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C program to calculate the Grade of Student using Percentage | Grading program in C

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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 :

Output :

student grading Program in C
Output of Student Grading Program 

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Venkatesh Macha

Hi Guys, I am Venkatesh Macha. I am a programmer, Presently working as a Software Engineer. I love Linux and Open Source. Apart from that, I love to explore new technologies and things.

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4 Responses

  1. sidvaldez says:
    July 2, 2018 at 11:07 am

    hey guys can u help me in this problem

    Reply
  2. Vimalan says:
    November 29, 2017 at 6:19 am

    Hey I need a favour. How do I do an algorithm, pseudocode and flowchart fo the attached program?

    #include
    #include // The Rounding Of Application //
    #include // To Copy The Score and Grade for sorting Purposes //

    main()
    {
    char MATRICS[26][31], temp4[10], temp5[10];
    int i, M, N, O, P, count;
    float MANUAL[26], CAD[26], TEST1[26], TEST2[26], TEST3[26], PROJ[26], j[26];
    float MAN[26], CADK[26], TEST[26], PRO[26];
    float sum, aveg;
    float temp1, temp2, temp, temp3;
    float add;
    int total1=0, total2=0, total3=0, total4=0, total5=0, total6=0, total7=0, total8=0;

    FILE *in; // declare input data file //
    in = fopen( “project1.txt”, “r” ); // assign input data file //

    printf(“\n \n”);

    while(!feof(in)) // read input datafile and display until end of file //
    {
    for(i=1; i<27; i=i+1)
    {
    fscanf(in ,"%s %f %f %f %f %f %f", &MATRICS[i], &MANUAL[i], &CAD[i], &TEST1[i], &TEST2[i], &TEST3[i], &PROJ[i]);
    MAN[i] = MANUAL[i]*.2;
    CADK[i] = CAD[i]*.1;
    TEST[i] = ((TEST1[i]*.1) + (TEST2[i]*.1) + (TEST3[i]*.2));
    PRO[i] = PROJ[i]*.3;
    j[i] = MAN[i] + CADK[i] + TEST[i] + PRO[i];

    }
    }

    for (M=1; M<26; M=M+1) // start sorting process //
    {
    for (i=1; i<26; i=i+1)
    {
    if (j[i] < j[i+1]) // "<" Indicates for Descending Order //
    {
    temp=j[i];
    j[i]=j[i+1];
    j[i+1]=temp;

    temp1=MAN[i];
    MAN[i]= MAN[i+1];
    MAN[i+1]= temp1;

    temp2=CADK[i];
    CADK[i]= CADK[i+1];
    CADK[i+1]= temp2;

    temp3=TEST[i];
    TEST[i]= TEST[i+1];
    TEST[i+1]= temp3;

    strcpy(temp4, MATRICS[i]);
    strcpy(MATRICS[i], MATRICS[i+1]);
    strcpy(MATRICS[i+1], temp4);

    }

    }
    }

    for(i=1; i=80)
    printf(” A”);

    else if(round(j[i])>=75 && (j[i]=70 && (j[i]=65 && (j[i]=60 && (j[i]=55 && (j[i]=50 && (j[i]<55))
    printf(" C");

    else
    printf(" FAIL");

    }
    for(i=1;i=80)
    {
    total1 += 1;
    }

    else if((j[i]>=75) && (j[i]=70) && (j[i]=65) && (j[i]=60) && (j[i]=55) && (j[i]=50) && (j[i]<55))
    {
    total7 += 1;
    }

    else
    {
    total8 += 1;
    }

    }

    printf("\n\n\n *****Statistics Of Results!!!*****");

    printf("\n\n Total Number Of Students: 26");

    for(i=1; i=80)
    printf(“\n Average Grade: A\n”);

    else if((aveg>=75) && (aveg=70) && (aveg=65) && (aveg=60) && (aveg=55) && (aveg=50) && (aveg<55))
    printf("\n Average Grade: C-");

    else
    printf("\nFAIL");

    printf("\n Number Of A : %i", total1);
    printf("\n Number Of A- : %i", total2);
    printf("\n Number Of B+ : %i", total3);
    printf("\n Number Of B : %i", total4);
    printf("\n Number Of B- : %i", total5);
    printf("\n Number Of C+ : %i", total6);
    printf("\n Number Of C : %i", total7);
    printf("\n Number Of 'Fail': %i", total8);

    fclose(in);
    }

    Reply
  3. Judy Nekoye says:
    June 12, 2017 at 6:50 pm

    Hey, kindly help me with this: Write a pseudocode that multiply positive integers x and y by using addition and subtraction only

    Reply
    • Venkatesh Macha says:
      June 13, 2017 at 1:07 am

      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.

      Reply

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