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Program to find the number of men initially
  • Last Updated : 15 Apr, 2021

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 <bits/stdc++.h>
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




<script>
 
// Javascript implementation of above approach.
 
// Function to return the
// number of men initially
function numberOfMen(D, m, d)
{
 
    var Men = (m * (D - d)) / d;
 
    return Men;
}
 
// Driver code
var D = 5, m = 4, d = 4;
document.write(numberOfMen(D, m, d));
 
</script>
Output: 
1

 

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