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railroad bridge

this is an easy one.

let time taken for man to run back and just escape = t1

time taken for man to run ahead and be hit as he exits tunnel = t2

speed of the train = x
speed of the running man = y

x*t2 = y*t1 + y*t2 + x*t1

x/y = (t2 + t1) / (t2 - t1)                        -----------  (1)


y*t1 / y*t2  =  (1/4 )  /  (3/4)

t2 = 3*t1.

use this in (1)

x = 2*y

hence the train is moving twice as fast as the man.


shiv reddy
Saturday, October 9, 2004

oops, i didnt notice that this question as already been beaten to death. sorry folks!


shiv reddy
Saturday, October 9, 2004

correct answer...what do it mean 'beaten to death'

Tuesday, December 28, 2004

Correct, but overly complicated. We know that a man can run 1/4 of a tunnel before the train enters it (that's what happens if he runs back). So if he runs forward, he'll be at 1/2 of the tunnel when the train enters. He will then run half a tunnel in a same time that it takes a train to go the whole tunnel. The man runs at 1/2 of the speed of the train.

Monday, January 3, 2005

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