Answer : Option A

Explanation :

Let the fraction = x

$MF#%\begin{align}
&\text{Then }x - \dfrac{1}{x} = \dfrac{9}{20}\\\\
&\Rightarrow \dfrac{x^2 - 1}{x} = \dfrac{9}{20}\\\\
&\Rightarrow x^2 - 1 = \dfrac{9x}{20}\\\\
&\Rightarrow 20x^2 - 20 = 9x\\\\
&\Rightarrow 20x^2 - 9x - 20 = 0\\\\
&x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\\\
&= \dfrac{9 \pm \sqrt{\left(-9\right)^2 - 4\times20\times\left(-20\right)}}{2 \times 20}\\\\
&= \dfrac{9 \pm \sqrt{81 + 1600}}{40}
= \dfrac{9 \pm 41}{40}=\dfrac{50}{40} \text{ Or } -\dfrac{32}{40}\\\\\\\\
&\text{Given that the fraction is positive. Hence }\\\\
&x = \dfrac{50}{40} = \dfrac{5}{4}\\\\
&\text{}\dfrac{1}{x} = \dfrac{4}{5}\\\\
&x + \dfrac{1}{x} = \dfrac{5}{4} + \dfrac{4}{5} = \dfrac{5 \times 5 + 4 \times 4}{20} = \dfrac{41}{20}\\\\
\end{align} $MF#%