The numerator of a fraction is 5 less than its denominator. If 3 is added to the numerator and denominator both, the fraction becomes $\dfrac{4}{5}$ . Find the original fraction.

Let the denominator of the fraction be x.

The numerator of a fraction is 5 less than its denominator

So, numerator = x - 5

According to the question,

when 3 is added to the numerator and denominator both, the fraction becomes $\dfrac{4}{5}$

$⇒ \dfrac{numerator + 3}{denominator + 3} = \dfrac{4}{5}\\[1em] ⇒ \dfrac{(x - 5) + 3}{x + 3} = \dfrac{4}{5}\\[1em] ⇒ \dfrac{x - 5 + 3}{x + 3} = \dfrac{4}{5}\\[1em] ⇒ \dfrac{x - 2}{x + 3} = \dfrac{4}{5}\\[1em]$

By cross multiplying, we get :

⇒ 5(x - 2) = 4(x + 3)

⇒ 5x - 10 = 4x + 12

⇒ 5x - 4x = 10 + 12

⇒ x = 22

Denominator = 22

Numerator = (x - 5)

= 22 - 5

= 17

Hence, the fraction is $\dfrac{17}{22}$.