SOLUTION: A jet flow at an average speed of 280 mph from Point X to Point Y. Because of head winds, the jet averaged only 260 miles on the return trip, the return trip took 10 minutes longer

Question 1089379: A jet flow at an average speed of 280 mph from Point X to Point Y. Because of head winds, the jet averaged only 260 miles on the return trip, the return trip took 10 minutes longer. How many hours was the flight from Point Y to Point X?
Please help. Thanks.
Found 3 solutions by josgarithmetic, MathTherapy, ikleyn:
Answer by josgarithmetic(39615)   (Show Source): You can put this solution on YOUR website!
One would expect equal distances both directions, so 260 MILES for both ways.

Speed for the return trip: 227.5 mph
Answer by MathTherapy(10551)   (Show Source): You can put this solution on YOUR website!
A jet flow at an average speed of 280 mph from Point X to Point Y. Because of head winds, the jet averaged only 260 miles on the return trip, the return trip took 10 minutes longer. How many hours was the flight from Point Y to Point X?
Please help. Thanks.

Correct answer:
By the way, you should've written 260 mph, and not 260 miles.
IGNORE the RUBBISH/NONSENSE the other person posted.
Answer by ikleyn(52772)   (Show Source): You can put this solution on YOUR website!
.

We are given: Average speed of 280 mph from X to Y. Average speed of 260 mph from Y to X. Time for the flight (Y --> X) is 12 minutes longer than that for (x --> Y). Let t be the time for the flight (Y --> X) (which is under the question). Then 280*(t-1/6) = 260*t. (1) This equation says that the distance was the same for both flights. Notice that = of an hour = 10 minutes. From (1) you have 280*t - 280/6 = 260*t ====> 280*t - 260t = -280/6 ====> 20*t = 280/6 ====> t = = = = hours. Answer. The flight from Y to X was hours.

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