A certain sum of money, invested for 5 years at 8% p.a. simple interest, earns an interest of ₹ 12,000. Find:

(i) the sum of money.

(ii) the compound interest earned by this money in two years at 10% p.a. compound interest.

(i) For simple interest

T = 5 years

R = 8%

S.I. = ₹ 12,000

Let the principal amount be $P$.

As we know,

$\text{S.I.} = \Big(\dfrac{P \times R\times T}{100}\Big)\\[1em] \Rightarrow 12,000 = \Big(\dfrac{P\times 8 \times 5}{100}\Big)\\[1em] \Rightarrow 12,000 = \Big(\dfrac{40P}{100}\Big)\\[1em] \Rightarrow 12,000 = \Big(\dfrac{2P}{5}\Big)\\[1em] \Rightarrow 12,000 = \dfrac{2P}{5}\\[1em] \Rightarrow P = \dfrac{5 \times 12,000}{2} \\[1em] \Rightarrow P = \dfrac{60,000}{2} \\[1em] \Rightarrow P = 30,000$

Hence, the principal amount = ₹ 30,000.

(ii) For compound interest

P = ₹ 30,000

R = 10%

n = 2 years

$\text{A} = P\Big[1 + \dfrac{R}{100}\Big]^n\\[1em] \Rightarrow\text{A} = 30,000\Big[1 + \dfrac{10}{100}\Big]^2\\[1em] = 30,000\Big[1 + \dfrac{1}{10}\Big]^2\\[1em] = 30,000\Big[\dfrac{10}{10} + \dfrac{1}{10}\Big]^2\\[1em] = 30,000\Big[\dfrac{(10 + 1)}{10}\Big]^2\\[1em] = 30,000\Big[\dfrac{11}{10}\Big]^2\\[1em] = 30,000\Big[\dfrac{121}{100}\Big]\\[1em] = \Big[\dfrac{36,30,000}{100}\Big]\\[1em] = 36,300$

And

$\text{C.I. = A - P}\\[1em] \Rightarrow C.I. = 36,300 - 30,000\\[1em] = 6,300$

Hence, the compound interest ₹ 6,300