Subj:     MATH PROB. - Train Bridge (S447b)           From: William Wu of U. C. Berkeley           At: http://www.ocf.berkeley.edu/~wwu/riddles/intro.shtml  Source: http://www.ocf.berkeley.edu/~wwu/riddles/easy.shtml#trainBridge  A man is 3/8's of the way across a train bridge, when he  hears the whistle of an approaching train behind him. It  turns out that he can run in either direction and just  barely make it off the bridge before getting hit.  If he  is running at 15 mph, how fast is the train traveling?  Assume the train travels at a constant speed, despite  seeing you on the tracks. =================================================================  Jack's solution: I did this very differently than your formal solution (mostly because I was looking for a way to do it in my head): If I run towards the train I cover 3/8 the length of the bridge just as the train arrives.  If I run the other way, I'm at the 3/8+3/8=3/4 point as the train arrives at the start.  Therefore I cover 1/4 of the bridge while the train covers the whole thing, so the train is moving 4x my speed => 60mph. Thank you Jack.  I could not imagine how you could do this problem without a solid knowledge of at least Algebra II. ==================================================================  My solution: Let  X = the width of the bridge          and    Y = the distance the train is from the bridge at the start          and    R = the rate of the train Distance equals Rate times Time  ( D = R x T )   or Time equals Distance divided by Rate  ( T = D/R ) Therefore Time 1 of train = time of man returning  ( Y / R  = 3/8 (X) / 15  )   and Time 2 of train = time of man going forward (  (Y + X)/R = 5/8 (X) / 15 ) That's two equations and three unknowns, but it solves. Cross multiplying yields 15 Y  = 3/8 X R   and  15 Y  +  15 X  =  5/8 X R Substitute the 15 Y from the first equation into the second yields 3/8 X R + 15 X = 5/8 X R Divide everywhere by X. 3/8 X + 15 = 5/8 X 15 = 1/4 X 60 mph = X The train is going 60 mph no matter the width of the bridge. The only reason I worked the problem was the site host's statement that the problem came from a 7th grade pre-algebra book.  I would be curious to see how anyone worked the problem without knowing Algebra II.