SUBTRACTION OF COMPOUND NUMBERS. 307. Subtraction of Compound Numbers is the pro cess of finding the difference between two similar compound numbers. 1. From 10 oz. 12 pwt. 20 gr. take 7 oz. 15 pwt. 16 gr. OPERATION oz. pwt. gr. 10 12 20 7 15 16 17 4 2 SOLUTION. We write the subtrahend under the minuend, placing similar units in the same column, and begin at the lowest denomination to subtract; 16 gr. subtracted from 20 gr. leaves 4 gr. which we write under the grains : 15 pwt. from 12 pwt. we cannot take; we will therefore take 1 oz. from the 10 oz., leaving 9oz.; 1oz. equals 20 pwt., which, added to 12 pwt. equals 32 pwt.; 15 pwt. subtracted from 32 pwt. equals 17 pwt., which we write under the pwt.; 7 oz. from 9 oz. (or, since it will give the same result, we may add 1oz. to 7 oz., and say 8 oz. from 10 oz.) leaves 2 oz. Hence the following Rule.-I. Write the subtrahend under the minuend, so that similar units stand in the same column. II. Begin with the lowest denomination and subtract each term of the subtrahend from the corresponding term of the minuend. III. If any term of the subtrahend exceeds the corre sponding term of the minuend, add to the latter as many units of that denomination as make one of the next higher, and then subtract; add 1 also to the next term of the subtrahend before subtracting. IV. Proceed in the same manner with each term to the last. Proof. The same as in the subtraction of simple numbers. NOTE. The pupil will notice that the general principle of addition and subtraction is the same as in simple numbers, the difference being in the Irregularity of the scale, the units themselves being expressed in the deci mal scale. 13. A farmer had 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 3lb. 16 pwt. less to his father; how much did he retain? Ans. 79 lb. 3 oz. 4 pwt. 8 gr. 15. Subtract 16 mi. 223 rd. 3 yd. 1 ft. 8 in. from 36 mi. Ans. 20 mi. 48 rd. 3 in. 271 rd. 3 yd. 1 ft. 11 in. SUPPLEMENTARY PROBLEMS. 16. Subtract $16 57 5 mills from $25 20 7 mills, and add 2 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. 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. --- OPERATION. d. 7 £ 6. 12 11 100 12 SOLUTION. We write the multiplier under the lowest denomination of the multiplicand, and begin at the right to multiply. 8 times 7 d. are 56 d., which, by reduction, we find equals 4s. and 8 d.; we write the 8d. under the pence, and reserve the 4s. to add to the next product: 8 times 11 s. are 88 s., which, added to the 4 s., equals 92s., which we find by reduction equals £4 and 12s.; we write the 12s. 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 denomi nation of the multiplicand. II. Begin with the lowest denomination, and multiply each term in succession as in simple numbers, reducing as in addition 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. 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 15cw 90 lb.; how much did he sell? Ans. 79 cwt. 50 lb. 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 he 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 he raise at the same rate on 48 acres? Ans. 2925 bu. 3pk. 14. If a pipe discharges 11 hhd. 40 gal. 2 qt. 1 pt. of water in an hour, how much will it discharge in 56 hours? Ans. 652 hhd. 7 gal. 15. A had 1000 A. of land; he 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 s. 6 d. into 5 equal parts, that is, take of it. SOLUTION.-We write the divisor at the left of the dividend, and begin at the highest denomination to divide. of £103 is £20 and £3 remaining; £3 equal 60 s., which added to 7 s. 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 6 d. Hence the following OPERATION. £ 8. d. 5)103 7 6 20 13 6 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. 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 Ans. 17 mi. 417 rd. the average distance each day? 12. If 26 casks contain 21 hhd. 11 gal. 2 qt. 1 pt., what is the capacity of each cask? CASE II. Ans. 51 gal. 1 qt. pt. 312. To divide one compound number by a similar one. 1. Divide £26 6 s. 2 d. by £4 15 s. 8 d. SOLUTION.-£26 6 s. 2 d. equais 6314 pence; £4 15 s. 8 d. equals 1148 pence; and dividing 6314d. by 1148 d. we obtain a quotient of 54. From this solution we have the following OPERATION. £26 68. 2d. 6314d. £4 15s. 8d.=1148 d. 1148)6314(5), Ans. 6314 |