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Minimum time required to fill a cistern using N pipes
• Last Updated : 13 Apr, 2021

Given the time required by a total of N+1 pipes where N pipes are used to fill the Cistern and a single pipe is used to empty the Cistern. The task is to Calculate the amount of time in which the Cistern will get filled if all the N+1 pipes are opened together.

Examples

```Input: n = 2,
pipe1 = 12 hours, pipe2 = 14 hours,
emptypipe = 30 hours
Output: 8 hours

Input: n = 1,
pipe1 = 12 hours
emptypipe = 18 hours
Output: 36 hours ```

Approach:

• If a pipe1 can fill a cistern in ‘n’ hours, then in 1 hour, the pipe1 will able to fill ‘1/n’ Cistern.
• Similarly If a pipe2 can fill a cistern in ‘m’ hours, then in one hour, the pipe2 will able to fill ‘1/m’ Cistern.
• So on…. for other pipes.

So, total work done in filling a Cistern by N pipes in 1 hours is

1/n + 1/m + 1/p…… + 1/z
Where n, m, p ….., z are the number of hours taken by each pipes respectively.

The result of the above expression will be the part of work done by all pipes together in 1 hours, let’s say a / b.
To calculate the time taken to fill the cistern will be b / a.

Consider an example of two pipes:

Time taken by 1st pipe to fill the cistern = 12 hours
Time taken by 2nd pipe to fill the cistern = 14 hours
Time taken by 3rd pipe to empty the cistern = 30 hours
Work done by 1st pipe in 1 hour = 1/12
Work done by 2nd pipe in 1 hour = 1/14
Work done by 3nd pipe in 1 hour = – (1/30) as it empty the pipe.
So, total work done by all the pipes in 1 hour is
=> ( 1 / 12 + 1/ 14 ) – (1 / 30)
=> ((7 + 6 ) / (84)) – (1 / 30)
=> ((13) / (84)) – (1 / 30)
=> 51 / 420
So, to Fill the cistern time required will be 420 / 51 i.e 8 hours Approx.

Below is the implementation of above approach:

## C++

 `// C++ implementation of above approach``#include ``using` `namespace` `std;` `// Function to calculate the time``float` `Time(``float` `arr[], ``int` `n, ``int` `Emptypipe)``{` `    ``float` `fill = 0;``    ``for` `(``int` `i = 0; i < n; i++)``        ``fill += 1 / arr[i];` `    ``fill = fill - (1 / (``float``)Emptypipe);` `    ``return` `1 / fill;``}` `// Driver Code``int` `main()``{``    ``float` `arr[] = { 12, 14 };``    ``float` `Emptypipe = 30;``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);` `    ``cout << ``floor``(Time(arr, n, Emptypipe)) << ``" Hours"``;` `    ``return` `0;``}`

## Java

 `// Java implementation of``// above approach``import` `java.io.*;` `class` `GFG``{``    ` `// Function to calculate the time``static` `float` `Time(``float` `arr[], ``int` `n,``                  ``float` `Emptypipe)``{``    ``float` `fill = ``0``;``    ``for` `(``int` `i = ``0``; i < n; i++)``        ``fill += ``1` `/ arr[i];` `    ``fill = fill - (``1` `/ (``float``)Emptypipe);` `    ``return` `1` `/ fill;``}` `// Driver Code``public` `static` `void` `main (String[] args)``{``    ``float` `arr[] = { ``12``, ``14` `};``    ``float` `Emptypipe = ``30``;``    ``int` `n = arr.length;``    ` `    ``System.out.println((``int``)(Time(arr, n,``                        ``Emptypipe)) + ``" Hours"``);``}``}` `// This code is contributed``// by inder_verma.`

## Python3

 `# Python3 implementation of``# above approach` `# Function to calculate the time``def` `Time(arr, n, Emptypipe) :` `    ``fill ``=` `0``    ``for` `i ``in` `range``(``0``,n) :``        ``fill ``+``=` `(``1` `/` `arr[i])` `    ``fill ``=` `fill ``-` `(``1` `/` `float``(Emptypipe))` `    ``return` `int``(``1` `/` `fill)`  `# Driver Code``if` `__name__``=``=``'__main__'``:``    ``arr ``=` `[ ``12``, ``14` `]``    ``Emptypipe ``=` `30``    ``n ``=` `len``(arr)``    ``print``((Time(arr, n, Emptypipe))``          ``, ``"Hours"``)` `# This code is contributed by``# Smitha Dinesh Semwal`

## C#

 `// C# implementation of``// above approach``using` `System;` `class` `GFG``{``    ` `// Function to calculate the time``static` `float` `Time(``float` `[]arr, ``int` `n,``                  ``float` `Emptypipe)``{``    ``float` `fill = 0;``    ``for` `(``int` `i = 0; i < n; i++)``        ``fill += 1 / arr[i];` `    ``fill = fill - (1 / (``float``)Emptypipe);` `    ``return` `1 / fill;``}` `// Driver Code``public` `static` `void` `Main ()``{``    ``float` `[]arr = { 12, 14 };``    ``float` `Emptypipe = 30;``    ``int` `n = arr.Length;``    ` `    ``Console.WriteLine((``int``)(Time(arr, n,``                             ``Emptypipe)) +``                                ``" Hours"``);``}``}` `// This code is contributed``// by inder_verma.`

## PHP

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## Javascript

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Output:
`8 Hours`

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