Multiple choice

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. 10

  2. 20

  3. 30

  4. 25
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let distance be D. Car 1 time = t, speed v1 = D/t. Car 2 time = t-1, speed v2 = D/(t-1). v2/v1 = t/(t-1) = 1 + 1/(t-1). To maximize percentage increase, minimize t. Given t >= 6, min t = 6. v2/v1 = 6/5 = 1.2. Increase is 20%.

AI explanation

Let the first car travel for exactly 6 hours, as this minimum time maximizes the second car's required percentage speed increase. Since the second car started an hour later, it travelled for 5 hours to reach the destination at the same time. Using the constant distance formula where speed is inversely proportional to time, the ratio of their speeds is 6:5. The percentage by which the second car's speed exceeds the first car's speed is (1 divided by 5) multiplied by 100, which equals 20.

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