Question

The sum of numerator and denominator of a fraction is $3$ less than twice the denominator. If each of the numerator and denominator is decreased by $1$, the fraction becomes $\frac{1}{2}$. Find the fraction.

Answer 1

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

Let the fraction be $\frac{x}{y}$

$x+y+3=2y$

(or) $x-y=-3$

(or) $x=y-3$ _____ (1)

Also , $\frac{x-1}{y-1}=\frac{1}{2}$

$x-1=\frac{y-1}{2}$

$x=\frac{y-1}{2}+1\Rightarrow 2x=y-1+2=y+1$

(or) $2x-y=1$ _____ (2)

substituting (1) in (2),

$2(y-3)-y=1$

$\Rightarrow 2y-6-y=1\Rightarrow y=7$

put $y=7$ in (1) we get $x=7-3=4$

$\therefore x=4,y=7$

$\therefore$ the fraction is $\frac{x}{y}=\frac{4}{7}$