Banking — Exercise 2

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

Answer

Given,

P = ₹1,000

n = 3 years = 3 x 12 = 36 months

r = 8%

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

$\therefore I = 1000 \times \dfrac{36 \times 37}{2 \times 12} \times \dfrac{8}{100} \\[1em] I = 1000 \times \dfrac{1332}{24} \times 0.08 \\[1em] I = 1000 \times 55.5 \times 0.08 \\[1em] I = ₹4,440$

Sum deposited = ₹1,000 x 36 = ₹36,000

Maturity value = Sum deposited + Interest = ₹36,000 + ₹4,440 = ₹40,440

Hence, the matured value is ₹40,440

Inderjeet opened a cumulative time deposit account with Punjab National Bank. He deposited ₹360 per month for 2 years. If the rate of interest be 7% per annum, how much did he get at the time of maturity?

Answer

Given,

P = ₹360

n = 2 years = 2 x 12 = 24 months

r = 7%

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

$\therefore I = 360 \times \dfrac{24 \times 25}{2 \times 12} \times \dfrac{7}{100} \\[1em] I = 360 \times \dfrac{600}{24} \times 0.07 \\[1em] I = 360 \times 25 \times 0.07 \\[1em] I = ₹630$

Sum deposited = ₹360 x 24 = ₹8,640

Maturity value = Sum deposited + Interest = ₹8,640 + ₹630 = ₹9,270

Hence, Inderjeet got ₹9,270 at the time of maturity.

Neema had a recurring deposit account in a bank and deposited ₹ 600 per month for $2\dfrac{1}{2}$ years. If the rate of interest was 10% per annum, find the maturity value of this account.

Answer

Given,

P = ₹600

n = $2\dfrac{1}{2}$ years = 2.5 years = 24 months + 6 months = 30 months

r = 10%

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

$\therefore I = 600\times \dfrac{30\times 31}{2 \times 12} \times \dfrac{10}{100} \\[1em] I = 600 \times \dfrac{930}{24} \times 0.1 \\[1em] I = 600 \times 38.75 \times 0.1 \\[1em] I = ₹2,325$

Sum deposited = ₹600 x 30 = ₹18,000

Maturity value = Sum deposited + Interest = ₹18,000 + ₹2,325 = ₹20,325

Hence, Neema got ₹20,325 at the time of maturity.

Sajal invests ₹600 per month for $2\dfrac{1}{2}$ years in a recurring deposit scheme of Oriental Bank of Commerce. If the bank pays simple interest at $6\dfrac{2}{3}$ % per annum, find the amount received by him on maturity.

Answer

Given,

P = ₹600

n = $2\dfrac{1}{2}$ years = 2.5 years = 24 months + 6 months = 30 months

r = 6 $\dfrac{2}{3}$ % = $\dfrac{20}{3}$

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

$\therefore I = 600\times \dfrac{30\times 31}{2 \times 12} \times \dfrac{\dfrac{20}{3}}{100}\\[1em] I = 600 \times \dfrac{930}{24} \times \dfrac{20}{300}\\[1em] I = 600 \times 38.75 \times 0.67\\[1em] I = ₹1,550$

Sum deposited = ₹600 x 30 = ₹18,000

Maturity value = Sum deposited + Interest = ₹18,000 + ₹1,550 = ₹19,550

Hence, Sajal got ₹19,550 at the time of maturity.

Mr. Richard has a recurring deposit account in a bank for 3 years at 7.5% per annum simple interest. If he gets ₹8,325 as interest at the time of maturity, find:

(i) The monthly deposit,

(ii) The maturity value.

Answer

(i) Given,

n = 3 year = 36 months

r = 7.5%

I = ₹8,325

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

$\therefore 8325 = P\times \dfrac{36\times 37}{2 \times 12} \times \dfrac{7.5}{100} \\[1em] 8325 = P \times \dfrac{1332}{24} \times 0.075 \\[1em] 8325 = P \times 4.1625 \\[1em] P =\dfrac{8325}{4.1625}\\[1em] P = ₹2,000$

Hence, monthly deposited = ₹ 2,000. Sum deposited = ₹2,000 x 36 = ₹72,000

(ii) Maturity value = Sum deposited + Interest = ₹72,000 + ₹8,325 = ₹80,325.

Hence, (i) Mr.Richard deposited ₹2,000 monthly (ii) Mr.Richard got ₹80,325 at the time of maturity.

Katrina opened a recurring deposit account with a Nationalised Bank for a period of 2 years, If the bank pays interest at 6% per annum and the monthly installment is ₹1,000 find:

(i) interest earned in 2 years,

(ii) matured value .

Answer

(i) Given,

P = ₹1,000

n = 2 years = 24 months

r = 6%

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

$\therefore I = 1000 \times \dfrac{24 \times 25}{2 \times 12} \times \dfrac{6}{100} \\[1em] I = 1000\times \dfrac{600}{24} \times 0.06 \\[1em] I = 1000 \times 25 \times 0.06 \\[1em] I = ₹1,500$

Hence, interest earned in 2 years = ₹1,500.

(ii) Sum deposited = ₹1,000 x 24 = ₹24,000

Maturity value = Sum deposited + Interest = ₹24,000 + ₹1,500 = ₹25,500.

Hence,(i)Interest earned by Katrina ₹1,500.(ii) Katrina got ₹25,500 at the time of maturity.

Ahmed has a recurring deposit account in a bank. He deposits ₹2,500 per month for 2 years. If he gets ₹66,250 at the time of maturity, find:

(i) the interest paid by the bank

(ii) the rate of interest.

Answer

Given,

P = ₹2,500

n = 2 years = 24 months

Maturity value = ₹66,250

Sum deposited = ₹2,500 x 24 = ₹60,000

Maturity value = Sum deposited + Interest

Interest = Maturity value - Sum deposited

∴ I = ₹66,250 - ₹60,000

I = ₹6,250

Let rate of interest be r %

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

$\therefore I = 2500 \times \dfrac{24 \times 25}{2 \times 12} \times r \\[1em] 6250 = 2500\times \dfrac{600}{24} \times\dfrac{r} {100} \\[1em] 6250\times 100= 2500 \times 25 \times r\\[1em] 625000 = 62500\times r\\[1em] r=\dfrac{625000}{62500}\\[1em] r=10\%$

Hence,(i) Interest earned by Ahmed ₹6,250 (ii) Rate of interest is 10% .

Mr. Gupta opened a recurring deposit account in a bank. He deposited ₹2,500 per month for 2 years. At the time of maturity he got ₹67,500. Find:

(i) the total interest earned by Mr. Gupta

(ii) the rate of interest per annum

Answer

Given,

P = ₹2,500

n = 2 years = 24 months

Maturity Value = ₹67,500

Sum deposited = ₹2,500 × 24 = ₹60,000

Maturity value = Sum deposited + Interest

Interest = Maturity value - Sum deposited

∴ I = ₹67,500 − ₹60,000 = ₹7,500

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

$\therefore I = 2500 \times \dfrac{24 \times 25}{2 \times 12} \times \dfrac{r}{100} \\[1em] 7500 = 62500 \times \dfrac{r}{100} \\[1em] r = \dfrac{7500 \times 100}{62500} \\[1em] r=12$

Hence, (i)Mr. Gupta earned ₹7,500 as interest.(ii)The rate of interest was 12% per annum.

Mr. Thomas has a 4 years cumulative time deposit account in Corporation Bank and deposits ₹650 per month. If he receives ₹36,296 at the time of maturity, find:

(i) the total interest earned by Mr. Thomas.

(ii) the rate of interest per annum.

Answer

Given,

P = ₹650

n = 4 years = 4 x 12 months = 48 months

Maturity Value = ₹36,296

Sum deposited = ₹650 × 48 = ₹31,200

Maturity value = Sum deposited + Interest

Interest = Maturity value - Sum deposited = ₹36,296 − ₹31,200 = ₹5,096

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

$\therefore I = 650 \times \dfrac{48 \times 49}{2 \times 12} \times \dfrac{r}{100} \\[1em] 5096 = 63700 \times \dfrac{r}{100} \\[1em] r = \dfrac{5096 \times 100}{63700} \\[1em] r =8$

(i)Mr. Thomas earned ₹5,096 as interest.

(ii) The rate of interest was approximately 8% per annum.

Tanvy has a recurring deposit account in a finance company for 1½ years at 9% per annum. If she gets ₹15,426 at the time of maturity, how much per month has been invested by her?

Answer

Given,

T = 1½ years = 18 months

r = 9%

Maturity Value = ₹15,426

Let monthly deposit be P

Sum deposited = P × 18 = 18P

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

$\therefore I = P \times \dfrac{18 \times 19}{2 \times 12} \times \dfrac{9}{100}\\[1em] = P \times \dfrac{342}{24} \times \dfrac{9}{100} \\[1em] = P \times \dfrac{57}{4} \times \dfrac{9}{100} \\[1em] = P \times \dfrac{513}{400}\\[1em] \text{Maturity Value} = 18P + \dfrac{513P}{400} \\[1em] = \dfrac{7200P + 513P}{400} \\[1em] = \dfrac{7713P}{400}\\[1em] \therefore 15426 = \dfrac{7713P}{400} \\[1em] P = \dfrac{15426 \times 400}{7713} = ₹800$

Hence, Tanvy deposited ₹800 per month.

Punam opened a recurring deposit account with Bank of Baroda for 1½ years. If the rate of interest is 6% per annum and the bank pays ₹11,313 on maturity, find how much Punam deposited each month?

Answer

Given,

n = 1½ years = 18 months

r = 6%

Maturity Value = ₹11,313

Let monthly deposit be P

Sum deposited = P × 18 = 18P

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

$\therefore I = P \times \dfrac{18 \times 19}{2 \times 12} \times \dfrac{6}{100}\\[1em] I= P \times \dfrac{342}{24} \times \dfrac{6}{100} \\[1em] I= P \times \dfrac{57}{4} \times \dfrac{6}{100} \\[1em] I= P \times \dfrac{342}{400}\\[1em] I = \dfrac{171}{200}P\\[1em]$

Maturity Value = Sum deposited + Interest

$Maturity Value = 18P + \dfrac{171P}{200} \\[1em] = \dfrac{3600P + 171P}{200} \\[1em] = \dfrac{3771P}{200}\\[1em] \therefore 11313 = \dfrac{3771P}{200} \\[1em] P = \dfrac{11313 \times 200}{3771} = ₹600$

Hence, Punam deposited ₹600 per month.

Kavita has a cumulative time deposit account in a bank. She deposits ₹600 per month and gets ₹6,165 at the time of maturity. If the rate of interest be 6% per annum, find the total time for which the account was held. (Hint: x² + 411x − 10x − 4110 = 0)

Answer

Given,

P = ₹600

Maturity Value = ₹6,165

r = 6% per annum

Let the number of months be 'x'.

Sum deposited = P × x = 600x

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

$I = 600 \times \dfrac{x(x+1)}{24} \times \dfrac{6}{100}\\[1em] I = 600 \times \dfrac{6x(x+1)}{2400} \\[1em] = \dfrac{3600x(x+1)}{2400} \\[1em] = \dfrac{3}{2} x(x+1)$

Maturity value = Sum deposited + Interest

$\Rightarrow 600x + \dfrac{3x(x + 1)}{2} = 6165 \\[1em] \Rightarrow \dfrac{1200x + 3x^2 + 3x}{2} = 6165 \\[1em] \Rightarrow 3x^2 + 1203x = 12330 \\[1em] \Rightarrow 3x^2 + 1203x - 12330 = 0 \\[1em] \Rightarrow 3(x^2 + 401x - 4110) = 0 \\[1em] \Rightarrow x^2 + 401x - 4110 = 0 \\[1em] \Rightarrow x^2 + 411x - 10x - 4110 = 0 \\[1em] \Rightarrow x(x + 411) - 10(x + 4110) = 0 \\[1em] \Rightarrow (x - 10)(x + 411) = 0 \\[1em] \Rightarrow x = 10 \text{ or } x = -411.$

Since the number of months cannot be negative

∴ x = 10 months.

Hence,total time for which the account was held = 10 months.

Kavita has a cumulative time deposit account in a bank. She deposits ₹800 per month and gets ₹16,700 as maturity value. If the rate of interest be 5% per annum, find the total time for which the account was held. (Hint: x² + 481x − 10020 = 0 ⇒ x² + 501x − 20x − 10020 = 0)

Answer

Given,

P = ₹800

Maturity Value = ₹16,700

r = 5%

Let the number of months be 'x'.

Sum deposited = P × x = 800x

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

$I = 800 \times \dfrac{x(x+1)}{24} \times \dfrac{5}{100}\\[1em] I = 800 \times \dfrac{5x(x+1)}{2400} \\[1em] = \dfrac{4000x(x+1)}{2400} \\[1em] = \dfrac{5}{3} x(x+1)$

Maturity Value = Sum deposited + Interest

$\Rightarrow 16700 = 800x + \dfrac{5}{3} x(x+1) \\[1em] \Rightarrow 50100 = 2400x + 5x(x+1) \\[1em] \Rightarrow 50100 = 2400x + 5x^2 + 5x \\[1em] \Rightarrow 5x^2 + 2405x - 50100 = 0 \\[1em] \Rightarrow 5(x^2 + 481x - 10020) = 0 \\[1em] \Rightarrow x^2 + 481x - 10020 = 0 \\[1em] \Rightarrow x^2 + 501x - 20x - 10020 = 0 \\[1em] \Rightarrow x(x + 501) - 20(x + 501) = 0 \\[1em] \Rightarrow (x - 20)(x + 501) = 0 \\[1em] \Rightarrow x = 20 \text{ or } x = -501$

Since the number of months cannot be negative

∴ x = 20 months

Hence, total time for which the account was held is 20 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%.

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.

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

Given,

n = 20 months

r = 9%

I = ₹441

Let the amount Rekha deposited each month be 'P'.

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

$\therefore 441 = P \times \dfrac{20 \times 21}{24} \times \dfrac{9}{100}\\[1em] 441 = P \times \dfrac{420}{24} \times \dfrac{9}{100}\\[1em] 441 = P \times 17.5 \times 0.09\\[1em] 441 = 1.575P\\[1em] P = \dfrac{441}{1.575}\\[1em] P=₹280$

Hence, Rekha deposited ₹280 each month.

Mr. Sonu has a 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

Given,

P = ₹750

n = 2 years = 24 months

Maturity Value = ₹19,125

Let the rate of interest be 'r' per annum

Sum deposited = P × n = 750 × 24 = ₹18,000

Maturity Value = Sum deposited + Interest

Interest = Maturity Value - Sum deposited

∴ I = ₹19,125 - ₹18,000 = ₹1,125

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

$1125 = 750 \times \dfrac{24 \times 25}{24} \times \dfrac{r}{100}\\[1em] 1125 = 750 \times 25 \times \dfrac{r}{100}\\[1em] 1125 = 18750 \times \dfrac{r}{100}\\[1em] 1125 = 187.5r\\[1em] r = \dfrac{1125}{187.5}\\[1em] r=6$

Hence, the rate of interest is 6% per annum.

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

(i) the total interest Salman earns.

(ii) the rate of interest.

Answer

Given,

P = ₹1,000

n = 2 years = 24 months

Maturity Value = ₹26,000

(i) The total interest Salman earns.

Sum deposited = P × n = 1,000 × 24 = ₹24,000

Maturity Value = Sum deposited + Interest

Interest = Maturity Value - Sum deposited

∴ I = ₹26,000 - ₹24,000 = ₹2,000

The total interest Salman earns is ₹2,000.

(ii) The rate of interest

I = ₹2,000

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

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

Hence, (i) The total interest Salman earns is ₹2,000.(ii) The rate of interest is 8% per annum.

Mrs. Rao deposited ₹ 250 per month in a recurring deposit account for a period of 3 years. She received ₹ 10,110 at the time of maturity. Find:

(i) the rate of interest.

(ii) how much more interest Mrs. Rao will receive if she had deposited ₹50 more per month at the same rate of interest and for the same time.

Answer

(i) Given,

Mrs. Rao deposited ₹ 250 per month in a recurring deposit account for a period of 3 years.

Total deposit = ₹ 250 × 3 × 12 = ₹ 9,000.

By formula,

Interest = Maturity value - Total deposit = ₹ 10,110 - ₹ 9,000 = ₹ 1,110.

Let rate of interest be r%.

Time (n) = 36 months

By formula,

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

Substituting values we get :

$\Rightarrow 1110 = \dfrac{250 \times 36 \times (36 + 1)}{2 \times 12} \times \dfrac{r}{100} \\[1em] \Rightarrow 1110 = \dfrac{9000 \times 37}{24} \times \dfrac{r}{100} \\[1em] \Rightarrow 1110 = \dfrac{333000}{24} \times \dfrac{r}{100} \\[1em] \Rightarrow r = \dfrac{24 \times 1110 \times 100}{333000} \\[1em] \Rightarrow r = \dfrac{2664000}{333000} \\[1em] \Rightarrow r = 8$

Hence, rate of interest = 8%.

(ii) If per month ₹ 50 more is deposited, then :

P = ₹ 250 + ₹ 50 = ₹ 300.

P = ₹ 300, r = 8%, n = 36 months

By formula,

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

Substituting values we get :

$\Rightarrow \text{Interest} = \dfrac{300 \times 36 \times (36 + 1)}{2 \times 12} \times \dfrac{8}{100} \\[1em] \Rightarrow \text{Interest} = \dfrac{10800 \times 37}{24} \times \dfrac{8}{100} \\[1em] \Rightarrow \text{Interest} = \dfrac{399600}{24} \times \dfrac{8}{100} \\[1em] \Rightarrow \text{Interest} = \dfrac{399600 \times 8}{24 \times 100} \\[1em] \Rightarrow \text{Interest} = \dfrac{3996}{3} \\[1em] \Rightarrow \text{Interest} = ₹1,332$

Additional Interest = New Interest - Old Interest

= ₹1,332 - ₹1,110

= ₹222.

Hence, Mrs. Rao would receive ₹222 more as interest if she had deposited ₹50 more per month at the same rate of interest and for the same time.

Mr. Anil has a recurring deposit account. He deposits a certain amount of money per month for 2 years. If he received an interest whose value is the double of the deposit made per month, then find the rate of interest.

Answer

Let deposit per month be P.

Given,

Time = 2 years = 24 months

Interest = 2 × Principle per month

By formula,

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

Substituting values we get :

$\Rightarrow 2P = \dfrac{P \times 24(24 + 1)}{24} \times \dfrac{R}{100} \\[1em] \Rightarrow 2P = \dfrac{P \times 25 \times R}{100} \\[1em] \Rightarrow R = \dfrac{2P \times 100}{P \times 25} \\[1em] \Rightarrow R = \dfrac{200}{25} \\[1em] \Rightarrow R = 8$

Hence, the rate of interest received by Mr. Anil = 8% .