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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.
Please post an explanation to solve the problem? Can`t figure it out.
400 meters
19 Oct 2010, 10:04
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jimmy86
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.
General Discussion
19 Oct 2010, 07:01
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jimmy86
This question comes under the concept of Relative speed.
Given A & B are 100m apart and B is moving in the same direction away from A
Therefore their relative speed (ie., speed of B with respect to speed of A) is 2 km/hr
Carefully observe that speeds of A & B are given in Km/hr, we need to convert into m/s
distance = speed * time
100 = (2 * 5/18) * t
=> t = 180 sec
Therefore distance travelled by B in 180 sec is ( 8 * 180) = 1440 m or 1.4 KM
OA please??
