Question

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?

Answer 1

Correct Answer: 10

Step 1: Define rates 

Let:

• Rate of each filling pipe = \( x \ \text{litres/hour} \)
• Rate of each draining pipe = \( y \ \text{litres/hour} \)

Step 2: First condition

From the first scenario: \[ \text{Capacity} = (6x - 5y) \times 6 \] (6 filling pipes and 5 draining pipes for 6 hours)

Step 3: Second condition

From the second scenario: \[ \text{Capacity} = (5x - 6y) \times 60 \] (5 filling pipes and 6 draining pipes for 60 hours)

Step 4: Equating capacities

\[ (6x - 5y) \times 6 = (5x - 6y) \times 60 \] \[ 6x - 5y = 50x - 60y \] \[ 44x = 55y \] \[ 4x = 5y \] \[ x = 1.25y \]

Step 5: Finding tank capacity

Substitute \( x = 1.25y \) into the first condition: \[ \text{Capacity} = (6x - 5y) \times 6 = (7.5y - 5y) \times 6 = 2.5y \times 6 = 15y \]

Step 6: Required time with 2 filling pipes and 1 draining pipe

Effective rate: \[ 2x - y = 2(1.25y) - y = 2.5y - y = 1.5y \] Time to fill: \[ \text{Time} = \frac{\text{Capacity}}{\text{Rate}} = \frac{15y}{1.5y} = 10 \ \text{hours} \]

✅ Final Answer: The required time is 10 hours.