Pipe A can fill a tank in 12 hours and pipe B can empty the tank in 18 hours. Both pipes are opened at 6 a.m. and after some time pipe B is closed and the tank is full at 8 p.m. At what time was the pipe B closed ?

Options

A. 10 a.m.
B. 8 a.m.
C. 9 a.m.
D. 11 a.m.

Explanation

To solve the problem step by step, we will first determine the rates at which pipes A and B work, then calculate how long pipe B was open before it was closed.

### Step 1: Determine the filling and emptying rates of the pipes.
- **Pipe A** can fill the tank in 12 hours. Therefore, in 1 hour, it fills:
\[ \text{Rate of Pipe A} = \frac{1 \text{ tank}}{12 \text{ hours}} = \frac{1}{12} \text{ tank/hour} \]
- **Pipe B** can empty the tank in 18 hours. Therefore, in 1 hour, it empties:
\[ \text{Rate of Pipe B} = \frac{1 \text{ tank}}{18 \text{ hours}} = \frac{1}{18} \text{ tank/hour} \]

### Step 2: Convert rates to a common unit.
To make calculations easier, we can find a common capacity for the tank. We can take the Least Common Multiple (LCM) of the times taken by both pipes:
- LCM of 12 and 18 is 36.

Thus, we can consider the tank capacity as 36 units.
- In 1 hour, Pipe A fills:
\[ \frac{36 \text{ units}}{12 \text{ hours}} = 3 \text{ units/hour} \]
- In 1 hour, Pipe B empties:
\[ \frac{36 \text{ units}}{18 \text{ hours}} = 2 \text{ units/hour} \]

### Step 3: Calculate the net effect when both pipes are open.
When both pipes are open, the net effect is:
\[ \text{Net Rate} = \text{Rate of A} - \text{Rate of B} = 3 \text{ units/hour} - 2 \text{ units/hour} = 1 \text{ unit/hour} \]

### Step 4: Determine the total time from 6 a.m. to 8 p.m.
From 6 a.m. to 8 p.m. is a total of:
\[ 8 \text{ p.m.} - 6 \text{ a.m.} = 14 \text{ hours} \]

### Step 5: Let \( x \) be the number of hours both pipes are open.
Let \( x \) be the number of hours both pipes are open before Pipe B is closed. After that, only Pipe A is working until the tank is full.

### Step 6: Calculate the amount of work done.
The work done by both pipes together for \( x \) hours is:
\[ \text{Work done by A and B} = 1 \text{ unit/hour} \times x \text{ hours} = x \text{ units} \]
After Pipe B is closed, Pipe A continues to fill the tank for the remaining \( 14 - x \) hours:
\[ \text{Work done by A alone} = 3 \text{ units/hour} \times (14 - x) \text{ hours} = 3(14 - x) \text{ units} \]

### Step 7: Set up the equation for the total work done.
Since the total work done must equal the capacity of the tank (36 units), we can write:
\[ x + 3(14 - x) = 36 \]

### Step 8: Solve the equation.
Expanding the equation:
\[ x + 42 - 3x = 36 \]
Combining like terms:
\[ -2x + 42 = 36 \]
Subtracting 42 from both sides:
\[ -2x = 36 - 42 \]
\[ -2x = -6 \]
Dividing by -2:
\[ x = 3 \]

### Step 9: Determine the time when Pipe B was closed.
Since Pipe B was open for \( x = 3 \) hours after 6 a.m., it was closed at:
\[ 6 \text{ a.m.} + 3 \text{ hours} = 9 \text{ a.m.} \]

### Final Answer:
Pipe B was closed at **9 a.m.**.

---