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by Tame the CAT » Sun Apr 15, 2007 4:33 am
Just want to see if I got the same answer as the many math-skilled folks here. My friend asked me this question

There is a high speed train (train A) that takes x hours to cover z miles and a regular speed train (train B) that takes y hours to cover the same distance. If train B leaves 2 hours earlier, how long would it take train A to catch up assuming they started from the same point?





My Ans -> t = 2x/(y-x)
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by gmatme » Mon Apr 16, 2007 11:12 am
Distance = rate x time
So, for train A : z = R1 * X
Train B : z = R2 * Y

Let t be the time travelled by train A, since train B starts 2 hr earlier, the time travelled by train B is (t+2). To catch up the train B, train A and B should travel the same distance,

R2 * (t+2) = R2 * (t)
z/Y * (t+2) = z/X * t
t+2/Y = t/X
t(x-y) = -2X
so, t = 2X/(Y-X)
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by Cybermusings » Wed Apr 18, 2007 9:19 am
There is a high speed train (train A) that takes x hours to cover z miles and a regular speed train (train B) that takes y hours to cover the same distance. If train B leaves 2 hours earlier, how long would it take train A to catch up assuming they started from the same point?

Relative speed in this case = (z/x - z/y)
Distance covered by B in these 2 hours = 2z/y
Thus time taken by A to cover this distance of 2z/y miles

= D/S = (2z/y) / (z/x - z/y)
=(2z/y) / [(yz-xz)/xy]
=2z/y * (xy/yz-xz)
=2/(y-z)
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by f2001290 » Tue Apr 24, 2007 1:28 am
I am getting 2x/(y-x)
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