Find the difference between compound interest on ₹8000 for $1\dfrac{1}{2}$ years at 10% p.a. when compounded annually and semi-annually.

Case 1:

When compounded annually,

rate for first year = 10%, rate for next $\dfrac{1}{2}$ year = 5%.

$A = P\Big(1 + \dfrac{r}{100}\Big)^n$

On substituting values,

$A = ₹8000\Big(1 + \dfrac{10}{100}\Big)\Big(1 + \dfrac{5}{100}\Big) \\[1em] = ₹8000\Big(\dfrac{110}{100}\Big)\Big(\dfrac{105}{100}\Big) \\[1em] = ₹8000 \times \dfrac{11}{10} \times \dfrac{21}{20} \\[1em] = ₹(40 \times 11 \times 21) \\[1em] = ₹9240.$

C.I. = A - P = ₹9240 - ₹8000 = ₹1240.

Case 2:

When compounded semi-annually,

rate = 5%, n = 3.

$A = P\Big(1 + \dfrac{r}{100}\Big)^n$

On substituting values,

$A = ₹8000\Big(1 + \dfrac{5}{100}\Big)^3 \\[1em] = ₹8000\Big(\dfrac{105}{100}\Big)^3 \\[1em] = ₹8000 \times \dfrac{21}{20} \times \dfrac{21}{20} \times \dfrac{21}{20} \\[1em] = ₹(21 \times 21 \times 21) \\[1em] = ₹9261.$

C.I. = A - P = ₹9261 - ₹8000 = ₹1261.

Difference between two C.I. = ₹1261 - ₹1240 = ₹21.