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MULTIPLICATION OF DECIMAL FRACTIONS.

171. 1. How much hay in 7 loads, each containing 236571 ewt. ?

OPERATION.
23'571 cwt. = 23571 1000ths of a cwt.
7

7

OPERATION.

675

150

Ans. 164'997 cwt. = 164997 1000ths of a cwt. We may here (TT 69) consider the multiplicand so many thousandths of a cwt., and then the product will evidently bé thousandths, and will be reduced to a mixed or whole number by pointing off 3 figures, that is, the same number as are in the multiplicand; and as either factor may be made the multiplier, so, if the decimals had been in the multiplier, the same number of places must have been pointed off for decimals. Hence it follows, we must always point of in the product as many places for decimals as there are decimal places in both factors. 2. Multiply (75 by '25.

In this example, we have 4 decimal

places in both factors; we must therefore 625

point off 4 places for decimals in the pro

duct. The reason of pointing off this 375

number may appear still more plain, if we consider the two factors as common or

vulgar fractions. Thus, 675 is 100, and 1875, Product.

“25 is zor: now, 70 X 100700

'1875, Ans. same as before. 3. Multiply '125 by '03.

Here, as the number of significant figures '125

in the product is not equal to the number of '03

decimals in both factors, the deficiency must

be supplied by prefixing ciphers, that is, '00375 Prod. placing them at the left hand. - The correct

ness of the rule may appear from the following process: '125 is 125, and '03 is

18 : now, X Too 203360 :00375, the same as before. These examples will be sufficient to establish the following

RULE. In the multiplication of decimal fractions, multiply as in whole numbers, and from the product point of so many figures for decimals as there are decimal places in the multiplicand and multiplier counted together, and, if there are not so many figures in the product, supply the deficiency by prefixing ciphers.

10

OPERATION,

EXAMPLES FOR PRACTICE. 4. At $5'47 per yard, what cost 8'3 yards of cloth?

Ans. $45'401. 5. At $07 per pound, what cost 26'5 pounds of rice?

Ans. $1'855. 6. If a barrel contain 1'75 cwt. of flour, what will be the weight of 63 of a barrel ?

Ans. 1'1025 cwt. 7. If a melon be worth $'09, what is cry of a melon worth?

Ans.
636

cents. 8. Multiply five hundredths by seven thousandths.

Product, '00035. 9. What is '3 of 116?

Ans. 34'8. 10. What is '85 of 3672 ?

Ans. 3121'2. 11. What is '37 of '0563 ?

Ans. '020831. 12. Multiply 572 by '58.

Product, 331676. 13. Multiply eighty-six by four hundredths. Product, 3'44. 14. Multiply ‘0062 by '0008.

15. Multiply forty-seven tenths by one thousand eighty-six hundredths.

Note. It has already been remarked, (Note 2, Avoirdupois Weight,) that the hundred weight of 112 lbs. is going out of use, and that 2000 pounds are considered a ton.-$10 per ton (of 2000 lbs.) is evidently 50 cents for 100 lbs. which is į cent, or 5 mills, for i lb. Hence, -Having the cost per ton to find the cost for a 100 lbs. or, for 1 lb., -GENERAL RULE : Divide the cost per ton by 2, and the quotient, remembering to remove the separatrix one figure from its natural place towards the left hand, will express the cost for a 100 lbs: or remove it three places, and it will express the cost for 1 lb.

16. At $13.50 per ton, what is that for 100 lbs.? for 1 lb.?

$13.50 - 2=675; then removing the separatrix one place to the Jeft, ('675 or '67) gives Ans. 67 cents, cost of 100 lbs. or removing it three places, ('0064) gives 64 mills, cost of i lb.

17. At $11'40 per ton, what costs 3476 lbs. of hay?

$11'40 - 2, &c. = 57 cents cost per hundred. We may now reduce the pounds (3476) to hundreds and decimals of a hundred, by pointing off the two right hand figures for decimals, thus, 34,76 hun. dreds. Then, 34'76 hundreds multiplied by 57 cents, the cost per hundred, pointing the product according to rule, Multiplication of Decimals, (34'76 X 57 = 19*8132) gives the Ans. $19-81 t.

18. At 37 cents per hundred what will be the cost of transporta. tion on 5863 lbs. that is, on 58'63 hundreds ? Ans. 21'98 +.

DIVISION OF DECIMAL FRACTIONS.

72. Multiplication is proved by division. We have seen, in multiplication, that the decimal places in the product must always be equal to the number of decimal places in the multiplicand and multiplier counted together. The multiplicand and multiplier, in

OPERATION

proving multiplication, become the divisor and quotient in division. It follows of course, in division, that the number of decimal places in the divisor and quotient, counted together, must always be equal to the number of decimal places in the dividend. This will still further appear from the examples and illustrations which follow.

1. If 6 barrels of flour cost $44718, what is that a barrel ?

By taking away, the decimal point, $44718 = 44718 mills, or 1000ths, which, divided by 6, the quotient is 7453 mills, = $7'453, the Answer. Or, retaining the decimal point, divide as in whole numbers.

As the decimal places in the divisor and 6)44718 quotient, counted together, must be equal to

the number of decimal places in the diviAns. 7'453 dend, there being no decímals in the divisor,

-therefore point off three figures for decimals in the quotient, equal to the number of decimals in the dividend, which brings us to the same result as before.

2. At $4'75 a barrel for cider, how many barrels may be bought for $31 ?

In this example, there are decimals in the divisor, and none in the dividend. $475 = 475 cents, and $31, by annexing two ciphers, =3100 cents; that is, reduce the dividend to parts of the same denomination as the divisor. Then, it is plain, as many times as 475 cents are contained in 3100 cents, so many barrels may be bought. 475)3100(695% barrels, the Ansuper ; that is, 6 barrels and 17 2850

barrel

But the remainder, 250, instead of being ex250 pressed in the form of a common fraction, may be

reduced to 10ths by annexing a cipher, which, in effect, is multiplying it by 10, and the division continued, placing the decimal point after the 6, or whole ones already obtained, to distinguish it from the decimals which are to follow. The points may be withdrawn or not from the divisor and dividend.

OPERATION. 4'75)31'00(6'526 + barrels, the answer; that is, 6 barrels and 2850

526 thousandths of another barrel.

By annexing a cipher to the first remainder, 2500 thereby, reducing it to loths, and continuing 2375 the division, we obtain from it'5, and a still

further remainder of 125, which, by annexing 1250 another cipher, is reduced to 100ths, and so 950

The last remainder, 150, is 15 of a thou3000 sandth part of a barrel, which is of so trifling 2850 a value as not to merit notice.

If now we count the decimals in the divi. 150 dend, (for every cipher annexed to the remain

der is evidently to be counted a decimal of the dividend,) we shall find them to be five, which corresponds with the number of decimal places in the divisor and quotient counted together,

on.

3. Under T 71, ex. 3, it was required to multiply '125 by '03; the product was ‘00375. Taking this product for a dividend, let it be required to divide ‘00375 by '125. One operation will prove the other. Knowing that the number of decimal places in the quotient and divisor, counted together, will be equal to the decimal places in the dividend, we may divide as in whole numbers, being careful to retain the decimal points in their proper places. Thus,

OPERATION The divisor, 125, in 375 goes 3 times, and '125)'00375('03 no remainder. We have only to place the 375 decimal point in the quotient, and the work is

done. There are five decimal places in the 000 dividend ; consequently there must be five in

the divisor and quotient counted together; and, as there are three in the divisor, there must be two in the quo tient; and, since we have but one figure in the quotient, the deficiency must be supplied by prefixing a cipher.

The operation by vulgar fractions will bring us to the same result. Thus, '125 is 125 and '00375 is

3750: now, 1000 1000 + 10% = 1338888 = 1 03, the same as before.

11 73. The foregoing examples and remarks are sufficient to establish the following

RULE. In the division of decimal fractions, divide as in whole numbers, and from the right hand of the quotient point of as many figures for decimals as the decimal figures in the dividend exceed those in the divisor, and if there are not so many figures in the quoticnt, supply the deficiency by prefixing.ciphers.

If at any time there is a remainder, or if the decimal figures in the divisor exceed those in the dividend, ciphers may be annexed to the dividend or the remainder, and the quotient carried to any necessary degree of exactness, but the ciphers annexed must be counted so many decimals of the dividend.

EXAMPLES FOR PRACTICE. 4. If $472'875 be divided equally between 13 men, how much will each one receive ?

Ans. $36-375. 5. At $75 per bushel, how many bushels of rye can be bought for $141 ?

Ans. 188 bushels. 6. At 12} cents per lb., how many pounds of butter may be bought for $37 ?

Ans. 296 lb. 7. At 6 cents apiece, how many oranges may be bought for $8 ?

Ans. 128 oranges. 8. If '6 of a barrel of flour cost $5, what is that per barrel ?

Ans. $86333 9. Divide 2 by 53'1.

Quot. '037 10. Divide 'oi2 by '005.

Quot. 24. 11. Divide three thousandths hy four hundredths. Quot. '075. 12. Divide eighty-six tenths by ninety-four thousandths. 13. How many times is '17 contained in 8 ?

REDUCTION OF COMMON OR VULGAR FRACTIONS

TO DECIMALS.

174. 1. A man has of a barrel of flour ; what is that expressed in decimal parts ?

As many times as the denominator of a fraction is contained in the numerator, so many whole ones are contained in the fraction. We can obtain no whole ones

in , because

the denominator is not contained in the numerator. We may, however, reduce the numerator to tenths, (TT 72, ex. 2,) by annexing a cipher to it, (which, in effect, is multiplying it by 10,) 'making 40 tenths, or 4'0. Then, as many times as the denominator, 5, is contained in 40, so many tenths are contained in the fraction. 5 into 40 goes 8 times, and no remainder.

Ans. '8 of a bushel. 2. Express of a dollar in decimal parts.

The numerator, 3, reduced to tenths, is 1 o, 30, which, divided by the denominator, 4, the quotient is 7 tenths, and a remainder of 2. This remainder must now be reduced to hundredths, by annexing another cipher, making 20 hundredths. Then, as many times as the denominator, 4, is contained in 20, so many hundredths also may be obtained. 4 into 20 goes 5 times, and no remainder. # of a dollar, therefore, reduced to decimals, is 7 tenths and 5 hundredths, that is, 675 of a dollar. The operation may be presented in form as follows:

Num.
Denom. 4) 3'0 ( 675 of a dollar, the Answer,

28

20
20

8. Reduce me to a decimal fraction.

The numerator must be reduced to hundredths, by annexing two ciphers, before the division can begin.

66 ) 4'00 ('0606 t, the Answer.

396

400 396

As there can be no tenths, a cipher must be placed in the quotient, in tenth's place.

Note. cannot be reduced exactly; for, however long the division be continued, there will still be a remainder.* It is sufficient

• Decimal figures, which continually repeat, like '06 in this example, are called Repetends, or Circulating Decimals. If only one figure repeats, as '3333 or 7777, &c., it is called a single repetend. "If two or more figures circulate alternately, as '060606, 234234234, &c., it is called a compound re

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