Pipes A and B when working together, can fill an empty tank in 24 hours. They work together for 8 hours and then B is 1 stopped but A continued to work. It took a total of 28 hours to fill the tank. How long would it take A to fill the empty tank alone?

Options

A. 30 hours
B. 29 hours
C. 28 hours
D. 31 hours

Answer

To solve the problem step by step, we will first define the rates at which pipes A and B can fill the tank, then use the information provided to find out how long it takes for pipe A to fill the tank alone.

### Step 1: Define the rates of work
Let:
- \( X \) = time taken by pipe A to fill the tank alone (in hours)
- \( Y \) = time taken by pipe B to fill the tank alone (in hours)

From the problem, we know that when both pipes A and B work together, they can fill the tank in 24 hours. Therefore, we can express their combined rate of work as:
\[ \frac{1}{X} + \frac{1}{Y} = \frac{1}{24} \]

### Step 2: Work done by A and B together for 8 hours
When A and B work together for 8 hours, the amount of work done can be calculated as:
\[ \text{Work done in 8 hours} = 8 \left( \frac{1}{X} + \frac{1}{Y} \right) = 8 \times \frac{1}{24} = \frac{1}{3} \]

This means that after 8 hours of working together, \( \frac{1}{3} \) of the tank is filled.

### Step 3: Work remaining after B stops
After 8 hours, B stops working, and A continues to fill the tank alone. The total time taken to fill the tank is 28 hours, which means A worked alone for:
\[ 28 - 8 = 20 \text{ hours} \]

### Step 4: Work done by A alone
The amount of work done by A in 20 hours can be expressed as:
\[ \text{Work done by A in 20 hours} = 20 \times \frac{1}{X} \]

### Step 5: Total work done
The total work done to fill the tank is equal to 1 (the whole tank):
\[ \text{Work done together} + \text{Work done by A alone} = 1 \]
Substituting the values we have:
\[ \frac{1}{3} + 20 \times \frac{1}{X} = 1 \]

### Step 6: Solve for \( X \)
Now, we can solve for \( X \):
\[ 20 \times \frac{1}{X} = 1 - \frac{1}{3} \]
\[ 20 \times \frac{1}{X} = \frac{2}{3} \]
\[ \frac{20}{X} = \frac{2}{3} \]
Cross-multiplying gives:
\[ 20 \times 3 = 2 \times X \]
\[ 60 = 2X \]
\[ X = \frac{60}{2} = 30 \]

### Conclusion
Thus, pipe A alone would take **30 hours** to fill the empty tank.

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