C - Exercise 3

In this exercise section of the tutorial we will write some C program.

Write a program in C to compute (a + b)2 where the value of a and b will be provided by the user

The formula of (a + b)2 is given below.

(a + b)2 = a2 + b2 + 2ab

Now we will convert this into a valid C expression.

Let x = (a + b)2
      = a2 + b2 + 2ab
      = (a * a) + (b * b) + (2 * a * b)

Now we will write the code.

/**
 * file: algebra.c
 * author: yusuf shakeel
 * date: 2010-12-15
 * description: algebra program
 */

#include <stdio.h>
int main(void)
{
  //variables
  float a, b, x;
  
  //user input
  printf("Enter value of a: ");
  scanf("%f", &a);
  
  printf("Enter value of b: ");
  scanf("%f", &b);
  
  //compute
  x = (a * a) + (b * b) + (2 * a * b);
  
  //output
  printf("Result = %f\n", x);
  
  printf("End of code\n");
  return 0;
}

Output

Enter value of a: 4
Enter value of b: 2
Result = 36.000000
End of code

Write a program in C to compute (a + b)3 where the value of a and b will be provided by the user

Try to solve this problem yourself.

Help:

Let x = (a + b)3
      = a3 + b3 + 3a2b + 3ab2

Write a program in C to compute Simple Interest based on user input

To find Simple Interest we will need the Principal, Rate of Interest and Time.

Click here to learn more about Simple Interest.

To keep things simple we will assume Time is in years and an integer value whereas, Principal and Rate of Interest are float type.

Formula for Simple Interest is given below.

SI = PRT/100
   = (P * R * T) / 100

where,
P = Principal
R = Rate of Interest
T = Time

Now, lets write the C code.

/**
 * file: simple-interest.c
 * author: yusuf shakeel
 * date: 2010-12-15
 * description: simple interest program
 */

#include <stdio.h>
int main(void)
{
  //variables
  float si, p, r;
  int t;
  
  //user input
  printf("Enter P: ");
  scanf("%f", &p);
  
  printf("Enter R: ");
  scanf("%f", &r);
  
  printf("Enter T: ");
  scanf("%d", &t);
  
  //compute
  si = (p * r * t) / 100;
  
  //output
  printf("SI = %f\n", si);
  
  printf("End of code\n");
  return 0;
}

Output

Enter P: 1000
Enter R: 10
Enter T: 2
SI = 200.000000
End of code