Question : Geeta runs $\frac{5}{2}$ times as fast as Babita. In a race, if Geeta gives a lead of 40 m to Babita, find the distance from the starting point where both of them will meet (correct up to two decimal places).
Option 1: 66.67 m
Option 2: 65 m
Option 3: 65.33 m
Option 4: 66 m
Correct Answer: 66.67 m
Solution :
Let the speed of Babita be $2x$.
Speed of Geeta = $\frac{5}{2} \times 2x = 5x$
Let the distance covered by Geeta be $y$ metres.
Distance covered by Babita = $(y - 40)$ metres
As time is constant, distance is directly proportional to speed.
$⇒\frac{2x}{5x}=\frac{y-40}{y}$
$⇒ 2y = 5y - 200$
$⇒ y = \frac{200}{3}$ = 66.67 m
$\therefore$ The distance from the starting point where both of them will meet is 66.67 m.
Hence, the correct answer is 66.67 m.
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