Chapter 3: Compound Interest (Stage 2) [Applications]

Exercise 3(A)

Question 1(a)

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

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

Question 1(b)

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

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

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.

Question 1(c)

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

₹ 615
₹ 630
₹ 600
₹ 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.

Question 1(d)

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

₹ 7260
₹ 4000
₹ 5000
₹ 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.

Question 1(e)

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

₹ 5000
₹ 6000
₹ 4500
₹ 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.

Question 1(f)

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

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.

Question 1(g)

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

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.

Question 2

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

Question 3

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.

Question 4

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

Question 5

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.

Question 6

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.

Question 7

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.

Question 8

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%.

Question 9

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.

Question 10

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.

Question 11

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%.

Question 12

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%.

Question 13

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

Question 14

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.

Question 15

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.

Question 16

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.

Question 17

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.

Question 18

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.

Question 19

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.

Exercise 3(B)

Question 1(a)

On ₹ 6000, the difference between C.I. and S.I. in 2 years and at 10% compound interest, compounded per year, is :

₹ 120
₹ 600
₹ 60
₹ 180

Answer

For S.I. :

Principal = ₹ 6000

Rate = 10%

Time = 2 years

S.I. = $\dfrac{6000 \times 10 \times 2}{100}$ = ₹ 1200.

For C.I. :

C.I. = A - P

$= P\Big(1 + \dfrac{r}{100}\Big)^n - P \\[1em] = 6000 \times \Big(1 + \dfrac{10}{100}\Big)^2 - 6000 \\[1em] = 6000 \times \Big(\dfrac{110}{100}\Big)^2 - 6000 \\[1em] = 6000 \times \Big(\dfrac{11}{10}\Big)^2 - 6000 \\[1em] = 6000 \times \dfrac{121}{100} - 6000 \\[1em] = 7260 - 6000 \\[1em] = ₹ 1260.$

Difference between C.I. and S.I. = ₹ 1260 - ₹ 1200 = ₹ 60.

Hence, Option 3 is the correct option.

Question 1(b)

₹ 10000 amounts to ₹ 12500 in one year. The rate of interest per year is :

15%
12.5%
20%
25%

Answer

Let rate of interest be r%.

Given,

P = ₹ 10000

A = ₹ 12500

n = 1 year

By formula,

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

Substituting values we get :

$\Rightarrow 12500 = 10000 \times \Big(1 + \dfrac{r}{100}\Big)^1 \\[1em] \Rightarrow \dfrac{12500}{10000} = \Big(1 + \dfrac{r}{100}\Big) \\[1em] \Rightarrow \dfrac{5}{4} = 1 + \dfrac{r}{100}\\[1em] \Rightarrow \dfrac{5}{4} - 1 = \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{5 - 4}{4} = \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{1}{4} = \dfrac{r}{100} \\[1em] \Rightarrow r = \dfrac{100}{4} = 25%.$

Hence, Option 4 is the correct option.

Question 1(c)

The C.I. on ₹ 16000 in 2 years at the rate of 20% per annum is :

₹ 19360
₹ 7040
₹ 23040
₹ 22400

Answer

Given,

P = ₹ 16000

n = 2 years

r = 20%

By formula,

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

C.I. = A - P

$C.I. = P\Big(1 + \dfrac{r}{100}\Big)^n - P \\[1em] = 16000 \times \Big(1 + \dfrac{20}{100}\Big)^2 - 16000 \\[1em] = 16000 \times \Big(\dfrac{120}{100}\Big)^2 - 16000 \\[1em] = 16000 \times \Big(\dfrac{6}{5}\Big)^2 - 16000 \\[1em] = 16000 \times \dfrac{36}{25} - 16000 \\[1em] = 23040 - 16000 \\[1em] = ₹ 7040.$

Hence, Option 2 is the correct option.

Question 1(d)

Simple interest, at the same rate for the same period as given above in part (c) is :

₹ 7040
₹ 6400
₹ 3200
₹ 1280

Answer

By formula,

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

Substituting values we get :

$S.I. = \dfrac{16000 \times 20 \times 2}{100} \\[1em] = ₹ 6400.$

Hence, Option 2 is the correct option.

Question 1(e)

The difference between C.I. and S.I. in 2 years as given above for parts (c) and (d) is :

₹ 640
₹ 3840
₹ 1920
₹ 1280

Answer

From part (c), we get :

C.I. = ₹ 7040

From part (d), we get :

S.I. = ₹ 6400

C.I. - S.I. = ₹ 7040 - ₹ 6400 = ₹ 640.

Hence, Option 1 is the correct option.

Question 2

The difference between simple interest and compound interest on a certain sum is ₹ 54.40 for 2 years at 8 percent per annum. Find the sum.

Answer

Given,

n = 2 years

r = 8%

Let sum of money be ₹ P.

C.I. = A - P

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

By formula,

$S.I. = \dfrac{P \times R \times T}{100} \\[1em] = \dfrac{P \times 8 \times 2}{100} \\[1em] = \dfrac{4P}{25}.$

Given,

Difference between S.I. and C.I. = ₹ 54.40

$\Rightarrow \dfrac{104P}{625} - \dfrac{4P}{25} = 54.40 \\[1em] \Rightarrow \dfrac{104P - 100P}{625} = 54.40 \\[1em] \Rightarrow \dfrac{4P}{625} = 54.40 \\[1em] \Rightarrow P = \dfrac{54.40 \times 625}{4} \\[1em] \Rightarrow P = ₹ 8500.$

Hence, sum = ₹ 8500.

Question 3

Pramod and Anand each lent the same sum of money for 2 years at 5% at simple interest and compound interest respectively. Anand received ₹ 15 more than Pramod. Find the amount of money lent by each and the interest received.

Answer

Let sum of money be ₹ x.

For Pramod :

P = ₹ x

Time = 2 years

Rate = 5%

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

Substituting values we get :

$S.I. = \dfrac{x \times 5 \times 2}{100} \\[1em] = \dfrac{x}{10}.$

For Anand :

P = ₹ x

Time (n) = 2 years

Rate (r) = 5%

C.I. = A - P

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

Given,

Anand received ₹ 15 more than Pramod.

∴ C.I. - S.I. = ₹ 15

$\Rightarrow \dfrac{41x}{400} - \dfrac{x}{10} = 15 \\[1em] \Rightarrow \dfrac{41x - 40x}{400} = 15 \\[1em] \Rightarrow \dfrac{x}{400} = 15 \\[1em] \Rightarrow x = 400 \times 15 = ₹ 6000. \\[1em] S.I. = \dfrac{x}{10} = \dfrac{6000}{10} = ₹ 600. \\[1em] C.I. = \dfrac{41x}{400} = \dfrac{41 \times 6000}{400} = ₹ 615.$

Hence, sum lent by each = ₹ 6000 and interest received by Pramod = ₹ 600 and Anand = ₹ 615.

Question 4

Simple interest on a sum of money for 2 years at 4% is ₹ 450. Find the compound interest on the same sum and at the same rate for 2 years.

Answer

Let the sum be ₹ x.

Given,

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

By formula,

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

Substituting values we get :

$\Rightarrow 450 = \dfrac{x \times 4 \times 2}{100} \\[1em] \Rightarrow x = \dfrac{450 \times 100}{8} \\[1em] \Rightarrow x = 225 \times 25 \\[1em] \Rightarrow x = ₹5625.$

By formula,

C.I. = A - P

$= P\Big(1 + \dfrac{r}{100}\Big)^n - P \\[1em] = 5625 \times \Big(1 + \dfrac{4}{100}\Big)^2 - 5625 \\[1em] = 5625 \times \Big(\dfrac{104}{100}\Big)^2 - 5625 \\[1em] = 5625 \times \Big(\dfrac{26}{25}\Big)^2 - 5625 \\[1em] = 5625 \times \dfrac{676}{625} - 5625 \\[1em] = 6084 - 5625 \\[1em] = ₹ 459.$

Hence, the compound interest = ₹ 459.

Question 5

Compound interest on a certain sum of money at 5% per annum for two years is ₹ 246. Calculate simple interest on the same sum for 3 years at 6% per annum.

Answer

Let sum of money be ₹ x.

By formula,

$\Rightarrow C.I. = A - P \\[1em] \Rightarrow 246 = P\Big(1 + \dfrac{r}{100}\Big)^n - P \\[1em] \Rightarrow 246 = x \times \Big(1 + \dfrac{5}{100}\Big)^2 - x \\[1em] \Rightarrow 246 = x \times \Big(\dfrac{105}{100}\Big)^2 - x \\[1em] \Rightarrow 246 = x \times \Big(\dfrac{21}{20}\Big)^2 - x \\[1em] \Rightarrow 246 = x \times \dfrac{441}{400} - x \\[1em] \Rightarrow 246 = \dfrac{441x}{400} - x \\[1em] \Rightarrow 246 = \dfrac{441x - 400x}{400} \\[1em] \Rightarrow 246 = \dfrac{41x}{400} \\[1em] \Rightarrow x = \dfrac{246 \times 400}{41} \\[1em] \Rightarrow x = ₹ 2400.$

For S.I. :

P = x = ₹ 2400

R = 6%

T = 3 years

By formula,

$\Rightarrow S.I. = \dfrac{P \times R \times T}{100} \\[1em] \Rightarrow S.I. = \dfrac{x \times 6 \times 3}{100} \\[1em] \Rightarrow S.I. = \dfrac{2400 \times 6 \times 3}{100} \\[1em] \Rightarrow S.I. = ₹ 432.$

Hence, simple interest = ₹ 432.

Question 6

A sum of money, invested at compounded interest, amounts to ₹ 19360 in 2 years and to ₹ 23425.60 in 4 years. Find the rate percent and the original sum of money.

Answer

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

By formula,

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

Given,

Sum amounts to ₹ 19360 in two years.

$\therefore P\Big(1 + \dfrac{r}{100}\Big)^2 = 19360$ ......(1)

Given,

Sum amounts to ₹ 23425.60 in four years.

$\therefore P\Big(1 + \dfrac{r}{100}\Big)^4 = 23425.60$ ......(2)

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

$\Rightarrow \dfrac{P\Big(1 + \dfrac{r}{100}\Big)^4}{P\Big(1 + \dfrac{r}{100}\Big)^2} = \dfrac{23425.60}{19360} \\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big)^2 = \dfrac{146.41}{121} \\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big)^2 = \Big(\dfrac{12.1}{11}\Big)^2 \\[1em] \Rightarrow 1 + \dfrac{r}{100} = \dfrac{12.1}{11} \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{12.1}{11} - 1 \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{12.1 - 11}{11} \\[1em] \Rightarrow r = \dfrac{1.1}{11} \times 100 \\[1em] \Rightarrow r = 10%.$

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

$\Rightarrow P\Big(1 + \dfrac{r}{100}\Big)^2 = 19360 \\[1em] \Rightarrow P\Big(1 + \dfrac{10}{100}\Big)^2 = 19360 \\[1em] \Rightarrow P\Big(\dfrac{110}{100}\Big)^2 = 19360 \\[1em] \Rightarrow P \times \Big(\dfrac{11}{10}\Big)^2 = 19360 \\[1em] \Rightarrow P \times \dfrac{121}{100} = 19360 \\[1em] \Rightarrow P = \dfrac{19360 \times 100}{121} \\[1em] \Rightarrow P = 160 \times 100 \\[1em] \Rightarrow P = ₹ 16000.$

Hence, sum of money = ₹ 16000 and rate of interest = 10%.

Question 7

A sum of money let out at C.I. at a certain rate per annum becomes three times of itself in 8 years. Find in how many years will the money become twenty-seven times of itself at the same rate of interest p.a.

Answer

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

By formula,

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

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 27 times.

$\Rightarrow P\Big(1 + \dfrac{r}{100}\Big)^n = 27P \\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big)^n = \dfrac{27P}{P} \\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big)^n = 27 \\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big)^n = 3^3 \\[1em]$

From equation (1)

$\Rightarrow \Big(1 + \dfrac{r}{100}\Big)^n = \Big[\Big(1 + \dfrac{r}{100}\Big)^8\Big]^3 \\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big)^n =\Big(1 + \dfrac{r}{100}\Big)^{24} \\[1em] \Rightarrow n = 24 \text{ years}.$

Hence, in 24 years money will becomes 27 times of itself.

Question 8

On what sum of money will compound interest (payable annually) for 2 years be the same as simple interest on ₹ 9430 for 10 years, both at the rate of 5 percent per annum ?

Answer

For S.I. :

P = ₹ 9430

T = 10 years

R = 5%

$S.I. = \dfrac{P \times R \times T}{100} \\[1em] = \dfrac{9430 \times 5 \times 10}{100} \\[1em] = ₹ 4715.$

Since, C.I. = S.I. = ₹ 4715

Let sum on which C.I. = ₹ 4715 for 2 years at 5% be ₹x.

$\Rightarrow C.I. = A - P \\[1em] \Rightarrow 4715 = P\Big(1 + \dfrac{r}{100}\Big)^n - P \\[1em] \Rightarrow 4715 = x \times \Big(1 + \dfrac{5}{100}\Big)^2 - x \\[1em] \Rightarrow 4715 = x \times \Big(\dfrac{105}{100}\Big)^2 - x \\[1em] \Rightarrow 4715 = x \times \Big(\dfrac{21}{20}\Big)^2 - x \\[1em] \Rightarrow 4715 = \dfrac{441x}{400} - x \\[1em] \Rightarrow 4715 = \dfrac{441x - 400x}{400} \\[1em] \Rightarrow 4715 = \dfrac{41x}{400} \\[1em] \Rightarrow x = \dfrac{4715 \times 400}{41} \\[1em] \Rightarrow x = 115 \times 400 \\[1em] \Rightarrow x = ₹ 46000.$

Hence, sum of money = ₹ 46000.

Question 9

Simple interest on a certain sum of money for 4 years at 4% per annum exceeds the compound interest on the same sum for 3 years at 5 percent per annum by ₹ 228. Find the sum.

Answer

Let sum of money be ₹ x.

For S.I. :

P = ₹ x

Time (T) = 4 years

Rate of interest (R) = 4%

By formula,

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

Substituting values we get :

$\Rightarrow S.I. = \dfrac{x \times 4 \times 4}{100} \\[1em] = \dfrac{4x}{25}.$

For C.I. :

P = ₹ x

Rate of interest (r) = 5%

Time (n) = 3 years

By formula,

$\Rightarrow C.I. = A - P \\[1em] \Rightarrow C.I. = P\Big(1 + \dfrac{r}{100}\Big)^n - P \\[1em] = x \times \Big(1 + \dfrac{5}{100}\Big)^3 - x \\[1em] = x \times \Big(\dfrac{105}{100}\Big)^3 - x \\[1em] = x \times \Big(\dfrac{21}{20}\Big)^3 - x \\[1em] = \dfrac{9261x}{8000} - x \\[1em] = \dfrac{9261x - 8000x}{8000} \\[1em] = \dfrac{1261x}{8000}.$

Given,

Simple interest on ₹ x for 4 years at 4% per annum exceeds the compound interest on the same sum for 3 years at 5 percent per annum by ₹ 228.

$\therefore \dfrac{4x}{25} - \dfrac{1261x}{8000} = 228 \\[1em] \Rightarrow \dfrac{1280x - 1261x}{8000} = 228 \\[1em] \Rightarrow \dfrac{19x}{8000} = 228 \\[1em] \Rightarrow x = \dfrac{228 \times 8000}{19} \\[1em] \Rightarrow x = ₹ 96000.$

Hence, sum of money = ₹ 96000.

Question 10

A certain sum of money amounts to ₹ 23400 in 3 years at 10% per annum simple interest. Find the amount of the same sum in 2 years at 10% p.a. compound interest.

Answer

Let sum of money be ₹ x.

Given,

It amounts to ₹ 23400 in 3 years at 10% per annum simple interest.

By formula,

$\Rightarrow A = P + S.I. \\[1em] \Rightarrow 23400 = x + \dfrac{x \times 10 \times 3}{100} \\[1em] \Rightarrow 23400 = x + \dfrac{3x}{10} \\[1em] \Rightarrow \dfrac{13x}{10} = 23400 \\[1em] \Rightarrow x = \dfrac{23400 \times 10}{13} \\[1em] \Rightarrow x = ₹ 18000.$

In case of compound interest :

By formula,

$A = P\Big(1 + \dfrac{r}{100}\Big)^n \\[1em] = 18000 \times \Big(1 + \dfrac{10}{100}\Big)^2 \\[1em] = 18000 \times \Big(\dfrac{110}{100}\Big)^2 \\[1em] = 18000 \times \Big(\dfrac{11}{10}\Big)^2 \\[1em] = 18000 \times \Big(\dfrac{121}{100}\Big) \\[1em] = ₹ 21780.$

Hence, amount on same sum in 2 years at 10% p.a. compound interest = ₹ 21780.

Question 11

Mohit borrowed a certain sum at 5% per annum compound interest and cleared this loan by paying ₹ 12600 at the end of the first year and ₹ 17640 at the end of the second year. Find the sum borrowed.

Answer

Let Mohit borrowed ₹ x and ₹ y.

By formula,

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

For ₹ 12600,

P = ₹ x

A = ₹ 12600

r = 5%

n = 1 year

Substituting values in formula we get :

$\Rightarrow 12600 = x \times \Big(1 + \dfrac{5}{100}\Big)^1 \\[1em] \Rightarrow 12600 = x \times \dfrac{105}{100} \\[1em] \Rightarrow x = \dfrac{12600 \times 100}{105} \\[1em] \Rightarrow x = ₹ 12000.$

For ₹ 17640,

P = ₹ y

A = ₹ 17640

r = 5%

n = 2 year

Substituting values in formula we get :

$\Rightarrow 17640 = y \times \Big(1 + \dfrac{5}{100}\Big)^2 \\[1em] \Rightarrow 17640 = y \times \Big(\dfrac{105}{100}\Big)^2 \\[1em] \Rightarrow 17640 = y \times \Big(\dfrac{21}{20}\Big)^2 \\[1em] \Rightarrow y = \dfrac{17640 \times 20^2}{21^2} \\[1em] \Rightarrow y = ₹ 16000.$

Total money borrowed = ₹ (x + y) = ₹ (12000 + 16000) = ₹ 28000.

Hence, sum borrowed = ₹ 28000.

Exercise 3(C)

Question 1(a-i)

If the letters have usual meanings, the formula for finding compound interest, compounded yearly for the given time is :

$1\dfrac{1}{2}$ years :

$P\Big[\Big(1 + \dfrac{r}{100}\Big)^{1\dfrac{1}{2}}- 1\Big]$
$P\Big(1 + \dfrac{r}{100}\Big)\Big(1 + \dfrac{r}{200}\Big)$ - P
$P\Big(1 + \dfrac{r}{200}\Big)^3$
$P\Big(1 + \dfrac{r}{100}\Big)^3\Big(1 + \dfrac{r}{200}\Big)- P$

Answer

Let sum = ₹ P.

Amount for first year :

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

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

Sum = ₹ $P\Big(1 + \dfrac{r}{100}\Big)$

A = $P\Big(1 + \dfrac{r}{100}\Big)\Big(1 + \dfrac{r}{2 \times 100}\Big)^{\dfrac{1}{2} \times 2} = P\Big(1 + \dfrac{r}{100}\Big)\Big(1 + \dfrac{r}{200}\Big)$

By formula,

C.I. = A - P = $P\Big(1 + \dfrac{r}{100}\Big)\Big(1 + \dfrac{r}{200}\Big) - P$

Hence, Option 2 is the correct option.

Question 1(a-ii)

If the letters have usual meanings, the formula for finding compound interest, compounded yearly for the given time is :

1 year :

$P\Big(1 + \dfrac{r}{100}\Big)^1 - P$
$P\Big(1 + \dfrac{r}{200}\Big)$ - P
$P\Big(1 + \dfrac{1}{200}\Big)^2$ - P
$P\Big(1 - \dfrac{r}{200}\Big)^2$ - P

Answer

By formula,

C.I. = A - P

$C.I. = P\Big(1 + \dfrac{r}{100}\Big)^1 - P$

Hence, Option 1 is the correct option.

Question 1(a-iii)

If the letters have usual meanings, the formula for finding compound interest, compounded yearly for the given time is :

$2\dfrac{1}{2}$ years :

$P\Big(1 + \dfrac{r}{100}\Big)^{\dfrac{5}{2}} - P$
$P\Big(1 + \dfrac{r}{100}\Big)^{\dfrac{5}{2}}$
$P\Big(1 + \dfrac{r}{100}\Big)^2\Big(1 + \dfrac{r}{200}\Big)$
$P\Big(1 + \dfrac{r}{100}\Big)^2\Big(1 + \dfrac{r}{200}\Big)$ - P

Answer

Amount for first two years :

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

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

Sum = ₹ $P\Big(1 + \dfrac{r}{100}\Big)^2$

A = $P\Big(1 + \dfrac{r}{100}\Big)^2\Big(1 + \dfrac{r}{2 \times 100}\Big)^{\dfrac{1}{2} \times 2} = P\Big(1 + \dfrac{r}{100}\Big)^2\Big(1 + \dfrac{r}{200}\Big)$

By formula,

C.I. = A - P = $P\Big(1 + \dfrac{r}{100}\Big)^2\Big(1 + \dfrac{r}{200}\Big) - P$

Hence, Option 4 is the correct option.

Question 1(b-i)

When interest is compounded half-yearly then the formula for C.I. for the given time is :

1 year :

$P\Big(1 + \dfrac{r}{200}\Big)^2 - P$
$P\Big(1 + \dfrac{r}{100}\Big)^2 - P$
$P\Big(1 + \dfrac{r}{200}\Big)^2$
$P\Big(1 + \dfrac{r}{100}\Big)^2$

Answer

When interest is compounded half-yearly.

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

C.I. = A - P

For 1 year :

$C.I. = P\Big(1 + \dfrac{r}{2 \times 100}\Big)^{1 \times 2} - P \\[1em] = P\Big(1 + \dfrac{r}{200}\Big)^2 - P$

Hence, Option 1 is the correct option.

Question 1(b-ii)

When interest is compounded half-yearly then the formula for C.I. for the given time is :

$1\dfrac{1}{2}$ years

$P\Big(1 + \dfrac{r}{200}\Big)^{\dfrac{3}{2}} - P$
$P\Big(1 + \dfrac{r}{200}\Big)^3 - P$
$P\Big(1 + \dfrac{r}{100}\Big)^3 - P$
$P\Big(1 + \dfrac{r}{100}\Big)^{\dfrac{3}{2}} - P$

Answer

When interest is compounded half-yearly.

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

C.I. = A - P

$C.I. = P\Big(1 + \dfrac{r}{2 \times 100}\Big)^{\dfrac{3}{2} \times 2} - P \\[1em] = P\Big(1 + \dfrac{r}{200}\Big)^3 - P$

Hence, Option 2 is the correct option.

Question 1(b-iii)

When interest is compounded half-yearly then the formula for C.I. for the given time is :

2 years :

$P\Big(1 + \dfrac{r}{100}\Big)^2 - P$
$P\Big(1 + \dfrac{r}{200}\Big)^2 - P$
$P\Big(1 + \dfrac{r}{200}\Big)^4 - P$
$P\Big(1 + \dfrac{r}{100}\Big)^4 - P$

Answer

When interest is compounded half-yearly.

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

C.I. = A - P

$C.I. = P\Big(1 + \dfrac{r}{2 \times 100}\Big)^{2 \times 2} - P \\[1em] = P\Big(1 + \dfrac{r}{200}\Big)^4 - P$

Hence, Option 3 is the correct option.

Question 1(c)

If ₹ 6000 earns C.I. = ₹ 1200 in 6 months; then the rate of interest per year is :

Answer

Let rate of interest be r%.

Given,

P = ₹ 6000

C.I. = ₹ 1200

By formula,

$\Rightarrow C.I. = P\Big(1 + \dfrac{r}{2 \times 100}\Big)^{n \times 2} - P \\[1em] \Rightarrow 1200 = 6000\Big(1 + \dfrac{r}{200}\Big)^{\dfrac{1}{2} \times 2} - 6000 \\[1em] \Rightarrow 1200 + 6000 = 6000\Big(1 + \dfrac{r}{200}\Big) \\[1em] \Rightarrow 7200 = 6000\Big(1 + \dfrac{r}{200}\Big) \\[1em] \Rightarrow \dfrac{7200}{6000} = \Big(1 + \dfrac{r}{200}\Big) \\[1em] \Rightarrow \dfrac{7200}{6000} - 1 = \dfrac{r}{200} \\[1em] \Rightarrow \dfrac{7200 - 6000}{6000} = \dfrac{r}{200} \\[1em] \Rightarrow \dfrac{1200}{6000} = \dfrac{r}{200} \\[1em] \Rightarrow r = \dfrac{1200}{6000} \times 200 \\[1em] \Rightarrow r = 40%$

Hence, Option 1 is the correct option.

Question 1(d-i)

On a certain sum, the S.I. for 2 years is ₹ 2400. If the rate of interest is 10% p.a., then :

C.I. for 1st year is :

₹ 1200
₹ 1320
₹ 1800
₹ 2640

Answer

Let sum of money be ₹ x.

Given,

S.I. = ₹ 2400

Time = 2 years

Rate = 10%

By formula,

$\Rightarrow S.I. = \dfrac{P \times R \times T}{100} \\[1em] \Rightarrow 2400 = \dfrac{x \times 10 \times 2}{100} \\[1em] \Rightarrow x = \dfrac{2400 \times 100}{10 \times 2} \\[1em] \Rightarrow x = ₹ 12000.$

By formula,

C.I. = A - P

$C.I. = P\Big(1 + \dfrac{r}{100}\Big)^1 - P \\[1em] = 12000\Big(1 + \dfrac{10}{100}\Big)^1 - 12000 \\[1em] = 12000\Big(1 + \dfrac{1}{10}\Big)^1 - 12000 \\[1em] = 12000 \times \Big(\dfrac{11}{10}\Big)^1 - 12000 \\[1em] = 12000 \times \dfrac{11}{10} - 12000 \\[1em] = 13200 - 12000 \\[1em] = ₹ 1200.$

Hence, Option 1 is the correct option.

Question 1(d-ii)

On a certain sum, the S.I. for 2 years is ₹ 2400. If the rate of interest is 10% p.a., then :

C.I. for 2nd year is :

₹ 1320
₹ 2640
₹ 1980
₹ 2400

Answer

For 2nd year :

P = 12000 + 1200 = ₹ 13200

By formula,

C.I. = A - P

$C.I. = P\Big(1 + \dfrac{r}{100}\Big)^1 - P \\[1em] = 13200\Big(1 + \dfrac{10}{100}\Big)^1 - 13200 \\[1em] = 13200 \times \dfrac{110}{100} - 13200 \\[1em] = 14520 - 13200 \\[1em] = ₹ 1320.$

Hence, Option 1 is the correct option.

Question 1(d-iii)

On a certain sum, the S.I. for 2 years is ₹ 2400. If the rate of interest is 10% p.a., then :

the sum is :

₹ 12000
₹ 26400
₹ 13200
₹ 24000

Answer

Let sum of money be ₹ x.

Given,

S.I. = ₹ 2400

Time = 2 years

Rate = 10%

By formula,

P = ₹ x = ₹ 12000.

Hence, Option 1 is the correct option.

Question 1(d-iv)

On a certain sum, the S.I. for 2 years is ₹ 2400. If the rate of interest is 10% p.a., then :

the amount in 2 years, at compound interest is :

₹ 14520
₹ 12000
₹ 12120
₹ 24000

Answer

Amount in 2 years = Principal + C.I. for 1st year + C.I. for 2nd year

= ₹ 12000 + ₹ 1200 + ₹ 1320

= ₹ 14520.

Hence, Option 1 is the correct option.

Question 2

If the interest is compounded half-yearly, calculate the amount when principal is ₹ 7400; the rate of interest is 5% per annum and the duration is one year.

Answer

Given,

P = ₹ 7400

r = 5% compounded half-yearly

n = 1 year

When rate of interest is compounded half-yearly,

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

Substituting values we get :

$A = 7400 \times \Big(1 + \dfrac{5}{2 \times 100}\Big)^{1 \times 2} \\[1em] = 7400 \times \Big(1 + \dfrac{1}{40}\Big)^2 \\[1em] = 7400 \times \Big(\dfrac{41}{40}\Big)^2 \\[1em] = 7400 \times \dfrac{1681}{1600} \\[1em] = ₹ 7774.63$

Hence, amount = ₹ 7774.63

Question 3

Find the difference between the compound interest compounded yearly and half-yearly on ₹ 10000 for 18 months at 10% per annum.

Answer

Given,

P = ₹ 10000

n = 18 months or 1.5 years

r = 10%

When interest is compounded yearly :

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

For 1st year :

$A = 10000 \times \Big(1 + \dfrac{10}{100}\Big)^1 \\[1em] = 10000 \times \dfrac{110}{100} \\[1em] = ₹ 11000.$

For next half-year :

₹ 11000 is the principal.

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

Substituting values we get :

$A = 11000 \times \Big(1 + \dfrac{10}{2 \times 100}\Big)^{\dfrac{1}{2} \times 2} \\[1em] = 11000 \times \Big(1 + \dfrac{1}{20}\Big)^1 \\[1em] = 11000 \times \dfrac{21}{20} \\[1em] = ₹ 11550.$

When rate of interest is compounded half-yearly :

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

Substituting values we get :

$A = 10000 \times \Big(1 + \dfrac{10}{2 \times 100}\Big)^{1.5 \times 2} \\[1em] = 10000 \times \Big(1 + \dfrac{1}{20}\Big)^3\\[1em] = 10000 \times \Big(\dfrac{21}{20}\Big)^3 \\[1em] = 10000 \times \dfrac{9261}{8000} \\[1em] = ₹ 11576.25$

Difference in C.I. between two cases = ₹ 11576.25 - ₹ 11550 = ₹ 26.25

Hence, difference between C.I. in two cases = ₹ 26.25

Question 4

A man borrowed ₹ 16000 for 3 years under the following terms :

20% simple interest for the first 2 years.

20% C.I. for the remaining one year on the amount due after 2 years, the interest being compounded half-yearly.

Find the total amount to be paid at the end of three years.

Answer

For S.I. :

P = ₹ 16000

Time = 2 years

Rate = 20%

S.I. = $\dfrac{P \times R \times T}{100} = \dfrac{16000 \times 2 \times 20}{100}$ = ₹ 6400.

Amount = P + S.I. = ₹ 16000 + ₹ 6400 = ₹ 22400.

For C.I. :

P = ₹ 22400

Time = 1 year

Rate = 20%

When interest is compounded half-yearly :

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

Substituting values we get :

$A = 22400\Big(1 + \dfrac{20}{2 \times 100}\Big)^{1 \times 2} \\[1em] = 22400 \times \Big(1 + \dfrac{1}{10}\Big)^2 \\[1em] = 22400 \times \Big(\dfrac{11}{10}\Big)^2 \\[1em] = 22400 \times \dfrac{121}{100} \\[1em] = 224 \times 121 \\[1em] = ₹ 27104.$

Hence, total amount to be paid = ₹ 27104.

Question 5

What sum of money will amount to ₹ 27783 in one and a half years at 10% per annum compounded half-yearly ?

Answer

Let sum of money be ₹ x.

Given,

Time = 1.5 years

Rate = 10% 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 :

$\Rightarrow 27783 = x \times \Big(1 + \dfrac{10}{2 \times 100}\Big)^{1.5 \times 2} \\[1em] \Rightarrow 27783 = x \times \Big(1 + \dfrac{1}{20}\Big)^3 \\[1em] \Rightarrow 27783 = x \times \Big(\dfrac{21}{20}\Big)^3 \\[1em] \Rightarrow 27783 = \dfrac{9261x}{8000} \\[1em] \Rightarrow x = \dfrac{27783 \times 8000}{9261} \\[1em] \Rightarrow x = ₹ 24000.$

Hence, sum of money = ₹ 24000.

Question 6

Ashok invests a certain sum of money at 20% per annum, compounded yearly. Geeta invests an equal amount of money at the same rate of interest per annum compounded half-yearly. If Geeta gets ₹ 33 more than Ashok in 18 months, calculate the money invested.

Answer

Given,

r = 20%

n = 18 months or 1.5 years

Let sum of money invested by both be ₹ x.

For Ashok interest is compounded annually :

For 1st year :

$A = P\Big(1 + \dfrac{r}{100}\Big)^n \\[1em] = x \times \Big(1 + \dfrac{20}{100}\Big)^1 \\[1em] = x \times \Big(\dfrac{120}{100}\Big)^1 \\[1em] = x \times \dfrac{6}{5} \\[1em] = \dfrac{6x}{5}.$

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

P = $\dfrac{6x}{5}$

$A = P\Big(1 + \dfrac{r}{100 \times 2}\Big)^{n \times 2} \\[1em] = \dfrac{6x}{5} \times \Big(1 + \dfrac{20}{100 \times 2}\Big)^{\dfrac{1}{2} \times 2} \\[1em] = \dfrac{6x}{5} \times \Big(1 + \dfrac{20}{200}\Big)^1 \\[1em] = \dfrac{6x}{5} \times \Big(\dfrac{220}{200}\Big) \\[1em] = \dfrac{66x}{50}.$

For Geeta interest is compounded half-yearly :

$A = P\Big(1 + \dfrac{r}{100 \times 2}\Big)^{n \times 2} \\[1em] = x \times \Big(1 + \dfrac{20}{100 \times 2}\Big)^{1.5 \times 2} \\[1em] = x \times \Big(1 + \dfrac{20}{200}\Big)^3 \\[1em] = x \times \Big(\dfrac{220}{200}\Big)^3 \\[1em] = x \times \Big(\dfrac{11}{10}\Big)^3 \\[1em] = x \times \dfrac{1331}{1000} \\[1em] = \dfrac{1331x}{1000}.$

Given, Geeta receives ₹ 33 more :

$\therefore \dfrac{1331x}{1000} - \dfrac{66x}{50} = 33 \\[1em] \Rightarrow \dfrac{1331x - 1320x}{1000} = 33 \\[1em] \Rightarrow \dfrac{11x}{1000} = 33 \\[1em] \Rightarrow x = \dfrac{33 \times 1000}{11} \\[1em] \Rightarrow x = ₹ 3000.$

Hence, sum invested by both = ₹ 3000.

Question 7

At what rate of interest per annum will a sum of ₹ 62500 earn a compound interest of ₹ 5100 in one year ? The interest is to be compounded half-yearly ?

Answer

Let rate of interest be r% per annum.

When interest is compounded half-yearly.

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

Substituting values we get :

$\Rightarrow C.I. = A - P \\[1em] \Rightarrow C.I. = P\Big(1 + \dfrac{r}{2 \times 100}\Big)^{n \times 2} - P \\[1em] \Rightarrow 5100 = 62500 \times \Big(1 + \dfrac{r}{200}\Big)^{1 \times 2} - 62500 \\[1em] \Rightarrow 5100 + 62500 = 62500 \times \Big(1 + \dfrac{r}{200}\Big)^2 \\[1em] \Rightarrow 67600 = 62500 \times \Big(1 + \dfrac{r}{200}\Big)^2 \\[1em] \Rightarrow \dfrac{67600}{62500} = \Big(1 + \dfrac{r}{200}\Big)^2 \\[1em] \Rightarrow \dfrac{676}{625} = \Big(1 + \dfrac{r}{200}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{26}{25}\Big)^2 = \Big(1 + \dfrac{r}{200}\Big)^2 \\[1em] \Rightarrow 1 + \dfrac{r}{200} = \dfrac{26}{25} \\[1em] \Rightarrow \dfrac{r}{200} = \dfrac{26}{25}- 1 \\[1em] \Rightarrow \dfrac{r}{200} = \dfrac{1}{25} \\[1em] \Rightarrow r = \dfrac{200}{25} = 8%.$

Hence, rate of interest = 8%.

Question 8

In what time will ₹ 1500 yield ₹ 496.50 as compound interest at 20% per year compounded half-yearly ?

Answer

Given,

C.I. = ₹ 496.50

P = ₹ 1500

Rate = 20%

A = P + C.I. = ₹ 1500 + ₹ 496.50 = ₹ 1996.50

Let time required be n years.

When interest is compounded half-yearly.

$\Rightarrow A = P\Big(1 + \dfrac{r}{2 \times 100}\Big)^{n \times 2} \\[1em] \Rightarrow 1996.50 = 1500 \times \Big(1 + \dfrac{20}{200}\Big)^{2n} \\[1em] \Rightarrow \dfrac{1996.50}{1500} = \Big(1 + \dfrac{20}{200}\Big)^{2n} \\[1em] \Rightarrow \dfrac{199650}{150000} = \Big(\dfrac{220}{200}\Big)^{2n} \\[1em] \Rightarrow \dfrac{1331}{1000} = \Big(\dfrac{11}{10}\Big)^{2n} \\[1em] \Rightarrow \Big(\dfrac{11}{10}\Big)^3 = \Big(\dfrac{11}{10}\Big)^{2n} \\[1em] \Rightarrow 2n = 3 \\[1em] \Rightarrow n = \dfrac{3}{2}.$

Hence, required time = $1\dfrac{1}{2}$ years.

Question 9

Calculate the C.I. on ₹ 3500 at 6% per annum for 3 years, the interest being compounded half-yearly.

Answer

When interest is compounded half-yearly.

$\Rightarrow A = P\Big(1 + \dfrac{r}{2 \times 100}\Big)^{n \times 2} \\[1em] \Rightarrow A = 3500 \times \Big(1 + \dfrac{6}{200}\Big)^{3 \times 2} \\[1em] \Rightarrow A = 3500 \times \Big(\dfrac{206}{200}\Big)^6 \\[1em] \Rightarrow A = 3500 \times (1.03)^6 \\[1em] \Rightarrow A = 3500 \times 1.194052 \\[1em] \Rightarrow A = ₹ 4179.18$

By formula,

C.I. = A - P = ₹ 4179.18 - ₹ 3500 = ₹ 679.18

Hence, C.I. = ₹ 679.18

Question 10

Find the difference between compound interest and simple interest on ₹ 12000 and in $1\dfrac{1}{2}$ years at 10% p.a. compounded yearly.

Answer

Calculating C.I. :

For 1st year :

P = ₹ 12000

Rate = 10%

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

Substituting values we get :

$A = 12000 \times \Big(1 + \dfrac{10}{100}\Big)^1 \\[1em] = 12000 \times \dfrac{110}{100} \\[1em] = ₹ 13200.$

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

P = ₹ 13200

Rate = 10%

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

Substituting values we get :

$A = 13200 \times \Big(1 + \dfrac{10}{200}\Big)^{\dfrac{1}{2} \times 2} \\[1em] = 13200 \times \dfrac{210}{200} \\[1em] = ₹ 13860.$

C.I. = A - P = ₹ 13860 - ₹ 12000 = ₹ 1860

Calculating S.I. :

P = ₹ 12000

Rate = 10%

Time = $1\dfrac{1}{2}$

$S.I. = \dfrac{P \times R \times T}{100} \\[1em] = \dfrac{12000 \times 10 \times \dfrac{3}{2}}{100} \\[1em] = ₹ 1800.$

Difference between C.I. and S.I. = ₹ 1860 - ₹ 1800 = ₹ 60.

Hence, required difference = ₹ 60.

Question 11

Find the difference between compound interest and simple interest on ₹ 12000 and in $1\dfrac{1}{2}$ years at 10% compounded half-yearly.

Answer

Calculating S.I. :

P = ₹ 12000

Rate = 10%

Time = $1\dfrac{1}{2}$

$S.I. = \dfrac{P \times R \times T}{100} \\[1em] = \dfrac{12000 \times 10 \times \dfrac{3}{2}}{100} \\[1em] = ₹ 1800.$

Calculating C.I. :

When interest is compounded half-yearly :

$A = P\Big(1 + \dfrac{r}{2 \times 100}\Big)^{n \times 2} \\[1em] = 12000 \times \Big(1 + \dfrac{10}{200}\Big)^{\dfrac{3}{2} \times 2} \\[1em] = 12000 \times \Big(1 + \dfrac{1}{20}\Big)^3 \\[1em] = 12000 \times \Big(\dfrac{21}{20}\Big)^3 \\[1em] = 12000 \times \dfrac{9261}{8000} \\[1em] = \dfrac{27783}{2} \\[1em] = ₹ 13891.50$

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

Difference between C.I. and S.I. = ₹ 1891.50 - ₹ 1800 = ₹ 91.50

Hence, required difference = ₹ 91.50

Exercise 3(D)

Question 1(a)

When the present population (P) of a certain locality increases by r% per year, the population in n years will be :

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

Answer

For growth :

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

Hence, Option 3 is the correct option.

Question 1(b)

The population of a town decreases by 10% in a particular year and then increases by 15% in the next year. The population at the end of two years is :

$\Big(1 + \dfrac{10}{100}\Big)^3\Big(1 + \dfrac{15}{100}\Big)$ times
$\Big(1 - \dfrac{10}{100}\Big)\Big(1 + \dfrac{15}{100}\Big)$ times
$\Big(1 - \dfrac{10}{100}\Big)\Big(1 - \dfrac{15}{100}\Big)$ times
$\Big(1 + \dfrac{10}{100}\Big)\Big(1 - \dfrac{15}{100}\Big)$ times

Answer

Given,

The population of a town decreases by 10% in a particular year and then increases by 15% in the next year.

Let present population be P.

$\text{Population after 2 years} = P \Big(1 - \dfrac{10}{100}\Big)\Big(1 + \dfrac{15}{100}\Big).$

Hence, Option 2 is the correct option.

Question 1(c)

When the cost of a machine decreases by r% per year; the cost of machine in 3 years is :

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

Answer

For depreciation :

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

So,

Value of machine after 3 years = Present value $\times \Big(1 - \dfrac{r}{100}\Big)^3$

Hence, Option 2 is the correct option.

Question 1(d)

On a certain sum the rate of C.I. is x% per annum for the first two years and y% per annum for the next three years. Then the amount after 5 years is :

$\Big(1 + \dfrac{x}{100}\Big)\Big(1 + \dfrac{y}{100}\Big)$ times
$\Big(1 + \dfrac{x}{100}\Big)^3\Big(1 + \dfrac{y}{100}\Big)^2$ times
$\Big(1 + \dfrac{x}{100}\Big)^2\Big(1 + \dfrac{y}{100}\Big)^3$ times
$\Big(1 + \dfrac{x}{100}\Big)^2\Big(1 - \dfrac{y}{100}\Big)^3$ times

Answer

By formula,

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

Given,

On a certain sum the rate of C.I. is x% per annum for the first two years and y% per annum for the next three years.

$\text{Amount after 5 years} = P\Big(1 + \dfrac{x}{100}\Big)^2\Big(1 + \dfrac{y}{100}\Big)^3.$

Hence, Option 3 is the correct option.

Question 1(e)

The cost (₹ x) of a machine increases by 20% in the first two years and then decreases by 25% in the next two years. Then the cost of machine becomes :

₹ $x \times \dfrac{120}{100} \times \dfrac{75}{100}$
₹ $x \times \Big(\dfrac{120}{100}\Big)^2 \times \dfrac{75}{100}$
₹ $x \times \Big(\dfrac{120}{100}\Big)^2 \times \Big(\dfrac{75}{100}\Big)^2$
₹ $x \times \Big(\dfrac{80}{100}\Big)^2 \times \Big(\dfrac{75}{100}\Big)^2$

Answer

According to question :

The cost (₹ x) of a machine increases by 20% in the first two years and then decreases by 25% in the next two years.

$\Rightarrow x \times \Big(1 + \dfrac{20}{100}\Big)^2 \times \Big(1 - \dfrac{25}{100}\Big)^2 \\[1em] \Rightarrow x \times \Big(\dfrac{120}{100}\Big)^2 \times \Big(\dfrac{75}{100}\Big)^2.$

Hence, Option 3 is the correct option.

Question 2

The cost of a machine is supposed to depreciate each year by 12% of its value at the beginning of the year. If the machine is valued at ₹ 44,000 at the beginning of 2008, find its value :

(i) at the end of 2009.

(ii) at the beginning of 2007.

Answer

(i) For depreciation :

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

The cost of machine depreciates at the beginning of the next year, or we can say that at end of each year.

So, Value at end of 2009 = Value at beginning of 2010.

$\text{Value at beginning of 2010 }= 44000 \times \Big(1 - \dfrac{12}{100}\Big)^2 \\[1em] = 44000 \times \Big(\dfrac{88}{100}\Big)^2 \\[1em] = 44000 \times \Big(\dfrac{22}{25}\Big)^2 \\[1em] = 44000 \times \dfrac{484}{625} \\[1em] = ₹ 34073.60$

Hence, value at the end of 2009 = ₹ 34073.60

(ii) Let value at beginning of 2007 be ₹ x.

After one year it becomes ₹ 44000.

For depreciation :

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

Substituting values we get :

$\Rightarrow 44000 = x \times \Big(1 - \dfrac{12}{100}\Big)^1 \\[1em] \Rightarrow 44000 = x \times \Big(\dfrac{88}{100}\Big) \\[1em] \Rightarrow 44000 = \dfrac{88x}{100} \\[1em] \Rightarrow x = \dfrac{44000 \times 100}{88} \\[1em] \Rightarrow x = ₹ 50000.$

Hence, value at beginning of 2007 = ₹ 50000.

Question 3

The value of an article decreased for two years at the rate of 10% per year and then in the third year it increased by 10%. Find the original value of the article, if its value at the end of 3 years is ₹ 40,095.

Answer

Let original value be ₹ x.

Given, it decreases by 10% for two years. So,

For depreciation :

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

$\text{Value after 2 years} = x \times \Big(1 - \dfrac{10}{100}\Big)^2 \\[1em] = x \times \Big(\dfrac{90}{100}\Big)^2\\[1em] = x \times \Big(\dfrac{9}{10}\Big)^2 \\[1em] = x \times \dfrac{81}{100} \\[1em] = \dfrac{81x}{100}.$

Given,

In the third year value is increased by 10%.

For growth :

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

$\text{Value after 1 year} = \dfrac{81x}{100} \times \Big(1 + \dfrac{10}{100}\Big)^1 \\[1em] = \dfrac{81x}{100} \times \dfrac{110}{100} \\[1em] = \dfrac{891x}{1000}.$

Given,

At the end of 3 years value is ₹ 40095.

$\therefore \dfrac{891x}{1000} = 40095 \\[1em] \Rightarrow x = \dfrac{40095 \times 1000}{891} \\[1em] \Rightarrow x = ₹ 45000.$

Hence, original value of article = ₹ 45000.

Question 4

According to a census taken towards the end of the year 2009, the population of a rural town was found to be 64000. The census authority also found that the population of this particular town had a growth of 5% per annum. In how many years after 2009 did the population of this town reach 74088 ?

Answer

Let in n years population reaches from 64000 to 74088.

For growth :

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

Substituting values we get :

$\Rightarrow 74088 = 64000 \times \Big(1 + \dfrac{5}{100}\Big)^n \\[1em] \Rightarrow \dfrac{74088}{64000} = \Big(\dfrac{105}{100}\Big)^n \\[1em] \Rightarrow \dfrac{9261}{8000} = \Big(\dfrac{21}{20}\Big)^n \\[1em] \Rightarrow \Big(\dfrac{21}{20}\Big)^3 = \Big(\dfrac{21}{20}\Big)^n \\[1em] \Rightarrow n = 3.$

In 3 years the population will reach 74088.

Question 5

The population of a town decreased by 12% during 1998 and then increased by 8% during 1999. Find the population of the town, at the beginning of 1998, if at the end of 1999 its population was 285120.

Answer

Let population in the beginning of 1998 be x.

Given,

The population of a town decreased by 12% during 1998 and then increased by 8% during 1999.

Population at the end of 1999 = 285120.

$\therefore 285120 = x\Big(1 - \dfrac{12}{100}\Big)\Big(1 + \dfrac{8}{100}\Big) \\[1em] \Rightarrow 285120 = x \times \dfrac{88}{100} \times \dfrac{108}{100} \\[1em] \Rightarrow 285120 = \dfrac{9504x}{10000} \\[1em] \Rightarrow x = \dfrac{285120 \times 10000}{9504} \\[1em] \Rightarrow x = 300000.$

Hence, population of town at beginning of 1998 = 300000.

Question 6

A sum of money, invested at compound interest, amounts to ₹ 16500 in 1 year and to ₹ 19965 in 3 years. Find the rate per cent and the original sum of money invested.

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 ₹ 16500 in 1 year.

$\Rightarrow A = P\Big(1 + \dfrac{r}{100}\Big)^n \\[1em] \Rightarrow 16500 = x \times \Big(1 + \dfrac{r}{100}\Big)^1 \\[1em] \Rightarrow 16500 = x\Big(1 + \dfrac{r}{100}\Big) ......(1)$

The sum of money, invested at compound interest, amounts to ₹ 19965 in 3 years.

$\Rightarrow A = P\Big(1 + \dfrac{r}{100}\Big)^n \\[1em] \Rightarrow 19965 = x \times \Big(1 + \dfrac{r}{100}\Big)^3 \\[1em] \Rightarrow 19965 = x\Big(1 + \dfrac{r}{100}\Big)^3 ......(2)$

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

$\Rightarrow \dfrac{19965}{16500} = \dfrac{x\Big(1 + \dfrac{r}{100}\Big)^3}{x\Big(1 + \dfrac{r}{100}\Big)} \\[1em] \Rightarrow \dfrac{121}{100} = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{11}{10}\Big)^2 = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big) = \dfrac{11}{10} \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{11}{10} - 1 \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{1}{10} \\[1em] \Rightarrow r = \dfrac{100}{10} = 10%.$

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

$\Rightarrow 16500 = x\Big(1 + \dfrac{10}{100}\Big) \\[1em] \Rightarrow 16500 = x \times \Big(\dfrac{110}{100}\Big) \\[1em] \Rightarrow x = \dfrac{16500 \times 100}{110} \\[1em] \Rightarrow x = ₹ 15000.$

Hence, rate percent = 10% and sum invested = ₹ 15000.

Question 7

The difference between C.I. and S.I. on ₹ 7500 for two years is ₹ 12 at the same rate of interest per annum. Find the rate of interest.

Answer

Given,

P = ₹ 7500

Time = 2 years

Let rate of interest be r%.

By formula,

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

Substituting values we get :

$S.I. = \dfrac{7500 \times r \times 2}{100} \\[1em] = 150r.$

By formula,

C.I. = A - P

$C.I. = P\Big(1 + \dfrac{r}{100}\Big)^n - P \\[1em] = 7500 \times \Big(1 + \dfrac{r}{100}\Big)^2 - 7500 \\[1em] = 7500 \times \Big(\dfrac{100 + r}{100}\Big)^2 - 7500 \\[1em] = 7500 \times \Big(\dfrac{10000 + r^2 + 200r}{10000}\Big) - 7500 \\[1em] = \dfrac{3}{4} \times (10000 + r^2 + 200r) - 7500 \\[1em] = 7500 + \dfrac{3r^2}{4} + 150r - 7500 \\[1em] = \dfrac{3r^2}{4} + 150r$

The difference between C.I. and S.I. on ₹ 7500 for two years is ₹ 12.

∴ C.I. - S.I. = ₹ 12

$\therefore \dfrac{3r^2}{4} + 150r - 150r = 12 \\[1em] \Rightarrow \dfrac{3r^2}{4} = 12 \\[1em] \Rightarrow r^2 = \dfrac{12 \times 4}{3} \\[1em] \Rightarrow r^2 = 16 \\[1em] \Rightarrow r = \sqrt{16} = 4%.$

Hence, rate of interest = 4%.

Question 8

A sum of money lent out at C.I. at a certain rate per annum becomes three times of itself in 10 years. Find in how many years will the money become twenty-seven times of itself at the same rate of interest p.a.

Answer

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

By formula,

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

Given,

₹ P becomes three times of itself in 10 years.

$\therefore 3P = P\Big(1 + \dfrac{r}{100}\Big)^{10} \\[1em] \Rightarrow \dfrac{3P}{P} = \Big(1 + \dfrac{r}{100}\Big)^{10} \\[1em] \Rightarrow 3 = \Big(1 + \dfrac{r}{100}\Big)^{10} .....(1)$

Let in n years money will become 27 times.

From equation (1)

$\Rightarrow \Big(1 + \dfrac{r}{100}\Big)^n = \Big[\Big(1 + \dfrac{r}{100}\Big)^{10}\Big]^3 \\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big)^n =\Big(1 + \dfrac{r}{100}\Big)^{30} \\[1em] \Rightarrow n = 30 \text{ years}.$

Hence, in 30 years money will becomes 27 times of itself.

Question 9

Mr. Sharma borrowed a certain sum of money at 10% per annum compounded annually. If by paying ₹ 19360 at the end of second year and ₹ 31944 at the end of the third year he clears the debt; find the sum borrowed by him.

Answer

Let sum borrowed be ₹ x.

By formula,

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

At the end of 2 years :

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

Given,

He pays back ₹ 19360 at the end of second year. So,

Principal for third year = ₹ $\Big(\dfrac{121x}{100} - 19360\Big)$

Amount after 3 year :

$A = \Big(\dfrac{121x}{100} - 19360\Big) \times \Big(1 + \dfrac{10}{100}\Big)^1 \\[1em] = \Big(\dfrac{121x}{100} - 19360\Big) \times \dfrac{110}{100} \\[1em] = \dfrac{1331x}{1000} - 21296.$

Given,

On giving ₹ 31944 at the end of the third year he clears the debt.

$\therefore \dfrac{1331x}{1000} - 21296 = 31944 \\[1em] \Rightarrow \dfrac{1331x}{1000} = 31944 + 21296 \\[1em] \Rightarrow \dfrac{1331x}{1000} = 53240 \\[1em] \Rightarrow x = \dfrac{53240 \times 1000}{1331} \\[1em] \Rightarrow x = ₹ 40000.$

Hence, sum borrowed = ₹ 40000.

Question 10

The difference between compound interest for a year payable half-yearly and simple interest on a certain sum of money lent out at 10% for a year is ₹ 15. Find the sum of money lent out.

Answer

Let sum of money lent out be ₹ x.

Calculating C.I. payable half-yearly :

P = ₹ x

Rate of interest = 10%

Time = 1 year

C.I. = A - P

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

Calculating S.I. :

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

Given,

Difference between compound interest for a year payable half-yearly and simple interest on ₹ x lent out at 10% for a year is ₹ 15.

$\therefore \dfrac{41x}{400} - \dfrac{x}{10} = 15 \\[1em] \Rightarrow \dfrac{41x - 40x}{400} = 15 \\[1em] \Rightarrow \dfrac{x}{400} = 15 \\[1em] \Rightarrow x = 15 \times 400 = ₹ 6000.$

Hence, sum of money lent out = ₹ 6000.

Question 11

The ages of Pramod and Rohit are 16 years and 18 years respectively. In what ratio must they invest money at 5% p.a. compounded yearly so that both get the same sum on attaining the age of 25 years ?

Answer

Let Pramod invest ₹ x and Rohit invest ₹ y.

By formula,

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

For Pramod :

P = ₹ x

r = 5%

n = 9 years (As it will take 9 years for Pramod to reach 25 years of age)

Substituting values we get :

$\text{Amount received by Pramod} = x\Big(1 + \dfrac{5}{100}\Big)^9 \\[1em] = x \times \Big(\dfrac{105}{100}\Big)^9 \\[1em] = x \times \Big(\dfrac{21}{20}\Big)^9.$

For Rohit :

P = ₹ y

r = 5%

n = 7 years (As it will take 7 years for Pramod to reach 25 years of age)

Substituting values we get :

$\text{Amount received by Rohit} = y\Big(1 + \dfrac{5}{100}\Big)^7 \\[1em] = y \times \Big(\dfrac{105}{100}\Big)^7 \\[1em] = y \times \Big(\dfrac{21}{20}\Big)^7.$

Since,

Amount received by both are equal.

$\Rightarrow x \times \Big(\dfrac{21}{20}\Big)^9 = y \times \Big(\dfrac{21}{20}\Big)^7 \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{\Big(\dfrac{21}{20}\Big)^7}{\Big(\dfrac{21}{20}\Big)^9} \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{1}{\Big(\dfrac{21}{20}\Big)^2} \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{20^2}{21^2} \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{400}{441}.$

Hence, ratio in which sum must be invested = 400 : 441.

Test Yourself

Question 1(a)

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

₹ 1,200
₹ 1,400
₹ 800
₹ 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.

Question 1(b)

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

₹ 1
₹ 41
₹ 00
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.

Question 1(c)

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:

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

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.

Question 1(d)

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:

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

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.

Question 1(e)

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

Both the statements are true.
Both the statements are false.
Statement 1 is true, and statement 2 is false.
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.

Question 1(f)

Statement 1: The population of a town in the year 2024 is x and it increases by 10% every year. The population of in the year 2021 was = 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%.

Both the statements are true.
Both the statements are false.
Statement 1 is true, and statement 2 is false.
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.

Question 1(g)

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):

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

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

Question 1(h)

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

Reason (R): A = $P\Big(1 - \dfrac{10}{100}\Big)^2$

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

So, assertion (A) is true.

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

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

⇒ P = A $\Big(1 + \dfrac{10}{100}\Big)^{-2}$

So, reason (R) is false.

∴ A is true, but R is false.

Hence, option 1 is correct option.

Question 2

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

Question 3

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.

Question 4

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.

Question 5

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.

Question 6

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.

Question 7

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.

Question 8

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.

Question 9

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.

Question 10

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.

Question 11

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

Question 12

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.

Chapter 3

Compound Interest (Stage 2) [Applications]

Class - 9 Concise Mathematics Selina

Exercise 3(A)

Question 1(a)

Question 1(b)

Question 1(c)

Question 1(d)

Question 1(e)

Question 1(f)

Question 1(g)

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Exercise 3(B)

Question 1(a)

Question 1(b)

Question 1(c)

Question 1(d)

Question 1(e)

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Exercise 3(C)

Question 1(a-i)

Question 1(a-ii)

Question 1(a-iii)

Question 1(b-i)

Question 1(b-ii)

Question 1(b-iii)

Question 1(c)

Question 1(d-i)

Question 1(d-ii)

Question 1(d-iii)

Question 1(d-iv)

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Exercise 3(D)

Question 1(a)

Question 1(b)

Question 1(c)

Question 1(d)

Question 1(e)

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8