A sum of money is invested at 10% per annum compounded half-yearly. If the difference of amounts at end of 6 months and 12 months is ₹ 189, find the sum of money invested.

Let sum of money invested be ₹ x.

When interest is compounded half-yearly :

A = $P\Big(1 + \dfrac{r}{2 \times 100}\Big)^{n \times 2}$

For first $\dfrac{1}{2}$ year :

$A = x \times \Big(1 + \dfrac{10}{2 \times 100}\Big)^{\dfrac{1}{2} \times 2} \\[1em] = x \times \Big(1 + \dfrac{1}{20}\Big) \\[1em] = x \times \dfrac{21}{20} \\[1em] = \dfrac{21x}{20}.$

For first 1 year :

$A = x \times \Big(1 + \dfrac{10}{2 \times 100}\Big)^{1 \times 2} \\[1em] = x \times \Big(1 + \dfrac{1}{20}\Big)^2 \\[1em] = x \times \Big(\dfrac{21}{20}\Big)^2 \\[1em] = \dfrac{441x}{400}.$

Given,

Difference of amounts at end of 6 months and 12 months is ₹ 189.

$\Rightarrow \dfrac{441x}{400} - \dfrac{21x}{20} = 189 \\[1em] \Rightarrow \dfrac{441x - 420x}{400} = 189 \\[1em] \Rightarrow \dfrac{21x}{400} = 189 \\[1em] \Rightarrow x = \dfrac{189 \times 400}{21} \\[1em] \Rightarrow x = ₹ 3600.$

Hence, sum invested = ₹ 3600.