A certain money is deposited every month for 8 months in a recurring deposit account at 12% p.a. simple interest. If the interest at the time of maturity is ₹ 36, the monthly instalment is :

1. ₹ 200

2. ₹ 1000

3. ₹ 100

4. ₹ 500

Given,

Time (n) = 8 months

Rate (r) = 12%

Interest = ₹ 36

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 :

$\Rightarrow 36 = P \times \dfrac{8 \times (8 + 1)}{2 \times 12} \times \dfrac{12}{100} \\[1em] \Rightarrow 36 = P \times \dfrac{8\times 9}{24} \times \dfrac{3}{25} \\[1em] \Rightarrow 36 = P \times \dfrac{24 \times 9}{24 \times 25} \\[1em] \Rightarrow P = \dfrac{36 \times 25 \times 24}{24 \times 9} \\[1em] \Rightarrow P = 4 \times 25 \\[1em] \Rightarrow P = ₹100.$

Hence, Option 3 is the correct option.