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 9, 2004 oops, i didnt notice that this question as already been beaten to death. sorry folks! --shiv shiv reddy Saturday, October 9, 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 3, 2005   Fog Creek Home