# PIPES AND CISTERNS

Quantitative Aptitude Study Notes for Bank & SSC
Exam
PIPES AND CISTERNS
You know that quantitative
aptitude section is most important in bank
exams in PO and
Clerk and for
other competitive exams because if you want good score in bank exam then you
have to score good in maths. In competitive exams the most important thing is
time management, if you know how to manage your time then you can do well in Bank Exams. That’s where maths shortcut
tricks and formula are comes into action. So continuously we are providing
shortcut tricks on different maths topics.
Today’s topic is PIPES AND CISTERNS. This is the one of the most important
topic in quantitative aptitude section in bank and ssc exam. You should know how
to calculate pipe and cisterns questions and answers in very short time for
bank exam. From this chapter around 1-2 questions are given in the SBI and IBPS exams. For this here we
are providing shortcut tricks and quicker
method
to calculate pipe and
cisterns
in very short time.

Pipes and Cisterns problems are
almost the same as those of Time and Work problems. Thus, if a pipe fills a
tank in 6 hrs, then the pipe fills 1/6 th of the tank in 1 hour. The only
difference with pipes and Cisterns problems is that there are outlets as well
as inlets. Thus, there are agents (the outlets) which perform negative work
too. The rest of the process is almost similar.
Inlet: A pipe connected with a tank (or
a cistern or a reservoir) is called an inlet, if it fills it.
Outlet: A pipe connected with a tank is
called an outlet, if it empties it.
DIRECT FORMULAE AND TIPS FOR PIPES AND CISTERNS QUESTIONS:
If we want to solve pipe and
cisterns questions or any other type of questions, then the first thing that we
need that is Formulas about that topic. So here is the list of formulas that is
used in time and distance quantitative topic.

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

ii.
If a pipe can empty a tank in y hours, then the part of the full
tank emptied in 1 hour = 1/y.

iii.
If a pipe can fill a tank in x hours and another pipe can empty
the full tank in y hours, then the net part filled in 1 hour, when both the
pipes are opened = (1/x – 1/y). Time taken to fill the tank, when both the
pipes are opened = xy / y-x

iv.
If a pipe can fill a tank in x hrs and another can fill the same
tank in y hrs, then the net part filled in 1 hr, when both the pipes are opened
= (1/x + 1/y).
Time taken to fill the tank = xy
/ y+x

v.
If a pipe fills a tank in x hrs and another fills the same tank in
y hrs, but a third one empties  the full
in z hrs, and all of them are opened together, the net part filled in 1 hr  = (1/x + 1/y + 1/z)
time taken to fill the tank = xyz
/ yz + xz – xy hrs.

vi.
A pipe can fill a tank in x hrs. Due to a leak in the bottom it is
filled in y hrs. If the tank is full, the time taken by the  leak to empty the tank = xy / y-x hrs.
Here, we are providing some of
the examples on pipes and cisterns
questions and their solutions according to bank
exam.
EXAMPLE 1:
Two pipes A and B can fill a tank in 36 hrs and 45 hrs respectively. If both
the pipes are opened simultaneously, how much time will be taken to fill the tank?

Solution: part filled by A alone
in 1 hour = 1/36
Part filled by B alone in 1 hour
= 1/45
Part filled by (A+B) in 1 hour =
(1/36 + 1/45)
= 9/180
= 1/20
Hence both the pies together will
fill the tank in 20 hours.
Direct method: by formula (iv)
Time taken = 36*45 /36+45
= 20 hrs.
EXAMPLE 2: A
pipe can fill a tank in 15 hrs. Due to a leak in the bottom, it is filled in 20
hours. If the tank is full, how much time will the leak take to empty it ?
Solution: Work done by the leak
in 1 hour = (1/15 – 1/20)
= 1/60
Direct method by formula (vi)
Required time = 15*20 / 20-15
= 60 hrs.
EXAMPLE 3:
Two pipes A and B can fill a tank in 24 minutes and 32 minutes respectively. If
both the pipes are opened simultaneously, after how much time should B be
closed so that the tank is full in 18 minutes?
Solution: let B be closed after x
minutes. Then, part filled by (A+B) in x min. + part filled by A in (18 – x)
min = 1
X ( 1/24 + 1/32) + (18-x) * 1/24
= 1
Or 7/96 + 18-x/24 = 1
Or 7x + 4(18- x)
= 96
Or 3x = 24
X = 8
So B should be closed after 8
min.
Direct formula:
Pipe B should be closed after (1 –
18/24) * 32
8 min.
EXAMPLE 4:
If three taps are opened together, a tank is filled in 12 hrs. One of the taps
can fill it in 10 hrs and another in 15 hrs. How does the third tap work?
Solution: We have to find the
nature of the third tap – whether it is a filler or a waster pipe.
Let it be a filler pipe which
fills in x hrs.
Then, 10*15*x / 10*15+10x+15x
= 12
Or, 150x = 150*12*25x*12
Or -150x  = 1800
X = – 12
-ve sign shows that the third
pipe is a waste pipe which vacates the tank in 12 hrs.
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