Compound Interest [Applications]

If P = sum lent, r = rate of interest per year, n = number of years and amounts =A, then :

$A = P\Big(1 + \frac{r}{100}\Big)^n$

Hence, Option 1 is the correct option.

If the letters used have usual meanings : r₁% and r₂% are rate of interests for two consecutive years then :

$A = P\Big(1 + \frac{r_1}{100}\Big)\Big(1 + \frac{r_2}{100}\Big)$

Hence, Option 2 is the correct option.

Given,

P = ₹ 6000

r = 5%

n = 2 years

By formula,

$$\begin{aligned}A = P\Big(1 + \frac{r}{100}\Big)^n \\[1em] = 6000\Big(1 + \frac{5}{100}\Big)^2 \\[1em] = 6000 \times \Big(\frac{105}{100}\Big)^2 \\[1em] = 6000 \times \frac{105}{100} \times \frac{105}{100} \\[1em] = 3 \times 21 \times 105 \\[1em] = ₹ 6615.\end{aligned}$$

C.I. = A - P = ₹ 6615 - ₹ 6000 = ₹ 615.

Hence, Option 1 is the correct option.

Given,

r = 10%

n = 2 years

Let sum of money lent be ₹ x.

P = ₹ x

A = ₹ 6050

By formula,

$A = P\Big(1 + \frac{r}{100}\Big)^n$

Substituting values we get :

$$\begin{aligned}\Rightarrow 6050 = x\Big(1 + \frac{10}{100}\Big)^2 \\[1em] \Rightarrow 6050 = x \times \Big(\frac{110}{100}\Big)^2 \\[1em] \Rightarrow 6050 = x \times \Big(\frac{11}{10}\Big)^2 \\[1em] \Rightarrow 6050 = \frac{121x}{100} \\[1em] \Rightarrow x = \frac{6050 \times 100}{121} \\[1em] \Rightarrow x = ₹ 5000.\end{aligned}$$

Hence, Option 3 is the correct option.

Given,

r = 10%

n = 1 year

C.I. = ₹ 550

Let sum of money lent out be ₹ x.

By formula,

C.I. = A - P

$$\begin{aligned}\Rightarrow C.I. = P\Big(1 + \frac{r}{100}\Big)^n - P \\[1em] \Rightarrow 550 = x(1 + \frac{10}{100})^1 - x \\[1em] \Rightarrow 550 = x + \frac{10x}{100} - x \\[1em] \Rightarrow 550 = \frac{x}{10} \\[1em] \Rightarrow x = 5500.\end{aligned}$$

Hence, Option 4 is the correct option.

Let rate of interest be r%.

Given,

P = ₹ 4000

A = ₹ 4600

n = 1 year

By formula,

$A = P(1 + \frac{r}{100})^n$

Substituting values we get :

$$\begin{aligned}\Rightarrow 4600 = 4000(1 + \frac{r}{100})^1 \\[1em] \Rightarrow \frac{4600}{4000} = 1 + \frac{r}{100} \\[1em] \Rightarrow \frac{23}{20} = 1 + \frac{r}{100} \\[1em] \Rightarrow \frac{23}{20} - 1 = \frac{r}{100} \\[1em] \Rightarrow \frac{23 - 20}{20} = \frac{r}{100} \\[1em] \Rightarrow r = \frac{3}{20} \times 100 \\[1em] \Rightarrow r = 15%.\end{aligned}$$

Hence, Option 1 is the correct option.

Let rate of interest per month be r%.

Given,

P = ₹ 4000

A = ₹ 5017.60

T = 2 months

By formula,

$A = P\Big(1 + \frac{r}{100}\Big)^n$

Substituting values we get :

$$\begin{aligned}\Rightarrow 5017.60 = 4000(1 + \frac{r}{100})^2 \\[1em] \Rightarrow \frac{5017.60}{4000} = \Big(1 + \frac{r}{100}\Big)^2 \\[1em] \Rightarrow \frac{501760}{400000} = \Big(1 + \frac{r}{100}\Big)^2 \\[1em] \Rightarrow \frac{50176}{40000} = \Big(1 + \frac{r}{100}\Big)^2 \\[1em] \Rightarrow \Big(\frac{224}{200}\Big)^2 = \Big(1 + \frac{r}{100}\Big)^2 \\[1em] \Rightarrow 1 + \frac{r}{100} = \frac{224}{200} \\[1em] \Rightarrow \frac{r}{100} = \frac{224}{200} - 1 \\[1em] \Rightarrow \frac{r}{100} = \frac{24}{200} \\[1em] \Rightarrow r =\frac{24}{200} \times 100 \\[1em] \Rightarrow r = 12%.\end{aligned}$$

Hence, Option 1 is the correct option.

Given,

P = ₹ 12000

n = 3 years

r = 5%

By formula,

$A = P\Big(1 + \frac{r}{100}\Big)^n$

Substituting values we get :

$$\begin{aligned}\Rightarrow A = 12000\Big(1 + \frac{5}{100}\Big)^3 \\[1em] \Rightarrow A = 12000\Big(1 + \frac{1}{20}\Big)^3 \\[1em] \Rightarrow A = 12000\Big(\frac{21}{20}\Big)^3 \\[1em] \Rightarrow A = 12000 \times \frac{9261}{8000} \\[1em] \Rightarrow A = ₹13891.50\end{aligned}$$

C.I. = A - P = ₹ 13891.50 - ₹ 12000 = ₹ 1891.50

Hence, amount = ₹ 13891.50 and compound interest = ₹ 1891.50

Given,

P = ₹ 15000

r₁ = 8%

r₂ = 10%

n = 2 years

By formula,

A = $P\Big(1 + \frac{r_1}{100}\Big)\Big(1 + \frac{r_2}{100}\Big)$

Substituting values we get :

$$\begin{aligned}A = 15000 \times \Big(1 + \frac{8}{100}\Big) \times \Big(1 + \frac{10}{100}\Big) \\[1em] = 15000 \times \frac{108}{100} \times \frac{110}{100} \\[1em] = 15000 \times \frac{27}{25} \times \frac{11}{10} \\[1em] = 60 \times 27 \times 11 \\[1em] = ₹17820.\end{aligned}$$

Hence, amount = ₹ 17820.

Given,

P = ₹ 6000

r₁ = 5%

r₂ = 8%

r₃ = 10%

n = 3 years

By formula,

A = $P\Big(1 + \frac{r_1}{100}\Big)\Big(1 + \frac{r_2}{100}\Big)\Big(1 + \frac{r_3}{100}\Big)$

Substituting values we get :

$$\begin{aligned}A = 6000 \times \Big(1 + \frac{5}{100}\Big) \times \Big(1 + \frac{8}{100}\Big) \times \Big(1 + \frac{10}{100}\Big) \\[1em] = 6000 \times \frac{105}{100} \times \frac{108}{100} \times \frac{110}{100} \\[1em] = 6000 \times \frac{21}{20} \times \frac{27}{25} \times \frac{11}{10} \\[1em] = \frac{30 \times 21 \times 27 \times 11}{25} \\[1em] = \frac{187110}{25} \\[1em] = ₹7484.40\end{aligned}$$

By formula,

C.I. = A - P = ₹ 7484.40 - ₹ 6000 = ₹ 1484.40

Hence, compound interest = ₹ 1484.40

Let sum of money be ₹ x.

Given,

P = ₹ x

r = 10%

n = 2 years

A = ₹ 5445

By formula,

A = $P\Big(1+ \frac{r}{100}\Big)^n$

Substituting values we get :

$$\begin{aligned}\Rightarrow 5445 = x \times \Big(1 + \frac{10}{100}\Big)^2 \\[1em] \Rightarrow 5445 = x \times \Big(\frac{110}{100}\Big)^2 \\[1em] \Rightarrow 5445 = x \times \Big(\frac{11}{10}\Big)^2 \\[1em] \Rightarrow 5445 = x \times \frac{121}{100} \\[1em] \Rightarrow x = \frac{5445 \times 100}{121} \\[1em] \Rightarrow x = ₹4500.\end{aligned}$$

Hence, sum of money = ₹ 4500.

Let sum of money be ₹ x.

Given,

n = 2 years

r = 5%

C.I. = ₹ 768.75

A = P + I = ₹ x + ₹ 768.75

By formula,

A = $P\Big(1 + \frac{r}{100}\Big)^n$

Substituting values we get :

$$\begin{aligned}\Rightarrow x + 768.75 = x \times \Big(1 + \frac{5}{100}\Big)^2 \\[1em] \Rightarrow x + 768.75 = x \times \Big(\frac{105}{100}\Big)^2 \\[1em] \Rightarrow x + 768.75 = x \times \Big(\frac{21}{20}\Big)^2 \\[1em] \Rightarrow x + 768.75 = x \times \frac{441}{400} \\[1em] \Rightarrow 400(x + 768.75) = 441x \\[1em] \Rightarrow 400x + 307500 = 441x \\[1em] \Rightarrow 441x - 400x = 307500 \\[1em] \Rightarrow 41x = 307500 \\[1em] \Rightarrow x = \frac{307500}{41} \\[1em] \Rightarrow x = ₹7500.\end{aligned}$$

Hence, sum of money = ₹ 7500.

Let sum of money be ₹ x.

Given,

n = 3 years

r = 10%

C.I. = ₹ 1655

A = P + I = ₹ x + ₹ 1655

By formula,

A = $P\Big(1 + \frac{r}{100}\Big)^n$

Substituting values we get :

$$\begin{aligned}\Rightarrow x + 1655 = x \times \Big(1 + \frac{10}{100}\Big)^3 \\[1em] \Rightarrow x + 1655 = x \times \Big(\frac{110}{100}\Big)^3 \\[1em] \Rightarrow x + 1655 = x \times \Big(\frac{11}{10}\Big)^3 \\[1em] \Rightarrow x + 1655 = x \times \frac{1331}{1000} \\[1em] \Rightarrow 1000(x + 1655) = 1331x \\[1em] \Rightarrow 1000x + 1655000 = 1331x \\[1em] \Rightarrow 1331x - 1000x = 1655000 \\[1em] \Rightarrow 331x = 1655000 \\[1em] \Rightarrow x = \frac{1655000}{331} \\[1em] \Rightarrow x = ₹5000.\end{aligned}$$

Hence, sum of money = ₹ 5000.

Given,

P = ₹ 6000

A = ₹ 6615

n = 2 years

Let rate of interest be r%.

By formula,

A = $P\Big(1 + \frac{r}{100}\Big)^n$

Substituting values we get :

$$\begin{aligned}\Rightarrow 6615 = 6000 \times \Big(1 + \frac{r}{100}\Big)^2 \\[1em] \Rightarrow \frac{6615}{6000} = \Big(1 + \frac{r}{100}\Big)^2 \\[1em] \Rightarrow \frac{441}{400} = \Big(1 + \frac{r}{100}\Big)^2 \\[1em] \Rightarrow \Big(\frac{21}{20}\Big)^2 = \Big(1 + \frac{r}{100}\Big)^2 \\[1em] \Rightarrow \frac{21}{20} = 1 + \frac{r}{100} \\[1em] \Rightarrow \frac{21}{20} - 1 = \frac{r}{100} \\[1em] \Rightarrow \frac{21 - 20}{20} = \frac{r}{100} \\[1em] \Rightarrow r = \frac{100}{20} = 5%.\end{aligned}$$

Hence, rate of interest = 5%.

Given,

A = ₹ 9856

r₁ = 10%

r₂ = 12%

n = 2 years

Let principal amount be ₹ x.

By formula,

A = $P\Big(1 + \frac{r_1}{100}\Big)\Big(1 + \frac{r_2}{100}\Big)$

Substituting values we get :

$$\begin{aligned}\Rightarrow 9856 = x\Big(1 + \frac{10}{100}\Big)\Big(1 + \frac{12}{100}\Big) \\[1em] \Rightarrow 9856 = x \times \frac{110}{100} \times \frac{112}{100} \\[1em] \Rightarrow 9856 = x \times \frac{11}{10} \times \frac{28}{25} \\[1em] \Rightarrow x = \frac{9856 \times 10 \times 25}{11 \times 28} \\[1em] \Rightarrow x = \frac{2464000}{308} = ₹8000.\end{aligned}$$

Hence, principal amount = ₹ 8000.

Given,

C.I. = ₹ 4240

r₁ = 10%

r₂ = 15%

n = 2 years

Let principal amount be ₹ x.

A = P + C.I. = ₹ x + ₹ 4240

By formula,

A = $P\Big(1 + \frac{r_1}{100}\Big)\Big(1 + \frac{r_2}{100}\Big)$

Substituting values we get :

$$\begin{aligned}\Rightarrow x + 4240 = x \times \Big(1 + \frac{10}{100}\Big)(1 + \frac{15}{100}) \\[1em] \Rightarrow x + 4240 = x \times \frac{110}{100} \times \frac{115}{100} \\[1em] \Rightarrow x + 4240 = x \times \frac{11}{10} \times \frac{23}{20} \\[1em] \Rightarrow x + 4240 = \frac{253x}{200} \\[1em] \Rightarrow 200(x + 4240) = 253x \\[1em] \Rightarrow 200x + 848000 = 253x \\[1em] \Rightarrow 253x - 200x = 848000 \\[1em] \Rightarrow 53x = 848000 \\[1em] \Rightarrow x = \frac{848000}{53} = ₹16000.\end{aligned}$$

Hence, principal amount = ₹ 16000.

Let sum of money be ₹ x and let rate of percent be r%.

A = ₹ 1.44x

n = 2 years

By formula,

A = $P\Big(1 + \frac{r}{100}\Big)^n$

Substituting values we get :

$$\begin{aligned}\Rightarrow 1.44x = x \times \Big(1 + \frac{r}{100}\Big)^2 \\[1em] \Rightarrow \frac{1.44x}{x} = \Big(1 + \frac{r}{100}\Big)^2 \\[1em] \Rightarrow 1.44 = \Big(1 + \frac{r}{100}\Big)^2 \\[1em] \Rightarrow (1.2)^2 = \Big(1 + \frac{r}{100}\Big)^2 \\[1em] \Rightarrow 1.2 = 1 + \frac{r}{100} \\[1em] \Rightarrow 1.2 - 1 = \frac{r}{100} \\[1em] \Rightarrow 0.2 = \frac{r}{100} \\[1em] \Rightarrow r = 0.2 \times 100 = 20%.\end{aligned}$$

Hence, rate of interest = 20%.

Let rate of interest be r%.

Given,

P = ₹ 4000

C.I. = ₹ 1324

A = P + C.I. = ₹ 4000 + ₹ 1324 = ₹ 5324

n = 3 years

By formula,

A = $P\Big(1 + \frac{r}{100}\Big)^n$

Substituting values we get :

$$\begin{aligned}\Rightarrow 5324 = 4000 \times \Big(1 + \frac{r}{100}\Big)^3 \\[1em] \Rightarrow \frac{5324}{4000} = \Big(1 + \frac{r}{100}\Big)^3 \\[1em] \Rightarrow \Big(1 + \frac{r}{100}\Big)^3 = \frac{1331}{1000} \\[1em] \Rightarrow \Big(1 + \frac{r}{100}\Big)^3 = \Big(\frac{11}{10}\Big)^3 \\[1em] \Rightarrow 1 + \frac{r}{100} = \frac{11}{10} \\[1em] \Rightarrow \frac{r}{100} = \frac{11}{10} - 1 \\[1em] \Rightarrow \frac{r}{100} = \frac{11 - 10}{10} \\[1em] \Rightarrow r = \frac{1}{10} \times 100 \\[1em] \Rightarrow r = 10%.\end{aligned}$$

Hence, rate of interest = 10%.

(i) For 2 years :

A = ₹ 6272

P = ₹ 5000

Let rate of interest be r%.

By formula,

A = $P\Big(1 + \frac{r}{100}\Big)^n$

Substituting values we get :

$$\begin{aligned}\Rightarrow 6272 = 5000 \times \Big(1 + \frac{r}{100}\Big)^2 \\[1em] \Rightarrow \frac{6272}{5000} = \Big(1 + \frac{r}{100}\Big)^2 \\[1em] \Rightarrow \frac{784}{625} = \Big(1 + \frac{r}{100}\Big)^2 \\[1em] \Rightarrow \Big(\frac{28}{25}\Big)^2 = \Big(1 + \frac{r}{100}\Big)^2 \\[1em] \Rightarrow \frac{28}{25} = 1 + \frac{r}{100} \\[1em] \Rightarrow \frac{r}{100} = \frac{28}{25} - 1 \\[1em] \Rightarrow \frac{r}{100} = \frac{28 - 25}{25} \\[1em] \Rightarrow r = \frac{3}{25} \times 100 \\[1em] \Rightarrow r = 12%.\end{aligned}$$

Hence, rate of interest per annum = 12%.

(ii) By formula,

A = $P\Big(1 + \frac{r}{100}\Big)^n$

Substituting values we get :

$$\begin{aligned}\Rightarrow A = 5000 \times \Big(1 + \frac{12}{100}\Big)^3 \\[1em] = 5000 \times \Big(\frac{112}{100}\Big)^3 \\[1em] = 5000 \times \Big(\frac{28}{25}\Big)^3 \\[1em] = 5000 \times \frac{21952}{15625} \\[1em] = \frac{109760000}{15625} \\[1em] = ₹7024.64\end{aligned}$$

Hence, amount in 3 years = ₹ 7024.64

Given,

P = ₹ 7000

A = ₹ 9317

r = 10%

Let in n years ₹ 7000 amount to ₹ 9317.

By formula,

A = $P\Big(1 + \frac{r}{100}\Big)^n$

Substituting values we get :

$$\begin{aligned}\Rightarrow 9317 = 7000 \times \Big(1 + \frac{10}{100}\Big)^n \\[1em] \Rightarrow \frac{9317}{7000} = \Big(\frac{110}{100}\Big)^n \\[1em] \Rightarrow \frac{1331}{1000} = \Big(\frac{11}{10}\Big)^n \\[1em] \Rightarrow \Big(\frac{11}{10}\Big)^3 = \Big(\frac{11}{10}\Big)^n \\[1em] \Rightarrow n = 3.\end{aligned}$$

Hence, in 3 years ₹ 7000 amounts to ₹ 9317 at 10 percent compound interest.

Given,

P = ₹ 4000

C.I. = ₹ 630.50

r = 5%

A = P + C.I. = ₹ 4000 + ₹ 630.50 = ₹ 4630.50

Let time taken be n years.

By formula,

A = $P\Big(1 + \frac{r}{100}\Big)^n$

Substituting values we get :

$$\begin{aligned}\Rightarrow 4630.50 = 4000 \times \Big(1 + \frac{5}{100}\Big)^n \\[1em] \Rightarrow \frac{4630.50}{4000} = \Big(\frac{105}{100}\Big)^n \\[1em] \Rightarrow \frac{9.261}{8} = \Big(\frac{2.1}{2}\Big)^n \\[1em] \Rightarrow \Big(\frac{2.1}{2}\Big)^3 = \Big(\frac{2.1}{2}\Big)^n \\[1em] \Rightarrow n = 3.\end{aligned}$$

Hence, time taken = 3 years.

Let share of A = ₹ x and share of B = ₹ 28730 - ₹ x.

For A :

P = ₹ x

r = 10%

n = 3 years

By formula,

Amount = $P\Big(1 + \frac{r}{100}\Big)^n$

Substituting values we get :

$$\begin{aligned}\Rightarrow \text{Amount } = x\Big(1 + \frac{10}{100}\Big)^3 \\[1em] = x \times \Big(\frac{110}{100}\Big)^3\\[1em] = x \times \Big(\frac{11}{10}\Big)^3\end{aligned}$$

For B :

P = ₹ (28730 - x)

r = 10%

n = 5 years

By formula,

Amount = $P\Big(1 + \frac{r}{100}\Big)^n$

Substituting values we get :

$$\begin{aligned}\Rightarrow \text{Amount } = (28730 - x) \times \Big(1 + \frac{10}{100}\Big)^5 \\[1em] = (28730 - x) \times \Big(\frac{110}{100}\Big)^5 \\[1em] = (28730 - x) \times \Big(\frac{11}{10}\Big)^5\end{aligned}$$

Since, amount received by A and B are equal.

$$\begin{aligned}\therefore x \times \Big(\frac{11}{10}\Big)^3 = (28730 - x) \times \Big(\frac{11}{10}\Big)^5 \\[1em] \Rightarrow x = (28730 - x) \times \Big(\frac{11}{10}\Big)^2 \\[1em] \Rightarrow x = (28730 - x) \times \frac{121}{100} \\[1em] \Rightarrow 100x = 121(28730 - x) \\[1em] \Rightarrow 100x = 3476330 - 121x \\[1em] \Rightarrow 100x + 121x = 3476330 \\[1em] \Rightarrow 221x = 3476330 \\[1em] \Rightarrow x = \frac{3476330}{221} \\[1em] \Rightarrow x = 15730.\end{aligned}$$

₹ (28730 - x) = ₹ (28730 - 15730) = ₹ 13000.

Hence, share of A and B are ₹ 15730 and ₹ 13000 respectively.

(i) Let share of John = ₹ x and share of Smith = ₹ 44200 - ₹ x.

For John :

P = ₹ x

r = 10%

n = 4 years (John has 4 years to reach 16 years of age)

By formula,

Amount = $P\Big(1 + \frac{r}{100}\Big)^n$

Substituting values we get :

$$\begin{aligned}\Rightarrow \text{Amount } = x\Big(1 + \frac{10}{100}\Big)^4 \\[1em] = x \times \Big(\frac{110}{100}\Big)^4\\[1em] = x \times \Big(\frac{11}{10}\Big)^4 ........(1)\end{aligned}$$

For Smith :

P = ₹ (44200 - x)

r = 10%

n = 2 years (Smith has 2 years to reach 16 years of age)

By formula,

Amount = $P\Big(1 + \frac{r}{100}\Big)^n$

Substituting values we get :

$$\begin{aligned}\Rightarrow \text{Amount } = (44200 - x) \times \Big(1 + \frac{10}{100}\Big)^2 \\[1em] = (44200 - x) \times \Big(\frac{110}{100}\Big)^2 \\[1em] = (44200 - x) \times \Big(\frac{11}{10}\Big)^2 ...........(2)\end{aligned}$$

Since, amount received by A and B are equal.

$$\begin{aligned}\therefore x \times \Big(\frac{11}{10}\Big)^4 = (44200 - x) \times \Big(\frac{11}{10}\Big)^2 \\[1em] \Rightarrow x \times \Big(\frac{11}{10}\Big)^2 = (44200 - x) \\[1em] \Rightarrow \frac{121}{100}x = (44200 - x) \\[1em] \Rightarrow 121x = 100(44200 - x) \\[1em] \Rightarrow 121x = 4420000 - 100x \\[1em] \Rightarrow 121x + 100x = 4420000 \\[1em] \Rightarrow 221x = 4420000 \\[1em] \Rightarrow x = \frac{4420000}{221} \\[1em] \Rightarrow x = ₹ 20000.\end{aligned}$$

₹ (44200 - x) = ₹ (44200 - 20000) = ₹ 24200.

Hence, share of John and Smith are ₹ 20,000 and ₹ 24,200 respectively.

(ii) Substituting value of x in equation (1), we get :

$$\begin{aligned}\Rightarrow 20000 \times \Big(\frac{11}{10}\Big)^4 \\[1em] \Rightarrow 20000 \times \frac{14641}{10000} \\[1em] \Rightarrow ₹ 29282.\end{aligned}$$

Since, both receive equal amount.

Hence, each receive ₹ 29282.

(i) Let the sum be ₹ x.

Given, S.I. = ₹ 6000 in 2 years at 10% rate of interest.

By formula,

S.I. = $\frac{P \times R \times T}{100}$

Substituting values we get :

$$\begin{aligned}\Rightarrow 6000 = \frac{x \times 10 \times 2}{100} \\[1em] \Rightarrow x = \frac{6000 \times 100}{10 \times 2} \\[1em] \Rightarrow x = 30000.\end{aligned}$$

Hence, sum = ₹ 30000.

(ii) By formula,

A = $P\Big(1 + \frac{r}{100}\Big)^n$

Substituting values we get :

$$\begin{aligned}A = 30000 \times \Big(1 + \frac{10}{100}\Big)^3 \\[1em] = 30000 \times \Big(\frac{110}{100}\Big)^3 \\[1em] = 30000 \times \Big(\frac{11}{10}\Big)^3 \\[1em] = 30000 \times \frac{1331}{1000} \\[1em] = 39930.\end{aligned}$$

Hence, amount due at the end of 3 years = ₹ 39930.

(iii) By formula,

C.I. = A - P = ₹ 39930 - ₹ 30000 = ₹ 9930.

Hence, compound interest earned in 3 years = ₹ 9930.

For S.I. :

Principal = ₹ 8000

Rate = 5%

Time = 2 years

S.I. = $\frac{8000 \times 5 \times 2}{100}$ = ₹ 800.

For C.I. :

C.I. = A - P

$$\begin{aligned}= P\Big(1 + \frac{r}{100}\Big)^n - P \\[1em] = 8000 \times \Big(1 + \frac{5}{100}\Big)^2 - 8000 \\[1em] = 8000 \times \Big(\frac{105}{100}\Big)^2 - 8000 \\[1em] = 8000 \times \Big(\frac{21}{20}\Big)^2 - 8000 \\[1em] = 8000 \times \frac{441}{400} - 8000 \\[1em] = 8820 - 8000 \\[1em] = ₹ 820.\end{aligned}$$

Difference between C.I. and S.I. = ₹ 820 - ₹ 800 = ₹ 20.

Hence, difference between C.I. and S.I. = ₹ 20.

For S.I. :

Principal = ₹ 6000

Rate = 10%

Time = 2 years

S.I. = $\frac{6000 \times 10 \times 2}{100}$ = ₹ 1200.

For C.I. :

C.I. = A - P

$$\begin{aligned}= P\Big(1 + \frac{r}{100}\Big)^n - P \\[1em] = 6000 \times \Big(1 + \frac{10}{100}\Big)^2 - 6000 \\[1em] = 6000 \times \Big(\frac{110}{100}\Big)^2 - 6000 \\[1em] = 6000 \times \Big(\frac{11}{10}\Big)^2 - 6000 \\[1em] = 6000 \times \frac{121}{100} - 6000 \\[1em] = 7260 - 6000 \\[1em] = ₹ 1260.\end{aligned}$$

Difference between C.I. and S.I. = ₹ 1260 - ₹ 1200 = ₹ 60.

Hence, Option 3 is the correct option.

Let rate of interest be r%.

Given,

P = ₹ 10000

A = ₹ 12500

n = 1 year

By formula,

A = $P\Big(1 + \frac{r}{100}\Big)^n$

Substituting values we get :

$$\begin{aligned}\Rightarrow 12500 = 10000 \times \Big(1 + \frac{r}{100}\Big)^1 \\[1em] \Rightarrow \frac{12500}{10000} = \Big(1 + \frac{r}{100}\Big) \\[1em] \Rightarrow \frac{5}{4} = 1 + \frac{r}{100}\\[1em] \Rightarrow \frac{5}{4} - 1 = \frac{r}{100} \\[1em] \Rightarrow \frac{5 - 4}{4} = \frac{r}{100} \\[1em] \Rightarrow \frac{1}{4} = \frac{r}{100} \\[1em] \Rightarrow r = \frac{100}{4} = 25%.\end{aligned}$$

Hence, Option 4 is the correct option.

Given,

P = ₹ 16000

n = 2 years

r = 20%

By formula,

A = $P\Big(1 + \frac{r}{100}\Big)^n$

C.I. = A - P

$$\begin{aligned}C.I. = P\Big(1 + \frac{r}{100}\Big)^n - P \\[1em] = 16000 \times \Big(1 + \frac{20}{100}\Big)^2 - 16000 \\[1em] = 16000 \times \Big(\frac{120}{100}\Big)^2 - 16000 \\[1em] = 16000 \times \Big(\frac{6}{5}\Big)^2 - 16000 \\[1em] = 16000 \times \frac{36}{25} - 16000 \\[1em] = 23040 - 16000 \\[1em] = ₹ 7040.\end{aligned}$$

Hence, Option 2 is the correct option.

By formula,

S.I. = $\frac{P \times R \times T}{100}$

Substituting values we get :

$$\begin{aligned}S.I. = \frac{16000 \times 20 \times 2}{100} \\[1em] = ₹ 6400.\end{aligned}$$

Hence, Option 2 is the correct option.

From part (c), we get :

C.I. = ₹ 7040

From part (d), we get :

S.I. = ₹ 6400

C.I. - S.I. = ₹ 7040 - ₹ 6400 = ₹ 640.

Hence, Option 1 is the correct option.

Given,

n = 2 years

r = 8%

Let sum of money be ₹ P.

C.I. = A - P

$$\begin{aligned}= P\Big(1 + \frac{r}{100}\Big)^n - P \\[1em] = P\Big(1 + \frac{8}{100}\Big)^2 - P \\[1em] = P \times \Big(\frac{108}{100}\Big)^2 - P \\[1em] = P \times \Big(\frac{27}{25}\Big)^2 - P\\[1em] = P \times \frac{729}{625} - P \\[1em] = \frac{729P}{625} - P \\[1em] = \frac{729P - 625P}{625} \\[1em] = \frac{104P}{625}.\end{aligned}$$

By formula,

$$\begin{aligned}S.I. = \frac{P \times R \times T}{100} \\[1em] = \frac{P \times 8 \times 2}{100} \\[1em] = \frac{4P}{25}.\end{aligned}$$

Given,

Difference between S.I. and C.I. = ₹ 54.40

$$\begin{aligned}\Rightarrow \frac{104P}{625} - \frac{4P}{25} = 54.40 \\[1em] \Rightarrow \frac{104P - 100P}{625} = 54.40 \\[1em] \Rightarrow \frac{4P}{625} = 54.40 \\[1em] \Rightarrow P = \frac{54.40 \times 625}{4} \\[1em] \Rightarrow P = ₹ 8500.\end{aligned}$$

Hence, sum = ₹ 8500.

Let sum of money be ₹ x.

For Pramod :

P = ₹ x

Time = 2 years

Rate = 5%

S.I. = $\frac{P \times R \times T}{100}$

Substituting values we get :

$$\begin{aligned}S.I. = \frac{x \times 5 \times 2}{100} \\[1em] = \frac{x}{10}.\end{aligned}$$

For Anand :

P = ₹ x

Time (n) = 2 years

Rate (r) = 5%

C.I. = A - P

$$\begin{aligned}C.I. = P\Big(1 + \frac{r}{100}\Big)^n - P \\[1em] = x \times \Big(1 + \frac{5}{100}\Big)^2 - x \\[1em] = x \times \Big(\frac{105}{100}\Big)^2 - x \\[1em] = x \times \Big(\frac{21}{20}\Big)^2 - x \\[1em] = x \times \frac{441}{400} - x \\[1em] = \frac{441x}{400} - x \\[1em] = \frac{441x - 400x}{400} \\[1em] = \frac{41x}{400}.\end{aligned}$$

Given,

Anand received ₹ 15 more than Pramod.

∴ C.I. - S.I. = ₹ 15

$$\begin{aligned}\Rightarrow \frac{41x}{400} - \frac{x}{10} = 15 \\[1em] \Rightarrow \frac{41x - 40x}{400} = 15 \\[1em] \Rightarrow \frac{x}{400} = 15 \\[1em] \Rightarrow x = 400 \times 15 = ₹ 6000. \\[1em] S.I. = \frac{x}{10} = \frac{6000}{10} = ₹ 600. \\[1em] C.I. = \frac{41x}{400} = \frac{41 \times 6000}{400} = ₹ 615.\end{aligned}$$

Hence, sum lent by each = ₹ 6000 and interest received by Pramod = ₹ 600 and Anand = ₹ 615.

Let the sum be ₹ x.

Given,

Simple interest on the sum of money for 2 years at 4% is ₹ 450.

By formula,

S.I. = $\frac{P \times R \times T}{100}$

Substituting values we get :

$$\begin{aligned}\Rightarrow 450 = \frac{x \times 4 \times 2}{100} \\[1em] \Rightarrow x = \frac{450 \times 100}{8} \\[1em] \Rightarrow x = 225 \times 25 \\[1em] \Rightarrow x = ₹5625.\end{aligned}$$

By formula,

C.I. = A - P

$$\begin{aligned}= P\Big(1 + \frac{r}{100}\Big)^n - P \\[1em] = 5625 \times \Big(1 + \frac{4}{100}\Big)^2 - 5625 \\[1em] = 5625 \times \Big(\frac{104}{100}\Big)^2 - 5625 \\[1em] = 5625 \times \Big(\frac{26}{25}\Big)^2 - 5625 \\[1em] = 5625 \times \frac{676}{625} - 5625 \\[1em] = 6084 - 5625 \\[1em] = ₹ 459.\end{aligned}$$

Hence, the compound interest = ₹ 459.

Let sum of money be ₹ x.

By formula,

$$\begin{aligned}\Rightarrow C.I. = A - P \\[1em] \Rightarrow 246 = P\Big(1 + \frac{r}{100}\Big)^n - P \\[1em] \Rightarrow 246 = x \times \Big(1 + \frac{5}{100}\Big)^2 - x \\[1em] \Rightarrow 246 = x \times \Big(\frac{105}{100}\Big)^2 - x \\[1em] \Rightarrow 246 = x \times \Big(\frac{21}{20}\Big)^2 - x \\[1em] \Rightarrow 246 = x \times \frac{441}{400} - x \\[1em] \Rightarrow 246 = \frac{441x}{400} - x \\[1em] \Rightarrow 246 = \frac{441x - 400x}{400} \\[1em] \Rightarrow 246 = \frac{41x}{400} \\[1em] \Rightarrow x = \frac{246 \times 400}{41} \\[1em] \Rightarrow x = ₹ 2400.\end{aligned}$$

For S.I. :

P = x = ₹ 2400

R = 6%

T = 3 years

By formula,

$$\begin{aligned}\Rightarrow S.I. = \frac{P \times R \times T}{100} \\[1em] \Rightarrow S.I. = \frac{x \times 6 \times 3}{100} \\[1em] \Rightarrow S.I. = \frac{2400 \times 6 \times 3}{100} \\[1em] \Rightarrow S.I. = ₹ 432.\end{aligned}$$

Hence, simple interest = ₹ 432.

Let original sum of money be ₹ P and rate of interest be r%.

By formula,

$A = P\Big(1 + \frac{r}{100}\Big)^n$

Given,

Sum amounts to ₹ 19360 in two years.

$\therefore P\Big(1 + \frac{r}{100}\Big)^2 = 19360$ ......(1)

Given,

Sum amounts to ₹ 23425.60 in four years.

$\therefore P\Big(1 + \frac{r}{100}\Big)^4 = 23425.60$ ......(2)

Dividing equation (2) by (1), we get :

$$\begin{aligned}\Rightarrow \frac{P\Big(1 + \frac{r}{100}\Big)^4}{P\Big(1 + \frac{r}{100}\Big)^2} = \frac{23425.60}{19360} \\[1em] \Rightarrow \Big(1 + \frac{r}{100}\Big)^2 = \frac{146.41}{121} \\[1em] \Rightarrow \Big(1 + \frac{r}{100}\Big)^2 = \Big(\frac{12.1}{11}\Big)^2 \\[1em] \Rightarrow 1 + \frac{r}{100} = \frac{12.1}{11} \\[1em] \Rightarrow \frac{r}{100} = \frac{12.1}{11} - 1 \\[1em] \Rightarrow \frac{r}{100} = \frac{12.1 - 11}{11} \\[1em] \Rightarrow r = \frac{1.1}{11} \times 100 \\[1em] \Rightarrow r = 10%.\end{aligned}$$

Substituting value of r in equation (1), we get :

$$\begin{aligned}\Rightarrow P\Big(1 + \frac{r}{100}\Big)^2 = 19360 \\[1em] \Rightarrow P\Big(1 + \frac{10}{100}\Big)^2 = 19360 \\[1em] \Rightarrow P\Big(\frac{110}{100}\Big)^2 = 19360 \\[1em] \Rightarrow P \times \Big(\frac{11}{10}\Big)^2 = 19360 \\[1em] \Rightarrow P \times \frac{121}{100} = 19360 \\[1em] \Rightarrow P = \frac{19360 \times 100}{121} \\[1em] \Rightarrow P = 160 \times 100 \\[1em] \Rightarrow P = ₹ 16000.\end{aligned}$$

Hence, sum of money = ₹ 16000 and rate of interest = 10%.

Let rate of interest be r% and sum of money be ₹ P.

By formula,

A = $P\Big(1 + \frac{r}{100}\Big)^n$

Given,

₹ P becomes three times of itself in 8 years.

$$\begin{aligned}\therefore 3P = P\Big(1 + \frac{r}{100}\Big)^8 \\[1em] \Rightarrow \frac{3P}{P} = \Big(1 + \frac{r}{100}\Big)^8 \\[1em] \Rightarrow 3 = \Big(1 + \frac{r}{100}\Big)^8 .....(1)\end{aligned}$$

Let in n years money becomes 27 times.

$$\begin{aligned}\Rightarrow P\Big(1 + \frac{r}{100}\Big)^n = 27P \\[1em] \Rightarrow \Big(1 + \frac{r}{100}\Big)^n = \frac{27P}{P} \\[1em] \Rightarrow \Big(1 + \frac{r}{100}\Big)^n = 27 \\[1em] \Rightarrow \Big(1 + \frac{r}{100}\Big)^n = 3^3 \\[1em]\end{aligned}$$

From equation (1)

$$\begin{aligned}\Rightarrow \Big(1 + \frac{r}{100}\Big)^n = \Big[\Big(1 + \frac{r}{100}\Big)^8\Big]^3 \\[1em] \Rightarrow \Big(1 + \frac{r}{100}\Big)^n =\Big(1 + \frac{r}{100}\Big)^{24} \\[1em] \Rightarrow n = 24 \text{ years}.\end{aligned}$$

Hence, in 24 years money will becomes 27 times of itself.

For S.I. :

P = ₹ 9430

T = 10 years

R = 5%

$$\begin{aligned}S.I. = \frac{P \times R \times T}{100} \\[1em] = \frac{9430 \times 5 \times 10}{100} \\[1em] = ₹ 4715.\end{aligned}$$

Since, C.I. = S.I. = ₹ 4715

Let sum on which C.I. = ₹ 4715 for 2 years at 5% be ₹x.

$$\begin{aligned}\Rightarrow C.I. = A - P \\[1em] \Rightarrow 4715 = P\Big(1 + \frac{r}{100}\Big)^n - P \\[1em] \Rightarrow 4715 = x \times \Big(1 + \frac{5}{100}\Big)^2 - x \\[1em] \Rightarrow 4715 = x \times \Big(\frac{105}{100}\Big)^2 - x \\[1em] \Rightarrow 4715 = x \times \Big(\frac{21}{20}\Big)^2 - x \\[1em] \Rightarrow 4715 = \frac{441x}{400} - x \\[1em] \Rightarrow 4715 = \frac{441x - 400x}{400} \\[1em] \Rightarrow 4715 = \frac{41x}{400} \\[1em] \Rightarrow x = \frac{4715 \times 400}{41} \\[1em] \Rightarrow x = 115 \times 400 \\[1em] \Rightarrow x = ₹ 46000.\end{aligned}$$

Hence, sum of money = ₹ 46000.

Let sum of money be ₹ x.

For S.I. :

P = ₹ x

Time (T) = 4 years

Rate of interest (R) = 4%

By formula,

S.I. = $\frac{P \times R \times T}{100}$

Substituting values we get :

$$\begin{aligned}\Rightarrow S.I. = \frac{x \times 4 \times 4}{100} \\[1em] = \frac{4x}{25}.\end{aligned}$$

For C.I. :

P = ₹ x

Rate of interest (r) = 5%

Time (n) = 3 years

By formula,

$$\begin{aligned}\Rightarrow C.I. = A - P \\[1em] \Rightarrow C.I. = P\Big(1 + \frac{r}{100}\Big)^n - P \\[1em] = x \times \Big(1 + \frac{5}{100}\Big)^3 - x \\[1em] = x \times \Big(\frac{105}{100}\Big)^3 - x \\[1em] = x \times \Big(\frac{21}{20}\Big)^3 - x \\[1em] = \frac{9261x}{8000} - x \\[1em] = \frac{9261x - 8000x}{8000} \\[1em] = \frac{1261x}{8000}.\end{aligned}$$

Given,

Simple interest on ₹ x for 4 years at 4% per annum exceeds the compound interest on the same sum for 3 years at 5 percent per annum by ₹ 228.

$$\begin{aligned}\therefore \frac{4x}{25} - \frac{1261x}{8000} = 228 \\[1em] \Rightarrow \frac{1280x - 1261x}{8000} = 228 \\[1em] \Rightarrow \frac{19x}{8000} = 228 \\[1em] \Rightarrow x = \frac{228 \times 8000}{19} \\[1em] \Rightarrow x = ₹ 96000.\end{aligned}$$

Hence, sum of money = ₹ 96000.

Let sum of money be ₹ x.

Given,

It amounts to ₹ 23400 in 3 years at 10% per annum simple interest.

By formula,

$$\begin{aligned}\Rightarrow A = P + S.I. \\[1em] \Rightarrow 23400 = x + \frac{x \times 10 \times 3}{100} \\[1em] \Rightarrow 23400 = x + \frac{3x}{10} \\[1em] \Rightarrow \frac{13x}{10} = 23400 \\[1em] \Rightarrow x = \frac{23400 \times 10}{13} \\[1em] \Rightarrow x = ₹ 18000.\end{aligned}$$

In case of compound interest :

By formula,

$$\begin{aligned}A = P\Big(1 + \frac{r}{100}\Big)^n \\[1em] = 18000 \times \Big(1 + \frac{10}{100}\Big)^2 \\[1em] = 18000 \times \Big(\frac{110}{100}\Big)^2 \\[1em] = 18000 \times \Big(\frac{11}{10}\Big)^2 \\[1em] = 18000 \times \Big(\frac{121}{100}\Big) \\[1em] = ₹ 21780.\end{aligned}$$

Hence, amount on same sum in 2 years at 10% p.a. compound interest = ₹ 21780.

Let Mohit borrowed ₹ x and ₹ y.

By formula,

A = $P\Big(1 + \frac{r}{100}\Big)^n$

For ₹ 12600,

P = ₹ x

A = ₹ 12600

r = 5%

n = 1 year

Substituting values in formula we get :

$$\begin{aligned}\Rightarrow 12600 = x \times \Big(1 + \frac{5}{100}\Big)^1 \\[1em] \Rightarrow 12600 = x \times \frac{105}{100} \\[1em] \Rightarrow x = \frac{12600 \times 100}{105} \\[1em] \Rightarrow x = ₹ 12000.\end{aligned}$$

For ₹ 17640,

P = ₹ y

A = ₹ 17640

r = 5%

n = 2 year

Substituting values in formula we get :

$$\begin{aligned}\Rightarrow 17640 = y \times \Big(1 + \frac{5}{100}\Big)^2 \\[1em] \Rightarrow 17640 = y \times \Big(\frac{105}{100}\Big)^2 \\[1em] \Rightarrow 17640 = y \times \Big(\frac{21}{20}\Big)^2 \\[1em] \Rightarrow y = \frac{17640 \times 20^2}{21^2} \\[1em] \Rightarrow y = ₹ 16000.\end{aligned}$$

Total money borrowed = ₹ (x + y) = ₹ (12000 + 16000) = ₹ 28000.

Hence, sum borrowed = ₹ 28000.

Let sum = ₹ P.

Amount for first year :

A = $P\Big(1 + \frac{r}{100}\Big)^1$

For next $\frac{1}{2}$ year :

Sum = ₹ $P\Big(1 + \frac{r}{100}\Big)$

A = $P\Big(1 + \frac{r}{100}\Big)\Big(1 + \frac{r}{2 \times 100}\Big)^{\frac{1}{2} \times 2} = P\Big(1 + \frac{r}{100}\Big)\Big(1 + \frac{r}{200}\Big)$

By formula,

C.I. = A - P = $P\Big(1 + \frac{r}{100}\Big)\Big(1 + \frac{r}{200}\Big) - P$

Hence, Option 2 is the correct option.

By formula,

C.I. = A - P

$C.I. = P\Big(1 + \frac{r}{100}\Big)^1 - P$

Hence, Option 1 is the correct option.

Amount for first two years :

A = $P\Big(1 + \frac{r}{100}\Big)^2$

For next $\frac{1}{2}$ year :

Sum = ₹ $P\Big(1 + \frac{r}{100}\Big)^2$

A = $P\Big(1 + \frac{r}{100}\Big)^2\Big(1 + \frac{r}{2 \times 100}\Big)^{\frac{1}{2} \times 2} = P\Big(1 + \frac{r}{100}\Big)^2\Big(1 + \frac{r}{200}\Big)$

By formula,

C.I. = A - P = $P\Big(1 + \frac{r}{100}\Big)^2\Big(1 + \frac{r}{200}\Big) - P$

Hence, Option 4 is the correct option.

When interest is compounded half-yearly.

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

C.I. = A - P

For 1 year :

$$\begin{aligned}C.I. = P\Big(1 + \frac{r}{2 \times 100}\Big)^{1 \times 2} - P \\[1em] = P\Big(1 + \frac{r}{200}\Big)^2 - P\end{aligned}$$

Hence, Option 1 is the correct option.

When interest is compounded half-yearly.

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

C.I. = A - P

$$\begin{aligned}C.I. = P\Big(1 + \frac{r}{2 \times 100}\Big)^{\frac{3}{2} \times 2} - P \\[1em] = P\Big(1 + \frac{r}{200}\Big)^3 - P\end{aligned}$$

Hence, Option 2 is the correct option.

When interest is compounded half-yearly.

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

C.I. = A - P

$$\begin{aligned}C.I. = P\Big(1 + \frac{r}{2 \times 100}\Big)^{2 \times 2} - P \\[1em] = P\Big(1 + \frac{r}{200}\Big)^4 - P\end{aligned}$$

Hence, Option 3 is the correct option.

Let rate of interest be r%.

Given,

P = ₹ 6000

C.I. = ₹ 1200

By formula,

$$\begin{aligned}\Rightarrow C.I. = P\Big(1 + \frac{r}{2 \times 100}\Big)^{n \times 2} - P \\[1em] \Rightarrow 1200 = 6000\Big(1 + \frac{r}{200}\Big)^{\frac{1}{2} \times 2} - 6000 \\[1em] \Rightarrow 1200 + 6000 = 6000\Big(1 + \frac{r}{200}\Big) \\[1em] \Rightarrow 7200 = 6000\Big(1 + \frac{r}{200}\Big) \\[1em] \Rightarrow \frac{7200}{6000} = \Big(1 + \frac{r}{200}\Big) \\[1em] \Rightarrow \frac{7200}{6000} - 1 = \frac{r}{200} \\[1em] \Rightarrow \frac{7200 - 6000}{6000} = \frac{r}{200} \\[1em] \Rightarrow \frac{1200}{6000} = \frac{r}{200} \\[1em] \Rightarrow r = \frac{1200}{6000} \times 200 \\[1em] \Rightarrow r = 40%\end{aligned}$$

Hence, Option 1 is the correct option.

Let sum of money be ₹ x.

Given,

S.I. = ₹ 2400

Time = 2 years

Rate = 10%

By formula,

$$\begin{aligned}\Rightarrow S.I. = \frac{P \times R \times T}{100} \\[1em] \Rightarrow 2400 = \frac{x \times 10 \times 2}{100} \\[1em] \Rightarrow x = \frac{2400 \times 100}{10 \times 2} \\[1em] \Rightarrow x = ₹ 12000.\end{aligned}$$

By formula,

C.I. = A - P

$$\begin{aligned}C.I. = P\Big(1 + \frac{r}{100}\Big)^1 - P \\[1em] = 12000\Big(1 + \frac{10}{100}\Big)^1 - 12000 \\[1em] = 12000\Big(1 + \frac{1}{10}\Big)^1 - 12000 \\[1em] = 12000 \times \Big(\frac{11}{10}\Big)^1 - 12000 \\[1em] = 12000 \times \frac{11}{10} - 12000 \\[1em] = 13200 - 12000 \\[1em] = ₹ 1200.\end{aligned}$$

Hence, Option 1 is the correct option.

For 2nd year :

P = 12000 + 1200 = ₹ 13200

By formula,

C.I. = A - P

$$\begin{aligned}C.I. = P\Big(1 + \frac{r}{100}\Big)^1 - P \\[1em] = 13200\Big(1 + \frac{10}{100}\Big)^1 - 13200 \\[1em] = 13200 \times \frac{110}{100} - 13200 \\[1em] = 14520 - 13200 \\[1em] = ₹ 1320.\end{aligned}$$

Hence, Option 1 is the correct option.

Let sum of money be ₹ x.

Given,

S.I. = ₹ 2400

Time = 2 years

Rate = 10%

By formula,

$$\begin{aligned}\Rightarrow S.I. = \frac{P \times R \times T}{100} \\[1em] \Rightarrow 2400 = \frac{x \times 10 \times 2}{100} \\[1em] \Rightarrow x = \frac{2400 \times 100}{10 \times 2} \\[1em] \Rightarrow x = ₹ 12000.\end{aligned}$$

P = ₹ x = ₹ 12000.

Hence, Option 1 is the correct option.

Amount in 2 years = Principal + C.I. for 1st year + C.I. for 2nd year

= ₹ 12000 + ₹ 1200 + ₹ 1320

= ₹ 14520.

Hence, Option 1 is the correct option.

Given,

P = ₹ 7400

r = 5% compounded half-yearly

n = 1 year

When rate of interest is compounded half-yearly,

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

Substituting values we get :

$$\begin{aligned}A = 7400 \times \Big(1 + \frac{5}{2 \times 100}\Big)^{1 \times 2} \\[1em] = 7400 \times \Big(1 + \frac{1}{40}\Big)^2 \\[1em] = 7400 \times \Big(\frac{41}{40}\Big)^2 \\[1em] = 7400 \times \frac{1681}{1600} \\[1em] = ₹ 7774.63\end{aligned}$$

Hence, amount = ₹ 7774.63

Given,

P = ₹ 10000

n = 18 months or 1.5 years

r = 10%

When interest is compounded yearly :

A = $P\Big(1 + \frac{r}{100}\Big)^n$

For 1st year :

$$\begin{aligned}A = 10000 \times \Big(1 + \frac{10}{100}\Big)^1 \\[1em] = 10000 \times \frac{110}{100} \\[1em] = ₹ 11000.\end{aligned}$$

For next half-year :

₹ 11000 is the principal.

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

Substituting values we get :

$$\begin{aligned}A = 11000 \times \Big(1 + \frac{10}{2 \times 100}\Big)^{\frac{1}{2} \times 2} \\[1em] = 11000 \times \Big(1 + \frac{1}{20}\Big)^1 \\[1em] = 11000 \times \frac{21}{20} \\[1em] = ₹ 11550.\end{aligned}$$

When rate of interest is compounded half-yearly :

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

Substituting values we get :

$$\begin{aligned}A = 10000 \times \Big(1 + \frac{10}{2 \times 100}\Big)^{1.5 \times 2} \\[1em] = 10000 \times \Big(1 + \frac{1}{20}\Big)^3\\[1em] = 10000 \times \Big(\frac{21}{20}\Big)^3 \\[1em] = 10000 \times \frac{9261}{8000} \\[1em] = ₹ 11576.25\end{aligned}$$

Difference in C.I. between two cases = ₹ 11576.25 - ₹ 11550 = ₹ 26.25

Hence, difference between C.I. in two cases = ₹ 26.25

For S.I. :

P = ₹ 16000

Time = 2 years

Rate = 20%

S.I. = $\frac{P \times R \times T}{100} = \frac{16000 \times 2 \times 20}{100}$ = ₹ 6400.

Amount = P + S.I. = ₹ 16000 + ₹ 6400 = ₹ 22400.

For C.I. :

P = ₹ 22400

Time = 1 year

Rate = 20%

When interest is compounded half-yearly :

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

Substituting values we get :

$$\begin{aligned}A = 22400\Big(1 + \frac{20}{2 \times 100}\Big)^{1 \times 2} \\[1em] = 22400 \times \Big(1 + \frac{1}{10}\Big)^2 \\[1em] = 22400 \times \Big(\frac{11}{10}\Big)^2 \\[1em] = 22400 \times \frac{121}{100} \\[1em] = 224 \times 121 \\[1em] = ₹ 27104.\end{aligned}$$

Hence, total amount to be paid = ₹ 27104.

Let sum of money be ₹ x.

Given,

Time = 1.5 years

Rate = 10% compounded half-yearly

When rate of interest is compounded half-yearly :

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow 27783 = x \times \Big(1 + \frac{10}{2 \times 100}\Big)^{1.5 \times 2} \\[1em] \Rightarrow 27783 = x \times \Big(1 + \frac{1}{20}\Big)^3 \\[1em] \Rightarrow 27783 = x \times \Big(\frac{21}{20}\Big)^3 \\[1em] \Rightarrow 27783 = \frac{9261x}{8000} \\[1em] \Rightarrow x = \frac{27783 \times 8000}{9261} \\[1em] \Rightarrow x = ₹ 24000.\end{aligned}$$

Hence, sum of money = ₹ 24000.

Given,

r = 20%

n = 18 months or 1.5 years

Let sum of money invested by each be ₹ x.

For Ashok interest is compounded annually :

For 1st year :

$$\begin{aligned}A = P\Big(1 + \frac{r}{100}\Big)^n \\[1em] = x \times \Big(1 + \frac{20}{100}\Big)^1 \\[1em] = x \times \Big(\frac{120}{100}\Big)^1 \\[1em] = x \times \frac{6}{5} \\[1em] = \frac{6x}{5}.\end{aligned}$$

For next $\frac{1}{2}$ year :

P = $\frac{6x}{5}$

$$\begin{aligned}A = P\Big(1 + \frac{r}{100 \times 2}\Big)^{n \times 2} \\[1em] = \frac{6x}{5} \times \Big(1 + \frac{20}{100 \times 2}\Big)^{\frac{1}{2} \times 2} \\[1em] = \frac{6x}{5} \times \Big(1 + \frac{20}{200}\Big)^1 \\[1em] = \frac{6x}{5} \times \Big(\frac{220}{200}\Big) \\[1em] = \frac{66x}{50}.\end{aligned}$$

For Geeta interest is compounded half-yearly :

$$\begin{aligned}A = P\Big(1 + \frac{r}{100 \times 2}\Big)^{n \times 2} \\[1em] = x \times \Big(1 + \frac{20}{100 \times 2}\Big)^{1.5 \times 2} \\[1em] = x \times \Big(1 + \frac{20}{200}\Big)^3 \\[1em] = x \times \Big(\frac{220}{200}\Big)^3 \\[1em] = x \times \Big(\frac{11}{10}\Big)^3 \\[1em] = x \times \frac{1331}{1000} \\[1em] = \frac{1331x}{1000}.\end{aligned}$$

Given, Geeta receives ₹ 33 more :

$$\begin{aligned}\therefore \frac{1331x}{1000} - \frac{66x}{50} = 33 \\[1em] \Rightarrow \frac{1331x - 1320x}{1000} = 33 \\[1em] \Rightarrow \frac{11x}{1000} = 33 \\[1em] \Rightarrow x = \frac{33 \times 1000}{11} \\[1em] \Rightarrow x = ₹ 3000.\end{aligned}$$

Hence, sum invested by each = ₹ 3000.

Let rate of interest be r% per annum.

When interest is compounded half-yearly.

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

Substituting values we get :

$$\begin{aligned}\Rightarrow C.I. = A - P \\[1em] \Rightarrow C.I. = P\Big(1 + \frac{r}{2 \times 100}\Big)^{n \times 2} - P \\[1em] \Rightarrow 5100 = 62500 \times \Big(1 + \frac{r}{200}\Big)^{1 \times 2} - 62500 \\[1em] \Rightarrow 5100 + 62500 = 62500 \times \Big(1 + \frac{r}{200}\Big)^2 \\[1em] \Rightarrow 67600 = 62500 \times \Big(1 + \frac{r}{200}\Big)^2 \\[1em] \Rightarrow \frac{67600}{62500} = \Big(1 + \frac{r}{200}\Big)^2 \\[1em] \Rightarrow \frac{676}{625} = \Big(1 + \frac{r}{200}\Big)^2 \\[1em] \Rightarrow \Big(\frac{26}{25}\Big)^2 = \Big(1 + \frac{r}{200}\Big)^2 \\[1em] \Rightarrow 1 + \frac{r}{200} = \frac{26}{25} \\[1em] \Rightarrow \frac{r}{200} = \frac{26}{25}- 1 \\[1em] \Rightarrow \frac{r}{200} = \frac{1}{25} \\[1em] \Rightarrow r = \frac{200}{25} = 8%.\end{aligned}$$

Hence, rate of interest = 8%.

Given,

C.I. = ₹ 496.50

P = ₹ 1500

Rate = 20%

A = P + C.I. = ₹ 1500 + ₹ 496.50 = ₹ 1996.50

Let time required be n years.

When interest is compounded half-yearly.

$$\begin{aligned}\Rightarrow A = P\Big(1 + \frac{r}{2 \times 100}\Big)^{n \times 2} \\[1em] \Rightarrow 1996.50 = 1500 \times \Big(1 + \frac{20}{200}\Big)^{2n} \\[1em] \Rightarrow \frac{1996.50}{1500} = \Big(1 + \frac{20}{200}\Big)^{2n} \\[1em] \Rightarrow \frac{199650}{150000} = \Big(\frac{220}{200}\Big)^{2n} \\[1em] \Rightarrow \frac{1331}{1000} = \Big(\frac{11}{10}\Big)^{2n} \\[1em] \Rightarrow \Big(\frac{11}{10}\Big)^3 = \Big(\frac{11}{10}\Big)^{2n} \\[1em] \Rightarrow 2n = 3 \\[1em] \Rightarrow n = \frac{3}{2}.\end{aligned}$$

Hence, required time = $1\frac{1}{2}$ years.

When interest is compounded half-yearly.

$$\begin{aligned}\Rightarrow A = P\Big(1 + \frac{r}{2 \times 100}\Big)^{n \times 2} \\[1em] \Rightarrow A = 3500 \times \Big(1 + \frac{6}{200}\Big)^{3 \times 2} \\[1em] \Rightarrow A = 3500 \times \Big(\frac{206}{200}\Big)^6 \\[1em] \Rightarrow A = 3500 \times (1.03)^6 \\[1em] \Rightarrow A = 3500 \times 1.194052 \\[1em] \Rightarrow A = ₹ 4179.18\end{aligned}$$

By formula,

C.I. = A - P = ₹ 4179.18 - ₹ 3500 = ₹ 679.18

Hence, C.I. = ₹ 679.18

Calculating C.I. :

For 1st year :

P = ₹ 12000

Rate = 10%

A = $P\Big(1 + \frac{r}{100}\Big)^n$

Substituting values we get :

$$\begin{aligned}A = 12000 \times \Big(1 + \frac{10}{100}\Big)^1 \\[1em] = 12000 \times \frac{110}{100} \\[1em] = ₹ 13200.\end{aligned}$$

For next $\frac{1}{2}$ year :

P = ₹ 13200

Rate = 10%

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

Substituting values we get :

$$\begin{aligned}A = 13200 \times \Big(1 + \frac{10}{200}\Big)^{\frac{1}{2} \times 2} \\[1em] = 13200 \times \frac{210}{200} \\[1em] = ₹ 13860.\end{aligned}$$

C.I. = A - P = ₹ 13860 - ₹ 12000 = ₹ 1860

Calculating S.I. :

P = ₹ 12000

Rate = 10%

Time = $1\frac{1}{2}$

$$\begin{aligned}S.I. = \frac{P \times R \times T}{100} \\[1em] = \frac{12000 \times 10 \times \frac{3}{2}}{100} \\[1em] = ₹ 1800.\end{aligned}$$

Difference between C.I. and S.I. = ₹ 1860 - ₹ 1800 = ₹ 60.

Hence, required difference = ₹ 60.

Calculating S.I. :

P = ₹ 12000

Rate = 10%

Time = $1\frac{1}{2}$

$$\begin{aligned}S.I. = \frac{P \times R \times T}{100} \\[1em] = \frac{12000 \times 10 \times \frac{3}{2}}{100} \\[1em] = ₹ 1800.\end{aligned}$$

Calculating C.I. :

When interest is compounded half-yearly :

$$\begin{aligned}A = P\Big(1 + \frac{r}{2 \times 100}\Big)^{n \times 2} \\[1em] = 12000 \times \Big(1 + \frac{10}{200}\Big)^{\frac{3}{2} \times 2} \\[1em] = 12000 \times \Big(1 + \frac{1}{20}\Big)^3 \\[1em] = 12000 \times \Big(\frac{21}{20}\Big)^3 \\[1em] = 12000 \times \frac{9261}{8000} \\[1em] = \frac{27783}{2} \\[1em] = ₹ 13891.50\end{aligned}$$

C.I. = A - P = ₹ 13891.50 - ₹ 12000 = ₹ 1891.50.

Difference between C.I. and S.I. = ₹ 1891.50 - ₹ 1800 = ₹ 91.50

Hence, required difference = ₹ 91.50

For growth :

Value after n years = Present value $\times \Big(1 + \frac{r}{100}\Big)^n = P \times \Big(1 + \frac{r}{100}\Big)^n.$

Hence, Option 3 is the correct option.

Given,

The population of a town decreases by 10% in a particular year and then increases by 15% in the next year.

Let present population be P.

$\text{Population after 2 years} = P \Big(1 - \frac{10}{100}\Big)\Big(1 + \frac{15}{100}\Big).$

Hence, Option 2 is the correct option.

For depreciation :

Value after n years = Present value $\times \Big(1 - \frac{r}{100}\Big)^n$

So,

Value of machine after 3 years = Present value $\times \Big(1 - \frac{r}{100}\Big)^3$

Hence, Option 2 is the correct option.

By formula,

A = $P\Big(1 + \frac{r}{100}\Big)^n$

Given,

On a certain sum the rate of C.I. is x% per annum for the first two years and y% per annum for the next three years.

$\text{Amount after 5 years} = P\Big(1 + \frac{x}{100}\Big)^2\Big(1 + \frac{y}{100}\Big)^3.$

Hence, Option 3 is the correct option.

According to question :

The cost (₹ x) of a machine increases by 20% in the first two years and then decreases by 25% in the next two years.

$$\begin{aligned}\Rightarrow x \times \Big(1 + \frac{20}{100}\Big)^2 \times \Big(1 - \frac{25}{100}\Big)^2 \\[1em] \Rightarrow x \times \Big(\frac{120}{100}\Big)^2 \times \Big(\frac{75}{100}\Big)^2.\end{aligned}$$

Hence, Option 3 is the correct option.

(i) For depreciation :

Value after n years = Present value $\times \Big(1 - \frac{r}{100}\Big)^n$

The cost of machine depreciates at the beginning of the next year, or we can say that at end of each year.

So, Value at end of 2009 = Value at beginning of 2010.

$$\begin{aligned}\text{Value at beginning of 2010 }= 44000 \times \Big(1 - \frac{12}{100}\Big)^2 \\[1em] = 44000 \times \Big(\frac{88}{100}\Big)^2 \\[1em] = 44000 \times \Big(\frac{22}{25}\Big)^2 \\[1em] = 44000 \times \frac{484}{625} \\[1em] = ₹ 34073.60\end{aligned}$$

Hence, value at the end of 2009 = ₹ 34073.60

(ii) Let value at beginning of 2007 be ₹ x.

After one year it becomes ₹ 44000.

For depreciation :

Value after n years = Present value $\times \Big(1 - \frac{r}{100}\Big)^n$

Substituting values we get :

$$\begin{aligned}\Rightarrow 44000 = x \times \Big(1 - \frac{12}{100}\Big)^1 \\[1em] \Rightarrow 44000 = x \times \Big(\frac{88}{100}\Big) \\[1em] \Rightarrow 44000 = \frac{88x}{100} \\[1em] \Rightarrow x = \frac{44000 \times 100}{88} \\[1em] \Rightarrow x = ₹ 50000.\end{aligned}$$

Hence, value at beginning of 2007 = ₹ 50000.

Let original value be ₹ x.

Given, it decreases by 10% for two years. So,

For depreciation :

Value after n years = Present value $\times \Big(1 - \frac{r}{100}\Big)^n$

$$\begin{aligned}\text{Value after 2 years} = x \times \Big(1 - \frac{10}{100}\Big)^2 \\[1em] = x \times \Big(\frac{90}{100}\Big)^2\\[1em] = x \times \Big(\frac{9}{10}\Big)^2 \\[1em] = x \times \frac{81}{100} \\[1em] = \frac{81x}{100}.\end{aligned}$$

Given,

In the third year value is increased by 10%.

For growth :

Value after n year = Present value $\times \Big(1 + \frac{r}{100}\Big)^n$

$$\begin{aligned}\text{Value after 1 year} = \frac{81x}{100} \times \Big(1 + \frac{10}{100}\Big)^1 \\[1em] = \frac{81x}{100} \times \frac{110}{100} \\[1em] = \frac{891x}{1000}.\end{aligned}$$

Given,

At the end of 3 years value is ₹ 40095.

$$\begin{aligned}\therefore \frac{891x}{1000} = 40095 \\[1em] \Rightarrow x = \frac{40095 \times 1000}{891} \\[1em] \Rightarrow x = ₹ 45000.\end{aligned}$$

Hence, original value of article = ₹ 45000.

Let in n years population reaches from 64000 to 74088.

For growth :

Value after n year = Present value $\times \Big(1 + \frac{r}{100}\Big)^n$

Substituting values we get :

$$\begin{aligned}\Rightarrow 74088 = 64000 \times \Big(1 + \frac{5}{100}\Big)^n \\[1em] \Rightarrow \frac{74088}{64000} = \Big(\frac{105}{100}\Big)^n \\[1em] \Rightarrow \frac{9261}{8000} = \Big(\frac{21}{20}\Big)^n \\[1em] \Rightarrow \Big(\frac{21}{20}\Big)^3 = \Big(\frac{21}{20}\Big)^n \\[1em] \Rightarrow n = 3.\end{aligned}$$

In 3 years the population will reach 74088.

Let population in the beginning of 1998 be x.

Given,

The population of a town decreased by 12% during 1998 and then increased by 8% during 1999.

Population at the end of 1999 = 285120.

$$\begin{aligned}\therefore 285120 = x\Big(1 - \frac{12}{100}\Big)\Big(1 + \frac{8}{100}\Big) \\[1em] \Rightarrow 285120 = x \times \frac{88}{100} \times \frac{108}{100} \\[1em] \Rightarrow 285120 = \frac{9504x}{10000} \\[1em] \Rightarrow x = \frac{285120 \times 10000}{9504} \\[1em] \Rightarrow x = 300000.\end{aligned}$$

Hence, population of town at beginning of 1998 = 300000.

Let original sum of money invested be ₹ x and rate of percent be r%.

By formula,

A = $P\Big(1 + \frac{r}{100}\Big)^n$

Given,

The sum of money, invested at compound interest, amounts to ₹ 16500 in 1 year.

$$\begin{aligned}\Rightarrow A = P\Big(1 + \frac{r}{100}\Big)^n \\[1em] \Rightarrow 16500 = x \times \Big(1 + \frac{r}{100}\Big)^1 \\[1em] \Rightarrow 16500 = x\Big(1 + \frac{r}{100}\Big) ......(1)\end{aligned}$$

The sum of money, invested at compound interest, amounts to ₹ 19965 in 3 years.

$$\begin{aligned}\Rightarrow A = P\Big(1 + \frac{r}{100}\Big)^n \\[1em] \Rightarrow 19965 = x \times \Big(1 + \frac{r}{100}\Big)^3 \\[1em] \Rightarrow 19965 = x\Big(1 + \frac{r}{100}\Big)^3 ......(2)\end{aligned}$$

Dividing equation (2) by (1), we get :

$$\begin{aligned}\Rightarrow \frac{19965}{16500} = \frac{x\Big(1 + \frac{r}{100}\Big)^3}{x\Big(1 + \frac{r}{100}\Big)} \\[1em] \Rightarrow \frac{121}{100} = \Big(1 + \frac{r}{100}\Big)^2 \\[1em] \Rightarrow \Big(\frac{11}{10}\Big)^2 = \Big(1 + \frac{r}{100}\Big)^2 \\[1em] \Rightarrow \Big(1 + \frac{r}{100}\Big) = \frac{11}{10} \\[1em] \Rightarrow \frac{r}{100} = \frac{11}{10} - 1 \\[1em] \Rightarrow \frac{r}{100} = \frac{1}{10} \\[1em] \Rightarrow r = \frac{100}{10} = 10%.\end{aligned}$$

Substituting value of r in equation (1), we get :

$$\begin{aligned}\Rightarrow 16500 = x\Big(1 + \frac{10}{100}\Big) \\[1em] \Rightarrow 16500 = x \times \Big(\frac{110}{100}\Big) \\[1em] \Rightarrow x = \frac{16500 \times 100}{110} \\[1em] \Rightarrow x = ₹ 15000.\end{aligned}$$

Hence, rate percent = 10% and sum invested = ₹ 15000.

Given,

P = ₹ 7500

Time = 2 years

Let rate of interest be r%.

By formula,

S.I. = $\frac{P \times R \times T}{100}$

Substituting values we get :

$$\begin{aligned}S.I. = \frac{7500 \times r \times 2}{100} \\[1em] = 150r.\end{aligned}$$

By formula,

C.I. = A - P

$$\begin{aligned}C.I. = P\Big(1 + \frac{r}{100}\Big)^n - P \\[1em] = 7500 \times \Big(1 + \frac{r}{100}\Big)^2 - 7500 \\[1em] = 7500 \times \Big(\frac{100 + r}{100}\Big)^2 - 7500 \\[1em] = 7500 \times \Big(\frac{10000 + r^2 + 200r}{10000}\Big) - 7500 \\[1em] = \frac{3}{4} \times (10000 + r^2 + 200r) - 7500 \\[1em] = 7500 + \frac{3r^2}{4} + 150r - 7500 \\[1em] = \frac{3r^2}{4} + 150r\end{aligned}$$

The difference between C.I. and S.I. on ₹ 7500 for two years is ₹ 12.

∴ C.I. - S.I. = ₹ 12

$$\begin{aligned}\therefore \frac{3r^2}{4} + 150r - 150r = 12 \\[1em] \Rightarrow \frac{3r^2}{4} = 12 \\[1em] \Rightarrow r^2 = \frac{12 \times 4}{3} \\[1em] \Rightarrow r^2 = 16 \\[1em] \Rightarrow r = \sqrt{16} = 4%.\end{aligned}$$

Hence, rate of interest = 4%.

Let rate of interest be r% and sum of money be ₹ P.

By formula,

A = $P\Big(1 + \frac{r}{100}\Big)^n$

Given,

₹ P becomes three times of itself in 10 years.

$$\begin{aligned}\therefore 3P = P\Big(1 + \frac{r}{100}\Big)^{10} \\[1em] \Rightarrow \frac{3P}{P} = \Big(1 + \frac{r}{100}\Big)^{10} \\[1em] \Rightarrow 3 = \Big(1 + \frac{r}{100}\Big)^{10} .....(1)\end{aligned}$$

Let in n years money will become 27 times.

$$\begin{aligned}\Rightarrow P\Big(1 + \frac{r}{100}\Big)^n = 27P \\[1em] \Rightarrow \Big(1 + \frac{r}{100}\Big)^n = \frac{27P}{P} \\[1em] \Rightarrow \Big(1 + \frac{r}{100}\Big)^n = 27 \\[1em] \Rightarrow \Big(1 + \frac{r}{100}\Big)^n = 3^3 \\[1em]\end{aligned}$$

From equation (1)

$$\begin{aligned}\Rightarrow \Big(1 + \frac{r}{100}\Big)^n = \Big[\Big(1 + \frac{r}{100}\Big)^{10}\Big]^3 \\[1em] \Rightarrow \Big(1 + \frac{r}{100}\Big)^n =\Big(1 + \frac{r}{100}\Big)^{30} \\[1em] \Rightarrow n = 30 \text{ years}.\end{aligned}$$

Hence, in 30 years money will becomes 27 times of itself.

Let sum borrowed be ₹ x.

By formula,

$A = P\Big(1 + \frac{r}{100}\Big)^n$

At the end of 2 years :

$$\begin{aligned}A = x \times \Big(1 + \frac{10}{100}\Big)^2 \\[1em] = x \times \Big(\frac{110}{100}\Big)^2 \\[1em] = x \times \Big(\frac{11}{10}\Big)^2 \\[1em] = \frac{121x}{100}.\end{aligned}$$

Given,

He pays back ₹ 19360 at the end of second year. So,

Principal for third year = ₹ $\Big(\frac{121x}{100} - 19360\Big)$

Amount after 3 year :

$$\begin{aligned}A = \Big(\frac{121x}{100} - 19360\Big) \times \Big(1 + \frac{10}{100}\Big)^1 \\[1em] = \Big(\frac{121x}{100} - 19360\Big) \times \frac{110}{100} \\[1em] = \frac{1331x}{1000} - 21296.\end{aligned}$$

Given,

On giving ₹ 31944 at the end of the third year he clears the debt.

$$\begin{aligned}\therefore \frac{1331x}{1000} - 21296 = 31944 \\[1em] \Rightarrow \frac{1331x}{1000} = 31944 + 21296 \\[1em] \Rightarrow \frac{1331x}{1000} = 53240 \\[1em] \Rightarrow x = \frac{53240 \times 1000}{1331} \\[1em] \Rightarrow x = ₹ 40000.\end{aligned}$$

Hence, sum borrowed = ₹ 40000.

Let sum of money lent out be ₹ x.

Calculating C.I. payable half-yearly :

P = ₹ x

Rate of interest = 10%

Time = 1 year

C.I. = A - P

$$\begin{aligned}C.I. = P\Big(1 + \frac{r}{2 \times 100}\Big)^{n \times 2} - P \\[1em] = x \times \Big(1 + \frac{10}{200}\Big)^{1 \times 2} - x \\[1em] = x \times \Big(\frac{210}{200}\Big)^2 - x \\[1em] = \frac{441x}{400} - x \\[1em] = \frac{441x - 400x}{400} \\[1em] = ₹ \frac{41x}{400}.\end{aligned}$$

Calculating S.I. :

$$\begin{aligned}S.I. = \frac{P \times R \times T}{100} \\[1em] = \frac{x \times 10 \times 1}{100} \\[1em] = ₹ \frac{x}{10}.\end{aligned}$$

Given,

Difference between compound interest for a year payable half-yearly and simple interest on ₹ x lent out at 10% for a year is ₹ 15.

$$\begin{aligned}\therefore \frac{41x}{400} - \frac{x}{10} = 15 \\[1em] \Rightarrow \frac{41x - 40x}{400} = 15 \\[1em] \Rightarrow \frac{x}{400} = 15 \\[1em] \Rightarrow x = 15 \times 400 = ₹ 6000.\end{aligned}$$

Hence, sum of money lent out = ₹ 6000.

Let Pramod invest ₹ x and Rohit invest ₹ y.

By formula,

A = $P\Big(1 + \frac{r}{100}\Big)^n$

For Pramod :

P = ₹ x

r = 5%

n = 9 years (As it will take 9 years for Pramod to reach 25 years of age)

Substituting values we get :

$$\begin{aligned}\text{Amount received by Pramod} = x\Big(1 + \frac{5}{100}\Big)^9 \\[1em] = x \times \Big(\frac{105}{100}\Big)^9 \\[1em] = x \times \Big(\frac{21}{20}\Big)^9.\end{aligned}$$

For Rohit :

P = ₹ y

r = 5%

n = 7 years (As it will take 7 years for Pramod to reach 25 years of age)

Substituting values we get :

$$\begin{aligned}\text{Amount received by Rohit} = y\Big(1 + \frac{5}{100}\Big)^7 \\[1em] = y \times \Big(\frac{105}{100}\Big)^7 \\[1em] = y \times \Big(\frac{21}{20}\Big)^7.\end{aligned}$$

Since,

Amount received by both are equal.

$$\begin{aligned}\Rightarrow x \times \Big(\frac{21}{20}\Big)^9 = y \times \Big(\frac{21}{20}\Big)^7 \\[1em] \Rightarrow \frac{x}{y} = \frac{\Big(\frac{21}{20}\Big)^7}{\Big(\frac{21}{20}\Big)^9} \\[1em] \Rightarrow \frac{x}{y} = \frac{1}{\Big(\frac{21}{20}\Big)^2} \\[1em] \Rightarrow \frac{x}{y} = \frac{20^2}{21^2} \\[1em] \Rightarrow \frac{x}{y} = \frac{400}{441}.\end{aligned}$$

Hence, ratio in which sum must be invested = 400 : 441.

Given, P = ₹ 1,000

R = 20%

n = 2 years

Using the formula, A = $P\Big(1 + \frac{R}{100}\Big)^n$

Substituting the values,

$$\begin{aligned}A = 1,000 \Big(1 + \frac{20}{100}\Big)^2\\[1em] = 1,000 \Big(1 + \frac{2}{10}\Big)^2\\[1em] = 1,000 \times \Big(\frac{12}{10}\Big)^2\\[1em] = 1,000 \times \frac{144}{100}\\[1em] = 1,440.\end{aligned}$$

Hence, option 4 is the correct option.

For S.I. :

P = ₹ 100

R = 10%

T = 2 years

$$\begin{aligned}I = \frac{P \times R \times T}{100} \\[1em] = \frac{100 \times 10 \times 2}{100} \\[1em] = \frac{2000}{100}\\[1em] = 20.\end{aligned}$$

For C.I. :

For 1st year :

P = ₹ 100

T = 1 year

R = 10%

$$\begin{aligned}I = \frac{P \times R \times T}{100}\\[1em] = \frac{100 \times 10 \times 1}{100}\\[1em] = \frac{1000}{100}\\[1em] = 10.\end{aligned}$$

Amount = P + I = ₹ 100 + ₹ 10 = ₹ 110

For 2nd year :

P = ₹ 110

R = 10%

T = 1 year

$$\begin{aligned}I = \frac{P \times R \times T}{100} \\[1em] = \frac{110 \times 10 \times 1}{100} \\[1em] = \frac{1100}{100} \\[1em] = 11.\end{aligned}$$

Amount = P + I = ₹ 110 + ₹ 11 = ₹ 121.

C.I. = Final amount - Initial principal = ₹ 121 - ₹ 100 = ₹ 21.

Difference between C.I. and S.I. = ₹ 21 - ₹ 20 = ₹ 1.

Hence, option 1 is the correct option.

Given, the principal amount = ₹ P.

The time period = $3\frac{1}{2} = \frac{7}{2}$ years.

The annual interest rate = r% .

Interest is compounded half-yearly.

The formula for C.I. when compounded half-yearly is I

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

Substituting the values, we get :

$$\begin{aligned}I = P \Big(1 + \frac{r}{2 \times 100}\Big)^{\frac{7}{2} \times 2} - P\\[1em] = P \Big(1 + \frac{r}{2 \times 100}\Big)^7 - P\end{aligned}$$

Hence, option 4 is the correct option.

Given, the principal amount = ₹ P, rate of interest r%.

For complete 3 years it will be calculated normally but for half year it will be calculated by taking 1 half year and half rate of interest.

A = $P\Big(1 + \frac{r}{100}\Big)^3 \times \Big(1 + \frac{r}{2 \times 100}\Big)^1$

C.I. = A - P

= $P\Big(1 + \frac{r}{100}\Big)^3 \times \Big(1 + \frac{r}{2 \times 100}\Big)^1 - P$

Hence, option 2 is the correct option.

Interest for a particular year = Amount in that year - Amount in previous year

Interest for 7th year = Amount in 7 years - Amount in 6 years

= $P\Big(1 + \frac{r}{100}\Big)^7 - P\Big(1 + \frac{r}{100}\Big)^6$

Thus, statement 1 is true and statement 2 is false.

Hence, option 3 is correct option.

The general formula for population growth with a fixed annual percentage increase is:

P_t = P_o $\Big(1 + \frac{r}{100}\Big)^t$

where, P_t = Population after t years

P₀ = Initial population

r = Annual growth rate (in percentage)

t = Time in years

Given, the population in 2024 is x, and the annual growth rate is 10%. To find the population in 2021, we need to go back 3 years (from 2024 to 2021).

Rearranging the formula to solve for P₀ = $\frac{P_t}{\Big(1 + \frac{r}{100}\Big)^t}$

⇒ P₀ = $\frac{x}{\Big(1 + \frac{10}{100}\Big)^3}$

So, statement 1 is false.

If the population increases from year 2021 to year 2024 at the rate of 10%.

Let P₀ be the population in 2021 and P_t be the population in 2024.

⇒ P_t = P₀ $\Big(1 + \frac{10}{100}\Big)^3$

⇒ P_t = P₀ (1 + 0.1)³

⇒ P_t = P₀ x 1.1³

⇒ P_t = 1.331P₀

The percentage decrease = $\frac{P_t - P_o}{P_t} \times 100$

$$\begin{aligned}= \frac{1.331P_o - P_o}{1.331P_o} \times 100\\[1em] = \frac{0.331P_o}{1.331P_o} \times 100\\[1em] = \frac{331}{1331} \times 100\\[1em] ≈ 24.87%\end{aligned}$$

So, statement 2 is false.

∴ Both the statements are false.

Hence, option 2 is correct option.

Let P be the principal amount, r% be rate of interest and t be the time.

By formula, A = P $\Big(1 + \frac{r}{100}\Big)^t$

A certain sum of money P let out at r% C.I. increased for first 5 years and then decreased for next five years at the same rate, then

A₁ = P $\Big(1 + \frac{r}{100}\Big)^5 \times \Big(1 - \frac{r}{100}\Big)^5$

A certain sum of money P let out at r% C.I. decreased on the same sum at the same rate (r%) during the first five years and then increase further next five years at the same rate, then

A₂ = P $\Big(1 - \frac{r}{100}\Big)^5 \times \Big(1 + \frac{r}{100}\Big)^5$

By rules of multiplication :

Since, $P \Big(1 + \frac{r}{100}\Big)^5 \times \Big(1 - \frac{r}{100}\Big)^5 = P \Big(1 - \frac{r}{100}\Big)^5 \times \Big(1 + \frac{r}{100}\Big)^5$

So, reason (R) is true.

Thus, amount will be same.

So, assertion (A) is true and reason (R) clearly explains assertion (A).

∴ Both A and R are true, and R is the correct reason for A.

Hence, option 3 is correct option.

Let P be the principal amount, r% be rate of interest and t be the time.

By formula, A = P $\Big(1 + \frac{r}{100}\Big)^t$

A = P $\Big(1 + \frac{10}{100}\Big)^2 = P \Big(1 + \frac{1}{10}\Big)^2 = (1 + 0.1)^2P = 1.1^2$ P = 1.21P

∴ Both A and R are true, and R is the correct reason for A.

Hence, option 3 is correct option.

Let sum of money be ₹ x.

Given,

Simple interest on sum of money for 2 years at 4% growth rate is ₹ 450.

By formula,

$$\begin{aligned}S.I. = \frac{P \times R \times T}{100} \\[1em] \Rightarrow 450 = \frac{x \times 4 \times 2}{100} \\[1em] \Rightarrow x = \frac{450 \times 100}{4 \times 2} \\[1em] \Rightarrow x = ₹ 5625.\end{aligned}$$

For C.I. :

P = ₹ 5625

n = 1 year

r = 4% compounded half-yearly

$$\begin{aligned}A = P\Big(1 + \frac{r}{2 \times 100}\Big)^{n \times 2} \\[1em] = 5625 \times \Big(1 + \frac{4}{200}\Big)^{1 \times 2} \\[1em] = 5625 \times \Big(\frac{204}{200}\Big)^2 \\[1em] = 5625 \times \Big(\frac{51}{50}\Big)^2 \\[1em] = 5625 \times \frac{2601}{2500} \\[1em] = ₹ 5852.25\end{aligned}$$

C.I. = A - P = ₹ 5852.25 - ₹ 5625 = ₹ 227.25

Hence, compound interest = ₹ 227.25

Given,

P = ₹ 10800

T = 2.5 years

r = 10%

For first 2 years :

$$\begin{aligned}A = P\Big(1 + \frac{r}{100}\Big)^n \\[1em] = 10800 \times \Big(1 + \frac{10}{100}\Big)^2 \\[1em] = 10800 \times \Big(\frac{110}{100}\Big)^2\\[1em] = 10800 \times \Big(\frac{11}{10}\Big)^2 \\[1em] = 10800 \times \frac{121}{100} \\[1em] = ₹ 13068.\end{aligned}$$

For next $\frac{1}{2}$ year :

P = ₹ 13068

$$\begin{aligned}A = P\Big(1 + \frac{r}{2 \times 100}\Big)^{n \times 2} \\[1em] = 13068 \times \Big(1 + \frac{10}{2 \times 100}\Big)^{\frac{1}{2} \times 2} \\[1em] = 13068 \times \Big(1 + \frac{1}{20}\Big)^1 \\[1em] = 13068 \times \frac{21}{20} \\[1em] = ₹ 13721.\end{aligned}$$

By formula,

C.I. = A - P = ₹ 13721 - ₹ 10800 = ₹ 2921.

Hence, compound interest = ₹ 2921.

(i) In depreciation :

Value after n years = Present value $\times \Big(1 - \frac{r}{100}\Big)^n$

$$\begin{aligned}\text{Value of machine after 2 years} = 97200 \times \Big(1 - \frac{10}{100}\Big)^2 \\[1em] = 97200 \times \Big(\frac{90}{100}\Big)^2 \\[1em] = 97200 \times \Big(\frac{9}{10}\Big)^2 \\[1em] = 97200 \times \frac{81}{100} \\[1em] = 972 \times 81 \\[1em] = ₹ 78732.\end{aligned}$$

Hence, value of machine after 2 years = ₹ 78732.

(ii) Let value of machine when it was purchased be ₹ x and its depreciate to ₹ 97200 in two years.

$$\begin{aligned}\therefore 97200 = x \times \Big(1 - \frac{10}{100}\Big)^2 \\[1em] \Rightarrow 97200 = x \times \Big(\frac{90}{100}\Big)^2 \\[1em] \Rightarrow 97200 = x \times \Big(\frac{9}{10}\Big)^2 \\[1em] \Rightarrow 97200 = x \times \frac{81}{100} \\[1em] \Rightarrow x = \frac{97200 \times 100}{81} \\[1em] \Rightarrow x = 120000.\end{aligned}$$

Hence, machine's value when it was purchased = ₹ 120000.

Let money lent by both be ₹ x.

For Anuj :

S.I. = $\frac{P\times R \times T}{100} = \frac{x \times 8 \times 2}{100} = \frac{4x}{25}$.

For Rajesh :

C.I. = A - P

$$\begin{aligned}C.I. = P\Big(1 + \frac{r}{100}\Big)^n - P \\[1em] = x \times \Big(1 + \frac{8}{100}\Big)^2 - x \\[1em] = x \times \Big(\frac{108}{100}\Big)^2 - x \\[1em] = x \times \Big(\frac{27}{25}\Big)^2 - x \\[1em] = x \times \frac{729}{625} - x \\[1em] = \frac{729x}{625} - x \\[1em] = \frac{729x - 625x}{625} \\[1em] = \frac{104x}{625}.\end{aligned}$$

Given,

Rajesh received ₹ 64 more than Anuj. So, it means Rajesh received ₹ 64 more than Anuj as interest.

$$\begin{aligned}\therefore \frac{104x}{625} - \frac{4x}{25} = 64 \\[1em] \Rightarrow \frac{104x - 100x}{625} = 64 \\[1em] \Rightarrow \frac{4x}{625} = 64 \\[1em] \Rightarrow x = \frac{625 \times 64}{4} \\[1em] \Rightarrow x = ₹ 10000.\end{aligned}$$

Calculating S.I. and C.I. :

$$\begin{aligned}S.I. = \frac{4x}{25} = \frac{4 \times 10000}{25} = ₹ 1600. \\[1em] C.I. = \frac{104x}{625} = \frac{104 \times 10000}{625} = ₹ 1664.\end{aligned}$$

Hence, sum of money lent = ₹ 10000 and interest received by Anuj = ₹ 1600 and by Rajesh = ₹ 1664.

For S.I. :

P = ₹ 4715

R = 5%

T = 5 years

S.I. = $\frac{P \times R \times T}{100} = \frac{4715 \times 5 \times 5}{100}$ = ₹ 1178.75

Given,

C.I. is four times the S.I.

∴ C.I. = 4 × 1178.75 = ₹ 4715

For C.I. :

Let P = ₹ x, n = 2 years, r = 5%

By formula,

C.I. = A - P

$$\begin{aligned}C.I. = P\Big(1 + \frac{r}{100}\Big)^n - P \\[1em] \Rightarrow 4715 = x \times \Big(1 + \frac{5}{100}\Big)^2 - x \\[1em] \Rightarrow 4715 = x \times \Big(\frac{105}{100}\Big)^2 - x \\[1em] \Rightarrow 4715 = x \times \Big(\frac{21}{20}\Big)^2 - x \\[1em] \Rightarrow 4715 = x \times \frac{441}{400} - x \\[1em] \Rightarrow 4715 = \frac{441x}{400} - x \\[1em] \Rightarrow 4715 = \frac{441x - 400x}{400} \\[1em] \Rightarrow 4715 = \frac{41x}{400} \\[1em] \Rightarrow x - \frac{4715 \times 400}{41} \\[1em] \Rightarrow x = ₹ 46000\end{aligned}$$

Hence, sum of money = ₹ 46000.

Let sum invested be ₹ x.

Amount after 1st year :

$$\begin{aligned}A = x \times \Big(1 + \frac{10}{100}\Big) \\[1em] = x \times \frac{110}{100} \\[1em] = \frac{11x}{10}.\end{aligned}$$

Amount after 2nd year :

$$\begin{aligned}A = x \times \Big(1 + \frac{10}{100}\Big)\Big(1 + \frac{15}{100}\Big) \\[1em] = x \times \frac{110}{100} \times \frac{115}{100} \\[1em] = \frac{253x}{200}.\end{aligned}$$

C.I. for 2nd year = Amount after 2 years - Amount after 1 year = $\frac{253x}{200} - \frac{11x}{10}$

Given,

Compound interest for the second year amounted to ₹ 4950.

$$\begin{aligned}\therefore \frac{253x}{200} - \frac{11x}{10} = 4950 \\[1em] \Rightarrow \frac{253x - 220x}{200} = 4950 \\[1em] \Rightarrow \frac{33x}{200} = 4950 \\[1em] \Rightarrow x = \frac{4950 \times 200}{33} \\[1em] \Rightarrow x = ₹ 30000.\end{aligned}$$

Hence, sum invested = ₹ 30000.

Let sum of money invested be ₹ x.

When interest is compounded half-yearly :

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

For first $\frac{1}{2}$ year :

$$\begin{aligned}A = x \times \Big(1 + \frac{10}{2 \times 100}\Big)^{\frac{1}{2} \times 2} \\[1em] = x \times \Big(1 + \frac{1}{20}\Big) \\[1em] = x \times \frac{21}{20} \\[1em] = \frac{21x}{20}.\end{aligned}$$

For first 1 year :

$$\begin{aligned}A = x \times \Big(1 + \frac{10}{2 \times 100}\Big)^{1 \times 2} \\[1em] = x \times \Big(1 + \frac{1}{20}\Big)^2 \\[1em] = x \times \Big(\frac{21}{20}\Big)^2 \\[1em] = \frac{441x}{400}.\end{aligned}$$

Given,

Difference of amounts at end of 6 months and 12 months is ₹ 189.

$$\begin{aligned}\Rightarrow \frac{441x}{400} - \frac{21x}{20} = 189 \\[1em] \Rightarrow \frac{441x - 420x}{400} = 189 \\[1em] \Rightarrow \frac{21x}{400} = 189 \\[1em] \Rightarrow x = \frac{189 \times 400}{21} \\[1em] \Rightarrow x = ₹ 3600.\end{aligned}$$

Hence, sum invested = ₹ 3600.

Given,

P = ₹ 86000

Rate of interest = 5%

Time = 2 years

Calculating S.I. :

S.I. = $\frac{P \times R \times T}{100} = \frac{86000 \times 5 \times 2}{100}$ = ₹ 8600.

Calculating C.I. :

C.I. = A - P

$$\begin{aligned}C.I. = P\Big(1 + \frac{r}{100}\Big)^n - P \\[1em] = 86000 \times \Big(1 + \frac{5}{100}\Big)^2 - 86000 \\[1em] = 86000 \times \Big(\frac{105}{100}\Big)^2 - 86000 \\[1em] = 86000 \times \Big(\frac{21}{20}\Big)^2 - 86000 \\[1em] = 86000 \times \frac{441}{400} - 86000 \\[1em] = 94815 - 86000 \\[1em] = ₹ 8815.\end{aligned}$$

Rohit's profit = Interest received by him - Interest paid by him

= ₹ 8815 - ₹ 8600 = ₹ 215.

Hence, Rohit's profit = ₹ 215.

Let sum of money be ₹ x.

For S.I. :

P = ₹ x

Rate = 5%

Time = 3 years

S.I. = $\frac{P \times R \times T}{100} = \frac{x \times 5 \times 3}{100} = \frac{3x}{20}$.

Given,

S.I. = ₹ 1200

$$\begin{aligned}\therefore \frac{3x}{20} = 1200 \\[1em] \Rightarrow x = \frac{1200 \times 20}{3} = ₹ 8000.\end{aligned}$$

When rate is compounded annually:

$$\begin{aligned}A = P\Big(1 + \frac{r}{100}\Big)^n \\[1em] = x \times \Big(1 + \frac{5}{100}\Big)^2 \\[1em] = 8000 \times \Big(\frac{105}{100}\Big)^2 \\[1em] = 8000 \times \Big(\frac{21}{20}\Big)^2 \\[1em] = ₹ 8820.\end{aligned}$$

C.I. = A - P = ₹ 8820 - ₹ 8000 = ₹ 820.

Hence, amount due = ₹ 8820 and C.I. = ₹ 820.

(a) Let rate percent be r%.

Given,

P = ₹ 6000

n = 1 year

A = ₹ 6720

By formula,

A = $P\Big(1 + \frac{r}{100}\Big)^n$

Substituting values we get :

$$\begin{aligned}\Rightarrow 6720 = 6000\Big(1 + \frac{r}{100}\Big)^1 \\[1em] \Rightarrow \frac{6720}{6000} = 1 + \frac{r}{100} \\[1em] \Rightarrow \frac{6720}{6000} - 1 = \frac{r}{100} \\[1em] \Rightarrow \frac{6720 - 6000}{6000} = \frac{r}{100} \\[1em] \Rightarrow \frac{720}{6000} = \frac{r}{100} \\[1em] \Rightarrow r = \frac{720 \times 100}{6000} \\[1em] \Rightarrow r = 12%.\end{aligned}$$

Hence, rate percent = 12%.

(b) By formula,

A = $P\Big(1 + \frac{r}{100}\Big)^n$

$$\begin{aligned}A = 6000 \times \Big(1 + \frac{12}{100}\Big)^2 \\[1em] = 6000 \times \Big(\frac{112}{100}\Big)^2 \\[1em] = 6000 \times \Big(\frac{28}{25}\Big)^2 \\[1em] = 6000 \times \frac{784}{625} \\[1em] = ₹ 7526.40\end{aligned}$$

Hence, amount at the end of 2 years = ₹ 7526.40

Let rate of interest be r% and sum of money be ₹ P.

By formula, A = $P\Big(1 + \frac{r}{100}\Big)^t$

Given,

₹ P becomes three times of itself in 8 years.

$$\begin{aligned}\therefore 3P = P\Big(1 + \frac{r}{100}\Big)^8 \\[1em] \Rightarrow \frac{3P}{P} = \Big(1 + \frac{r}{100}\Big)^8 \\[1em] \Rightarrow 3 = \Big(1 + \frac{r}{100}\Big)^8 .....................(1)\end{aligned}$$

Let in n years money becomes 81 times.

$$\begin{aligned}\Rightarrow P\Big(1 + \frac{r}{100}\Big)^n = 81P\\[1em] \Rightarrow \Big(1 + \frac{r}{100}\Big)^n = \frac{81P}{P}\\[1em] \Rightarrow \Big(1 + \frac{r}{100}\Big)^n = 81\\[1em] \Rightarrow \Big(1 + \frac{r}{100}\Big)^n = 3^4\\[1em] \Rightarrow \Big(1 + \frac{r}{100}\Big)^n = \Big[\Big(1 + \frac{r}{100}\Big)^8\Big]^4 \text{ [From (1)]}\\[1em] \Rightarrow \Big(1 + \frac{r}{100}\Big)^n = \Big(1 + \frac{r}{100}\Big)^{32}\\[1em]\end{aligned}$$

⇒ n = 32.

Hence, in 32 years money will becomes 81 times of itself.

Let rate of interest be r% and sum of money be ₹ P.

By formula, A = $P\Big(1 + \frac{r}{100}\Big)^t$

Given,

₹ P becomes twice of itself in 8 years.

$$\begin{aligned}\therefore 2P = P\Big(1 + \frac{r}{100}\Big)^8 \\[1em] \Rightarrow \frac{2P}{P} = \Big(1 + \frac{r}{100}\Big)^8 \\[1em] \Rightarrow 2 = \Big(1 + \frac{r}{100}\Big)^8 .....................(1)\end{aligned}$$

Let the money become four times in n years.

$$\begin{aligned}\Rightarrow P\Big(1 + \frac{r}{100}\Big)^n = 4P\\[1em] \Rightarrow \Big(1 + \frac{r}{100}\Big)^n = \frac{4P}{P}\\[1em] \Rightarrow \Big(1 + \frac{r}{100}\Big)^n = 4\\[1em] \Rightarrow \Big(1 + \frac{r}{100}\Big)^n = 2^2\\[1em] \Rightarrow \Big(1 + \frac{r}{100}\Big)^n = \Big[\Big(1 + \frac{r}{100}\Big)^8\Big]^2 \text{ [From (1)]}\\[1em] \Rightarrow \Big(1 + \frac{r}{100}\Big)^n = \Big(1 + \frac{r}{100}\Big)^{16}\\[1em]\end{aligned}$$

⇒ n = 16.

Hence, in 16 years money will becomes 4 times of itself.

Given:

Total money = ₹1,68,200

Rate = 5% p.a (compound interest)

Younger son = 16 years → will get money after 5 years

Elder son = 18 years → will get money after 3 years

Let the amount to be distributed to younger son be '₹ x' and to elder son to be '₹ y'

∴ x + y = 168200.......(1)

By formula, A = $P\Big(1 + \frac{r}{100}\Big)^n$

Amount after 5 years for younger son (at the age of 21),

A = $x\Big(1 + \frac{5}{100}\Big)^5$

= x(1 + 0.05)⁵

Amount after 3 years for elder son (at the age of 21),

A = $y\Big(1 + \frac{5}{100}\Big)^3$

= y(1 + 0.05)³

Since both will have same amount at age 21:

∴ x(1 + 0.05)⁵ = y (1 + 0.05)³

⇒ x(1.05)² = y

⇒ x × 1.1025 = y.....(2)

Substituting the value of y from equation (2) in equation (1) :

⇒ x + 1.1025x = 168200

⇒ 2.1025x = 168200

⇒ x = $\frac{168200}{2.1025}$

⇒ x = 80,000.

Substituting the value of x in equation (1),

⇒ 80,000 + y = 1,68,200

⇒ y = 1,68,200 - 80,000

⇒ y = 88,200.

Hence, sum should be divided as ₹ 80,000 and ₹ 88,200.

Given,

Principal (P) = ₹ 4,000

Rate (r) = 6% per year

Time (n) = 1 year

Formula for compound interest (when compounded N times per year) :

A = $P\Big(1 + \frac{r}{100 × N}\Big)$^n×N

(a) Quarterly means 4 times in a year.

So, N = 4

By formula,

$$\begin{aligned}\Rightarrow A = P\Big(1 + \frac{r}{100 × N}\Big)^{n×N} \\[1em] \Rightarrow A = 4000\Big(1 + \frac{6}{100 × 4}\Big)^{1×4} \\[1em] \Rightarrow A = 4000\Big(1 + \frac{6}{400}\Big)^4\\[1em] \Rightarrow A = 4000 (1 + 0.015)^4 \\[1em] \Rightarrow A = 4000 (1.015)^4 \\[1em] \Rightarrow A = 4000 × 1.06136 \\[1em] \Rightarrow A \approx 4245.45\end{aligned}$$

Hence, amount = ₹ 4,245.45.

(b) From question:

Compounded every 4 months = ₹ 4,244.83.

Compounded half-yearly = ₹ 4,243.60

Compounded quarterly = ₹ 4,245.45

∴ As the number of times interest is compounded increases, the amount also increases.

Hence, when the number of times the interest is compounded in one year is more, the C.I accrued is also more.

(i) Given,

Units collected by IRCS in 2024-25 = 30,000

Increase rate = 15% per year

From 2024-25 to 2026-27 = 2 years

By formula, A = $P\Big(1 + \frac{r}{100}\Big)$ⁿ

$$\begin{aligned}\Rightarrow A = 30000 \Big(1 + \frac{15}{100}\Big)^2 \\[1em] \Rightarrow A = 30000(1 + 0.15)^2 \\[1em] \Rightarrow A = 30000(1.15)^2 \\[1em] \Rightarrow A = 30000 × 1.3225 \\[1em] \Rightarrow A = 39,675 \text { units}\end{aligned}$$

Hence, in 2026-27 number of units collected = 39,675 units.

(ii) Given,

Initial (P) = 30,000

Final (A) = 39,930

Time (n) = 3 years

By formula, A = $P\Big(1 + \frac{r}{100}\Big)$ⁿ

$$\begin{aligned}\Rightarrow 39930 = 30000 \Big(1 + \frac{r}{100}\Big)^3 \\[1em] \Rightarrow \frac{39930}{30000} = \Big(1 + \frac{r}{100}\Big)^3 \\[1em] \Rightarrow 1.331 = \Big(1 + \frac{r}{100}\Big)^3 \\[1em] \Rightarrow 1 + \frac{r}{100} = \sqrt[3]{1.331} \\[1em] \Rightarrow 1 + \frac{r}{100} = 1.1 \\[1em] \Rightarrow \frac{r}{100} = 0.1 \\[1em] \Rightarrow r = 100 × 0.1 \\[1em] \Rightarrow r = 10%.\end{aligned}$$

Hence, rate of increase = 10%.