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.

Ans. 113 A.

sq.

ft. 144 P. 1 sq. yd. 7 sq. ft.

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 bad as much as he sold A remaining; how much had he at first? Ans. 72hhd. 34 gal. 1 qt.

15. What is the sum of 126 yr. 10 mo. 5 wk. 17 h., 236 yr. 9 mo. 2 wk. 7 da. 18 hr. 41 min., 425 yr. 8 mo. 4 wk. 3 da. 20 h. 16 min., 198 yr. 7 mo. 6 wk. 19 h. 52 min., 385 yr. 5 wk. -40 min.? Ans. 1373 yr. 3 mo. 3 wk. 6 da. 4 h. 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 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

2

17

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 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 dec 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 mother 17 lb. 10 oz. 15 pwt. 20 gr., and 3 lb. his father; how much did he retain?

271 rd. 3 yd. 1 ft. 11 in.

gold, sent his

16 pwt. less to

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. 16. Subtract $16 577 5 mills from $25 20 74 mills, and add 2 eagles and 25 dimes to the result.

Ans. $31.2065.

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. 3qt. 1 pt., and then after putting in 12 gal. he found it lacked 16 gal. 1 pt. of containing €3 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.

£

12

OPERATION.

S.

d.

11

7

8

8

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 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 Ronvenient 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.

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; 2s. equal 24 d., which added to 6 d. equals 30 d.; of 30 d. is 6 d. Hence the following

OPERATION.

£ s. 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 Aumbers