Compound Interest

For first year :

P = ₹ 25,000

T = 1 year

R = 8%

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

$= \frac{25000 \times 8 \times 1}{100}$ = ₹ 2,000.

Amount = P + I = ₹ 25,000 + ₹ 2,000 = ₹ 27,000.

For second year :

P = ₹ 27,000

T = 1 year

R = 8%

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

$= \frac{27000 \times 8 \times 1}{100}$ = ₹ 2160.

Amount = P + I = ₹ 27,000 + ₹ 2,160 = ₹ 29,160.

Compound interest = Final amount - Initial principal

= ₹ 29,160 - ₹ 25,000 = ₹ 4,160.

Hence, compound interest = ₹ 4,160 and amount = ₹ 29,160.

For Rohit,

P = ₹ 62,500

T = 2 year

R = 10% per annum simple interest

Interest Rohit pays to Arun:

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

$= \frac{62500 \times 10 \times 2}{100}$ = ₹ 12,500.

For Kunal,

For first year :

P = ₹ 62,500

T = 1 year

R = 10% per annum compounded annually

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

$= \frac{62500 \times 10 \times 1}{100}$ = ₹ 6,250.

Amount = P + I = ₹ 62,500 + ₹ 6,250 = ₹ 68,750.

For second year :

P = ₹ 68,750

T = 1 year

R = 10% per annum compounded annually

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

$= \frac{68750 \times 10 \times 1}{100}$ = ₹ 6,875.

Amount = P + I = ₹ 68,750 + ₹ 6,875 = ₹ 75,625.

Compound interest = Final amount - Initial principal

= ₹ 75,625 - ₹ 62,500 = ₹ 13,125.

∴ Interest Kunal pays to Rohit = ₹ 13,125

Rohit's profit = Compound interest received from Kunal - Simple interest paid to Arun = ₹ 13,125 - ₹ 12,500 = ₹ 625.

Hence, Rohit's profit in the transaction at the end of two years = ₹ 625.

(i) Given,

P = ₹ 10,000

T = 3 year

Amount at the end of first year = ₹ 11,200

Interest in the first year = Amount - Principal

= ₹ 11,200 - ₹ 10,000 = ₹ 1,200.

So, for 1 year interest equals to ₹ 1,200 on ₹ 10,000. Let rate of interest be R%. Substituting values we get :

$$\begin{aligned}\Rightarrow I = \frac{P \times R \times T}{100} \\[1em] \Rightarrow 1200 = \frac{10000 \times R \times 1}{100} \\[1em] \Rightarrow 1200 = 100 \times R \\[1em] \Rightarrow R = \frac{1200}{100} \\[1em] \Rightarrow R = 12%.\end{aligned}$$

Hence, the rate of interest per annum = 12% p.a.

(ii) Given,

For second year :

P = ₹ 11,200

T = 1 year

R = 12%

Interest accrued in the second year,

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

$= \frac{11200 \times 12 \times 1}{100}$

= ₹ 1,344.

Hence, the interest accrued in the second year = ₹ 1,344.

(iii) For third year,

P = ₹ 11,200 + ₹ 1,344 = ₹ 12,544

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

$= \frac{12544 \times 12 \times 1}{100}$

= ₹ 1,505.28

Amount at the end of the third year = P + I = ₹ 12,544 + ₹ 1,505.28 = ₹ 14,049.28

Hence, the amount at the end of the third year = ₹ 14,049.28.

For first year :

P = ₹ 22,500

T = 1 year

R = 10 %

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

$= \frac{22500 \times 10 \times 1}{100}$ = ₹ 2,250.

Amount = P + I = ₹ 22,500 + ₹ 2,250 = ₹ 24,750.

Amount payed at end of first year = ₹ 11,250.

Amount left at beginning of second year = ₹ 24,750 - ₹ 11,250 = ₹ 13,500.

For second year :

P = ₹ 13,500

R = 10%

T = 1 year

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

$= \frac{13500 \times 10 \times 1}{100}$ = ₹ 1,350.

Amount = P + I = ₹ 13,500 + ₹ 1,350 = ₹ 14,850.

Amount payed at end of second year = ₹ 12,550.

Amount left at beginning of third year = ₹ 14,850 - ₹ 12,550 = ₹ 2,300

For third year :

P = ₹ 2,300

R = 10%

T = 1 year

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

$= \frac{2300 \times 10 \times 1}{100}$ = ₹ 230.

Amount due at the end of third year = P + I = ₹ 2,300 + ₹ 230 = ₹ 2,530.

Hence, the amount outstanding at the end of the third year = ₹ 2,530.

For first year :

P = ₹ 15,000

T = 1 year

R = 12%

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

$= \frac{15000 \times 12 \times 1}{100}$ = ₹ 1,800.

Amount = P + I = ₹ 15,000 + ₹ 1,800 = ₹ 16,800.

Amount payed at end of first year = ₹ 4,400.

Amount left at beginning of second year = ₹ 16,800 - ₹ 4,400 = ₹ 12,400.

For second year :

P = ₹ 12,400

R = 12%

T = 1 year

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

$= \frac{12400 \times 12 \times 1}{100}$ = ₹ 1,488.

Amount = P + I = ₹ 12,400 + ₹ 1,488 = ₹ 13,888.

Amount payed at end of second year = ₹ 4,400.

Amount left at beginning of third year = ₹ 13,888 - ₹ 4,400 = ₹ 9,488.

Hence, amount left at beginning of third year = ₹ 9,488.

For first year :

P = ₹ 16,000

T = 1 year

R = 10 %

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

$= \frac{16000 \times 10 \times 1}{100}$ = ₹ 1,600.

Amount = P + I = ₹ 16,000 + ₹ 1,600 = ₹ 17,600.

Amount payed at end of first year = ₹ 5,600.

Amount left at beginning of second year = ₹ 17,600 - ₹ 5,600 = ₹ 12,000.

For second year :

P = ₹ 12,000

R = 12%

T = 1 year

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

$= \frac{12000 \times 12 \times 1}{100}$ = ₹ 1,440.

Amount = P + I = ₹ 12,000 + ₹ 1,440 = ₹ 13,440.

Hence, amount outstanding at end of second year = ₹ 13,440.

For first year :

P = ₹ 30,000

T = 1 year

R = 10%

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

$= \frac{30000 \times 10 \times 1}{100}$ = ₹ 3,000.

Amount = P + I = ₹ 30,000 + ₹ 3,000 = ₹ 33,000.

For second year :

P = ₹ 33,000

T = 1 year

R = 10%

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

$= \frac{33000 \times 10 \times 1}{100}$ = ₹ 3,300

Amount = P + I = ₹ 33,000 + ₹ 3,300 = ₹ 36,300

For next 4 months :

P = ₹ 36,300

T = 4 months = $\frac{4}{12}$ year = $\frac{1}{3}$ year

R = 10%

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

$= \frac{36300 \times 10 \times \frac{1}{3}}{100}$

$= \frac{363000}{300}$ = ₹ 1,210.

Amount = P + I = ₹ 36,300 + ₹ 1,210 = ₹ 37,510.

Hence, final amount = ₹ 37,510.

For first year :

P = ₹ 31,250

T = 1 year

R = 8%

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

$= \frac{31250 \times 8 \times 1}{100}$ = ₹ 2,500.

Amount = P + I = ₹ 31,250 + ₹ 2,500 = ₹ 33,750.

For second year :

P = ₹ 33,750

T = 1 year

R = 8%

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

$= \frac{33750 \times 8 \times 1}{100}$ = ₹ 2,700.

Amount = P + I = ₹ 33,750 + ₹ 2,700 = ₹ 36,450.

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

P = ₹ 36,450

T = $\frac{1}{2}$ year

R = 8%

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

$= \frac{36450 \times 8 \times \frac{1}{2}}{100}$

$= \frac{291600}{200}$ = ₹ 1,458.

Amount = P + I = ₹ 36,450 + ₹ 1,458 = ₹ 37,908.

Hence, final amount = ₹ 37,908.

For first year :

P = ₹ 15,000

T = 1 year

R = 8%

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

$= \frac{15000 \times 8 \times 1}{100}$ = ₹ 1,200.

Amount = P + I = ₹ 15,000 + ₹ 1,200 = ₹ 16,200.

For second year :

P = ₹ 16,200

T = 1 year

R = 9%

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

$= \frac{16200 \times 9 \times 1}{100}$ = ₹ 1,458.

Amount = P + I = ₹ 16,200 + ₹ 1,458 = ₹ 17,658.

Compound interest = Final amount - Initial principal

= ₹ 17,658 - ₹ 15,000 = ₹ 2,658.

Hence, final amount = ₹ 17,658 and compound interest = ₹ 2,658.

For first year :

P = ₹ 25,000

T = 1 year

R = 8%

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

$= \frac{25000 \times 8 \times 1}{100}$ = ₹ 2,000.

Amount = P + I = ₹ 25,000 + ₹ 2,000 = ₹ 27,000.

For second year :

P = ₹ 27,000

T = 1 year

R = 9%

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

$= \frac{27000 \times 9 \times 1}{100}$ = ₹ 2,430.

Amount = P + I = ₹ 27,000 + ₹ 2,430 = ₹ 29,430.

For third year :

P = ₹ 29,430

T = 1 year

R = 10%

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

$= \frac{29430 \times 10 \times 1}{100}$ = ₹ 2,943.

Amount = P + I = ₹ 29,430 + ₹ 2,943 = ₹ 32,373.

Compound interest = Final amount - Initial principal

= ₹ 32,373 - ₹ 25,000 = ₹ 7,373.

Hence, final amount = ₹ 32,373 and compound interest = ₹ 7,373.

For first year :

P = ₹ 2,40,000

T = 1 year

R = 10%

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

$= \frac{240000 \times 10 \times 1}{100}$ = ₹ 24,000.

Income tax deducted = 20% of Interest

= $\frac{20}{100} \times 24000$ = ₹ 4,800

Interest after deduction = ₹ 24,000 - ₹ 4,800 = ₹ 19,200.

Amount = P + I = ₹ 2,40,000 + ₹ 19,200 = ₹ 2,59,200.

For second year :

P = ₹ 2,59,200

T = 1 year

R = 10%

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

$= \frac{259200 \times 10 \times 1}{100}$ = ₹ 25,920.

Income tax deducted = 20% of Interest

= $\frac{20}{100} \times 25920$ = ₹ 5,184.

Interest after deduction = ₹ 25,920 - ₹ 5,184 = ₹ 20,736.

Amount = P + I = ₹ 2,59,200 + ₹ 20,736 = ₹ 2,79,936.

Hence, final amount received at the end of 2 years = ₹ 2,79,936.

Given,

Rate = 12%

Half yearly rate (R) = $\frac{\text{Rate}}{2} = \frac{12}{2}$ = 6%

For first half year :

P = ₹ 10,000

T = 1 half year

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

$= \frac{10000 \times 6 \times 1}{100}$ = ₹ 600.

Amount = P + I = ₹ 10,000 + ₹ 600 = ₹ 10,600.

For second half year :

P = ₹ 10,600

T = 1 half year

Half yearly rate = 6%

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

$= \frac{10600 \times 6 \times 1}{100}$ = ₹ 636.

Amount = P + I = ₹ 10,600 + ₹ 636 = ₹ 11,236.

Compound interest = Final amount - Initial principal

= ₹ 11,236 - ₹ 10,000 = ₹ 1,236.

Hence, final amount = ₹ 11,236 and compound interest = ₹ 1,236.

Given,

Rate = 15%

Half yearly rate (R) = $\frac{\text{Rate}}{2} = \frac{15}{2}$ % = 7.5%

Time = $1\frac{1}{2}$ year = $\frac{3}{2} \times 2$ = 3 half-year.

For first half year :

P = ₹ 64,000

T = 1 half year

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

$= \frac{64,000 \times 7.5 \times 1}{100}$ = ₹ 4,800

Amount = P + I = ₹ 64,000 + ₹ 4,800 = ₹ 68,800

For second half year :

P = ₹ 68,800

Half yearly rate (R) = 7.5%

T = 1 half year

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

$= \frac{68800 \times 7.5 \times 1}{100}$ = ₹ 5,160.

Amount = P + I = ₹ 68,800 + ₹ 5,160 = ₹ 73,960.

For third half year :

P = ₹ 73,960

Half yearly rate (R) = 7.5%

T = 1 year

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

$= \frac{73960 \times 7.5 \times 1}{100}$ = ₹ 5,547.

Amount = P + I = ₹ 73,960 + ₹ 5,547 = ₹ 79,507.

Compound interest = Final amount - Initial principal

= ₹ 79,507 - ₹ 64,000 = ₹ 15,507.

Hence, final amount = ₹ 79,507 and compound interest = ₹ 15,507.

(i) Given,

The simple interest on a sum of money for 2 years at 10% p.a. is ₹ 1700.

I = ₹ 1,700

T = 2 year

R = 10%

Let sum of money be ₹ P.

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

Substituting values we get :

$$\begin{aligned}\Rightarrow 1700 = \frac{P \times 10 \times 2}{100} \\[1em] \Rightarrow 1700 = \frac{P \times 20}{100}\\[1em] \Rightarrow 1700 = \frac{P}{5} \\[1em] \Rightarrow P = 1700 \times 5 = 8,500.\end{aligned}$$

Hence, the sum of money = ₹ 8,500

(ii) Given,

For first half year :

P = ₹ 8,500

R = 10%

Half yearly rate = $\frac{Rate}{2} = \frac{10}{2}$ = 5%

T = 1 half year

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

$= \frac{8500 \times 5 \times 1}{100} = ₹ 425$

Amount = P + I = ₹ 8,500 + ₹ 425 = ₹ 8,925.

For second half year :

P = ₹ 8,925

Half yearly rate = 5%

T = 1 half year

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

$= \frac{8925 \times 5 \times 1}{100} = ₹ 446.25$

Amount = P + I = ₹ 8,925 + ₹ 446.25 = ₹ 9,371.25

Compound interest = Final Amount - Initial Pincipal

= ₹ 9,371.25 - ₹ 8,500

= ₹ 871.25

Hence, compound interest = ₹ 871.25

Given,

P = ₹ 10,000

n = 2 years

r = 8%

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow A = 10000\Big(1 + \frac{8}{100}\Big)^2 \\[1em] \Rightarrow A = 10000\Big(\frac{100+8}{100}\Big)^2 \\[1em] \Rightarrow A = 10000\Big(\frac{108}{100}\Big)^2 \\[1em] \Rightarrow A = 10000\Big(\frac{27}{25}\Big)^2 \\[1em] \Rightarrow A = 10000 \times \frac{729}{625} \\[1em] \Rightarrow A = ₹ 11,664\end{aligned}$$

Compound interest = Final amount - Initial principal

= ₹ 11,664 - ₹ 10,000 = ₹ 1,664

Hence, amount = ₹ 11,664 and compound interest = ₹ 1,664.

Given,

P = ₹ 64,000

n = 3 years

r = $7\frac{1}{2}$% = 7.5%

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow A = 64000\Big(1 + \frac{7.5}{100}\Big)^3 \\[1em] \Rightarrow A = 64000\Big(\frac{100+7.5}{100}\Big)^3 \\[1em] \Rightarrow A = 64000\Big(\frac{107.5}{100}\Big)^3 \\[1em] \Rightarrow A = 64000 \times \Big(\frac{21.5}{20}\Big)^3 \\[1em] \Rightarrow A = 64000 \times \frac{9938.375}{8000} \\[1em] \Rightarrow A = ₹ 79,507.\end{aligned}$$

Compound interest = Final amount - Initial principal = ₹ 79,507 - ₹ 64,000 = ₹ 15,507

Hence, amount = ₹ 79,507 and compound interest = ₹ 15,507.

Given,

P = ₹ 12,000

r₁ = 10%

r₂ = 11%

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 :

$$\begin{aligned}\Rightarrow A = 12000 \times \Big(1 + \frac{10}{100}\Big) \times \Big(1 + \frac{11}{100}\Big) \\[1em] = 12000 \times \frac{110}{100} \times \frac{111}{100} \\[1em] = 12000 \times \frac{11}{10} \times \frac{111}{100} \\[1em] = \frac{12000 \times 11 \times 111}{10 \times 100} \\[1em] = ₹ 14,652.\end{aligned}$$

Hence, final amount = ₹ 14,652.

Given,

P = ₹ 25,000

r₁ = 8%

r₂ = 9%

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 :

$$\begin{aligned}\Rightarrow A = 25000 \times \Big(1 + \frac{8}{100}\Big) \times \Big(1 + \frac{9}{100}\Big) \times \Big(1 + \frac{10}{100}\Big) \\[1em] = 25000 \times \frac{108}{100} \times \frac{109}{100} \times \frac{110}{100} \\[1em] = 25000 \times \frac{27}{25} \times \frac{109}{100} \times \frac{11}{10}\\[1em] = \frac{25000 \times 27 \times 109 \times 11}{25 \times 100 \times 10} \\[1em] = 27 \times 109 \times 11 \\[1em] = ₹ 32,373.\end{aligned}$$

Compound interest = Final amount - Initial principal

= ₹ 32,373 - ₹ 25,000

= ₹ 7,373.

Hence, amount = ₹ 32,373 and compound interest = ₹ 7,373.

Given,

P = ₹ 7,500

n = 2 years 8 months

= 2 $\frac{8}{12}$ years = 2 $\frac{2}{3}$ years

r = 10%

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow A = 7500\Big(1 + \frac{10}{100}\Big)^2 \Big(1 + \frac{\frac{2}{3} \times 10}{100}\Big) \\[1em] \Rightarrow A = 7500\Big(\frac{100+10}{100}\Big)^2 \Big(1 + \frac{20}{300}\Big) \\[1em] \Rightarrow A = 7500\Big(\frac{110}{100}\Big)^2 \Big(\frac{300 + 20}{300}\Big) \\[1em] \Rightarrow A = 7500\Big(\frac{11}{10}\Big)^2 \Big(\frac{320}{300}\Big)\\[1em] \Rightarrow A = 7500 \times \frac{121}{100} \times \frac{32}{30}\\[1em] \Rightarrow A = ₹ 9,680\end{aligned}$$

C.I. = A - P = ₹ 9,680 - ₹ 7,500 = ₹ 2,180

Hence, amount = ₹ 9,680 and compound interest = ₹ 2,180.

Given,

I = ₹ 7,500

T = 3 years

R = 8% p.a. simple interest

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

$$\begin{aligned}\Rightarrow 7500 = \frac{P \times 8 \times 3}{100} \\[1em] \Rightarrow 7500 = \frac{P \times 24}{100} \\[1em] \Rightarrow P = \frac{7500 \times 100}{24} \\[1em] \Rightarrow P = ₹ 31,250.\end{aligned}$$

Let's calculate compound interest for this principal, rate of interest and time.

By formula,

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

$$\begin{aligned}\Rightarrow A = 31250\Big(1 + \frac{8}{100}\Big)^3 \\[1em] \Rightarrow A = 31250\Big(\frac{100+8}{100}\Big)^3 \\[1em] \Rightarrow A = 31250\Big(\frac{108}{100}\Big)^3 \\[1em] \Rightarrow A = 31250\Big(\frac{27}{25}\Big)^3 \\[1em] \Rightarrow A = 31250 \times \frac{19683}{15625} \\[1em] \Rightarrow A = ₹ 39366.\end{aligned}$$

Compound interest = Final amount - Initial principal

= ₹ 39,366 - ₹ 31,250 = ₹ 8,116.

Hence, compound interest = ₹ 8,116.

Given,

Principal (P) = ₹ 16,000

Time (n) = 1 year

Rate (r) = 15% 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 :

$$\begin{aligned}\Rightarrow A = 16000 \times \Big(1 + \frac{15}{2 \times 100}\Big)^{1 \times 2} \\[1em] \Rightarrow A = 16000 \times \Big(\frac{200+15}{200}\Big)^2 \\[1em] \Rightarrow A = 16000 \times \Big(\frac{215}{200}\Big)^2 \\[1em] \Rightarrow A = 16000 \times \Big(\frac{43}{40}\Big)^2 \\[1em] \Rightarrow A = 16000 \times \Big(\frac{1849}{1600}\Big) \\[1em] \Rightarrow A = \frac{16000 \times 1849}{1600} = ₹ 18,490.\end{aligned}$$

Compound interest = Amount - Principal = ₹ 18,490 - ₹ 16,000 = ₹ 2,490

Hence, amount = ₹ 18,490 and compound interest = ₹ 2,490.

Given,

Principal (P) = ₹ 1,25,000

Time (n) = 1 $\frac{1}{2}$ years = 1.5 years

Rate (r) = 12% 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 :

$$\begin{aligned}\Rightarrow A = 125000 \times \Big(1 + \frac{12}{2 \times 100}\Big)^{1.5 \times 2} \\[1em] \Rightarrow A = 125000 \times \Big(1 + \frac{3}{50}\Big)^3 \\[1em] \Rightarrow A = 125000 \times \Big(\frac{50 + 3}{50}\Big)^3 \\[1em] \Rightarrow A = 125000 \times \Big(\frac{53}{50}\Big)^3 \\[1em] \Rightarrow A = 125000 \times \Big(\frac{148877}{125000}\Big) \\[1em] \Rightarrow A = \frac{125000 \times 148877}{125000} = ₹ 1,48,877\end{aligned}$$

Compound interest = Amount - Principal

= ₹ 1,48,877 - ₹ 1,25,000 = ₹ 23,877

Hence, amount = ₹ 1,48,877 and compound interest = ₹ 23,877.

(i) Given,

P = ₹ 12,500

n = 1 half year

Amount = ₹ 13,000

When rate of interest is compounded half-yearly :

By formula,

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

For first half year:

$$\begin{aligned}\Rightarrow 13000 = 12500\Big(1 + \frac{R}{2 \times 100}\Big)^{0.5 \times 2} \\[1em] \Rightarrow \frac{13000}{12500}= \Big(1 + \frac{R}{200}\Big)^1 \\[1em] \Rightarrow 1.04 - 1 = \Big(\frac{R}{200}\Big) \\[1em] \Rightarrow 0.04 = \Big(\frac{R}{200}\Big) \\[1em] \Rightarrow R = 0.04 \times 200 \\[1em] \Rightarrow R = 8%\end{aligned}$$

Hence, Rate of interest = 8% p.a .

(ii) Given,

P = ₹ 12,500

n = 1.5 years

R = 8%

Let's calculate compound interest:

When rate of interest is compounded half-yearly :

By formula,

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

$$\begin{aligned}\Rightarrow A = 12500 \Big(1 + \frac{8}{2 \times 100}\Big)^{2 \times 1.5} \\[1em] \Rightarrow A = 12500 \Big(\frac{200 + 8}{200}\Big)^3 \\[1em] \Rightarrow A = 12500 \Big(\frac{208}{200}\Big)^3 \\[1em] \Rightarrow A = 12500 \Big(\frac{26}{25}\Big)^3 \\[1em] \Rightarrow A = 12500 \times \frac{17576}{15625} \\[1em] \Rightarrow A = ₹ 14,060.80 \approx ₹ 14,061\end{aligned}$$

Hence, compound interest = ₹ 14,061.

(i) Given,

I = ₹ 900

R = 12%

T = 1 year

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

$$\begin{aligned}\Rightarrow 900 = \frac{P \times 12 \times 1}{100}\\[1em] \Rightarrow 900 = \frac{P \times 12}{100}\\[1em] \Rightarrow P = \frac{900 \times 100}{12} \\[1em] \Rightarrow P = ₹ 7,500.\end{aligned}$$

Hence, principal = ₹ 7,500.

(ii) When rate of interest is compounded half-yearly :

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow A = 7500 \Big(1 + \frac{12}{2 \times 100}\Big)^{2 \times 1} \\[1em] \Rightarrow A = 7500 \Big(\frac{200 + 12}{200}\Big)^2 \\[1em] \Rightarrow A = 7500 \Big(\frac{212}{200}\Big)^2 \\[1em] \Rightarrow A = 7500 \Big(1.06\Big)^2 \\[1em] \Rightarrow A = 7500 \times 1.1236 \\[1em] \Rightarrow A = ₹ 8,427\end{aligned}$$

By formula,

Compound interest = Amount - Principal = ₹ 8,427 - ₹ 7500 = ₹ 927.

Hence, compound interest = ₹ 927.

Let sum of money be ₹ x.

Given,

P = ₹ x

r = 10%

n = 2 years

A = ₹ 18,150

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow 18150 = x \times \Big(1 + \frac{10}{100}\Big)^2 \\[1em] \Rightarrow 18150 = x \times \Big(\frac{110}{100}\Big)^2 \\[1em] \Rightarrow 18150 = x \times \Big(\frac{11}{10}\Big)^2 \\[1em] \Rightarrow 18150 = x \times \frac{121}{100} \\[1em] \Rightarrow x = \frac{18150 \times 100}{121} \\[1em] \Rightarrow x = ₹ 15,000.\end{aligned}$$

Hence, sum of money = ₹ 15,000.

Let sum of money be ₹ x.

Given,

P = ₹ x

r = 10%

n = 3 years

A = ₹ 93,170

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow 93170 = x \times \Big(1 + \frac{10}{100}\Big)^3 \\[1em] \Rightarrow 93170 = x \times \Big(\frac{110}{100}\Big)^3 \\[1em] \Rightarrow 93170 = x \times \Big(\frac{11}{10}\Big)^3 \\[1em] \Rightarrow 93170 = x \times \frac{1331}{1000} \\[1em] \Rightarrow x = \frac{93170 \times 1000}{1331} \\[1em] \Rightarrow x = ₹ 70,000.\end{aligned}$$

Hence, sum of money = ₹ 70,000.

Let sum of money be ₹ x.

Given,

P = ₹ x

n = 2 years

r = 8%

C.I. = ₹ 7,488

A = P + I = ₹ x + ₹ 7,488

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow x + 7488 = x \times \Big(1 + \frac{8}{100}\Big)^2 \\[1em] \Rightarrow x + 7488 = x \times \Big(\frac{108}{100}\Big)^2 \\[1em] \Rightarrow x + 7488 = x \times \Big(\frac{27}{25}\Big)^2 \\[1em] \Rightarrow x + 7488 = x \times \frac{729}{625} \\[1em] \Rightarrow 625(x + 7488) = 729x \\[1em] \Rightarrow 625x + 4680000 = 729x \\[1em] \Rightarrow 729x - 625x = 4680000 \\[1em] \Rightarrow 104x = 4680000 \\[1em] \Rightarrow x = \frac{4680000}{104} \\[1em] \Rightarrow x = ₹ 45,000.\end{aligned}$$

Hence, sum of money = ₹ 45,000.

Given,

n = 2 years

r = 12%

Let sum of money be ₹ P.

C.I. = A - P

$$\begin{aligned}C.I. = P\Big(1 + \frac{r}{100}\Big)^n - P \\[1em] = P\Big(1 + \frac{12}{100}\Big)^2 - P \\[1em] = P \times \Big(\frac{112}{100}\Big)^2 - P \\[1em] = P \times \Big(\frac{28}{25}\Big)^2 - P\\[1em] = P \times \frac{784}{625} - P \\[1em] = \frac{784P}{625} - P \\[1em] = \frac{784P - 625P}{625} \\[1em] = \frac{159P}{625}.\end{aligned}$$

By formula,

T = 2 years

$$\begin{aligned}S.I. = \frac{P \times R \times T}{100} \\[1em] = \frac{P \times 12 \times 2}{100} \\[1em] = \frac{6P}{25}.\end{aligned}$$

Given,

Difference between S.I. and C.I. = ₹ 216

$$\begin{aligned}\Rightarrow \frac{159P}{625} - \frac{6P}{25} = 216 \\[1em] \Rightarrow \frac{159P - 150P}{625} = 216 \\[1em] \Rightarrow \frac{9P}{625} = 216 \\[1em] \Rightarrow P = \frac{216 \times 625}{9} \\[1em] \Rightarrow P = 24 \times 625 \\[1em] \Rightarrow P = ₹ 15,000.\end{aligned}$$

Hence, sum = ₹ 15,000.

Given,

n = 3 years

r = 10 %

Let sum of money be ₹ P.

C.I. = A - P

$$\begin{aligned}C.I. = P\Big(1 + \frac{r}{100}\Big)^n - P \\[1em] = P\Big(1 + \frac{10}{100}\Big)^3 - P \\[1em] = P \times \Big(\frac{110}{100}\Big)^3 - P \\[1em] = P \times \Big(\frac{11}{10}\Big)^3 - P\\[1em] = P \times \frac{1331}{1000} - P \\[1em] = \frac{1331P}{1000} - P \\[1em] = \frac{1331P - 1000P}{1000} \\[1em] = \frac{331P}{1000}.\end{aligned}$$

Calculating S.I.,

R = 10%

T = 3 years

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

Given,

Difference between S.I. and C.I. = ₹ 558

$$\begin{aligned}\Rightarrow \frac{331P}{1000} - \frac{3P}{10} = 558 \\[1em] \Rightarrow \frac{331P - 300P}{1000} = 558 \\[1em] \Rightarrow \frac{31P}{1000} = 558 \\[1em] \Rightarrow P = \frac{558 \times 1000}{31} \\[1em] \Rightarrow P = 18 \times 1000 \\[1em] \Rightarrow P = ₹ 18,000.\end{aligned}$$

Hence, the sum = ₹ 18,000.

Let sum of money lent out be ₹ x.

Calculating C.I. payable half-yearly :

P = ₹ x

r = 10%

n = 1 year

C.I. = A - P

When rate of interest is compounded half-yearly :

By formula,

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

C.I. = A - P

Substituting values we get :

$$\begin{aligned}C.I. = P\Big(1 + \frac{r}{2 \times 100}\Big)^{n \times 2} - P \\[1em] = x \times \Big(1 + \frac{10}{200}\Big)^{1 \times 2} - x \\[1em] = x \times \Big(\frac{210}{200}\Big)^2 - x \\[1em] = x \times \Big(\frac{21}{20}\Big)^2 - x \\[1em] = \frac{441x}{400} - x \\[1em] = \frac{441x - 400x}{400} \\[1em] = ₹ \frac{41x}{400}.\end{aligned}$$

Calculating S.I. :

T = 1 year

R = 10%

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

Given,

Difference between compound interest for a year payable half-yearly and simple interest on ₹ x lent out at 10% for a year is ₹ 360.

$$\begin{aligned}\therefore \frac{41x}{400} - \frac{x}{10} = 360 \\[1em] \Rightarrow \frac{41x - 40x}{400} = 360 \\[1em] \Rightarrow \frac{x}{400} = 360 \\[1em] \Rightarrow x = 360 \times 400 = ₹ 1,44,000.\end{aligned}$$

Hence, the sum = ₹ 1,44,000.

Given,

P = ₹ 6,250

A = ₹ 7,290

n = 2 years

Let rate of interest be r%.

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow 7290 = 6250 \times \Big(1 + \frac{r}{100}\Big)^2 \\[1em] \Rightarrow \frac{7290}{6250} = \Big(1 + \frac{r}{100}\Big)^2 \\[1em] \Rightarrow \frac{729}{625} = \Big(1 + \frac{r}{100}\Big)^2 \\[1em] \Rightarrow \Big(\frac{27}{25}\Big)^2 = \Big(1 + \frac{r}{100}\Big)^2 \\[1em] \Rightarrow \frac{27}{25} = 1 + \frac{r}{100} \\[1em] \Rightarrow \frac{27}{25} - 1 = \frac{r}{100} \\[1em] \Rightarrow \frac{27 - 25}{25} = \frac{r}{100} \\[1em] \Rightarrow \frac{2}{25} = \frac{r}{100} \\[1em] \Rightarrow r = \frac{200}{25} = 8%.\end{aligned}$$

Hence, rate of interest = 8% p.a.

Given,

P = ₹ 3,000

A = ₹ 3,993

n = 3 years

Let rate of interest be r%.

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow 3993 = 3000 \times \Big(1 + \frac{r}{100}\Big)^3 \\[1em] \Rightarrow \frac{3993}{3000} = \Big(1 + \frac{r}{100}\Big)^3 \\[1em] \Rightarrow \frac{1331}{1000} = \Big(1 + \frac{r}{100}\Big)^3 \\[1em] \Rightarrow \Big(\frac{11}{10}\Big)^3 = \Big(1 + \frac{r}{100}\Big)^3 \\[1em] \Rightarrow \frac{11}{10} = 1 + \frac{r}{100} \\[1em] \Rightarrow \frac{11}{10} - 1 = \frac{r}{100} \\[1em] \Rightarrow \frac{11 - 10}{10} = \frac{r}{100} \\[1em] \Rightarrow \frac{1}{10} = \frac{r}{100} \\[1em] \Rightarrow r = \frac{100}{10} = 10%.\end{aligned}$$

Hence, rate of interest = 10% p.a.

Given,

P = ₹ 5,120

Rate = $12\frac{1}{2}$% = 12.5%

A = ₹ 7,290

Let time required be n years.

$$\begin{aligned}\Rightarrow A = P\Big(1 + \frac{r}{100}\Big)^{n } \\[1em] \Rightarrow 7290 = 5120 \times \Big(1 + \frac{12.5}{100}\Big)^{n} \\[1em] \Rightarrow \frac{7290}{5120} = \Big(\frac{100 + 12.5}{100}\Big)^{n} \\[1em] \Rightarrow \frac{729}{512} = \Big(\frac{112.5}{100}\Big)^{n} \\[1em] \Rightarrow \frac{729}{512} = \Big(\frac{1125}{1000}\Big)^{n} \\[1em] \Rightarrow \frac{729}{512} = \Big(\frac{9}{8}\Big)^{n} \\[1em] \Rightarrow \Big(\frac{9}{8}\Big)^3 = \Big(\frac{9}{8}\Big)^{n} \\[1em] \Rightarrow n = 3\end{aligned}$$

Hence, required time = 3 years.

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 ₹ 7,260 in 2 years.

$$\begin{aligned}\Rightarrow A = P\Big(1 + \frac{r}{100}\Big)^n \\[1em] \Rightarrow 7260 = x \times \Big(1 + \frac{r}{100}\Big)^2 \\[1em] \Rightarrow 7260 = x\Big(1 + \frac{r}{100}\Big)^2 ......(1)\end{aligned}$$

The sum of money, invested at compound interest, amounts to ₹ 7,986 in 3 years.

$$\begin{aligned}\Rightarrow A = P\Big(1 + \frac{r}{100}\Big)^n \\[1em] \Rightarrow 7986 = x \times \Big(1 + \frac{r}{100}\Big)^3 \\[1em] \Rightarrow 7986 = x\Big(1 + \frac{r}{100}\Big)^3 ......(2)\end{aligned}$$

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

$$\begin{aligned}\Rightarrow \frac{7986}{7260} = \frac{x\Big(1 + \frac{r}{100}\Big)^3}{x\Big(1 + \frac{r}{100}\Big)^2} \\[1em] \Rightarrow \frac{1331}{1210} = \Big(1 + \frac{r}{100}\Big) \\[1em] \Rightarrow \Big(1 + \frac{r}{100}\Big) = \frac{1331}{1210} \\[1em] \Rightarrow \frac{r}{100} = \frac{1331}{1210} - 1 \\[1em] \Rightarrow \frac{r}{100} = \frac{1331 - 1210}{1210} \\[1em] \Rightarrow \frac{r}{100} = \frac{121}{1210} \\[1em] \Rightarrow r = \frac{100 \times 121}{1210} = 10%.\end{aligned}$$

Hence, rate percent = 10% p.a.

Given,

P = 15625

R = 4% p.a.

n = 3 years

By formula,

Population after n years = $P \times \Big(1 + \frac{r}{100}\Big)^n$

Substituting the values in formula,

$$\begin{aligned}\text{Population after 3 years }= 15625 \times \Big(1 + \frac{4}{100}\Big)^3 \\[1em] = 15625 \times \Big(\frac{100 + 4}{100}\Big)^3 \\[1em] = 15625 \times \Big(\frac{104}{100}\Big)^3 \\[1em] = 15625 \times \Big(\frac{26}{25}\Big)^3 \\[1em] = 15625 \times \frac{17576}{15625} \\[1em] = 17576.\end{aligned}$$

Hence, the number of inhabitants of the town after 3 years = 17576.

(i) Given,

P = 36300

R = 10% p.a.

n = 2 years

By formula,

Population after n years = $P \times \Big(1 + \frac{r}{100}\Big)^n$

Substituting the values in formula,

$$\begin{aligned}\text{Population after 2 years } = 36300 \times \Big(1 + \frac{10}{100}\Big)^2 \\[1em] =36300 \times \Big(\frac{100 + 10}{100}\Big)^2 \\[1em] =36300 \times \Big(\frac{110}{100}\Big)^2 \\[1em] =36300 \times \Big(\frac{11}{10}\Big)^2 \\[1em] =36300 \times \frac{121}{100} \\[1em] =43923.\end{aligned}$$

Hence, population of the town after 2 years = 43923.

(ii) Given,

P = 36300

R = 10% p.a.

n = 2 years

By formula,

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

Substituting the values in formula,

$$\begin{aligned}\text{Population 2 years ago }= \frac{36300}{\Big(1 + \frac{10}{100}\Big)^2} \\[1em] = \frac{36300}{\Big(\frac{100+10}{100}\Big)^2} \\[1em] = \frac{36300}{\Big(\frac{110}{100}\Big)^2} \\[1em] = \frac{36300}{\Big(\frac{11}{10}\Big)^2} \\[1em] = \frac{36300}{\frac{121}{100}} \\[1em] = \frac{36300 \times 100} {121} \\[1em] = 30000.\end{aligned}$$

Hence, population of the town 2 years ago = 30000.

(i) Given,

P = 176400

R = 5% p.a.

n = 2 years

By formula,

Population after n years = $P \times \Big(1 + \frac{r}{100}\Big)^n$

Substituting the values in formula,

$$\begin{aligned}\text{Population after 2 years}=176400 \times \Big(1 + \frac{5}{100}\Big)^2 \\[1em] =176400 \times \Big(\frac{100 + 5}{100}\Big)^2 \\[1em] =176400 \times \Big(\frac{105}{100}\Big)^2 \\[1em] =176400 \times \Big(\frac{21}{20}\Big)^2 \\[1em] =176400 \times \frac{441}{400} \\[1em] =194481\end{aligned}$$

Hence, population of the town after 2 years = 194481.

(ii) Given,

P = 176400

R = 5% p.a.

n = 1 year

By formula,

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

Substituting the values in formula,

$$\begin{aligned}\text{Population one year ago} =\frac{176400} {\Big(1 + \frac{5}{100}\Big)} \\[1em] = \frac{176400} {\Big(\frac{105}{100}\Big)} \\[1em] = \frac{176400} {\frac{21}{20}} \\[1em] = \frac{176400 \times 20} {21} \\[1em] = 168000\end{aligned}$$

Hence, population of the town before 1 year = 168000.

Given,

P = 50000

r₁ = 5%

r₂ = 8%

r₃ = 10%

By formula,

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

Substituting the values in formula,

$$\begin{aligned}\text{Present population} = 50000 \times \Big(1 + \frac{5}{100}\Big) \times \Big(1 + \frac{8}{100}\Big) \times \Big(1 + \frac{10}{100}\Big) \\[1em] = 50000 \times \Big(\frac{100 + 5}{100}\Big) \times \Big(\frac{100 + 8}{100}\Big) \times \Big(\frac{100 + 10}{100}\Big) \\[1em] = 50000 \times \Big(\frac{105}{100}\Big) \times \Big(\frac{108}{100}\Big) \times \Big(\frac{110}{100}\Big) \\[1em] = 50000 \times \Big(\frac{21}{20}\Big) \times \Big(\frac{27}{25}\Big) \times \Big(\frac{11}{10}\Big) \\[1em] = \frac{50000 \times 21 \times 27 \times 11}{5000} \\[1em] = 10 \times 21 \times 27 \times 11 \\[1em] = 62370.\end{aligned}$$

Hence, present population of the city = 62370.

Given,

P = 2916 quintals

R = 12.5% p.a.

n = 2 year

By formula,

Wheat production before n years = $\frac{P}{\Big(1 + \frac{r}{100}\Big)^n}$

Substituting the values in formula,

$$\begin{aligned}\text{Wheat production before 2 years }= \frac{2916}{\Big(1 + \frac{12.5}{100}\Big)^2} \\[1em] = \frac{2916}{\Big(\frac{100 + 12.5}{100}\Big)^2} \\[1em] = \frac{2916}{\Big(\frac{112.5}{100}\Big)^2} \\[1em] = \frac{2916}{\frac{12656.25}{10000}} \\[1em] = \frac{2916 \times 10000}{12656.25} \\[1em] = 2304\end{aligned}$$

Hence, farmer's annual production of wheat 2 years ago = 2304 quintals.

Given,

P = 64000

Net growth rate (R) = Birth rate - Death rate

= 11.7% - 4.2% = 7.5%

n = 3 years

By formula,

Population after n years = $P \times \Big(1 + \frac{r}{100}\Big)^n$

Substituting the values in formula,

$$\begin{aligned}\text{Population after 3 years} = 64000 \times \Big(1 + \frac{7.5}{100}\Big)^3 \\[1em] = 64000 \times \Big(\frac{100 + 7.5}{100}\Big)^3 \\[1em] = 64000 \times \Big(\frac{107.5}{100}\Big)^3 \\[1em] = 64000 \times \Big(\frac{43}{40}\Big)^3 \\[1em] = 64000 \times \frac{79507}{64000} \\[1em] = 79507\end{aligned}$$

Hence, the population of the town after 3 years = 79,507.

Given,

Present height (P) = 162 cm

R = 20% p.a.

n = 2 years

By formula,

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

Substituting the values in formula,

$$\begin{aligned}\text{Height before 2 years }=\frac{162}{\Big(1 + \frac{20}{100}\Big)^2} \\[1em] = \frac{162}{\Big(\frac{100 + 20}{100}\Big)^2} \\[1em] = \frac{162}{\Big(\frac{120}{100}\Big)^2} \\[1em] = \frac{162}{\Big(\frac{6}{5}\Big)^2} \\[1em] = \frac{162}{\frac{36}{25}} \\[1em] = \frac{162 \times 25}{36} \\[1em] = 112.5\end{aligned}$$

Hence, the height of the tree when the tree was planted was 112.5 cm.

Given,

P = 4000

r₁ = 6%

r₂ = 5%

Given,

The population increased by 6% in first year and decreased by 5% in second year.

By formula,

Population after n years = $P \times \Big(1 + \frac{r}{100}\Big) \times \Big(1 - \frac{r}{100}\Big)$

Substituting the values in formula,

$$\begin{aligned}\text{Population after 2 years }= 4000 \times \Big(1 + \frac{6}{100}\Big) \times \Big(1 - \frac{5}{100}\Big) \\[1em] = 4000 \times \Big(\frac{100 + 6}{100}\Big) \times \Big(\frac{ 100 - 5}{100}\Big) \\[1em] = 4000 \times \Big(\frac{106}{100}\Big) \times \Big(\frac{95}{100}\Big) \\[1em] = 4000 \times \Big(\frac{53}{50}\Big) \times \Big(\frac{19}{20}\Big) \\[1em] = \frac{4000 \times 53 \times 19}{50 \times 20} \\[1em] = 4 \times 53 \times 19 \\[1em] = 4028.\end{aligned}$$

Hence, the present population of the village = 4028.

Given,

P = 13125000

r₁ = 10%

r₂ = 8%

r₃ = 12%

$$\begin{aligned}\therefore \text{Count of bacteria after 3 years} = 13125000 \times \Big(1 + \frac{10}{100}\Big) \times \Big(1 - \frac{8}{100}\Big) \times \Big(1 + \frac{12}{100}\Big) \\[1em] = 13125000 \times \Big(\frac{100 + 10}{100}\Big) \times \Big(\frac{100 - 8}{100}\Big) \times \Big(\frac{100 + 12}{100}\Big) \\[1em] = 13125000 \times \Big(\frac{110}{100}\Big) \times \Big(\frac{92}{100}\Big) \times \Big(\frac{112}{100}\Big) \\[1em] = 13125000 \times \Big(\frac{11}{10}\Big) \times \Big(\frac{23}{25}\Big) \times \Big(\frac{28}{25}\Big) \\[1em] = \frac{13125000 \times 11 \times 23 \times 28}{10 \times 25 \times 25} \\[1em] = 14876400\end{aligned}$$

Hence, the count of bacteria after 3 hours = 14876400.

Given,

Initial Production = 40000

Production after 2 years = 57600

n = 2 years

Let the rate of growth per annum be r.

By formula,

Production after n years = $P \times \Big(1 + \frac{r}{100}\Big)^n$

Substituting the values in formula,

$$\begin{aligned}\Rightarrow 57600 = 40000 \times \Big(1 + \frac{r}{100}\Big)^2 \\[1em] \Rightarrow \frac{57600}{40000} = \Big(1 + \frac{r}{100}\Big)^2 \\[1em] \Rightarrow \frac{576}{400} = \Big(1 + \frac{r}{100}\Big)^2 \\[1em] \Rightarrow \Big(\frac{24}{20}\Big)^2 = \Big(1 + \frac{r}{100}\Big)^2 \\[1em] \Rightarrow \frac{24}{20} = 1 + \frac{r}{100} \\[1em] \Rightarrow \frac{24}{20} - 1 = \frac{r}{100} \\[1em] \Rightarrow \frac{r}{100} = \frac{24 - 20}{20} \\[1em] \Rightarrow \frac{r}{100} = \frac{4}{20} \\[1em] \Rightarrow r = \frac{4 \times 100}{20} \\[1em] \Rightarrow r = 20%\end{aligned}$$

Hence, the rate of growth per annum = 20%.

Given,

P = ₹ 5,00,000

r₁ = 5%

r₂ = 10%

r₃ = 12%

Given,

In the first year, Amit incurred a loss of 5%, in second year, he earned a profit of 10% which in the third year rose to 12%.

Substituting the values in formula,

$$\begin{aligned}\text{Value after 3 years} = 500000 \times \Big(1 - \frac{5}{100}\Big) \times \Big(1 + \frac{10}{100}\Big) \times \Big(1 + \frac{12}{100}\Big) \\[1em] = 500000 \times \Big(\frac{100 - 5}{100}\Big) \times \Big(\frac{100 + 10}{100}\Big) \times \Big(\frac{100 + 12}{100}\Big) \\[1em] = 500000 \times \Big(\frac{95}{100}\Big) \times \Big(\frac{110}{100}\Big) \times \Big(\frac{112}{100}\Big) \\[1em] = 500000 \times \Big(\frac{19}{20}\Big) \times \Big(\frac{11}{10}\Big) \times \Big(\frac{28}{25}\Big) \\[1em] = \frac{500000 \times 19 \times 11 \times 28}{20 \times 10 \times 25} \\[1em] = ₹ 5,85,200\end{aligned}$$

Net profit = Value after 3 years - Investment

= ₹ 5,85,200 - ₹ 5,00,000

= ₹ 85,200

Hence, the net profit for the entire period of three years = ₹ 85,200.

(i) Given,

Present value of machine (V) = ₹ 64,800

R = 10%

n = 2 years

By formula,

Value of machine after n years = ₹ $\Big[V \times \Big(1 - \frac{r}{100}\Big)^n \Big]$

Substituting the values in formula,

$$\begin{aligned}\text{Value of machine after 2 years} = 64800 \times \Big(1 - \frac{10}{100}\Big)^2 \\[1em] =64800 \times \Big(\frac{100 - 10}{100}\Big)^2 \\[1em] =64800 \times \Big(\frac{90}{100}\Big)^2 \\[1em] =64800 \times \Big(\frac{9}{10}\Big)^2 \\[1em] =64800 \times \frac{81}{100} \\[1em] =\frac{64800 \times 81}{100} \\[1em] = ₹ 52,488\end{aligned}$$

Hence, value of machine after two years = ₹ 52,488.

(ii) Given,

Present value of machine (V) = ₹ 64,800

R = 10%

n = 2 years

By formula,

Value of machine n years ago = ₹ $\frac{V}{\Big(1 - \frac{r}{100}\Big)^n}$

Substituting the values in formula,

$$\begin{aligned}\text{Value of machine 2 years ago } = \frac{64800}{\Big(1 - \frac{10}{100}\Big)^2} \\[1em] =\frac{64800}{\Big(\frac{100 - 10}{100}\Big)^2} \\[1em] =\frac{64800}{\Big(\frac{90}{100}\Big)^2} \\[1em] =\frac{64800}{\Big(\frac{9}{10}\Big)^2} \\[1em] =\frac{64800}{\frac{81}{100}} \\[1em] =\frac{64800 \times 100}{81} \\[1em] =₹ 80,000\end{aligned}$$

Hence, value of machine two years ago = ₹ 80,000.

(i) Given,

V = ₹ 20,000

R = 15%

n = 1 year

By formula,

Value of refrigerator after n years = ₹ $\Big[V \times \Big(1 - \frac{r}{100}\Big)^n \Big]$

Substituting the values in formula,

$$\begin{aligned}\text{Present value }=20000 \times \Big(1 - \frac{15}{100}\Big)^1 \\[1em] =20000 \times \Big(\frac{100 - 15}{100}\Big) \\[1em] =20000 \times \Big(\frac{85}{100}\Big) \\[1em] =20000 \times \frac{17}{20} \\[1em] =\frac{20000 \times 17}{20} \\[1em] = ₹ 17,000\end{aligned}$$

Hence, present value of refrigerator = ₹ 17,000.

(ii) Given,

V = ₹ 17,000

R = 15%

n = 1 year

By formula,

Value of refrigerator after n years = ₹ $\Big[V \times \Big(1 - \frac{r}{100}\Big)^n \Big]$

Substituting the values in formula,

$$\begin{aligned}\text{Value after 1 year }=17000 \times \Big(1 - \frac{15}{100}\Big)^1 \\[1em] =17000 \times \Big(\frac{100 - 15}{100}\Big) \\[1em] =17000 \times \Big(\frac{85}{100}\Big) \\[1em] =17000 \times \frac{17}{20} \\[1em] =\frac{17000 \times 17}{20} \\[1em] =₹ 14,450\end{aligned}$$

Hence, value of refrigerator after 1 year = ₹ 14,450.

(i) Given,

Value of machine at the end of the year 2015 (V) = ₹ 57,500

R = 8%

n = 1 year

By formula,

Value of machine n years ago = ₹ $\frac{V}{\Big(1 - \frac{r}{100}\Big)^n}$

Substituting the values in formula,

$$\begin{aligned}\text{Value of machine at the end of the year 2014 }=\frac{57500}{\Big(1 - \frac{8}{100}\Big)^1} \\[1em] =\frac{57500}{\frac{100 - 8}{100}} \\[1em] =\frac{57500}{\frac{92}{100}} \\[1em] =\frac{57500}{\frac{23}{25}} \\[1em] =\frac{57500 \times 25}{23} \\[1em] =₹ 62,500\end{aligned}$$

Hence, value of machine at the end of the year 2014 = ₹ 62,500.

(ii) Given,

Value of machine at the end of the year 2015 (V) = ₹ 57,500

R = 8%

n = 1 year

By formula,

Value of machine after n years = $\Big[V \times \Big(1 - \frac{r}{100}\Big)^n \Big]$

Substituting the values in formula,

$$\begin{aligned}\text{Value of machine at the end of 2016}=57500 \times \Big(1 - \frac{8}{100}\Big)^1 \\[1em] =57500 \times \Big(\frac{100 - 8}{100}\Big) \\[1em] =57500 \times \Big(\frac{92}{100}\Big) \\[1em] =57500 \times \frac{23}{25} \\[1em] =\frac{57500 \times 23}{25} \\[1em] =₹ 52,900.\end{aligned}$$

Hence, value of machine at the end of the year 2016 = ₹ 52,900.

Given,

Present value of machine (V) = ₹ 62,500

R = $16\frac{2}{3} = \frac{48 + 2}{3} = \frac{50}{3}%$

n = 3 years

By formula,

Value of machine n years ago = ₹ $\frac{V}{\Big(1 - \frac{r}{100}\Big)^n}$

Substituting the values in formula,

$$\begin{aligned}\text{Value of machine 3 years ago}=\frac{62500}{\Big(1 - \frac{\frac{50}{3}}{100}\Big)^3} \\[1em] =\frac{62500}{\Big(1 - \frac{50}{3 \times 100}\Big)^3} \\[1em] =\frac{62500}{\Big(\frac{300 - 50}{300}\Big)^3} \\[1em] =\frac{62500}{\Big(\frac{250}{300}\Big)^3} \\[1em] =\frac{62500}{\Big(\frac{25}{30}\Big)^3} \\[1em] =\frac{62500}{\frac{15625}{27000}} \\[1em] =\frac{62500 \times 27000}{15625} \\[1em] =₹ 1,08,000\end{aligned}$$

Hence, its purchase price = ₹ 1,08,000.

Given,

P = ₹ 3,750

n = 2 years

r = 8%

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow A = 3750 \times \Big(1 + \frac{8}{100}\Big)^2 \\[1em] \Rightarrow A = 3750 \times \Big(\frac{100 + 8}{100}\Big)^2 \\[1em] \Rightarrow A = 3750\times \Big(\frac{108}{100}\Big)^2 \\[1em] \Rightarrow A = 3750 \times \Big(\frac{27}{25}\Big)^2 \\[1em] \Rightarrow A = 3750 \times \frac{729}{625} \\[1em] \Rightarrow A = ₹ 4,374\end{aligned}$$

Compound interest = Final amount - Initial principal

= ₹ 4374 - ₹ 3750 = ₹ 624.

Hence, option 3 is correct option.

Given,

P = ₹ 46,875

n = 1 year

r = 4%

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow A = 46875 \times \Big(1 + \frac{4}{100}\Big)^1 \\[1em] \Rightarrow A = 46875 \times \Big(\frac{100 + 4}{100}\Big) \\[1em] \Rightarrow A = 46875 \times \Big(\frac{104}{100}\Big) \\[1em] \Rightarrow A = 46875 \times \frac{26}{25} \\[1em] \Rightarrow A = ₹ 48,750.\end{aligned}$$

Compound interest = Final amount - Initial principal

= ₹ 48,750 - ₹ 46,875 = ₹ 1,875.

Hence, option 3 is correct option.

Given,

P = ₹ 10,000

n = 3 years

r = 9%

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow A = 10000 \times \Big(1 + \frac{9}{100}\Big)^3 \\[1em] \Rightarrow A = 10000 \times \Big(\frac{100 + 9}{100}\Big)^3 \\[1em] \Rightarrow A = 10000\times \Big(\frac{109}{100}\Big)^3 \\[1em] \Rightarrow A = 10000 \times \frac{1295029}{1000000} \\[1em] \Rightarrow A = ₹ 12,950.29\end{aligned}$$

Hence, option 1 is correct option.

Given,

P = ₹ 16,000

r₁ = 8%

r₂ = 12%

A = ₹ 21,384

Let the interest for third year be r₃.

By formula,

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

Substituting the values in formula,

$$\begin{aligned}\Rightarrow 21384 = 16000 \times \Big(1 + \frac{8}{100}\Big) \times \Big(1 + \frac{12}{100}\Big) \times \Big(1 + \frac{r_3}{100}\Big) \\[1em] \Rightarrow 21384 = 16000 \times \Big(\frac{100 + 8}{100}\Big) \times \Big(\frac{100 + 12}{100}\Big) \times \Big(1 + \frac{r_3}{100}\Big) \\[1em] \Rightarrow 21384 = 16000 \times \Big(\frac{108}{100}\Big) \times \Big(\frac{112}{100}\Big) \times \Big(1 + \frac{r_3}{100}\Big) \\[1em] \Rightarrow 21384 = 16000 \times \Big(\frac{27}{25}\Big) \times \Big(\frac{28}{25}\Big) \times \Big(1 + \frac{r_3}{100}\Big) \\[1em] \Rightarrow 21384 = \frac{16000 \times 27 \times 28}{25 \times 25} \times \Big(1 + \frac{r_3}{100}\Big) \\[1em] \Rightarrow 21384 \times 25 \times 25= 16000 \times 27 \times 28 \times \Big(1 + \frac{r_3}{100}\Big) \\[1em] \Rightarrow \frac{21384 \times 25 \times 25}{16000 \times 27 \times 28} = \Big(1 + \frac{r_3}{100}\Big) \\[1em] \Rightarrow \frac{21384 \times 625}{16000 \times 756} = \Big(1 + \frac{r_3}{100}\Big) \\[1em] \Rightarrow \frac{13365000}{12096000} = \Big(1 + \frac{r_3}{100}\Big) \\[1em] \Rightarrow 1 + \frac{r_3}{100} = 1.10 \\[1em] \Rightarrow \frac{r_3}{100} = 1.10 - 1 \\[1em] \Rightarrow r_3 = 0.10 \times 100\\[1em] \Rightarrow r_3 = 10 %\end{aligned}$$

Hence, option 2 is correct option.

For first six moths :

P = ₹ 5,000

T = 6 months = 0.5 year

R = 12%

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

$= \frac{5000 \times 12 \times 0.5}{100}$ = ₹ 300.

Amount = P + I = ₹ 5,000 + ₹ 300 = ₹ 5,300.

Amount payed at end of six months = ₹ 1,800.

Amount left at beginning of second six months = ₹ 5,300 - ₹ 1,800 = ₹ 3,500.

For next six months :

P = ₹ 3,500

R = 12%

T = 6 months = 0.5 year

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

$= \frac{3500 \times 12 \times 0.5}{100}$ = ₹ 210.

Amount = P + I = ₹ 3,500 + ₹ 210 = ₹ 3,710.

Amount payed at end of second six months = ₹ 1,800.

Amount left at beginning of third six months = ₹ 3,710 - ₹ 1,800 = ₹ 1,910

For next six months :

P = ₹ 1,910

R = 12%

T = 6 months = 0.5 year

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

$= \frac{1910 \times 12 \times 0.5}{100}$ = ₹ 114.6

Amount due at the end of third year = P + I = ₹ 1,910 + ₹ 114.6 = ₹ 2,024.60

Hence, option 1 is correct option.

Given,

P = ₹ 8,000

T = 1 year

r = 10%

For the first year,

By formula,

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

$= \frac{8000 \times 10 \times 1}{100}$ = ₹ 800.

Amount = P + I = ₹ 8,000 + ₹ 800 = ₹ 8,800.

Amount at beginning of second year = ₹ 8,800.

P = ₹ 8,800

T = 1 year

r = 10%

For the first year,

By formula,

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

$= \frac{8800 \times 10 \times 1}{100}$ = ₹ 880.

The compound interest for the second year is ₹ 880

Hence, option 2 is correct option.

Given,

P = ₹ 6,000

r₁ = 10%

r₂ = 12%

n = 2 years

By formula,

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

Substituting the values in formula,

$$\begin{aligned}\Rightarrow A = 6000 \times \Big(1 + \frac{10}{100}\Big) \times \Big(1 + \frac{12}{100}\Big) \\[1em] \Rightarrow A = 6000 \times \Big(\frac{100 + 10}{100}\Big) \times \Big(\frac{100 + 12}{100}\Big) \\[1em] \Rightarrow A = 6000 \times \Big(\frac{110}{100}\Big) \times \Big(\frac{112}{100}\Big) \\[1em] \Rightarrow A = 6000 \times \Big(\frac{11}{10}\Big) \times \Big(\frac{28}{25}\Big) \\[1em] \Rightarrow A = \frac{6000 \times 11 \times 28}{10 \times 25} \\[1em] \Rightarrow A = ₹ 7,392.\end{aligned}$$

Hence, option 4 is correct option.

Given,

P = ₹ 6,000

n = 1 year

A = ₹ 6,720

Let the rate of interest be r,

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow 6720 = 6000 \times \Big(1 + \frac{r}{100}\Big)^1 \\[1em] \Rightarrow \frac{6720}{6000} = 1 + \frac{r}{100} \\[1em] \Rightarrow 1 + \frac{r}{100} = 1.12 \\[1em] \Rightarrow \frac{r}{100} = 1.12 - 1 \\[1em] \Rightarrow r = 0.12 \times 100 \\[1em] \Rightarrow r = 12%\end{aligned}$$

Hence, option 3 is correct option.

Given,

P = ₹ 8,640

n = 3 year

r = 8%

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow A = 8640 \times \Big(1 + \frac{8}{100}\Big)^3 \\[1em] \Rightarrow A = 8640 \times \Big(\frac{100 + 8}{100}\Big)^3 \\[1em] \Rightarrow A = 8640 \times \Big(\frac{108}{100}\Big)^3 \\[1em] \Rightarrow A = 8640 \times \Big(\frac{27}{25}\Big)^3 \\[1em] \Rightarrow A = 8640 \times \frac{19683}{15625} \\[1em] \Rightarrow A = ₹ 10,883.9\end{aligned}$$

Compound interest = Final amount - Initial principal

= ₹ 10,883.9 - ₹ 8,640

= ₹ 2,243.9

Given,

P = ₹ 7,400

r = 5%

n = 1 year

Given,

When interest is compounded half-yearly.

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow A = 7400 \times \Big(1 + \frac{5}{2 \times 100}\Big)^{1 \times 2} \\[1em] \Rightarrow A = 7400 \times \Big(\frac{200 + 5}{200}\Big)^2 \\[1em] \Rightarrow A = 7400 \times \Big(\frac{205}{200}\Big)^2 \\[1em] \Rightarrow A = 7400 \times \Big(\frac{41}{40}\Big)^2 \\[1em] \Rightarrow A = 7400 \times \frac{1681}{1600} \\[1em] \Rightarrow A = 7774.625\end{aligned}$$

Compound interest = Final amount - Initial principal

= ₹ 7774.625 - ₹ 7400

= ₹ 374.625 ≈ ₹ 374.63

Hence, option 2 is correct option.

Given,

I = ₹ 340

R = 4%

T = 2 years

Let sum of money be ₹ P.

By formula,

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

$$\begin{aligned}\Rightarrow 340 = \frac{P \times 4 \times 2}{100} \\[1em] \Rightarrow 340 = \frac{P \times 8}{100} \\[1em] \Rightarrow P = \frac{340 \times 100}{8} \\[1em] \Rightarrow P = ₹ 4,250.\end{aligned}$$

Given, interest is compounded half-yearly.

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow A = 4250 \Big(1 + \frac{4}{2 \times 100}\Big)^{2 \times 1} \\[1em] \Rightarrow A = 4250 \Big(\frac{200 + 4}{200}\Big)^2 \\[1em] \Rightarrow A = 4250 \Big(\frac{204}{200}\Big)^2 \\[1em] \Rightarrow A = 4250 \Big(1.02\Big)^2 \\[1em] \Rightarrow A = 4250 \times 1.0404 \\[1em] \Rightarrow A = ₹ 4,421.7\end{aligned}$$

Compound interest = Amount - Principal = ₹ 4,421.7 - ₹ 4,250 = ₹ 171.70.

Hence, option 3 is correct option.

Given,

I = ₹ 246

R = 5%

n = 2 years

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

Compound interest = Amount - Principal

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

$$\begin{aligned}\Rightarrow 246 = P \Big(1 + \frac{5}{100}\Big)^{2} - P \\[1em] \Rightarrow 246 = P \Big(\frac{100 + 5}{100}\Big)^2 - P \\[1em] \Rightarrow 246 = P \Big(\frac{105}{100}\Big)^2 - P \\[1em] \Rightarrow 246 = P [(1.05)^2 - 1] \\[1em] \Rightarrow 246 = P [1.1025 - 1] \\[1em] \Rightarrow 246 = 0.1025P \\[1em] \Rightarrow \frac{246}{0.1025} = P\\[1em] \Rightarrow P = ₹ 2,400.\end{aligned}$$

For calculating Simple interest,

P = ₹ 2,400

R = 6%

T = 3 years

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

$$\begin{aligned}\Rightarrow I = \frac{2400 \times 6 \times 3}{100} \\[1em] \Rightarrow I = \frac{43200}{100} \\[1em] \Rightarrow I = ₹ 432\end{aligned}$$

Hence, option 1 is correct option.

Given,

A = ₹ 6,050

R = 10%

n = 2 years

Let amount he needs to keep be ₹ P.

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

Substituting values we get :

$$\begin{aligned}\Rightarrow 6050 = P \Big(1 + \frac{10}{100}\Big)^{2} \\[1em] \Rightarrow 6050 = P \Big(\frac{100 + 10}{100}\Big)^2 \\[1em] \Rightarrow 6050 = P \Big(\frac{110}{100}\Big)^2 \\[1em] \Rightarrow 6050 = P (1.1)^2 \\[1em] \Rightarrow 6050 = P \times 1.21 \\[1em] \Rightarrow \frac{6050}{1.21} = P \\[1em] \Rightarrow P = ₹ 5,000. \\[1em]\end{aligned}$$

Hence, option 2 is correct option.

By formula,

Given,

T = 2 years

r = 5%

Let sum of money be ₹ P.

By formula,

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

By formula,

C.I. = A - P

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

Given,

Difference between S.I. and C.I. = ₹ 12

$$\begin{aligned}\Rightarrow \frac{41P}{400} - \frac{P}{10} = 12 \\[1em] \Rightarrow \frac{41P - 40P}{400} = 12 \\[1em] \Rightarrow \frac{P}{400} = 12 \\[1em] \Rightarrow P = 400 \times 12 \\[1em] \Rightarrow P = ₹ 4,800.\end{aligned}$$

Hence, option 3 is correct option.

Given,

A = ₹ 26,620

P = ₹ 20,000

n = 3 years

Let rate of interest be r.

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

$$\begin{aligned}\Rightarrow 26620 = 20000 \Big(1 + \frac{R}{100}\Big)^{3} \\[1em] \Rightarrow \frac{26620}{20000} = \Big(1 + \frac{R}{100}\Big)^3 \\[1em] \Rightarrow \frac{1331}{1000} = \Big(1 + \frac{R}{100}\Big)^3 \\[1em] \Rightarrow \Big(\frac{11}{10}\Big)^3 = \Big(1 + \frac{R}{100}\Big)^3 \\[1em] \Rightarrow \frac{11}{10} = \Big(1 + \frac{R}{100}\Big) \\[1em] \Rightarrow \frac{11}{10} - 1 = \frac{R}{100} \\[1em] \Rightarrow \frac{11-10}{10} = \frac{R}{100} \\[1em] \Rightarrow \frac{1}{10} = \frac{R}{100} \\[1em] \Rightarrow \frac{100}{10} = R \\[1em] \Rightarrow R = 10%\end{aligned}$$

Hence, option 4 is correct option.

Given,

Amount = ₹ 5,832

Principal = ₹ 5,000

R = 8%

Let time be n years.

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

$$\begin{aligned}\Rightarrow 5832 = 5000 \Big(1 + \frac{8}{100}\Big)^{n} \\[1em] \Rightarrow \frac{5832}{5000} = \Big(\frac{100 + 8}{100}\Big)^n \\[1em] \Rightarrow 1.1664 = \Big(\frac{108}{100}\Big)^n \\[1em] \Rightarrow 1.1664 = (1.08)^n \\[1em] \Rightarrow (1.08)^2 = (1.08)^n \\[1em] \Rightarrow n = 2\end{aligned}$$

Hence, option 1 is correct option.

Given,

V = ₹ 8,000

n = 3 years

R = 10%

Value of machine after n years = $\Big[V \times \Big(1 - \frac{r}{100}\Big)^n \Big]$

$$\begin{aligned}\text{Value of machine after 3 years }= 8000 \Big(1 - \frac{10}{100}\Big)^{3} \\[1em] = 8000 \Big(\frac{100 - 10}{100}\Big)^3 \\[1em] = 8000 \Big(\frac{90}{100}\Big)^3 \\[1em] = 8000 (0.9)^3 \\[1em] = 8000 \times 0.729 \\[1em] = ₹ 5,832.\end{aligned}$$

Value of machine after 3 years = ₹ 5,832

Hence, option 2 is correct option.

Given,

P = 200000

r₁ = 10% p.a.

r₂ = 15% p.a.

By formula,

Population after two years = $P \times \Big(1 + \frac{r_1}{100}\Big) \times \Big(1 + \frac{r_2}{100}\Big)$

Substituting the values in formula,

$$\begin{aligned}\text{Population after two years }=200000 \times \Big(1 + \frac{10}{100}\Big) \times \Big(1 + \frac{15}{100}\Big) \\[1em] =200000 \times \Big(\frac{100 + 10}{100}\Big) \times \Big(\frac{100 + 15}{100}\Big) \\[1em] =200000 \times \Big(\frac{110}{100}\Big) \times \Big(\frac{115}{100}\Big) \\[1em] =200000 \times \Big(\frac{11}{10}\Big) \times \Big(\frac{23}{20}\Big) \\[1em] =200000 \times 1.10 \times 1.15\\[1em] =253000.\end{aligned}$$

Hence, option 1 is correct option.

Given,

Initial value (P) = ₹ 10,000

Depreciated value (A) = ₹ 7,744

R = 12%

Value of machine after n years = $\Big[V \times \Big(1 - \frac{r}{100}\Big)^n \Big]$

$$\begin{aligned}\Rightarrow 7744 = 10000 \Big(1 - \frac{12}{100}\Big)^{n} \\[1em] \Rightarrow 7744 = 10000 \Big(\frac{100 - 12}{100}\Big)^n \\[1em] \Rightarrow 7744 = 10000 \Big(\frac{88}{100}\Big)^n \\[1em] \Rightarrow \frac{7744}{10000} = \Big(\frac{88}{100}\Big)^n \\[1em] \Rightarrow \Big(\frac{88}{100}\Big)^2 = \Big(\frac{88}{100}\Big)^n \\[1em] \Rightarrow n = 2.\end{aligned}$$

Hence, option 1 is correct option.

Given,

Initial value (P) = ₹ 25,000

Depreciated value (A) = ₹ 19,360

n = 2 years

Let rate of depreciation be R.

Value of machine after n years = $\Big[V \times \Big(1 - \frac{R}{100}\Big)^n \Big]$

$$\begin{aligned}\Rightarrow 19360 = 25000 \Big(1 - \frac{R}{100}\Big)^{2} \\[1em] \Rightarrow 19360 = 25000 \Big(1 - \frac{R}{100}\Big)^2 \\[1em] \Rightarrow \frac{19360}{25000} = \Big(1 - \frac{R}{100}\Big)^2 \\[1em] \Rightarrow \frac{484}{625} = \Big(1 - \frac{R}{100}\Big)^2 \\[1em] \Rightarrow \Big(\frac{22}{25}\Big)^2 = \Big(1 - \frac{R}{100}\Big)^2 \\[1em] \Rightarrow \frac{22}{25} = 1 - \frac{R}{100} \\[1em] \Rightarrow \frac{R}{100} = 1 - \frac{22}{25} \\[1em] \Rightarrow \frac{R}{100} = \frac{25 - 22}{25} \\[1em] \Rightarrow \frac{R}{100} = \frac{3}{25} \\[1em] \Rightarrow R = \frac{3 \times 100}{25} \\[1em] \Rightarrow R = 3 \times 4\\[1em] \Rightarrow R = 12%\end{aligned}$$

Hence, option 3 is correct option.

1. Given,

Population (in 2015) = 20000

r = 4% p.a.

n = 1 year

By formula,

Population in 2016 = $P \times \Big(1 + \frac{r}{100}\Big)^n$

Substituting the values in formula,

$$\begin{aligned}\text{Population in 2016 } = 20000 \times \Big(1 + \frac{4}{100}\Big)^1 \\[1em] =20000 \times \Big(\frac{100 + 4}{100}\Big)^1 \\[1em] =20000 \times \Big(\frac{104}{100}\Big)^1 \\[1em] =20000 \times (1.04)\\[1em] =20800.\end{aligned}$$

Hence, option (a) is correct option.

2. Given,

Population (in 2016) = 20800

r = 5% p.a.

n = 1 year

By formula,

Population in 2017 = $P \times \Big(1 + \frac{r}{100}\Big)^n$

Substituting the values in formula,

$$\begin{aligned}\text{Population in 2017} = 20800 \times \Big(1 + \frac{5}{100}\Big)^1 \\[1em] =20800 \times \Big(\frac{100 + 5}{100}\Big)^1 \\[1em] =20800 \times \Big(\frac{105}{100}\Big)^1 \\[1em] =20800 \times (1.05)\\[1em] =21840.\end{aligned}$$

The difference in population at the end of the years 2017 and 2015 = 21840 - 20000 = 1840.

Hence, option (d) is correct option.

3. Given,

Population (in 2017) = 21840

r (decrease) = 5% p.a.

n = 1 year

By formula,

Population in 2018 = $P \times \Big(1 - \frac{r}{100}\Big)^n$

Substituting the values in formula,

$$\begin{aligned}\text{Population in 2018}=21840 \times \Big(1 - \frac{5}{100}\Big)^1 \\[1em] =21840 \times \Big(\frac{100 - 5}{100}\Big)^1 \\[1em] =21840 \times \Big(\frac{95}{100}\Big)^1 \\[1em] =21840 \times (0.95)\\[1em] =20748.\end{aligned}$$

Hence, option (c) is correct option.

4. Given,

P = ₹ 20,000

r₁ = 4%

r₂ = 5%

r₃ = 5% (decrease)

By formula,

Population at the end = $P \times \Big(1 + \frac{r_1}{100}\Big) \times \Big(1 + \frac{r_2}{100}\Big) \times \Big(1 - \frac{r_3}{100}\Big)$

Substituting the values in formula,

$$\begin{aligned}\Rightarrow \text{Population at the end of year 2018} = 20000 \times \Big(1 + \frac{4}{100}\Big) \times \Big(1 + \frac{5}{100}\Big) \times \Big(1 - \frac{5}{100}\Big) \\[1em] = 20000 \times \Big(\frac{100 + 4}{100}\Big) \times \Big(\frac{100 + 5}{100}\Big) \times \Big(\frac{100 - 5}{100}\Big) \\[1em] = 20000 \times \Big(\frac{104}{100}\Big) \times \Big(\frac{105}{100}\Big) \times \Big(\frac{95}{100}\Big)\end{aligned}$$

Hence, option (b) is correct option.

5. Given,

A (population in 2019) = 21,163

P (population in 2018) = 20,748

n = 1 year

By formula,

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

Substituting the values in formula,

$$\begin{aligned}\Rightarrow 21163 = 20748 \times \Big(1 + \frac{R}{100}\Big)^1 \\[1em] \Rightarrow \frac{21163}{20748} = \Big(1 + \frac{R}{100}\Big)^1 \\[1em] \Rightarrow 1.02 = \Big(1 + \frac{R}{100}\Big)^1 \\[1em] \Rightarrow 1.02 - 1 = (\frac{R}{100}) \\[1em] \Rightarrow 0.02 = (\frac{R}{100}) \\[1em] \Rightarrow 0.02 \times 100 = R \\[1em] \Rightarrow R = 2%\end{aligned}$$

Hence, option (a) is correct option.

Given,

P = ₹ 1,00,000

A = ₹ 1,21,000

n = 1 year

By formula,

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

Substituting the values in formula,

$$\begin{aligned}\Rightarrow 121000 = 100000 \times \Big(1 + \frac{r}{100}\Big)^1 \\[1em] \Rightarrow \frac{121000}{100000} = 1 + \frac{r}{100} \\[1em] \Rightarrow \frac{r}{100} = 1.21 - 1 \\[1em] \Rightarrow r = 0.21 \times 100 \\[1em] \Rightarrow r = 21%\end{aligned}$$

Lets calculate the amount for 2 years with this rate of interest.

By formula,

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

Substituting the values in formula,

$$\begin{aligned}\Rightarrow A = 100000 \times \Big(1 + \frac{21}{100}\Big)^2 \\[1em] \Rightarrow A = 100000 \times \Big(\frac{100 + 21}{100}\Big)^2 \\[1em] \Rightarrow A = 100000 \times \Big(\frac{121}{100}\Big)^2 \\[1em] \Rightarrow A = 100000 \times (1.21)^2 \\[1em] \Rightarrow A = 100000 \times 1.4641 \\[1em] \Rightarrow A = 146410.\end{aligned}$$

Assertion (A) is true.

Formula,

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

Reason is true, as it is the formula to calculate the amount for compound interest when the interest is compounded annually.

Hence, option 3 is correct option.

Given,

P (two years ago) = 24,000

R = 5% p.a.

n = 2 year

By formula,

Population after n years = $P\Big(1 - \frac{r}{100}\Big)^n$

Substituting the values in formula,

$$\begin{aligned}\text{Present population }=24000 \times \Big(1 - \frac{5}{100}\Big)^2 \\[1em] =24000 \times \Big(\frac{100 - 5}{100}\Big)^2 \\[1em] =24000 \times \Big(\frac{95}{100}\Big)^2 \\[1em] =24000 \times (0.95)^2\\[1em] =24000 \times 0.9025\\[1em] =21,660\end{aligned}$$

Assertion is False

Reason is also false as we do not use that formula to calculate the growth in population. We use,

Population growth after n years at R% with initial Population as P is calulated using formula : $P \times \Big(1 + \frac{R}{100}\Big)^n$

Hence, option 4 is correct option.

Let annual installment be ₹ x.

Annual installment 1 (Paid at the end of 1st year) : This installment will earn interest for 3 years.

$$\begin{aligned}\Rightarrow x + x \times \frac{10}{100} \times 3 \\[1em] \Rightarrow x + \frac{3x}{10} \\[1em] \Rightarrow \frac{13x}{10} \\[1em] \Rightarrow 1.3x\end{aligned}$$

Annual installment 2 (Paid at the end of 2nd year) : This installment will earn interest for 2 years.

$$\begin{aligned}\Rightarrow x + x \times \frac{10}{100} \times 2 \\[1em] \Rightarrow x + \frac{2x}{10} \\[1em] \Rightarrow \frac{12x}{10} \\[1em] \Rightarrow 1.2x\end{aligned}$$

Annual installment 3 (Paid at the end of 3rd year) : This installment will earn interest for 1 year.

$$\begin{aligned}\Rightarrow x + x \times \frac{10}{100} \times 1 \\[1em] \Rightarrow x + \frac{x}{10} \\[1em] \Rightarrow \frac{11x}{10} \\[1em] \Rightarrow 1.1x\end{aligned}$$

Annual installment 4 (Paid at the end of 4th year) : This installment will not earn any interest.

⇒ x

The sum of these values must be equal to debt value of ₹ 4,600.

⇒ 1.3x + 1.2x + 1.1x + x = 4,600

⇒ 4.6x = 4,600

⇒ x = $\frac{4,600}{4.6}$

⇒ x = ₹ 1,000.

Hence, Option 1 is correct option.

Let the sum be ₹ P.

Given,

I = $\frac{P}{2}$

T = 6 years

R = 8%

By formula,

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

Substituting values we get :

$\Rightarrow \frac{P}{2} = \frac{P \times 8 \times 6}{100}$

By doing this calculation we will not obtain the sum, we need to have the interest to calculate the sum.

Hence, Option 4 is correct option.

Let the principal amount be ₹ P.

For 1 year :

R = 10%

T = 1 year

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

Substituting values we get :

$\Rightarrow I = \frac{P \times 10\times 1}{100} = \frac{P}{10}$

A = $P + \frac{P}{10} = \frac{11P}{10}$

For 2nd year :

Principal = $\frac{11P}{10}$

R = 10%

T = 1 year

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

$$\begin{aligned}I = \frac{\frac{11P}{10} \times 10 \times 1}{100} \\[1em] = \frac{11P \times 10}{1000} \\[1em] = \frac{11P}{100}.\end{aligned}$$

Given,

Interest for 2nd year = ₹ 132

$$\begin{aligned}\therefore \frac{11P}{100} = 132 \\[1em] \Rightarrow P = \frac{132 \times 100}{11} \\[1em] \Rightarrow P = 1,200.\end{aligned}$$

Hence, Option 3 is correct option.

Given,

V = ₹ 5 crores

R = 2%

Price of land increases by 2% quarterly.

n = There are 4 Quarters in a year = 4

By Formula,

Value after n years = $V \times \Big(1 + \frac{R}{100}\Big)^n$

By substituting values,

$\text{Value after 1 year}= 5 \times \Big(1 + \frac{2}{100}\Big)^4$ crores.

Hence, Option 1 is correct option.

Given,

C.I = ₹ 2,100

n = 2 years

R = 10%

By Formula,

C.I = Amount - Principal

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

Substituting the values in formula

$$\begin{aligned}\Rightarrow 2100 = P \times \Big(1 + \frac{10}{100}\Big)^2 - P\\[1em] \Rightarrow 2100 = P \times \Big(\frac{100 + 10}{100}\Big)^2 - P\\[1em] \Rightarrow 2100 = P \times \Big(\frac{110}{100}\Big)^2 - P\\[1em] \Rightarrow 2100 = P \times (1.10)^2 - P\\[1em] \Rightarrow 2100 = 1.21P - P \\[1em] \Rightarrow 2100 = 0.21P \\[1em] P = \frac{2100}{0.21} \\[1em] P = 10,000\end{aligned}$$

To calculate the simple interest on the same principal.

T = 2 years

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

Substituting the Values,

I = $\frac{10000 \times 10 \times 2}{100}$ = ₹ 2,000

Hence, Option 3 is correct option.

Given,

Present value of machine (V) = ₹ 6,25,000

r = 20%

Depreciated value = ₹ 6,25,000 - ₹ 3,69,000 = ₹ 2,56,000

By formula,

Value of machine after n years = ₹ $V \times \Big(1 - \frac{r}{100}\Big)^n$

Substituting the values in formula,

$$\begin{aligned}\Rightarrow 256000 = 625000 \times \Big(1 - \frac{20}{100}\Big)^n\\[1em] \Rightarrow \Big(1 - \frac{20}{100}\Big)^n=\frac{256000}{625000} \\[1em] \Rightarrow \Big(1 - \frac{1}{5}\Big)^n=\frac{256}{625} \\[1em] \Rightarrow \Big(\frac{4}{5}\Big)^n=\Big(\frac{4}{5}\Big)^4 \\[1em] \Rightarrow n = 4\end{aligned}$$

Hence, It will take 4 years for the price of the car to depreciate.

Given,

P = 6,31,680

r₁ = 4%

r₂ = 6%

n = 1 year

By formula,

Population n years ago = $\frac{P}{\Big(1 - \frac{r_1}{100}\Big) \times \Big(1 - \frac{r_2}{100}\Big)}$

Substituting the values in formula,

$$\begin{aligned}\text{Population 2 years ago}= \frac{631680}{\Big(1 - \frac{4}{100}\Big) \times \Big(1 - \frac{6}{100}\Big)} \\[1em] = \frac{631680}{\Big(\frac{100 - 4}{100}\Big) \times \Big(\frac{100 - 6}{100}\Big)} \\[1em] = \frac{631680}{\Big(\frac{96}{100}\Big) \times \Big(\frac{94}{100}\Big)} \\[1em] = \frac{631680 \times 100 \times 100}{96 \times 94} \\[1em] = \frac{6316800000}{9024} \\[1em] = 700000.\end{aligned}$$

Hence, the population of the town 2 years ago was 700000.