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

The Correct Option is B

Given:

• Usual speed = \( s \)
• Usual time = \( t \)
• Due to reduction in speed to \( \frac{s}{3} \), the train is 30 minutes (i.e., \( \frac{1}{2} \) hour) late.

Step 1: Total Distance

Total distance = \( s \cdot t \)

Step 2: Time loss due to slow speed

The initial speed is \( \frac{s}{3} \). Let’s consider time for this part is 5 minutes = \( \frac{1}{12} \) hour.

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

Step 3: Remaining Distance and Required Speed

Let remaining time be \( \frac{11}{12} \) hour (assuming usual time is 1 hour).

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

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

Instead, using proportion method (as done in your reasoning):

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