# Program to find the number of men initially

• Last Updated : 30 May, 2022

A certain number of men can do a certain piece of work in D days. If there were m more men engaged in the work then the work can be done in d days less. The task is to find how many men were there initially.
Examples:

Input: D = 5, m = 4, d = 4
Output: 1
Input: D = 180, m = 30, d = 20
Output: 240

Approach: Let the initial number of men be M and days be D
Amount of work completed M men in D days will be M * D
i.e. Work Done = M * D …(1)
If there are M + m men then the same amount of work is completed in D – d days.
i.e. Work Done = (M + m) * (D – d) …(2)
Equating equations 1 and 2,

M * D = (M + m) * (D – d)
M * D = M * (D – d) + m * (D – d)
M * D – M * (D – d) = m * (D – d)
M * (D – (D – d)) = m * (D – d)
M = m * (D – d) / d

Below is the implementation of the above approach:

## C++

 `// C++ implementation of above approach.``#include ``using` `namespace` `std;` `// Function to return the``// number of men initially``int` `numberOfMen(``int` `D, ``int` `m, ``int` `d)``{` `    ``int` `Men = (m * (D - d)) / d;` `    ``return` `Men;``}` `// Driver code``int` `main()``{``    ``int` `D = 5, m = 4, d = 4;` `    ``cout << numberOfMen(D, m, d);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.util.*;` `class` `GFG``{``    ` `// Function to return the``// number of men initially``static` `int` `numberOfMen(``int` `D, ``int` `m, ``int` `d)``{` `    ``int` `Men = (m * (D - d)) / d;` `    ``return` `Men;``}` `// Driver code``public` `static` `void` `main(String args[])``{``    ``int` `D = ``5``, m = ``4``, d = ``4``;` `    ``System.out.println(numberOfMen(D, m, d));` `}``}``// This code is contributed by Arnab Kundu`

## Python3

 `# Python implementation of above approach.` `# Function to return the``# number of men initially``def` `numberOfMen(D, m, d):``    ``Men ``=` `(m ``*` `(D ``-` `d)) ``/` `d;` `    ``return` `int``(Men);`  `# Driver code``D ``=` `5``; m ``=` `4``; d ``=` `4``;` `print``(numberOfMen(D, m, d));`  `# This code contributed by Rajput-Ji`

## C#

 `//  C# implementation of the approach``using` `System;` `class` `GFG``{``    ` `// Function to return the``// number of men initially``static` `int` `numberOfMen(``int` `D, ``int` `m, ``int` `d)``{` `    ``int` `Men = (m * (D - d)) / d;` `    ``return` `Men;``}` `// Driver code``public` `static` `void` `Main()``{``    ``int` `D = 5, m = 4, d = 4;` `    ``Console.WriteLine(numberOfMen(D, m, d));` `}``}` `// This code is contributed by anuj_67..`

## Javascript

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Output:

`1`

Time Complexity: O(1)

Auxiliary Space: O(1)

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