Question

In a fraction, if the numerator is decreased by $1$ and the denominator is increased by $1$, then the resulting fraction is $\frac{1}{4}$. Instead if the numerator is increased by $1$ and the denominator is decreased by $1$, then the resulting fraction is $\frac{2}{3}$. Find the absolute difference of the numerator and the denominator of the fraction.

• A. $4$
• B. $5$
• C. $6$
• D. None of these

Answer 1

The correct option is A $4$

Let the fraction be $\frac{x}{y}$
As per the given statements, $\frac{x-1}{y+1}=\frac{1}{4}$
$\Rightarrow 4x-4=y+1$
$\Rightarrow 4x-y=5$ --- $\left(1\right)$
Also, $\frac{x+1}{y-1}=\frac{2}{3}$
$\Rightarrow 3x+3=2y-2$
$\Rightarrow 3x-2y=-5$ --- $\left(2\right)$

Multi[plying eqn $\left(1\right)$ by $2$ and subtracting eqn $\left(2\right)$, we get

$2(4x-y)-(3x-2y)=2\left(5\right)-(-5)$

$\Rightarrow 5x=15\Rightarrow x=3$

Substituting this value of $x$ in eqn $\left(1\right)$,

$4\left(3\right)-y=5$

$\Rightarrow y=7$

$\therefore x=3;y=7$
So, the absolute difference between the numerator and denominator $=7-3=4$