Pipes A and B can fill an empty cistern in 78 minutes and 39 minutes, respectively. Both Pipe A and Pipe B are opened together. After how much time should Pipe A be turned off so that the empty cistern is completely filled in a total of 26 minutes?

Correct option is A

Given:

Pipe A fills the tank in 78 minutes

Pipe B fills the tank in 39 minutes

Both are opened together

Total time to fill the tank = 25 minutes

Find: After how many minutes Pipe A should be turned off (using LCM method)

Formula Used:

Work done = Efficiency × Time

Solution:

Total Work = LCM(78, 39) = 78 units

Pipe A efficiency = $$\frac{78}{78}$$​ = 1 unit/min

Pipe B efficiency = $$\frac{78}{ 39}$$​ = 2 units/min

Then Pipe B works for full 26 minutes.

Work done by B in 26 minutes = 26 × 2 = 52 unit 

Remaining work = 78 - 52 = 26 unit 

Time taken by A to complete 26 unit work;

= $$\frac{26}{1}$$ = 26 min.

Thus, Pipe A should be turned off after 26 minutes.