Banking

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.

If Vijay opened a recurring deposit account in a bank and deposited ₹800 per month for 1½ years, then the total money deposited in the account is

1. ₹11400
2. ₹14400
3. ₹13680
4. none of these

Answer

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

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

∴ Option 2 is the correct option.

Mrs. Asha Mehta deposit ₹250 per month for one year in a bank’s recurring deposit account. If the rate of (simple) interest is 8% per annum, then the interest earned by her on this account is

1. ₹65
2. ₹120
3. ₹130
4. ₹260

Answer

Here,
P = money deposited per month = ₹250,
n = number of months for which the money is deposited = 1 x 12 = 12,
r = simple interest rate percent per annum = 8

Using the formula:

$I = P \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{r}{100} \text{, we get} \\[0.7em] I = \Big( 250 \times \dfrac{12 \times 13}{2 \times 12} \times \dfrac{8}{100} \Big) \\[0.5em] \enspace\medspace = \text{₹130}$

Interest earned by Mrs. Asha Mehta = ₹130

∴ Option 3 is the correct option.

Mr. Sharma deposited ₹500 every month in a cumulative deposit account for 2 years. If the bank pays interest at the rate of 7% per annum, then the amount he gets on maturity is

1. ₹875
2. ₹6875
3. ₹10875
4. ₹12875

Answer

Here,
P = money deposited per month = ₹500,
n = number of months for which the money is deposited = 2 x 12 = 24,
r = simple interest rate percent per annum = 7

Using the formula:

$I = P \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{r}{100} \text{, we get} \\[0.7em] I = \Big( 500 \times \dfrac{24 \times 25}{2 \times 12} \times \dfrac{7}{100} \Big) \\[0.5em] \enspace\medspace = \text{₹875}$

Using the formula:

$MV = P \times n + I \text{, we get} \\ MV = (500 \times 24) + 875 \\ \qquad\medspace = 12000 + 875 \\ \qquad\medspace = \text{₹12875}$

The amount Mr. Sharma will get at the time of maturity = ₹12875

∴ Option 4 is the correct option.

Radha deposited ₹ 400 per month in a recurring deposit account for 18 months. The qualifying sum of money for the calculation of interest is :

1. ₹ 3600

2. ₹ 7200

3. ₹ 68,400

4. ₹ 1,36,800

Answer

Since, Radha deposits ₹ 400 per month in a recurring deposit account for 18 months, thus the amount deposited in first month will earn interest for 18 months, the amount deposited in second month will earn interest for 17 months and so on.

$\text{Qualifying sum }= ₹ 400 \times (18 + 17 + 16 + ........ + 1) \\[1em] = ₹ 400 \times \dfrac{18(18 + 1)}{2} \\[1em] = ₹ 400 \times 9 \times 19 \\[1em] = ₹ 68,400.$

Hence, Option 3 is the correct option.

A person deposit ₹P, every month for n months at R percent per annum, simple interest in a recurring deposit account.

Assertion (A): The maturity value is more than total amount deposited by the person.

Reason (R): Maturity value includes an interest equal to $\dfrac{\text{P} \times \text{n(n + 1)} \times \text{R}}{2400}$

1. Assertion (A) is true, but Reason (R) is false.

2. Assertion (A) is false, but Reason (R) is true.

3. Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

4. Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).

Answer

Given,

Amount deposited per month = ₹ P

Time = n months

Rate of interest = R%

The total amount deposited is the monthly deposit multiplied by the number of months.

Total deposited = P x n

The interest earned is given by the formula = $\text{P} \times\dfrac{\text{n(n + 1)}}{2 \times 12} \times \dfrac{\text{R}}{100}$

The maturity value is the total deposited amount plus the interest earned.

Maturity value = P x n + $\text{P} \times\dfrac{\text{n(n + 1)}}{2 \times 12} \times \dfrac{\text{R}}{100}$

The maturity value includes the total deposited amount plus the interest earned, which is a positive value.

∴ The maturity value is more than total amount deposited by the person.

So, assertion (A) is true.

By formula,

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

= $\dfrac{\text{P} \times \text{n(n + 1)} \times \text{R}}{2400}$

So, both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

Jena has a cumulative deposit account in Indian Bank. She deposit ₹ 500 per month for 2 years. Bank pays interest at the rate of 6% p.a.

Assertion (A): Money received by Jena at maturity is ₹ 12,750.

Reason (R): Maturity value = money deposit + interest.

1. Assertion (A) is true, but Reason (R) is false.

2. Assertion (A) is false, but Reason (R) is true.

3. Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

4. Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).

Answer

Given,

P = ₹ 500/month

n = 2 years or 24 months

r% = 6%

By formula,

M.V. = money deposit + interest

So, reason (R) is true.

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

Substituting the values, we get :

$M.V. = 500 \times 24 + 500 \times \dfrac{24(24 + 1)}{2 \times 12} \times \dfrac{6}{100}\\[1em] = 1200 + 500 \times \dfrac{24 \times 25}{24} \times \dfrac{6}{100}\\[1em] = 1200 + 500 \times 25 \times \dfrac{6}{100}\\[1em] = 1200 + 500 \times \dfrac{6}{4}\\[1em] = 1200 + 125 \times 6\\[1em] = 1200 + 750\\[1em] = ₹ 12,750$

So, both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

Rohana has a recurring deposit account in State Bank of India. She deposits ₹ 800 per month for $2\dfrac{1}{2}$ years. Bank pays interest at the rate of 5% p.a.

Assertion (A): Interest earned by Rohana in $2\dfrac{1}{2}$ years is ₹ 1,650.

Reason (R): Interest earned = $P \times \dfrac{\text{n(n + 1)}}{2 \times 12} \times \dfrac{\text{r}}{100}$

where, P = monthly instalment, n = number of installments and r = rate of interest p.a.

1. Assertion (A) is true, but Reason (R) is false.

2. Assertion (A) is false, but Reason (R) is true.

3. Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

4. Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).

Answer

Given,

P = ₹ 800/month

n = $2\dfrac{1}{2}$ years or 30 months

r% = 5%

By formula,

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

So, reason (R) is true.

Substituting the values, we get :

$I = 800 \times \dfrac{30(30 + 1)}{2 \times 12} \times \dfrac{5}{100}\\[1em] = 800 \times \dfrac{30 \times 31}{24} \times \dfrac{5}{100}\\[1em] = 800 \times \dfrac{10 \times 31}{8} \times \dfrac{5}{100}\\[1em] = 10 \times 31 \times 5\\[1em] = ₹ 1,550.$

So, assertion (A) is false.

Thus, Assertion (A) is false, but Reason (R) is true.

Hence, option 2 is the correct option.

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 2½ 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 Salman 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 4½ 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.

Mr. 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%.