If the difference between the reciprocal of a positive proper fraction and the fraction itself be $$\frac{9}{20}$$, then the fraction is :

Solution(By Examveda Team)

Let the fraction be $$\frac{a}{1}$$
Then,
$$\eqalign{ & \Leftrightarrow \frac{1}{a} - a = \frac{9}{{20}} \cr & \Leftrightarrow \frac{{1 - {a^2}}}{a} = \frac{9}{{20}} \cr & \Leftrightarrow 20 - 20{a^2} = 9a \cr & \Leftrightarrow 20{a^2} + 9a - 20 = 0 \cr & \Leftrightarrow 20{a^2} + 25a - 16a - 20 = 0 \cr & \Leftrightarrow 5a\left( {4a + 5} \right) - 4\left( {4a + 5} \right) = 0 \cr & \Leftrightarrow \left( {4a + 5} \right)\left( {5a - 4} \right) = 0 \cr & \Leftrightarrow a = \frac{4}{5}\,\,\,\,\,\,\,\,\left[ {\because a \ne - \frac{5}{4}} \right] \cr} $$