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Chapter 3

Compound Interest [Applications] — Exercise 3(C)

Class - 9 Concise Mathematics Selina

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 :

1121\dfrac{1}{2} years :

  1. P[(1+r100)112−1]P\Big[\Big(1 + \dfrac{r}{100}\Big)^{1\dfrac{1}{2}}- 1\Big]

  2. P(1+r100)(1+r200)P\Big(1 + \dfrac{r}{100}\Big)\Big(1 + \dfrac{r}{200}\Big) - P

  3. P(1+r200)3P\Big(1 + \dfrac{r}{200}\Big)^3

  4. P(1+r100)3(1+r200)−PP\Big(1 + \dfrac{r}{100}\Big)^3\Big(1 + \dfrac{r}{200}\Big)- P

Answer

Let sum = ₹ P.

Amount for first year :

A = P(1+r100)1P\Big(1 + \dfrac{r}{100}\Big)^1

For next 12\dfrac{1}{2} year :

Sum = ₹ P(1+r100)P\Big(1 + \dfrac{r}{100}\Big)

A = P(1+r100)(1+r2×100)12×2=P(1+r100)(1+r200)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(1+r100)(1+r200)−PP\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 :

  1. P(1+r100)1−PP\Big(1 + \dfrac{r}{100}\Big)^1 - P

  2. P(1+r200)P\Big(1 + \dfrac{r}{200}\Big) - P

  3. P(1+1200)2P\Big(1 + \dfrac{1}{200}\Big)^2 - P

  4. P(1−r200)2P\Big(1 - \dfrac{r}{200}\Big)^2 - P

Answer

By formula,

C.I. = A - P

C.I.=P(1+r100)1−PC.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 :

2122\dfrac{1}{2} years :

  1. P(1+r100)52−PP\Big(1 + \dfrac{r}{100}\Big)^{\dfrac{5}{2}} - P

  2. P(1+r100)52P\Big(1 + \dfrac{r}{100}\Big)^{\dfrac{5}{2}}

  3. P(1+r100)2(1+r200)P\Big(1 + \dfrac{r}{100}\Big)^2\Big(1 + \dfrac{r}{200}\Big)

  4. P(1+r100)2(1+r200)P\Big(1 + \dfrac{r}{100}\Big)^2\Big(1 + \dfrac{r}{200}\Big) - P

Answer

Amount for first two years :

A = P(1+r100)2P\Big(1 + \dfrac{r}{100}\Big)^2

For next 12\dfrac{1}{2} year :

Sum = ₹ P(1+r100)2P\Big(1 + \dfrac{r}{100}\Big)^2

A = P(1+r100)2(1+r2×100)12×2=P(1+r100)2(1+r200)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(1+r100)2(1+r200)−PP\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 :

  1. P(1+r200)2−PP\Big(1 + \dfrac{r}{200}\Big)^2 - P

  2. P(1+r100)2−PP\Big(1 + \dfrac{r}{100}\Big)^2 - P

  3. P(1+r200)2P\Big(1 + \dfrac{r}{200}\Big)^2

  4. P(1+r100)2P\Big(1 + \dfrac{r}{100}\Big)^2

Answer

When interest is compounded half-yearly.

A = P(1+r2×100)n×2P\Big(1 + \dfrac{r}{2 \times 100}\Big)^{n \times 2}

C.I. = A - P

For 1 year :

C.I.=P(1+r2×100)1×2−P=P(1+r200)2−PC.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 :

1121\dfrac{1}{2} years

  1. P(1+r200)32−PP\Big(1 + \dfrac{r}{200}\Big)^{\dfrac{3}{2}} - P

  2. P(1+r200)3−PP\Big(1 + \dfrac{r}{200}\Big)^3 - P

  3. P(1+r100)3−PP\Big(1 + \dfrac{r}{100}\Big)^3 - P

  4. P(1+r100)32−PP\Big(1 + \dfrac{r}{100}\Big)^{\dfrac{3}{2}} - P

Answer

When interest is compounded half-yearly.

A = P(1+r2×100)n×2P\Big(1 + \dfrac{r}{2 \times 100}\Big)^{n \times 2}

C.I. = A - P

C.I.=P(1+r2×100)32×2−P=P(1+r200)3−PC.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 :

  1. P(1+r100)2−PP\Big(1 + \dfrac{r}{100}\Big)^2 - P

  2. P(1+r200)2−PP\Big(1 + \dfrac{r}{200}\Big)^2 - P

  3. P(1+r200)4−PP\Big(1 + \dfrac{r}{200}\Big)^4 - P

  4. P(1+r100)4−PP\Big(1 + \dfrac{r}{100}\Big)^4 - P

Answer

When interest is compounded half-yearly.

A = P(1+r2×100)n×2P\Big(1 + \dfrac{r}{2 \times 100}\Big)^{n \times 2}

C.I. = A - P

C.I.=P(1+r2×100)2×2−P=P(1+r200)4−PC.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 :

  1. 40%

  2. 15%

  3. 20%

  4. 24%

Answer

Let rate of interest be r%.

Given,

P = ₹ 6000

C.I. = ₹ 1200

By formula,

⇒C.I.=P(1+r2×100)n×2−P⇒1200=6000(1+r200)12×2−6000⇒1200+6000=6000(1+r200)⇒7200=6000(1+r200)⇒72006000=(1+r200)⇒72006000−1=r200⇒7200−60006000=r200⇒12006000=r200⇒r=12006000×200⇒r=40\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 :

  1. ₹ 1200

  2. ₹ 1320

  3. ₹ 1800

  4. ₹ 2640

Answer

Let sum of money be ₹ x.

Given,

S.I. = ₹ 2400

Time = 2 years

Rate = 10%

By formula,

⇒S.I.=P×R×T100⇒2400=x×10×2100⇒x=2400×10010×2⇒x=₹12000.\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(1+r100)1−P=12000(1+10100)1−12000=12000(1+110)1−12000=12000×(1110)1−12000=12000×1110−12000=13200−12000=₹1200.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 :

  1. ₹ 1320

  2. ₹ 2640

  3. ₹ 1980

  4. ₹ 2400

Answer

For 2nd year :

P = 12000 + 1200 = ₹ 13200

By formula,

C.I. = A - P

C.I.=P(1+r100)1−P=13200(1+10100)1−13200=13200×110100−13200=14520−13200=₹1320.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 :

  1. ₹ 12000

  2. ₹ 26400

  3. ₹ 13200

  4. ₹ 24000

Answer

Let sum of money be ₹ x.

Given,

S.I. = ₹ 2400

Time = 2 years

Rate = 10%

By formula,

⇒S.I.=P×R×T100⇒2400=x×10×2100⇒x=2400×10010×2⇒x=₹12000.\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.

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 :

  1. ₹ 14520

  2. ₹ 12000

  3. ₹ 12120

  4. ₹ 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(1+r2×100)n×2P\Big(1 + \dfrac{r}{2 \times 100}\Big)^{n \times 2}

Substituting values we get :

A=7400×(1+52×100)1×2=7400×(1+140)2=7400×(4140)2=7400×16811600=₹7774.63A = 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(1+r100)nP\Big(1 + \dfrac{r}{100}\Big)^n

For 1st year :

A=10000×(1+10100)1=10000×110100=₹11000.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(1+r2×100)n×2P\Big(1 + \dfrac{r}{2 \times 100}\Big)^{n \times 2}

Substituting values we get :

A=11000×(1+102×100)12×2=11000×(1+120)1=11000×2120=₹11550.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(1+r2×100)n×2P\Big(1 + \dfrac{r}{2 \times 100}\Big)^{n \times 2}

Substituting values we get :

A=10000×(1+102×100)1.5×2=10000×(1+120)3=10000×(2120)3=10000×92618000=₹11576.25A = 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. = P×R×T100=16000×2×20100\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(1+r2×100)n×2P\Big(1 + \dfrac{r}{2 \times 100}\Big)^{n \times 2}

Substituting values we get :

A=22400(1+202×100)1×2=22400×(1+110)2=22400×(1110)2=22400×121100=224×121=₹27104.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(1+r2×100)n×2P\Big(1 + \dfrac{r}{2 \times 100}\Big)^{n \times 2}

Substituting values we get :

⇒27783=x×(1+102×100)1.5×2⇒27783=x×(1+120)3⇒27783=x×(2120)3⇒27783=9261x8000⇒x=27783×80009261⇒x=₹24000.\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 by each.

Answer

Given,

r = 20%

n = 18 months or 1.5 years

Let sum of money invested by each be ₹ x.

For Ashok interest is compounded annually :

For 1st year :

A=P(1+r100)n=x×(1+20100)1=x×(120100)1=x×65=6x5.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 12\dfrac{1}{2} year :

P = 6x5\dfrac{6x}{5}

A=P(1+r100×2)n×2=6x5×(1+20100×2)12×2=6x5×(1+20200)1=6x5×(220200)=66x50.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(1+r100×2)n×2=x×(1+20100×2)1.5×2=x×(1+20200)3=x×(220200)3=x×(1110)3=x×13311000=1331x1000.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 :

∴1331x1000−66x50=33⇒1331x−1320x1000=33⇒11x1000=33⇒x=33×100011⇒x=₹3000.\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 each = ₹ 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(1+r2×100)n×2A = P\Big(1 + \dfrac{r}{2 \times 100}\Big)^{n \times 2}

Substituting values we get :

⇒C.I.=A−P⇒C.I.=P(1+r2×100)n×2−P⇒5100=62500×(1+r200)1×2−62500⇒5100+62500=62500×(1+r200)2⇒67600=62500×(1+r200)2⇒6760062500=(1+r200)2⇒676625=(1+r200)2⇒(2625)2=(1+r200)2⇒1+r200=2625⇒r200=2625−1⇒r200=125⇒r=20025=8\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.

⇒A=P(1+r2×100)n×2⇒1996.50=1500×(1+20200)2n⇒1996.501500=(1+20200)2n⇒199650150000=(220200)2n⇒13311000=(1110)2n⇒(1110)3=(1110)2n⇒2n=3⇒n=32.\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 = 1121\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.

⇒A=P(1+r2×100)n×2⇒A=3500×(1+6200)3×2⇒A=3500×(206200)6⇒A=3500×(1.03)6⇒A=3500×1.194052⇒A=₹4179.18\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 1121\dfrac{1}{2} years at 10% p.a. compounded yearly.

Answer

Calculating C.I. :

For 1st year :

P = ₹ 12000

Rate = 10%

A = P(1+r100)nP\Big(1 + \dfrac{r}{100}\Big)^n

Substituting values we get :

A=12000×(1+10100)1=12000×110100=₹13200.A = 12000 \times \Big(1 + \dfrac{10}{100}\Big)^1 \\[1em] = 12000 \times \dfrac{110}{100} \\[1em] = ₹ 13200.

For next 12\dfrac{1}{2} year :

P = ₹ 13200

Rate = 10%

A = P(1+r2×100)n×2P\Big(1 + \dfrac{r}{2 \times 100}\Big)^{n \times 2}

Substituting values we get :

A=13200×(1+10200)12×2=13200×210200=₹13860.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 = 1121\dfrac{1}{2}

S.I.=P×R×T100=12000×10×32100=₹1800.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 1121\dfrac{1}{2} years at 10% compounded half-yearly.

Answer

Calculating S.I. :

P = ₹ 12000

Rate = 10%

Time = 1121\dfrac{1}{2}

S.I.=P×R×T100=12000×10×32100=₹1800.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(1+r2×100)n×2=12000×(1+10200)32×2=12000×(1+120)3=12000×(2120)3=12000×92618000=277832=₹13891.50A = 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

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