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Train speed increase for return journey

CAT 2020 Slot 1 · Quantitative Ability · Medium · Speed, Time & Distance

This is a medium Quantitative Ability question from the CAT 2020 Slot 1 paper. It tests Speed, Time & Distance. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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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. A.

    58

  2. B.

    67

  3. C.

    50

  4. D.

    61

Answer

B

Explanation

Let the normal speed be ss and scheduled time be tt minutes.

When speed becomes s3\frac{s}{3}, time taken becomes 3t3t. Given 3t−t=30  ⟹  2t=30  ⟹  t=153t - t = 30 \implies 2t = 30 \implies t = 15 minutes.

On the return journey:

  • Total scheduled time available =15= 15 minutes.
  • The train travels for 5 minutes at usual speed ss. Remaining scheduled distance corresponds to 15−5=1015 - 5 = 10 minutes of travel at usual speed.
  • The train stops for 4 minutes, leaving 15−5−4=615 - 5 - 4 = 6 minutes to complete the remaining distance.

Let the new speed be s′s'. To cover the remaining distance: s′×6=s×10  ⟹  s′=106s=53ss' \times 6 = s \times 10 \implies s' = \frac{10}{6}s = \frac{5}{3}s

Percentage increase in speed: (5/3−11)×100%=23×100%=66.67%≈67%\left(\frac{5/3 - 1}{1}\right) \times 100\% = \frac{2}{3} \times 100\% = 66.67\% \approx 67\%

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