Management Aptitude Test Sep 2011 Paper

(3) Let total required distance = d
Speed of train = x
and time taken = t
Then, according to first condition,
$\frac{150}{\mathit{v}}+\frac{\mathit{d}-150}{\frac{3}{5}\mathit{v}}$ = t + 8 .......(i)
According to the second condition,
$\frac{410}{\mathit{v}}+\frac{\mathit{d}-410}{\frac{3}{5}\mathit{v}}$ = t + 4 .......(ii)
SubtractingEq. (ii) from Eq. (i),
$-\frac{360}{\mathit{v}}+\frac{360}{\frac{3}{5}\mathit{v}}$ = 4
⇒ v = $\frac{-360+600}{4}=\frac{240}{4}$ = 60 km/h
∴ Time t = $\frac{\mathit{d}}{60}$
Now, from Eq. (i),
$\frac{150}{60}+\frac{\mathit{d}-150}{\frac{3}{5}\times 60}$ = $\frac{\mathit{d}}{60}$ + 8
⇒ $\frac{\mathit{d}-150}{36}$ - $\frac{\mathit{d}}{60}$ = 8 - $\frac{5}{2}$
⇒ $\frac{5\left(\mathit{d}-150\right)-3\mathit{d}}{36\times 5}=\frac{11}{2}$
⇒ 2d - 750 = $\frac{11}{2}$ x 36 x 5 = 990
⇒ d = $\frac{1740}{2}$ = 870 km