SOLUTION: An airplane takes 5 hours to travel a distance of 4200km 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

Algebra ->  Equations -> SOLUTION: An airplane takes 5 hours to travel a distance of 4200km 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      Log On


   

Question 1012409: An airplane takes 5 hours to travel a distance of 4200km 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?
Found 2 solutions by macston, fractalier:
Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
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R=rate of plane; W=rate of wind
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Against wind:(5 hours)(R-W)=4200km
R-W=840
R=W+840
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With wind: (4 hours)(R+W)=4200km
R+W=1050
(W+840)+W=1050
2W=210
W=105
ANSWER 1: Rate of wind = 105 km/hr.
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R=W+840
R=105+840
R=945
ANSWER 2: Rate of plane in still air = 945 km/hr.
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CHECK:
(5 hours)(R-W)=4200
(5hrs)(945-105)=4200
(5hrs)(840)=4200
4200=4200
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(4 hrs)(R+W)=4200
(4 hrs)(945+105)=4200
(4 hrs)(1050)=4200
4200=4200
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Answer by fractalier(6550) About Me  (Show Source):
You can put this solution on YOUR website!
Call its rate in still air, r.
Call the rate of the wind, w.
In general rate times times equal distance.
We can then write and solve this system:
4(r + w) = 4200
5(r - w) = 4200
Let's divide the first by 4 and the second by 5 and get
r + w = 1050
r - w = 840
Now add equations and get
2r = 1890
r = 945 mph
Thus w = 105 mph