Subject: Plane Problem Date: Sat, 06 Mar 1999 18:22:17 PST Math Central: I'm having difficulty with this question about a plane that left New York. It's a secondary level question. A plane left New York and headed East to its destination 3600 miles away across the Atlantic. On the way back its speed was boosted by a 50 mph tail wind and it arrived an hour early. What was its normal speed? If you could help that would be great! Thanks! B.M.R. Hi, An fact that connects the two trips is that the usual time is 1 h greater than the shorter time. So we know that Usual Time - This Time = 1 h. We know that Distance = Speed x Time so we know that Time = Dist/Speed = 3600/Speed. If we let the usual speed be x mi/h (x > 0), then the new speed is x + 50 mi/h. So Usual Time = 3600/x and New Time = 3600/(x + 50). Thus we know that 3600/x -  3600/(x + 50)  =  1. You can now solve for x. If I were going to solve for x  I would multiply both sides of the equation by  x(x + 50) to "eliminate" the denominators. This leads to a quadratic equation that can be solved by factoring or by using the quadratic formula. You will likely find that one of the roots is negative. If so, then it is inadmissible since x > 0. Cheers Jack Go to Math Central To return to the previous page use your browser's back button.