Two pipes A and B can fill an empty cistern in 1.8 and 2.7 hours, respectively. Pipe C can drain the entire cistern in 4.5 1 hours when no other pipe is in operation. Initially when the cistern was empty Pipe A and Pipe C were turned on. After a few hours Pipe A was fumed off and Pipe B was turned on instantly. In all it took 5.S hours to fill the cistern. For how many hours was Pipe B turned on?

Options

A. 5
B. 4.5
C. 3
D. 2.7

Answer

To solve the problem, we need to determine how long Pipe B was turned on while filling the cistern. Let's break down the solution step by step.

### Step 1: Determine the filling and draining rates of the pipes
- **Pipe A** can fill the cistern in 1.8 hours. Therefore, its rate of filling is:
\[ \text{Rate of A} = \frac{1 \text{ cistern}}{1.8 \text{ hours}} = \frac{5}{9} \text{ cisterns per hour} \]

- **Pipe B** can fill the cistern in 2.7 hours. Therefore, its rate of filling is:
\[ \text{Rate of B} = \frac{1 \text{ cistern}}{2.7 \text{ hours}} = \frac{10}{27} \text{ cisterns per hour} \]

- **Pipe C** can drain the cistern in 4.5 hours. Therefore, its rate of draining is:
\[ \text{Rate of C} = \frac{1 \text{ cistern}}{4.5 \text{ hours}} = \frac{2}{9} \text{ cisterns per hour} \]

### Step 2: Set up the equations for the filling process
Let \( x \) be the time (in hours) that Pipe A was on, and \( y \) be the time (in hours) that Pipe B was on. According to the problem, the total time taken to fill the cistern is 5.5 hours:
\[ x + y = 5.5 \]

### Step 3: Calculate the total amount of water filled
During the time Pipe A and Pipe C are on, the effective filling rate is:
\[ \text{Effective rate of A and C} = \text{Rate of A} - \text{Rate of C} = \frac{5}{9} - \frac{2}{9} = \frac{3}{9} = \frac{1}{3} \text{ cisterns per hour} \]
Thus, the amount of water filled by Pipe A and drained by Pipe C in \( x \) hours is:
\[ \text{Water filled by A and C} = \left(\frac{1}{3}\right) x \]

When Pipe B is turned on, the effective filling rate becomes:
\[ \text{Effective rate of B and C} = \text{Rate of B} - \text{Rate of C} = \frac{10}{27} - \frac{2}{9} = \frac{10}{27} - \frac{6}{27} = \frac{4}{27} \text{ cisterns per hour} \]
Thus, the amount of water filled by Pipe B and drained by Pipe C in \( y \) hours is:
\[ \text{Water filled by B and C} = \left(\frac{4}{27}\right) y \]

### Step 4: Set up the equation for the total water filled
The total amount of water filled in the cistern is 1 (the full cistern):
\[ \left(\frac{1}{3}\right) x + \left(\frac{4}{27}\right) y = 1 \]

### Step 5: Substitute \( y \) from the first equation into the second equation
From \( x + y = 5.5 \), we can express \( y \) as:
\[ y = 5.5 - x \]
Substituting this into the water filled equation:
\[ \left(\frac{1}{3}\right) x + \left(\frac{4}{27}\right)(5.5 - x) = 1 \]

### Step 6: Solve for \( x \)
Now, multiply through by 27 to eliminate the fractions:
\[ 9x + 4(5.5 - x) = 27 \]
Distributing the 4:
\[ 9x + 22 - 4x = 27 \]
Combine like terms:
\[ 5x + 22 = 27 \]
Subtract 22 from both sides:
\[ 5x = 5 \]
Divide by 5:
\[ x = 1 \]

### Step 7: Find \( y \)
Substituting \( x = 1 \) back into the equation \( y = 5.5 - x \):
\[ y = 5.5 - 1 = 4.5 \]

### Conclusion
Pipe B was turned on for **4.5 hours**.