In a recurring deposit account for 2 years, the total amount deposited by a person is ₹ 9600. If the interest earned by him is one-twelfth of his total deposit, then find :

(a) the interest he earns

(b) his monthly deposit

(c) the rate of interest

(a) Given,

Interest earned by man = One-twelfth of the total deposit

= $\dfrac{1}{12} \times 9600$

= ₹ 800.

Hence, the interest earned = ₹ 800.

(b) Given,

In a recurring deposit account for 2 years (or 24 months), the total amount deposited by a person is ₹ 9600.

Money deposited per month = $\dfrac{9600}{24}$ = ₹ 400.

Hence, monthly deposit = ₹ 400.

(c) By formula,

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

Substituting values we get :

$\Rightarrow 800 = \dfrac{400 \times 24 \times (24 + 1)}{2 \times 12} \times \dfrac{r}{100} \\[1em] \Rightarrow 800 = \dfrac{400 \times 24 \times 25}{24} \times \dfrac{r}{100} \\[1em] \Rightarrow 800 = 400 \times 25 \times \dfrac{r}{100} \\[1em] \Rightarrow 800 = 4 \times 25 \times r \\[1em] \Rightarrow 800 = 100 \times r \\[1em] \Rightarrow r = \dfrac{800}{100} \\[1em] \Rightarrow r = 8\%.$

Hence, rate of interest = 8%.