₹ 16,000 invested at 10% p.a. compounded semi-annually amounts to ₹ 18,522. Find the time period of investment.

Given:

P = ₹ 16,000

R = 10%

A = ₹ 18,522

Let the time be $n$.

Since the interest is compounded half-yearly,

$\text{A} = P\Big[1 + \dfrac{r}{2 \times 100}\Big]^{2\times n}\\[1em] \Rightarrow 18,522 = 16,000\Big[1 + \dfrac{10}{200}\Big]^{2\times n}\\[1em] \Rightarrow \dfrac{18,522}{16,000} = \Big[1 + \dfrac{10}{200}\Big]^{2\times n}\\[1em] \Rightarrow \dfrac{18,522}{16,000} = \Big[1 + \dfrac{1}{20}\Big]^{2\times n}\\[1em] \Rightarrow \dfrac{18,522}{16,000} = \Big[\dfrac{20}{20} + \dfrac{1}{20}\Big]^{2\times n}\\[1em] \Rightarrow \dfrac{18,522}{16,000} = \Big[\dfrac{(20 + 1)}{20}\Big]^{2\times n}\\[1em] \Rightarrow \dfrac{18,522}{16,000} = \Big[\dfrac{21}{20}\Big]^{2\times n}\\[1em] \Rightarrow \dfrac{9,261}{8,000} = \Big[\dfrac{21}{20}\Big]^{2\times n}\\[1em] \Rightarrow \Big[\dfrac{21}{20}\Big]^3 = \Big[\dfrac{21}{20}\Big]^{2\times n}\\[1em]$

Hence, $2 \times n = 3$

$n = \dfrac{3}{2}$ years

$n = 1\dfrac{1}{2}$ years

Hence, the time period of investment is $1\dfrac{1}{2}$ years.