SOLUTION: a canoe can travel 15 mi/hr 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
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Question 118432: a canoe can travel 15 mi/hr 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 Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! a canoe can travel 15 mi/hr 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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Let the speed of the current be "c"
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Downstream DATA:
Distance = 20 miles ; rate = 15+c ; time = d/r = 20/(15+c) hrs
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Upstream DATA;
Distance = 10 miles ; rate = 15-c ; time = d/r = 10/(15-c) hrs
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EQUATION:
time up = time down
10/(15-c) = 20/(15+c)
1/(15-c) = 2/(15+c)
15+c = 2(15-c)
15+c = 30-2c
3c = 15
c = 5 mph ( speed of the current )
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Cheers,
Stan H.
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