Compound Interest (Stage 1) [Basic Concepts]

For a particular year the simple interest at 10% is ₹ 800. The compound interest for the next year at the same rate is :

1. ₹ 880

2. ₹ 800

3. ₹ 720

4. ₹ 968

Answer

Given,

Interest = ₹ 800

T = 1 year

R = 10%

Let money on which interest is ₹ 800 be P.

By formula,

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

Substituting values we get :

$\Rightarrow 800 = \dfrac{P \times 10 \times 1}{100} \\[1em] \Rightarrow P = \dfrac{800 \times 100}{10} = 8000.$

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

Hence, Option 1 is the correct option.

The compound interest on ₹ 5000 at 10% per annum and in 6 months amounts to :

1. ₹ 5500

2. ₹ 250

3. ₹ 600

4. ₹ 2500

Answer

Given,

P = ₹ 5000

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

R = 10%

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

Hence, Option 2 is the correct option.

The compound interest on ₹ 5000 at 10% per annum and in one year compounded half-yearly is :

1. ₹ 1050

2. ₹ 525

3. ₹ 512.50

4. ₹ 5512.50

Answer

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

For first half-year :

P = ₹ 5000

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

R = 10%

By formula,

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

Substituting values we get :

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

For second half-year :

P = ₹ 5000 + ₹ 250 = ₹ 5250

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

R = 10%

By formula,

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

Substituting values we get :

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

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

Hence, Option 3 is the correct option.

A sum of ₹ 20,000 is lent at 12% compound interest compounded yearly. The compound interest accrued in the second year will be :

1. ₹ 4800

2. ₹ 288

3. ₹ 2688

4. ₹ 5088

Answer

For first year :

P = ₹ 20,000

R = 12%

T = 1 year

By formula,

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

Substituting values we get :

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

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

For second year :

P = ₹ 22400

R = 12%

T = 1 year

By formula,

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

Substituting values we get :

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

Hence, Option 3 is the correct option.

During the year 2022, the interest accrued at the rate of 5% is ₹ 1250. The compound interest accrued at the same rate during the year 2023 is :

1. ₹ 1312.50

2. ₹ 62.50

3. ₹ 6250

4. ₹ 2000

Answer

Let principal for first year be ₹ P.

For first year :

Interest = ₹ 1250

R = 5%

By formula,

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

Substituting values we get :

$\Rightarrow 1250 = \dfrac{P \times 5 \times 1}{100} \\[1em] \Rightarrow P = \dfrac{1250 \times 100}{5} \\[1em] \Rightarrow P = 25000.$

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

For second year :

P = ₹ 26250

R = 5%

T = 1 year

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

Hence, Option 1 is the correct option.

Rates of interest for two consecutive years are 10% and 12% respectively. The percentage increase during these two years is :

1. 22%

2. 23.2%

3. 123.2%

4. 122%

Answer

Let initial principal be ₹ x.

For first year :

P = x

R = 10%

T = 1 year

By formula,

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

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

For second year :

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

R = 12%

T = 1 year

By formula,

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

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

Compound interest = Final Amount - Initial Principal

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

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

$= \dfrac{\dfrac{223x}{1000}}{x} \times 100 = \dfrac{223x}{1000x} \times 100$ = 23.2%

Hence, Option 2 is the correct option.

₹ 16,000 is invested at 5% compound interest compounded per annum. Use the table, given below, to find the amount in 4 years.

Answer

For 2nd year :

P = ₹ 16800

R = 5%

T = 1 year

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

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

For 3rd year :

P = ₹ 17640

R = 5%

T = 1 year

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

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

For 4th year :

P = ₹ 18522

R = 5%

T = 1 year

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

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

For 5th year :

P = ₹ 19488.10

R = 5%

T = 1 year

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

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

Hence, amount in 4 years = ₹ 19448.10

Calculate the amount and the compound interest on ₹ 8000 in $2\dfrac{1}{2}$ years at 15% per annum.

Answer

For first year :

P = ₹ 8000

T = 1 year

R = 15%

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

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

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

For second year :

P = ₹ 9200

T = 1 year

R = 15%

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

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

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

For next half year :

P = ₹ 10580

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

R = 15%

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

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

Calculate the amount and the compound interest on :

₹ 4600 in 2 years when the rates of interest of successive years are 10% and 12% respectively.

Answer

For first year :

P = ₹ 4600

T = 1 year

R = 10%

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

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

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

For second year :

P = ₹ 5060

T = 1 year

R = 12%

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

$= \dfrac{5060 \times 12 \times 1}{100}$ = ₹ 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

Meenal lends ₹ 75000 at C.I. for 3 years. If the rate of interest for the first two years is 15% per year and for the third year it is 16%, calculate the sum Meenal will get at the end of third year.

Answer

For first year :

P = ₹ 75000

T = 1 year

R = 15%

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

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

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

For second year :

P = ₹ 86250

T = 1 year

R = 15%

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

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

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

For third year :

P = ₹ 99187.50

T = 1 year

R = 16%

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

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

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

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

Calculate the amount and the compound interest on ₹ 16000 in 3 years, when the rates of interest for successive years are 10%, 14% and 15% respectively.

Answer

For first year :

P = ₹ 16000

T = 1 year

R = 10%

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

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

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

For second year :

P = ₹ 17600

T = 1 year

R = 14%

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

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

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

For third year :

P = ₹ 20064

T = 1 year

R = 15%

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

$= \dfrac{20064 \times 15 \times 1}{100}$ = ₹ 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

Calculate the compound interest for the second year on ₹ 8000 invested for 3 years at 10% per annum.

Answer

For first year :

P = ₹ 8000

T = 1 year

R = 10%

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

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

For second year :

P = ₹ 8800

T = 1 year

R = 10%

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

Hence, C.I. for second year = ₹ 880.

Find the compound interest correct to the nearest rupee, on ₹ 2400 for $2\dfrac{1}{2}$ years at 5 percent per annum.

Answer

For first year :

P = ₹ 2400

T = 1 year

R = 5%

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

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

For second year :

P = ₹ 2520

T = 1 year

R = 5%

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

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

For next half-year :

P = ₹ 2646

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

R = 5%

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

A borrowed ₹ 2500 from B at 12% per annum compound interest. After 2 years, A gave ₹ 2936 and a watch to B to clear the account. Find the cost of the watch.

Answer

For first year :

P = ₹ 2500

T = 1 year

R = 12%

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

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

For second year :

P = ₹ 2800

T = 1 year

R = 12%

I = $\dfrac{P \times R \times T}{100} = \dfrac{2800 \times 1 \times 12}{100}$ = ₹ 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.

How much will ₹ 50000 amount to in 3 years compounded yearly, if the rates for the successive years are 6%, 8% and 10% respectively.

Answer

For first year :

P = ₹ 50000

T = 1 year

R = 6%

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

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

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

For second year :

P = ₹ 53000

T = 1 year

R = 8%

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

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

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

For third year :

P = ₹ 57240

T = 1 year

R = 10%

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

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

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

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

Govind borrows ₹ 18000 at 10% simple interest. He immediately invests the money at 10% compound interest compounded half-yearly. How much money does Govind gain in one year ?

Answer

Calculating simple interest :

P = ₹ 18000

R = 10%

T = 1 year

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

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

Calculating compound interest :

For 1st half year :

P = ₹ 18000

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

R = 10%

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

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

For 2nd half-year :

P = ₹ 18900

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

R = 10%

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

Find the compound interest on ₹ 4000 accrued in three years, when the rate of interest is 8% for the first year and 10% per year for the second and the third years.

Answer

For first year :

P = ₹ 4000

T = 1 year

R = 8%

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

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

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

For second year :

P = ₹ 4320

T = 1 year

R = 10%

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

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

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

For third year :

P = ₹ 4752

T = 1 year

R = 10%

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

$= \dfrac{4752 \times 10 \times 1}{100}$ = ₹ 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

The difference between C.I. and S.I. in one year on ₹ 5000 at the rate of 10% per annum is :

1. ₹ 00

2. ₹ 500

3. ₹ 5500

4. ₹ 5250

Answer

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.

₹ 2000 is saved during the year 2022 and deposited in a bank at the beginning of year 2023 at 8% compound interest. During 2023, ₹ 3000 more is saved and deposited in the same bank at the beginning of 2024 and at the same rate of interest. The C.I. earned upto the end of 2024 is :

1. ₹ 560

2. ₹ 572.80

3. ₹ 5400

4. ₹ 400

Answer

At beginning of 2023 :

P = ₹ 2000

R = 8%

T = 1 year

I = $\dfrac{P \times R \times T}{100} = \dfrac{2000 \times 8 \times 1}{100}$ = ₹ 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 = $\dfrac{P \times R \times T}{100} = \dfrac{5160 \times 8 \times 1}{100}$ = ₹ 412.80

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

Hence, Option 2 is the correct option.

₹ 1000 is borrowed at 10% per annum C.I. If ₹ 300 is repaid at the end of each year, the amount of loan outstanding at the end of 2nd year is :

1. ₹ 880

2. ₹ 610

3. ₹ 580

4. ₹ 484

Answer

For first year :

P = ₹ 1000

R = 10%

T = 1 year

I = $\dfrac{P \times R \times T}{100} = \dfrac{1000 \times 10 \times 1}{100}$ = ₹ 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 = $\dfrac{P \times R \times T}{100} = \dfrac{800 \times 10 \times 1}{100}$ = ₹ 80.

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

Hence, Option 1 is the correct option.

The difference between C.I. and S.I. in 2 years on ₹ 4000 at 10% per annum is :

1. ₹ 840

2. ₹ 800

3. ₹ 400

4. ₹ 40

Answer

For S.I. :

P = ₹ 4000

R = 10%

T = 2 years

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

For C.I. :

For 1st year :

P = ₹ 4000

T = 1 year

R = 10%

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

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

For 2nd year :

P = ₹ 4400

R = 10%

T = 1 year

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

₹ 10 is the difference between C.I. and S.I. in 2 years and at 5% per annum. The principal amount is :

1. ₹ 4400

2. ₹ 4100

3. ₹ 4000

4. none of these

Answer

Let principal amount be ₹ x.

For S.I. :

P = ₹ x

R = 5%

T = 2 year

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

For C.I. :

For first year :

P = ₹ x

R = 5%

T = 1 year

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

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

For second year :

P = ₹ $\dfrac{21x}{20}$

R = 5%

T = 1 year

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

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

C.I. = Final amount - Initial Principal

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

Given,

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

$\Rightarrow \dfrac{41x}{400} - x = 10 \\[1em] \Rightarrow \dfrac{41x - 40x}{400} = 10 \\[1em] \Rightarrow \dfrac{x}{400} = 10 \\[1em] \Rightarrow x = 4000.$

Hence, Option 3 is the correct option.

Calculate the difference between the simple interest and compound interest on ₹ 4000 in 2 years at 8% per annum compounded yearly.

Answer

For S.I. :

P = ₹ 4000

T = 2 years

R = 8%

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

For C.I. :

For 1st year :

P = ₹ 4000

T = 1 year

R = 8%

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

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

For 2nd year :

P = ₹ 4320

T = 1 year

R = 8%

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

A sum of money is lent at 8% per annum compound interest. If the interest for the second year exceeds that for the first year by ₹ 96, find the sum of money.

Answer

Let sum of money be ₹ x.

For first year :

P = ₹ x

T = 1 year

R = 8%

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

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

For 2nd year :

P = ₹ $\dfrac{27x}{25}$

T = 1 year

R = 8%

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

Given,

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

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

Hence, sum of money = ₹ 15000.

A man invests ₹ 5600 at 14% per annum compound interest for 2 years. Calculate :

(i) the interest for the first year.

(ii) the amount at the end of the first year.

(iii) the interest for the second year, correct to the nearest rupee.

Answer

(i) For first year :

P = ₹ 5600

R = 14%

T = 1 year

I = $\dfrac{P \times R \times T}{100} = \dfrac{5600 \times 14 \times 1}{100}$ = ₹ 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 = $\dfrac{P \times R \times T}{100} = \dfrac{6384 \times 14 \times 1}{100}$ = ₹ 893.76

Hence, interest for the second year = ₹ 894.

A man saves ₹ 3000 every year and invests it at the end of the year at 10% compound interest. Calculate the total amount of his savings at the end of the third year.

Answer

Savings at the end of every year = ₹ 3000

For 2nd year :

P = ₹ 3000

R = 10%

T = 1 year

⇒ I = $\dfrac{P \times R \times T}{100} = \dfrac{3000 \times 10 \times 1}{100}$ = ₹ 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 = $\dfrac{P \times R \times T}{100} = \dfrac{6300 \times 10 \times 1}{100}$ = ₹ 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.

A man lends ₹ 12500 at 12% for the first year, at 15% for the second year and at 18% for the third year. If the rates of interest are compounded yearly; find the difference between the C.I. of the first year and the compound interest for the third year.

Answer

For first year :

P = ₹ 12500

R = 12%

T = 1 year

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

Amount = ₹ 12500 + ₹ 1500 = ₹ 14000

For second year :

P = ₹ 14000

R = 15%

T = 1 year

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

Amount = ₹ 14000 + ₹ 2100 = ₹ 16100

For third year :

P = ₹ 16100

R = 18%

T = 1 year

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

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

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

A man borrows ₹ 6000 at 5 percent C.I. per annum. If he repays ₹ 1200 at the end of each year, find the amount of the loan outstanding at the beginning of the third year.

Answer

For first year :

P = ₹ 6000

T = 1 year

R = 5 %

I = $\dfrac{P \times R \times T}{100} = \dfrac{6000 \times 5 \times 1}{100}$ = ₹ 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 = $\dfrac{P \times R \times T}{100} = \dfrac{5100 \times 5 \times 1}{100}$ = ₹ 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.

A man borrows ₹ 5000 at 12 percent compound interest payable every six months. He repays ₹ 1800 at the end of every six months. Calculate the third payment he has to make at the end of 18 months to clear the entire the loan.

Answer

For 1st half-year

P = ₹ 5000

R = 12%

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

I = $\dfrac{P \times R \times T}{100} = \dfrac{5000 \times 12 \times \dfrac{1}{2}}{100}$ = ₹ 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 = $\dfrac{1}{2}$ year

I = $\dfrac{P \times R \times T}{100} = \dfrac{3500 \times 12 \times \dfrac{1}{2}}{100}$ = ₹ 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 = $\dfrac{1}{2}$ year

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

Amount = ₹ 1910 + ₹ 114.60 = ₹ 2024.60

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

On a certain sum of money, the difference between the compound interest for a year, payable half-yearly, and the simple interest for a year is ₹ 180. Find the sum lent out, if the rate of interest in both cases is 10% per annum.

Answer

Let sum of money be ₹ x.

For S.I. :

P = ₹ x

R = 10%

T = 1 years

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

For C.I. :

For first half-year :

P = ₹ x

R = 10%

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

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

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

For second year :

P = ₹ $\dfrac{21x}{20}$

R = 10%

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

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

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

C.I. = Final amount - Initial principal

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

Given,

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

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

Hence, sum lent out = ₹ 72000.

A manufacturer estimates that his machine depreciates by 15% of its value at the beginning of the year. Find the original value of machine, if it depreciates by ₹ 5355 during the second year.

Answer

Let original value of machine be ₹ x.

For first year :

P = ₹ x

R (of depreciation) = 15%

T = 1 year

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

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

For second year :

P = ₹ $\dfrac{17x}{20}$

R (of depreciation) = 15%

T = 1 year

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

Given,

Depreciation in second year = ₹ 5355

$\therefore \dfrac{51x}{400} = 5355 \\[1em] \Rightarrow x = \dfrac{5355 \times 400}{51} \\[1em] \Rightarrow x = 105 \times 400 = 42000.$

Hence, original value of machine = ₹ 42000.

A man borrows ₹ 10000 at 5% per annum compound interest. He repays 35% of the sum borrowed at the end of the first year and 42% of the sum borrowed at the end of the second year. How much must he pay at the end of the third year in order to clear the debt ?

Answer

For first year :

P = ₹ 10000

R = 5%

T = 1 year

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

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

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

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

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

For second year :

P = ₹ 7000

R = 5%

T = 1 year

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

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

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

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

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

For third year :

P = ₹ 3150

R = 5%

T = 1 year

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

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

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

₹ 300 and ₹ 360 are the compound interest for two consecutive years. The rate of interest is :

1. 1.2%

2. 12%

3. 120%

4. 20%

Answer

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 = $\dfrac{100 \times I}{P \times T} = \dfrac{100 \times 60}{300 \times 1}$ = 20%.

Hence, Option 4 is the correct option.

A certain sum of money amounts to ₹ 5000 at the end of 5th year and to ₹ 6000 at the end of 6th year. The rate of interest is :

1. 120%

2. 20%

3. 1.2%

4. 12%

Answer

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

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

By formula,

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

Hence, Option 2 is the correct option.

At the end of 2020, the compound interest amounted to ₹ 3850 at 10% C.I. The C.I. on the same sum and at same rate amounted at the end of 2019 was :

1. ₹ 3500

2. ₹ 4235

3. ₹ 3181

4. ₹ 3182

Answer

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 = $\dfrac{100 \times I}{P \times T}$

Substituting values we get :

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

Hence, Option 1 is the correct option.

A sum of money, lent out at C.I. amounts to ₹ 4500 in 6 years. If rate of C.I. is 12%, the same money will amount in 7 years to rupees :

1. 540

2. 5040

3. 5400

4. 4725

Answer

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 = $\dfrac{100 \times I}{P \times T}$

Substituting values we get :

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

Hence, Option 2 is the correct option.

For two consecutive years, a sum lent out at C.I. earns ₹ 600 and ₹ 690 respectively. The rate of C.I. is :

1. 12%

2. 18%

3. 15%

4. 5%

Answer

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 = $\dfrac{100 \times I}{P \times T} = \dfrac{100 \times 90}{600 \times 1}$ = 15%.

Hence, Option 3 is the correct option.

For two consecutive years, a sum of money lent out at C.I. amounts to ₹ 2400 and ₹ 2760 respectively. The rate of interest is :

1. 5%

2. 15%

3. 18%

4. 10%

Answer

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

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

By formula,

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

Hence, Option 2 is the correct option.

A certain sum amounts to ₹ 5292 in two years and ₹ 5556.60 in three years, interest being compounded annually. Find :

(i) the rate of interest

(ii) the original sum.

Answer

(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 = $\dfrac{I \times 100}{P \times T} = \dfrac{264.60 \times 100}{5292 \times 1} = \dfrac{100}{20}$ = 5%.

Hence, rate of interest = 5%.

(ii) Let original sum be ₹ x.

For 1st year :

P = ₹ x

R = 5%

T = 1 year

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

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

For second year :

P = ₹ $\dfrac{21x}{20}$

R = 5%

T = 1 year

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

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

Given,

Amount after 2 years = ₹ 5292

$\therefore \dfrac{441x}{400} = 5292 \\[1em] \Rightarrow x = \dfrac{5292 \times 400}{441} = 4800.$

Hence, original sum = ₹ 4800.

Mohit invests ₹ 8000 for 3 years at a certain rate of interest, compounded annually. At the end of one year, it amounts to ₹ 9440. Calculate :

(i) the rate of interest per annum.

(ii) the amount at the end of the second year.

(iii) the interest accrued in the third year.

Answer

(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 = $\dfrac{I \times 100}{P \times T} = \dfrac{1440 \times 100}{8000 \times 1} = \dfrac{144}{8}$ = 18%.

Hence, rate of interest = 18%.

(ii) For second year :

P = ₹ 9440

R = 18%

T = 1 year

I = $\dfrac{P \times R \times T}{100} = \dfrac{9440 \times 18 \times 1}{100} = \dfrac{169920}{100}$ = ₹ 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 = $\dfrac{P \times R \times T}{100} = \dfrac{11139.20 \times 18 \times 1}{100} = \dfrac{200505.60}{100}$ = ₹ 2,005.06

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

The compound interest, calculated yearly, on a certain sum of money for the second year is ₹ 1089 and for the third year it is ₹ 1197.90. Calculate the rate of interest and the sum of money.

Answer

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 = $\dfrac{100 \times I}{P \times T} = \dfrac{100 \times 108.9}{1089 \times 1}$ = 10%.

Let sum of money be ₹ x.

For first year :

P = ₹ x

R = 10%

T = 1 year

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

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

For second year :

P = ₹ $\dfrac{11x}{10}$

R = 10%

T = 1 year

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

Given,

C.I. for 2nd year = ₹ 1089

$\therefore \dfrac{11x}{100} = 1089 \\[1em] \Rightarrow x = \dfrac{1089 \times 100}{11} = 9900.$

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

A sum is invested at compound interest compounded yearly. If the interest for two successive years be ₹ 5700 and ₹ 7410, calculate the rate of interest.

Answer

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 = $\dfrac{100 \times I}{P \times T} = \dfrac{100 \times 1710}{5700 \times 1}$ = 30%.

Hence, the rate of interest = 30%.

The cost of a machine depreciated by ₹ 4000 during the first year and by ₹ 3600 during the second year. Calculate :

(i) the rate of depreciation.

(ii) the original cost of the machine.

(iii) its cost at the end of third year.

Answer

(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 = $\dfrac{100 \times I}{P \times T} = \dfrac{100 \times 400}{4000 \times 1}$ = 10%.

Hence, rate of depreciation = 10%.

(ii) Let cost of machine be ₹ x.

Given,

Depreciation in first year = ₹ 4000

Depreciation % = 10%

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

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

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

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

Ramesh invests ₹ 12800 for three years at the rate of 10% per annum compound interest. Find :

(i) the sum due to Ramesh at the end of the first year.

(ii) the interest he earns for the second year.

(iii) the total amount due to him at the end of third year.

Answer

(i) For first year :

P = ₹ 12800

R = 10%

T = 1 year

I = $\dfrac{P \times R \times T}{100} = \dfrac{12800 \times 10 \times 1}{100}$ = ₹ 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 = $\dfrac{P \times R \times T}{100} = \dfrac{14080 \times 10 \times 1}{100}$ = ₹ 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 = $\dfrac{P \times R \times T}{100} = \dfrac{15488 \times 10 \times 1}{100}$ = ₹ 1548.80

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

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

A certain sum of money is put at compound interest, compounded half-yearly. If the interest for two successive half-years are ₹ 650 and ₹ 760.50; find the rate of interest.

Answer

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

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

By formula,

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

Hence, the rate of interest = 34%.

Geeta borrowed ₹ 15000 for 18 months at a certain rate of interest compounded semi-annually. If at the end of six months it amounted to ₹ 15600; calculate :

(i) the rate of interest per annum.

(ii) the total amount of money that Geeta must pay at the end of 18 months in order to clear the account.

Answer

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

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

By formula,

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

Hence, the rate of interest = 8%.

(ii) For 2nd half-year :

P = ₹ 15600

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

R = 8%

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

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

For 3rd half-year :

P = ₹ 16224

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

R = 8%

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

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

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

₹ 8000 is lent out at 7% compound interest for 2 years. At the end of the first year ₹ 3560 are returned. Calculate :

(i) the interest paid for the second year.

(ii) the total interest paid in two years

(iii) the total amount of money paid in two years to clear the debt.

Answer

(i) For first year :

P = ₹ 8000

R = 7%

T = 1 year

I = $\dfrac{P \times R \times T}{100} = \dfrac{8000 \times 7 \times 1}{100}$ = ₹ 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 = $\dfrac{P \times R \times T}{100} = \dfrac{5000 \times 7 \times 1}{100}$ = ₹ 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.

Find the sum invested at 10% compounded annually, on which the interest for the third year, exceeds the interest of the first year by ₹ 252.

Answer

Let sum of money be ₹ x.

For first year :

P = ₹ x

R = 10%

T = 1 year

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

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

For second year :

P = ₹ $\dfrac{11x}{10}$

R = 10%

T = 1 year

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

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

For third year :

P = ₹ $\dfrac{121x}{100}$

R = 10%

T = 1 year

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

Given,

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

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

Hence, sum = ₹ 12000.

A man borrows ₹ 10000 at 10% compound interest compounded yearly. At the end of each year, he pays back 30% of the sum borrowed. How much money is left unpaid just after the second year ?

Answer

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

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

For first year :

P = ₹ 10000

R = 10%

T = 1 year

I = $\dfrac{P \times R \times T}{1000} = \dfrac{10000 \times 10 \times 1}{100}$ = ₹ 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 = $\dfrac{P \times R \times T}{100} = \dfrac{8000 \times 10 \times 1}{100}$ = ₹ 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.

A man borrows ₹ 10000 at 10% compound interest compounded yearly. At the end of each year, he pays back 20% of the amount for that year. How much money is left unpaid just after the second year ?

Answer

For first year :

P = ₹ 10000

R = 10%

T = 1 year

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

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

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

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

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

For second year :

P = ₹ 8800

R = 10%

T = 1 year

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

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

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

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

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

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

Certain sum is lent at 10% compound interest per annum. If the interest accrued during the year 2024 was ₹1,331 then the interest accrued during the year 2022, was:

1. ₹ 1,210

2. ₹ 1,100

3. ₹ 1,464.10

4. ₹ 1,610.51

Answer

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.

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

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

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

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

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

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

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

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

$\Rightarrow \dfrac{\text{121P}}{1000} = 1331\\[1em] \Rightarrow \text{P} = \dfrac{1000 \times 1331}{121}\\[1em] \Rightarrow \text{P} = 11,000$

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

Hence, option 2 is correct option.

During 2023, the population of a small village was 45,000 which increased every year by 5%. The population during the year 2024 was:

1. 45,000 + 5% of 45,000

2. 45,000 - 5% of 45,000

3. 45,000 x 5% of 45,000

4. 45,000 ÷ 5% of 45,000

Answer

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.

In how many years, will a sum of money double itself at 10% C.I.:

1. 5 years

2. 10 years

3. 8 years

4. none of these

Answer

Given, rate of interest = 10%

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

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

₹ P doubles itself in n years.

A = 2P

Substituting values in formula, we get :

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

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.

The cost of a machine depreciates every year by 10%; the percentage decrease during two years will be:

1. 20%

2. 18%

3. 19%

4. 21%

Answer

Let initial value of machine be ₹ x.

For first year :

P = ₹ x

Rate of depreciation = 10%

n = 2 years

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

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

By formula,

$\text{Percentage depreciated }= \dfrac{\text{Total depreciation}}{\text{Initial value}} \times 100\\[1em] = \dfrac{\dfrac{19x}{100}}{x} \times 100\\[1em] = \dfrac{19}{100} \times 100\\[1em] = 19$

Hence, option 3 is the correct option.

Assertion (A): On a certain sum and at a certain rate,

C.I. for 3rd year = Amount of 3rd year - Amount of 2nd year

Reason (R): A in 3 years = Principal + C.I. for 3 years and A in 2 years = Principal + C.I. for 2 years

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

1. A is true, but R is false.

2. A is false, but R is true.

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

4. Both A and R are true, and R is the incorrect reason for A.

Answer

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.

Assertion (A): Interest for 5th year = Interest in 5 years - Interest in 4 years

Reason (R): Interest of 5th year = Amount in 5 years - Amount in 4 years

1. A is true, but R is false.

2. A is false, but R is true.

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

4. Both A and R are true, and R is the incorrect reason for A.

Answer

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.

Statement 1: Rate of C.I. accrued in 3rd year

= $\dfrac{\text{Amount of 3 years - Amount of 2 years}}{\text{Amount in 2 years}} \times 100$

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

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

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

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

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

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.

Statement 1: If P is the sum invested for 2 years at 20% rate of interest

= ₹ P x $\dfrac{20}{100} \times \dfrac{20}{100}$

Statement 2: Interest accrued in 2 years = ₹ (P x $\dfrac{120}{100} \times \dfrac{120}{100} - P)$

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

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

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

Thus, Statement 1 is false.

C.I. = A - P

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

Thus, Statement 2 is true.

Hence, option 4 is correct option.

What sum will amount to ₹ 6,593.40 in 2 years at C.I., if the rates are 10 percent and 11 percent for the two successive years ?

Answer

Let sum be ₹ x.

For first year :

P = ₹ x

R = 10%

T = 1 year

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

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

For second year :

P = ₹ $\dfrac{11x}{10}$

R = 11%

T = 1 year

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

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

Given,

Amount after two years = ₹ 6593.40

$\therefore \dfrac{1221x}{1000} = 6593.40 \\[1em] \Rightarrow x = \dfrac{6593.40 \times 1000}{1221} \\[1em] \Rightarrow x = 5.4 \times 1000 = 5400.$

Hence, required sum = ₹ 5400.

The value of a machine depreciated by 10% per year during the first two years and 15% per year during the third year. Express the total depreciation of the machine, as percent, during the three years.

Answer

Let initial value of machine be ₹ x.

For first year :

P = ₹ x

R = 10%

T = 1 year

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

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

For second year :

P = ₹ $\dfrac{9x}{10}$

R = 10%

T = 1 year

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

Value at end of second year = P - Depreciation

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

For third year :

P = ₹ $\dfrac{81x}{100}$

R = 15%

T = 1 year

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

Value at end of third year = P - Depreciation

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

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

$= x - \dfrac{1377x}{2000} \\[1em] = \dfrac{2000x - 1377x}{2000} \\[1em] = \dfrac{623x}{2000}.$

Percent depreciated

$\dfrac{\text{Total depreciation}}{\text{Initial value}} \times 100 \\[1em] = \dfrac{\dfrac{623x}{2000}}{x} \times 100 \\[1em] = \dfrac{623x}{x \times 2000} \times 100 \\[1em] = \dfrac{623}{20} \\[1em] = 31.15$

Hence, total depreciation of machine = 31.15%.

Rachna borrows ₹ 12000 at 10 per cent per annum interest compounded half-yearly. She repays ₹ 4000 at the end of every six months. Calculate the third payment she has to make at the end of 18 months in order to clear the entire loan.

Answer

For 1st half-year

P = ₹ 12000

R = 10%

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

I = $\dfrac{P \times R \times T}{100} = \dfrac{12000 \times 10 \times \dfrac{1}{2}}{100}$ = ₹ 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 = $\dfrac{1}{2}$ year

I = $\dfrac{P \times R \times T}{100} = \dfrac{8600 \times 10 \times \dfrac{1}{2}}{100}$ = ₹ 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 = $\dfrac{1}{2}$ year

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

Amount = ₹ 5030 + ₹ 251.50 = ₹ 5281.50

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

On a certain sum of money, invested at the rate of 10 percent per annum compounded annually, the interest for the first year plus the interest for the third year is ₹ 2,652. Find the sum.

Answer

Let the sum of money be ₹ x.

For first year :

P = ₹ x

R = 10%

T = 1 year

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

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

For second year :

P = ₹ $\dfrac{11x}{10}$

R = 10%

T = 1 year

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

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

For third year :

P = ₹ $\dfrac{121x}{100}$

R = 10%

T = 1 year

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

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

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

Hence, sum of money is ₹ 12000.

During every financial year, the value of a machine depreciates by 12%. Find the original cost of a machine which depreciates by ₹ 2640 during the second financial year of its purchase.

Answer

Let original value of machine be ₹ x.

For first year :

P = ₹ x

R (of depreciation) = 12%

T = 1 year

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

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

For second year :

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

R (of depreciation) = 12%

T = 1 year

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

Given,

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

$\therefore \dfrac{1056x}{10000} = 2640 \\[1em] \Rightarrow x = \dfrac{2640}{1056} \times 10000 \\[1em] \Rightarrow x = 2.5 \times 10000 \\[1em] \Rightarrow x = 25000.$

Hence, original cost of machine = ₹ 25000.

Find the sum on which the difference between the simple interest and the compound interest at the rate of 8% per annum compounded annually be ₹ 64 in 2 years.

Answer

Let sum be ₹ x.

For S.I. :

P = ₹ x

R = 8%

T = 2 years

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

For C.I. :

For first year :

P = ₹ x

R = 8%

T = 1 year

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

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