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denominator. Now, if we wish to bring a fraction already in its lowest terms to an equivalent fraction having some power of 10 for its denominator, it can only be done by multiplying its numerator and denominator by some integer; and it is impossible to obtain any power of 10 by multiplication only, from a number which contains any prime factor, other than 2 or 5.

Reduction of Terminating Decimals to Fractions. 25. Remembering (Art. 1) that any given terminating decimal may be considered as derived from an integer by diminishing it 10, or 10', 109, &c. fold, we have

•345 345 ; 103. Hence, .345 is the value of the ratio 345 : 103; and we

345 345 have, also, 345

103 1000 The following rule is, therefore, clear :

RULE.—Make a numerator of the integer, formed by taking away the decimal point; and for a denominator put 1, followed by as many ciphers as there are given decimal figures. Ex. 1.- •625

8. 3236 319_*_5

318. ·0025 To ovo

ado.

6 2 5

5 X 125
8 X 125

3.95 =

20 * 5

2 5 400 X 25

Reduction of Recurring or Circulating Decimals to

Fractions. 26. There are two kinds, and it is convenient to treat them separately.

(1.) Pure circulating decimals, where the whole of the decimal figures recur. RULE. —Take

away the decimal point and the dots, make a numerator of the integer thus obtained, and place under this as denominator as many nines as there are recurring figures. The following example will make this rule clear. Ex.-Reduce 207 to a fraction.

The value of the decimals is evidently not altered by writing it .207207. Let us remove the decimal point three

places to the right, or, what is the same thing (Art. 1), let us multiply the given decimal by 1000.

We then get 207.207 as the value of 1000 times the given decimal. Now the number 207 207 includes the integer 207, and the given decimal -207; and it therefore follows that the integral part 207 is (1000 – 1), or 999 times the value of the given decimal. Hence, dividing it by 999, we get

207

207 • 999 97. (2.) Mixed circulating decimals.Where part only of the figures recurs. RULE.-Take

away the decimal point and the dots, subtract from the integer thus obtained the integer formed by the figures which do not recur, and make a numerator of the result. Then, for a denominator, place underneath as many nines as there are recurring figures, followed by as many ciphers as there are figures which do not recur.

We sħall make this rule clear by the following example:Ex.-Reduce .24573 to a fraction.

Let us remove the decimal point two places to the right, thus, by Art. 1, multiplying the given decimal by 100; we,

100 times the value of -24573 24.573. Now, by case (1) above, 24.573 = 2437% ; or reducing to an improper fraction, and noticing that 24 x 999 =. 24000 – 24, we have

24:573 24000 – 24 + 573 Hence, dividing this result by 100, we get .24573 = 2 4 5 7 3 – 24 ; 100 =

(Art. 24.) Ex. 1.-428571 = 43887! Ex. 2.—2709

then get

2 4 5 7 3

999

24,

999

- 24

2 4 573

99900

999

348 27. In arithmetical operations involving circulating decimals, and, indeed, any decimals having a large number of decimal figures, it is generally sufficient to obtain a result correct to a given number of decimals.

3 X 1 4 2 8 5 7
7 X 1 4 2 8 57
2682 149 X 2 X 9
9900

550 X 2 X 9

2709 27

9900

1. In addition and subtraction we obtain this result most easily by using in our operation one or two more figures of the given decimals than are required in the result.

Ex. 1.-Add together (correct to five places) the following:

•3026, 6.7294, •016, .4163729.

•3026026 6.7294444 .0166666 -4163729

7.4650865. Ans. 746508. NOTE.- If our object is to obtain a decimal of five places which shall give the approximate sum of the given decimals we must write 746509 as the approximate sum; for 7.46509 is nearer to the true value of 7•465086 &c. than 7'46508. The general rule is to increase by 1 the decimal figure at which we stop, when the next figure is 5 or above 5.

Ex. 2.— Find the difference (correct to six places) of 3.0745 and 4.263, and express the approximate difference by a decimal of five places.

4.26326326
3.07454545

1.18871781 Hence the difference correct to six places is 1.188717, and the required approximate difference 1.18872.

2. In multiplication and division of circulating decimals it is generally preferable to reduce the given decimals to fractions, bring out the result in a fractional form, and afterwards reduce this to a decimal.

Ex. VII. 1. Express as a decimal the sum of the following fractions:

1,I1, 3, 7, 7. 2. Reduce to fractions the following decimals :

•35, 026, 16, .142857, 16, 4285714. 3. Find the value (correct to six places) of 237 + 3.816 6.0235 + 4.29 ·002 + 1.374.

+

+

+

+

+ &c.

+

4. Add together •62, 037, 2.476i, 58106, 7, +375.
5. Find the product and quotient of 3.54 by 4.3.
6. Simplify the expression-

(4.6 x 428571 – 2.2 x •36) (1 – .16). In the next six examples the dots are signs of multiplication, and you are required to give the values of the expressions correct to six places. 1 1 1

1 7. 1

i 1.2 1.2.3 1.2.3.4 1 1

1

1 8.

3 3.33 5.35 7.37
1 1

1

1 9. 1.2 1.2.3 1.2.3.4 1.2.3.4.5

1 1 1 10. 1

+
87

86
1
1 1
1

1 11. 16

+
5 3.58 5.55 7.57

* &c.

239 1 3.2393 1 1 1 1 1

1 12. 1-2 3.4 4.5 5.6

6.7

+ &c.

+

+ &c.

+

+

+ &c.

87

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[ocr errors]

+ &c.

}

+

+

[blocks in formation]

CHAPTER III:

APPLICATION OF THE PRECEDING ARTICLES TO CONCRETE

QUANTITIES.

To find the Value of a Fraction of a Concrete Quantity.

28. RULE.—Multiply the concrete quantity by the numerator of the fraction, and divide the product by the denominator.

(2 tons, 3 cwt. 21 lbs.) x 3

6 tons, 9 cwt. 2 qrs. 7 lbs.

7

7

£. 8.

81

Ex. 1.–Find the value of of 2 tons, 3 cwt. 21 lbs. of 2 tons, 3 cwt. 21 lbs.

18 cwt. 2

qrs.

1 lb. Ans. Ex. 2.—Find the value of 33 of £12. 6s. 2 d.

d. 3 times £12. 6s. 24d. (£12. 6s. 27d.) x 3 =

36 18 6
(£12. 6s. 2 d.) x 2
of £12. 6s. 21d.

9
£24. 12s. 41d.

2 14

9 Hence, adding, the value required = £39 13 37

Ex. 3.–Find the value of of 4 miles + 3 of 3 fur. + at of 8 poles.

Mile. Fur. Poles. Yards. (4 x 3) miles of 4 miles

1 5 28 34 y of 3 fur.

0 2 13. 18

40 poles 1 of 8 poles

0 0 1 Hence, adding, the value required 2 0 3 NOTE.—The addition of the yards is thus effected :(37 + 18 + 441) yds. (8 + 6+36*142) yds. (8 + 131) yds.

931 yds. 1 pole,(31 + Áf) yds. = 1 pole, (3 + 1**) yds. = 1 pole, 44% yds.

12 miles

21 fur.

9

(3 x 7) fur.

9
(8 x 5) poles

2 1

21

41%

Ex. VIII. Find the values of

1. of £1; 3 of 1s.; of 12s.; of £3. 2. f of £5; 4 of a guinea; 1 of 2s. 6d.; of a crown. 3. 3} of £1. 12s.; 2} of £3. 10s.; 7141 of £3. 4s. 51. 4. of 1 ton; of 1 qr.; of 1 stone; if of 9 lbs. 5. 3 mile; 4 of 3 fur.; of 15 poles ; 34 of 24 of 12 yds. 8 2

53 6.

lunar month;

of 10 h. 15' 12". 20

10 7. 12 lbs. 3 oz. 7 dwt. 5 grs, x 34; 10 oz. 4 gr. ; 16.

leap year ; 12

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