Problems related to Pipes and Cistern are the same as those of time and work.

Inlet pipe:- A pipe which fills up the tank is known as inlet.
Outlet pipe:- A pipe which empties a tank is known as outlet.

Important Points

1. If a pipe can fill a tank in x hours, then the part of tank filled in 1 hour =	1
   	x

   Example:

   A pipe can fill the tank in 5 hours, then the tank filled in 1 hour =	1
   	5

   2. If a pipe can fill 1/x part in 1 hour, then it can fill the whole tank in x hours.

   Example: If a pipe can fill 1/3 part of a tank in 1 h, then it can fill the whole tank in 3 h.
   3. Time taken to fill a tank is taken positive and time taken to empty a tank is taken negative.

Important Formulas

Rule 1: If a pipe can fill a tank in x hours and an another pipe can fill the same tank in y hours, then the

Example: Two pipes A and B can fill a tank in 9 and 18 hours, respectively. If both the pipes are opened simultaneously, how much time will be taken to fill the tank ?
Solution: Part filled by A in 1 hour = 1/9
Part filled by B in 1 hour = 1/18

Part filled by ( A + B ) in 1 hour =	1	+	1	=	(2 + 1)	=	3	=	1
	9		18		18		18		6

Hence, both the pipes together will fill the tank =	1	= 6 hours
	1
	6

By Formula,
Here, x = 9 hours and y = 18 hours

=	(9 × 18)	=	9 × 18	= 6 hours
	9 + 18		27

Rule 2: If a pipe fill a tank in x hours, and an another pipe can empty in y hours, then the

Example: A pipe can fill a tank in 3 hours, while another pipe can empty it in 4 hours. If both the pipes are opened simultaneously, how much time will be taken to fill the tank ?

Solution:
Part filled by 1st pipe in 1 hours = 1/3
Part filled by 2nd pipe in 1 hours = 1/4

Part filled in 1 h by both pipes =	1	=	1	=	(4 - 3)	=	1
	3		4		12		12

Hence, the tank will be filled completely in = 12 hours

By Formula,

Here, x = 3 hours and y = 4 hoours

=	(3 × 4)
	(4 - 3)

Example: A pipe can fill a tank in 3 hours and because of a leakage it is taking 3.5 hours to fill the tank. Find the time in which the leakage can drain all the water when it is full.
Solution: Let the lick drain all the water in B min.
Part filled by A in 1 min = 1/3
Part filled by ( A + B ) in 1 min = 1/3.5
Part drain by lick in 1 min = 1/B
According to the question,

=	1	-	1	=	1
	3		B		3.5

or,	1	-	1	=	1
	3		3.5		B

or,	7	-	6	=	1
	21		21		B

or,	1	=	1
	21		B

∴ B = 21 min
So, the leakage will drain all water in 21 min.

Rule 3: If three pipes can fill a tank separately in x, y and z hours respectively, then part of tank filled in 1 hour by all three pipes 1/x + 1/y + 1/z and total

Example: Three pipes A, B and C can fill a tank separately in 4 hours, 5 hours and 6 hours respectively. Find the time taken by all the three pipes to fill the tank when the pipes are opened together.
Solution:-
Part filled by A in 1 hour = 1/4
Part filled by B in 1 hour = 1/5
Part filled by C in 1 hour = 1/6

Part filled by ( A + B + C ) =	1	+	1	+	1
	4		5		6

Part filled by ( A + B + C ) =	( 15 + 12 + 10 )
	60

=	37
	60

∴ Time taken to fill the tank =	37
	60

=  1	23
	27

By Formula,

Here, x = 4 hours , y = 5 hours and z = 6 hours

=	(4 × 5 × 6)
	(5 × 6 + 4 × 6 + 4 × 5)

=	120
	(30 + 24 + 20)

=	60
	37

=  1	23	= hours
	37

Rule 4: If two taps A and B, which can fill a tank, such that efficiency of A is x times of B and takes t min less/more than B to fill the tank, then

Example: If tap A can fill a tank in 4 times faster than tap B and takes 24 min less than tap B to fill the tank. If both the taps are opened simultaneously, then find the time taken to fill the tank.
Solution: Let the time taken by pipe A to fill the tank be x min. Then, time taken by pipe B to fill the tank be 4x min.
According to the question,
4x - x = 24
or, 3x = 24 or, x = 24/3 = 8
∴ Time taken by pipe A = 8 min
Time taken by pipe B = 4 × 8 = 32 min
Time taken to fill the tank by pipe A in 1 min = 1/8
Time taken to fill the tank by pipe B in 1 min = 1/32

Time taken to fill the tank by pipes in 1 min =	1	+	1
	8		32

=	(4 + 1)
	32

=	5	min.
	32

So, time taken by both pipes to fill the tank working together =	5	min.
	32

By Formula,
Here, x = 4 and t = 24

Time taken to fill the tank by both pipes together =	xt	min.
	(x ² - 1)

=	( 4 × 24 )
	(42 - 1)

= 96/15
= 32/5 min

Capacity of tank

=		c	-   y		=	c
		x				z

or,	c	-	c	=   y
	x		z

or, c	1	-	1	=   y
	x		z

or, c		z -	x		=   y
			xz

Where, c = capacity of tank
y = admit water

x = time required to empty
z = new time required to empty

Example: A cistern had leakage which can empty in 4 hours. A pipe which admit 10 L of water per minute into the cistern is turn on and now the cistern is empty in 12 h, Find the capacity of tank the tank.
Solution:- Let the capacity of the tank be c litres.
According to the question,

=	c	- (10 × 60) =	c
	4		12

or,	c	-	c	=  (10 × 60)
	4		12

or, 3c -	c	=  600
	12

or,	2c	=  600
	12

or,	c	=  600
	6

or, c = 600 × 6
∴ Capacity = 3600 L

By Formula,
Here, x = 4 hours , y = 10 L per minute = 10 × 60 = 600 L per hour and z = 12 hours

∴ c   =	xyz
	(z - x)

=	4 × 600 × 12
	(12 - 4)

=	4 × 600 × 12
	8

= 3600 L

Example: A pipe can fill a tank in 12 min and B can fill the same tank in 16 min. Both pipes are open together. But 4 min before the tank is full, one pipe A is closed. How much time will it take to fill the tank.
Solution: Let pipe B open for x min.
then, pipe A open for ( x - 4 ) min.
Part filled by pipe A in 1 min = 1/12 min
Part filled by pipe A in ( x - 4 ) min = ( x - 4 )/12
Part filled by Pipe B in 1 min = 1/16
Part filled by pipe B in x min = x/16

=	(x - 4)	+	x	= 1
	12		16

or,	4 ( x - 4 ) + 3x	= 1
	48

or, 4x - 16 + 3x 48
or, 7x = 48 + 16
or, 7x = 64
or, x = 64/7

∴ Time required to fill the tank =	67	min.
	7

Example: Pipe A can fill a tank in 8 hours , and pipe B in 6 hours. If both the pipes are open and after 2 hours , pipe A is closed then how much time is required to fill the tank by pipe B?
Solution: Time taken by pipe A = 8 hours
Part filled by pipe A in 1 hour = 1/8

Part filled by pipe A in 2 hour =	1	× 2 =	1
	8		4

Time taken by pipe B = 6 hours
Part filled by pipe B in 1 hour = 1/6

Part filled by pipe B in 2 hour =	1	× 2 =	1
	6		3

Part filled by both ( A + B ) in 2 hour =		1	+	1
		4		3

=	( 3 + 4 )
	12

= 7/12

Remaining part = 1 -	7	=	5
	12		12

=	1
	12
	1
	6

=	(5 × 6)
	12

= 5/2
= 2.5 hour