Pipes and Water Tank in maths is another form of time and work-based questions. All the time and work-related problems such as work is done or time is taken to fill the water tank are included in the Pipes and Water Tank concept. The particulars connected to the water tank and it can be used to fill the water tank as well as to empty the water tank are called ‘inlet’ and ‘outlet’ pipes. Water Tanks are also called Cisterns.
Inlet Pipe: If the pipe is used to fill the tank, then it is called an ‘Inlet Pipe’.
Outlet Pipe: when the pipe which is attached to the tank is used to empty the tank that is called an ‘Outlet Pipe’.
The work done by an inlet pipe is considered positive work done by the pipe. The work done by an outlet pipe is considered as negative work done by the pipe or we can call it leakage. Time taken to fill or empty the tank is taken as work is done.
Do Check: Problems on Time and Work
Pipes and Cisterns Formulas
To easily solve the problems on pipes and water tanks or cisterns, we need to know the below-mentioned formulas. They are
- If a tank filled in ‘a’ hours by pipe, then the part of the tank filled in 1 hour is equal to \(\frac{ 1 }{ a }\).
Example: a tank filled in ‘4’ hours by a pipe, then the part of the tank filled in 1 hour is \(\frac{ 1 }{ 4 }\). - If a pipe can take the ‘b’ hours of time to empty the tank, then the empty part of the tank in 1 hour is equal to \(\frac{ 1 }{ b }\).
Example: a pipe can empty a tank in 6 hours, then the tank is emptied in 1 hour is \(\frac{ 1 }{ 6 }\) - If the two pipes are working at a time. That means, inlet pipe fills the full tank in ‘a’ hours and the outlet pipe empties the full tank in ‘b’ hours at a time. So the final part of the tank filled in 1 hour is equal to \(\frac{ 1 }{ a }\) – \(\frac{ 1 }{ b }\) (where a > b).
- So, the time taken to fill the tank is (where a > b).
Example: a pipe can fill the tank in ‘8’ hours and another pipe can empty the tank in ‘6’ hours. Then the net amount of part filled in 1 hour is equal to \(\frac{ 1 }{ 8 }\) – \(\frac{ 1 }{ 6 }\).Time taken to fill the tank is \(\frac{ (8) (6) } { 8-6 }\).= 24 hours. - If the two pipes are working at a time. That means, inlet pipe fills the full tank in ‘a’ hours and the outlet pipe empties the full tank in ‘b’ hours at a time. So the final part of the empty tank in 1 hour is equal to \(\frac{ 1 }{ b }\) – \(\frac{ 1 }{ a }\) (where b > a).
- So, the time taken to empty the full tank is equal to\(\frac{ ab }{ b-a }\) (where b > a).
Example: a pipe can fill the tank in ‘5’ hours and a pipe can empty the tank in ’10’ hours. So the part of the empty tank in 1 hour is equal to \(\frac{ 1 }{ 10 }\) – \(\frac{ 1 }{ 5 }\). Time taken to empty the tank is \(\frac{ (10)(5) }{ 10-5 }\) = 10 hours. - If the two pipes are used to fill the tank at a time. The first pipe takes ‘a’ hours and the other pipe takes ‘b’ hours of time to fill the tank. Then the net amount of part filled by two pipes is \(\frac{ 1 }{ a }\) + \(\frac{ 1 }{ b }\).
Example: two pipes are filled a tank by taking the times ‘2’ hours and ‘4’ hours. So the net amount of part filled by two pipes is \(\frac{ 1 }{ 2 }\) + \(\frac{ 1 }{ 4 }\) = \(\frac{ 3 }{ 4 }\). - The total time is taken to fill the full tank by two pipes ‘a’ and ‘b’ is \(\frac{ ab }{ a+b }\). If a = 2 and b = 4, then the total time taken to fill the tank is = \(\frac{ (4)(2) }{ 4+2 }\) = \(\frac{ 4 }{ 3 }\).
- Same as above, two pipes are working as outlet pipes. One pipe can empty the tank in ‘a’ hours and another pipe can empty the tank in ‘b’ hours. Then the net amount of part emptied in 1 hour is equal to \(\frac{ 1 }{ a }\) + \(\frac{ 1 }{ b }\).
Some More Important Formulas on Pipes and Water Tank Concept
- If a pipe can fill the tank in ‘a’ hours, another pipe can fill the tank in ‘b’ hours and third pipe can empty the tank in ‘c’ hours. The final part filled in 1 hour is equal to \(\frac{ 1 }{ a }\) + \(\frac{ 1 }{ b }\) – \(\frac{ 1 }{ c }\).
- The total time taken to fill the tank is \(\frac{ abc }{ bc+ac-ab }\)
Example: Two pipes are taken time to fill the tank is 2 hours and 6 hours. Another pipe can take 5 hours of time to empty the tank. So the net amount of part filled in 1 hour is \(\frac{ 1 }{ 2 }\) + \(\frac{ 1 }{ 6 }\) – \(\frac{ 1 }{ 5 }\).The total time taken to fill the tank is \(\frac{ (2)(6)(5) }{ (6)(5)+(2)(5)-(2)(6) }\) = \(\frac{ 60 }{ 40-12 }\) = \(\frac{ 60 }{ 28 }\) = \(\frac{ 15 }{ 7 }\). - If three pipes can fill the tank in ‘a’ hours, ‘b’ hours, and ‘c’ hours individually, then the time taken by all three pipes together to fill the water tank is \(\frac{ abc }{ bc+ac+ab }\) hours.
Example: three pipes are taken 2 hours, 4 hours, 6 hours of time to fill the cistern. If we open all the three pipes at a time, then the total time to fill the cistern is = \(\frac{ abc }{ bc+ac+ab }\) = \(\frac{ (2)(4)(6) }{ (4)(6)+(2)(6)+(2)(4) }\) = \(\frac{ 48 }{ 24+12+8 }\) = \(\frac{ 48 }{ 44 }\)hours. - If the two pipes A and B can fill the cistern in ‘a’ and ‘b’ min and after some time pipe B is closed. So, the cistern is filled in ‘x’ min and that is equal to {b[1 – x/a]}min.
Example: two pipes A and B fill the water tank in ‘6’ and ‘3’ min and after how much time pipe B closed to fill the water tank in 2 min. That is {b[1 – x/a]} = {3[1 – ]} = {3[]} = 2min. - If two pipes A and B can fill the cistern in time ‘X’ together. If the time taken by pipe A alone is greater than X (a > X) and the time taken by pipe B alone is greater than X (b > X), then the total time X = √ab.
Example: Two pipes A and B are opened together and fills the water tank in time ‘X’ min. Pipe A alone can fill the tank in 100 min which is greater than ‘X’ min. Pipe B alone can fill the tank in 36 min which is greater than ‘X’ min. So, the total time taken by the X = √(100)(36) = (10)(6) = 60 min. - If we have a number of pipes to fill the cistern and the number of pipes to empty the cistern.
Then the net amount of part to be filled by the pipes = (sum of the inlet pipes) + (sum of the outlet pipes).
The net amount of part emptied by the pipes = (sum of the inlet pipes) – (sum of the outlet pipes).
Problems on Pipes and Water Tank
In this concept, we are providing the number of problems on pipes and water tanks with solutions. Learn and practice the below problems and get the grip on pipes and cistern topics.
Example 1.
A water tank can be filled by a pipe in 10 hours and another pipe can empty the water tank in 15 hours. If the two pipes are opened simultaneously, then find the time to fill the tank completely.
Solution:
As per the given information, water can be filled by a pipe in 10 hours. That is a = 10 hours.
Another pipe can empty the water tank in 15 hours. That is b = 15 hours.
Total time to fill the tank = \(\frac{ ab }{ b-a }\) (where b > a).
By substituting the a and b values in the above equation. We will get \(\frac{ ab }{ b-a }\) = \(\frac{ (10)(15) }{ 15-10 }\) =\(\frac{ 150 }{ 5 }\) = 30 hours.
Therefore, by working two pipes simultaneously, the water tank can be filled out in 30 hours.
Example 2.
If a pipe can empty a tank in 20 hours, then find the time to empty the 3/4th of the part of the cistern?
Solution:
As per the given information, a pipe can empty the tank in 20 hours. That is a = 20 hours
Time to empty the 3/4th of the part of cistern = \(\frac{ 3 }{ 4 }\)(20) = 3 x 5 = 15.
Therefore, a pipe can empty the \(\frac{ 3 }{ 4 }\)th of the part of the cistern in 15 hours.
Example 3.
A pipe leakage can empty the full tank in 6 hours. Pipe B can fill the tank in 12 hours and another pipe can fill the tank in 8 hours. If we open three pipes at a time, then calculate the total time to fill the tank?
Solution:
As per the details, we have three pipes.
Pipe A can empty the tank in 6 hours. That is c = 6 hours.
Pipe B can fill the tank in 12 hours. That is a = 12 hours.
Pipe C can fill the tank in 8 hours. That is b = 8 hours.
Total time to fill the tank = \(\frac{ abc }{ bc + ca – ab }\).
Substitute the a, b, and c values in the above equation. Then we will get
= \(\frac{ (6) (12) (8) }{ (12) (8) + (8) (6) – (6) (12) }\)
= \(\frac{ 576 }{ 96+ 48 – 72 }\)
= \(\frac{ 576 }{ 72 }\)= 8 hours.
So, the total time to fill the tank is 8 hours.
Example 4.
Pipe A can fill a cistern in 4 hours but the leakage pipe B can empty the tank in 2 hours. Then find the time to empty the full cistern?
Solution:
As per the given information, Pipe A can fill the cistern in 4 hours. That is a = 4 hours.
Pipe B can empty the cistern in 2 hours. That is b = 2 hours.
Total Time to empty the cistern = \(\frac{ a b}{ a – b }\) (where a > b).
Substitute the a and b values in the above equation. Then we will get
\(\frac{ (4) (2) }{ 4 – 2 }\)= 4 hours.
So, the total time to empty the cistern is equal to 4 hours.
Example 5.
Three pipes are connected to the water tank. Pipe A, Pipe B, and Pipe C will take 8 hours, 12 hours, and 10 hours of time to fill the water tank individually. If we open three pipes at a time, then what is the time to fill the water tank?
Solution:
As per the given details, we have three pipes such as A, B, and C which are connected to the water tank.
Pipe A can fill the water tank in 8 hours. (a = 8 hours).
Pipe B can fill the water tank in 12 hours. (b = 12 hours).
Pipe C can fill the water tank in 10 hours. (c = 10 hours).
If three pipes are open at a time,
Total time to fill the water tank =\(\frac{ abc }{ ab + bc + ca}\)
Substitute the a, b, and c values in the above equation. Then we will get
=\(\frac{ (8) (12) (10) }{ (8) (12) + (12) (10) + (10) (8) }\)= \(\frac{ 960 }{ 96 + 120 + 80}\)
=\(\frac{ 960 }{ 296 }\) = 3 hours.
Therefore, a Water tank is filled in 3 hours time by three pipes.
FAQs on Pipes and Water Tank Concept
1. What is the use of inlet and outlet pipes?
Every cistern has two pipes. They are inlet and outlet pipes. To fill the cistern or water tank, an inlet pipe is used. An outlet pipe is used to empty the water tank or cistern.
2. If the Water Tank is filled in ‘X’ hours, then what is the amount partly filled in 1 hour?
If the water tank or cistern is filled in ‘X’ hours, then the water tank is filled in 1 hour is\(\frac{ 1 }{ X }\).
3. Pipe A and Pipe B can fill the water tank in ‘X’ and ‘Y’ hours. If we open the two pipes at a time, then what is the total time to fill the Water tank?
Pipe A can fill the water tank in ‘X’ hours. Pipe B can fill the water tank in ‘Y’ hours. So, the total time to fill the water tank or cistern by two pipes is\(\frac{ XY }{ X+Y }\).
4. Pipe A and Pipe B can fill the water tank in ‘X’ and empty the water tank in ‘Y’ hours. If we open the two pipes at a time, then what is the Total time to empty the Water tank?
Pipe A can fill the water tank in ‘X’ in hours and pipe B can empty the water tank in ‘Y’ hours. If the two pipes are open at a time, then the total time taken by the pipes to empty the water tank is \(\frac{ XY }{ X-Y }\) (where X > Y) and \(\frac{ XY }{ Y-X }\) (where Y > X).