Lab Assignment 3 Write a program using the structured programming approach that

Question

Lab Assignment 3

Write a program using the structured programming approach that determines a student's grade. It reads three test scores (between 0 and 100) and calculates the average score and converts it to a letter grade. Average Score Calculation: Average Score = (Test Score 1 + Test Score 2 + Test Scores 3) /3 Grade Conversion Rules: rule a. If the average score is 90 or more the grade is 'A'. rule b. If the average score is 70 or more and less than 90 then check the third test score. If the third score is 90 or more the grade is 'A' otherwise the grade is 'B'. rule c. If the average score is 50 or more and less than 70 then check the average of the second and third scores. If the average of the last two is 70 or more, the grade is 'C' otherwise it is a 'D' rule d. If the average score is less than 50 then the grade is 'F'. Rounding Rule: Midpoint Rounding Calculate the grade average as a double. Round up to the next int if the fractional part is .5 or greater, otherwise truncate the fraction by casting to an int. The algorithm is: Add .5 to the average and cast the result to an int. Example: average = (int)(average+0.5);

To solve this problem, we can identify IPO and develop the Pseudocode. 1. Input: three test scores 2. Processing: Calculate the average and convert it to letter grade according to the conversion rules. 3. Output: Display the letter grade 4. Variables: int test1, test2, test2 – double average, average2 – char letterGrade 5. Constants: none Detailed Pseudocode: 1. Prompt user to enter first test score 2. Read user input and store value into variable test1 3. Prompt user to enter second test score 4. Read user input and store value into the variable test2 5. Prompt user to enter third test score 6. Read user input and store value into the variable test3 7. average = (test1 + test2 + test3) / 3.0 //use 3.0 to avoid integer division average = (int) (average +0.5) //apply rounding rule 8. average2 = (test2+test3) / 2.0 //use 2.0 to avoid integer division 1/16/2017 COSC 1436 Programming Fundamentals I C# average2 = (int) (average2+0.5) //apply rounding rule 9. if (average >=90) letterGrade=’A’ else if (average >=70) if (test3 >= 90) letterGrade=’A’ else letterGrade=’B’ else if (average >=50) if (average2>=70) letterGrade=’C’ else letterGrade=’D’ else letterGrade=’F’ 10. Display letterGrade Once the above detailed pseudocode is developed, you can map it line by line into C# code.

I keep testing this code project but my output keeps getting errors :

using System;

namespace GradeCalculatorNew
{
   class MainClass
   {
       public static void Main (string[] args)
       {
           int test1,test2,test3;
           double average,average2;
           char letterGrade;
           Console.WriteLine("Enter test score1");
           test1=Convert.ToInt32(Console.ReadLine());
           Console.WriteLine("Enter test score2");
           test2=Convert.ToInt32(Console.ReadLine());
           Console.WriteLine("Enter test score3");
           test3=Convert.ToInt32(Console.ReadLine());
           average=(test1+test2+test3)/3.0;
           average=(int)(average+0.5);
           if(average>=90)
           letterGrade='A';
           else if(average>=70)
           {
               if(test3>=90)
               letterGrade='A';
               else
               letterGrade='B';
           }
           else if(average>=50)
           {
               average2=(test2+test3)/2.0;
               average2=(int)(average2+0.5);
               if(average2>=70)
               letterGrade='C';
               else
               letterGrade='D';
           }
           else
               letterGrade='F';
           Console.WriteLine("Grade of Student is {0}",letterGrade);
       }
   }
}

My teacher keeps saying my if statements are not presented but my else statements are I don't know where she is talking about?

Explanation / Answer

Hi

I do not see major issues with this code. it is working fine. Did minor changs annd highlighted the same

using System;

namespace GradeCalculatorNew
{
class MainClass
{
public static void Main (string[] args)
{
int test1,test2,test3;
double average,average2;
char letterGrade;
Console.WriteLine("Enter test score1");
test1=Convert.ToInt32(Console.ReadLine());
Console.WriteLine("Enter test score2");
test2=Convert.ToInt32(Console.ReadLine());
Console.WriteLine("Enter test score3");
test3=Convert.ToInt32(Console.ReadLine());
average=(test1+test2+test3)/3.0;
average=(int)(average+0.5);
if(average>=90)
letterGrade='A';
else if(average>=70 && average<90)
{
if(test3>=90)
letterGrade='A';
else
letterGrade='B';
}
else if(average>=50 && average<70)
{
average2=(test2+test3)/2.0;
average2=(int)(average2+0.5);
if(average2>=70)
letterGrade='C';
else
letterGrade='D';
}
else
letterGrade='F';
Console.WriteLine("Grade of Student is {0}",letterGrade);
}
}
}

Output:

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