SOLUTION: Solve the problem and show the work. Jim can run 5 miles per hour on level ground on a still day. One windy day, he runs 10 miles with the wind, and in the same amount of time run

Question 83704: Solve the problem and show the work.
Jim can run 5 miles per hour on level ground on a still day. One windy day, he runs 10 miles with the wind, and in the same amount of time runs 7 miles against the wind. What is the rate of the wind.
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
Let r=rate of wind
Distance(d) equals Rate(r) times Time(t) or d=rt; r=d/t and t=d/r
time it takes to run 10 mi with the wind:
t=10/(5+r)
time it take to run 7 mi against the wind:
t=7/(5-r)
Now we are told that these two times are the same, so:
10/(5+r)=7/(5-r) multiply both sides by(5+r)(5-r) to get rid of fractions. In other words, cross multiply:
10(5-r)=7(5+r) get rid of parens
50-10r=35+7r subtract 7r and also 50 from both sides
50-50-10r-7r=35-50+7r-7r collect like terms
-17r=-15 divide both sides by -17
r=0.882 mph-------------------------------------------rate of wind
ck
10/(5+0.882)=7/(5-0.882)
10/5.882=7/4.118
~1.7=~1.7
Hope this helps-----ptaylor

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