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A train travelled at one$-$thirds of its usual speed, and hence reached the destination $30$ minutes after the scheduled time. On its return journey, the train initially travelled at its usual speed for $5$ minutes but then stopped for $4$ minutes for an emergency. The percentage by which the train must now increase its usual speed so as to reach the destination at the scheduled time, is nearest to

1. $58$
2. $67$
3. $61$
4. $50$

Let the usual speed and the usual time taken to travel the distance be $s$ and $t,$ respectively.

We know that, $\boxed{\text{Speed} = \frac{\text{Distance}}{\text{Time}}}$

$\Rightarrow \text{Speed} \propto \frac{1}{\text{Time}} \quad [\because \text{Distance constant}]$

We get, $\boxed{\frac{s_{1}}{s_{2}} = \frac{t_{2}}{t_{1}}}$

$\begin{array}{cc} \underline{\text{Speed}} & \underline{\text{Time}} \\ S & t \\ \frac{1}{3}S & t + 30 \\ \frac{1}{3}S & 3t \end{array}$

Here, $3t = t + 30$

$\Rightarrow 2t = 30$

$\Rightarrow \boxed{ t = 15 \; \text{minutes}}$

$\begin{array}{cc} \underline {\text{Speed}} & \underline{\text{Time}} \\ S & 10\;\text{minutes} \\ \frac{10}{6}S & 6\;\text{minutes} \\ \frac{5}{3}S & 6 \;\text{minutes}\end{array}$

Required percentage $= \dfrac{\left(\frac{5}{3}s – s \right)} {s} \times 100 \%$

$\qquad = \frac{2}{3} \times 100 \% = 66.66 \% \simeq 67 \%$

Correct Answer$: \text{B}$

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