Determine the rate of interest for a sum that becomes $\dfrac{216}{125}$ times of itself in $1\dfrac{1}{2}$ years, compounded semi-annually.

Let rate of interest per annum be r% per annum, i.e. $\dfrac{r}{2}$% half-yearly.

n = $1\dfrac{1}{2}$ years or 3 half-years.

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

Let principal be P,

∴ A = $\dfrac{216}{125}P$.

Putting values in formula we get,

$\Rightarrow \dfrac{216}{125}P = P\Big(1 + \dfrac{\dfrac{r}{2}}{100}\Big)^3 \\[1em] \Rightarrow \dfrac{216}{125} = \Big(1 + \dfrac{r}{200}\Big)^3\\[1em] \Rightarrow \Big(\dfrac{6}{5}\Big)^3 = \Big(1 + \dfrac{r}{200}\Big)^3\\[1em]$

Taking cube root on both sides,

$\dfrac{6}{5} = 1 + \dfrac{r}{200} \\[1em] \dfrac{6}{5} - 1 = \dfrac{r}{200} \\[1em] \dfrac{6 - 5}{5} = \dfrac{r}{200} \\[1em] \dfrac{1}{5} = \dfrac{r}{200} \\[1em] r = \dfrac{200}{5} \\[1em] r = 40\%.$

Hence, rate of interest = 40% per annum.