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A tank is fitted with pipes, some filling it and the rest draining it. All filling pipes fill at the same rate, and all draining pipes drain at the same rate. The empty tank gets completely filled in $6$ hours when $6$ filling and $5$ draining pipes are on, but this time becomes $60$ hours when $5$ filling and $6$ draining pipes are on. In how many hours will the empty tank get completely filled when one draining and two filling pipes are on?

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 $:10$
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