A tank has two outlet pipes, A and B, which together take 6 hours to empty a full tank when they are opened simultaneously. The tank was initially half-full and both the outlets were opened, after an hour an inlet pipe was also opened. If the inlet alone can fill the empty tank in 4 hours, how much time will it now take to fill the tank completely?

8 hrs

The correct answer is Option (2) → 8 hrs

Let the tank capacity be 1 unit.

Rate of A and B together: $\frac{1}{6}$ tank per hour (emptying)

Initial tank: half full = $\frac{1}{2}$ unit

After 1 hour with only outlets A and B open:

Amount emptied = $\frac{1}{6} \cdot 1 = \frac{1}{6}$

Remaining in tank = $\frac{1}{2} - \frac{1}{6} = \frac{1}{3}$ unit

Inlet pipe fills in 4 hours ⇒ rate of inlet = $\frac{1}{4}$ tank per hour (filling)

Net rate with inlet and outlets open:

Net rate = $\frac{1}{4} - \frac{1}{6} = \frac{3-2}{12} = \frac{1}{12}$ tank per hour

Time required to fill remaining $\frac{2}{3}$ units (since emptying outlets reduce tank further?):

Actually, remaining tank = $\frac{1}{3}$ unit, we need to reach full tank = 1 unit ⇒ amount to be filled = $1 - \frac{1}{3} = \frac{2}{3}$ units

Time = $\frac{\text{Amount to fill}}{\text{Net rate}} = \frac{2/3}{1/12} = \frac{2}{3} \cdot 12 = 8$ hours