# URI Online Judge Solution 1006 Average 2 – Solution in C, C++, Java, Python and C#

### URI Online Judge Solution 1006 Average 2 – Solution in C, C++, Java, Python and C#

URI Online Judge Solution 1006 Average 2 | Beginner
URI Problem Link -1006 Average 2 – https://www.urionlinejudge.com.br/judge/en/problems/view/1006

Problem Name: 1006 Average 2 solution
Problem Number : URI – 1006 Average 2 code
Online Judge : URI Online Judge Solution
Category: Beginner
Solution Language : C,C plus plus, java, python, c#(c sharp)
URI 1006 problem screenshot:

See a video if you have some time –
URI 1006 solution Video –

### URI Solution 1006 Average 2 Code in C:

`#include <stdio.h>int main(){   double a, b, c, media;   scanf("%lf%lf%lf", &a, &b, &c);   media = (a/10 * 2) + (b/10 * 3) + (c/10 * 5);   printf("MEDIA = %.1lfn", media);   return 0;}`

### URI Solution 1006 Average 2 Code C++/ URI 1006 Average 2 solution in CPP:

`#include <iostream>#include <iomanip>using namespace std;int main(){    double a, b, c, media;    cin >> a >> b >> c;    media = (a/10 * 2) + (b/10 * 3) + (c/10 * 5);    cout << "MEDIA = "<< fixed << setprecision(1) << media << endl;    return 0;}`

### URI Solution 1006 Average 2 Code in java / URI 1006 Average 2 solution in Java:

`import java.util.Scanner;public class Main{    public static void main(String[] args){        double a, b, c, med;        Scanner sc =new Scanner(System.in);           a =sc.nextDouble();           b =sc.nextDouble();           c =sc.nextDouble();         med = (a/10 * 2) + (b/10 * 3) + (c/10 * 5);         String media = String.format("MEDIA = %,.1f", med);        System.out.print(media +"n");    }}`

### URI Solution 1006 Average 2 Code / URI 1006 Average 2 solution in  Python:

`a = float(input())b = float(input())c = float(input())media = (a/10 * 2) + (b/10 * 3) + (c/10 * 5);print("MEDIA = %0.5f" %media)`

### URI Solution 1006 Average 2 Code / URI 1006 Average 2 solution in  C# (C Sharp):

`using System;class URI {     static void Main(string[] args) {            double a, b, c;            a = Convert.ToDouble(Console.ReadLine());            b = Convert.ToDouble(Console.ReadLine());            c = Convert.ToDouble(Console.ReadLine());            Console.WriteLine("MEDIA = " + ((a/10 * 2) + (b/10 * 3) + (c/10 * 5)).ToString("0.00000"));            Console.ReadKey();    }}`

Demonstration:

Since in the question, it has told that each grade has grown from 0 to 10, so devide it with 10 and multiplication into with their weight. Just main line is this,

`media = (a/10 * 2) + (b/10 * 3) + (c/10 * 5);`

Just implement this in coding. Since having any problem just put a comment below. Thanks

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