## NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Ex 8.3

- Class 8 Maths Comparing Quantities Exercise 8.1
- Class 8 Maths Comparing Quantities Exercise 8.2
- Class 8 Maths Comparing Quantities Exercise 8.3
- Comparing Quantities Class 8 Extra Questions

**NCERT Solutions for Class 8 Maths Chapter 8 Comparing Quantities Exercise 8.3**

Ex 8.3 Class 8 Maths Question 1.

Calculate the amount and compound interest on

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

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

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

(d) â‚¹ 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) â‚¹ 10,000 for 1 year at 8% per annum compounded half yearly.

Solution:

(a) Given:

P = â‚¹ 10,800, n = 3 years,

CI = A – P = â‚¹ 15,377.35 – â‚¹ 10,800 = â‚¹ 4,577.35

Hence amount = â‚¹ 15,377.34 and CI = â‚¹ 4,577.34

(b) Given: P = â‚¹ 18,000, n = 2\(\frac { 1 }{ 2 }\) years = \(\frac { 5 }{ 2 }\) years

R = 10% p.a.

The amount for 2\(\frac { 1 }{ 2 }\) years, i.e., 2 years and 6 months can be calculated by first calculating the amount to 2 years using CI formula and then calculating the simple interest by using SI formula.

The amount for 2 years has to be calculated

Total CI = â‚¹ 3780 + â‚¹ 1089 = â‚¹ 4,869

Amount = P + I = â‚¹ 21,780 + â‚¹ 1,089 = â‚¹ 22,869

Hence, the amount = â‚¹ 22,869

and CI = â‚¹ 4,869

(c) Given: P = â‚¹ 62,500, n = 1\(\frac { 1 }{ 2 }\) years = \(\frac { 3 }{ 2 }\) years per annum compounded half yearly

= \(\frac { 3 }{ 2 }\) Ã— 2 years = 3 half years

R = 8% = \(\frac { 8 }{ 2 }\) % = 4% half yearly

CI = A – P = â‚¹ 70,304 – â‚¹ 62,500 = â‚¹ 7,804

Hence, amount = â‚¹ 70304 and CI = â‚¹ 7804

(d) Given: P = â‚¹ 8,000, n = 1 years R = 9% per annum compounded half yearly

Since, the interest is compounded half yearly n = 1 Ã— 2 = 2 half years

CI = A – P = â‚¹ 8,736.20 – â‚¹ 8,000 = â‚¹ 736.20

Hence, the amount = â‚¹ 8736.20 and CI = â‚¹ 736.20

(e) Given: P = â‚¹ 10,000, n = 1 year and R = 8% pa compounded half yearly

Since the interest is compounded half yearly n = 1 Ã— 2 = 2 half years

CI = A – P = â‚¹ 10,816 – â‚¹ 10,000 = â‚¹ 816

Hence the amount = â‚¹ 10,816 and Cl = â‚¹ 816

Ex 8.3 Class 8 MathsÂ Question 2.

Kamala borrowed â‚¹ 26,400 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? (Hint: Find amount for 2 years with interest is compounded yearly and then find SI on the 2nd

year amount for \(\frac { 4 }{ 12 }\) years).

Solution:

Given:

P = â‚¹ 26,400

R = 15% p.a. compounded yearly

n = 2 years and 4 months

Amount after 2 years and 4 months = â‚¹ 34,914 + â‚¹ 1745.70 = â‚¹ 36,659.70

Hence, the amount to be paid by Kamla = â‚¹ 36,659.70

Ex 8.3 Class 8 MathsÂ Question 3.

Fabina borrows â‚¹ 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?

Solution:

For Fabina: P = â‚¹ 12,500, R = 12% p.a. and n = 3 years

Difference between the two interests = â‚¹ 4500 – â‚¹ 4137.50 = â‚¹ 362.50

Hence, Fabina pays more interest by â‚¹ 362.50.

Ex 8.3 Class 8 MathsÂ Question 4.

I borrowed â‚¹ 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?

Solution:

Given: P = â‚¹ 12,000, R = 6% p.a., n = 2 years

Difference between two interests = â‚¹ 1483.20 – â‚¹ 1440 = â‚¹ 43.20

Hence, the extra amount to be paid = â‚¹ 43.20

Ex 8.3 Class 8 MathsÂ Question 5.

Vasudevan invested â‚¹ 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?

Solution:

(i) Given: P = â‚¹ 60,000, R = 12% p.a. compounded half yearly

Hence, the required amount = â‚¹ 67416

Ex 8.3 Class 8 MathsÂ Question 6.

Arif took a loan of â‚¹ 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:

(i) Given: P = â‚¹ 80,000

R = 10% p.a.

n = 1\(\frac { 1 }{ 2 }\) years

Since the interest is compounded annually

Difference between the amounts = â‚¹ 92,610 – â‚¹ 92,400 = â‚¹ 210

Ex 8.3 Class 8 MathsÂ Question 7.

Maria invested â‚¹ 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 third year.

Solution:

(i) Given: P = â‚¹ 8,000, R = 5% p.a.

and n = 2 years

Hence, interest for the third year = â‚¹ 441

Ex 8.3 Class 8 MathsÂ Question 8.

Find the amount and the compound interest on â‚¹ 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?

Solution:

Given: P = â‚¹ 10,000, n = 1\(\frac { 1 }{ 2 }\) years

R = 10% per annum

Since the interest is compounded half yearly

Total interest = â‚¹ 1,000 + â‚¹ 550 = â‚¹ 1,550

Difference between the two interests = â‚¹ 1,576.25 – â‚¹ 1,550 = â‚¹ 26.25

Hence, the interest will be â‚¹ 26.25 more when compounded half yearly than the interest when compounded annually.

Ex 8.3 Class 8 MathsÂ Question 9.

Find the amount which Ram will get on â‚¹ 4,096, if he gave it for 18 months at 12\(\frac { 1 }{ 2 }\) per annum, interest being compounded half yearly.

Solution:

Given: P = â‚¹ 4,096, R = 12\(\frac { 1 }{ 2 }\) % pa, n = 18 months

Hence, the required amount = â‚¹ 4913

Ex 8.3 Class 8 MathsÂ 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?

Solution:

(i) Given: Population in 2003 = 54,000

Rate = 5% pa

Time = 2003 – 2001 = 2 years

Population in 2003 = Population in 2001

Ex 8.3 Class 8 MathsÂ 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.

Solution:

Given: Initial count of bacteria = 5,06,000

Rate = 2.5% per hour

n = 2 hours

Number of bacteria at the end of 2 hours = Number of count of bacteria initially

Thus, the number of bacteria after two hours = 5,31,616 (approx).

Ex 8.3 Class 8 MathsÂ Question 12.

A scooter was bought at â‚¹ 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.

Solution:

Given: Cost price of the scooter = â‚¹ 42,000

Rate of depreciation = 8% p.a.

Time = 1 year

Final value of the scooter

Hence, the value of scooter after 1 year = â‚¹ 38,640.

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