A man was walking through a tunnel. He is 1/4 of the way through when he hears a train approaching the tunnel from behind him.

If he turns and runs back, he will make it out of the tunnel just as the train is entering it (and will save himself by a hair). If he goes forward to the far end of the tunnel, he will also just barely make it, emerging from the tunnel just as the train is about to catch up to him.

If the man's running speed is 7 miles per hour, how fast is the train moving?

(from techInterview.org)

Let d=the distance from the train to the entrance of the tunnel.  Let L =the length of the tunnel.  Let S=the speed of the train.

The time it takes for the train to reach the entrance of the tunnel is: d/S.  This is equal to the time it takes the man to reach the same spot: .25L/7=L/28.  So we have: d/S=L/28.

The time it takes for the train to reach the end of the tunnel is: (d+L)/S which is equal to the time it takes for the man to reach the end of the tunnel: .74L/7=3L/28.  So we have: (d+L)/S=3L/28.

Subtract these equations to obtain: L/S=2L/28=L/14

Thus: S=14

Posted by lmnop on 2005-01-17 22:24:45