# Concentration of juice after mixing n glasses in equal proportion

Given an array arr[] where arr[i] is the concentration of juice in ith glass. The task is to find the concentration of the resultant mixture when all the glasses are mixed in equal proportions.

Examples:

Input: arr[] = {10, 20, 30}
Output: 20

Input: arr[] = {0, 20, 20}
Output: 13.3333

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Since the juices are mixed in equal proportions so the resultant concentration will be the average of all the individual concentrations. Therefore the required answer would be sum(arr) / n where n is the size of the array.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the concentration ` `// of the resultant mixture ` `double` `mixtureConcentration(``int` `n, ``int` `p[]) ` `{ ` `    ``double` `res = 0; ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``res += p[i]; ` `    ``res /= n; ` `    ``return` `res; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 0, 20, 20 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``cout << mixtureConcentration(n, arr); ` `} `

## Java

 `// Java implementation of the approach ` ` `  `class` `GFG ` `{ ` `     `  `// Function to return the concentration ` `// of the resultant mixture ` `static` `double` `mixtureConcentration(``int` `n, ``int` `[]p) ` `{ ` `    ``double` `res = ``0``; ` `    ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``res += p[i]; ` `    ``res /= n; ` `    ``return` `res; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main (String[] args)  ` `{ ` ` `  `    ``int` `[]arr = { ``0``, ``20``, ``20` `}; ` `    ``int` `n = arr.length; ` `    ``System.out.println(String.format(``"%.4f"``, ` `                        ``mixtureConcentration(n, arr))); ` `} ` `} ` ` `  `// This code is contributed by chandan_jnu `

## Python3

 `# Python3 implementation of the approach ` `     `  `# Function to return the concentration ` `# of the resultant mixture ` `def` `mixtureConcentration(n, p): ` ` `  `    ``res ``=` `0``; ` `    ``for` `i ``in` `range``(n): ` `        ``res ``+``=` `p[i]; ` `    ``res ``/``=` `n; ` `    ``return` `res; ` ` `  `# Driver code ` `arr ``=` `[ ``0``, ``20``, ``20` `]; ` `n ``=` `len``(arr); ` `print``(``round``(mixtureConcentration(n, arr), ``4``)); ` ` `  `# This code is contributed  ` `# by chandan_jnu `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `     `  `// Function to return the concentration ` `// of the resultant mixture ` `static` `double` `mixtureConcentration(``int` `n, ``int` `[]p) ` `{ ` `    ``double` `res = 0; ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``res += p[i]; ` `    ``res /= n; ` `    ``return` `Math.Round(res,4); ` `} ` ` `  `// Driver code ` `static` `void` `Main() ` `{ ` `    ``int` `[]arr = { 0, 20, 20 }; ` `    ``int` `n = arr.Length; ` `    ``Console.WriteLine(mixtureConcentration(n, arr)); ` `} ` `} ` ` `  `// This code is contributed by chandan_jnu `

## PHP

 ` `

Output:

```13.3333
```

Time Complexity: O(N)

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Improved By : Chandan_Kumar