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Program to find the profit or loss when CP of N items is equal to SP of M items

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Given N   and M   denoting that the Cost Price of N articles is equal to the Selling Price of M articles. The task is to determine the profit or Loss percentage.

Examples: 

Input:  N = 8, M = 9
Output: Loss = -11.11%

Input:  N = 8, M = 5
Output: Profit = 60%

Formula:-
 


Below is the implementation of the above approach:

C++

// C++ implementation of above formula
#include <bits/stdc++.h>
using namespace std;
 
// Function to calculate
// Profit or loss
void profitLoss(int N, int M)
{
    if (N == M)
        cout << "No Profit nor Loss";
    else {
 
        float result = 0.0;
 
        result = float(abs(N - M)) / M;
 
        if (N - M < 0)
            cout << "Loss = -" << result * 100 << "%";
        else
            cout << "Profit = " << result * 100 << "%";
    }
}
 
// Driver Code
int main()
{
    int N = 8, M = 9;
 
    profitLoss(N, M);
 
    return 0;
}

                    

Java

// Java implementation of above formula
public class GFG{
 
    // Function to calculate
    // Profit or loss
    static void profitLoss(int N, int M)
    {
        if (N == M)
            System.out.print("No Profit nor Loss");
        else {
     
            float result = 0;
     
            result = (float)(Math.abs(N - M))/ M;
     
            if (N - M < 0)
                System.out.print("Loss = -" + result * 100 +"%");
            else
                System.out.print("Profit = " + result * 100 + "%");
        }
    }
     
    // Driver Code
     public static void main(String []args){
          
        int N = 8, M = 9;
     
        profitLoss(N, M);
         
     }
     // This code is contributed by ANKITRAI1
}

                    

Python3

# Python 3 implementation of above formula
 
# Function to calculate Profit or loss
def profitLoss(N, M):
    if (N == M):
        print("No Profit nor Loss")
    else:
        result = 0.0
 
        result = float(abs(N - M)) / M
 
        if (N - M < 0):
            print("Loss = -",'{0:.6}' .
                   format(result * 100), "%")
        else:
            print("Profit = ",'{0:.6}' .
                   format(result * 100), "%")
 
# Driver Code
if __name__ == '__main__':
    N = 8
    M = 9
 
    profitLoss(N, M)
 
# This code is contributed by
# Sanjit_Prasad

                    

C#

// C# implementation of above formula
using System;
 
class GFG
{
// Function to calculate
// Profit or loss
static void profitLoss(int N, int M)
{
    if (N == M)
        Console.Write("No Profit nor Loss");
    else
    {
        float result = 0;
 
        result = (float)(Math.Abs(N - M))/ M;
 
        if (N - M < 0)
            Console.Write("Loss = -" +
                           result * 100 +"%");
        else
            Console.Write("Profit = " +
                           result * 100 + "%");
    }
}
 
// Driver Code
static public void Main ()
{
    int N = 8, M = 9;
 
    profitLoss(N, M);
}
}
 
// This code is contributed by ajit.

                    

PHP

<?php
// PHP implementation of above formula
 
// Function to calculate
// Profit or loss
function profitLoss($N, $M)
{
    if ($N == $M)
        echo "No Profit nor Loss";
    else
    {
 
        $result = 0.0;
 
        $result = (abs($N - $M)) / $M;
 
        if ($N - $M < 0)
            echo "Loss = -",
                  $result * 100, "%";
        else
            echo "Profit = ",
                  $result * 100, "%";
    }
}
 
// Driver Code
$N = 8;
$M = 9;
profitLoss($N, $M);
 
// This code is contributed
// by Sach_Code
?>

                    

Javascript

<script>
 
// Javascript implementation of above formula
 
// Function to calculate
// Profit or loss
function profitLoss(N, M)
{
    if (N == M)
        document.write( "No Profit nor Loss");
    else
    {
        var result = 0.0;
 
        result = (Math.abs(N - M)) / M;
 
        if (N - M < 0)
            document.write("Loss = -" +
              result.toFixed(6) * 100 + "%");
        else
            document.write("Profit = " +
               result.toFixed(6) * 100 + "%");
    }
}
 
// Driver Code
var N = 8, M = 9;
 
profitLoss(N, M);
 
// This code is contributed by rutvik_56
 
</script>

                    

Output: 
Loss = -11.1111%

 

Time Complexity: O(1)

Auxiliary Space: O(1)



Last Updated : 11 Jul, 2022
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