Class 8 NCERT Solutions Chapter 8 Comparing Quantities Exercise 8.3 (original) (raw)

Last Updated : 23 Jul, 2025

In this section, we explore Chapter 8 of the Class 8 NCERT Mathematics textbook, which focuses on Comparing Quantities. This chapter introduces students to concepts like percentages, profit and loss, discounts, and simple interest. Exercise 8.3 specifically deals with problems related to calculating simple interest, profit, loss, and discounts, helping students apply these concepts in real-life situations.

Class 8 NCERT Solutions - Chapter 8 Comparing Quantities - Exercise 8.3

This section provides detailed solutions for Exercise 8.3 from Chapter 8 of the Class 8 NCERT Mathematics textbook. These solutions are designed to help students understand and solve problems related to comparing quantities, including calculations of simple interest, profit and loss, and discounts.

**Question 1. Calculate the amount and compound interest on

****(i) Rs 10,800 for 3 years at 12**\mathbf{\frac{1}{2}} ****% per annum compounded annually.**

****(ii) Rs 18,000 for 2**\mathbf{\frac{1}{2}} **years at 10% per annum compounded annually.

****(iii) Rs 62,500 for 1**\mathbf{\frac{1}{2}} **years at 8% per annum compounded half-yearly.

****(iv) Rs 8,000 for 1 year at 9% per annum compounded half-yearly.**

****(v) Rs 10,000 for 1 year at 8% per annum compounded half-yearly.**

**Solution:

****(i) Given values are,**

P = Rs 10,800

R = 12 \frac{1}{2} % per annum = \frac{25}{2} %

T = 3 Years

As it is compounded annually then, n = 3 times.

We have,

**A = P (1 + \mathbf{\frac{R}{100}} ****)** n

A = 10,800 (1+ \frac{25}{2*100} )3

A = 10,800 (1+ \frac{1}{8} )3

A = 10,800 (\frac{9}{8} )3

**A = Rs 15,377.34

**CI = A - P

CI = 15,377.34 - 10,800

**CI = Rs 4,577.34

**Hence, the amount = Rs 15,377.34 and

**Compound interest = Rs 4,577.34

****(ii) Given values are,**

P = Rs 18,000

R = 10 % per annum

T = 2\frac{1}{2} Years

As it is compounded annually then, n = 2\frac{1}{2} times.

We have,

**A = P (1 + \mathbf{\frac{R}{100}} ****)** n

A = 18,000 (1+ \frac{10}{100} )2½

What we will do here is Firstly we know 2\frac{1}{2} Years is 2 years and 6 months which 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 :

A = 18,000 (1+ \frac{1}{10} )2

A = 18,000 (\frac{11}{10} )2

**A = Rs 21,780

**CI = A - P

CI = 21,780 - 18,000

**CI = Rs 3,780

Now, The amount for \frac{1}{2} year has to be calculated:

New P is equal to the amount after 2 Years. Hence,

P = Rs 21,780

R = 10 % per annum

T = \frac{1}{2} year

SI = \frac{PRT}{100}

SI = \frac{21,780 × 10 × \frac{1}{2}}{100}

SI = \frac{21,780 × 10 × 1}{200}

**SI = Rs 1,089

**Hence, the Total amount = A + SI

= 21,780 + 1,809

= Rs 22,869

**Total compound interest = CI + SI

= 3,780 + 1,809

= Rs 4,869

****(iii) Given values are,**

P = Rs 62,500

R = 8 % per annum hence 4% Half Yearly

T = 1\frac{1}{2} Years

As it is compounded Half yearly then, n = 3 times. (1\frac{1}{2} Years contains 3 half years)

We have,

**A = P (1 + \mathbf{\frac{R}{100}} ****)** n

A = 62,500 (1+ \frac{4}{100} )3

A = 62,500 (1+ \frac{1}{25} )3

A = 62,500 (\frac{26}{25} )3

**A = Rs 70,304

**CI = A - P

CI = 70,304 - 62,500

**CI = Rs 7,804

**Hence, the amount = Rs 70,304 and

**Compound interest = Rs 7,804

****(iv) Given values are,**

P = Rs 8,000

R = 9 % per annum hence, \frac{9}{2} % Half Yearly

T = 1 Year

As it is compounded Half yearly then, n = 2 times. (1 Year contains 2 half years)

We have,

**A = P (1 + \mathbf{\frac{R}{100}} ****)** n

A = 8,000 (1+ \frac{9}{2*100} )2

A = 8,000 (1+ \frac{9}{200} )2

A = 8,000 (\frac{209}{200} )2

**A = Rs 8,736.20

**CI = A - P

CI = 8,736.20 - 8,000

**CI = Rs 736.20

**Hence, the amount = Rs 8,736.20 and

**Compound interest = Rs 736.20

****(v) Given values are,**

P = Rs 10,000

R = 8 % per annum hence, 4% Half Yearly

T = 1 Year

As it is compounded Half yearly then, n = 2 times. (1 Year contains 2 half years)

We have,

**A = P (1+ \mathbf{\frac{R}{100}} ****)** n

A = 10,000 (1+ (\frac{4}{100} ))2

A =10,000 (1+ (\frac{1}{25} ))2

A = 10,000 (\frac{26}{25} )2

**A = Rs 10,816

**CI = A - P

CI = 10,816- 10,000

**CI = Rs 816

**Hence, the amount = Rs 10,816 and

**Compound interest = Rs 816

**Question 2. Kamala borrowed Rs 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?

**Solution:

Here, Given values are,

P = Rs 26,400

R = 15 % per annum

T = 2 Years and 4 months, which is 2 \frac{1}{3} years

As it is compounded annually then, n = 2\frac{1}{3} times

We have,

**A = P (1 + \mathbf{\frac{R}{100}} ****)** n

A = 26,400 (1 + (\frac{15}{100} )2(1/3)

What we will do here is Firstly 2 years and 4 months which 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:

A = 26,400 (1+ (\frac{15}{100} )2

A = 26,400 (1+ (\frac{3}{20} )2

A = 26,400 (\frac{23}{20} )2

**A = Rs 34,914

Now, The amount for (1/3) year (4 months) has to be calculated :

New P is equal to the amount after 2 Years. Hence,

P = Rs 34,914

R = 15 % per annum

T = \frac{1}{3} year

SI = \frac{P × R × T}{100}

SI = \frac{(34,914 × 15 × (1/3)}{100}

SI = \frac{(34,914 × 15 × 1)}{300}

**SI = 1,745.70

Hence, the Total amount = **A + SI

= 34,914 + 1,745.70

= Rs 36,659.70

**Hence, the amount to be paid by Kamla = ₹ 36,659.70

**Question 3. Fabina borrows Rs 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:

Let's see each case

**Fabina Case: at simple interest

P = 12,500

R = 12% per annum

T = 3 Years

SI = \frac{(P × R × T)}{100}

SI = \frac{(12,500 × 12 × 3)}{100}

**SI = Rs 4,500

**Radha Case: at compound interest

P = 12,500

R = 10% per annum

T = 3 Years

As it is compounded annually then, n = 3 times

We have,

**A = P (1 + \mathbf{\frac{R}{100}} ****)** n

A = 12,500 (1 + (\frac{10}{100} ))3

A =12,500 (1 + \frac{1}{10} )3

A = 12,500 (\frac{11}{10} )3

A = Rs 16,637.5

**CI = A - P

CI = 16,637.5 - 12,500

**CI = 4,137.5

Clearly we can see that **Fabina paid more interest, and she paid

4,500 - 4,137.5 = **Rs 362.5 more than Radha

**Solution:

Lets see each case First

**At simple interest

P = 12,000

R = 6% per annum

T = 2 Years

SI = \frac{(P × R × T)}{100}

SI = \frac{(12,000 × 6 × 2)}{100}

SI = Rs 1,440

**At compound interest

P = 12,000

R = 6% per annum

T = 2 Years

As it is compounded annually then, n = 2 times

We have,

**A = P (1 + \mathbf{\frac{R}{100}} ****)** n

A = 12,000 (1+ (\frac{6}{100} ))2

A =12,000 (1+ (\frac{3}{50} ))2

A = 12,000 (\frac{53}{50} )2

**A = Rs 13,483.2

CI = A - P

CI = 13,483.2 - 12,000

**CI = 1,483.2

Clearly we can see that,

1,483.2 - 1,440 = Rs 43.2

Hence, **the extra amount to be paid = ₹ 43.20

**Question 5. Vasudevan invested Rs 60,000 at an interest rate of 12% per annum compounded half-yearly. What amount would he get

****(a) after 6 months?**

****(b) after 1 year?**

**Solution:

Let's see each case

****(a)**

P = 60,000

R = 12% per annum (6% Half yearly)

T = 6 Months

As it is compounded Half Yearly then, n = 1 times (as 6 months is 1 half year)

We have,

**A = P (1 + \mathbf{\frac{R}{100}} ****)** n

A =60,000 (1+ (\frac{6}{100} ))1

A =60,000 (1+ (\frac{3}{50} ))1

A = 60,000 (\frac{53}{25} )1

**A = Rs 63,600

**He would get Rs 63,600 after 6 Months.

****(b)**

P = 60,000

R = 12% per annum (6% Half yearly)

T = 1 Year

As it is compounded Half Yearly then, n = 2 times (as 1 Year is 2 half year)

We have,

**A = P (1 + \mathbf{\frac{R}{100}} ****)** n

A = 60,000 (1+ (\frac{6}{100} ))2

A = 60,000 (1+ (\frac{3}{50} ))2

A = 60000 (\frac{53}{25} )2

**A = Rs 67,416

**He would get Rs 67,416 after 1 Year.

**Question 6. Arif took a loan of Rs 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

****(a) compounded annually.**

****(b) compounded half-yearly.**

**Solution:

Let's see each case

****(a) Compounded Annually**

P = 80,000

R = 10% per annum

T = 1\frac{1}{2} Year

As it is compounded annually then, n = 1 \frac{1}{2} times

We have,

**A = P (1 + \mathbf{\frac{R}{100}} ****)** n

A = 80,000 (1 + (\frac{10}{100} )1½

What we will do here is Firstly we know 1\frac{1}{2} Years is 1 year and 6 months which can be calculated by first calculating the amount to 1 year using CI formula and then calculating the simple interest by using SI formula.

The amount for 1 years has to be calculated :

A = 80,000 (1+ (\frac{10}{100} ))1

A = 80,000 (1+ (\frac{1}{10} )1

A = 80,000 (\frac{11}{10} )1

**A = Rs 88,000

Now, The amount for \frac{1}{2} Year (6 months) has to be calculated :

New P is equal to the amount after 1 Year. Hence,

P = Rs 88,000

R = 10 % per annum

T =\frac{1}{2} Year

SI = \frac{(P × R × T)}{ 100}

SI = \frac{(88,000 × 10 × \frac{1}{2})}{100}

SI = \frac{(88,000 × 10 × 1)}{200}

**SI = 4,400

Hence, the Total amount = **A + SI

= 88,000 + 4,400

= Rs 92,400

****(b) Compounded Half-yearly**

P = 80,000

R = 10% per annum (5 % Half Yearly)

T = 1\frac{1}{2} Year

As it is compounded annually then, n = 3 times (as 1\frac{1}{2} Year is 3 half year)

We have,

**A = P (1 + \mathbf{\frac{R}{100}} ****)** n

A = 80,000 (1+ (\frac{5}{100} )3

A = 80,000 (1+ (\frac{1}{20} )3

A = 80,000 (\frac{21}{20} )3

**A = Rs 92,610

**Hence, the Total amount = Rs 92,610

**Question 7. Maria invested Rs 8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find

****(a) The amount credited against her name at the end of the second year.**

****(b) The interest for the 3rd year.**

**Solution:

Let's see each case

Here,

P = 8,000

R = 5% Per annum

****(a) The amount credited against Maria's name at the end of the second year.**

T = 2 Year

As it is compounded annually then, n = 2 times

We have,

**A = P (1 + \mathbf{\frac{R}{100}} ****)** n

A = 8,000 (1+ (\frac{5}{100} ))2

A = 8,000 (1+ (\frac{1}{20} ))2

A = 8,000 (\frac{21}{20} )2

A = Rs 8,820

**Hence, the amount credited against Maria's name at the end of the second year = Rs 8,820

****(b) The interest for the 3rd year.**

T = 3 Year

As it is compounded annually then, n = 3 times

We have,

**A = P (1+ \mathbf{\frac{R}{100}} ****)** n

A = 8,000 (1+ (\frac{5}{100} ))3

A = 8,000 (1+ (\frac{1}{20} ))3

A = 8,000 (\frac{21}{20} )3

A = Rs 9,261

**The interest for the 3rd year = Amount after 3 years - Amount after 2 Years

= 9,261 - 8,820

= Rs 441

**Another Solution for (b)

As we can calculate interest of 3rd year by having 2nd Year Amount as P.

P = 8,820

R = 5% per annum

T = 1 Year (2nd to 3rd year)

**SI = \frac{(P × R × T)}{100}

SI = \frac{(8,820 × 5 × 1)}{100}

**SI = Rs 441

**The interest for the 3rd year = Rs 441

**Question 8. Find the amount and the compound interest on Rs 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:

Let's see each cases

**Compounded Annually

P = 10,000

R = 10% per annum

T = 1\frac{1}{2} Year

As it is compounded annually then, n = 1 \frac{1}{2} times

We have,

**A = P (1 + \mathbf{\frac{R}{100}} ****)** n

A = 10,000 (1 + (\frac{10}{100} )1½

What we will do here is Firstly we know 1½ Years is 1 year and 6 months which can be calculated by first calculating the amount to 1 year using CI formula and then calculating the simple interest by using SI formula.

The amount for 1 year has to be calculated:

A = 10,000 (1 + \frac{10}{100} )1

A = 10,000 (1+ \frac{1}{10} )1

A = 10,000 (\frac{11}{10} )1

A = Rs 11,000

CI = A - P

CI = 11,000-10,000

**CI = 1,000

Now, The amount for \frac{1}{2} Year (6 months) has to be calculated :

New P is equal to the amount after 1 Year. Hence,

P = Rs 11,000

R = 10 % per annum

T =\frac{1}{2} Year

SI = \frac{(P × R × T)}{100}

SI = \frac{(11,000 × 10 × \frac{1}{2})}{100}

SI = \frac{11,000 × 10}{200}

**SI = 550

**Hence, the Total Interest (compounded annually)= CI + SI

= 1,000 + 550

= Rs 1,550

**Compounded Half-yearly

P = 10,000

R = 10% per annum (5 % Half Yearly)

T = 1\frac{1}{2} Year

As it is compounded annually then, n = 3 times (as 1\frac{1}{2} Year is 3 half year)

We have,

**A = P (1 + \mathbf{\frac{R}{100}} ****)** n

A = 10,000 (1 + (\frac{5}{100} )3

A = 10,000 (1+ \frac{1}{20} )3

A = 10,000 (\frac{21}{20} )3

A = Rs 11,576.25

CI = A - P

CI = 11,576.25 - 10,000

CI = 1,576.25

**Hence, the Total Interest (compounded Half Yearly) = Rs 11576.25

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

**Hence, the interest will be Rs 26.25 more when compounded half-yearly than the interest when compounded annually.

Question 9. Find the amount which Ram will get on Rs 4096, if he gave it for 18 months at 12 \frac{1}{2} % per annum, interest being compounded half-yearly.**

**Solution:

**Let's see this case

P = Rs 4,096

R = 12 \frac{1}{2} % per annum (\frac{25}{4} % Half yearly)

T = 18 Months = 1\frac{1}{2} Year

As it is compounded Half yearly then, n = 3 Times

We have,

**A = P (1 + \mathbf{\frac{R}{100}} ****)** n

A = 4,096 (1+ (\frac{\frac{25}{4}}{100} )3

A = 4,096 (1+ \frac{25}{400} )3

A = 4,096 (1+ (\frac{1}{16} )3

A = 4,096 (\frac{17}{16} )3

**A = Rs 4,913

**Ram will get the amount = Rs 4,913

**Question 10. The population of a place increased to 54,000 in 2003 at a rate of 5% per annum

****(a) find the population in 2001.**

****(b) what would be its population in 2005?**

**Solution:

**Here,

P = 54,000 (in 2003)

R = 5% per annum

****(a) Population in 2001**

T = 2 Years (back)

n = 2

**Population in 2003 = Population in 2001 (1 + \mathbf{\frac{R}{100}} ****)** n

54,000 = P1 (1+(\frac{5}{100} ))2

54,000 = P1 (\frac{21}{20} )2

54,000 = P1 (\frac{441}{400} )

P1 = 54,000 (\frac{441}{400} )

P1 = 48,979.59

P1 = 48,980 (approx.).

**Population in 2001 was 48,980 (approx.).

****(b) Population in 2005**

T = 2 Years

n = 2

We have,

**A = P (1 + \mathbf{\frac{R}{100}} ****)** n

A = 54,000 (1+ \frac{5}{100} )2

A = 54,000 (1+ (\frac{1}{20} )2

A = 54,000 (\frac{21}{20} )2

**A = 59,535

**Population in 2005 will be 59,535

**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:

**Here,

P = 5,06,000

R = 2.5% per hour

T = 2 hours

We have,

**A = P (1 + \mathbf{\frac{R}{100}} ****)** n

A = 5,06,000 (1+ \frac{2.5}{100} )2

A = 5,06,000 (1+ \frac{25}{1000} )2

A = 5,06,000 (1+ \frac{1}{40} )2

A = 5,06,000 (\frac{41}{40} )2

A = 5,31,616.25

**A = 5,31,616 (approx.)

**Bacteria at the end of 2 hours = 5,31,616 (approx.)

**Question 12. A scooter was bought at Rs 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.

**Solution:

**Here,

P = 42,000

R = 8% per annum (depreciated)

T = 1 Year

We have,

**A = P (1 + \mathbf{\frac{R}{100}} ****)** n

A = 42,000 (1- \frac{8}{100} )1 (negative sign because the price is reduced)

A = 42,000 (1- (\frac{2}{25} )1

A = 42,000 (\frac{23}{25} )1

**A = Rs 38,640

**The value of scooter after one year will be = Rs 38,640

Summary:

Exercise 8.3 in Chapter 8, "Comparing Quantities," from the Class 8 NCERT Mathematics textbook focuses on the concepts of percentage increase and decrease, profit and loss calculations, and simple interest. Students learn how to apply formulas for percentage change, calculate profits or losses in transactions, and determine simple interest on various amounts. This exercise enhances students' ability to solve real-life problems related to financial transactions and percentage-based comparisons.