Find the difference between compound interest and simple interest on ₹ 12000 and in $1\dfrac{1}{2}$ years at 10% p.a. compounded yearly.

Calculating C.I. :

For 1st year :

P = ₹ 12000

Rate = 10%

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

Substituting values we get :

$A = 12000 \times \Big(1 + \dfrac{10}{100}\Big)^1 \\[1em] = 12000 \times \dfrac{110}{100} \\[1em] = ₹ 13200.$

For next $\dfrac{1}{2}$ year :

P = ₹ 13200

Rate = 10%

A = $P\Big(1 + \dfrac{r}{2 \times 100}\Big)^{n \times 2}$

Substituting values we get :

$A = 13200 \times \Big(1 + \dfrac{10}{200}\Big)^{\dfrac{1}{2} \times 2} \\[1em] = 13200 \times \dfrac{210}{200} \\[1em] = ₹ 13860.$

C.I. = A - P = ₹ 13860 - ₹ 12000 = ₹ 1860

Calculating S.I. :

P = ₹ 12000

Rate = 10%

Time = $1\dfrac{1}{2}$

$S.I. = \dfrac{P \times R \times T}{100} \\[1em] = \dfrac{12000 \times 10 \times \dfrac{3}{2}}{100} \\[1em] = ₹ 1800.$

Difference between C.I. and S.I. = ₹ 1860 - ₹ 1800 = ₹ 60.

Hence, required difference = ₹ 60.