# URI Online Judge Solution 1005 Average 1 – Solution in C, C++, Java, Python and C#

### URI Online Judge Solution 1005 Average 1 – Solution in C, C++, Java, Python and C#

URI Online Judge Solution 1005 | Beginner
Problem Name: URI 1005 Average 1 code
Problem Number : URI – 1005 Average 1 Solution
Online Judge : URI Online Judge Solution
Category: Beginner
Solution Language : C,C plus plus, java, python, c#(c sharp)
Problem’s Screenshot:

### See a 4 minutes video – URI online judge solution 1005 in C programming :

If you have a little more time, then see the video, – URI 1005 solution video

### URI Solution 1005 Code in C / URI 1005 solution in C:

`#include <stdio.h>int main(){    float x,y;    scanf("%f %f", &x, &y);    printf("MEDIA = %.5fn", (x*3.5+y*7.5)/(3.5+7.5));    return 0;}`

### URI Solution 1005 Code / URI 1005 solution in CPP:

`#include <cstdio>int main(){ double a,b,m; scanf("%lf %lf", &a, &b); m = (a/11 * 3.5) + (b/11 * 7.5); printf("MEDIA = %.5lfn", m); return 0;}`

### URI Solution 1005 Code / URI 1005 solution in Java:

`import java.util.Scanner;public class URI_1005 {    public static void main(String[] args){  float a, b, m;  Scanner sc = new Scanner(System.in);  a = sc.nextFloat();  b = sc.nextFloat();  m = (float) ((a/11 * 3.5) + (b/11 * 7.5));  System.out.printf("MEDIA = %.5fn", m);    }}`

### URI Solution 1005 Code / URI 1005 solution in  Python:

`nota1 = float(input())nota2 = float(input())media = (((nota1 * 3.5) + (nota2 * 7.5)) / 11)print("MEDIA = %0.5f" %media)`

### URI Solution 1005 Code / URI 1005 solution in  C# (C Sharp):

`using System;class URI {     static void Main(string[] args) {            double a, b;            a = Convert.ToDouble(Console.ReadLine());            b=Convert.ToDouble(Console.ReadLine());            Console.WriteLine("MEDIA = " + ((a*3.5 + b*7.5)/(3.5+7.5)).ToString("0.00000"));            Console.ReadKey();    }}`

Demonstration:

This is a simple problem.
Steps to solve URI 1005 Average solution in C:

First take the value of a and b –

` scanf("%f %f", &x, &y);`

Second Get the median and print –

`printf("MEDIA = %.5fn", (x*3.5 + y*7.5) / (3.5 + 7.5));`

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

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