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Aptitude

Introduction to Pipes and Cisterns

Tips and Tricks Of HCF and LCM
Inlet pipe is the pipe for filling the Cistern while Outlet Pipe is for emptying the cistern.Suppose a  pipe can fill a tank in x hours, Then, the pipe can fill same tank in 1 hour = 1n \frac{1}{n} parts . Let us look at some of the Tips And Tricks for Pipes and Cisterns that are given below.
Tips and Tricks Of HCF and LCM

Pipes and Cisterns Tips and Tricks & Shortcuts

  • Here, are quick and easy tips and tricks for you to solve Pipes and Cisterns questions quickly, easily, and efficiently in competitive exams.

Tips and tricks and shortcuts of  Pipes and Cisterns questions

  • Pipes and Cisterns questions on are almost like Time and Work questions.
  • In pipes and cistern problems, the portion of the tank are filled or emptied is calculated and it is relatively same as the amount of work done in time and work problems.
  • Similarly, time taken to fill or empty a tank completely or to a desired level is the time taken to do a piece of work.
Type 1: Calculate time taken to fill a tank by two or more pipes

Question 1. Pipes A and B can fill a tank in 5 and 6 hours respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together, then the tank will be filled in:

Options

A. $1\frac{13}{17}$ hrs

B. $2\frac{8}{11}$ hrs

C. $3\frac{9}{17}$ hrs

D. $4\frac{1}{2}$ hrs

Solution :

$\eqalign{ & {\text{Net}}\,{\text{part}}\,{\text{filled}}\,{\text{in}}\,{\text{1}}\,{\text{hour}} \cr & = {\frac{1}{5} + \frac{1}{6} - \frac{1}{{12}}} = \frac{{17}}{{60}} \cr & \therefore {\text{The}}\,{\text{tank}}\,{\text{will}}\,{\text{be}}\,{\text{full}}\,{\text{in}}\,\frac{{60}}{{17}}\,{\text{hours}} \cr & i.e.,\,3\frac{9}{{17}}\,{\text{hours}} \cr} $

Correct option: A

Type 2 : Tips and Tricks and Shortcuts for Pipes and Cisterns Problems related to Calculate time taken to fill a tank having a hole or leakage

Question 1. A pump can fill a tank with water in 2 hours. Because of a leak, it took 2\frac{1}{2} hours to fill the tank. The time taken to empty the full tank due to leakage (in hours) is :

Options

A. \frac{10}{3} hrs

B. \frac{3}{10} hrs

C. \frac{4}{5} hrs

D. \frac{1}{2} hrs

Solution :

Work done by the leak in 1 hour
$\eqalign{& = {\frac{1}{2} - \frac{2}{5}} = \frac{3}{{10}} \cr}$
∴ Leak will empty the tank in \frac{10}{3} hours

Correct option: A

Type 3: Calculate Time Taken When Pipes Are Opened For Different Periods

Question 1: Two pipes A and B can fill a tank in 15 minutes and 20 minutes respectively. Both the pipes are opened together but after 4 minutes, pipe A is turned off. What is the total time required to fill the tank?

Options

A. 10 min. 20 sec.

B. 11 min.

C. 14 min. 40 sec.

D. 12 min. 30 sec.

Solution :

$\eqalign{ & {\text{Part}}\,{\text{filled}}\,{\text{in}}\,{\text{4}}\,{\text{minutes}} \cr & = 4\left( {\frac{1}{{15}} + \frac{1}{{20}}} \right) = \frac{7}{{15}} \cr & {\text{Remaining}}\,{\text{part}} = {1 - \frac{7}{{15}}} = \frac{8}{{15}} \cr & {\text{Part}}\,{\text{filled}}\,{\text{by}}\,B\,{\text{in}}\,{\text{1}}\,{\text{minute}} = \frac{1}{{20}} \cr & \therefore \frac{1}{{20}}:\frac{8}{{15}}::1:x \cr & x = {\frac{8}{{15}} \times 1 \times 20} \cr & \,\,\,\,\,\, = 10\frac{2}{3}\,\min \cr & \,\,\,\,\,\, = 10\min .\,40\,\sec . \cr & \therefore {\text{The}}\,{\text{tank}}\,{\text{will}}\,{\text{be}}\,{\text{full}}\,{\text{in}}\, \cr & = {4\min . + 10\min . +\, 40\sec .} \cr & = 14\min .\,40\sec . \cr} $

Correct option: C