NIACL Model Paper 10 with solutions for free online practice

(3) Let the time taken by B be x days.
∴ Time taken by A = (x-5) days.
According to the question,
$\frac{1}{\mathit{x}-5}+\frac{1}{\mathit{x}}=\frac{9}{100}$
⇒ $\frac{\mathit{x}+\mathit{x}-5}{\mathit{x}\left(\mathit{x}-5\right)}=\frac{9}{100}$
=> $\frac{2\mathit{x}-5}{\mathit{x}\left(\mathit{x}-5\right)}=\frac{9}{100}$
=> $9{\mathit{x}}^2-45\mathit{x}=200\mathit{x}-500$
=> $9{\mathit{x}}^2-245\mathit{x}+500=0$
=> $9{\mathit{x}}^2-225\mathit{x}-20\mathit{x}+500=0$
⇒ $9\mathit{x}\left(\mathit{x}-25\right)-20\left(\mathit{x}-25\right)=0$
=> $\left(\mathit{x}-25\right)\left(9\mathit{x}-20\right)=0$ =>x = 25 because ≠ $\frac{20}{9}$