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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 traveled 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

- $30$
- $25$
- $10$
- $20$

1 vote

Given that,

Two cars travel the same distance starting at $10 \; \text{am}$ and $11:00 \; \text{am}$, respectively, on the same day.

For the highest possible value of the percentage by which the speed of the second car could exceed that of the first car the time taken by both cars should be minimum as their difference is constant.

If the first car takes the hours to reach the destination $ = 6 \; \text{hours}$

The second car started one hour late. And, they reached their common destination at the same point of time.

So, the second car takes the hours to reach the destination $ = 5 \; \text{hours}$

Using this, when the distance is constant $: \boxed{\text{S} \; \propto \; \frac{1}{ \text{T}}}$

$\begin{array} {c c c} \text{Time} & \text{Speed} \\ \text{First car = 6 hours} & 5 \; \frac{\text{Km}} {\text{h}} \\ \text{Second car = 5 hours} & 6 \; \frac{\text{Km}} {\text{h}} \end{array}$

$\therefore$ The highest possible value of the percentage by which the speed of the second car could exceed that of the first car $ = \left(\frac{6-5}{5} \right) \times 100 = \frac{1}{5} \times 100 = 20\%.$

Correct Answer $: \text{D}$

Two cars travel the same distance starting at $10 \; \text{am}$ and $11:00 \; \text{am}$, respectively, on the same day.

For the highest possible value of the percentage by which the speed of the second car could exceed that of the first car the time taken by both cars should be minimum as their difference is constant.

If the first car takes the hours to reach the destination $ = 6 \; \text{hours}$

The second car started one hour late. And, they reached their common destination at the same point of time.

So, the second car takes the hours to reach the destination $ = 5 \; \text{hours}$

Using this, when the distance is constant $: \boxed{\text{S} \; \propto \; \frac{1}{ \text{T}}}$

$\begin{array} {c c c} \text{Time} & \text{Speed} \\ \text{First car = 6 hours} & 5 \; \frac{\text{Km}} {\text{h}} \\ \text{Second car = 5 hours} & 6 \; \frac{\text{Km}} {\text{h}} \end{array}$

$\therefore$ The highest possible value of the percentage by which the speed of the second car could exceed that of the first car $ = \left(\frac{6-5}{5} \right) \times 100 = \frac{1}{5} \times 100 = 20\%.$

Correct Answer $: \text{D}$