Assuming that no pajrment is made until Dec. 1, James Maun is entitled to interest On $700 for 8 mo. = $700 x 8 or $5600 for 1 month. Ou 1400 for 6 mo. = $400 x 6 or $^400 for 1 mouth. Ilence he is entitled to the use of $8000 for 1 month. 2. $8000 -^ IICOO = 5, the number of months (642—4) which must be counted back from Dec. 1 to find the equated time, which is July 1. Hence the bill can be equitably settled in one payment July 1, 1876. 2. Bought merchandise as follows : Feb. 3, 1875, $S80 ; A])ril 18, $520 ; May 18, $260 ; and Aug. 12, $350, each item on interest from date. What must be the date of a note fcjr the sum of the items bearing interest which will equitably settle the bill ? 3. A man purchased a farm May 23, 1876. for $8600, on which he paid ^2600, and was to pay the balance, without interest, as follows : Aug. 10, 1876, $2500 : Jan. 4, 1877, $1500 ; and June 14, 1877, $2000. Afterwards it was agreed that the whole should be settled in one payment. At what date must the payment be made ? ( ma- sum is the 5 ma- gives worth , 1876, t what J date of J reduced 1, James must be Hence $380; :h item lote for luitably which lerest, as- \d June whole ^ust the H Q UA T 1 y OF PA Y M EXTS. 271 Find the date at which a note bearing interest can be given as an eq uitable settlement for the amount of oacli of tlie fol- lowing bills, each item being on interest from the date of purchase : 4. Purchased as foUows : July 9, 1876, $380 Sept. 13, " $270 Nov. 24, " $840 Dec. 29, " $260. 6. Purchased as follows : April 17, 1877, $186 ; June 24, " $250; Sept. 13, " $462. 6. Purchased as follows May 5, 1876, $186 Aug. 10, " $230 Oct. 15, " $170 Dec. 20, " $195. 7. Purchased as follows Aug. 25, 1877, $280; Oct. 10, " $193; Dec. 18, " $290. 8. Sold A. Williams the following bills of goods : July 10, $2300, on 6 mo. credit ; Aug. 15, $900, on 5 mo. ; and Oct. 13, $830, on 7 mo. What must be the date of note for the three amounts, bearing interest, which will equitably settle the account. 9. Find the average term of credit on goods purchased as follows: Mar. 23, $700, on 95 da. credit; May 17, $480, on 45 da. ; Aug. 25, $690, on 60 da. ; and Oct. 2, $380 on 35 da. 045. Prob. II. — To settle equitably an account con- taining both debit and credit items. Find the amount equitably due at the latest maturity of either the debit or credit side of the following account, and the equated time of paying the balance : JDr. B. Whitney. ( 7r. 1877. 1877. Mar. 17 To mdse. . . . $400 Apr. 13 By cash .... $200 May 10 " at 4 mo. aso Juno 15 " draft at 30 da. 250 Ang. 7 " •' at 2 mo. 540 I!/ :«! Ill i i! ,1 1^! ':!« i^ 272 BUSIXESS ARITHMETIC. Before examining the following solntion, study carefully the three propoBitionB under (64 1 ). SOLUTION BY PBODCCT METHOD. I •■'> ' Dtie. Mar. 17. 400 Sept. 10. 380 Oct. 7. 5^ Total debt, $1320 Total paid, 450 Balance, |870 ytmt. Days. Products. 204 = 27 = 81600 $91860 5&400 Paid. Amt. Days. Products. Apr. 13. 200 X 177 = 35400 July 15. 250 X 84 = 21000 450 56400 Amt. whose Int. for 1 da. is due to Cred. Amt. whose Int. for 1 da. is due to Debt. $35460 Bal. whose Int. for 1 da. is due to Cred. f Explanation.— Assuming Oct. 7, the latest maturity on either side of the account, as the date of settlement, the creditor is entitled to interest on each item of the debit side, and the debtor on each item of the credit side to this date (641). Hence, we find, according to (642—3), the amount whose interest for 1 day both creditor and debtor are entitled to Oct. 7. 2. The creditor being entitled to the most interest, we subtract the amount whose interest for 1 day the debtor is entitled to from the cred- itor's amount, leaving $35460, the amount whose interest for 1 day the creditor is still entitled to receive. 8. We find the sum of the debit and credit items, and subtract the latter from the former, leaving $870 yet unpaid. This, with $6.80, the interest on $35460, is the amount equitably due Oct. 7, equal $876.80. 4. According to (641—4), $35460 + $870 = 40??, the number of days jrevious to Oct. 7 when the debt can be discharged by paying the balance, $870, in cash, or by a note bearing interest. Hence the equated time of paying the balance is Aug. 27. Tli3 following points regarding the foregoing solution should be carefully studied : 1. In the given example, the sum of the debit is greater than the sum of the credit items ; consequently the balance on the account is due to the creditor. But the balance of interest being also due him, it is evident that to settle the account equitably he should be paid the $870 before the assumed date of settlement. Hence the equated time of paying the balance must be before Oct. 7. 2. Had the balance of interest been on the credit side. It is evident the debtor would be entitled to keep the balance on the account until the e three 'Products. 35400 21000 56400 e to Cred. 3 to Debt. to Cred. er Bule of interest the credit i— 3), the jutitled to ibtract the 1 the cred- 1 day the the latter le interest ;r of days |e balance, }d time of Bhould lie sum of liie to the Ident that ]>efore the le balauce ^ident the until the EQUATION OF PAYMEXTS. 273 Interest upon it would be equal the Interest due him. Tlence the equated time of paying the balance woakl be after Oct. 7. 3. Had the balance of the account been on the credit side, the creditor would be overpaid, and hence the balance would be due to the debtor. Now in case the balance of interest is also on the credit side and due to the debtor, it is evident that to settle the account equitably the debtor should be paid the amount of the balance before the a^ssumed date of set- tlement. Hence the equated time would be before Oct. 7. In case the balance of interest ia on the debtor ^^i(le, it is evident that while the creditor has been overpaid on the account, he is entitled to a balance of interest, and consequently should keep the amount he has been overpaid until the interest uiwu it would be equal to the interest due him. Hence the equated time would be afitr Oct. 7. 4. The interest method given (642) can be used to mlvautage in finding the equated time when the time is long between the maturity of the items and the assumed date of settlement. In case this method of solution is adopted, the foregoing conditions are equally applicable. From these illustrations we obtain the following HTJLE. 646. /. Find the maturity of each item on the debit and credit side of the accovnt. II. AsfiJfme as the date of settlement the latest maturity on either side of the account, and find, in the manner indicated abfwe, tlie number of days from this date to the maturity of each, on both sides of the account. III. Multiply each debit and credit item by the number of daps from its maturity to the date of settlement, and divide the bal- ance of the debit and credit products by the balance of the debit and credit items ; the quotient is the number (fdays the equated time is from the assumed date of settlement. IV. In case the balance of items and balance of interest are both on the same side of the account, subtract this number of days from, the assumed date of settlement, but add this number of days in ease they are on opposite sides ; the result thus obtainsd is th€ equated time. M frw '■\ if w 274 BUSINESS ARITHMETIC, EXAMPLES FOR PRACTICE. 647. 1. Find the face of a note and the date from which it must bear interest to settle equitably the following account : Dr. jAifES Hand in acct. vyith P. Anstead. Cr. 1876. 1876. Jan. 7 To mdee. on 3 mo. $430 Mar. 15 By draft at 90 da. $500 May 11 «♦ " " 2 mo. 390 May 17 *' cash .... 280 June 6 " " 5 mo. 570 Aug. 9 *' mdse. on 30 da. 400 2. Equate the following account, and find the cash payment Dec. 7.1876: Dr. William Henderson. CV, 1876. 1876. Mar. 83 To mdse. on 46 da. $470 Apr. 16 By cash .... $4!K) May 16 44 4i ti 25 da. 380 June 25 " mdee. on 30 da. 650 Aug. 7 " " " 36 da. 590 July 13 " draft at 60 da. 2C0 8. Find the equated time of paying the balance on the follow- ing account : Dr. Hugh Quthbib. Or. 1876. 1876. Jan. 13 To mdse. on 60 da. $840 Feb. 15 By note at 60 da. $700 Mar. 24 " '• 40 da. 580 Apr. 17 " cash . . . • 460 June 7 " " " 4 mo. 360 June 9 '' draft at 30 da. 1150 July 14 " " 80 da. 730 1 4. I purchased of Wm. Rodgfers, March 10, 1876, $930 worth of goods ; June 23, $G80 ; and paid, April 3, |870 cash, and gave a note May 24 on 30 days for $500. What must be the date of a note bearing interest that will equitably settle the balance ? i,! biicli it nt: Cr. $500 280 I. I 400 Btynient Cr. — ■ $4W a. 650 2C0 s follow- er. $700 4(jO 1150 , $930 10 cash, lUst be settle HE VIEW, 275 REVIEW AND TEST QUESTIONS. 048. 1. Define Simple, Compound, and Annual Interest. 2. Illustrate by an example every step in the six per cent method. 3. Show that 12% may be used as conveniently as 6J^, and write a rule for finding the interest for months by this method. 4. Explain the method of finding the e:vnct interest of any sum for any given time. Give reasons for each step in the process. 5. Show by an example the difference between true and hiink discount. Give reasons for your answer. 6. Explain the method of finding the present worth. 7. Explain how the face of a note is found when the pro- ceeds are given. Illustrate each step in the process. 8. Define Erchange, and state the difference between Inland and Foreign Exchan<]^e. 9. State the difference in the three biUs in a Set of Ex- change. 10. What is meant by Par of Exchange ? 11. State the various methods of Inland Exchange, and illus- trate each by an example. 12. Illustrate the method of finding the cost of a draft when exchange is at a discount and brokerage allowed. Give reasons for each step. 13. State the methods of Foreign Exchange. 14. Illustrate by an example the difference between Direct and Indirect exchan^\ 15. Define Equation of Payments, an Account, Equated Time, and Term of Credit. 16. Illustrate the Interest Method of finding the Equated Time when there are but debit items. 17. State when and why you count forward from the assumed date of settlement to find the equated time. i If i i w RATIO. PREPARATORY PROPOSITIONS. 04:9« Tico numbers are compared and their relation deter- mined by dividing the first by the second. For example, the relation of $8 to $4 is determined thus, |8+$4 = 2. Observe, the quotient 2 indicates that for every one dollar in the $4, there are two dollars in the |8. I) "J '•■< ,1 .»(, Be particular to observe the following: 1 . Wlien the greater of two numbers is compared with the less, the relation of the numbers is expressed either by the relation of an integer or of a mixed number to i\\e iinit 1, that is, by an improper fraction whose denominator is 1. Thup, 20 compared with 4 gives 20+4 = 5 ; that ie, for every 1 in the 4 there are 5 in the 20, Hence the relation of 20 to 4 is that of the integer 5 to the nnit 1, expressed fractionally thus, |. Again, 29 compared with 4 gives 29+4 = 7^ ; that is, for every 1 in the 4 there are 7^ in 29. Hence, the relation of 29 to 4 is that of the mixed num- hs the same relation. PROBLEMS ON BATIO. 602. Since every ratio is either a proper or improper frac- tion, the principles of reduction discussed in (224) apply to the reduction of ratios. The wording of tlic principles must l)e slightly modified thus : Prin. I. — TJie terms of a ratio must each represent units of the same kind. Pit IN. II. — Multiplying both terms of a ratio by tJie same number does not change t?ie value of the ratio. Prin. III. — Dividing both terms of a ratio by the same num- ber does not change the value of the ratio. For the illustration of these principles refer to (224). 063. Prob. I.— To find the ratio between two given numbers. Ex. 1. Find the ratio of $56 to $84. Solution.— Since, accordin": to (649), two numbers are compared by divii'.in^ the first by the eecoud, we divide |i3G by $84, giving $56 + |84 = l\ J that ie, $56 is 2' of |84. Hence the ratio of $56 to $84 is ||. f ii m 280 Lf . , ■t^' I' f r^ 1 .■ i -"■ : , ' ! ^ ;«- - ^*t. ^^B' u^u.::-' f BUSINESS ARITHMETIC, Ex. 2. Find th.* ratio of 1 yd. 2 ft. to 1 ft. 3 In. Solution.— 1. Since, according to (051), only numbers of the Bamc dcnoniliiatlon can bo coniimrotl, wo reduce both termH to Inchec, jjivlnjr 6U In. and 15 in. 2. Dividing 60 in. by 16 in. we havo 60 in. -«- 16 in. = 4 ; that it), 60 in. it) 4 timuH 15 in. llcnco Ibo ratio of 1 yd. 2 ft. to 1 It. 3 lu. it) ;. EXAMPLES FOR PRACTICE. «04. Find tlio ratio 1. Of 143 yd. to 305 yd. 3. Of 73 A. to 3G5 A. 2. Of $512 lo $250. 4. Of 082 da. to 21)40 da. 5. Of £41 58. Od. to £2 38. 6d. 6. Of 20 T. cwt. 93 lb. to 25 cwt. 43 lb. 5 oz. 0(J5. P«OB. II.— To reduce a ratio to its simplest terms. Reduce the ratio Y ^o i^^ simplest tenns. Solution.— Since, according to (602— III), the value of the ratio V i.s not cbanped by dividing both terms by the pame number, we divide the antecedent 15 and the consequent 9 by 3, their greatest common divisor, giving — "*" = -. But having divided 15 and 9 by their greatest common ^*'9+33 divisor, the quotients T> and 3 must be prime to each other. Ilence (600) 1 are the simplest terms of the ratio V* EXAMPLES FOR PRACTICE. COG. Reduce to its simplest terms 1. The ratio 21 : 50. 2. The ratio : 9. 3. The ratio ^. 4. The ratio 05 : 85. 5. The ratio |f f . 0. The ratio 195 : 39. Express in its simplest terms the ratio (see 516) 7. Of I ft. to 2 yd. 8. Of 90 T. to 50 T. 9. Of 3 pk. 5 qt. to 1 bu. 2 pk. 10. Of8s. 9d. to£l. RATIO. 281 1 667. Prob. III.— To find a number that has a given ratio to a given number. How many dollars are g of |72 f BoLVTiON.— The fhiction } denotes the ratio of the required nambor to ITS; uamely, for every $8 iu f79 there are $5 in the required uumbur. Connequontly we divide the $73 by $8, and maltiply $5 by the quotient. Honce, first step, 173 -»• $8 = 9; second step, $5x9 = 945, the required number. ObservSt that this problem is the same as Pbob. VIII (400), and Prob. n (963)* Compare this solution with the solution iu each of these problems. i EXAMPLES FOR PRACTICE. 668. Solve and explain each of the following examples, regarding the fraction in every case as a ratio. 1. A man owning a farm of 248 acres, sold /^ of it ; how many acres did he sell 1 2. How many days are -^ of 360 days ? 8. James has $706 and John has } as much ; how much has John? 4. Mr. Jones has a quantity of flour worth $3140 ; part of it being damaged he sells the whole for ^ of its value ; how much does he receive for it? 5. A man's capital is $4500, and he gains ^ of his capital ; how much does he gain ? 669. Prob. IV.— To find a number to which a given number has a given ratio. $42 are j of how many dollars 7 Solution.— The fraction | denotes the ratio of $43 to the required number ; namely, for every |7 In $43 there are $4 in the required number. Consequently we divide the $43 by $7 and multiply |4 by the quotient. Hence, first step^ t4S -»- $7 = 6 ; teoond step, $4x6 = $94, the required number. Observe, that this problem is vhe same as Pbob. IX (491). Compare the solutions and notice the points of difference. 19 i,' * 282 BUSINESS ARITHMETIC, ,1 EXAMPLES FOR PRACTICE. 670. Solve and explain each of the following examples, regarding the fraction in every case as a ratio. 1. I received |75, which is f of my wages; how much is still due? 2. 96 acres are |f of how many acres 1 3. James attended school 117 days, or ^^ of the term ; how many days in the term? 4. Sold my house for |2150, which was ^f of what I paid for it ; how much did I lose ? 5. 48 cd. 3 cd. ft. of wood is ^j of what I bought; how much did I buy % 6. Henry reviewed 249 lines of Latin, or f of the term's work ; how many lines did he read during the term ? 7. Mr. Smith's expenses are f of his income. He spends ^^1500 per year ; what is his income ? 8. A merchant sells a piece of cloth at a profit of $2.o0, which is ^\ of what it cost him ; how much did he pay for it ? 9. 4 gal. 3 qt. 1 pt. are /j of how many gallons? 10. 7 yards and 2 ft, are f of how many yards ? 671. Prob. v.— To find a number to -which a giren number has the same ratio that two other given numbers have to each other. To how many dollars have $18 tlie same ratio that 6 yd. have to 15 yd. ? SoLtrnoN.— 1. We find by (663—1) the ratio of (5 yd. to 15 yd., which ^ r*f = i« a^xording to (659). 8. SiBoe f denotev the ratio of ikab fid to tito reqoired wuDber, the |18 mnat be the antecedent ; benoe we have, aooordiAg to <6^0). Jin^t s(ej), 418 -I- $3 = 9 ; aeoond et^, $5x9:? $46, the reqaired number. Ob^erre, that in this problem we have the antecedent of a ratio ^ren to And the coneaqueut. In the following we have the consequeot given to find the antecedent R A TIO. 283 mples, lucb is tt ; how ,t I paid it; how e term's e spends of $2.50, y for it? a given lumbers It 6 yd. [yd. , which r, ihe |18 lo piTcn to [t given to 672. Prob. VI. — To find a number that has the same ratio to a given number that two other given numbers have to each other. How many acres have the same ratio to 12 acres that $56 have to $84? Solution.— 1. We find by (663—1) the ratio of $56 to $84, which is IS = S, according to (660). 2. Since i denotes the ratio of the required number to 12 acres, the 12 acres must be the consequent; hence we have, according to (666".,//>'< step, 12 acr. + 8 acr. = 4 ; second stsp, 2 acr. x 4 = 8 acres, the required number. fiXAMPLES FOR PRACTICE. 673. The following are applications of Prob. V and VI. 1. If 12 bu. of wheat coat $16, what will 42 bu. cost? Regarding the solution of examples of this kind, observe that the price or rate per unit is assumed to be the same for each of the quantities given. Thus, since the 12 bn. cost $15, the price per bushel or unit Is $1.25, and the example aslis for the cost of 42 bu. at this price per bushel. Cuu- eequently whatever part the 12 bu. are of 42 bu., the $15, the cost of 12 bu., must be the same part of the cost of 42 bu. Hence we find the ratio of 12 bu. to 42 bu. and solve the example by Prob. V. 2. If a man earn $18 in 2 weeks, how much will he earn in 52 weeks ? 3. What will 16 cords of wood cost, if 2 cords cost $9 ? 4. If 24 bu. of wheat cost $18, what will 30 bu. cost ? 5. If 24 cords of wood cost $60, what will 40 cords cost ? 6. Bought 170 pounds of butter for $51 ; what would 680 pounds cost, at the same price ? 7. At the rate of 16 yards for $7, how many yards of cloth can be bought for $100 ? 8. Two numbers are to each other as 10 to 15, and the less number is 329 ; what is the greater? P'; ■> <. J' ^'' PROPORTION f* DEFIinTIONS. 674, A Proportioti is an equality of ratios, the tenns of the ratios being expressed. ThuH the ratio J is equal to the ratio U ; hence i = y is a proportion, and is read, The ratio of 3 to 5 is equal to the ratio of 12 to 30, or 3 is to 5 as 12 is to 20. 075. The equality of two ratios constituting a proportion is indicated either by a double colon (: :) or by the sign (=). Thus, I = T*s, or 8 : 4 = 9 : 19, or 8 : 4 : : 9 : 12. 076. A Simple Proportion is an expression of the equality of two simple ratios. Thus, t\ = II, or 8 : 12 : : 82 : 48, or 8 : 12 = 32 : 48 is a simple propor- tion. Hence a simple proportion contains four terms. C77. A Compound Proportion is an expression of the equality of a compound (057) and a simple ratio (650). 2:3) Thus, " > : : 48 : 60, or S X J = 43, is a compound proportion. It is 6:5) read. The ratio 2 into 6 is to 3 into 5 as 4S is to 60. 078. A Proportional is a number used as a term in a pro{)ortion. Tims in the simple proportion 3 : 5 : : 6 : 15 the numbers 2, 5, 6, and 15 are its terms ; hence, each one of these numbers is called a proportional, and the four numbers together are called proportionals. Wh<^n three numbers form a proportion, one of them is repeated. Thus, 32 : 8 : : 8 : 8. SIMPLE PROPORTION. 285 •me of >ortion, is to 5 rtion is of tlie propor- of tlie 1. It IB in a \, and 15 )rtion!iK Thus, 670. A Mean Proportional is a number that is the Conaequent of ono and the Antecedent of the other of the two ratios forming a proportion. Tbas in the proportion 4 : 8 : : 8 : 16, the namber 8 iB the consequent of the first ratio and the antecedent of the second ; hence is a mean propar- tional. (>80. The Aiifeceflents of a proportion are the first and third terms, and the Consequents are the second and fourth terms. 081. The Extremes of a proportion arc its first and fourth terms, and the Means are its second and third terms. SIMPLE PROPORTION. PREPARATORY STEPS. 082. The following preparatory steps should be i^erfectly mastered before applying proportion in the solution of problems. The solution of each example under Step I should be given in full, as shown in ((571 and 072), and Step II and III should be illustrated by the pupil, in the manner shown, by a number of examples. . C583. Step I. — Find by Prob. V and Yl.in ratio, the miss- ing term in tJiefolloicing proportions : The required term is represented by the letter x. 1. 24 : 60 : : r : 15. 4. 2 yd. : 8 in. : : a; : 3 ft. 4 in. 2. 6 : 42 ; 3. 84 : X : 5 : X. 21 : 68. 5. 5 bu. 2 pk. : 3 pk. : : ./• : 4 bu. 6. a? : £3 28. : . 49 T. : 18 cwt. Step II. — Show that the product of tlie extremes of a propor- tion is eqiuU to the product of the means. Thus the proportion 2 : 3 : : 6 : 9 expressed fractionally gives 1 = J. ^86 BUSINESS ARITHMETIC. El ■ -J Now if both tenne of tliis equality be multiplied by 3 and by 9, the consoqucntB of the given ratios, the equality U not changed; hence, iix9x8 0x3x9 _ _ Cancelling (1 76) the factor 3 in the left-hand term and 9 in the right-hand term we have 2x9 = 6x3. Bat 2 and 9 are the extremes of the proportion and 6 and 3 are the means ; hence the truth of the proposition. I ■ ■| 1. 14 : 3 : : 3- : 13. 4. 2. 27 : c : : 9 : 5. 6. 3. aj : 24 : : 7 : 8. 6. Step III. — 8?u>w that, since the product of the extremes is equal to the product of the means ^ any term of a proportion can he found when the other three are known. Thus in the proportion 8 : » : : 9 : 15 we have known the two extremes 3 and 15 and the mean 9. Bat by Step II, 3 x 15, or 45, U equal to 9 times the required mean ; hence 45 + 9 = 6, the required mean. In the eanic manner an* one of the terms may be found ; hence the truth of the proposition. Find by this method the missing term in the following : $13 : T : : 5 yd. : 3 yd. 64 cwt. \x : : $120 : $15. 128 bu. : 3 pk. : : x : ^1.25. Solution by Simple Proportion, 684-. The quantities considered in problems that occur in practical business are so related that when certain conditions are assumed as invariable, they form ratios that must be equal to each other, and hence can be stated as a proportion thus, If 4 yd. of cloth cost $10, what will 18 yd. cost ? Observe, that in this example the price per yard is assumed to be invariable, that is. tlie price is the same in both cases ; consequently whatever part the 4 yd. are of the 18 yd., the $10 are the same part of the cost of the 18 yd., hence the ratio of the 4 yd. to the 18 yd. is equal the ratio of the flO to the required cost, giving the proportion 4 yd. : 18 yd. : : $10 : %x. SIMPLE PROPORTION, 287 9, the hence, id term are the truth of Ion can sxtrcmes equal to . In the th of the ng S15. ^1.25. )ccur in Editions )e equal thus, issumed cases ; [the $10 itio of to the lO : |2- EXAMPLES FOR PRACTICE. 685. Examine carefully the following proportions, and state what must be considered in each case as invariable, and why, in order that the proportion may be correct. 1. The number of units bought in oue case is to ' The number 1 f The cost ] of unit? bought in ■as - in the first is to . another case . case The 2 Principal in one case The number of men that can do a piece of work in one case The Principal in another case The interest in the first case f 8. The number " r The 1 of men number that can do of days the same ■ • • the • work in second another case . . work , ' The cost in the second case. The interest in the second case. The number of dajs the first work. Why is the second ratio of this proportion made the ratio of the number of days the second work to the number of days the first work f Illustrate this arrangement of the terms of the ratio by other examples. In solving examples by simple proportion, the following course should be pursued : /. Represent the required term by x, and make it the last extreme or consequent of the second ratio in the proportion. II. Find the term in the example thaJt is of the same denom- ination as the required term, and make it the second mean or the antecedent of the second ratio of the proportion. III. Determine, hy inspecting carefuUy the condi'.ions giren in the example, whetTier x, the required term of the ratio now expressed^ must he greater or less than the given term. IV. If T, the required term of the ratio expressed, must he greater than the given term, m^ike the greater of the remaining terms in the example the consequent of the first ratio of the proportion ; if less, make it the antecedent. % 288 BUSINESS ARITHMETIC, m0 l:\ '« * t^ V. When the proportion w stated, find the required term either as shown in (67 1) or in (672). Observe, that in either way of finding the required term, any flictor that is common to the given extreme and either of the given means should t>e cancelled, as shown in (175)* 4. If 77 pounds of sugar cost $8.25, what will 84 pounds cost? 5. How many bushels of wheat would be required to make 39 barrels of flour, if 15 bushels will make 3 barrels? 6. If 6 men pat up 73 feet of fence in 3 days, how many feet will they put up in 33 days ? 7. I raised 245 bushels of com on 7 acres of land ; how many bushels groW on 2 acres? 8. What will 168 pounds of salt cost, if Sj^ pounds cost 37^ cents? 9. If 25 cwt. of iron cost $84.50, what will 24| cwt. cost? 10. Paid $2225 for 18 cows, and sold them for $2675 ; what should I gain on 120 cows at the same rate ? 11. If 5 lb. 10 oz. of tea cost $5.25 ; what will 7 lb. 8 oz. cost? 12. My horse can travel 2 mi. 107 rd. in 20 minutes ; how far can he travel in 2 hr. 20 min. ? 18. If a piece of cloth containing 18 yards is worth $10.80, what are 4 yards of it worth ? 14. If 18 gal. 3 qt. 1 pt. of water leaks out of a cistern in 4 hours, how much will leak out in 36 hours ? 15. Bought 28 yards of cloth for $20; what price per yard would give me a gain of $7.50 on the whole ? 16. My annual income on U.S. 6%'s is $337.50 when gold is at 112^ ; what would it be if jyold were at $125 ? 17. If I lend a man $69.60 for 8i^ months, how long should he lend me $17.40 to counterbalance it ? 18. If 10 bu. of apples cost $7.50, what will 60 bu. cost? Ans. $45. 19. If a board 13 ft. long cast a shadow of 10 ft., what will be the height of a tree which casts a shadow of 115 ft. ? Ans. 149 It. 6 in. COMPOUND PROPORTION, 289 km rthflt lid be »undB make y fe«t ; how s cost >8t? : wliat COMPOUND PROPOKTION". PREPARATORY STEPS. 686. Step. I. — A compound ratio U reduced to a simple one hy multiplying the antecedents together for an antecedent and the consequents for a consequent (657). ThuB, the componnd ratio ] 4 ! o [ J" reduced to a elmple ratio by m'iltiplyiiJg the antecedents 6 and 4 together, and the consequents 7 and 3. Expressing the ratios fractionally we have f x j = |f = ; (665). Obsefve, that any factor that Is common to any antecedent and con- sequent may be cancelled before the terms are multiplied. Reduce the following compound ratios to simple ratios in their simplest terms. 1. 9 : 25 15 : 18 28 : 50 3 : 7 2. ''8^1 35 115 J 8. 16 : 9 27 : 15 28 8 Step II. — A compound proportion is reduced to a simple proportion by reducing the compound ratio to a simple ratio. (8*9) Thus, in the componnd proportion ] » , 4 r : : 24 : 18, the compound ratio 2 X ; is equal the simple ratio ; ; substituting this in the proportion for the compound ratio we have the simple proportion 4 : 3 : : 24 : 18. Observe, that when a compound proportion is reduced to a simple pro- portion, the missing term is found according to (671), or (673). Find the missing term in the following : (24 : 15) I. < 7 : 16 [• ( 25 : 21 ) 24 : 15 7 . 16 25 : 21 23 12 8 32 36 9 ( 23 : 12 ) i 8 : 32 >■ ( 36 : 9 ) 40 : X, 75 : X, 2. (28 : 7) i 9 : 36 [• ( 50 : 10 ) 28 : X, 32 : X. 13 ■i. 'vrf 290 BUSINESS ARITHMETIC, Solution by Cotnpound Proportion. 087. The following preparatory propositions should be carefully studied, and the course indicated observed in solving problems involving comi)ound proportion. Prop. I. — There are one or more conditions in every example incolviug proportion, which must he regarded as iavuriablc in order that a solution may be given, thus, If 9 lioraes can sabsitjt on 60 bu. of oats for SO day^, how long can 6 burses subsist ou 70 bu. ? In this example there are two conditions that must be considered as invariable iu order to give a solution : 1. The fact that each horse subsists on the same quantity of oats each dsjy. 2. The fact that each bushel of oats contains the same amount of food. Prop. II. — To solve a problem involving a compound pro- portion, the ffect of each ratio, which forms the compound Toiio, on the Required term, must he considered separately, thus : If 5 men can build 40 yards of a fence in 12 days, how many yards can 8 men build in 9 days ? 1. We observe that the invariable conditions in this example are (1.) That each man in both cases does the same amount of toork in the same time. (8.) That the sam£ amount of work is required in each case to build one yard of the fence. 2. We determine by examining the problem how the required term is affected by the relations of the given term, thus : (1.) We observe that the 5 men in 12 days can build 40 yards. Now since each man can build the same extent of the fence in one day, it is evi- dent that if the 8 men work 12 days the same as the 5 men, the 40 yards built by the 5 men in 12 days must have the same ratio to the number of Id be alving cample 'iable ong can iered as of oats amount d pro- npaund irately, many tample * in the lUd one luired \. Now is> evi- |0 yards iber of COMPOUND PB O PORTION. 291 yards that can be built by the 8 men in 12 days as 5 men have to 8 men ; hence the proportion 5 men : 8 men : : 40 yards : x yards. This proportion will give the number of yards the 8 men can build in 12 days. (i.) We now observe that the 8 men work only 9 days ; and since they can do the same amount of work each day, the work done in 12 days must have the same ratio to the work they can do in 9 days that 12 days? have to 9 days. Hence we have the compound proportion 5 men : 8 men (..„,, We find from this proportion, according to (686—11), ttiat the 8 men can build 48 yards of fence in 9 days. EXAMPLES FOR PRACTICE. 688. 1. If 12 men can saw 45 cords of wood in 3 days, working 9 hours a day, how much can 4 men saw iu 18 days^ working 12 hours a day ? 2. If it cost $88 to hire 12 horses for 5 days, what will it cost to hire 10 horses for 18 days ? 3. When the charge for carrying 20 centals of grain 50 miles is $4.50, what is the charge for carrying 40 centals- 100 miles ? 4. If 28 horses consume 240 bushels of corn in 112 days, how many bushels wiU 12 horses consume in lOG days. 5. The average cost of keeping 25 soldiers 1 year is $3000 ; what would it cost to keep 139 soldiers 7 years ? 6. 64 men dig a ditch 72 feet long, 4 feet wide, and 2 feet deep, ill 8 days ; how long a ditch, 2| feet wide and 1^ feet deep, can 96 men dig in 60 days ? 7. If 1 pound of thread makes 3 yards of linen, 1} yard wide, how many pounds would make 45 yards of linen, 1 yard wide ? 8. If it requires 8400 yd. of cloth 1| yd. wide to clothe 8500 8 )ldierH, how many yards | wide will clothe 6720 ? Ill I- %t:: PARTNERSHIP u ' m if It- i DEFINITIONS. 080. A Partnership is au association of two or more persons lor the transaction of business. The persons associated are called partners, and the Association is called a Vomitany^ Firm^ or House. CM)0. The Cajntal is the money or other property invested in the business. The Capital is aleo called the Investment or Joint-stock of the Company. 091. The Assets or Effects of a Company are the proiM5rty of all kinds belonging to it, together with all the amounts due to it. G92. The Liabilities of a Company are its debts. PREPARATORY PROPOSITIONS. 093. Prop. I. — The profits and the losses of a comr pany are divided among the partners, according to the 'vcdue of eojch man's investment at the time the division is made. Observe carefully the following regarding this proposition : Since the use of money or property ie iteelf value, it ia evident that the value of an investment at any time after it is made, depends first upon the amount invested, second on the length of the time the investment has been made, and third the rate of interest. Thus the value of an investment of $500 at the time it is made is just P A RT y E R s n IP . 293 $500 ; bnt at tfu end of ^ yearn, reckontn •: -((^ainor)- S ■< wtOH f invented 1 i loaa ) ' or Each ),*a gain loaa. II. By Percentage ilius : Find what per cent (493) the whole gain or low in of the whole capital invested, and take the same per c^ent of emh man's investment as his share of the gain or loss. II. By FractioHB thus : Find tcJiat fractional part each man's investment is 8. 1. Three men hire a pasture for $136.50 ; A \)\ii% in 16 cows for 8 weeks, B puts in 6 cows for 12 weeks, and C the same number for 8 weeks ; what should each man pay ? 2. A and B engage in business ; A puts in $1120, for 5 months and B $480 for 8 months ; they gain $354 ; what is each man's share of the gain ? 3. The joint capital of a company wa« $7800, which was docbled at the end of the year. A put in \ for 9 mo., B ^ for 8 mo., and C the remainder for 1 year. What is each one's stock at the end of the year ? 4. A and B formed a partnership Jan. 1, 1876. A put in ^^6000 and at the end of 8 mo. $900 more, and at the end of 10 mo. drew out $800 ; B put in $9000 and 8 mo. after $1500 more, and drew out $600 Dec. 1 ; at the end of the year the net profits were $8900. Find the share of each. 5. Jan. 1, 1875, three persons began business. A put in $1200, B put in $500 and May t $800 more, C put in $700 and July 1 $400 more ; at the end of the year the profits were $875 ; how shall it be divided ? m;^ iir r- H -■ v.- 1' H4 ALLIGATION. ALLIGATION MEDIAL. 699* Alligation Medial is the process of finding the mean or average price or quality of a mixture composed of several ingredients of different prices or qualities. ■I i^ & -^ EXAMPLES FOR PRACTICE. TOO. 1. A grocer mixed 7 lb. of coffee worth 30 ct. a pound with 4 lb. @ 25 ct. and 10 lb. (ib 32 ct. ; in order that he may neither gain nor lose, at what price must he sell the mixture? 7 lb. @ 30 ct. 4 lb. @ 25 ct. 10 lb. @ 32 ct. 21 lb. = 16.30 ^ 21 = = $2.10 = 1.00 = 3.20 $6.30 30 ct. Solution.— 1. Since the valne of each kind of coffee is not changed hy mixing, we find the value of the entire mixture by finding the value of each kind at the given price, and taking the Bum of these values as shown in illus- tration. 2. Having found that the 21 lb. of coffee are worth at the given prices ♦6.30, it is evident that to realize this amount fk-om the sale of the 21 lb. at a uniform price per pound, he must get for each pound ,V of $6.30 ; hence, $6.30 -i- 31 = 30 cents, the selling price of the mixture. 2. A wine merchant mixes 2 gallons of wine worth $1.20 a gallon with 4 gallons worth $1.40 a gallon, 4 gallons worth $.90 and 8 gallons worth $.80 a gallon ; what is the mixture worth per gallon ? 3. A grocer mixes 48 lb. of sugar at 17 ct. a pound with 68 lb. at 13 ct. and 94 lb. at 11 ct. ; what is a pound of the mixture worth? ALL IQ A Tl N. 297 emean several 4. A merchant purchased 60 gallons of molasses at 00 ct. per gallon and 40 gallons at 25 cents, which he mixed with 8 gallons of water. He sold the entire mixture so as to gain 20 per cent on the original cost ; what was his selling price per gallon ? 5. A goldsmith melts together 6 ounces of gold 22 carats fino, 30 ounces 20 carats fine, and 12 ounces 14 carats fine ; liuw many carats fine is the mixture? 6. A farmer mixes a quantity of barley at 90 cents a bushel with oats at 37 cents a bushel and rye at 65 cents a bushel. Find the price of the mixture. 50 ct. a that he sell the value of inged by [he entire of each iking the in illus- jn prices 21 lb. at : hence. |tll.20 a wortli lixture ^d with of the ALLIGATION ALTERNATE. 701. Alligation Alternate is the process of finding the proportional quantities of ingredients of different prices or qualities that must be used to form any required mixture, when the price or quality of the mixture is given. PREPARATORY PROPOSITIONS. 702, Prop. I. — In forming any mixture, it is assumed that the value of the entire mixture must be equal to the aggregate value of its ingredients at their given prices. Thus, if 10 pounds of tea at 45 ct. and 5 pound? at 60 ct. be mixed, the value of the mixture must be the value of the 10 pounds plus the vatae of the 5 pounds at the given prices, Avhloh is equal $4.50 + $3.00 ■-- $7.50. Hence there is neither gain nor loss in forming a mixture. Prop. II. — The price of a mixture must he less than tiie highest and greater tJian the lowest price of any ingredient used informing the mixture. Thus, if sugar at 10 ct. and at 15 rt. per pound be mixed. It it» evident the price of the mixture must be lens than I.** cents and greater than 10 cents ; that u, It must be some price between 10 and 15 cents. wr 298 BUSIXESS ARITHMETIC, ILLUSTRATION OF PROCESS. 703. If tea at 50 ct., 60 ct., 75 ct., and 90 ct. per pound be mixed and sold at 66 ct. per })ound ; bow mucb of eacb kind of tea can be put in tbe mixture ? First Step in Solution, We find the jmin or loss on one unit of each injjredient thns : 66 ct. 66 ct. (1.) ] . ( 75 ct. — 66 ct. = 9 ct. loss. ^ '' < 90 ct. - 66 ct. = 24 ct. loss. 56 ct. = 10 ct. gain. 60 ct. = 6 ct. gaiu. Second Step in Solution, We now lake an ingredient on which there i>* a gain, and one on which there is a loss, and ascertain how much of each must be put in tbe mixture to make the gain and loss equal ; thus : Producino Gain. Gaotbd and Lost. Pbodttcino Loss. (1.) 9 lb. at 10 ct. per lb. gain. = 90 ct. = 10 lb. at 9 ct. per lb. lojs. (2.) 4 lb. at 6 ct. per lb. gain. = 24 ct. = 1 lb. at 24 ct. per lb. low. Hence the mixture must contain 9 lb. at 56 ct. per pound, 10 lb. at 75 ct. per poHnd, 4 lb. at 60 ct. per pound, and 1 lb. at 90 ct. per pound. 704. Observe carefully the following : 1. The gain and loss on any two ingredients may be balanced by assuming any amount as the sum gained and lost. Thus, ins^tead of taking 90 cents, as in (1) in the above solution, as the amount gained and lost, we might take 360 cents ; and dividing 360 cents by 10 cents would give 36, the number of pounds of 56 ct. ton that would gain this sum. Again, dividing 360 cents by 9 cents would give 40, the number of pounds of 75 ct. tea that would lose this sum. 2. To obtain intefjrnl proportional parts the amount -sumed must be a multiple of the gain and loss on one unit of the ingredients balanced, and to obtain the least integral propor- tional parts it must be the least common multiple. und be kind of I ALLIGATION, 299 3. \Mien a number of ingredients are given on which there is a gain and also on which tliere is a loss, they may be balanced with each other in several ways ; hence a series of diflferent mixtures may be formed as follows : Taking the foregoing example we have A Second Mixture thus: PBODUCiNa Gain. Gained and Lost. Producing Loss. (1.) 24 lb. at 10 ct. per II). gain. = 240 ct. = 10 lb. at 24 ct. per lb. loss. (2.) 9 lb. at 6 ct. per lb. gain. = 54 ct. = 6 lb. at 9 ct. per lb. loss. Ilence the mixture is composed of 24 lb. @ 56 ct., 9 lb. @ 60 ct., 10 lb. @ 90 ct., and 6 lb. ® 75 ct. on which e mixture Loss. tr lb. loH. T lb. lo». at 75 ct. )alanced kn, as the 360 cents lut would ire 40, the •sumed of the Ipropor- A TJiird Mixture thus : Producing Gain. Gained and Lost. Pboducino Loss. (1.) 9 lb. at 10 ct. per lb. gain. = 90 ct. = 10 lb. at 9 ct. per lb. loss. (2.) 24 lb. at 10 ct. per lb. gain. = 240 ct. = 10 lb. at 24 ct. per lb. loss. (3.) 9 lb. at 6 ct. per lb. gain. = 54 ct. = 6 lb. at 9 ct. per lb. loss. Observe, that in (1) and (2) we have balanced the loss on the T.'i ct. and 90 ct. tea by the gain on the 56 ct. tea; hence we have 9 lb. + 24 lb., or 33 lb. of the 56 ct. tea in the mixture. Observe, also, that in (3) we have balanced the gain on the 60 ct. tea by a loss on the 75 ct. tea ; hence we have 10 lb. + 6 lb., or 16 lb. of the 75 ct. tea in the mixture. Hence the mixture is composed of 33 lb. @ 56 ct., 9 lb. @ 60 ct., 16 lb. % 75 ct., and 10 lb, @ 90 ct. 4. Mixtures may be fonned as follows : /. Take any pair of ingredients, one giving a gain and the other a loss, and find the gain and loss on one unit of each. II. Assume the least common multiple of the gain and loss on one unit as the amount gained and lost, by putting t?ie two ingredients in the mixture. III. Divide the amount thus assumed hy the gain and then by the loss on one unit; the results will be respectively the i^*;:'f :•, -i-.f 300 BUSI^'£SS ARITHMETIC, ^f'^^i M number of units of each ingredient thtt must he in the mixture that the gain and loss viay balance ea^h other. IV. Proceed in the same manner with other ingredients; the results will be tlie proportional parts. EXAMPLES FOR PRACTICE. 705. 1. A man wishes to mix sufficieut water with mo- lasses worth 40 cents a gallon to make the mixture worth 24 cents a gallon ; what amount must he take of each 'I 2. How much sugar at 10, 9, 7, and 5 ct. will produce a mixture worth 8 cents u pound ? 3. A merchant desires to mix flour worth ^6, $7^, and $10 a barrel so as to sell the mixture at $9 ; what proportion of each kind can he use V 4. A jeweller has gold 16, 18, 22, and 24 carats fine; how much of each must he use to form gold 20 car»is fine? 5. A farmer has wheat worth 40, 55, 80, and 90 cents a bushel ; how many bushels of each must bo mixed with 270 @ 40 ct. to form a mixture worth 70 cents a bushel? Examples like this where the quantity of one or more ingredients is limited may be solved thus : First, we find the gain or loss on one unit as in (703). Second, we balance the whole gain or loss on an ingredient where the quantity is limited, by using any ingredient giving an opposite result thus : Producing Gain. Gaiked and Lost. Producing Loss. (1.) 270 bu. at 30 ct. per ba. gain. =$81.00=405 bu. at 20 ct. per bu. loss. (2.) 2 bu. at 15 ct. per bu. gain'= .30= 3 bu. at 10 ct. per ba. loss. *»72 bu. + 408 bu.=680 bu. in mixtun. Observe,, the pfaln on the 270 bu. may be balanced with the other injEjre- dlent that produces a loss, or with both incredieuts that produce a loss, and these may be put in the mixture in different proportions ; hence a series of different mixtures may thus be formed. mixture its; the ALLIQ ATIOX, 301 ith mo- ' B worth oduce a [id $10 a L of each ne; li3w cents a dth 270 )r more redient giving jOSS. or bu. loss, her bu. loss. in mixtun. [er Inffre- te a loss, hence a 6. I wish to mix vinegar worth 18, 21, and 27 cents a gallon with 8 gallons of water, making a mixture worth 25 cents a gallon ; how much of each kind of vinegar can I use V 7. A merchant having good flour worth $7, $0, and ^12 a barrel, and 240 barrels* of a poorer quality worth $5 a barrel, wishes to sell enough of each kind to realize an average ])rice of $10 a barrel on the entire quantity sold. How many barrels of each kind can he sell? 8. A man bought u lot of sheep at an average price of $2 apiece, lie paid lor 50 of them $2.50 per head, a«id for the rest $1.50, $1.75, and s3.25 i)er head ; how many sheej) could there be in the lot at each price ? 9. A milkman mixes milk worth 8 cents a (juartnyith water, making 24 quarts worth 6 cents a quart ; how much water did he use ? Examples like this, where the quantity of the mixture la limited, may be solved thus : Solution. — 1. We find, according to (T02), the smallest proportional parts that can be ubed, namely, 3 quarts of milk and 1 quart of water, making a mixture of 4 quarts, 2. Now, since in 4 qt. of the mixture there are 3 qt. of milk and 1 ql. of water, in 94 qt. there must be as many times 3 qt. of milk and 1 qt. of water as 4 qt. are contained times in 24 qt. Consequently we have as the Jirst step 24 qt. -*- 4 qt. =r 6, t^ecoml itfep 8 qt. x ♦» = 18 qt. and 1 qt. x 6 = 6 qt. Hence in 24 qt. of the mixture there are 18 qt. of milk and 6 qt. of water. 10. A jeweler melts together gold 14, 18, and 24 carats fine, 80 as to make 240 oz. 22 carats fine; how much of each kind did it require? 11. A grocer has four kinds of coffee worth 20, 25, 35, and 40 cents a pound, from which he fills an order for 135 pounds worth 32 cents -^ pound ; how may he form tlio mixture ? 12. I wish to fill an order for 224 lb. of sugar at 12 cents, by forming a mixture from 8, 10, and 16 cent sugar; how much of each must I take ? 13. How much candy at 35, 39, and 47 ct. will produce a mixture worth 20 cents a pound ? 'M 'A':i:'^ ■■:■-, i INVOLUTION. DEFINITIONS. 706. A Power of a number is either the number itself or the product obtained by taking the number two or more times as a factor. Thus 35 is the prodnct of 6 x 5 or of 5 taken twice, ae a factor : hence 25 1& a power of 5. 707. An Exjtonent is a number written at the right and a little above a number to indicate : (1.) The number of times the given number is taken as a foctor. Thus in 7"* the 8 indicates that the 7 i^ taken 3 timeb as a fiuitor ; hence 7' = 7x7x7 = 343. (2.) The degree of the power or the order of the power with reference to the other powers of the given number. Thus, in 5* the 4 indicates that the given power is the fourth power of 5, and hence there are three powers of 5 below 5* ; namely, 5, 5', and 5\ 708. The Square of a number is its second power, so called because in finding the superficial contents of a given square we take the second power of the number of linear units in one of its sides (395). 709. The Cube of a number is its third power, so called because in finding the cubic contents of a given cube we take the third power of the number of linear units in one of its- t^-^.gea (403). 710. Involution is the process of finding any required power of a given number. tseif or e times ence 25 i& iglit and ir. Thus !nce 7* = rerence to Is that the towers of called WiTe we one of called Kve take le of its- required I NVO L UTI X, PROBLEMS IN INVOLUTION. 303 711. Prob. I. — To find any power of any given number. 1. Find the fourth power of 17. Solution.— Since according to (706) the fourth power of 17 is the product of 17 talcen as a factor 4 times, we have 17 x 17 x 17 >( 17 = 83521, the required power. 2. Find the square of 294. Of 386. Of 497. Of 253. 3. Find the second power of 48. Of 65. 01 432. 4. Find the third power of 5. Of J. Of ,;,. Of .8. 5. Find the cube of 63. Of 25. Of 76. Of 392. Observe, any power of a fraction it? found by involving each of its terms separately to the required power (256;. Find the required power of the following : 10. (.3?)'. 12. (.71)^ 454. 2372. 8. (H>'.^ 11. (.25)*. 13. (.l/jj)4. 14. .0303^ 15. (.005;)«. 712. Pbob. II. — To find the exponent of the product of two or more powers of a given number. 1. Find the exponent of product of 7^ and 7^ Solution.— Since 7* = 7 x 7 x 7 and 7' = 7 x 7, the product of 7* and 7' must be (7 X 7 X 7) X (7 X 7), or 7 taken as a foctor as many times as the sum oi the exponents 3 and 2. Hence to find the exponent of the product of two or more powers of a given number, we take the sum of the given exponents. Find the exponent of the product 3. OfCD'^xd)*. 8. Of 35* X 353. 4. Of 18* X 18'. 5. of(*rx(jr. 6. Of(T-'V)*x(A)». 7. Of23'x23^ 7«. 8. Of (7*)«. Observe, (7*)* = 7* x 7* = 7*><« Hence the required exponent is the product of the given exponents. 9. Of(12»)*. 10. Of(9«)\ 11. Of (16Y. 13. Of[(i)»]*. I n !!l lif EVOLUTION. DEFINITIONS. 713. A Root of a numl)er is either the number itself or one of the equal factors or into which it can be resolved, ThuB, since 7 x 7 = 49, the factor 7 is a root of 49. 714:» The Second or Square Root is one of the two equal factors of a number. Thus, 5 is the square root of 25. 715. The Third or Cube Roof is one of the three equal factors of a number. Thus, 2 is the cube root of 8. 716. The Radical or Root Sign is V, or a fractional exponent. When the siffo, ^, 1b used, the degree or name of the root is indicated by a small figure written over the sign ; when the fractional exponent is used, the denominator indicates the name of the root ; thus, /?/9 or 9* indicates that the second or square root is to be found. yy/27 or 27» indicates that the third or cube root is to be found. Any required root is expressed in the same manner. The index is usually omitted when the square root is required. 717. A Perfect Power is a number whose exact root can be found. 718. An Imperfect Power is a number whose exact root cannot be found. The indicated root of an imperfect power is called a eurd ; thus ^5. 719. Evolution is the process of finding the roots of numbers. r itself or ed. of the two »t of 25. three equal fractional iB indicated exponent is ound. nd. e index is 3xact root lose exact las Vs. roots of nvoLUTioy^, 305 SQUABE BOOT. PREPARATORY PROPOSITIONS. 720. Prop. I. — Any perfect second power may be represented to the eye by a square, and the number of units in the side of such square will represent the second or square root of tlie given power. For example, if 25 is the given power, we can suppose the number represents 25 small squares and arrange them thus : 1. Since 25 = 5 x 5, we can arrange the 35 squares 5 in a row, making 5 rows, and hence forming a equarc as shown in the illustration. 2. Since the side of the square is 5 units, it represents the pquare root of 25, the given power ; hence the truth -of the proposition. 731. Prop. II. — Any number being given, by suppoHvg it to represent small squares, we can find by arranging these squares in a large square the largest perfect second power the given number contains, and hence its square root. For example, if we take 83 as the given number and suppose it to represent 83 small squares, we can proceed thus : 1. We can take any number of the 83 squares, as 36, that we know will form a perfect square (Prop. I), and arrange them in a square, as t^bown in (1), leaving 47 of the 83 squares yet to be disposed of. 2. We can now place a row of squares on two adja- cent sides of the square in (1) and a square in the comer, and still have a perfect square as shown in (2). 8. Observe, that in putting one row of small squares on each of two adjacent sides of the ^(\xi&re first formed, we must use twice as many squares as there are units in the side of the square. 4. Now since it takes twice 6 or 12 squares to put one row on each of two a4Jacent sides, we can put on f!> ^ r I" tfelH 3 _ - -1- fe fc P s. i\tm - -.}: r p.t.t=j i_"i: (1) Bf 4_i 1 1 WW 306 BUSINESS ARITHMETIC, m ' f^^ 1 11 It ■^1 i « ,1 (8) 6x8=18. 8'=9. as many rows as 12 is contained times in 47, the number of squares remain- ing. Hence we can put on 8 rows as shown in (8) and have 11 squares still remaining. 5. Again, having pnt 8 rows of squares on each of two adjacent sides, it talces 3 x 8 or 9 squares to fill the corner thus formed, as shown in (3), leaving only 2 of the 11 squares. Hence, the square in (3) represents the greatest r-* perfect power in 83, namely 81 ; and 9, the number so of units in its side, represents the square root of 81. •■o 6. Now observe that the length of the side of the square in (3) is 6+3 units, and that the number of Hmall squares may be represented in terms of 6 + 3; thus. 6^ = 36. (1.) (6+3)« = Q' + S^ + ttoiceQxd = 36+9 + 36 = 81. Again, suppose 5 units had been taken as the side of the first square, the number of small squares would be represented thus : (2.) (5 + 4)'^ = S' + ^' + ticke 5x4: = 25 + 16 + 40 = 81. In the same manner it may be shown that the square of the sum of any two numbers expressed in terms of the numbers, is the square of each of the numbers plus twice their product. Hence the square of any number may be expressed in terms of its tens and units ; thus 57 = 50+7 ; hence, (3.) 572 = (50 + 7)"^ = 50^ + r + tmce 50x7 = 3249. This may also be shown by actual multiplication. Tims, in multiplying 57 by 57 we have, Jlrst, 57 x 7 = 7 x 7+50 x 7 = 7' + 50 x 7 ; we have, second^ 67x50 = 50x7+50x50 = 50x7+50'; hence, 57 » = QO^+V+ twice 50x1. Find, by constructing a diagram as above, the square root of each of the following : Observe, that when the number is large enough to give tens in the root* we can take as the side of the first square weconstnict the greatest number of tens whose square can be taken out of the given number. 1. Of 144. 2. Of 196. 3. Of 289. 4. Of 520. 5. Of 729. 6. Of 1089. 7. Of 1125. 8. Of 584. 9. Of 793. 10. Of 1054. 11. Of 2760. 12. Of 3832. 9 remain- laretf tttill ) on each quares to ;), leaving 3 greatest le number root ol SI. ide of the lumber of Qsof6+3; E VOL UTION. 307 square, the mm of any of each of of its ten& 7S22. Prop. Ill— TAd square of any number must, wn- tain twice as many figures as the number, or twice as many less one. This proposition may be shown thus : 1. Observe, the Bqaare of either of the digits 1, 2, 8, is expressed by one figure, and the t -= ^ .^ r ~ :f -3 ^ s ^ == -~ ss :i=l :?. -~ r -~ ^ -3 :^ ^ ^ := =a: ^ =ri = P =e ^ ^ •# !?*^ * ^ S ^ s ^ ^ m g ^ =^^ m. (a ) 1 i_ :1( X) Id step. id Step \ 336(10 10* B 10 X 10 s 100 (1) T. diviaAQ % 3=30) 135 ( 5 foo^ 15 S0k5= 6x6: <»>n:rsh 135 S II Explanation.— 1. We observe, as ebown in (a), tbat 1 ten is tbe lar^et nnmber of tens wboso sqnara is con- tained in 335. Hence in let step we subtract 10*=100 from 836, leaving 136. 3. Having formed a square wliose side is 10 units, we observe, as sbown 1q {b) and (c), that it will take tu^ce ten to put one row on two acyaoent eidesi. Hence the Trial Divisor is 10 x 2 = sa 8. We observe tbat SO is contained 6 times In 125, but if we add 6 unite to the aide of the square (a) we will not have enough left for the comer ((f), hence we add 5 unite. 4. Having added 6 units to the side of the square (a), we observe, as shown in (6) and (r 600x3 = 1300)106489 •} i 1300x80=96000 * ^ M 80x80= 6400 f = 3 103400 688 required root. 'TmRS Step, 1 (1) 2Wa;(fi«i4or 680x3 = 1360) 4089 • "j ( 1360x3=4080} 8x8= 9 4089 Bxplanation.— 1. We place a point over every second figure beginning with the units, and thus find, according to (722), that the root must have three places. Hence the first figure of the root expresses hundreds. 835(10 . m =20)125 (5 root 15 = 1* observe, as B the Iftr^eat aare is con- 1st «lep we leaving 18S. luare wlioee re, as pbown B add 6 un\ts or the corner observe, as jed by 5 plus step. \ated. EVOL UTION, 3oa 8. We ob^erro that the square of 600 is the fnreatcst second poNor of buudrcd-i cuuuiiued iu 4(kM8:). ileuce iu the tirel rtep we 8ublract 600 X GOO -^ 360(M)0 from 460^t8*», ioavin^ 106480 3. Wu now duubit' the 600, the rout foaud, for a trial dlvimor, accordiug to (744—2). DlvidiuK 106489 by 1200 we And, accordluK to (7>i4— 2), that we cau odd 80 to tbe root. For tUiit additlou wu une. u» hUowu Iu (2), teomU step, 1200 ^m - mm aud 80 X 80 - 0400(7*4-4). muklii),' in all 10^100. Subtracting 10^100 from 1»^0, we have atill rvmainiug 40W. 4. We aguJu double 680, the root found, for a trial divisor, according to (f 24— 3), and proceed in the same manner as before, as shown In third vtep. obtierve, from the foregoing it is evident that one digit in the root is equivalent to two digits in the square, and convorHoly, two dlfrltt* In the . »quare give one digit in the root. Flence by dividing thu giv«n number into i>erio ^^^ 68x2 = 130) 408J» THIRD STEi . ^ ^2^ 1363 ^ 3 ^ 40^9 anired root, beginning )t must have Irede. Explanation.— 1. Observe, in the^rs^ step we luiow that the square 600 must occupy the fifth and sixth place (721). lleuce the ciphers are omitted. 2. Observe, that in (1), second step, we use 6 instead of 600, tlius dividing the divisor by 100 ; hence we reject the tens and uniti firom the right of the dividend (131). 3. Observe, also. In (2), second step, we unite in one three operations. Instead of multiplying 12 by 80, the part of the root found by dividing 1064 by 12, we multiply the 12 flfst by 10 by annexing the 3 to it (82), and having annexed the 8 we multiply the result by 8, which gives ua the- prodnct of 12 by 80, plus the square of 8. But the square of 8, written, as it is, in the third and fourth place, is the square of 80. Hence by annexing the 8 and writing the result an we do, we have united in one three operations ; thus, 128 x 8 = 12 x 80 -t- 80*. mam 310 n Ua I N E S S A R I TUM E Tl C, mM m I From these illustrations we have the following 727. liuLK. — /. ISeparate the number into periods of two figures rarh, by 2d(icin(/ a point over ecery second figure, begin- ning with the units figure. II. Mud the greatest square in the left-hand period and pface its root on the right. (Subtract this square from the ptrioil and annex to the remainder the next period for a dicidend. III. Double the j)art of the root found for a tnal divisor, and find flow many times this divisor is contained in the dividend, omitting tlie right-hand figure. Annex the quotient thus found both to the root and to the divisor. Multiply the ditisor thus completed by the figure of the root last obtained, and subtract the product from tJie dividend. IV. Tf there are more periods, continue the operation in the same manner as before. In applying this rule be particular to observe : 1. When there is a remainder after the last period hae been need, annex periodf" of ciphers and continue the root to as many decimal places as may be required. 2. We eeparatc a number into periods of two figure? by beginning at the nnltt* place and prococdin/j; to the left if the number is an integer, and to the ri{;ht if a decimal, and to the right and left if both. 3. Mix'-d numbers and fractions are reduced to decimals before extract- ins: the root. But in case the numerator and the denotninator are perfect powers, or the detwminator alone, the root may be more readily fornied by extracting the root of each term separately. Tims 80 on. I ^ _ V49 _ 7 81 V^i 9' and Extract the square root 1. Of ^W 3. Of 2. Of ,Vj. 4. Of f Jf . 9. 540131000000. 10. 191810713444. 35 04 V35 V^S -7_ = -^v,-, and V64 S 6. Of iff. 7 Of 72p 6. Of ,ViV- Q Of 3 ^ 1 Ans. 739000. Ans. 43T9G2. is of two re, begin - riod nnd from the- iod for a Ivisor, find I dividend, huH found icisor thua ubtract the Uion in the upcd, annex laces as may inninir at the tc'Rer, aud to eforc extract- r are perfect y foriped by Vi^^-, and 8 ' 7. Of ,Vo^. 8. Of ^Vj- 9000. 79G2. E VO L UT I ON, EXAMPLES FOR PRACTICE. 728. Extract the square root 311 1. Of 3481. 2 Of40»0. 8. Of 7509. 4. Of2l-'()9. T). Of 0210. C. or 8049. 7. Of?i2}- 9. Of.022.j, 10 Of ' "''* 11. or .5770, 12. or .2804 18. Of 137041. 14. Of 4100.25. 15. Of 708427.50. 10. or 28022.70. 17. Of 57.1530. 18. Of 474.8041. Find tlie square root to three decimal i)laces : 19. Of 32. 20. Of 59. 21. or 7. 22. or .93. 23. or .8. 24. Of .375. 25. Of 14.7. 20. or 80.2. 27. Of 5.973. 28. Of I. 29. or/g. 30. or A„ Perform the operations indicated in the following : 34. ^/558009-<-(4J^)i. 31. ^^889-^1024. 82. V^209 + V225. 33. V!SF§ ^ ^2209. 35. 70890^-^^2130. 30. (SS?S)^x 131376^. 87. What is the length of a square floor containing 9025 square feet of lumber? 38. How many yards in one of the equal sides of a square acre? 39. A square garden contains 237109 square feet ; how many reet in one or its sides? 40. A triangular field contains 1900.24 P. What is the length of one side of a square field of equal area ? 41. An orchard containing 9210 trees is planted in the form or a square, each tree an equal distance from another ; how many trees in each row? 42. Find the square root of 2, of 5, and of 11, to 4 decimal places. 43. Find the square root of 2, j'^j,and of \l, to 3 decimal places. ll ^*^Tf :^^'^ 312 B COSINESS ARITHMETIC, M ..i CUBE ROOT. PREPARATORY PROPOSITIONS. 729. Prop. I.— Any perfect third power may be repre^ sented to the eye by a cube, and the number of units in the side of such cube will represent the third or cube root of the given power. Represent to the eye by a cabe 343. 1. We can suppose the number 343 to repre- sent small cubes, and we can take 2 or more of these cubes and arrange them iu a row, Ub bhovvu 2. Having formed a row of 5 cube?, as shown m (1), we can arrange 5 of these rows side by side, as shown in (2), forming a square slab coiituiiiiug 5x5 small cubes, or as many small cubes as the square of the number of units in the side of the slab. 3. Placing 5 such slabs together, as shown in (3), we form a cube. Now. since each blab con- tains 5x5 small cubes, and since 5 slabs are placed together, the cube in (3) contain** 5x5x5, or 125 small cubes, and hence represents the third power 125, and each edge of the cube represents to the eye 5, the cube root of 125. We have now remaining yet to be disposed of »43-125, or 218 small cubes. 4. Now, observe, that to enlarge the cube in (3) so that it may contain the 343 small cubes, wo must build the same number of tiers of small cubes upon each of three adjacent sides, as shown in (4). Observe, also, that a slab of small cubes to cover one side of the cube in (3) must contain 5x5 or 25 small cubes, as shown iu (4), or as many small cubes as the square of the number of unite in one edge of the cube in (3). Hence, to find the number of cubes necessary to put one slab on each of three sides of the cube in (3), we multiply the square of its edge by 3, giving 6' X 3 = 5 X 5 X 3 — 75 small cubes. be repre' )i the side tot of the a43 to repre- t or more of jw, as tiUowu ep, as shown 8 side by side, abcouiaiuiug cubes as the he side of the , as Bhown in jach blub cou- 5 elabsj are ktains 5x5^5, iente the third jbc represents je disposed of |the cube in (3) lall cubes, wo [tiers of small Ides, as shown Imall cubes to must contain iu (4), or as le edge of the Llab on each of Ige by 3, giving' £ VOL UTI N. 313 (5) 5. Having found that 75 small cubes will put one tier on each of three adjacent sides of the cube in (3), we divide 218, the number of small cubes yet remaining, by 75, and find how many such tiers we can form. Thus, 218+75 = 2 and 68 remaining. Hence we can put 2 tiers on each of three adjacent sides, as shown in (5), and have 68 email cubes remaining. 6, Now, observe, that to complete this cube we must fill euch of the three comers formed by building on three adjacent sides. Examine cartfuUy (6) and observe that to fill one of these three comers we require as many email cubes as is expressed by the square of the number of tiers added, multiplied by the number of units in the side of the cube to which the addition is made. Hence we require 2' x 5 or 20 small cubes. And to fill the three corners we require 3 times 2^ x 5 or 60, leaving 68—60 or 8 of the small cubes. ffi) (7) 7. Examine again (5) and (6) Andobserve that when the three comers are filled we require to complete the cube as shown in (7), another cube whose side contains as many units as there are units added to the side of the cube on which we have built. Consequently we require 2* or 3x2x2 = 8 small cubes. Hence we have formed a cube containing 343 small cubes, and any one of its edges repre- sents to the eye 5+2 or 7 units, the cube root of 343. Prom these illustrations it will be seen that the steps in finding the cube root of 343 may be stated thus : "tt: First Step. 343 125 Skoond Step. We assume that 343 represents small cubes and take 5 as the length of the side of a large cube formed from these. Hence we subtract the cube of 5 = r 1. We observe it takes 5' x 3 = 75 to put oiie tier on three adjacent Hides. Hence we can put on 75)218(2 2. We have now found that we can add 2 units to the Bide of the cube. Hence to add this we require (1) For the 3 sides of the cube 5« x 2 x 3= 150 ^ (2) For the 8 corners thus formed 2' x 5 x 3= (3) For the cube in the comer last formed 2*: J= 60 )- = 218 »= 8J u Hence the cube root of 343 is 5+2 = 7. 21 ^<''^ 314 BUSINESS ARITHMETIC, u • [.f ■] ■^ 730« Observe^ that the number of small cubes in the cube (7) in the foregoing Illustrations, are expressed in terms of 6+2; namely, the num- ber of units in the side of the first cube formed, plus the number of tiers added in enlarging this cube ; thus : <7) (B) m II (5 + 2)8 5» + 5«x2x3 + 2«x5x3 + 2» In this manner it may be shown that the cube of the sum of any two numbers is equal to the cube of each number plus 3 times the square of the first multiplied by the second number, plus 3 times the square of the second multiplied by the^r^^ number. Hence the cube of any number may be expressed in terms of its tens and units ; thus, 74 = 70+4; hence, (70+4)» = 70» +3 times 70" x 4+3 times 4" x 70+4* = 405224. Solve each of the following examples, by applying the fore- going illustrations : 1. Find the side of a c abe which contains 739 small cubes, taking 6 units as the side of the first cube formed. 2. How many must be added to 9 that the sum may be the cube root of 4096? Of 2197? 0/2744? 8. Take 20 units as the side of the first cube formed, and find the side of the cube that contains 15625 cubic units. 4. Find the jube root of 1368. Of 3405. Of 2331. Of 5832. 5. ExpresB the cabe of 54, of 72, of 95, of 123, of 274, in terms of the tens and units of each number. 6. Express the cube of 83 in terms of 80+3. be (7) in the y, the nnm- iber of tiers (fi) EVOLUTION, 315 II 5x3+2» um of any two e square of tbe re of the second 18 of its tens 405224. |ing the fore- small cubes, may be the jed, and find Its. \\. Of 5832. IS3, of 274, in 731. Prop. II. — The cube of any mimbcr must contain three times as many places as the number^ or three times aa jnany less one or two places. This proposition may be shown thus : 1. Observe, V = 1, 2» = 8, 3' = 27, 4* = 64, 5» = 125, and 9» = 7»; hence the cube of 1 and 2 is expressed each by one figure, the cube of 3 and 4 each by two figures, and any number from 5 to 9 Inclusive each by three figures. 2. Observe, also, that for every cipher at the right of a number there must (82) be three ciphers at the right of its cube ; thus, 10* = 1,000, 100' = 1,000,000. Hence the cube of tens can occupy no place lower than thousands, the cube of hundreds no place lower than millions, and bo on with higher orders. 3. From the foregoing we have the following : (1.) Since the cube of 1 or 2 contains one figure, the cube of 1 or 2 tens must contain /o?/r places ; of 1 or 2 hundreds, seven places, and so on with higher orders. (2.) Since the cube of 3 or 4 contains two figures, the cube of 3 or 4 tens must contain ^t'e places ; of 3 or 4 hundreds, eight places, and so on with higher orders. (3.) Since the cube of any number from 5 to 9 inclusive contains three places, the cube of any number of tens from 5 to 9 tens inclusive must con- tain six places ; of liuudreds, from 5 to 9 hundred inclusive, ?iin€ places, and so on with higher orders ; hence the truth of the proposition. Hence also the following : 7322. / If any nuTnber be separated into periods of th/ree figures ea/ih, beginning with the units place, the number of periods mil be equal to the number of pLices in tlie cube root of the greatest perfect third power which the given number contains. II. The cube of units contains no order higher than hundreds. III. The cube of tens contains no order lower than thousands nor higher than hundred thousands, the cube of hundreds no order lower than millions nor ftigh^r than hundred millions, and 80 on with higher orders. ;t! ^iilll// 316 B^/SI^ESS ARITHMETIC, !<■' :j ILLUSTRATION OF PROCESS. 733. Solution ivith every Operation Indicated^ Find the cube root of 92345408. FlKST Stbp. 400* = 400x400x400 = 9234M08(400 64000000 (1) Trial divisor 400» x 8 = 480000 ) 28M5408 ( 50 Second Step. < / 40O' x 50 x 3 = 24000000 \ (2) "I 50"^ X 400 X 3 = 3000000 >• = 27125000 ( 50» = 125000 J Thtpo \y ( (1) TricU divisor 450' x 3 = 607500 ) 12204C8 ( _2 I I 450" X 3 X 2 = 1215000 \ Hoot 453 ( (2; ■{ 2» X 450 x 3 = 5400 ( 2» = 8 \ = 1220408 . f Explanation.— -t. vV' j ...z'. a period over every third fljjure begln- iiiDg with the units, and thus fled, according to (732), that the root must have three places. Hence the first figure of the root expressee huudrcds. 2. We observe that 400 is the greatest number whose cube is contained k) the given number. Subtracting 400* = 64000000 from 92345408, we have 88345408 remaining. 3. We find a trial divisor, according to (729—4), by taking 3 times the square of 400, as shown in (1), second step. Dividing by this divisor, according to (729—5), the result is that we find we can add 50 to the root already found. Observe, the root now found is 400+50, and that according to (730), (400 + 50)» = 400» + 400"x50x3 + 50''x400x3+50». We have already subtracted 400* = 640000 from the given number. Hence we have only now to subtract, 400'x50x3 + 50»x400x3+50» = 27126000, as shown in (2), second step, leaving 1220408. 5. We find another trial divisor and proceed in the same manner to find the next figure of the root, as shown in the third 8tQ3. EVOL UTION, 317 .dicatcd* [08(400 WO 408 ( 50 )000 34C8 ( _2 JiOOt 452 10408 i figure begin- that the root root expresses )e iB contained i&408, we have tag 3 times the ly this divisor, Id 50 to the root ig to (730), given number. manner to find 734. Contracted Solution of tJie foregoing Example, First Step. Second Step. 4» = 4x4x4 = 0234M08(452 64 i(l) Trial divisor 40* x 3 = 41 / 40» X 5 X 3 = eiooo \ (2) ^ 5» X 40 X 3 = 3000 > = ( 5* = 125 ) Trial divisor 450* x 3 = 607500 ) 1220408 Thibd Stsp. ni) Trials } 1 460* ( (2) j 2« X X 2 X 3 = 1216000 \ 450 X 8= MOOJ- = a»= 8) 1220406 Explanation.— 1. Observe, in the first step, we know that the cube of 400 must occupy the seventh and eighth places (732—111). Ilence the ciphers are omitted. 2. Observe, also, that no part of the cube of hundreds and tens is found below thousands (732— III). We therefore, in finding the number of tens in the root, disregard, as shown in second step, the right-hand period in the given number, and consider the hundreds and tens in the root as tens and units respectively. Hence, in general, whatever number of places there are in the root, we ■ w^ (T? ■«■■" 318 BUSINESS ARITHMETIC, r : . if . IV. fMytractfrom the dividend P> times the square of the root before found, considered as tens, multiplied by the figure last found, plus 3 times the square of the figure last found, multi- plied by tTie root before found, plus the cube of t?ie figure last found, and to the remainder annex the next period, if any, for a new dividend. V. Jf there are nune figures in the root, find in the same manner trial divisors and proceed as before. In applying this rule be particular to observe : ■- 1. In dividing by the Trial Divisor the quotient may be larger than the required figure in the root, on account of the addition to be made, as ehovvn in (729—6) second step. In Buch case try a figure 1 leps than the quotient found. S. When there is a remainder after the last period has been used, annex periods of ciphers and continue the root to as many decimal places as may be required. 8. We separate a number Into periods of three figures by beginning at the units place and proceeding to the left if the number is an integer, and to the right if a decimal, and to the right and left if both. 4. Mixed numbers and ft'actions arc reduced to decimals before extract* ing the root. But in case the numerator and the denominator are perfect third powers, or the denominator alone, the root m8.y be more readily found by extracting the root of each term separately. EXAMPLES FOR PRACTICE. 736. Find the cubic root of 1. 729. 2. 216. 3. 2197. 4. 1331. 5. 10648. 6. 4096. 7 518 8. 6859. 10. un- 11. 438976. 12. 250047. 13. 47045881. 14. 24137569. 15. 113.379904. Of 86. 17 . Find, to two decimal places, the cube root of 11. Of 84. Of 235. Of^V Of^^. Of 75.4. Of 6.7. 18. Find to three decimal places the cube root of 3. Of 7. Of .5. Of .04. Of 009. Of 2.06. EVOL UTION, 319 I of the, root figure last mnd, multi- g figure last , if any, for in the same larger than the to be made, as 1 lesB than the een used, annex a\ places as may by beginning at B an integer, and l8 before extract- Inator are perfect be more readily 13. 47045881. 14. 24137569. 15. 113.379904. 16. t^VtA^- of 11. 0136. |6.7. ot ol 3. 0£ 7. 19. Find the sixth root of 4096. Observe, the sixth root may be found by extracting ^r«< the square root, then the cube root of the result. For example, V^M = 64; hence, 4096 = 64 x 64. Now, if we extract the cube root of 64 we will have one of the three equal factors of 64, and hence one of the six equal factors or sixth root of 4096. Thus, Mr represents the Constant or Common Difference. 4. A' represents the Number qf Terms. 5. S represents the Sum of all the Terms. a series of manner that 8 is co)i»to^T^^- 3Ties of num- ner that the tant. numTsers of ire called the \rence of an ^een any two or ilfiiWi- xt obtained by *ogre8sion ling one. Progression Ing one. 746. Any three of these quantities being given, the other two may be found. This may be shown thus : Taking 7 as the first term of an increasing series, and 5 the constant difference, the series may he written in two forms ; thus : Isi Term. (1) t (2) 7 M Term. IS 7+ (5) Sd Term. 17 ^ \ 7 + (5+5) Uh Term. 22, and so on. 7+(5+5+5) Observe, in (2), each term is composed of the first term 7 plus as many times the constant difi'erence 5 as the nnmber of the term less 1. Thas, for example, the ninth term in this series would be 7+5 ;< (9—1) = 47. Hence, tram the manner in which each term is composed, we have the following formulae or rules : 1. A= L-J)x{N-\). Read, The first term is equal to the last term, minus the common difference multiplied by the number qf terme less 1. 2. X = ^ + 2>x(2 = T^-A {The last term is equal to the first term, plus the common difference mul- tiplied by the number qf terms less 1. (The common difference is equal to the last term minus the first term^ divided by the number of terms less 1. •:r /■ f 322 BUsry^ESS arithmetic. il m r 1 4. N— - _ +1. {The number of teryuH is equai to (fir last term minus the first tertn, iJlvidid by the common cUference^ plus l. Observe, that io a decreasing ecricp, tho flret term is the largest ani tbu last term the stnaUest iu the McrU^B. Uence, to make the above formuliu apply to a dccrcanlnt,' Hories, we mu'»t place L where A la, and -i whore /> la, and read tho formulee accordingly. 747. To show how to find the sum of a series let (1.) 4 7 10 18 16 19 (3.) 19 _16_ 18_10 7__4 (8.) 28 + 28 + 28 + 23 + 23 + 28 = twice the sum of the termj. be an arithmetical series. be the same series reversed. Now, observe, that in (3), which is equal to twice the Bum of the ^cric?, each term is equal to the first term plus the last term ; hence, a=\ot{A + L)^y. Read, The sum qf the terms of an arithmetical series is equal to one-haHf qf the sum qf thfi first and last term, multiplied by the number qf terms. EXAMPLES FOR PRACTICE. 748. 1. The first round of an upright ladder is 12 inches from the ground, and the nineteenth 246 inches ; how far apart are the rounds ? 2. The first term of an arithmetical progression is 4, the common difference 2 ; what is the 12th term? 3. Weston travelled 14 miles the first day, increasing 4 miles each day ; how far did he travel the 15th day, and how many miles did he travel in all the first 12 days? 4. The tenth term of an arithmetical progression is 190, the common difference 20 ; what is the first term ? 5. The first term of an arithmetical series of 100 terms is 150, and the last term 1338 ; what is the common difference ? 6. The amount of $360 for 7 years at simple interest was $486 ; what was tho yearly interest ? PROGRESSION, 32a lenn, dlvidtd », plus 1. argest ani tbo ibove formulif tnd ^l where /-< let netical series, series reversed. im of Ibe terms. tin of the ecrlc?, ice, r an arithmetiiol qf the sum of the led by the number er is 12 inches ; Low far apart Bsion is 4, the [ay, increasing [the 15th day, all the first 3sion is 190, the of 100 terms the common Lie interest was 7. A merchant bought 10 pieces of cloth, giving 10 cents for the first and $12.10 for the last, the several prices forming an arithmetical series ; find the cost of the cloth. 8. What is the sum of the first 1000 numbers in their natural order? 0. How many less strokes are made daily by a clock which strikes the hours from 1 to 12, than by one which strikes from 1 to 24 ? 10. A man set out on a journey, going 6 miles the first day, increasing the distance 4 miles each day. The last day he went 50 miles ; how long and how far did he travel ? GEOMETBICAIi PROGRESSION. 749. There are Jive quantities considered in geometrical progression, which we denote by lette-s in the same manner as in arithmetical progression ; thus : 1. A = First Term. 3. X = Last Term. B. Jt = Conatant Ratio. 4. A^ = Number of Terms. 6. A = the Sam < ull the terms. 750. Any t?iree of these quantities being given, the ot7ier two may be found. This may be shown thus : Taking 3 as the first term and 2 as the constant ratio or maltiplier, the series may be written in three forms ; thus : 1st Term. (1.) 3 2d Term. 6 8x2 8x3 Sd Term. 12 lUh Term. 84 6f.h Term. 48 (8.) 8 (3.) 8 3x(2x2) 8x8* 8x(3x3x2) 8xa» 3x(2;<2x2xS> 8x2* Observe, in (8), each term is composed of the first term, 3, multiplied by the constant multiplier 2, raised to the power indicated by the number of the term less 1. Thus, for example, the seventh term would be3x 2'-' =3 X 2« = 192. m f.ff stmm m Is Mr .'■s: ■■■! I, :-:i I v » ■' i 324 BUSIiVIJSS ARITHMETIC, Hence, from the manner in which each tenn is composed, we have the followiug formulae or rules : r T?ie first term is equal to the last term^ di- \. A = ^^n_i. Read, -I vided by the constant rnultijTlier raised to the 1 2^ower indicated l/y the nunUfer of terms less 1. r The last term is equal to the first term, muUi- %. L = A yi K°-'' Read, •{ i^ied try the constant multiplier raised to the I power indicated try the number of terins Cess 1. n— I /"T ■ ij 8. «= V:i 4. JM-I = ^. Read, Read, The constant multiplier is equal to the root, whose index is indicated by the number of terms less one, of the quotient of the last term divided by the first. The number of terms less ont. is equal to the exponent of tfie power to which the common multiplier must 6e reused to be equal to the quotient of the last term divided by tlie first. 751. To show how to find the sum of a geometrical series, we take a series whose common multiplier is known ; thus : ^^ = 5 + 15 + 45 + 185 + 405. 'i I ■. t Multiplying each term in this series by 3, the common multiplier, we will have 3 times the sum. (1.) ;S'x3 = 5x3 + 15x3 + 45x3 + 135x3 + 405x3, or (2.)'Sx3 = lo +45 +135 +405 +405x3. Subtracting the sum of the series from this result as expressed in (2), we liave, /Sx3= 15 + 45 + 135 + 405 + 405x3 S = 5 + 15 + 45 + 135 + 405 ^^x2 = 405x3-5 Now, observe, in this remainder -S x 2 is S x (ft - i), and 406 x 3 is Jj X R, and 5 i& A. Ilence, S x {H — i) = L x R — A. And since M — \ times the Sum is equal to X > iJ - ^, we have. « = Tj X R^ A R-1 ' Read, The sum of a geometrical series is eqtial to the difference between the last term mvlfiplied by the ratio and the first term, diridfd by the L ratio minus 1. ►sed, we have last term^ di- raised to the )f terms lt«s 1. St term, mtUti- r raised to the of terms less 1. Ml to the root, fhe nunvber of f the last term V is eqvcU to the ch the common be eqval to the i by tJie first. etrical series, wn ; thus : 1 multiplier, we )5 X 3, or )5 X 3. |;xpres8ed In (2), |3 and 406 X 3 is \A. And since is equal to \term multiplied ». diridtd by tlu PROGRESSION, EXAMPLES FOR PRACTICE. 325 752. 1. The first term of a geometrical progression is 1, and the ratio 2 ; what is the 12th term ? 2. The first term of a geometrical progression is 3, the ratio 4 ; what is the 8th term ? 3. Tlie extremes of a geometrical progression are 2 and 1458, and the ratio 3 ; what is the sum of all the terms ? 4. The extremes are 4 and 2916, and the ratio 3 ; what is the number of terms ? 5. A man, coming from Winnipeg to the Province of Ontario, travelled days ; the first day he went 5 miles, and doubled the distance each day ; his last day's ride was 160 miles ; how far did he travel ? 6. The first term is 3, the seventeenth 196608 ; what is the sum of all the terms ? 7. The first term of a geometrical progression is 4, the 7th term is 2916 ; what is the ratio and the sum of the series ? 8. Supposing an engine, on the Intercolonial Railway be- tween Quebec and Halifax, should start at a speed of 3 miles an hour, and the speed could be doubled each hour until it equalled 96 miles, how far would it have moved in all, and how many hours would it be in motion 7 ANNUITIES. 75;j. An Annuity is a fixed sum of money, payable annually, or at the end of any equal periods of time. 754. The Atnount or Final Value of annuity is the sum of all the payments, each payment being increased by its interest from the time it is due until the annuity ceases. 755. The Present Worth of an annuity is such a sum of money as will amount, at the given rate per cent, in the given time, to the Amount or Final Value of the annuity. 1 II IS, ■■> iO iT L. 326 B USIXESS ARITHMETIC. 756. An Annuity at Simple Interest forms an arith- metical progressUm whose common difference is the interest on the given annuity for one interval of time. Thns an annaity of $400 for 4 years, at 7^ eimple interest, gives the fol- lowing progresBion : f; 1 r '■{ m : i ■ ■ - let Term, id Term, a.) $400 $400 +($38) (3.) $400 $428 Sd Term. $400 +($28 + 128) $466 lUh Term. $400 -r ($28 +$28 +128), or Observe, there le no interest on the laet payment ; hence it forms the M Term. The payment before the last bears one year's interest, hence forms the Sd Term ; and so on with the other terms. Hence all problems in annuities at simple interest are solved by arith' meticcU progression. 757. An Annuity at Compound Interest forms a geometrical progression whose common multiplier is represented by the amount oi$t for one interval of time. Thus an annuity of $300 for 4 years, at Z% compound interest, gives the following progression : Ist Term. $900 id Term. $900x1.06 Sd Term. $900x1.06x1.06 kth Term. $900x1.06x1.06x1.06 Observe carefully the following : (1.) The last payment bears no interest, and hence forms the 1st Term of the progression. (2.) The payment before the last, when not paid until the annuity ceases, bears interest for one year; hence its amount is $800x1.06 and forms the id Term. (3.) The second payment before the last, bears interest when the annuity ceases, for two years ; hence its amount at compound interest ie ♦900 X 1.06, the amount for one year, multiplied by 1.06, equal $300 x 1.06 x 1.06, and forms the Sd Term^ and so on with other terms. Hence all problems in annuities at compound interest are solved by geometrical progression. r RO G R Essioy, 327 l! ms an nritli- B interest on :, gives the fol- h Term. 28 +$28 +128), or $484. ;nce it forms the s interest, hence solved byarf/A- ^rest forms a r is represented aterest, gives the \kth Term. 1.06x1.06x1.06 18 the 1st Tenn of intil the annnlty lis $300 X 1.06 and llerest when the jpoond interest is pqual $300 X 1.06 X Bt are eolved by EXAMPLES FOR PRACTICE. 758. 1. A father deposits $150 annually for the benefit of his son, beginning with his 12fh birthday ; what will be the amount of the annuity on his 2l8t birthday, allowing simple interest at 6% ? 2. What is the amount of an annuity of $200 for 6 years at 7fc simple interest ? 3. What is the amount of an annuity of $400 for 4 years at 7%, comi)ound interest ? 4. What is the present worth of an annuity of $000 for 5 years at 8^ , simple interest ? 5. What is the present worth of an annuity of $700 at B%, simple interest, for 10 years? 6. What is the present worth of an annuity of $100 for 6 years at 6%, compound interest ? 7. What is the present worth of an annuity of $350 for 9 years at 6%, compound interest? 8. What is the amount of an annuity of $600 &t I'/c, com- pound interest, for 12 years ? 9. At what rate % will $100 amount to $119.1016 in 3 years, at compound interest ? This example and the fonr following should be solved by applying the formulffi for geometrical progression on page 394. 10. At what rate % will $1000 amount to $1500.73 in 6 years, compound interest ? 11. What sum at compound interest 8 years, at 7%, will amount to $4295.465 ? 13. The amount of a certain sum of money for 12 years, at 7% compound interest, was $1126.096; what was the original Hum? 13. In how many years will $20 smount to $23.82032, at 6% compound interest ? '.!"■ 'J Urn m f 1 m IM MENSURATION k*2 f ' 'I » "^ i GENERAL DEFINITIONS. 759* A Line ie that which has only IcDgth. 7 GO. A Straight Line is a Une which has the same direction at every point. 701. A Curved Utie la a line which changes its direction at every point. 76iS. Parallel Line* are lines which have the same direction. 703. An Angle is tlic opening between two lines which meet in a common point, called the veitex. Angles are of three Itinde , thas : (1) (2) (8) (4) 1 5 1 c m J9 1 Honiz ONTAL.. " A c r — c 7'wo Right Angles. One Right Angle. Obtuse Angle. Acute Angle. 7G4:. When a line meets another line, making, as shown in (1), two equal angles, each angle is a Right Angle f and the lines are said to bo perpendicular to each other. 7GJ^. An Obtuae Angle, as shown in (3), is greater than a right angle, and an Acute Angle^ bl9 shown In (4), is less than a right angle. Angles are read by using letters, the letter at the vertex being always read in the middle. Thus, in (8), we read, the angle BAG or CAB. 7C$0. A Plane is a surface such that if any two points in it be joined by a straight line, every point of that line will be in the surface. MENSURATIOX. 329 me direction at tB direction at ; direction. wWch meet in a (4) Acute Angle. lown In (1), two are eaid to be sr than a right right angle. being always CAB. In It be joined bface. 7G7« A J'lnne Tiynve \a a plane bounded cither by straight or curved lines, or by Oi-w curved line. 70S. A Polygon \» a plane figure bounded by straight linen. It is named by the number of sides in its boundary ; thus : Trigon. TBtragon, Pentagon. BeKogon, and m> on. Observe, that a regular polygon Is one that has all Its sides and all its angles equal, and that the Baae of a polygon la the aide on which it stands. 760. A Trigon is a ^Are^-sided polygon. It is usually called a TYiangle on account of having three angles. Triangles are of three kinds, thus : ^ D c Ac Sight-ana^ Triangle. Acute-angled TYiangle. Obtuse-angled Trian^ Observe^ a right-angled triangle has okx right angle, an acntc-angled triangle has thbeb acute angles, and au obtuse-angled triangle has om obtuse angle. Observe, also, as ehovm in (9) and (3), that the Altitude of a triangle is the perpendicular distance from one of its angles to the side opposite. 770. An £!quilateral Triangle is a triangle whose three sides are equal 771* An l80»eele» Triangle has two of Us sides eqaaL 772. A Scalene Triangle luu all of its sides unequal. 773. A Tetragon is a four-sided polygon. It is usually called a Quadrilateral. 22 m ■An M' '1 ir: ill ^m -yrnmavKxm 330 D[/SI.yESS ARITHMETIC, Qoadrilaterals are of three kinds, thus ; *"■ Yh> ) ParaUelogram. Trapezoid. Trapezium. f> S 1. ^/ : \ si < ' '.I Observe, that a Parallelogram has its opposite sides parallel, that a Trapezoid has only two sides parallel, and that a Trapezium has no sides parallel. Observe, also, that the Diagonal of a qoadrilateral, as shown in (1), (3) and (3), is a line joining any two opposite angles. 774. A Parallelogram is a quadrilateral which has its opposite Bides parallel. Parallelograms are of four kinds, thus : (I) (2) (3) ffi . J* »| -|C Bm Square. Rectangle. A — -F A R/iomboid. Rhombns. Observe, that a Square has all its sides equal and all its angles right angles, that a Rectangle has its opposite sides equal and all its angles right angles, that a Rhomboid has its opposite ^c^ equal and its angles a8t |fiO. villbeused in pavlnjf 785* Prob. Vf.—When Vie base and perpendicular are given in a iide 126 feet, and what I right-angled triangle, to find the other side : Extract the square root of Ihs mm qf the squares of the base and perpendicular. ,ulredtoboardupthe Ige of the roof H feet llmension are given^jo U divide by the given The reason of this rule and the one in Prob. V will be seen by examin- ing the diagram in the margin. sq. lo* Observe, that the square on the side AB opposite the right angle, contains as many small squaren as the sum of the small squares in the squares on the base AC and the perpendicular BC. This i:^ ebown by geometry to be true of all right-angled triai)<4les. Hence, by extracting the square root of the sum of the squares of the base and perpendicular of a right-angled triangle, we have the length of the Bide opposite the right angle. •a 1 *■ I w. The side opposite the right angle is called the Hypothenuse. and altitude 1 tdl Find the hypothenuse of a right-anjjled triangle [ ^ Mn stones, each 2 fl 1. Whose base is 15 ft. and perpendicular 36 ft. t distance from the rida 2. Whose base is 40 ft. and perpendicular 16 ft. 8. A tree 104 ft. high stands upon the bank of a stream 76 feet wide ; rhat is the distance of a man upon the opposite baulc from a raven upon the a triangle are given,tov of the tree f gides subtract each sia 4. What is the length of the shortest rope by which a horse may be tied emainders together ; tl^ a post in the middle of a field 20 rods square, and yet be allowed to graze ' on every part of it f 6. A and B start ftt)m one comer of a field a mile square, travelliner at the Ivrhoee base la 8** **• Hme rate ; A follows the fence around the field, and B proceeds directly tross to the opposite comer ; when B reaches the comer, how far will Is 90, 35 feet. | be trom A ? li m^r ' r ik ]-, tij ••:i ■,'l 334 BUSINESS ARITHMETIC, 780. Proij. V,— ]rA«n /A« 6a«« or perj)endicular is to de found; Extract the tquure root qfthe difference Itetiveen the square qf the hypothe- nme and the square of the given aide. Flud tbo bane of a rigbt-unglcd triangle 1. WhoHe perpendicular is 20 feet and hypothenase 45 feet. 2. Whot>e hypothenuHe is 40 feet and perpendicular 15 feet. 3. Bunker Hill monument is 230 feet lii^b ; a man 360 feet from the base Bhot a bird hovering above the top ; the man wan 423 feet from the bird ; how far was the bird Arom the top of the monument f 4. The lower endu of two opposite rafters are 48 feet apart and the length of each rafter is 80 feet ; what is the elevation of the ridge above the eaves ? 5. A ladder 36 feet long reaches from the middle of the street to a win- dow 28 feet high ; how wide is the street f FBOBLEMS ON QUADBILATERALS. 787. PnoB. \l.—To find tlie area qf a parallelogram : Multiply the base by the altitude. 1. now many acres in a piece of land in the form of a parallelogram, whose base is 9.86 ch. and altitude 7.5 ch. ? 2. Find the area of a parallelogram whose base is 3 it. 9 in. and altitude 7 ft. 8 in. ; whose altitude is 2 yd. 5 in. and base 3 yd. 6 in. 3. The base of a rhombus is 9 A. 8 in. and its altitude 3 ft. ; how many square feet in its surface ? 4. How many square feet in the roof of a building 86 ft. long, and whose rafters are each 16 ft. 6 in. long ? 788. Prob. vn — To find the area qf a trapezoid: Multiply one^ half of the sum qf the parallel sides by the altitude. Find the area of a trapezoid 1. Whose parallel sides are 8 and 11 inches and altitude 6 inches. 2. Whose parallel sides are 15 and 25 feet and altitude 11 feet. 3. One side of a field measures 47 rods, the side opposite and parallel to it measures 39 rods, and the distance between the two sides is 15 rods ; how much is it worth at $40 per acre f 4. How many square feet in a board 1 ft. 4 in. wide, one side of which i» 82 ft. long and the other side 34 feet long f to be found ; )f the hypothe- t. St. it from the base from the bird; t apart and the the ridge above street to a win- BALS. \m: Multiply the a parallelogram, in. and altitude 3 ft. ; how many long, and whose ; ; Multiply one^ inches, feet. and parallel to ^es is 15 rods; Bide of which i» M Ey SUR A TIO N, 335 789. Pbob. WU..— To find the area of a trapezium : Multiply the diagoml by half the turn of the perpe^ndicukvra to U from the opposite angles. Refer to diagram (3> in (773) and And the area of a trapesiom 1. Whose diagonal ii* IG feet and perpendiculars to this diagonal 7 feet and 6 feet. 2. Whose diagonal is 45 in. and perpendiculars to this diagonal 11 inches and 9 inches. S. How many acres in a field in the form of a trapezium whose diagonal is \ mi. and the perpendiculars to this diagonal 5 ch. and 6 ch.? 4. Whose diagonal is 37 ft. 6 in. and perpendiculars to this diagonal 7 ft. 4 in. and 8 ft. 8 in. 790. Pbob. VS..— To find the diameter of a circle: Divide the cir- eumference by S.1U16. To find the circumference : Multiply the diameter by S.1U16. 1. Find the circumference of a circle whose diameter is 14 inches; whose radius is 9 inches. 2. Find the diameter of a circle whose circumference is 94.248 Inches ; whose ciicumference is 78-54 feet. 3. Ilow many miles does the earth pass over in its revolution around the sun, its distance flrom the sun being 95,000,000 miles ? 4. What will it cost to fence a circular park 8 rods in diameter, at $4.80 per rod? 791. Pbob. X.— 7b JInd the area of a circle: Multiply \ of it* diameter by the drcumference ; or^ Multiply the square qf its diameter by .785k. 1. Wliat is the area of the largest circular plot that can be cut from a field 135 foet square ? How much must be cut ofi" at the comers in making this plot ? How much less will it cost to fence this than the square, at $2.50 a rod ? 2. What is the area of a circle whose diameter in 20 feet? Whoee diameter is 42 inches ? Whose circumference is 157.08 feet ? 3. The distance around a circular park is 1 2 miles. How many acres does it contain ? 4. How many square yards arc contained in a circle, who!«e diameter is Si feet? Ans. 1.069. ^1 'SI' PM '•/',; i tK' I 336 BUSINESS ARITHMETIC, J /' 7012. Pbob. XI.— To find the diameter when the area qf a circle is given: Extract the tgtuire root qf the quotient qf the area divided by .786k. Observe, that nvben the diameter Is found, the drcnmflDrcnce can be found by multiplyint; the diameter by 8.141(i (700). 1. What is the circumference of a circle whose area is 103.9884 square feet? 2. What is the diameter of a circle whose area is 50.S666 sq. ft. t 8. The area of a circular lot is 19.635 square rods; what la its diameter ? 4. The area of a circle is 118.0976 sq. in. ; what is its circumference t 6. What is the radius of a circle whose area is 804.349U sq, in. ? 6. How many rods of fence will be required to enclose a circle whofle area is 2Mfi square rods 7 •'.^ J *-r FBOBLEMS ON SOLIDS OB VOLUMES. 703. A Solid or Volume has three dimensions: length, breadth, and thickness. The boundaries of a solid are planes. They are called /(UM, and their intereectiouB edgee. 704. A Prism is a solid or volume having two of its faces equal and parallel polygons, and its other faces parallelograms. Obeerve, a prism is named by the number of sides in Its equal and parallel foces or bases ; thus : (1) Triangular I^ism. Quadrangular Prism. Pent m- Observe, a Prism whose parallel faces or bases are p. ^Uelogrf a, as shown in (3), is called a Parol letopipedon. Observe, also, that the Altittide of a prism is the perpendicular aietance between ita bases. MENSURATIOX. 337 "■ea qf a circle area divided ierence can be 3a la 163.9884 1. ft.f ; what la ita inference f in.r i circle whose jUMES. ngth, breadth, lOM, and their icee equal and elogr id, aa alar Uietance 795. A CyHnder, as ehown In (1), is a round solid or volume having Uoo equal and parallel circles as its ba»e$. 1. Obeerve, that the altitude of a cylin- der is the perpendicular distance between the two circles forming Its b coived to be (generated by revolving a rectangle about one of its sides. TOO. A Sphere, au Mhown in (3), ia a solid or volume bounded by a curved snrfoce, such that all points in it are equally distafit fh>m a point within, called the centre. 797. The Diameter of a sphere la a line, as CD in (S), passing through its centre and terminating at both ends in the surface. 798* The Radius of a sphere is a Use drawn f^om the centre to any point in the surflice. 799. A Pyratnid, as shown in (1), is a solid or volume having as its base any polygon, and as its other/aces triangles, which meet in a com* mon point called the vertex. Pyramid. Frustum. Cone. Frustum, HOO. A Cone^ as shown in (8), is a solid or volume whose base is a circle and whose convex surface tapers uniformly to a point, called the vertex. 1. Obfterve, that the Attitude of a pyramid or cone is the perpendicular distance between the vertex and the bnse. 2. Ob/'crtr, also, that the Slant Height of a pyramid is the perpen- diculjir (lintance between the vertex and one of the sides of theiutse; a 1 of a cone the distance between the vertex and the circumference of tJ base. 15 " ..'I m _ illL Jill % : V 338 BUSINESS ARITHMETIC. - <^ 801. A frufiiwtn of a pyramid or cone, as phown in (2) and (4), is the part which remains after cuttiug off the top by a plane parallel to^ the base. 80t2. Prob. XII.— To find the convex surface qf a priinn or cylinder : Multiply the perim etkb of the base by the altitude. To find tht ENTIRE auRFACH add the area qf the bases. The reasion of this rule may be shown thus : (1) (2) (3) G F e Observe, that if the three faces of the prism in (1) are marked out side by side, as shown in (2), we have a rectangle which is equal to the convex surface in the prism. Observe, also, that the surfocc of the cylinder in (8) may be conceived as spread out, as shown in (2) ; hence the reason of the rule. Find the area of the convex surface 1. Of a cylinder whose altitude is 4 ft. 9 in. and the circumference of its base 7 ft. 8 in. 2. Of a prism whose altitude is 8 feet, and its base a triangle, the sides of whose base measures 4 ft., 3 ft., 3 ft. 6 in. 3. Of a prism whose altitude is 9 inches, and its base a hexagon, each sido of which is 2} inches. 4. Find the entire surface of a cylinder 9 ft. high, the diameter of whose base is 8 ft. 5. Find the entire surfiiicc of a parallelopipedon 9 ft long, 5 (1. 6 in. wide, and 3 ft high. 803. Prob. 'SilW.— To find the volume qf any prism or cylinder: Multiply the area of the base by the altttude. 1. Wtiat > the volume of a triangular prism whose altitude is 28 ft., and the sides of its bnsc 6 ft., 7 ft, 5 ft. respectively. %. Find the volume of a triangular prism wuose altitude is 15 ft. and the tides of the base each 4 ft. in (2) and (4), ane parallel to nn or cylinder : (3) bed out side by to the convex c conceiTcd as iference of its gle, the side? lexagon, each eter of whose ft. 6 in. wide, or cylinder : i(< 28 n., and 5 ft. and the M E S SU RATION, 339 3. What is the volume of a parallelopipedon 15 ft. long, 12 ft. high, 10 ft. wide ? 4. Find the contents of a cylinder whose altitude is 19 ft. and the diame- ter of its base 4 ft. 5. A log is 90 ft. long and its diameter is 16 in. ; how many cubic feet does it contain t i\. What k the value of a piece of timber 15 in. square and 50 feet long, at 40 cents a cubic foot ? 804- . Prob. XrV.— To find the convex mrface of a jryramid or cone : MiUti[dy the perimeter of the base by one-haif the slant height. To find the entire surface, add the area of the base. Find the convex surflace of a cone 1. Whose slant height is 15 feet, and the diameter of the base 10 feet. 2. Whose base is 19 in. in circumference, and the slant height 12 inchest. Find the convex surface of a pyramid 3. Whose base is 3 ft. 6 in. square, and the slant height 6 ft. 4. Whose slant height is 19 ft, and the base a triangle whose sides are 12, 14, 8 ft. Find the entire surflace of a pyramid 6. Whose slant height is 56 in., and its baao a triangle each of whoso sides is 6 in. 6. Whose slant height is 45 feet, and the base a rectangle 7 ft. long and 8 ft. wide. Find the entire surfbcc of a cone 7. Whose slant height is 42 feet, and the circumference of the base 31.416 ft. 8. Whose slant height is 75 in., and the diameter of the base 6 inches. 80R, Prob. XV.— To find the volume qf a pyramid or cone : Multiply the urea of the btise by one-third the altitude. Find the volume of a cone 1. Whose altitude is 24 feet, and the circnmferencc of the base 6.2832 feet. •! 11 t i" II r'> ^40 nusry^Ess a rithmei^i c. r'1^1 .*.' 8. Whose altitude is 12 it, and the diameter of the baee 4 ft. Find (he volame of a pyramid 3. Whose altitude is 15 feet, and its base 4 feet square. 4. Whose altitude is 45 ft., and its base a rectangle 15 feet by 16 feet. 5. Whose altitude is 18 inches, and the base a triangle 8 inches on each side. m •( :;/■ 800. Prob. XVI.— 7b find the convex surface cf a frwitum of a pyramid or cone : Multiply the gum of the i)erimeter.'< or circumferences by one ha^ the slant height. To find the entire surface^ add the area of both the bates. 1. What is the convex surface of a fhistum of a cone whose slant height is 9 inches, and the circumference of the lower base 17 inches, and of the upper base 18 inchei* ? 2. What is the convex surftu:e of a frustum of a triangular pjrramid whose slant height is 6 feet, each side of the greater base 3 feet, and of the less base 2 feet? 3. Find the entire surftice of a frustum of a pyramid whose slant height is 14 feet, and its bases triangles, each side of the larger base being 8 feet, and of the smaller base 6 feet. 4. Find the entire surfiMse of a frustum of a cone whose slant height is 27 feet, the circumference of the greater base being 87.6992 feet, and of the less base 31.416 feet. 807. Pbob. XVn.— Tto find Vie volume of a frustum of a pyramid or cone: To the mm of the areas of both banes add the square root of their product and multiply the result by one-third of the altitude. 1. How many cubic feet in a frurtum of a cone whose altitude Is 9 feet, the diameters of its ba^s 8 feet and 6 feet * 2. Find the volume of a frustum of a square p3rr«mid whose altitude is 6 feet, and each side of the lower base 16 feet, and of the upper base 12 feet. 8. How many cubic feet In a section of a tree-trunk 90 feet long, the diameter of the lower base being IR inches, and of the upper base 12 inches ? 4. One of the big trees of California is 32 feet in diameter at the foot of the tree ; how many cubic feet in a section of this tree 9U feet high, the upper base being 20 feet in diameter i 15 feet by i iuches on -wttum of a cutnferenceii slant height , and of the lar pyramid , and of the elant height )eiiig 8 feet. It height is and of the a ityramid the square ird of the le is 9 feet, pe altitude upper base feet long, ipper base ho foot of high, the 3IEXSURA TION, 341 5. A granite rock, whose form is a fhistum of a triangular pyramid, is 40 feet high, the Bides of tlic lower base being 30 feel tach, and of tl»e upper base 16 feet each. How many cubic feet in the rock. 808. Prob. XVUl.— To find the surface of a spJiere : Multiply the diameter by the circumference qf a great circle of the given sphere. 1. What is the surface of a globe 9 inches in diameter ? 2. Find the surfoce of a sphere whose diameter is 8 feet. a How many square feet In the surface of a sphere 45 feet in diameter ? 4. How noAny square inches in the surface of a globe 5 inchuH iu diameter ? 5. What is the surface of a globe whose radius is 1 ft. 6 in. f 80t). Pbob. 'KlX.—To find the volume qf a sphere, surface by one-sixth of the diameter. MultliAy the 1 How many cubic yards in a sphere who^e diameter is 3 yards ? 2. Find the volume of a sphere whot^c diameter is 20 iuches ? 3. How many cubic feet in a globe 9 iuchcB in diameter? 4. Find the solid contents of a globe 2 ft. 6 in. in diameter. 5. Find the volume of a globe whose radius is 4 inches. 8 lO. Pbob. XX.— To find the capacity qf casks in galloiui : MuUijdy the number of inches in the length by the square of the number qf' inchts in the mean diameter^ and this product by .0031*. Observe^ that the mean diameter is found (nearly) by adding to tltc head diameter %, or if the staves are but slightly curved, { of the dilTcreuc-e between the head and bung diameters. The process of finding the capacity of casks is called Gauging, 1. How many gallons will a cask hold whose head diameter is 21 inches, bung diameter 90 inches, and length 42 inches ? 2. How many gallons in a cask whose head diameter is 20, bung diameter 26 inches, and its length 80 inches ? 8. A cask slightly curved is 40 inches long, its head diameter being 22 inches, and its bung diamet4ir 27 inches ; how many gallons will it hold? 4. What is the volume of a ca^k whose diameters are 18 and 24 inches respectively, and the length 32 inches ? i I REVIEW AND TEST EXAMPLES. !, 811. In using this set of review and test examples, the following suggestions should be carefully regarded : 1. Tlie examples cover all the important subjects in arith- metic, and are designed as a test of the pupil's strength in solving difficult problems and of his knowledge of principles and processes. 2. The teacher should require the pupil to master the thought expressed in each example before attempting a solution. To do this he must notice carefully the meaning of each sentence, and especially the technical terms peculiar to arith- metic ; he must also locate definitely the business relations involved. 3. When the soluticms are given in class, the teacher should require the pupils to state clearly : (a). Wluit u given and wJuit is required in each example. (6). The relations of the given quantities from which what is required can he found. (c). The xtejis thtt must he taken in their order, and the pro- cesses that must he used to nhtuin the required remit. In making these three statements, no set form should Oe used ; each pupil should be left free to pursue his own course and give liis own solution. Clearness, accuracy, and brevity should be the only conditions imposed. The pupil should be allowed to make re)>eated efforts until the precise point of his failure is made apparnt. ^Vlien the work is written on a slate, paper, or blackboard, neatness and a logical order in arranging the steps in the solution should be invariably required. PLES. samples, the i: sets in arith- I strength in of principles master the ittempting a ning of each iliar to arith- less relations acher should rample. which tchat is and the pro- l. m should oe s own course and brevity m1 should be point of his blackboard, steps in the REV I E W E XA M PLES, 343 1. A gentleman held a note for .11643.20, payable in 8 mo., without interest. lie discounted the note at 8% for ready cash, and invested the proceeds in stock at $104 per share. How many shares did he purchase? Ans. 15 shares. 2. Three daughters, Mary, Jane, and Ellen, are to share an estate of $80000, in the proportion of ^, i|, and \, respectively ; but Ellen dies, and the whole amount is to be divided in a proper proportion between the other two. What share does each receive ? Ans. Mary, $48,000 ; Jane, $32,000. 3. What must be the dimensions of a rectangular bin that will hold 350 bushels of grain, if its length is twice its width, and its width twice its depth ? Ana. Length, 15.5 + ft. ; width, 7. 75 4- ft.; depth, 3.87 f ft. 4. A Montreal merchant bought 800 barrels of flour at $7 per barrel, and sent it to Halifax, paying 9% of tlie cost for freight and other charges ; his agent sold it at an advance of 25% on the original cost and charged 3% commission. What was the net gain ? Ah8. $686. 5. What sum invested in railroad stock paying 7% annually willyieldaquarterly dividend of $325.50? Ana. $18,600. 6. A, B, and C together can dig a ditch in 4 days. A can dig it alone in 10 days ; B can dig it alone in 12 days. How long will it take C to do the work alone ? Ana. 15 days. 7. A person owning 7^ acres in the form of a rectangle 3 times as long as it is wide, wishes to tether his horw to a stake by the shortest rope that will allow him to graze upon any part of the field. What is the length of rope required ? Ana. 31.02 +itl, 8. A cubical block contains 64 cubic feet ; what is the dis- tance from one corner to the oppf)site diagonal corner ? Ana. 6.92+ feet. 9. A farmer bought a horse, wagon, and plough for $134 ; the horse cost I as much as the wagon, and the plough J as much as the horse. What was the cost of each ? Ana. Horse, {^70; wagon, $50; plough, $14. i u Ill ■■' m^ y ■r It : . t' ( ! 1! 3M REVIEW EXAMPLES, 10. A grocer mixed 15 pounds of Hyson tea with 9 pounds of Gunpowder tea, and sold it at $.96 j)er pound, thus gainini;^ 25% on the original cost. If a pound of the Gunpowder cost IG cents more than a pound of the Hyson, what was the cost of each per pound ? An%. Gunpowder, $.86 ; Hysou, ^.70. 11. A farmer has a cornfield whose width is to its length as 8 tu 4, and contains 4f acres. The hills of corn, supposing them to occupy only a mathematical point, are 2 feet apart, and no hill is nearer the fence than 3 feet. What must he pay a man to hoe his com, at the rate of $.50 per day, if he hoes 750 hiUa in a day ? Am. $34.33,\. 12. The duty at 20% ad valorem on a quantity of tea in chests, each weighing^ 75 pounds gross, and invoiced at $.70 per pound, was $6,552, tare heing 4%. How many chests were Imported ? Ans. 650. 13. A room is 22 feet long, 18 feet wide, and 14 feet high. What is the distance from one of the lowest corners to the opposite upper comer? Ans. Q\. OS + ft. 14. A farmer sold 85 sheep at $2, $2.20 and $2.80 per head, and thus realized an average price of $2.40 per head. What number of each did the lot contain ? Ans. 17 at $2 ; 84 at $f .20 ; 84 at $2.80. 15. If the ratio of increase of a certain c\o]) is 3, and a man begins by planting 5 bushels, using all the crop for seed the next year, and so on ; what will be his crop the seventh year V Ans. 10,935 bushels. 10. A can do a piece of work in 4 J days that requires B G days and C 9 days to do the same amount of work. In how many days can they do it working together ? Ans. 2 days. 17. A father divided his property among his wife and four sons, directing that his wife should have $8 as often as the oldest son $6, the second eon $3 as often as the wife $5, the youngest son $12 as often as the third $14, the third sou i^o as often as the oldest $7. The youngest son received S4,500; what was the value of the father's property? Ans. $32,780. 1 9 pounds of bus gaininjor ipowder cost was the cost ysou, ^.70. its length as rn, supposing 2 feet apart, I must he pay ly, if he hoes J. $34.23,^. tity of tea in roiced at $.T0 many chests Ans. 65a 1 14 feet high, corners to tho s. 31.08 + ft. i3.80 per head, head. What |34at$2.80. 3, and a man for seed the Kventh yearV )35 bushels. lat requires B rork. In how [w«. 2 days. ife and four often as the le wife $5, the lird sou 4^5 at» ceived S4,500; ^18. $32,780. REVIEW AXD TEST EXAMPLES. 345 18. If 72 men dig a trench 20 yd. long, 1 ft. G in. broad, and 4 ft. deep in 3 days of 10 hours each, how many men would be required to dig a trench 30 yd. long, 2 ft. 3 in. broad, and 5 ft. deep in 15 days of U hours eacli t Aii». 45 men. 19. Saniuol Wells paid 3^ times as much for a house as for a baru ; had the bam cost him 6 % more, and the house 8 % more, the whole cost would have been $7260. What was the actual cost ? Ans. $G,750. 20. Change ^^5 of — to a simple fraction, and re- 1 -f- 3 + i In- duce to lowest terms. Ana. 21. A person sells out .*4,500 of 4% stock at 95, and invests the proceeds in bank stock at 80, which pays an annual divi- dend of 2| % . How much ia the gain or loss per annum ? Am. $37.50 loss. 22. James Oriswold bought f of a ship ; but the property having fallen in value 8^, he sells 14% of his share for $2700. What was the value of the ship at first ? Ans. $25,000. 23. A gentleman willed to the youngest of his five sons $2000, to the next a sum greater by one-half, and so on, the eldest receiving $10,125, thus disposing of his entire estate. What was the gentleman's estate worth? Ans. $20,375. 24. I sent $7847 to my agent in New Orleans, who pur- chased sugar at an average price of $16 \>et barrel ; he charged 3| % commission. How many barrels did he buy? I Ans. 475. 25. Bought 3,000 bushels of wheat at $1.50 per bu.shel. What must I ask per bushel that I may fall 20% on the asking price and still make 16%, allowing 10 'r of the sales for bad debts? Ans. $241,;. 26. Henry Swift has $0,000 worth of 5% stock: but not being satisfitnl ^\^th his inc-<»me, lu* sells at 96 and invests in stock paying 4J%, which pves him an income greater by $45.60. At what price did he purchase the latter strxik? ^„ Ans. At 75. li r '.* ■ ,, u. ,e ^ ■ i ■ ( V . 4 ^ ■ ; u • ■: " ^1 j ll : 1 i * ! 34G i?i&r/-Bir A^D TEST EXAMPLES, 27. A drover bought a number of horses, cows, and sheep for $3,900. For every horse he paid $75, for each cow he paid f as much as for a horse, and for each sheep ^ as much an for a cow. He bought 3 times as many sheep as cows, and twice as many cows as horses ; how many did he buy of each ? Ans. 20 horses ; 40 cows ; 120 sheep. 28. I shipped to my agent in Buffalo a quantity of flour, which he immediately sold at $7.50 per barrel. I then in- structed him to purchase goods for me at a commission of 8^^ ; he charged me 4% commission for selling, and received as his whole commission $800. How many barrels of flour did I send him? Ans. 1,472. 29. Adam Gesner gave his note for $1,250, and at the end of 3 years 4 months and 21 days paid off the note, which then amounted to $1504.375 ; reckoning only simple interest, what was the rate fo ? Ana. 30. A hound in pursuit of a fox runs 6 rods while tlie fox runs 3 rods, but the fox had 60 rods the start. How far must the hound run before he overtakes the fox ? Ans. 150 rods. 31. A man divided his property, amounting to $15,000, among his three sons, in such a manner that their shares put at 6 % simple interest should all amount to the same sum when they were 21 years old ; the ages of the children were respec- tively 6 yr., 9 yr., and 13 yr. What was the share of each ? Ans. Eldest, $5683.082+ ; second, $4890.094+ ; youngest, $4426.822 + . ♦ a 32. A certain garden is 12f rods long, and 9^ rods wide. At 2^ cents per cubic foot, what will it cost to dig a ditch ^ around it that shall be 3 J feet wide and 4 feet deep ? ^^ Ans. $258. 03|. 33. A farmer sells a merchant 40 bushels of oats at $.60 pe bushel and makes 20 % ; the merchant sells the farmer 4 yard of broadcloth at $3.75 per yard, 15 yards of calico at 8 cent per yard, and 40 yards of cotton cloth at 12 cents per ya i.^ i LE8» wB, and Bbeep each cow lie ) \ as much as ) as cows, and B buy of each 1 J ; 120 sheep. antity of flour, •el. I then in- commission of ig, and received rrels of flour did Ans. 1,472. REVIEW A. YD TEST EXAMPLES. 347 and makes a profit of 25 ^ . Which gains the more by tbf ' t rade, uud how much ? Ans. Merchant gains |.20. 34. A triangular cornfield consisting of 146 rows, has 437 hills in the longest row, and 2 in the shortest ; how many corn hills in the field ? 35. What is the value of r 44| of .056 - 3.04 of t\ 1 L (8-2.4) + |of3| J ' Ans. 82047 hills. + 2 285, 561 Ans. i\. 36. Bought 60 barrels of flour at $8.50 per barrel, but on account of its having been damaged, one-half of it was sold at a loss of 10%, and the remainder at |9 per barrel. W^hat % was lost by the operation ? Ans. 2^ % . 37. A room 22 feet long, 16 feet wide, and 9 feet high, con- pds while the fox I i^jng 4 windows, each of which is 5 J feet high and 3 feet wide ; ^ and at the end the note, which f simple interest, Ans. Qfo. .t. How far must Ans. 150 rods. nting to $15,000, t their shares put le same sum when iren were respec share of each? td, $4890.094+ ; id 9i rods wide, lost to dig a dit<" let deep? Ans. $258.03|. lof oats at $.60 pe 4 yard the farmer calico at I12 cents per yai also two doors, 7 feet in height and 3^ feet in width. The base-boards are | of a foot wide. What will it cost to plaster and paper the room, if the plastering cost 16 cents per square yard and the papering to cents ? Ans. $21.20^. 38. Two persons, A and B, each receive the same salary. A pends 76^ % of his money, and B spends as much as would il 46A % of what both received. At the end of the year they both together have left $276.25 ; what part of it belongs to A, and what to B? Ans. A, $199.75 ; B, $76.50. 39. Two persons 280 miles apart travel toward each other ntil they meet, one at the rate of 6 miles jier hour, the other t the rate of 8 miles per hour. Ilow far docs each travel ? Ans. First, 120 miles ; second, 160 miles. 40. James Welch has a debt in Ottawa amounting to 4489.32. For what sum must a note be drawn at 90 days, that hen discounted at 6% at an Ottawa bauk, will just pay the bt? Ans. $4560. i . I 1 ! « 4 ! ill 348 i? J? r / A' ir and test examples, 41. I went to the store to buy carpeting, and found that any one of three pieces, width rebpectivuly 1^, 1^, and 2 J yards, would exactly fit my room without cutting anything from the width of the carpet. What is the width of my room ? Aji8. 22^ feet. 42. If 10 horses in 25 days consume 3^ tons of hay, how long will 6j^ tons last horsoH, 12 cows, and 8 sheep, if each cow consumes | as much as a horse, and each sbecp | as much as a cow ? Au8. 25 days. 43. The distance between the opposite corners of a square field is GO rods ; how many acres in the field ? Ans. 11 A. 40 sq. rd. 44. At |225 per ton, what is the cost of 17 cwt. 2 qr. 21 lb. of sugar? Ana. $199.2371. 45. A drover bought 12 sheep at $6 per head ; how many must he buy at $9 and $15 per head, that he may sell them all at $12 per head and lose nothing ? Ans. 1 at $9, 25 at $15. 46. Three men bought a field of grain in circular form con- taining 9 A., for which they paid $192, of which the first man paid !?48. the second $64. tlie third $80. They agreed to take their shares in the form of rini^s ; the first man mowing around r,he field until he got his share, then the second, and so on. What depth of ring must each man mow to get his share of the grain'' Ans. 1st man, 2.80+ rd. ; 2d man, 4.73 + rd. ; 3d man, 13.81 + rd. 47. A young man inherited an estate and spent 15% of it during the first year, and 30% of the remainder during the second year, when he had only $9401 left. How much money did he inherit ? Ans. $15800. 48. Mr. Webster bought a house for $6750, on a credit of 10 months ; after keeping it 4 months, he sold it for $7000 on a credit of 8 months. Money being worth 6%, what was his net cash gain at the time of the sale ? Ane. $177.37 + . O^:; r L E S, nd that any id 2^ yards, ig from the aiV 22 J feet. ly, how long if each cow 18 much as a . 25 days. of a square 40 sq. rd. I qr. 21 lb. of $199^37^. ; how many sell them all 25 at |15. IT form con- le first man reed to take wing around and so on. sliare of the J.80+ rd. ; 73+ rd. ; 3.81 + rd. t 15% of it during the luch money $15800. a credit of T $7000 on lat was his 177.37 + . REVIEW AND TEST EXAMPLES, 349 49. A and B can do a piece of work in 18 days ; A can do | 418 much as B. In how many days can each do it alone? Ana. A, 40J days ; B, 32f days. 50. If ? of a farm is worth $7524 at $45 i^er acre, how many acres in the whole I'urm V Ana. 195,^^ A. 51. A person paid $1450 for two building lots, the price of one being 45^ that of the other ; he sold the cheai)er lot at a gain of C0%, and the dearer one at a loss of 25%. What % did he gain or lose on the whole transaction ? Ana. \^\ fc gain. 52. A certain sum of money, at 8% compound interest for 10 years, amounted to $2072.568. What was the amount at interest? Ana. $960. 53. There are two church towers, one 120 feet high, and the other 150 feet. A certain object upon the ground between thum is 125 feet from the top of the first and 160 feet from the top of the second ; how far apart are their tope ? Ana. 95.50+ feet. 54. A farmer sold to a merchant 80 bushels of wheat nt $1 90 l^er bushel, 70 bushels of barley at $1.10, and 176 bushels of oats at $.75. He took in payment a note for 5 months, and immediately got it discounted at bank at 6% ; how much money did he receive ? iln«. $351 . 06 + . 55. There is a pile of 100 railroad ties, which a man is required to carry, one by one, and place in their proper places, 3 feet apart ; supposing the first to be laid 3 feet from the pile, how far will the man travel in placing them all ? Am. 30300 feet. 56. Sound travels at the rate of 1142 feet a second. If a gun be discharged at a distance of 4| miles, how much time will elapse, after seeing the flash, before the report is heard ? Ana. 20J?? sec. 57. If a company of 480 men have provisions for 8 months, how many men must be sent away at the end of 6 montlis, that the remaining provisions may last 6 months longer ? Alia. 320 men. !i;l« 850 REVIEW AJSD TEST EXAMPLES, .^4 v ' ■ i i K V i f r' ,1^ 58. The first fear a man was In btutineBs ho cleared $300, and each year his profit increased by a commun difference ; the fourteenth year he made |950. How much did he make the third year ? Ana. $400. 69. What number is that, which being increased by ^, |,aQd 1 of itself, and diminished by 25, equals 391 ¥ Ans. 180. 60. At what time between 5 and will the hour and minute hands of a clock be together? Ans. 27^ min. past five. 61. A field whose length is to its width as 4 to 3, contains 2 A. 2 R. 82 rd. ; what are its dimeuHions ? Ans. Length, 24 rd. ; width, 18 rd. 62. Three persons formed a partnership with a capital of $4600. The first man's stock was in trade 8 months and gained $752 ; the second man's stock was in trade 12 months, and gained $600 ; and the third man had his stock in 16 months, and gained $640. What was each man's stuck ? Ans. First, $2350; second, $1250 ; third, $1000. 68. How many thousand shingles, 18 inches long and 4 in. wide, lying ^ to the weather, are required to shingle the roof of a building 54 feet long, with rafters 22 feet long, the first row of shingles being double ? Ans. l^^. 64. Employed an agent who charges 4% commission to col- lect a bill of $550. He succeeded in obtaining only 85^ ; how much did I receive ? Ans. $448.80. 65. A and B entered into partnership and gained $4450.50. A put in enough capital to make his gain 15% more than B's ; what was each man's share of the gain ? Ans. A, $2880.50 ; B, $2070. 66. A building is 75 feet long and 44 feet wide, and the elevation of the roof is 14 feet. How many feet of boards will be required to cover the roof, if the rafters extend 2 feet beyond the plates, and the boarding projects 1\ feet at each end, and ^ allowed for waste ? Ans. 547497+ feet. i'LES, REVIEW AND TEST EXAMPLES, 351 cleared $300, difference ; the 1 be make the Am. $400. Bed by f \, and Am. 180. )ur and minute lin. past five. V to 8, contains width, 18 rd. itb a capital of Qths and gained 2 months, and k in 16 months, \ third, $1000. long and 4 in. lingle the roof long, the first Ana. 14|8. nission to ool- Qly 85%; how rut. $448.80. ined $4450.50. Lore than B's ; ; B, $2070. nride, and the >f boards will )xtend 3 feet feet at eacb 4,97+ feet. 67. A circular court is laid with 19 rows of flat stono^, each row forming a complete circle ; the outside row is 89 inches wide, and the width of each row diminishes 2 inches as it uears the centre. What is the width of the innermost row Y Ana. 3 inches. a of fi of AJ^ 68. Reduce . , . .7^ ^ • simple fraction, and take the 4offofU result from the sum of 10}, ^^j^, and 7}|. Ana. 8Jj{. 69. Bought 75 yards of cloth nt 10% less than the first cost, and sold it at 10% more than the first cost and gained $25. What was the first cost per yard t Ana. |1.66f . 70. A grain merchant bought 7500 bushels of com at $1.85 per bushel, 5450 bushels of oats at $.80, 3250 bushels of barley at $.95, paid $225 for freight and $170 for storage ; he immo- diately sold it at an advance of 20 % on the entire cost, on a credit of 6 months. What % did be gain at the time of the sale, money being worth 8% ? Ana. 15+ %. 71. A farmer employs a number of meu and 8 boys ; he pays the boys .$.05 and the men $1.10 per day. The amount that he paid to all was as much as if each had received $.92 i)er day : how many men were employed? Ana. 12 men. 72. S. Howard can mow 6 acres in 4 days, and hib son can mow 7 acres in 5 days. How long will it take them both to mow 49i^ acres ? Ana. 11 ^ days. 78. I lent a friend $875, which he kept 1 year and 4 months. Some time afterward I borrowed of him $350 ; how long must I keep it to balance the favor ? Ana. ^ yr. 4 mo. 74. Find the difference between the surface of a floor 80 ft. 9 in. long and 66 ft. 6 in. broad, and the sum of the surfaces of three others, the dimensions of each of which are exactly one- third of those of the other. Ana. 891 sq. yd. 7 sq. ft. 12 sq. in. 75. A tree broken off 24 feet from the ground rests on the stump, the top touching the ground 30 feet from the foot of the tree. What was the height of the iree ? Ana. 62.41 + ft. X; / \ }:% '^:' 352 BEVJEW AND TEST EXAMPLES, ' )'■ ' ' "'* 1 ,M! I ( I i I,! Pf» 76. Two persons entered into partnership for trading. A put in $245 for 375 days and rec€ived ^ of the gain ; the num- ber of dollars that B put in was equrl to the number of days it was employed in trade. What 'vas B's capital ? Aii8. $350. 77. How many square feet of boards 1^ inches thick will lie reciuired to make a box, open at the top, whoso inner dimen- sions are 6 feet long, 4 feet wide, and 3 feet deep ? AhH. 88,'u sq. ft. 78. A farmer having 80 acres of l.ind, worth $5.") an aero, wishes to buy enough more at $50 and .$G5, respectively, so that the value of his land shall averr%'t $60 an acre. How much of each must he buy? Am. 1 A. at $50 ; 82 A. at $65. 79. What is the amount of an annuity of $700 for 8 years, at 6% comi'ound inU^rest ? Am. $6928.22J. 80. What must be the price of stock yielding 5} 'v , that will yield the same profit as 4^% stock at 96? Am. 112. 81. A jjerson after sptuiding \ and J of his money an.l $20, had $80 loft. What had he at first ? Am. $240. 82. James Hari>er has a large jewelry store, which with its contents he insun^s in the C'itiz«'ns' Insurance Company for * of its estimated value, at 3}^. This Company immediately insures \ of its risk in the Phoenix Company, at 2 J '/c . After two years and a half, the store and Its contents w«'ro destrr)yed by fire, when it was found tliat the PhoBnix Company lost $2025 more than the Citizons' Compaay. Reckoning 6% simple ' iterest on the premiums that the owner |>aid, what would 1)e his entire loBsV Ant. $78815.75. 88. A drover sold 42 cows and 34 oxen f.)r $3374, receiving $21 i»er lu'ad more for the oxen than for the cows. What aid he receive for each jMir head ? Anit. $35 for cows ; $56 for oxen. 84. A certain room is 27 ft. 5 In. long, 14 ft. 7 in. wide, and 12 ft. 10 in. high. How much paper ; of a yard w'uh; will he required to cover the walls ? Ans. 136 yd. 2 ft. 8 in. LES, trading. A \ ; the num- 3r of days it iii8. $350. hick will 1)© uncr dimen- ^8,^ sq. ft. jiS.") an iicro, ipectively, so . How much I A. at $63. or 8 year«, at $0928. 22 J. %, that will Am. 112. ,oy un.l $20, Am. $240. licli with its 'ompuny for immediately After two roye*! by fire, $25)25 moro pie " itrnist Ixi his rntire ^78815.75. r4, rorciving What did for cows; for oxen. |n. wide, and i'mV'! will 1)6 2 ft. 8 in. REVIEW AND TEST EXAMPLES, 353 85. The area of a triangular field is A. 36 rd. ; the base is (U rods. Whnt is the p<}rpendicular distance from the base to the angle opposite ? Aiia. 31 i rfxls, 86. If the width of a building is 50 feet, and the length of the rafters 30 feet, what will it cost to board the gable ends, at $.18 per square yard ? Ana. $16.58 -♦- . 87. What is the solidity of the largest ball that can be cut out of a cubical bUx:k whose sides are 6 inches stpiareV Am. 113.0976 cu. in. 88. A privateer took a prize ivorth £.348 15s., which was to be divided among 1 captain, 3 mates, and 27 privates, so that a private; should have one share, a mate twice as much as a private, and the captain 6 times as much as a mate. What was the share of each? Am. Private, £7 ISs. ; mate, £15 10s. ; captain, £93. 89. The' width of a certain building is 38 feet, and the elevation of the roof is 16 feet ; how many square feet of boards will be required to cover the gable ends? Am. 608 s '^:* 354 BEVIEW AND TEST EXAMPLES, Li i i I 94 At what time between 10 and 11 o'clock will the hands be directly opposite? Arm. 21^ min. past ten. 95. How much better is it to invest |15000 in 6% stock, at a discount of 25%, than to loan the same sum at 7% simple interest? Ans. $150. 96. What is the present worth of an annuity of $550 for 6 years, at 8;^ simple interest? Ans. $2675.675 + . jjof9J + 4|of J\ 1.30 97. What is the value of(« 5-4f ') .005 Ans. i^oV- 08. If 36 men working 8 hours a day for 16 days can dig a trench 72 yd. long, 18 ft. wide, and 12 ft. deep, iu liow many days will 32 men, working 12 hours a day, dig a trench 04 yd. long, 27 ft. wide, and 18 ft. deep ? Ans. 24 days. 99. Bought a piece of broadcloth at $2.75 per yard. At what price shall it be marked that I may sell it at 5 % less than the marked price and still make 20% profit ? Ans. $3.47i'5. 100. A man hired a mechanic for 85 days, on condition th -, for every day he worked he should receive $1.75, and to: every day he was absent he should forfe't $2.50. At the end of the time he received $40 ; how many days did he work ? Ans. 30 days. 101. If stock bought at 25;^ premium pay 7J;^ on the in- vestment, what % will it pay if bought at 4% discount? Ans. 9g%. 102. The interest on a note for 2 yr. 3 mo. 18 da., at 8%, was $155.02 ; what was the face of the note ? Ans. $842.50. 103. Tlie distance on the road anmnd a certain park is 17 miles. If three persons start from the same point on the road at the same time and travel in the same direction around tlio park, how far will oach have to travel before they all come together, if the first travels 5 miles an hour, the second 6, and the third 7 miles an hour? Am. First, 85 mi. ; second, 102 mi. ; third, 110 mi LE8. ill the hands k. past ten. 6% stock, at It 7 % simple Ans. $150. of $550 for i675.675 + . OG .005 Ans. ^oV- ays can dig a in how many trench 04 yd. ris. 24 days. )er yard. At it at 5% less ? u. |3.47i'ft. londition th .75, and to: At the end he work? IS. 30 days. ^f on the in- icount ? Ans. 9g%. , at 8%, was | . What was his gain ? Aii8. .i;553.a9-»-. 121. Two men form a partnership for trading ; A's capital is $3500, B's $4800. At the end of 7 months, how much must A put in that he may receive ^ of the year's gain ? Ans. $3120. 122. A maft having lost 25% of his capital, is worth exactly as much as another who has just gained 15% on his capital ; the second man's capital was originally $9000. What was the first man's capital V Ans. $13800. 123. A merchant imported 18 barrels of syrup, each contain- ing 42 gallons, invoiced at $.95 per gallon ; paid .*>b5 for freight ami a duty of 30%. What % will he gain by selling the whole lor $1171.459"^ Am. 15%. lamess, $56. I b^ock 25 rods and Herbert ■)ni the same block before rt, times. 'I 1 , » : ANSWERS. «• tJH Hj ). •t ! ii The answers to oral exercises and the more simple examples have been omitted. The answers for the exercises taken from the Arithmetical Tables, commence on page 376. Art. 39. /. 1213. 2. 11526. 3. vm. 4. 5726. 5. 322633. 6. 1543. 7. $4030. 8. $795. .9. $2617. 10. $1330. Art. 45. 1. $3553.66. 2. 16217.66. S. $1004.94. 4. IU06.75. 5. $75.38. 6. $312.09. S. $607.65. 9. $17931. 10 !?(Uy.90. 11. $5746.62. le ?1751.'32. 13. $228.20. 14. $8985.60. 15. $70.11. Art. 01. 1. 100 years. 2. $UH6. 3. 6269. 4. (i892 feet 5. 3502. 6. 6940. S. $550. iy. $41874. /ry. $1074. 11. $4820. i4^. $1634. 13. $245. i4. $1136. 75. 1005 ba. 16. $262.64. 17. $166.38. 18. $11.06. /!). $95.12. m $901.92. 21. $272.59. Art. 81. 0. $1995. 7. $639. ^. i?;:i882. .9. 63360 feet. iO. $2556. 11. 1516 far. i.'. $241. !//?. $311. />^. $76. Art. 90. i. $8:;()0. $24630. .tl498. $72. 522270 gal. J. 4. 5. n. 14136 bu. 7. 142692 da. 8. 8946 trees, ry. $2373. 10. $9048. ii. 3.')857586. 12. $1086800. 13, $11.26. /4. $72.25. 15. $638.27. 16. $6253.04; 17. $21.99 gain. 18. $69.42. 19. $2112; $345.60 ga. 20. $32.33 gain. Art. 113. /. 281 ; 302. 2. $986. /. 210 hours. 4. 147 barrels. 5. 121 montlis 6. 26 we«kB. 7. 86400 arc. cV. 134 baskets 0. 480. /(y. 47. //. 59 dozen. 12. $178. 13. m sli«M«p. i.^. 76 barrels. Art. 120. 7.205. 2.2X», 3. 440. 4. 315. 5. 882. 6. 3158. 7. 1002. S. 137. .9.445. /^. 963. 11. 4455. /J. 375. iJ. 4144. 14. 560. /.5. 4661. 16. 2247. /7. 108. 75. 176. 7.9. 276. 20. 5362. 2U 7967. .?i?. 90807. ?J. 1234. U. 4(;834. ;;A'J. 3147. 2C. 864;J. .^7. 40367. /. 440. 815. 832. 8158. 7. 1002. S. 137. .9.445. 0. 962. i. 44')5. J. 375. b. 4144. U. 560. .5. 4561. l6\ 2247. 7. 108. [5. 176. ;.9. 2T«. [^A 6862. ;. 7967. b. 90807. b. 12:54. 4r»H34. 8147. 8.56. 4. :j;767.3«. r>. 120502.50. a. :j;658.56. 7, ♦632.20. 5. $756. 9. $598.58. JU. $;i31.89. Art. 147. S. $13. 'i $18000. «>. ^o, I . ♦♦*•'• 4. $7. .V. $73. 5. $7'. .V*. $7J $52. ' 2 Art. 148. $1260. .?. $1148. //. *8172. 5. 31104 railea. a. $2496. 7. ♦162. Art. I4». :?. 85 barn^l.s. 3. 73 horse-s. 4. 17 weeks. 5. \^\\\ acres. C. 327 thousand. Art. 1 50. ?. .{.T.l T^iinds. 3. 184 huxes. 4. 4900 pounds. .5. 776 baskets. 6*. 168 days. 7. 168 cords. Art. 151. 1. 25 pounds. J. 712 corda. 3. $2625. 4. $3;«24. 5. $426l 6. $470 gahL 7. $1241. 8. $504. 9. 60; 13. W. $2096 ; $1861. //. 4714: 4262. 12. $4028 . $2466. 13. 50 days. /4. $576. 75. $56. /6\ 149|f acres ; 14811 acres. /7. 6 bushels. Art. 173. /. 3, 7, 2, 5, 3. .'. 3, 7. 2. 5. 3. i, 3. 3, 7. 11. 4. 2. 3. 7, 11. 5. 2. 2, 2. 3. 8. 113. 6. 2, 5. 7. 3. 7. 7. 7. 7. 81>. *. 2. 3. 11, J, J. ^', «>, o, 0, O, < ) 7. /^>. 3, 11, 2. 5, 7, 13. //. 5. 2. 5, 2, 2, 73. I /?. 5, 5, 13, 59. 13. 2, 2, 4007. 74. 2. 5, 5, 5, 5. in. 3,3,31, 37. I m. 3, 5, 2, 2, 2, 2, 2. 7. ' 77. 2, 5. 2, 5. 2, 2. 2, 2, ' 2. 2. 7^. 5. 377, 43. /'. 2. 5, 2, 5, 2, 5. "'. 2.5.3. 7. 11. 1 21. 2, 3. 7, 11, 17. 359 22. 2, 5, 6, 109. »3. 5, 2, 2, 2, 229. 2Jt. 3, 2, 2, 2, 2. 3, 8, 8, 8, 3. S5. 2, 5, 11, 73. Sii. 6, 7, 7. 17. .?7. 2,5,2,5,5,5,5,5. i?6. 2,5,2,5,2.5,3.8. 8.3. f9. 2,5,2,5,2,5,2,2, 2, «, 2, 3. SO. 2, 5, 2, 5, 5, 7, 13. SU 5, 5, 5. 5, 3, 3, 3. 52. 2,5,11,13,18. 53, 3, 5, 2, 3, 8, 7, 13. Art. 177. i. 16^. f.^' 2. \\. S. 20. 5.200. 4. 15U. 5. 166j. c'y. 20. i(>. 2. 77. 4 barrels. 12. 42. 7J. 6 shillings. U. 80 weeks. 7.5. 80 loads. 76. 915 IM) unda 77. 5 boxes. Art. 185. 1. 5. 7^.34. 2. 15. 77. ;^. .y. 21. 12. 35. 4. 39. 7.?. 46. r>. 18. 74. 5. 6-. 4. 15. 8. 7. 17. 7ft\ 12. S. 81. 77. 24. i/. 75. 7.V. 46. Art. 192. 7. 21. 0'. 5. ;?. 15. 7.4. ^. 22. cV. 3. 4.^. .'/. 91. 5. 35. 10.^ w ly H' 111 \' p ■ Iff 1 , t fi 360 Art. 199. 1. 3 feet. ^. 8. 3. 12 inches. 4. 80 feet. 5. 13 barrels. G. 14 feet. 7. 14 yards. 8. $88; 214 acres. 105 acres. Art. 20«. i. 00 cents. ^. 2520. ^. 80 quarts. 4. $7i>H. 5. U'tH rows. 6. 30 feet. 7. 1050 feet. Art. a 10. J. 330. i:'. 182. 5. 390. 4. 2145. 5. 5005. 0. 402. 7. 0000. 5. 2548. Art. 237. S. fiS". i6-. i^y,K 17. \\\ Art. 240. 4. '6^i\. ANSWERS, 5, G. 7. S. 'J. 10. 11. U. IJ. /4. 15. IG. 17. IS. 12||. 4. 57 *i ore 8er|. 80,Vff. 89 jV «3h. 64J2. 16 B ffOff" 1 I* „7 Art. 7. O. -r' 11. I?. 24G. /^. ^|. 15. 10 17 18. \ '• Ik. . HI Art. 249. 6. G. 7. 5. 'J. 10. 11. if; f^ u. tfC' BO «<» 4B . T2fl > m- mi 4Hfl . T20 » ^■5 •iJ [J i^. « OH . DA • Stttl » m Ml m HI IHO 6'. ii. 1^. iJ. 14. 15. 10. 17. 16'. /.''. :. OOi pounds, Art. 202. '//. 35!. U. ISi'o- iJ. 50. i6\ 72. 17. 10, ni- Art. 204. 1. 7867; 61 A; 75U; 9; 90. 2. $206; $777J : $3780 ; $7:T85. 3. $300. 4. 142Hi{ acres. 5. $580. G. 836 j^s yds. Art. 2G7. ^' HI; nu: 7. ?• Art. 270. 10. 953 i. //. 959|. 12. 44571 1 . 7^. 1534. i4. 74800. 15. 73000. Art. 275. / 1^' ^' «• 4. t T * ^'. li "w 4B0 '• 11*1 IF' v r. s 10. 6 tl 1 • "a .'. • i^. iJ^ . 50. . 72. \rt. 204. . 7867; 61A; 75U; 9; 90. . $266; $7775 ; $11780 ; $77785. . laoo. . 142|;{ acres, . $580. . 835A yds. lit. 2G7. 1^51 I IS* .rt. 270. 953 i. 9592. 44571 I . 1534. 74800. 73600. At. ti75. n 1 r- .\. 4 «• 8 IR" , 1 3 450 1 " -' hi .1 • M 5- ANSWERS, i4. ■ i^. 1380104. f i. 126? £3. 500it. i?4 29 ct. S5. 69rV tone. J6\ $10^f. S7, 2H. ^'! 8540. tJi- 7. ^- I hi' i(^' lif. V IS- 53. 825^. 54. ' ' 35. 36. 37. 11295. Art. 285. i. 2H. 4^ iifif. 5. 20Ja. 6\ 48071. 7". 18 |. 8. 50 . 9. 4»i f /O. 81 f. 11 5f /^. iLo.t. mil 16. 6 ^?'. |2U; |22| $68|. ^«?- mil Art. 282. ^4. 1|. 26. i|- :S?A Mil. ^^.' m'i. 31. im^. Art. 205. 7 1 • A- ft • -»• f • f » g » 8|.2| 2J 3 ' 8 • 3 2 ?!. ^.^ 7 ' 7 ' 7 • •^' ao • 9* 7 * 7 * 7 ' 7' 7' 7* ^' 'nio t TTnF » JUL • «o • Too » Too » 361 100 • 100* 100 ' 100 ' 217,V 100 • Art. 296. .J. 100' 100' 100' "^' Toy 100* 10' 10' 10' OJJ 10' S' tWo;t'. $289|. ^(;. 47 J j miles; 38! X miles. '■> Too » ^' 100 '100' .'/. $8113,',. 22. 656Ar. _ . „.. . S3. $ior>oo. J? . il*J .Uj. 11800. 26. 123? corrls. i'7. Increased by A. 7611 100 ' 24 H'. n . : I m 362 eS. 4|}| Ibfl. ; SI. 2U days. SS. 15 feet. 3S. 1106|f$yd& 5^. $289i{|. ^5. 18 U acres. S6. %rms\. Art. 320. 1. .6. ;?. .4875. 3. .36. 4. .84875. 5. .4625. G. .712. 7. .063125. 8. .5008. 9. .3571f m .42301g. 11. .3538 A i.?. .1^571 13. .0352 14. .0879, i5. .1917tW. 16. .niSsYi- 27. .019047. 18. .42a57i. i5. .5367142^. m .^6410^. fSL .S46153. ;?5. .33^92307. SS. ..^02156862- 7450986. f4. .863. S5. .430. i?(>. 8.tl42a5. 57. 82 4076923. 28. 24.32i42857. ANSWERS, i: Art. 332. 1. H. ^. |. 5. g i^. iHi- Art. 335. i?. A. Art. 338. 2. M. r _i 1 !• a 1 JO. II. 'OT. i:\ v^. /4. V. 15.^. Art. 341. i. 158.9656. J?. 572.877. .9. i58a26. 4. 492.4198. 5. 105.9817. 6. 16.62220. 7. 115.2549a 8. $121.11. Art. 343. 1. 82.0445. ^. 782.48. 3. $87.26. 4. 87.124. 5. $16.65. 6. 17.49608. 7. 874.960401. 8. 6.08995. 9. .600091. 10. .0894097. 11. $72,091. if. 94.1881. IS. 17.42. /4. 5.552. 15. $170.75. i6. $788.0225. Art. 346. /. 4.958. 2. 85.77. 3. 217.496. 4. $58,555. 5. .$4075.26. 0. 60.285618. 7. 286 025901. 8. .029418. 5. .80288. 70. .00748. //. .0000352. U. .000072. /.?. .(X)3045. /4. .001535. /.I. .00000101. in. 24.17J. i7. 1344.13^. IS. $1.06375. 19. .46075. *o. .0068. il. .00088575. S2. $7.47891. 23. $685.6289. 24. $588.2114 25. 22:.900| gr. 2G. 2.85644 yd. 27. 52.16681 in. 28. $271.8017. 2U. 1908.75. SO. $5260.888 •(- 31. $46070. ?. $88.88|. 33. $4.96. 34. $862.90. Art. 347. 1. .85. 2. .75. .J. .625. 4. .46875. 5. .825. 6". .1125. 7. .tS. 5. .^1428lJ. 9. .8^. /r;. .15. i/. .&0769S. ii'. .07954. Art. :)48. i. 20.5652 + . 2. 46.2857 + . J. 8.0731 +. 4. 8.8235 + . .';. .8040 + . (I. .8604 + .- 7. 5.4844 + . 5. 29.8661+. 9. 9.0009 + . Art. ,352. /. 6.854«; 38.3680. 16075. iX)088576. ^7.47891. |;685.6289. 1(583.2114. •2:.tt0«| gr. J.85644 yd. 32.10681 in. H271.8017. 1»08.75. 15260.8884- 'i46070. id.88|. 4.96. ;62.90. rt. 347. .85. .75. .626. .46875. .825. .1125. .'71428b. .8^. .15. .^709i^. 079M. t, :i48. 120.5052 + . 46.2857 + . 8.0721 + . 13.8235 + . .8040 + . .8604 + .- i.4844 + . 59.8001 +. .1)009 + . Irt. 352. 1854«; }8.3080. f. 17.0709. 5. 1.0979. 4. 1960.5945. 6. .7583. (;. 6.08JJ3. 7. 160.6284. 5. 142857.142a 9. 41710. 20. 1.4743. J I. 161.2496. 1?. 73.3743. 13. 89.5901. 14. |5.8:m2. 1,5. 101.1879 yd. in. 5.1309. 17. 17!>.78. 27700 pounds. 7. 103240 grains. 47 ix)unds. //. 217339 graina 12. 175393 pounds. Li. 5749 dwt. 14. 588S000 ounces. 15. 7000 grains. iC. 39377 dwt. Art. 300. 5. 438 T. 4 cwt. 451b. G. lb. 9 oz. 8 dwt 12 gr. 7. 1017 lb. 2 ot. 8. 16 T. 14 cwt. 77 lb. 8 oa. P. 48 lb. 6 oz. 4 dr. 2 sc. 7 gr. 10. 35 lb. 9 oz. 17 dwt 11. 173 T. 8 cwt 13. 1504 dr. 1 sc. 15 gr. IS. 92 lb. 9 oz. 1 dwt. 11 gr. 14. 27 T. cwt. 4 oz. 15. 1011). 3oz. Idwt 21gr.;13i''^lb. 10 lb. 8 oz. 2 sc 5gr. 16. 70 lb. 3 oz \ dwt 23i gr. 363 17. T,V, lb. IS. 17 lb. oz. 4 dr. 1 80. 21 gr. 19. 51b. 310 -0 D2 8gr. Art. 372, /. 32 2 gr. 2. 11 cwt. 11 lb 1} oz. J. 8 oz. 14 dwt. 13,Vgr. 4. 9.0 oz. 5. 58 34 T)i 14? gr. 6". 17 cwt. 7. 14 dwt. 14.4 gr. S. £>2 10.4 gr. 9. 15 lb 10 oz. 10. 58 lb. 5J oz. 11. 7 lb. 5 oz. dwt. 10 gr. 12. 8 cwt. 70 lb. 13. 13 T. 14 t wt. 28 lb. 9> oz. 14. lb. 5 3 11 33 14.4 gr. 15. 8 oz. 10 dwt. 17 004 gr. 16. 18 cwt. 71 lb. 8.2 oz. 17. lib. 15 dwt. 5gr. 18. 102 lb. 1 oz. 19. 80 lb. 9 oz 13 dwt 8 gr. ; lb.80|9 35 31. Art. 375. 0. 11000ft T. A'off '''• T. . I 10. gj^lb.; ^^- sffino '»*• > sift II*' 1-'- ^ T. ; ,«„ t. : TTsW ^' • TnOO ^ •» s T iJ. A lb. ; rt^5 cwt. '' I I! ■^^ IMAGE EVALUATION TEST TARGET {MT-3) I 1.0 I.I M 12.0 12.2 1.8 1.25 1.4 1.6 ■• 6" ► Photographic Sciences Corporation 23 WEST MAIN STREET WEBSTER, N.Y. 14S80 (716) 872-4503 V iV 40^ ^\ .<-*u 9> V ^^ #; c^ ■'^ 4if C C/j o »' \i 364 ANSWERS. H .1 1 h I.. J 'J- m ff! |fel . yu^.. Art. 378. i. IH lb. 2. /A T. 5. m cwt. 4. tJS- lb. 5. Ill lb. 6. ^A lb. 7. if lb. 5. tWs lb. 5- Hie lb. 10. H- 12. 1. i.?. .05375. /^. .122|. 15. .4455 T. 26. .9868+ lb. 17. .1138|. i<9. .81875 lb. 19. .634 T. ^0. .8121527J lb. 21. .7361 j lb. ^2. .71498 + . 23. .5944f Art. 380. 1. 74 T. 3 qr. 4 lb. 10 oz. 2. 33 lb. 2 oz. 1 sc. 12 gr. S. 1 lb. 3 oz. 18 dwt. 20f gr. 4. 49 cwt. 2 qr. 6 lb. 5. 5 cwt. 20 lb. 12^ oz. 6\ 4 T. 3 cwt. 9 lb. 10 oz. 7. Ib.l4 53 34 32 14 gr. 8. 10 oz. 4 dwt. 9.0 gr. 5. lb. 40 §10 31. 10. 87 lb. 2 oz. 12 dwt. 18 gr. 11. 10 T. 11 cwt 42 lb. 15 oz. 12. 8 lb. 11 oz. 19 dwt. 15.2 gr. 13. 80 lb. .5 oz. 14. IT. 14 cwt. 921b. 14 oz. Art. 382. 1. 11 lb. 4 oz. 13 dwt. 12 gr. ;?. lb. 5 59 3 6 32 3. UT. 18cwt.lqr. 21 lb. 3 oz. 4. 5 lb. 10 oz. 7 dwt. 5. 1 T. 18 cwt. 62 lb. 8 oz. G. 5 6 3 7 10 gr. 7. 4 lb. 6 oz. 11 dwt. 12 gr. .?. 7 T. 7 cwt. 45 lb. 9. 9 oz. 8 dwt. 16.8 gr. 10. 3 cwt. 64 lb. 12 oz. 11. lb. 7 55 3 6 31 5gr. 12. lb.3 511 3132 10 gr. 13. 6 T. 12 cwt. 2 qr. 141b. Art. 384. 1. 3 T. 9 cwt. 14 lb. 13 oz. ; 4 T. 12 cwt. 19 lb. 12 oz. ; 24 T. 4 cwt. 8 lb. 11 oz. 2. 15 lb. 2 o2. 10 dwt. 20 gr. ; 261b. 7 oz. 8 dwt. 23 gr. 3. 18 T. 17 cwt. 521b. 4. 7 cwt. 66 lb. 5 oz. 5. 62 lb. 4 oz. 7 dr. 15 gr. 6. 1 cwt. 19 lb. 12| oz. 7. 25 lb. 14 oz. ^. 3 T. 6 cwt. 92 lb. 9. 54 lb. 12 oz. 10. 1 cwt. 7 lb. 14f oz. 11. 6 T. 3 cwt. 12 lb. 12. 32 11.12 gr. 13. 2 cwt. 59 lb. 6.72 oz. 14. 9 cwt. 72 lb. 3f oz. 15. 8H oz. 16. 50 lb. 17. 19.44 gr. IS. 78T.3cwt.161b. 19. 40 T. 5 cwt. 92 lb. Art. 385. 1. 4 lb. a oz. 16 dwt. 13| gr. 2. 4 T. 17 cwt. 3 qr. 21 lb. 8 oz. ; IT. 19 cwt. 181b. 9| oz. 1 qr. 1 T. 4 cwt. 24 lb. 2 oz. 16 cwt. 1 qr. 7 lb. 12 oa. 5. 3 1 5 gr. 4.510 3710 gr.; 57 32 6fgr.; 53 35 3igr.; 5. 6. 7. 8. 9. 10. 11. 5l323213|gr. 1 cwt. 10 lb. IJfloz. 34, I bo: 35 boxes. 25 5ii,v ; 1920. 52 3 5 31. 12. i2413|f ; lOf- Art. 389. i. 21.86611+ mi. cwt. 19 lb. l^ oz. I lb. 14 oz. T. 6 cwt. 92 lb. t lb. 12 oz. cwt. 7 lb. 4|oz. T. 3 cwt. 12 lb. )2 11.12 gr. cwt. 59 lb. ;.72 oz. cwt. 72 lb. J| oz. II oz. 10 lb. .9.44 gr. '8T.3cwt.l61b. to T. 5 cwt. 92 lb. Art. 385. Ub.aoz.l6dwt. 13? gr. 4T. 17 cwt. 3 qr. 21 lb. 8 oz. ; LT. 19 cwt. 181b. 9|oz.; L T. 4 cwt. 1 qr. 24 lb. 2 oz. 6 cwt. 1 qr. 7 lb. I 12 oa. ' 1 5gr. 10 3 710gr.; 7 3 2 6| gr. ; 3 3 5 3^ gr. ; I3 23213igr. cwt. 10 lb. {do: xes. >5. )11,^^ ; 1920. 2 36 31. 413|i; lOf. ,rt. 389. 51.&65U+ ml* 2. .5280 yd. J. 1386 in. 4. 2100 1. 5. 84f in. e. 5.5f 1. 7. 1087.5 mi. S. 4.3866+ statute mi. 9. 2544 in. 10. 245 statute mi. //. 165751. 1^. 233 rd. 3 yd. 10.8 in. J3. .000482+ mi. 14. .4646. 15. 2 mL 48 ch. 3 rd. 24 1. 1.08 in. 16. 66 ch. 2 rd. 7 1. 1.76 in. 17. 10 mi. 14 ch. 5 1. 18. 7 mi. 258 rd, 2 yd. 2H ft. 19. 10a5.916 St. mi. 20. 2\ degrees ; 155.625 St. mi. SI. w- SS. .66 ft. 33. 145.83 St. mi. 24. .(m6. 25. .036 mi Art. 398. 1. 4017.2 sq. yd. 2. 15488000 sq. yd. 3. 180680 sq. ft. 4. 8000 sq. I. 5. 448000 sq. 1. 6. 19200 sq. ch. 7. .00516528+ A. 8. .0001367 + sq. mi. 9. 284 A. 71 P. 3 sq. yd. 3 sq. ft. 36 sq. in. 10. 21.78 sq. ft. ANSWERS, 11. 25 sq. mi. 457 A. 1 sq. ch. 6 P. 535fc sq. 1. 12. 81 P. 27 sq. yd. 7 sq. ft. 67.68 sq. in. IS. 4 A. 68 P. 6 sq. ft. 32H in. 14. 1 Tp. 12 8q. mi. 188 A. 9 sq. ch. IP. 15. 4 sq. mi. 319 A. 9 sq. ch. 8 P. 458^ sq. 1. IB. 9 sq. ch. 15 P. 624 sq. 1. 35 sq. mi. 639 A. 9 sq. ch. 15 P. 624 sq. 1. 17. 140 A. 6 sq. ch. 13 P. 89f sq. I. 18. 5 P. 28 sq. yd. 6 sq. ft. 108 sq.in. 19. 9 P. 5 sq. yd. 4 sq. ft. 108 sq. in. 20. 5 P. 29 sq. yd. 1 sq. ft. 18 sq. in. 21. 2 A. 3 sq. ch. 1 P. 376 sq. 1. 22. 18 sq. yd. 1 sq. ft. 572 sq. in. 23. 1 P. 2 sq. yd. 8 sq. ft. 14.4 sq. in. 24. 309i yds. 25. 384 boards. 26. 2230 sq. ft. 27. $620. 28. $485,275. 29. 1096.98 tiles. 30. 11352 shingles. 31. $500.76. 32. 1200 stones; $2120.725. 33. $50.53. Alt. 406. i. 167616 cu. in. 365 2. 2.43 cu. ft. 3. 867456 cu. in. 4. 32.6592 cu. in. 5. 18 cu. ft. 1080 cu. in. 6'. .0296 cu. yd. 7. .02*> cu. yd. 8. 9 cu. ft. 648 cu. in. 9. 4 cu. yd. 8 cu. ft. 432 cu. in. 10. 58 cu. yd. 21 cu. ft. 1664 cu. in. 11. 3 cu. yd. 18 cu. ft. 12. 29 cu. yd. 22 cu. ft. 576 cu. in. 13. 5cu.yd.l4cu. ft. 1080 cu. in. 14. 13 ft. 5 J in. 15. 48 ft. 16. 827851 cu. yd. 17. 92808U cu. ft. 18. 4 ft. 6 in. 19. mil cu. ft. 20. 598JHCU. yd. 21. $233.6244 + . 22. 87|«pch. Art. 407. 1. 336 cd. ft. 2. 1584 cu. ft. 3. 4608 cu. ft. 4. 20.79 cu. ft. 5. 91? cu. ft. 6. .005703125 cd. 7. 2.6337448 + cu. yd. 8. .03232 pch 9. 134.S2098 + . 10. .1111. 11. 2cd.3cd.ft.8cu. ft. 972 cu. in. 12. 16 cd. 7 cd. ft. 15 cu. ft. 345^ cu. in. 366 AJVSWEBS. '1,1 If pi m tn 13. 91 1 cd. 14. $53.0578J. i5. 8 ft. 6f in. 10. 108,*Tpch. 17, mOti cd. i- Hi Alt. 410. 1. 26 B. ft. 2. 35 B. ft. 3. 24 B. ft. 4. 24 B. ft. 5. 28 B. ft. 2 B. in. 6. 42 B. ft. 6 B. in. 7. 7 B. ft. 6 B. in. 8. 42 B, ft. 6 B. in. 9. 29 B. ft. 2 B. in. 10. 18 B. ft. 9 B. in. 11. 40 B. ft. 6 B. in. le. 44 B. ft. 4 B. in. 13. 32 B. ft. 7 B. in. U. 13 B. ft. 6 B. in. 15. 504 B. ft. 16. 36 ft. 17. 31 ft. 2| in. 18. $70.98. 19. *24 80625. go. $85,008. Art. 412. 1. 32 gi. : 64 gi. ; 56 gi.; 28 gi.; 120 gi. e. 128 pt.; 58 pt.; 110 pt. 5. O. 6 f 5 5 f 3 2 TTV15. 4. 2 bu. 3 pk. 1 pt. ; 4 bu. 2 pk. 3 qt. 1 pt.; 9 bu. 7 q,t. 6 bu. 1 pk. 6 qt. 1 pt. 11 bu. 2 qt. 6. 871Hbbl. 6. 369 1 J bush. 7. Cong. .13777- 66 + . 8. 1463f^ bu. .9. 56|f^ bbl. 10. Cong. 7.46972 + . 11. .875 gal. IS. .064453125. 13. .00236 + . 14. 1523. 21 f. 15. $424.52. Alt. 413. 1. 125.958+ bu. ^. 1244111 cu. ft. 3. 1282l| cu. ft. 4. 505 A cu. ft. 5. 2462.431 cu. ft. 6. 6011 cu. ft. 7. 1001.475 + CU. ft. 8. 7.085+ ft. 9. 144^*^ oz. Art. 417. 1. 66 yr. 6 mo. 13 da. 10 hr. S. 22 yr. 3 mo. 14 da. 14 br. 3. 99 yr. 4 mo. 19 da. 4. 118 yr. 5 mo. 17 da. 7 hr. 5. 7 yr.- 9 mo. 1 da. 6. Feb. 22, 1732. Art. 422. 1. 385740". 2. 30600'. 3. 19663 '. 5. .025 cir. 6. 9 s. 28' 48'. 7. A ^ 'ir. ; i Cir. ; i Cir. 5. 64" 17' 8|". 9. 2 sextants; 2i ; 2| ; 4^ ; m 10. 33 sex. 47°. 11. 2 s. 12". 12. 883 yr. 8 mo. 20 da. 13. 94 doz. ; 165 doz. 14. 1264 doz.; 2280 doz. 15. 29| doz.; 327| doz. ; 141|^ doz. IG. 13320 sheets. 17. .333. Ai-t. 427. 1. 2736 far. 2. $83,745. 3. £.0375. 4. 34000 mills. 5. £.002916|. G. .015. 7. 12s. 6d. 5. 16s. 9d. 2.4 far. 9. £1713 88. lid. 10. $457,024. 11. £89 lis. 9id. 12. $10461.564. 13. $1172.0965}; 4914 marks 45.08+ pfennig; 6073 fr. 3 ct. 8.9+ m. 14. .417375. 15. .4632. IG. 647 f r. 6 dc. 6 ct. 8.3+ m. o 3. 4- 6. 6. 7. Art. 440. 38.1024 Kg. 33.5627 Ton. 33.8304 HI. 2.2185 L. 68.0494 St. 13274.16 A. 250 cu. in. ANSWFBS. 367 If eextauts; 2^; J sex. 47°. s. 12 . 33 yr. 8 mo. 30 da. 4 doz. ; 165 doz. 264 doz.; 2280 doz. 9| doz.; 327| loz. ; 141} doz. 3320 sheets. 333. I.rt. 427. 5736 far. B83.745. E.0375. 34000 mills. . £.002916|. .015. 128. 6d. 16s. 9d. 2.4 far. £1713 88. lid. S457.024. £89 Us. 8id. $10461.564. $1172.0965} ; 4914 marks 45.08+ pfennig; 6073 fr. 3 ct. 8.9+ m. .417375. .4632. 647 f r. 6 dc. 6 ct. 8.3+ m. Art. 440. 38.1024 Kg. 33.5627 Ton. 33.8304 HI. 2.2185 L. 68.0494 St. 13274.16 A. 250 cu. in. 8. 38 sq. yd. 9. 7.7353+ bu. 10. 652960 gr. n. 36.8965+ od. J2. 1176.15+ cu.ft IS. $.024624. U. $1.41975. 15. $14.5675 + . 16. $1,357 + . 17. 8340 dg. 83.4 Dg. 18. 8400 L. 840000 cl. 19. 790.75+ mi. eo. $960. ei. $74.00048 + . Art. 449. 7. 153 sq. ft. 8' 9" 3'" 8"" 8. 212 sq. ft. 9' 11" 2'" S"". 9. 146 sq. ft. 2' 1" W" S"". 10. 376 sq. ft. 2' 2' 8'" 6"". 11. 236 sq. ft. 9' 11" 2'" 9"". ' 12. 915 sq. ft. 8". IS. 156 cu. ft. r 7" 7'" 14' 216 cu. ft. '8' 11" 2'" 9"". Art. 466. i. 8400. 2. 76000. 3. 573. 4. 38097. 5. 897.52. 6. 30084. 7. 3426000. \8. 720000000. 5.463000000. Art. 457. 1. 300800. 2. 18018. S. 350000. 4. 1673800. 5. 24473.6. 6. 11500000. 7. 27360000. 8. 414000. 9. 3141. Art. 458. 1. 6628122. 2. 385605. 3. 87546366. 4. 53111520. 5. 839516040. 6. 2585632. I 7. 7349272. I 8. 62767170. 9. 4388206. 4. .013. 5. .008. 6'. .0004. 7. .674*. 8. 2.7041*. 9. .013921. Art. 461. Art. 453. 1. 10 min. 45| sec. 5. 3 hr. 58 min. 1 sec 3. 54 min. 16Y*y sec. 4- 2 hr. 6 min. 59 sec. 6. 10hr.53min.65|j sec. 6. 6 hr. 40 min. 10 sec. 7. 107° 19' 48 j". 8. 8hp.38min.24wc 9. 90° 2' 30". 10. 43° 45'. Art. 459. 1. 87.36. 2. 43.72. 3. 7.903. 4. .02397. 5. .05236. 6. .0006934. 7. .0054. 8. .00007. 9. .0072. Art. 460. /. 183.8. 2. 6.54625. 3. .1098i. Art. 463. 1. 103. 2. 30. 3. 28. 4' 25. 5. 112. 6. u. 7. 50. 8. 80. 9. 158f. Art. 468. '. 15300. 2. 111875. 3. 596| 4. 48700. 5. 78000. e. 9655|. 7. 8000. 8. 14725. 9. 1090. 10. 14.6. /e. $980; $1470; $735, 17. $425 ; $550 ; $612.50; $5334. 18. $600; 1315 ; $910; $700; 19. $630; $675; $650. 20. *1112; $1251 : $1042.50; $973. 21. $112; $128: $149,331. 22. $1110. 23. $371.25 ; $309.37^; $433.12 i; 24. $522; ■ $543.75; $606; $761.25. 25. $548. $657.60, $1315.20; $1534.40. 26. $255; $510; $408. ■ \ ■ f * r * '1 ■ ■ ■»' rii,'l:l a ;i Ti H '• 368 Art. 469. i. 217.74. ^. 6025. S. 1952.8. ^. 2619. J. 8.1584. €. 1.9708. 7. 11.388. S. 1.8434. 5. 3.62552. iO. 219.288. 15. 358 16 A. 179.08 A. 268.62 A. 8.954 A. ; 71.632 A.; 537.24 A. ; 35.816 A. 16. 6083.2 bu. ; 2541.6 bu. 17. 7926 yd. ; 18494 yd. ; 10568 yd. ; 792.6 vd. ; 1056.8 yd. ; 2377.8 yd. ; 1849.4 yd. 18. $1725 ; $1971f ; $2300; $3450; $1863 ; $2116 ; $2380.50. Art. 483. 10. $14444^. 11. $87. 12. $56. IS. $35.57|. U. $5.68U. 15. $17,791. 16. $4656.88f . 17. $143.46f . 18. $52.66|. 19. $19.1 ItV. SO. $940.95. ANSWERS, o. Art. 484. 6. $78.66 + . 7. $1.4826. 8. $.3811. d. $4,375. 10. $15.50. U. $.2758 + . 12. $2.75. Art. 485. 1. $4.05. $7812 ; $12361U. $668.25 ; $7G7.88 ; $816.48. I^. $1163.35. 5. $465.75 ; $465.75. 6. $3.4546. 7. $1346.2295. 8. $273.12525. 9. $379.64024. Art. 486. 5. 51 rd. 6. 626 bu. ; 516H bu. ; 630 bu. ; 5341^ bu. ; 468y^ bu, 7. 274 yd. 8. 476 bu. 9. 285 lb. 10. 395 yd. 11. 217 A. Art. 487. 1. 30 yd. 2. 96 ft. 3. 38| bu. ; 40 bu. 4. 37T.; 5. 483yd. * 6. 9 ft. 7. 53 od. ^.«9. Art. 488. 6. $1886. 7. $685.50. *. $847. 09|. 9. $5075. iO. $120.25. 11. $40455. Art. 489. ^- * ; f ; ^ f 4. I- 5. 18f. 10. -j^ yr. ; \l1 898 ^4. 1 ; f i5. }|, or .68. Art. 490. ;?. 20.4; 77.76; 604. ^. 42 ; 84 ; 329 ; 905|. -^$60; $641.36f : 5. 22.96. 6. 78i A. ; 82 bu. 7. $58.88; 15.80 lb. 8. 25.02 lb. 9. 15 men. 10. 1.96 yd. 11. 13.251 mi. 1^. 91.50. 13. $6.45. 14. $837.3. 15. $347.75. 16. 120.7 yd. 17. 244 lb. 15. 269.86 A. 19. 84 ct. ;?0. $1173 ; $4 26 per yd. Art. 491. 1. 108. ;?. 324 yd. 3. $45 ; $63. 4. $1323. 5. $23680. Art. 492. 7.200. 5.400. 9. 1200. 10. 800. 11. 2400 yd. 12. 700 bu. 1^. 20. 14. $9H. 15. 6U pk. 16. 14 ft. 17. 2k 15.4. 15>. 42|. ^6?. $36. 21. 874. ^^.550. ^^- 10 yd. 2k. 8. ^5. 3100 lb. 26. $1.50. ^7. $35500. 28. $75. J?5. 1. .ro! $412500. 31. $75. 5iF. $1350. 33. $83381. ^4. 300 A. 35. $8000. Art. 494. 7. 20%. 8. 16|^. r. 24i lb. ?. 259.86 A. h 84 ct. ?. $1173 ; (4 25 per yd. Art. 491. 1. 108. 2. 324 yd. 9. |45 ; $63. ^. $1323. 5. $23580. Art. 492. 7.200. S. 400. 9. 1200. 9. 800. 1. 2400 yd. 2. 700 bu. S. 20. 4. $9H. 7.2\. ?. 4. ^42|. ^ $36. m ANSWERS, 10 yd. 8. 3100 lb. $1.50. $35500. $75. 1. $412500. $75. $1350. $8338i. 300 A. $8000. Lrt. 494. 20%. 16|^. 9, 716 fe' 10. iii%. 11. 16}%. 12. 29}%. 13. 85f % ; 55%; 10?%. U- 95 15. 44 26. 42U%. i7. 261%. 18. liH%. i9. 14||^%. ^0. 39i|% ^/. 8%. J?^. 76A%. 23. 6l||%. ;?4. 10%. 25. 331%. ^e. 13}%. ^7. 6im%. 28. 15|%. Art. 496. . 1 . Art. 668. i. 46A. ;?. 150 da. ^. f 597. ^. $2386.40. J. $2100. Art. 670. 1. $60. ;^. 104 A. 3. 130 da. 4. $678}f • 5. 177 cd. 3 cd. ft. €. 415 lines. 7. $2000. S. $13.50. ^. 17gal.3qt.lpt. Art. 673. ^. $72. -*. $27. ANSWBJtS, 5. $100. 6-. $204. 7. 2284 yd. 8. 498i. Art. 683. 7. 6. e. 35. 5. 272. 4. 10 yd. 5. 29i bu. 6'. £168 158. 6|d. 1. 56. ^. 15. ^. 21. 4. $7f . 5. 8 cwt. 6. 213i. Art. 685. 4- $9. 5. 195 bu. 6'. 803 ft. 7. 70 bu. 5. 183.655. 10. $3000. ii. $7. i^. 16 mi. 109 rd. IS. $2.40. 14. 169 gal. 3 qt. 1 pt. 15. $.98-ft-. 16. $375. i7. 2 yr. 10 mo. Art. 686. ^. V. i. 48. 2. 23^. Art. 688. i. 120 cd. 2. 264. 5. $18. 4. 180 bu. 5. $116760. 6. 1728 ft. 7. 12 lb. 8. 23040 yd. Art. 694. 1. $405,125. ^. $992. S. $3222.26}. 4. $2400 for 4 mo. J. $800 for 4 mo. Art. 606. 1. s. $2382.545, A'b share ; $1737.272, B's share ; $1340.181, C's share. $1661.538, A's share ; $1107.692, B's share ; $1550.769, C's share. 3. $2814.128, A's share ; $1644.112, B's share ; $8431.758, C's share. 4. $900, 1st district ; $700. 2d district ; 600, 3d district; 400, 4th district. 1178.947. ^tna ; $1547.363, Home ; $2394.736, Mu. tual. 4. Art. 698. 1. $70,451, A's share. $39 .629, B's share; $36,419, C's share. w^ ■-„.( 8. Obu. 16760. 28 ft. lb. 040 yd. .rt. 694. ^.126. 192. !222.26f. 1400 for 4 mo. iOO for 4 mo. .rt. 606. 1382.545, 'b share; 737.272, 's share ; 340.181, 's share. 661.638, 's share ; 107.692, s share ; 50.769, share. 14.128, share; .112. share ; 1.768, share. 1st district; 2d district ; jO, 3d district ; , 4th district. 8.947, iBtna ; 7.363, Home ; 4.736, Mu. ). 698. ANSWERS, n, A's re. 829, B'b share; 119, C'b shure. S. f:!0, A's share; $144, B's share. 0. $1940, A's stock ; 13510, B's stock ; $7150, C's stock. 4. $8666.06 + , A's share ; $5288.98 + , B's share. . 6. $385.10 + , A's profit ; $288.56 + , B's profit ; $251.32 + , C's profit. Art. TOO. 5. $1.00. 5. $.13^. 4. $-31}. 6. 18} carats. Art. 705. 1. 3 gal. of mo. to 2 gal. of water. 2. 1, 3, 2, 1 lb. 5. 1, 2, and 6 bbl. 4. 1 part of each. 6. 2, 1, and 109 gal. 7. 2, 2, 604, and 240 bbl. 8. 50, 50, 5, and 1 sheep. 10. 30, 30, and 180 oz. il. 18, 27, 27, and 631b. i^. 44t,89|, and89if lb. Art. 711. e. 86436 : 148996 ; 247009; 64009. 5.2804; 4225; 186624 ^* TijsJ Fnr » TTnnF » .512. 5. 250047; 15625: 438976 : 60286288. 6. 4100625. 7. 66169. 8. 9. r*fJir- 10. .039304. 11. .00390625. 12. m- 13. .00028561. 14. .00091204. 16. .000000166375. Art. 712. 2. 7. 3. 7. 4. 9. 6. 15. 6. 18. 7. 12. 9. 12. 10. 30. 11. 24. 12. 12. Art. 728. 1. 59. 2. 64. 3. 87. 4- 53. 5, 96. 6. 93. 5'. I*. S. If. 9. .15. m. u. 11. .76. 12. .48. 13. 371. 14> 64.5. 15. 876.6. 16. 167.4 17. 7.56. 18. 21.79. 19. 5.656 + . 20. 7.681 + . 375 SI. 2.646+. 22. .964 + . 23. .894 + . 24- .612+. 25. 3.834 + . 26. 9.284 + . 27. 2.443 + . 28. .881 + . 29. .404 + . 30. .346 + ' 31. 51. 32. 62. 33. 88*. 34' 2656. 36. 6. 36. 354906 + . 37. 95 ft. 38. 69.57+ yd. 39. 487 ft. 40. 44842+ rd. 41. 96 trees. 42. 1.4142; 2.2860; 3.3166. 43. .654; .852; .785, Art. 736. m flt m / f^j^(i I ■ IV' 374 i5. 1.442+ ; 1.913+ ; .798+ ; .8414 ; .208+ ; 1.272 +. 19. 4. m 9. Sfl. 87. ;?^. 76.86 + . 183. 8. ;?i 439 ft. 25. 2730JI sq. ft. ;?6. 78.3+ inches. £7. 196.9+ inches. 2S. 32 feet. ;?P. 26.9+ feet. SO. 436 feet. Art. 748. 1. 18 in. ;?. 26. S. 70 ; 432. 4. 10. 5. 12. 6. $18. 7. $97.60. 5. 600500. 9. 144. iO. 12 days ; 886 mi. Art. 752. 1. 2048. 2. 49152. 5. 2186. 4. 7. 5. 815 mi. 6. 393213. 7. 8; 4372. 8. 189 mi. ; 6. Art. 758. 1. $1905. S. $1410. 5. $1775.9772. 4. $2485.714+. ANSWERS. 5. $5288.88 + . 6. $491.73 + . 7. $2380.59 + . 8. $8944.226. 9. 6%. 10. 7%. ii. $2500. 12. $500. i^. 3 years. Art. 782. i. 3 rd. ch. ^. or^ sq i?. 53l| sq. ft. S. 112.292 sq 4. 168 sq. ch. 5. 2160 stones ; $367.50. 6. 510 sq. ft. Art. 783. 1. 22 rd. ^. 10 ft. 3. 16 ft. 8 in. 4. 6 yd. 6. 42 rd. 6. 4 rd. 4 yd. 1 ft. 9 in. 7. 20 ft. Art. 784, 1. 612.87 sq. in. ;^. 150 sq. ft. 3. .924+ A. 4. 692.82+ sq. ft. 5. $292.68 + . Art. 785. 1. 39 ft. 2. 43.08+ ft. 3. 128.80+ ft. 4. 1414+ rd. 5. 187.45+ rd. Art. 786. 1. 40.31 + ft. S. 37.08+ ft. 3. 2.10+ ft. 4. 18 ft. ^. 42 ft. Art. 787. i. 7 A. 8 sq. ch. 15 P. 125 sq. 1. 2. 28 sq. ft. 108 sq. in. ; 6 sq. yd. 6 sq. ft 138 sq. in. 3. 29 sq. ft. 4. 2805 sq. ft. Art. 788. 1. 57 sq. in. 2. 220 sq. ft 3. $161.25. 4. 44 sq. ft. Art. 789. 1. 104 sq. ft. 2. 450 sq. in. 3. 11 A. 4. 800 sq. ft. Art. 790. 1. 8.9824 in.; 66.5488 in. 2. 30 in ; 25 ft. 3. 596904000 mi. 4. $45,239 + . Art. 791. 1. 14313.915 sq.ft.; 3911.085 sq. ft; $17.55+. 2. 314.16 sq. ft ; 1385.4456 sq. in.; 1963.50 sq. ft. 3. 79.5727+ A. Art. 792. 1. 48.9824 ft 2. 8 ft. 3. 5rd. . 2.10+ ft. . 18 ft. . 42 ft. Art. 787. . 7 A. 8 sq. ch. 16 P. 125 sq. 1. 28 sq. ft. 108 eq. in. ; 6 sq. yd. 6 sq. ft 188 sq. in. 29 sq. ft. 2805 sq. ft. Art. 788. 67 sq. in. 220 sq. ft $161.26. 44 sq. ft. Art. 789. 104 sq. ft. 450 sq. in. 11 A. 300 sq. ft. Art. 790. 3.9824 in.; 66.5488 in. 30 in ; 26 ft. 596904000 mi. 146.239 + . Art. 791. 1431.3.915 sq.ft.; }911.086sq. ft; M7.554. J14.16sq. ft; [385.4456 sq. in.: L963.50 sq. ft. r9.5727+ A. Vrt. 792. 3.9824 ft ft rd. -#. 87.6992 in. S. 16 in. G. 62.882 rd. Art. 802. i 36,4710 489069 502072 484149 614932 E, F,G,U,I. 301125 ai7139 3iX)680 431760 441629 449362 F,G, H, I, J. 511291 571434 . B, C, D, E. C,D,E,F. D, E, F,G. E,F, G,H. F, G, H, I. G, H, I, J. 1 11735 6942 23565 21414 11080 11562 42710 2 8310 15408 68:>00 31518 10152 43064 23010 3 23730 57384 45184 45402 43776 43240 5a37t 4 27880 46128 46008 60183 41-^74 a5180 55713 5 43875 8758 68274 mU5 69568 41784 86823 6 27476 26073 48433 73560 17622 47930 17607 7 67872 24215 759a3 35024 227:34 631«50 44766 8 55848 58710 62872 77382 4V,m 85)472 78165 9 67179 47808 875)12 40134 48279 7175)2 29247 10 894-^8 60046 46312 39480 5:38C2 88712 61047 11 60752 53523 87912 33495 70704 59948 45192 [. ;E,F,o, B,I,J. 11520;} 232651 201473 1816S8 673778 1830&4 219732 QIOM 2070:M Illi:i4 16232!> tier one. 3, H, I. H, I, J. 2562 2710 30&4 5010 3240 4;i74 3180 6713 5784 5823 2930 2607 7160 4705 8^172 616.") 7793 2247 2712 7047 3W8 5192 n,l. )62 G, H, I, J. 42710 m 23010 MO 58:nt ;80 55713 m 8e82;i )30 17607 m 44765 72 78165 t>2 29247 12 61047 »48 45192 ANS WERS. Multiplicand six figures, multiplier one. 379 A,B,C,D,B,P. B, C,D,E,F,G. c,D,E,r,o,H. D, E,r,Q,H,I. E,P,G, U,I,J. 1 117:35«)5 1041414 3771080 2141562 692710 2 1108200 2911518 3410152 4io;jow 1523010 3 3165184 6()9!>402 5083776 3243240 4378374 4 3;«60(» 51i>0183 4601274 2675180 4815713 5 4388274 4379M5 6069568 a52l7»4 7820823 6 48()»433 6953560 6227622 4597930 587607 7 7635!Ki3 3875024 50fi27;i4 3503160 18947G5 8 &382872 8807382 5501fM)2 JM3JM72 53881«i5 9 8<)37912 8586134 8838279 6151792 2069247 10 7886312 4281>480 ^173802 3158712 8071017 11 3037912 416^195 8530704 3349948 6285192 Art. 89, Multiplicand five figures, multiplier three. A, B, C, D, E. B, C, D, B, F. C, D, E, F, O. D, E, F, 0,H. E, F, G, H, I. 1 19997292 18224326 11925914 4M05130 5306082 2 7812528 24964200 41432958 4X392832 lftl48184 3 30300024 6.*«92864 38805882 57015456 38675160 4 42270628 3379;i448 66ftll003 46&47784 66318380 5 33691778 801915:34 147M455 5«i230768 5(H)60904 6 57!)06513 379a360;i 26155710 72554862 1752(M70 t 6(5601755 416073S3 50158044 43175954 82892860 8 778701()0 75155712 541.54022 77129442 5a311432 9 5.>474128 47156952 87529»14 743593W 4()7frl372 10 93a5;«(» 41077142 4544*360 67595752 50578:392 11 52327483 53172:332 89761395 2992:3024 62615508 Multiplicand six figures, multiplier five. A, B, C, D, E, F. B, C, D, E, F, G. C, D, B, F, O, H. D, E, F, O, H, I. 1 9042318325 29574555914 24704210975 18119756082 2 13^9639200 21761400958 552<»90248:32 26558882184 3 22816228864 73a50041882 3777753t>456 44623739160 4 48843909448 4;3752(}()6003 450127297t^ 58161088880 5 42:3801875»4 60605755455 69768171768 11496276904 6 S3272295603 7aS109:35710 3OI001 48862 ^484771 1470 T 8302719<5383 38071142044 72554010954 2<)215897860 8 47700798174 95598232758142 81315762360 27707044752 62040262392 11 65994081332 41008041395 &4813207024 45360431506 J 1 "t; t ,1^1 Si ■^il nil 380 ANSWERS. Art. 111. Dividend three figures, diioiaoi' one. A, B, C. B, 0, D. 0, D, E. D, K, P. E, F, O. F, G, H. O, H,I. H, I, J. 1 56 96 173t 863 168» 55? 243' 92 2 53 41 51| 159J 41S 107f 183 82A 3 OS 154 82 66} 127? 9t>? 96J 226S 4 288 192 111 283 68 1603 at? 28 5 138 85} 20!)J 41 S 126:1 »)!» 76J 424 6 53 88 70? 185i 89 84i 184 i 48| 1131 7 159 403 329 97,^ 121? 165 119; 8 143 105? m im 42; 169J 82 103 9 1«)7 134^ 143 363 2S4J 43 439i 113? 10 237 242 Ml 15Hi w» 76? asj 155i 11 74 232 40? 171?, 1981 242 114 141.] Dividend five fig wes, divisor one. A,B,C,]),B. B, 0, D, E,F. C,D,E,F,0. D, B, r,«,H. E,r,a,n,i. F, G, H, I, J. 1 3373J 858611 144918 7721 g 28748} 6217 3 4718J 61C«i 51524 ;)017? 12516i 12582;; 3 18082 13209 49271 11596J 7096S 12893 4 9614 25616J 8568 14160.1 7069? 10(5845 5 192091 7597;-! 13968.1 18760 150761 2(5924 6 (KWO? 4985 J 16422J 11584^ 6684-J 8428J 7 31829 6097,^ 9407 l!M98i 106191 99131 . 8 10659i- 13184'i 3012S 12769J mb^ 94863 9 13143 11786 J 107341 3265-I 4(i9ii0] 5M1{^ 10 10538?, 16158?, 10694J 6790} 887n 7828. 11 8467? 18571 J 9582 21492 9SM7;i 161411 Art. 121. Dividend four figures, divisor two. A, B, 0,D. B, 0, D, B. 0, D, B^. D, B, F, G. E, F, G, H. F, G, n, I. O, H, I, J. 1 45U 91H 14952 80-5 137* J 58SI 202|S 152?! ft 49i',1 ffi^S 4i;i 180A ^58^^? 103? S 3 85J* 14635 mi 61 '^1 llfiJJ 78*5 932 S 4i awA lOTt-H 108.^3 223?~J 5Tl';i' 143.U 79}J 5 112jf2 8i;i 170U as,"* 121;:? 216/, 74 6 51(8 e7i* 65? J 17411 70>.i aQjis 153J T 1291t 39Ji' 263,^:t 87?? 116fJ 130J? 1121 1 8 ISlJi mn 32r.' 171,'s 40J? 151?!- 71? 9' 146 12913 112?^ 34H I9in 42JI 325H • lO 225? 5 167,*. 9US 125^, 67^1 372 i 11 TOa 1971? 38,-j 168 J S i75i: 210J3 103S w one. G, U, I. H, I, J. aWl 92 i83 82* 96J 226i a>? 28 76? 424 184i 48i 119; 113J 82 103 439i 113? IB* 155i 114 141:; ANSWERS, Dividend six figures, divisor three. 381 A,B,C,D,E,F. B, C,D, E,F,Q. C,D,E,F,G,II. D,E,r,o,n,i. E, F, O, H, I,J. 1 222HI 289l:3 lOOOj-H 1001 „' J? looi::; 2 611.12? 933AV 655i?J 510^33 3 616,Vt 2412 j?| 826?SI 1774i!|5 8-145 J i 1295iJ: 4 1!M5:JS law.; 51 w>:.i? 730.U*; o 905hV's 1378: ;.i 870»n 600;! Sj 2413ij; 2991 ii; 3152il 6 113:i,,V. 331 rVs 915iS| 7 687*13 ISTOj^.-V 978;;¥S m^At ^206^,?? 8 1460^", 62iai 553111 mtn r,76?«s 9 saira 2455ii| 516iU 504} U 597U5 510'is loi2«;f 10 32374VS 2190,St 930} IS 11 1248,V, 1283yy, 535? il 24&4,^j'k 1280J-I Art. 194. Examples with two numbers. »,n,i. F,G,H,I,J. 48} 6217 16', 12582.^, W>a 12893 m 10(584; T»)1 2(i924 ^i 8428J 'JJ 9913? 5:? 94862 !9» 5541? u 7822 7* • 11 16141]; lor two. U n, I. a, H, I, J. e;j 202IS 152?,| w?S 81 r, 932S 3.H 79,^J 6/, 74 )j$ 153J )iS 112J us 71,! 325^ 37S m 103^ 1 6 2 22 88 22 7 8 56 16 48 168 2 6 6 18 6 42 8 9 9 99 63 27 3 36 132 24 20 44 9 70 10 60 60 «» 4 60 40 20 35 5 10 44 44 44 44 132 5 132 132 66 66 66 11 24 24 W 84 36 6 3r 105 12 3 45 1 12 156 26 26 91 39 Art. 195. Examples with two nurnbers. 1 6 6 2 2 4 18 T 6 35 1 8 21 6 2 12 6 24 6 4 18 8 6 85 3 3 9 18 3 12 12 8 60 5 3 9 6 10 2A 4 60 6 4 12 24 4 42 1 3 lO 12 20 24 28 12 12 5 4 3 12 6 9 3 11 12 ^ 2 28 21 27 6 4 35 28 12 21 15 Art. 196. Examples with two numbers. 1 6 36 2S 16 44 im 6 2 105 3 3 68 9 2 12 30 8 6 7 3 7 6 TO 16 12 48 210 3 \% 132 las 6 22 2ftl 8 66 5 ;« 77 8 66 4 20 15 20 12 35 195 9 42 10 24 60 m 90 o 12 8 4 6 6 24 1 lo 12 4 2 28 3 8 |J*.'' m 382 ANS WE Its, m 11^ it I Art. 197, Examples with three numbers. 1 2 2 22 22 T 8 8 16 24 2 6 6 6 6 8 9 9 9 9 3 12 12 4 4 9 10 10 60 ao 4 20 20 5 5 10 44 44 44 44 •■> 132 66 66 66 11 24 12 M 12 6 36 3 3 3 12 26 26 13 13 Art. 198. Examples with three numbers. 1 6 6 2 2 4 18 6 2 35 1 8 21 3 2 12 6 8 6 1 3 T 6 35 1 3 3 6 3 12 12 4 6 1 3 8 6 6 3 1 3 6 4 4 3 4 6 1 3 9 6 10 24 4 12 6 6 4 1 4 6 3 3 10 12 4 2 28 3 3 ^'1 f ■ Art. 211. Examples with two numbers, according to 194. 1 190 2772 1232 176 792 3 1680 2520 216 756 252 3 2772 792 5280 5280 3960 4 120 120 840 2100 6825 5 264 792 2772 4158 4752 6 420 420 5460 16380 4095 T 3360 560 1341 1344 1680 8 6930 10395 2079 6237 12474 9 1050 840 1200 3600 5010 10 660 1320 1408 4620 2640 11 3360 10080 2520 1260 3780 la 312 1560 1820 1092 2457 Art. 211. Exa/mples with two numbers, accmding to 195. 1 240 1260 5544 4752 1848 1386 2 1008 13860 792 4320 4620 5544 3 1260 39()0 1320 3360 7700 51480 4 660 1320 3960 4620 103950 34320 5 4620 9240 12860 11012 20790 102960 6 8360 740 1680 2496 4368 32760 7 3168 2520 33»}4 44352 9072 83160 8 6980 3150 237H0 69300 45360 20790 9 4620 770 2640 23100 7920 55440 10 1848 5280 5544 4620 18860 23760 11 2184 6210 32760 5160 3276 7020 her9. mbers. 16 a4 9 9 60 au 44 44 M 1« 13 " 1 1 3 24 2 8 21 3 8 3 6 1 3 6 4 12 B 28 3 8 ^•ding to 194:. sioo H58 1260 1092 792 252 3960 6825 4752 4095 1680 12474 5040 2640 3780 2457 A XSWERS. 383 Art. 211. Examples with two numbers, according to 15^C5. 1 1280 396 1848 5280 440 798 2 240 840 360 3780 2100 8190 3 2772 792 792 3(i'.KiO 5940 15M 4 420 »K) &10 5-160 1575 585 5 1056 8140 11088 147M »]2W 18480 6 13HtiO 815 41580 36036 28;J5 38610 7 3:J60 1400 1680 4800 5040 2520 8 306 13860 2376 2772 124740 4752 9 840 16800 5040 2100 5040 3780 10 1716 17160 17160 4004 60060 61776 Art. 211. Examples with three numbers, according to 1?>7. 1 140 210 84 48 48 2 90 90 90 72 72 3 1512 4;« 48 48 168 4 1440 1440 910 14256 (;;)36 7 594 6930 39()0 41580 55440 7128 792 8 2376 18480 41580 462C 5280 792 2376 q '^ '■;•! •1 *?l '?' a^ /.. , 3 tj. ■ 'i \'^'^ ■• I '^fcf A.\ •