COMPOUND INTEREST EXAMPLES | EXERCISE 8.3 CLASS 8 MATHS

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Compound Interest Examples | Compound Interest Formula

Compound interest is the interest calculated on the previous year’s amount (A = P + I)
• Amount when interest is compounded annually

A = P\left(1+\frac{r}{100}\right)^t
where, P is principal, r is rate of interest, t is time period.

• Amount when interest is compounded half yearly

A = P\left(1+\frac{r/2}{100}\right)^{2t}
where, r/2 is half yearly rate and
2t = number of ’half-years’

CI = A - P

EXERCISE 8.3
Application of Compound Interest Formula

QUESTION 1

Calculate the amount and compound interest on

(a) ₹ 10,800 for 3 years at 12\frac{1}{2} % per annum compounded annually.

(b) ₹ 18,000 for 2\frac{1}{2} years at 10% per annum compounded annually.

(c) ₹ 62,500 for 1\frac{1}{2} years at 8% per annum compounded half yearly.

(d) ₹ 8,000 for 1 year at 9% per annum compounded half yearly.
(You could use the year by year calculation using SI formula to verify).

(e) ₹ 10,000 for 1 year at 8% per annum compounded half yearly.

ANSWER

(a) P = ₹ 10,800
t = 3 years,
r = 12\frac{1}{2} % per annum compounded annually,

Amount = P\left(1+\frac{r}{100}\right)^t

= ₹ 10800\left(1+\frac{25}{200}\right)^3

= ₹ 10800\left(1+\frac{1}{8}\right)^3

= ₹ 10800\left(\frac{8+1}{8}\right)^3

= ₹ 10800\left(\frac{9}{8}\right)^3

= ₹ 10800\left(\frac{729}{512}\right)^3

= ₹ 15,377.34

Compound Interest = A – P
= ₹ (15377.34 – 10800)
= ₹ 4,577.34

(b) P = ₹ 18,000
t = 2\frac{1}{2} years,
r = 10 % per annum compounded annually,

Amount = P\left(1+\frac{r}{100}\right)^t

Amount for 2 years = ₹ 18000\left(1+\frac{10}{100}\right)^2

= ₹ 18000\left(1+\frac{1}{10}\right)^2

= ₹ 18000\left(\frac{11}{10}\right)^2

= ₹ 18000\times \frac{121}{100}

= ₹ 180\times 121

= ₹ 21,780

CI for 2 years = ₹ (21,780 – 18,000)
= ₹ 3,780

Amount for 2 years will act as Principal for next \frac{1}{2} years,

SI on ₹ 21,780 = ₹ \frac{21780\times 10\times \frac{1}{2}}{100}

= ₹ 1,089

Total CI = ₹ (3,780 + 1,089)
= ₹ 4,869

Total Amount = ₹ (18,000 + 4,868)
= ₹ 22,869.

(c) P = ₹ 62,500,

t = 1\frac{1}{2} years,

r = 8% per annum compounded half yearly,

Since interest is compounded half yearly, so there will be two conversion periods in a year,

therefore, t_h = 2\times \frac{3}{2} = 3 years

and r_h = \frac{8\%}{2} = 4\% per half-year,

Now,

Required Amount = P\left(1+\frac{r_h}{100}\right)^{t_h}

= ₹ 62,500\left(1+\frac{4}{100}\right)^3

= ₹ 62,500\left(1+\frac{1}{25}\right)^3

= ₹ 62,500\left(\frac{26}{25}\right)^3

= ₹ 62,500\times \frac{26\times 26\times 26}{25\times 25\times 25}

= ₹ 4\times 17,576

= ₹ 70,304

CI = A – P

= ₹ (70,304 – 62,500)

= ₹ 7,804

(d) P = ₹ 8,000,

t = 1 year,

r = 9% per annum compounded half yearly,

Since interest is compounded half yearly, so there will be two conversion periods in a year,

therefore, t_h = 2\times 1 = 2 years

and r_h = \frac{9}{2}\% per half-year,

Now,

Required Amount = P\left(1+\frac{r_h}{100}\right)^{t_h}

= ₹ 8,000\left(1+\frac{9}{2\times 100}\right)^2

= ₹ 8,000\left(\frac{209}{200}\right)^2

= ₹ 8,000\times\frac{209\times 209}{200\times 200}

= ₹ \frac{43,681}{5}

= ₹ 8,736.20,

CI = A – P

= ₹ (8,736.20 – 8,000)

= ₹ 736.20

(e) P = ₹ 10,000,

t = 1 years,

r = 8% per annum compounded half yearly,

Since interest is compounded half yearly, so there will be two conversion periods in a year,

therefore, t_h = 2\times 1 = 2 years

and r_h = \frac{8\%}{2} = 4\% per half-year,

Now,

Required Amount = P\left(1+\frac{r_h}{100}\right)^{t_h}

= ₹ 10,000\left(1+\frac{4}{100}\right)^2

= ₹ 10,000\left(1+\frac{1}{25}\right)^2

= ₹ 10,000\left(\frac{26}{25}\right)^2

= ₹ 10,000\times \frac{26\times 26}{25\times 25}

= ₹ 16\times 676

= ₹ 10,816

CI = A – P

= ₹ (10,816 – 10,000)

= ₹ 816.

QUESTION 2

Kamala borrowed ₹ 26,400 from a Bank to buy a scooter at a rate of 15% p.a. compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan?

(Hint: Find A for 2 years with interest is compounded yearly and then find SI on the 2nd year amount for \frac{4}{12} years).

ANSWER

Principal, P = ₹ 26,400

Rate, r = 15% p.a.,

Time period, t = 2 years 4 months
= 2\frac{4}{12} years = 2\frac{1}{3} years

Amount for 2 years = ₹ 26,400\left(1+\frac{15}{100}\right)^2

= ₹ 26,400\left(1+\frac{3}{20}\right)^2

= ₹ 26,400\left(\frac{23}{20}\right)^2

= ₹ 26,400\times \frac{529}{400}

= ₹ 66\times 529

= ₹ 34,914

So, CI for two years = ₹ (34,914 – 26,400)

= ₹ 8,514

Amount after 2 years will serve as Principal for next \frac{1}{3} years,

Simple Interest for next \frac{1}{3} years = \frac{34,914\times 15\times \frac{1}{3}}{100}

= ₹ \frac{34,914\times 15\times}{3\times 100}

= ₹ \frac{11,638\times 3\times}{20}

= ₹ \frac{5,819\times 3\times}{10}

= ₹ 1,745.70

Therefore, total CI = ₹ (8,514 + 1,745.70) = ₹ 10,259.70

Hence, amount Kamala has to pay at the end of 2 years and 4 months to clear the loan = ₹ (26,400 + 10,259.70)
= ₹ 36,659.70.

QUESTION 3

Fabina borrows ₹ 12,500 at 12% per annum for 3 years at simple interest and Radha borrows the same amount for the same time period at 10% per annum, compounded annually. Who pays more interest and by how much?

ANSWER

Fabina and Radha borrows same amount for the same time period,

Principal, P = ₹ 12,500,

time period, t = 3 years,

Rate at which Fabina borrowed money, r_f = 12\% per annum,

Since Fabina borrowed the money at simple interest,

therefore, SI to be paid by Fabina = ₹ \frac{12,500\times 12\times 3}{100}

= ₹ 12,5\times 12\times 3

= ₹ 4,500

Rate at which Radha borrowed money, r_f = 10\% per annum,

Since Radha borrowed the money at compound interest,

So, Amount to be paid by Radha = ₹ 12,500\left(1+\frac{10}{100}\right)^3

= ₹ 12,500\times \left(\frac{11}{10}\right)^3

= ₹ 12,500\times \times \frac{1331}{1000}

= ₹ 16,637.50

Compound interest to be paid by Radha = ₹ 16,637.50 – 12,500
= ₹ 4,137.50

Comparing interests to be paid by Fabina and Radha,

₹ 4,500 > ₹ 4,137.50,
and ₹ 4,500 – ₹ 4,137.50 = ₹ 362.50

Hence, Fabina has to pay more interest than Radha by ₹ 362.50.

[ Compound Interest Examples | Ncert Solutions Exercise 8.3 Class 8 Maths ]

QUESTION 4

I borrowed ₹ 12,000 from Jamshed at 6% per annum simple interest for 2 years. Had I borrowed this sum at 6% per annum compound interest, what extra amount would I have to pay?

ANSWER

P = ₹ 12,000

r = 6% per annum,

t = 2 years,

SI = \frac{P\times r\times t}{100}

= ₹ \frac{12,000\times 6\times 2}{100}

= ₹ 120\times 12

= ₹ 1,440

Amount = P + SI

= ₹ (12,000 + 1,440)

= ₹ 13,440,

Had the money borrowed at compound interest,

Amount to be paid = ₹ 12,000 \left(1+\frac{6}{100}\right)^2

= ₹ 12,000\left(\frac{106}{100}\right)^2

= ₹ 12,000 \times \frac{11,236}{10000}

= ₹ 13,483.20

Thus, extra amount to pay had the money borrowed at compound interest = ₹ (13,483.20 – 13,440)
= ₹ 43.20

QUESTION 5

Vasudevan invested ₹ 60,000 at an interest rate of 12% per annum compounded half yearly. What amount would he get
(i) after 6 months?
(ii) after 1 year?

ANSWER

P = ₹ 60,000

r = 12% per annum,

Since interest is compounded half yearly, so there will be two conversion period in a year,

(i) r_h = \frac{12\%}{2} per half year,

t_h = 2\times \frac{1}{2} half year

= 1 half year

\therefore \, A = P\left(1+\frac{r_h}{100}\right)^{t_h}

= ₹ 60,000\left(1+\frac{6}{100}\right)^1

= ₹ 60,000\times \frac{106}{100}

= ₹ 60,0\times 106

= ₹ 63,600

Thus, amount Vasudevan will get after 6 months = ₹ 63,600.

(ii) r_h = \frac{12\%}{2} per half year,

t_h = 2\times 1 half year

= 2 half year

\therefore \, A = P\left(1+\frac{r_h}{100}\right)^{t_h}

= ₹ 60,000\left(1+\frac{6}{100}\right)^2

= ₹ 60,000\times \frac{106\times 106}{10000}

= ₹ 6\times 11,236

= ₹ 67,416

Thus, amount Vasudevan will get after 1 years = ₹ 67,416.

QUESTION 6

Arif took a loan of ₹ 80,000 from a bank. If the rate of interest is 10% per annum, find the difference in amounts he would be paying after 1\frac{1}{2} years if the interest is
(i) compounded annually.
(ii) compounded half yearly.

ANSWER

P = ₹ 80,000

r = 10% per annum

t = 1\frac{1}{2} years = 1 and \frac{1}{2} years

(i) When Interest is compounded annually,

A for 1st year = P\left(1+\frac{r}{100}\right)^t

= ₹ 80,000\left(1+\frac{10}{100}\right)^1

= ₹ 80,000\times \frac{11}{10}

= ₹ 88,000

Now, simple interest for next \frac{1}{2} years on new P (= ₹ 88,000),

SI = ₹ \frac{88,000\times 10\times \frac{1}{2}}{100}

= ₹ 4,400

So, Amount for 1\frac{1}{2} years = ₹ (88,000 + 4,400)
= ₹ 92,400

(ii) When Interest is compounded half-yearly,

r_h = \frac{10}{2}% per half year = 5% per half year,

t_h = 2\times \frac{3}{2} half year = 3 half year,

A = P\left(1+\frac{r_h}{100}\right)^{t_h}

= ₹ 80,000\left(1+\frac{5}{100}\right)^3

= ₹ 80,000\left(\frac{21}{20}\right)^3

= ₹ 80,000\times \frac{9,261}{8,000}

= ₹ 92,610

\therefore Amount for 1\frac{1}{2} years if the interest is compounded half yearly = ₹ 92,610

So, the difference between amounts when the interest is calculated annually and when the interest is calculated half yearly,

= ₹ (92,610 – 92,400)
= ₹ 210

[ Compound Interest Examples | Ncert Solutions Exercise 8.3 Class 8 Maths ]

QUESTION 7

Maria invested ₹ 8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find
(i) The amount credited against her name at the end of the second year.
(ii) The interest for the 3rd year.

ANSWER

P = ₹ 8,000

r = 5% per annum,

(i) Amount for second year = ₹ 8,000\left(1+\frac{5}{100}\right)^2

= ₹ 8,000\left(\frac{21}{20}\right)^2

= ₹ 8,000\left(\frac{441}{400}\right)

= ₹ 20\times 441

= ₹ 8,820

(ii) Principal for 3rd year = Amount for 2nd year = ₹ 8,820,

Interest for 3rd year = ₹ \frac{8,820\times 5}{100}

= ₹ \frac{882}{2}

= ₹ 441

QUESTION 8

Find the amount and the compound interest on ₹ 10,000 for 1\frac{1}{2} years at 10% per annum, compounded half yearly. Would this interest be more than the interest he would get if it was compounded annually?

ANSWER

P = ₹ 10,000

r = 10% per annum,

t = 1\frac{1}{2} years,

Since,the interest is compounded half-yearly,

r_h = \frac{10}{2}% per half year
= 5% per half year,

t_h = 2\times \frac{3}{2} half year
= 3 half year,

A = P\left(1+\frac{r_h}{100}\right)^{t_h}

= ₹ 10,000\left(1+\frac{5}{100}\right)^3

= ₹ 10,000\left(\frac{21}{20})^3

= ₹ 10,000\times \frac{9,261}{8,000}

= ₹ 11,576.25

CI = 11,576.25 – 10,000 = ₹ 1,576.25

If the Interest is compounded annually,

Interest for 1st year = \frac{P\times r}{100}

= ₹ \frac{10,000\times 10}{100})^1

= ₹ 1000

Amount for 1st year = ₹ (10,000 + 1000)
= ₹ 11,000

Now, the interest for next \frac{1}{2} years on new P (= ₹ 11,000),

SI = ₹ \frac{11,000\times 10\times \frac{1}{2}}{100}

= ₹ 550

So, total interest if interest is calculated annually = ₹ (1000 + 550)
= ₹ 1,500

Comparing the interests in two cases,

₹ 1,576.25 > ₹ 1,500

Therefore, the interest on the sum compounded half yearly is more than the interest compounded annually.

[ Compound Interest Examples | Ncert Solutions Exercise 8.3 Class 8 Maths ]

QUESTION 9

Find the amount which Ram will get on ₹ 4096, if he gave it for 18 months at 1 12\frac{1}{2} % per annum, interest being compounded half yearly.

ANSWER

P = ₹ 4,096

t = 18 months = \frac{3}{2} years

r = \frac{25}{2}%

Since interest is compounded half yearly,

t_h = 2\times \frac{3}{2} half years
= 3 half years

r_h =  \frac{25}{2\times 2} half years
= \frac{25}{4} half years

A = P\left(1+\frac{r_h}{100}\righg)^{t_h}

= ₹ 4,096\left(1+\frac{25}{400}\right)^3

= ₹ 4,096\left(1+\frac{1}{16})^3

= ₹ 4,096\left(\frac{17}{16})^3

= ₹ 4,096\times \frac{4,913}{4,096}

= ₹ 4,913

\therefore Ram will get = ₹ 4,913 after 18 months.

QUESTION 10

The population of a place increased to 54,000 in 2003 at a rate of 5% per annum
(i) find the population in 2001.
(ii) what would be its population in 2005?

ANSWER

Population in 2003 (P_{2003}) = 54,000

r = 5% per annum,

(i) Let, the population of the place in 2001 be x,

t = 2 years,

Using Compound Interest formula,

P_{2003} = P_{2001}\left(1+\frac{r}{100}\right)^t

\implies 54,000 = x\left(1+\frac{5}{100}\right)^2

\implies 54,000 = x\left(1+\frac{1}{20}\right)^2

\implies 54,000 = x\left(\frac{21}{20}\right)^2

\implies x = 54,000\times \frac{400}{441}

\implies x = 48,980

\therefore population of the place in 2001 was 48,980.

(ii) Let, the population of the place in 2005 be y,

t = 2 years,

Using Compound Interest formula,

P_{2005} = P_{2003}\left(1+\frac{r}{100}\right)^t

\implies y = 54,000\left(1+\frac{5}{100}\right)^2

\implies y = 54,000\left(1+\frac{1}{20}\right)^2

\implies y = 54,000\left(\frac{21}{20}\right)^2

\implies y = 54,000\times \frac{441}{400}

\implies y = 135\times 441

\implies y = 59,535

\therefore population of the place in 2005 will be 59,535.

[ Compound Interest Examples | Ncert Solutions Exercise 8.3 Class 8 Maths ]

QUESTION 11

In a Laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially 5, 06,000.

ANSWER

Initial count = 5,06,000,

rate of increase = 2.5% per hour,

t = 2 hours,

Using compound interest formula,

Bacterial count after 2 hours = 5,06,000\left(1+\frac{25}{1000}\right)^2

= 5,06,000\left(1+\frac{1}{40}\right)^2

= 5,06,000\left(\frac{41}{40}\right)^2

= 5,06,000\times \frac{1681}{1600}

= 5,31,616.25

QUESTION 12

A scooter was bought at ₹ 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.

ANSWER

Initia value = ₹ 42,000

rate of depreciation = 8% per annum,

t = 1 year,

Using compound interest formula,

Value of the scooter after one year = ₹ 42,000\left(1-\frac{8}{100}\right)^1

= ₹ 42,000\left(1-\frac{2}{25}\right)

= ₹ 42,000\times \frac{23}{25}

= ₹ 1,680\times 23

= ₹ 38,640

\therefore Value of the scooter after one year = ₹ 38,640.


NCERT SOLUTIONS CLASS 8 MATHS CHAPTER 8 COMPARING QUANTITIES

Check NCERT Solutions of other exercises from Class 8 Maths Chapter 8 Comparing Quantities by clicking the links given below.

• Exercise 8.1
• Exercise 8.2
• Exercise 8.3

We hope Compound Interest Examples | Ncert Solutions Exercise 8.3 Class 8 Maths has helped you understand how to solve Compound Interest Problems.

Write in the comment section for any error or any solution related queries from the exercise. [ Compound Interest Examples | Ncert Solutions Exercise 8.3 Class 8 Maths ]

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