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

18 hours

Step-by-step explanation:

Let the volume of the pool be x. Since pipe A filled the  pool in 6 hours, the rate of pipe A = x / 6.

Let the rate of pipe b be y, Hose A filled the pool alone for the first 2 hours, this means that the volume filled in the 2 hours is x/6(2 hours) and the two hoses, working together, then finished filling the pool in another 3 hours for the 3 hours the volume filled is x/6(3) + y(3).  hence the total time is:

x/6(2) + x/6(3) + y(3) = x

x/3 + x/2 + 3y = x

Multiply through by 6:

2x + 3x + 18y = 6x

5x + 18y = 6x

18y = x

y = x/18

The rate of pipe B is x/18, this means it would take pipe B 18 hours to full the pool alone

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step-by-step explanation:

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