Compound Interest [Applications] — Exercise 3(A)

If P = sum lent, r = rate of interest per year, n = number of years and amounts =A, then :

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

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

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

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

Answer

If P = sum lent, r = rate of interest per year, n = number of years and amounts =A, then :

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

Hence, Option 1 is the correct option.

If the letters used have usual meanings : r₁% and r₂% are rate of interests for two consecutive years then :

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

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

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

4. $P = A\Big(1 - \dfrac{r_1}{100}\Big)\Big(1 - \dfrac{r_2}{100}\Big)$

Answer

If the letters used have usual meanings : r₁% and r₂% are rate of interests for two consecutive years then :

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

Hence, Option 2 is the correct option.

Compound interest on ₹ 6000 in 2 years at 5% per annum is :

1. ₹ 615

2. ₹ 630

3. ₹ 600

4. ₹ 690

Answer

Given,

P = ₹ 6000

r = 5%

n = 2 years

By formula,

$A = P\Big(1 + \dfrac{r}{100}\Big)^n \\[1em] = 6000\Big(1 + \dfrac{5}{100}\Big)^2 \\[1em] = 6000 \times \Big(\dfrac{105}{100}\Big)^2 \\[1em] = 6000 \times \dfrac{105}{100} \times \dfrac{105}{100} \\[1em] = 3 \times 21 \times 105 \\[1em] = ₹ 6615.$

C.I. = A - P = ₹ 6615 - ₹ 6000 = ₹ 615.

Hence, Option 1 is the correct option.

A sum of money, lent out at 10% C.I. compounded yearly becomes ₹ 6050 in 2 years. The sum lent is :

1. ₹ 7260

2. ₹ 4000

3. ₹ 5000

4. ₹ 7320.50

Answer

Given,

r = 10%

n = 2 years

Let sum of money lent be ₹ x.

P = ₹ x

A = ₹ 6050

By formula,

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

Substituting values we get :

$\Rightarrow 6050 = x\Big(1 + \dfrac{10}{100}\Big)^2 \\[1em] \Rightarrow 6050 = x \times \Big(\dfrac{110}{100}\Big)^2 \\[1em] \Rightarrow 6050 = x \times \Big(\dfrac{11}{10}\Big)^2 \\[1em] \Rightarrow 6050 = \dfrac{121x}{100} \\[1em] \Rightarrow x = \dfrac{6050 \times 100}{121} \\[1em] \Rightarrow x = ₹ 5000.$

Hence, Option 3 is the correct option.

On a certain sum, the compound interest accrued in one year is ₹ 550. If the rate of interest is 10%, the sum is :

1. ₹ 5000

2. ₹ 6000

3. ₹ 4500

4. ₹ 5500

Answer

Given,

r = 10%

n = 1 year

C.I. = ₹ 550

Let sum of money lent out be ₹ x.

By formula,

C.I. = A - P

$\Rightarrow C.I. = P\Big(1 + \dfrac{r}{100}\Big)^n - P \\[1em] \Rightarrow 550 = x(1 + \dfrac{10}{100})^1 - x \\[1em] \Rightarrow 550 = x + \dfrac{10x}{100} - x \\[1em] \Rightarrow 550 = \dfrac{x}{10} \\[1em] \Rightarrow x = 5500.$

Hence, Option 4 is the correct option.

₹ 4000 amounts to ₹ 4600 in one year at compound interest compounded yearly. The rate of interest is :

1. 15%

2. 12%

3. 10%

4. 20%

Answer

Let rate of interest be r%.

Given,

P = ₹ 4000

A = ₹ 4600

n = 1 year

By formula,

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

Substituting values we get :

$\Rightarrow 4600 = 4000(1 + \dfrac{r}{100})^1 \\[1em] \Rightarrow \dfrac{4600}{4000} = 1 + \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{23}{20} = 1 + \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{23}{20} - 1 = \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{23 - 20}{20} = \dfrac{r}{100} \\[1em] \Rightarrow r = \dfrac{3}{20} \times 100 \\[1em] \Rightarrow r = 15$

Hence, Option 1 is the correct option.

₹ 4000 amounts to ₹ 5017.60 in two months at compound interest compounded per month. The rate of interest per month is :

1. 12%

2. 15%

3. 10%

4. 20%

Answer

Let rate of interest per month be r%.

Given,

P = ₹ 4000

A = ₹ 5017.60

T = 2 months

By formula,

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

Substituting values we get :

$\Rightarrow 5017.60 = 4000(1 + \dfrac{r}{100})^2 \\[1em] \Rightarrow \dfrac{5017.60}{4000} = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{501760}{400000} = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{50176}{40000} = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{224}{200}\Big)^2 = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow 1 + \dfrac{r}{100} = \dfrac{224}{200} \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{224}{200} - 1 \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{24}{200} \\[1em] \Rightarrow r =\dfrac{24}{200} \times 100 \\[1em] \Rightarrow r = 12$

Hence, Option 1 is the correct option.

Find the amount and the compound interest on ₹ 12000 in 3 years at 5%; interest being compounded annually.

Answer

Given,

P = ₹ 12000

n = 3 years

r = 5%

By formula,

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

Substituting values we get :

$\Rightarrow A = 12000\Big(1 + \dfrac{5}{100}\Big)^3 \\[1em] \Rightarrow A = 12000\Big(1 + \dfrac{1}{20}\Big)^3 \\[1em] \Rightarrow A = 12000\Big(\dfrac{21}{20}\Big)^3 \\[1em] \Rightarrow A = 12000 \times \dfrac{9261}{8000} \\[1em] \Rightarrow A = ₹13891.50$

C.I. = A - P = ₹ 13891.50 - ₹ 12000 = ₹ 1891.50

Hence, amount = ₹ 13891.50 and compound interest = ₹ 1891.50

Calculate the amount, if ₹ 15000 is lent at compound interest for 2 years and the rates for the successive years are 8% p.a. and 10% p.a. respectively.

Answer

Given,

P = ₹ 15000

r₁ = 8%

r₂ = 10%

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 :

$A = 15000 \times \Big(1 + \dfrac{8}{100}\Big) \times \Big(1 + \dfrac{10}{100}\Big) \\[1em] = 15000 \times \dfrac{108}{100} \times \dfrac{110}{100} \\[1em] = 15000 \times \dfrac{27}{25} \times \dfrac{11}{10} \\[1em] = 60 \times 27 \times 11 \\[1em] = ₹17820.$

Hence, amount = ₹ 17820.

Calculate the compound interest accrued on ₹ 6000 in 3 years, compounded yearly, if the rates for the successive years are 5%, 8% and 10% respectively.

Answer

Given,

P = ₹ 6000

r₁ = 5%

r₂ = 8%

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 :

$A = 6000 \times \Big(1 + \dfrac{5}{100}\Big) \times \Big(1 + \dfrac{8}{100}\Big) \times \Big(1 + \dfrac{10}{100}\Big) \\[1em] = 6000 \times \dfrac{105}{100} \times \dfrac{108}{100} \times \dfrac{110}{100} \\[1em] = 6000 \times \dfrac{21}{20} \times \dfrac{27}{25} \times \dfrac{11}{10} \\[1em] = \dfrac{30 \times 21 \times 27 \times 11}{25} \\[1em] = \dfrac{187110}{25} \\[1em] = ₹7484.40$

By formula,

C.I. = A - P = ₹ 7484.40 - ₹ 6000 = ₹ 1484.40

Hence, compound interest = ₹ 1484.40

What sum of money will amount to ₹ 5445 in 2 years at 10% per annum compound interest ?

Answer

Let sum of money be ₹ x.

Given,

P = ₹ x

r = 10%

n = 2 years

A = ₹ 5445

By formula,

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

Substituting values we get :

$\Rightarrow 5445 = x \times \Big(1 + \dfrac{10}{100}\Big)^2 \\[1em] \Rightarrow 5445 = x \times \Big(\dfrac{110}{100}\Big)^2 \\[1em] \Rightarrow 5445 = x \times \Big(\dfrac{11}{10}\Big)^2 \\[1em] \Rightarrow 5445 = x \times \dfrac{121}{100} \\[1em] \Rightarrow x = \dfrac{5445 \times 100}{121} \\[1em] \Rightarrow x = ₹4500.$

Hence, sum of money = ₹ 4500.

On what sum of money will the compound interest for 2 years at 5 per cent per annum amount to ₹ 768.75 ?

Answer

Let sum of money be ₹ x.

Given,

n = 2 years

r = 5%

C.I. = ₹ 768.75

A = P + I = ₹ x + ₹ 768.75

By formula,

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

Substituting values we get :

$\Rightarrow x + 768.75 = x \times \Big(1 + \dfrac{5}{100}\Big)^2 \\[1em] \Rightarrow x + 768.75 = x \times \Big(\dfrac{105}{100}\Big)^2 \\[1em] \Rightarrow x + 768.75 = x \times \Big(\dfrac{21}{20}\Big)^2 \\[1em] \Rightarrow x + 768.75 = x \times \dfrac{441}{400} \\[1em] \Rightarrow 400(x + 768.75) = 441x \\[1em] \Rightarrow 400x + 307500 = 441x \\[1em] \Rightarrow 441x - 400x = 307500 \\[1em] \Rightarrow 41x = 307500 \\[1em] \Rightarrow x = \dfrac{307500}{41} \\[1em] \Rightarrow x = ₹7500.$

Hence, sum of money = ₹ 7500.

Find the sum on which the compound interest for 3 years at 10% per annum amounts to ₹ 1655.

Answer

Let sum of money be ₹ x.

Given,

n = 3 years

r = 10%

C.I. = ₹ 1655

A = P + I = ₹ x + ₹ 1655

By formula,

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

Substituting values we get :

$\Rightarrow x + 1655 = x \times \Big(1 + \dfrac{10}{100}\Big)^3 \\[1em] \Rightarrow x + 1655 = x \times \Big(\dfrac{110}{100}\Big)^3 \\[1em] \Rightarrow x + 1655 = x \times \Big(\dfrac{11}{10}\Big)^3 \\[1em] \Rightarrow x + 1655 = x \times \dfrac{1331}{1000} \\[1em] \Rightarrow 1000(x + 1655) = 1331x \\[1em] \Rightarrow 1000x + 1655000 = 1331x \\[1em] \Rightarrow 1331x - 1000x = 1655000 \\[1em] \Rightarrow 331x = 1655000 \\[1em] \Rightarrow x = \dfrac{1655000}{331} \\[1em] \Rightarrow x = ₹5000.$

Hence, sum of money = ₹ 5000.

At what rate per cent per annum will ₹ 6000 amount to ₹ 6615 in 2 years when interest is compounded annually ?

Answer

Given,

P = ₹ 6000

A = ₹ 6615

n = 2 years

Let rate of interest be r%.

By formula,

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

Substituting values we get :

$\Rightarrow 6615 = 6000 \times \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{6615}{6000} = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{441}{400} = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{21}{20}\Big)^2 = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{21}{20} = 1 + \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{21}{20} - 1 = \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{21 - 20}{20} = \dfrac{r}{100} \\[1em] \Rightarrow r = \dfrac{100}{20} = 5$

Hence, rate of interest = 5%.

What principal will amount to ₹ 9856 in two years, if the rates of interest for successive years are 10% and 12% respectively?

Answer

Given,

A = ₹ 9856

r₁ = 10%

r₂ = 12%

n = 2 years

Let principal amount be ₹ x.

By formula,

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

Substituting values we get :

$\Rightarrow 9856 = x\Big(1 + \dfrac{10}{100}\Big)\Big(1 + \dfrac{12}{100}\Big) \\[1em] \Rightarrow 9856 = x \times \dfrac{110}{100} \times \dfrac{112}{100} \\[1em] \Rightarrow 9856 = x \times \dfrac{11}{10} \times \dfrac{28}{25} \\[1em] \Rightarrow x = \dfrac{9856 \times 10 \times 25}{11 \times 28} \\[1em] \Rightarrow x = \dfrac{2464000}{308} = ₹8000.$

Hence, principal amount = ₹ 8000.

On a certain sum, the compound interest in 2 years amounts to ₹ 4240. If the rates of interest for successive years are 10% and 15% respectively, find the sum.

Answer

Given,

C.I. = ₹ 4240

r₁ = 10%

r₂ = 15%

n = 2 years

Let principal amount be ₹ x.

A = P + C.I. = ₹ x + ₹ 4240

By formula,

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

Substituting values we get :

$\Rightarrow x + 4240 = x \times \Big(1 + \dfrac{10}{100}\Big)(1 + \dfrac{15}{100}) \\[1em] \Rightarrow x + 4240 = x \times \dfrac{110}{100} \times \dfrac{115}{100} \\[1em] \Rightarrow x + 4240 = x \times \dfrac{11}{10} \times \dfrac{23}{20} \\[1em] \Rightarrow x + 4240 = \dfrac{253x}{200} \\[1em] \Rightarrow 200(x + 4240) = 253x \\[1em] \Rightarrow 200x + 848000 = 253x \\[1em] \Rightarrow 253x - 200x = 848000 \\[1em] \Rightarrow 53x = 848000 \\[1em] \Rightarrow x = \dfrac{848000}{53} = ₹16000.$

Hence, principal amount = ₹ 16000.

At what rate per cent compound interest, does a sum of money become 1.44 times of itself in 2 years ?

Answer

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

A = ₹ 1.44x

n = 2 years

By formula,

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

Substituting values we get :

$\Rightarrow 1.44x = x \times \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{1.44x}{x} = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow 1.44 = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow (1.2)^2 = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow 1.2 = 1 + \dfrac{r}{100} \\[1em] \Rightarrow 1.2 - 1 = \dfrac{r}{100} \\[1em] \Rightarrow 0.2 = \dfrac{r}{100} \\[1em] \Rightarrow r = 0.2 \times 100 = 20$

Hence, rate of interest = 20%.

At what rate per cent will a sum of ₹ 4000 yield ₹ 1324 as compound interest in 3 years ?

Answer

Let rate of interest be r%.

Given,

P = ₹ 4000

C.I. = ₹ 1324

A = P + C.I. = ₹ 4000 + ₹ 1324 = ₹ 5324

n = 3 years

By formula,

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

Substituting values we get :

$\Rightarrow 5324 = 4000 \times \Big(1 + \dfrac{r}{100}\Big)^3 \\[1em] \Rightarrow \dfrac{5324}{4000} = \Big(1 + \dfrac{r}{100}\Big)^3 \\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big)^3 = \dfrac{1331}{1000} \\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big)^3 = \Big(\dfrac{11}{10}\Big)^3 \\[1em] \Rightarrow 1 + \dfrac{r}{100} = \dfrac{11}{10} \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{11}{10} - 1 \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{11 - 10}{10} \\[1em] \Rightarrow r = \dfrac{1}{10} \times 100 \\[1em] \Rightarrow r = 10$

Hence, rate of interest = 10%.

A person invests ₹ 5000 for three years at a certain rate of interest compounded annually. At the end of two years this sum amounts to ₹ 6272. Calculate :

(i) the rate of interest per annum.

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

Answer

(i) For 2 years :

A = ₹ 6272

P = ₹ 5000

Let rate of interest be r%.

By formula,

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

Substituting values we get :

$\Rightarrow 6272 = 5000 \times \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{6272}{5000} = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{784}{625} = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{28}{25}\Big)^2 = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{28}{25} = 1 + \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{28}{25} - 1 \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{28 - 25}{25} \\[1em] \Rightarrow r = \dfrac{3}{25} \times 100 \\[1em] \Rightarrow r = 12$

Hence, rate of interest per annum = 12%.

(ii) By formula,

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

Substituting values we get :

$\Rightarrow A = 5000 \times \Big(1 + \dfrac{12}{100}\Big)^3 \\[1em] = 5000 \times \Big(\dfrac{112}{100}\Big)^3 \\[1em] = 5000 \times \Big(\dfrac{28}{25}\Big)^3 \\[1em] = 5000 \times \dfrac{21952}{15625} \\[1em] = \dfrac{109760000}{15625} \\[1em] = ₹7024.64$

Hence, amount in 3 years = ₹ 7024.64

In how many years will ₹ 7000 amount to ₹ 9317 at 10 per cent per annum compound interest ?

Answer

Given,

P = ₹ 7000

A = ₹ 9317

r = 10%

Let in n years ₹ 7000 amount to ₹ 9317.

By formula,

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

Substituting values we get :

$\Rightarrow 9317 = 7000 \times \Big(1 + \dfrac{10}{100}\Big)^n \\[1em] \Rightarrow \dfrac{9317}{7000} = \Big(\dfrac{110}{100}\Big)^n \\[1em] \Rightarrow \dfrac{1331}{1000} = \Big(\dfrac{11}{10}\Big)^n \\[1em] \Rightarrow \Big(\dfrac{11}{10}\Big)^3 = \Big(\dfrac{11}{10}\Big)^n \\[1em] \Rightarrow n = 3.$

Hence, in 3 years ₹ 7000 amounts to ₹ 9317 at 10 percent compound interest.

Find the time, in years, in which ₹ 4000 will produce ₹ 630.50 as compound interest at 5 percent p.a. interest being compounded annually.

Answer

Given,

P = ₹ 4000

C.I. = ₹ 630.50

r = 5%

A = P + C.I. = ₹ 4000 + ₹ 630.50 = ₹ 4630.50

Let time taken be n years.

By formula,

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

Substituting values we get :

$\Rightarrow 4630.50 = 4000 \times \Big(1 + \dfrac{5}{100}\Big)^n \\[1em] \Rightarrow \dfrac{4630.50}{4000} = \Big(\dfrac{105}{100}\Big)^n \\[1em] \Rightarrow \dfrac{9.261}{8} = \Big(\dfrac{2.1}{2}\Big)^n \\[1em] \Rightarrow \Big(\dfrac{2.1}{2}\Big)^3 = \Big(\dfrac{2.1}{2}\Big)^n \\[1em] \Rightarrow n = 3.$

Hence, time taken = 3 years.

Divide ₹ 28730 between A and B so that when their shares are lent out at 10 per cent compound interest compounded per year, the amount that A receives in 3 years is same as what B receives in 5 years.

Answer

Let share of A = ₹ x and share of B = ₹ 28730 - ₹ x.

For A :

P = ₹ x

r = 10%

n = 3 years

By formula,

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

Substituting values we get :

$\Rightarrow \text{Amount } = x\Big(1 + \dfrac{10}{100}\Big)^3 \\[1em] = x \times \Big(\dfrac{110}{100}\Big)^3\\[1em] = x \times \Big(\dfrac{11}{10}\Big)^3$

For B :

P = ₹ (28730 - x)

r = 10%

n = 5 years

By formula,

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

Substituting values we get :

$\Rightarrow \text{Amount } = (28730 - x) \times \Big(1 + \dfrac{10}{100}\Big)^5 \\[1em] = (28730 - x) \times \Big(\dfrac{110}{100}\Big)^5 \\[1em] = (28730 - x) \times \Big(\dfrac{11}{10}\Big)^5$

Since, amount received by A and B are equal.

$\therefore x \times \Big(\dfrac{11}{10}\Big)^3 = (28730 - x) \times \Big(\dfrac{11}{10}\Big)^5 \\[1em] \Rightarrow x = (28730 - x) \times \Big(\dfrac{11}{10}\Big)^2 \\[1em] \Rightarrow x = (28730 - x) \times \dfrac{121}{100} \\[1em] \Rightarrow 100x = 121(28730 - x) \\[1em] \Rightarrow 100x = 3476330 - 121x \\[1em] \Rightarrow 100x + 121x = 3476330 \\[1em] \Rightarrow 221x = 3476330 \\[1em] \Rightarrow x = \dfrac{3476330}{221} \\[1em] \Rightarrow x = 15730.$

₹ (28730 - x) = ₹ (28730 - 15730) = ₹ 13000.

Hence, share of A and B are ₹ 15730 and ₹ 13000 respectively.

A sum of ₹ 44200 is divided between John and Smith, 12 years and 14 years old respectively, in such a way that if their portions be invested at 10 percent per annum compound interest, they will receive equal amounts on reaching 16 years of age.

(i) What is the share of each out of ₹ 44200?

(ii) What will each receive, when 16 years old ?

Answer

(i) Let share of John = ₹ x and share of Smith = ₹ 44200 - ₹ x.

For John :

P = ₹ x

r = 10%

n = 4 years (John has 4 years to reach 16 years of age)

By formula,

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

Substituting values we get :

$\Rightarrow \text{Amount } = x\Big(1 + \dfrac{10}{100}\Big)^4 \\[1em] = x \times \Big(\dfrac{110}{100}\Big)^4\\[1em] = x \times \Big(\dfrac{11}{10}\Big)^4 ........(1)$

For Smith :

P = ₹ (44200 - x)

r = 10%

n = 2 years (Smith has 2 years to reach 16 years of age)

By formula,

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

Substituting values we get :

$\Rightarrow \text{Amount } = (44200 - x) \times \Big(1 + \dfrac{10}{100}\Big)^2 \\[1em] = (44200 - x) \times \Big(\dfrac{110}{100}\Big)^2 \\[1em] = (44200 - x) \times \Big(\dfrac{11}{10}\Big)^2 ...........(2)$

Since, amount received by A and B are equal.

$\therefore x \times \Big(\dfrac{11}{10}\Big)^4 = (44200 - x) \times \Big(\dfrac{11}{10}\Big)^2 \\[1em] \Rightarrow x \times \Big(\dfrac{11}{10}\Big)^2 = (44200 - x) \\[1em] \Rightarrow \dfrac{121}{100}x = (44200 - x) \\[1em] \Rightarrow 121x = 100(44200 - x) \\[1em] \Rightarrow 121x = 4420000 - 100x \\[1em] \Rightarrow 121x + 100x = 4420000 \\[1em] \Rightarrow 221x = 4420000 \\[1em] \Rightarrow x = \dfrac{4420000}{221} \\[1em] \Rightarrow x = ₹ 20000.$

₹ (44200 - x) = ₹ (44200 - 20000) = ₹ 24200.

Hence, share of John and Smith are ₹ 20,000 and ₹ 24,200 respectively.

(ii) Substituting value of x in equation (1), we get :

$\Rightarrow 20000 \times \Big(\dfrac{11}{10}\Big)^4 \\[1em] \Rightarrow 20000 \times \dfrac{14641}{10000} \\[1em] \Rightarrow ₹ 29282.$

Since, both receive equal amount.

Hence, each receive ₹ 29282.

The simple interest on a certain sum of money at 10% per annum is ₹ 6000 in 2 years. Find :

(i) the sum

(ii) the amount due at the end of 3 years and at the same rate of interest compounded annually.

(iii) the compound interest earned in 3 years.

Answer

(i) Let the sum be ₹ x.

Given, S.I. = ₹ 6000 in 2 years at 10% rate of interest.

By formula,

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

Substituting values we get :

$\Rightarrow 6000 = \dfrac{x \times 10 \times 2}{100} \\[1em] \Rightarrow x = \dfrac{6000 \times 100}{10 \times 2} \\[1em] \Rightarrow x = 30000.$

Hence, sum = ₹ 30000.

(ii) By formula,

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

Substituting values we get :

$A = 30000 \times \Big(1 + \dfrac{10}{100}\Big)^3 \\[1em] = 30000 \times \Big(\dfrac{110}{100}\Big)^3 \\[1em] = 30000 \times \Big(\dfrac{11}{10}\Big)^3 \\[1em] = 30000 \times \dfrac{1331}{1000} \\[1em] = 39930.$

Hence, amount due at the end of 3 years = ₹ 39930.

(iii) By formula,

C.I. = A - P = ₹ 39930 - ₹ 30000 = ₹ 9930.

Hence, compound interest earned in 3 years = ₹ 9930.

Find the difference between compound interest and simple interest on ₹ 8000 in 2 years and at 5% per annum.

Answer

For S.I. :

Principal = ₹ 8000

Rate = 5%

Time = 2 years

S.I. = $\dfrac{8000 \times 5 \times 2}{100}$ = ₹ 800.

For C.I. :

C.I. = A - P

$= P\Big(1 + \dfrac{r}{100}\Big)^n - P \\[1em] = 8000 \times \Big(1 + \dfrac{5}{100}\Big)^2 - 8000 \\[1em] = 8000 \times \Big(\dfrac{105}{100}\Big)^2 - 8000 \\[1em] = 8000 \times \Big(\dfrac{21}{20}\Big)^2 - 8000 \\[1em] = 8000 \times \dfrac{441}{400} - 8000 \\[1em] = 8820 - 8000 \\[1em] = ₹ 820.$

Difference between C.I. and S.I. = ₹ 820 - ₹ 800 = ₹ 20.

Hence, difference between C.I. and S.I. = ₹ 20.