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October 26, 2014

October 26, 2014

Posted by **Skylar** on Saturday, November 5, 2011 at 10:18pm.

Please use the work equation

- Math -
**MathMate**, Sunday, November 6, 2011 at 12:20amThe 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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