5 Types of Pipes and Cisterns Problems For Bank Exams

1_Pipes_Cisterns

Dear Reader, below you will find 5 very important types of pipes and cistern problems. You have to apply a lot of common sense and logic to solve these problems.

At the end of the tutorial, you will find a short online practice test. This will help to test yourself if you have learned well.

Type I: Calculate Time Taken to Fill a Tank By 2 or More Pipes

This is the easiest type of pipes and cisterns problems. In this type, you will be given time taken by each pipe to individually fill a tank. You then have to find the time taken to fill the tank when all the pipes are opened together. Below is your example.

Example Question 1: Two pipes P and Q can fill a tank in 30 and 42 hours respectively. If both the pipes are opened simultaneously, how much time will be taken to fill the tank?

Solution:
You know that the pipe P takes 30 hours to fill the tank. You have to first calculate the portion of the tank filled in 1 hour. You can use the direct proportion table as shown below.

Hours Portion of the tank
30 1 (1 represents full tank)
1 ?

In 1 hour, the portion of the tank filled by pipe P = 1 x 1/30 = 1/30 … value 1
By the same logic, in 1 hour, the portion of the tank filled by pipe Q =
1/42 … value 2

In 1 hour, the portion of the tank filled by both pipes P and Q together = value 1 + value 2
= 1/30 + 1/42 = 12/210 = 2/35

You can find the time taken by both the pipes together to fill the entire tank using direct proportion table as shown below.

Hours Portion of the tank
1 2/35
? 1 (1 represents full tank)

Therefore, pipes P and Q can fill the tank in 1 x 1 / (2/35) = 35/2 hrs = 17 hours and 30 minutes.

Type II: Calculate Time Taken to Fill a Tank With Leakage

This is an extension to type 1. You have to find the time taken to fill tanks with leakages. Below example will help you to understand better.

Example Question 2: In a school, to fill a cistern, the authorities use two taps namely A and B. A can fill the tank in 20 hours and B can fill in 28 hours. The pipes are opened simultaneously and it is found that due to leakage in the cistern, it takes 20 minutes more to fill the cistern. If the cistern is full, then find the time taken by the leak to empty it.

Solution:
You have to solve this problem in 3 parts.

Part 1: Assume That There is No Leakage
Portion of the tank filled by pipe A in 1 hour = 1/20 … value 1
Portion of the tank filled by pipe B in 1 hour = 1/28 … value 2
Portion of the tank filled by pipes A and B together in 1 hour = value 1 + value 2
= (1/20 + 1/28)
= 12/140
= 6/70

Therefore, time taken to fill full tank by pipes A and B together = 1/(6/70)
= 70/6 hours … value 3

Note: Till the last step, every step is the same as in example 1.

Part 2: Include Leakage
In the question, it is given that due to leakage, the pipes will take extra 20 minutes (or 1/3 hours) to fill the tank.
(20 minutes = 20/60 hours = 1/3 hours)
Therefore, time taken to fill full tank by the pipes A and B together when there is leakage
= value 3 got in part 1 + 1/3 hours
= 70/6 + 1/3
= 72/6
= 12 hours

Part 3: Calculate Portion of Tank Emptied by Leakage Alone
Portion of tank filled by 2 pipes together with leakage in 1 hour = 1/12 … value 4
From part 1, you know that
Portion of the tank filled by 2 pipes together without leakage 1 hour =
6/70 … value 5

Portion of the tank emptied by leakage alone in 1 hour can be found by subtracting value 4 from value 5
Portion of the tank emptied by leakage alone in 1 hour = 6/70 – 1/12
= (36 – 35) /420
= 1/420
Therefore, leakage alone will empty the cistern in 420 hours.

Type III: Equations Based Pipes and Cistern Problems

In this type, you have to form equations based on conditions given in the question. Solving those equations will help you to find the answer. Here is your example.

Example Question 3: A tank can be filled in 10 hours by two taps Tap1 and Tap2. Tap2 is twice as fast as tap1. How much time will tap1 alone take to fill the tank?

Solution:
Assume that the tank can be filled by tap1 in X hours.
tap2 is twice as fast as tap1. Therefore, Then Tap2 can fill it in X/2 hours.

(Some candidates may make a mistake and write 2X hours instead of X/2 hours. If you have the same doubt, here is your explanation. When speed increases time decreases. Therefore, if tap2 is 2 times faster than tap1, then the time taken by tap2 will be half of that of tap1.)

Portion of the tank filled by tap1 in 1 hour = 1/X … value 1
Portion of the tank filled by tap2 in 1 hour = 2/X … value 2
Portion of the tank filled by both the taps in 1 hour = value 1 + value 2
= 1/X + 2/X … value 3

In the question, you can see that the taps together take 10 hours to fill the tank.
Therefore,
Portion of the tank filled by both the taps in 1 hour = 1/10 … value 4

As you can see, both the values 3 and 4 are same. Therefore, you can equate these values to find X.
i.e., 1/X + 2/X = 1/10
3/X = 1/10
X = 30 hours
So, tap1 alone will take 30 hours to fill the tank.

Type IV: Calculate Time Taken When Pipes Are Opened For Different Periods

In all the above types, all the pipes were open for the full duration. But what if a pipe is closed when other pipes are still open? For scenarios like this, you have to learn this type. Below example will help you.

Example Question 4: In a five-star hotel, there are three inlets namely P, Q and R, which can fill the tank in 10 hours. After working together for 5 hours, R is closed and inlets P and Q fill the remaining part in 12 hours. Find the time taken by R alone to fill the tank.

Solution:
From the question, you know that the 3 pipes (if opened together) will take 10 hours to fill the entire tank.
Portion of the tank filled by the pipes P, Q and R in 1 hour = 1/10

Part 1: First 5 Hours When All 3 Pipes Are Open
Portion of the tank filled by the pipes P, Q and R in 5 hours = 5 x 1/10 = ½
Remaining portion of tank to be filled = 1 – ½ = ½

Part 2: Second 5 Hours When Only P And Q Are Open
P and Q take 12 hours to fill the remaining part
So, portion of the tank filled by the pipes P and Q in second 5 hours = 1/2
Portion of the tank filled by P and Q in 1 hour can be found using the direct proportion table shown below.

Hours Portion of the tank
12 ½
1 ?

Portion of the tank filled by P and Q in 1 hour = 1 x ½ / 12 = 1/24

Part 3: Calculate Time Taken by R Alone to Fill the Tank
At the start of the solution, we found the below value.
Portion of the tank filled by the pipes P, Q and R in 1 hour = 1/10 …value 1
In part 2, we found the below value.
Portion of the tank filled by P and Q in 1 hour = 1/24 … value 2

If you subtract value 2 from value 1, you will get the portion of the tank filled by the pipe R in an hour.
Therefore, portion of the tank filled by R in 1 hour = 1/10 – 1/24 = 7/120
Therefore, time is taken by R to fill the entire tank = 1/(7/120) = 120/7 hours
= 17 hours 8 minutes (approximately)

Type V: Calculate Number of Pipes

Below is an example of this type.

Example Question 5: In a college hostel, the water tank is fitted with 6 pipes, in which some of them are used to fill the tank and others to drain the tank. Each of the filling pipes can fill the tank in 6 hours and each of the draining pipes can drain the tank in 4 hours. If all the pipes are opened together, the tank will drain in 4 hours. How many of the pipes are fill pipes?

Solution:
Assume that there are X fill pipes.
In the question, you can see that the total number of pipes is 6.
Therefore, will be (6-X) drain pipes.

Part 1: Calculate Values for Fill Pipes
Each fill pipe (alone) will take 6 hours to fill the tank.
So, portion of the tank filled by each fill pipe in 1 hour = 1/6
Therefore, portion of the tank filled by X fill pipes in 1 hour = X x 1/6 =
X/6 … value 1

Part 2: Calculate Values for Drain Pipes
Each drain pipe (alone) will take 4 hours to drain the tank.
So, portion of the tank drained by each drain pipe in 1 hour = 1/4
Therefore, portion of the tank drained by (6-X) drain pipes in 1 hour =
(6-X)/4 … value 2

Part 3: Final Solution
If all the pipes are opened together, then in 1 hour 1/4 of the tank gets drained.
Therefore, you can form the below equation.

Portion of the tank filled by X fill pipes in 1 hour – Portion of the tank drained by (6-X) drain pipes in 1 hour = -1/4

Note: Since the water is draining out you have to use “-” (minus) sign before 1/4

If you substitute value 1 (from part 1) and value 2 (from part 2) in above equation, you will get
X/6 – (6-X)/4 = -1/4
4X – 36X +6X / 24 = -1/4
10x -36 = -6
10X = 30
X =3
Therefore, there are 3 fill pipes.

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