Question 201939: Hi Please help me answer this hard word problem.
An Airplane whose speed in still air is 530mi/h carries enough fuel for 10 hours of flight. On a certain flight it flies against a wind of 30 mi/h. On the return flight it travels with a wind of 30 mi/h. How far can the plane fly without refueling?
Thank you Very much,
Leo
Answer by scott8148(6628)   (Show Source):
You can put this solution on YOUR website! during the flight (against the wind) the speed is 500mph (530-30)
during the return (with the wind) the speed is 560mph (530+30)
the distance (out and back) is the same for both trips
if "t" is the time for the flight, then "10-t" is the time for the return
d = r * t, so, ___ 500 * t = 560 * (10 - t)
500t = 5600 - 560t
1060t = 5600
t = 5600 / 1060 = 280 / 53
substituting ___ d = 500 * 280 / 53 = 2641.5 mi (approx) one way
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