Banking

Given,

P = ₹1,000

n = 3 years = 3 x 12 = 36 months

r = 8%

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

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

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

Given,

P = ₹360

n = 2 years = 2 x 12 = 24 months

r = 7%

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

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

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.

Given,

P = ₹600

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

r = 10%

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

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

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.

Given,

P = ₹600

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

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

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

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

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.

(i) Given,

n = 3 year = 36 months

r = 7.5%

I = ₹8,325

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

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

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.

(i) Given,

P = ₹1,000

n = 2 years = 24 months

r = 6%

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

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

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.

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

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

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

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

$$\begin{aligned}\therefore I = 2500 \times \frac{24 \times 25}{2 \times 12} \times \frac{r}{100} \\[1em] 7500 = 62500 \times \frac{r}{100} \\[1em] r = \frac{7500 \times 100}{62500} \\[1em] r=12%\end{aligned}$$

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

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

$$\begin{aligned}\therefore I = 650 \times \frac{48 \times 49}{2 \times 12} \times \frac{r}{100} \\[1em] 5096 = 63700 \times \frac{r}{100} \\[1em] r = \frac{5096 \times 100}{63700} \\[1em] r =8%\end{aligned}$$

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

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

Given,

T = 1½ years = 18 months

r = 9%

Maturity Value = ₹15,426

Let monthly deposit be P

Sum deposited = P × 18 = 18P

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

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

Hence, Tanvy deposited ₹800 per month.

Given,

n = 1½ years = 18 months

r = 6%

Maturity Value = ₹11,313

Let monthly deposit be P

Sum deposited = P × 18 = 18P

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

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

Maturity Value = Sum deposited + Interest

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

Hence, Punam deposited ₹600 per month.

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

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

Maturity value = Sum deposited + Interest

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

Since the number of months cannot be negative

∴ x = 10 months.

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

Given,

P = ₹800

Maturity Value = ₹16,700

r = 5%

Let the number of months be 'x'.

Sum deposited = P × x = 800x

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

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

Maturity Value = Sum deposited + Interest

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

Since the number of months cannot be negative

∴ x = 20 months

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

Let rate of interest be r%.

Given,

P = ₹ 600/month

n = 2 years or 24 months

M.V. = ₹ 15600

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow 15600 = 600 \times 24 + 600 \times \frac{24 \times (24 + 1)}{2 \times 12} \times \frac{r}{100} \\[1em] \Rightarrow 15600 = 14400 + 6 \times 25 \times r \\[1em] \Rightarrow 15600 - 14400 = 6 \times 25 \times r \\[1em] \Rightarrow 150r = 1200 \\[1em] \Rightarrow r = \frac{1200}{150} = 8%.\end{aligned}$$

Hence, rate of interest = 8%.

Let time be n months.

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow 1040 = 2000 \times \frac{n(n + 1)}{24} \times \frac{8}{100} \\[1em] \Rightarrow 1040 = 2000 \times \frac{n(n + 1)}{300} \\[1em] \Rightarrow n(n + 1) = \frac{1040 \times 300}{2000} \\[1em] \Rightarrow n(n + 1) = 156 \\[1em] \Rightarrow n^2 + n - 156 = 0 \\[1em] \Rightarrow n^2 + 13n - 12n - 156 = 0 \\[1em] \Rightarrow n(n + 13) - 12(n + 13) = 0 \\[1em] \Rightarrow (n - 12)(n + 13) = 0 \\[1em] \Rightarrow n - 12 = 0 \text{ or } n + 13 = 0 \\[1em] \Rightarrow n = 12 \text{ or } n = -13.\end{aligned}$$

Since, no. of months cannot be negative.

∴ n = 12.

Hence, total time for which the account was held = 12 months.

Given,

n = 20 months

r = 9%

I = ₹441

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

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

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

Hence, Rekha deposited ₹280 each month.

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

$$\begin{aligned}1125 = 750 \times \frac{24 \times 25}{24} \times \frac{r}{100}\\[1em] 1125 = 750 \times 25 \times \frac{r}{100}\\[1em] 1125 = 18750 \times \frac{r}{100}\\[1em] 1125 = 187.5r\\[1em] r = \frac{1125}{187.5}\\[1em] r=6%\end{aligned}$$

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

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

$$\begin{aligned}\therefore 2000 = 1000 \times \frac{24 \times 25}{24} \times \frac{r}{100}\\[1em] 2000 = 1000 \times 25 \times \frac{r}{100}\\[1em] 2000 = 250r\\[1em] r = \frac{2000}{250}\\[1em] r=8%\end{aligned}$$

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

(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,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow 1110 = \frac{250 \times 36 \times (36 + 1)}{2 \times 12} \times \frac{r}{100} \\[1em] \Rightarrow 1110 = \frac{9000 \times 37}{24} \times \frac{r}{100} \\[1em] \Rightarrow 1110 = \frac{333000}{24} \times \frac{r}{100} \\[1em] \Rightarrow r = \frac{24 \times 1110 \times 100}{333000} \\[1em] \Rightarrow r = \frac{2664000}{333000} \\[1em] \Rightarrow r = 8%.\end{aligned}$$

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,

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

Substituting values we get :

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

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.

Let deposit per month be P.

Given,

Time = 2 years = 24 months

Interest = 2 × Principle per month

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow 2P = \frac{P \times 24(24 + 1)}{24} \times \frac{R}{100} \\[1em] \Rightarrow 2P = \frac{P \times 25 \times R}{100} \\[1em] \Rightarrow R = \frac{2P \times 100}{P \times 25} \\[1em] \Rightarrow R = \frac{200}{25} \\[1em] \Rightarrow R = 8%.\end{aligned}$$

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

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

Hence, Option 2 is the correct option.

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.

Maturity value = Sum deposited + Interest

Sum deposited = P × n

Maturity value = P × n + Interest

Hence, Option 1 is the correct option.

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 × $\frac{n(n + 1)}{2}$

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 = $\frac{1}{12}$ years

Final formula,

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

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

Hence, Option 4 is the correct option.

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

Hence, Option 2 is the correct option.

Given:

Monthly deposit = P

Rate = R

Time = n

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

Hence, Option 3 is the correct option.

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.

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.

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.

In first case :

P = ₹ 1000

r = 5%

n = 12 months

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

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

In second case :

P = ₹ x (Let)

r = 4%

n = 12 months

Interest = ₹ 325

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

Hence, Option 2 is the correct option.

Given,

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

Amount deposited = ₹ 11,700

Amount deposited per month

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

$\frac{11700}{18}$

= ₹ 650.

Hence, Option 1 is the correct option.

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 × $\frac{18(18 + 1)}{2}$

= ₹ 400 × 9 × 19

= ₹ 68,400.

Hence, Option 3 is the correct option.

According to Assertion:

Given,

P = ₹1,600

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

r = 10%

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

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

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.

According to Assertion:

The maturity amount is ₹25,000.

Given,

P = ₹1,000

n = 2 years = 24 months

r = 6%

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

Substituting values we get :

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

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

So, Reason (R) is true.

Hence, Option 4 is the correct option.

Given,

Time (n) = 2 years or 24 months

Rate = r% (let)

P = ₹ x/month

I = ₹ x

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow x = \frac{x \times 24 \times (24 + 1)}{2 \times 12} \times \frac{r}{100} \\[1em] \Rightarrow x = \frac{x \times 24 \times 25}{24} \times \frac{r}{100} \\[1em] \Rightarrow 1 = \frac{r}{4} \\[1em] \Rightarrow r = 4%.\end{aligned}$$

Hence, rate of interest = 4%.

(i) Let money be deposited for n months.

By formula,

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

Substituting values we get :

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

Since, no. of months cannot be negative.

∴ n = 18.

Hence, number of monthly installments = 18.

(ii) By formula,

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

Substituting values we get :

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

Hence, total interest paid = ₹ 1026.

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

= ₹ 600 × 18

= ₹ 10800.

Hence, total amount deposited by Amit = ₹ 10800.

1.Given,

I = 2000

R = 8%

n = 2 years = 24 months

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

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%

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

Since the number of months cannot be negative.

∴ n = 12 months

Hence, Option (c) is the correct option.