Two pipes A and B can fill a tank in 24 minutes and 32 minutes respect...
Solution:
Let's assume that B is closed after x minutes.
Step 1: Find the work done by A in 18 minutes.
We know that A can fill the tank in 24 minutes.
Therefore, in 1 minute, A can fill 1/24th of the tank.
In 18 minutes, A can fill 18/24th of the tank.
Hence, the work done by A in 18 minutes = 18/24 = 3/4.
Step 2: Find the work done by B in (18 - x) minutes.
We know that B can fill the tank in 32 minutes.
Therefore, in 1 minute, B can fill 1/32nd of the tank.
In (18 - x) minutes, B can fill (18 - x)/32th of the tank.
Hence, the work done by B in (18 - x) minutes = (18 - x)/32.
Step 3: Find the total work done by both the pipes in 18 minutes.
Since both the pipes are opened simultaneously, they will fill the tank together.
Therefore, in 1 minute, both the pipes can fill (1/24 + 1/32)th of the tank.
In 18 minutes, both the pipes can fill (18/24 + 18/32)th of the tank.
Hence, the total work done by both the pipes in 18 minutes = (18/24 + 18/32) = 27/32.
Step 4: Equate the total work done by both the pipes with the work required to fill the tank.
Since the tank is full, the total work done by both the pipes should be equal to the work required to fill the tank.
Therefore, 27/32 = 1
Step 5: Find x.
Now, we need to find x such that B can be closed after x minutes and the tank can still be filled in 18 minutes.
We know that the work done by A in 18 minutes = 3/4.
We also know that the work done by B in (18 - x) minutes = (18 - x)/32.
Therefore, the work done by both the pipes in x minutes = 27/32 - 3/4 - (18 - x)/32.
Simplifying this equation, we get:
x/24 - (18 - x)/32 = 1/32
Solving for x, we get:
x = 8
Therefore, B should be closed after 8 minutes so that the tank can be filled in 18 minutes.
Hence, the correct answer is option B.