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

• A. 30
• B. 25
• C. 10
• D. 20

Answer 1

The Correct Option is D

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%.