Pipe A can fill an empty cistern in 5 hours while Pipe B takes 6.25 hours to fill it. The two pipes are opened 6 simultaneously when the cistem is empty, but Pipe B is turned off after some time as a result of which the cistem is illed in 3.4 hours. For how many hours was Pipe B open?

Options

A. `1.5`
B. 1
C. 2
D. `2.5`

Answer

To solve the problem step by step, we will first determine the rates at which each pipe can fill the cistern, then set up an equation based on the information given in the problem.

### Step 1: Determine the rates of filling for each pipe.
- Pipe A can fill the cistern in 5 hours. Therefore, the rate of Pipe A is:
\[ \text{Rate of Pipe A} = \frac{1}{5} \text{ cisterns per hour} \]

- Pipe B can fill the cistern in 6.25 hours. Therefore, the rate of Pipe B is:
\[ \text{Rate of Pipe B} = \frac{1}{6.25} = \frac{1}{\frac{25}{4}} = \frac{4}{25} \text{ cisterns per hour} \]

### Step 2: Set up the equation based on the total filling time.
Let \( x \) be the number of hours Pipe B is open. Since both pipes are open for 3.4 hours, we can express the total work done as:
\[ \text{Work done by Pipe A in 3.4 hours} + \text{Work done by Pipe B in } x \text{ hours} = 1 \text{ cistern} \]
This can be written as:
\[ \left( \frac{1}{5} \times 3.4 \right) + \left( \frac{4}{25} \times x \right) = 1 \]

### Step 3: Calculate the work done by Pipe A.
Calculating the work done by Pipe A:
\[ \frac{1}{5} \times 3.4 = \frac{3.4}{5} = 0.68 \text{ cisterns} \]

### Step 4: Substitute and simplify the equation.
Now substitute this value into the equation:
\[ 0.68 + \frac{4}{25}x = 1 \]

### Step 5: Solve for \( x \).
Subtract 0.68 from both sides:
\[ \frac{4}{25}x = 1 - 0.68 \]
\[ \frac{4}{25}x = 0.32 \]

Now, multiply both sides by \( \frac{25}{4} \) to solve for \( x \):
\[ x = 0.32 \times \frac{25}{4} \]
\[ x = 0.32 \times 6.25 = 2 \]

### Conclusion
Pipe B was open for **2 hours**.