Reema runs $$\frac{5}{4}$$ times as fast as Rekha. In view of this Reema allows Rekha a lead of 50 metres to Rekha in a friendly race competition. What is the distance from the starting point where both Reema and Rekha meet?

Let Rekha’s speed be $$v \text{ m s}^{-1}$$.
Reema runs $$\frac{5}{4}$$ times as fast, so Reema’s speed is $$\frac{5v}{4} \text{ m s}^{-1}$$.

At the start of the race Rekha is given a head-start (lead) of 50 m. Hence, when the stop-watch starts Reema is at the starting line while Rekha is already 50 m ahead.

Suppose both runners meet after $$t$$ seconds. In that time:

• distance run by Reema $$= \frac{5v}{4}\,t$$
• distance run by Rekha $$= v\,t$$

Because Reema had to make up the 50 m lead, the distance she covers exceeds Rekha’s by exactly 50 m:

$$\frac{5v}{4}\,t = vt + 50$$

Simplify:

$$\left(\frac{5v}{4} - v\right)t = 50$$
$$\frac{v}{4}\,t = 50$$
$$t = \frac{200}{v}$$

Total distance from the starting point to the meeting point equals the distance run by Reema:

$$\text{Distance} = \frac{5v}{4}\,t = \frac{5v}{4}\left(\frac{200}{v}\right) = 5 \times \frac{200}{4} = 250 \text{ m}$$

Hence, they meet 250 m from the starting line.

Option A which is: 250 m