SOLUTION: Rowing with the current, a canoeist paddled 10 mi in 2 h. Against the current, the canoeist could paddle only 6 mi in the same amount of time.
Find the rate of the canoest in c
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Find the rate of the canoest in c
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Question 1051863: Rowing with the current, a canoeist paddled 10 mi in 2 h. Against the current, the canoeist could paddle only 6 mi in the same amount of time.
Find the rate of the canoest in calm water and the rate of the current. Answer by ikleyn(52781) (Show Source):
You can put this solution on YOUR website! .
Rowing with the current, a canoeist paddled 10 mi in 2 h. Against the current, the canoeist could paddle only 6 mi in the same amount of time.
Find the rate of the canoest in calm water and the rate of the current.
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Let u be the the canoe rate in still water, in mph.
Let v be the rate of the current.
Then your governing equations are
= u + v (1) (the rate paddling with the stream)
= u - v (2) (the rate paddling against the srteam)
Simplify:
u + v = 5, (3)
u - v = 3. (4)
To solve, add the equations (3) and (4). You will get
2u = 5 + 3 ---> 2u = 8 ---> u = 4.
Thus you found the canoe rate in still water. It is 4 mph.
Then from (3) v = 5 - u = 5-4 = 1.
Answer. the canoe rate in still water is 4 mph. The rate of the current is 1 mph.
