Question
A 50m long platoon is marching ahead. The last person in the platoon wants to give a letter to the first person leading the platoon. So while the platoon is marching he runs ahead, reaches the first person and hands over the letter to him and without stopping he runs and comes back to his original position. In the mean time the whole platoon has moved ahead by 50m. How much distance (approximately) did the last person cover in that time. Assuming that he ran the whole distance with uniform speed
Answer: Option A
:
A
The last person covered 120.71 meters.
It is given that the platoon and the last person moved with uniform speed. Also, they both moved for the identical amount of time. Hence, the ratio of the distance they covered - while person moving forward and backward - are equal. Let's assume that when the last person reached the first person, the platoon moved X meters forward.
Thus, while moving forward the last person moved (50+X) meters whereas the platoon moved X meters.
Similarly, while moving back the last person moved [50-(50-X)] = X meters whereas the platoon moved (50-X) meters.
Now, as the ratios are equal,
[50+X]X=X[50−X]
[50+X]×[50−X]=X×X
Solving, X=35.355 meters
Thus, total distance covered by the last person
=[50+X]+X
=2×X+50
=2×[35.355]+50
=120.71 metres.
=120 m (Approximately)
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:
A
The last person covered 120.71 meters.
It is given that the platoon and the last person moved with uniform speed. Also, they both moved for the identical amount of time. Hence, the ratio of the distance they covered - while person moving forward and backward - are equal. Let's assume that when the last person reached the first person, the platoon moved X meters forward.
Thus, while moving forward the last person moved (50+X) meters whereas the platoon moved X meters.
Similarly, while moving back the last person moved [50-(50-X)] = X meters whereas the platoon moved (50-X) meters.
Now, as the ratios are equal,
[50+X]X=X[50−X]
[50+X]×[50−X]=X×X
Solving, X=35.355 meters
Thus, total distance covered by the last person
=[50+X]+X
=2×X+50
=2×[35.355]+50
=120.71 metres.
=120 m (Approximately)
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