A fraction becomes 13 when 1 is subtracted from the numerator and it becomes 14 when 8 is added to its denominator.Form equation to find the fraction.( Assume 'x' as the numerator and 'y' as the denominator. )

Let 'x' be the numerator and 'y' be the denominator.As per condition 1,x-1y=13-- (1)As per condition 2,xy+8=14 -- (2)

A fraction becomes 13 when 1 is subtracted from the numerator and it becomes 14 when 8 is added to its denominator.Form equation to find the fraction.( Assume 'x' as the numerator and 'y' as the denominator. )

Let 'x' be the numerator and 'y' be the denominator.As per condition 1,x-1y=13-- (1)As per condition 2,xy+8=14 -- (2)

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A fraction becomes $\frac{1}{3}$ when 1 is subtracted from the numerator and it becomes $\frac{1}{4}$ when 8 is added to its denominator.
Form equation to find the fraction.
( Assume 'x' as the numerator and 'y' as the denominator. )

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$\frac{x - 1}{y} = \frac{1}{3}, \frac{x}{y + 8} = \frac{1}{4}$

$\frac{x - 1}{y - 1} = \frac{1}{3}, \frac{x + 8}{y + 8} = \frac{1}{4}$

$\frac{x - 1}{y + 8} = \frac{1}{3}, \frac{x + 1}{y + 8} = \frac{1}{4}$

$\frac{x - 1}{y - 8} = \frac{1}{3}, \frac{x + 1}{y + 8} = \frac{1}{4}$

answer is A.

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Detailed Solution

Let 'x' be the numerator and 'y' be the denominator.

As per condition 1,

$\frac{x - 1}{y} = \frac{1}{3}$ -- (1)

As per condition 2,

$\frac{x}{y + 8} = \frac{1}{4}$ -- (2)

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A fraction becomes 13 when 1 is subtracted from the numerator and it becomes 14 when 8 is added to its denominator.Form equation to find the fraction.( Assume 'x' as the numerator and 'y' as the denominator. )

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A fraction becomes $\frac{1}{3}$ when 1 is subtracted from the numerator and it becomes $\frac{1}{4}$ when 8 is added to its denominator.
Form equation to find the fraction.
( Assume 'x' as the numerator and 'y' as the denominator. )