Mo2men wrote:Hoses A and B spout water at different constant rates, and hose A can fill a certain pool in 6 hours. Hose A filled the pool alone for the first 2 hours and the two hoses, working together, then finished filling the pool in another 3 hours. How many hours would it have taken hose B, working alone, to fill the entire pool?
A 18
B 15
C 12
D 6
E 3
Let the pool = 36 gallons.
Since A takes 6 hours to fill the 36-gallon pool, A's rate = w/t = 36/6 = 6 gallons per hour.
In the first 2 hours, the amount of water generated by A = rt = 6*2 = 12 gallons.
Since A and B together take 3 hours to fill the remaining 24 gallons of the pool, the combined rate for A and B = w/t = 24/3 = 8 gallons per hour.
B's rate = (combined rate for A and B) - (A's rate) = 8-6 = 2 gallons per hour.
At a rate of 2 gallons per hour, the time for B to fill the 36-gallon pool = w/r = 36/2 = 18 hours.
The correct answer is
A.
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