Geeta runs 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).
Show answer & explanation
Correct answer: A
Explanation
The correct answer is option 1 i.e. 66.67 m
We are given that Geeta runs 5/2 times as fast as Babita
Thus,
Let Speed of Babita = x
Thus, Speed of Geeta = 5/2x
Since the time taken by both Geeta and Babita is the same they will meet each other at the same time.
Thus,
⇒ tG = tb
if the distance traveled by Babita is D meters.
Then Geeta will travel (D + 40) m. (as Geeta has given 40 m lead to Babita)
[ (D + 40) is the total distance between the start point and the meeting point of Geeta and Babita. ]
We know,
Time = Distance/Speed
Thus,
⇒ tg = (D + 40)/(5/2x) ⇒ tb = (D)/x
thus,
⇒ (D + 40)/(5/2x) = (D)/x
⇒ 5D = 2D + 40
⇒ 3D = 80
⇒ D = 80/3
⇒ Thus, the total distance between the starting point and the meeting point is
⇒ D + 40
⇒ 80/3 + 40
⇒ (80 + 40)/3
⇒ 66.67 m (approximately).