Question 459758
AN AIRCRAFT FLEW 4 HOURS WITH THE WIND. THE RETURN TRIP TOOK 5 HOURS AGAINST THE WIND. IF THE SPEED OF OF THE PLANE IN STILL AIR IS 384 MILES PER HOUR MORE THAN THE WIND SPEED, FIND THE SPEED OF THE WIND SPEED AND THE SPEED OF THE PLANE IN STILL AIR, 
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With wind DATA:
time = 4 hrs ; distance = x miles ; rate = d/t = x/4 mph
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Against wind DATA:
time = 5 hrs ; distance = x miles ; rate = d/t = x/5 mph
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Let wind speed be "w".
Let speed of the plane in still air = "p".
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Equation:
p + w = x/4
p - w = x/5
p = w + 384
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Substitute for "p" and solve for w and x:
w + 384 + w = x/4
w + 384 - w = x/5
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2w + 384 = x/4
384 = x/5
x = 1920 miles
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Solve for "w":
2w + 384 = 1920/4
2w = 96
wind speed = 48 mph
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Solve for "p":
p = w + 384
plane speed in still air = 48+384 = 432 mph
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Cheers,
Stan H.
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