Pipe A can fill the tank in 4 hours, while pipe B can fill it in ’6 hours working separately. Pipe C can empty whole the tank in 4 hours. He opened the pipe A and B simultaneously to fill the empty tank. He wanted to adjust his alarm so that he could open the -pipe C when it was half-filled, but he mistakenely adjusted his alarm at a time when his tank would be `3//4` th filled. What is the time difference between both the cases, to fill, the tank fully

Options

A. 48 min
B. 54 min
C. 30 min
D. none of these

Answer

To solve the problem, we will calculate the time taken to fill the tank in two scenarios: when the tank is filled to half and when it is filled to three-fourths.

### Step 1: Determine the filling rates of pipes A, B, and C

- Pipe A can fill the tank in 4 hours. Therefore, its rate of filling is:
\[ \text{Rate of A} = \frac{1}{4} \text{ tank/hour} \]

- Pipe B can fill the tank in 6 hours. Therefore, its rate of filling is:
\[ \text{Rate of B} = \frac{1}{6} \text{ tank/hour} \]

- Pipe C can empty the tank in 4 hours. Therefore, its rate of emptying is:
\[ \text{Rate of C} = \frac{1}{4} \text{ tank/hour} \]

### Step 2: Calculate the combined filling rate of pipes A and B

The combined rate of pipes A and B when they are working together is:
\[ \text{Combined Rate of A and B} = \frac{1}{4} + \frac{1}{6} \]

To add these fractions, we find a common denominator (which is 12):
\[ \text{Combined Rate of A and B} = \frac{3}{12} + \frac{2}{12} = \frac{5}{12} \text{ tank/hour} \]

### Step 3: Calculate the time taken to fill half the tank

To fill half the tank (which is 0.5 of the tank), we use the combined rate of A and B:
\[ \text{Time to fill half the tank} = \frac{0.5}{\frac{5}{12}} = 0.5 \times \frac{12}{5} = \frac{6}{5} \text{ hours} = 1.2 \text{ hours} \]

### Step 4: Calculate the time taken to fill three-fourths of the tank

To fill three-fourths of the tank (which is 0.75 of the tank), we again use the combined rate of A and B:
\[ \text{Time to fill three-fourths of the tank} = \frac{0.75}{\frac{5}{12}} = 0.75 \times \frac{12}{5} = \frac{9}{5} \text{ hours} = 1.8 \text{ hours} \]

### Step 5: Calculate the remaining filling time after reaching three-fourths

After the tank is three-fourths filled, we need to fill the remaining one-fourth (0.25 of the tank) with all three pipes (A, B, and C):
\[ \text{Combined Rate of A, B, and C} = \frac{1}{4} + \frac{1}{6} - \frac{1}{4} = \frac{5}{12} - \frac{3}{12} = \frac{2}{12} = \frac{1}{6} \text{ tank/hour} \]

Now, we calculate the time to fill the remaining one-fourth:
\[ \text{Time to fill remaining one-fourth} = \frac{0.25}{\frac{1}{6}} = 0.25 \times 6 = 1.5 \text{ hours} \]

### Step 6: Calculate total time for both scenarios

- **Total time for filling half the tank**:
\[ \text{Total Time (half)} = 1.2 \text{ hours} \]

- **Total time for filling three-fourths of the tank**:
\[ \text{Total Time (three-fourths)} = 1.8 + 1.5 = 3.3 \text{ hours} \]

### Step 7: Calculate the difference in time

Now, we find the difference in time taken to fill the tank fully in both cases:
\[ \text{Time Difference} = 3.3 \text{ hours} - 1.2 \text{ hours} = 2.1 \text{ hours} = 2 \text{ hours} + 6 \text{ minutes} \]

### Final Answer

The time difference between both cases to fill the tank fully is **2 hours and 6 minutes**.

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