SOLUTION: A six passenger plane cruises at 180 mph in calm air. If the plane flies 7 miles with the wind in the same amount of time as it flies 5 miles against the wind, then what is the win

Question 77448: A six passenger plane cruises at 180 mph in calm air. If the plane flies 7 miles with the wind in the same amount of time as it flies 5 miles against the wind, then what is the wind speed?
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
Distance (d)= rate(r) times time (t) or d=rt; t=d/r
Let x=rate of speed of wind
Speed of plane with wind =180+x
Distance travelled with wind is 7 mi
Time to travel 7 mi with wind is 7/(180+x)-----Time #1
Speed of plane against wind=180-x
Distance travelled against wind is 5 mi
Time to travel 5 mi against wind is 5/(180-x)----Time #2
Now we are told that Time #1 = Time #2 so our equation to solve is:
7/(180+x)=5/(180-x) Multiply both sides by (180+x)(180-x) to get rid of the fractions:
7(180+x)(180-x)/(180+x)=5(180+x)(180-x)/(180-x) Simplify by cancelling
7(180-x)=5(180+x) get rid of parens
1260-7x=900+5x subtract 1260 and also 5x from both sides
1260-1260-7x-5x=900-1260+5x-5x collect like terms
-12x=-360 divide both sides by -12
x=30 mph--------------------speed of the wind
CK
t=d/r
7/(180+30)=5/(180-30)
7/210=5/150
1/30=1/30
Hope this helps----ptaylor

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