A certain sum of money invested at CI triples itself in 8 year interest being payable annually. In how many years will it be 81 times?

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

By formula, A = $P\Big(1 + \dfrac{r}{100}\Big)^t$

Given,

₹ P becomes three times of itself in 8 years.

$\therefore 3P = P\Big(1 + \dfrac{r}{100}\Big)^8 \\[1em] \Rightarrow \dfrac{3P}{P} = \Big(1 + \dfrac{r}{100}\Big)^8 \\[1em] \Rightarrow 3 = \Big(1 + \dfrac{r}{100}\Big)^8 …………………(1)$

Let in n years money becomes 81 times.

$\Rightarrow P\Big(1 + \dfrac{r}{100}\Big)^n = 81P\\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big)^n = \dfrac{81P}{P}\\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big)^n = 81\\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big)^n = 3^4\\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big)^n = \Big[\Big(1 + \dfrac{r}{100}\Big)^8\Big]^4 \text{ [From (1)]}\\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big)^n = \Big(1 + \dfrac{r}{100}\Big)^{32}\\[1em]$

⇒ n = 32.

Hence, in 32 years money will becomes 81 times of itself.