One pipe can fill the tank in 18 minutes while another pipe can empty the completely filled tank in 19 minutes. If both the pipes are operated together on empty tank, how long (in minutes) will it take to fill one-third of the tank?​

Correct option is A

Given:
- Pipe 1 can fill the tank in 18 minutes.
- Pipe 2 can empty the tank in 19 minutes.
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Formula Used:
 Rate of Filling = $$\frac{ 1 }{ Time \ taken \ by \ Pipe 1}$$​​
 Rate of Emptying = $$\frac{ 1 }{ Time \ taken \ by \ Pipe 2}$$​​
 Net Rate of Filling = Rate of Filling by Pipe 1 - Rate of Emptying by Pipe 2
Time Taken to Fill One-third of the Tank = $$\frac{Required\ Portion }{ Net \ Rate \ of \ Filling}$$​​

Solution:
Rate of Filling by Pipe 1 = $$\frac{1 }{ 18}$$​​
Rate of Emptying by Pipe 2 = $$\frac{1 }{ 19}$$​​
The Net Rate of Filling when both pipes are working together is:
Net Rate of Filling = $$\frac{1 }{ 18 }- \frac{1 }{ 19 }= \frac{(19 - 18) }{ (18 \times 19)} = \frac{1 }{ 342}$$​​
Thus, the Net Rate of Filling is 1 / 342 of the tank per minute.

To fill one-third of the tank, the time taken is:
Time = $$\frac{(\frac{1}{3}) }{ (\frac{1}{342})} =\frac{ 342 }{ 3 }$$​= 114 minutes
It will take 114 minutes to fill one-third of the tank

Alternative Method: 

Pipe 1 can fill the tank in 18 minutes.
Pipe 2 can empty the tank in 19 minutes. 

Capacity of tank = LCM of 18 , 19 = 342

Efficiency of first Pipe  =  $$\frac{342}{18} = 19$$unit / minutes ​

Efficiency of second Pipe   = $$\frac{342}{19} = - 18$$ unit/minutes 

Time Taken to Fill One-third of the Tank  = $$\frac{1}{3} \times \frac{342}{(19-18)}$$  = 114 Minutes ​