Posted by Skylar on Saturday, November 5, 2011 at 10:18pm.
Two hoses are connected to a swimming pool. Working together, they can fill the pool in 4 hr. The larger hose, working alone, can fill the pool in 6 hr less time than the smaller one. How long would it take the smaller one, woking alone, to fill the pool?
Please use the work equation

Math  MathMate, Sunday, November 6, 2011 at 12:20am
The large one can fill in x hours,
then the small one in (x+6) hours.
During each hour, working together, the two hoses can fill 1/x+1/(x+6) of the pool, or (2x+6)/(x(x+6)).
Since using both hoses, it fills the pool in 4 hours, each hour fills 1/4 of the pool, therefore:
1/x + 1/(x+6) = 1/4
(2x+6)/(x(x+6)) = 1/4
4(2x+6)=x(x+6)
x² 2x 24 = 0
Solve for x (reject the negative root).
The large one will fill in x hours alone, and the small one, x+6 hours.
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