Understanding the Pipe and Cistern Problem

This question involves analyzing the work rates of two pipes filling a cistern and a drain pipe emptying it. The filling pipes have varying efficiencies over time, while the drain pipe works at a constant rate. We need to find out how long it takes the drain pipe alone to empty a full cistern.

The key concept here is work rate. If a pipe can fill a cistern in $T$ hours, its filling rate is $\frac{1}{T}$ of the cistern per hour. Similarly, if a drain pipe can empty a cistern in $T$ hours, its draining rate is $\frac{1}{T}$ of the cistern per hour (often treated as a negative rate when working with filling pipes).

Calculating Full Capacity Rates

• Pipe 1 (P1) at full speed fills the cistern in 1 hour.
• Full capacity rate of P1 = $\frac{1}{1} = 1$ cistern per hour.
• Pipe 2 (P2) at full speed fills the cistern in 2 hours.
• Full capacity rate of P2 = $\frac{1}{2}$ cistern per hour.

Analyzing Varied Efficiency Rates Hour by Hour

The rates of Pipe 1 and Pipe 2 change over the 5 hours they work together. Let's calculate their rates for each hour:

Pipe 1 Rates:

• Hour 1: $\frac{1}{12}$ of usual capacity. Rate = $\frac{1}{12} \times 1 = \frac{1}{12}$ cistern/hour.
• Hour 2: $\frac{1}{9}$ of usual capacity. Rate = $\frac{1}{9} \times 1 = \frac{1}{9}$ cistern/hour.
• Hour 3: $\frac{1}{6}$ of usual capacity. Rate = $\frac{1}{6} \times 1 = \frac{1}{6}$ cistern/hour.
• Hour 4: $\frac{1}{4}$ of usual capacity. Rate = $\frac{1}{4} \times 1 = \frac{1}{4}$ cistern/hour.
• Hour 5: $\frac{1}{3}$ of usual capacity. Rate = $\frac{1}{3} \times 1 = \frac{1}{3}$ cistern/hour.

Pipe 2 Rates:

• Hour 1: $\frac{1}{12}$ of usual capacity. Rate = $\frac{1}{12} \times \frac{1}{2} = \frac{1}{24}$ cistern/hour.
• Hour 2: $\frac{1}{9}$ of usual capacity. Rate = $\frac{1}{9} \times \frac{1}{2} = \frac{1}{18}$ cistern/hour.
• Hour 3: $\frac{1}{6}$ of usual capacity. Rate = $\frac{1}{6} \times \frac{1}{2} = \frac{1}{12}$ cistern/hour.
• Hour 4: $\frac{1}{4}$ of usual capacity. Rate = $\frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$ cistern/hour.
• Hour 5: $\frac{1}{3}$ of usual capacity. Rate = $\frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$ cistern/hour.

Calculating Combined Work of Filling Pipes Over 5 Hours

We calculate the total work done by Pipe 1 and Pipe 2 together over the 5 hours by summing their individual work (rate $\times$ time for each hour, but time is 1 hour for each step):

Total work by P1 and P2 in 5 hours = (P1 Rate Hour 1 + P2 Rate Hour 1) + (P1 Rate Hour 2 + P2 Rate Hour 2) + (P1 Rate Hour 3 + P2 Rate Hour 3) + (P1 Rate Hour 4 + P2 Rate Hour 4) + (P1 Rate Hour 5 + P2 Rate Hour 5)

• Hour 1 combined rate: $\frac{1}{12} + \frac{1}{24} = \frac{2}{24} + \frac{1}{24} = \frac{3}{24} = \frac{1}{8}$
• Hour 2 combined rate: $\frac{1}{9} + \frac{1}{18} = \frac{2}{18} + \frac{1}{18} = \frac{3}{18} = \frac{1}{6}$
• Hour 3 combined rate: $\frac{1}{6} + \frac{1}{12} = \frac{2}{12} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4}$
• Hour 4 combined rate: $\frac{1}{4} + \frac{1}{8} = \frac{2}{8} + \frac{1}{8} = \frac{3}{8}$
• Hour 5 combined rate: $\frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2}$

Total work done by P1 and P2 over 5 hours = $\frac{1}{8} + \frac{1}{6} + \frac{1}{4} + \frac{3}{8} + \frac{1}{2}$

To sum these fractions, find a common denominator, which is 24:

Total work = $\frac{3}{24} + \frac{4}{24} + \frac{6}{24} + \frac{9}{24} + \frac{12}{24} = \frac{3+4+6+9+12}{24} = \frac{34}{24} = \frac{17}{12}$ of the cistern.

So, in 5 hours, the two filling pipes together filled $\frac{17}{12}$ of the cistern.

Including the Drain Pipe's Effect

All three pipes (Pipe 1, Pipe 2, and the Drain Pipe) work concurrently for 5 hours to fill the empty cistern. Let the rate of the drain pipe be $d$ (as a fraction of the cistern emptied per hour). The work done by the drain pipe in 5 hours is $-5d$.

The total work done by all three pipes to fill the cistern (which means filling 1 whole cistern) is the sum of the work done by the filling pipes and the drain pipe:

(Work by P1 and P2 in 5 hours) + (Work by Drain Pipe in 5 hours) = 1 (full cistern)

$\frac{17}{12} - 5d = 1$

Solving for the Drain Pipe's Rate

Now, we solve the equation for $d$:

$\frac{17}{12} - 1 = 5d$

$\frac{17}{12} - \frac{12}{12} = 5d$

$\frac{5}{12} = 5d$

$d = \frac{5}{12 \times 5} = \frac{5}{60} = \frac{1}{12}$ cistern per hour.

The drain pipe drains $\frac{1}{12}$ of the cistern every hour.

Time for Drain Pipe to Empty the Cistern Alone

If the drain pipe drains $\frac{1}{12}$ of the cistern in 1 hour, the time it takes to drain the entire cistern (1 unit of work) is:

Time = $\frac{\text{Total Work}}{\text{Rate}} = \frac{1}{\frac{1}{12}} = 1 \times 12 = 12$ hours.

Therefore, it will take the drain pipe 12 hours to empty the filled cistern by itself.

Sum of combined filling rates over 5 hours: $\frac{1}{8} + \frac{1}{6} + \frac{1}{4} + \frac{3}{8} + \frac{1}{2} = \frac{17}{12}$.

Let drain rate be $d$ cistern/hour. Total work = Work done by fillers - Work done by drainer.

In 5 hours, total work = $\frac{17}{12} - 5d$.

Since the cistern is filled in 5 hours, total work = 1.

$\frac{17}{12} - 5d = 1$

$5d = \frac{17}{12} - 1 = \frac{5}{12}$

$d = \frac{1}{12}$ cistern/hour.

Time for drain pipe alone = $\frac{1}{\text{Rate}} = \frac{1}{\frac{1}{12}} = 12$ hours.

Revision Table: Pipe and Cistern Concepts

Additional Information: Pipe and Work Problems

Pipe and cistern problems are a type of work problem. They often involve calculating rates at which pipes fill or empty tanks. The core idea is that the total work (filling or emptying the cistern, which is usually considered '1' unit of work) is equal to the rate of work multiplied by the time taken.

When multiple pipes work together, their rates are combined. Filling rates add up, and draining rates subtract from the filling rates. If a cistern is filled, the total work done is +1. If a cistern is emptied, the total work done is -1 (or just 1 if we are only considering the time to empty from full).

Problems often introduce complexities like varying rates, pipes starting/stopping at different times, or leaks (which act like drain pipes). Solving these requires carefully calculating the work done by each pipe during the specific time intervals it is active and summing up the work to equal the total change in the cistern's volume.