Question 1179659: Flying with the wind, a plane flew 1,120 mi in 4 h. Against the wind, the plane required 7 h to fly the same distance. Find the rate of the plane in calm air and the rate of the wind.
Found 2 solutions by mananth, MathTherapy:
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website! Plane speed =x mph in still air
wind speed =y mph
against wind x-y 7.00 hours
with wind x+y 4.00 hours
Distance = same= 1120 miles
t=d/r
1120 / ( x - y )= 7.00
1120.00 /( x + y) = 4.00
1 x -1 y = 160.00 ....................1
1120 / ( x+ 1 y )= 4.00
1120.00 /( x - 1 y) = 4.00
1.00 x + 1.00 y = 280.00 ...............2
Multiply (1) by 1.00
Multiply (2) by 1.00
we get
1 x + 1 y = 160
1 x - 1 y = 280
2 x = 440
/ 2
x = 220 mph
plug value of x in (1)
1 x -1.00 y = 160.00
220 -1.00 y = 160.00
- y = 160 -220
- y = -60
y = 60 mph
Answer by MathTherapy(10551)  (Show Source):
You can put this solution on YOUR website!
Flying with the wind, a plane flew 1,120 mi in 4 h. Against the wind, the plane required 7 h to fly the same distance. Find the rate of the plane in calm air and the rate of the wind.
That MUMBO-JUMBO method by the other person is RIDICULOUS! I hope you're not planning on using it. If not, then read on.
Let the speed in still air, and speed of wind be S, and W, respectively
Then we get the following system of TOTAL-SPEED equations:  ====> 
Now, add the 2 equations and solve for S to get the plane's speed in still air.
Afterwards, substitute the value of S into either eq (i) or (ii) and solve for W to get the wind-speed.
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