Question 201939
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