A man gets ₹ 1,404 as interest at the end of one year. If the rate of interest is 12% per annum in R.D. account, the monthly installment is :

1. ₹ 1200

2. ₹ 1800

3. ₹ 2400

4. ₹ 3600

Given,

Interest = ₹ 1,404

Rate of interest (r) = 12%

Time (n) = 12 months

Let monthly installment be ₹ P.

By formula,

Interest = $P \times \dfrac{n(n + 1)}{2 \times 12} \times \dfrac{r}{100}$

Substituting values we get :

$\Rightarrow 1404 = P \times \dfrac{12 \times (12 + 1)}{2 \times 12} \times \dfrac{12}{100} \\[1em] \Rightarrow 1404 = P \times \dfrac{12 \times 13}{2 \times 12} \times \dfrac{12}{100} \\[1em] \Rightarrow P = \dfrac{1404 \times 100 \times 2 \times 12}{12 \times 13 \times 12} \\[1em] \Rightarrow P = \dfrac{3369600}{1872} \\[1em] \Rightarrow P = ₹ 1800.$

Hence, Option 2 is the correct option.