Program to find the quantity after mixture replacement

Given a container that has X liters of milk. Y liters of milk is drawn out and replaced with Y liters of water. This operation is done Z times. The task is to find out the quantity of milk left in the container.
Examples:
Input: X = 10 liters, Y = 2 liters, Z = 2 times Output: 6.4 liters Input: X = 25 liters, Y = 6 liters, Z = 3 times Output: 10.97 liters
Formula:-
Below is the required implementation:
C++
// C++ implementation using above formula#include <bits/stdc++.h>using namespace std;// Function to calculate the Remaining amount.float Mixture(int X, int Y, int Z){ float result = 0.0, result1 = 0.0; // calculate Right hand Side(RHS). result1 = ((X - Y) / (float)X); result = pow(result1, Z); // calculate Amount left by // multiply it with original value. result = result * X; return result;}// Driver Codeint main(){ int X = 10, Y = 2, Z = 2; cout << Mixture(X, Y, Z) << " litres"; return 0;} |
Java
// Java code using above Formula.import java.io.*;class GFG{// Function to calculate the // Remaining amount. static double Mixture(int X, int Y, int Z) { double result1 = 0.0, result = 0.0; // calculate Right hand Side(RHS). result1 = ((X - Y) / (float)X); result = Math.pow(result1, Z); // calculate Amount left by // multiply it with original value. result = result * X; return result; } // Driver Codepublic static void main(String[] args){ int X = 10, Y = 2, Z = 2; System.out.println((float)Mixture(X, Y, Z) + " litres");}}// This code is contributed // by Naman_Garg |
Python 3
# Python 3 implementation using # above formula# Function to calculate the# Remaining amount.def Mixture(X, Y, Z): result = 0.0 result1 = 0.0 # calculate Right hand Side(RHS). result1 = ((X - Y) / X) result = pow(result1, Z) # calculate Amount left by # multiply it with original value. result = result * X return result# Driver Codeif __name__ == "__main__": X = 10 Y = 2 Z = 2 print("{:.1f}".format(Mixture(X, Y, Z)) + " litres")# This code is contributed by ChitraNayal |
C#
// C# code using above Formula.using System;class GFG{// Function to calculate the // Remaining amount. static double Mixture(int X, int Y, int Z) { double result1 = 0.0, result = 0.0; // calculate Right hand Side(RHS). result1 = ((X - Y) / (float)X); result = Math.Pow(result1, Z); // calculate Amount left by // multiply it with original value. result = result * X; return result; } // Driver Codepublic static void Main(){ int X = 10, Y = 2, Z = 2; Console.WriteLine((float)Mixture(X, Y, Z) + " litres");}}// This code is contributed // by Akanksha Rai(Abby_akku) |
PHP
<?php // PHP implementation of above formula// Function to calculate the// Remaining amount. function Mixture($X, $Y, $Z) { $result = 0.0; $result1 = 0.0; // calculate Right hand Side(RHS). $result1 = (($X - $Y) / $X); $result = pow($result1, $Z); // calculate Amount left by // multiply it with original value. $result = $result * $X; return $result; } // Driver Code $X = 10;$Y = 2;$Z = 2;echo Mixture($X, $Y, $Z), " litres"; // This code is contributed // by Sanjit_Prasad?> |
Javascript
<script>// Javascript implementation using above formula// Function to calculate the Remaining amount.function Mixture(X, Y, Z){ var result = 0.0, result1 = 0.0; // calculate Right hand Side(RHS). result1 = ((X - Y) / X); result = Math.pow(result1, Z); // calculate Amount left by // multiply it with original value. result = result * X; return result;}// Driver Codevar X = 10, Y = 2, Z = 2;document.write( Mixture(X, Y, Z).toFixed(1) + " litres");</script> |
Output:
6.4 litres
Time Complexity: O(log(n))
Auxiliary Space: O(1)
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