Pipe A and B can fill a tank in 15 hours and 20 hours respectively. Pipe C can empty the tank in 25 hours. Pipes A, B and C works together but pipe C is closed after 10 hours, then time taken by pipe A and B to fill the remaining tank is:

2 hours

The correct answer is Option (2) → 2 hours **

Let rates: A = $\frac{1}{15}$ tank/hr, B = $\frac{1}{20}$ tank/hr, C (emptying) = $-\frac{1}{25}$ tank/hr.

Combined rate while all three work: $r_{1}=\frac{1}{15}+\frac{1}{20}-\frac{1}{25}=\frac{20+15-12}{300}=\frac{23}{300}$ tank/hr.

Work done in first 10 hr: $W_{1}=10\cdot r_{1}=10\cdot\frac{23}{300}=\frac{23}{30}$ tank.

Remaining tank = $1-\frac{23}{30}=\frac{7}{30}$ tank.

After C is closed, A and B together: $r_{2}=\frac{1}{15}+\frac{1}{20}=\frac{20+15}{300}=\frac{35}{300}=\frac{7}{60}$ tank/hr.

Time to fill remaining: $t=\frac{\text{remaining}}{r_{2}}=\frac{\frac{7}{30}}{\frac{7}{60}}=\frac{7}{30}\cdot\frac{60}{7}=2\ \text{hours}.$

Answer: 2 hours