Nikita invests ₹ 6,000 for two years at a certain rate of interest compounded annually. At the end of the first year, it amounts to ₹ 6,720. The rate of interest p.a. is:

1. 8%

2. 10%

3. 12%

4. 14%

Given,

P = ₹ 6,000

n = 1 year

A = ₹ 6,720

Let the rate of interest be r,

By formula,

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

Substituting values we get :

$\Rightarrow 6720 = 6000 \times \Big(1 + \dfrac{r}{100}\Big)^1 \\[1em] \Rightarrow \dfrac{6720}{6000} = 1 + \dfrac{r}{100} \\[1em] \Rightarrow 1 + \dfrac{r}{100} = 1.12 \\[1em] \Rightarrow \dfrac{r}{100} = 1.12 - 1 \\[1em] \Rightarrow r = 0.12 \times 100 \\[1em] \Rightarrow r = 12\%$

Hence, option 3 is correct option.