Pipes and Cistern

Why Are The Pipes And Cistern Problems Important?

Questions based on pipes and cistern is a very common and repetitive topic, from which questions  are asked frequently in the general aptitude sections for various competitive exams like bank exams, SSC, RRB, Insurance, Olympiads and others. 

Candidates who are preparing themselves for the upcoming exams must know that topics like these are quite important and easy to understand. They are easy to score in. Just get through the logic and you will be able to get those very important marks to add on to your total score.

The Concept Of Pipes And Cistern: 

The concept is based on work and time. It includes statement problems like time taken to empty or fill the reservoir.       

Make sure you are perfectly understanding the terminologies like inlet, outlet, work (don’t consider it as how work is defined in physics, its a general term used in mathematics logics) etc.

Try to understand the formulas, through your own logic, make diagrams, solve many questions, to grab the correct way of solving these types of pipes and cistern problems.

Extremely Important Concepts And Formulas To Solve The Problems On Pipes And Cistern:

1) Part of reservoirs filled or emptied in one hour:

• Let us imagine that ‘A’ hours are required to fill up a tank or reservoir, thus 1/A will be the part that will be  filled in 1 hr 

• Let us imagine to empty the tank that ‘A’ hours is needed, thus 1/A will be the part emptied in 1 hour 

2) Now let us consider if both inlet and outlet pipes are on:

Net amount of water in reservoir will be logically = (Sum of work done by Inlets/water in) – (Sum of work done by Outlets/water out)

           Case 1: When outlet is more than the inlet: 

If a tank can be filled in “A” hours by an inlet pipe  and emptied by  another outlet pipe in “B” hours. When both the pipes are opened at the same time, then the net part of the tank is filled in 1 hr :

= {(BA) / (B-A)}

Case 2: When inlet is more than the outlet:

If an inlet pipe can fill a tank in “A” hours and an outlet pipe can empty the same tank in “B” hours. When both the pipes are opened at the same time, then the net part of the tank filled in 1 hr

= {(BA) / (A-B)}

NOTE: If the inlet and outlet are at the same pace, net work is 0, and logically, it will take infinite time to empty the tank, logically, the tank will never be emptied.

When Two Inlets Are Given:

One inlet can fill the tank in “A” hr and the other inlet can fill the same tank in “B” hrs, if both the inlets are opened at the same time, the time taken to fill the whole tank 

= {(AB) / (A+B)}= 1/ {1/A + 1/B}

When Two Inlet And One Outlet Pipe Is Given:

If two inlet pipes take “A” and “B” hours respectively to fill a tank of water and a third outlet pipe is opened which takes “C” hours to empty the tank, then the time taken to fill the tank 

= 1/ { (1/A)+(1/B)+(1/C)} 

 And the net part of the tank filled in 1 hr = (1/A)+(1/B)-(1/C)

Will The Formulas Be Enough To Solve The Problem?

Just rot learning won’t help, try to understand the formulas, through your own logic, make diagrams, solve many questions, to grab the correct way of solving these types of pipes and cistern problems.

Make sure you are perfectly understanding the terminologies like inlet, outlet, work (don’t consider it as how work is defined in physics, its a general term used in mathematics logics), etc.

Though you should practise a lot, search problems on pipes and cistern on google.

Here Are Two Sample Questions For Your Reference- 

(You will find the same or similar question all around the web and reference books. These are two famous questions)

Q.1 A tank is filled by 3 pipes with uniform flow. Operating simultaneously, the first 2 pipes fill the tank at the same time in which the tank is filled by the third pipe alone. The tank is filled faster by 5 hours through the 2nd pipe than the 1st pipe and the 3rd pipe does it slower by 4 hours. The time required by the 1st pipe is?

1. 15 hours

2. 6 hours

3. 30 hours

4. 10 hours

Answer: (1) 15 hours

Solution:

Suppose, To fill the tank the 1st pipe alone takes x hours 

Then, 2nd and 3rd pipes will take (x-5) and (x-9) hours respectively to fill the tank.

Therefore, (1/x) + {1/(x-5)} = {1/(x-9)}

[(x-5+x)/{x(x-5)}] = {1/(x-9)}

(2x-5)(x-9) = x(x-5)

X2 – 18x + 45 = 0

(x-15)(x-3) = 0

X = 15 [neglecting x=3 ] 

Q 2. A tank can be filled by Two pipes X and Y in 20 minutes and 15 minutes respectively. Both the pipes are opened simultaneously but after 4 minutes, pipe Y is turned off. What is the total time required to fill the tank? 

1. 10 min. 20 sec

2. 12 min. 30 sec

3. 11 min. 45 sec

4. 14 min. 40 sec

Answer: (4) 14 min. 40 sec

Solution:

Part filled in 4 minutes = 4{(1/15) + (1/20)} = 7/15 

Remaining part = {1 – (7/15)} = 8/15 

Part filled by B in 1 minute = 1/20 

(1/20) : (8/15) :: 1 : x 

x = {(8/15) x 1 x 20} = 10 (⅔) min = 10 min. 40 sec

The tank will be full in (4 min. + 10 min. + 40 sec.) = 14 min. 40 sec.