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MACKINLAY. 1871. 1 i Lntered, according to Act of Parliament of Canada, in the year 1871 By a. & W. Mackinlay, ' In the office of the Minister of Agriculture, at Ottawa. m "KOVA SCOTIA PRINTING COMPANY," CORNER SACKVILLE AND GRANVILLE STREETS, HALIFAX, N. 8. Mi # i V n, PBEFACE. The " Elementary Arithmetic " is intended to occupy an intermediate position, coming between the concrete and the advanced stages, and is adapted for the junior classes in our common schools, for securing the mental development, as well as the accuracy and expedition in calculation of the pupils between seven and eleven years of age. The plan consists of such a delineation of the principles that the pupils are enabled, by induction, to form the appro- priate rules. After the accuracy of their knowledge is tested by a few mental exercises, the examples are reduced to practice on the bla'^kboard or slate. A number of self-testing exercises to many of the rules are introduced, which avIU save the teacher much labour, and be of benefit to the pupils. The definitions and tables have been interspersed through the work, thereby rendering them more available to the student. The plan jDursued in the rule of P'-actice, is, we think, well calculated to exercise the reflective powers of the young, the examples and illustrations having been carefully selected, rising from the easy to the more difficult. After Practice, Proportion is introduced, in a way not usually found in works of the kind ; and several operations generally included under Interest and other rules, are group- ed together, by which means the pupils are enabled to solve all questions where ratio is involved. VI. PREFACE. Under each rule will be found a large number of well graded exercises, many of which liavc been selected trom real occurrences in business. The compiler has availed himself of the best works in the New and the Old World, viz., Dr. llobinson's, edited by Fish, Dr. Thomson's, Greenleaf 's, Barnard Smith's, Currie's, Hay's and othei-s, but especially that of Dr. Robinson. Note.— Tn this Work, £ 3. d. mean Sterling Money; $ and cfs. mean Canada Currency. m m ell L'al is. ••"V 4. THE ELEMENTARY ARITHMETIC. DEFINITIONS. !• Anything which can be multiplied, divided or measureu is called Quantity. Thus, lines, weight, time, number, &c., are quantities. S. Aritliinctic is the science of number, and teaches how to represent numbers by symbols or signs, and the various methods of using these in calculation. 3. ]Viiinbei*s are expressions for one or more units. Thus, the words one, two, three, four, &c., or the characters 1, 2, 3, 4, &c., are expressions by which we indicate how many single things, or units, are to be taken. 4. Numbers are divided into two classes, Abstract and Concrete or denominate. If the units represented have no reference to any particular object, as when we say seven and two are nine, they are called abstract numbei-s. If the units referred to have reference to partictdar objects, as tivo days, seven men, &c., they are called concrete or denominate numbers. NOTATION AND NUMERATION. Art. 1. I¥otatioii is the writing or expressing of num- bers by characters ; and ]¥iiincratioii is the reading of numbers expressed by characters. a. Two systems of notation are in general use — the Roman ^n^ i\iQ Arahict 8 NOTATION AND NUMERATION. The Roman IVotalioii «. Employs seven capital lettc^rs to express numbers, ill us, Letters— I V X L C D M Values— (;ne, five, ten, fifUj, one five one hundred, hundred, thousand. Bycoml)Ining these letters, the ancient Komans] formed the tollowin The nought or cipher is also called nothinq or zero. The cipher has, of itself, no value, but is used to indicate the order ot the significant figures which precede it. The ten Arabic characters are the Alphabet of Arithmetic- and bv TxpreS '^'"^ "^'"'^^"^ '' certain Unciples, all numbei-s ca. bl u o t ai 2 m ol 1^ ti NOTATION AND NUMERATION. 9 5. To facilitate tlie readincr of laro^o numhors thov are divided into periods of three figures each, begimiing at the right-liaud side, according to the following IVuiiieratlon Table. ( i-* Units, Period I. Units -| to Tens, ( CO Hundreds. ( >p- ITnits of Thousands, " II. Thousands -! ot Tens of Thousands, ( oi Hundreds of Thousands, ( -J Units of INIillions, « HI. Millions ' cc Tens of Millions, ( « Hundreds of ISIillions. r ^ Units of Billions, « IV. Billions ) :i Tens of Billions, ( '^ Hundreds of Billions. r ^ Units of Trillions, " V. Trillions ) ;^ Tens of Trillions, ( ^ Hundreds of Trillions- ^ ^ Units of Quadrillions, " VI. Quadrillions ... 3 |^^ Tens of Quadrillions, ( ^ Hundreds of Quadrillions. 0. Figures occupying different places in a number, as units, tens, hundreds, &c., are said to express different orderB of units. Simple units are called units of the/r5« order. Tens " " second " Hundreds " " third " Thousands " " fourth " and so on. Thus, 327 contains 3 units of the third order, 2 units of the second order, and 7 units of the first order. Exercises for the Slate. Write and read the following numbers : 1. One unit of the third order, four of the second. 2. Eight units of the fifth order, three of the second. 3. Two units of the seventh order, five of the sixth, three of the fourth, nine of the third, eight of the first. 10 NOTATION ANDNUMEUATION. 4. Four units oftho t(Mith order, h\x of tho (Mj^'litli, four of tlic seventh, three of the fifth, seven of the fourth, nine of tho Bocond, one of tlie fii-st order. 7. PrinolplcM of IVotafion and IViiiiioralloii. 1st. Figures have two values. Simple and Local. The Nlmpl© Value of a fi<^urc is its value when taken alone. Thus, 3, 4, 5. The I^mral Value of a fijrure is its value when used Avith another fi^nu'e or fi^^ures in the sanu^ number. Thus, in 472 the simpli! values of the several finjui-es are 4, 7, and 2 ; but the local value of the 4 is 400 ; of the 7 is 7 tens, or 70 ; and of the 2 is 2 units. Note.— When a llgure occupies the first place, its simple and local values are the same. 2nd. A dijrit or fijrure, if used in the second place, expresses tens ; in the third place, hundreds ; in the fourth place, thousands ; and so on. 3rd. As 10 units make 1 ten, 10 tens 1 hundred, 10 hundreds 1 thousand, and 10 units of any order, or in any place, make 1 unit of the next higher order, we readily sec that the Arabic form of notation is based on the followin**- GENERAL LAWS. 1 i: The diflferent orders of units increase from right to left, in a ten-fold ratio. II. Every removal of a figure one place to the left, in- creases its local value ten-fold ; and every removal of a figure one place to the right, diminishes its local value to one-tenth of its previous value. Thus, 6 is 6 units. 60 is 10 times 6 units. 600 is 10 times 6 tens. 6000 is 10 times 6 hundreds. 4th. Every period contains three fijrnres, (units, tens, and hundreds,) except the left hand period, which sometimes contains only one or two figures, (units, or units and tens.) RULE FOR NOTATION. • ^' 4. ^fP^^>^^ ^1 the left hand, write the figures belong- ing to the highest period. c^wub II. Write the hundreds, tens, and units of each suc- Sn ^'/A?^/^?^'^ '"* *^-^''' "^^4?^' placing a cipher wherever an order of figures 18 wanting. i: X t V o 1> f IC coi ADDITION. 11 t t RULE FOR NUMERATION. ^hu SoP^'-rate tho number into periods of three flaurea each, commencing at the right hand. ""^-o uKuroa »ih. 3°^^^!^^^^^^^ >*^^ ^°f* *»'^^di read off the number of SSme ot theVeriod.'" °"°*^ ^''"^^ Bcparately. and addke «. Until tlie pupil can write numlxTs readily, it may be well for him to write several periods of eiphers/point them on, and over each period write its name. 'J'hus, Trillions, Billions, Millions, Thousands, Units. 000,000,000,000,000 And then write the given numbers in their appropriate ExerclMcs for llie Nlatc. Express the following numbei-s by figures : 1. Thirty-six. 2. Three hundred and thirty-six. 3. Five thousand, three hundred and thirty-six. 4. Fourteen thousand, two hundred and forty-seven 5. Four hundred and fifty thousand, and fii'ty-nine.' 6. Ninety-six thousand and four. 7. Nine hundred thousand, and ninety. 8. Sixty-one billions, four millions, and ninety-seven. Point off, and read the following numbei-s : 9. 489 14. 3786 10. 586 15. 20900 11. 4070 16. 57631 12. 307 17. 37000 13. 10010 18. 94000554 24. Write seven millions and thirty-six. 103705^0()'f '"'"'^'''^ ""^ ""'^ '''■'' contained in the number 19. 2987654300 20. 4783006001 21. 3456789012 22. 6830428301 23. 7932643162 ADDITION. Explanatory Exei*ci!«cs. O. 1. John gave 5 dollars for a vest, and doUai-s for a coat ; how many dollars did he pay for both *? 12 ADDITION. Analysis. — Ho gave as many dollars as 5 dollars and 9 dollai-s, which are 14 dollars. 2. A farmer sold a lamb for 3 dollars, and a calf for 4 dollars ; how many dollai-s did he receive for both ? ^ 3. John got 3 apples from his mother, 2 apples from his sister, and 1 ajjple from his brother ; how manv apples did he get altogether V 4. How many are 4 and 5 ? 4 and 7 ? 3 and 6 ? 5. How many are 5 cents, 6 cents, and 7 cents ? 10. From the preceding operations we perceive that Additlou is the process of uniting several numbers into one equivalent number. 11. The Sum or Amount is the result obtained by the process of addition. Note.— Concrete ruinbers, that is numbers of objects, cannot be added tof^^cthor unless the objects are of the same kind. Thus, ^ granunars and 5 geographies cannot be added togethe •. If, however, we drop the distinctive names of the objects, and use in their stead a more general term, which will include the several kinds in one class, the addition can be performed. Thus, if we consider geographies and gr-.mmars merely as ioo^v. we raav sav 4 grammars (books) and G geographies (books) are 10 books. This principle applies to all opera- t'ons with concrete numbers. 12. The sign -{-, is called plm, which signifies more. When placed betveen tAvo numbers, it denotes that they are to be ad First row 3456 Second ' 3456 Same as first row. Third " 6912== Sum of second and first Fourth " 10368 z=z Sum of third and second. Fifth " 17280 z= Sum of fourth and third. Ans. 41472 rr: Sum of all the rows. 16 ADDITION. (^>) (10) (11) (12) (la) (14) 632781 547182 987006 875871 76 7808 6r. 171 621 531 432 135 252 801 (34) (35) (36) (37) (38) (15) (16) (!') (18) (If)) (20) (21) 6 r. 1233 4581 6543 7632 8901 9342 1899 1234584 2781099 3765789 4572171 5706018 (22) (23) (24) (25) (26) (27) (28) 6r, 10y872 234531 901827 728109 879102 512361 987642 240357897 304578927 457028973 758203434 987645312 \ 4 (I) (2) (3) (4) (5) (6) 45 extended 54 153 162 549 1089 8r. 8 r, 6 6 5 4 SHOW THAT = 18 extended 8 r. » 36 '* = 90 " = 72 " = 261 " ' 531 " 8 r. 6 r. 6 r. 5 r. 4 r. 4- 27 -f 18 + 63 -j- 90 + 288 -f 558 extended 8 r. 8r. 6r. 6r. 5r, 4r. SECTION ly. \'-J''"^ *^^^ sum of 1247 + 91679 + 27-f- 1987 -f 1800 o wi . Ans. 98536. 2. What IS the sum of 250120 -4- 30402 -4- 7850 -L '%'''± '^1 + %045. ^ its. '^2846t 408; sftVaft' ' '''' '''' '''' '''' '''ir'7^2? 9ri'7,(!f^ ^?%f}^L^^^^^^ 250763, 7561, 830654,"293fo6; 2.37104, and 316725. Ans. 4251988. o. innd the sum of 629405, 7629, 31000401 263019 1300512, 390217, and 13268. ' ''Tn^Ss'oLlI' Af^t'' rT"" ^""""^ ^^^^ "^^^^^^'^ *« ^'^ elflest son, to the next 406O, to the next 6750 to the next 8000, and to the younges 7276 , how much did he give to all. Ans. 31551 dollai-s. 7. A merchant on settling up his business, found he owed one creditor 176 dollars, another 841 dolla^, another 1356 dollars, another 2370 dollars, another 840 dollars; what wa. the amount of his debts? a^.^ nr^oo ^^i « y\..A +1,^ „ r.; ^ ,, . , ^"^- •^•^83 dollars. «r,d"fiftv"^,V r 'T "'/ 1 i^^iowmg numbers ; seven hundred and fifty-wx, four hundred and twenty-five, six hundred and ? 4 SUBTRACTION. 17 t iirty-three, five Imiulrcd and forty-one, nine hundred and sixty-nmo . . . ^ Ans.3324. .. Add to;jethcr six, sixty-five, six hundred and fifty-five, three housand six hundred and fifty-fi^ e, twenty-six thousand thiee liundred and hfty-nine. Ans. 30740. 10 A man willed his estate to his wife, two sons and four daughters. To his dau.:,hters he gave 2030 ^ P \ 24 MULTIPLICATION. Verify the following — (3) 47 X 2 = 10 X 2 + 28 X 2 (4) 59 X 2 = 27 X 2 + 32X 2 (5) 75 X 2 = 49 X 2 + 2« X 2 (G) 124 X 2 = 50 X 2 + 68 X 2 (7) 309 X 2 = 24G X 2 + 123 X 2 [,) 003 X 2 = 431 X 2 + 232 X 2 (9) 984 X 2 = 015 X 2 + 309 X 2 (10) 190 X 2 = 94 X 2 + 102 X 2 NoTic. — Instead of 2 as multiplier take successively 3, 4, 5, 0, 7, 8, 9, 10, 11, and 12 as multipliers, using the exercises in the section. — Thus, (10) 196 X 9 = 94 X 9 + 102 X 9, &c. 11. What will be the cost of 344 cords of wood at 4 dol- lars a cord '? Ans. 1376 dollars. 12. In one day are 8G400 seconds ; how many seconds in 7 days ? Ans. G04800 seconds. 1 3. In one bushel there are 256 gills ; how many gills are there in 12 bushels? Ans. 3072 gills CASE II. 35. When the multiplier is a composite number, none of whose factors are greater than 12. »«. A Composite Nitiiiber is one that may be pro- duced by nuiltlplying together two or more numbers. Thus, 18 is a composite number, since 6X3zz;18; or9X2::=18; or 3 X 3 X 2 = 18. 37. The Component Factors of a number are the several numbers Avhich, multiplied together, produce the given number. Thus, the component factors of 16 are 4 and 4, (4 X 4 = 16); or, 8 and 2, (8 X 2 = 16) ; or, 2 and 2 and 2 and 2, (2X2X2X2== 16). Note. — The pupil must not confound the factors with the parti of a number. Ihus, i\\Q factors of Avhich 14 is composed are 7 and 2, (7X2 = 14); Avhile the parts of which 14 is composed are 8 and 6 (8 _+ = 14), or, 10 and 4, (10 + 4 = 14). The factors are multi- pUed, while the pa7'ts are added. Example 2. — "VYhat will 36 cows cost, at 196 dollars each ? Multiplicand 196 cost of 1 cow. 1st factor 4 2nd factor Product 784 cost of 4 cows. 9 7056 cost of 36 cows. Analyst s. — The factors of 36 are 4 and 9. If Ave multiply the cost of 1 cow by 4, we obtain the cost of 4 cows I and by multi- plying the cost of 4 cows by 9, we obtain J MULTIPLICATION. 25 of i the cost of 9 times 4 cows, or 36 cows, the number boucht. iienoe we have the fbllowiiu/ m?r^^^toJs.^^^*'"**® *^® composite number into two or and'th^t^'nliyHJ^nf^® multiplicand by one of these factors, So^i^.^ product by another, and so on until all the factors P?oduct?eSui?ed'.^°°"'''^"'^' *^^ '^«* product wiUbe'thl Find the product of (1) 123G456 X 15 (2) 23450 79 X IG (3) 40 71320 X 18 (4) 7235409 X 21 (5) 9876519 X 24 (6) 8297568 X 27 (7) 9726354 X 35 SECTION II. (8) 87645231 X 32 (9) 18765432 X 35 (10) 33236775 X ?>^ (11) 21876543 X 42 (12) 54670104 X 44 (13) 32336775 X 54 (14) 68206986 X 55 16. AVhat will 573 oxen cost, at 63 dollars each ? _ Ti. Ans. 36099 dollars. 16. \\ an army consume 1645 pounds of bread in a day, how mucli will they consume in 96 days ? ,_ TT ^'^^' 157920 pounds. 17. How many are 84 times six hundred and four thous- and, seven hundred and fifty-six? An-'. 50799504. 18. A merchant bouorht 145 pieces of broadcloth, each piece containing 48 yards, at 4 dollars a yard ; how much did the whole cost ? Ans. 27840 dollars. CASE III. as. When the multiplier consists of two or more figures. Example 3.— Multiply 646 by 29. Multiplicand 646 Multiplier 29 5814 y tiinea the multiplicand. 1292 20 times the multiplicand. Product 1 8 734 29 times the multiplicand. Analysis. — Writing the mul- tiplicand and mul- tiplier as in Case I, we first multiply each figure of the multiplicand by the unit figure of ihe multipher, exactljr as in Case I. We then multiply by the 2 tens. 2 tens times 6 units, or 6 times 2 tens, are 12 tens, equal to 1 hundred, and 2 tens; w« place the two tens 26 MULTIPLICATION. under tlic tens' place in the product already o})tainc(L 2 tens times 4 tt'iis are 8 hundreds, and 1 hundred of the last pro(hi(t adivisor is the number divided by. 45. ^ The Cluoticnt is the result obtained by the process of division and shows how many times the divisor is contained in the dividend. Note.— 1. When the dividend does not contain the divisor an exaet number of tunes, the part of the dividend left is ca led the remainder, and it must l)e less than the divisor. the 'sam: "^^^^r^^llS^. " ''^""" '''' '' '''' ^'^^^^^^"^' '' '« ^^^^^^^ ^' 3. When there is no remainder the division is said to be com2>kte. 4«. The sign, -f-, placed between two numbers, denotes division, and shows that the number on the left is to be divided by the number on the right. Thus, 39 -^ 3, is read 39 aivKicjl hi/ 3. Division is often indicated by writing the dividend above and the divisor helow a short horizontal line. Thus 4a CASE I. 47. When the dwixor does not exceed 12. Example 1.— How many times is 3 contained in 936 '? OPERATION. Dividend. Divisor 3)936 Quotient 312 Analysis— After writing the divi- sor on the left of the dividend, with a line between them, we begin at the left hand and say: 3 is contained in 9 hundreds, 3 hundreds times, and write 3 in hundreds' place in the quotient : 80 DIVISION. then 3 is contained in 3 tens 1 ten times, and write 1 in tens' place in the quotient ; then 3 is contained in G units 2 units times ; and writing the 2 in units' place in the quotient, we have the entire quotient, 312. 2. HoAV many times is 4 contained in 1684 ? Analysis. — As we cannot divide 1 tliousand by 4, we take the 1 thousand and the 6 hun- dreds together, and say, 4 is contained in 16 OPERATION. 4)1684 421 hundreds 4 hundieds times, which we write in hundreds' place in the quotient ; then 4 is con- tained in 8 tens 2 tens times, which we write in the tens' place in the quotient ; and 4 is contained in 4 units 1 unit time, which we write in the units' place in the quotient, and we have the entire quotient, 421. 3. How many times is 7 contained in 2835 ? opi:ration. Analysis. — Beginning as in the last ex- 7)2835 ample, we say, 7 is contained in 28 hundreds — — 4 hundreds times, which we write in the hun- 405 dreds' place in the quotient ; then, 7 is contained in 3 tens no times, and we write a cipher in the tens' place in the quotient ; and taking the 3 tens and 5 units together, 7 is contained in 35 units 5 units times, which we write in the units' place in the quotient, and we have the entire quotient, 405. 4. How many times is 8 contained in 987 ? operation. 8)987 123 or 123f Analysis. — Here 8 is contained in 9 hundreds 1 hundred times, and 1 hun- dred, or 10 tens, over, which, united to 3 Rem. the 8 tens, make 18 tens ; 8 in 18 tens, 2 tens times and 2 tens, or 20 units, over, which, united to the 7 units, make 27 units ; 8 in 27 units 3 units times and The 3 which is left after performing the divi- be divided by 8 ; but the method of doing reach fractions ; so we 3 units over. sion, should so cannot be explained until we merely indicate the division by placing the divisor under the dividend, thus, |. (46). The entire (Quotient is written 123f, which may be read, one hundred and twenty-three and three-eif/htJis, or one hundred and twenty-three and a remninder of three. From the foregoing examples and illutitratious, we deduce the following DIVISION. 31 in BULE. I. "Write the divisor at the left of the dividend, with a line between them. II. Beginning at the left hand, find how many times the divisor is contained in the fewest number of figures of the dividend that will contain it, and write the result under the dividend. III. If there be a remainder after dividing any figure, regard it as prefixed to the figure of the next lower order in the dividend, and divide as before. IV. Should any figure or part of the dividend be less than the divisor, write a cipher in thp quotient, and prefix the number to the figure of the next lower order m the dividend, and divide as before. V. If there be a remainder after dividing the last figure, place it over the divisor at the right hand of the quotient. Mental Exercises. 1. If 4 casks of lime cost 12 dollars, what Is the cost of 1 cask? 2. If a man perform a certain piece of work In 30 clays, how long will it take 5 men to do the same ? How long will it take 6 men ? How long will it take 7 men ? 3. If 24 pounds of tea can be purchased for 12 dollars, how much can be bought for 1 dollar ? How much for 9 dollai-s ? How much for 5 dollars ? 4. Gave OG cents for 6 pounds of raisins ; what cost 1 pound ? What cost 7 pounds ? 5. A man gave 15 dollars for 3 barrels of apples ; what was the cost of each barrel ? What would 5 barrels cost at the same rate ? Exercises for tlie Slate. SECTION I. 42240-1-2,4,6,8, 10,11 14784 — 3, 7, 11, 2,4, 8 76032-^4, 3, 2, 8, 9, 11 (4) 1209GO-I-5, 7,6,4,8 (5) 30888 -:- 9, 3, 8 (6) 13608 -^ 7, 3, 9 (7) 34668 ~ 6, 9, 3 (8) 363285 ~ 5, 9, 3 snow THAT (9) 869 -i- 3 = 24- 42 men, each man will receive as many dollars as 42 is contained in 5376 dollars. 42 may be resolved into tiie factors 6 and 7; and we may suppose the 42 men divided into six jrroups of 7 men each ; divid- ir;^ the 53 76 by 6, the number of groups, we liave 896, the number of dollars to be given to each o-roup ; and dividing 896 by 7, the number of n)en in each group, we have 128, the number of dollars that each man will receive. Hence, RULE. Divide the dividend by one of the factors, and the quotient thus obtained by another, and so on if there ^®,i^°]^^ 1^^^ two factors, until every factor has been made a divisor. The last quotient will be the quotient SECTION III. 1. 2. 3. 4. 5. 6, Dlv Div Div Div Div Div de de de de 5 by 1 5 = 3 X by 16 = 4 X 4 by 45 = 5 X i) by 56 z= 8 y 7 de 3948767388 by 108 — 3 X 4 X 9 de 3176823672 by 132 = 12 X U 98.-> 768545 68 7698464 931684770 945328G08 Quotients. 65717903. 42981154. 20704106. 16880868. 36562661. 24066846. DIVISION. aa 49, To find the true remainder. Example 2.— Divide 1143 by C4, usiu^ tlie factors 2, 8, and 4, and find the true remainder. OPERATION, 2)1143 8)571 1 rem. 4)71 3X2 = 17 3 X 8 X 2=48 55 true rem. ANALYsis.-Divid* ing 1143 by 2 wc g€t a remainder of 1 undivided, which be- ing a part of the giv- en dividend must also be a i)art of the truti remainder.' — And in dividing the first quotient by 8, we get a remainder 3, which we multiply by 2, the first divisor, to bring it to the same name, or units, as the first remainder, and in dividing by 4, we have a remainder of 3, which we multiply by 8 and 2, the preceding divisors, in order to bring it also to the same name as the first remainder. Adding the three partial re- mainders, we obtain 55, the true remainder. Hence the KUIjIj, I. Multiply each partial remainder, except the first, by all tbe preceding divisors. II. Add the several products with the first remaindert and the sum will be the true remainder, IsToTE. — For other methods see Advanced Arithmetic. SECTION IV. 1. 234567 • -^ 18 6. 751113 • 63 11. 23456781 — 216 2. 345G72 ■ -^ 27 7. 804024 4- 72 12. 83456712 -f- 225 3. 427311 ■ 4- 36 8. 887625 ^ 81 13. 40107645 -f 432 4. 453672- -^ 45 9. 999999 -f- 99 14. 57763323 — 441 5. 672345 - ^ 54 10. 723456 ~- 108 15. 68960286 -^504 SI -^TION V. 1. 958768461 —■ 27 Ans. 35509948. 2. 726894784 ' 32 (( 22715462. 3. 729368465 ' 35 (( 20839099. • 4. 675487368 ' 36 u 18763538. 5. 945328608 -j- 56 u 16880868. 6. 1796842688 • 64 u 2S075667. 7. 897684192 ' 72 IL 12467836. 8. 910364312 -j- 88 U 10345049. 9. 39187G7388 -j- 108 a odud2uo1. 10. 31768 2 23672 -^ 132; (i 2:4066846. r1 ' 4 M PIV19IOK. 24 24 CASE III. 00. To rftvicTtf hy a number consisting of several Jigvres. NoTB.— To illustrate the method of operation more cleftrly, w« will tftke »n example usually pertormed by Short Division. 1. How many times is 6 contained in 564, OPERATION. Analysis. — As & is not contained in 5 hun- e)564(94 dreds, we take 5 and 6 as one nnmber, and consider how many times 6 is contained in this partial dividend, 56 tens, and find that it is contained 9 tens times, and a remainder. To find this remainder, we multiply the divisor, 6, -— by the quotient figure, 9 tens, and subtract tbe product, 54 tens, from the partial dividend, 56 tens, and there remain 2 tens. To this remainder we bring down the 4 units, and consider the 24 units the second partial dividend. Then, 6 is contained in 24 units 4 units times. Multiplying and subtracting as before, we find that nothing remains, and we have for the entire quotient, 94. J. How many times is 23 contained in 4807 ? operation. Analysis. — We first find how Pivkor. Dividend. Quotieni. many times 25 is contained in 48> 23 ) 4807 ( 209 the least number of figures that will contain 23, and place the result in the quotient on the right of the dividend. We then multiply the divisor, 23, by the quotient figure, 2, and subtract the product, 46^ from the part of the dividend used, and to the remainder bring down the next figure of the dividend, which is 0, mak- ing 20, for the second partial dividend. Then, since 23 is contained in 20 no times, we place a cipher in the quotient, and bring down the next figure of the dividend, making a third partial dividend, 207 ; 23 is contained in 207, 9 times : multiplying and subtracting as before, nothing remains, and we have for the entire quotient, 209. Notes.— 1. When the process of dividing is performed mentally, and the results only are written, as in Case I, Uie operation is termed iSAort Division. . . . ^ ^, *• • 2. When the whole process of division is written, the operation is l«nned Long Dividon. From the preceding illustarations we derive the following general 46* 207 207 DIVISION. , 85 «.v?^' ,Pi.^i<^®.*^e least number of the left hand flgures in JJlS.?'"'''*?^? that wiU contain the divisor one or moS «^K*i*v ^ ^i!*?® *^® quotient at the right of the dirldend, with a line between them. «w«**, ♦^««?\v?^^^*^5^y x*^,® divisor by this quotient figure, sub- tract the product from the partial dividend used, and to the remainder bring down the next figure of the d/vidend. T,™ K?i^* V® ^^ ^,ff°r®» ^^**1 *11 **^« figures of the dividend have been brought down and divided. *««»"« ^iYa^"? *^^ partial dividend wiU not contain the divisor. figure of the dividend, and divide as before. ^/^t- ¥ **?f re ^e. a remainder afier dividing aU the figures of the dividend, it must be written in th© quotient, with the divisor underneath. wv*«««, w^ba Note.— 1. If any remainder be equal to, <5r ffreater than the divisor, the quotient figure is too small, and must be increased. *u J} ^^^ .P!;^?."^' of ^he divisor by the quotient figure be oreater than the partial dividend, the quotient figure is too larae, and must be diminished. " SECTION VI. (1) 79865379-r-702 (2) 81136863-T-801 (3) 90909963^-117 (4) 23659245-^126 (5) 37018764-i-135 (6) 53146827 -f-459 (7) 61327548 -f-558 (8) 128713536-r-567 (9) 123456789-4-576 (10) 987654321^585 (11) 709005474-5-882 (12) 407049570-J-918 (13) 981234567-*-891 (14) 900664200-^909f (W) 11177711 1-*-900» 1. 2. 3. 4. 5. 6. SECTION VII. Divide 556^804464 by 7346. Divide 1 74 70 7 1 255 by 6483. Divide 8287864532 by 8594. Divide 35365114332 by 98846. Divide 520090972776 by 654321. Divide 7428927415293 by 8496427. Divide 936864880704 by 987654. Ans. 756984. Ans. 269485, Ans. 964378, Ans. 376843. Ans. 794858. Ans. 874859. Ans. 948576. 8 The number of post offices in the United State, in 1853 was 22320, and the revenue of this department wa« 5937120 dollars; what was the average revenue of each ^^^^ ^ J . . ^"8- 266 doUaro. 9. A bag containing three hundred and twenty-four nuti waa divided among nine boys ; how many did each boy get ? , /v -r.. , , . Ans. 86, 10. Find the 17th part of 6508. Ans. 324. 1 1 . How many miles an hour does a train go which traveli 1692 miles in 47 hours ? Ans. 86. 12. A gentleman left £5000. By his will he dire<5t«d that after paying hia debts, amomntinf to iSfi, th« rMl If«? mrisWiV, •hould be divided equally among his seven eliildren : t,hai was the shai-e of *jach V T" ' .. r 7?. 1 Q Tl 1 ^ /. , Alls. ib/o-. Arf. Uie prodnct of two numbers is 31383450, and one of the numbers is 4050 5, what i. th« other number ? Ans. 7 749, CASE IV. M. To chviile bi/ 10, 100, 1000, &c. Example 1. -Divide 486 acres of land equally amonj. 10 men ; how many acres wdl each have '? "^ ^ l?oT4^8trr''''* *. ANALTsis.^Acjording to «he decimal sys^ i\0)48\6 tern of notation if we remove a figure one — ^ place toward the left by annexing a cipher, 48 6 rem. its value is increased ten fold, or 5 muKiplied 41. ,y. V' J'o on tlie contrary, by cutting off or taJcing away the nght "liand figure of a number, each of 'tl^ figures IS removed one place toward the right and conse^ guently reduced to one-tenth its former value, or divTled lot), if three, we divide by /OOO, and so oS. ilencc the ;^an?.* ^ "" ^''^'^^ ^^ "" ''"'''^''' ^*^'''''*^ ^'>^'^^* ^" ^^^ ^»^^»' Example 1.— Divide 587618 by 400 OPERATION. 4j00)5876|I8 1469 18 rem. Analysis.— In this example we resolve 400 into the factors, 4 and 100, and divide first by 100, by cutting off the two right hand figures of the dividend, .nd a e..,.,,, ,, W ^^^^^^ "^ttf ^ 1469 for a quotient; and the entire quotient is 1469^^ 53. When there is a remainder afler dividijx^by the ^gnificant figures, it must be prefixed to the figu^S cut off from the dividend to give the true remainder. I. 2. 3. 4. SECTION VIH. Divide 48600 by 100. Divide 596 73 by 1000. Divide 34716 by yOO Divide 1789S0 by 10. Ans. 486' Ans. 59 rem. 673 or b(\3J.i . Ans. 38 rem. 516 or 38|||. Ans. 17898. DIVISION. 37 ■ 6. Divide 47321046 by 45000. Am. 1051, rem. 26046 6. Divide 1047634 by 2400. Ans. 436, rem. 1234 Or 436i?'''i 7. Tlie sum of 40000 dollars is paid to 1600 men ; what does each man receive ? Ans. 25 dollai-s. ». llie circumference of the earth at the equator is 24898 miles. How many hours would a train of cars require to travel that distance, going at the rate of 60 miles an hour ? Ans. 41414. SECTION VIII. To one, annex as many ciphers as you please. From this subtract any number. To the two numbers thus formed prefix two figures whose sum is less than tliC proposed divisor by one, then divide by the proposed divisor. Example 1. — To 1, annex 5 ciphers. Thus, 100000 From this subtract any number (say) 54321 (a) 45679 (b) Take any divisor, as 9. To (a) and (h) prefix two figure! hose sum = 9 less 1, /. e. to 8. Say 6 and 2, then— 9)6,54321 9)2,45679 ( 72702| (a) ^""'"^^^'j 272971 (-JJ 72702f 27297| Sum of do. 100000 For Long Division take, say 54. Prefix as before. 54)27,54321(5100511 54)26,456 79(48994A 270 216 54 « 485 54 432 321 536 270 486 a 507 486 ANSWERS. (a) 51005-1^ 219 W 48994^ 216 ' H Sum of do. lOOOQO A 56 I! DIVISION. MULTIPLICATION AND DIVISION BY FRACTIONAL NUMBERS. Example 1.— Multiply 1483 bv 123|. OPERATION. Analysis.— Here we multiply 1483 by 123 in the usual way; but before adding the partial products wc find the 5 eighths of 1483, namely 926|, and write it u^der the partial products, as in addition, then adding the four lines we obtain the required product. 1483 123} 4449 2966 1483 926| 1833351 We multiply by | (or any other fraction) by multiplying the given number by the upper number of the given fraction and dividing the product by the lower. Thus, 1483 V 5 (the upper figure) z=z 7415 — 8 (the lower figure) z= 9261, Example 2.— Divide 1234 by 4|. ^ Analysis.— We first bring both divisor and dividend to the same name as the given fraction— that is (in this instance) to fourths, then proceed as in division. OPERATION. 4f)1234 4 4 19 )4936(259i4 38* • 113 95 186 171 (2) (3) (4) (5) (6) (7) (8) 8 Exercises 18947632 X H 46738479 X 4 94327865 X 304- 29768342 X lo| 29648732 X 2006U 43796284 — 64 49625483 -i- 304- 76587938 -i- 148^ for the Slate. Ans. 104211976 3038001134 28534179164- 317528981|. 59502309784A- 6737889U 164051lj^^ 6911479,8^V PROMISCUOUS EXERniSFs IN '^V^^ T>r>T:'npT.TXTr. ^j, lol' y^^.u'^oo?^ contains 60 pupils, a second 83, a third 125, a fourth 234, a fiflh 672, andV sixth 1003 ; how many pnpils are there m the six schools ? Ans. 3177. I " PRIME NUMBBUfl. 80 4. 5. 6. 91?' ,J^|,^^y^«" 100 miles long, the Forth 11 5, the Thame. 215 the Shannon 224, and the Wrn 240; what wouldX 3 A^ hat i;' r' r r^^^ '^ .^'^^ ^" ^ ^-- «^ "-j^- o. unat IS the ditterence between 8964 and 14,382 ? Tfi /• ■ Ans. 5418. Two factors are 57682 and 8493 ; what is their product ? TT„ 11. Ans. 489893226. How much less is 7289 than 8723 ? Ans 1434 19 ^.o '''T.^ ''^^f' of drawers; in each chest there are 12 drawers, and m each drawer there are placed 12 dollars • how many dollars are there altogether in the chests ? ' ^ J.^ Multiply 94836 by 768, and divide^'t^he prodttTy Q ' -p^^ .1 - , ^ Ans. 7903. o. J?rom the sum of 189649, 283726 5428QS 9A»^(i7 by 847G2, and divide the product by 9418. Ans. IMMSeO. J. A man commenced business when 22 years old and l^tired at the age of seventy with a fortune 0/48768 dollar Required how much he cleared on an average each year ? 10. A wood of 6723 trees is to be thtned^ cullTe ^* Ans. 5976. PRIME NUMBERS. 54. A Prime ]¥umber is one that cannot be resolved mto two or more integral factors; thus 7, 3, 11, &c are />nme because they are not divisible by any number greatS than 1, without a remainder. ^ greater «5- To find the prime factors of any composite number. Example 1.— What are the prime factors of 30 ? OPERATION. ANALY8is.-We divide the given number by 2, the least prime factor; this gives an odd number for the quotient, divisible by the prime factor, 3, and obtain the quotient 5 ; this being a prime number, the division cannot be carried anv further. T^»« r]i*ir,'o^>» ^^a au- i_ , tient, 2, 3 and 5, are all the prime factora of r n w „ *^® S^^®^ number, 30. Hence the proof 2 X 3 X 5 X 1 = 30. 2 30 15 40 GREATEST COMMON MEASURE. KULB. Divide the given number by any prime factor j divide tlie quotient in the same manner, and so continue the division until the quotient is a prime number. The several divisors and the last quotient will be the prime factors required. 1. 2. 3. What are the prime factors of 9, 12, 15, 16 and 18 ? Wliat are the prime factors of .'}9, 26, 34, 38 and 42 ? What are th'i \mnui factors of 65, 85, 95, 105 and 115 ? ExcrciNCN for tli© Slate. Find the prime factors of the following numbei's and prove he results. ri) 15 (5) 39 (9) 57 (13) 85 (2) 18 (6) 42 (10) 69 (14) 91 (15) 99 (16) 108 ;i) 15 (5) 39 (9) 57 ;2) 18 (6) 42 (10) 69 (3) 24 (7) 45 (11) 78 (4)36 (8)49 (12)88 (17) 120 (18) 144 (19) 714 (20) 836 (21) 1492 (22) 8032 (23) 4604 (24) 1728 GREATEST COiMAlON MEASURE. 50. A Common Divisor of two or more numbers is a number that will exactly divide each of them. 57. The Crreat^st Common DlTisor of two or more numbers is the greatest number that will exactly divide each of them. Numbers prime to each other are such as have no common divisor. Note.— A common divisor is called a common measure ; and the greatest common divisor, the ^'reatest common measure. The latter IS usually indicated by the initial letters G. c. M. 58. To find the greatest common measure of two numbers. lix. — Find the greatest common measure of 105 and 165. OPERATION. 105)165(1 105 60)105(1 60 45)60(1 45 15)45(3 45 Analysis.— Here we divide the greater number, 165, by the less, 105, and thus obtain a remainder, 60, which we now make a divisor, and 105, the former divisor, the divi- dend, and wso on. When the re- mainder, 15, is used as a divisor it leaves no remainder, and is therefore tiio greatest common measure re- quire ^ Hence, 4 i GREATEST COMMON MEASUllE. 41 RULE. I. Divide the greater number by the leai. II. Divide the preceding divisor by the last remainder, and so on till nothing remains. The last divisor will bo the greatest common measure. 50. To find the greatest common measure of three or more given numbers. RUIjE. I. Find the greatest common measure of any two of the given numbers, by the last rule. II. Then, that of the common divisor thus obtained and of another of the given numbers, and so on through all the given numbers. III. The last common divisor found will be the greatest common measure of all the given numbers. £xcrci«cfi ibr the Slate. SECTION I. Find tlic ^eatest common measure of (1) 12 and 18. (2) 21 and 28. (3) 39 and 52. (4) 42 and 77. (5) 28 and 12G. Ans. 6 7 13 7 14 (6) 1024 and 2240. Ans. 64 (7) 1624 and 14500. 116 (8) 714 and 1176. 42 (9) 21671 and 22111. prime (10) 11 256 and 19899. 201 11. What is the greatest common divisor of 72, 120, 240, and 384 ? Ans. 24. 12. What is the greatest common measure of 300, 525, 225, and 375 ? Ans. 75. Example 2. — Find the greatest common measure of 42, 63, and 105. OlERATION. 42 — 2X3X7 prime factors. 63 = 3 X 3 X 7 " " 105 — 3 X 5 X 7 " " The factors common to the three given numbers are 3 and 7. Therefore 3 X 7 z= 21 , the greatest common measure. Hence, RULE. I. Resolve each number into its prime factors. II. Select those which are common to all the numbers, and their product will be their greatest common measure. SECTION II. Find the greatest common measure of (!) 12, 36, 60 and 72. Ans. 12 (2) 18, 24, 30, 36 and 42. 6 (ij) 50, i2(i, 72, 216. 18 (4) 32, 80 and 256. 16 (5) 200, 625, and 150. Ans. 25 (6) 252, 630, 1134 and 1386. 126 (7) 28, 140 and 280. ^8 (8) 468 And 1184. 4 I 42 LEAST COMMON MULTIPLE. LEAST COMMON MULTIPLE. •O. A Multiple is a number exactly divisible bv & given number; thus 16 is a multiple of 4. "''''''''^® ^^ » blp^lw f^ ^*'™"**^» Multiple is a number exactly divisi- bP^f;. J^'!i'^^*M* Commou Multiple is the least num- ber exactly divisible by two or more given numbers • thus 24 IS the least common multiple of 2, 4, 6, and 8 ' i^-^n'^ulL;^.'^ ''"''^^' ^^""^^ ^"^'^>'^ ^/'-^- -^ore A^^nTe^.^ 1— Find the lea^t common multiple of 12, 30, OPERATION. A VAT vara TU« 12 — 3 V 2 V 9 r.r.:«,« ^ i 1 ANALYSIS. — Ihe num- 30 = 3 o , P""^«/actors. ber cannot be less than 66, 42 = 3X2X7 « r^"^ ?* "'"'^ ^«"*^^" 66 ; 66 = 3 X 2 V T 1 « J^"''® '* "^"^^ ^<^"*ain the DO __ d X i X 11 « factors of 66, viz., 3 X 2 X 11 fa^tol,^of V? I ^1 ^ ^ ^ =t^^^^'. ^"^- ^^ have all the prime factors of 66, and also the prime factors of 42, except the factor 7. Annexing 7 to the serie. .f factors, ^ ^ ^ , 3 X 2X11 X 7 3X2X11X7X5X2 be^^'^.nrt.''" *tP"™« '■»««<'« of ea^ih of the given num- bers , and hence the product of the series of factor, ;.„ common multiple of the given numbel *"*°" " * «factor°rf„'^o5f!" series can be omitted without omitting .^te^t°rrm«ii;ro?t"&-T^^^^ J^rom thi« illustration we deduce the following B-TJT.in T -o •■__ ,. factora7' "' -^-^^^^^o ^ae suren numbers into their prime DECIMALS. 43 II. Take all the prime factors of the largest number, and such prime factors of the other numbers as are not found in the largest number, and their product wiU be the least common multiple. Note.— For other methods see Advanced Arithmetic. Find the least common multiple of the following numbers. 7, 85 and 98. Ans. 490 4, 9, 6 and 8. 72. 8, 15, 77 and 385. 9240. 12, 15, 42 and 60. 420- 21, 35 and 42. 210- 4, 16, 20, 48, 60 and 72. 720- 5, 10, 15, 20, 25, 30, 35 and 40. 4200- 3, 6, 9, 12, 48, 21, 24 and 16. 1008- 15, 12, 128, 30, 16, 4, 320 and 96. 1920 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30 and . , ^"s. 1441440. What IS the smallest sum of money for which I could purchase an exact number of books, at 5 dollars, or 3 dollars, or 4 dollars, or 6 dollars each ? Ans. 60 dollars. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 32. 11. DECIMALS. 64. Beelinal Fractions are the decimal divisions of a unit ; thus a unit is divided into ten equal parts called tenths ; each of these tenths is divided into ten other equal parts called hundredths; and so on. Since the denominators of decimal fractions increase and decrease by the scale of 10, the same as simple numbers, in writing decimals the denomi* nators are generally omitted. 65- In simple numbeid the unit 1, is the starting point of notation and numeration ; and so also is it in decimals. ««. The Decimal Point is a period, C) which must always be placed before the left hand figure of the decimal. Thus, Y®^ is expressed .6 tWt " .567 «y. The names of the different orders of decimals, or places below units, may be easily learned from the following 44 NOTATION AND NUMERATION OT DECIMALS. Decimal Table. i H o 2 i a 'o Ph I— H ,-C -tJ OS m a> i2 S-^rB 25 -a 2 fl 3=5 000000000-000000000 -t-3 1 • r:} 0) -M (h s -V) o r^ s • 1— ( w ;^ en G S:2 ^ 'TJ T3 o Cl, a, Pn'Ei'ai'PH'PH'EH'PH 1— (.cCOS By examining this table wc see that Tenths are expressed by one figure. Hun(h-edtlis " two figures. Thousandths " three figures. 68. Every cipher on the left hand of a decimal reduces It to one-tenth its previous value. Thus, .5 is 5 tenths, .05 is 5 hundredths, and .005 is 5 thousandths. Ciphers on the right do not alter the value, for .5, .50, .500 are the same as ^-^, {>^^, ^^^^^ and these are all equal. NOTATIOX AND NUMERATION OF DECIMALS. 6©« Rule for decimal notation. r.hlr,J^l*f *^® decimals as a whole number, placing cl- frue lo^al valul.°^''^''^ *° ^'""^ ^^°^ significaAt^figu?! fts II. Place the decimal point before the first figure. 70. Rule for decimal numeration. T^ne^Ik deiSmTnaTor. '"°^ **^^ ^^""'"^^^ P°^^*' *° ^«*^^- th"num''e?at'(?r? ^"'^^^^^ *.*^® decimal point, to determine TM'Loi* «?l^^ *^® decimal as a whole number, giving it the mean or denommation of the right hand figure. i 1 ucos 05 is .500 ; ci- ) its ter- Line the 1. 2. 3. 4. 5. 6, ADDITION OF DECIMALS. Exercises for flic JSlatc. Write 2G5 ten thousandths. Write six liundred and thirteen thousandtlis. Write 3G5 thousands, and 4 billionths. Write seven hundred thousandths. Write one hundred, and 2 tenths. E,ead the following numbers : 1.265 4.0005 «.0007 "8-898 i/.2006 1267.9876543 .5967 119.3200 3.0000678 4G.7325 .5000 123.45607890 4$ ADDITION OF DECIMALS. Tl. Example 1.— Add 3 tenths, 45 hundredths, U tenths, and 365 thousandths. OPERATION. Analysis. — As in simple numbers, we write .3 -45 1.6 .365 2.715 the numbers so that units shall stand under units, tenths under tenths, hundredths under hun- dredths, &c. This brings the decimal points directly under each other. Commencing at the right hand we add each column, and carry as ia whole numbers, and in th« result we place a point between the units and tenths, or directly andcr the decimal point in the numbers added Hence the BULE. I. Write the numbers so that thetieeimal point* snail stand directly under each otlier. II. Add as in whole mimlaers, and place tTie decimal point, in theTeSv'lt, directly under the points in the num- bers added. Iflcntal ^Exercises. 1. Add .6 and .06 ; 10 and .01 ; 3.6 and 3.007; .8 and .3. 2. Add 6 hundredths and 56 thousandths; .06 and .056. 5. Add 20 cents and 156 cents; .20 and 1.56. 4. Add 256 dollars and 3 dollars and 25 cents; 25<5 4- I -f .25. ^ Exercises fior the Slate. SECTION I. (1)27.655 +71.784 +98.687 +84.769. (2) 210.373 + 376.458 + 843.847 + 591.738 + 456.153. (3) 26.3756 + 74.5673 + 5b.8948 -4- 74.7355 4- .55.1052- (4) 254.172 + 888.627 + 568.296 + 756 939+531.704. (5) 214-735 + C07.434 + fi65.75S + 456.376 + 730.242. •3' . ft u SUBTKAOTIOK OF DKCIMAL*. SECTION II. Add 25.7, 8.389, 23.056. Ans 57 145 Add 36.258, 2.0675, 382.45. Ans.* 420.7755. Add32.764,5.78, 16.0037 and49.3046. Ana. 103.8523. Add 1152.01, 14.11018, 152348.21, 9.000083. A ij .v^ «« ^^^- 153528.330263. Add 37.03, 0.521, .9, 1000, 4000.0004. Ans. 5038.4514. What is the sum of twenty-six, and twenty-six hun- dredths; seven tenths; six, and eighty-three thousandths; four, and four thousandths ? Ans. 37.047. 7. How many yards in three pieces of cloth, the first piece containing 18.375 yards, the second piece 41.625 yards, and the third piece 35.5 yards ? Ans. 95.5 yard*. I. 2. 3. 4. A. 6. 79. SUBTBACTION OF DECIMALS. Example l.—From 81.63 take 27.85. OPBRATI027. 31.63 27.85 S.78 Ex. 2.— Prom 3.86 74 take 1.36. OPERATIOX. 9.8674 1.36 9.5074 Ex. 3. — From 15.36 take 8.1234 OPERATION. 15.36 8.1234 Analysis.— In each of these three examples, we write the subtrahend un- der the minuend, placing units under units, tenths under tenths, &c. Com- mencing at the right hand we subtract M m whole numbers, and in the remain- ders we place the decimal points directly under those in the numbers above. In the second example the number of deci- mal places in the minuend is greater than the number in the subtrahend, and in the third example less. In both cases, we reduce both minuend and sub- trahend to the same name, or number of decimal places, by annexing ciphers ; or we suppose them to be annexed before perfon..ing the subtraction.— Hence, 7.2366 ^BULIB. Plao« the numbers as in addition, sub+^ae*^^ •- *- i MULTIPLICATION OF DSCIMALS. 47 nientol exercises. 1. From five tenths take forty-nine hundredths. 2. From .63 take .496 ; 2.19 take .63 ; .5 take .005. 3. From .16 take .006; 12.34 take 2.345; 100 take .001- 4. From one take two hundredths. 5. From 3.10 dollars take 75 cents; 3.10 take .7«. Exercises fbr tlie 81«,te« SKCTIONT I. 1 . From 20.34 2. From 40,68 3. From 16.272 4. From 6.21083 take 4€ 13.56 27,12 10.848 4.3392 5. From 52.0704 6. From 430.2816 7. From 2603.52 8. From 933.9607 tftk« S4.71SC " 286.8544 ** 1735.6t ^ 655.973* Find tlie value of — SECTION II. (1) 111.1116—22.22222. (2) 279.00906—117.916. <3) 8.135—2.6875. (4) 627.4—91.7469 Ans. 88.88938 161.09306 5.4475 5S5.653I (6) 21.004— .75 <«) 714.0— .916 (7) 2— .298 (8) 1000— .001 Ans. 20.254 713.084 1.702 909.99t 7S. by .5 OPERA TlOIf. .25 ,6 MULTIPLICATION OF DEaMALS. Example.— What is the product of .25 multiplied .125 Analysis.-— We perform the multiplication the same as in whole numbers. Since the multiplicand is 25 hundredths, and the multi- plier 5 tenths, and huadredtha multiplied by tenths give thousandths, and thousandths being expressed by three figures, we must have three places of decimals in the product. Hence we see the product contains as many decimal places as are contained in both multiplicand and multiplier. Hence, mJIJE. Multiply as in whole nnml)eT8, and from tli« right hand of the product point ofT as many figures for decimals as there are decimal places in both factors. Note 1.— If there are not as many figures in the product as there are decimals ia both factors, supply the deficiency by jare^xin^ cipher*. 2.— To muitiplv by 10, 100, 1000, &c., remove the decimal point m many places to the right as there are ciphen on the right of the mul tiplien •» t 4B DIVISION . OF DECIMAL!?. lUental Exercises. 1. If a man cjir* i-eap .i>6 of an acre in a day, how much can he reap in .5 ofa day V 2. If 1 pound of coffee cost .a of a dollar^ what will 4 younds cost 'i 3. Add 3.G -f- .26 + .00(> -f- 3.006, and multiply the pro- duct by .8 4. From 3.606 take 1.4, and multip' r ? result by .09 5. If 1 ton oi' hay cost 8.75 dollarjiy >-' ;ill .25 ofa ton «ogt? Hxercises r»p tlie S^late^ SECTION I- • Multiply and add together the products of— (1) 1234.5e789 by 78.91 and 21.09 f (6) by .550.8 and 449.2 (2) 345.789612 by 35.79 and 64.21 '^ " (.3) 40&.783089 by 60^09 and 39.91 (4) 2492.67339 by 42.82 and 57.18 (5) 5063.48001 by .99 and 99.01 (7) by 900.9 and 99.1' (8) by 428..6 and 571.4 (9) by 624.8 and 375.2 (10) by 99.73 and .27 SECTiai? ir. Find the protluct of — iTL) .132X-2"41 Aus. .081812 ^2) .23 X .009 .00207 ^3) 21.71()X2.06 44.734!»() ^4) 11.111 X 9.7116 107.9055876 (r5) .2X.7X-06X-004X.1 .00000336 (fi) .0006 X .fW)0r2 Ans. .000800072 (7) 8.0004 X .004 .0320016 (8) 164.023 X 12.88 2112.61624 (9) 178.006 X 100.001 17800.778006 (10) 43.1 X .6 X 100. X .01 25.86 11. Multiply fotn- hundred^ and four thousandths by thirty and three hundredths. Ans. 1 20 1 2-. 1 20 1 2^. 12. If a cord of wood be- -vrorth 2.37 bushels of wheat,, how many bushels of wheat must be given for 9.58 cords of «oo^ P^"'^^'^' ^e., are the,^ in 27s.? in 2««. ^ £6^7s'^TnS^5;?^"^^'"^^ -^5 7-? in ofL'^HrXei'^''"''' '^- '^^•' -^^^ -as the cost pounds'^''"" ''' ^^^''^'""-^^ ^« P--^- In 690s. how manj 7. What cost 85 pairs of gloves at 7 pence per pair 7 Exercises for the Slate. SECTION I. iieduce to FarOimgs. £ 8. D. £ (7) 129 3 (13) 3974 (8) 103 12 n ' - (9) 354 10 loi (10) 530 17 2l (H) 531 2 3 8. (14) 1009 15 (15) 4983 16 (16) 5993 11 (17) 5221 4 ,. ^ „ <^^2^ ^^^ 7^3|((18) 5575 15 .^ 21. In95guintasf7s ^rrr^^^^^^ ,Ans. 284079 guineas, 1 /s. i)|(l., how many fartliings ? 22. R«duc<3 £15 15« RA f^ • ^"s. 96615. 23. Eeduce £l I4s 1 . fj^l^^^es. Ans. 631. eauce ^1. 14s. 9d. to three pences. Aus. 1259. SECTION II. Reduce to Pounds. ^7^7^.^^ 'f.'-- I.f ">!??!« >-'f pence. (1) 17448 far. (2) 43632 (3) 138657 (4) 156113 (5) U22^2 u u (( (7) 78536 (8) 198786 " (9) 302547 " (10) 103753 " a (12) 21600 (13) 99393 V --4<.;o peneeu i) 17066o « REDUCTION or DECIMAL CURRENCY. 53 Refluec (16) 197421 far. to slillHngs. (17) 171504 halfpence " (18) 7r>6 shillinjjja to guineas. (19) 4536 three pences (i (20) 6480 far. to crowns. (21) 11340 pence " (22) 2700 " " (23) 2160 lialf pence " 24. How many pounds, shIDiriigs, &c., are there in 367841 farthings ? Ans. £383 3s. 4|d. 25. In 1059120 pence how many sovereipjns ? Ans. 4413. 26. A farmer, during the year, sold 1367 quarts of milk at 3 pence per quart, what did it all amount to ? Ans. £17 Is. 9d. REDUCTION OF DECIMAL CURRENCY. 83. A I>€(ciiiial Currency is a currency whose denom- inations increase in a ten-fold ratio, and each denomination is one-tenth the value of the next higher. The currency of the Dominion of Canada, the United States, France, Barbadoes and some others of the Wind- ward Islands, and Demerara, is decimal. 84. CANADA CURRENCY. TABLE. 10 Mills (m) make 1 Cent, marked Ct. or C. 10 Cents " 1 Dime, " (I 10 Dimes " 1 Dollar, " $. Note 1. — It is usual in writing dollars and cents, to place the sign ($) of dollars in front of the sum, and a point ( . ) between the dollars and cents. Thus, tifty.six dollars, four dimes, six cents, and five mills would be written $56,465, or $56.46^, and read 56 dollars and 46^ cents. 2. If the sum consists of dollars, and a number of cents less than ten, there must be a cipher between the dollars and cents in place or dimes. Thus, 5 dollars and 4 cents must be written $5.04. 85. By examining the above table we see that 10 mills make 1 cent, and 100 cents, or 1000 mills one dollar; hence, 86. To change dollars to cents, multiply by 100 ; that is, annex two ciphers. To change dollars to mills, annex three ciphers. To change cents to mills, annex one cipher. To change dollars and cents to cents, or dollars, cents and mills to mills, remove the decimal point and the sign $. 54 1. 2. S. 4. 5. RKDUCTION OF DECIMAL CURRWTCY Exercises for the Nlat4s. ""^'l/.'t'r^^^ Ans. 19600. »32otom.lk " 1325000. »i.4b to cents. " 146 56 centa to mills. « figo." $19,425 to mills. " 19425. 1. 2. 3. 4. — ■^ * «* ^i#« Exercises for the Slate. Change 1967 cents to dollars a -. , " 1432 mills to « An8.Sl9.67. In 84567 mills how many dollars ? a""' f^'^^^. ^ ^duce 3195 mills to doCt'dlents. t^1:f^|: ^e^al tt^L^'Zy 'op'Sn" L" ^'fr^^ P"-^P^« ^s multiplication, &c., mav ^be niJ?' ^^^d^tion, subtraction, manner as upin deciml. P^^^™^^ "PO" it in the same' *•• Accounts are kenf In »♦« i- TABLE, fixAMPLE.-Ileduce £5 10s. l^d. to Canada currency oPERATiox. Analysis <5- ^ Unhand nt~"'^ P^""^«' ship- s' P^",""? ^^® composed of farthings multiplying bv 20 12 and 4, reduces the whole amount to farthings = 5285 farthingr A„d since one farthing is equal^ A^f a Canadian cent. 5285 farthinlJtre operation. £5 10s IJd = 5285 far. 73 15855 36995 that w. Si 9.6 7. $1,432. 34.567. I3.19J. iple as action, ' same lillincfs tid ^ 1 and ■y ^ shil- 3d of , 12 ntto And kof are t\ REDUCTION OP DECIMAL CUnnENCY. fi5 RULE. Reduoe pounds, nhillins^s and pence sterling to farthings, and multiply by 73 and divide by 144. The quo- tient will be the equivalent in Canada ourrenoy. Note 1. — In a final remainder reckon over j^ as a cent, les-s than | reject. Note 2. — When there are only pounds in the exercise multiply by 483 2-3, the number of Canadian cents in a pound sterling. See Appendix II. mental Exerciseti. ^ 1, How many Canadian cents are there in a three-penny piece ? in a four-penny piece ? in a sixpence ? in a shilling ? 2. How many Canadian dollars and cents are there in 2s, or a florin ? in 5 florins ? in 5s, or a crown ? in 10 crowns ? in 3 florins -|- 2 crowns ? 3. How many Canadian dollars and cents are there in 10s, or a half-sovereign ? in £l, or a sovereign ? in 10 sov- ereigns? in £l Is, or a guinea? in 2 guineas -|- 3 half-sov- ereigns ? Exercises for the Slate. Reduce the following to Canadian currency :- (1) £1 3 (2) £11 11 (3) £44 15 (4) £26 18 (5)£115 16 llf (6) £110 11 lli (7) £365 4 5} Ans. $5.73 $56.35 $217.94 $131.11 $rH)3.80 $538.26 $1777.41 (8) (9) (10) (11) (12) (13) (14) £27 6 £26 16 £10 11 £25 £82 £64 £5 7 h 4f Ans $133.01 $130.60 $51.44 $121.67 $399.07 $311.47 $24.33 •!• To reduce Canadian currency to pounds, Sfc, Stg. BUIjE. Reduce the dollars and cents to farthings by multiplying by 144 and dividing by 73. Reduce the farth- ings to pounds, shillings and pence. See Appendix ii. Example. — Reduce $110.12^ to pounds, &c., stg OPERATION. $110.12J X 144 = 1585800 -^ 73 = 4)21723 farthings 12)5430-}- I 2,0)45,2 -f 6 22 -}- 12 Ans. £22 12s 6| Note.— For exercises under this rule the pupil may prove those^of the former one. «•■ n\ 56 il i ! REDUCTiON OF LINEAR OB LONG MEASURE. REDUCTION OF LINEAR OR LONG MEASURE. ®®- LONG MEASURE — TABLE. 12 Inches make 1 Foot marked ft. 3 haat u I Yard " / lll^'f' T. , " 11^«^1. Pole or Perch" ri'or« 40 Rods or Perches " 1 Furlong " fur 8 Furlongs u i Mile ^ ^^'Z' nJ ?J.'^' . " 1 League " lea 69i Miles (nearly) " 1 Degree " deg. or EXAMPLES. L ]lllF?- ' "• ^ '"• ''"'' I ?■ ?<'d™« 53 73 inches ti many inches ? OPERATION. 18 po. 0yd. 1ft. Gin. H 90 9 99 z= yds. in 18 po. 3 ^ 298 = ft. in 18 po. 1 ft. 12 poles, &c. '•^■^^ii^-d. ^OPERATION. 12)5383 3)447 ft. 9 inches. H) 149 yds. P 2 11 )298 27po.Jyd.zr:lft.6in + 9 in 27 po. yd. 2 ft. 3 in 3582 = in. in 18 po. 1 ft. 6 in. Mental Exercises. in 12 R.7[uT^ '"''^^' ^'''' ^^'^''^ "' 3 ft. ? in 5 ft. ? in 10 ft. ? 9 yds. fT:ir;z. f '' '" *''^^ ^^ ' >''^- ' - ^ >'^«- ^ - fur'-? in 72 r7fu'r f "^" ^'^ ''"^ ^" ' "^^^^« ^ - ^ - ^ 4. In 100 inches how many yards, feet and inches ? iron Jill ooTtI ' ''' "^'^ "'"^ ^""'-^^ ^^'^" ^ y^^' ' ft- «f Exercises for tlie Slate. (1) Reduce 71280 in. to fur. (2) '• 3564 in. to po. (3) " 63.360 yds. to miles. ^•< "-1U iii. to iniies. (5) " 190080 ft. to miles. I (10) " 74 m. tfur (6) Reduce 36 po. 3 ft. to inches fsi " ^5 "^- ? P«- 1 y^^- to yds. (8) 27 m. 1 po. 3i vd. to fR^t- (y) " 72 m. 13 po. i .yd. oyds'. 1 po- i yd. to yds ICEDUCTION OF LINEAR OR LONG MEASURE. 57 11. In 9768042 inches how many miles? Ans. 154 m. 1 fur. 13 po. 3 yds. 12. In 897682 yards how many miles V Ans. 510 m. fur. 14 po. 5 yds. 13. Reduce 103 m. 5 fur. 32 po. 5 yds. to feet. Ans. 547683. CLOTH M ASURE — TABLE. 93. 2\ Inches make 1 Nail. 4 Nails " 1 Quarter, qr. 4 Quarters " 1 Yard, 1 yd. 5 Quarters " 1 English ell. 6 Quarters " 1 French ell. 3 Quarters " 1 Flemish ell. EXAMPLES. 1. Reduce 27 yards 3^qr. to inches. OPERATION. 27 yds. 3 qr. 4 111 :=z qrs. in 27 yds. 3 qr. 4 444 = nls. in 27 yds. 3 qr. _H 888 111 999 2. Reduce 153 nails to yds, &c. OPERATION. 4)153 4)38 qrs. 1 nl. 9 yds. 2 qrs. 1 nl. :in. in 27 yds. 3 qr. Mental Exercises. 1. How many inches are there in 3 nls. ? in 2 qr. 1 nl. ? in 2 yds. 1 nl. ? in 5 qrs. ? 2. HoAV many quarters are there In 5 yds. ? in 3 yds. 3 qrs. ? in 6 yds. 2 qrs. ? 3. How many yards arc there in 5 qrs. ? in 17 nls. ? in 123 nls.? in 196 qi-s. ? 4. How many inches are there in 4 English ells ? m 5 Flemish ells? in 19 French ells? 6 What is tha cost of 8 French ells at 2 cents per inch ? if i ill 58 ItEDUCTION OF SQUARE MEA8URB. (UVA o. *^**'**''«» ^w 'ards opp;ration. 135 ac. 3ro. 15 po. 4 543ro. in 135 ac. 3 ro. 40 to acres. OPERATIOX. 30^)261414 4 4 21735po.inl35ac.3ro.15po I Cental Exercises 121 " 11)1045656 11) 95059 7^ 95 4|0) 864)1 7)7 ~^^^ 4)216 ro. Ipo. "77 [yds. 54 ac. ro. 1 po. 23| ■.'1 REDUCTION OF CUBIC OR SOLID MEASURE. 59 2. How many acres are there in 880 poles ? in 160 poles ? in 320 poles? in 1240 poles? 3. At $4 per acre what will 920 poles of land cost ? 4. Find the cost of 12 yards 3 feet at 7 dimes per foot. Exercises for the Slate. (1 ) Reduce 126 ac. 4 po. 5 yds. to yds. (2) " 162ac.5po.l04yd8.toyds. (3) " 9 po. 9 in. to inches. (4) " 90 ac. 18 yds. to yards. 9. (5) Reduce 1411380 in. to poles. (6) " 304983 yds.to acres. (7) 94ac.2ro. lpo.5|yds. to yds. (8) " 697104 yds.to acres. In 36 ac. 3 ro. 28 po. 5 yds., how many feet ? Ans. 1608498. 10. Reduce 29 ac. 3 ro. 38 po. 15^ yds. 8 feet to inches. Ans. 188122032. 11. In 6463 76^ feet how many acres ? Ans. 14 ac. 3 ro. 14 po. 6 ydi . 1 foot. \ REDUCTION OF CUBIC OR SOLID MEASURE. 05» SOLID MEASURE — TABLE. 1728 Cubic inches make 1 Cubic foot, marked cm. yi; 27 Cubic feet " 1 Cubic yard, " cu.yd. 40 Cubic feet of rough or ' 50 Cubic feet of hewn timber 42 Cubic feet of timber 128 Cubic feet 5 Cubic feet « 1 Ton. 1 Ton. 1 Cord of fire wood. 1 Barrel bulk. Exercises for tlie Slate* 1. In 125 cu. ft. 840 cu. in. how many cu. in. ? Ans. 216840. 2. Reduce 5224 cubic feet to cords. Ans. 40 \^. 3. In a pile of wood 60 feet long, 20 feet wide, and 15 feet high, how many cords ? Ans. 140|. 4. A cellar is 32 feet long, 24 feet wide, and 6 feet deep, how much did it cost to dig it at 15 cents a cubic yard ? Ans. $25.60. 5. In a school-room 30 feet long, 20 feet wide and 10 feet high, with 50 pupils each breathing 10 cubic feet of air in one minute, in how long time will mey breathe as mnch as the room contains ? Ans. 12 min. II 60 REDUCTION OF CUBIC OR SOLID MEASURE. oo. MEASURE OF CAPACITY- -f ABLE. 4 Gills (g) make 1 Pint, marked pt. 2 Pints '' 1 Quart, " qt. 4 Quarts " 1 Gallon, " gal. 2 Gallons " 1 Peck, " pk. 4 Pecks " 1 Bushel, " hush. 36 Bushels " 1 Chaldron " clial.\ EXAMPLES. 1. Kcduce 27 bus. 1 pk. 1 gal. 1 qt. 1 pint to pints. OPERATION. 27 bus. 1 pk. 1 gal. 1 qt. 1 pt. 4 109 pks. 2 219 gals. 4 677 qts. 2 2. Keduce 594 gills to galw Ions. 4)594 OPERATION. 2)148 pts. 2 gills. 4) 74 qts. pts. 18 gals. 2 qts. Opts. 2 gills 1 755 pints. Note,— As Liquid and Dry Measure are similarly divided, the above table and examples will answer both. (See Nova Scotia Table-book, pages 24 and 25.) Mental Exercises. 1. How many gills are there in 4 pts. ? in 3 qts. 3 pts. ? in 6 qts. 3 pts. 1 gdl ? 2. How many quarts are there in 6 gals. ? in 3 gals. 2 qts. ? in 2 pks. 1 qt. ? 3. How many gallons are there in 8 qts. ? in 8 pts. ? in 24 pts. ? in 38 qts. ? 4. What will be the cost of 7 gak ' qts of burning fluid at 15 cents a quart ? Exercisefs for ithe Slate. Ci) Reduce ID gals. 1 pt. to gills. ,2) « 11 pks. 1 gal. 1 qt. 3 gil. to gills. (3) « 3 bus. 1 gal. 1 gill to gills. (4) «' 2 bus. 1 pk. 3 qt, 3 gils. to gills. (5) Rp.duce ll>42 bus. 1 qt. to qts. (G) " 2880 gills to pks. (7) " 18432 gills to bus. (8) « 694 qts. to bush. . 5 BEDUCTION OF WEIGHTS. n 0. In 4983265 gills how many qiiai;ts ? Ans, 622908 qts. 1 gill. 10. Reduce 126 bus. 3 pks. 1 pt. to pints. Ans. 8113. 11. Reduce 1467896 quarts to chaldrons ? Ans. 1274 ch. 7 bus. 3 pks. 12. An innkeeper lK>ught 50 bushels of oats at 65 cents a bushel, and retailed them at 25 cents a peck ; how much did lie make on the lot ? Ans. $17.50- REDUCTION OF WEIGHTS, 97» TROY WEIGHT — TABLE. 24 Grains make 1 Pennyweight, 1 dwt. 20 Pennyweights " 1 Ounce, 1 oz. 12 Ounces " 1 Pound, 1 lb. This weight is used in weighing the precious metals and! atones ; also in scientific investigations. EXAMPLES. 1. Reduce 31 lbs, 10 oz. 8 dwts. 1 2 grs. to grains. OPERATION. 81 lbs. lOoz. 8dwt. 12grs. 12 382 oz. 20 7648 dwt. 24 80604 15296 2. Reduce 28197 dwt. to lbs. OPERATION. 210)281917 12)1409 oz. 17 dwt. 117 lbs. 5 oz. 17 dwt. 183564 grains. mental Hxerciiies. 1. How mar y grains are there in 5 dwts.? in 6 dwts. 7 grains ? in 15 dwts. 3 grs. ? 2. How many ounces are there in 120 dwt. ? in 200 dwt. ? in 240 dwts. ? 8. What will a gold chain weighing 9 dwt. 15 grs. cost at 3 cents a grain ? I ■ REDUCTION OF WEIGHTS. 4, What is the value of a silver cup, weighing 5 oz. 4 dwts. at 15 cents per pennyweight? 5. In 5 ingots of gold, each weighing 9 oz. 6 dwt. how many dwts. ? Exercises for the Slate. (1) Reduce 9 oz. 12 dwt. 18 grs. to grs. (2) (3) (4) « 1 lb. 1 oz. 19 dwts. to grs. 1 lb. 3 oz. 9 dwt. to grs. 20 lbs, lOoz. ISdwts.to dwts. (5) Reduce 207396 grs. to lbs. (6) " 4338 dwts. to lbs. (7) " 155520 grs. to lbs. (8) " 17280 dwts. to lbs. 9. Reduce 37 lbs. 11 oz. 19 dwts. to dwts. Ans. 9119 dwts. 10. Reduce 87 lbs. 19 grs* to grains. Ans. 501139. 11. Reduce 578096 grains to pounds. Ans. 100 lbs. 4 oz. 7 dwts. 8 grs. 12. A miner had 14 lbs. 10 oz. 18 dwt. of gold dust : how much was it worth at 75 cents a dwt. ? Ans. $2683.50. 1 f' M 99* APOTHECARIES WEIGHT — TABLE. 20 3 8 12 Grains make Scruples Drams Ounces " u (( 3 3 1 Scruple, 1 Dram, 1 Ounce, 1 Pound, Note. Apothecaries and Physicians mix their medicine by this weight, but they buy and sell by Avoirdupois. 1 sc. or 1 dr. or 1 oz. or 1 lb. or lb 1 (2) (3) (4) Exereises for the Slate. Raduoe 9 lbs. 1 oz. 1 dr. to grs. " 18 lbs. 6 dre. to scr. ♦* 36 lbs. 1 scr. 16 grs. to grs. *• 45 lbs. 2 scr. 6 grs. to grs. (5) Reduce 63 lbs. 1 dr. 3 m. to gn. (6) " 84 lbs. 7 oz. 7 drs. to gri. (7) « 207360 grains to lbs. (8) * 259200 grains to lbs. 10. Reduce 47tt). 6^. 43. to scruples. Ans. 13692 scr. 11. How many pounds of medicine would a physician use in 365 days, if he averaged daily 5 prescriptions of 20 graini each? Ans. 61b. 4|. 13. •9. AVOIRDUPOIS WEIGHT — TABLK. 16 Drams make 1 Ounce, marked 1 oz. 16 Ounces 28 Pounds 4 Quarters 20 Hundredweight 1 Pound, " 1 lb. 1 Quarter, " 1 qr. 1 Hundredwdght 1 cwt. 1 Ton, " 1 ton. ^Ki RBDUCTION OF WEIGHTS. y NEW SYSTEM OF WEIGHT. The different units are the same as in the old system, thus 16 Drams make 1 Ounce, marked 1 oz. 16 Ounces " 1 Pound, " 1 lb. 25 Pounds " 1 Quarter, " 1 qr. 4 Quarters " 1 Hundredweight 1 cwt. 20 Hundredweight " 1 Ton, " 1 ton. Note. — The old system of weight is called long, and the new tyt- tem short weight. EXAMPLES. 1. Reduce 81 cwt. 2qrs. 25 lbs., long weight, to pounds. OPERATION. 81 cwt. 2 qrs. 25 lbs. S26 qrs. 28 2633 652 9153 IH. Or, 81 cwt. 2 qrs. 25 lbs. 8100= 81 X 100 972 = 81 X 12 66 = pounds m 2 qrs. 25 =z " given. 9153 (( required. 2. Reduce 72 cwt. 2 qrs. 22 lbs., short weight,to pounds. OPERATION. 72 cwt. 2 qr. 22 lbs. 4 290 qrs. 25 1472 580 7272 lbs. Or, 72 cwt. 2 qrs. 22 lbs 7200 =: pounds in 72 cwt. 50= " " 2 qrs. 22 = " giv en. 7272 = " required. IHental Exercises. 1. How many ounces are there in 3 lbs. ? in 5 lbs. 10 oz. ? 6 lbs. 13 oz.? 2. In 3 cwt. 5 lbs. short weight, how many pounds ? How many ounces ? 3. What will 1 ton 5 cwt. of hay cost, if 5 cwt. cost $3 ? 4. What will 2 cwt. 12 lbs., short weight, of beef cost at r%i\ir\Tft « ^^. J o iJVU !l\~i. 5. If 8 ounces of tea cost 40 cents, what is the cost of 2 lbs. ? %m u REDUCTION OF TIME. Exercises for tlie Slate. 1. Reduce 8 cwt. 2 qi-s. 10 lbs. 4 oz. 12 drs., long weight, 1 ton 2 cwt. 3 qrs. 7 lbs. 9 oz. 13 drs., long weight, 22 tons 13 cwt. 1 qr. 5 lbs. 9 oz., long weight. to drs 2. " to drs. 3. " to drs. 4. " 25 tons 2 cwt. 1 qr. 13 oz., long weight, to oz. 5. " 42 tons 14 cwt. 2 qrs. 3 lbs. 5 oz., short weight, to ounces. . -i ^ x j 7 cwt. 1 qr. 4 lbs. 7 oz. 5 drs., short weight, to drs. 6939 drams to pounds. 1032228 drams to cwt., long weight. 3 qrs. 15 lbs. 15 oz. 15 drs., long weight, to drs. Ans. 25599 drs. 94 tons 19 cwt. 2 qrs. 24 lbs. 10 oz. 15 drs., long weight, to drams. Ans. 54468783. 11. " 493865 lbs. to tons, long weight. Ans. 220 tons 9 c. 2 qr. 1 lb. 12. " 204250 oz. to cwt., short weight. Ans. 127 cwt. 2 qr. 15 lb. 10 oz. S. 7. 8. 9. 10. u u u it a 100. REDUCTION OF TIME. TABLE. 1 Second is written thus : 1" 60 Seconds make 1 Minute, marked 1'. 60 Minutes 24 Hours 7 Days 28 Days 28, 29, 30, or 31 Days " 1 Calendar month. 12 Calendar months " 1 Year 365 Days - 1 Hour, " " 1 Day, " " 1 Week, " 1 Lunar month. 1 Jir. 1 day. 1 wk. u 1 Common year. 366 Days « 1 Leap year. mental Exercises. 1. How many seconds arc there in 3 hrs. ? in 4 hrs. 20' ? in 5 hrs. 9" V n • , i 2. Hov: i.ianv hours are there in 4 days 5 hrs. .'' in 2 wks, 3 days U hrs. 'i . , o • ok i o 3. Hov many weeks are there in 72 days i m 85 days i 'n 63 d-tr^ ? EEDDCTION OP TIME, 6» ) 4. How many days are there from April 15 th to August loth inclusive ? Exercises Tor the Slate* REDUCE (6) 365 dys. 5 hrs. 48 min. 45 sec. to «ewf. s ors., 2nd. 1 cwt= 2 nrs. 14 lbs., 3rd, 1 ewt. 3 qr., and the 4th, 2 ewt. qr. 14 lbs. ;^wliat wag the weight of tlie whole ? I COMPOUND ADDITIOH. ef Exercises fbr the Slaiie* aination denomi- monina- aation. £ s. d. £ s. d. £ s. d. £ s. d. 2 16 9 2 7 8 2 10 7f 29 9 104 8 17 6 3 14 5 7 16 10 25 18 44 8 18 5 9 10 7 9 1# 9i 76 16 ll} 9 5 11 9 2 10 8 15 8 94 14 8 9 ^'^ 1 (6) lbs. oz. dr. cwt. qr. lb» (8) tons. cwt. qr» 3 10 54 7 13 4l 33 10 7 31 2 23 3 17 2 37 8 13 27 1 16 I 13 6 12 8; J. 78 12 8 49 8 5 8 3 4 9 6 . 65 14 5 57 3 12 6 12 1 5 13 5 . 26 6 10 79 2 6 7 13 2 5 18 4 . 81 13 8 50 3 20 4 11 3 4 16 6 rs. 11 14 7 11 32 16 2 17 2 oz. awt. g 35 12 5 (10) oz. dr. scr. 35 5 2 yds. ft. in. 35 2 10 yd. qrs. nl» 38 2 3 64 17 19 38 2 1 34 6 45 1 2 48 16 11 75 6 69 2 8 37 S 65 18 4 47 7 2 42 1 11 72 8 1 61 13 23 89 4 1 35 2 7 42 2 2 98 ly 14 52 1 2 60 1 8 67 3 1 66 1 6 42 S (13) m. fur. po. 36 6 33 jmr. po. yds. 35 26 34 (15) ac. ro. po. 37 1 35 (16) ac. ro. po. 24 8 7 67 4 16 74 35 2i 25 2 18 76 1 38 63 5 9 57 17 5 68 1 36 16 2 23 28 6 25 46 8 4^ 34 3 15 63 3 19 84 2 8 65 14 3 46 1 13 40 34 35 4 31 12 22 0^ 50 1 17 1 1 51 7 15 83 31 1 63 3 22 49 1 87 •J i. xiuCl tau sum ui oiio. o ox. i-xuvii., oo luz. ivui. O dwt., 761b. 4 oz. 12dwt., 38 lb. 8oz. lOdwt., 83 lb. 11 oz 18 dwt., 6 7 lb. 5oz. 7 dwt. I:' J f a €S COMPOUND SPBTR4CTI0W. 18. Find the snm of 37 69 yds. qrs.nla. yds. qr«- nls. (5) 3 15 (6) 4 19 (7) 5 17 (8) 6 18 yds. ft. (11) 44 2 (12) 70 9 m. fur. po, (17) 57 2 28 (18) 61 6 18 (19) 44 6 33 (20) 16 4 4 ac. ro. po. ac. ro. po. bus. pks. (21) 74 1 20 — 44 2 20 (25) 74 1 (22) 44 3 35 — 26 3 37 (26) 83 (23)284 1 15—170 2 17 (27)602 3 (24) 131 3 12|— 79 15| (28) 301 3 tt> ! 3 3 grs. ib I (29) 114 11 7 2 10 — 68 11 (30) 73 8 2 — 44 2 (31) 90 2 5 15 — 54 1 2 3 — 47 3 3 1 — 67 2 1 3^—380 2 3 21— 41 3 1 3 2f po. d. £ s. d, 64—2 5 3| 10|— 2 19 lli 7|— 3 10 6| 5|— 4 3 0| in. yds. ft. in 9i— 26 2 104 Ol— 42 5 54- . yds. m. fur. po. yds 3^34 3 9 1 1 —37 26 5 41—26 7 12 If Oj— 9 7 10 4 gals. bus. pks. gals 1—44 2 1 1—49 3 1 0^—361 2 14 1 —181 1 3 3 grs. 7 2 14 3 1 16 4 2 5 32. From 546 lbs. 10 oz. 2 dwt. 8 grs. take 397 lbs. 11 oz. 15 dwt. 14 grs. Ans. 148 lbs. 10 oz. 6 dwt. 18 grs. 33. From 486 years take 895 years 8 mo. 3 wks. 5 days. Ans. 90 yrs. 3 mo. 2 dnys. 34. From 310 tons 13 cwt. 2 qrs., long' weight, take 77 tone 13ewt. 1 qr. 14 lbs. four times. Ani. 0. COMPOUND MULTIPLICATION. 71 35. From 481 acres 1 ro. 18 po. 11 yds. take 120 ac. 1 ro. 14 po. 18 yds. four times. h Ans. 0. 36. What is the difference betwec* 198 m. 7 fur. 25 po. 2 yd. 1 ft. 10 in. and 300 miles ? Ans. 101 m. 14 po. 2 yd. 2ft. 8 in. 37. A person having 63 gallons of wine, drank, on an average, for five years, mcluding two leap years, one gill of wine a day ; how much remained ? Ans. 5 gals. 3 qts. 1 pt. 1 gill. 38. A man having dug from a trench 126 cub. yds. 16 cub. ft., from a cistern 18 cu. yd. 18 cu. ft. 196 cu. in., and from other places 126 cu. yd. 26 cu. ft., was paid for 196 cu. yd. 26 cu. ft. 1714 cu. in. ; how much remained unpaid ? Ans. 75 cub. yd. 6 cub. ft. 210 cub. in. COMPOUND MULTIPLICATION. 105. CompouiKl Multiplication is the method of multiplying a quantity consisting of several denominations by a given number. 106. To Multiply a Compound Number. CASE I. 107. When the multiplier is under 12. Example 1. — A man sold 6 lots of land, each lot con- taining 4 ae. 2 ro. 14 po. : how much land is there in all ? OPERATION. ac. ro. po. 4 2 14 e 27 2 4 Analysis.— In 6 lots there are 6 times as much land as in 1 lot. We write the multi- pher under the lowest denomination of the multiplicand, and proceed thus ; 6 times 14 po. are 84 poles, equal to 2 ro. 4 po. ; and we write the 4 po. under the number multiplied, and carry the 2 ro. to the next product. Then, 6 times 2 ro. are 12 ro., and 2 ro. added make 14 ro., equal to 3 ac. 2 ro. ; and we write the 2 ro. under the number multiplied. Again, 6 times 4 ac. are 24 ac, and 3 ac. added make 27 ac, which we write under the number multiplied. Prom the above example and illustration we deduce the following general rule : 72 COMPOUND MULTIPLICATION. m m BULB. I. Write the multiplier under the lowest denom- ination of the multiplicand. II. Multiply as in »imple numbers, and earry as in addi- tion oi compound numbers. Mental Exercises. 1. Find the cost of 5 lbs. of tea at 3s. 9d. per pound. 2. What will 9 lbs. of coffee cost at Is. 6d. per pound ? . 3. What will 36 pairs of stockings cost at 3s. lid. per pair ? 4. How many acres are there in four fields each contain- ing 2 ac. 3 ro. 10 po.? 5. If a tailor requires 3 yds. 1 qr. 1 nl. of cloth to make a coat, how many yards must he have to make five coats of the same size ? Exercises for the Slate. SECTION I. Example.— Multiply £l 2s. 9^d. by 4, and £8 7s. 2|d. by 4. OPERATION. OPERATION. £ s. d. £ s. d. £ s. d. 12 9^ 8 17 4 4 23 Test f 4 11 1 (35 8 11 £4 11 1 £35 8 11 40 Multiply each of the following couplets by 2, 3, 4/5, 6, 7, 8, 9, 10, 11, 12. Multiplying them all first by 2, then all by 3, then all by 4, &c., testing them as above. s. (1) 2 (2) 3 (5) 6 a. s. 3 and 17 4 and 16 5| and 15 9| and 12 8^ and 13 d. 9 8 64 2| H ac. ro. po. yds. ac. ro. po. yds.i (11) 2 3 21 16 and 7 18 I44 (12) 5 3 24 19 and 4 15 Hi' (13) 3 2 17 3 and 6 1 22 274 1 (14) 6 27 15 and 3 3 12 154| £ s. d. £ s. d. (6) 4 3 9^ and 5 16 2k (7) 3 12 84 and 6 7 34 (8) 8 19 111 and 1 04 (9) 5 17 6jand 4 2 5} (10) 6 13 9|and 3 6 24 yds. qrs. nls. yds. qrs. iils.'j (15) 3 3 3 and 6 1 (10) 7 2 1 and 2 1 3 (17) 8 I 1 and 1 2 3 (18) 9 2 li and 1 2^ CASE II. 108. When the Multiplier is a Composite number. Example.— What is the weight of 42 bundles of hay each weighing 3 cwt. 2 qrs. 1 2 lbs, (short weight) ? COMPOUND MULTIPLICATION. 73 OPERATION, cwt. qr. lbs. 3 2 12 6 21 2 22 weight of 6 bundles. 7 Analysis. — Multiply- ing the weight of 1 bundle by 6, we obtain the weight of 6 bundles, and the weight of 6 bundles mul- tiplied by 7, gives the weight of 42 bundles. 152 4 weight of 42 bundles. SECTION II. Example.— Multiply £46 13s. lO^d., and £53 6s. l^d., by 48. OPEPvATION, OPERATION. £ s. d. £, s. d. £ s. d. 46 13 10^ 53 6 1^ 6 12 280 3 3 8 639 13 6 4 Test P241 6 ^^^^' I 2558 14 £2241 6 £2558 14 £4800 Multiply each of the folloAving couplets by 14, 16, 18, 20, 21, 22, 24, 27, 28, 30, 32, 36, 40, 42, 45, 48, 50, 54, 56, 60, 64, 72, 81, 96, testing the products as above. £ s. d. £ s. d. \ lbs. oz. dr. lbs. oz. dr, (1) 89 13 6|andlO 6* 5i I (4) 19 14 14 and 80 1 2 (2) 72 14 3h and 27 5 8^ (3) 36 10 114 and 63 9 0| (long WEIGHT.) tons cwt. qrs. lbs. tons cwt. qrs. lbs (7) 83 15 3 27 and 16 4 1 (8) 72 16 2 22^ " 27 3 1 5^ (9) 91 18 3 Hi " 8 10 16^ (10)54 15 2 27|" 45 4 1 04 Multiply each of the above by 100, 110, 120, 121, 132, 144, using two factors, and by 112, 144,420, 441, 504, using three factors. CASE III. 100. When the multiplier cannot he reduced to factors. Example. — How many bushels of oats in 47 barrels, each containing 3 bus. 1 pk. ? (4) 89 15 (6) 72 13 11 and 10 34 arH 27 2 5 121 (short weight.) cwt. qrs. (11) 72 3 (12) 91 1 (13) 12 3 (14) 87 1 lbs. cwt. qrs. 22 and 27 24 "82 19^ " 87 22i " 12 2 lb«. 3 1 ^ 2i 74 COMPOUND MULTIPLICATION. I ■im OPERATION. 47= (5 X 9) + 2 bus. pks. 3 1X2 5 16 1 in 5 barrels. 9 346 1 in 45 barrels. 6 2 " 2 " Analysis.— Multiplying the contents of 1 barrel by 5, ^nd the resulting product by 9, we have the contents of 45 barrels, which is the composite number next less than the given prime number 47. Next mul- tiplying the contents of 1 barrel by 2, we have the contents of 2 barrels, which added to the contents of 45 barrels, gives us the contents of 45 -{- 2i=r 47 barrels. 152 3 in 47 barrels. SECTION III. 67 7fi H^.^of t7°f.^^^,?o"^'^^"- '°"P^^*^ ^y 19' 29, 31, 43. 783 I'OO 'qn' «'««'^' ]\^' ^^[' ^^^' ^^^' 1^9' 1^4' 345,461, section '■^'"^*' ""' ^" *^^ preceding qts. pte. bus. pks. 3 1 and 864 2 1 and 364 2 1 1 and 700 3 bus. (1) 135 (2) 635 (3) 299 pks. 3 1 pal. 1 1 gal. 1 qts. 1 2 pts. 1 1 1 ■ SECTION IV. fion. T L- ''''^ ''f^Pl-^^' ^'J" """"y «^ *he preceding sec- tions Take anij multipher. Prefix to the couplet any two numbers whose sum is one less than the multiplier chosen Multiply both the multiplicands thus-formed bv the niuW pher chosen, and add the products. ' Example l.--Takethe couplet £l6 13s. 9d. and £83 6s. 3d. lake 8 as multiplier. Prefix to the couplet 7, (8 — n Then multiply by 8. A — 4 S 3>v ^* ?!fr7.''''!.' OPERAT^^N.^ ^41^ 13 9 £383 Q 3 ^ 8 £3333 10 £3333 10 £3066 10 £3066 10 £6400 = 82 X 100 o^'^'^f ^'^^^ 2-— Take the couplet 196 cwt. 2ars. 27 lbs. nnd 1x^7 -^VIq \^^,«- l«"^\^^ight' and 48 as multiplier. Pre"- nx 47, _ (29 -f 18), and multiply as before. COMPOUND MULTIPLICATION. 75 OPERATION. •- OPERATION. wt. 29,196 qrs. lbs. 2 27 6 cwt. qi'S. lbs 18,803 1 1 6 175180 1 22 112819 2 6 8 8 1401443 2 8 902556 1 20 cwt. 1401443 qrs. 2 Iba. 8 902556 1 20 2304000 = 482 y^ 1000 SECTION V. Find the value of — 1. 37 tons 13 cwt. 3 qrs. 12 lbs., long weight, X 6 Ans. 226 tons 3 cwt. 16 Iba. 2. 39 m. 7 fur. 28 po. 4 yds. X 6. Ans. 239 m. 6 fur. 12 po. 2 yd. 8. 92 yd. 3 qr. 1 nl. 2 In. X 765. Ans. 71044 yd. qr. 1 nl. 4. 27 y. 54 days 15 h. 29 m. X 921. Ans. 25004 y. 323 d. 4h. 9 m. 5. If 1 acre of land produce 45 bus. 3 pks. 6 qts. 1 pt. of corn, how much will 64 acres produce ? Ans. 2941 bus. 6. If $80 purchase 4 ac. 3 ro. 26 po. 20 sq. yd. 3 sq. ft. of land, how much will $4800 buy ? Ans. 295 ac. 10 sq. yd. 7. What will 16 tons of hay cost at £3 19s. 64d. per ton ? Ans. £63 12s. 8d. 8. Wliat is the cost of 8 bus. 3 pks. of beans at 5^ per quart ? Ans. £6 8s. 4d. 9. If 1 pt. 3 gills of wine fill 1 bottle, how much will be required to fill a great gross of bottles of the same capacity ? Ans. 378 gals. Windsor, March 17th, 1866. Bo't. of J. C. Smith & Co. at $0.11 $ 10. Mr. C. Clarke, 25 lbs. Sugar, 5 lbs. Tea, 4 gals. Molasses, 30 yds. White Cotton, " .62i u .49 .27 Beceived payment, J. C. Smith & Co per John Ne woomb. $16.93^ 76 COMPOUXD DIVISION. «r^^' ^ _ Halifax, March 19th, 1866. William Joxes, Esq., T r^ To W. P. DuFFUs, Dr. Jan. 1. To 15 lbs. Tea, at 50c. s Dec. 6. " 25 lbs. Su^rar, at 10c. Feb. 5. " 1 bbl. Flour, at $9.50, Mar. 14. " 26 yds. Grey Homespun, at 62^0. $35.75 Truro, Feb. 22nd, 1866. Bought of S. Johnson. at 6^(1. £ " 2s. 7id. " Is. 9d. " 7s. 6d. 12. Mr. James Crowe, 1 7 lbs. Sugar, 3^ lbs. Tea, 13 lbs. Coffee, 3 gals. Burning Fluid, 15 lbs. Brown Soap, £3 9 3|- ^f^h ^4''*''"' -^^^ ^ ^*^' 1866-Mr. Andrew Bryden, bought of John Fraser & Co., 1 7^ yds. superfine cloth at 22s. 6d, per yd 27|. yds. drab cloth at 12s. 8d, 34^ drugget at 7s. lOd., 18^ yds. broad cloth at 17s. 4d., 29f yds. serge at 2s. lOd. 1^ TT IV -n , « , Ans. £70 4s. 7id. .f f Y ^ ''''' l""^' ?^'''^^ 1866.-Mr. James Scott, bought of John Young, 24 yds. white cotton, at 27 cents per ya'rd, m yds. flannel at $0.45, 26^ yds. shalloon at $0.37, 54 yds broad cloth at $4.75, 15 yds. broad cloth at $1.82, 27 Vds lining cotton at 7^ cents. Ans. $ 78.53|; COMPOUND DIVISION. 111. Compound BiTision is the method of dividin^r a quantity consisting of several denominations. '^ 118. Compound division is divided into two cases— 1st. When the divisor is an Abstract number. 2nd. When the divisor is a Compound number. CASE I. ExAMFLJZ.^ir 6 acres of land produce 153 bush(;l. 3 pks. 3 qts ot oats, how much will 1 acre produce ? ^ii*. COMPOUND DIVISION. 77 OPEaATION. bus. pks. qts. pts. 6)153 3 3 25 2 4 1 Analysis. — One acre will produce ^ as much as 6 acres. Writing the divisor on the left of the dividend, we divide 153 bus. by 6, and obtain a quotient of 25 bus., and a remainder of 3 bus. We write the 25 bus. under the denom- ination of bnshels, and reduce the 3 bus. to pecks, making 12 pecks, and the 3 peeks of the dividend added make 15 pecks. Dividing 15 pks. by 6, we obtain a quotient of 2 pks. and a remainder of 3 pks. ; writing the 2 pecks under the ©rder of peek«, we next reduce 3 pl^ to quarts, adding the 3 qts. of the dividend, making 27 qt«-, w^hich beiiag divided by 6 gives a quotient of 4 qts. and a remainder of 3 qts. Writing the 4 qts. under the order of quarts, and reducing the re- mainder, 3 qts., to pints, we have 6 pints, which divided by 6 give a quotient of i pt^ which we write under the order of pints, and the work is finished. Example f. — When 96 acres produee 2731J bush. 1 pL 5 qts. of grain, what wHl 1 acre prodaice ? OPEKATIOy- 1)118. pks. gaL qte. 98)2739 1 S '27 bua. 196 779 373(3 pkiu ~79 2 When the divisor is large and not a composite number, we di- vide by long division, ae -showiL in the operation. Fr©m these examples we foria the foBowiai^ rule J ti ; ?tl . 158(1 gai. 98 m, 245(2 qta. 196 •vl 49 2 58(1 pt. Jjas. 27 bu. Z pks. 1 g^. 2 Qt. Sl ^ T8 COMPOUND DIVISION. m^^H* o^«P*^^^T? *^® highest denomination, as in simple nJoJie*rclS^to%ffeTi^^^ »e?lSred^® *^''^''*^ **'*'*^ quotients wiU be the quotient we^mirsw;.7.hf *''lK'''i-'"-^.^^"r ^"*^ i8 a compwtVe number, ▼e may shorten the work by dividing by the factors. pence P"^ '"''''^ '"^*'' ^* ^'^' ^"""^ ^^' "^^^ ^^ ^°"g^* ^^^ ^^^ fo/js. fr? """""^ ""^'^^ '"^^'' ^^ ^^- P^"" ^^- "^^^^ ^^ ^"&^^* w4ht^o/i'bSV' '^ ""^^^'^ '' ^^^- ' *^^-' "^^* ™ *^« what JL U^w^l^'l^^STtxf '' ^^*- ' ^^-^^ ^^"^ ^^^^^^^) Exercises f©r the Slate. SECTION I. Answers to. be tested as in Reduction ascending. (1) £ 1& 16 r2^ 109 1 824 4 858 10 &04 1515 2 '7) 1513 2 {8) 2521 4 (9^ 1488 1 7 (105 1624 4 £ 7947 ie40 2927 6121 4636 21624 25055 48483 12 80886 13 46690 13 6 6 2 4 3 4 6 8-1-14 1U-M4 4|-T--18 7-^-20 0|-^27 -f-96 4i-M21 ~-128 4 -~176 —216 '.Si COMPOUND DIVISION. 75> In the following ible by 9. tons. cwt. 0) 82 (2) 101 (3) (4) 181 2 1631 18 (5) 72036 1 80163 SECTION II. exercises the remainders (if any) are divi- qrs. lbs. oz. drs. (long weight.) 27 3 8-^45, 81 and 171 2 3 11 4- '>4, 63 and 162 1 13 15 -i- 243, 423 and 432 2 8 10 15 -i- 621, 162 and 261 1 27 10 9-1. 765, 675 and 999 5 2 7-1 4302, 5904 and 904S lbs. (7) 46 8) 326 9) 7908 I oz. 5 4 7 lbs. (10) 29 (11) 9876 (12) 305511 yrs. '20) 353 '21) 1278 '22) 7877 '23) 3274 dwt. grs. 11 10 9 2 21 miles, fur. 13) 887 3 *14) 2662 3 l5) 4644 3 'l6) 59816 1 dys. hrs. (17) 1314 (18) 32626 10 (19) 32627 22 mo. 6 1 18, 27 and 36 126, 261 and 396 576, 729 and 891 oz. 3 1 drs. scr. grs. 0-^ 90, 126 and 207 e 1 4 -1 45, 369 and 639 4 2 8-1 702, 837 and 909 po. yds. ft. 30 2 11 i 2 34 1 18 5 m. 9 ~-621, 54 and 702 8 _1 207, 594 and 94« 9 -1 846, 468 and 711 -1 333, 649 and 27 mm. see. 2 42-^45, 72, 81 and 99 8 24 — 612, 711, 549 and 279 4 21-1324,981, 117 and 819 wks. dys. hrs. min. see- 183 6 46 48-7- 63 and 117 199 10 37 12-^972 and 711 4 17 34 48 -7- 567 and 766 1 4 10 10 48 -f- 576 and 667 i 6ECTIOK III. Take any couplet^-as £l34 6s. 84d. and £865 iSs. Sjd.— name any number as divisor — say 17 — then prefix to the ao B COMPOUND DiyiSIOir. «hosen— as and 9, and proceed as in the followin^r (a)17)9.134 J ai ,.>) 17)7^5 '3 .U 61 124 ft 20 lOG 12 5 4i £ 8. D. 22 (a) 537 & 314-^^ 17 (b) 462 13 8|-(-{| 106 10«> 45 34 ]] 20 233 221 12 12 147 136 11 4 46 34 5 1000 12 Tlie teacher will dictate a list of divisors gra^hmllj ri.m^ m difficulty. Prefix to the foflowing < upletTtwo munberf loee sina isone less than the divisor .l.nln Ja .t-!/^" .^ff -.„„^ *.v...^ Lw tuu loiiowing ( unlets two numhpri* TvXT- "'"" '."'' *''^" "'« '^'^- of cloth"at t* ed! per yd., 20^ ds. white cotton at 35 cents per yard • what change did I get out of £5 ? aHiL llf of which IS made up of one hundred and sIk equal ZS there is a c ear gain of «209f,.S0 at the end oTthe veS' How much Should Mr. P. receive ? Ans.lsK VULGAR OR COMMON FRACTIONS. DcfiiiUlons, ^rotation and Nuinerattoii. .oi^* ^ """.i''!' ?'"''"'' '■'*o '' •"l""' parts, one of the Darts i. called one Ihml, two of the parte two tUrk ^ if a unit be divided into 4 equal parts, one of the parts is VULGAR OR COMMON FRAOTIOXS. 85 ., what will IS. $456.37 ns. $39.60. iig a field ns. 9 days. 6d. each, 9 15s. 8d. of getting ch money :8 lis. 8d. m. $72.70. should 11 ns. 4s. 6d. at 7s. 6d. 'd; what 18s. 8fd. iiuch will if nearly. s divided 20 poles 20 farms, us. 2 pks. . $17.20. e capital I shai^ he year, . $39.60. NS. n. of these parts ij8 parts 13 called one fourth^ two of the parts Uvo fourths, three of the parts three fourths, &c. The parts are expressed by figures ; thus, One half is written ^ One third Two thirds u n One fourth is written \ Two fourths " | Three fourths " f Hence we see that the parts into which a unit is divided take their name and their value from the number of equal parts into which the unit is divided. Thus, if we divide an apple into three equal parts, the parts are called thii^ds ; if into 4 equal parts, fourths, &c.; and each fourth is less in value than each third, and the greater the number of parts the less the value of each. When a unit is divided into any number of equal parts, one or more such parts is a fractional part of the whole num- ber, and is called ?i fraction. Hence, 115. A Fraction is one or more of the equal parts of a unit. 116. To write a fraction we require two integers, one to express the number of parts into which the whole number is divided, and the other to express the number of parts taken. Thus, if one orange be divided into 5 equal parts, the parts are called fifths, and three of these parts are called three fifths of an orange. Tiiese may be written 3 the number of parts taken. 5 the number of parts into which, the orange is divided. 117. The Denominator is the number below the line. It denominates or names the parts ; and It shows how many parts are equal to a unit. 118. The Wumorator is the number above the line. It numerates or numbers the parts ; and It show* how many parts are taken or expressed by the fraction. 119. The Terms of a fraction are the numerator and denominator taken together. 130. Fractions indicate division, the numerator answering to the dividend, and the denominator to the divisor. Henctt, S6 VULGAR OR COMMON FRACTIONS. 1. 2. 3. 4. 5. 6. Exercises In WotaMo™ and 3lumevaM«i5. Jixpress the following fractions bv figures ■— 1. seven eighths. ' Three Iwenty-ffths. Twenty-seven nineti/sixths. Seven one hundred aid tii-enty-sevenlhs. 1 wo hundred and tmr four hundred and fifiy-thirds Ivme hundred one thousand and fifty-fourths. 188. To analyze a fraction is to desio-nate and de<,<.rihp j^^numerator and donom.nator. Thus, f is anal^'ed Stf. 4 t^;£:rtst:t^' '''^* "- -'* -^ ^^^^^d int„ ^i^:z^^:\^^^'^-^^^ it is ine value of the fraction is the quotient of 3 -^ 4, or f Read atid analyse the foUmnng fractions ■— p'p *■ '""' '"'" distinguished as Proper and /m- i-.ino'Sof ' A """ ''^. T' ■•'•"^■'^ """"'™'-' - •- than o. tSproS|i"-i: one .hose numerator equals 180. ?RTW "t' l'"""!*'"^* "f Fraction*. '80 ¥ 3. tient of the ntioia. 'fty-tUrds. s. nd describe J zed as fol- livided into aken ; it is md divisor. ■4, orf and /»!- s less than iter equals ^ssed by a t obtained r the laws i REDUCTION OP FRACTIONS. REDUCTION OF FRACTIONS. 87 CASE I. 127- To reduce fractions to their lowest terms. A fraction is in its lowest terms when its numerator and denominator are prime to each other; that is, when both terms have no common divisor. Example. — Reduce the fraction ff to its lowest terms. FIRST OPERATION. A N A L Y s I s. — Dividing both *-|^- :=. ^\^ ■=. 4 Ans. terms of a fraction by the same number does not alter the value of the fraction or quotient (130, Prin. III.,) hence, we divide both terms of || by 3, both terms of the result, ^f, by 2. As the terms of | are prime to each other, the lowest terms of |-| are ^. We have, in effect, cancelled all the factors common to the numerator and denominator. SECOND OPERATION. In this operation we have divided 6)||z:r.|, Ans. the terms of the fraction by the great- est common divisor, (57,) and thus performed the reduction at a single division. Hence the RUIiE. I. Cancel or reject all factors common to both, numerator and denominator. Or, II. Divide both terms by their greatest common measure, or divisor. Mental Exercities. Reduce the followino; fractions to their lowest terms : — 6.8. and 4il. |;II;M; A? A; ^; H; ^f^ ^t; i^V^ if; m 1^6' 1. 2. 3. 4. 6. Exercises fop the Slate. 15 5 675 ¥T¥ Ans. |i I H 6. 7. 8. 9. 10. aojio T94"0 m 684_ ITT? Ans. 1^ fi aultiplies the frac- idea the the frac- is of the ue of the CASE II. 138. To reduce an improper fraction to a luhole or mixed number. Example. — Reduce ^^ to a whole or mixed number. OPERATION. ANALYSis.-Since 7 sevf^nths Ajf^ z=z'd2 -^ 7 zzz 44, Ans. equal 1, 32 sevenths are equal to as many times 1 as 7 is con- tained in 32, which is 4| times. Hence the following — 88 i n I REDUCTION OF FRACTIONS. the result i, a whole mmtT'"'""'"' ^""'^"y ^'""''^ «he numerator, reduced'jL"r"i::S. t™r'"« ''■■^'"""•^ ">o rrao«„„, should be 1 ti Montal Exercises. 24 halv^:7 "'"^' "'^^^^ ^^""g^ -« - 12 halves ? 16 halves ^ to whole or mitfj'n ' iLlJ ' ^' ^'' W, W, W, ^m. 3. 2. 3. 4. 5. 6. 7. Tn US F^'^''^''*^^ «>i- the Slate. Tn S ""5 "" r^"'^'' ^^^ ^^">^ "»«nt'^« ? in ijx of a bushel, how many bushels? In ^f^ of a dollar, how many dollars? ^ -^ of a ton, how many tons 9 Reduce i^3_t to a mixed number. Reduce ^VV^ to a mixed number. Change ^f 1 6^ to a whole number 'iVS -i^ / Ans. 16| 23| . 187| 22 7032 129 T 7 CASE III. ^Iven denominator' "" '"^''^' "^^'^^^ ^^ « /^'^^^''^n ^a.m^ a ExAMPLE.^^Jeduce 15 bushels to sevenths of . T. t, , ANATv<*r« G- ''v^"^"s ot a bushel. - =e™ml7, fr,, v,"" '\^ '""'''«' ""''•e are In practice we mult?riv ?? ^''™nt''« = -H^. busKels, by 7 ,?o 'P '",' *''« . ""'-iber of , . taking heVo'dict fo.? *"°""'«'tor, and fS£Si^?-?^StSf-f^-a " 1 • Mental Exercises. 1. g^;^ "ce 25 bushels to 4ths of a bushel I' r ^f.^f,^^^^«to4thsofayard V ' 3. In 56 dollars how many lOth's of a dollar ? ' OPERATION. 15 7 ^Ans. 2. 3. 4. 5. 6. f^^'^^^i'^Vf^'^Stsm^S-Xir^,,^ -.^ng REDUCTION OF FRACTIONS. 89 nator. le numerator, 15 should be 16 halves? 18 thirds? 1^ ' IQ-} Ana. 16^ 23| . 187| 22 7032 having a )ushel. here are times 7 ^=^^. fiiber of or, and rator of iiinator, lenom- which w rlting :. A man distributed 3 dollars among some poor persons,^ ing 1 of a dollar to each ; how many persons received the 4 givi money V Exercises for the Slate. Change 126 to a fraction whose denominator shall be Ans. -affi. Reduce 145 pounds to 16ths of a pound. Ans. ^'j^ Change 3(55 to the form of a fraction. 1. 13. 2. 3. 4. 5. 23. In 196 gallons how many Sths ? Ans. ^^ Change 187 to a fraction whose denominator shall be Ans. ^F CASE IV. 130^ To reduce a mixed number to an improper fraction. Example. — In &\ dollars, how many eighths of a dollar ? Analysis. — Since in 1 dollar there are 8 eighths, in 6 dollars there are 6 times 8 eighths, 01 48 eighths, and 48 eighths -|- 1 eighth =49 eighths, or ^. From this we derive the following OPERATION 8 49 T KULE. Multiply the whole number by the denominator of the fraction; to the product add the numerator, and under the sum write the denomii:ator. Mental Exeri'dses. 1. How many times \ are in 6| ? in 5*^ ? in 18f ? in 16f ? 2. How many times ^i^ are in SjV? in 8,3 ? in 15 A ? in 22^80? 3. In \Q\ how many thirds ? 4. In 9y^^ how many twelfths ? 5. Reduce 20| to an improper fraction. 6. How do you change a whole number to a fraction having a required denominator ? 7. How do you change a mixed number to an improper fraction ? ^ Exercises for tlic !§Iafe. Reduce the folio winjr T>-ixed numbers to im|,.oper fractions. 1. 71-1 2. 161|i 3. 27ia 4. 39M 5. 126ifT 6. 567.4t Ans. 'i-?-^ 6_4tjJL 33A 81 7. 8. 9 10. 11. 12. 216^ 131|i 156H 1111 AV 1 2S4|M Ans. -&||^ ^853. 90 jiiii I 'fi s lii 1 i i ii f i J'jii BEDUOTION OF FRACTIONS. thev ,r.ay also be rod„m7 fo W. r ^^ /'"^""^ ^>^ ^'^^^^'^on, and all the higher terms mult ^JZ^^J^^^^^^^^- I ExAMPLE.-Heduce « to n iVn ,' , ^""'''^'"'^ t^'^»is. OPERATION. '' "iNir vt^^'\'r^^^" clenominator is 24 24-^-6 = 4 . ANALYSIS.— We first divide 24 ftl' iXi = h '"^"-""^ clenominator, by 6 til' t ^ A 4 - ij nominator of the m;en fraJ-tinn f" The division shows Z^^T^^^ ^'^Y f^^^™ ^ obtain |o, tll^^S^^tS:^ ^^ ^ O^O^. IIi;) JJ:^ In I of 1 r, ^^'»*»* Exereisc«. Tn I i ] . '''''''y tenths ? In 1 of 1 1 "^ '''''''y twentieths ? In ' Jf ^'^ '"^">' thirty-sixths ? In J« of , r '"^"^ ^°"^<^^^"ths ? n. of 1 how many one hundred and eightieths ? Exercises for the Slate Heduec | to a fraction .hose df„'or„;tor U 2«4. 1. 2. 3. 4. o. 1. 2. 3. 4. I^educe H to a fraction who.e denominator is^5 J deduce ^U to a fraction whose denominator is^^lsl^ Reduce I to a fraction whoseMenominator is^^e'oo^^^^ A Coiniiion JOenominatoi. • i to two or more fractions. Thus 4 is ?, ^^"^minator common tor of I, 8 and |. ■"-''"' 4 ^'^ the common denomina- ExAMPLE.— Reduce 4 ht.i fi f^ Red nator. 1. 2. 3. 4. 5. u REDUCTION OP FRACTIONS. 91 ator. by division, Itiplicatiori ; )west terms. inator is 24. de 24, the 6) the de- 'action, to this term 6. at 4 is the of 6. We III-,) and e denomi- • ms of the leths ? 16 5 1 264. A.ns. ! 51. Ans is 3488 denom- ommon lomina- ator. irms of of the faction by the denominator of the first, (136.) This must reduce each fraction to the same denominator, for each new denominator will be il.o product of the given denominators. Hence the RULE. Multiply the terms of each fraction by the de- nominators of Itiply allth e other fra'tions. m^ m^ m m^ m^ m Note.— Mixed numbers must first be reduced to improper fractions. Exercises for tlie Slate. Reduce to equivalent fractions having a common denomi- nator. 1. i,|,|andf Ans. Ill, Ifl, III, ^ i, j\, and f tV' i and f . I, 2|, I and f 1|, -^Tj and 4. CASE VII. 133. To reduce fractions to the least common denominator. The licast Common Denominator of two or more fractions, is the least common denominator to which they can all be reduced, and it must be the least common multiple of the lowest denominators. Example. — Reduce ^, | and | to the least common de- nominator. 2. 3. 4. 5. iTT' ttt 8 60, ^ 8 4 8 T¥' TIT OPERATIOX. 4 8 3 2 4 3X4X2 24 OR, 6 = 2X3 4 = 2X2 '8 — 2X2X2 Therefore 2X2X2X3 = 24 Analysis.— We find the least common multi- ple of the given denom- inators, v^hicli is 24. — This must be the least common denominator to which the fractions can be reduced. We then divide this least common multiple, 24, by the denominator of the ^iven fraction, and Since 24-^-6=:4.-.^Xi = -!^ multiplying each term " 24 -^ 4 := 6 .-. I X I = H ^^ *'^^* fraction by the " 24 -i- 8 = 3 .'. 4 X # ~ M 51^^^'^^«n*' (13«0 we ¥ ^ ¥ . f? have the answer.-- Ilence the jRUXiE. I. Find the least common multiple of the a:iven I 92 ADDITION OF FRACTIONS. /li?;; JP^^l^® *^^^ common denominator by each of the eiven 5^^S^'''''*°''^- ''^.^ ^H.^^iPly each numerator by the corJes^ ?ators?^ quotient. The products wiU be the new n^mel fra^dons. ^' ^^'""""^ """^bcrs must first be reduced to improper 2. If the several fractions are not in their lowest terms, they should be reduced to their lowest terms before applying the rule. Exercises for Keduce the following to their 1- li^' iV 1^ and ^. hhh h tjV and {^ 2. the Slate. least common denominator, ^"s. ^\\, ^, -11^, ^1^ 3. 4. 6. 6. 7. 8. 9. 10. h f iir» and ^^. h TT' H and 5f . f I' ^' i' i and ^. 7|, 5 j-\, 7, and 8. foi i¥(T'and^f t\' ^' II' and 4 J. H' 2^, 3i, 5|, and f t\' ^i' If and 5. ^'^»«-T¥ir»i%V-mHf,i%,ii^ F6' ^H' ¥¥^5 ^^ ^ ^ ii M' H' W M' II' F6' /iT' A' /e W' W' W' W ■§^' ft' I T^A' ik' I^A' 41 fi if W, V/' II If ^¥-' ih W ADDITION OF FRACTIONS. CASE I. 184. To add fractions wing a common denominator. Example.— What is the sum of |, |, | and |- ? 1 -L i ^ 8 Tt^'^'Ts''- .. A ANALYsis.-Since t-hfH-ir-hi = Y = H' Ans. the given fractions • ^ _ , . have a common de- nominator, 9, their sum may be found by adding their nume- rators, 1, 2, 3, and 7, and placing the sum, 13, over the com- mon denominator. We thus obtain J/ - i^ the required sum. Hence the » a^ m vi comm''o?-d^nom*nlt''o"r!^'"^*°"*^ *^^ ^'^^^ *^^ «^^ °^^^ t^« '^': '.^ ADDITION OF FRACTIONS. as improper 1. 2. 3. 4. 5. and ^^. £xcreli»es for the Slate. Add ;V. ^, A, tV. Add J, ^'^, 1^, ^7^, and |f Add ■-^, ^, ^^, ^o_^, ^1 and ^J. Find the sum oi^, ^t^-, ^i and ||. Find the sum of ^j^/^, ^/^^ Mi and ^|f Ans. 3^ 2f If CASE II. 199. T > add fractions having different denominators, ExAMPi :.— What is the sum of | and | V ^ _Fi ST OPERATION. ANALYSIS.— In f-f-TF — I "hlf — H=^1H Ans. whole numbers we can add like num- bers only, o those of the same unit value ; so in fractions we can add the numerators when they have a common denomi- nator, but iiot otherwise. As | and |- have not a common denommator, we first reduce them to a common denominator, (132 or 133) and then add the numerators, 36 -|- 35 := 71, the same as whole numbers, and place the sum over the com- mon denom'nator. Analysis. — Since it is easier to perform addition when the num- bers are in columns, we therefore place the new numerators as in addition of simple numbers and write the common denominator at the side. From the above examples we have the following RULE. I. Reduce the fractions to a common or to their least common denominator. k"oix common denomx^naSfr!^''""' *^^ ^"^^^ '^^ "^"^ ^""^^ **^« Note.— If the amount be an improper fraction, reduce it to a whole or a mixed number. SECOND Ol {ATION. 1=35} ^' ^- ^' ^^' li == 1|| Ans. 1. 2. 3. 4. 5. Exereises far the Siate. Add ^-, |, I, I and ^% Add |, |, f and y\. Add ^\=V' A' i\ «nd jV- Add I, H, i|. If and |f Add *, A. U. II. U. II and 11. •-•- iv- ii- i.i- ii-- 4 5 ii- i C14401 ~io van i 94 SUBTRACTION OF FRACTIONS. CASE in. 130. To add mixed numbers Example.— Add 31, 5f , and 7.\ OPERATION. ^ 34 == 8) 5| =:12^16L. CM. ^V= O or CD. ANALYsis.-Thesumofthe fractions, ^, I, and ^, is IX; the sum of the integers 3, 5, and 7, is 15 : and the sum of both fractions ami integei-s is 16i\. Hence the followinjr— 1. 2. 3. 4. 5. Ans. 19^ 505^1 273|| ■* 16j<^ Ans. tlS?JfdT^ei?sums!'*°*^*'^« *^^ ^^*^8^«" separately, and Exercises For the Slate. Add 5J 3J, 4| and 6}. Find the sum of |, 1^, io|, and 5. Find the sum of 126i, I83f, and 196^%. What is the sum of 3i, 126|, and 144A. -.r^n Bought 5 lots oflandcontaining 127 acres j^.. „ ^^JJ /n fheTSts^^^^ ^^^^^' ^"^ ^* --^ '^""^^^^^^^^^ 136X dofkrf ^fxl^^'l ?""fl^* 1''^ ^"«^^^« of'^wh'eat'iS i 7^1 *^7^^^^' 3,6 ^i bushels of barley for 21 9| dollars 'iOfiil Sm h1Vu?an/T '"'^^ t''>r^ ^^^^ man/Lsh&gS «ia he buy, and how much did he pay for the whole ? Ans I l^OOU bushels. 1 592|| dollars. SUBTRACTION OF FRACTIONS. CASE I. Ji.XAMPLE,-~Fiom ^tj. take ^8^. OPERATION. AwATvoTc c t 8 -_ 7-a ^ . -^N^LYsis— Since the TT — iV — \if = ^ = I given fractions have a com- mon denominator, 1 0, we find feUBTRACTIuN OF FRACTIO. 95 rence by suhtracting 3, the less numerator, from 7, 1 ■ ' the di the crre.rer, ami write the remainder, 4, over the unmon d( >n. nator, 10. We thus obtain ^ =: ^, the required dif- fert ice. Hence the tbllowing — B. LE. Subtract the im^ Jr of the subtrahend from the numi i-ator of thn a i^, and place the difference over the common denoi lator. 1. 2. 8. 4. 5. 6. I^^xerelses for the Slato. Fr ui I ke |- From ^ take j^. From \^ take ^*j. From ^Vg- take ^^. From jYff take ^y From ill take ^^. CASE II. Ans. ^ A 138. To subtract fractions having different denominators. Example. — From f take f . OPERATION. = If - It ^ ^-/^ = ii, Ans. OR, f =:35> 1=24;^ 56 C. D. tt' Ans. Analysis. — As in whole num- bers we subtract like numbers on- ly, or those hav- injrthe same unit value, so, we can subtract fractions only when they have a common denominator. As | and f have not a com- mon denominator, we first reduce them to a common denom- inator, and then subtract the less numerator, 24, from the greater numerator, 35, and write the difference, 11, over the common denominator, 56. We thus obtain ^, the required difference. Hence the following — KULE. Beduce the fractions to a common denominator and subtract as in the former rule. Exercises for tlie Slate. 1. From I take |. 2. From ^ take ^. 3. From ^ take ^. Ans. ^ fi I ^ — ^ ,b. IMAGE EVALUATION TEST TARGET (MT-S) / ^/ 7 ^ o ^ .<$- &5 fe ^ 1.0 I.I 1.25 M 1.8 1-4 IIIIII.6 P> <^ ^% /M A -y ^M w om Phote)graphic Sdences Corporafon 1> ^ v w^^' 23 WEST MAIN STREET WEBSTER, N.Y. 14580 (716) 872-4503 fo g 5 ^ 96 SUBTRACTION OP FRACTIONS. 4. From 1^ take ^. 5. From || take ^V^. m CASE III. 18©, To subtract Example. — What OPERATION. 18i = 18^«^ n= 7^ OR, 10 H 18|=3> 12 CD. leaves 10. We thus Hence the foUowing- mixed numbers. is the difference between 18:^ and 7 J. Analysis. — We first reduce the fractional parts, ^ and ^, to a common denominator, 12. Since we cannot take ^^ fi'om j^^, we add 1 = i| to ^^, which makes l|, and ^^ from \^ leaves \^. Again, having added 1 to the upper number, we must add 1 to the lower number, so that the dif- ference between the t\(ro numbers may not be altered ; and adding 1 to 7 we have 8, which taken from 18, obtain 10^^ the difference required. — RULE.— Reduce the fractional parts to a common denom- inator, and then subtract the fractional and integral parts separately. Or, We may reduce the mixed numbers to improper fractions, and subtract the less from the greater by the usual method. 1. 2. 3. 4. 6. 6. Ans. 3| Exercises for the Slate. From 8| take 5|. From 27| take 10^^. 8j^ From 5^ take 4f. | From 27 take 18^. 8( F.om 3^ take 1,^^. 2^ From a barrel of Kerosene oil containing 56 J gallons 27^ gallons were drawn ; how many gallons remained ? Ans. 28f 7. If flour, which cost $6 J per barrel, be sold for S7| per barrel, what will be the gain per barrel ? Ans. $|- 8. From the sum of 5^, 31 and 8^^ take the sum of 2-|^, 7i and ^. Ans. 6-^^ 9. What fraction added to ^ will make ^f ? Ans. y|^ 10. A man having 368|^ dollars, paid $100X for a horse, S25A for a set of harness, $X for a whip, and $175^^ for a waggon ; how much had he tefl ? Ans. md-^ 9 6 1. 2. 3. 4. .5. 6. 7 If 42 43| MCTLTIPLICATION OF FRACtlONS. St MULTIPLICATION OF FRACTIONS. CASE I. 140. To multiply a fraction hy an integer. Example 1. — If 1 yard of cloth cost £|, how much Trill 7 yds. cost ? OPERATION AxALYSis, — Sincc 1 yd. cost 3 f X 7 =: ^ =: 5^ Ans. fourths of one pound, 7 yds. will cost 7 times 3 fourths of one pounds or 21 fourths^ equal to £5^. A fraction is multiplied by multiplying its numerator, (19Q.) Example 2. — If 1 pound of Tea cost ^^ of a dollar, how much wiH 4 lbs. cost ? Analysis. — Since 4, the mul- tiplier, is a factor of 20, the denom- inator, of the multiplicand, we per- form the multiplication by dividing the denominator, 20, by the mm- tiplier, 4, and we have 4 =: 1^ doUai-s. A fraction is multiplied by dividing its denominator, (l!86). Hence the following — RUIjIj. Multiply the numerator of the fraction by ths whole number, and write the product over the numerator, Diride the denominator b j the whole number, when thia can be done without a remainder. operation. 6 d 1. 2. 3. 4. .6. 6. H 7 li 42 43| Exercises for the Slate. Multiply I by 6. Multiply 1^ by 9. Multiply ^ by 5. Multiply X by 84. Multiply ^ by 55. Multiply el by 7. operation. or, H = ^ i^ X 7 = iji = 43| 43f Analysis. — In multi- plying a mixed number, we first multiply the frac- tional part, and then the integer, and add the two products, or we reduce the mixed number to an improper fraction, and tiien multiply it. li Ki MtrLTTiTLrcATtcwr Of rRAcno^!9. 7. Multiply l-7|ly &, 8. Multiply ^Vt by 7. 9. Multipiy 16| by 16?. 10. Multiply {U by 544. Jl. If 1' ton of hay cost $ Afa. 85 f 1 9 6- 266 404 JAj ^^ae wiS T2 tons cost. ? Ans. $105| 12. What win 14 yds. c^silk cest at Ij dollars per yard / Aas. $26| CASE II. 141. TV muUipiy a vfhoie mmnhtr hif a fraction. E3CAMPI.E.-— A* 83- dollars a» acre, how mijch will | of afi tcre cost ? opekatio^n:. 83* p¥iee of 1 acre*. 3 1)249 =: cost of ? aefiesr. 49| = " I of aft aerc. A5f ALTSI8. — MwKiply' ing the price of 1 acre by 3, we have the price of 8> aeres ; and f.3 ^ of 3 acres' is the safiae as f of 1 acre, we divide the cost of £P acres by 5, and we have the eost of ^ of an acre. — ' jKence the following— RULEI, Multiply tHe given dumber "by tie fiumerator* and divide the product by the denominator. NoTS. — When the denominator is exactly contained in the given dumber, it will be found easier to tirst divide by it, and then multi- ply the quotient by the numerator. 1. 2. 8. 4. 5. e. 2f6 H 120 Kxefeises for tlae I81at«« Multiply 4 by ^.• Multiply 165 by ^. Multiply 457 by ^^. What is ^ of 4261, What is ^ of 1644, Multiply 26 by 5|. OFERATION. Or 5| = 26^X^ = ^^ 139| Ans. =:|of26 Ans. 2f 20* 266 J^ hQQ^- -9 959 139|, An», AlTALYSIS. — We BEiuItiply by the inte- ger and fraction se- parately, and add the products ; or reduce the mixed number to an improper fraction, and then multiply by it MtTLTIPLlCATlON OY FRACTIONS. 9^ 7. Multiply 83 by 7f Ani. 59 7| 8. Multiply 45 by 8f 375 9. Multiply 156 by f|. 108 10. If a man walk 16 miles in one day, how many will he travel in 1 1 2| days ? Ans. 1 798 11. At 18 dollars per ton, what is the cost of 18| tons of hay ? Ans. $338 CASE HI. 14a. To mnltiply a fraction by a fraction. Example 1. — At | of a dollar per yard, how much will | of a yard cost ? OPERATION. Analysis. — Since 1 }Xi = AX3z=^ Ans. yard cost | of a dollar, ^ of a yard will cost ^ of |> which is ^ of a dollar ; and as ^ of a yard costs ^ of a dol- lar, J of a yard will cost 3 times as much, or A X 3 = t|^ It wfll read^' ^® ^^^^ *^^* ^^ ^^^® multiplied together the two numerators, 3 and 3, for a new numerator, and the two denominators, 8 and 4, for a new denominator, as shown in the whole work of the operation. Hence for multipUcation of fractions we have this general B,UI, anti reduce thi:« fraction lio a decimal.. !Exereises for i^e INlate. Reduce 1 . £ 7». 4 |d. to the dcci kkiI of £ 1 . 2. 3. 4k. 5. 6. 7. 8. D. 30. 10s. Ofd. to the decimal of £l. Ans. .£.3!7 £.503125 3 pks. 1.12 qt. to the decimal of a bashel. .785 bi*. lOoz. 13 dwt. 9 grs. to the decimal of 1 lb. Troy. Aaa. .889«M)25 lb-. 2 oz. 1"3 (Jwt. to the deeimal of 1 fb. .22083 lb. 4 Va. 2 SG. to the decimal of 1 oz. 48.083 oz. 4 da. 18 hrs. to the decimal of 1 week. .67857142 wk. 2| iiKihes to the decimal of 2\ miles. 3^ acres to the decimal of 3f sq. yards. ^ of a crown to tlie decimal of 2 la. .000015 -j^ 5212.30769^ JL488M3aS FROJiTTSCUOXJS EXETlCnS*ES. HI l?ROMTSCUOUS EXERCISES IN THE PRECEDING RULE&. 1. Reduce ^, |, | and 6 t© fraotlons havmg a conimoB denominator. -^i^s. -1^, ^-g-^ ^f , \-^ 2. What is the value of .75 of an ell English ? Ans. 3 qr. 3 nails. 3. Add 4^, 3^, 5|, I of 8^, and ^1- Ans. 15|^ 4. What number multiplied by 4 will produce 1141^^? Ans. 3043| ;S. If the dividend be f a«id the (|uatierLt J, what is the divisor ? ^^^- * 6- If 3^0 ®^ *^ barrel of flour cost $2.34, what will be the cost of a whole barrel. Ans. $7.8® 7. If *,he smaller of two fraetiofts be f^, aad theia- differ- ence -^-g, what k the greaafcer ? Ans. ^ 5. Find the diffttrence between f -Qf 6fj and -| >« ""i*«» fio-nrA of ttie product for shillings. PRACTICE. 117 \\ Exerdiies for tlie Slato. Find the value of (1) 426 at £7 83. and at £2 123. (2) 446 " X4 3s. and " £5 ITs. (3) 642 " £b 7s. and " £i 138. (4) 741 " .£6 93. and " £3 lis. (5) 684 " £9 13s. and " £0 78. (6) 563 at £6 78. and at £3 13«. (7) 851 " £8 13s. and " £1 7s. (8) 754 " £6 17s. and " £3 38. (9) 694 " £4 15s. and " £5 58. (10) 339 " £5158. aad " £4 58. Ans. £484 19s. £999 15s. S17301.60. S23 720.80. CASE IV. 160. To find the value of any number of articles^ when the price is given in shillings and pence, or in pounds, shillings and pence. Example 1. — If 1 yard cost 16s. 3d., what will 127 yards cost at the same rate ? OPERATION. . 11. 183 at £2 13s. 12. 129 " £7 153. 13- 486 " £8 18s. 14. 596 " £9 19s. 127 at 16s. 3d. per yard. 10s. Is ^ of £1 63 10 = price at £0 10 6s. is 4 of 10s. I 31 15 0= " 5 Is. 3d. is ^ of 5s.! 7 18 9= " 13 £103 3 9 = price at £0 16 3 Example 2. — Find the price of 187 yards at £2 ISs. 4d. per yard. OPERATION. 187 at £2 13s. 4d. 2 10s. is i of £1 3s. 4d. IS J^ of 10s. 374 = price at £2 per yard. 93 10 z=: " 10 " 31 3 4 ::= " 3 4 " £498 13 4 = price at £2 13 4 From the foregoing we have the following BUTjB.— Multiply the quantity by the pounds, if any, and ta! liquet parts for the shillings and pence. 118 PRACTICE. Vlnereimem for tlic Slate. (1) 132 (2) 156 (3) 999 (4) 3fi5 (6) 831 (6) 144 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. at 3fl. '• 3s " 18s. " 12s. " 17s. " lis. 2436 2739 4938 9852 3482 9584 7947 543 296 568 496 624 9(1. and 4(1. and 4d. and Od. and 5d. and 7d. and at 15s. at 10s. at 15s. at 15s. at 19s. at lis. at 18s. at £l at £2 at £2 at £3 at £8 at 10s. 3d. " 16s. 8d. " Is. 8d. " 7s. 6d. " 2s 7d. " 8s. 5d. lOd. 7^cl. IIH ll^d. 6fd. Ojd. 8s. 8d. 13s. 4d. 18s. 4d. 19s. 8|d. 18s. llfd. (7) 127 at 68. 7id. andatl4s. 4id. (8) 395 " 12s. 2Jd. and " 78. 9id. (9)987"12s. lid. and "7s. lOJd. (10) 1118 at 14s. Sid. and at 5s. 3|d. (11) 5639 " 189. 4id. and " Is. 7id. (12) 3017 " 16s. 2|d. and " 3s. 94d. Ans. £1827 Os. Od. £1483 12s. 6d. £3857 16s. 3d. £7850 16s. 3d. £3471 2s. 4id. £5540 15s. Od. £7160 lis. 6|d. £778 6s. Od. £789 6s. 8(L £1656 133. 4d. £1976 58. Od. £4689 68. Id. CASE V. 161. To find the price when the quantity contains a fraction. Example. — What is the valuo of 136| yards of cloth at 17s. 6d. per yard? operation. 136| at 17s. ed.per yard. £136 12 6 =: price at £ 1 per yard. 10s. =: ^ ^1 5s. r=^ lOs, 2s. 6d. :=: \ 5s. 68 6 3 =: price at lOs. " 34 3 li = price at 5s. " 17 1 6| =: price at 2s. 6d. " £119 10 11}= price at 17s. 6d." SECOND OPERATION. 136| at 17s. 6d. lOs. zr:^of£l 68 ~ price of 136 at 10s. 5s. — h of 10s. 34 — " " 5s. . 6d.— ^ of 5s. 17 — " " 2s. 6d. 1 of 17s. 6d. £ 10 ni= 1 119 1 /\ 1 1 1 lit — Note.— The price of 1 may also be found by means of aliquot parts. PRACTICE. THIRD OPERATION. 136| at 17s. 6(L 8 119 109. = ^ of£l 6s. = I of 10s. 23. 6d.= | of 53. 1093 546 10 273 5 136 12 6 Dividing by 8)956 7 6 £119 10 Hi 1 78. 6d. OR, 1361 at 8 i 1093 ^ 109 6 9 2 2 1 2 H 2 2:^ of£l of 2s. of 2d. £119 10 Hi BT DOLLARS AND CENTS. 1464 at $3.50 cw. 146.625 3.5 733125 439875 1464 at $3.50 146f 4 = i of 1 1=1 of! 21.00 140.0. 350. 1.75 .4375 $513.1875 $513.1875 Exercises fojr tlic Slate. 1. 284^ at 10s. Ans. £142 5s. Od. 2. 354^ " 13s. 4d. ^236 6s. 8d. 3. 968f " 15s. 6d. $2923.04 4. 279^ " $4.90. ' $1368.50. 120 PRACTICE. ,1 \ 6. 512|iitS11.56f 6. 849^ " $15.7l|. 7. 440fff at£2 12s. l^d. 8. 678f at$19.5G|. 9. 427^ " £5 19s. 7^(1. 10. 651^^ " $15.75. 11. 542 ji^ " £2 6s. 9(1. 12. 491^ " £2 5s. 6(1. S5768.88 $1334».40j^. £1149 Is. Od. $11316.87^V £2558 128. lid. $10264.27^ £1267 17s. llfd. £1117 14s. 6d. CASE VI. loa. To find the value of a comp^u id quantity when the price of a unit of tie quantity is given in dollars and cents. Example 1. — Find the value of 126 cwt. 3 qrs. 14 lbs. (long weight) at $14.62^ per cwt. OPERATION. 126 cwt. 3 qrs. 14 lbs. at $14,625 126 2 qrs. 1 qr. 14 lbs. ^ of 1 cwt. ^ of 2 qrs. I of 1 (jr. 1842.75 7.3125 3.65625 1.828125 u price of 126 cwt 2 qrs. 1 (^r. 14 lbs. $.1855.546875 = priceofl 26 cwt.,&c. Example 2.— What will 13 cwt. 2 qrs, 15 lbs. (short weight) of oatmeal cost, at $3.75 per cwt.? OPERATION. 13 cwt. 2 qrs. 15 lbs. at $: .75 per 100 lbs. 13 $48.75 = price of 13 cwt. 2 qrs. = i^ of 1 cwt. 1.875 = " 2 qrs. lOlbs. =:^of2qrs. .375 = " 10 lbs. 5lbs. = ^of lOlbs. .1875= " 5 lbs. $51.1875 = price of 18 cwt., &c. PRACTICE. 131 OR, 13 cwfc. 2 qrs. 15 lbs. = 13.65 cwt. at $3.75 3.75 68 25 955 5 4095 $51.1875 = price as before. NoTK.— In calculating, it will often be found more convenient to reduce the lower denoniinationa to a decimal of a higher, and multi- ply as in decimals. Example 3.— Find the price of 14 ac. 3 ro. 35 po. at $22.16^ per acre. OPEUATION. 14 ac. 3 ro. 35 po. at $22,162 per acre 14 2ro. 1 ro. 20 po. 10 po. 5 po. ^ of 1 ac. I of 2 ro. ; ^ of 1 ro. : I of 20 po. Jof lOpo. 310.268 —r- price of 14 ac. 11.081 «= 2ro. 5.5405 — 1 ro. 2.77025 20 po. 1.385125 x= 10 po. .6925625 =-= 5 po. $331.7374375 = price of 14 ac, &c. OR, 14 ac. 3ro. 35 po. = 14Uac. = 14.96875 ac. at $22.16^ 22.16^ 299375 8981250 1496875 2993750 2993750 $331.7374375 Ans. as before. From these illustrations we deduce the following general RULE. Multiply the price by the integral part of the auantity, then separate the remainder into ahquot parts ol i oi Ziie quantity v.-iiust; ia-.c i= sivc^ii x^^ ^^•^•^^-. ^ — each other, as the case may require. Or, 122 PRACTICE. Reduce the quantity to a decimal of the same denomina- tion as the quantity wnose rate is given, and multiply as in decimals. Exercises for tlie Slat«. Answers. $2490.90 $1515.616G-|- $268.58109.3 £406 16s. l|d. £109 9s. 8.46-j- £397 18s. l^d. $725,025 $3318.5267 482.03395 1901.883375 £371 i7s. 2i|| $2854.196-f- $33,206^ $75.51094- $408.5317-1- $1 66 1.94|-J $3777.72^^V $6993.69^7^ $5697.90| $2793.82,^^ 103. To find the value of a compound quantity ivhen the price of a unit of the quantity is given in pounds, shillirys and pence. Example 1. — Fiad the price of 3 cwt. 2 qrs. 4 lbs., long •weight, of flour, at £l per cwt. Analysis. — Since 1 cwt. costs £l, 8 cwt. will cost 3 times as much, or £3. Again 1 qr. will cost ^ of £l, or 5s., and 2 qrs. will cost 2 times as much, or 10s. Lastly, 1 lb. will cost ^^ of .'^s. or 2i|^d., and 4 lb. will cost 4 times 2|d., or 8|d. 1. cwt. qrs. lbs, (long weight.) 163 3 14 at $15.20. 2. 3. 115 2 17at$13.10|. 18 3 21 at $14.18}. 4. 136 2 27 at £2 19s. 6d. 6. 18 3 24^ at £5 15s. 5^d. 6. « (short weight.) 181 3 15 at £2 3s. 9d. 7. 165 2 22 at $4.37f 8. 172 3 18 at $19.19. 9. Ill 1 1 at $4.33:^. 10. ac. ro. po. 121 3 14 at $15.61. 11. 136 2 19 at £2 14s. 5^d. 12. 183 1 38^ at $15.55^. 13. yds. qrs, nls. 15 3 1 at $2.10. 14. 16 2 3 at $4.52|. 15. 28 3 3^ at $14.10|. 16. tons, cwt, qrs, 113 12 3at$14.62i. 17. 165 13 2at$22.80|. 18. 567 2 3 at $12.33|. 19. 384 19 34 at $14.80. 20. 144 18 3| at $19,27^. CASE VIL OPERATION. cwt. qrs. lbs. 3 2 4 5 2| £3 10s. 8|d. practice:. 123 Example 2.— Find the value of 16 cwt. 3 qrs. 14 lbs., long weight, at £2 13s. 6d. per cwt. OPERATION. cwt. qi-s. lbs. 16 3 5 14 JlTT £16 17 6 = price at £l per cwt. 2 10s. = ^of£l 2s. 6d. == ^ of 10s. Is. = ^^ of 10s. 33 15 8 8 2 2 16 : 9 : 10^: price at £2 " 10 " 2 6 " 10 £45 1 6 cwt. 3 qrs. OR, 14 lbs. at £2 13 6 16 2 qrs. = ^ of 1 cwt 1 qr. == ^ of 2 qrs. 14 lbs. ^= I of 1 qr. 42 16 1 6 13 6 == 9 = 4| = 8i = price at £2 13 6 cwt. qrs. lbs. price of 16 " 2 "010 14 <( £45 2 9S= price of 16 5 14 From these examples we deduce the following general KUIili. Find the value of the quantity, if any, of which the rate is given, by Compound Multiplication, then sepa- rate the remainder of the quantity into aliquot parts, as in the former rule. Or, Find the price of the given quantity at £1, by one of the following rules, then multiply the result by the pounds, if any, in the price and separate the shilUngs and pence into aliquot parts. RULES. In calculating tlie price of 1. Hundreds^ quarters and pounds, long weight, at S.I per cwt., multiply tlie pounds by 2| for pence, and the quartei-s by 5 for shillings. 3. Hundreds, quarters and pounds, short weight, at £l per cwt., multiply the pounds by 2f for pence, and the quarters by 5 for shillingg. m 124 PRACTICE. [i: 8. Tons, hundreds and quarters, at £l per f on, take the tons and hundreds as pounds and shillings, and multiply the quarters by 3 for pence. 4. ^cres, roods and poles, at £l per acre, multiply the poles by U tor pence, and the roods by 5 for shillings. 5. Yards, quarters and nails, at £1 per yard, take each quarter at 5s. and each nail at Is. 3d. «. Oz.,dwts. and grains, Troy weight, at £l per ounce, take the ounces as pounds, the pennyweights as shillings, and half the grains as pence. 164. In calculating by means of aliquot parts, it will often be more convenient to use the decimal form ot remain- der instead of the common fractional. It will be sufficient to carry the decimals to two places, as in the following example. Example 3.— What will 126 ac. 3 ro. 15 po. cost at £2 lis. 3d. per acre ? OPERATION. J 26 3 15 at £2 lis. 3d. 6 1^ £126 16 10.50 == price at £l per acre. 2 10s. = ^of£l is. 3d. -=^ of 10s. 253 13 9.00 63 8 5.25 7 18 6.66 : price at £2 per acre " 10 " " 13 " £325 9(91-=priceat£2 11 3peracre) NoTK.— In working by this method the penny is supposed to^be divided into 100 equal parts. Hence .25d. = i, .50d. = i, .75d. — f In valuing the decimal in the answer we consider to which of these tt is nearest and value it accordingly. General Exercises. 1. 18967 at $15.01. 2. 13468 at £2 18s. 3. 1768 at £9 13s. 4. 1476 at £11 15s. 6. 1367 at £3 19s. 6d. 6. 387 at $14.83^. 7. 1429 at $18.62. 8. 148^ atSll.lO^. Ans. $284694.67 $190078.38 £17061 4s. Od. $84402.60 £5433 16s. 6d. $26607.98 $1646.81| / PRACTICE. 125 9. 367^ at £11 13s. 6d. $20859.44 10. 463| at $18.18f $8430.56| 11. 5191^ at £1 Os. 6d. £532 18s. 3^d. 12. 345^1 at $G.72f cwt. qrs. lbs. (long weight,) $2325.5890625 13. 15 3 16 at£0 13s. 6d. per cwt. £10 14s. 6.642-- 14. 14 2 24^ at £3 18s. 6d. " £57 15s. 5.062d. 15. 19 3 23 at $15.62^ " (short weight.) $311.80^ nearly. 16. 17 3 15 at £3 15s. 6d. " £67 lis. 5.4d. 17. 19 3 14 at $18.61^ " $370.20^ 18. 23 3 11 at$12.32| " $294.07^ nearly. 19. 26 2 17iatl9s. 7^d. " £26 3s. 5.96d. 20. 136 2 10^ at £3 16s. « £519 Is. 9.48d. 21. 48 1 27 at $7.87^ " tons. cwt. qrs. $382,095 22. 11 13 3 at £5 16s. 3d. per ton. £67 18s. 8.06d. 23. 14 17 2 at $18.88 " $280.84 24. 13 14 1 at $27.33 " $374.7626 25. 18 19 3^ at £2 19s. 7^d. " £56 12s. 6.05d. 26. 84 3 2^ at £11 3s. 4|d. « yd, qrs. nls. £940 3s. 3.4d. 27. 15 3 1 at $2,18 per yd. $34.47125 28. 18 2 3 at$11.16 " $208.55^ 29. 15 1 2 at 13s. 9id. " £^0 12s. 0.56d. 30. 25 3 2 at 18s. lid. " £24 9s. 5.62d. 81. 27 3 1^ at $4.16^ " $115,969 32. ac. ro. po. 126 3 14 at £2 19s. 8d. per acre. £378 7s. ll||d. 33. 384 1 27 at $18.55 « $7130.96- _ 34. 361 2 19 at $18.27^ « $6608.58- ■> 35. 84 1 37^ at $10.19 " $860,895- — 36. 172 1 15 at£l 18s. 9d. " oz. dwt. grs. £333 18s. 3|^d. 37. 14 12 9 at $1.62 per oz. $23,682-}- 38. 17 3 19 at $18.50 " $318,007-- 39. 12 13 20 at £2 3s. 6^d. " £27 12s. 7.39d. 40. 15 11 16^at£l 19s. 6d. " g-al. qts. pis. £30 15s. 6.993d.-- 41. 13 3 1 at 13s. 6d. per gal. £9 7s. 3|d. 42. 18 3 Oat $1.10 " $20.62^ 43. 27 1 i at $14.16 " $387.63 44. 9 1 at l8. 9d. p«r quart. $15.7 5 Hll 126 PROPORTION. PROPORTION. 165. In the foregoing exercises on the Rules of Practice there are apparently only two terms given, the price and quantity ; but in each there are really three things given. Taking the last exercise as an example, it may be written thus : — If 1 quart of oil cost Is. 9d., what is the cost of 37 quarts V Analysis. — Here the price of a certain quantity is given, and we wish to know the price of so many times that quantity. 37 quarts are 37 times 1 quart, therefore the price of 37 quarts will be 37 times the price of 1 quart; that is Is. 9d. X 37 = £3 4s. 9d., or $15.75. Example 2.— K 6 lbs. of tea cost 18s. 9d., what is the cost of 48 lbs. ? Analysis.— Here the price of 6 lbs is given, and we wish to know the price of 48 lbs. 48 lbs. are 8 times 6 lbs., there- fore the price of 48 lbs. will be 8 times the price of 6 lbs. ; that is 18. 9d. X 8= £7 10s. Questions of this sort, in which the quantity whose price is sought in so many times the quantity whose price is given, arc generally solved by Multiplication. In all such questions there are three numbers given, two being of the same kind, and the third of a different kind ; hence it is sometimes called the " Rule of Three." A fourth quantity is in all cases sought, which is of the the same kind as the third given. Exercises for the Slate. 1. If 5 yards cost £9, what will 20 yards cost ? 2. If 3 yds. cost £2 5s. 5^d., what will 24 yards cost ? 3. How much must be paid for 32 yds., if 4 yards cost £6 16s. 4d. 4. If a man walk 81 miles in 3 days, how far will he walk in 1 5 days ? 5. If 2 quarts cost $1.53, what cost 2 gallons? G. The wa'n'P. of 8 men amount to £7 Gs. 5^d, what will the wages of 128 men amount to ? 7. If ^ lb. of tea cost 22^ cents, what cost 8 Ibi. ? FROFcmTicrnr. 127 S. How many yds. of cloth at 3g. 6d. are worth 27 yds. at 14s. per yard V 9. If 8 yards cost £2 8s,, what is the price of 2 yards ? Analysis. — Here the quantity whose price is sought is an even part of that whose price is given. Since 2 yards is the fourth part of 8 yards, the price of 2 yards will be the fourth part of that of 8 yanls. Now ^ of £2 8s.==£0 12s., which is the answer required. Such questions are solved by Division. 10. If lbs. of butter cost $1.62, what will S lbs. cost ? Jl. If 32 cwt. cost £72, what cost 4 cwt. ? 12. If &6 sheep cost £73 4s. what will 7 cost ? 1 3. If the school tax on $ 1 6 7 3. 1 2 is $i) , what will it be on $418.28 ? 14. How lon^ will 36 laboui*ers take to dig^ a trench which 3 2 mfen can dig in 2 7 days ? 1 5. A firnt expended £ 1 90 1 Ts. Gd. in 75 days, what will be the expenses for 25 days ? 16. If 8 yards cost £4 123., what will 13 yards cost ? Analysis. — Here the quantity whose price is sought nei- ther contains, nor is contained in, the quantity whose price i» given, an even number of times. We therefore find the price of 1 yard as an intermediate step, the number 1 befng m both quantities. Thus, since 8 yds. cost £4 12&., I yd. cost | of £4 12s.; and since 1 yd. cost^ of £4 12s., 13 yds. cost^* of £4 12s. ; that is, £7 9s. 6d. Such exercises are solved by Division and Multiplication combined. 17. If 7 articles cost 15s. 9d., what is the cost of 4 ? 18. If 11 tons of hay cost £37 9s. lOd., what is the cost of 8 tons ? 19. If a man walk 21 miles in 7 hours, how far wil he walk in 9 hours '? 20. A boy earns Ss 6d. in 3 daya^ in what time will he earn ^9 18s. ? 21. If 18 sheep are worth 3 cows, how many sheep are worth 21 cows ? •22. What will 34 iheep cost at the rate of £368 2s. 9d. fer 153 sheep ? ^ 128 PROPORTION. t 23. If 18 lbs. of rice cost 67^ cents, how many pounds can be purchased for 13s. 6d. ? 24. How many yards of cloth may be had for $64.80, when 12 yards cost £3 12s. ? 10 -i. In the preceding exercises we found what multiple or part the quantity whose price was given, or the price whose quantity was given, was of that required, — and multi- plied the remaining term by the result. Thus, in the first exercise, dividing 20 by 5, we obtain 4 as quotient, then multiplying £9 by 4, we have £36 for the answer. The question might have been asked thus : — What sum of money will contain £9 as often as 20 yards contains 5 yards ? Ans. £36 The number of times that one number is contained in another is called the ratio of the two numbers ; thus the ratio of 5 to 20 is 4, and of 9 to 36 is 4. 167. Rati© is the comparison with each other of two numbers of the same kind. 168. The Terms are the two numbers compai-ed. 169. The Antececlent is the Jirst term. 170. The Consequent is the second term. 171. Ratio is expressed in two ways — 1st. — By placing two points, or a colon ( : ) between the numbers compared, writing the divisor before the points, and the dividend after the points. Thus, the ratio 5 to 7 is written 6:7; the ratio of 6 to 12 is written 6:12. 2nd. — In the form of a fraction. Thus, the ratio of 8 to 7 is I ; the ratio of 5 to 9 is |. Note.— In British publications the antecedent is put for the nume- rator and the consequent for the denominator; but the above form, which is that used in France, and in many parts of the United States, is more readily understood by beginners, because the Jirst term of a proportion is always used as a divisor. It also renders the inversion of the fraction unnecessary when that form of ratio is used. 173. A Simple Ratio consists of a single couplet as 4 : 12. 173. A Compound Ratio is the product of two or more simple ratios. Thus, The simple ratio of 4 to 8 is 2 The simple ratio of 3 to 9 is 3 ? The compound ratio of these is 12 to 72 6 i*]xepORTieN. iff 1L7^* Ih comparing numbers with each other, they mustt he of the same kind, and of the same denomination. Thus, ishiilings have a ratio to shillings. A foot has a ratio to a .yard; for one is three tunes as long as the other"; hut a foot %as not properly a ratio to an hour, for one cannot be said td be longer or shorter than the other. Note. — When questions are solved hy a direct Application of the •elementary rules, they are said to be worked by analysis. In thift •case of the iprevious exercises, it is merely finding the ratio t)t the two given terms of the sanio name, and multiplying the tliird by tlws Result. FiJTcrclses lor (lae 81«tte. 1. What is the ratio of 5 to 27 ? 2. What is the ratio of 32 to 8 ? 5. What is -the ratio of 4 oz. to 8 lbs. ? Ans. 4 oz. : 3 lbs. == 4 oz. Ans. 9 i 48 61. = ^ Required the ratios of the following n-umbers — 1. 2. 5. 4. 7 to 14 9 to 36 108 to 18 136 to 17 20 ft. to 40 yds. 60 m. to 4 fur. 45 bus. to 3 qts» 3s. to 16 shillings 5. 6 lbs. to 18 lbs. ' 9. 6. 28 lbs. to 4 lbs. 10. 7. -9 0z. to 63 lbs., 11. 8. 17 yds. to 68 yds. j 12. 13. Which is the greater, the ratio of 9 to 63, or that ot « to 72 ? 14. Which is the greater, the ratio -of 120 to 85, ©i that ^f 240 to 1 70 ? 15. What is the ratio compounded of 8 : 10 aiid 20 : 16 ? Ans. 1 16. What is the ratio compounded of 35 : 40, and 60 : 75 and 21: 19? Ans. i|^ 1 7. What is the ratio of 19 lbs. 5 oz» 8 dwts. to 58 W 4 oz. 4 dwts. Ahs. I^ 18. If the antecedent be | n nd the ratio ^, what is the con- sequent ? Ans. X 19. If the antecdent be 14.5 and the ratio S, what is the consequent ? Ans. 43.5 20. What sum of money will contain £6 10s. as often as 32 yards contain 8 yards ? Ans. £26 21. How many acres of land will have the same ratio to 7 ac, that £16 has to £112 ? Ans. 49 ac. 22. How many yards of cloth will have the same ratio to 3 yds. 2 qrs. 2 nls.', that £2 16s. 3d. has to £9 16s. lO^d. ? Ans. 12 yds. 2 qrs. 3 nls. ^ ISt) PROPORTfOJSr. -(- 28 What number compared with 8 will form a ratio equal to that of 4 to G? Ans. 15^ IT'S. When the ratio of two numbers is equal to that of two- other numbers, they are said to be proportional. Thus, the ratio oi' 4 to 6 is equal to the ratio of 8 to 12 ; and the four numbers are on that account said to be projiartwnal, or to form a simple proportion. ITO. Proportion is usually indicated by placing a double colon (: :) between tl;e two ratios. Thus, 4 : 6 : : 8 : 12, and are read, As 4 is to 6 so is 8 to 12. 177. Smce each ratio consists of two terms, every propor- tion must consist of at least /oi/r terms. 178. The Extremes are the first and foui'th terms. The Means are the second and third terms. 179. In every proportion the product of the extremes i» equal to the product of the means. Thus, in the proportion 4 : 8 : : 5 : 10 we have 4 X 10 = 5 X 8- 180. From the preceding principles and itatrations, it follows that, any three terms of a proportion being given, the fourth may readily be found by the following RULE. I. Divide the product of the extremes by one of the means, and the quotient will be the other mean. Or, II Divide the product of the means by one of the ex- tremes, and the quotient wiU be the other extreme. Note.— When the first and second terms are not both of the samer same they must be reduced. The fourth term is always the same aa the third term. Exercises foi= tlie Slate. Find the term not given in each of the following proportions r Ans. 12a 903 144 yd. Is. 8^d. 100 yd. 1. 48 : 20 : : ( ) : 50. 2. 42:70::3:( ). 3. 16:129::112:( ). 4. 48yd.:( ): :$67.25 : $201.75. 5. 17 yd. : 221 yd. : : ( ) : £l Is. ll^d. 6. ( ):160yd. : :8s. 5id. :13s. fid. OS. ^wll. ( 'd. : : 187 yd. £1 17s. l^d. 8. A = ( )--i-i I SIMPLE ritO PORTION. 131 ) equal Lns. 12 ; of two us, the le four , or to double 12, and prcpor- s. The ernes I» )portiort tions, it ven, the y one of . Or, the ex- the samer c same a^ portions r Ans. 12a 5 903 144 yd. Is. 8id. 100 yd. 17s, Ud. SIMPLE PROPORTION. OPERATIOX. yd. yd. ■As, 8 : 20 : : $d^ 20 8)1920 $240 Ans. 181. Simple Proportion is an equaiity of two simple ratios, and consists of four terms, any three of which being given, the fourth may readily be found. Example 1.— If 8 yds. of cloth cost $96, how much witt 20 yds. cost at the same rate ? Analysis. — Since 8 yards liavethe same ratio to 20 yds. as S96, the cost of the former has to the cost of the latter, we have the first three terms of a propor- tion given, namely one of the pitr ernes and the two means. Now to arrange the given num- bers in the order of a propor- tion, or state the question^ we make $96 the third term, because it is of the same kind, and has the same ratio to the required answer, or fourth term, as the first has to the second. From the nature ©f the question, since the answer will be more than f 96, or the third terra, the second term must be larger than th« first; we therefore put 20, the larger number, for the second term, and 8, the smaller, for the first term, and then the pro- duct of the means divided by the given extreme, gives the required extreme. (180.) Example 2. — If 50 men consume a certain quantity of flour in 20 daj's, how long would it take 35 men to consume a like quantity ? Analysis. — Having stated the question as in the last example, we perceive that the first and second terms have a common factor, 5, we therefore cancel it, Avhich leaves 10 and 7 as the new ratio. A^ain the factor 16 is common to the nrst and last terms, and we cancel it also, then multiplying 7 by 2 we have the answer as b«fore. OPERATION. men men day* 35 :: 20 20 Ag 50 50)700 14 Ans. Ag OR, 10 7 2 14 as before. .132' SIMPLE PitoroiiTiOii^r. t I Exercises for the Slate. J^OTE.— The piipil .sliouul write out eatu of the Ibllbwing exerciser ife words which will embrace the ^ven terma. This will greatly tacil-' i»«te his progress, and render hhn familiar mfh orie of fiiis moit i:h - ^rtant agents ot tho bcienoe of calculation. 1. % 3. 4. 5. 8i 7'. 8. 9. 10. H. 13 yds. : 143 yds. :: 3s. 4,^(1. : 39 yds. : 4li% yds. ^ : £ 1 Is. 1 l^Jd. :■ 8s. 5^d. : i:s. 6d.: :50yds.-: 13s. Gd.: £2 17s. 4^d. : : 68 yda. :■ 48 men : 1 2 men : : 20 days : 5 bu. : 470 bu. : : £3 38. : 136 ewtr. :-M* ewtv :-: S:>;8^. Ahs. £1 17s. l^d.' £i2 3s. Od. 80 yds.. 289 yds. 5 days- £296 2s. Od.- 15s. 2^d.. £13 18. 5^d. : £95 8s. 6fd. : : 165 tons : 1131 tons.. 144 days: 89 days:: £60 15s.: £^7 lOs. ll:id.. $41.87 : £58 1 9s. 63d. : : 34 yeai-s. : 233 years.. 9 ac. 2 ro. 38 po. : 14 ac. 2 ro. i7 |)o. : : ^8.45. Ans. $12.67|: Hi. 27' ac. r rO. 8 po. r 1^6 ae, 3 m. 24 po. : : £22 3s. 7^d. : Ane. $;66.83' 18. £14 Gs. 11| : $27.92^ : : 19 yds. 2 qj-s. 3nls. Ans. 7 yds. 3 qrs. 2 nls,. U. 2 days : 8 years :: $1.10 : Ah«. £124 6s. Gfd.- 15. 6 weeks : &8 years : : £4 15^. 4^ : An*. £2S10 7s. 8d. M. 2 oz. » dwt. 21 grs : 4 oz. S 7 dwt 18 grs, : : £ 1 2s. 9^d. Ans. $11.0i> 18®. From the preceding illustrations and principles, we tfeduce the following gertcraJ tlUIjE. I. Write for the third t6rm that number whicl*. •s of the same name as the required fourth term. It. Of the other two numbers, write the larger for th© »»econd term, and the smaller for the first, when the answer should exceed the third term ; but write ivhe less for thrt second term, arid the greater for the first, when the answer should' be less than the third term. ,?^^\' Multiply the second and third terms together, and divide their product by the first. Note. — To shorten the teorh factors common to the Jirti and second lerwM, w to the first and third terms, may be cancelled, Exercises for tlie Slate. If I get 60 yardls of clofh for S486.6G#, how many yard"^ ^ci lui- ^^v i Ans. 24 yards* 2. If 36 men ©ara $192 in a week, what will 72 men earn in th« saioe time ? Ans. $384 1. «_i:ii T i rr.. o J rt o SIMPLE PROPORTION. 133 S. ir a railway train can run 525 miles in 15 lioui"s, how fur would it run in 7 horn's ? Ans. 245 miles. 4. It' a grass field maintain 34 cows for 6 months, how long ■will it maintain 51 eows? Ans. 4 months. 5. If 1 7 evvt. be bought foi* £ 14, how many may be boui^ht for !$ 1 1 (5 .80 ? Ans. 20 cwt. 1 G lbs. 6. If 59 cwts. cost $196, how many CAvt. may be bought for $140 ? Ans. 42 cwt Ifi lbs. 7. A silversmith pays £144 for 19 lbs. of silver, how much ought he to get for £234 ? Ans. 30 lbs. 10 oz. 10 dwt 8. A lump of gold weighing 154 oz. costs $2258.14, what ■will be the weight of a nugget which costs £290 ? Ans. 96 oz. 5 dwt. 9. I bought 24 cwt. of sugar at £52 16s., required the price of 16 cwt. ? Ans. £35 4s. 10. The wages of 6 men amount to $18, required the ■wages of 9 men V Ans. $27 11. Three score of sheep cost £66 16b. 8d., what will 36 sheep cost ? Ans. $195.16 12. A truckman charges $15.47^ for 84 miles, how much is that for 56 miles ? Ans. £^ lis. 7d. 13. If 4^ yds. cost £2 168. 3d., what will 9 yds. cost at the same rate? Ans. $27.38 14. A snail travels at the rate of 16 po. 2 yds. 2 ft. 9 in. in 3 houi*s, how far will he have gone in 2 days, travelling night and day ? Ans. 6 fur. 24 po. 2 yds. 2 ft. 15. A school-room containing 120 pupils is 92 yds. 2 ft. in area, how much is that for each pupil ? Ans. 6 ft. 132 in. 16. If 24f barrels of fish cost 39.27^, what will 8^ barrels •ost? Ans. $13.09^ 17. If 2 J tons of coal cost $13.33, required the price of 19{ tons? Ans. £l9 Is. 6d. 18. A person saves each week as much money as buys a square pole of ground, in what time will he be able to purchase a farm containing 21 ac. 7 po. ? Ans. 64 yrs. 39 wks. 19. If 2 yds. 2 qrs. cost 16s. 7^d., what will 12 yds. 2 qrs. cost? Ans. 20.23 20. A boy who lives 455 yds. from the school goes to it in 6 min. 30 sec, how long would he take to go, if he ■were 2 miles 6 fur. 26 po. 1 yd. from it ? Ans. 1 h. 11 min. 12 sec. 21. A chest of tea weighing 3 qrs. 22 lb. 15 oz., long: wt., cost $121.43, what will 5 chests, each oz. cost? jighing 1 qr. 27 lbs. 13 Ans. £65 2s. 3^d. 134 SIMI'LE PROI'OHTION. 22. If a man mow 6 ac. 2 ro. 8G po. of barley in 5 days 8 hours, woi'kin*; 10 hours a day, in what time would lie mow IG ac. 3 ro. 10 po. Ans. 1 ! la. 5 ho. i 23. If 1 3 cwt. (^r. 9 Ihs., long weiglit, cost .i:22 lis. 5.3d., what will 20 cwt. 8 (jrs. 20 lb. cost ? Ans. HiiC) 7s. 2d. 24. A farmer draws a net i)rofit of £ 23 1 7s. 2|d. fnmi 2 nc. 1 7 po. ; how niucli should he receive at the same rate from 3H acres 3 ro. 32 po. V Ans. ^2147.28 25. If 8;J bushels of corn cost $4.20, what will be the cost of 13^ bushels at the same rate ? Ans. $6.48 2G. If 1 f yds. of cotton cloth cost S0.10^"tj, how niany yds. can be bought for $100? Ans. l^ yds. 27. If 15| bu. of clover seed cost $156-j, what Avill \) bu. 2 pk . 2 f qt. cost ? Ans. S i) 5 . 7 5 28. If I of a barrel of apples cost $j\, how many can be bought for $f 1| V Ans. | of a barrel. 29. A butcher selling meat sells 1A\1 oz. for a pound ; how much does he cheat a custonjer who bu)S of him to the amount of $30? Ans. $2.46^^ 30. If I pay $6 for the loan of $100 for 1 year, what should I pay for $493 ? Ans. $29.58 31. If I borrow $2000, and keep it 1 year 4 mo., how long should I lend $240 as an cqivalent for the favour? Ans. 2 yr. 9^ mo. 32. If ^ of I of 4 ac. cost \ of i\ of $140, what is the cost of II acres? . Ans. $36| 33. If I pay $4| to a person for buying $100 worth of goods for me, what should I pay for buying $189.75 worth ? Ans. $7.82f nearly. 34. If a merchant makes a reduction of 1 penny in each shillino-s' worth of goods sold, how much is that in £l00? Ans. £h 6s. 8d. 35. An insolvent debtor fails for $2000, of which he is able to pay only SGGO, how much is that in each dollar, and how much will a person receive whose claim is $900 ? Ans. $0.43 and $387 36. If £100 gain £3 in one year, what will £256 10s. 6d. gain in the same time? Ans. £7 13s. lid. nearly. 37. Find the interest of £ 126 for one year at £5 per cent. Ans. £6 6s. Note.— In this exercise there are apparently only two tenns. £5 per cent, however, just means £5 for £100. The above may titere- fore be written thus :— 1 f SIMPLE PIlOrOllTION. i 135 If£lOO{jrain £5 in one year, how much will £12G gain in the same time V .•m. Find tiie interest of £12G 14s. 6(1. for 1 year at 8 J per cent. Analysis. — Hero, and in all similnr caws, the first term bc'iiin; 100, we ni^kc no fonual statement Ixit merely multiply the second term bv the third an at 8V «i 42 4 10 1013 16 £10,56 10 20 1 1 ,20 12 2,50 2 1,00 £10 lis. 2|d., Ans. OR, £ £ s. D. £ 12 £12G 14 G-i-12 = £10 11 2.Vasbelbre. Here the third term is con tained exactly 12 times in 100, we therefore cancel it. Divid- 'sng the second term by 12 we obtain the answer. 39. Find the interest of $186 for 1 year at 8 per cent Analysis. — Here, dividing the first and thii'd terms by 100 we have the quotients 1 and .08. We therefore multi})ly the second term by .08, and obtain the required interest. In a sim- ilar manner avc may find the interest for one year at any given per cent. Write out and solve the following exercises — 40. Find the interest of £186 10s. for 1 year at G^ per cent. Ans. £11 13s. l|d. 41. At 5| per cent., what is the interest of £196 16s. 8d. fori year? Ans* £10 Is. 9^d. OPERATION. df* dt» (ft -tp ^j2:3 Into 4:2) extremes. means. That IS 6X4:3X2:: 12X3; for 6 X 4X3 = 12 X 3X2 Note.— Compound proportion is chiefly applied to the solution of questions Avhich would xQ(\Vi\.T^ two or mwe statements \n simple pro- portion. Example 1. — If 8 men can reap 32 acres in 6 days, how many acres can 12 men reap in 24 days ? STATEMENT. ANALYSIS. — In this CX- As 8 men : 12 men 1 . . og ap ^"^P^® ^^ ^^ supposed that 8 6 days : 15 days ^ • • ^^ *^^- men can reap 32 acres in 6 days ; this being the case, it is asked or demanded how many acres 12 men can reap in 15 days. The question may therefore be divided into two parts, supposition and demand. In order to state the question in the form of a proportion, we take from the supposition that quantity, 32 acres, whica is of the same kind as the answer required, and place it for the third term. Then, taking the next number, 8 men, in the sup- position, and 12 men, tlie corresponding number in the de- mand, and considering these with reference to the third term only, as in simple proportion, we find the answer is to exceed COMPOUND PROPORTION. 137 the third term, and therefore place 12 men for the second term and 8 for the first. Ai^ain, comparing tlie remaining quantity, 6 days, in the supposition with the corresponding quantity, 1 5 days, in the demand with reference to the third term, 32 acres, we observe that if the time be increased the number of acres will also be increased ; we therefore place 15 days in the second term and the 6 days in the first, and the question is stated. Analysis. — Since the product of the antecedents has the same ratio to the product of the consequents, as 32 has to the answer, (Art. l'^'^), we multiply 8 by 6 and 12 by 15 to form a simple ratio. The remainder of the work is the same as simple pro- portion. OPERATION. As 8 6 ; ]lh^^ 48 :180 32 3G0 540 acres. 48)5760(120 Ans. 48 •• 96. 96 Example 2. — If 12 horses can plough 11 acres in 5 days, how many horses can plough 33 acres in 18 days '? Dividing the question into supposition and demand we have 1 2 horses 11 acres 5 days ? 11 acres 18 days As 1 1 acres : 33 acres 18 days : 5 days 198 :165 165 X 12 |::'12 horses. = 10 horses. 198 Stating and working as in the former example we obtain 10 horses for the answer. BY CANCELLATION. 3 1 As ,1^ : ^ .18: %]■■ n 5 X 2=10 as before. Here 1 1 is a common factor of the fii-st and second terms, we therefore cancel it. Again, 3 being a common factor of 3 and 18, we divide each (3 and 18) by it, and set dofwn the 138 COMPOUND PROPORTION. quotients 1 and 6. For similar reasons we omit G and write 2 instead of 12. We then multiply 5 and 2 together and find the answer as before. From these examples and illustrations we have the following RUIiE. I. Take from the supposition that number which is of the same kind as the answer required, and place it for the third term. II. Take the remaining numbers in pairs, one from the supposition and a corresponding one from the demand, and arrange them as in Simple Proportion. III. Finally, multiply together all the second and third terms, divide the result by the product of the first terms, and the qviotient will be the fourth term or answer. Note. -When the first term has factors which are common to the second or third terms, cancel the factors which are common^ then divide the product of those remaining in the second and third terms by the pro- duct of those remainimj in thefrst, and the quotient will be the answer. Exercisefi for the Slatc« 1. If 18 masons can build a wall 120 feet long in 3 days, in what time will 2i men build a wall 480 feet long ? Ans. 9 days. 2. If the wages for 8 men for 12 days be $64, what will be the wages of 1 men for 6 days ? Ans: $40 3. If $1 00 gain S4 of interest in 12 months, how much will $G0 gain in 15 months? Ans. $3 4. If £100 gain £5 of interest in 10 months, how much would £250 gain in 8 months ? Ans. £10 5. The wages of 8 men for 4 days are $19.50, what will be the wages of 12 men for 2 days ? Ans. SI 4. 6 2^ 6. If 12 reapers cut 71 ac. 2 ro. 8 po. in 6 days, how many acres will 8 reapers cut in 10 days ? Ans. 79 ac. 2 ro. 7. If IG horses in 9 days plough 110 acres, how many acres will 27 horses plough in G days. Ans. 123 ac. 3 ro. 8. If 208 families consume 6 cwt. of tea in 42 weeks, how much will 63 families consume m a year. Ans. 2| cwt. 9. If 18 men plant 29 ac. 2 ro. 26 1 po. of potatoes with the spade in 15 days, how many men would plant 17 ac. 3 ro. 8 po. in 6 days. Ans. 27 men. 10. If 69 yards of cloth 3 qrs. wide, make 24 pairs of trou- sers, how many pairs will 301 yds. 3 (jrs. 2 nls., which is 1 yard wide, make V Ans. 140 pairs. 11. If a man walk 1 70 miles in 6 days, walking 15 houn' a day, how many miles will he walk in 5 days, walking 1 2 hours a-day ? Ans. 113 miles 2 fur. 26 po. 3| yds. percentage; 139 12. If 18 reapers cut 30 acres of barley in 6 days, working 1 liours a-day, how many reapers will it take to cut 40 acres in 4 days, working 12 hours a-day ? Ans. 30 reapers. 13. If IG men earn $(j2.40 in 18 days, how many men will it take to earn $140.40 in 24 days ? Ans. 27 men. 14. If a family of 8 pei'sons spend $200 in 9 months, how much will 18 persons spend in 12 months ? Ans. $600 15. If 15 men working 12 hours a-day, can hoe 60 acres in 20 days, how long will it take 30 boys working 10 hours a-day, to hoe 96 acres, 6 men being equal to 10 boys ? Ans. 32 days. 16. If 125 men can make an embankment 100 yards long, 20 feet wide, and 4 feet high in 4 days, working J 2 hours a-day, how many men must be employed to make an embankment 1000 yards long, 16 feet wide, and 6 feet high, in 3 days, work- ing 10 hours a-day ? Ans. 2400 men. 17. A log of wood 60 feet long, 4 broad, 2 thick cost $1 28, what would be the price of one 45 feet long, 3} broad, and 2 J thick? Ans. $115.50. 18. If 42^ yards of cloth, which is 18 in. wide, cost $238. 83^-, what will 118^ yards of yard-wide cloth of the same quality cost? Ans. ifSl329.04. 19. If 400 men can make a canal which is to be a mile long, 40 feet broad, and 12 feet deep, in 20 days, working 8 hours a day, what length of canal, 30 feet wide and 16 deep, could 300 men make in 45 days, working 10 hours a day ? Ans. 2 miles 35 po. 20. Forty men engaged to finish a road, which was to be a mile long, in 60 days, but after three-fourths of it was done they left off. How many men would it take to finish the re- mainder in 6 days ? Ans. 100 men. 21. If 5 horses require as much oats as 8 ponies, and 120 bushels last 12 ponies for 64 days, how long may 25 horses be kejDt for $165 when oats are selling at $0.55 per bushel ? Ans. 48 days. 22. If $250 gain $30 in 2 years, what wnll be the interest of $ 7 5 for 5 years ? A ns. $ 2 25 23. If $100 gain $5 In 1 year, what will be the interest of $575 for 3} years ? Ans. 1 00.62^ 24. What will be the interest of £ 125 for 4 years, if £l 50 will gain £lO 10s. in 1 year ? Ans. £35 25. If £100 gain £3 10s. in 1 year, what will £375 gain in 3 years and 8 months ? Ans. £48 2s. 6d. 140 COMPOUND PROPORTION. hi Ilk 26. If $100 gain $4.50 In 1 year, what $426. 6G| s;n\n from June 15th, 1865, to Sept. 18th, 1865 ? Ans. $4.99 27. If £100 o-ain £4 in 365 days, what will be the gain on £690 10s. 6(1. for 85 days ? Ans. £0 8s. 7^d. 28. Find the interest of $2737.50 for 56 days at 3^ per cent. Ans. $14.70 Note. — The pupil may suppose that the full number of terms are not given in this exercise : but it will be readily seen that 3^ per cent is in reality $3| for the loan or interest of $100 for one year or 365 days. The above question may be written thus :— If $100 gain $3^ in 365 days, how much will $2737.50 gain in 56 days ? Note. — The terms pe7^ cent, interest, cf c, have not been explained in the preceding pages: but as the illustrations of percentage in gen- eral depend on proportion, the pupil should, at this stage, be made acquainted with the principles involved. This will enable him to solve almost every question relating to per centage without consider- ing them under any special rule. Write out and solve the following exercises — 29. Find the interest of £812 6s. 8d. for 7 years 3 months at 5 per cent. Ans. £294 9s. 5d. 30. Lent $2400 for 4 months, and received $24.60 for interest ; what was the rate per cent ? Ans. 3.07|- 31. Find the interest of $3311.50 for 292 days at 2^ per cent. Ans. $66.23 32. What is the interest of £660 for 8 months at 4^ per cent? Ans. £19 16s. 33. The value of a share in a railway is $300, and the half- yearly dividend is $16.80 ; required the rate per cent? Ans. ll|^p. c. 34. Bought $6000 worth of goods, and at the end of 70 days sold them for $6200, what was tlie gain per cent ? Ans. 1 7^j- p. c. 35. A person havirtg borrowed a certain sum of money at 5 per cent., at the end of 3 months paid $15, the amount of interest then due ; how much did lie borrow ? Ans. $1200 36. A j.erson having mortgaged his property, pays $40 of interest every three months ; for what amount was the mort- eajie drawn, interest being charged at 6 per cent ? ^ ^ « b ^ ^^g $2666.661 37. Dec. 18th, 1865—1 borrowed $6866.46. with which 1 purchased Hour at $6.66 a barrel. March 17th, 1866—1 sold the flour for $7.3 7^ a barrel, cash. How much did 1 gain by the transaction, interest being reckoned at 6 per cent V i Ans. $636.71^ fiPEKCENTA.G'E. lU PEECENTAOE. 18 1. Pcp Cent, is a term derived from the Latin words per centum^ and signifies hi/ the hundred, or hundredths, that is, ■a certain number of parts -of each i)ne hundred parts, of what- ever denomination. Thus, hy 4 per cent., is ajieaBt -$4 of .every $100, 4 bushels for every 100 busliek, &c. Therefore, 4 per cent equals 4 hundredtlis = .04 = -j^-g- =x ~^^ = ^. 6 per cent equals .08 = ^l^ = ^. 185. Percentage is such a part ©f a number as indi- «cated by the per cent. 18«. Tlie B»g»e ©f percentage is the number on which ithe percentage is computed. 187. Since per cent, is any numT3er of hundredths, it if (usually expressed in the form of a decimal ; but it may be expressed either as a deoimal or a common fraction as in the lfollowi«ig tabk. jSoTjs. — In business, -per cent is usually ir dicated by fb« sign %, Decimals 1 ,pGr cent. = .01 •2 per cent. = .€2 4 per «ent. = .04 t> per cent. = .05 ^ per cent. = .06 7 per cent. =* .07 10 por cent. = .1 12^ per cent. = .125 TABLE. Common fi-actkm. Lowest terms. it 142 COMPOUND PROPORTION. 4. Express decimally | per cent. ; f per cent. ; | per cenl. 5. Express in the form of common fractions, in their lowest terms, 6 per cent. ; 5 per cent. ; 33| per cent. ; 31^ per cent. ; 113 per cent. ; 18| per cent. CASE I. 188. To find the percentage of mnj number. ExAMPLr: man Laving 125 boshels of wheat, sold 25 .ntity, how much did he sell ? Analysis. — Since 25 per cent, is -A^ ==.25, he sold .25 X 125 bus., or 125 bush. X .25 = 1 3| bhshels. Or, 25 per cent, is ,2^ == i and i of 1 25 = 31 :^. Hence th© following — per cent, of t, OPERATION. 125 .25 625 250 31.25 =51| RULE. Multiply the given number or quantity, by the 7ate per cent., expressed decimaUy, and point off as m deci- mals. Or, Take such a part of the given number as the number ex- pres&^ng the rate is part of 100. Exercises. 1. 2. 3. A.. 5. 6. 7. 8. 9. Ans. S847 $106.08 $4.11/^ 132.7725 bus. 26.95 miles. What is 5 per cent, of $18940 ? What is 8|- per cent, of $1248 ? What is 7i per cent, of $56.75? What le C^ pftr opnt. of 1967 bus. ? What is 9| per cent, of 275- miles ? What is 25 per cent, of | ? 25 per cent. =^^=^, and | X ? = iV An&; What is I per cent, of $2526.40 ? Ans. $6.3 IS. What is I per cent, of $75,000 ? $250.00 A farmer having 1500 sheep, sold 25 per cent, of them ; iiow many did he sell ? Ans. 3 75 sheep. 10. A merchant imported 1500 boxes of oranges, and 12^ per cent, of them decayed ; how many boxes did he lose, and how many had he left ? Ans. 187.5 lost. 1312.5 saved. itta. CASE II. 77h iJnfT n-iliftt nfir r^nt- f>nfi numher is o-f another: Example. — A man having purchased a horse for $170, sold him for $1 7 less \ what per cent, of his money did he lose ? PERCENTA.aE. 14^ ■1% )47 Ans. OPERATION. Analysis. — We mul- 17-^1 70 = .10=10 per cent. tiply the base by the rate OR, per cent, to obtain the ^^ = ^^=x.lO= 10 per cent. percentage (188) ; con- versely, we divide the per rentage by the base to obtain the rate. Or, since $1 70 is 10() per cent, of his money, SI 7 is y^j, equal to ^^ of 100 per cent., which is 10 per cent. Hence tlie following — RUIjE. Divide the per centage by the base, and the quotient will be the rate per cent., expressed decrmally. Oi:, Take such a part of 100 as the per centage is part of the base. Exercises Tor flic Slate* 1. AVhat per cent, of $9876 is $24^0 ? Aiis. 25 2. What per cent, of $7G56 is S957 ? Ans. 12^ 3. What per eent. of 4 tons 16 cwt. is 3 tons. 12 cwt '? Ans. 75 per cent 4. What per cent, of 6 bushels 1 peck is 4 bushels 2 pecks 6 quarts ? Ans. 75 per cent 5. A roan having 9^0 acres of land, sold \ of it at one time, and ^ of the remainder at anothuer time ; what per cent, remained unsold ? Aas. .33| pci* cent CASE KI. 190. To find a nurnker when a certain per cent, of k is given. ExAMFLE. — A man sold 31 1 bushels of wheat, being 25 per cent, of all he had ; how much had be at first ? OPERATION. 31.25 bushels -i- .25 OR, 3U _ _. 12S 125 25 4 X 10<5 - X 100 100 125 Analysis. — We are here required to find the base, of which 81 1 bush- els is the percentage. — Now, percen-tage equal* base multiplied by the rate per cent. ; convei*sely, base €quak percentage divided by the rate per cent. Or, 31^ bushels is 25 per cent, of all he had; -^^ of 31 ^ bushels, or i|| equals 1 per cent, of all he had, and 100 times 1|| equals 100 per cent, of all he had. Hence the following — RULE. Divide the percentage by the rate per cent., ex- pressed decimally, and the quotient will be the base, or number reqnired. Or, Take as many times 100 as the percentage is times the pate per cent u^ FERCTENTAGB;. Exercises Tor tlic State. 24 13 8 per cent, of what number ? Abs. 30iJ» 2. 42 is 7 per cent, of what aumber j^ 600 3. 39^ is- 5 per cent, of what number ? 790 4. A mao, owning 30 per cent, of a shoe factory, sells 33 J jffer cent, of his share for $U1-L275, what is the value of the •idiok factory.?. ' ' Aks. Hil2.7a t t is i i I 30i)» 790 33* ■ the". 2.7.a APPEMIIX I ' KEY TO THE SELF-TESTING EXERCISES- ADDITION. All the exercises given in this Rule as self-testing are formed as shown in section 3. To test the sum- of any number of rows orKnes we may use any of the three following m^thocb. 1st. As the first line of each exercise is a mmltiple of 9,. the Slim of any niitiiber of lines must also be a multiple of 9 ; therefore casting the 9's out ofthe sum, if the work is correct, there will be no excess.. If there be an error ia any of the lines it may also be de- tected' by casting out the 9*s ia the same manner. 2nd. If the exercise is composed of 5 rows,, the sum of all the rows will be 12 times the fii'st line. If composed of 6 rows it will be 20 tiines the first line, and so on as may be seen in the following exannples. (1) (2> 1467 First line =s.l times 1467 First line = 1 1467 Second " = 1 " 14P7 Second " = 1 2934 Third " = 2 " tet Hne; 2934 Third " =r2 ^01 Fourth " =3 7335 Fifth " =.5 u. « u 37604 Sum =12 times 1st line. 4401 Fou;-th " = 3 7335 Fifth «* =5 11736 Sixth " - 8 29340- Surffl' = 20 times Ist line. &rd. The siwn of a req-uired number of lines added to the i'rst line will be equal ta the line that is two more than the; required number of lines. Thus let 6 be the required num- ber of lines. Tlie sum of six lines added to the first line will .1 + ®f lines. ^:i TitllXi Vit^r" The sum of eleven, lines added to the first line will gjive the; 134Ji Iiii£. 146 APPENDIX. Example. — Find the sum of 162 extended to 8 rows, and test the result bj the tenth lino. OPERATION. Istl inc J 162 2nd a 1G2 3rd ti 324 4th K 486 5th (. 810 <5th 1( 1296 7th it, 2106 8th i( 3402 9th " 5508 8748 zz: sum of eight lineg. 162 — first line. Tenth line 8910 8910 r= line that is two more than tli« required number of lines, i. e., (8 X 2) 10th line. Note. — As soon as the pupil fully understands the principles of addition he should be required to test his work as above, and thus £acilitate his progress. SUBTRACTION. The exercises under this rule are to bo worked b^r the pupil as shown in the following example. 18717 minuend. 12478 subtrahend. 6239 difference. 6239 difference betwen 2'1 and Sdllne. Analysis. — We first take th« sub- trahend from the minuend, then this difference from the subtrahend. If the two last lines are alike, the work is eorrect. f^ MULTIPLICATION. Section 1. — The test of the exercises in this section may be seen from the construction of each. SECiiON 2. — In the exercises in this section the teaclj^r will observe that every line in the working, aad every pro- duct, is a multiple of nine, and by adding the digits in any line or prxjduct he caw ascertain if it is ocwrect. APPENDIX. 147 Sections 3, 4 and 5. — Tlie manner of testing the exercises in tliese sections may be readily seen from their construction. DIVISION. Skction 1. — Each dividend is a multiple of its divisor, consequently, if worked correctly there will be no remainders. Sections 4 and 6. — ^In the exercises under these sections each dividend is a multiple of nine, also each divisor, and the remainders, if any, are divisible by 9, and each dividend is divisible by all the divisors given with remainders as above. These sections, therefore, contain 841 exercises. ADDITIO^T OF DECIMALS. Increase each figu 'e of the second line by unity, and pre- fix the first figure of the exercise. The effect of ft occurring in the second line should be particularly noted. Note. — The second line n:\ay be varied at pleasure. SUBTRACTION OF DECIMALS. Same as Simple Subtraction. MULTIPLICATION OF DECIMALS. Same as Section 3 of Simple Multiplication. DIVISIO:^ OF DECIMALS. Tlie quotients ai-e without remainders, and each is a mul- tiple of &. REDUCTION DESCENDING. Tlie answers to all the exercises given in Reduction de- scending are to be tested by the sum of the digits, which, if "correct,'^will be found to coatain some multiple of ^ withouti any excess. I4;s jLrT»i':»rnx- REDUCTION ASCENDINO. (1) (2) RoUiKie 15270 ]iencG to pwmJis. llodiKu; ilU^Ci? fur. to poiuiAt 12)15270 4)3112G7 2,0) 1 2 7,2s. 6(L £6S 12 6 12)77816|(!. («) «,0)648,48. Sd. .£824 4 8} Reduce 28197 dwte. to Ibe. 2,0)28197 12)1409 17 U7 lbs. 5 oz. 17d.wt. For «xepclses like examj)les (1) and (2) test the pounds by the sum of the digits, then double the two right hand figures, calling the units pence, and the other figures shil- lings. Thus .£36 7s. 2d. Here thenumlDer of pounds z= 36. Test the pounds by tl>e sum of tl>e digits. Then 3