SOLUTION: Solve the Problem: A boat can go 33 mph in still water. It takes as long to go 300 miles upstream as it does to go downstream 360 miles. How fast is the current?

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Question 187405: Solve the Problem: A boat can go 33 mph in still water. It takes as long to go 300 miles upstream as it does to go downstream 360 miles. How fast is the current?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let c = the speed of the current
Let b = the speed of the boat in still water
For upstream:
d%5Bu%5D+=+%28b+-+c%29%2At%5Bu%5D
For downstream:
d%5Bd%5D+=+%28b+%2B+c%29%2At%5Bd%5D
given:
b+=+33 mi/hr
d%5Bu%5D+=+300 mi
d%5Bd%5D+=+360 mi
t%5Bu%5D+=+t%5Bd%5D (I'll call them both t)
Now I can rewrite the equations
(1) 300+=+%2833+-+c%29%2At
(2) 360+=+%2833+%2B+c%29%2At
This is 2 equations and 2 unknowns,
so it's solvable
(1) 300+=+%2833+-+c%29%2At
(1) 300+=+33t+-+ct
(2) 360+=+%2833+%2B+c%29%2At
(2) 360+=+33t+%2B+ct
I'll add (1) and (2)
(1) 300+=+33t+-+ct
(2) 360+=+33t+%2B+ct
(3) 660+=+66t
t+=+10 hr
Plug this back into (1) or (2)
(2) 360+=+%2833+%2B+c%29%2A10
(2) 36+=+33+%2B+c
(2) c+=+3 mi/hr
The speed of the current is 3 mi/hr
check answer:
(1) 300+=+%2833+-+3%29%2At
(1) 300+=+30t
(1) t+=+10
(2) 360+=+%2833+%2B+3%29%2At
(2) 360+=+36t
(2) t+=+10
OK