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 lossvoid Loss(int SP, int P){ float loss = 0; loss = (2 * P * P * SP) / float(100 * 100 - P * P); cout << "Loss = " << loss;}// Driver Codeint 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 lossstatic 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 Codepublic 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 lossstatic 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 Codestatic 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> |
Loss = 474.725
Time Complexity: O(1)
Auxiliary Space: O(1)
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