Pipe A alone can fill a tank in 8 h. Pipe S alone can fill it in 6 h. If both the pipes are opened and after 2 h pipe A is closed, then the other pipe will fill the tank in

Options

A. 4h
B. `2 (1)/(2) h`
C. `6h`
D. `3 (1)/(2)` h

Answer

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

### Step 1: Determine the filling rates of both pipes.
- Pipe A can fill the tank in 8 hours. Therefore, in 1 hour, it fills:
\[ \text{Rate of Pipe A} = \frac{1}{8} \text{ tanks/hour} \]
- Pipe S can fill the tank in 6 hours. Therefore, in 1 hour, it fills:
\[ \text{Rate of Pipe S} = \frac{1}{6} \text{ tanks/hour} \]

### Step 2: Find the combined filling rate of both pipes.
- When both pipes are opened, their combined rate is:
\[ \text{Combined Rate} = \frac{1}{8} + \frac{1}{6} \]
- To add these fractions, we need a common denominator. The least common multiple (LCM) of 8 and 6 is 24. Thus, we convert the rates:
\[ \frac{1}{8} = \frac{3}{24}, \quad \frac{1}{6} = \frac{4}{24} \]
- Now, adding these:
\[ \text{Combined Rate} = \frac{3}{24} + \frac{4}{24} = \frac{7}{24} \text{ tanks/hour} \]

### Step 3: Calculate how much of the tank is filled in 2 hours.
- If both pipes are opened for 2 hours, the amount filled is:
\[ \text{Amount filled in 2 hours} = 2 \times \frac{7}{24} = \frac{14}{24} = \frac{7}{12} \text{ of the tank} \]

### Step 4: Determine the remaining part of the tank.
- The total capacity of the tank is 1 (or 24 units if we consider it in units). Therefore, the remaining part of the tank is:
\[ \text{Remaining} = 1 - \frac{7}{12} = \frac{5}{12} \text{ of the tank} \]

### Step 5: Calculate the time taken by Pipe S to fill the remaining part.
- After 2 hours, Pipe A is closed, and only Pipe S is filling the tank. The rate of Pipe S is:
\[ \text{Rate of Pipe S} = \frac{1}{6} \text{ tanks/hour} \]
- To find the time taken to fill the remaining \(\frac{5}{12}\) of the tank, we use the formula:
\[ \text{Time} = \frac{\text{Remaining amount}}{\text{Rate of Pipe S}} = \frac{\frac{5}{12}}{\frac{1}{6}} \]
- Simplifying this:
\[ \text{Time} = \frac{5}{12} \times 6 = \frac{5 \times 6}{12} = \frac{30}{12} = \frac{5}{2} \text{ hours} = 2.5 \text{ hours} \]

### Final Answer:
The time taken by Pipe S to fill the remaining part of the tank is **2.5 hours**.

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