Two pipes A and B can fill a cistern in 36 minutes and 45 minutes respectively. These two pipes are opened simultaneously in the cistern. After what time should A be closed so that the cistern is filled in 25 minutes?

Correct option is D

Given:
Pipe A can fill the cistern in 36 minutes
Pipe B can fill the cistern in 45 minutes
Both pipes are opened together
The cistern is to be filled in 25 minutes

Concept Used:

The rate of work of each person is the reciprocal of the time they take to complete the work. example,
A complete work in x day,
Rate of work in one day = $$\frac{1}{x}$$​ day

Solution:

Let A is to be closed after x minutes, and B runs for full 25 minutes.

Let the total work (i.e., filling the tank) = 1 unit

Work done by A in 1 min = $$\frac{1}{36}$$​​

Work done by B in 1 min = $$\frac{1}{45}$$

Total Work = Work by A in x min + Work by B in 25 min

$$1 = \frac{x}{36} + \frac{25}{45}$$​​

​$$1 = \frac{x}{36} + \frac{5}{9}$$

$$\frac{x}{36} = 1 - \frac{5}{9} = \frac{9-5}{9} = \frac{4}{9}$$

$$x = \frac{4}{9} \times 36 = 16$$ min

So, Pipe A should be closed after 16 minutes.

Thus, the correct answer is (d).

Alternative Method:

The cistern filled by pipe B in 25 minutes = 25 $$\times$$​ 4 = 100
Remaining capacity of cistern = 180 - 100 = 80
So, the remaining capacity must be filled by A,
Time taken by pipe A = $$\frac{80}{5}$$​ = 16 minutes