AP State Syllabus AP Board 8th Class Maths Solutions Chapter 5 Comparing Quantities Using Proportion Ex 5.3 Textbook Questions and Answers.

## AP State Syllabus 8th Class Maths Solutions 5th Lesson Comparing Quantities Using Proportion Exercise 5.3

Question 1.

Sudhakar borrows ₹ 15000 from a bank to renovate his house. He borrows the money at 9% p.a. simple interest over 8 years. What are his monthly repayments’?

Solution:

P = 15,000

R = 9%

T = 8 years

A = ₹ 25800

∴ His monthly payment = \(\frac{25800}{8 \times 12}\)

= ₹268.75

∴ Monthly he has to pay = ₹268.75

Question 2.

A TV was bought at a price of ₹ 21000. After 1 year the value of the TV was depreciated by 5% (Depreciation means reduction of the value due to use and age of the item). Find the value of the TV after 1 year.

Solution:

The C.P. of T.V = ₹ 21,000.

After 1 year its value

= 21000 – 5% of 21000

=21000 – \(\frac { 5 }{ 100 }\) × 21000

= 21000 – 1050

= ₹19,950

Question 3.

Find the amount and the compound interest on ₹ 8000 at 5% per annum, for 2 years

compounded annually.

Solution:

P = ₹8000

R = 5%

The interest is compounded every year.

Then 2 time periods wII be occurred.

∴ n = 2

∴ Amount (A) = ₹8820

C.I = A – P

= 8820 – 8000 = ₹ 820

Question 4.

Find the amount and the compound interest on ₹ 6500 for 2 years, compounded annually, the rate of interest being 5% per annum during the first year and 6% per annum during the second year.

Solution:

P = ₹ 6500

R = 5%

T = 1 years

∴ \(\frac{\mathrm{PTR}}{100}=\frac{6500 \times 5 \times 1}{100}\) = 325

∴ A = P + I = 6500 + 325 = 6825

∴ P = 6825

(At the begining of 2,id year A=P)

R = 6%

T = 1 year

∴ \(\frac{\mathrm{PTR}}{100}=\frac{6825 \times 6 \times 1}{100}\) = 409.5

∴ A = P + I = 6825 + 409.5

∴ Amount = ₹ 7234.50

C.I. = A – P

= ₹ 7234.50 – 6500

= ₹734.50

Question 5.

Prathibha borrows ₹47000 from a fmance company to buy her first car. The rate of simple interest is 17% and she borrows the money over a 5 year period. Find: (a) How much

amount Prathibha should repay the finance company at the end of five years. (b) her equal

monthly repayments.

Solution:

P = ₹ 47000

R = 17%

T =5 years

∴ I = \(\frac{\mathrm{PTR}}{100}=\frac{47000 \times 5 \times 17}{100}\)

= ₹ 39,950

a) Amount to be paid

A = P + I

= 47000 + 39,950

= 86950

∴ Amount to be pay = ₹ 86950

b) In monthly equal instalments she has to pay

= 149.1

= ₹ 1450 (approx)

Question 6.

The population of Hyderabad was 68,09,000 in the year 2011. If it increases at the rate of 4.7% per annum. What will be the population at the end of the year 2015.

Solution:

The population of Hyderabad

= 68,09,000

If every year increase in 4.7%.

Then the population of the city in 2015

= 68,09,000 ( 1 + \(\frac{4.7}{100}\) )^{4}

100 J

[ ∵ P = 6809000, R = 4.7 %, n = 4(2015 -2011)]

= 68,09,000 x \(\frac{104.7}{100} \times \frac{104.7}{100} \times \frac{104.7}{100} \times \frac{104.7}{100}\)

= 81,82,199

Question 7.

Find Compound interest paid when a sum of ₹ 10000 is invested for 1 year and 3 months at 8\(\frac{1}{2}\) % per annum compounded annually.

Solution:

P = ₹10,000; R = 8\(\frac { 1 }{ 2 }\) % = \(\frac { 17 }{ 2 }\)%

T = 1 year

= 50 × 17 = 850

∴ I = ₹ 850

∴ A = P + I = 10,000 + 850

A = 10,850

∴ P = 10,850; R = \(\frac { 17 }{ 2 }\)% % ; T = 3 months

= ₹ 230.50

∴ Compound Interest

= 850 + 230.50

= ₹ 1080.50

Question 8.

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.

Solution:

P = ₹ 80,000; R = 10%;

T = 1 year

∴ \(\frac{\mathrm{PTR}}{100}\) = \(\frac{80000 \times 10 \times 1}{100}\)

= ₹8000

∴ A = P + I = 80000 + 8000

= ₹ 88,000

Interest on 6 months :

P = 88,000 ; R = 10% ; T = 6 Months

= \(\frac { 1 }{ 2 }\) year

i) The amount to be paid after 1 year 6 months = P + I

= 88000 + 4400

A_{1} = ₹ 92,400

ii) He has to pay compounded on

every 6 months in 1 \(\frac { 1 }{ 2 }\) years

∴ 3 time periods will be occurred.

∴ n = 3

R = \(\frac { 10 }{ 2 }\) = 5% P = ₹ 80,000

A_{2} = ₹ 92610

∴ Difference between the amounts = A_{2} – A_{1} = 92610 – 92400 = ₹ 210

Question 9.

I borrowed ₹ 12000 from Prasad at 6°/o per annum simple interest for 2 years. Had

I borrowed this sum at 6% per annum compounded annually, what extra amount would

I have to pay9

Solution:

Sum borrowed from Prasad

P = ₹ 12000

T = 2 years;

R = 6%

= ₹144O

A = P + I

A_{1} = P + I = 12000 + 1440

= ₹13440

12000 + 1440 , = ₹ 13440

∴ He has to pay the amount after 2 years at the rate of 6% on C.I.

P = ₹12,000; R = 6%; n = 2 years

A_{2} = ₹13483.2

∴ The difference between the C.I and S.I = 13483.2 – 13440

= ₹ 43.20

Question 10.

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

Solution:

No. of bacteria in a laboratory = 5,06,000

If they are increased at the rate of 2.5% per hour then their number after 2 hours

Question 11.

Kamala borrowed ₹ 26400 from a bank to buy a scooter at a rate of 15% per annum compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan?

Solution:

Kanala borrowed from bank = ₹ 26400

Rah of interest (R) =15%

n = 2 years

After 4 rpnths the amount will be ₹ 34914

∴ P = 34914; R = 15%; T = 4 months

T = \(\frac { 4 }{ 12 }\) year

= \(\frac { 1 }{ 3 }\) year

∴ \(I=\frac{P T R}{100}=\frac{34914 \times 15 \times \frac{1}{3}}{100}\)

= ₹1745.7

∴ Kamala has to pay the amount after 2 years and 4 months to the bank = 34914 + 1745.7

= ₹36659.7

Question 12.

Bharathi borrows an amount of ₹ 12500 at 12% per annum for3 years at a simple interest and Madhuri borrows the same amount for the same time period at 10% per annum, compounded annually. Who pays more interest and by how much?

Solution:

Bharathi borrowed the sum

P = ₹12500

R = 12%

T = 3 years

S. I (I) = \(\frac { PTR }{ 100 }\)

= \(\frac{12500 \times 12 \times 3}{100}\)

= 125 × 36

= 4500

After 3 years she has to pay

(A_{1})= P + I

= 12500 + 4500 .

A_{1} = ₹17,000

Madhuri has to pay the amount on

A_{2} = 16637.5

∴ A_{1} > A_{2}

A_{1} – A_{2} = 17000 – 16637.5

= ₹ 362.5

∴ Bharathi has to pay ₹ 362.5 more than Madhuri.

Question 13.

Machinery worth ₹ 10000 depreciated by 5%. Find its value after 1 year.

Solution:

The value of machinery after 1 year on 5% depreciation

= 95 × 100

= ₹ 9500

Question 14.

Find the population of a city after 2 years which is at present 12 lakh, if the rate of increase is 4%.

Solution:

Present population of a city = 12,00,000 If its population increases at the rate of 4%, then the population after 2 years

= 120 × 104 × 104

= 12,97,920

Question 15.

Calculate compound interest on ₹ 1000 over a period of 1 year at 10% per annum, if interest is compounded quarterly?

Solution:

compounded quarterly then 4 time periods will be there in 1 year.

∴ n = 4

C.I. on ₹ 1000 over a period of 1 year at

10% per annum A = P (1 + \(\frac{\mathrm{R}}{100}\) )^{n}

P = 1000; n = 4; R = \(\frac{10}{4}=\frac{5}{2}\) %

= ₹ 1103.81

A = ₹ 1103.81

C.I. for 1 year

= 1103.81 – 1000

= ₹ 10.81