Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

MULTIPLICATION & DIVISION OF COMPOUND NUMBERS.

T 41. 1. A man bought 2 yards of cloth, at 1 s. 6 d. per yard; what was the cost?

2. If 2 yards of cloth cost 3 shillings, what is that per Jard?

3. A man has three pieces of cioth, each measuring 10 yds. 3 qrs.; how many yards in the whole?

4. If 3 equal pieces of cloth contain 32 yds. 1 qr., how much does each piece contain?

5. A man has five bottles, each containing 2 gal. 1 qt. 1 pt.; how much wine do they all contain?

6. A man has 11 gal. 3 qts. 1 pt. of wine, which he would divide equally into five bottles; how much must he put into each bottle?

7. How many shillings are 3-times 8 d.?
3 x 10 d. ?
7 X 6 d.?

4 X 7 d.?

d.?

2 X 3 qrs.?

8 d.?
6 d. ?

8. How much is one third
of 2 s. 6 d. ?
To of 7 s. 6 d. ?

[ocr errors]
[ocr errors]

5 X 2 qrs.?

of 2 shillings?

of 2 s. 4 d.?

of 14 d.?

OPERATION.

£.

s. d. qr.

1 5 8 3 price of 1 yard.
6 number of yards.

3 x 9 d.?

10 X

of 2 s 3 of 3 s of 21 d. ?

9. At 1£. 5 s. 8 d. per 10. If 6 yards of cloth cost yard, what will 6 yards of 7£. 14 s. 4 d., what is the cloth cost? price per vard?

Here, as the numbers are large, it will be most convenient to write them down before multiplying and dividing.

OPERATION.
£. s. d. qr.

6)7 14 4 2 cost of 6 yards.

Ans. 7 14 4 2 cost of 6 yards.

1583 price of 1 yard. Proceeding after the man

6 times 3 qrs. are 18 qrs.ner of short division, 6 is con4 d. and 2 qrs. over; we set tained in 7£. 1 time, and 1£. down the 2 qrs.; then, 6 timesjover; we write down the 8 d. are 48 d., and 4 to carry quotient, and reduce the re makes 52 d. = 4 s. and 4 d. mainder (1.) to shillings, over, which we write down; (20 s.,) which, with the given again, 6 times 5 s. are 30 s. shillings, (14 s.,) make 34 s. ¡

and 4 to carry makes 34 s. 16 in 34 s. goes 5 times, and 4 s. 1. and 14 s. over; 6 times over; 4 s. reduced to pence 1£. are 6£., and 1 to carry = 48 d., which, with the makes 7£., which we write given pence, (4 d.,) make 52 down; and it is plain, that the d.; 6 in 52 d. goes 8 times, and united products arising from 4 d. over; 4 d. = 16 qrs., the several denominations is which, with the given qrs. the real product arising from (2) 18 qrs. ; 6 in 18 qrs. goes the whole compound number. 3 times; and it is plain, that the united quotients arising from the several denominations, is the real quotient arising from the whole compound number.

11. Multiply 3£. 4 s. 6 d. by 7.

13. What will be the cost 14. At 2£. 12 s. 6 d. for 5 of 5 pairs of shoes at 10 s. 6 d. pairs of shoes, what is that a a pair? pair?

15. In 5 barrels of wheat, each containing 2 bu. 3 pks. 6 qts., how many bushels?

16. If 14 bu. 2 pks. 6 qts. of wheat be equally divided into 5 barrels, how many bushels will each contain ?

18. If 9 coats contain 39

12. Divide 22£. 11 s. 6 d. by 7.

17. How many yards of cloth will be required for 9 yds. 3 qrs. 3 na., what does 1 coats, allowing 4 yds. 1 qr. coat contain? 3 na. to each ?

19. In 7 bottles of wine, 20. If 5 gal. 1 gill of wine each containing 2 qts. 1 pt. 3 be divided equally into 7 bot gills, how many gallons? tles, how much will each con tain?

21. What will be the 22. If 8 silver cups weigh weight of 8 silver cups, each 3 lb, 9 oz. 1 pwt. 16 grs., what weighing 5 oz. 12 pwt. 17 is the weight of each?

grs.?

23. How much sugar in 12 24. If 119 cwt. 1 qr. of su hogsheads, each containing gar be divided into 12 hogs cwt. 3 qrs. 21 lb. ? heads, how much will each hogshead contain?

25. In 15 loads of hay, each 26. If 15 teams be loaded weighing 1 T. 3 cwt, 2 qrs., with 17 T. 12 cwt. 2 qrs. of how many tons? hay, how much is that to each team?

When the multiplier, or divisor, exceeds 12, the operations of multiplying and dividing are not so easy, unless they be composite numbers; in that case, we may make use of the component parts, or factors, as was done in simple numbers.

=

Thus 15, in the example 15 being a composite numabove, is a composite number ber, and 3 and 5 its compoproduced by the multiplica-nent parts, or factors, we may tion of 3 and 5, (3 × 5 divide 17 T. 12 cwt. 2 qrs. by 15.) We may, therefore, one of these component parts, multiply 1 T. 3 cwt. 2 qrs. by or factors, and the quotient one of those component parts, thence arising by the other, or factors, and that product by which will give the true the other, which will give the answer, as already taught, true answer, as has been al-(¶ 20.)

ready taught, (¶ 11.)

T.

1

A

3

OPERATION.

crot. gr.

One factor,

3 2

3 one of the factors. The other factor, 5) 5 17 2

Ans. 1 3 2

10 2

5 the other factor.

17

12 2 the answer.

27. What will 24 barrels of flour cost, at 2£. 12 s. 4 d. a barrel?

29. What will 112 lb. of Bugar cost, at 74 d. per lb. ?

Note. 8, 7, and 2, are fac-lb. ? tors of 112.

33. What will 139 yards of cloth cost, at 3 £. 6 s. 5 d. per yard?

139 is not a composite number. We may, however, decompose this number thus, 139 100+30+ 9.

OPERATION. T. cut. qr. 3) 17 12 2

28. Bought 24 barrels of flour for 62 £. 16 s.; how much was that per barrel ?

30. If 1 cwt. of sugar cest 3 £. 7 s. 8 d., what is that per

31. How much brandy in 32. Bought 84 pipes of 84 pipes, each containing 112 brandy, containing 9468 gal. gal. 2 qts. 1 pt. 3 g.? 1 qt. 1 pt.; how much in a pipe?

34. Bought 139 yards of cloth for 461 £. 11 s. 11 d.; what was that per yard?

When the divisor is such a
number as cannot be produced
by the multiplication
numbers, the better way is to

We
Te may now multiply the divide after the manner of

price of 1 yard by 10, which long division, setting down will give the price of 10 yards, the work of dividing and re and this product again by 10, ducing in manner as fol which will give the price of lows: 100 yards.

£.

417

We may then multiply the 139) 461 price of 10 yards by 3, which will give the price of 30 yards, and the price of 1 yard by 9, which will give the price of 9 yards, and these three products, added together, will evidently give the price of 139 yards; thus:

£. S. d.

3

6

33

4

332 1

99 12 29 17

5 price of 1 yd.

10

2 price of 10 yds.

10

44

20

S. d.

57

12

11

891 (6 s.

834

695 ( 5 d.

695

11 (3£.

The divisor, 139, is contained in 461 £. 3 times, 8 price of 100 yds. (3£.,) and a remainder of 6 price of 30 yds. 44., which must now be 9 price of 9 yds. reduced to shillings, multi461 11 11 price of 139 yds. in the given shillings, (11 s.,) plying it by 20, and bringing Note. In multiplying the making 891 s., in which the price of 10 yards (33£. 4 s. divisor is contained 6 times, 2 d.) by 3, to get the price of (6 s.,) and a remainder of 30 yards, and in multiplying 57 s., which must be reduced the price of 1 yard (3£. 6 s. to pence, multiplying it by 12, 5 d.) by 9, to get the price of and bringing in the given 9 yards, the multipliers, 3 and pence, (11 d.,) together mak9, need not be written down, ing 695 d., in which the dibut may be carried in the visor is contained 5 times, mind. (5 d.,) and no remainder.

The several quotients, 3, 6 s., 5 d., evidently make the

answer.

The processes in the foregoing examples may now be pre sented in the form of a

RULE for the Multiplication of RULE for the Division of Com Compound Numbers. pound Numbers. I. When the multiplier does I. When the divisor does not exceed 12, multiply suc-not exceed 12, in the manner cessively the numbers of each of short division, find how denomination, beginning with many times it is contained in the least, as in multiplication the highest denomination, unof simple numbers, and carry der which write the quotient, as in addition of compound and, if there be a remainder,, numbers, setting down the reduce it to the next less dewhole product of the highest nomination, adding thereto the denomination. number given, if any, of that denomination, and divide as before; so continue to do through all the denominations, and the several quotients will be the answer.

II. If the multiplier exceed II. If the divisor exceed 12, 12, and be a composite num- and be a composite, we may diber, we may multiply first by vide first by one of the comone of the component parts, ponent parts, that quotient by that product by another, and another, and so on, if the comso on, if the component parts ponent parts be more than be more than two; the last two; the last quotient will be product will be the product re- the quotient required.

quired.

III. When the multiplier III. When the divisor exexceeds 12, and is not a com-ceeds 12, and is not a composite, multiply first by 10, posite number, divide after the and this product by 10, which manner of long division, setwill give the product for 100; ting down the work of diand if the hundreds in the mul-viding and reducing.

tiplier be more than one, multiply the product of 100 by the number of hundreds; for the tens, multiply the product of 10 by the number of tens; for the units, multiply the multiplicand; and these several products will be the product required.

1.

« ΠροηγούμενηΣυνέχεια »