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Shortcut Tricks to Solve - Pipes and Cisterns

SBI PO - IBPS PO - SSC CGL

Shortcut Rules to Solve Problems on Pipes and Cisterns

Effective for IBPS PO - SBI PO Exam

Dear Reader,

Here we will start a series of Quantitative Aptitude Shortcut Tricks for your upcoming SBI - IBPS - SSC and Other Government Competitive Exams. We will try to cover up all topics of the quantitative Aptitude Sections from which question was generally asked.

Note: The page may takes sometime to load the Quantitative formula's. If you face any problem just comment below the posts.

Trick - 1

  • 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 hours, when both the pipes are opened = $\left( \frac{1}{x}-\frac{1}{y} \right)$ 

: Time (T) taken to fill the tank, when both the pipes are opened
$=\frac{xy}{y-x}$
Trick - 2 

  • If a tap can fill a ${{x}_{1}}$ part of the cistern in  ${{t}_{1}}$ min and  ${{x}_{2}}$ part in ${{t}_{2}}$ min, then following expression is obtained

$\frac{{{t}_{1}}}{{{x}_{1}}}=\frac{{{t}_{2}}}{{{x}_{2}}}$
Trick - 3
  • A tap A can empty a cistern in x hours and the other tap B can empty is in y hours. If both emptying taps are opened together, then the time taken to empty the full cistern is given by 

 $\left( \frac{xy}{x+y} \right)hrs.$
Trick- 4
  • If a pipe can fills a tank in x hours and another fills the same tank in y hours, but a third one empties the full tank in z hours, and all of them are opened together, the net part filled in 1 hour

$=\left[ \frac{1}{x}+\frac{1}{y}-\frac{1}{z} \right]$
:time taken to fill the tank 
$=\left[ \frac{xyz}{yz+xz-xy} \right]hrs$
Trick - 5
  • Two pipes A and B can fill a cistern in x hours and y hours respectively. There is also an outlet C. If all the three pipes are opened together, the tank is full in T hrs, then the time taken by C to empty the full tank is 

$\left[ \frac{xyT}{yT+xT-xy} \right]hrs.$ 
Trick - 6

  • Two pipes A and B can fill a tank in x min and y min respectively. If  both the pipes are opened simultaneously, then the time after which pipe B should be closed, so that the tank is full in t min, is 

$\left[ y\left( 1-\frac{t}{x} \right) \right]\min utes$

Questions for Practice

Q1. A pipe can fill a tank in 10 hours and another pipe can empty it in 12 hours. If both the pipes are opened, find the time in which tank is filled. 

Q2. A pipe can fill $\frac{1}{4}$ of cistern in 16 min. In how many min, it can fill $\frac{3}{4}$
 
of the cistern ?


Q3. A pipe can empty a tank in 10 hours and another pipe can empty it in 5 hours. If both the pipes are opened simultaneously, find the time in which a full tank is emptied.

Q4. Pipe A can fills a tank in 20 hours while pipe B alone can fill it in 30 hours and pipe C can empty the full tank in 40 hours. If all the pipes are opened together, how much time will be needed to make the tank full ?

Q5. Two pipes A and B can fill a cistern in 1 hours and 75 mins respectively. There is also an outlet C. If all the three pipes are opened together, the tank is full in 50 mins. How much time will be taken by C to empty the full tank ?  


Q6. Two pipes A and B can fill a tank in 24 min and 32 min 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 ? 

Answers


Answer 1. 60 hrs
Answer 2. 48 minutes
Answer 3. $3\frac{1}{3}$ hrs 
Answer 4. $17\frac{1}{7}$ hrs 
Answer 5. 100 minutes 
Answer 6. 8 minutes
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