₹ 4000 amounts to ₹ 5017.60 in two months at compound interest compounded per month. The rate of interest per month is :

1. 12%

2. 15%

3. 10%

4. 20%

Let rate of interest per month be r%.

Given,

P = ₹ 4000

A = ₹ 5017.60

T = 2 months

By formula,

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

Substituting values we get :

$\Rightarrow 5017.60 = 4000(1 + \dfrac{r}{100})^2 \\[1em] \Rightarrow \dfrac{5017.60}{4000} = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{501760}{400000} = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{50176}{40000} = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{224}{200}\Big)^2 = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow 1 + \dfrac{r}{100} = \dfrac{224}{200} \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{224}{200} - 1 \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{24}{200} \\[1em] \Rightarrow r =\dfrac{24}{200} \times 100 \\[1em] \Rightarrow r = 12\%.$

Hence, Option 1 is the correct option.