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Question 320409: Zelon Pharmaceuticals maintains a courier plane to transport company executives back and forth between the company headquarters and the main manufacturing plant 250 miles away. Today the pilot flew from headquarters to the plant, picked up the Chief of Operations, and returned to headquarters. On the first leg of the trip, the pilot flew against a headwind averaging 10 miles per hour. On the return leg, the average tailwind was 30 miles per hour. If the pilot spent a total of four hours in the air, what would the average speed of the courier plane be with no wind? Round your answer to the nearest mile per hour.
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website! Let the speed in still air be x mph
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250/x-10 +250/x+30 =4
(x-10)(x+30) is the LCM
250(x+30)+250(x-10)/ (x-10(x+30)=4
250x+7500+250x-2500 = 4(x-10)(x+30)
500x+5000=4(x^2+30x-10x-300)
500x+5000=4x^2+120x-40x-1200
4x^2-420x-6200=0
x^2-105x-1550=0
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Find the roots of the equation
x1, x2
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x1= +420 +sqrt((-420)^2-4*4(-6200)))/2*4
=118.1 mph
x2=+420 -sqrt((-420)^2-4*4(-6200)))/2*4
x2=-13.1
Ignore the negative value.
so the speed in still air = 118.1 mph
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CHECK
(250/118.1-10)+(250/118.1+30)= 4.004
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