Given that, all filling pipes fill at the same rate and all draining pipes drain at the same rate.

Let the rate of each filling pipe be $`f\text{' liters/hr}$ similarly, the rate of each draining pipe be $`d\text{’ liters/hr}.$

Six filling pipes and five draining pipes filled a tank in $6$ hours. Then the capacity of tank $ = (6f – 5d) \times 6 \; \text{hours} \quad \longrightarrow(1) $

Similarly, five filling pipes and six draining pipes filled same tank in $60$ hours. Then the capacity of tank $ = (5f - 6d) \times 60 \;\text {hours} \quad \longrightarrow(2) $

Equating the equation $(1)$ and $(2),$ because capacity of the tank should be the same.

$ (6f – 5d)6 = (5f- 6d)60 $

$ \Rightarrow 6f – 5d = 50f - 60d $

$ \Rightarrow 6f – 50f = -60d + 5d $

$ \Rightarrow -44f = -55d $

$ \Rightarrow 4f = 5d $

$ \Rightarrow \boxed {\frac {f}{d} = \frac {5}{4}} $

The total capacity of tank $ = ( 6f - 5d ) \times 6 $

$ = ( 6 \times \frac{5d}{4} - 5 d) \times 6 $

$ = \left(\frac{15d-10d}{2}\right) \times 6 $

$ = 15d$ litres

$ \therefore$ Number of hours in which one draining and two filling pipes completely fill the tank $ = \frac {15d}{2f – 1d} $

$ \qquad = \frac {15d}{2 \left( \frac{5d}{4} \right) – d} $

$ \qquad = \frac{30d}{5d-2d} $

$ \qquad = \frac{30d}{3d} = 10$

Hence, in the $10$ hours, the empty tank gets completely filled when one draining and two filling pipes are on.

Correct Answer $:10$