Banking — Exercise 2

Anushka deposits ₹1000 every month in a recurring deposit account for 3 years at 8% interest per annum. Find the matured value.

Answer

Here,
P = money deposited per month = ₹1000,
n = number of months for which the money is deposited = 3 x 12 = 36,
r = simple interest rate percent per annum = 8

Using the formula:

$I = P \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{r}{100} \text{, we get} \\[0.7em] I = \Big( 1000 \times \dfrac{36 \times 37}{2 \times 12} \times \dfrac{8}{100} \Big) \\[0.5em] \enspace\medspace = \text{₹4440}$

Using the formula:

$MV = P \times n + I \text{, we get} \\ MV = (1000 \times 36) + 4440 \\ \qquad\medspace = 36000 + 4440 \\ \qquad\medspace = \text{₹40440}$

∴ The amount Anushka will get at the time of maturity = ₹40440.

Sonia had a recurring deposit account in a bank and deposited ₹600 per month for 2½ years. If the rate of interest was 10% p.a., find the maturity value of this account.

Answer

Here,
P = money deposited per month = ₹600,
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:

$I = P \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{r}{100} \text{, we get} \\[0.7em] I = \Big( 600 \times \dfrac{30 \times 31}{2 \times 12} \times \dfrac{10}{100} \Big) \\[0.5em] \enspace\medspace = \text{₹2325}$

Using the formula:

$MV = P \times n + I \text{, we get} \\ MV = (600 \times 30) + 2325 \\ \qquad\medspace = 18000 + 2325 \\ \qquad\medspace = \text{₹20325}$

∴ The maturity value of Sonia's account = ₹20325.

Kiran deposited ₹200 per month for 36 months in a bank’s recurring deposit account. If the banks pays interest at the rate of 11% per annum, find the amount she gets on maturity?

Answer

Here,
P = money deposited per month = ₹200,
n = number of months for which the money is deposited = 36,
r = simple interest rate percent per annum = 11

Using the formula:

$I = P \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{r}{100} \text{, we get} \\[0.7em] I = \Big( 200 \times \dfrac{36 \times 37}{2 \times 12} \times \dfrac{11}{100} \Big) \\[0.5em] \enspace\medspace = \text{₹1221}$

Using the formula:

$MV = P \times n + I \text{, we get} \\ MV = (200 \times 36) + 1221 \\ \qquad\medspace = 7200 + 1221 \\ \qquad\medspace = \text{₹8421}$

∴ The amount Kiran will get at the time of maturity = ₹8421.

Haneef has a cumulative bank account and deposits ₹600 per month for a period of 4 years. If he gets ₹5880 as interest at the time of maturity, find the rate of interest per annum.

Answer

Here,
P = money deposited per month = ₹600,
n = number of months for which the money is deposited = 4 x 12 = 48

Let the rate of interest be r% per annum, then by using the formula:

$I = P \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{r}{100} \text{, we get} \\[0.7em] I = \Big( 600 \times \dfrac{48 \times 49}{2 \times 12} \times \dfrac{r}{100} \Big) \\[0.5em] \enspace\medspace = 588r$

According to the given,

$588r = 5880 \\[0.5em] \Rightarrow r = \dfrac{5880}{588} \\[0.5em] \Rightarrow r = 10$

∴ Rate of (simple) interest = 10% p.a.

David opened a Recurring Deposit Account in a bank and deposited ₹300 per month for two years. If he received ₹7725 at the time of maturity, find the rate of interest.

Answer

Here,
P = money deposited per month = ₹300,
n = number of months for which the money is deposited = 2 x 12 = 24

Let the rate of interest be r% per annum, then by using the formula:

$I = P \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{r}{100} \text{, we get} \\[0.7em] I = \Big( 300 \times \dfrac{24 \times 25}{2 \times 12} \times \dfrac{r}{100} \Big) \\[0.5em] \enspace\medspace = 75r$

Total money deposited by David = ₹300 x 24 = ₹7200

∴ The amount of maturity = total money deposited + interest
= 7200 + 75r

According to the given,

$7200 + 75r = 7725 \\[0.5em] \Rightarrow 75r = 7725 - 7200 \\[0.5em] \Rightarrow 75r = 525 \\[0.5em] \Rightarrow r = \dfrac{525}{75} \\[0.5em] \Rightarrow r = 7$

∴ Rate of (simple) interest = 7% p.a.

Mr. Gupta opened a recurring deposit account in a bank. He deposited ₹2500 per month for two years. At the time of maturity he got ₹67500. Find:

(i) the total interest earned by Mr. Gupta.
(ii) the rate of interest per annum.

Answer

(i) Here,
P = money deposited per month = ₹2500,
n = number of months for which the money is deposited = 2 x 12 = 24

∴ Total money deposited by Mr. Gupta = ₹(2500 x 24) = ₹60000

Money Mr. Gupta gets at the time of maturity = ₹67500

∴ Total interest earned by Mr. Gupta = ₹67500 - ₹60000 = ₹7500

(ii) Let the rate of interest be r% per annum, then by using the formula:

$I = P \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{r}{100} \text{, we get} \\[0.7em] 7500 = \Big( 2500 \times \dfrac{24 \times 25}{2 \times 12} \times \dfrac{r}{100} \Big) \\[0.5em] 7500 = 625r \\[0.5em] \Rightarrow r = \dfrac{7500}{625} \\[0.5em] \Rightarrow r = 12$

∴ Rate of (simple) interest = 12% p.a.

Shahrukh opened a Recurring Deposit Account in a bank and deposited ₹800 per month for 1½ years. If he received ₹15084 at the time of maturity, find the rate of interest per annum.

Answer

Here,
P = money deposited per month = ₹800,
n = number of months for which the money is deposited = 1 x 12 + 6 = 18

Let the rate of interest be r% per annum, then by using the formula:

$I = P \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{r}{100} \text{, we get} \\[0.7em] I = \Big( 800 \times \dfrac{18 \times 19}{2 \times 12} \times \dfrac{r}{100} \Big) \\[0.5em] \enspace\medspace = 114r$

Total money deposited by Shahrukh = ₹800 x 18 = ₹14400

∴ The amount of maturity = total money deposited + interest
= 14400 + 114r

According to the given,

$14400 + 114r = 15084 \\[0.5em] \Rightarrow 114r = 15084 - 14400 \\[0.5em] \Rightarrow 114r = 684 \\[0.5em] \Rightarrow r = \dfrac{684}{114} \\[0.5em] \Rightarrow r = 6$

∴ Rate of (simple) interest = 6% p.a.

Om has recurring deposit account and deposits ₹ 750 per month for 2 years. If he gets ₹ 19,125 at the time of maturity, find the rate of interest.

Answer

Let the rate of interest be r%.

Given,

P = ₹ 750/month

n = 2 years or 24 months

M.V. = ₹ 19,125

By formula,

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

Substituting values we get :

$\Rightarrow 19125 = 750 \times 24 + 750 \times \dfrac{24(24 + 1)}{2 \times 12} \times \dfrac{\text{r}}{100}\\[1em] \Rightarrow 19125 = 18000 + 750 \times \dfrac{24 \times 25}{24} \times \dfrac{\text{r}}{100}\\[1em] \Rightarrow 19125 = 18000 + 750 \times 25 \times \dfrac{\text{r}}{100}\\[1em] \Rightarrow 19125 = 18000 + 750 \times \dfrac{\text{r}}{4}\\[1em] \Rightarrow 750 \times \dfrac{\text{r}}{4} = 19125 - 18000 \\[1em] \Rightarrow 750 \times \dfrac{\text{r}}{4} = 1125\\[1em] \Rightarrow \text{r} = \dfrac{1125 \times 4}{750}\\[1em] \Rightarrow \text{r} = \dfrac{4500}{750} \\[1em] \Rightarrow \text{r} = 6$

Hence, rate of interest = 6%.

Rekha opened a recurring deposit account for 20 months. The rate of interest is 9% per annum and Rekha receives ₹441 as interest at the time of maturity. Find the amount Rekha deposited each month.

Answer

Here,
n = number of months for which the money is deposited = 20,
r = interest rate per annum = 9

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

Using the formula:

$I = P \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{r}{100} \text{, we get} \\[0.7em] I = \Big( x \times \dfrac{20 \times 21}{2 \times 12} \times \dfrac{9}{100} \Big) \\[0.5em] \enspace\medspace = ₹1.575x$

According to the given,

$1.575x = 441 \\[0.5em] \Rightarrow x = \dfrac{441}{1.575} \\[0.5em] \Rightarrow x = 280 \\[0.5em]$

∴ The monthly installment = ₹280

Mohan has a recurring deposit account in a bank for 2 years at 6% p.a. simple interest. If he gets ₹1200 as interest at the time of maturity, find

(i) the monthly installment.
(ii) the amount of maturity.

Answer

Here,
n = number of months for which the money is deposited = 2 x 12 = 24,
r = interest rate per annum = 6

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

Using the formula:

$I = P \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{r}{100} \text{, we get} \\[0.7em] I = \Big( x \times \dfrac{24 \times 25}{2 \times 12} \times \dfrac{6}{100} \Big) \\[0.5em] \enspace\medspace = ₹1.5x$

According to the given,

$1.5x = 1200 \\[0.5em] \Rightarrow x = \dfrac{1200}{1.5} \\[0.5em] \Rightarrow x = 800 \\[0.5em]$

∴ The monthly installment = ₹800

(ii) Total amount deposited by Mohan = ₹(800 x 24) = ₹19200

∴ Amount of maturity = total amount deposited + interest
= ₹19200 + ₹1200
= ₹20400

Mr. R.K. Nair gets ₹6455 at the end of one year at the rate of 14% per annum in a recurring deposit account. Find the monthly installment.

Answer

Here,
n = number of months for which the money is deposited = 1 x 12 = 12,
r = interest rate per annum = 14

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

Using the formula:

$I = P \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{r}{100} \text{, we get} \\[0.7em] I = \Big( x \times \dfrac{12 \times 13}{2 \times 12} \times \dfrac{14}{100} \Big) \\[0.5em] \enspace\medspace = ₹\dfrac{91}{100}x$

Total money deposited by Mr. Nair = ₹12x

∴ The amount of maturity = total money deposited + interest

$= ₹12x + ₹\dfrac{91}{100}x \\[0.5em] = ₹\dfrac{1291}{100}x$

According to the given,

Amount of maturity = ₹6455

$\Rightarrow \dfrac{1291}{100}x = 6455 \\[0.5em] \Rightarrow \dfrac{x}{100} = \dfrac{6455}{1291} \\[0.5em] \Rightarrow x = 5 \times 100 \\[0.5em] \Rightarrow x = 500$

∴ The monthly installment = ₹500

Suhani has a recurring deposit account in a bank of ₹2000 per month at the rate of 10% p.a. If she gets ₹83100 at the time of maturity, find the total time for which the account was held.

Answer

Here,
P = money deposited per month = ₹2000,
r = simple interest rate percent per annum = 10

Let the account be held for n months

Using the formula:

$I = P \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{r}{100} \text{, we get} \\[0.7em] I = \Big( 2000 \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{10}{100} \Big) \\[0.5em] \enspace\medspace = \dfrac{25n(n+1)}{3}$

Total money deposited by suhani = ₹(2000 x n) = ₹2000n

∴ Amount of maturity = total amount deposited + interest

$= ₹2000n + \dfrac{25n(n+1)}{3} \\[0.5em] = ₹\dfrac{6000n + 25n(n+1)}{3} \\[0.5em] = ₹\dfrac{25n^2 + 6025n}{3}$

According to the given,

$\dfrac{25n^2 + 6025n}{3} = 83100 \\[0.5em] \Rightarrow 25n^2 + 6025n - 249300 = 0 \\[0.5em] \Rightarrow n^2 + 241n - 9972 = 0 \\[0.5em] \Rightarrow n^2 + 277n -36n - 9972 = 0 \\[0.5em] \Rightarrow n(n + 277) -36(n + 277) = 0 \\[0.5em] \Rightarrow (n + 277)(n - 36) = 0 \\[0.5em] \Rightarrow n = -277, 36 \\[0.5em] \text{ (but n cannot be negative)} \\[0.5em] \Rightarrow n = 36$

∴ The account was held for 36 months i.e. 3 years.