A sum of money, invested at compounded interest, amounts to ₹ 19360 in 2 years and to ₹ 23425.60 in 4 years. Find the rate percent and the original sum of money.

Let original sum of money be ₹ P and rate of interest be r%.

By formula,

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

Given,

Sum amounts to ₹ 19360 in two years.

$\therefore P\Big(1 + \dfrac{r}{100}\Big)^2 = 19360$ ……(1)

Given,

Sum amounts to ₹ 23425.60 in four years.

$\therefore P\Big(1 + \dfrac{r}{100}\Big)^4 = 23425.60$ ……(2)

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

$\Rightarrow \dfrac{P\Big(1 + \dfrac{r}{100}\Big)^4}{P\Big(1 + \dfrac{r}{100}\Big)^2} = \dfrac{23425.60}{19360} \\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big)^2 = \dfrac{146.41}{121} \\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big)^2 = \Big(\dfrac{12.1}{11}\Big)^2 \\[1em] \Rightarrow 1 + \dfrac{r}{100} = \dfrac{12.1}{11} \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{12.1}{11} - 1 \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{12.1 - 11}{11} \\[1em] \Rightarrow r = \dfrac{1.1}{11} \times 100 \\[1em] \Rightarrow r = 10\%.$

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

$\Rightarrow P\Big(1 + \dfrac{r}{100}\Big)^2 = 19360 \\[1em] \Rightarrow P\Big(1 + \dfrac{10}{100}\Big)^2 = 19360 \\[1em] \Rightarrow P\Big(\dfrac{110}{100}\Big)^2 = 19360 \\[1em] \Rightarrow P \times \Big(\dfrac{11}{10}\Big)^2 = 19360 \\[1em] \Rightarrow P \times \dfrac{121}{100} = 19360 \\[1em] \Rightarrow P = \dfrac{19360 \times 100}{121} \\[1em] \Rightarrow P = 160 \times 100 \\[1em] \Rightarrow P = ₹ 16000.$

Hence, sum of money = ₹ 16000 and rate of interest = 10%.