Question

If 1÷3 be added to the numerator of a certain fraction, the fraction is increased by 1÷15 and if 1÷4 be taken from the denominator the fraction becomes 8÷19 Find the fraction.

Answer 1

$Let \: a \: fraction = \frac{x}{y}$

/* According to the problem given */

If 1÷3 be added to the numerator of a certain fraction, the fraction is increased by 1÷15

$\frac{x+\frac{1}{3}}{y} = \frac{x}{y} + \frac{1}{15}$

$\implies \frac{x}{y} + \frac{1}{3y} =\frac{x}{y} + \frac{1}{15}$

$\implies \frac{1}{3y} = \frac{1}{15}$

$\implies 3y = 15$

/* Divide each term by 3 ,we get */

$\implies y = 5 \: ---(1)$

/*if 1÷4 be taken from the denominator the fraction becomes 8÷19 */

$\frac{x}{y - \frac{1}{4}} = \frac{8}{19}$

$\implies \frac{x}{5- \frac{1}{4}} = \frac{8}{19}\: [ From \: (1)]$

$\implies \frac{x}{ \frac{20-1}{4}} = \frac{8}{19}$

$\implies \frac{x}{\frac{19}{4}} = \frac{8}{19}$

$\implies x = \frac{8}{19}\times \frac{19}{4}$

$\implies x = \frac{8}{4}$

$\implies x = 2\: --(2)$

Therefore.,

$\red { Required \: fraction \: \Big(\frac{x}{y}\Big)} \green {= \frac{2}{5}}$

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