​Pipes A and B can fill a tank in 6 hours and 15 hours, respectively. Pipe C is a drain pipe. When all the three pipes are opened together for 6 hours, then 65% of the tank is filled. Initially, pipes A and C are opened together for 8 hours and then C is closed and B is opened. Pipes A and B together will fill the remaining part of tank in __________ hours.​

Correct option is B

Given:
Pipe A can fill the tank in 6 hours → A’s rate = 1/6 tank/hour
Pipe B can fill the tank in 15 hours → B’s rate = 1/15 tank/hour
Pipe C is a drain pipe (let its rate = –1/C)
When A, B, and C are opened together for 6 hours, 65% = 65/100 = 13/20 of the tank is filled.

Solution:

Step 1: Let Pipe C’s rate = –1/x
Together for 6 hours:
A + B + C = $$\left( \frac{1}{6} + \frac{1}{15} - \frac{1}{x} \right)$$​tank/hour
In 6 hours, this fills 13/20 of the tank:

​$$6 \times \left( \frac{1}{6} + \frac{1}{15} - \frac{1}{x} \right) = \frac{13}{20} \\\ \\6 \times \left( \frac{5 + 2 - \frac{30}{x}}{30} \right) = \frac{13}{20} \\\ \\\frac{42 - \frac{180}{x}}{30} = \frac{13}{20} \\\ \\20 \times \left(42 - \frac{180}{x}\right) = 390 \\840 - \frac{3600}{x} = 390 \\\ \\\frac{3600}{x} = 450 \\\ \\x = \frac{3600}{450} = 8$$

So, Pipe C empties the tank in 8 hours → C’s rate = –1/8

Step 2: Initial phase – A and C opened for 8 hours
A’s rate = 1/6
C’s rate = –1/8
Combined rate = (1/6 – 1/8) = (4 – 3)/24 = 1/24

In 8 hours:
→ (1/24) × 8 = 1/3 of the tank filled

​$$\text{Remaining work} = 1 - \frac{1}{3} = \frac{2}{3} \\\text{Combined rate of A and B} = \frac{1}{6} + \frac{1}{15} = \frac{5 + 2}{30} = \frac{7}{30} \\\ \\\text{Time} = \frac{\frac{2}{3}}{\frac{7}{30}} = \frac{2}{3} \times \frac{30}{7} = \frac{60}{21} = \frac{20}{7}$$

​$$\frac{20}{7} = 2 \frac{6}{7} \text{ hours}$$​​

Final Answer: (B)​​