algebra (23)
posted by Anita .
How do you set up this problem?
A boater travels 16 miles per hour on the water on a still day. During one particular windy day, he finds that he travels 48 miles with the wind behind him in the same amount of time that he travels 16 miles into the wind. Find the rate of the wind.

this one is similar to the previous two questions you asked and were helped by bobpursley.
Remember Time = Distance/Rate, so..
Time for first case = 48/(16+x)
time for second case = 16/(16x)
but the times were the same
48/(16+x) = 16/(16x)
crossmultiply and solve for x
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