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.
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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 |
3 | Chapter 3 | Understanding Quadrilaterals |
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.
Answer 9: Let “x” be the initial amount he had initially.
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
- a) 20%
- b) 25%
- c) 22%
- 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
- a) 50%
- b) 40%
- c) 30%
- 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”.
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- 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.
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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
$\begin{array}{l}\mathrm{Selling}\text{price of washing machine}=5760\\ \mathrm{Two}\text{successive discounts are 15\% and 10\%.}\\ \text{Let marked price of washing machine}=\mathrm{x}\\ \mathrm{S}.\mathrm{P}.\text{of washing machine after first discount}=\mathrm{x}\left(\frac{100\u02c6\u201915}{100}\right)\\ \text{}=\frac{85\mathrm{x}}{100}\\ \text{}=\frac{17\mathrm{x}}{20}\\ \mathrm{S}.\mathrm{P}.\text{of washing machine after second discount}=\frac{17\mathrm{x}}{20}\left(\frac{100\u02c6\u201910}{100}\right)\\ \text{}=\frac{17\mathrm{x}}{20}\u2014\frac{90}{100}\\ \text{}=\frac{153\mathrm{x}}{200}\\ \mathrm{Then},\text{according to condition}\\ \text{}\frac{153\mathrm{x}}{200}=\mathrm{Rs}.5760\\ \text{}\mathrm{x}=5760\u2014\frac{200}{153}\\ \text{}\mathrm{x}=7529.40\left(\mathrm{approx}\right)\\ \mathrm{Thus}\text{, the marked price of washing machine is}7529.40.\end{array}$
Q.2 Find the compound interest on 1,60,000 for 2 years at 10% per annum when compounded semi-annually.
Marks:2
Ans
$\begin{array}{l}\mathrm{Principal}=\text{160000,}\\ \text{rate = 10\% per annum = 5\% per half year}\\ \text{Time = 2 years = 4 half years.}\\ \mathrm{Amount}\text{=}\left\{160000\text{}\u2014\text{}{\left(1+\frac{5}{100}\right)}^{4}\right\}\\ =\text{160000}\u2014\frac{21}{20}\u2014\frac{21}{20}\text{}\u2014\frac{21}{20}\text{}\u2014\frac{21}{20}\\ =\text{}1,94,481\\ \mathrm{Compound}\text{}\mathrm{Interest}=\text{}1,94,481\text{}\u02c6\u2019\text{}1,60,000\\ \text{= 34,481}\end{array}\text{}$
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
$\begin{array}{l}\mathrm{A}=\mathrm{P}{\left(1+\frac{\mathrm{r}}{100}\right)}^{\mathrm{n}}\\ =2000{\left(1+\frac{10}{100}\right)}^{2}\\ =2000\u2014\frac{110}{100}\text{}\u2014\text{}\frac{110}{100}\text{}=\text{}2420\\ \mathrm{Amount}\text{}=\text{}2420\\ \mathrm{Compound}\mathrm{Interest}\text{}=\mathrm{Amount}\u2013\mathrm{Principal}\\ =2420\u20132000=420\end{array}$
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
$\begin{array}{l}\text{}5\%\mathrm{of}300\mathrm{is}\frac{5}{100}\u2014300=15\\ \mathrm{buying}\mathrm{price}\mathrm{of}\mathrm{Bedsheet}\mathrm{is}300+15=315\end{array}$
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
$\begin{array}{l}\text{C.P. of watch for Rakesh}=\text{800}\\ \text{S.P. of watch for Rakesh}=\text{}\text{1000}\\ \text{Profit on watch to Rakesh}=\text{1000}\u02c6\u2019800\\ \text{}=200\\ \text{}\mathrm{Rate}\text{of Profit}=\frac{200}{1000}\u2014100\\ \text{}=20\%\\ \text{C.P. of car for Mukesh}=\text{}\text{4,00,000}\\ \text{S.P. of car for Mukesh}=\text{}\text{4,20,000}\\ \text{Profit on car for Mukesh}=\text{}\text{}\left(\text{4,20,000}\u02c6\u2019\text{4,00,000}\right)\\ \text{Profit on car for Mukesh}=\text{}20,000\\ \text{}\mathrm{Rate}\text{of Profit}=\frac{20,000}{\text{4,00,000}}\u2014100\\ \text{}=5\%\\ \mathrm{So},\text{Rakesh made a better sale.}\end{array}$
Important Questions for Class 8 Maths
FAQs (Frequently Asked Questions)
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