Cancelling 3 from the 15 of the numerator and from the 9 of the denominator gives 5 in place of the former and 3 in place of the latter. As no further cancellation can be made, we multiply the remaining factors toge 1 X 5 3X2

ther, which gives

5

=

6

for the result.

The following is a convenient form of writing the work:

(h.) Reduce the following to their lowest terms:

86. To find a Fractional Part of a Number.

1. What is of 715?

1ST SOLUTION.

of 715 equals 5 times

of 715.

dividing 715 by 6, is 119, and 5 times 119 is 5955. 5955.

2D SOLUTION. -§ of 715 of 3575 is 595g. Hence, & of 715

of 715, found by Hence, & of 715 —

=

of 5 times 715. 5 times 715

=

5955, as before.

(a.) As of 715 equals 5 times

of 715, or 1 of 5 times 715, so of any number equals 5 times of that number, or of 5 times that number; .07 of any number equals 7 times .01 of that number, or .01 of 7 times that number, etc. Hence

(b.) To get a fractional part of a number, we either divide that number by the denominator of the fraction and multiply the quotient by the numerator, or we multiply the number by the numerator of the fraction and divide the product by the denominator. (c.) What is

1ST SOLUTION.-.009 of 32.45 equals 9 times .001 of 32.45. .001 of 32.45, found by removing the decimal point 3 places to the left (see 14), is .03245, and 9 times .03245 = .29205.

=

=

2D SOLUTION.-.009 of 32.45 .001 of 9 times 32.45. 9 times 32.45 = 292.05, and .001 of this, found by removing the decimal point 3 places further to the left, is .29205, as before.

*The 1.3 should be read as an improper fraction, as should all other numbers similarly situated.

87. Reduction of Compound Fractions to Simple Ones.

(a.) To reduce a compound fraction to a simple one is merely to find a fractional part of a fraction.

1. Reduce the compound fraction off to a simple fraction, i. e. find of fz.

SOLUTION.

of may be found by writing, and then making 9 a factor of the numerator and 10 a factor of the denominator (see 83, ex. 15), cancelling and reducing as in the written work below:

2. Reduce of 42 of 15% to a simple fraction.

SOLUTION. Writing 1, or its equal, 7, as the number of which the fractional part is to be obtained, we find 2 of 2 by making 49 a factor of the numerator and 54 a factor of the denominator; and of this result by making 11 a factor of the numerator and 21 a factor of the denominator. The written work would take the following form:

(b.) These solutions show that, to reduce a compound fraction to a simple one, we first reduce the mixed numbers, if there are any, to improper fractions, then make all the numerators of the compound fraction factors of the new numerator, and all the denominators factors of the new denominator, and then cancel and reduce.

(c.) Reduce each of the following compound fractions to simple

(a.) To reduce a vulgar fraction to an equivalent decimal, we have only to get the part of 1 which the given fraction indicates, carrying out the work to the required number of decimal places.

(b.) Practically, the reduction is made by annexing zeroes to the numerator and dividing by the denominator.

NOTE.-In most cases it will be sufficiently exact to carry out the division to 3 or 4 decimal places, but sometimes it is necessary to earry it out much further.

(c.) Reduce the following to decimal fractions, carrying out the division, when only approximate values can be obtained, to 6 places of decimals.

1. What is the value of of a mile in units of lower denomi nations, i. e. in fur. rd. yd. etc.?

1ST SOLUTION.-1⁄2 of a mile

12

=

12

of 7 miles, which is found by dividing

7 m. 0 fur. 0 rd. 0 yd. 0 ft. 0 in. by 12, as in compound division.

3D SOLUTION. There must be 8 times as many furlongs as miles, or in

of a mile, 8 times

rods as furlongs, or in

fur.

=

4 fur. But there must be 40 times as many fur. 40 times rods 26 rods. But there must

be 5 or times as many yards as rods, or in & of a rod 1 times yd. 3 yd. But there must be three times as many feet as yards, or in of a yard, 3 times ft. 2 ft. Hence, of a mile 4 fur. 26 rd. 3 yd..

2 ft.

=

(a.) Reduce each of the following to units of a lower denomination:

10. Reduce .5135 of a pound Troy to units of a lower denomination.

SOLUTION. There must be 12 times as many ounces as pounds, or, in this case, 12 times .5135 of an ounce = 6.1620 oz. 6.162 oz. But there must be 20 times as many pennyweights as ounces, or, in .162 of an ounce, 20 times .162 of a pennyweight = 3.240 dwt. 3.24 dwt. But there must be 24 times as many grains as pennyweights, or, in .24 of a pennyweight, 24 times .24 of a grain Hence, .5135 of a lb.

=

1=3

6 oz. 3 dwt. 5.76 gr.

5.76 gr.

.5135

12

6.1620

20

3.240

24

5.76

NOTE. It will be seen that, practically, the reductions in the last example, as well as in those immediately preceding it, were made by multiplying the fractional part of each denomination by the number which expresses how many units of the next lower denomination are equal to a unit of the denomination of the fraction.

(b.) Reduce each of the following to lower denominations: