Programming Example 2: Final Mark Computation

Problem Statement

Two examination papers are written at the end of the course. The final mark is either the average of the two papers, or the average of the two papers and the class record mark (all weighted equally), whichever is the higher. The program should reads in the class record mark and the marks of the papers, computes the average, and shows PASS (>= 50%) or FAIL (< 50%).

Solution

! -------------------------------------------------------------
! Two examination papers are written at the end of the course.
! The final mark is either the average of the two papers, or
! the average of the two papers and the class record mark (all
! weighted equally), whichever is the higher.  The program
! should reads in the class record mark and the marks of the
! papers, computes the average, and shows PASS (>= 50%) or
! FAIL (< 50%).
! -------------------------------------------------------------

PROGRAM  FinalMark
   IMPLICIT  NONE

   REAL    :: Mark1, Mark2         ! the marks of the papers
   REAL    :: Final                ! the final marks
   REAL    :: ClassRecordMark      ! the class record mark

   REAL, PARAMETER :: PassLevel = 50.0  ! the pass level

   READ(*,*)  ClassRecordMark, Mark1, Mark2

   Final = (Mark1 + Mark2) / 2.0
   IF (Final <= ClassRecordMark) THEN
      Final = (Mark1 + Mark2 + ClassRecordMark) / 3.0
   END IF

   WRITE(*,*)  'Class Record Mark : ', ClassRecordMark
   WRITE(*,*)  'Mark 1            : ', Mark1
   WRITE(*,*)  'Mark 2            : ', Mark2
   WRITE(*,*)  'Final Mark        : ', Final

   IF (Final >= PassLevel) THEN
      WRITE(*,*)  'Pass Status       : PASS'
   ELSE
      WRITE(*,*)  'Pass Status       : FAIL'
   END IF

END PROGRAM  FinalMark

Program Input and Output

• If the input to class record mark, the mark of the first paper, and the mark of the second paper are 40.0, 60.0 and 43.0, the average is (60.0 + 43.0)/2 = 51.5, which is larger than the class record mark (40.0). Therefore, this student has a final mark 51.5 and receives a PASS status.

  40.0  60.0  43.0
  
  Class Record Mark : 40.
  Mark 1            : 60.
  Mark 2            : 43.
  Final Mark        : 51.5
  Pass Status       : PASS
  

• If the input to class record mark, the mark of the first paper, and the mark of the second paper are 60.0, 45.0 and 43.0, then the average is (45.0 + 43.0)/2 = 44.0, which is less than the class record mark (60.0). Therefore, this student's new mark is the average of his marks and the class record mark, (60.0 + 45.0 + 43.0)/3.0 = 49.33333. Since this new mark is less than 50.0, this student receives a FAIL status.

  60.0  45.0  43.0
  
  Class Record Mark : 60.
  Mark 1            : 45.
  Mark 2            : 43.
  Final Mark        : 49.3333321
  Pass Status       : FAIL
  

Discussion

• The READ statement reads in values for ClassRecordMark, Mark1 and Mark2.
• The average of papers is computed and stored in Final
• If this final mark is less than or equal to the class record mark, a new final mark is computed as the average of the two marks and the class record mark. Here, the IF-THEN-END form is used. You can use the logical IF form; but, this line could be too long. As mentioned earlier, the IF-THEN-END IF is preferred.
• The four WRITE(*,*) statements display the input and the computed final mark.
• Finally, the IF-THEN-ELSE-END IF determines if the final mark is a pass or a fail.