jimmy86 wrote:

A and B are 100 meters apart. A starts moving towards B at the same time B starts moving in the same direction away from A. If the Speed of B is 8 km/hr and that of A is 10 km/hr , how far will B have gone before he is overtaken?

Please post an explanation to solve the problem? Can`t figure it out.

First approach:

To catch up to B, A must travel 100 meters more than B.

Rate for A = 10000 m/hr, Rate for B = 8000 m/hr.

Distance A = r*t = 10000t

Distance B = r*t = 8000t

Thus, 10000t = 8000t + 100

2000t = 100

t = 100/2000 = 1/20 hours.

Distance for B = r*t = (1/20)*8000 = 400 meters.

Second approach:

Plug in the answer choices, which would represent the distance traveled by B.

B = 400 meters

Time for B = d/r = 400/8000 = 1/20 hours.

Distance for A = r*t = 10000*(1/20) = 500 meters.

A-B = 500-400 = 100 meters.

Since A travels 100 meters more than B, A will catch up to B. Thus, we would have found the correct answer.

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