Banking — Analytical & Application Based Questions

A man opened a recurring deposit account in a branch of PNB. The man deposits certain amount of money per month such that after 2 years, the interest accumulated is equal to his monthly deposits. Find the rate of interest per annum that the bank was paying for the recurring deposit account.

Answer

Given,

Time (n) = 2 years or 24 months

Rate = r% (let)

P = ₹ x/month

I = ₹ x

By formula,

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

Substituting values we get :

$\Rightarrow x = \dfrac{x \times 24 \times (24 + 1)}{2 \times 12} \times \dfrac{r}{100} \\[1em] \Rightarrow x = \dfrac{x \times 24 \times 25}{24} \times \dfrac{r}{100} \\[1em] \Rightarrow 1 = \dfrac{r}{4} \\[1em] \Rightarrow r = 4$

Hence, rate of interest = 4%.

Amit deposited ₹ 600 per month in a recurring deposit account. The bank pays a simple interest of 12% p.a. Calculate the:

(i) number of monthly installments Amit deposits to get a maturity amount of ₹ 11826?

(ii) total interest paid by the bank.

(iii) total amount deposited by him.

Answer

(i) Let money be deposited for n months.

By formula,

M.V. = P × n + $\dfrac{P \times n(n + 1)}{2 \times 12} \times \dfrac{r}{100}$

Substituting values we get :

$\Rightarrow 11826 = 600 \times n + \dfrac{600 \times n(n + 1)}{2 \times 12} \times \dfrac{12}{100} \\[1em] \Rightarrow 11826 = 600n + \dfrac{600(n^2 + n)}{200} \\[1em] \Rightarrow 11826 = 600n + 3(n^2 + n) \\[1em] \Rightarrow 11826 = 600n + 3n^2 + 3n \\[1em] \Rightarrow 3n^2 + 603n = 11826 \\[1em] \Rightarrow 3(n^2 + 201n) = 11826 \\[1em] \Rightarrow n^2 + 201n = \dfrac{11826}{3} \\[1em] \Rightarrow n^2 + 201n = 3942 \\[1em] \Rightarrow n^2 + 201n - 3942 = 0 \\[1em] \Rightarrow n^2 + 219n - 18n - 3942 = 0 \\[1em] \Rightarrow n(n + 219) - 18(n + 219) = 0 \\[1em] \Rightarrow (n - 18)(n + 219) = 0 \\[1em] \Rightarrow n - 18 = 0 \text{ or } n + 219 = 0 \\[1em] \Rightarrow n = 18 \text{ or } n = -219.$

Since, no. of months cannot be negative.

∴ n = 18.

Hence, number of monthly installments = 18.

(ii) By formula,

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

Substituting values we get :

$\text{Interest } = \dfrac{600 \times 18(18 + 1)}{2 \times 12} \times \dfrac{12}{100} \\[1em] = \dfrac{600 \times 18 \times 19}{200} \\[1em] = 3 \times 18 \times 19 \\[1em] = \text{₹ 1026}.$

Hence, total interest paid = ₹ 1026.

(c) Total amount deposited by Amit = P × n

= ₹ 600 × 18

= ₹ 10800.

Hence, total amount deposited by Amit = ₹ 10800.