Compound Interest [Basic Concepts] — Test Yourself

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): Amount in 3 years = Principal + C.I. of 3 years and Amount in 2 years = Principal + C.I. of 2 years

⇒ Amount in 3 years - Amount in 2 years = C.I. in 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): At compound interest, interest of 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, then

Amount in 2 years = ₹ 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 = $x + \dfrac{2x}{25} = \dfrac{27x}{25}$.

For second year :

P = ₹ $\dfrac{27x}{25}$

R = 8%

T = 1 year

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

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

C.I. = Final amount - Initial Principal

$= \dfrac{729x}{625} - x \\[1em] = \dfrac{729x - 625x}{625} \\[1em] = \dfrac{104x}{625}.$

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

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

Hence, required sum = ₹ 10000.

A sum of ₹ 13500 is invested at 16% per annum compound interest for 5 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 = ₹ 13500

R = 16%

T = 1 year

I = $\dfrac{P \times R \times T}{100} = \dfrac{13500 \times 16 \times 1}{100}$ = ₹ 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 = $\dfrac{P \times R \times T}{100} = \dfrac{15660 \times 16 \times 1}{100}$ = ₹ 2505.6

Hence, interest for second year = ₹ 2506.

Saurabh invests ₹ 48000 for 7 years at 10% per annum compound interest. Calculate :

(i) the interest for the first year.

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

(iii) the interest for the third year.

Answer

(i) For first year :

P = ₹ 48000

R = 10%

T = 1 year

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

Hence, interest for first year = ₹ 4800.

(ii) For second year :

P = ₹ 48000 + ₹ 4800 = ₹ 52800

R = 10%

T = 1 year

I = $\dfrac{P \times R \times T}{100} = \dfrac{52800 \times 10 \times 1}{100}$ = ₹ 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 = $\dfrac{P \times R \times T}{100} = \dfrac{58080 \times 10 \times 1}{100}$ = ₹ 5808.

Hence, interest for third year = ₹ 5808.

Ashok borrowed ₹ 12000 at some rate per cent compound interest. After a year, he paid back ₹ 4000. If compound interest for the second year be ₹ 920, find :

(i) the rate of interest charged

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

Answer

(i) Let rate of interest be x%.

For first year :

P = ₹ 12000

R = x%

T = 1 year

I = $\dfrac{P \times R \times T}{100} = \dfrac{12000 \times x \times 1}{100}$ = 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 = $\dfrac{P \times R \times T}{100} = \dfrac{(8000 + 120x) \times x \times 1}{100}$.

Given,

Interest for second year = ₹ 920

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

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.

On a certain sum of money, let out at C.I., interests for first, second and third years are ₹ 1500; ₹ 1725 and ₹ 2070 respectively. Find the rate of interest for the (i) second year (ii) third year.

Answer

(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

= $\dfrac{\text{Interest earned on ₹ 1500 in second year}}{\text{Interest earned in first year}} \times 100 = \dfrac{225}{1500} \times 100$ = 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

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

Hence, interest for the third year = 20%.