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Concentration of juice after mixing n glasses in equal proportion
  • Last Updated : 27 Apr, 2021

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 

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 <bits/stdc++.h>
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[0]);
    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




<?php
// PHP implementation of the approach
     
// Function to return the concentration
// of the resultant mixture
function mixtureConcentration($n, $p)
{
    $res = 0;
    for ($i = 0; $i < $n; $i++)
        $res += $p[$i];
    $res /= $n;
    return $res;
}
 
// Driver code
$arr = array( 0, 20, 20 );
$n = count($arr);
print(round(mixtureConcentration($n, $arr), 4));
 
// This code is contributed by chandan_jnu
?>

Javascript




<script>
 
// Javascript implementation of the approach
 
// Function to return the concentration
// of the resultant mixture
function mixtureConcentration(n, p)
{
    var res = 0;
    for(var i = 0; i < n; i++)
        res += p[i];
         
    res /= n;
    return res;
}
 
// Driver Code
var arr = [ 0, 20, 20 ];
var n = arr.length;
document.write(mixtureConcentration(n, arr).toFixed(4));
 
// This code is contributed by Ankita saini
 
</script>
Output: 



13.3333

 

Time Complexity: O(N)
 

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