Swimming Pool

Algebra Level 4

A swimming pool is fitted with three pipes with constant flow rates. The first two pipes operating simultaneously can fill the pool in the same amount of time that it takes for the third pipe alone to fill the pool. The second pipe fills the pool five hours faster than the first pipe and four hours slower than the third pipe.

How many hours would it take for the first pipe to fill the pool alone?


The answer is 15.

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Ved Sharda by ved sharda , 20, India

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1 solution

Ved Sharda
Jan 21, 2016

Let 'V' be the volume of the pool & let 'x' be the number of hours required by the SECOND pipe alone to fill the pool. Then the parts of the pool filled by the FIRST , SECOND & THIRD pipes in one hour are respectively V x + 5 \large{\dfrac{V}{x+5}} , V x \large{\dfrac{V}{x}} & V x − 4 \large{\dfrac{V}{x-4}}

It is given that V x + 5 + V x \large{\dfrac{V}{x+5} + \dfrac{V}{x}} = V x − 4 \large{\dfrac{V}{x-4}}

Since V is not equal to zero ,

we have 1 x + 5 + 1 x \large{\dfrac{1}{x+5} + \dfrac{1}{x}} = 1 x − 4 \large{\dfrac{1}{x-4}}

or x 2 x^2 - 8x - 20 = 0

or (x-10)(x+2) = 0

This gives x=10 or x=-2 , but x, being the number of hours , can't be negative.

Therefor, x=10

so the FIRST pipe fill the pool in 15 hours

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