There are two equivalent way. Suppose the item costs C units (of whatever currency), and suppose the tax rate is R %. You calculate the total tax = C*R/100 currency units so total costs = C + C*R/100 Alternatively, you calculate the tax multiplier, = (1 + R/100) and then the total cost is C*(1 + R/100). It is simple to show the two arrive at the same answer.
100 Runs in a Century
100 coins in a row 100 cents in a rand 100 cubits in a rood
100 coins in a row or 100 cents in a rand or 100 cubits in a rood.
The amount of interest earned on an investment of C, for y years at r per cent is C*y*r/100.
Rs 2000. The amount a after n years of an amount c with interest rate r % is: a = c(1 + r/100)^n After 3 & 6 years of the question there is: Rs 4000 = c(1 + r/100)^3 Rs 8000 = c(1 + r/100)^6 Dividing the second by the first gives: Rs 8000 ÷ Rs 4000 = (c(1 + r/100)^6) ÷ (c(1 + r/100)^3) → 2 = (1 + r/100)^3 Using the first equation and substituting for (1 + r/100)^3 found above: Rs 4000 = c(1 + r/100)^3 = 2c → c = Rs 2000 Alternatively: As it is compound interest, the saved interest also attracts interest; the amount doubles in the 3 years from year 3 to year 6, so it doubles every 3 years. Therefore initially it must be half the value at 3 years → it is ½ × Rs 4000 = Rs 2000.
p = r - c r - c = p r - c - r = p - r -(-c) = -(p) c = -p
Let c= original cost Let p= percent reduced by Let r= cost after reduction Equation: c=p*(100%-r)
The simple interest, I, on a deposit of C, at an interest rate of r% per year, deposited for y years, is calculated as I = C*(r/100)*y
If an amount C is invested for n years with an interest rate of r%, then the amount of interest earned is C*n*r/100
The freezing point of R-12 ( Dichlorodifluoromethane, a form of Freon) is −157.7 °C.
#include<stdio.h> #include<conio.h> void main() { int a[3][3],b[3][3],c[3][3],r,c; for(r=0;r<=3;r++) { for(c=0;c<3;c++) { printf("\n enter the value="); scanf("%d%d",&a[r][c],&b[r][c]); } } printf("\n first matrix=\n"); for(r=0;r<=3;r++) { for(c=0;c<3;c++) { printf("%d\t",a[r][c]); } printf("\n"); } printf("\n scond matrix=\n"); for(r=0;r<=3;r++) { for(c=0;c<3;c++) {printf("%d\t",b[r][c]); } printf("\n"); } printf("\n sum of given matrix=\n"); for(r=0;r<=3;r++) { for(c=0;c<3;c++) { c[r][c]=a[r][c]+b[r][c]; printf("%d\t",c[r][c]); } printf("\n"); } getch(); }