Pipes and Cistern Short Tricks : How to Solve Problems in Easy way

The problems often asked on pipes and cistern have similarity with problems on Time and Work. Hence the approach to solve the problems became very easy if you have good command on concept of Time and Work.

So before moving forward I would recommended you to read The Article on Time and Work Short Tricks.

Before going through I would like to provide some terminology for pipe and cistern –

  • An inlet pipe fill the tank/cistern.
  • An outlet pipe or waste pipe empties a cistern.
  • Time – Stands for time taken for filling or emptying.
  • When a cistern is filled completely, then amount of work done (filling) = 1

Pipe and Cistern Short Tricks

Let’s move on some basic concept for Pipes and Cistern Short Tricks :

Concept 1. –

If an inlet pipe can fill a cistern in ‘x’ hours, then

Pipe and Cistern with Short Tricks

Concept 2. –

If an outlet pipe can empty a cistern in ‘y’ hours, then

Pipe and Cistern Short Tricks

Concept 3. –

Net work done in 1 hours = (filling work in 1 hour) – (Empty work in 1 hour)

Pipe and Cistern

If W is ( – ) ve, then cistern is emptied.

Concept 4. –

short tricks for Pipe and Cistern

Concept 5. –

If more than one inlet pipe or more than one out let pipes are fitted, then

pipe and tank short cut
aptitude shortcut

Time taken to fill or Empty = 1/part filled or emptied in 1 hour

Concept 6. –

If one inlet pipe can fill in t1 hours and one outlet pipe can empties it in t2 hours, then part of cistern filled or emptied in 1 hours = (1/t1 – 1/t2)

Note : Chain rule will work here same as Time and Work problem.

Problem on Leakage:

Two fill pipes can separately fill a cistern, say, in ‘x’ hours and ‘y’ hours respectively, but due to leak it takes ‘P’ hours extra to fill the cistern. Now both pipes are closed and the fill cistern can be emptied through the leak in ‘T’ hours,

competition math short cut

If there is only one fill pipe, then above relation can be reduces to

Empty time by leak, T

aptitude short tricks

Also read – Easiest way to find out the square of any number

Lets try to solve some question of Pipe and Cistern

Q 1. A cistern is filled in 9 hours and it takes 10 hours when there is a leak in its bottom. If the cistern is full, in which time shall the leak empty it?

(A) 90 h                                                (B) 94 h

(C) 92 h                                                 (D) 91 h

Q 2. Two filling pipes can fill a cistern in 10 and 12 min. respectively and when the waste pipe is open, they can together fill it in 15 min. The waste pipe can empty the full cistern in –

(A) 15 min.                                          (B) 37 min.

(C) 60/7 min.                                      (D) 15/2 min.

Q 3. A cistern can be filled by two pipes A and B in 15/2 hours. Both of them are open for 5/2 hours and then B is closed. A alone now requires 20/3 hours more to fill cistern. How long would A take to fill the cistern working alone?

(A) 8 h                                                   (B) 10 h

(C) 9 h                                                   (D) 11 h

Q 4. A cistern has three pipes A, B and C the pipes A and B can fill it in 4 and 5 hours respectively and pipe C can empty it in 2 hours. If the pipes are opened in order at 1, 2 and 3 a.m. respectively, when will the cistern be empty?

(A) 3 p.m.                                            (B) 4 p.m.

(C) 7 a.m.                                             (D) 6 a.m.


Answer Key

Q. 1Q. 2Q. 3Q. 4


Explanation : By using Pipes and Cistern Short Tricks you can solve the above question easily –

Ans 1. Use T = x(x+p) / p = 9 * 10/ 1 = 90 hours.

Ans 2. From formula – 1/Empty time = 1/10 + 1/12 – 1/15 = 7/60 hence answer is 60/7 hours.

Ans 3. Let the filling time for A and B are Fa and Fb then both are open 5/2 hours,

Part filled = 5/2 * [ 1/Fa + 1/Fb] = 5/2 x (20/15)

Next A is open for 20/3 hours, part filled = 20/3 x [1/Fa]

So cistern is filled so, [1/3] + 20/3 x [1/Fa] = 1  ⇒ Fa = 10 hours.

Ans 4.  Upto 3 a.m. both pipes fill the tank = 2/4 + 1/5 = 7/10 th part.

Remaining 3/10 part is filled in x hours, then

(1/4 + 1/5 – 1/2)*X = 3/10  ⇒ X = 6 hours. So the cistern is filled at 1am + 6 hours = 7 a.m.

Leave comment if you have any confusion.

Thank you.. good Luck  😎