12_Pipes___Cisterns

Pipes & Cisterns - Aptitude Mastery Guide

Category: Quantitative Aptitude
Generated on: 2025-07-15 09:19:08
Source: Aptitude Mastery Guide Generator

Pipes & Cisterns: A Comprehensive Guide for Aptitude Success

This guide aims to provide a complete understanding of Pipes & Cisterns problems, a common topic in quantitative aptitude tests. We’ll cover the foundational concepts, essential formulas, shortcut tricks, solved examples, and practice problems to help you master this area.

1. Foundational Concepts

Pipes and Cisterns problems are essentially an application of the concept of Time and Work. Instead of people working, we have pipes filling or emptying a tank (cistern). The core idea revolves around understanding the rate at which a pipe fills or empties the tank.

Inlet: A pipe that fills the tank.
Outlet: A pipe that empties the tank.
Rate of Filling/Emptying: The amount of tank filled or emptied in a unit of time (usually 1 hour or 1 minute).

Why the time and work analogy?

Think of a pipe filling a tank as a person doing work. The tank represents the total work to be done. The rate of filling (or emptying) is analogous to the person’s efficiency. Therefore, all the principles of Time and Work directly apply here.

Key Concept:

If a pipe can fill a tank in ‘x’ hours, then the part of the tank filled in 1 hour is 1/x. Similarly, if a pipe can empty a tank in ‘y’ hours, the part of the tank emptied in 1 hour is 1/y.

When multiple pipes are working together, their rates are combined. Inlet pipes contribute positively (filling), while outlet pipes contribute negatively (emptying).

2. Key Tricks & Shortcuts

This section is the heart of this guide! We’ll explore techniques to solve Pipes & Cisterns problems quickly and efficiently.

Trick 1: Combined Rate (Filling and Emptying):

If a pipe fills a tank in ‘x’ hours and another pipe empties it in ‘y’ hours, then the net part filled in 1 hour is (1/x - 1/y). The time taken to fill the tank completely is 1 / (1/x - 1/y) = xy / (y - x) hours. (Remember: y > x for the tank to fill eventually).
- When to use: When you have one or more pipes filling and one or more pipes emptying the tank simultaneously.
Trick 2: Work Done in a Specific Time Interval:

If pipes A and B can fill a tank in ‘x’ and ‘y’ hours respectively, and they are opened alternately for 1 hour each, starting with pipe A, then:
- Part of tank filled in 2 hours = (1/x + 1/y)
- Calculate how many complete cycles of 2 hours are required to almost fill the tank.
- Calculate the remaining part and the time taken by the next pipe (in this case A) to fill it.
- When to use: Problems where pipes are opened and closed alternately.
Trick 3: Proportionality and Ratios:

If pipe A is ‘k’ times faster than pipe B, then the ratio of time taken by A to B is 1:k. This can be very helpful in problems involving comparison of efficiencies.
- When to use: Problems comparing the efficiency or speed of different pipes.
Trick 4: Assumption Method (Efficiency-Based):

Assume the total capacity of the tank to be a multiple of the individual times taken by each pipe to fill or empty the tank. This makes calculations with fractions easier. For example, if pipe A takes 10 hours and pipe B takes 15 hours, assume the total capacity of the tank to be LCM(10, 15) = 30 units. This means pipe A fills 3 units/hour, and pipe B fills 2 units/hour.
- When to use: Most problems involving multiple pipes and different filling/emptying times. This simplifies calculations significantly.
Trick 5: Percentage to Fraction Conversion (Vedic Maths):

Quickly convert percentages to fractions to speed up calculations. Knowing common conversions (e.g., 12.5% = 1/8, 16.66% = 1/6, 33.33% = 1/3, 66.66% = 2/3, 75% = 3/4) is crucial. For example, if a pipe fills 16.66% of a tank in 1 hour, it can fill the entire tank in 6 hours.
- Vedic Maths Connection: Vedic Maths emphasizes quick mental calculations. Knowing percentage-to-fraction conversions is a key skill.
- When to use: Whenever percentages are involved in the problem.
Trick 6: Using LCM for Total Work (Tank Capacity):

This is an extension of the assumption method. Instead of just assuming, use the Least Common Multiple (LCM) of the given times as the “total work” (or tank capacity). This avoids fractions and simplifies calculations.
- When to use: This is generally applicable and often the most efficient method for many Pipes & Cisterns problems.
Trick 7: Negative Work for Emptying Pipes:

Always treat the work done by emptying pipes as negative. This helps in correctly calculating the combined rate of filling or emptying.
- When to use: In all problems that involve both filling and emptying pipes.

3. Essential Formulas & Rules

Formula/Rule	Description
Work Done = Rate x Time	Fundamental formula. Amount of tank filled/emptied = (Rate of filling/emptying) * Time
Rate = Work Done / Time	Rate of filling/emptying = (Amount of tank filled/emptied) / Time
Time = Work Done / Rate	Time taken to fill/empty the tank = (Amount of tank filled/emptied) / (Rate of filling/emptying)
Combined Rate (Filling)	If two pipes fill a tank in x and y hours respectively, the combined rate is (1/x + 1/y)
Combined Time (Filling)	Time taken by two pipes to fill the tank together = xy / (x + y)
Combined Rate (Filling & Emptying)	If a pipe fills a tank in x hours and another empties it in y hours, the combined rate is (1/x - 1/y) (y > x)
Combined Time (Filling & Emptying)	Time taken to fill the tank when filling and emptying pipes are open = xy / (y - x) (y > x)
Efficiency Ratio = Time Ratio (Inversely Prop.)	If A is k times faster than B, then the ratio of time taken by A to B is 1:k. Efficiency and time are inversely proportional.

4. Detailed Solved Examples

Example 1: Basic Problem (Using LCM and Combined Rate)

Two pipes, A and B, can fill a tank in 20 minutes and 30 minutes respectively. If both pipes are opened simultaneously, how long will it take to fill the tank?

Solution:

Method: LCM Method and Combined Rate
Step 1: Find the LCM of 20 and 30. LCM(20, 30) = 60. Assume the total capacity of the tank is 60 units.
Step 2: Calculate the rate of each pipe.
- Pipe A fills 60/20 = 3 units per minute.
- Pipe B fills 60/30 = 2 units per minute.
Step 3: Calculate the combined rate.
- Combined rate = 3 + 2 = 5 units per minute.
Step 4: Calculate the time to fill the tank.
- Time = Total capacity / Combined rate = 60 / 5 = 12 minutes.

Answer: It will take 12 minutes to fill the tank.

Example 2: Filling and Emptying (Using Combined Rate with Negative Work)

A tank can be filled by pipe A in 25 minutes and by pipe B in 30 minutes. An outlet pipe C can empty the full tank in 40 minutes. If all three pipes are opened simultaneously, how much time will it take to fill the tank?

Solution:

Method: LCM Method, Combined Rate with Negative Work.
Step 1: Find the LCM of 25, 30, and 40. LCM(25, 30, 40) = 600. Assume the capacity of the tank is 600 units.
Step 2: Calculate the rate of each pipe.
- Pipe A fills 600/25 = 24 units per minute.
- Pipe B fills 600/30 = 20 units per minute.
- Pipe C empties 600/40 = 15 units per minute.
Step 3: Calculate the net rate (remember to use a negative sign for the emptying pipe).
- Net rate = 24 + 20 - 15 = 29 units per minute.
Step 4: Calculate the time to fill the tank.
- Time = Total capacity / Net rate = 600 / 29 = 20 20/29 minutes.

Answer: It will take 20 20/29 minutes to fill the tank.

Example 3: Alternate Filling (Using Work in a Specific Time Interval)

Two pipes, A and B, can fill a tank in 24 minutes and 30 minutes respectively. If A opens first, and they are opened alternately for 1 minute each, how long will the tank take to fill?

Solution:

Method: Work Done in a Specific Time Interval.
Step 1: Calculate the work done by each pipe in 1 minute.
- Pipe A fills 1/24 of the tank in 1 minute.
- Pipe B fills 1/30 of the tank in 1 minute.
Step 2: Calculate the work done in 2 minutes (one cycle of A and B).
- Work done in 2 minutes = 1/24 + 1/30 = (5 + 4)/120 = 9/120 = 3/40 of the tank.
Step 3: Find out how many cycles of 2 minutes are needed to fill almost the entire tank.
- (3/40) * 13 = 39/40 (Almost the entire tank in 26 minutes)
Step 4: Calculate remaining work and time required.
- Remaining work = 1 - 39/40 = 1/40
- In the 27th minute, pipe A opens and fills 1/24 of the tank. Since 1/24 > 1/40, pipe A will fill the remaining part in less than a minute.
- Time taken by A to fill the remaining 1/40 = (1/40) / (1/24) = 24/40 = 3/5 minute = 36 seconds.
Step 5: Calculate the total time.
- Total time = 26 minutes + 36 seconds = 26 minutes 36 seconds.

Answer: It will take 26 minutes and 36 seconds to fill the tank.

Example 4: Efficiency Based (Proportionality and Ratios)

Pipe A is twice as fast as pipe B. If both pipes together can fill a tank in 20 minutes, how long will pipe B alone take to fill the tank?

Solution:

Method: Proportionality and Ratios
Step 1: Establish the relationship between their efficiencies.
- Since A is twice as fast as B, the ratio of their efficiencies (work rates) is A:B = 2:1.
Step 2: Let the efficiencies be 2x and x.
Step 3: Calculate the combined efficiency.
- Combined efficiency = 2x + x = 3x
Step 4: Since they fill the tank in 20 minutes together, the total work (tank capacity) is (3x) * 20 = 60x.
Step 5: Calculate the time taken by pipe B alone.
- Time = Total work / Efficiency of B = 60x / x = 60 minutes.

Answer: Pipe B alone will take 60 minutes to fill the tank.

5. Practice Problems (Graded Difficulty)

[Easy]

A pipe can fill a tank in 15 hours. How much of the tank will it fill in 3 hours?

[Easy]

Pipe A can fill a tank in 8 hours, and Pipe B can fill the same tank in 12 hours. If both pipes are opened together, how long will it take to fill the tank?

[Medium]

A cistern has two inlets, A and B, which can fill it in 12 minutes and 15 minutes respectively. There is also an outlet C. If A, B, and C are opened together, the cistern is filled in 20 minutes. How much time will C take to empty the full cistern?

[Medium]

Two pipes A and B can fill a tank in 36 minutes and 45 minutes respectively. Another pipe C can empty the tank in 30 minutes. First A and B are opened. After 7 minutes, C is also opened. In what time will the tank be full?

[Hard]

A tank has a leak which would empty it in 8 hours. A tap is turned on which admits 6 liters a minute into the tank, and it is now emptied in 12 hours. What is the capacity of the tank?

[Hard]

Two pipes A and B can fill a tank in 20 minutes and 30 minutes respectively. If pipe A is kept open all the time and pipe B is opened for every alternate minute, how long would it take for the tank to be filled?

[Hard]

A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. Find the time required by the first pipe to fill the tank.

6. Advanced/Case-Based Question

A tank is fitted with three pipes. The first two pipes can fill the tank in 10 hours and 12 hours respectively, while the third pipe can empty the tank in 20 hours. Initially, the first pipe is opened at 6 am. Then, the second pipe is opened at 7 am. Finally, the third pipe is opened at 8 am. At what time will the tank be completely filled? Also calculate what percentage of the tank was already filled by 8 am. This requires you to track the filling rates at different time intervals, accounting for when each pipe is operational.