Ramesh deposits ₹ 2400 per month in a recurring deposit account deposit scheme of a bank for one year. If he gets ₹ 1248 as interest at the time of maturity, find the rate of interest. Also, find the maturity value of this deposit.

Given,

Monthly installment (P) = ₹ 2400

Time (n) = 12 months

Interest = ₹ 1248

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 :

$\Rightarrow 1248 = 2400 \times \dfrac{12 \times 13}{2 \times 12} \times \dfrac{r}{100} \\[1em] \Rightarrow 1248 = 12 \times 13 \times r \\[1em] \Rightarrow r = \dfrac{1248}{12 \times 13} \\[1em] \Rightarrow r = 8\%.$

Maturity value = P × n + Interest

= ₹ 2400 × 12 + ₹ 1248

= ₹ 28800 + ₹ 1248

= ₹ 30048.

Hence, rate of interest is 8% and maturity value = ₹ 30048.