C++ program that will display if a students is pass or not in his exam.
Object:
Write a C++ program that will display if a students is pass or not in his exam. (50% or more is pass). If the student is Pass than your program should display which letter the student has obtained:
- 85% or more E for excellent
- 75% or more but less than 85% O for Outstanding
- 65% or more but less than 75% G for good
- Less than 65% S for satisfactory
If however the student is Fail (below 50% marks) your program should display. Whether the student should Resit or Redo depending on the following criteria. – 33% or more Resit in exam – Less than 33% Redo course.
Code:
#include<iostream>
#include<conio.h>
using namespace std;
void main()
{
int per;
cout<<“Enter a Student Percentage: “;
cin>>per;
if(per>=85 && per <=100)
{
cout<<“\nExcellent Work.”;
}
else if(per>=75 && per<85)
{
cout<<“\nOutstanding Work.”;
}
else if(per>=65 && per<75)
{
cout<<“\nGood Work.”;
}
else if(per>=50 && per<65)
{
cout<<“\nSatisfactory.”;
}
else if(per<50)
{
cout<<“\nBad Performance.”<<endl;
if(per<50 && per>=33)
{
cout<<“\nYou can Resit in Exam.”;
}
else
{
cout<<“\nPlease Redo your course!!”;
}
}
else
{
cout<<“\nInvalid Entry Please Enter Valid Percentage!!!”;
}
getch();
}
#include<iostream> #include<conio.h> using namespace std; void main() { int per; cout<<"Enter a Student Percentage: "; cin>>per; if(per>=85 && per <=100) { cout<<"\nExcellent Work."; } else if(per>=75 && per<85) { cout<<"\nOutstanding Work."; } else if(per>=65 && per<75) { cout<<"\nGood Work."; } else if(per>=50 && per<65) { cout<<"\nSatisfactory."; } else if(per<50) { cout<<"\nBad Performance."<<endl; if(per<50 && per>=33) { cout<<"\nYou can Resit in Exam."; } else { cout<<"\nPlease Redo your course!!"; } } else { cout<<"\nInvalid Entry Please Enter Valid Percentage!!!"; } getch(); }
Output:
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