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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