Question

A swimming pool is filled with three pipes with uniform flow. The first two pipes operating simultaneously, fill the pool at the same time during which the pool is filled by the third pipe alone. The second pipe fills the pool five hours faster than the first pipe and four hours slower than the third pipe. Find the time required by each pipe to fill the pool separately.

Answer 1

Let $V$ be the volume of the pool and $x$ the number of hours required by the second pipe alone to fill the pool. Then, the first pipe takes $(x+5)$ hours, while the third pipe takes $(x-4)$ hours to fill the pool. So, the parts of the pool filled by the first, second and third pipes in one hour are respectively

$\frac{V}{x+5},\frac{V}{x},\frac{V}{x-4}$

Let the time taken by the first and second pipes to fill the pool simultaneously be $t$ hours.
Then, the third pipe also takes the same time to fill the pool

$\therefore (\frac{V}{x+5}+\frac{V}{x})t=$ Volume of the pool

Also, $\frac{V}{x-4}t=$ Volume of the pool

$\Rightarrow (\frac{V}{x+5}+\frac{V}{x})t=\frac{V}{x-4}t\Rightarrow \frac{1}{x+5}+\frac{1}{x}=\frac{1}{x-4}$

$\Rightarrow (2x+5)(x-4)=x^2+5x\Rightarrow x^2-8x-20=0\Rightarrow x^2-10x+2x-20=0\Rightarrow (x-10)(x+2)=0$

$\Rightarrow x=10orx=-2$

But $x$ cannot be negative. So $x=10$

hence the timings required by first, second and third pipes to fill the pool individually are $15$ hours, $10$ hours and $6$ hours respectively.