​A and B are emptying pipes and C is a filling pipe. Pipes A and B can empty a full tank in 15 hours and 18 hours, respectively. When all three pipes are opened together, it takes $$1\frac{1}{4}$$ hours to empty $$\frac{1}{9}$$​ part of the tank. In how many hours can pipe C alone fill $$\frac{2}{3}$$​ part of the tank?

Correct option is A

Given:

Pipe A can empty the tank in 15 hours.

Pipe B can empty the tank in 18 hours.

Pipe C is a filling pipe.

When all three pipes are open together, it takes $$1\frac{1}{4}$$​ hours to empty $$\frac{1}{9}$$​ of the tank.

Formula Used:

Total time = $$\frac{\text{Total work }}{\text{efficiency}}$$​​

Rate of Pipe A = $$\frac{-1}{15}$$​ (since it is emptying).

Rate of Pipe B =$$\frac{-1}{18}$$​ (since it is emptying).

Rate of Pipe C = $$\frac{1}{x}$$​ (filling pipe, where x is the time it takes for C to fill the tank).

Solution:

Combined rate for all pipes together =$$\frac{-1}{9} \div 1 \frac{1}{4}$$​ (as they empty$$\frac{1}{9}$$​of the tank in$$1\frac{1}{4}$$​ hours).

​$$1\frac{1}{4} = \frac{5}{4}$$​ hours.

Combined rate = $$\frac{-1}{9} \div \frac{5}{4} = \frac{-1}{9} \times \frac{4}{5} = \frac{-4}{45}$$​ (emptying rate).

​$$\frac{-1}{15} + \frac{-1}{18} + \frac{1}{x} = \frac{-4}{45}.$$​​

$$\frac{-1}{15} + \frac{-1}{18} = \frac{-18}{270} - \frac{15}{270} = \frac{-33}{270} = \frac{-11}{90}.$$​

​$$\frac{1}{x} = \frac{-4}{45} - \frac{-11}{90} = \frac{-8}{90} + \frac{11}{90} = \frac{3}{90} = \frac{1}{30}.$$ 

x whole work complete 30 days work is 90 and work efficiency is 3 So, 90$$\times\frac23$$ = 60

Total time = $$\frac{60}3$$ = 20days 

​Alternate Method: 

A+B+C = $$\frac91\times1\frac14= \frac91\times\frac54= \frac{45}4$$​ 

C efficiency = (A + B) - (A+ B+ C) = (6+5)-8 = 11-8 = 3 

Total work = 90

But c fill $$\frac23$$​= $$90\times\frac23$$ = 60 

C take total time to fill $$\frac23$$ tank = $$\frac{60}3$$ 

= 20h