<PRE><B>Question 925262</B>
Hank goes 10 mph in still water. He notices one day he goes 4 miles upstream in the same amount of time he goes downstream. What is the approximate speed of that day?
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Upstream DATA:
distance = 4 miles ; rate = 10-c ; time = d/r = 4/(10-c) hrs
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Downstream DATA:
distance = 4 miles ; rate = 10+c ; time = d/r = 4/(10+c) hrs
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Equation:
4/(10-c) = 4/(10+c)
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Comment:: Either the speed of the current is zero or
his distance downstream is different then his distance
upstream. Please check your problem statement.
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Cheers,
Stan H.
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