Maximum Speed Difference Between Two Cars

CAT 2019 Slot 1 · QA · Medium · Speed, time and distance

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Two cars travel the same distance starting at 10:00 am and 11:00 am, respectively, on the same day. They reach their common destination at the same point of time. If the first car travelled for at least 6 hours, then the highest possible value of the percentage by which the speed of the second car could exceed that of the first car is

  1. A.

    20

  2. B.

    10

  3. C.

    30

  4. D.

    25

Answer

A

Explanation

Let time taken by first car be t1≥6t_1 \ge 6 hours. Time taken by second car t2=t1−1t_2 = t_1 - 1. Ratio of speeds: S2S1=t1t1−1=1+1t1−1\frac{S_2}{S_1} = \frac{t_1}{t_1 - 1} = 1 + \frac{1}{t_1 - 1}. To maximize S2S1\frac{S_2}{S_1}, we minimize t1t_1. Minimum t1=6t_1 = 6. Max S2S1=1+15=1.20\frac{S_2}{S_1} = 1 + \frac{1}{5} = 1.20, which is 20%20\% excess.

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