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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. Find the fraction.
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Let the fraction be $\frac{x}{y}$ [$\frac{1}{2}$ mark]
$\frac{x - 1}{y} = \frac{1}{3} \Rightarrow 3x - y = 3 \dots (i)$ [$\frac{1}{2}$ mark]
$\frac{x}{y + 8} = \frac{1}{4} \Rightarrow 4x - y = 8 \dots (ii)$ [$\frac{1}{2}$ mark]
On solving the equations $(i)$ and $(ii)$, we get $x = 5, y = 12$ [$1$ mark]
Required fraction is $\frac{5}{12}$ [$\frac{1}{2}$ mark]
$\frac{x - 1}{y} = \frac{1}{3} \Rightarrow 3x - y = 3 \dots (i)$ [$\frac{1}{2}$ mark]
$\frac{x}{y + 8} = \frac{1}{4} \Rightarrow 4x - y = 8 \dots (ii)$ [$\frac{1}{2}$ mark]
On solving the equations $(i)$ and $(ii)$, we get $x = 5, y = 12$ [$1$ mark]
Required fraction is $\frac{5}{12}$ [$\frac{1}{2}$ mark]