SOLUTION: togo flew his ultralight plane to a landing field 30 mi away. With the wind, the flight took 3/5h. Returning against the wind, the flight took 5/6h. Find the rate of the plane in s

Question 161535: togo flew his ultralight plane to a landing field 30 mi away. With the wind, the flight took 3/5h. Returning against the wind, the flight took 5/6h. Find the rate of the plane in still air and the rate of the wind.
Found 2 solutions by checkley77, gonzo:
Answer by checkley77(12844) (Show Source): You can put this solution on YOUR website!
D=RT
30/.6=r
r=50 mph with the wind.
30/(5/6)=r
5r/6=30
5r=30*6
5r=180
r=180/5
r=36 mph against the wind.
(50-36)/2=14/2=7 is the wind speed.
50-7=43 speed of the plane.
36+7=43 ditto.

Answer by gonzo(654) (Show Source): You can put this solution on YOUR website!
togo flies 30 miles in 3/5 of an hour with the wind.
togo flies 30 miles in 5/6 of an hour against the wind.
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his rate of speed with the wind was 30 / (3/5) = 30*5/3 = 150/3 = 50 mph.
his rate of speed against the wind was 30 / (5/6) = 30*6/5 = 6*6 = 36 mph.
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3/5 of an hour times 50 mph = 30 miles.
5/6 of an hour times 36 mph = 30 miles.
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mph looks good.
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going he was with the wind.
his rate of speed then was the airplane speed plus the wind speed.
coming back he was against the wind.
his rate of speed then was the airplane speed minus the wind speed.
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50 mph = A + W
36 mph = A - W
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looks like A + W - (A - W) should equal 50 - 36.
removing parentheses, equation becomes
A + W - A + W = 14.
combining like terms, the equation becomes
2*W = 14.
dividing both sides by 2, equation becomes
W = 7.
if W = 7, then A can be found as follows:
50 = A + 7
A = 43.
likewise,
36 = A - 7
A = 43.
since the airplane speed is the same, it looks like the answer is good.
solving in the original equations, we get
3/5 * (43 + 7) = 30
3/5 * 50- = 30
30 = 30
good.
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5/6 * (43 - 7) = 30
5/6 * 36 = 30
30 = 30
good.
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answer is W = 7 and A = 43.
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