Two pipes running together can fill a tank in $11\dfrac{1}{9}$ minutes. If one pipe takes 5 minutes more than the other to fill the tank, find the time in which each pipe would fill the tank.

The tank is filled by the two pipes together in $11\dfrac{1}{9}$ minutes i.e. in $\dfrac{100}{9}$ minutes,

∴ the part of tank filled in one minute = $\dfrac{9}{100}$

Let the time taken by two pipes to fill tank separately be x minutes and (x + 5) minutes.

∴ the part of tank filled by the first pipe in one minute = $\dfrac{1}{x}$ and

the part of tank filled by the second pipe in one minute = $\dfrac{1}{x + 5}$

According to the question

$\dfrac{1}{x} + \dfrac{1}{x + 5} = \dfrac{9}{100} \\[1em] \Rightarrow \dfrac{x + 5 + x}{x(x + 5)} = \dfrac{9}{100} \\[1em] \Rightarrow \dfrac{2x + 5}{x^2 + 5x} = \dfrac{9}{100} \\[1em] \Rightarrow 100(2x + 5) = 9(x^2 + 5x) \\[1em] \Rightarrow 200x + 500 = 9x^2 + 45x \\[1em] \Rightarrow 9x^2 + 45x - 200x - 500 = 0 \\[1em] \Rightarrow 9x^2 - 155x - 500 = 0 \\[1em] \Rightarrow 9x^2 - 180x + 25x - 500 = 0 \\[1em] \Rightarrow 9x(x - 20) + 25(x - 20) = 0 \\[1em] \Rightarrow (9x + 25)(x - 20) = 0 \\[1em] \Rightarrow x = -\dfrac{25}{9} \text{ or } x = 20$

Since, time cannot be negative hence, x ≠ $-\dfrac{25}{9}$

∴ x = 20 , x + 5 = 25.

Time taken by each pipe is 20 minutes and 25 minutes.