Solving the Cistern Filling Problem with Multiple Pipes

This question involves understanding the concept of work rate, where each pipe contributes to filling or draining a cistern at a certain speed. The total work done is filling one entire cistern. We need to figure out how long Pipe B was operating given the sequence of operations and the total time taken.

Understanding the Rates of Filling and Draining Pipes

First, let's determine the rate of work for each pipe. The rate is the fraction of the cistern filled or drained per hour.

• Pipe A fills the cistern in 32 hours. Its filling rate is $\text{Rate}_A = \frac{1}{32}$ of the cistern per hour.
• Pipe B fills the cistern in 48 hours. Its filling rate is $\text{Rate}_B = \frac{1}{48}$ of the cistern per hour.
• Pipe C drains the cistern in 64 hours. Its draining rate is $\text{Rate}_C = -\frac{1}{64}$ of the cistern per hour (negative sign indicates draining).

Combined Rates in Different Phases

The operation happens in two phases:

• Phase 1: Pipe A and Pipe C are on.
• Phase 2: Pipe B and Pipe C are on.

Let's calculate the combined rate for each phase:

• Combined Rate in Phase 1 (A and C): Since Pipe A is filling and Pipe C is draining, their combined rate is $\text{Rate}_{A+C} = \text{Rate}_A + \text{Rate}_C$. $\text{Rate}_{A+C} = \frac{1}{32} + (-\frac{1}{64}) = \frac{1}{32} - \frac{1}{64}$ To subtract these fractions, we find a common denominator, which is 64. $\text{Rate}_{A+C} = \frac{2}{64} - \frac{1}{64} = \frac{2-1}{64} = \frac{1}{64}$ of the cistern per hour.
• Combined Rate in Phase 2 (B and C): In this phase, Pipe B is filling and Pipe C is draining. The combined rate is $\text{Rate}_{B+C} = \text{Rate}_B + \text{Rate}_C$. $\text{Rate}_{B+C} = \frac{1}{48} + (-\frac{1}{64}) = \frac{1}{48} - \frac{1}{64}$ To subtract these fractions, we find the Least Common Multiple (LCM) of 48 and 64. LCM(48, 64) = 192. $\text{Rate}_{B+C} = \frac{1 \times 4}{48 \times 4} - \frac{1 \times 3}{64 \times 3} = \frac{4}{192} - \frac{3}{192} = \frac{4-3}{192} = \frac{1}{192}$ of the cistern per hour.

Setting Up the Equations

Let $t_1$ be the number of hours Pipe A and Pipe C were turned on (duration of Phase 1). Let $t_2$ be the number of hours Pipe B and Pipe C were turned on (duration of Phase 2).

The total time taken to fill the cistern is 112 hours. So,

$\qquad t_1 + t_2 = 112 \quad \text{(Equation 1)}$

The total work done (filling 1 cistern) is the sum of the work done in Phase 1 and Phase 2. Work done = Rate $\times$ Time

Work done in Phase 1 = $\text{Rate}_{A+C} \times t_1 = \frac{1}{64} \times t_1 = \frac{t_1}{64}$

Work done in Phase 2 = $\text{Rate}_{B+C} \times t_2 = \frac{1}{192} \times t_2 = \frac{t_2}{192}$

Since the total work is filling the entire cistern (which represents 1 unit of work), we have:

$\qquad \frac{t_1}{64} + \frac{t_2}{192} = 1 \quad \text{(Equation 2)}$

Solving for the Time Pipe B Was On

We need to find the value of $t_2$, which is the number of hours Pipe B was turned on. We can use the system of equations we derived. From Equation 1, we can express $t_1$ in terms of $t_2$:

$\qquad t_1 = 112 - t_2$

Now substitute this expression for $t_1$ into Equation 2:

$\qquad \frac{112 - t_2}{64} + \frac{t_2}{192} = 1$

To eliminate the denominators, multiply the entire equation by the LCM of 64 and 192, which is 192.

$\qquad 192 \times \left( \frac{112 - t_2}{64} \right) + 192 \times \left( \frac{t_2}{192} \right) = 192 \times 1$

$\qquad 3 \times (112 - t_2) + t_2 = 192$

Distribute the 3:

$\qquad 336 - 3t_2 + t_2 = 192$

Combine the $t_2$ terms:

$\qquad 336 - 2t_2 = 192$

Subtract 192 from both sides and add $2t_2$ to both sides:

$\qquad 336 - 192 = 2t_2$

$\qquad 144 = 2t_2$

Divide by 2 to find $t_2$:

$\qquad t_2 = \frac{144}{2}$

$\qquad t_2 = 72$

So, the duration of Phase 2 was 72 hours. Pipe B was turned on during Phase 2.

Final Answer

Pipe B was turned on for 72 hours.

Revision Table: Key Concepts in Pipe and Cistern Problems

Additional Information: Tips for Solving Pipe and Cistern Problems

Solving problems involving pipes filling or draining cisterns often relies on the concept of work rate, similar to time and work problems. Here are some useful tips:

• Always determine the rate of each pipe first. Remember filling rates are positive, and draining rates are negative.
• Identify the different phases of operation and the pipes active in each phase.
• Calculate the combined rate for each phase by summing the individual rates (algebraically).
• If the problem involves a sequence of operations, consider the time spent in each phase. Let variables represent these times.
• The total work done is the sum of the work done in each phase. If the cistern is filled, the total work is 1.
• Set up equations based on the total time and the total work done.
• Use algebraic methods (substitution, elimination) to solve the system of equations for the unknown time variable(s).
• Pay attention to units (usually hours and cisterns).
• Make sure the calculated rates and times make sense in the context of the problem (e.g., a combined rate of filling and draining should be positive if the cistern is eventually filled).