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

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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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//  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..

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Output:

1


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Improved By : andrew1234, vt_m, Rajput-Ji