Pipes `A` and `B` can fill a tank in `8` hours and `12` hours, respectively. `C` is an outlet pipe. When all the three are opened together, the tank is filled in `13 1/3` hours. `C` alone can empty the full tank in-

Options

A. `6 1/2` hours
B. `7 1/2` hours
C. `8` hours
D. `6` hours

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Determine the rates of pipes A and B
- Pipe A can fill the tank in 8 hours, so its rate is:
\[ \text{Rate of A} = \frac{1}{8} \text{ tank per hour} \]
- Pipe B can fill the tank in 12 hours, so its rate is:
\[ \text{Rate of B} = \frac{1}{12} \text{ tank per hour} \]

### Step 2: Determine the total time taken when all pipes are opened
- When all three pipes (A, B, and C) are opened together, the tank is filled in \( 13 \frac{1}{3} \) hours, which can be converted to an improper fraction:
\[ 13 \frac{1}{3} = \frac{40}{3} \text{ hours} \]
- Therefore, the combined rate of A, B, and C is:
\[ \text{Combined Rate} = \frac{1}{\frac{40}{3}} = \frac{3}{40} \text{ tank per hour} \]

### Step 3: Set up the equation for the combined rates
- The equation for the combined rates of A, B, and C is:
\[ \text{Combined Rate} = \text{Rate of A} + \text{Rate of B} - \text{Rate of C} \]
- Substituting the known values:
\[ \frac{3}{40} = \frac{1}{8} + \frac{1}{12} - \text{Rate of C} \]

### Step 4: Find a common denominator for A and B
- The least common multiple (LCM) of 8 and 12 is 24. Therefore, we convert the rates:
\[ \frac{1}{8} = \frac{3}{24}, \quad \frac{1}{12} = \frac{2}{24} \]
- Now, substituting these into the equation:
\[ \frac{3}{40} = \frac{3}{24} + \frac{2}{24} - \text{Rate of C} \]
- Simplifying the right side:
\[ \frac{3}{40} = \frac{5}{24} - \text{Rate of C} \]

### Step 5: Isolate Rate of C
- Rearranging gives:
\[ \text{Rate of C} = \frac{5}{24} - \frac{3}{40} \]

### Step 6: Find a common denominator for the fractions
- The LCM of 24 and 40 is 120. Converting both fractions:
\[ \frac{5}{24} = \frac{25}{120}, \quad \frac{3}{40} = \frac{9}{120} \]
- Now substituting back:
\[ \text{Rate of C} = \frac{25}{120} - \frac{9}{120} = \frac{16}{120} = \frac{2}{15} \]

### Step 7: Determine the time taken by C alone to empty the tank
- Since Rate of C is \( \frac{2}{15} \) tank per hour, the time taken by C to empty the tank is the reciprocal:
\[ \text{Time taken by C} = \frac{1}{\frac{2}{15}} = \frac{15}{2} = 7.5 \text{ hours} \]

### Final Answer
C alone can empty the full tank in **7.5 hours** or **7 hours and 30 minutes**.

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