Question:
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
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
Updated On: Jul 28, 2025
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- 25
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- 20
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The Correct Option is D
Solution and Explanation
Since both cars cover the same distance, we can use the inverse relationship between speed and time for constant distance:
\[ \frac{V_2}{V_1} = \frac{T_1}{T_2} = \frac{6}{5} \]
So, the second car's speed is: \[ \frac{6}{5} \times 100\% = 120\% \] of the first car’s speed.
Therefore, the second car is \[ 120\% - 100\% = 20\% \] faster than the first car.
Answer:
The second car’s speed exceeds the first car’s speed by 20%.
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