Banking — Chapter Test

Dhruv deposits ₹600 per month in a recurring deposit account for 5 years at the rate of 10% per annum (simple interest). Find the amount he will receive at the time of maturity.

Answer

Here,
P = money deposited per month = ₹600,
n = number of months for which the money is deposited = 5 x 12 = 60,
r = simple interest rate percent per annum = 10

Using the formula:

$I = P \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{r}{100} \text{, we get} \\[0.7em] I = \Big( 600 \times \dfrac{60 \times 61}{2 \times 12} \times \dfrac{10}{100} \Big) \\[0.5em] \enspace\medspace = \text{₹9150}$

Using the formula:

$MV = P \times n + I \text{, we get} \\ MV = (600 \times 60) + 9150 \\ \qquad\medspace = 36000 + 9150 \\ \qquad\medspace = \text{₹45150}$

∴ The amount Dhruv will get at the time of maturity = ₹45150.

Ankita started paying ₹400 per month in a 3 years recurring deposit. After six months her brother Anshul started paying ₹500 per month in a $2\dfrac{1}{2}$ years recurring deposit. The bank paid 10% p.a. simple interest for both. At maturity who will get more money and by how much?

Answer

For Ankita,
P = money deposited per month = ₹400,
n = number of months for which the money is deposited = 3 x 12 = 36,
r = simple interest rate percent per annum = 10

Using the formula:

$I = P \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{r}{100} \text{, we get} \\[0.7em] I = \Big( 400 \times \dfrac{36 \times 37}{2 \times 12} \times \dfrac{10}{100} \Big) \\[0.5em] \enspace\medspace = \text{₹2220}$

Using the formula:

$MV = P \times n + I \text{, we get} \\ MV = (400 \times 36) + 2220 \\ \qquad\medspace = 14400 + 2220 \\ \qquad\medspace = \text{₹16620}$

The amount Ankita will get at the time of maturity = ₹16620.

For Anshul,
P = money deposited per month = ₹500,
n = number of months for which the money is deposited = 2 x 12 + 6 = 30,
r = simple interest rate percent per annum = 10

Using the formula:

$I = P \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{r}{100} \text{, we get} \\[0.7em] I = \Big( 500 \times \dfrac{30 \times 31}{2 \times 12} \times \dfrac{10}{100} \Big) \\[0.5em] \enspace\medspace = \text{₹1937.50}$

Using the formula:

$MV = P \times n + I \text{, we get} \\ MV = (500 \times 30) + 1937.50 \\ \qquad\medspace = 15000 + 1937.50 \\ \qquad\medspace = \text{₹16937.50}$

The amount Anshul will get at the time of maturity = ₹16937.50.

Difference in maturity amount = 16937.50 - 16620 = 317.50

∴ Anshul will get ₹317.50 more than Ankita at maturity.

Salman deposits ₹ 1,000 every month in a recurring deposit account for 2 years. If he receives ₹ 26,000 on maturity, find :

(a) total interest he earns

(b) the rate of interest.

Answer

(a) Given,

Salman deposits ₹ 1000 every month in a recurring deposit account for 2 years.

Total deposit = ₹ 1,000 × 2 × 12 = ₹ 24,000.

By formula,

Total interest earned = Maturity value - Total deposit

= ₹ 26,000 - ₹ 24,000

= ₹ 2,000.

Hence, interest earned = ₹ 2,000.

(b) Let rate of interest be r%. Time (n) = 24 months

By formula,

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

Substituting values we get :

$\Rightarrow 2000 = 1000 \times \dfrac{24 \times 25}{2 \times 12} \times \dfrac{r}{100} \\[1em] \Rightarrow 2000 = 1000 \times 25 \times \dfrac{r}{100} \\[1em] \Rightarrow 2000 = 250r \\[1em] \Rightarrow r = \dfrac{2000}{250} \\[1em] \Rightarrow r = 8$

Hence, rate of interest earned = 8%.

Mr. Chaturvedi has a recurring deposit account in a Bank for $4\dfrac{1}{2}$ years at 11% p.a. (simple interest). If he gets ₹101418.75 at the time of maturity, find the monthly installment.

Answer

Here,
n = number of months for which the money is deposited = 4 x 12 + 6 = 54,
r = interest rate per annum = 11

Let the monthly installment be ₹x, then P = ₹x.

Using the formula:

$I = P \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{r}{100} \text{, we get} \\[0.7em] I = \Big( x \times \dfrac{54 \times 55}{2 \times 12} \times \dfrac{11}{100} \Big) \\[0.5em] \enspace\medspace = ₹\dfrac{1089}{80}x$

Total money deposited by Mr. Chaturvedi = ₹54x

∴ The amount of maturity = total money deposited + interest

$= ₹54x + ₹\dfrac{1089}{80}x \\[0.5em] = ₹\dfrac{5409}{80}x$

According to the given,

$\dfrac{5409}{80}x = 101418.75 \\[0.5em] \Rightarrow \dfrac{5409}{80}x = \dfrac{10141875}{100} \\[0.5em] \Rightarrow x = \dfrac{10141875 \times 80}{100 \times 5409} \\[0.5em] \Rightarrow x = 1500 \\[0.5em]$

∴ The monthly installment = ₹1500

Rajiv Bhardwaj has a recurring deposit account in a bank of ₹600 per month. If the bank pays simple interest of 7% p.a. and he gets ₹15450 as maturity amount, find the total time for which the account was held.

Answer

Here,
P = money deposited per month = ₹600,
r = simple interest rate percent per annum = 7

Let the account be held for n months

Using the formula:

$I = P \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{r}{100} \text{, we get} \\[0.7em] I = \Big( 600 \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{7}{100} \Big) \\[0.5em] \enspace\medspace = \dfrac{7n(n+1)}{4}$

Total money deposited by Rajiv Bhardwaj = ₹(600 x n) = ₹600n

∴ Amount of maturity = total amount deposited + interest

$= ₹600n + \dfrac{7n(n+1)}{4} \\[0.5em] = ₹\dfrac{2400n + 7n(n+1)}{4} \\[0.5em] = ₹\dfrac{7n^2 + 2407n}{4}$

According to the given,

$\dfrac{7n^2 + 2407n}{4} = 15450 \\[0.5em] \Rightarrow 7n^2 + 2407n - 61800 = 0 \\[0.5em] \Rightarrow 7n(n - 24) + 2575(n - 24) = 0 \\[0.5em] \Rightarrow (n - 24)(7n + 2575) = 0 \\[0.5em] \Rightarrow n = 24, -\dfrac{2575}{7} \\[0.5em] \text{ (but n cannot be negative)} \\[0.5em] \Rightarrow n = 24$

∴ The account was held for 24 months i.e. 2 years.

In a recurring deposit account for 2 years, the total amount deposited by a person is ₹ 9600. If the interest earned by him is one-twelfth of his total deposit, then find :

(a) the interest he earns

(b) his monthly deposit

(c) the rate of interest

Answer

(a) Given,

Interest earned by man = One-twelfth of the total deposit

= $\dfrac{1}{12} \times 9600$

= ₹ 800.

Hence, the interest earned = ₹ 800.

(b) Given,

In a recurring deposit account for 2 years (or 24 months), the total amount deposited by a person is ₹ 9600.

Money deposited per month = $\dfrac{9600}{24}$ = ₹ 400.

Hence, monthly deposit = ₹ 400.

(c) By formula,

Rate of interest = $\dfrac{\text{Interest earned}}{\text{Amount invested}} \times 100$

$= \dfrac{800}{9600} \times 100$ %.

Hence, rate of interest = $8\dfrac{1}{3}$ %.

Suresh has a recurring deposit account in a bank. He deposits ₹2000 per month and the bank pays interest at the rate of 8% per annum. If he gets ₹1040 as interest at the time of maturity, find in years total time for which the account was held.

Answer

Let time be n months.

By formula,

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

Substituting values we get :

$\Rightarrow 1040 = 2000 \times \dfrac{n(n + 1)}{24} \times \dfrac{8}{100} \\[1em] \Rightarrow 1040 = 2000 \times \dfrac{n(n + 1)}{300} \\[1em] \Rightarrow n(n + 1) = \dfrac{1040 \times 300}{2000} \\[1em] \Rightarrow n(n + 1) = 156 \\[1em] \Rightarrow n^2 + n - 156 = 0 \\[1em] \Rightarrow n^2 + 13n - 12n - 156 = 0 \\[1em] \Rightarrow n(n + 13) - 12(n + 13) = 0 \\[1em] \Rightarrow (n - 12)(n + 13) = 0 \\[1em] \Rightarrow n - 12 = 0 \text{ or } n + 13 = 0 \\[1em] \Rightarrow n = 12 \text{ or } n = -13.$

Since, no. of months cannot be negative.

∴ n = 12.

Hence, total time for which the account was held = 12 months.

Sameer has a recurring deposit account and deposits ₹ 600 per month for 2 years. If he gets ₹ 15600 at the time of maturity, find the rate of interest earned by him.

Answer

Let rate of interest be r%.

Given,

P = ₹ 600/month

n = 2 years or 24 months

M.V. = ₹ 15600

By formula,

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

Substituting values we get :

$\Rightarrow 15600 = 600 \times 24 + 600 \times \dfrac{24 \times (24 + 1)}{2 \times 12} \times \dfrac{r}{100} \\[1em] \Rightarrow 15600 = 14400 + 6 \times 25 \times r \\[1em] \Rightarrow 15600 - 14400 = 6 \times 25 \times r \\[1em] \Rightarrow 150r = 1200 \\[1em] \Rightarrow r = \dfrac{1200}{150} = 8$

Hence, rate of interest = 8%.