C program to calculate the Grade of Student using Percentage | Grading program in C
Description:
Find the grade of a student by reading marks or by taking Percentage. Here we are assuming greater than 80 percentage or marks as grade A, if the Marks are between 80-60 Grade is B, if the marks are between 60-40 grade is C, If student got marks below 40 it means he is Failed.
Algorithm :
Step 1 : start
Step 2 : read marks or Percentage
Step 3 : if marks >= 80 then grade =A, go to step 7
Step 4 : if marks >= 60 and marks <=80 then grade = B, go to step 7
Step 5 : if marks >=40 and marks <=60 then grade = C go to step 7
Step 6 : display failed
Step 7 : display grade.
Step 8 : stop.
Flow Chart :
Program :
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#include <stdio.h> int main(void) { int num; printf("Enter your mark "); scanf("%d",&num); printf(" You entered %d Marks n", num); // printing outputs if(num >= 80){ printf(" You got A grade n"); // printing outputs } else if ( num >=60){ // Note the space between else & if printf(" You got B grade n"); } else if ( num >=40){ printf(" You got C grade n"); } else if ( num < 40){ printf(" You Failed in this exam n"); printf(" Better Luck Next Time n"); } return 0; } |
Output :
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| Output of Student Grading Program |
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Hey, kindly help me with this: Write a pseudocode that multiply positive integers x and y by using addition and subtraction only
Hi Judy,
#include <stdio.h>
int main(void) {
int x, y, i, mul=0;
printf("Enter x and y values : ");
scanf("%d%d",&x,&y);
for(i=1;i<=y;i++)
{
mul = mul + x;
}
printf("Multiplication is : %d \n",mul);
return 0;
}
Pseudocode:
1) Take the input from user and store them in x and y.
2) Now create a for loop, That will loop from 1 to the y.
3) For each iteration we are going to add x to mul. Mul is variable which is initially 0 and we are adding x to mul. Means we are
multiplying the value.
4) This process will repeat until we reach the y. Once we reach, Loop will break and Multiplication value is stored in variable mul
5) display the mul value on the console.
cheers,
Venkey.