C Tutorial
This c program is used to find the depreciation amount after few years.
Current Value = (principal amount * power(1-(rate per annum/ 100), year)) Depreciation Amount = Principal Amount - Current Value
#include<stdio.h>
#include<math.h>
void main()
{
int p, r, y;
double current, depreciation;
printf("Enter principal (p): ");
scanf("%d", &p);
printf("Enter rate per annum (r): ");
scanf("%d", &r);
printf("Enter number of years (y): ");
scanf("%d", &y);
current = (double)p*(pow((1.0-(((double)r)/100)), y));
printf("Current Total Value after %d years: %lf\n", y, current);
depreciation = p-current;
printf("Depreciation Amount after %d years: %lf\n",y, depreciation);
}
Output:
$ cc depreciation.c -o a.out -lm $ ./a.out Enter principal (p): 7500 Enter rate per annum (r): 4 Enter number of years (y): 2 Current Total Value after 2 years: 6912.000000 Depreciation Amount after 2 years: 588.000000
Total Value = (current amount / power(1-(rate per annum/ 100), year)) Depreciation Amount = Total Value - current
#include<stdio.h>
#include<math.h>
void main()
{
int p, r, y;
double total, depreciation;
printf("Enter principal (p): ");
scanf("%d", &p);
printf("Enter rate per annum (r): ");
scanf("%d", &r);
printf("Enter number of years (y): ");
scanf("%d", &y);
total = (double)p/(pow((1.0-(((double)r)/100)), y));
printf("Total Value %d years ago: %lf\n", y, total);
depreciation = total - p;
printf("Depreciation Amount after %d years: %lf\n",y, depreciation);
}
Output:
$ cc depreciation.c -o a.out -lm $ ./a.out Enter principal (p): 7500 Enter rate per annum (r): 4 Enter number of years (y): 2 Total Value 2 years ago: 8138.020833 Depreciation Amount after 2 years: 638.020833
C Tutorial
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