Question:medium

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

Updated On: Jan 15, 2026
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The Correct Option is B

Solution and Explanation

Provided:

  • Standard speed = \( s \)
  • Standard time = \( t \)
  • Speed reduced to \( \frac{s}{3} \), resulting in a 30-minute (\( \frac{1}{2} \) hour) delay.

Step 1: Total Distance

Total distance = \( s \cdot t \)

Step 2: Time lost due to reduced speed

Initial speed is \( \frac{s}{3} \). Assuming time for this segment is 5 minutes (\( \frac{1}{12} \) hour).

Distance covered in this time: \( \frac{s}{3} \cdot \frac{1}{12} = \frac{s}{36} \)

Step 3: Remaining Distance and Required Speed

Assuming standard time is 1 hour, remaining time is \( \frac{11}{12} \) hour.

To compensate for the delay, the required speed \( v \) must cover the remaining distance:

\[ v = \frac{\text{Remaining distance}}{\text{Time left}} = \frac{s \cdot t - \frac{s}{36}}{t - \frac{1}{12}} \]

Alternatively, using the proportion method:

The last \( \frac{2st}{3} \) distance must be covered in \( \frac{2t}{5} \) time (6 minutes).

Required speed: \[ = \frac{\frac{2st}{3}}{\frac{2t}{5}} = \frac{5s}{3} \]

Step 4: Calculate % Increase in Speed

\[ \text{Increase} = \frac{\frac{5s}{3} - s}{s} \cdot 100 = \frac{2s}{3s} \cdot 100 = \frac{2}{3} \cdot 100 = 66.\overline{6}\% \]

Final Answer: 67%

Correct Option: (B)

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