Management Aptitude Test Sep 2020 Paper

Let the distance covered be $\mathit{D}\mathit{k}\mathit{m}$. and speed of train be $\mathit{x}\mathit{k}\mathit{m}∕\mathit{h}$.
$\therefore$ According to the question,
$\frac{\mathit{D}}{\mathit{x}}-\frac{\mathit{D}}{\mathit{x}+4}=\frac{30}{60}\Rightarrow \frac{\mathit{D}\left(\mathit{x}+4-\mathit{x}\right)}{\mathit{x}\left(\mathit{x}+4\right)}=\frac{1}{2}$
$\Rightarrow 8\mathit{D}={\mathit{x}}^2+4\mathit{x}$ . . . (i)

$\Rightarrow \mathit{D}\left[\frac{\mathit{x}-\mathit{x}+2}{\mathit{x}\left(\mathit{x}-2\right)}\right]=\frac{1}{3}$
$\Rightarrow 6\mathit{D}={\mathit{x}}^2-2\mathit{x}$ . . . (ii)
From Eqs. (i) and (ii), we get
$\frac{{\mathit{x}}^2+4\mathit{x}}{8}=\frac{{\mathit{x}}^2-2\mathit{x}}{6}$
$\Rightarrow 6{\mathit{x}}^2+24\mathit{x}=8{\mathit{x}}^2-16\mathit{x}\Rightarrow 2{\mathit{x}}^2-40\mathit{x}=0$
$\therefore 2\mathit{x}\left(\mathit{x}-20\right)=0$
$\therefore \mathit{x}=20\mathit{k}\mathit{m}∕\mathit{h}$
Hence, distance covered $=\frac{{\left(20\right)}^2-2\times 20}{6}=60\mathit{k}\mathit{m}$