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)
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)
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)
Or 3x = 24
X = 8
So B should be closed after 8 min.
Pipe B should be closed after (1 – 18/24) * 32
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
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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