Understanding Pipe and Cistern Problems

Pipe and cistern problems are similar to work and time problems. Instead of individuals doing work, pipes fill or empty a tank (cistern). The key is to determine the rate at which each pipe performs the task.

• A filling pipe has a positive rate (adds to the tank).
• An emptying pipe has a negative rate (removes from the tank).
• The rate is usually expressed as the fraction of the tank filled or emptied per unit of time (e.g., per minute).

Calculating Individual Pipe Rates

If a pipe can fill or empty a tank in 't' units of time, its rate is $\frac{1}{t}$ of the tank per unit of time.

• Pipe A fills the tank in 43.2 minutes.
• Rate of Pipe A = $\frac{1}{43.2}$ of the tank per minute.
• Pipe B fills the tank in 108 minutes.
• Rate of Pipe B = $\frac{1}{108}$ of the tank per minute.

Calculating Combined Rate

When multiple pipes work together, their individual rates are added to find the combined rate. Since Pipe C is an emptying pipe, its rate will be subtracted when combining.

• Pipes A, B, and C together fill the tank in 54 minutes.
• Combined rate of A, B, and C = $\frac{1}{54}$ of the tank per minute.

Let the rate of Pipe C be $R_C$ (which will be negative since it's emptying). The combined rate is the sum of individual rates:

\( \text{Rate of A} + \text{Rate of B} + \text{Rate of C} = \text{Combined Rate of A, B, C} \)

\( \frac{1}{43.2} + \frac{1}{108} + R_C = \frac{1}{54} \)

Finding the Rate of Pipe C

Now, we solve for $R_C$:

\( R_C = \frac{1}{54} - \frac{1}{43.2} - \frac{1}{108} \)

To perform this calculation, it's helpful to find a common denominator or convert decimals to fractions. Let's work with fractions:

\( 43.2 = \frac{432}{10} = \frac{216}{5} \)

So, \( \frac{1}{43.2} = \frac{5}{216} \).

Now, find a common denominator for 54, 108, and 216. The least common multiple (LCM) is 216.

• \( \frac{1}{54} = \frac{1 \times 4}{54 \times 4} = \frac{4}{216} \)
• \( \frac{1}{108} = \frac{1 \times 2}{108 \times 2} = \frac{2}{216} \)
• \( \frac{1}{43.2} = \frac{5}{216} \) (as calculated above)

Substitute these values back into the equation for $R_C$:

\( R_C = \frac{4}{216} - \frac{5}{216} - \frac{2}{216} \)

\( R_C = \frac{4 - 5 - 2}{216} \)

\( R_C = \frac{-3}{216} \)

\( R_C = -\frac{1}{72} \)

The rate of Pipe C is $-\frac{1}{72}$ of the tank per minute. The negative sign confirms it's an emptying pipe, emptying $\frac{1}{72}$ of the tank per minute.

Relating Rate to Capacity

We are given that Pipe C empties at a rate of 3 litres per minute. This means that the fraction of the tank emptied per minute by C corresponds to 3 litres.

\( \frac{1}{72} \text{ of the tank capacity} = 3 \text{ litres} \)

Let the total capacity of the tank be $V$ litres.

\( \frac{1}{72} \times V = 3 \)

To find the total capacity $V$, multiply both sides by 72:

\( V = 3 \times 72 \)

\( V = 216 \)

The capacity of the tank is 216 litres.

Step-by-Step Calculation Summary

1. Calculate the filling rate of Pipe A: $R_A = \frac{1}{43.2}$ tank/min.
2. Calculate the filling rate of Pipe B: $R_B = \frac{1}{108}$ tank/min.
3. Calculate the combined filling rate of A, B, and C: $R_{ABC} = \frac{1}{54}$ tank/min.
4. The combined rate is $R_{ABC} = R_A + R_B + R_C$.
5. Solve for the rate of Pipe C: $R_C = R_{ABC} - R_A - R_B = \frac{1}{54} - \frac{1}{43.2} - \frac{1}{108}$.
6. Convert rates to fractions with a common denominator: $\frac{4}{216} - \frac{5}{216} - \frac{2}{216} = -\frac{3}{216} = -\frac{1}{72}$ tank/min.
7. The rate of Pipe C is $-\frac{1}{72}$ tank/min, meaning it empties $\frac{1}{72}$ of the tank per minute.
8. This rate is given as 3 litres/minute. So, $\frac{1}{72}$ of the tank capacity = 3 litres.
9. Let capacity be V. $\frac{1}{72} \times V = 3$.
10. Solve for V: $V = 3 \times 72 = 216$ litres.

Checking the rates: Rate of A + Rate of B + Rate of C = $\frac{5}{216} + \frac{2}{216} - \frac{3}{216} = \frac{5+2-3}{216} = \frac{4}{216} = \frac{1}{54}$, which matches the combined rate. This confirms our calculation for $R_C$.

Revision Table: Pipe and Cistern Concepts

Additional Information: Work, Time, and Rate

The relationship between Work, Rate, and Time is fundamental to solving these problems. The basic formula is:

\( \text{Work} = \text{Rate} \times \text{Time} \)

In pipe and cistern problems:

• The 'Work' is usually filling or emptying one full tank (considered as '1 unit' of work or the total capacity).
• The 'Rate' is the fraction of the tank filled or emptied per unit of time.
• The 'Time' is the duration the pipe(s) operate.

If a pipe completes a job (fills a tank) in time T, its rate is 1/T per unit time. If multiple pipes work together, their combined rate is the sum (or difference for emptying pipes) of their individual rates. If the combined rate is R and they complete the job in time T, then $R \times T = 1$ (where 1 represents the full tank as the total work unit).

When a problem gives the rate in actual volume per unit time (like 3 litres/minute for pipe C), we can set the fraction of the tank's volume corresponding to that rate equal to the given volume. For example, if pipe C empties 1/72 of the tank per minute, and this amount is 3 litres, then (1/72) * Total Capacity = 3 litres, allowing us to find the Total Capacity.