SOLUTION: problem: An airplane made a trip of 200 miles against the wind in 1 hour. returning with the wind, the trip took only 50 minutes. Find the speed of the plane in calm air and the sp

Question 853229: problem: An airplane made a trip of 200 miles against the wind in 1 hour. returning with the wind, the trip took only 50 minutes. Find the speed of the plane in calm air and the speed of the wind

For this I'm guessing I have to find two equations that relate speed, time and distance:
I used Y for the speed of the wind and X for the speed of the airplane
X-Y = 200m/h
X+Y = 200m x 50min
Found 2 solutions by JulietG, lwsshak3:
Answer by JulietG(1812) (Show Source): You can put this solution on YOUR website!
Let X be the airplane's speed
Let Y be the wind's speed (assuming it is constant)
Here's what we know from the problem:
X - Y = 200
X + Y = 240 (60min/50 min * 200)
Add together
2X = 440
X = 220
If the airplane's speed is 220mph, then the wind is 20mph (Subtract 20 heading into it, add 20 traveling with it).
.
That's exactly what you had, you just needed to go one step further.
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
An airplane made a trip of 200 miles against the wind in 1 hour. returning with the wind, the trip took only 50 minutes. Find the speed of the plane in calm air and the speed of the wind
***
let x=speed of plane in calm air
let y=speed of wind
x+y=speed of plame with the wind
x-y=speed of plame against the wind
distance/speed=travel time
50 min=5/6 hr
..
2 equations:
200/(x-y)=1(against the wind)
200/(x+y)=5/6(with the wind)
..
x-y=200
5x+5y=1200
..
5x-5y=1000
5x+5y=1200
add
10x=2200
x=220
y=x-200=20
speed of plane in calm air=220 mph
speed of wind=20 mph

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