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.
Show SolutionHide Solution↓
(a) Time = 2 years = 24 months
r = 8%
P = ₹ 300 for Vinay
By formula,
I = $\frac{P \times n(n + 1)}{2} \times \frac{r}{12 \times 100}$
Substituting values, we get :
$$\begin{aligned}\Rightarrow I = \frac{300 \times 24(24 + 1)}{2} \times \frac{8}{1200} \\[1em] = \frac{300 \times 24(25)}{2} \times \frac{8}{1200} \\[1em] = 300 \times 300 \times \frac{8}{1200} \\[1em] = 300 \times 2 \\[1em] = ₹ 600.\end{aligned}$$
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 = $\frac{P \times n(n + 1)}{2} \times \frac{r}{12 \times 100}$
Substituting values, we get :
$$\begin{aligned}\Rightarrow I = \frac{P \times 24(24 + 1)}{2} \times \frac{8}{1200} \\[1em] \Rightarrow 1400 = P \times 300 \times \frac{8}{1200} \\[1em] \Rightarrow 1400 = 2P \\[1em] \Rightarrow P = \frac{1400}{2} = ₹ 700.\end{aligned}$$
Hence, Rohit deposits ₹700 per month.