SOLUTION: a plane which can fly 520mph in still air flies for 3 hours against a wind and for 2 hours with the same wind. The total distance it covers is 2565 miles. find rate of the wind.

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Question 166202: a plane which can fly 520mph in still air flies for 3 hours against a wind and for 2 hours with the same wind. The total distance it covers is 2565 miles. find rate of the wind.
Found 2 solutions by Mathtut, MathTherapy:
Answer by Mathtut(3670) About Me  (Show Source):
You can put this solution on YOUR website!
lets call s the rate of the plane which is 520mph d=2565
now with the formula distance(d)=rate*time or d=rt lets break this trip down into 2 segments. We will call the speed of the wind w.
d=(s-w)3+(s+w)2---->2565=(520-w)3+(520+w)2--->2565=1560-3w+1040+2w
w=35 after combining like terms and solving for w.
so the rate of the wind 35mph

Answer by MathTherapy(10719) About Me  (Show Source):
You can put this solution on YOUR website!
a plane which can fly 520mph in still air flies for 3 hours against a wind and for 2 hours with the same wind.
The total distance it covers is 2565 miles. find rate of the wind.
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Speed, in still air: 520 mph

Let wind-speed be W
Average speed, against the wind: 520 - W
Average speed, with the wind: 520 + W 

Distance travelled against the wind, in 3 hours: 3(520 - W) = 1,560 - 3W
Distance travelled with the wind, in 2 hours: 2(520 + W) = 1,040 + 2W

With the TOTAL DISTANCE travelled being 2,565 miles, we get the following TOTAL DISTANCE equation:
1,560 - 3W + 1,040 + 2W = 2,565 
      - 3W + 2,600 + 2W = 2,565
              - 3W + 2W = 2,565 - 2,600
                    - W = - 35
   Speed of wind, or