In a recurring deposit account, John deposits ₹ 500 per month for 24 months. If the interest he earns is one-tenth of his total deposit, the rate of interest is :

1. 4.8%

2. 9.6%

3. 7.2%

4. 3.2%

Deposit per month (P) = ₹ 500

Time (n) = 24 months

Total deposit = ₹ 500 × 24 = ₹ 12000

Given,

Interest earned is one-tenth of total deposit.

Interest = $\dfrac{1}{10} \times 12000$ = ₹ 1200.

Let rate of interest be r%.

By formula,

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

Substituting values we get :

$\Rightarrow 1200 = 500 \times \dfrac{24 \times (24 + 1)}{2 \times 12} \times \dfrac{r}{100} \\[1em] \Rightarrow 1200 = 500 \times \dfrac{24 \times 25}{24} \times \dfrac{r}{100} \\[1em] \Rightarrow 1200 = 5 \times 25 \times r \\[1em] \Rightarrow r = \dfrac{1200}{125} \\[1em] \Rightarrow r = 9.6\%.$

Hence, Option 2 is the correct option.