SOLUTION: a twin engine aircraft can fly 720 miles from a city "a" to a city "b" in 4 hours with the wind and make a return in 6 hours against the wind. what is the speed of the wind?

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Question 511994: a twin engine aircraft can fly 720 miles from a city "a" to a city "b" in 4 hours with the wind and make a return in 6 hours against the wind. what is the speed of the wind?
Answer by oberobic(2304) About Me  (Show Source):
You can put this solution on YOUR website!
d = r*t is the fundamental distance equation.
.
Fly to B:
r = airspeed + wind speed = (a + w) mph
t = 4 hr
d = 720 mi
.
Return to A:
r= airspeed -wind speed = (a - w) mph
t = 6 hr
d = 720 mi
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6a - 6w = 720
4a + 4w = 720
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12a - 12w = 1440
12a + 12w = 2160
-----------------
0a -24w = -720
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w = 30 = wind speed
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substitute w=30 to find airspeed
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6a - 6(30) = 720
6a = 720 + 180
6a = 900
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a = 150 = airspeed
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Check to see if it can cover the distances in the correct times.
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(150-30)*6 = ??
120*6 = 720
correct
.
(150+30)*4 = ??
(180)*4 = 720
correct
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Answer: The wind speed is 30 mph.
.
Done.