Question

Two pipes A and B can fill a tank in 12 minutes and 15 minutes respectively while a third pipe C can empty the full tank in 20 minutes. All the three pipes are opened in the beginning but pipe C is closed 6 minutes before the tank is filled. In what time will the tank is full?

• A. 15 minutes
• B. 10 minutes
• C. 7 minutes
• D. 17 minutes

Hint:

All three pipes fill \(\frac{1}{10}\) per minute. A and B alone fill \(\frac{3}{20}\) per minute for the last 6 minutes.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
In pipe problems we find the part of the tank each pipe fills in one minute. Filling pipes give a positive part and an emptying pipe gives a negative part.

Step 2: Find the one minute work.
Pipe A fills \(\frac{1}{12}\), pipe B fills \(\frac{1}{15}\), pipe C empties \(\frac{1}{20}\) of the tank per minute.
All three together: \[ \frac{1}{12} + \frac{1}{15} - \frac{1}{20} = \frac{5 + 4 - 3}{60} = \frac{6}{60} = \frac{1}{10} \]
A and B alone: \[ \frac{1}{12} + \frac{1}{15} = \frac{9}{60} = \frac{3}{20} \]

Step 3: Set up the equation.
Let the total time be \(T\) minutes. All three pipes work for \(T - 6\) minutes and only A and B work for the last 6 minutes.
\[ \frac{T - 6}{10} + 6 \times \frac{3}{20} = 1 \]
\[ \frac{T - 6}{10} + 0.9 = 1 \]
\[ T - 6 = 1 \]
\[ T = 7 \]

Step 4: Check the options.
In 7 minutes, the first minute gives \(\frac{1}{10}\) and the last 6 minutes give \(\frac{18}{20} = 0.9\). The total is 1, so the tank is exactly full. Options 15, 10 and 17 minutes are too long, because even with C open the tank fills in 10 minutes, and closing C only makes it faster.

Final Answer:
The tank is full in 7 minutes, which is option 3. \[ \boxed{7 \text{ minutes}} \]