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Banking

Exercise 2

Question 1

Mrs. Goswami 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×n(n+1)2×12×r100, we getI=(1000×36×372×12×8100) =₹4440I = 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×n+I, we getMV=(1000×36)+4440 =36000+4440 =₹40440MV = P \times n + I \text{, we get} \\ MV = (1000 \times 36) + 4440 \\ \qquad\medspace = 36000 + 4440 \\ \qquad\medspace = \text{₹40440}

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

Question 2

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×n(n+1)2×12×r100, we getI=(600×30×312×12×10100) =₹2325I = 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×n+I, we getMV=(600×30)+2325 =18000+2325 =₹20325MV = 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.

Question 3

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×n(n+1)2×12×r100, we getI=(200×36×372×12×11100) =₹1221I = 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×n+I, we getMV=(200×36)+1221 =7200+1221 =₹8421MV = 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.

Question 4

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×n(n+1)2×12×r100, we getI=(600×48×492×12×r100) =588rI = 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⇒r=5880588⇒r=10588r = 5880 \\[0.5em] \Rightarrow r = \dfrac{5880}{588} \\[0.5em] \Rightarrow r = 10

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

Question 5

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×n(n+1)2×12×r100, we getI=(300×24×252×12×r100) =75rI = 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⇒75r=7725−7200⇒75r=525⇒r=52575⇒r=77200 + 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.

Question 6

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×n(n+1)2×12×r100, we get7500=(2500×24×252×12×r100)7500=625r⇒r=7500625⇒r=12I = 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.

Question 7

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×n(n+1)2×12×r100, we getI=(800×18×192×12×r100) =114rI = 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⇒114r=15084−14400⇒114r=684⇒r=684114⇒r=614400 + 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.

Question 8

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×n(n+1)2×12×r100, we getI=(x×20×212×12×9100) =₹1.575xI = 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⇒x=4411.575⇒x=2801.575x = 441 \\[0.5em] \Rightarrow x = \dfrac{441}{1.575} \\[0.5em] \Rightarrow x = 280 \\[0.5em]

∴ The monthly installment = ₹280

Question 9

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×n(n+1)2×12×r100, we getI=(x×24×252×12×6100) =₹1.5xI = 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⇒x=12001.5⇒x=8001.5x = 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

Question 10

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×n(n+1)2×12×r100, we getI=(x×12×132×12×14100) =₹91100xI = 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. R.K. Nair = ₹12x

∴ The amount of maturity = total money deposited + interest

=₹12x+₹91100x=₹1291100x= ₹12x + ₹\dfrac{91}{100}x \\[0.5em] = ₹\dfrac{1291}{100}x

According to the given,

Amount of maturity = ₹6455

⇒1291100x=6455⇒x100=64551291⇒x=5×100⇒x=500\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

Question 11

Samita 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.

Hint:

Let the account be held for n months, then

2000n+2000×n(n+1)2×12×10100=831002000 n + 2000 \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{10}{100} = 83100

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×n(n+1)2×12×r100, we getI=(2000×n(n+1)2×12×10100) =25n(n+1)3I = 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 Samita = ₹(2000 x n) = ₹2000n

∴ Amount of maturity = total amount deposited + interest

=₹2000n+25n(n+1)3=₹6000n+25n(n+1)3=₹25n2+6025n3= ₹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,

25n2+6025n3=83100⇒25n2+6025n−249300=0⇒n2+241n−9972=0⇒n2+277n−36n−9972=0⇒n(n+277)−36(n+277)=0⇒(n+277)(n−36)=0⇒n=−277,36 (but n cannot be negative)⇒n=36\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.

Multiple Choice Questions

Question 1

If Sharukh 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 Shahrukh = ₹800 x 18 = ₹14400

∴ Option 2 is the correct option.

Question 2

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×n(n+1)2×12×r100, we getI=(250×12×132×12×8100) =₹130I = 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.

Question 3

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×n(n+1)2×12×r100, we getI=(500×24×252×12×7100) =₹875I = 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×n+I, we getMV=(500×24)+875 =12000+875 =₹12875MV = 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.

Chapter Test

Question 1

Mr. 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×n(n+1)2×12×r100, we getI=(600×60×612×12×10100) =₹9150I = 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×n+I, we getMV=(600×60)+9150 =36000+9150 =₹45150MV = P \times n + I \text{, we get} \\ MV = (600 \times 60) + 9150 \\ \qquad\medspace = 36000 + 9150 \\ \qquad\medspace = \text{₹45150}

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

Question 2

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×n(n+1)2×12×r100, we getI=(400×36×372×12×10100) =₹2220I = 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×n+I, we getMV=(400×36)+2220 =14400+2220 =₹16620MV = 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×n(n+1)2×12×r100, we getI=(500×30×312×12×10100) =₹1937.50I = 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×n+I, we getMV=(500×30)+1937.50 =15000+1937.50 =₹16937.50MV = 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.

Question 3

Shilpa has a 4 year recurring deposit account in Bank of Maharashtra and deposits ₹800 per month. If she gets ₹48200 at the time of maturity, find

(i) the rate of (simple) interest,
(ii) the total interest earned by Shilpa

Answer

Here,
P = money deposited per month = ₹800,
n = number of months for which the money is deposited = 4 x 12 = 48

∴ Total money deposited by Shilpa = ₹(800 x 48) = ₹38400

Money Shilpa gets at the time of maturity = ₹48200

∴ Total interest earned by Shilpa = ₹48200 - ₹38400 = ₹9800

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

I=P×n(n+1)2×12×r100, we get9800=(800×48×492×12×r100)9800=784r⇒r=9800784⇒r=12.5I = P \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{r}{100} \text{, we get} \\[0.7em] 9800 = \Big( 800 \times \dfrac{48 \times 49}{2 \times 12} \times \dfrac{r}{100} \Big) \\[0.5em] 9800 = 784r \\[0.5em] \Rightarrow r = \dfrac{9800}{784} \\[0.5em] \Rightarrow r = 12.5

(i) Rate of (simple) interest = 12.5% p.a.

(ii) Total interest earned by Shilpa = ₹9800

Question 4

Mr. Chaturvedi has a recurring deposit account in Grindlay’s 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×n(n+1)2×12×r100, we getI=(x×54×552×12×11100) =₹108980xI = 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+₹108980x=₹540980x= ₹54x + ₹\dfrac{1089}{80}x \\[0.5em] = ₹\dfrac{5409}{80}x

According to the given,

540980x=101418.75⇒540980x=10141875100⇒x=10141875×80100×5409⇒x=1500\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

Question 5

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×n(n+1)2×12×r100, we getI=(600×n(n+1)2×12×7100) =7n(n+1)4I = 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+7n(n+1)4=₹2400n+7n(n+1)4=₹7n2+2407n4= ₹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,

7n2+2407n4=15450⇒7n2+2407n−61800=0⇒7n(n−24)+2575(n−24)=0⇒(n−24)(7n+2575)=0⇒n=24,−25757 (but n cannot be negative)⇒n=24\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.