Compound Interest [Applications] — Test Yourself

The amount of ₹ 1,000 in 2 years and at 20% compound interest compounded per year is:

1. ₹ 1,200

2. ₹ 1,400

3. ₹ 800

4. ₹ 1,440

Answer

Given, P = ₹ 1,000

R = 20%

n = 2 years

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

Substituting the values,

$A = 1,000 \Big(1 + \dfrac{20}{100}\Big)^2\\[1em] = 1,000 \Big(1 + \dfrac{2}{10}\Big)^2\\[1em] = 1,000 \times \Big(\dfrac{12}{10}\Big)^2\\[1em] = 1,000 \times \dfrac{144}{100}\\[1em] = 1,440.$

Hence, option 4 is the correct option.

The difference between C.I. and S.I. at 10% in 2 years on ₹ 100 is:

1. ₹ 1

2. ₹ 41

3. ₹ 00

4. none of these

Answer

For S.I. :

P = ₹ 100

R = 10%

T = 2 years

$I = \dfrac{P \times R \times T}{100} \\[1em] = \dfrac{100 \times 10 \times 2}{100} \\[1em] = \dfrac{2000}{100}\\[1em] = 20.$

For C.I. :

For 1st year :

P = ₹ 100

T = 1 year

R = 10%

$I = \dfrac{P \times R \times T}{100}\\[1em] = \dfrac{100 \times 10 \times 1}{100}\\[1em] = \dfrac{1000}{100}\\[1em] = 10.$

Amount = P + I = ₹ 100 + ₹ 10 = ₹ 110

For 2nd year :

P = ₹ 110

R = 10%

T = 1 year

$I = \dfrac{P \times R \times T}{100} \\[1em] = \dfrac{110 \times 10 \times 1}{100} \\[1em] = \dfrac{1100}{100} \\[1em] = 11.$

Amount = P + I = ₹ 110 + ₹ 11 = ₹ 121.

C.I. = Final amount - Initial principal = ₹ 121 - ₹ 100 = ₹ 21.

Difference between C.I. and S.I. = ₹ 21 - ₹ 20 = ₹ 1.

Hence, option 1 is the correct option.

A certain sum of money (₹ P) is lent for $3\dfrac{1}{2}$ years at r% C.I. compounded half yearly. The interest accrued will be:

1. $P\Big(1 + \dfrac{r}{100}\Big)^{\dfrac{7}{2}} - P$

2. $P\Big(1 + \dfrac{r}{100}\Big)^3 \times \Big(1 + \dfrac{r}{2 \times 100}\Big)^1 - P$

3. $P\Big(1 + \dfrac{r}{2 \times 100}\Big)^{\dfrac{7}{2}} - P$

4. $P\Big(1 + \dfrac{r}{2 \times 100}\Big)^7 - P$

Answer

Given, the principal amount = ₹ P.

The time period = $3\dfrac{1}{2} = \dfrac{7}{2}$ years.

The annual interest rate = r% .

Interest is compounded half-yearly.

The formula for C.I. when compounded half-yearly is I

= P $\Big(1 + \dfrac{r}{2 \times 100}\Big)^{n \times 2} - P$

Substituting the values, we get :

$I = P \Big(1 + \dfrac{r}{2 \times 100}\Big)^{\dfrac{7}{2} \times 2} - P\\[1em] = P \Big(1 + \dfrac{r}{2 \times 100}\Big)^7 - P$

Hence, option 4 is the correct option.

A certain sum of money (₹ P) is lent for $3\dfrac{1}{2}$ years at r% C.I. compounded yearly. The interest accrued will be:

1. $P \Big(1 + \dfrac{r}{100}\Big)^{\dfrac{7}{2}} - P$

2. $P\Big(1 + \dfrac{r}{100}\Big)^3 \times \Big(1 + \dfrac{r}{2 \times 100}\Big)^1 - P$

3. $P\Big(1 + \dfrac{r}{2 \times 100}\Big)^{\dfrac{7}{2}} - P$

4. $P\Big(1 + \dfrac{r}{2 \times 100}\Big)^7 - P$

Answer

Given, the principal amount = ₹ P, rate of interest r%.

For complete 3 years it will be calculated normally but for half year it will be calculated by taking 1 half year and half rate of interest.

A = $P\Big(1 + \dfrac{r}{100}\Big)^3 \times \Big(1 + \dfrac{r}{2 \times 100}\Big)^1$

C.I. = A - P

= $P\Big(1 + \dfrac{r}{100}\Big)^3 \times \Big(1 + \dfrac{r}{2 \times 100}\Big)^1 - P$

Hence, option 2 is the correct option.

Statement 1: $P\Big(1 + \dfrac{r}{100}\Big)^7 - P\Big(1 + \dfrac{r}{100}\Big)^6$ = Interest accrued in 7th year

Statement 2: C.I. accrued in 7 years = P + $P\Big(1 + \dfrac{r}{100}\Big)^7$ and C.I. accrued in 6 years = P + $P\Big(1 + \dfrac{r}{100}\Big)^6$

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

Interest for a particular year = Amount in that year - Amount in previous year

Interest for 7th year = Amount in 7 years - Amount in 6 years

= $P\Big(1 + \dfrac{r}{100}\Big)^7 - P\Big(1 + \dfrac{r}{100}\Big)^6$

Thus, statement 1 is true and statement 2 is false.

Hence, option 3 is correct option.

Statement 1: The population of a town was x in the year 2024 which increased by 10% every year. The population in the year 2021 was equal to x $\Big(1 - \dfrac{10}{100}\Big)^3$

Statement 2: If the population increases from year 2021 to year 2024 at the rate of 10%, then corresponding decrease from 2024 to 2021 is 10 x 3%.

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

The general formula for population growth with a fixed annual percentage increase is:

P_t = P_o $\Big(1 + \dfrac{r}{100}\Big)^t$

where, P_t = Population after t years

P₀ = Initial population

r = Annual growth rate (in percentage)

t = Time in years

Given, the population in 2024 is x, and the annual growth rate is 10%. To find the population in 2021, we need to go back 3 years (from 2024 to 2021).

Rearranging the formula to solve for P₀ = $\dfrac{P_t}{\Big(1 + \dfrac{r}{100}\Big)^t}$

⇒ P₀ = $\dfrac{x}{\Big(1 + \dfrac{10}{100}\Big)^3}$

So, statement 1 is false.

If the population increases from year 2021 to year 2024 at the rate of 10%.

Let P₀ be the population in 2021 and P_t be the population in 2024.

⇒ P_t = P₀ $\Big(1 + \dfrac{10}{100}\Big)^3$

⇒ P_t = P₀ (1 + 0.1)³

⇒ P_t = P₀ x 1.1³

⇒ P_t = 1.331P₀

The percentage decrease = $\dfrac{P_t - P_o}{P_t} \times 100$

$= \dfrac{1.331P_o - P_o}{1.331P_o} \times 100\\[1em] = \dfrac{0.331P_o}{1.331P_o} \times 100\\[1em] = \dfrac{331}{1331} \times 100\\[1em] ≈ 24.87$

So, statement 2 is false.

∴ Both the statements are false.

Hence, option 2 is correct option.

Assertion (A): A certain sum of money (P) let out at r% C.I. increased for first 5 years and then decreased for next five years at the same rate is same as decrease on the same sum at the same rate (r%) during the first five years and then increase further next five years at the same rate.

Reason (R):

$P\Big(1 + \dfrac{r}{100}\Big)^5 \times \Big(1 - \dfrac{r}{100}\Big)^5$ is same as $P\Big(1 - \dfrac{r}{100}\Big)^5 \times \Big(1 + \dfrac{r}{100}\Big)^5$.

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 rate of interest and t be the time.

By formula, A = P $\Big(1 + \dfrac{r}{100}\Big)^t$

A certain sum of money P let out at r% C.I. increased for first 5 years and then decreased for next five years at the same rate, then

A₁ = P $\Big(1 + \dfrac{r}{100}\Big)^5 \times \Big(1 - \dfrac{r}{100}\Big)^5$

A certain sum of money P let out at r% C.I. decreased on the same sum at the same rate (r%) during the first five years and then increase further next five years at the same rate, then

A₂ = P $\Big(1 - \dfrac{r}{100}\Big)^5 \times \Big(1 + \dfrac{r}{100}\Big)^5$

By rules of multiplication :

Since, $P \Big(1 + \dfrac{r}{100}\Big)^5 \times \Big(1 - \dfrac{r}{100}\Big)^5 = P \Big(1 - \dfrac{r}{100}\Big)^5 \times \Big(1 + \dfrac{r}{100}\Big)^5$

So, reason (R) is true.

Thus, amount will be same.

So, assertion (A) is true and reason (R) clearly explains assertion (A).

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

Hence, option 3 is correct option.

Assertion (A): A = P $\Big(1 + \dfrac{10}{100}\Big)^2$ = 1.21P

Reason (R): $\Big(1 + \dfrac{10}{100}\Big)^2$ = (1.1)² = 1.21

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 rate of interest and t be the time.

By formula, A = P $\Big(1 + \dfrac{r}{100}\Big)^t$

A = P $\Big(1 + \dfrac{10}{100}\Big)^2 = P \Big(1 + \dfrac{1}{10}\Big)^2 = (1 + 0.1)^2P = 1.1^2$ P = 1.21P

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

Hence, option 3 is correct option.

Simple interest on a sum of money for 2 years at 4% growth rate is ₹ 450. Find compound interest on the same sum and at the same rate for 1 year, if the interest is reckoned half-yearly.

Answer

Let sum of money be ₹ x.

Given,

Simple interest on sum of money for 2 years at 4% growth rate is ₹ 450.

By formula,

$S.I. = \dfrac{P \times R \times T}{100} \\[1em] \Rightarrow 450 = \dfrac{x \times 4 \times 2}{100} \\[1em] \Rightarrow x = \dfrac{450 \times 100}{4 \times 2} \\[1em] \Rightarrow x = ₹ 5625.$

For C.I. :

P = ₹ 5625

n = 1 year

r = 4% compounded half-yearly

$A = P\Big(1 + \dfrac{r}{2 \times 100}\Big)^{n \times 2} \\[1em] = 5625 \times \Big(1 + \dfrac{4}{200}\Big)^{1 \times 2} \\[1em] = 5625 \times \Big(\dfrac{204}{200}\Big)^2 \\[1em] = 5625 \times \Big(\dfrac{51}{50}\Big)^2 \\[1em] = 5625 \times \dfrac{2601}{2500} \\[1em] = ₹ 5852.25$

C.I. = A - P = ₹ 5852.25 - ₹ 5625 = ₹ 227.25

Hence, compound interest = ₹ 227.25

Find the compound interest to the nearest rupee on ₹ 10800 for $2\dfrac{1}{2}$ years at 10% per annum.

Answer

Given,

P = ₹ 10800

T = 2.5 years

r = 10%

For first 2 years :

$A = P\Big(1 + \dfrac{r}{100}\Big)^n \\[1em] = 10800 \times \Big(1 + \dfrac{10}{100}\Big)^2 \\[1em] = 10800 \times \Big(\dfrac{110}{100}\Big)^2\\[1em] = 10800 \times \Big(\dfrac{11}{10}\Big)^2 \\[1em] = 10800 \times \dfrac{121}{100} \\[1em] = ₹ 13068.$

For next $\dfrac{1}{2}$ year :

P = ₹ 13068

$A = P\Big(1 + \dfrac{r}{2 \times 100}\Big)^{n \times 2} \\[1em] = 13068 \times \Big(1 + \dfrac{10}{2 \times 100}\Big)^{\dfrac{1}{2} \times 2} \\[1em] = 13068 \times \Big(1 + \dfrac{1}{20}\Big)^1 \\[1em] = 13068 \times \dfrac{21}{20} \\[1em] = ₹ 13721.$

By formula,

C.I. = A - P = ₹ 13721 - ₹ 10800 = ₹ 2921.

Hence, compound interest = ₹ 2921.

The value of a machine, purchased two years ago, depreciates at the annual rate of 10%. If its present value is ₹ 97200, find :

(i) its value after 2 years.

(ii) its value when it was purchased.

Answer

(i) In depreciation :

Value after n years = Present value $\times \Big(1 - \dfrac{r}{100}\Big)^n$

$\text{Value of machine after 2 years} = 97200 \times \Big(1 - \dfrac{10}{100}\Big)^2 \\[1em] = 97200 \times \Big(\dfrac{90}{100}\Big)^2 \\[1em] = 97200 \times \Big(\dfrac{9}{10}\Big)^2 \\[1em] = 97200 \times \dfrac{81}{100} \\[1em] = 972 \times 81 \\[1em] = ₹ 78732.$

Hence, value of machine after 2 years = ₹ 78732.

(ii) Let value of machine when it was purchased be ₹ x and its depreciate to ₹ 97200 in two years.

$\therefore 97200 = x \times \Big(1 - \dfrac{10}{100}\Big)^2 \\[1em] \Rightarrow 97200 = x \times \Big(\dfrac{90}{100}\Big)^2 \\[1em] \Rightarrow 97200 = x \times \Big(\dfrac{9}{10}\Big)^2 \\[1em] \Rightarrow 97200 = x \times \dfrac{81}{100} \\[1em] \Rightarrow x = \dfrac{97200 \times 100}{81} \\[1em] \Rightarrow x = 120000.$

Hence, machine's value when it was purchased = ₹ 120000.

Anuj and Rajesh each lent the same sum of money for 2 years at 8% simple interest and compound interest respectively. Rajesh received ₹ 64 more than Anuj. Find the money lent by each and interest received.

Answer

Let money lent by both be ₹ x.

For Anuj :

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

For Rajesh :

C.I. = A - P

$C.I. = P\Big(1 + \dfrac{r}{100}\Big)^n - P \\[1em] = x \times \Big(1 + \dfrac{8}{100}\Big)^2 - x \\[1em] = x \times \Big(\dfrac{108}{100}\Big)^2 - x \\[1em] = x \times \Big(\dfrac{27}{25}\Big)^2 - x \\[1em] = x \times \dfrac{729}{625} - x \\[1em] = \dfrac{729x}{625} - x \\[1em] = \dfrac{729x - 625x}{625} \\[1em] = \dfrac{104x}{625}.$

Given,

Rajesh received ₹ 64 more than Anuj. So, it means Rajesh received ₹ 64 more than Anuj as interest.

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

Calculating S.I. and C.I. :

$S.I. = \dfrac{4x}{25} = \dfrac{4 \times 10000}{25} = ₹ 1600. \\[1em] C.I. = \dfrac{104x}{625} = \dfrac{104 \times 10000}{625} = ₹ 1664.$

Hence, sum of money lent = ₹ 10000 and interest received by Anuj = ₹ 1600 and by Rajesh = ₹ 1664.

Calculate the sum of money on which the compound interest (payable annually) for 2 years be four times the simple interest on ₹ 4715 for 5 years, both at the rate of 5 percent per annum.

Answer

For S.I. :

P = ₹ 4715

R = 5%

T = 5 years

S.I. = $\dfrac{P \times R \times T}{100} = \dfrac{4715 \times 5 \times 5}{100}$ = ₹ 1178.75

Given,

C.I. is four times the S.I.

∴ C.I. = 4 × 1178.75 = ₹ 4715

For C.I. :

Let P = ₹ x, n = 2 years, r = 5%

By formula,

C.I. = A - P

$C.I. = P\Big(1 + \dfrac{r}{100}\Big)^n - P \\[1em] \Rightarrow 4715 = x \times \Big(1 + \dfrac{5}{100}\Big)^2 - x \\[1em] \Rightarrow 4715 = x \times \Big(\dfrac{105}{100}\Big)^2 - x \\[1em] \Rightarrow 4715 = x \times \Big(\dfrac{21}{20}\Big)^2 - x \\[1em] \Rightarrow 4715 = x \times \dfrac{441}{400} - x \\[1em] \Rightarrow 4715 = \dfrac{441x}{400} - x \\[1em] \Rightarrow 4715 = \dfrac{441x - 400x}{400} \\[1em] \Rightarrow 4715 = \dfrac{41x}{400} \\[1em] \Rightarrow x - \dfrac{4715 \times 400}{41} \\[1em] \Rightarrow x = ₹ 46000$

Hence, sum of money = ₹ 46000.

A sum of money was invested for 3 years, interest being compounded annually. The rates for successive years were 10%, 15% and 18% respectively. If the compound interest for the second year amounted to ₹ 4950, find the sum invested.

Answer

Let sum invested be ₹ x.

Amount after 1st year :

$A = x \times \Big(1 + \dfrac{10}{100}\Big) \\[1em] = x \times \dfrac{110}{100} \\[1em] = \dfrac{11x}{10}.$

Amount after 2nd year :

$A = x \times \Big(1 + \dfrac{10}{100}\Big)\Big(1 + \dfrac{15}{100}\Big) \\[1em] = x \times \dfrac{110}{100} \times \dfrac{115}{100} \\[1em] = \dfrac{253x}{200}.$

C.I. for 2nd year = Amount after 2 years - Amount after 1 year = $\dfrac{253x}{200} - \dfrac{11x}{10}$

Given,

Compound interest for the second year amounted to ₹ 4950.

$\therefore \dfrac{253x}{200} - \dfrac{11x}{10} = 4950 \\[1em] \Rightarrow \dfrac{253x - 220x}{200} = 4950 \\[1em] \Rightarrow \dfrac{33x}{200} = 4950 \\[1em] \Rightarrow x = \dfrac{4950 \times 200}{33} \\[1em] \Rightarrow x = ₹ 30000.$

Hence, sum invested = ₹ 30000.

A sum of money is invested at 10% per annum compounded half-yearly. If the difference of amounts at end of 6 months and 12 months is ₹ 189, find the sum of money invested.

Answer

Let sum of money invested be ₹ x.

When interest is compounded half-yearly :

A = $P\Big(1 + \dfrac{r}{2 \times 100}\Big)^{n \times 2}$

For first $\dfrac{1}{2}$ year :

$A = x \times \Big(1 + \dfrac{10}{2 \times 100}\Big)^{\dfrac{1}{2} \times 2} \\[1em] = x \times \Big(1 + \dfrac{1}{20}\Big) \\[1em] = x \times \dfrac{21}{20} \\[1em] = \dfrac{21x}{20}.$

For first 1 year :

$A = x \times \Big(1 + \dfrac{10}{2 \times 100}\Big)^{1 \times 2} \\[1em] = x \times \Big(1 + \dfrac{1}{20}\Big)^2 \\[1em] = x \times \Big(\dfrac{21}{20}\Big)^2 \\[1em] = \dfrac{441x}{400}.$

Given,

Difference of amounts at end of 6 months and 12 months is ₹ 189.

$\Rightarrow \dfrac{441x}{400} - \dfrac{21x}{20} = 189 \\[1em] \Rightarrow \dfrac{441x - 420x}{400} = 189 \\[1em] \Rightarrow \dfrac{21x}{400} = 189 \\[1em] \Rightarrow x = \dfrac{189 \times 400}{21} \\[1em] \Rightarrow x = ₹ 3600.$

Hence, sum invested = ₹ 3600.

Rohit borrows ₹ 86000 from Arun for two years at 5% per annum simple interest. He immediately lends out this money to Akshay at 5% compound interest compounded annually for the same period. Calculate Rohit's profit in transaction at the end of two years.

Answer

Given,

P = ₹ 86000

Rate of interest = 5%

Time = 2 years

Calculating S.I. :

S.I. = $\dfrac{P \times R \times T}{100} = \dfrac{86000 \times 5 \times 2}{100}$ = ₹ 8600.

Calculating C.I. :

C.I. = A - P

$C.I. = P\Big(1 + \dfrac{r}{100}\Big)^n - P \\[1em] = 86000 \times \Big(1 + \dfrac{5}{100}\Big)^2 - 86000 \\[1em] = 86000 \times \Big(\dfrac{105}{100}\Big)^2 - 86000 \\[1em] = 86000 \times \Big(\dfrac{21}{20}\Big)^2 - 86000 \\[1em] = 86000 \times \dfrac{441}{400} - 86000 \\[1em] = 94815 - 86000 \\[1em] = ₹ 8815.$

Rohit's profit = Interest received by him - Interest paid by him

= ₹ 8815 - ₹ 8600 = ₹ 215.

Hence, Rohit's profit = ₹ 215.

The simple interest on a certain sum of money for 3 years at 5% per annum is ₹ 1200. Find the amount due and the compound interest on this sum of money at the same rate and after 2 years, interest is reckoned annually.

Answer

Let sum of money be ₹ x.

For S.I. :

P = ₹ x

Rate = 5%

Time = 3 years

S.I. = $\dfrac{P \times R \times T}{100} = \dfrac{x \times 5 \times 3}{100} = \dfrac{3x}{20}$.

Given,

S.I. = ₹ 1200

$\therefore \dfrac{3x}{20} = 1200 \\[1em] \Rightarrow x = \dfrac{1200 \times 20}{3} = ₹ 8000.$

When rate is compounded annually:

$A = P\Big(1 + \dfrac{r}{100}\Big)^n \\[1em] = x \times \Big(1 + \dfrac{5}{100}\Big)^2 \\[1em] = 8000 \times \Big(\dfrac{105}{100}\Big)^2 \\[1em] = 8000 \times \Big(\dfrac{21}{20}\Big)^2 \\[1em] = ₹ 8820.$

C.I. = A - P = ₹ 8820 - ₹ 8000 = ₹ 820.

Hence, amount due = ₹ 8820 and C.I. = ₹ 820.

Nikita invests ₹ 6000 for two years at a certain rate of interest compounded annually. At the end of first year it amounts to ₹ 6720. Calculate :

(a) the rate percent (i.e. the rate of growth)

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

Answer

(a) Let rate percent be r%.

Given,

P = ₹ 6000

n = 1 year

A = ₹ 6720

By formula,

A = $P\Big(1 + \dfrac{r}{100}\Big)^n$

Substituting values we get :

$\Rightarrow 6720 = 6000\Big(1 + \dfrac{r}{100}\Big)^1 \\[1em] \Rightarrow \dfrac{6720}{6000} = 1 + \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{6720}{6000} - 1 = \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{6720 - 6000}{6000} = \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{720}{6000} = \dfrac{r}{100} \\[1em] \Rightarrow r = \dfrac{720 \times 100}{6000} \\[1em] \Rightarrow r = 12$

Hence, rate percent = 12%.

(b) By formula,

A = $P\Big(1 + \dfrac{r}{100}\Big)^n$

$A = 6000 \times \Big(1 + \dfrac{12}{100}\Big)^2 \\[1em] = 6000 \times \Big(\dfrac{112}{100}\Big)^2 \\[1em] = 6000 \times \Big(\dfrac{28}{25}\Big)^2 \\[1em] = 6000 \times \dfrac{784}{625} \\[1em] = ₹ 7526.40$

Hence, amount at the end of 2 years = ₹ 7526.40

A certain sum of money invested at CI triples itself in 8 year interest being payable annually. In how many years will it be 81 times?

Answer

Let rate of interest be r% and sum of money be ₹ P.

By formula, A = $P\Big(1 + \dfrac{r}{100}\Big)^t$

Given,

₹ P becomes three times of itself in 8 years.

$\therefore 3P = P\Big(1 + \dfrac{r}{100}\Big)^8 \\[1em] \Rightarrow \dfrac{3P}{P} = \Big(1 + \dfrac{r}{100}\Big)^8 \\[1em] \Rightarrow 3 = \Big(1 + \dfrac{r}{100}\Big)^8 .....................(1)$

Let in n years money becomes 81 times.

$\Rightarrow P\Big(1 + \dfrac{r}{100}\Big)^n = 81P\\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big)^n = \dfrac{81P}{P}\\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big)^n = 81\\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big)^n = 3^4\\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big)^n = \Big[\Big(1 + \dfrac{r}{100}\Big)^8\Big]^4 \text{ [From (1)]}\\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big)^n = \Big(1 + \dfrac{r}{100}\Big)^{32}\\[1em]$

⇒ n = 32.

Hence, in 32 years money will becomes 81 times of itself.

A certain sum of money doubles itself at a given rate in 8 years compounded yearly. In how many years will it be four times at the same rate compounded yearly ?

Answer

Let rate of interest be r% and sum of money be ₹ P.

By formula, A = $P\Big(1 + \dfrac{r}{100}\Big)^t$

Given,

₹ P becomes twice of itself in 8 years.

$\therefore 2P = P\Big(1 + \dfrac{r}{100}\Big)^8 \\[1em] \Rightarrow \dfrac{2P}{P} = \Big(1 + \dfrac{r}{100}\Big)^8 \\[1em] \Rightarrow 2 = \Big(1 + \dfrac{r}{100}\Big)^8 .....................(1)$

Let the money become four times in n years.

$\Rightarrow P\Big(1 + \dfrac{r}{100}\Big)^n = 4P\\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big)^n = \dfrac{4P}{P}\\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big)^n = 4\\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big)^n = 2^2\\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big)^n = \Big[\Big(1 + \dfrac{r}{100}\Big)^8\Big]^2 \text{ [From (1)]}\\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big)^n = \Big(1 + \dfrac{r}{100}\Big)^{16}\\[1em]$

⇒ n = 16.

Hence, in 16 years money will becomes 4 times of itself.

Mr. Sharma wants to divide ₹ 1,68,200 between his two sons who are 16 years and 18 years old respectively, in such a way that the sum invested at the rate of 5% p.a compound interest annually will give the same amount to each when they attain the age of 21 years. How much should he divide the sum ?

Answer

Given:

Total money = ₹1,68,200

Rate = 5% p.a (compound interest)

Younger son = 16 years → will get money after 5 years

Elder son = 18 years → will get money after 3 years

Let the amount to be distributed to younger son be '₹ x' and to elder son to be '₹ y'

∴ x + y = 168200.......(1)

By formula, A = $P\Big(1 + \dfrac{r}{100}\Big)^n$

Amount after 5 years for younger son (at the age of 21),

A = $x\Big(1 + \dfrac{5}{100}\Big)^5$

= x(1 + 0.05)⁵

Amount after 3 years for elder son (at the age of 21),

A = $y\Big(1 + \dfrac{5}{100}\Big)^3$

= y(1 + 0.05)³

Since both will have same amount at age 21:

∴ x(1 + 0.05)⁵ = y (1 + 0.05)³

⇒ x(1.05)² = y

⇒ x × 1.1025 = y.....(2)

Substituting the value of y from equation (2) in equation (1) :

⇒ x + 1.1025x = 168200

⇒ 2.1025x = 168200

⇒ x = $\dfrac{168200}{2.1025}$

⇒ x = 80,000.

Substituting the value of x in equation (1),

⇒ 80,000 + y = 1,68,200

⇒ y = 1,68,200 - 80,000

⇒ y = 88,200.

Hence, sum should be divided as ₹ 80,000 and ₹ 88,200.