Question

A fraction becomes $\frac{1}{4}$ when $1$ is subtracted from the numerator and it becomes $\frac{1}{5}$ when $7$ is added to its denominator. Find the fraction.

Answer 1

Let numerator be $x$ & denominator be $y,y\ne 0$

Case:1

$\Rightarrow \frac{x-1}{y}$ = $\frac{1}{4}$

$\Rightarrow 4x-4–y=0$---- (1)

Case:2

$\Rightarrow \frac{x}{(y+7)}$ = $\frac{1}{5}$

$\Rightarrow 5x–y–7=0$ ---- (2)

Equation (2) – equation (1)

$5x–y–7-4x+4+y=0$

$\Rightarrow x-3=0$

$\Rightarrow x=3$

Now, lets substitute $x=3$ in $4x-4–y=0$

$\Rightarrow 4\left(3\right)-4–y=0$

$\Rightarrow 8–y=0$

$\Rightarrow y=8$

Hence required fraction is $\frac{3}{8}$