SOLUTION: A jet plane flying with the wind flew 2400 miles in 4 hours. Against the wind, the plane flew only 2000 miles in 4 hours. Find the rate of the plane in calm air and the rate of the

Question 307439: A jet plane flying with the wind flew 2400 miles in 4 hours. Against the wind, the plane flew only 2000 miles in 4 hours. Find the rate of the plane in calm air and the rate of the wind.
Found 3 solutions by rapaljer, mananth, Alan3354:
Answer by rapaljer(4671)   (Show Source): You can put this solution on YOUR website!
Let x = rate of the plane in still air.
Let y = rate of the wind.

Rate WITH the wind =
Rate AGAINST the wind =

x+y=600
x-y=500

Add the equations together:
2x=1100
x=550 mph = speed of the plane in still air.

Now, x+y=600, where x=550
550+y=600
y=50 mph = speed of the wind.

Dr. Robert J. Rapalje, Retired
Seminole State College of Florida
Answer by mananth(16946)   (Show Source): You can put this solution on YOUR website!
A jet plane flying with the wind flew 2400 miles in 4 hours. Against the wind, the plane flew only 2000 miles in 4 hours. Find the rate of the plane in calm air and the rate of the wind.

let the the rate of plane instill air be x mph
& the rate of wind be y mph
.
with wind the rate will be x+y mph
Against wind the rate will be x-y mph
.
Time taken with wind will be 2400 / x+y...... ( time = distance / rate)
Time taken against wind will be 2000/ x-y
.
2400 /x+y = 4
2000/ x-y =4
.
4(x+y)=2400
4(x-y)=2000
.
4x+4y=2400
4x-4y=2000
.
Add the two equations
.
4x+4y+4x-4y= 4400
8x= 4400
x= 550 mph the rate of the plane
.
4x+4y=2400
plug the value of x
4*550+4y= 2400
2200 +4y= 2400
4y= 2400-2200
4y=200
y=50 mph the rate of the wind
.
.
Ananth

Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
2400/4 = 600 mph
2000/4 = 500 mph
The airspeed is the average, (600+500)/2 = 550 mph
The windspeed is the difference, 550 - 500 = 50 mph
Don't make it a thesis.

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