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two hoses when connected to a swimming pool can fill together in four hours. if the larger hose alone is used, it can fill the pool in 6 hours less than the smaller hose. how long will it the smaller hose to fill the swimming pool alone?-- ^-^

Sagot :

Mlcparra16idn
We let the time taken when ....
the larger hose alone is used be x - 6
and the smaller hose as x.

This would make their rates [tex] \frac{1}{x-6} [/tex] and [tex] \frac{1}{x} [/tex] respectively.

Take note that:
1 / rate = time 
so,
1 / rate of both hoses = 4
rate of both hoses = 1/4

[tex] \frac{1}{x-6} + \frac{1}{x} = \frac{1}{4}\\ \frac{2x-6}{x^2-6x} = \frac{1}{4} \\ 8x-24=x^2-6x \\ 0=x^2-14x+24 \\ 0=(x-12)(x-2)[/tex]

This would mean that x = 12 or 2 but x cannot be 2 since that would make the time taken for the larger hose negative.

Therefore it will take the smaller hose 12 hours to fill the swimming pool alone.
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