Pipes A and B can fill an empty cistern in 78 minutes and 24 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 20 minutes?

Correct option is B

Given:

Pipe A can fill the tank in = 78 minutes

Pipe B can fill the tank in = 24 minutes

Both pipes are opened together initially

Total time to completely fill the tank = 20 minutes

We need to find: After how many minutes should Pipe A be turned off so that the tank fills in exactly 20 minutes.

Formula Used:

Work done = Rate × Time

Solution:

Let Pipe A be turned off after x minutes.

So,

In x minutes, work done by Pipe A =$$\frac{x}{78}$$​​

In all 20 minutes, work done by Pipe B =$$\frac{20}{24} = \frac{5}{6}$$​​

According to the question:

​$$\frac{x}{78} + \frac{5}{6} = 1$$​

​$$\frac{x}{78} = 1 - \frac{5}{6} = \frac{1}{6}$$​

x $$= \frac{78}{6}$$​= 13 minutes

Alternate Method:

Assume total work = LCM of 78 and 24 = 312 units 

Pipe A fills the tank in 78 min => A’s efficiency =$$\frac{312}{78}$$​ = 4 units/min

Pipe B fills the tank in 24 min => B’s efficiency =$$\frac{312}{24}$$​= 13 units/min

Let A be turned off after x minutes

Then:

A works for x min => Work done by A = 4x

B works for full 20 min => Work done by B = 13 × 20 = 260

Total work done = 4x + 260 = 312

4x = 312 - 260 = 52

x =$$\frac{52}{4}$$​= 13 minutes