At what rate per cent compound interest, does a sum of money become 1.44 times of itself in 2 years ?

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

A = ₹ 1.44x

n = 2 years

By formula,

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

Substituting values we get :

$\Rightarrow 1.44x = x \times \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{1.44x}{x} = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow 1.44 = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow (1.2)^2 = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow 1.2 = 1 + \dfrac{r}{100} \\[1em] \Rightarrow 1.2 - 1 = \dfrac{r}{100} \\[1em] \Rightarrow 0.2 = \dfrac{r}{100} \\[1em] \Rightarrow r = 0.2 \times 100 = 20\%.$

Hence, rate of interest = 20%.