Question

The numerator of the rational number is less than its denominator by $3$. If the numerator becomes three times and the denominator is increased by $20$, the new number becomes $\frac{1}{8}$. Find the original number.

Answer 1

Let the denominator be $x$.

Thus, numerator will be $x-3$.

As per the question, we have

numerator is $3$ times i.e., $3(x-3)$ and denominator is increased by $20$ i.e., $x+20$.

The new number thus formed is $\frac{1}{8}$.

Therefore, $\frac{3(x-3)}{x+20}=\frac{1}{8}$

$\Rightarrow 24(x-3)=1(x+20)$

$\Rightarrow 24x-72=x+20$

$\Rightarrow 23x=92$

$\Rightarrow x=4$

Denominator is $4$.

Thus, numerator is $x-3=4-3=1$

Therefore, the fraction is $\frac{1}{4}$.