Pipes and Cisterns

The topic of Pipes and Cisterns is a fundamental part of quantitative aptitude tests, especially in competitive exams like the RRB ALP. It deals with problems involving filling and emptying tanks by pipes, each operating at a different rate. These problems often test your understanding of rates of work and how to combine them. It's a direct application of the principles learned in the "Time and Work" chapter, but with a focus on filling and emptying rather than just completing a task.

In essence, a cistern is a tank or reservoir, and pipes are used to fill or empty it. The core idea is to determine the time taken to fill or empty the cistern, or to find the rate at which it is being filled or emptied, considering multiple pipes working simultaneously or sequentially.

Understanding the Basics: Rates of Work

Just like in Time and Work problems, we associate a rate with each pipe. The rate represents how much of the cistern the pipe can fill or empty in one unit of time (usually one hour).

If a pipe can fill a cistern in 'x' hours, then in 1 hour, the portion of the cistern it fills is 1/x.

If a pipe can empty a cistern in 'y' hours, then in 1 hour, the portion of the cistern it empties is 1/y.

When dealing with filling pipes, their rates are positive. When dealing with emptying pipes, their rates are negative, as they are removing water from the cistern.

Scenario 1: Only Filling Pipes

Suppose pipe A can fill a cistern in $x_1$ hours, and pipe B can fill the same cistern in $x_2$ hours. If both pipes are opened simultaneously, we want to find the time it takes for them to fill the cistern together.

Rate of pipe A = $\frac{1}{x_1}$ (portion of cistern filled per hour)

Rate of pipe B = $\frac{1}{x_2}$ (portion of cistern filled per hour)

When both pipes are open, their rates add up.

Combined rate of A and B = Rate of A + Rate of B = $\frac{1}{x_1} + \frac{1}{x_2}$

To find the total time taken to fill the cistern together, we take the reciprocal of the combined rate.

Time taken (together) = $\frac{1}{\frac{1}{x_1} + \frac{1}{x_2}} = \frac{1}{\frac{x_2 + x_1}{x_1 x_2}} = \frac{x_1 x_2}{x_1 + x_2}$ hours.

This formula is identical to the one used for two people working together in Time and Work.

Example 1:

Pipe A can fill a cistern in 10 hours. Pipe B can fill the same cistern in 15 hours. If both pipes are opened at the same time, how long will it take to fill the cistern?

Rate of Pipe A = $\frac{1}{10}$ per hour.

Rate of Pipe B = $\frac{1}{15}$ per hour.

Combined Rate = $\frac{1}{10} + \frac{1}{15} = \frac{3 + 2}{30} = \frac{5}{30} = \frac{1}{6}$ per hour.

Time taken together = $\frac{1}{\text{Combined Rate}} = \frac{1}{\frac{1}{6}} = 6$ hours.

Alternatively, using the formula: Time = $\frac{10 \times 15}{10 + 15} = \frac{150}{25} = 6$ hours.

Scenario 2: Filling and Emptying Pipes Working Together

This is where the concept of negative rates becomes crucial. Suppose pipe A fills a cistern in $x_1$ hours, and pipe B empties the same cistern in $y_1$ hours. If both pipes are opened simultaneously, we want to find the time taken to fill the cistern.

Rate of filling pipe A = $\frac{1}{x_1}$ (portion of cistern filled per hour)

Rate of emptying pipe B = $-\frac{1}{y_1}$ (portion of cistern emptied per hour)

When both pipes are open, the net rate of filling is the sum of their rates.

Net rate of filling = Rate of A + Rate of B = $\frac{1}{x_1} + (-\frac{1}{y_1}) = \frac{1}{x_1} - \frac{1}{y_1}$

The time taken to fill the cistern is the reciprocal of this net rate.

Time taken to fill = $\frac{1}{\frac{1}{x_1} - \frac{1}{y_1}} = \frac{1}{\frac{y_1 - x_1}{x_1 y_1}} = \frac{x_1 y_1}{y_1 - x_1}$ hours.

Important Note: This formula is valid only if the filling rate ($1/x_1$) is greater than the emptying rate ($1/y_1$), meaning $y_1 > x_1$. If the emptying rate is greater than the filling rate ($x_1 > y_1$), the cistern will never be filled; instead, it will be emptied if it has some water initially. The net rate would be negative, indicating emptying.

Example 2:

A tap can fill a cistern in 8 hours. Another tap can empty it in 5 hours. If both taps are opened, how long will it take to fill the cistern?

Time for filling = 8 hours. Rate of filling = $\frac{1}{8}$ per hour.

Time for emptying = 5 hours. Rate of emptying = $-\frac{1}{5}$ per hour.

Net rate of filling = $\frac{1}{8} - \frac{1}{5} = \frac{5 - 8}{40} = -\frac{3}{40}$ per hour.

Since the net rate is negative, the cistern will not be filled; it will be emptied. The question implies that we are looking for the time to fill it *if it were possible*. In such competitive exam scenarios, if the emptying rate is higher, the answer is often interpreted as "the cistern will never be filled" or the question is framed as "how long to empty if it was full initially". Let's rephrase the question to make it solvable for filling.

Example 2 (Revised for Filling):

Tap A can fill a cistern in 5 hours. Tap B can empty it in 8 hours. If both taps are opened, how long will it take to fill the cistern?

Rate of filling (Tap A) = $\frac{1}{5}$ per hour.

Rate of emptying (Tap B) = $-\frac{1}{8}$ per hour.

Net rate of filling = $\frac{1}{5} - \frac{1}{8} = \frac{8 - 5}{40} = \frac{3}{40}$ per hour.

Time taken to fill = $\frac{1}{\text{Net Rate}} = \frac{1}{\frac{3}{40}} = \frac{40}{3}$ hours.

$\frac{40}{3}$ hours = $13 \frac{1}{3}$ hours = 13 hours and 20 minutes.

Scenario 3: Multiple Filling and Emptying Pipes

When there are more than two pipes, some filling and some emptying, we simply sum up all their rates. Filling pipes have positive rates, and emptying pipes have negative rates.

Let pipes $P_1, P_2, ..., P_n$ fill the cistern in $x_1, x_2, ..., x_n$ hours respectively.

Let pipes $Q_1, Q_2, ..., Q_m$ empty the cistern in $y_1, y_2, ..., y_m$ hours respectively.

Rate of $P_i = \frac{1}{x_i}$

Rate of $Q_j = -\frac{1}{y_j}$

Net rate of filling = $(\frac{1}{x_1} + \frac{1}{x_2} + ... + \frac{1}{x_n}) - (\frac{1}{y_1} + \frac{1}{y_2} + ... + \frac{1}{y_m})$

Time taken to fill = $\frac{1}{\text{Net Rate}}$ (assuming Net Rate is positive).

Example 3:

Three pipes A, B, and C can fill a cistern in 10, 12, and 15 hours respectively. Pipe D can empty it in 8 hours. If all pipes are opened, how long will it take to fill the cistern?

Rate of A = $\frac{1}{10}$

Rate of B = $\frac{1}{12}$

Rate of C = $\frac{1}{15}$

Rate of D = $-\frac{1}{8}$

Total filling rate (A, B, C) = $\frac{1}{10} + \frac{1}{12} + \frac{1}{15}$

LCM of 10, 12, 15 is 60.

Total filling rate = $\frac{6}{60} + \frac{5}{60} + \frac{4}{60} = \frac{6+5+4}{60} = \frac{15}{60} = \frac{1}{4}$ per hour.

Rate of emptying (D) = $\frac{1}{8}$ per hour.

Net rate of filling = (Total filling rate) - (Rate of emptying) = $\frac{1}{4} - \frac{1}{8}$

Net rate = $\frac{2}{8} - \frac{1}{8} = \frac{1}{8}$ per hour.

Time taken to fill = $\frac{1}{\text{Net Rate}} = \frac{1}{\frac{1}{8}} = 8$ hours.

Scenario 4: Pipes Opened Sequentially or at Different Times

In these problems, we need to carefully track the state of the cistern at different points in time.

Example 4: A cistern can be filled by pipe A in 4 hours and by pipe B in 6 hours. Both pipes are opened together for 2 hours, and then pipe B is closed. How long will pipe A take to fill the remaining part of the cistern?

Rate of A = $\frac{1}{4}$ per hour.

Rate of B = $\frac{1}{6}$ per hour.

Combined rate of A and B = $\frac{1}{4} + \frac{1}{6} = \frac{3+2}{12} = \frac{5}{12}$ per hour.

In the first 2 hours, the portion filled is: Combined Rate × Time = $\frac{5}{12} \times 2 = \frac{10}{12} = \frac{5}{6}$ of the cistern.

Remaining part of the cistern to be filled = $1 - \frac{5}{6} = \frac{1}{6}$.

Now, only pipe A is working. Pipe A fills at a rate of $\frac{1}{4}$ per hour.

Time taken by A to fill the remaining $\frac{1}{6}$ = $\frac{\text{Remaining Part}}{\text{Rate of A}} = \frac{\frac{1}{6}}{\frac{1}{4}} = \frac{1}{6} \times 4 = \frac{4}{6} = \frac{2}{3}$ hours.

$\frac{2}{3}$ hours = $\frac{2}{3} \times 60$ minutes = 40 minutes.

So, pipe A will take $\frac{2}{3}$ hours (or 40 minutes) to fill the remaining part.

Scenario 5: Cistern is Emptied

If the net rate is negative, it means the cistern is emptying. The question might ask how long it takes to empty a full cistern.

Example 5: Pipe A can fill a cistern in 6 hours. Pipe B can empty it in 4 hours. If the cistern is full and both pipes are opened, how long will it take to empty the cistern?

Rate of filling (A) = $\frac{1}{6}$ per hour.

Rate of emptying (B) = $-\frac{1}{4}$ per hour.

Net rate = $\frac{1}{6} - \frac{1}{4} = \frac{2 - 3}{12} = -\frac{1}{12}$ per hour.

The negative sign indicates emptying. The rate of emptying is $\frac{1}{12}$ of the cistern per hour.

Time taken to empty = $\frac{1}{\text{Rate of emptying}} = \frac{1}{\frac{1}{12}} = 12$ hours.

Working with Capacity (Alternative Method)

Sometimes, it's easier to assume a total capacity for the cistern, usually a number that is a multiple of the times taken by all pipes. This avoids fractions.

Let the capacity of the cistern be the LCM of the times taken by the pipes.

Example 6 (using capacity): Pipe A can fill a cistern in 10 hours. Pipe B can fill the same cistern in 15 hours. How long will it take for both to fill the cistern?

LCM of 10 and 15 is 30. Let the capacity of the cistern be 30 units.

In 1 hour, Pipe A fills $\frac{30 \text{ units}}{10 \text{ hours}} = 3$ units/hour.

In 1 hour, Pipe B fills $\frac{30 \text{ units}}{15 \text{ hours}} = 2$ units/hour.

When both pipes work together, their combined filling rate is $3 + 2 = 5$ units/hour.

Time taken to fill 30 units at a rate of 5 units/hour = $\frac{30 \text{ units}}{5 \text{ units/hour}} = 6$ hours.

This method is particularly useful when dealing with multiple pipes and complex scenarios.

Example 7 (using capacity with emptying):

Tap A can fill a cistern in 5 hours. Tap B can empty it in 8 hours. If both taps are opened, how long will it take to fill the cistern?

LCM of 5 and 8 is 40. Let the capacity of the cistern be 40 units.

Rate of filling (Tap A) = $\frac{40 \text{ units}}{5 \text{ hours}} = 8$ units/hour.

Rate of emptying (Tap B) = $\frac{40 \text{ units}}{8 \text{ hours}} = 5$ units/hour (this is the rate of removal).

Net rate of filling = Rate of filling - Rate of emptying = $8 - 5 = 3$ units/hour.

Time taken to fill 40 units at a net rate of 3 units/hour = $\frac{40 \text{ units}}{3 \text{ units/hour}} = \frac{40}{3}$ hours.

This matches the result obtained using fractions.

Key Takeaways and Formulas

The core concept is that rates add up. For filling pipes, rates are positive; for emptying pipes, rates are negative.

Formula Summary:

• If pipe A fills in $x$ hours, its rate is $\frac{1}{x}$.
• If pipe B empties in $y$ hours, its rate is $-\frac{1}{y}$.
• If pipes A and B fill a cistern in $x_1$ and $x_2$ hours respectively, time to fill together = $\frac{x_1 x_2}{x_1 + x_2}$.
• If pipe A fills in $x_1$ hours and pipe B empties in $y_1$ hours, time to fill together = $\frac{x_1 y_1}{y_1 - x_1}$ (provided $y_1 > x_1$).
• If the net rate is negative, the cistern empties.

Practice Problems

1. Two pipes A and B can fill a tank in 12 hours and 18 hours respectively. If they are opened simultaneously, in how much time will the tank be filled?

2. A pipe can fill a cistern in 9 hours. Due to a leak in the bottom, it takes 10 hours to fill the cistern. In how many hours can the leak empty the cistern?

3. Pipe A can fill a cistern in 5 hours. Pipe B can fill it in 6 hours. Pipe C can empty it in 10 hours. If all three pipes are opened, how long will it take to fill the cistern?

4. A cistern is filled by three pipes A, B, and C in 12, 20, and 24 minutes respectively. How long will it take to fill the cistern if all three pipes are opened?

5. A tap can fill a cistern in 4 hours. Another tap can empty it in 6 hours. If the cistern is half full and both taps are opened, how long will it take to empty the cistern?

6. Pipe A and B can fill a tank in 3 hours and 4 hours respectively. Pipe C can empty it in 2 hours. If all three pipes are opened at 7 am, at what time will the tank be filled?

Solutions to Practice Problems:

1. Rate A = 1/12, Rate B = 1/18. Combined rate = 1/12 + 1/18 = (3+2)/36 = 5/36. Time = 36/5 = 7.2 hours.
2. Rate of pipe = 1/9. Rate of pipe - leak = 1/10. Rate of leak = 1/9 - 1/10 = (10-9)/90 = 1/90. Time to empty = 90 hours.
3. Rate A = 1/5, Rate B = 1/6, Rate C = -1/10. Net rate = 1/5 + 1/6 - 1/10 = (6+5-3)/30 = 8/30 = 4/15. Time = 15/4 = 3.75 hours.
4. Rate A = 1/12, Rate B = 1/20, Rate C = 1/24. LCM(12, 20, 24) = 120. Capacity = 120 units. Rate A = 10 units/min, Rate B = 6 units/min, Rate C = 5 units/min. Combined rate = 10+6+5 = 21 units/min. Time = 120/21 = 40/7 minutes.
5. Rate fill = 1/4, Rate empty = -1/6. Net rate = 1/4 - 1/6 = (3-2)/12 = 1/12. It takes 12 hours to fill the whole cistern. So, it takes 12 hours to fill half. The question asks to empty if half full. If it's half full, and net rate is filling, it won't empty. Let's assume it was full and both taps opened. Time to empty = 12 hours. If it's half full, and net rate is filling, it will take 6 hours to fill the other half. The question is tricky. If it's half full, and both taps are opened, the net rate is 1/12 (filling). So it will continue to fill. It will never empty. Re-reading: "how long will it take to empty the cistern?". This implies the net rate must be negative. Let's swap rates. Pipe A empties in 4 hrs, Pipe B fills in 6 hrs. Net rate = 1/6 - 1/4 = -1/12. Time to empty = 12 hrs. If cistern is half full, it takes 6 hrs to empty. Assuming the question meant Pipe A empties in 4 hours and Pipe B fills in 6 hours.
6. Rate A = 1/3, Rate B = 1/4, Rate C = -1/2. Net rate = 1/3 + 1/4 - 1/2 = (4+3-6)/12 = 1/12. Time to fill = 12 hours. If started at 7 am, it will be filled at 7 pm.