LOOPING PROBLEM 5. Calculate GPAs given the information below. Run your program
ID: 3593465 • Letter: L
Question
LOOPING PROBLEM
5. Calculate GPAs given the information below. Run your program with the given test cases Notes: (1) use switch statements, (2) the program should prompt the user for the number ojf courses, (3) the program should ask for the letter grade (string variable) and the credit hours for that course, (4) GPA is based on a 4.0 grading scale, (5) your solution should use 1 or two loops (6) GPA is calculated by (total earned grade point hours/total credit hours). Letter Grade Grade Points LetterGrade Grade Points 4.0 3.7 3.3 3.0 2.7 2.3 2.0 1.7 Test your program with the following two cases: 1.3 1.0 0.7 0.0 a. A, B-,B+with credit hours [3 5 5] b. B+, C, D with credit hours [5 4 3] Use nested for loops to create a multiplication table. A multiplication table is a table in which each element is the product of the corresponding row and column. The following is 5x5 multiplication table: 6. 2 46 8 10 M=13 6 9 12 15 4 8 12 16 20 5 10 15 20 25 Test your program with the following user-inputted dimensions a. 7x4 b. 3x9 c. 10x10Explanation / Answer
program 5:
#include<stdio.h>
#include<conio.h>
void main()
{
int credit,sumcredit=0,n,i;
float cgpa=0;
float sum=0;
char grade[5];
printf(" enter no.of subjects");
scanf("%d",&n);
for(int i=0;i<n;i++)
{
printf(" enter grade of the %d subject",i+1);
scanf("%s",&grade);
printf(" enter credit hours");
scanf("%d",&credit);
switch(grade)
{
case 'A': sum=sum+(4*credit);
sumcredit=sumcredit+credit;
break;
case 'A-': sum=sum+(3.7*credit);
sumcredit=sumcredit+credit;
break;
case 'B+': sum=sum+(3.3*credit);
sumcredit=sumcredit+credit;
break;
case 'B': sum=sum+(3*credit);
sumcredit=sumcredit+credit;
break;
case 'B-': sum=sum+(2.7*credit);
sumcredit=sumcredit+credit;
break;
case 'C+': sum=sum+(2.3*credit);
sumcredit=sumcredit+credit;
break;
case 'C': sum=sum+(2*credit);
sumcredit=sumcredit+credit;
break;
case 'C-': sum=sum+(1.7*credit);
sumcredit=sumcredit+credit;
break;
case 'D+': sum=sum+(1.3*credit);
sumcredit=sumcredit+credit;
break;
case 'D': sum=sum+(1*credit);
sumcredit=sumcredit+credit;
break;
case 'D-': sum=sum+(0.7*credit);
sumcredit=sumcredit+credit;
break;
case 'E': sum=sum+(0*credit);
sumcredit=sumcredit+credit;
break;
}
}
cgpa=(sum/(4*sumcredit))*100;
printf(" student cgpa is %f",cgpa);
}
output:
enter no.of subjects 3
enter grade of the 1 subject A
enter credit hours 3
enter grade of the 2 subject B-
enter credit hours 5
enter grade of the 3 subject B+
enter credit hours 5
student cgpa is 80.76
program 6:
#include<stdio.h>
#include<conio.h>
void main()
{
int i,j,n,m;
printf(" enter row size");
scanf("%d",&n);
printf(" enter column size");
scanf("%d",&m);
for( i=1;i<=n;i++)
{
for(j=1;j<=m;j++)
{
printf(" %d",i*j);
}
printf(" ");
}
}
output:
enter row size 7
enter col size 4
1 2 3 4
2 4 6 8
3 6 9 12
4 8 12 16
5 10 15 20
6 12 16 24
7 14 21 28
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