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

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 in order 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 (cost) 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