4 Ways to Solve Pipes and Cisterns Problems Easily
In every competitive exam, aptitude section is very important and scoring section. Pipes and Cisterns is very difficult to prepare and getting some confusion to solve the Pipes and Cisterns problems, due to this confusion lots of applicants lose their marks in aptitude section. Here we are providing the 4 simple ways to solve Pipes and Cisterns problems easily at our website. Candidates who are going to prepare and appear for the competitive bank exam or any govt exam, those can download the Pipes and Cisterns problems with examples through online from our website. Applicants can follow the following Pipes and Cisterns easy problems with simple formulas.
4 Ways to Solve Pipes and Cisterns Problems:
Step 1: 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.
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?
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)|
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 (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.
Step 2: Calculate Time Taken to Fill a Tank With Leakage in Pipes and Cisterns:
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.
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)
Therefore, time taken to fill full tank by pipes A and B together = 1/ (6/70)
= 70/6 hours … value 3
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
= 12 hours
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
Therefore, leakage alone will empty the cistern in 420 hours.
Step 3: Equations Based Pipes and Cisterns Problems:
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?
Assume that the tank can be filled by tap1 in X hours.
Tap 2 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.
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.
Step 4: Calculate Time Taken When Pipes and Cisterns Are Opened For Different Periods:
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.
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
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
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|
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)