Management Aptitude Test Dec 2013 Paper

(b) Let the distance covered be D km

and speed of train be x km/h.
∴ According to the question,
$\frac{\mathit{D}}{\mathit{x}}-\frac{\mathit{D}}{\mathit{x}+4}=\frac{30}{60}$
⇒ $\frac{\mathit{D}\left(\mathit{x}+4-\mathit{x}\right)}{\mathit{x}\left(\mathit{x}+4\right)}=\frac{1}{2}$
⇒ 8D = ${\mathit{x}}^2$ + 4x    .....(i)
and
$\frac{\mathit{D}}{\mathit{x}-4}-\frac{\mathit{D}}{\mathit{x}}=\frac{20}{60}$
⇒ $\mathit{D}\left[\frac{\left(\mathit{x}+2-\mathit{x}\right)}{\mathit{x}\left(\mathit{x}-2\right)}\right]=\frac{1}{3}$
⇒6D = ${\mathit{x}}^2$ - 2x    .....(ii)
From Eqs. (i) and (ii), we get
$\frac{{\mathit{x}}^2+4\mathit{x}}{8}$ = $\frac{{\mathit{x}}^2-2\mathit{x}}{6}$
⇒ 6${\mathit{x}}^2$ + 24x = 8${\mathit{x}}^2$ - 16x
⇒ 2${\mathit{x}}^2$ - 40x = 0
∴ 2x(x - 20) = 0
∴ x = 20 km/h
Hence, distance covered = $\frac{{\left(20\right)}^2-2\times 20}{6}$ = 60 km