Pipes A and B are emptying pipes and can empty a tank in 6 hours and 16 hours, respectively. C is a filling pipe.All the three pipes were opened together. They took 80 minutes to empty 5/18th of the tank. Pipe C alone can fill the tank in `:`

Options

A. 40 hours
B. 48 hours
C. 36 hours
D. 42 hours

Answer

To solve the problem step by step, we need to find out how long pipe C takes to fill the tank given the rates at which pipes A and B empty the tank.

### Step 1: Determine the rates of pipes A and B
- Pipe A can empty the tank in 6 hours. Therefore, its rate of emptying is:
\[ \text{Rate of A} = \frac{1}{6} \text{ tank/hour} \]

- Pipe B can empty the tank in 16 hours. Therefore, its rate of emptying is:
\[ \text{Rate of B} = \frac{1}{16} \text{ tank/hour} \]

### Step 2: Calculate the combined rate of pipes A and B
To find the combined rate of pipes A and B, we add their rates:
\[ \text{Combined rate of A and B} = \frac{1}{6} + \frac{1}{16} \]

To add these fractions, we need a common denominator. The least common multiple of 6 and 16 is 48.
\[ \frac{1}{6} = \frac{8}{48}, \quad \frac{1}{16} = \frac{3}{48} \]
So,
\[ \text{Combined rate of A and B} = \frac{8}{48} + \frac{3}{48} = \frac{11}{48} \text{ tank/hour} \]

### Step 3: Determine the time taken to empty part of the tank
We know that pipes A and B, along with pipe C, took 80 minutes to empty \(\frac{5}{18}\) of the tank. First, convert 80 minutes to hours:
\[ 80 \text{ minutes} = \frac{80}{60} = \frac{4}{3} \text{ hours} \]

### Step 4: Calculate the total rate when all pipes are open
Let the rate of pipe C (which fills the tank) be \(r_C\) (in tank/hour). The combined rate when all three pipes are open is:
\[ \text{Combined rate} = \frac{11}{48} - r_C \]

Using the time taken to empty \(\frac{5}{18}\) of the tank, we can set up the equation:
\[ \left(\frac{11}{48} - r_C\right) \cdot \frac{4}{3} = -\frac{5}{18} \]

### Step 5: Solve for \(r_C\)
First, simplify the left side:
\[ \frac{4}{3} \cdot \left(\frac{11}{48} - r_C\right) = -\frac{5}{18} \]

Multiply both sides by 3 to eliminate the fraction:
\[ 4 \cdot \left(\frac{11}{48} - r_C\right) = -\frac{5}{6} \]

Now distribute the 4:
\[ \frac{44}{48} - 4r_C = -\frac{5}{6} \]

Convert \(\frac{44}{48}\) to \(\frac{11}{12}\) for easier calculations:
\[ \frac{11}{12} - 4r_C = -\frac{5}{6} \]

Convert \(-\frac{5}{6}\) to twelfths:
\[ -\frac{5}{6} = -\frac{10}{12} \]

Now, we have:
\[ \frac{11}{12} - 4r_C = -\frac{10}{12} \]

### Step 6: Isolate \(r_C\)
Rearranging gives:
\[ 4r_C = \frac{11}{12} + \frac{10}{12} = \frac{21}{12} \]
\[ r_C = \frac{21}{48} = \frac{7}{16} \text{ tank/hour} \]

### Step 7: Calculate the time for pipe C to fill the tank
If pipe C fills the tank at a rate of \(\frac{7}{16}\) tank/hour, the time taken by pipe C to fill the entire tank is:
\[ \text{Time} = \frac{1 \text{ tank}}{r_C} = \frac{1}{\frac{7}{16}} = \frac{16}{7} \text{ hours} \]

### Final Answer
Pipe C alone can fill the tank in \(\frac{16}{7}\) hours, or approximately 2 hours and 17 minutes.