Banking

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% .

A recurring deposit is also known as:

1. maturity deposit

2. cumulative time deposit

3. regular saving deposit

4. investment fund deposit

Answer

In recurring deposit, the deposits and interest accumulate over a fixed time period

Hence, Option 2 is the correct option.

In a recurring deposit (R.D.):

1. a person gets the same interest every month

2. a person gets the same maturity amount every year

3. a person deposits the same amount every month

4. the government deposits an amount equal to the interest every year.

Answer

Recurring deposit (RD) is a type of savings account where you deposit a fixed amount of money regularly.

Hence, Option 3 is the correct option.

In a recurring deposit, the maturity value is given by:

1. $(P \times n) + I$

2. $P \times n \times I$

3. $\dfrac{P \times n \times I}{100}$

4. $\dfrac{(P \times n) + I}{100}$

Answer

Maturity value = Sum deposited + Interest

Sum deposited = P × n

Maturity value = P × n + Interest

Hence, Option 1 is the correct option.

₹ P is deposited for n number of months in a recurring deposit account which pays interest at the rate of r% per annum. The nature and time of interest calculated is :

1. compound interest for n number of months

2. simple interest for n number of months

3. compound interest for one month

4. simple interest for one month

Answer

In a Recurring Deposit (RD), the interest is calculated using the concept of Equivalent Monthly Principal.

The first installment stays in the bank for $n$ months, the second for n - 1 months, and the last for 1 month. To simplify this, we use the sum of natural numbers formula to find the total "month-units" of interest:

Total monthly principal = P × $\dfrac{n(n + 1)}{2}$

Because we have converted the entire duration into an equivalent principal for just one month, the time (T) used in the standard S.I. formula is :

T = $\dfrac{1}{12}$ years

Final formula,

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

The interest is simple in nature, and it is calculated on the equivalent principal for one month.

Hence, Option 4 is the correct option.

If Ramesh Kumar has an R.D. in a post office, he has to deposit:

1. an amount only once

2. the same amount every month

3. a decreasing amount every month

4. an increasing amount every month

Answer

In a recurring deposit a person deposits the same amount every month.

Hence, Option 2 is the correct option.

In an R.D., the maturity value is the sum of the total amount deposited and the interest. If P is the amount deposited every month for n months and R is the rate of interest, then interest I is equal to:

1. $P \times \dfrac{n}{12} \times \dfrac{R}{100}$

2. $P \times \dfrac{n(n - 1)}{12} \times \dfrac{R}{100}$

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

4. $P \times \dfrac{n}{2 \times 12} \times \dfrac{R}{100}$

Answer

Given:

Monthly deposit = P

Rate = R

Time = n

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

Hence, Option 3 is the correct option.

Mohit opened a Recurring deposit account in a bank for 2 years. He deposits ₹1,000 every month and receives ₹25,500 on maturity. The interest he earned in 2 years is:

1. ₹13,500

2. ₹3,000

3. ₹24,000

4. ₹1500

Answer

Given:

P = ₹1000

n = 24 months

Maturity value = ₹25,500

Sum deposited = P × n =1000 × 24 = ₹24,000

Maturity Value = Sum deposited + Interest

Interest = Maturity Value - Sum deposited

∴ I = 25,500 - 24,000 = ₹1,500

Hence, Option 4 is the correct option.

Naveen deposits ₹800 every month in a recurring deposit account for 6 months. If he receives ₹4,884 at the time of maturity, then the interest he earns is:

1. ₹84

2. ₹42

3. ₹24

4. ₹284

Answer

Given:

P = ₹800

n = 6 months

Maturity Amount= ₹4,884

Sum deposited = P × n = 800 × 6 = ₹4,800

Maturity Value = Sum deposited + Interest

Interest = Maturity Value - Sum deposited

∴ I = ₹(4,884 - 4,800) = ₹84

Hence, Option 1 is the correct option.

Mr. Anuj deposits ₹ 500 per month for 18 months in a recurring deposit account at a certain rate. If he earns ₹570 as interest at the time of maturity, then his matured amount is:

1. ₹(500 x 18 + 570)

2. ₹(500 x 19 + 570)

3. ₹(500 x 18 x 19 + 570)

4. ₹(500 x 9 x 19 + 570)

Answer

Given,

Monthly deposit = ₹500

Number of months = 18

Interest earned = ₹570

By formula,

Matured amount = Total deposit + Interest

= Monthly deposit x number of months + Interest

= ₹(500 x 18 + 570)

Hence, option 1 is the correct option.

Anwesha intended to open a Recurring Deposit account of ₹ 1000 per month for 1 year in a Bank, paying a 5% per annum rate of simple interest. The bank reduced the rate to 4% per annum. How much must Anwesha deposit monthly for 1 year so that her interest remains the same?

1. ₹ 12325

2. ₹ 1250

3. ₹ 1200

4. ₹ 1000

Answer

In first case :

P = ₹ 1000

r = 5%

n = 12 months

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

$= 1000 \times \dfrac{12 \times (12 + 1)}{2 \times 12} \times \dfrac{5}{100} \\[1em] = 1000 \times \dfrac{12 \times 13}{2 \times 12} \times \dfrac{1}{20} \\[1em] = 25 \times 13 \\[1em] = ₹ 325.$

In second case :

P = ₹ x (Let)

r = 4%

n = 12 months

Interest = ₹ 325

$\therefore 325 = x \times \dfrac{12 \times 13}{2 \times 12} \times \dfrac{4}{100} \\[1em] \Rightarrow 325 = x \times \dfrac{13}{2} \times \dfrac{1}{25}\\[1em] \Rightarrow x = \dfrac{325 \times 25 \times 2}{13} \\[1em] \Rightarrow x = \dfrac{16250}{13} = \text{₹ 1250.}$

Hence, Option 2 is the correct option.

Rahul deposited ₹ 11,700 in a recurring deposit account for $1\dfrac{1}{2}$ years. The amount deposited by him per month is :

1. ₹ 650

2. ₹ 780

3. ₹ 6,500

4. ₹ 7,800

Answer

Given,

Time = $1\dfrac{1}{2}$ years = 18 months

Amount deposited = ₹ 11,700

Amount deposited per month

= $\dfrac{\text{Total amount deposited}}{\text{Total number of months}}$

$\dfrac{11700}{18}$

= ₹ 650.

Hence, Option 1 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.

Qualifying sum = ₹ 400 × (18 + 17 + 16 + ……..+ 1)

= ₹400 × $\dfrac{18(18 + 1)}{2}$

= ₹ 400 × 9 × 19

= ₹ 68,400.

Hence, Option 3 is the correct option.

Assertion (A) : Sunidhi deposits ₹1,600 per month in a bank for $1\dfrac{1}{2}$ years in a recurring deposit account at 10% p.a. She gets ₹31,080 on maturity.

Reason (R): Maturity value is given by MV = (P x n) - S.I.

1. Both A and R are true, and R is the correct explanation of A.

2. Both A and R are true, but R is not the correct explanation of A.

3. A is true, but R is false.

4. A is false, but R is true.

Answer

According to Assertion:

Given,

P = ₹1,600

n = $1\dfrac{1}{2}$ years = 18 months

r = 10%

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

$\therefore I = 1600\times \dfrac{18\times 19}{2 \times 12} \times \dfrac{10}{100} \\[1em] I = 1600 \times \dfrac{342}{24} \times 0.1 \\[1em] I = 1600 \times 14.25 \times 0.1 \\[1em] I = ₹2,280$

Sum deposited = ₹1,600 x 18 = ₹28,800

Maturity value = Sum deposited + Interest = ₹28,800 + ₹2,280 = ₹31,080

So, Assertion (A) is true.

According to Reason:

Maturity value is given by MV = (P x n) - S.I.

But,

Maturity value = Sum deposited + Interest

Sum deposited = P × n

Maturity value = (P × n) + Interest

So, Reason (R) is false.

Hence, Option 3 is the correct option.

Assertion (A): Pawandeep opened a recurring deposit account in a bank for a period of 2 years. If the bank pays interest at the rate of 6% p.a. and the monthly instalment is ₹1,000, then the maturity amount is ₹25,000.

Reason (R): For a recurring deposit account, we compute the interest using the following formula:

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

1. Both A and R are true, and R is the correct explanation of A.

2. Both A and R are true, but R is not the correct explanation of A.

3. A is true, but R is false.

4. A is false, but R is true.

Answer

According to Assertion:

The maturity amount is ₹25,000.

Given,

P = ₹1,000

n = 2 years = 24 months

r = 6%

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

Substituting values we get :

$\therefore I = 1000\times \dfrac{24\times 25}{2 \times 12} \times \dfrac{6}{100} \\[1em] I = 1000 \times \dfrac{600}{24} \times 0.06\\[1em] I = 1000 \times 25 \times 0.06 \\[1em] I = ₹1500$

Sum deposited = ₹1,000 x 24 = ₹24,000

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

The given Maturity amount = ₹25,500

So, Assertion(A) is false.

For a recurring deposit account, we compute the interest using the following formula:

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

So, Reason (R) is true.

Hence, Option 4 is the correct option.

A man opened a recurring deposit account in a branch of PNB. The man deposits certain amount of money per month such that after 2 years, the interest accumulated is equal to his monthly deposits. Find the rate of interest per annum that the bank was paying for the recurring deposit account.

Answer

Given,

Time (n) = 2 years or 24 months

Rate = r% (let)

P = ₹ x/month

I = ₹ x

By formula,

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

Substituting values we get :

$\Rightarrow x = \dfrac{x \times 24 \times (24 + 1)}{2 \times 12} \times \dfrac{r}{100} \\[1em] \Rightarrow x = \dfrac{x \times 24 \times 25}{24} \times \dfrac{r}{100} \\[1em] \Rightarrow 1 = \dfrac{r}{4} \\[1em] \Rightarrow r = 4$

Hence, rate of interest = 4%.

Amit deposited ₹ 600 per month in a recurring deposit account. The bank pays a simple interest of 12% p.a. Calculate the:

(i) number of monthly installments Amit deposits to get a maturity amount of ₹ 11826?

(ii) total interest paid by the bank.

(iii) total amount deposited by him.

Answer

(i) Let money be deposited for n months.

By formula,

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

Substituting values we get :

$\Rightarrow 11826 = 600 \times n + \dfrac{600 \times n(n + 1)}{2 \times 12} \times \dfrac{12}{100} \\[1em] \Rightarrow 11826 = 600n + \dfrac{600(n^2 + n)}{200} \\[1em] \Rightarrow 11826 = 600n + 3(n^2 + n) \\[1em] \Rightarrow 11826 = 600n + 3n^2 + 3n \\[1em] \Rightarrow 3n^2 + 603n = 11826 \\[1em] \Rightarrow 3(n^2 + 201n) = 11826 \\[1em] \Rightarrow n^2 + 201n = \dfrac{11826}{3} \\[1em] \Rightarrow n^2 + 201n = 3942 \\[1em] \Rightarrow n^2 + 201n - 3942 = 0 \\[1em] \Rightarrow n^2 + 219n - 18n - 3942 = 0 \\[1em] \Rightarrow n(n + 219) - 18(n + 219) = 0 \\[1em] \Rightarrow (n - 18)(n + 219) = 0 \\[1em] \Rightarrow n - 18 = 0 \text{ or } n + 219 = 0 \\[1em] \Rightarrow n = 18 \text{ or } n = -219.$

Since, no. of months cannot be negative.

∴ n = 18.

Hence, number of monthly installments = 18.

(ii) By formula,

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

Substituting values we get :

$\text{Interest } = \dfrac{600 \times 18(18 + 1)}{2 \times 12} \times \dfrac{12}{100} \\[1em] = \dfrac{600 \times 18 \times 19}{200} \\[1em] = 3 \times 18 \times 19 \\[1em] = \text{₹ 1026}.$

Hence, total interest paid = ₹ 1026.

(c) Total amount deposited by Amit = P × n

= ₹ 600 × 18

= ₹ 10800.

Hence, total amount deposited by Amit = ₹ 10800.

Case study: Joseph has a recurring deposit account in a bank for two years at the rate of 8% per annum simple interest.

1. If at the time of maturity Joseph receives ₹2,000 as interest, then the monthly instalment is:
(a) ₹1,200
(b) ₹600
(c) ₹1,000
(d) ₹1,600

2.The total amount deposited in the bank is:
(a) ₹25,000
(b) ₹24,000
(c) ₹26,000
(d) ₹23,000

3.The amount Joseph receives on maturity is:
(a) ₹27,000
(b) ₹25,000
(c) ₹26,000
(d) ₹28,000

4. If the monthly instalment is ₹100 and the rate of interest is 8%, in how many months Joseph will receive ₹52 as interest?
(a) 18
(b) 30
(c) 12
(d) 6

Answer

1.Given,

I = 2000

R = 8%

n = 2 years = 24 months

$I = P \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{R}{100}\\[1em] \therefore I = P \times \dfrac{24 \times 25}{2 \times 12} \times \dfrac{8}{100}\\[1em] I = P \times \dfrac{600}{24} \times 0.08\\[1em] I = P \times 25 \times 0.08\\[1em] 2000 = P \times 2 \\[1em] P=\dfrac{2000}{2}\\[1em] P =₹ 1,000$

Hence, Option (c) is the correct option.

2. Given,

P = ₹1000

n = 24 months

Total deposit = P x n

Total deposit = 1000 x 24

Total deposit= ₹24,000

Hence, Option (b) is the correct option.

3. Given:

Total deposit = ₹24,000

Interest = ₹2,000

Maturity amount = Total Deposit + Interest

Maturity amount = 24000 + 2000

Maturity amount= ₹ 26,000

Hence, Option (c) is the correct option.

4. Given:

I = ₹52

P = ₹100

r = 8%

$I = P \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{R}{100} \\[1em] \therefore I = 100 \times \dfrac{n(n+1)}{2 \times 12} \times \dfrac{8}{100} \\[1em] 52 = \dfrac{n(n+1)}{24}\times 8 \\[1em] 52 = \dfrac{n(n+1)}{3} \\[1em] n(n+1) =52 \times 3 \\[1em] n(n+1) = 156 \\[1em] n^2+n-156 = 0 \\[1em] n^2+13n - 12n-156 = 0 \\[1em] n(n+13) - 12(n+13) = 0 \\[1em] (n+13)(n-12) = 0 \\[1em] n=-13 \text{ or } n=12$

Since the number of months cannot be negative.

∴ n = 12 months

Hence, Option (c) is the correct option.