Question

A tank can be filled by two pipes A and B in 18 minutes and 24 respectively. Another tap C can empty the full tank in 36 mintues. If the tap C is opened 6 minutes after the pipes A and B are opened, the tank will become full in a total of :

• A. 6 minutes
• B. 12 minutes
• C. 18 minutes
• D. 36 minutes

Answer 1

The Correct Option is B

To solve the problem of determining when the tank will become full, we consider the rates at which the pipes fill or empty the tank. Let the total capacity of the tank be 1 unit.

Pipe A fills the tank in 18 minutes, so pipe A's rate of filling is 1/18 of the tank per minute.

Pipe B fills the tank in 24 minutes, so pipe B's rate is 1/24 of the tank per minute.

Tap C empties the tank in 36 minutes, so tap C's rate is 1/36 of the tank per minute.

The combined rate of filling for A and B is (1/18 + 1/24). To calculate this, find a common denominator:

$\frac{1}{18}+\frac{1}{24}=\frac{4+3}{72}=\frac{7}{72}$

Thus, the filling rate by A and B is 7/72 of the tank per minute.

Since tap C is opened 6 minutes later, during the first 6 minutes only A and B are working:

6 minutes * 7/72 = 7/12 of the tank filled.

After 6 minutes, tap C is opened, introducing a negative rate. The net rate when A, B, and C are all open is:

Net rate = (7/72) - (1/36).

Simplifying using a common denominator:

$\frac{7}{72}-\frac{1}{36}=\frac{7}{72}-\frac{2}{72}=\frac{5}{72}$

Thus, A, B, and C together fill 5/72 of the tank per minute after 6 minutes.

Let x be the time taken after 6 minutes when all three pipes are operating:

x * 5/72 = 5/12, because 5/12 tank still needs to be filled.

Simplifying gives x = 5/12 * 72/5 = 12 minutes.

The total time is 6 minutes (initial) + 12 minutes = 18 minutes. Thus, the correct choice is 12 minutes.