A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the second and third pipe can fill a tank together?

(Pipe & Cistern)

Let, the first pipe can alone fill the tank in x hour

$\therefore$2nd and 3rd pipe will take (x-5) and (x - 9) hours to fill the tank.

According to the question

$$\begin{gathered} \frac{1}{x}+\frac{1}{x-5}=\frac{1}{x-9} \\ =\frac{(x-5)+x}{x(x-5)}=\frac{1}{x-9} \\ =(2x-5) (x-9) =x(x-5) \\ =2x^2-18x-5x+45=x^2-5x \\ =x^2-18x+45=0 \\ =x^2-15x-3x+45=0 \\ =x(x-15)-3(x-15)=0 \\ (x-15) (x-3)=0 \\ Now x-15=0 [x=3 is not acceptable] \\ \therefore x=15 \\ The first pipe will take 15 minutes to fill the \tan k. \end{gathered}$$

$\therefore$The 2nd pipe can fill the tank in = 15 - 5 = 10 hours

And the 3 pipe can fill the tank in = 15 - 9 = 6 hours

2nd and 3rd pipe can fill in 1 hour = $\frac{1}{10}+\frac{1}{6}=\frac{3+5}{30}=\frac{8}{30}=\frac{4}{15}part$

$$\begin{gathered} Now, \frac{4}{15} part of the \tan k is filled by 2nd and 3rd pipe in 1 hour \\ \therefore 1 or complete part of the \tan k is filled by 2nd and 3rd pipe in = \frac{15}{4}=3.75 hours \\ \therefore 2nd and 3rd pipe can fill the \tan k in = 3 hours 45 minutes. \left(answer\right) \end{gathered}$$