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SUBTRACTION OF COMPOUND NUMBERS.

146. Subtraction of Compound Numbers is the process of finding the difference between two compound numbers. Ex. 1. From 617£. lis. 8d. take 18l£. 15s. 5d.

Ans. 435£. 16s. 3d.

Operation. Having placed the less number un

£ 8- d- der the greater, pence under pence, Min. 6 17 11 8 shillings under shillings, &c, we beSub. 18 1 15 5 gin with pence, thus: 5d. from 8d. P 4. Q ^ T~r Q leaves 3d., which we set under the Kem. 4oo lb o co1umn of pence. As we cannot take 15s. from lis., we add 20s. = l£. to the lis., making 31s., and then subtract the I5s. from it, and set the remainder, 16s., under the column of shillings. Then, having added l£. = 20s. to the 18l£., to compensate for the 20s. added to the lis. in the minuend, we subtract the pounds as in subtraction of simple numbers, and obtain 435£. for the remainder, and as the result complete, 435£. 16s. 3d.

Rule. Write the less compound number under the greater, so that units of the same denomination shall stand in the same column. Subtract as in subtraction o f simple numbers.

If any number in the subtrahend is larger than that above it, add to the upper number as many units as make one of the next higher denomination before subtracting, and curry one to the next lower number before subtracting it.

Proof. — The proof is the same as in simple subtraction,

Examples.
2. 3.

£ s. d. £ s. d.

871634; ,617 115$

19 17 9$ 18115 8$

67 1 8 5% ~~

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same point in the heavens, how far will the Moon have gained on the Sun in 24 hours? Ans. 12° 11' 27".

37. A farmer raised 136bu. of wheat; if he sells 49bu. 2pk. 7qt. lpt., how much has he remaining?

Ans. 86bu. lpk. Oqt. lpt.

38. If from a stick of timber containing 2T. 18ft. 1410in. there be taken 38ft. 1720in., how much will be left?

Ans. IT. 19ft. 1418in.

MULTIPLICATION OF COMPOUND NUMBEES.

HIi Multiplication of Compound Numbers is the process of taking a compound number any proposed number of times.

Ex. 1. What will 6 bales of cloth cost, at 7£. 12s. 7d. per bale? Ans. 45£. los. 6d.

Operation. Having written the multiplier under £. s. d. lowest denomination of the multiMultiplicand 7 12 7 p1icand, we multiply thus: 7d. X 6 Multiplier 6 = 42d. = 3s. 6d. We write the 6d.

Product 45 F5 6 un,ler the number multiplied, and reserve the 3s. to be added to the product of the shillings. Then, 12s. X 6 = 72s., and 3s. (carried) = 75s. = 3£. 15s. We write the 15s. under the column of shillings, and reserve the 3£. to be added to the product of the pounds. Again, 7£. X 6 = 42£., and 3£.. (carried) = 45£. This, placed under the column of pounds, gives 4o£. 15s. 6d.

Rule. Multiply each denomination of the compound number as in multiplication of simple numbers, and carry as in addition of compound numbers.

Proof. — Write down by themselves the several products obtained by multiplying each denomination of the multiplicand by the multiplier, and these partial products added together will equal the entire product, if the work be right. (Art. 60.)

Note. — Going a second time carefully over the work is a good way of testing its accuracy. On learning Division of Compound Numbers, the pupil will find that rule a better method of proving multiplication of compound numbers.

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2£. 8s. 9£d. x 8 = 19£. 10s. 4d.

Note. — The answers to the following examples may be found in corresponding numbers of examples in Division of Compound Numbers.

3. 4.

T. cwt , qr. lb. oz. dr. lb. oz. pwt. gr.

611931715 15 7111415 9 5

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