Question

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

• A. 58
• B. 67
• C. 61
• D. 50

Answer 1

Answer: (B)

67

Answer 2

The correct answer is (B) : $67$

Let the usual speed of the train be s and time taken at that speed be $'t'$.

Given by travelling at $s/3$, it reached $30$ min late.

Hence the usual time:

Distance travelled = $s×t$

Distance travelled in the first $5$ min = $\frac{s×t}{3}.D$

Distance to be travelled in the last $6$ min = $\frac{2st }{ 3}$

Required speed to cover that distance on time = $\frac{\frac{2st}{3}}{\frac{2t}{5}}$ i.e., $\frac{5s'}{3}$

Hence the percentage increase in its speed = $\bigg(\frac{2}{3}\bigg)×100$ i.e., $66\frac{2}{3}\%$ or $67\%$