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

  1. 58
  2. 67
  3. 61
  4. 50
Option: 2
Solution:

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 \( = s \times t\)

Distance travelled in the first \(5min = s \times t/3.\) D

Distance to be travelled in the last \(6min = 2st/3\)

Required speed to cover that distance on time \( = \frac{{2st/3}}{{2t/5}}\) i.e., \(\frac{{5s}}{3}\)

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

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