A water tank has one inlet pipe and one outlet pipe. The inlet pipe can fill the tank completely in 2 hours when the outlet pipe is closed, and the outlet pipe can empty the full tank in 6 hours when the inlet pipe is closed. There is also a leakage which can drain water at half the rate of the outlet pipe. How long will it take to fill the empty tank when the inlet, outlet and leakage are all operating together?

Introduction / Context:
This is a pipes and cisterns problem, a standard variation of time and work questions. Here, filling and emptying actions happen simultaneously because the inlet pipe, outlet pipe, and a leakage are all functioning together. You must convert each pipe and the leakage into a rate of work expressed as a fraction of the tank filled or emptied per hour and then combine these rates algebraically to find the net effect.

Given Data / Assumptions:

• The inlet pipe fills the tank in 2 hours when working alone.
• The outlet pipe empties a full tank in 6 hours when working alone.
• The leakage drains water at half the rate of the outlet pipe.
• All rates are constant over time.
• The tank is initially empty, and all three are open together.

Concept / Approach:
When multiple pipes and leakages act on the same tank, the net rate is the algebraic sum of their individual rates. Filling rates are taken as positive, and emptying rates are taken as negative. Once the net rate of change of the volume of water is found, we use the formula Time = Total work / Net rate. Here, total work is one full tank, which we treat as 1 unit of work.

Step-by-Step Solution:
Let the capacity of the tank be 1 unit. Inlet pipe rate = 1 / 2 tank per hour (filling). Outlet pipe rate = 1 / 6 tank per hour (emptying). Leakage rate = half of outlet rate = (1 / 6) / 2 = 1 / 12 tank per hour (emptying). Net rate when all three act together = filling rate − outlet rate − leakage rate. Net rate = 1 / 2 − 1 / 6 − 1 / 12. Convert to common denominator 12: 1 / 2 = 6 / 12, 1 / 6 = 2 / 12, 1 / 12 = 1 / 12. Net rate = (6 / 12) − (2 / 12) − (1 / 12) = 3 / 12 = 1 / 4 tank per hour. Time taken to fill 1 tank = 1 / (1 / 4) = 4 hours.

Verification / Alternative check:
You can think in terms of actual tank fractions per hour: in 4 hours the inlet alone would fill 2 tanks, but during these 4 hours the outlet alone would empty 4/6 = 2/3 of a tank, and the leakage alone would empty 4 * (1 / 12) = 1/3 of a tank. Total emptied is 2/3 + 1/3 = 1 tank. Net effect is 2 tanks filled minus 1 tank emptied = 1 full tank effectively filled in 4 hours, which matches our answer.

Why Other Options Are Wrong:
3 hours corresponds to a net rate of 1/3 tank per hour, which would imply the combined emptying effect is weaker than it actually is. 5 or 6 hours would mean the net rate is slower than 1/4 tank per hour, which contradicts the given strong inlet pipe of 1/2 tank per hour even after considering the outlet and leakage.

Common Pitfalls:
Students often forget to include the leakage as a separate emptying agent or misunderstand “half of the rate” and use an incorrect value. Another mistake is treating both the outlet and the leakage as positive when they actually reduce the water level. Always remember to subtract emptying rates from filling rates when calculating net effect.

Final Answer:
The empty tank will be completely filled in 4 hours when all three are operating together.