Question 1089379
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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 {{{1/6}}} = {{{1/6}}} 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 = {{{280(20*6)}}} = {{{14/6}}} = {{{7/3}}} = {{{2}}}{{{1/3}}} hours.


<U>Answer</U>.  The flight from Y to X was  {{{2}}}{{{1/3}}} hours.
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