geluh
Answered

Makakuha ng mabilis at tumpak na mga sagot sa iyong mga katanungan sa Imhr.ca, ang pinakamahusay na Q&A platform. Kumuha ng mga sagot mula sa mga eksperto nang mabilis at eksakto mula sa aming dedikadong komunidad ng mga propesyonal. Tuklasin ang komprehensibong mga solusyon sa iyong mga tanong mula sa mga bihasang propesyonal sa iba't ibang larangan sa aming platform.

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 :

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.