Correct option is A
Given:
- Pipe A can fill the tank in 40 minutes.
- Pipe B can fill the tank in 60 minutes.
- Both pipes are opened together, and the tank should be filled in 30 minutes.
Solution:
We will use the LCM method to calculate the time after which Pipe B should be turned off.
Let the time for which Pipe B is open be 't' minutes.
The work done by Pipe A in 1 minute = 1/40
The work done by Pipe B in 1 minute = 1/60
We need to calculate 't' such that the total work done in 30 minutes equals 1 (the whole tank).
Work done by Pipe A in 30 minutes = 30 × (1/40) = 30/40 = 3/4
Now, let Pipe B be open for 't' minutes. Work done by Pipe B in 't' minutes = t × (1/60) = t/60
The total work done should be 1 (i.e., the tank should be full):
Work done by Pipe A + Work done by Pipe B = 1
minutes
Pipe B should be turned off after 15 minutes so that the tank is filled in 30 minutes.