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)

(Shouldn't this problem have gone in "Just Math"?)

Assume the length of the tunnel is 28*d miles. He can return to the beginning in d hours and meet the train, or he can travel on and have the train catch up to him in 3*d hours. The train will go through the length of the tunnel (28*d miles) in (3*d) - d = 2*d hours, so its speed is [(28*d)miles/(2*d)hours] = 14 mph

Posted by TomM on 2002-05-26 15:36:03