Question 427553
A canoe can travel 15 mph in still water. 
Going with the current it can travel 20 miles in the same time that it takes to travel 10 miles against the current. 
Find the rate of the current
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With-current DATA:
distance = 20 miles ; rate = 15+c mph; time = d/r = 20/(15+c) hrs
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Against-current DATA:
distance = 10 miles ; rate = 15-c mph; time = d/r = 10/(15-c) hrs
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Equation:
time = time
20/(15+c) = 10/(15-c)
Cross-multiply to get:
20(15-c) = 10(15+c)
300-20c = 150+10c
30c = 150
c = 5 mph (speed of the current)
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Cheers,
Stan H.
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