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common or vulgar fractions. Thus, 75 is 7%, and ‘25 is : now, % × 20% == '1875, Ans. same as be

1875

fore.

3. Multiply '125 by '03.

OPERATION.

'125

'03

'00375 Prod.

Here, as the number of significant figures in the product is not equal to the number of decimals in both factors, the deficiency must be supplied by prefixing ciphers, that is, placing them at the left hand. The correctness of the rule may appear from the following process: 125 is, and ‘03 is To: now, 25 X TOU =1000000375, 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 off so many figures for deci mals 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.

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 175 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 "7 of a melon worth? Ans. 6 cents.

8. Multiply five hundredths by seven thousandths.

Product, '00035.

Ans. 34'8. Ans. 3121'2. Ans. '020831. Product, 331'76.

Product, 3'44.

14. Multiply '0062 by '0008.

15. Multiply forty-seven tenths by one thousand eightysix hundredths.

9. What is '3 of 116?
10. What is '85 of 3672?
11. What is '37 of '0563?

12. Multiply 572 by '58.

13. Multiply eighty-six by four hundredths.

16. Multiply two hundredths by eleven thousandths. 17. What will be the cost of thirteen hundredths of a ton of hay, at 11 a ton?

18. What will be the cost of three hundred seventy-five thousandths of a cord of wood, at $2 a cord?

19. If a man's wages be seventy-five hundredths of a dollar a day, how much will he earn in 4 weeks, Sundays excepted?

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 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 $44'718, what is that a barrel?

By taking away the decimal point, $44'718 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.

OPERATION.

6)44'718

Ans. 7'453

As the decimal places in the divisor and quotient, counted together, must be equal to the number of decimal places in the dividend, there being no decimals 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.

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2. At $475 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 di

vidend 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(624g barrels, the Answer; that is, 6 barrels and 2g of another barrel.

2850

250

But the remainder, 250, ir.stead of being expressed 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 cr not from the divisor and dividend.

OPERATION.

475)31'00(6'526+ barrels, the Answer; that is, 6 bar rels and 526 thousandths of another barrel.

2850

By annexing a cipher to the first remainder, thereby reducing it to 10ths, and continuing the division, we obtain from it '5, and a still further remainder of 125, which, by annexing another cipher, is reduced to 100ths, and so on.

The last remainder, 150, is % of . a thousandth part of a barrel, which

2500

2375

1250

950

3000

2850

150

is of so trifling a value, as not to merit notice.

If now we count the decimals in the dividend, (for every cipher annexed to the remainder 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.

3. Under ¶ 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,

The divisor, 125, in 375 goes 3 times, and no remainder. We have only to place the decimal point in the quotient, and the work is done. There are five decimal places in the 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 quotient; and, since we have but one figure in the quotient, the deficiency must be supplied by prefixing a cipher.

375

The operation by vulgar fractions will bring us to the same result. Thus, 125 is, and 00375 is Too: now, co ÷ 788% = 73138885 T8='03, the same as before.

125
000

*

OPERATION.

'125) '00375('03

375

000

¶ 73. The foregoing ex-mples 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 off 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 quotient, 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? Aus. $36'375.

5. At $5 per bushel, how many bushels of rye can be bought for $141 ? Ans. 188 bushels.

6. At 124 cents per lb., how many pounds of butter may be bought for $37? Ans. 296 ib.

7. At 64 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 barAns. $8333 +. Quot. '037 +

rel?

9. Divide 2 by 53'1.

10. Divide '012 by '005.
11. Divide three thousandths by four hundredths.

Quot. 2'4

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.

¶ 74. 1. A man has ĝ 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, (¶ 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 28, 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, '75 of a dollar.

The operatic n may be presented in form as follows :—

Num.

Denom. 4 ) 3′0 (75 of a dollar, the Answer.

28

20

20

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