There are three pipes connected with a tank. The first pipe can fill 1/2 part of the tank in 1 h, second pipe can fill 1/3 part of the tank in 1 h. Third pipe is connected to empty the tank. After opening all the three pipes, 7/12 part of the tank can be filled in 1 h, then how long will third pipe take to empty the full tank?

Options

A. 3h
B. 4h
C. 5h
D. 6h

Answer

To solve the problem step by step, we will analyze the information given about the three pipes and calculate the time taken by the third pipe to empty the full tank.

### Step 1: Determine the filling rates of the first two pipes.
- The first pipe (let's call it Pipe A) can fill 1/2 of the tank in 1 hour. Therefore, its filling rate is:
\[ \text{Rate of Pipe A} = \frac{1}{2} \text{ tank/hour} \]

- The second pipe (let's call it Pipe B) can fill 1/3 of the tank in 1 hour. Therefore, its filling rate is:
\[ \text{Rate of Pipe B} = \frac{1}{3} \text{ tank/hour} \]

### Step 2: Combine the filling rates of the first two pipes.
To find the combined filling rate of Pipes A and B:
\[ \text{Combined Rate of A and B} = \frac{1}{2} + \frac{1}{3} \]
To add these fractions, we need a common denominator. The least common multiple of 2 and 3 is 6:
\[ \text{Combined Rate of A and B} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6} \text{ tank/hour} \]

### Step 3: Determine the net filling rate when all three pipes are open.
It is given that when all three pipes are opened, they fill 7/12 of the tank in 1 hour. Therefore, the net filling rate when all pipes are open is:
\[ \text{Net Rate} = \frac{7}{12} \text{ tank/hour} \]

### Step 4: Set up the equation to find the rate of the third pipe.
Let the rate of the third pipe (which empties the tank) be denoted as \( C \) (in tank/hour). The net filling rate can be expressed as:
\[ \text{Net Rate} = \text{Rate of A} + \text{Rate of B} - \text{Rate of C} \]
Substituting the known values:
\[ \frac{7}{12} = \frac{5}{6} - C \]

### Step 5: Solve for \( C \).
First, convert \( \frac{5}{6} \) to a fraction with a denominator of 12:
\[ \frac{5}{6} = \frac{10}{12} \]
Now substitute:
\[ \frac{7}{12} = \frac{10}{12} - C \]
Rearranging gives:
\[ C = \frac{10}{12} - \frac{7}{12} = \frac{3}{12} = \frac{1}{4} \text{ tank/hour} \]

### Step 6: Calculate the time taken by the third pipe to empty the tank.
The time taken by the third pipe to empty the full tank can be calculated using the formula:
\[ \text{Time} = \frac{\text{Capacity of the tank}}{\text{Rate of C}} \]
Assuming the capacity of the tank is 1 tank unit:
\[ \text{Time} = \frac{1}{\frac{1}{4}} = 4 \text{ hours} \]

### Final Answer:
The third pipe will take **4 hours** to empty the full tank.

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