SOLUTION: A boat can go 20 miles against a current in the same time that it can go 60 miles with the current. The current is 4 mph. Find the rate of the boat in still water.

Algebra ->  Rate-of-work-word-problems -> SOLUTION: A boat can go 20 miles against a current in the same time that it can go 60 miles with the current. The current is 4 mph. Find the rate of the boat in still water.      Log On


   

Question 1068136: A boat can go 20 miles against a current in the same time that it can go 60 miles with the current. The current is 4 mph. Find the rate of the boat in still water.
Found 2 solutions by josgarithmetic, stanbon:
Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
                 speed     time       distance
AGAINST          r-4          t         20
WITH             r+4          t         60


20%2F%28r-4%29=60%2F%28r%2B4%29
Solve for r.

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A boat can go 20 miles against a current in the same time that it can go 60 miles with the current. The current is 4 mph. Find the rate of the boat in still water.
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With Current DATA:
rate = b+4 mph ; time = t hrs ; distance 60 miles
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Against Current DATA:
rate = b-4 mph ; time = t hrs ; distance = 20 miles
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Equations:
bt + 4t = 60
bt - 4t = 20
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8t = 40
time = 5 hrs
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rate in still water ::
(b+4)5 = 60
b+4 = 12
Ans: boat = 8 mph
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Cheers,
Stan H.
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