A tank has two inlet pipes which can fill the empty tank in 12 hours and 15 hours working alone and one outlet pipe which can empty the full tank in 8 hours working alone. The inlet pipes are kept open for all the time hut the outlet pipe was opened after 2 hours for one hour and then again closed for 2 hours then once again opened for one hour. This pattern of outlet pipe continued till the tank got completely filled. In how many hours the tank has been filled, working on the given pattern?

Options

A. a)8 hours 24 minutes
B. b)10 hours 15 minutes
C. c)9 hours 10 minutes
D. d)9 hours 6 minutes

Answer

To solve the problem step by step, we need to analyze the rates of the inlet and outlet pipes and how they work together over time.

### Step 1: Determine the rates of the pipes
1. **Inlet Pipe A** can fill the tank in 12 hours. Therefore, its rate is:
\[ \text{Rate of A} = \frac{1}{12} \text{ tank/hour} \]

2. **Inlet Pipe B** can fill the tank in 15 hours. Therefore, its rate is:
\[ \text{Rate of B} = \frac{1}{15} \text{ tank/hour} \]

3. **Outlet Pipe C** can empty the tank in 8 hours. Therefore, its rate is:
\[ \text{Rate of C} = -\frac{1}{8} \text{ tank/hour} \]

### Step 2: Find the combined rate of the inlet pipes
The combined rate of pipes A and B when they are both open is:
\[ \text{Combined Rate of A and B} = \frac{1}{12} + \frac{1}{15} \]
To add these fractions, we find a common denominator, which is 60:
\[ \text{Combined Rate of A and B} = \frac{5}{60} + \frac{4}{60} = \frac{9}{60} = \frac{3}{20} \text{ tank/hour} \]

### Step 3: Analyze the filling and emptying pattern
The pattern of operation is as follows:
- **For the first 2 hours**, both A and B are open:
\[ \text{Water filled in 2 hours} = 2 \times \frac{3}{20} = \frac{3}{10} \text{ tank} \]

- **For the next 1 hour**, A and B are open, and C is also open:
\[ \text{Water filled in 1 hour} = \left(\frac{3}{20} - \frac{1}{8}\right) \text{ tank} \]
To subtract, we find a common denominator (40):
\[ \frac{3}{20} = \frac{6}{40}, \quad -\frac{1}{8} = -\frac{5}{40} \]
Thus,
\[ \text{Water filled in 1 hour} = \frac{6}{40} - \frac{5}{40} = \frac{1}{40} \text{ tank} \]

### Step 4: Calculate the total water filled in a cycle
In one complete cycle of 3 hours (2 hours of filling and 1 hour of emptying), the total water filled is:
\[ \text{Total water filled in 3 hours} = \frac{3}{10} + \frac{1}{40} \]
Finding a common denominator (40):
\[ \frac{3}{10} = \frac{12}{40} \]
Thus,
\[ \text{Total water filled in 3 hours} = \frac{12}{40} + \frac{1}{40} = \frac{13}{40} \text{ tank} \]

### Step 5: Determine how many cycles are needed to fill the tank
Let’s find out how many cycles are needed to fill the tank:
- The tank is 1 full unit, so we need:
\[ \text{Number of cycles} = \frac{1}{\frac{13}{40}} = \frac{40}{13} \approx 3.08 \text{ cycles} \]
This means we need 3 complete cycles and a part of the 4th cycle.

### Step 6: Calculate the total time for 3 complete cycles
Each cycle takes 3 hours, so for 3 cycles:
\[ \text{Time for 3 cycles} = 3 \times 3 = 9 \text{ hours} \]

### Step 7: Calculate the remaining water to fill in the 4th cycle
After 3 cycles, the amount of water filled is:
\[ \text{Water filled} = 3 \times \frac{13}{40} = \frac{39}{40} \text{ tank} \]
Remaining water to fill:
\[ \text{Remaining water} = 1 - \frac{39}{40} = \frac{1}{40} \text{ tank} \]

### Step 8: Calculate the time to fill the remaining water
In the 4th cycle, for the first 2 hours, A and B will fill:
\[ \text{Water filled in 2 hours} = 2 \times \frac{3}{20} = \frac{3}{10} \text{ tank} \]
Since \(\frac{3}{10} = \frac{12}{40}\), this is more than \(\frac{1}{40}\). Therefore, we only need a fraction of this time.

To find the time to fill \(\frac{1}{40}\):
\[ \text{Time needed} = \frac{\frac{1}{40}}{\frac{3}{20}} = \frac{1}{40} \times \frac{20}{3} = \frac{1}{6} \text{ hours} = 10 \text{ minutes} \]

### Final Calculation
Total time taken to fill the tank:
\[ \text{Total time} = 9 \text{ hours} + \frac{1}{6} \text{ hours} = 9 \text{ hours} + 10 \text{ minutes} \]

Thus, the tank is completely filled in **9 hours and 10 minutes**.