# Important Questions for CBSE Class 8 Maths Chapter 8 Comparing Quantities

Chapter 8 of Class Maths introduces students to Comparing Quantities. The concept of comparing quantities sets up the basis for many higher Maths topics and is a practically essential skill to have to solve real-life problems. Hence, enough practice is required to build quick mental Maths skills, which will help students in the examinations, enabling them to save enough time on calculations.

Extramarks is the best study adviser for students and helps them with comprehensive online study solutions from Class 1 to Class 12. Our  Maths experts have prepared various NCERT solutions to help students in their studies and exam preparation. Students can refer to our Important Questions Class 8 Maths Chapter 8 to practise exam-oriented questions. We have collated questions from various sources such as NCERT textbooks and exemplars, CBSE sample papers, CBSE past year question papers, etc. Students can prepare well for their exams and tests by solving various chapter questions from our comparing quantities class 8 extra questions.

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## Get Access to CBSE Class 8 Maths Extra Questions 2022-23 with Chapter-Wise Solutions

You can also find CBSE Class 8 Maths Chapter-by-Chapter Important Questions here:

### CBSE Class 8 Maths Important Questions

Sr No. Chapters Chapters Name
1 Chapter 1 Rational Numbers
2 Chapter 2 Linear Equations in One Variable
4 Chapter 4 Practical Geometry
5 Chapter 5 Data Handling
6 Chapter 6 Squares and Square Roots
7 Chapter 7 Cubes and Cube Roots
8 Chapter 8 Comparing Quantities
9 Chapter 9 Algebraic Expressions and Identities
10 Chapter 10 Visualising Solid Shapes
11 Chapter 11 Mensuration
12 Chapter 12 Exponents and Powers
13 Chapter 13 Direct and Inverse Proportions
14 Chapter 14 Factorisation
15 Chapter 15 Introduction to Graphs
16 Chapter 16 Playing with Numbers

## Comparing Quantities Class 8 Extra Questions with Solutions

Our in-house Maths faculty experts have collected an entire list of Important Questions Class 8 Maths Chapter 8 by referring to various sources. For each question, the team experts have  prepared a step-by-step explanation that will help students understand the concepts used in each question. Also, the questions are chosen in a way that would cover full chapter topics. So by practising from our question bank, students will be able to revise the chapter and understand their strong and weak points. And improvise by further focusing on weaker sections of the chapter.

Given below are a few of the questions and answers from our question bank of Maths Class 8 Chapter 8 Important Questions:

### Question 1: Find the rate of discount given on a shirt whose selling price is ₹1092 after deducting a discount of ₹208 on its marked price.

Answer 1: we know that the SP = ₹1092

The discount = ₹208

By using the formula market price = SP + Discount

= 1092 + 208 = ₹1300

∴ The discount% = (discount/market price) × 100

= (208/1300) × 100 = 16%

### Question 2: A scooter was bought at the cost of ₹ 42,000. Its value drops down to the rate of 8% per annum. Find its value after one year.

Answer 2:Principal = Cost price = ₹ 42,000

Depreciation rate = 8% of ₹ 42,000 per year

= ( Principal x Rate x Time period)/100

= (42000 x 8 x 1)/100

= ₹ 3360

Therefore , the value after 1 year = ₹ 42000 − ₹ 3360 = ₹ 38,640.

### Question 3: A particular football team won 10 matches out of all the total number of matches they played. If their winning percentage was 40 %, how many matches did they play?

Answer 3: Let the total number of matches played be x.

The team won around 10 matches, and the team’s winning percentage was 40%.

40/100 × x = 10

40x = 10 × 100

40x = 1000

x = 1000/40

= 100/4

= 2

Hence, the team played 25 matches.

### Question 4:In a particular school, there are 456 girls. Compute the total number of students if 24% of the total students are boys.

Answer 4: The total number of students is 100.

It is given that 24% are boys.

So, total number of boys = 24% of 100 = 24

Therefore, the total number of girls will be = (100-24)%, i.e. 76%.

So, the total number of girls = 76% of 100 = 76

But, it is given that there are 456 girls.

Now, for 76 girls, the total number of students is 100

For 1 girl, the total number of students will be 100/76

Therefore, for 456 girls, the total number of students will be = 100/76 × 456 = 45600/76 = 600

Therefore, the total number of students = 600.

### Question 5: I borrowed an amount ₹ 12000 from Jamshed at a rate of 6% per annum simple interest for 2 years. Had I borrowed this sum at a rate of 6% per annum compound interest, what is the extra amount I would have to pay?

Answer 5: Principle = ₹ 12000

Rate = 6% per annum

Time period = 2 years

Simple Interest = (P x R x T)/100

= (12000 x 6 x 2)/100

= ₹ 1440

To find the compound interest,

the amount (A) has to be calculated

A = P(1 + R/100)n

= 12000(1 + 6/100)2

= 12000(106/100)2

= 12000(53/50)2

= ₹ 13483.20

∴ Compound Interest = A − P

= ₹ 13483.20 − ₹ 12000

= ₹ 1,483.20

Compound Interest − Simple Interest = ₹ 1,483.20 − ₹ 1,440

= ₹ 43.20

Hence, the extra amount to be paid is ₹ 43.20.

### Question 6: Find the cost price when:

SP = ₹34.40 and Gain = 7 ½ %

Answer 6: Cost price = (100 / (100+ Gain %)) × SP

= (100/ (100+ (15/2))) × 34.40

= (100/ (215/2)) × 34.40

= (200/215) × 34.40

=32

∴ the cost price is 32

### Question 7:₹ 62500 for 1½ years at 8% per annum compounded half yearly.

Answer 7:Principal (P) = ₹ 62,500

Rate = 8% per annum or 4% per half-year

Number of years = 1½

There will be 3 and half years in 1½ years

Amount, A = Principle (1 + Rate/100)time period

= 62500(1 + 4/100)3

= 62500(104/100)3

= 62500(26/25)3

= ₹ 70304

Compound Interest = A – P = ₹ 70304 – ₹ 62500 = ₹ 7,804

### Question 8:In a certain laboratory, the bacteria count in a particular experiment increased at 2.5% per hour. Find the bacteria at the end of exactly 2 hours if the count was initially 5,06,000.

Answer 8:The initial count of bacteria = 5,06,000

Bacteria at the end of exactly 2 hours = 506000(1 + 2.5/100)2

= 506000(1 + 1/40)2

= 506000(41/40)2

= 531616.25

Therefore, the bacteria count at the end of 2 hours will be 5,31,616 (approx.).

### Question 9:A person shops and spends 75% of his money. If he is now left with Rs. 600, find out how much he had in the beginning.

As per the given question, the person spent 75% of Rs.x and is left with Rs. 600.

So, the amount he spent = x – 600

=> 75% of x = x – 600

=> (75/100) × x = x – 600

=> 75x = 100x – 60,000

=>25x = 60,000

Or, x = 2400.

Thus, the person had Rs. 2400 initially.

### Question 10:A student scored 150 out of 200 in the subject maths and got 120 marks out of 180 in the subject science. In which subject did the student perform better?

Answer 10: Express the following marks in the form of ratios.

For the subject maths,the ratio = 150/200 = 3/4

For the subject science, the ratio = 120/180= 2/3

Here, the ratio 3/4 shows 75% (3/4× 100 = 75) and the ratio 2/3 shows 66.6% (2/3× 100 = 66.6).

Therefore, the student performed better in the maths exam than in his science exam.

### Question 11: The population of a particular place increased to 54000 in 2003 at 5% per annum

(i) find the population of the place in the year 2001

(ii) what would be the population of the place in the year 2005?

Answer 11: (i) Population in the year 2003 = 54,000

54,000 = (Population in the year 2001) (1 + 5/100)2

54,000 = (Population in the year 2001) (105/100)2

Population in the year 2001 = 54000 x (100/105)2

= 48979.59

Thus, the population of a place in the year 2001 was approximately 48,980

(ii) Population in the year 2005 = 54000(1 + 5/100)2

= 54000(105/100)2

= 54000(21/20)2

= 59,535

Hence, the population of a place in the year 2005 would be 59,535.

### Question 12:Compute the amount and compound interest on the principal amount ₹ 10,800 for 3 years at 12½ % per annum compounded annually.

Answer 12:Principal (P) = ₹ 10,800

Rate (R) = 12½ % = 25/2 % (annual)

Number of years (n) = 3

Amount (A) = P(1 + R/100)n

= 10800(1 + 25/200)3

= 10800(225/200)3

= 15377.34375

= ₹ 15377.34 (approximately)

Compound interest = A – P

= ₹ (15377.34 – 10800) = ₹ 4,577.34

### Question 13:Evaluate the ratio of 5 m to 20 km.

Answer 13: First, make both the units the same.

So, we must convert 20 km to the equivalent metre, i.e. “m”.

=> 20km = (20 × 1000)m

Therefore, the ratio of 5 m to 20 km = 5/20000 = 1:4000

### Question 14: The marked price of a particular water cooler is ₹4650. The shopkeeper proposes an off-season discount of 18% on it. Find its selling price.

Answer 14: we know that the marked price = ₹4650

The discount = 18%

And the Discount in amount = 18% of the market price

= (18/100) × 4650

= ₹ 837

By using the formula, SP = marked price – Discount

= 4650 – 837 = ₹ 3,813

### Question 15: Arun bought a new pair of skates at a sale where the Discount is given 20%. If the amount Arun pays is ₹ 1,600, find out the marked price.

Answer 15: Let the marked price be x

Discount percent = Discount/Marked Price x 100

20 = Discount/x × 100

Discount = 20/100 × x

= x/5

Also,

Discount = Marked price – Sale price

x/5 = x – ₹ 1600

x – x/5 = 1600

4x/5 = 1600

x = 1600 x 5/4

= 2000

Therefore, the marked price was ₹ 2000.

### Question 16: How much did she have in the beginning if Chameli had ₹600 left after spending 75% of her money?

Answer 16: Let the amount of money which Chameli had, in the beginning, be x

After spending amount 75% of ₹x, she was left with ₹600

So, (100 – 75)% of x = ₹600

Or, 25% of x = ₹600

25/100 × x = ₹600

x = ₹600 × 4

= ₹2,400

Hence, Chameli had ₹2,400 in the beginning.

### Question 17:Express 25% and 12% as decimals.

Answer 17: 25% = 25/100 = 0.25

12% = 12/100 = 0.12

### Question 18: A man got a 10% increase in his salary. If his new salary amounted to ₹1,54,000, what was his original salary?

Answer 18: Let the original salary be x

The new salary is ₹1,54,000

Original salary + Increment = New Salary

The increment is of 10% of the original salary

So, (x + 10/100 × x) = 154000

x + x/10 = 154000

11x/10 = 154000

x = 154000 × 10/11

= 140000

Thus, the original salary was ₹1,40,000.

### Question 19: A particular cell phone was marked at 40% above the cost price, and a discount of 30% was given on its marked price. Find out the gain or loss percent made by the shopkeeper.

Answer 19: Let us consider the CP of goods as x

The market price of the goods when goods marked above the 40% CP is

Market price = x + (40x/100) = 140x/100 = 1.4x

So the Discount = 30%

Discount amount = 30% of 1.40x = 0.42x

∴ The SP = Market price – Discount

= 1.4x – 0.42x= 0.98x

Since SP is less than CP, it’s a loss

We know that Loss = CP – SP

= x – 0.98x = 0.02x

∴ Loss % = (Loss × 100)/ CP

= (0.02x × 100)/ x

= 2%

### Question 20: Express 45% and 78% as a fraction.

Answer 20: 45% = 45/100 = 9/20

78% = 78/100 = 39/50

### Question 21: Obtain the amount Ram will get on ₹ 4,096, and he gave it for 18 months at 12½ per annum, interest being compounded half-yearly.

Answer 21: Principal = ₹ 4,096

Rate = 12½ per annum = 25/2 per annum = 25/4 per half-year

Time period = 18 months

There will be exactly 18 months in 3 half years

Thus, amount A = P(1 + R/100)n

= 4096(1 + 25/(4 x 100))3

= 4096 x (1 + 1/16)3

= 4096 x (17/16)3

= ₹ 4913

Hence, the required amount is ₹ 4,913.

### Question 22: A bookseller sells a book for ₹100, gaining ₹20. His gain percentage is

1. a) 20%
2. b) 25%
3. c) 22%
4. d) None of these

Answer 22: we know that the SP = ₹100

Gain = ₹20

By using the formula CP = SP – Gain

= 100 – 20

= 80

∴ Gain% = (Gain × 100) /CP

= (20 × 100) / 80

= 25%

### Question 23: Find out the amount and the compound interest on ₹ 10,000 for 1½ years at the rate of 10% per annum, compounded half yearly. Could this interest be more than the interest he would get if it were compounded annually?

Answer 23: P = ₹ 10,000

Rate = 10% per annum = 5% per half-year

Time period = 1½ years

There will 3 half years in 1½ years

Amount, A = P(1 + R/100)n

= 10000(1 + 5/100)3

= 10000(105/100)3

= ₹ 11576.25

Compound interest = Amount − Principal

= ₹ 11576.25 − ₹ 10000

= ₹ 1,576.25

The amount for 1 year and 6 months can be calculated by first calculating the amount for 1 year using the compound interest formula and then calculating the simple interest for 6 months on the amount obtained at the end of 1 year.

Amount, A = P(1 + R/100)n

= 10000(1 + 10/100)1

= 10000(110/100)

= ₹ 11000

By taking ₹ 11,000 as the principal, the S.I. for the next ½ year will be calculated as

Simple Interest = (P x R x T)/100

= (11000 x 10 x ½)/100

= ₹ 550

So, the interest for particularly the first year = ₹ 11000 − ₹ 10000 = ₹ 1,000

Therefore, Total compound interest = ₹ 1000 + ₹ 550 = ₹ 1,550

So the actual difference between two interests = 1576.25 – 1550 = 26.25

Thus, the interest would be 26.25 more when compounded half yearly than the interest when compounded annually.

### Question 24: Maria invested ₹ 8,000 in a particular business. She would eventually be paid interest at 5% per annum compounded annually. Find out

(i) The amount credited against her name, particular at the end of the second year

(ii) The interest for the third year

Answer 24: (i) P = ₹ 8,000

R = 5% per annum

n = 2 years

Amount, A = P(1 + R/100)n

= 8000(1 + 5/100)2

= 8000(105/100)2

= ₹ 8820

(ii) The interest for the next year, i.e. the third year, has to be calculated. By taking ₹

8,820 as principal, the Simple Interest for the next year will be calculated.

Simple Interest = (P x R x T)/100

= (8820 x 5 x 1)/100

= ₹ 441

Question 31: Vasudevan invested ₹ 60000 at an interest rate of 12% per annum compounded half yearly. What amount would he get?

(i) after 6 months?

(ii) after 1

Answer 31:(i) P = ₹ 60,000

Rate = 12% per annum = 6% per half-year

n = 6 months = 1 half-year

Amount, A = Principle(1 + Rate/100)time period

= 60000(1 + 6/100)1

= 60000(106/100)

= 60000(53/50)

= ₹ 63,600

(ii) There are almost 2 and half years in 1 year

So, time period = 2

Amount, A = P(1 + R/100)n

= 60000(1 + 6/100)2

= 60000(106/100)2

= 60000(53/50)2

= ₹ 67,416

### Question 25:Fabina borrows ₹ 12,500 at the rate of 12% per annum for 3 years at simple interest, and Radha borrows the same amount for the same period at 10% per annum, compounded annually. Who pays more interest, Fabina and Radha and by how much?

Answer 25: Interest paid by Fabina = (Principle x Rate x Time period)/100

= (12500 x 12 x 3)/100

= 4500

Amount paid by Radha at the end of 3 years = A = Principle(1 + Rate/100)time period

Amount = 12500(1 + 10/100)3

= 12500(110/100)3

= ₹ 16637.50

Compound Interest = Amount – Principal  = ₹ 16637.50 – ₹ 12500

= ₹ 4,137.50

The interest paid by Fabina is ₹ 4,500 and by Radha is ₹ 4,137.50

Therefore, Fabina pays more interest

₹ 4500 − ₹ 4137.50 = ₹ 362.50

Therefore, Fabina will have to pay ₹ 362.50 more than Radha.

### Question 26:Kamala borrowed ₹ 26400 from a particular 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 pay off the loan?

(Hint: Find A for 2 years with interest compounded yearly and then find S.I. on the 2nd year amount for 4/12 years.)

Answer 26:Principal (P) = ₹ 26,400

Rate (R) = 15% per annum

Number of years (n) = 2 4/12

The amount for 2 years and 4 months can be calculated by first calculating the amount for 2

years using the compound interest formula, and then calculating the simple interest for 4 months on the amount obtained at the end of 2 years

First, the amount for 2 years has to be calculated

Amount, A = Principal (1 + Rate/100)time period

= 26400(1 + 15/100)2

= 26400(1 + 3/20)2

= 26400(23/20)2

= ₹ 34914

By taking ₹ 34,914 as the principal amount, the S.I. for the next 1/3 years can be calculated.

Simple Interest = (34914 × 1/3 x 15)/100

= ₹ 1745.70

Interest for the first two years

= ₹ (34914 – 26400)

= ₹ 8,514

And interest for the next 1/3 year = ₹ 1,745.70

Total Compound Interest = ₹ (8514 + ₹ 1745.70) = ₹ 10,259.70

Amount = P + C.I. = ₹ 26400 + ₹ 10259.70 = ₹ 36,659.70

### Question 27: Oranges are purchased at 5 for ₹10 and sold at 6 for ₹15. His gain percentage is

1. a) 50%
2. b) 40%
3. c) 30%
4. d) 25%

Answer 27: we know that the CP of 1 orange = ₹10/5 = ₹2

SP of 1 orange = ₹15/6 = ₹2.5

Since SP is more than CP, it’s a Gain

Gain = SP – CP

=2.5 – 2

= 0.5

∴ Gain% = (Gain × 100) /CP

= ((0.5) × 100) / 2

= 25%

## Benefits Of Solving Important Questions Class 8 Maths Chapter 8

Practice is the key to scoring 100%  in Maths. The foundation of fundamental apprehension is the Maths taught in  Classes 8, 9, and 10 and it’s important to step up their learning experience and eliminate “maths phobia” among students. . We recommend students access  Extramarks to obtain important questions in Class 8 Maths Chapter 8. By systematically solving questions and going through the required solutions, students will get  the confidence to solve any tough  questions in the given chapter ”Comparing Quantities”.

### Below are a few benefits of frequently solving comparing quantities class 8 important questions:

•  Our  Maths  subject experts have carefully assembled the most important questions in  Class 8 Maths Chapter 8 by inspecting many past years’  question papers.
• The questions and solutions provided are based on the latest CBSE syllabus and as per CBSE guidelines. So the students can completely count on it.
• The questions covered in our set of important questions in Class 8 Maths Chapter 8 are entirely based on several topics covered in the chapter ”Comparing Quantities”. It is suggested that students revise and clear all their doubts before solving all these important questions.
• By practicing class 8 maths chapter 8 extra questions students will get a general idea of how the paper will be prepared. Practising questions similar to the exam questions would help the students gain confidence, score 100%  marks in their examinations, and set their own benchmark. .

Extramarks believes in incorporating joyful learning experiences through its own repository of resources. Extramarks provides comprehensive learning solutions for students from Class 1 to Class 12. We have other study resources on our website, along with important questions and solutions.To get good grades in exams students must practice a lot of questions and stick to a study schedule and follow it religiously to come out with flying colours. Students can also click on the below-mentioned links and access some of these resources:

• NCERT books
• CBSE Revision Notes
• CBSE syllabus
• CBSE sample papers
• CBSE past  year’s question papers
• Important formulas
• CBSE extra questions

Q.1 A washing machine was sold for 5760 after giving successive discounts of 15% and 10% respectively. What was the marked price

Marks:4
Ans

$\begin{array}{l}\mathrm{Selling}\text{price of washing machine}=5760\\ \mathrm{Two}\text{successive discounts are 15% and 10%.}\\ \text{Let marked price of washing machine}=\mathrm{x}\\ \mathrm{S}.\mathrm{P}.\text{of washing machine after first discount}=\mathrm{x}\left(\frac{100ˆ’15}{100}\right)\\ \text{}=\frac{85\mathrm{x}}{100}\\ \text{}=\frac{17\mathrm{x}}{20}\\ \mathrm{S}.\mathrm{P}.\text{of washing machine after second discount}=\frac{17\mathrm{x}}{20}\left(\frac{100ˆ’10}{100}\right)\\ \text{}=\frac{17\mathrm{x}}{20}—\frac{90}{100}\\ \text{}=\frac{153\mathrm{x}}{200}\\ \mathrm{Then},\text{according to condition}\\ \text{}\frac{153\mathrm{x}}{200}=\mathrm{Rs}.5760\\ \text{}\mathrm{x}=5760—\frac{200}{153}\\ \text{}\mathrm{x}=7529.40\left(\mathrm{approx}\right)\\ \mathrm{Thus}\text{, the marked price of washing machine is}7529.40.\end{array}$

Q.2 Find the compound interest on 1,60,000 for 2 years at 10% per annum when compounded semi-annually.

Marks:2
Ans

$\begin{array}{l}\mathrm{Principal}=\text{160000,}\\ \text{rate = 10% per annum = 5% per half year}\\ \text{Time = 2 years = 4 half years.}\\ \mathrm{Amount}\text{=}\left\{160000\text{}—\text{}{\left(1+\frac{5}{100}\right)}^{4}\right\}\\ =\text{160000}—\frac{21}{20}—\frac{21}{20}\text{}—\frac{21}{20}\text{}—\frac{21}{20}\\ =\text{}1,94,481\\ \mathrm{Compound}\text{}\mathrm{Interest}=\text{}1,94,481\text{}ˆ’\text{}1,60,000\\ \text{= 34,481}\end{array}\text{}$

Q.3 Find the amount on a principal of 2000 for 2 years at 10% per annum compounded annually. Also find the compound interest.

Marks:1
Ans

$\begin{array}{l}\mathrm{A}=\mathrm{P}{\left(1+\frac{\mathrm{r}}{100}\right)}^{\mathrm{n}}\\ =2000{\left(1+\frac{10}{100}\right)}^{2}\\ =2000—\frac{110}{100}\text{}—\text{}\frac{110}{100}\text{}=\text{}2420\\ \mathrm{Amount}\text{}=\text{}2420\\ \mathrm{Compound}\mathrm{Interest}\text{}=\mathrm{Amount}–\mathrm{Principal}\\ =2420–2000=420\end{array}$

Q.4 When 5% sale tax is added on the purchase of a bedsheet of 300, find the buying price or the cost price of the bedsheet.

Marks:1
Ans

$\begin{array}{l}\text{}5%\mathrm{of}300\mathrm{is}\frac{5}{100}—300=15\\ \mathrm{buying}\mathrm{price}\mathrm{of}\mathrm{Bedsheet}\mathrm{is}300+15=315\end{array}$

Q.5 Rakesh bought a watch for 800 and sold it for 1000. Mukesh bought a car for 4,00,000 and sold it for 4,20,000. Who made a better sale, Rakesh or Mukesh

Marks:4
Ans

$\begin{array}{l}\text{C.P. of watch for Rakesh}=\text{800}\\ \text{S.P. of watch for Rakesh}=\text{}\text{1000}\\ \text{Profit on watch to Rakesh}=\text{1000}ˆ’800\\ \text{}=200\\ \text{}\mathrm{Rate}\text{of Profit}=\frac{200}{1000}—100\\ \text{}=20%\\ \text{C.P. of car for Mukesh}=\text{}\text{4,00,000}\\ \text{S.P. of car for Mukesh}=\text{}\text{4,20,000}\\ \text{Profit on car for Mukesh}=\text{}\text{}\left(\text{4,20,000}ˆ’\text{4,00,000}\right)\\ \text{Profit on car for Mukesh}=\text{}20,000\\ \text{}\mathrm{Rate}\text{of Profit}=\frac{20,000}{\text{4,00,000}}—100\\ \text{}=5%\\ \mathrm{So},\text{Rakesh made a better sale.}\end{array}$

## Important Questions for Class 8 Maths

### 1. What are the formulas in ”Comparing Quantities” Class 8?

Following are the Important formulas for comparing Quantities Class 8 Maths:

• Cost Price= Selling Price – Profit
• Cost Price= Selling Price + Loss
• Selling Price=Cost Price + Profit
• Selling Price=Cost Price – Loss
• Profit= Selling Price – Cost Price
• Loss= Cost Price – Selling Price
• Profit %= Profit/ cost Price x 100
• Loss %= Loss/ Cost Price x 100

### 2. How can I score well in Class 8 Maths examinations?

Maths is a subject which requires a lot of practice. To score well in Maths, one must have a strong conceptual understanding of the chapter, be good in calculations, practice questions regularly, give mock tests from time to time, get feedback and avoid silly mistakes. Regular practice with discipline, working diligently and conscientiously towards your goal, will definitely ensure a 100% in your exams.

### 3. What can I get from the Extramarks website?

Extramarks is one of the best educational platforms and it has its own archive of educational resources, which assists students in acing their exams. You can get all the NCERT-related material like NCERT solutions, solved exemplar solutions, NCERT-based mock tests, CBSE revision notes, and  Class 8 Maths Chapter 8 important questions on the Extramarks website. Apart from this, you can get comprehensive guidance from our subject experts and doubt-clearing sessions once you sign up on our official website for any study resources.

### 4. How many total chapters will students study in Class 8 Maths?

There are 16 chapters in Class 8 Maths. The list of chapters is given below:

• Chapter 1- Rational Numbers
• Chapter 2 – Linear Equations in One Variable
• Chapter 3 – Understanding Quadrilaterals
• Chapter 4 – Practical Geometry
• Chapter 5 – Data Handling
• Chapter 6 – Square and Square Roots
• Chapter 7 – Cube and Cube Roots
• Chapter 8 – Comparing Quantities
• Chapter 9 – Algebraic Expressions and Identities
• Chapter 10 – Visualising Solid Shapes
• Chapter 11- Mensuration
• Chapter 12 – Exponents and Powers
• Chapter 13 – Direct and Inverse Proportions
• Chapter 14 – Factorisation
• Chapter 15 – Introduction to Graphs
• Chapter 16 – Playing with Numbers