₹ 4000 amounts to ₹ 4600 in one year at compound interest compounded yearly. The rate of interest is :

1. 15%

2. 12%

3. 10%

4. 20%

Let rate of interest be r%.

Given,

P = ₹ 4000

A = ₹ 4600

n = 1 year

By formula,

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

Substituting values we get :

$\Rightarrow 4600 = 4000(1 + \dfrac{r}{100})^1 \\[1em] \Rightarrow \dfrac{4600}{4000} = 1 + \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{23}{20} = 1 + \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{23}{20} - 1 = \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{23 - 20}{20} = \dfrac{r}{100} \\[1em] \Rightarrow r = \dfrac{3}{20} \times 100 \\[1em] \Rightarrow r = 15\%.$

Hence, Option 1 is the correct option.