## 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

where, P is principal, r is rate of interest, t is time period.

• Amount when interest is compounded half yearly

where, r/2 is half yearly rate and
2t = number of ’half-years’

### EXERCISE 8.3Application of Compound Interest Formula

QUESTION 1

Calculate the amount and compound interest on

(a) ₹ 10,800 for 3 years at % per annum compounded annually.

(b) ₹ 18,000 for years at 10% per annum compounded annually.

(c) ₹ 62,500 for 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.

(a) P = ₹ 10,800
t = 3 years,
r = % per annum compounded annually,

Amount =

= ₹

= ₹

= ₹

= ₹

= ₹

= ₹ 15,377.34

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

(b) P = ₹ 18,000
t = years,
r = 10 % per annum compounded annually,

Amount =

Amount for 2 years = ₹

= ₹

= ₹

= ₹

= ₹

= ₹ 21,780

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

Amount for 2 years will act as Principal for next years,

SI on ₹ 21,780 = ₹

= ₹ 1,089

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

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

(c) P = ₹ 62,500,

t = 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, years

and per half-year,

Now,

Required Amount =

= ₹

= ₹

= ₹

= ₹

= ₹

= ₹ 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, years

and per half-year,

Now,

Required Amount =

= ₹

= ₹

= ₹

= ₹

= ₹ 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, years

and per half-year,

Now,

Required Amount =

= ₹

= ₹

= ₹

= ₹

= ₹

= ₹ 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 years).

Principal, P = ₹ 26,400

Rate, r = 15% p.a.,

Time period, t = 2 years 4 months
= years = years

Amount for 2 years = ₹

= ₹

= ₹

= ₹

= ₹

= ₹

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

= ₹ 8,514

Amount after 2 years will serve as Principal for next years,

Simple Interest for next years =

= ₹

= ₹

= ₹

= ₹ 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?

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, per annum,

Since Fabina borrowed the money at simple interest,

therefore, SI to be paid by Fabina = ₹

= ₹

= ₹ 4,500

Rate at which Radha borrowed money, per annum,

Since Radha borrowed the money at compound interest,

So, Amount to be paid by Radha = ₹

= ₹

= ₹

= ₹ 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?

P = ₹ 12,000

r = 6% per annum,

t = 2 years,

SI =

= ₹

= ₹

= ₹ 1,440

Amount = P + SI

= ₹ (12,000 + 1,440)

= ₹ 13,440,

Had the money borrowed at compound interest,

Amount to be paid = ₹

= ₹

= ₹

= ₹ 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?

P = ₹ 60,000

r = 12% per annum,

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

(i) per half year,

half year

= 1 half year

= ₹

= ₹

= ₹

= ₹ 63,600

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

(ii) per half year,

half year

= 2 half year

= ₹

= ₹

= ₹

= ₹ 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 years if the interest is
(i) compounded annually.
(ii) compounded half yearly.

P = ₹ 80,000

r = 10% per annum

t = years = 1 and years

(i) When Interest is compounded annually,

A for 1st year =

= ₹

= ₹

= ₹

Now, simple interest for next years on new P (= ₹ 88,000),

SI = ₹

= ₹ 4,400

So, Amount for years = ₹ (88,000 + 4,400)
= ₹ 92,400

(ii) When Interest is compounded half-yearly,

% per half year = 5% per half year,

half year = 3 half year,

A =

= ₹

= ₹

= ₹

= ₹

Amount for 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.

P = ₹ 8,000

r = 5% per annum,

(i) Amount for second year = ₹

= ₹

= ₹

= ₹

= ₹

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

Interest for 3rd year = ₹

= ₹

= ₹

QUESTION 8

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

P = ₹ 10,000

r = 10% per annum,

t = years,

Since,the interest is compounded half-yearly,

% per half year
= 5% per half year,

half year
= 3 half year,

A =

= ₹

= ₹

= ₹

= ₹

CI = 11,576.25 – 10,000 = ₹ 1,576.25

If the Interest is compounded annually,

Interest for 1st year =

= ₹

= ₹

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

Now, the interest for next years on new P (= ₹ 11,000),

SI = ₹

= ₹ 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 % per annum, interest being compounded half yearly.

P = ₹ 4,096

t = 18 months = years

r = %

Since interest is compounded half yearly,

half years
= 3 half years

half years
= half years

A =

= ₹

= ₹

= ₹

= ₹

= ₹

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?

Population in 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,

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,

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.

Initial count = 5,06,000,

rate of increase = 2.5% per hour,

t = 2 hours,

Using compound interest formula,

Bacterial count after 2 hours =

=

=

=

=

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.

Initia value = ₹ 42,000

rate of depreciation = 8% per annum,

t = 1 year,

Using compound interest formula,

Value of the scooter after one year = ₹

= ₹

= ₹

= ₹

= ₹

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

## NCERT SOLUTIONS CLASS 8 MATHS CHAPTER 8 COMPARING QUANTITIES

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