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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

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

also,

y*t1 / y*t2  =  (1/4 )  /  (3/4)

so,
t2 = 3*t1.


use this in (1)

x = 2*y

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

--shiv

shiv reddy
Saturday, October 09, 2004

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

--shiv

shiv reddy
Saturday, October 09, 2004

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

michael
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.

Hobbit
Monday, January 03, 2005

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