Compound Interest — Exercise 2(B)

Calculate the amount and the compound interest on ₹ 10,000 for 2 years at 8% p.a., compounded annually.

Answer

Given,

P = ₹ 10,000

n = 2 years

r = 8%

By formula,

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

Substituting values we get :

$\Rightarrow A = 10000\Big(1 + \dfrac{8}{100}\Big)^2 \\[1em] \Rightarrow A = 10000\Big(\dfrac{100+8}{100}\Big)^2 \\[1em] \Rightarrow A = 10000\Big(\dfrac{108}{100}\Big)^2 \\[1em] \Rightarrow A = 10000\Big(\dfrac{27}{25}\Big)^2 \\[1em] \Rightarrow A = 10000 \times \dfrac{729}{625} \\[1em] \Rightarrow A = ₹ 11,664$

Compound interest = Final amount - Initial principal

= ₹ 11,664 - ₹ 10,000 = ₹ 1,664

Hence, amount = ₹ 11,664 and compound interest = ₹ 1,664.

Calculate the amount and the compound interest on ₹ 64,000 for 3 years at $7\dfrac{1}{2}$% per annum, compounded annually.

Answer

Given,

P = ₹ 64,000

n = 3 years

r = $7\dfrac{1}{2}$% = 7.5%

By formula,

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

Substituting values we get :

$\Rightarrow A = 64000\Big(1 + \dfrac{7.5}{100}\Big)^3 \\[1em] \Rightarrow A = 64000\Big(\dfrac{100+7.5}{100}\Big)^3 \\[1em] \Rightarrow A = 64000\Big(\dfrac{107.5}{100}\Big)^3 \\[1em] \Rightarrow A = 64000 \times \Big(\dfrac{21.5}{20}\Big)^3 \\[1em] \Rightarrow A = 64000 \times \dfrac{9938.375}{8000} \\[1em] \Rightarrow A = ₹ 79,507.$

Compound interest = Final amount - Initial principal = ₹ 79,507 - ₹ 64,000 = ₹ 15,507

Hence, amount = ₹ 79,507 and compound interest = ₹ 15,507.

How much will ₹ 12,000 amount to in 2 years at compound interest, the rates of interest for successive years being 10% and 11% respectively ?

Answer

Given,

P = ₹ 12,000

r₁ = 10%

r₂ = 11%

n = 2 years

By formula,

A = $P\Big(1 + \dfrac{r_1}{100}\Big)\Big(1 + \dfrac{r_2}{100}\Big)$

Substituting values we get :

$\Rightarrow A = 12000 \times \Big(1 + \dfrac{10}{100}\Big) \times \Big(1 + \dfrac{11}{100}\Big) \\[1em] = 12000 \times \dfrac{110}{100} \times \dfrac{111}{100} \\[1em] = 12000 \times \dfrac{11}{10} \times \dfrac{111}{100} \\[1em] = \dfrac{12000 \times 11 \times 111}{10 \times 100} \\[1em] = ₹ 14,652.$

Hence, final amount = ₹ 14,652.

Calculate the amount and the compound interest on ₹ 25,000 for 3 years, the rates of interest for the successive years being 8%, 9% and 10%, compounded annually.

Answer

Given,

P = ₹ 25,000

r₁ = 8%

r₂ = 9%

r₃ = 10%

n = 3 years

By formula,

A = $P\Big(1 + \dfrac{r_1}{100}\Big)\Big(1 + \dfrac{r_2}{100}\Big)\Big(1 + \dfrac{r_3}{100}\Big)$

Substituting values we get :

$\Rightarrow A = 25000 \times \Big(1 + \dfrac{8}{100}\Big) \times \Big(1 + \dfrac{9}{100}\Big) \times \Big(1 + \dfrac{10}{100}\Big) \\[1em] = 25000 \times \dfrac{108}{100} \times \dfrac{109}{100} \times \dfrac{110}{100} \\[1em] = 25000 \times \dfrac{27}{25} \times \dfrac{109}{100} \times \dfrac{11}{10}\\[1em] = \dfrac{25000 \times 27 \times 109 \times 11}{25 \times 100 \times 10} \\[1em] = 27 \times 109 \times 11 \\[1em] = ₹ 32,373.$

Compound interest = Final amount - Initial principal

= ₹ 32,373 - ₹ 25,000

= ₹ 7,373.

Hence, amount = ₹ 32,373 and compound interest = ₹ 7,373.

Find the amount and the compound interest on ₹ 7,500 for 2 years 8 months at 10% p.a., compounded annually.

Answer

Given,

P = ₹ 7,500

n = 2 years 8 months

= 2 $\dfrac{8}{12}$ years = 2 $\dfrac{2}{3}$ years

r = 10%

By formula,

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

Substituting values we get :

$\Rightarrow A = 7500\Big(1 + \dfrac{10}{100}\Big)^2 \Big(1 + \dfrac{\dfrac{2}{3} \times 10}{100}\Big) \\[1em] \Rightarrow A = 7500\Big(\dfrac{100+10}{100}\Big)^2 \Big(1 + \dfrac{20}{300}\Big) \\[1em] \Rightarrow A = 7500\Big(\dfrac{110}{100}\Big)^2 \Big(\dfrac{300 + 20}{300}\Big) \\[1em] \Rightarrow A = 7500\Big(\dfrac{11}{10}\Big)^2 \Big(\dfrac{320}{300}\Big)\\[1em] \Rightarrow A = 7500 \times \dfrac{121}{100} \times \dfrac{32}{30}\\[1em] \Rightarrow A = ₹ 9,680$

C.I. = A - P = ₹ 9,680 - ₹ 7,500 = ₹ 2,180

Hence, amount = ₹ 9,680 and compound interest = ₹ 2,180.

If simple interest on sum of money for 3 years at 8% per annum is ₹ 7,500, find the compound interest on the same sum for the same period at same rate.

Answer

Given,

I = ₹ 7,500

T = 3 years

R = 8% p.a. simple interest

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

$\Rightarrow 7500 = \dfrac{P \times 8 \times 3}{100} \\[1em] \Rightarrow 7500 = \dfrac{P \times 24}{100} \\[1em] \Rightarrow P = \dfrac{7500 \times 100}{24} \\[1em] \Rightarrow P = ₹ 31,250.$

Let's calculate compound interest for this principal, rate of interest and time.

By formula,

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

$\Rightarrow A = 31250\Big(1 + \dfrac{8}{100}\Big)^3 \\[1em] \Rightarrow A = 31250\Big(\dfrac{100+8}{100}\Big)^3 \\[1em] \Rightarrow A = 31250\Big(\dfrac{108}{100}\Big)^3 \\[1em] \Rightarrow A = 31250\Big(\dfrac{27}{25}\Big)^3 \\[1em] \Rightarrow A = 31250 \times \dfrac{19683}{15625} \\[1em] \Rightarrow A = ₹ 39366.$

Compound interest = Final amount - Initial principal

= ₹ 39,366 - ₹ 31,250 = ₹ 8,116.

Hence, compound interest = ₹ 8,116.

Calculate the amount and compound interest on ₹ 16,000 for 1 year at 15% per annum, compounded half yearly.

Answer

Given,

Principal (P) = ₹ 16,000

Time (n) = 1 year

Rate (r) = 15% compounded half-yearly

When rate of interest is compounded half-yearly :

By formula,

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

Substituting values we get :

$\Rightarrow A = 16000 \times \Big(1 + \dfrac{15}{2 \times 100}\Big)^{1 \times 2} \\[1em] \Rightarrow A = 16000 \times \Big(\dfrac{200+15}{200}\Big)^2 \\[1em] \Rightarrow A = 16000 \times \Big(\dfrac{215}{200}\Big)^2 \\[1em] \Rightarrow A = 16000 \times \Big(\dfrac{43}{40}\Big)^2 \\[1em] \Rightarrow A = 16000 \times \Big(\dfrac{1849}{1600}\Big) \\[1em] \Rightarrow A = \dfrac{16000 \times 1849}{1600} = ₹ 18,490.$

Compound interest = Amount - Principal = ₹ 18,490 - ₹ 16,000 = ₹ 2,490

Hence, amount = ₹ 18,490 and compound interest = ₹ 2,490.

Find the amount and compound interest on ₹ 1,25,000 for $1\dfrac{1}{2}$ years at 12% per annum, compounded half yearly.

Answer

Given,

Principal (P) = ₹ 1,25,000

Time (n) = 1 $\dfrac{1}{2}$ years = 1.5 years

Rate (r) = 12% compounded half-yearly

When rate of interest is compounded half-yearly :

By formula,

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

Substituting values we get :

$\Rightarrow A = 125000 \times \Big(1 + \dfrac{12}{2 \times 100}\Big)^{1.5 \times 2} \\[1em] \Rightarrow A = 125000 \times \Big(1 + \dfrac{3}{50}\Big)^3 \\[1em] \Rightarrow A = 125000 \times \Big(\dfrac{50 + 3}{50}\Big)^3 \\[1em] \Rightarrow A = 125000 \times \Big(\dfrac{53}{50}\Big)^3 \\[1em] \Rightarrow A = 125000 \times \Big(\dfrac{148877}{125000}\Big) \\[1em] \Rightarrow A = \dfrac{125000 \times 148877}{125000} = ₹ 1,48,877$

Compound interest = Amount - Principal

= ₹ 1,48,877 - ₹ 1,25,000 = ₹ 23,877

Hence, amount = ₹ 1,48,877 and compound interest = ₹ 23,877.

A sum of ₹ 12,500 is deposited for $1\dfrac{1}{2}$ years, compounded half yearly. It amounts to ₹ 13,000 at the end of first half year. Find:

(i) The rate of interest

(ii) The final amount. Give your answer correct to the nearest rupee.

Answer

(i) Given,

P = ₹ 12,500

n = 1 half year

Amount = ₹ 13,000

When rate of interest is compounded half-yearly :

By formula,

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

For first half year:

$\Rightarrow 13000 = 12500\Big(1 + \dfrac{R}{2 \times 100}\Big)^{0.5 \times 2} \\[1em] \Rightarrow \dfrac{13000}{12500}= \Big(1 + \dfrac{R}{200}\Big)^1 \\[1em] \Rightarrow 1.04 - 1 = \Big(\dfrac{R}{200}\Big) \\[1em] \Rightarrow 0.04 = \Big(\dfrac{R}{200}\Big) \\[1em] \Rightarrow R = 0.04 \times 200 \\[1em] \Rightarrow R = 8$

Hence, Rate of interest = 8% p.a .

(ii) Given,

P = ₹ 12,500

n = 1.5 years

R = 8%

Let's calculate compound interest:

When rate of interest is compounded half-yearly :

By formula,

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

$\Rightarrow A = 12500 \Big(1 + \dfrac{8}{2 \times 100}\Big)^{2 \times 1.5} \\[1em] \Rightarrow A = 12500 \Big(\dfrac{200 + 8}{200}\Big)^3 \\[1em] \Rightarrow A = 12500 \Big(\dfrac{208}{200}\Big)^3 \\[1em] \Rightarrow A = 12500 \Big(\dfrac{26}{25}\Big)^3 \\[1em] \Rightarrow A = 12500 \times \dfrac{17576}{15625} \\[1em] \Rightarrow A = ₹ 14,060.80 \approx ₹ 14,061$

Hence, compound interest = ₹ 14,061.

The simple interest on a sum of money at 12% per annum for 1 year is ₹ 900. Find :

(i) the sum of money and

(ii) the compound interest on this sum for 1 year, payable half-yearly at the same rate.

Answer

(i) Given,

I = ₹ 900

R = 12%

T = 1 year

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

$\Rightarrow 900 = \dfrac{P \times 12 \times 1}{100}\\[1em] \Rightarrow 900 = \dfrac{P \times 12}{100}\\[1em] \Rightarrow P = \dfrac{900 \times 100}{12} \\[1em] \Rightarrow P = ₹ 7,500.$

Hence, principal = ₹ 7,500.

(ii) When rate of interest is compounded half-yearly :

By formula,

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

Substituting values we get :

$\Rightarrow A = 7500 \Big(1 + \dfrac{12}{2 \times 100}\Big)^{2 \times 1} \\[1em] \Rightarrow A = 7500 \Big(\dfrac{200 + 12}{200}\Big)^2 \\[1em] \Rightarrow A = 7500 \Big(\dfrac{212}{200}\Big)^2 \\[1em] \Rightarrow A = 7500 \Big(1.06\Big)^2 \\[1em] \Rightarrow A = 7500 \times 1.1236 \\[1em] \Rightarrow A = ₹ 8,427$

By formula,

Compound interest = Amount - Principal = ₹ 8,427 - ₹ 7500 = ₹ 927.

Hence, compound interest = ₹ 927.

What sum of money will amount to ₹ 18,150 in 2 years at 10% per annum, compounded annually?

Answer

Let sum of money be ₹ x.

Given,

P = ₹ x

r = 10%

n = 2 years

A = ₹ 18,150

By formula,

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

Substituting values we get :

$\Rightarrow 18150 = x \times \Big(1 + \dfrac{10}{100}\Big)^2 \\[1em] \Rightarrow 18150 = x \times \Big(\dfrac{110}{100}\Big)^2 \\[1em] \Rightarrow 18150 = x \times \Big(\dfrac{11}{10}\Big)^2 \\[1em] \Rightarrow 18150 = x \times \dfrac{121}{100} \\[1em] \Rightarrow x = \dfrac{18150 \times 100}{121} \\[1em] \Rightarrow x = ₹ 15,000.$

Hence, sum of money = ₹ 15,000.

What sum of money will amount to ₹ 93,170 in 3 years at 10% per annum, compounded annually?

Answer

Let sum of money be ₹ x.

Given,

P = ₹ x

r = 10%

n = 3 years

A = ₹ 93,170

By formula,

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

Substituting values we get :

$\Rightarrow 93170 = x \times \Big(1 + \dfrac{10}{100}\Big)^3 \\[1em] \Rightarrow 93170 = x \times \Big(\dfrac{110}{100}\Big)^3 \\[1em] \Rightarrow 93170 = x \times \Big(\dfrac{11}{10}\Big)^3 \\[1em] \Rightarrow 93170 = x \times \dfrac{1331}{1000} \\[1em] \Rightarrow x = \dfrac{93170 \times 1000}{1331} \\[1em] \Rightarrow x = ₹ 70,000.$

Hence, sum of money = ₹ 70,000.

On what sum of money will the compound interest for 2 years at 8% per annum be ₹ 7,488?

Answer

Let sum of money be ₹ x.

Given,

P = ₹ x

n = 2 years

r = 8%

C.I. = ₹ 7,488

A = P + I = ₹ x + ₹ 7,488

By formula,

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

Substituting values we get :

$\Rightarrow x + 7488 = x \times \Big(1 + \dfrac{8}{100}\Big)^2 \\[1em] \Rightarrow x + 7488 = x \times \Big(\dfrac{108}{100}\Big)^2 \\[1em] \Rightarrow x + 7488 = x \times \Big(\dfrac{27}{25}\Big)^2 \\[1em] \Rightarrow x + 7488 = x \times \dfrac{729}{625} \\[1em] \Rightarrow 625(x + 7488) = 729x \\[1em] \Rightarrow 625x + 4680000 = 729x \\[1em] \Rightarrow 729x - 625x = 4680000 \\[1em] \Rightarrow 104x = 4680000 \\[1em] \Rightarrow x = \dfrac{4680000}{104} \\[1em] \Rightarrow x = ₹ 45,000.$

Hence, sum of money = ₹ 45,000.

The difference between the simple interest and the compound interest on a sum of money for 2 years at 12% per annum is ₹ 216. Find the sum.

Answer

Given,

n = 2 years

r = 12%

Let sum of money be ₹ P.

C.I. = A - P

$C.I. = P\Big(1 + \dfrac{r}{100}\Big)^n - P \\[1em] = P\Big(1 + \dfrac{12}{100}\Big)^2 - P \\[1em] = P \times \Big(\dfrac{112}{100}\Big)^2 - P \\[1em] = P \times \Big(\dfrac{28}{25}\Big)^2 - P\\[1em] = P \times \dfrac{784}{625} - P \\[1em] = \dfrac{784P}{625} - P \\[1em] = \dfrac{784P - 625P}{625} \\[1em] = \dfrac{159P}{625}.$

By formula,

T = 2 years

$S.I. = \dfrac{P \times R \times T}{100} \\[1em] = \dfrac{P \times 12 \times 2}{100} \\[1em] = \dfrac{6P}{25}.$

Given,

Difference between S.I. and C.I. = ₹ 216

$\Rightarrow \dfrac{159P}{625} - \dfrac{6P}{25} = 216 \\[1em] \Rightarrow \dfrac{159P - 150P}{625} = 216 \\[1em] \Rightarrow \dfrac{9P}{625} = 216 \\[1em] \Rightarrow P = \dfrac{216 \times 625}{9} \\[1em] \Rightarrow P = 24 \times 625 \\[1em] \Rightarrow P = ₹ 15,000.$

Hence, sum = ₹ 15,000.

The difference between the simple interest and the compound interest on a sum of money for 3 years at 10% per annum is ₹ 558. Find the sum.

Answer

Given,

n = 3 years

r = 10 %

Let sum of money be ₹ P.

C.I. = A - P

$C.I. = P\Big(1 + \dfrac{r}{100}\Big)^n - P \\[1em] = P\Big(1 + \dfrac{10}{100}\Big)^3 - P \\[1em] = P \times \Big(\dfrac{110}{100}\Big)^3 - P \\[1em] = P \times \Big(\dfrac{11}{10}\Big)^3 - P\\[1em] = P \times \dfrac{1331}{1000} - P \\[1em] = \dfrac{1331P}{1000} - P \\[1em] = \dfrac{1331P - 1000P}{1000} \\[1em] = \dfrac{331P}{1000}.$

Calculating S.I.,

R = 10%

T = 3 years

$S.I. = \dfrac{P \times R \times T}{100} \\[1em] = \dfrac{P \times 10 \times 3}{100} \\[1em] = \dfrac{3P}{10}.$

Given,

Difference between S.I. and C.I. = ₹ 558

$\Rightarrow \dfrac{331P}{1000} - \dfrac{3P}{10} = 558 \\[1em] \Rightarrow \dfrac{331P - 300P}{1000} = 558 \\[1em] \Rightarrow \dfrac{31P}{1000} = 558 \\[1em] \Rightarrow P = \dfrac{558 \times 1000}{31} \\[1em] \Rightarrow P = 18 \times 1000 \\[1em] \Rightarrow P = ₹ 18,000.$

Hence, the sum = ₹ 18,000.

The difference between the compound interest for 1 year, compounded half-yearly and the simple interest for 1 year on a certain sum of money at 10% per annum is ₹ 360. Find the sum.

Answer

Let sum of money lent out be ₹ x.

Calculating C.I. payable half-yearly :

P = ₹ x

r = 10%

n = 1 year

C.I. = A - P

When rate of interest is compounded half-yearly :

By formula,

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

C.I. = A - P

Substituting values we get :

$C.I. = P\Big(1 + \dfrac{r}{2 \times 100}\Big)^{n \times 2} - P \\[1em] = x \times \Big(1 + \dfrac{10}{200}\Big)^{1 \times 2} - x \\[1em] = x \times \Big(\dfrac{210}{200}\Big)^2 - x \\[1em] = x \times \Big(\dfrac{21}{20}\Big)^2 - x \\[1em] = \dfrac{441x}{400} - x \\[1em] = \dfrac{441x - 400x}{400} \\[1em] = ₹ \dfrac{41x}{400}.$

Calculating S.I. :

T = 1 year

R = 10%

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

Given,

Difference between compound interest for a year payable half-yearly and simple interest on ₹ x lent out at 10% for a year is ₹ 360.

$\therefore \dfrac{41x}{400} - \dfrac{x}{10} = 360 \\[1em] \Rightarrow \dfrac{41x - 40x}{400} = 360 \\[1em] \Rightarrow \dfrac{x}{400} = 360 \\[1em] \Rightarrow x = 360 \times 400 = ₹ 1,44,000.$

Hence, the sum = ₹ 1,44,000.

At what rate per cent per annum compound interest will ₹ 6,250 amount to ₹ 7,290 in 2 years?

Answer

Given,

P = ₹ 6,250

A = ₹ 7,290

n = 2 years

Let rate of interest be r%.

By formula,

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

Substituting values we get :

$\Rightarrow 7290 = 6250 \times \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{7290}{6250} = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{729}{625} = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{27}{25}\Big)^2 = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{27}{25} = 1 + \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{27}{25} - 1 = \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{27 - 25}{25} = \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{2}{25} = \dfrac{r}{100} \\[1em] \Rightarrow r = \dfrac{200}{25} = 8$

Hence, rate of interest = 8% p.a.

At what rate per cent per annum will ₹ 3,000 amount to ₹ 3,993 in 3 years, the interest being compounded annually?

Answer

Given,

P = ₹ 3,000

A = ₹ 3,993

n = 3 years

Let rate of interest be r%.

By formula,

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

Substituting values we get :

$\Rightarrow 3993 = 3000 \times \Big(1 + \dfrac{r}{100}\Big)^3 \\[1em] \Rightarrow \dfrac{3993}{3000} = \Big(1 + \dfrac{r}{100}\Big)^3 \\[1em] \Rightarrow \dfrac{1331}{1000} = \Big(1 + \dfrac{r}{100}\Big)^3 \\[1em] \Rightarrow \Big(\dfrac{11}{10}\Big)^3 = \Big(1 + \dfrac{r}{100}\Big)^3 \\[1em] \Rightarrow \dfrac{11}{10} = 1 + \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{11}{10} - 1 = \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{11 - 10}{10} = \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{1}{10} = \dfrac{r}{100} \\[1em] \Rightarrow r = \dfrac{100}{10} = 10$

Hence, rate of interest = 10% p.a.

In what time will ₹ 5,120 amount to ₹ 7,290 at $12\dfrac{1}{2}$% per annum, compounded annually?

Answer

Given,

P = ₹ 5,120

Rate = $12\dfrac{1}{2}$% = 12.5%

A = ₹ 7,290

Let time required be n years.

$\Rightarrow A = P\Big(1 + \dfrac{r}{100}\Big)^{n } \\[1em] \Rightarrow 7290 = 5120 \times \Big(1 + \dfrac{12.5}{100}\Big)^{n} \\[1em] \Rightarrow \dfrac{7290}{5120} = \Big(\dfrac{100 + 12.5}{100}\Big)^{n} \\[1em] \Rightarrow \dfrac{729}{512} = \Big(\dfrac{112.5}{100}\Big)^{n} \\[1em] \Rightarrow \dfrac{729}{512} = \Big(\dfrac{1125}{1000}\Big)^{n} \\[1em] \Rightarrow \dfrac{729}{512} = \Big(\dfrac{9}{8}\Big)^{n} \\[1em] \Rightarrow \Big(\dfrac{9}{8}\Big)^3 = \Big(\dfrac{9}{8}\Big)^{n} \\[1em] \Rightarrow n = 3$

Hence, required time = 3 years.

A certain sum of money amounts to ₹ 7,260 in 2 years and to ₹ 7,986 in 3 years, interest being compounded annually. Find the rate per cent per annum.

Answer

Let original sum of money invested be ₹ x and rate of percent be r%.

By formula,

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

Given,

The sum of money, invested at compound interest, amounts to ₹ 7,260 in 2 years.

$\Rightarrow A = P\Big(1 + \dfrac{r}{100}\Big)^n \\[1em] \Rightarrow 7260 = x \times \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow 7260 = x\Big(1 + \dfrac{r}{100}\Big)^2 ......(1)$

The sum of money, invested at compound interest, amounts to ₹ 7,986 in 3 years.

$\Rightarrow A = P\Big(1 + \dfrac{r}{100}\Big)^n \\[1em] \Rightarrow 7986 = x \times \Big(1 + \dfrac{r}{100}\Big)^3 \\[1em] \Rightarrow 7986 = x\Big(1 + \dfrac{r}{100}\Big)^3 ......(2)$

Dividing equation (2) by (1), we get :

$\Rightarrow \dfrac{7986}{7260} = \dfrac{x\Big(1 + \dfrac{r}{100}\Big)^3}{x\Big(1 + \dfrac{r}{100}\Big)^2} \\[1em] \Rightarrow \dfrac{1331}{1210} = \Big(1 + \dfrac{r}{100}\Big) \\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big) = \dfrac{1331}{1210} \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{1331}{1210} - 1 \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{1331 - 1210}{1210} \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{121}{1210} \\[1em] \Rightarrow r = \dfrac{100 \times 121}{1210} = 10$

Hence, rate percent = 10% p.a.