Compound Interest [Basic Concepts] — Exercise 2(A)

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

$= \dfrac{\dfrac{232x}{1000}}{x} \times 100 = \dfrac{232x}{1000x} \times 100$ = 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 borrowed 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