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13. A farmer bad 200 bu, of wheat and sold 28 bu. 2 pk. 5 qt. 1 pt. to one man, and as much more to another; how much remained ?

Ans. 142 bu. 2 pk. 5 qt. 14. A California miner having 112 lb. of gold, sent his mother 17 lb. 10 oz. 15 pwt. 20 gr., and 3 lb. 16 pwt. less to his father ; how much did be retain ?

Ans. 79 lb. 3 oz. 4 pwt. 8 gr. 15.

Subtract 16 mi. 223 rd. 3 yd. 1ft. 8 in. from 36 mi. 271 rd. 3 yd. 1 ft. 11 in.

Ans. 20 mi. 48 rd. 3 in. SUPPLEMENTARY PROBLEMS.

To be omitted unless otherwise directed. 16. Subtract $16 574 54 mills from $25 20% 74 mills, and add 3 bagles and 25 dimes to the result.

Ans. $31.206%. 17. Subtract 125 A. 37 P. 29 sq. yd. 4 sq. ft. 140 sq. in. from 240 A. 85 P. 16 sq. yd. 96 sq. in.

Ans. 115 A. 47 P. 16 sq. yd. 6 sq. ft. 136 sq. in. 18. A man bought a cask of molasses, from which there leaked away 11 gal. 3 qt. 1 pt., and then after putting in 12 gal, he found it lacked 16 gal. 1 pt. of containing 63 gal.; how much was in at first.

Ans. 46 gal. 3 qt.

OPERATION.

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7

MULTIPLICATION OF COMPOUND NUMBERS.

308. Multiplication of Compound Numbers is the process of finding the product when the multiplicand is a compound number.

1. Multiply £12 11 s. 7 d. by 8. Solution.-We write the multiplier under the lowest denomination of the multiplicand, and begin

£

d. at the right to multiply. 8 times 7d. are 56 d., 12 11 which, by reduction, we find equals 4s. and 8 d.; we

8 write the 8d. under the pence, and reserve the 48.

100 12 8 to add to the next product : 8 times 11 s. are 88 s., which, added to the 4 s., equals 92 s., which we find by reduction equals £4 and 12 s.; we write the 12 s. under the shillings, and reserve the £4 to add to the next product; 8 times £12 are £96, plus the £4, equals £100, which we write under the pounds. Hence the following

Rule.-I. Write the multiplier under the lowest denomination of the multiplicand.

II. Begin with the lowest denomination, and multiply each term in succession as in simple numbers, reducing as in ad dition of compound numbers.

Proof.—The same as in multiplication of simple numbers NOTE.-If the multiplier is a large composite number, it will be more convenient to multiply by its factors.

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8. Multiply 12 L. 2 mi. 232 rd. by 12.

Ans. 154 L. 2 mi. 224 rd. 9. Multiply 23 ch. (36 bu.) 18 bu. 2 pk. 7 qt. 1 pt. by 13.

Ans. 305 ch. 27 bu. 2 pk. 1 qt. 1 pt. 10. A farmer sold 5 loads of hay, each containing 150W 90 lb.; how much did he sell ?

Ans. 79 cwt. 501h.

11. Multiply 13 yr. 10 mo. 3 wk. 5 da. by 15, using the fac tors of the multiplier. Ans. 208 yr. 7 mo. 3 wk. 5 da.

12. If a man walk 17 mi. 300 rd. in each of 21 days, how far will be walk in all ?

Ans. 376 mi, 220 rd. 13. If a farmer raise 60 bu. 3 pk. 6 qt. 1 pt. of grain on one acre, how much can be raise at the same rate on 48 acres ?

Ans. 2925 bu. 3pk. 14. If a pipe discharges 11 bhd. 40 gal. 2 qt. 1 pt. of water in an hour, bow much will it discharge in 56 hours ?

Ans. 652 bhd. 7 gal. 16. Å had 1000 A. of land; be sold B 96 A. 150 P., and C 4 times as much; how much remained ? Ans. 515 A. 50 P.

16. A farmer raised 4000 bu. of grain; he sold 50 bu. 3 pk. 7 qt. to A, 7 times as much to B, and to C 6 times as much as to A and B together; how much remained ?

Ans. 1145 bu. 3 pk.

DIVISION OF COMPOUND NUMBERS. 309. Division of Compound Numbers is the process of finding the quotient when the dividend is a compound number.

310. There are two cases :1st. To divide a compound number into equal parts.' 2d. To divide one compound number by a similar one.

CASE I. 311. To divide a compound number into a number of equal parts.

1. Divide £103 7 8. 6 d. into 5 equal parts, that is, take } of it.

SOLUTION.- We write the divisor at the left of OPERATION. the dividend, and begin at the highest denomi

£ 8. d. nation to divide. of £103 is £20 and £3 re

5)1037 6 maining; £3 equal 60 s., which added to 7 s.

20 13 6 equals 67 s.; } of 67 s. is 13 s. and 2 s. remaining; 28. equal 24 d., which added to 6 d. equals 30 d.; } of 30 d. is 6d. Hence the following

Rule.-I. Begin with the highest denomination of the dividend and divide each term in succession, as in simple numbers

II. When there is a remainder, reduce it to the next lower denomination, add it to the term of that denomination, and divide the result as before.

III Proceed in the same manner until all the terms are divided.

Proof.—The same as in division of simple numbers. Note.-When the divisor is a large number and composite, the factors being not greater than 12, it is perhaps more convenient to divide by the factors.

WRITTEN EXERCISES.
(3)

(4)
d. lb. oz. pwt. gr.

T. cwt. lb. 4)61 18

4

6)76 10 14 12 7)112 16 66 15 9 7

16 2 38 (5) (6)

(7) cwt. lb. oz. bhd. gal. qt. pt. gi. mi. rd. yd. ft. 8)125 94 12 9)108 42 2 1 2 11)120 313 3 2

8.

8. Five sons share 112 A. 144 P. 24 sq. yd. of land; how much does each receive ? Ans. 22 A. 92 P. 29 sq yd.

9. If 9 farmers raise 1137 bu. 3 pk. 4 qt. 1 pt. of grain, what is the average amount raised by each?

Ans. 126 bu. 1 pk. 5 qt. 1 pt. 10. A miner sends 37 lb. 10 oz. 17 pwt. 16 gr. of gold to his 8 sisters; how much does each receive ?

Ans. 4 lb. 8 oz. 17 pwt. 5 gr. 11. A man walked 376 mi. 276 rd. in 22 days; what was the average distance each day? Ans. 17 mi. 41 7 rd.

12. If 26 casks contain 21 hhd. 11 gal. 2 qt. 1 pt., what is the capacity of each cask? Ans. 51 gal. 1 qt. H pt.

CASE II. 312. To divide one compound number by a similar one. 1. Divide £26 6 8. 2 d. by £4 15 8. 8 d.

OPERATION. SOLUTION.-£26 68. 2 d. equais 6314 £26 68. 2d.=6314d. pence; £4 15 8. 8 d. equals 1148 pence; and dividing 6314d. by 1148 d. we obtain

£4 158. 8d=1148 d. & quotient of 54. From this solution we 1148)6314(54, Ans. have the following

6314

Rule.-Reduce both dividend and divisor to the lowest denomination mentioned in either, and then divide as in simple numbers.

Proof.-The same as in division of simple numbers. NOTE.—The division may also be made without reducing to the lowest denomination, and this will be shorter when the quotient is integral.

WRITTEN EXERCISES. 2. Divide £48 7 8. 4 d. by £6 11 d.

Ans. 8. 3. 69 bu. 3 pk. 6 qt. by 6 bu. 3 pk. 6 qt. Ans. 1024

4. If a man feeds his horse 1 pk. 6 qt. of oats a day, how long will 3 bu. 2 gt. last him ?

Ans. 7 days. 5. How many demijohns, each containing 2 gal. 3 qt. 1 pt., can be filled from a tank holding 71 gal. 3 qt. 1 pt. of wine?

Ans. 25. 6. A drove of cattle ate 6 T. 15 cwt. 87 lb. of hay in a week; bow long will 33 T. 19cwt. 35 lb. last them ?

Ans. 5 weeks.

yr.

DIFFERENCE BETWEEN DATES.

CASE I. 313. To find the difference of time between two dates.

1. Washington was born Feb. 22d, 1732, and died Dec. 14th, 1799; what was his age ?

SOLUTION.—Dates are expressed in the number of OPERATION. the year, the month, and the day; hence the date of

mo. da his birth is 1732 yr. 2 mo. 22 da., and the date of his

1799 12 14 death is 1799 yr. 12 mo. 14 da.; and the difference of

1732 2 22 these two dates will equal his age, which we find to be 67 yr. 9 mo. 22 da.

67 9 22 Rule.- Write the number of the year, month, and day of the earlier date under the year, month, and day of the later date, and take the difference of the numbers.

N :TE.- In this method we reckon 30 days to the month; wted greater accuracy is required, we reckon the actual pumber of days in each month. The exact time between two dates is found by the table, Art. 296.

WRITTEN EXERCISES. 2. What is the difference in time from Dec. 12th, 1850, tu Jan. 5th, 1860 ?

Ans. 9 yr. 23 da

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