# Program to find the number of men initially

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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..

Output:

1

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Improved By : andrew1234, vt_m, Rajput-Ji

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