Question

Find the diffraction between C.I and S.I on sum of $Rs5000$ for $2$ years at $8\%$ per annum payable yearly.

Answer 1

Let us find the CI

$A=5000(1+\frac{\frac{R}{n}}{100}{)}^{nt}$

where, R = rate % = 8

n = Number of times interest is compounded per year=1 (as it is compounded once a year)

t = number of years=2

$CI=A-P=5000(1+\frac{8}{100}{)}^2-5000=(\frac{108}{100}{)}^{2}5000-5000=\frac{11664-10000}{10000}\ast 5000=\frac{1664}{10000}\ast 5000=\mathrm{832...........}\left(1\right)$

Now, $SI=\frac{PRT}{100}=\frac{5000\ast 8\ast 2}{100}=\mathrm{800...............}\left(2\right)$

hence, $CI-SI=832-800=32$