# Program to find the quantity after mixture replacement

• Difficulty Level : Basic
• Last Updated : 23 Jul, 2022

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 ``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 Code``int` `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 Code``public` `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 Code``if` `__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 Code``public` `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

 ``

## Javascript

 ``

Output:

`6.4 litres`

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

My Personal Notes arrow_drop_up