A sum of money, invested at compound interest, amounts to ₹ 16500 in 1 year and to ₹ 19965 in 3 years. Find the rate per cent and the original sum of money invested.

Let original sum of money invested be ₹ x and rate of percent be r%.

By formula,

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

Given,

The sum of money, invested at compound interest, amounts to ₹ 16500 in 1 year.

$\Rightarrow A = P\Big(1 + \dfrac{r}{100}\Big)^n \\[1em] \Rightarrow 16500 = x \times \Big(1 + \dfrac{r}{100}\Big)^1 \\[1em] \Rightarrow 16500 = x\Big(1 + \dfrac{r}{100}\Big) ……(1)$

The sum of money, invested at compound interest, amounts to ₹ 19965 in 3 years.

$\Rightarrow A = P\Big(1 + \dfrac{r}{100}\Big)^n \\[1em] \Rightarrow 19965 = x \times \Big(1 + \dfrac{r}{100}\Big)^3 \\[1em] \Rightarrow 19965 = x\Big(1 + \dfrac{r}{100}\Big)^3 ……(2)$

Dividing equation (2) by (1), we get :

$\Rightarrow \dfrac{19965}{16500} = \dfrac{x\Big(1 + \dfrac{r}{100}\Big)^3}{x\Big(1 + \dfrac{r}{100}\Big)} \\[1em] \Rightarrow \dfrac{121}{100} = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{11}{10}\Big)^2 = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big) = \dfrac{11}{10} \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{11}{10} - 1 \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{1}{10} \\[1em] \Rightarrow r = \dfrac{100}{10} = 10\%.$

Substituting value of r in equation (1), we get :

$\Rightarrow 16500 = x\Big(1 + \dfrac{10}{100}\Big) \\[1em] \Rightarrow 16500 = x \times \Big(\dfrac{110}{100}\Big) \\[1em] \Rightarrow x = \dfrac{16500 \times 100}{110} \\[1em] \Rightarrow x = ₹ 15000.$

Hence, rate percent = 10% and sum invested = ₹ 15000.