Pipes A and B can fill a tank in 12 hours and 18 hours, respectively. Pipe C is an emptying pipe. When all the three pipes are opened together for 8 hours, then 26/45 part of the tank is filled up. What part of the tank will be filled up by A and C together in 9 hours?

Correct option is B

Given:

Pipe A fills the tank in 12 h, pipe B in 18 h, and pipe C is an emptying pipe.
When A + B + C work together for 8 h, $$\frac{26}{45}$$​ of the tank is filled.
Find the part filled by A + C in 9 h.

Solution:

Rates:

A = $$\frac{1}{12}$$​,  B = $$\frac{1}{18}$$​,  C = -x

Combined work in 8 h:

​$$8\left(\frac{1}{12}+\frac{1}{18}-x\right) = \frac{26}{45} \\ \ \\\frac{1}{12}+\frac{1}{18}=\frac{5}{36}\\ \ \\8\left(\frac{5}{36}-x\right)=\frac{26}{45}\\ \ \\\frac{40}{36}-8x=\frac{26}{45}\\ \ \\\frac{10}{9}-\frac{26}{45}=8x\\ \ \\\frac{50-26}{45}=8x \Rightarrow \frac{24}{45}=8x\\ \ \\x=\frac{1}{15}$$​​

So, rate of A + C:

​$$\frac{1}{12}-\frac{1}{15}=\frac{5-4}{60}=\frac{1}{60}$$​​

Work done in 9 h:

​$$9\times\frac{1}{60}=\frac{3}{20} \times 100 = 15\%$$

Alternate Solution(Exam-hall Method): ​