SOLUTION: Please help me set up this problem: "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 ag

Question 427553: Please help me set up this problem:
"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."
Found 2 solutions by stanbon, mananth:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
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
-----
With-current DATA:
distance = 20 miles ; rate = 15+c mph; time = d/r = 20/(15+c) hrs
----
Against-current DATA:
distance = 10 miles ; rate = 15-c mph; time = d/r = 10/(15-c) hrs
--------------
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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Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
boat speed =15 mph
current speed =x mph
Speed upstream= 15-x
Speed downstream=15+x
Distance upstream= 10 miles
Distance downstream= 20 miles
t=d/r time upstream = time downstream
10/(15-x )=20/(15 +x)
10(15+x)=20(15-x)
150+10x=300-20x
20x-10x=150
10 x = 150
/ 10
x =15mph current speed
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