Rohit and Vinay both opened a recurring deposit account in a bank for 2 years at 8% simple interest. Vinay deposited ₹ 300 per month. On maturity, Rohit’s interest was ₹ 800 more than Vinay’s interest. Find:

(a) interest earned by Vinay.

(b) sum deposited by Rohit every month.

(a) Time = 2 years = 24 months

r = 8%

P = ₹ 300 for Vinay

By formula,

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

Substituting values, we get :

$\Rightarrow I = \dfrac{300 \times 24(24 + 1)}{2} \times \dfrac{8}{1200} \\[1em] = \dfrac{300 \times 24(25)}{2} \times \dfrac{8}{1200} \\[1em] = 300 \times 300 \times \dfrac{8}{1200} \\[1em] = 300 \times 2 \\[1em] = ₹ 600.$

Hence, interest earned by Vinay = ₹ 600.

(b) Rohit’s interest is ₹800 more than Vinay’s.

Interest earned by Rohit = ₹600 + ₹800 = ₹1400.

Let Rohit deposit ₹P per month.

By formula,

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

Substituting values, we get :

$\Rightarrow I = \dfrac{P \times 24(24 + 1)}{2} \times \dfrac{8}{1200} \\[1em] \Rightarrow 1400 = P \times 300 \times \dfrac{8}{1200} \\[1em] \Rightarrow 1400 = 2P \\[1em] \Rightarrow P = \dfrac{1400}{2} = ₹ 700.$

Hence, Rohit deposits ₹700 per month.