SOLUTION: A pilot can fly a plane at 75 mph in calm air. A recent trip of 300 mi flying with the wind and 300 mi returning against the wind took 9 h. Find the rate of the wind.

Question 719418: A pilot can fly a plane at 75 mph in calm air. A recent trip of 300 mi flying with the wind and 300 mi returning against the wind took 9 h. Find the rate of the wind.
Found 2 solutions by mananth, graphmatics:
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
Plane speed = 75 mph
wind speed = x mph
0.60
against current = 75 - x
with current = 75 + x

Total distance = 300 miles Total Time= 9.00 hours
Time upstream + time downstream =

300 /( 75 - x )+ 300 /( 75 + x )= 9
300 ( 75 + x )+ 300 ( 75 - x )= 9.00 ( 5625 - 1 X^2 )
22500 + 300 x + 22500 -300 x = 50625 - 9 X^2
45000 - -50625 = -9 X^2
-5625 = -9 X^2
-9 X^2= -5625
X^2 = 625
X = +/- 25

wind speed 25 mph

Answer by graphmatics(170) (Show Source): You can put this solution on YOUR website!
Let x be the speed of the wind. Then 75+x is the speed of the plane flying with the wind and 300/(75+x) is the amount of time the plane flew with the wind. Similarity 75-x is the speed of the plane flying against the wind and 300/;(75-x) is the amount of time the plane flew against the wind. As we are given that the amount of time the plane flew with and against the wind is 9 hours we have that
300/(75+x)+300/(75-x)=9
(75+x)*(75-x)*300/(75+x) + (75+x)*(75-x)*300/(75-x)=(75+x)*(75-x)*9
(75-x)*300 + (75+x)*300=(75+x)*(75-x)*9
22500-300*x + 22500+300*x=(5625-x^2)*9
45000=50625-9*x^2
9*x^2-5625=0
(3*x)^2-75^2=0
(3*x-75)*(3*x+75)=0
3*x-75=0, 3*x+75=0
x=25, x=-25
As the speed of the wind must be positive we know it is 25.

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