Pipes A and B can fill an empty cistern in 86 minutes and 43 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 42 minutes?

Correct option is D

Given:

Pipe A fills the cistern in 86 minutes, so its filling rate is $$\frac{1}{86}$$ cisterns per minute.

Pipe B fills the cistern in 43 minutes, so its filling rate is $$\frac{1}{43}$$ cisterns per minute.

The total time to fill the cistern is 42 minutes.

Formula Used:

Rate =$$\frac{\text{Work}}{\text{Time}}$$​​

Solution:

Let t be the time for which Pipe A is open. Pipe B works for the entire 42 minutes.

In t minutes, Pipe A fills $$\frac{t}{86}$$​ of the cistern.

In 42 minutes, Pipe B fills $$\frac{42}{43}$$​ of the cistern.

The total work done must add up to 1:

​$$\frac{t}{86} + \frac{42}{43} = 1$$​

​$$\frac{t}{86} = 1 - \frac{42}{43} = \frac{1}{43}$$​​

t =$$\frac{86}{43} =2 \text{ minutes}$$​

Thus, Pipe A should be turned off after 2 minutes.

Alternate Method:

 LCM of 86 and 43 is 86 (since 43 is a factor of 86), so total work = 86 units.

Pipe A's 1 min work =$$\frac{ 86}{86}$$​= 1 unit/min

Pipe B's 1 min work = $$\frac{86}{43}$$​= 2 units/min

Let Pipe A run for x minutes with B. Then, A is turned off, and only B runs for (42 - x) minutes.

 Work done by A and B in x minutes = 3x units (since together they do 1 + 2 = 3 units/min)

Work done by B alone in remaining (42 - x) minutes = 2(42 - x) units

For the cistern to be full: Total work = Work by both + Work by B alone

3x + 2(42 - x) = 86
3x + 84 - 2x = 86
x + 84 = 86
x = 2 minutes

Thus, Pipe A should be turned off after 2 minutes.