SOLUTION: When an airplane flies with a given wind, it can travel 3500km in 5 hours. When the same airplane flies in opposite direction against the wind it takes 7 hours to fly the same pldi

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Question 1035209: When an airplane flies with a given wind, it can travel 3500km in 5 hours. When the same airplane flies in opposite direction against the wind it takes 7 hours to fly the same pldistance. Find the air speed of the plane and the speed of the wind
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
So typical a type of travel rate exercise that this solution will be all in symbols, and will work for ALL of this type of exercise.


Known variables, h,d,k
Time with wind, h
Time against wind, k
Distance one way for trip, d

Unknown variables, r,w
Air speed for plane absent any wind, r
Speed of wind, w 
                 RATE       TIME    DISTANCE
WITHWIND         r+w          h       d
AGAINSTWIND      r-w          k       d

Starting System of Equations: system%28%28r%2Bw%29h=d%2C%28r-w%29k=d%29
Solve this system for r and w.

system%28rh%2Bwh=d%2Crk-wk=d%29
rh=d-wh
r=%28d-wh%29%2Fh-----* and you might return to this formula later.
-
%28%28d-wh%29%2Fh%29k-wk=d
%28dk-whk%29%2Fh-wk=d
dk-whk-whk=dh
-2whk=dh-dk
2whk=dk-dh

highlight%28w=%28dk-dh%29%2F%282hk%29%29

Use w to determine value for r.
r=%28d-wh%29%2Fh
r=%28d-%28%28dk-kh%29%2F%282hk%29%29h%29%2Fh
r=%28d-%28dk-kh%29%2F%282k%29%29%2Fh
r=d%2Fh-%28dk-hk%29%2F%282kh%29, LCD is 2kh;
r=%28d%2Fh%29%282k%29%2F%282k%29-%28dk-hk%29%2F%282kh%29
r=%282dk-%28dk-hk%29%29%2F%282kh%29
r=%282dk-dk%2Bhk%29%2F%282kh%29
r=%28dk%2Bhk%29%2F%282kh%29
highlight%28r=%28d%2Bh%29%2F%282h%29%29

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

When an airplane flies with a given wind, it can travel 3500km in 5 hours. When the same airplane flies in opposite direction against the wind it takes 7 hours to fly the same pldistance. Find the air speed of the plane and the speed of the wind
Let speed, in still air be S, and windspeed, W

Then S+%2B+W+=+3500%2F5_____S + W = 700 ------ eq (i)

Also, S+-+W+=+3500%2F7_____S – W = 500 ------ eq (ii)

2S = 1,200 -------- Adding eqs (ii) & (i)

S, or speed in still air = highlight_green%28matrix%281%2C4%2C+%221%2C200%22%2F2%2C+or%2C+600%2C+%22km%2Fh%22%29%29%29



600 + W = 700 -------- Substituting 600 for S in eq (i)

W, or windspeed = 700 – 600, or highlight_green%28matrix%281%2C2%2C+100%2C+%22km%2Fh%22%29%29