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Loss when two items are sold at same price and same percentage profit/loss

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  • Difficulty Level : Basic
  • Last Updated : 09 Jun, 2022
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Given the Selling price i.e ‘SP’ of the two items each. One item is sold at ‘P%’ Profit and other at ‘P%’ Loss. The task is to find out the overall Loss.
Examples: 
 

Input: SP = 2400, P = 30%  
Output: Loss = 474.725

Input: SP = 5000, P = 10%
Output: Loss = 101.01

 

Approach:
 

How does the above formula work? 
For profit making item : 
With selling price (100 + P), we get P profit. 
With selling price SP, we get SP * (P/(100 + P)) profit
For loss making item : 
With selling price (100 – P), we get P loss. 
With selling price SP, we get SP * (P/(100 – P)) loss
Net Loss = Total Loss – Total Profit 
= SP * (P/(100 – P)) – SP * (P/(100 + P)) 
= (SP * P * P * 2) / (100*100 – P*P)
Note: The above formula is applicable only when the Cost price of both the items are different. If CP of both the items are same then, in that case, there is ‘No profit No loss’.
Below is the implementation of the above approach
 

C++




// C++ implementation of above approach.
#include <bits/stdc++.h>
using namespace std;
 
// Function that will
// find loss
void Loss(int SP, int P)
{
 
    float loss = 0;
 
    loss = (2 * P * P * SP) / float(100 * 100 - P * P);
 
    cout << "Loss = " << loss;
}
 
// Driver Code
int main()
{
    int SP = 2400, P = 30;
 
    // Calling Function
    Loss(SP, P);
 
    return 0;
}


Java




// Java implementation of above approach.
class GFG
{
 
// Function that will
// find loss
static void Loss(int SP, int P)
{
 
    float loss = 0;
 
    loss = (float)(2 * P * P * SP) / (100 * 100 - P * P);
 
    System.out.println("Loss = " + loss);
}
 
// Driver Code
public static void main(String[] args)
{
    int SP = 2400, P = 30;
 
    // Calling Function
    Loss(SP, P);
}
}
 
// This code has been contributed by 29AjayKumar


Python3




# Python3 implementation of above approach.
 
# Function that will find loss
def Loss(SP, P):
     
    loss = 0
    loss = ((2 * P * P * SP) /
            (100 * 100 - P * P))
    print("Loss =", round(loss, 3))
 
# Driver Code
if __name__ == "__main__":
 
    SP, P = 2400, 30
 
    # Calling Function
    Loss(SP, P)
 
# This code is contributed by Rituraj Jain


C#




// C# implementation of above approach.
class GFG
{
 
// Function that will
// find loss
static void Loss(int SP, int P)
{
 
    double loss = 0;
 
    loss = (double)(2 * P * P * SP) / (100 * 100 - P * P);
 
    System.Console.WriteLine("Loss = " +
                            System.Math.Round(loss,3));
}
 
// Driver Code
static void Main()
{
    int SP = 2400, P = 30;
 
    // Calling Function
    Loss(SP, P);
}
}
 
// This code has been contributed by mits


PHP




<?php
// PHP implementation of above approach.
 
// Function that will find loss
function Loss($SP, $P)
{
     
    $loss = 0;
    $loss = ((2 * $P * $P * $SP) /
          (100 * 100 - $P * $P));
    print("Loss = " . round($loss, 3));
}
 
// Driver Code
$SP = 2400;
$P = 30;
 
// Calling Function
Loss($SP, $P);
 
// This code is contributed by mits
?>


Javascript




<script>
 
// javascript implementation of above approach.
// Function that will
// find loss
    function Loss(SP , P) {
 
        var loss = 0;
 
        loss =  (2 * P * P * SP) / (100 * 100 - P * P);
 
        document.write("Loss = " + loss.toFixed(3));
    }
 
    // Driver Code
     
        var SP = 2400, P = 30;
 
        // Calling Function
        Loss(SP, P);
 
// This code contributed by gauravrajput1
 
</script>


Output

Loss = 474.725

Time Complexity: O(1)
Auxiliary Space: O(1)


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