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

Correct Answer: 25

Let the distance travelled by both cars be $D$.

Let the time taken by the first car be $t_1$ hours. Since it starts at 10:00 am and the second car starts at 11:00 am, and both reach the destination at the same time, the second car travels for 1 hour less than the first car.

Thus, the time taken by the second car is $t_2=t_1-1$ hours.

We are given that the first car travelled for at least 6 hours, which means:
$t_1\ge 6$

Let $s_1$ and $s_2$ be the speeds of the first and second cars, respectively. Since they travel the same distance:
$s_1\times t_1=s_2\times t_2=D$

Therefore, the ratio of their speeds is:
$\frac{s_2}{s_1}=\frac{t_1}{t_2}=\frac{t_1}{t_1-1}$

The percentage by which the speed of the second car exceeds that of the first car is:
$P=\left(\frac{s_2}{s_1}-1\right)\times 100=\left(\frac{t_1}{t_1-1}-1\right)\times 100=\frac{1}{t_1-1}\times 100$

To find the highest possible value of this percentage, we need to maximize $P$, which means we need to minimize $t_1$.

Since the minimum value of $t_1$ is 6 hours:
$P_{\text{max}}=\frac{1}{6-1}\times 100=\frac{1}{5}\times 100=20\%$

Wait, looking at the options and the question, let's verify if the question says "at least 6 hours". If it travelled for at least 6 hours, then $t_1\ge 6$, so the maximum value of $\frac{1}{t_1-1}$ is indeed at $t_1=6$, giving $20\%$. Thus option 4 (which is 20) is the correct answer.