Compound Interest [Basic Concepts]

Given,

Interest = ₹ 800

T = 1 year

R = 10%

Let money on which interest is ₹ 800 be P.

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow 800 = \frac{P \times 10 \times 1}{100} \\[1em] \Rightarrow P = \frac{800 \times 100}{10} = 8000.\end{aligned}$$

We know that,

S.I. and C.I. for first year are equal.

C.I. for first year = ₹ 800

For second year :

P = ₹ 8000 + ₹ 800 = ₹ 8800

R = 10%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{8800 \times 10 \times 1}{100}$ = ₹ 880.

Hence, Option 1 is the correct option.

Given,

P = ₹ 5000

T = 6 months or $\frac{1}{2}$ years

R = 10%

Interest = $\frac{P \times R \times T}{100} = \frac{5000 \times 10 \times \frac{1}{2}}{100}$ = ₹ 250.

Hence, Option 2 is the correct option.

Since, time is 1 year so interest will be compounded half-yearly twice.

For first half-year :

P = ₹ 5000

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

R = 10%

By formula,

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

Substituting values we get :

Interest = $\frac{5000 \times \frac{1}{2} \times 10}{100}$ = ₹ 250.

For second half-year :

P = ₹ 5000 + ₹ 250 = ₹ 5250

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

R = 10%

By formula,

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

Substituting values we get :

Interest = $\frac{5250 \times \frac{1}{2} \times 10}{100}$ = ₹ 262.50

Total compound interest = ₹ 250 + ₹ 262.50 = ₹ 512.50

Hence, Option 3 is the correct option.

For first year :

P = ₹ 20,000

R = 12%

T = 1 year

By formula,

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

Substituting values we get :

Interest = $\frac{20000 \times 12 \times 1}{100}$ = ₹ 2400.

Amount = P + I = ₹ 20,000 + ₹ 2400 = ₹ 22400

For second year :

P = ₹ 22400

R = 12%

T = 1 year

By formula,

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

Substituting values we get :

Interest = $\frac{22400 \times 12 \times 1}{100}$ = ₹ 2688.

Hence, Option 3 is the correct option.

Let principal for first year be ₹ P.

For first year :

Interest = ₹ 1250

R = 5%

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow 1250 = \frac{P \times 5 \times 1}{100} \\[1em] \Rightarrow P = \frac{1250 \times 100}{5} \\[1em] \Rightarrow P = 25000.\end{aligned}$$

Amount = P + I = ₹ 25000 + ₹ 1250 = ₹ 26250

For second year :

P = ₹ 26250

R = 5%

T = 1 year

Interest = $\frac{P \times R \times T}{100} = \frac{26250 \times 5 \times 1}{100}$ = ₹ 1312.50

Hence, Option 1 is the correct option.

Let initial principal be ₹ x.

For first year :

P = x

R = 10%

T = 1 year

By formula,

I = $\frac{P \times R \times T}{100} = \frac{x \times 10 \times 1}{100} = \frac{x}{10}$.

Amount = P + I = x + $\frac{x}{10} = \frac{11x}{10}$

For second year :

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

R = 12%

T = 1 year

By formula,

I = $\frac{P \times R \times T}{100} = \frac{\frac{11}{10}x \times 12 \times 1}{100} = \frac{132x}{1000}$.

A = P + I = $\frac{11x}{10} + \frac{132x}{1000} = \frac{1100x + 132x}{1000} = \frac{1232x}{1000}$.

Compound interest = Final Amount - Initial Principal

= $\frac{1232x}{1000} - x = \frac{1232x - 1000x}{1000} = \frac{232x}{1000}$.

Percentage increase = $\frac{\text{Compound interest}}{\text{Initial principal}} \times 100$

$= \frac{\frac{232x}{1000}}{x} \times 100 = \frac{232x}{1000x} \times 100$ = 23.2%

Hence, Option 2 is the correct option.

For 2nd year :

P = ₹ 16800

R = 5%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{16800 \times 5 \times 1}{100}$ = ₹ 840.

A = P + I = ₹ 16800 + ₹ 840 = ₹ 17640.

For 3rd year :

P = ₹ 17640

R = 5%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{17640 \times 5 \times 1}{100}$ = ₹ 882.

A = P + I = ₹ 17640 + ₹ 882 = ₹ 18522.

For 4th year :

P = ₹ 18522

R = 5%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{18522 \times 5 \times 1}{100}$ = ₹ 926.10.

A = P + I = ₹ 18522 + ₹ 926.10 = ₹ 19448.10

For 5th year :

P = ₹ 19488.10

R = 5%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{19448.10 \times 5 \times 1}{100}$ = ₹ 972.405.

A = P + I = ₹ 19448.10 + ₹ 972.405 = ₹ 20420.505

Year	Initial amount (₹)	Interest (₹)	Final amount (₹)
1st	16000	800	16800
2nd	16800	840	17640
3rd	17640	882	18522
4th	18522	926.10	19448.10
5th	19448.10	972.405	20420.505

Hence, amount in 4 years = ₹ 19448.10

For first year :

P = ₹ 8000

T = 1 year

R = 15%

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

$= \frac{8000 \times 15 \times 1}{100}$ = ₹ 1200.

Amount = P + I = ₹ 8000 + ₹ 1200 = ₹ 9200.

For second year :

P = ₹ 9200

T = 1 year

R = 15%

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

$= \frac{9200 \times 15 \times 1}{100}$ = ₹ 1380.

Amount = P + I = ₹ 9200 + ₹ 1380 = ₹ 10580.

For next half year :

P = ₹ 10580

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

R = 15%

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

$= \frac{10580 \times 15 \times \frac{1}{2}}{100}$ = ₹ 793.50

Amount = P + I = ₹ 10580 + ₹ 793.50 = ₹ 11373.50

Compound interest = Final amount - Initial principal

= ₹ 11373.50 - ₹ 8000 = ₹ 3373.50

Hence, amount = ₹ 11373.50 and compound interest = ₹ 3373.50

For first year :

P = ₹ 4600

T = 1 year

R = 10%

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

$= \frac{4600 \times 10 \times 1}{100}$ = ₹ 460.

Amount = P + I = ₹ 4600 + ₹ 460 = ₹ 5060.

For second year :

P = ₹ 5060

T = 1 year

R = 12%

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

$= \frac{5060 \times 12 \times 1}{100}$ = ₹ 607.2.

Amount = P + I = ₹ 5060 + ₹ 607.2 = ₹ 5667.20

Compound interest = Final amount - Initial principal

= ₹ 5667.20 - ₹ 4600 = ₹ 1067.20

Hence, compound interest = ₹ 1067.20 and amount = ₹ 5667.20

For first year :

P = ₹ 75000

T = 1 year

R = 15%

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

$= \frac{75000 \times 15 \times 1}{100}$ = ₹ 11250.

Amount = P + I = ₹ 75000 + ₹ 11250 = ₹ 86250.

For second year :

P = ₹ 86250

T = 1 year

R = 15%

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

$= \frac{86250 \times 15 \times 1}{100}$ = ₹ 12937.50.

Amount = P + I = ₹ 86250 + ₹ 12937.50 = ₹ 99187.50

For third year :

P = ₹ 99187.50

T = 1 year

R = 16%

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

$= \frac{99187.50 \times 16 \times 1}{100}$ = ₹ 15870.

Amount = P + I = ₹ 99187.50 + ₹ 15870 = ₹ 115057.50

Hence, at the end of third year Meenal will get ₹ 115057.50

For first year :

P = ₹ 16000

T = 1 year

R = 10%

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

$= \frac{16000 \times 10 \times 1}{100}$ = ₹ 1600.

Amount = P + I = ₹ 16000 + ₹ 1600 = ₹ 17600.

For second year :

P = ₹ 17600

T = 1 year

R = 14%

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

$= \frac{17600 \times 14 \times 1}{100}$ = ₹ 2464

Amount = P + I = ₹ 17600 + ₹ 2464 = ₹ 20064

For third year :

P = ₹ 20064

T = 1 year

R = 15%

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

$= \frac{20064 \times 15 \times 1}{100}$ = ₹ 3009.60

Amount = P + I = ₹ 20064 + ₹ 3009.60 = ₹ 23073.60

Compound interest = Final amount - Initial principal

= ₹ 23073.60 - ₹ 16000 = ₹ 7073.60

Hence, amount = ₹ 23073.60 and compound interest = ₹ 7073.60

For first year :

P = ₹ 8000

T = 1 year

R = 10%

I = $\frac{P \times R \times T}{100} = \frac{8000 \times 1 \times 10}{100}$ = ₹ 800.

Amount = P + I = ₹ 8000 + ₹ 800 = ₹ 8800.

For second year :

P = ₹ 8800

T = 1 year

R = 10%

I = $\frac{P \times R \times T}{100} = \frac{8800 \times 1 \times 10}{100}$ = ₹ 880.

Hence, C.I. for second year = ₹ 880.

For first year :

P = ₹ 2400

T = 1 year

R = 5%

I = $\frac{P \times R \times T}{100} = \frac{2400 \times 1 \times 5}{100}$ = ₹ 120.

Amount = P + I = ₹ 2400 + ₹ 120 = ₹ 2520.

For second year :

P = ₹ 2520

T = 1 year

R = 5%

I = $\frac{P \times R \times T}{100} = \frac{2520 \times 1 \times 5}{100}$ = ₹ 126.

Amount = P + I = ₹ 2520 + ₹ 126 = ₹ 2646.

For next half-year :

P = ₹ 2646

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

R = 5%

I = $\frac{P \times R \times T}{100} = \frac{2646 \times \frac{1}{2} \times 5}{100}$ = ₹ 66.15

Amount = P + I = ₹ 2646 + ₹ 66.15 = ₹ 2712.15

Compound interest = Final amount - Initial principal

= ₹ 2712.15 - ₹ 2400 = ₹ 312.15 ≈ ₹ 312.

Hence, compound interest = ₹ 312.

For first year :

P = ₹ 2500

T = 1 year

R = 12%

I = $\frac{P \times R \times T}{100} = \frac{2500 \times 1 \times 12}{100}$ = ₹ 300

Amount = P + I = ₹ 2500 + ₹ 300 = ₹ 2800.

For second year :

P = ₹ 2800

T = 1 year

R = 12%

I = $\frac{P \times R \times T}{100} = \frac{2800 \times 1 \times 12}{100}$ = ₹ 336

Amount = P + I = ₹ 2800 + ₹ 336 = ₹ 3136.

Given,

After 2 years, A gave ₹ 2936 and a watch to B to clear the account. Let cost of watch be ₹ x.

∴ 3136 = 2936 + x

⇒ x = 3136 - 2936 = ₹ 200.

Hence, cost of watch = ₹ 200.

For first year :

P = ₹ 50000

T = 1 year

R = 6%

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

$= \frac{50000 \times 6 \times 1}{100}$ = ₹ 3000.

Amount = P + I = ₹ 50000 + ₹ 3000 = ₹ 53000.

For second year :

P = ₹ 53000

T = 1 year

R = 8%

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

$= \frac{53000 \times 8 \times 1}{100}$ = ₹ 4240

Amount = P + I = ₹ 53000 + ₹ 4240 = ₹ 57240

For third year :

P = ₹ 57240

T = 1 year

R = 10%

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

$= \frac{57240 \times 10 \times 1}{100}$ = ₹ 5724

Amount = P + I = ₹ 57240 + ₹ 5724 = ₹ 62964

Hence, ₹ 50000 will amount to ₹ 62964 in 3 years

Calculating simple interest :

P = ₹ 18000

R = 10%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{18000 \times 10 \times 1}{100}$ = 1800.

Amount Govind needs to return = P + I = ₹ 18000 + ₹ 1800 = ₹ 19800.

Calculating compound interest :

For 1st half year :

P = ₹ 18000

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

R = 10%

I = $\frac{P \times R \times T}{100} = \frac{18000 \times 10 \times \frac{1}{2}}{100}$ = 900.

Amount = P + I = ₹ 18000 + ₹ 900 = ₹ 18900.

For 2nd half-year :

P = ₹ 18900

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

R = 10%

I = $\frac{P \times R \times T}{100} = \frac{18900 \times 10 \times \frac{1}{2}}{100}$ = 945.

Amount Govind will get back = P + I = ₹ 18900 + ₹ 945 = ₹ 19845.

Gain = Amount Govind will get back - Amount Govind will return

= ₹ 19845 - ₹ 19800 = ₹ 45.

Hence, Govind will gain ₹ 45 in one year.

For first year :

P = ₹ 4000

T = 1 year

R = 8%

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

$= \frac{4000 \times 8 \times 1}{100}$ = ₹ 320.

Amount = P + I = ₹ 4000 + ₹ 320 = ₹ 4320.

For second year :

P = ₹ 4320

T = 1 year

R = 10%

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

$= \frac{4320 \times 10 \times 1}{100}$ = ₹ 432

Amount = P + I = ₹ 4320 + ₹ 432 = ₹ 4752

For third year :

P = ₹ 4752

T = 1 year

R = 10%

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

$= \frac{4752 \times 10 \times 1}{100}$ = ₹ 475.2

Amount = P + I = ₹ 4752 + ₹ 475.2 = ₹ 5227.20

Compound interest = Final amount - Initial principal

= ₹ 5227.20 - ₹ 4000 = ₹ 1227.20

Hence, compound interest = ₹ 1227.20

For one year and interest being compounded annually,

Simple interest = Compound interest

∴ Difference between S.I. and C.I. = 0.

Hence, Option 1 is the correct option.

At beginning of 2023 :

P = ₹ 2000

R = 8%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{2000 \times 8 \times 1}{100}$ = ₹ 160.

Amount = ₹ 2000 + ₹ 160 = ₹ 2160.

Since, ₹ 3000 is saved during 2023.

So, at beginning of 2024,

P = ₹ 2160 + ₹ 3000 = ₹ 5160

R = 8%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{5160 \times 8 \times 1}{100}$ = ₹ 412.80

C.I. earned upto 2024 = ₹ 412.80 + ₹ 160 = ₹ 572.80

Hence, Option 2 is the correct option.

For first year :

P = ₹ 1000

R = 10%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{1000 \times 10 \times 1}{100}$ = ₹ 100.

Amount = P + I = ₹ 1000 + ₹ 100 = ₹ 1100.

₹ 300 is repaid at end of each year.

Amount left at end of first year = ₹ 1100 - ₹ 300 = ₹ 800.

For second year :

P = ₹ 800

R = 10%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{800 \times 10 \times 1}{100}$ = ₹ 80.

Amount = P + I = ₹ 800 + ₹ 80 = ₹ 880.

Hence, Option 1 is the correct option.

For S.I. :

P = ₹ 4000

R = 10%

T = 2 years

I = $\frac{P \times R \times T}{100} = \frac{4000 \times 10 \times 2}{100}$ = ₹ 800.

For C.I. :

For 1st year :

P = ₹ 4000

T = 1 year

R = 10%

I = $\frac{P \times R \times T}{100} = \frac{4000 \times 10 \times 1}{100}$ = ₹ 400.

Amount = P + I = ₹ 4000 + ₹ 400 = ₹ 4400

For 2nd year :

P = ₹ 4400

R = 10%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{4400 \times 10 \times 1}{100}$ = ₹ 440.

Amount = P + I = ₹ 4400 + ₹ 440 = ₹ 4840.

C.I. = Final amount - Initial principal = ₹ 4840 - ₹ 4000 = ₹ 840

Difference between C.I. and S.I. = ₹ 840 - ₹ 800 = ₹ 40.

Hence, Option 4 is the correct option.

Let principal amount be ₹ x.

For S.I. :

P = ₹ x

R = 5%

T = 2 year

I = $\frac{P \times R \times T}{100} = \frac{x \times 5 \times 2}{100} = \frac{x}{10}$.

For C.I. :

For first year :

P = ₹ x

R = 5%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{x \times 5 \times 1}{100} = \frac{x}{20}$.

Amount = P + I = $x + \frac{x}{20} = \frac{21x}{20}$

For second year :

P = ₹ $\frac{21x}{20}$

R = 5%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{\frac{21x}{20} \times 5 \times 1}{100} = \frac{21x}{400}$.

Amount = P + I = $\frac{21x}{20} + \frac{21x}{400} = \frac{420x + 21x}{400} = \frac{441x}{400}$.

C.I. = Final amount - Initial Principal

= $\frac{441x}{400} - x = \frac{441x - 400x}{400} = \frac{41x}{400}$.

Given,

Difference between S.I. and C.I. = ₹ 10

$$\begin{aligned}\Rightarrow \frac{41x}{400} - x = 10 \\[1em] \Rightarrow \frac{41x - 40x}{400} = 10 \\[1em] \Rightarrow \frac{x}{400} = 10 \\[1em] \Rightarrow x = 4000.\end{aligned}$$

Hence, Option 3 is the correct option.

For S.I. :

P = ₹ 4000

T = 2 years

R = 8%

S.I. = $\frac{P \times R \times T}{100} = \frac{4000 \times 8 \times 2}{100}$ = ₹ 640.

For C.I. :

For 1st year :

P = ₹ 4000

T = 1 year

R = 8%

I = $\frac{P \times R \times T}{100} = \frac{4000 \times 8 \times 1}{100}$ = ₹ 320.

Amount = P + I = ₹ 4000 + ₹ 320 = ₹ 4320

For 2nd year :

P = ₹ 4320

T = 1 year

R = 8%

I = $\frac{P \times R \times T}{100} = \frac{4320 \times 8 \times 1}{100}$ = ₹ 345.60

C.I. = ₹ 320 + ₹ 345.60 = ₹ 665.60

Difference between C.I. and S.I. = C.I. - S.I. = ₹ 665.60 - ₹ 640 = ₹ 25.60

Hence, difference between C.I. and S.I. = ₹ 25.60

Let sum of money be ₹ x.

For first year :

P = ₹ x

T = 1 year

R = 8%

I = $\frac{P \times R \times T}{100} = \frac{x \times 8 \times 1}{100} = \frac{2x}{25}$.

A = P + I = $x + \frac{2x}{25} = \frac{25x + 2x}{25} = \frac{27x}{25}$.

For 2nd year :

P = ₹ $\frac{27x}{25}$

T = 1 year

R = 8%

I = $\frac{P \times R \times T}{100} = \frac{\frac{27x}{25} \times 8 \times 1}{100} = \frac{216x}{2500}$.

Given,

Interest for 2nd year exceeds interest for first year by ₹ 96

$$\begin{aligned}\therefore \frac{216x}{2500} - \frac{2x}{25} = 96 \\[1em] \Rightarrow \frac{216x - 200x}{2500} = 96 \\[1em] \Rightarrow \frac{16x}{2500} = 96 \\[1em] \Rightarrow x = \frac{96}{16} \times 2500 \\[1em] \Rightarrow x = 15000.\end{aligned}$$

Hence, sum of money = ₹ 15000.

(i) For first year :

P = ₹ 5600

R = 14%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{5600 \times 14 \times 1}{100}$ = ₹ 784

Hence, interest for the first year = ₹ 784.

(ii) Amount after first year = Principal for first year + Interest

= ₹ 5600 + ₹ 784 = ₹ 6384.

Hence, amount at the end of first year = ₹ 6384.

(iii) For second year :

P = ₹ 5600 + ₹ 784 = ₹ 6384

R = 14%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{6384 \times 14 \times 1}{100}$ = ₹ 893.76

Hence, interest for the second year = ₹ 894.

Savings at the end of every year = ₹ 3000

For 2nd year :

P = ₹ 3000

R = 10%

T = 1 year

⇒ I = $\frac{P \times R \times T}{100} = \frac{3000 \times 10 \times 1}{100}$ = ₹ 300

A = ₹ 3000 + ₹ 300 = ₹ 3300

Savings at end of second year = ₹ 3000

For third year :

P = 3000 + 3300 = ₹ 6300

R = 10%

T = 1 year

⇒ I = $\frac{P \times R \times T}{100} = \frac{6300 \times 10 \times 1}{100}$ = ₹ 630

A = ₹ 6300 + ₹ 630 = ₹ 6930

Savings at end of second year = ₹ 3000

Amount at the end of 3rd year

A = ₹ 6930 + ₹ 3000 = ₹ 9930.

Hence, amount at the end of third year = ₹ 9930.

For first year :

P = ₹ 12500

R = 12%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{12500 \times 12 \times 1}{100}$ = ₹ 1500.

Amount = ₹ 12500 + ₹ 1500 = ₹ 14000

For second year :

P = ₹ 14000

R = 15%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{14000 \times 15 \times 1}{100}$ = ₹ 2100

Amount = ₹ 14000 + ₹ 2100 = ₹ 16100

For third year :

P = ₹ 16100

R = 18%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{16100 \times 18 \times 1}{100}$ = ₹ 2898.

Difference between interest of first year and third year = ₹ 2898 - ₹ 1500 = ₹ 1398.

Hence, difference between interest of first year and third year = ₹ 1398.

For first year :

P = ₹ 6000

T = 1 year

R = 5 %

I = $\frac{P \times R \times T}{100} = \frac{6000 \times 5 \times 1}{100}$ = ₹ 300

Amount = ₹ 6000 + ₹ 300 = ₹ 6300

Amount payed at end of first year = ₹ 1200

Amount left at beginning of second year = ₹ 6300 - ₹ 1200 = ₹ 5100.

For second year :

P = ₹ 5100

R = 5%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{5100 \times 5 \times 1}{100}$ = ₹ 255

Amount = ₹ 5100 + ₹ 255 = ₹ 5355

Amount payed at end of second year = ₹ 1200

Amount left at beginning of third year = ₹ 5355 - ₹ 1200 = ₹ 4155.

Hence, amount left at beginning of third year = ₹ 4155.

For 1st half-year

P = ₹ 5000

R = 12%

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

I = $\frac{P \times R \times T}{100} = \frac{5000 \times 12 \times \frac{1}{2}}{100}$ = ₹ 300.

Amount = P + I = ₹ 5000 + ₹ 300= ₹ 5300.

Money paid at the end of 1st half year = ₹ 1800

Balance money for 2nd half-year = ₹ 5300- ₹ 1800 = ₹ 3500.

For 2nd half-year

P = ₹ 3500

R = 12%

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

I = $\frac{P \times R \times T}{100} = \frac{3500 \times 12 \times \frac{1}{2}}{100}$ = ₹ 210.

Amount = ₹ 3500 + ₹ 210 = ₹ 3710

Money paid at the end of 2nd half-year = ₹ 1800

Balance money for 3rd half-year = ₹ 3710 - ₹ 1800 = ₹ 1910

For 3rd half-year

P = ₹ 1910

R = 12%

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

Interest = $\frac{P \times R \times T}{100} = \frac{1910 \times 12 \times \frac{1}{2}}{100}$ = ₹ 114.60

Amount = ₹ 1910 + ₹ 114.60 = ₹ 2024.60

Hence, amount to be paid at the end of 18 months = ₹ 2024.60

Let sum of money be ₹ x.

For S.I. :

P = ₹ x

R = 10%

T = 1 years

I = $\frac{P \times R \times T}{100} = \frac{x \times 10 \times 1}{100} = \frac{x}{10}$.

For C.I. :

For first half-year :

P = ₹ x

R = 10%

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

I = $\frac{P \times R \times T}{100} = \frac{x \times 10 \times \frac{1}{2}}{100} = \frac{x}{20}$.

Amount = P + I = $x + \frac{x}{20} = \frac{21x}{20}$.

For second year :

P = ₹ $\frac{21x}{20}$

R = 10%

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

I = $\frac{P \times R \times T}{100} = \frac{\frac{21x}{20} \times 10 \times \frac{1}{2}}{100} = \frac{21x}{400}$.

Amount = P + I = $\frac{21x}{20} + \frac{21x}{400} = \frac{420x + 21x}{400} = \frac{441x}{400}$.

C.I. = Final amount - Initial principal

= $\frac{441x}{400} - x = \frac{41x}{400}$.

Given,

Difference between C.I. and S.I. = ₹ 180

$$\begin{aligned}\therefore \frac{41x}{400} - \frac{x}{10} = 180 \\[1em] \Rightarrow \frac{41x - 40x}{400} = 180 \\[1em] \Rightarrow \frac{x}{400} = 180 \\[1em] \Rightarrow x = 72000.\end{aligned}$$

Hence, sum lent out = ₹ 72000.

Let original value of machine be ₹ x.

For first year :

P = ₹ x

R (of depreciation) = 15%

T = 1 year

Depreciation = $\frac{P \times R \times T}{100} = \frac{x \times 15 \times 1}{100} = \frac{3x}{20}$.

New value = P - Depreciation = $x - \frac{3x}{20} = \frac{17x}{20}$.

For second year :

P = ₹ $\frac{17x}{20}$

R (of depreciation) = 15%

T = 1 year

Depreciation = $\frac{P \times R \times T}{100} = \frac{\frac{17x}{20} \times 15 \times 1}{100} = \frac{51x}{400}$.

Given,

Depreciation in second year = ₹ 5355

$$\begin{aligned}\therefore \frac{51x}{400} = 5355 \\[1em] \Rightarrow x = \frac{5355 \times 400}{51} \\[1em] \Rightarrow x = 105 \times 400 = 42000.\end{aligned}$$

Hence, original value of machine = ₹ 42000.

For first year :

P = ₹ 10000

R = 5%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{10000 \times 5 \times 1}{100}$ = ₹ 500

Amount = P + I = ₹ 10000 + ₹ 500 = ₹ 10500

35% of the sum borrowed is repaid at the end of the first year.

Sum repaid = $\frac{35}{100} \times 10000$ = ₹ 3500

Sum left = Amount - Sum repaid = ₹ 10500 - ₹ 3500 = ₹ 7000

For second year :

P = ₹ 7000

R = 5%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{7000 \times 5 \times 1}{100}$ = ₹ 350

Amount = P + I = ₹ 7000 + ₹ 350 = ₹ 7350

42% of the sum borrowed is repaid at the end of the second year.

Sum repaid = $\frac{42}{100} \times 10000$ = ₹ 4200

Sum left = Amount - Sum repaid = ₹ 7350 - ₹ 4200 = ₹ 3150

For third year :

P = ₹ 3150

R = 5%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{3150 \times 5 \times 1}{100}$ = ₹ 157.50

Amount = P + I = ₹ 3150 + ₹ 157.50 = ₹ 3307.50

Hence, amount to be paid after the end of third year = ₹ 3307.50

Difference between C.I. of two successive years = ₹ 360 - ₹ 300 = ₹ 60

∴ ₹ 60 is the interest of one year on ₹ 300.

By formula,

Rate of interest = $\frac{100 \times I}{P \times T} = \frac{100 \times 60}{300 \times 1}$ = 20%.

Hence, Option 4 is the correct option.

Difference between amounts of two successive years = ₹ 6000 - ₹ 5000 = ₹ 1000

∴ ₹ 1000 is the interest of one year on ₹ 5000.

By formula,

Rate of interest = $\frac{100 \times I}{P \times T} = \frac{100 \times 1000}{5000 \times 1}$ = 20%.

Hence, Option 2 is the correct option.

Let C.I. in 2019 be ₹ x.

Difference in C.I. of successive years = ₹ 3850 - ₹ x.

So,

∴ ₹ (3850 - x) is the interest of one year on ₹ x.

By formula,

Rate of interest = $\frac{100 \times I}{P \times T}$

Substituting values we get :

$$\begin{aligned}\Rightarrow 10 = \frac{100 \times (3850 - x)}{x \times 1} \\[1em] \Rightarrow 10x = 100(3850 - x) \\[1em] \Rightarrow x = 10(3850 - x) \\[1em] \Rightarrow x = 38500 - 10x \\[1em] \Rightarrow 11x = 38500 \\[1em] \Rightarrow x = \frac{38500}{11} \\[1em] \Rightarrow x = 3500.\end{aligned}$$

Hence, Option 1 is the correct option.

Let money amounts to ₹ x.

Difference between amounts of two successive years = ₹ x - ₹ 4500

∴ ₹ (x - 4500) is the interest of one year on ₹ 4500.

By formula,

Rate of interest = $\frac{100 \times I}{P \times T}$

Substituting values we get :

$$\begin{aligned}\Rightarrow 12 = \frac{100 \times (x - 4500)}{4500 \times 1} \\[1em] \Rightarrow 12 \times 45 = x - 4500 \\[1em] \Rightarrow 540 = x - 4500 \\[1em] \Rightarrow x = 540 + 4500 \\[1em] \Rightarrow x = 5040.\end{aligned}$$

Hence, Option 2 is the correct option.

Difference between C.I. of two successive years = ₹ 690 - ₹ 600 = ₹ 90

∴ ₹ 90 is the interest of one year on ₹ 600.

By formula,

Rate of interest = $\frac{100 \times I}{P \times T} = \frac{100 \times 90}{600 \times 1}$ = 15%.

Hence, Option 3 is the correct option.

Difference between amounts of two successive years = ₹ 2760 - ₹ 2400 = ₹ 360

∴ ₹ 360 is the interest of one year on ₹ 2400.

By formula,

Rate of interest = $\frac{100 \times I}{P \times T} = \frac{100 \times 360}{2400 \times 1}$ = 15%.

Hence, Option 2 is the correct option.

(i) Given,

Amount in two years = ₹ 5292

Amount in three years = ₹ 5556.60

Difference between the amounts of two successive years

= ₹ 5556.60 - ₹ 5292 = ₹ 264.60

∴ ₹ 264.60 is the interest of one year on ₹ 5292.

By formula,

Rate of interest = $\frac{I \times 100}{P \times T} = \frac{264.60 \times 100}{5292 \times 1} = \frac{100}{20}$ = 5%.

Hence, rate of interest = 5%.

(ii) Let original sum be ₹ x.

For 1st year :

P = ₹ x

R = 5%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{x \times 5 \times 1}{100} = \frac{x}{20}$.

Amount = P + I = $x + \frac{x}{20} = \frac{21x}{20}$.

For second year :

P = ₹ $\frac{21x}{20}$

R = 5%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{\frac{21x}{20} \times 5 \times 1}{100} = \frac{21x}{400}$.

Amount = P + I = $\frac{21x}{20} + \frac{21x}{400} = \frac{420x + 21x}{400} = \frac{441x}{400}$.

Given,

Amount after 2 years = ₹ 5292

$$\begin{aligned}\therefore \frac{441x}{400} = 5292 \\[1em] \Rightarrow x = \frac{5292 \times 400}{441} = 4800.\end{aligned}$$

Hence, original sum = ₹ 4800.

(i) Given,

Mohit invests ₹ 8000 (P)

Amount at end of one year = ₹ 9440

Interest = Amount - P = ₹ 9440 - ₹ 8000 = ₹ 1440.

∴ ₹ 1440 is the interest of one year on ₹ 8000.

By formula,

Rate of interest = $\frac{I \times 100}{P \times T} = \frac{1440 \times 100}{8000 \times 1} = \frac{144}{8}$ = 18%.

Hence, rate of interest = 18%.

(ii) For second year :

P = ₹ 9440

R = 18%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{9440 \times 18 \times 1}{100} = \frac{169920}{100}$ = ₹ 1699.20

Amount = P + I = ₹ 9440 + ₹ 1699.20 = ₹ 11,139.20

Hence, amount at the end of second year = ₹ 11,139.20

(iii) For third year :

P = ₹ 11,139.20

R = 18%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{11139.20 \times 18 \times 1}{100} = \frac{200505.60}{100}$ = ₹ 2,005.06

Hence, interest accrued in third year = ₹ 2,005.06

Difference between C.I. of two successive years = ₹ 1197.90 - ₹ 1089 = ₹ 108.9

∴ ₹ 108.9 is the interest of one year on ₹ 1089.

By formula,

Rate of interest = $\frac{100 \times I}{P \times T} = \frac{100 \times 108.9}{1089 \times 1}$ = 10%.

Let sum of money be ₹ x.

For first year :

P = ₹ x

R = 10%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{x \times 10 \times 1}{100} = \frac{x}{10}$.

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

For second year :

P = ₹ $\frac{11x}{10}$

R = 10%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{\frac{11x}{10} \times 10 \times 1}{100} = \frac{11x}{100}$.

Given,

C.I. for 2nd year = ₹ 1089

$$\begin{aligned}\therefore \frac{11x}{100} = 1089 \\[1em] \Rightarrow x = \frac{1089 \times 100}{11} = 9900.\end{aligned}$$

Hence, rate of interest = 10% and sum of money = ₹ 9900.

Difference between C.I. of two successive years = ₹ 7410 - ₹ 5700 = ₹ 1710

∴ ₹ 1710 is the interest of one year on ₹ 5700.

By formula,

Rate of interest = $\frac{100 \times I}{P \times T} = \frac{100 \times 1710}{5700 \times 1}$ = 30%.

Hence, the rate of interest = 30%.

(i) Difference between depreciation in value between the first and second years is ₹ 4,000 - ₹ 3,600 = ₹ 400

So, the depreciation of one year on ₹ 4,000 = ₹ 400

By formula,

Rate of depreciation = $\frac{100 \times I}{P \times T} = \frac{100 \times 400}{4000 \times 1}$ = 10%.

Hence, rate of depreciation = 10%.

(ii) Let cost of machine be ₹ x.

Given,

Depreciation in first year = ₹ 4000

Depreciation % = 10%

$\therefore \frac{x \times 1\times 10}{100} = 4000$

x = 40000.

Hence, original cost of machine = ₹ 40000.

(iii) Value of machine at beginning of third year = Original value - Depreciation in first and second years

= ₹ 40000 - (₹ 4000 + ₹ 3600)

= ₹ 40000 - ₹ 7600

= ₹ 32400.

For third year :

P = ₹ 32400

T = 1 year

Depreciation % = 10%

Depreciation = $\frac{P \times R \times T}{100} = \frac{32400 \times 10 \times 1}{100}$ = ₹3240.

Value at the end of third year = ₹ 32400 - ₹ 3240 = ₹ 29160.

Hence, value of machine at the end of third year = ₹ 29160.

(i) For first year :

P = ₹ 12800

R = 10%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{12800 \times 10 \times 1}{100}$ = ₹ 1280.

Amount = P + I = ₹ 12800 + ₹ 1280 = ₹ 14080.

Hence, sum due at the end of first year = ₹ 14080.

(ii) For second year :

P = ₹ 14080

R = 10%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{14080 \times 10 \times 1}{100}$ = ₹ 1408.

Hence, interest for second year = ₹ 1408.

(iii) Amount at end of second year = P + I = ₹ 14080 + ₹ 1408 = ₹ 15488.

For third year :

P = ₹ 15488

R = 10%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{15488 \times 10 \times 1}{100}$ = ₹ 1548.80

Amount = P + I = ₹ 15488 + ₹ 1548.80 = ₹ 17036.80

Hence, amount due at end of third year = ₹ 17036.80

Difference between C.I. of two successive half-years = ₹ 760.50 - ₹ 650 = ₹ 110.50

∴ ₹ 110.50 is the interest of $\frac{1}{2}$ year on ₹ 650.

By formula,

Rate of interest = $\frac{100 \times I}{P \times T} = \frac{100 \times 110.50}{650 \times \frac{1}{2}} = \frac{22100}{650}$ = 34%.

Hence, the rate of interest = 34%.

(i) Difference between C.I. of two successive half-years = ₹ 15600 - ₹ 15000 = ₹ 600

∴ ₹ 600 is the interest of $\frac{1}{2}$ year on ₹ 15000.

By formula,

Rate of interest = $\frac{100 \times I}{P \times T} = \frac{100 \times 600}{15000 \times \frac{1}{2}} = \frac{120}{15}$ = 8%

Hence, the rate of interest = 8%.

(ii) For 2nd half-year :

P = ₹ 15600

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

R = 8%

I = $\frac{P \times R \times T}{100} = \frac{15600 \times 8 \times \frac{1}{2}}{100}$ = ₹ 624.

Amount = P + I = ₹ 15600 + ₹ 624 = ₹ 16224.

For 3rd half-year :

P = ₹ 16224

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

R = 8%

I = $\frac{P \times R \times T}{100} = \frac{16224 \times 8 \times \frac{1}{2}}{100}$ = ₹ 648.96

Amount = P + I = ₹ 16224 + ₹ 648.96 = ₹ 16872.96

Hence, amount needed to pay at the end of 18 months = ₹ 16872.96

(i) For first year :

P = ₹ 8000

R = 7%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{8000 \times 7 \times 1}{100}$ = ₹ 560.

Amount = P + I = ₹ 8000 + ₹ 560 = ₹ 8560.

Amount paid back at end of first year = ₹ 3560

Amount left = ₹ 8560 - ₹ 3560 = ₹ 5000.

For second year :

P = ₹ 5000

R = 7%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{5000 \times 7 \times 1}{100}$ = ₹ 350.

Amount = P + I = ₹ 5000 + ₹ 350 = ₹ 5350

Hence interest paid in second year = ₹ 350.

(ii) Total interest paid in two years = ₹ 350 + ₹ 560 = ₹ 910.

Hence, total interest paid in two years = ₹ 910.

(iii) Amount of money paid in two years to clear the debt = Amount at end of 2nd year + Money paid back at end of first year

= ₹ 5350 + ₹ 3560 = ₹ 8910.

Hence, total amount of money paid in two years to clear the debt = ₹ 8910.

Let sum of money be ₹ x.

For first year :

P = ₹ x

R = 10%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{x \times 10 \times 1}{100} = \frac{x}{10}$

Amount = P + I = $x + \frac{x}{10} = \frac{11x}{10}$

For second year :

P = ₹ $\frac{11x}{10}$

R = 10%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{\frac{11x}{10} \times 10 \times 1}{100} = \frac{11x}{100}$

Amount = P + I = $\frac{11x}{10} + \frac{11x}{100} = \frac{110x + 11x}{100} = \frac{121x}{100}$.

For third year :

P = ₹ $\frac{121x}{100}$

R = 10%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{\frac{121x}{100} \times 10 \times 1}{100} = \frac{121x}{1000}$.

Given,

Interest for the third year exceeds the interest of the first year by ₹ 252.

$$\begin{aligned}\therefore \frac{121x}{1000} - \frac{x}{10} = 252 \\[1em] \Rightarrow \frac{121x - 100x}{1000} = 252 \\[1em] \Rightarrow \frac{21x}{1000} = 252 \\[1em] \Rightarrow x = \frac{252 \times 1000}{21} = 12000.\end{aligned}$$

Hence, sum = ₹ 12000.

30% of sum borrowed = $\frac{30}{100} \times 10000$ = ₹ 3000.

So, at the end of each year ₹ 3000 is returned back.

For first year :

P = ₹ 10000

R = 10%

T = 1 year

I = $\frac{P \times R \times T}{1000} = \frac{10000 \times 10 \times 1}{100}$ = ₹ 1000

Amount = P + I = ₹ 10000 + ₹ 1000 = ₹ 11000.

Amount left to pay at end of first year = ₹ 11000 - ₹ 3000 = ₹ 8000.

For second year :

P = ₹ 8000

R = 10%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{8000 \times 10 \times 1}{100}$ = ₹ 800

Amount = P + I = ₹ 8000 + ₹ 800 = ₹ 8800.

Amount left to pay at end of second year = ₹ 8800 - ₹ 3000 = ₹ 5800.

Hence, amount left to pay after second year = ₹ 5800.

For first year :

P = ₹ 10000

R = 10%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{10000 \times 10 \times 1}{100}$ = ₹ 1000

Amount = P + I = ₹ 10000 + ₹ 1000 = ₹ 11000.

Amount paid back = 20% of the amount for that year

= $\frac{20}{100} \times 11000$ = ₹ 2200

Amount left to pay at end of first year = ₹ 11000 - ₹ 2200 = ₹ 8800.

For second year :

P = ₹ 8800

R = 10%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{8800 \times 10 \times 1}{100}$ = ₹ 880

Amount = P + I = ₹ 8800 + ₹ 880 = ₹ 9680.

Amount paid back = 20% of the amount for that year

= $\frac{20}{100} \times 9680$ = ₹ 1936

Amount left to pay at end of second year = ₹ 9680 - ₹ 1936 = ₹ 7744.

Hence, amount left to pay after second year = ₹ 7744.

Given, Rate of interest = 10% per annum

Interest accrued during the year 2024 = ₹1,331

Let ₹ P be the principal amount at the beginning of 2022.

$$\begin{aligned}\text{Interest for the year 2022} = \frac{\text{P} \times \text{R} \times \text{T}}{100} \\[1em] = \frac{\text{P} \times 10 \times 1}{100} \\[1em] = \frac{\text{P}}{10}\end{aligned}$$

Amount at end of first year, 2022 = P + I

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

$$\begin{aligned}\text{Interest for the year 2023 }= \frac{\frac{\text{11P}}{10} \times 10 \times 1}{100}\\[1em] = \frac{\text{11P}}{100}\end{aligned}$$

Amount at end of second year, 2023 = P + I

= $\frac{\text{11P}}{10} + \frac{\text{11P}}{100} = \frac{\text{110P + 11P}}{100} = \frac{\text{121P}}{100}$

$$\begin{aligned}\text{Interest for the year 2024 }= \frac{\frac{\text{121P}}{100} \times 10 \times 1}{100}\\[1em] = \frac{\text{121P}}{1000}\end{aligned}$$

It is given, the interest accrued in 2024 = ₹1,331

$$\begin{aligned}\Rightarrow \frac{\text{121P}}{1000} = 1331\\[1em] \Rightarrow \text{P} = \frac{1000 \times 1331}{121}\\[1em] \Rightarrow \text{P} = 11,000\end{aligned}$$

Interest for year 2022 = $\frac{P}{10} = \frac{11,000}{10}$ = 1,100.

Hence, option 2 is correct option.

Given, the population of a small village = 45,000

Rate of increase = 5%

Increased population = 5% of 45,000

The population during the year 2024 = the population of 2023 + increased population

= 45,000 + 5% of 45,000

Hence, option 1 is correct option.

Given, rate of interest = 10%

Let sum of money be ₹ P and time be n years.

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

₹ P doubles itself in n years.

A = 2P

Substituting values in formula, we get :

$$\begin{aligned}\Rightarrow 2P = P\Big(1 + \frac{10}{100}\Big)^n \\[1em] \Rightarrow 2 = (1 + 0.1)^n \\[1em] \Rightarrow 2 = (1.1)^n \text{ ............(1)}\end{aligned}$$

Substituting, n = 5 in R.H.S. of equation (1),

(1.1)⁵ = 1.61

Thus, n ≠ 5.

Substituting, n = 10 in R.H.S. of equation (1),

(1.1)¹⁰ = 2.59

Thus, n ≠ 10.

Substituting, n = 8 in R.H.S. of equation (1),

(1.1)⁸ = 2.14

Thus, n ≠ 8.

Hence, option 4 is correct option.

Let initial value of machine be ₹ x.

For first year :

P = ₹ x

Rate of depreciation = 10%

n = 2 years

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

Depreciation = Initial value - Final value = $x - \frac{81x}{100} = \frac{19x}{100}$.

By formula,

$$\begin{aligned}\text{Percentage depreciated }= \frac{\text{Total depreciation}}{\text{Initial value}} \times 100\\[1em] = \frac{\frac{19x}{100}}{x} \times 100\\[1em] = \frac{19}{100} \times 100\\[1em] = 19%\end{aligned}$$

Hence, option 3 is the correct option.

We know that,

A in 3 years = Principal + C.I. for 3 years .............(1)

A in 2 years = Principal + C.I. for 2 years .............(2)

Subtracting equation (2) from (1), we get :

A in 3 years - A in 2 years = Principal + C.I. for 3 years - (Principal + C.I. for 2 years)

A in 3 years - A in 2 years = C.I. for 3 years - C.I. for 2 years

A in 3 years - A in 2 years = C.I. for 3rd year

∴ 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 the rate of interest and n be time.

We know that,

⇒ C.I. = Amount - Principal

As we know that Principal of 5th year = Amount of 4th year

⇒ C.I.₅ = A₅ - A₄

⇒ Interest of 5th year = Amount in 5 years - Amount in 4 years

So, reason (R) is true.

Thus, proved

⇒ Interest for 5th year = Amount in 5 years - Amount in 4 years

⇒ Interest for 5th year = P + Interest in 5 years - (P + Interest in 4 years)

⇒ Interest for 5th year = Interest in 5 years - Interest in 4 years

So, assertion (A) is true.

∴ 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 the rate of interest and n be time.

Rate of C.I. accured = $\frac{\text{C.I. accrued in a particular year}}{\text{Amount in previous year}} \times 100%$

⇒ Rate of C.I. accured in 3rd year = $\frac{\text{Amount of 3 years - Amount of 2 years}}{\text{Amount in 2 years}} \times 100%$

So, statement 1 is true.

⇒ Amount of 3 years = Amount of 2 years + C.I.

Amount at the end of 3 years = Amount at the end of 2nd year + Interest on it.

So, statement 2 is true.

∴ Both the statements are true.

Hence, option 1 is correct option.

Let ₹ P be the principal amount, R be the rate of interest and n be time.

$$\begin{aligned}A = P\Big(1 + \frac{R}{100}\Big)^n \\[1em] = P\Big(1 + \frac{20}{100})^2 \\[1em] = P\Big(\frac{120}{100}\Big)^2 \\[1em] = P \times \frac{120}{100} \times \frac{120}{100}.\end{aligned}$$

Thus, Statement 1 is false.

C.I. = A - P

= $P \times \frac{120}{100} \times \frac{120}{100} - P$

Thus, Statement 2 is true.

Hence, option 4 is correct option.

Let sum be ₹ x.

For first year :

P = ₹ x

R = 10%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{x \times 10 \times 1}{100} = \frac{x}{10}$

Amount = P + I = $x + \frac{x}{10} = \frac{11x}{10}$.

For second year :

P = ₹ $\frac{11x}{10}$

R = 11%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{\frac{11x}{10} \times 11 \times 1}{100} = \frac{121x}{1000}$

Amount = P + I = $\frac{11x}{10} + \frac{121x}{1000} = \frac{1100x + 121x}{1000} = \frac{1221x}{1000}$.

Given,

Amount after two years = ₹ 6593.40

$$\begin{aligned}\therefore \frac{1221x}{1000} = 6593.40 \\[1em] \Rightarrow x = \frac{6593.40 \times 1000}{1221} \\[1em] \Rightarrow x = 5.4 \times 1000 = 5400.\end{aligned}$$

Hence, required sum = ₹ 5400.

Let initial value of machine be ₹ x.

For first year :

P = ₹ x

R = 10%

T = 1 year

Depreciation = $\frac{P \times R \times T}{100} = \frac{x \times 10 \times 1}{100} = \frac{x}{10}$.

Value at end of 1 year = P - Depreciation = $x - \frac{x}{10} = \frac{9x}{10}$.

For second year :

P = ₹ $\frac{9x}{10}$

R = 10%

T = 1 year

Depreciation = $\frac{P \times R \times T}{100} = \frac{\frac{9x}{10} \times 10 \times 1}{100} = \frac{9x}{100}$.

Value at end of second year = P - Depreciation

= $\frac{9x}{10} - \frac{9x}{100} = \frac{90x - 9x}{100} = \frac{81x}{100}$.

For third year :

P = ₹ $\frac{81x}{100}$

R = 15%

T = 1 year

Depreciation = $\frac{P \times R \times T}{100} = \frac{\frac{81x}{100} \times 15 \times 1}{100} = \frac{243x}{2000}$.

Value at end of third year = P - Depreciation

= $\frac{81x}{100} - \frac{243x}{2000} = \frac{1620x - 243x}{2000} = \frac{1377x}{2000}$.

Total depreciation = Initial value - Value at end of third year

$$\begin{aligned}= x - \frac{1377x}{2000} \\[1em] = \frac{2000x - 1377x}{2000} \\[1em] = \frac{623x}{2000}.\end{aligned}$$

Percent depreciated

$$\begin{aligned}\frac{\text{Total depreciation}}{\text{Initial value}} \times 100 \\[1em] = \frac{\frac{623x}{2000}}{x} \times 100 \\[1em] = \frac{623x}{x \times 2000} \times 100 \\[1em] = \frac{623}{20} \\[1em] = 31.15%\end{aligned}$$

Hence, total depreciation of machine = 31.15%.

For 1st half-year

P = ₹ 12000

R = 10%

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

I = $\frac{P \times R \times T}{100} = \frac{12000 \times 10 \times \frac{1}{2}}{100}$ = ₹ 600.

Amount = P + I = ₹ 12000 + ₹ 600= ₹ 12600.

Money paid at the end of 1st half year = ₹ 4000

Balance money for 2nd half-year = ₹12600- ₹4000 = ₹8600.

For 2nd half-year

P = ₹ 8600

R = 10%

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

I = $\frac{P \times R \times T}{100} = \frac{8600 \times 10 \times \frac{1}{2}}{100}$ = ₹ 430.

Amount = ₹ 8600 + ₹ 430= ₹ 9030

Money paid at the end of 2nd half-year = ₹ 4000

Balance money for 3rd half-year = ₹ 9030- ₹ 4000 = ₹ 5030

For 3rd half-year

P = ₹ 5030

R = 10%

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

Interest = $\frac{P \times R \times T}{100} = \frac{5030 \times 10 \times \frac{1}{2}}{100}$ = ₹ 251.50

Amount = ₹ 5030 + ₹ 251.50 = ₹ 5281.50

Hence, amount to be paid at the end of 18 months = ₹ 5281.50

Let the sum of money be ₹ x.

For first year :

P = ₹ x

R = 10%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{x \times 10 \times 1}{100} = \frac{x}{10}$.

Amount = P + I = x + $\frac{x}{10} = \frac{11x}{10}$.

For second year :

P = ₹ $\frac{11x}{10}$

R = 10%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{\frac{11x}{10} \times 10 \times 1}{100} = \frac{11x}{100}$.

Amount = P + I = $\frac{11x}{10} + \frac{11x}{100} = \frac{110x + 11x}{100} = \frac{121x}{100}$.

For third year :

P = ₹ $\frac{121x}{100}$

R = 10%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{\frac{121x}{100} \times 10 \times 1}{100} = \frac{121x}{1000}$.

Given, interest of first year plus the interest of third year is ₹ 2652.

$$\begin{aligned}\Rightarrow \frac{x}{10} + \frac{121x}{1000} = 2652 \\[1em] \Rightarrow \frac{100x + 121x}{1000} = 2652 \\[1em] \Rightarrow \frac{221x}{1000} = 2652 \\[1em] \Rightarrow x = \frac{2652 \times 1000}{221} = 12000.\end{aligned}$$

Hence, sum of money is ₹ 12000.

Let original value of machine be ₹ x.

For first year :

P = ₹ x

R (of depreciation) = 12%

T = 1 year

Depreciation = $\frac{P \times R \times T}{100} = \frac{x \times 12 \times 1}{100} = \frac{12x}{100}$

Value at end of first year = P - Depreciation = $x - \frac{12x}{100} = \frac{88x}{100}$.

For second year :

P = $\frac{88x}{100}$

R (of depreciation) = 12%

T = 1 year

Depreciation = $\frac{P \times R \times T}{100} = \frac{\frac{88x}{100} \times 12 \times 1}{100} = \frac{1056x}{10000}$

Given,

Machine depreciates by ₹ 2640 during the second financial year of its purchase.

$$\begin{aligned}\therefore \frac{1056x}{10000} = 2640 \\[1em] \Rightarrow x = \frac{2640}{1056} \times 10000 \\[1em] \Rightarrow x = 2.5 \times 10000 \\[1em] \Rightarrow x = 25000.\end{aligned}$$

Hence, original cost of machine = ₹ 25000.

Let sum be ₹ x.

For S.I. :

P = ₹ x

R = 8%

T = 2 years

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

For C.I. :

For first year :

P = ₹ x

R = 8%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{x \times 8 \times 1}{100} = \frac{2x}{25}$.

Amount = P + I = $x + \frac{2x}{25} = \frac{27x}{25}$.

For second year :

P = ₹ $\frac{27x}{25}$

R = 8%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{\frac{27x}{25} \times 8 \times 1}{100} = \frac{54x}{625}$.

Amount = P + I = $\frac{27x}{25} + \frac{54x}{625} = \frac{675x + 54x}{625} = \frac{729x}{625}$.

C.I. = Final amount - Initial Principal

$$\begin{aligned}= \frac{729x}{625} - x \\[1em] = \frac{729x - 625x}{625} \\[1em] = \frac{104x}{625}.\end{aligned}$$

Given, difference between S.I. and C.I. = ₹ 64

$$\begin{aligned}\Rightarrow \frac{104x}{625} - \frac{4x}{25} = 64 \\[1em] \Rightarrow \frac{104x - 100x}{625} = 64 \\[1em] \Rightarrow \frac{4x}{625} = 64 \\[1em] \Rightarrow x = \frac{64}{4} \times 625 \\[1em] \Rightarrow x = 16 \times 625 = 10000.\end{aligned}$$

Hence, required sum = ₹ 10000.

(i) For first year :

P = ₹ 13500

R = 16%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{13500 \times 16 \times 1}{100}$ = ₹ 2160.

Hence, interest for first year = ₹ 2160.

(ii) By formula,

Amount = P + I = ₹ 13500 + ₹ 2160 = ₹ 15660.

Hence, amount at end of first year = ₹ 15660.

(iii) For second year :

P = ₹ 15660

R = 16%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{15660 \times 16 \times 1}{100}$ = ₹ 2505.6

Hence, interest for second year = ₹ 2506.

(i) For first year :

P = ₹ 48000

R = 10%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{48000 \times 10 \times 1}{100}$ = ₹ 4800.

Hence, interest for first year = ₹ 4800.

(ii) For second year :

P = ₹ 48000 + ₹ 4800 = ₹ 52800

R = 10%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{52800 \times 10 \times 1}{100}$ = ₹ 5280.

Amount = P + I = ₹ 52800 + ₹ 5280 = ₹ 58080

Hence, amount at the end of second year = ₹ 58080.

(iii) For third year :

P = ₹ 58080

R = 10%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{58080 \times 10 \times 1}{100}$ = ₹ 5808.

Hence, interest for third year = ₹ 5808.

(i) Let rate of interest be x%.

For first year :

P = ₹ 12000

R = x%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{12000 \times x \times 1}{100}$ = 120x.

Amount = P + I = ₹ 12000 + ₹ 120x.

Given amount paid back after a year = ₹ 4000

Amount left = ₹ 12000 + ₹ 120x - ₹ 4000 = ₹ 8000 + ₹ 120x

For second year :

P = ₹ 8000 + ₹ 120x

R = x%

T = 1 year

I = $\frac{P \times R \times T}{100} = \frac{(8000 + 120x) \times x \times 1}{100}$.

Given,

Interest for second year = ₹ 920

$$\begin{aligned}\therefore \frac{(8000 + 120x) \times x \times 1}{100} = 920 \\[1em] \Rightarrow (8000 + 120x)x = 92000 \\[1em] \Rightarrow 8000x + 120x^2 = 92000 \\[1em] \Rightarrow 40(200x + 3x^2) = 40(2300) \\[1em] \Rightarrow 3x^2 + 200x = 2300 \\[1em] \Rightarrow 3x^2 + 200x - 2300 = 0 \\[1em] \Rightarrow 3x^2 + 230x - 30x - 2300 = 0 \\[1em] \Rightarrow x(3x + 230) - 10(3x + 230) = 0 \\[1em] \Rightarrow (x - 10)(3x + 230) = 0 \\[1em] \Rightarrow x - 10 = 0 \text{ or } 3x + 230 = 0 \\[1em] \Rightarrow x = 10 \text{ or } 3x = -230 \\[1em] \Rightarrow x = 10 \text{ or } x = -\frac{230}{3}.\end{aligned}$$

Since, rate of interest cannot be negative.

Hence, rate of interest = 10%.

(ii) We know that :

For second year :

P = ₹ 8000 + ₹ 120x = ₹ 8000 + ₹ 120 × 10 = ₹ 9200.

I = ₹ 920

Amount at end of second year = P + I = ₹ 9200 + ₹ 920 = ₹ 10120.

Hence, the amount of debt at end of second year = ₹ 10120.

(i) Interest earned in first year = ₹ 1500

Interest earned in second year = ₹ 1725

Interest earned on ₹ 1500 in second year = Interest earned in second year - Interest earned in first year

= ₹ 1725 - ₹ 1500 = ₹ 225.

Rate of interest for second year

= $\frac{\text{Interest earned on ₹ 1500 in second year}}{\text{Interest earned in first year}} \times 100 = \frac{225}{1500} \times 100$ = 15%.

Hence, interest for the second year = 15%.

(ii) Interest earned in second year = ₹ 1725

Interest earned in third year = ₹ 2070

Interest earned on ₹ 1725 in third year = Interest earned in third year - Interest earned in second year

= ₹ 2070 - ₹ 1725 = ₹ 345.

Rate of interest for second year

= $\frac{\text{Interest earned on ₹ 1725 in third year}}{\text{Interest earned in second year}} \times 100 = \frac{345}{1725} \times 100$ = 20%.

Hence, interest for the third year = 20%.