ADDITION OF COMPOUND NUMBERS. 306. Addition of Compound Numbers is the process of finding the sum of two or more similar compound numbers. 1. Find the sum of £7 6 s. 8d.; £5 9 s. 3a; £14 13 8. 10 d.; and £11 19 s. 11 d. 7 OPERATION. £ S. d. 6 8 5 9 3 14 13 10 11 19 11 8 SOLUTION.-We write the numbers so that similar units shall stand in the same column, and begin at the right to add. 11 d. plus 10 d., plus 3 d., plus 8 d., are 32 d., which by reduction we find equals 2 s. and 8 d.; we write the 8 d. in the pence column, and reserve the 2s. to add to the column of shillings: 2 s. plus 19s., plus 13 s., plus 9s., plus 6 s., are 49s., which by reduction we find equals £2 and 9 s.; we write the 9 s. in the column of shillings, and reserve the £2 to add to the column of pounds; £2 plus £11, plus £14, plus £5, plus £7, equal £39, which we write under the pounds. Hence the following 39 9 Rule.-I. Write the compound numbers so that similar units stand in the same column. II. Begin with the lowest denomination and add each column separately, placing the sum, when less than a unit of the next higher denomination, under the column added. III. When the sum equals one or more units of the next higher denomination, reduce it to this denomination, write the remainder under the column added, and add the quotient obtained by reduction to the next column. IV. Proceed in the same manner with all the columns to the last, under which write the entire sum. Proof. The same as in addition of simple numbers. NOTE.-IL writing, if any places are wanting supply them with a cipher 12. Find the sum of 132 sq. yd. 8 sq. ft. 120 sq. in., 246 sq. yd. 7 sq. ft. 137 sq. in., 546 sq. yd. 3 sq. ft. 129 sq. in., 765 sq. yd. 6 sq. ft. 105 sq. in., 382 sq. yd. 5 sq. ft. 126 sq. in. Ans. 2074 sq. yd. 6 sq. ft. 41 sq. in. 13. Find the sum of 16 A. 104 P. 18 sq. yd. 7 sq. ft., 25A. 116 P. 28 sq. yd. 8 sq. ft., 18 A. 139 P. 17 sq. yd. 6 sq. ft., 27 A. 106 P. 30 sq. yd. 8 sq. ft., 24 A. 155 P. 26 sq. yd. 5 sq. ft. Ans. 113 A. 144 P. 1 sq. yd. 7 sq. ft. SUPPLEMENTARY EXAMPLES. To be omitted unless otherwise directed. 14. A vintner sold to A 5 hhd. 59 gal. 3 qt. 1 pt. of wine, to B 20 hhd. 45 gal. 2 qt., to C 39 hhd. 58 gal. 1 pt., and had as much as he sold A remaining; how much had he at first? Ans. 72 hhd. 34 gal. 1 qt. 15. What is the sum of 126 yr. 10 mo. 5 wk. 17 hr., 236 yr. 9 mo. 2 wk. 7 da. 18 hr. 41 min., 425 yr. 8 mo. 4 wk. 3 da. 20 hr. 16 min., 198 yr. 7 mo. 6 wk. 19 hr. 52 min., 385 yr. 5 wk. 40 min. ? Ans. 1373 yr. 3 mo. 3 wk. 6 da. 4 hr. 29 min. 16. Find the sum of 144 cu. yd. 18 cu. ft. 1329 cu. in., 275 cu. yd. 25 cu. ft. 1076 cu. in., 382 cu. yd. 17 cu. ft. 1521 cu. in., 420 cu. yd. 20 cu. ft. 1507 cu. in., 367 cu. yd. 21 cu. ft. 1473 cu. in. Ans. 1591 cu. yd. 23 cu. ft. 1722 cu in. SUBTRACTION OF COMPOUND NUMBERS. 307. Subtraction of Compound Numbers is the process 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 2 17 4 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 Hence the following Oz. 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 corresponding 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 num bers. 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. 271 rd. 3 yd. 1 ft. 11 in. Ans. 20 mi. 48 rd. 3 in. SUPPLEMENTARY EXAMPLES. To be omitted uniess otherwise directed. 16. Subtract $16 577 5 mills from $25 20 73 mills, and add 2 eagles 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 had a hogshead 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 at. 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. £ 8. d. 12 11 100 7 12 8 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 4 s. and 8 d.; we write the 8 d. 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 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 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 15cwt 90 lb.: how much did he sell? Ans. 79 cwt 50 lb. |