Calculating Pipe Rates

First, determine the filling/draining rate for each pipe:

• Pipe A fills the cistern in 18 hours. Rate (A) = $\frac{1}{18}$ cistern/hour.
• Pipe B fills the cistern in 27 hours. Rate (B) = $\frac{1}{27}$ cistern/hour.
• Pipe C drains the cistern in 45 hours. Rate (C) = $-\frac{1}{45}$ cistern/hour (negative sign indicates draining).

Analyzing the Filling Process

The process occurs in two phases:

• Phase 1: Pipes A and C are turned on. Let this phase last for $t_1$ hours.
• Phase 2: Pipe A is turned off, and Pipe B is turned on instantly. Pipe C remains on. Let this phase last for $t_2$ hours.

The total time to fill the cistern is given as 55 hours. Therefore:

$t_1 + t_2 = 55 \text{ hours}$

Setting Up the Work Equation

The total work done is filling 1 cistern. The work done is the sum of the work done by each pipe in its respective phase.

Combined rate in Phase 1 (A + C) = Rate(A) + Rate(C) = $\frac{1}{18} - \frac{1}{45} = \frac{5 - 2}{90} = \frac{3}{90} = \frac{1}{30}$ cistern/hour.

Combined rate in Phase 2 (B + C) = Rate(B) + Rate(C) = $\frac{1}{27} - \frac{1}{45} = \frac{5 - 3}{135} = \frac{2}{135}$ cistern/hour.

The total work equation is:

$(\text{Rate A + Rate C}) \times t_1 + (\text{Rate B + Rate C}) \times t_2 = 1$

$ \left( \frac{1}{30} \right) t_1 + \left( \frac{2}{135} \right) t_2 = 1 $

Solving for Pipe B's Duration ($t_2$)

We know $t_1 = 55 - t_2$. Substitute this into the work equation:

$ \frac{1}{30} (55 - t_2) + \frac{2}{135} t_2 = 1 $

Multiply the equation by the Least Common Multiple (LCM) of the denominators (30 and 135), which is 270:

$ 270 \times \frac{1}{30} (55 - t_2) + 270 \times \frac{2}{135} t_2 = 270 \times 1 $

$ 9 (55 - t_2) + 2 \times 2 t_2 = 270 $

$ 495 - 9 t_2 + 4 t_2 = 270 $

$ 495 - 5 t_2 = 270 $

Rearrange to solve for $t_2$:

$ 5 t_2 = 495 - 270 $

$ 5 t_2 = 225 $

$ t_2 = \frac{225}{5} $

$ t_2 = 45 \text{ hours} $

Pipe B was turned on for 45 hours.