SOLUTION: An airplane takes 5 hours to travel a distance of 4400km against the wind. The return trip takes 4 hours with the wind. What is the rate of the plane in still air and what is the

Question 500983: An airplane takes 5 hours to travel a distance of 4400km against the wind. The return trip takes 4 hours with the wind. What is the rate of the plane in still air and what is the rate of the wind?
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
Plane speed =x km/h
Wind speed =y km/h
against wind x-y 5.00 hours
with windx+y 4.00 hours

Distance = same= 4400 miles
t=d/r
4400 / ( x - y )= 5.00
5 ( x - y ) = 4,400.00
5 x -5 y = 4400 ....................1
4400 / ( x + y )= 4.00
4.00 ( x + y ) = 4400
4.00 x + 4.00 y = 4400 ...............2
Multiply (1) by 4.00
Multiply (2) by 5.00
we get
20 x + -20 y = 17600
20 x + 20 y = 22000
40 x = 39600
/ 40
x = 990 km/h

plug value of x in (1)
5 x -5 y = 4400
4950 -5 y = 4400
-5 y = 4400 -4950
-5 y = -550
y = 110 km/h

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