   Railroad Bridge Unless I misread this, it's a pretty simple algebra problem. Given: The train is X feet from the mouth of the tunnel. The length of the tunnel is Y feet, so the man is 0.25Y into the tunnel. The speed of the man (Sm) is unknown. The speed of the train (St) is unknown. Set up two equations. Equation #1 relates the man traveling back to the mouth of the tunnel in the same time the train travels there. (0.25Y / Sm) = (X /St) The reason I am dividing distance (feet) by speed (feet/sec) is so I end up with seconds. Essentially I am equating the time for both the man and train to reach the mouth of the tunnel to be the same. Equation #2 relates the man traveling to the end of the tunnel in the same time the train gets there. (0.75Y / Sm) = ((X+Y) / St) Solve the first equation for X and plug it into the second where the Y's nicely drop out along the way. The answer is the train is traveling twice as fast as the man. Jason Wednesday, April 6, 2005 Reached the same conclusion through a far less "structered" approach: - if the train reaches the start of the tunnel at the same time it takes the man to run 1/4 of the tunnel, then, as the man starts at the 1/4 position, if he had run in the same direction of the train, he would be at the middle. - if it takes the same amount of time for the train to go all the tunnel distance as the man to go half the tunnel, well, then our train is going twice as fast. We also now that the train is half the tunnel distance away from the start of the tunnel. And that the name of this puzzle perhaps should be "railroad tunnel" ;) Rui Brandão Sunday, April 10, 2005 Recent Topics Fog Creek Home