The numerator of a fraction is 3 less than its denominator. If 2 is added to both the numerator and denominator, the sum of the new fraction and the original fraction is $1\dfrac{9}{20}$. Find the new fraction.

Let the fraction be $\dfrac{a}{b}$.

It is given that the numerator of a fraction is 3 less than its denominator.

⇒ a = b - 3

And, If 2 is added to both the numerator and denominator, the sum of the new fraction and the original fraction is $1\dfrac{9}{20}$

$\Rightarrow \dfrac{a}{b} + \dfrac{a + 2}{b + 2} = 1\dfrac{9}{20}\\[1em] \Rightarrow \dfrac{b - 3}{b} + \dfrac{(b - 3) + 2}{b + 2} = \dfrac{29}{20}\\[1em] \Rightarrow \dfrac{b - 3}{b} + \dfrac{b - 1}{b + 2} = \dfrac{29}{20}\\[1em] \Rightarrow \dfrac{(b - 3) \times (b + 2)}{b \times (b + 2)} + \dfrac{(b - 1) \times b}{(b + 2) \times b} = \dfrac{29}{20}\\[1em] \Rightarrow \dfrac{b^2 - 3b + 2b - 6}{b^2 + 2b} + \dfrac{b^2 - b}{b^2 + 2b} = \dfrac{29}{20}\\[1em] \Rightarrow \dfrac{(b^2 - 3b + 2b - 6) + (b^2 - b)}{b^2 + 2b} = \dfrac{29}{20}\\[1em] \Rightarrow \dfrac{b^2 - b - 6 + b^2 - b}{b^2 + 2b} = \dfrac{29}{20}\\[1em] \Rightarrow \dfrac{2b^2 - 2b - 6}{b^2 + 2b} = \dfrac{29}{20}\\[1em] \Rightarrow 20(2b^2 - 2b - 6) = 29(b^2 + 2b)\\[1em] \Rightarrow 40b^2 - 40b - 120 = 29b^2 + 58b\\[1em] \Rightarrow 40b^2 - 40b - 120 - 29b^2 - 58b = 0\\[1em] \Rightarrow 11b^2 - 98b - 120 = 0\\[1em] \Rightarrow 11b^2 - 110b + 12b - 120 = 0\\[1em] \Rightarrow 11b(b - 10) + 12(b - 10) = 0\\[1em] \Rightarrow (b - 10)(11b + 12) = 0\\[1em] \Rightarrow (b - 10) = 0 \text{ or } (11b + 12) = 0\\[1em] \Rightarrow b = 10 \text{ or } b = -\dfrac{12}{11}\\[1em]$

As b cannot be fraction. So, b = 10.

When b = 10, a = b - 3 = 10 - 3 = 7

The fraction = $\dfrac{a}{b} = \dfrac{7}{10}$

New fraction = $\dfrac{a + 2}{b + 2} = \dfrac{7 + 2}{10 + 2} = \dfrac{9}{12} = \dfrac{3}{4}$

Hence, the fraction = $\dfrac{3}{4}$.