Mathematics
Grade
– 3xy
– 5x
y+
xy
11
1
4
a
0
a:b
%
Professor Vassen Naëck
- Head Curriculum Implementation,Textbook Development and Evaluation
MATHEMATICS PANEL
Savila Thapermall-Ramasawmy
Asifa Salehmohamed
Nandita Vandana Baijnath
Dilshad Beebee Codobaccus
Annelise Lafrance
Nishta Sunanda Rajinundun
Iswaree Ramsaran
Rakesh Teeluckdarry
Luchmun Useree
- Coordinator, Lecturer, MIE
- Lecturer, MIE
- Educator
- Educator
- Educator
- Educator
- Educator
- Educator
- Educator
Design
Nishi Manic
Sanjna Kathapermall
Keshrajsing Lokhun
- Graphic Designer, MIE
- Graphic Designer, MIE
- Graphic Designer
© Mauritius Institute of Education (2021)
ISBN: 978-99949-44-57-6
Acknowledgements
The Mathematics textbook panel wishes to thank the following:
- Mathematics Education Department (MIE) for proof reading and vetting.
- Komal Reshma Gungapersand (Lecturer, MIE) for proof reading.
- Yesha Devi Mahadeo-Doorgakant (Lecturer, MIE) for proof reading.
Consent from copyright owners has been sought. However, we extend our apologies to those we might have overlooked.
All materials should be used strictly for educational purposes.
ii
Foreword
As the first cohort to embark on the Nine Year Continuous Basic Education at secondary level, we are
pleased to offer you a new series of Grade 7 textbooks. These textbooks have been designed in line
with the National Curriculum Framework (NCF) and syllabi for Grades 7, 8 and 9 − documents that
offer a comprehensive notion of learning and teaching with regard to each subject area. These may be
consulted on the MIE website, www.mie.ac.mu.
This set of textbooks aims at providing a smooth transition from Grade 6 so that learners gradually
get initiated into the requirements of secondary schooling. As per the philosophy propounded by the
NCF, the content and pedagogical approach, as well as the activities, have been crafted to allow for an
incremental and continuous improvement of the learners’ cognitive skills. The content is contextual and
based on the needs of the Mauritian learners. Care has been taken to provide the basics that should
help every student develop key competencies, knowledge, skills, attitudes and values that would make
him or her a successful learner for the grades beyond. The comments and suggestions of a variety of
stakeholders have been kept in mind. We are especially appreciative of those made by educators who
have been part of our validation panels, and whose suggestions emanate from long standing experience
and practice in the field.
The production of this series of textbooks, within a relatively short period of time, has been a challenge
to the writers who have invested a considerable amount of time, effort and energy into the process.
I would therefore wish to thank all those who have been part of the whole process for the time they
devoted to it and their perseverance. The panel coordinators are to be commended for their leadership
and insistence on maintaining the standard and quality of the textbooks, while ensuring that the
objectives of the National Curriculum Framework are translated in the content.
Every endeavour involves a number of dedicated, hardworking and able staff whose contribution
needs to be acknowledged. Professor Vassen Naëck, Head Curriculum Implementation and Textbook
Development and Evaluation, had the very demanding task of ensuring that all panel leaders are
adequately guided with respect to the objectives of the NCF, while ascertaining that the instruction
designs are appropriate for the age group being targeted. I also have to acknowledge the efforts of the
Graphic Artists and the Graphic Unit for putting in much hard work to ensure that MIE publications have
a distinctive quality that singles them out. My thanks go equally to the support staff who have worked
hard to ensure that every one receives the necessary support and work environment that is conducive
to a creative endeavour.
I am thankful to all those who provided the support, both within and outside the MIE, and to the Ministry
of Education, Human Resources, Tertiary Education and Scientific Research for giving us the opportunity
to be part of the whole reform process.
Dr O Nath Varma
Director
Mauritius Institute of Education
iii
Preface
Mathematics is found at the core of the curriculum and empowers us with the requisite knowledge,
skills, attitudes and values. The tools and habits of mind that we acquire in learning Mathematics
prepare us not only for academic and vocational pursuits but also to be functional citizens in the
society. Thus, the Grade 7 Mathematics textbook sets the ground to lay a solid foundation in the
five content areas of the National Curriculum Framework (2017 - Mathematics), namely Numbers,
Algebra, Measures, Geometry and Probability & Statistics.
In designing the textbook, the Mathematics panel has ensured that students experience a smooth
transition from Grade 6 to Grade 7. In each chapter, we review and consolidate concepts learnt at
the primary level before extending and introducing new ideas. Concepts have been developed
systematically through activities and ‘illustrative’ examples followed by graded exercises to cater
for different learners. The textbook provides a range of conceptually-rich activities and problems to
provide students with varied opportunities to make a robust start in their first year of secondary
schooling.
The Grade 7 Mathematics textbook is premised on sense making and reasoning. Throughout the
textbook, we have included activities to engage students in their mathematical learning. There is
also a strong focus on the key components of the Framework for Mathematics Curriculum (Grades
7-9). An important feature of the textbook is the explicit connection made with real life situations for
students to see the applications of Mathematics and to arouse their interest and motivation to learn
the subject.
As a significant component of meaningful Mathematics learning, problem solving has also been
given much attention in the Grade 7 textbook through the provision of rich mathematical tasks
and investigations. In addition, the integration of online resources and digital tools, such as the
mathematical software GeoGebra, encourages learners to use ICT to further explore, investigate
and solve mathematical problems. It is hoped that the varied features of the textbook will allow
for numerous opportunities to create a positive learning environment where students can develop
and enhance their problem solving skills and develop positive attitudes such as perseverance and
commitment.
The Mathematics panel has made a conscious effort to make Mathematics accessible to each and
every student. We hope that the approaches and resources included in the textbook will enable the
21st century learners to successfully pave their first steps into secondary school Mathematics and to
experience a rich learning journey.
The Mathematics Panel
iv
Icons
EXERCISE
The exercise icon provides you with questions to practise.
RECALL
The recall icon reviews ideas and concepts covered in previous grades.
STOP AND THINK
This icon encourages you to pause and reason mathematically.
DID YOU KNOW
This icon provides you with interesting mathematical facts.
FIND OUT
This symbol indicates concepts that you may want to further investigate.
CHECK THIS LINK
This section provides you with links to online resources.
CHECK THAT YOU CAN:
KEY TERMS
Activity
Investigate
This icon indicates the prior knowledge and understanding you
would require.
This icon highlights key mathematical vocabulary.
This icon provides you with active learning opportunities to better
understand the Mathematics you are learning.
This encourages you to make better sense of what you are learning.
Summary
This provides you with the key ideas and main points in each chapter.
Note:
This provides you with additional information.
Caution:
This is a warning sign that alerts you to be careful about common
mistakes and misunderstandings.
GeoGebra
NOTE TO TEACHER
This encourages the use of the interactive dynamic tool, GeoGebra, to represent, explore and solve mathematical problems.
This provides additional information to the teacher.
v
Table of contents
NUMBERS I
Chapter 1
Chapter 2
Chapter 3
Chapter 4
Numbers, Factors and Multiples
Integers
Order of operations
Fractions and decimals
GEOMETRY I
Chapter 5
Chapter 6
Angles
66
Polygons84
MEASUREMENT I
Chapter 7
Length, Perimeter and Area
1
13
32
40
96
Chapter 8
Chapter 9
Chapter 10
Percentage117
Ratio and Proportion
124
Indices132
NUMBERS II
Chapter 11
Chapter 12
Chapter 13
Chapter 14
Money138
Time146
Speed155
Mass164
MEASUREMENT II
Algebraic expressions
Chapter 15
and algebraic equations170
ALGEBRA I
NUMBERS III
GEOMETRY II
ALGEBRA II
Chapter 22
vi
Chapter 16
Patterns and Sequences
Chapter 17
Coordinates194
Chapter 18
Symmetry207
Chapter 19
Geometrical Constructions
Chapter 20
Reflection229
Chapter 21
Sets240
Statistics257
STATISTICS
187
221
1
Chapter 1 - Numbers, Factors and Multiples
NUMBERS, FACTORS AND MULTIPLES
Learning
Objectives
By the end of this chapter, you should be able to:
• demonstrate an understanding of different types of numbers, common divisibility tests,
factors and multiples.
• find multiples and prime factors of a given number.
• express the factors of a given number in index form.
• find the H.C.F. and L.C.M. of 2 or 3 numbers.
• solve word problems involving H.C.F and L.C.M.
Numbers are everywhere and are so importantly linked
to our lives. When a baby is born, his birth date and
time are the very first numbers connected to his life.
CHECK THAT YOU CAN:
• Perform multiplication and
division of numbers.
Types of numbers
There are different types of numbers such as even, odd,
prime, composite, square and triangular numbers.
Even Numbers
Even numbers are all numbers which are exactly
divisible by 2. They end with 0, 2, 4, 6 or 8.
Examples of even numbers: 8, 12, 24, 36, 50.
Odd Numbers
Odd numbers are numbers which leave a remainder
of 1 when divided by 2. They end with 1, 3, 5, 7 or 9.
Examples of odd numbers: 11, 23, 67, 105, 219.
Prime Numbers
A prime number is a number that has only two distinct
factors: 1 and itself.
Examples of prime numbers: 7, 13, 29, 37.
Composite Numbers
A composite number is a number that has more than
two factors.
Examples of composite numbers: 4, 6, 32, 100, 450.
4 has three factors, namely 1, 2 and 4.
6 has four factors, namely 1, 2, 3 and 6.
KEY TERMS
•
•
•
•
•
•
•
•
•
•
•
•
Even numbers
Odd numbers
Prime numbers
Composite numbers
Triangular numbers
Square numbers
Factors
Multiples
Prime factors
Index and Expanded notations
Factor tree
H.C.F and L.C.M.
STOP AND THINK
Is 2 a prime number?
NOTE TO TEACHER
Prompt students to ‘realise’ that 2
is the only prime number which
is even.
Note: 0 and 1 are neither prime
nor composite numbers.
1
Chapter 1 - Numbers, Factors and Multiples
Square Numbers
When a whole number is multiplied by itself, the resulting product is called a square number.
For example, 1 × 1 = 1, 2 × 2 = 4, 3 x 3 = 9, 9 × 9 = 81. So, 1, 4, 9 and 81 are called square numbers.
We can represent square numbers through a square array of dots.
For example,
1
4
9
Triangular Numbers
Triangular numbers are numbers that can make a triangular dot pattern as shown below.
Fig.1
Fig. 2
Fig.3
Fig. 4
Fig. 5
Figure
1
2
3
4
5
Number of dots
1
1+2
1+2+3
1+2+3+4
1+2+3+4+5
Triangular numbers
1
3
6
10
15
Thus, the triangular numbers are 1, 3, 6, 10, 15 and so on .
This pattern can be continued to obtain more triangular numbers.
EXERCISE 1.1
CHECK THIS LINK
http://www.sheppardsoftware.com/
mathgames/numbers/fruit_shoot_
prime.htm
1. (a) Circle the even numbers from the list below:
4,
248,
7,
345,
1 023,
2 124,
10 321,
20 240
(b) Tick the correct box for the following statement:
True
False
The numbers 43, 7, 15, 8, 17, 61, 109 are all odd numbers.
(c) List the square numbers between 30 and 90.
(d) Circle the triangular numbers in the list below:
32,
28,
41,
45,
52,
55,
65,
63
2. Circle the correct answer.
(a) Which one of the following groups of numbers is prime?
A. 2, 3, 5, 7, 9, 12
B. 2, 3, 5, 7, 11, 13
C. 2, 4, 6, 8, 10
D. 1, 2, 3, 5, 7, 11
(b) Which one of the following groups of numbers is composite?
A. 2, 3, 4, 6, 8
2
B. 9, 11, 13, 15
C. 4, 6, 8, 9, 10
D. 17, 19, 23, 29
Chapter 1 - Numbers, Factors and Multiples
Factors and Multiples
Without realising it, we use factors and multiples in our daily life.
For example, when you need to change a 100-rupee note with a friend or at the school canteen,
you probably ask for four 25-rupee notes or ten 10-rupee coins. So, 4, 10 and 25 are a few of the
factors of 100.
Moreover, did you realise that when you are buying fruits or other items you are using multiples?
For example, if a pencil costs Rs 5 and you are buying 4 pencils, it will cost you Rs (5 × 4) = Rs 20.
Therefore 20 is a multiple of both 4 and 5.
Example 1
What are the factors of 12?
Solution
Method 2 - The factor rainbow of 12
Method 1
1 and 12 are factors of 12
2 and 6 are factors of 12
3 and 4 are factors of 12
12 = 1 × 12
=2×6
=3×4
1 2 3 4 6 12
If we multiply
• the 1st and last factors, we get 12 (1 × 12)
• the 2nd and 5th factors, we get 12 (2 × 6)
• the 3rd and 4th, the two middle factors,
we get 12 (3 × 4)
Rewriting all the factors of 12, we have 1, 2, 3, 4, 6 and 12.
Example 2
Find the factors of (a) 64, (b) 250.
Solution
(a)
(b)
1
2
4
8
16
32
64
The factors of 64 are 1, 2, 4, 8, 16, 32 and 64.
Note: 64 = 8 x 8
1
2
5
10
25
50
125
250
The factors of 250 are 1, 2, 5, 10, 25, 50,
125 and 250.
POINTS TO REMEMBER:
•
•
•
•
•
1 and the number itself are always factors of the number.
A prime number has only 2 distinct factors: 1 and the number itself.
Every number is exactly divisible by its factors.
Every factor of a number is less than or equal to that number.
The number of factors of a number is finite.
3
Chapter 1 - Numbers, Factors and Multiples
EXERCISE 1.2
1. Complete the following by writing “TRUE” or
“FALSE” for each statement below:
(a) 378 is exactly divisible by 3.
______
(b) 136 is exactly divisible by 6.
______
(c) 120 is not exactly divisible by 10. ______
(d) 1 104 is exactly divisible by 8.
______
2. Which one of the following statements is correct?
A. 8 is a factor of 116
B. 9 is a factor of 423
C. 15 is a factor of 235
D. 6 is not a factor of 36
RECALL
Divisibility tests
A number is
divisible by:
(b) 42
(c) 72
(d) 30
(e) 102
(f) 90
(g) 105
(h) 63
(i) 121
(j) 225
4. Consider the following numbers:
12, 16, 18, 20, 24, 28, 33.
(a) Write down all the numbers having 3 as a factor.
(b) Write down all the numbers having 4 as a factor.
(c) (i) Write down all the numbers which have both
3 and 4 as factors.
(ii) Hence, give another possible factor that is
common to 3 and 4.
5. In the list below, circle the numbers that are
factors of 108:
3,
12,
6,
5,
54,
8,
7,
if its last digit
32, 450, 5 324
is 0,2,4,6 or 8
3
522
if the sum of
since
the digits in
5+2+2 =9
the number is
is divisible
divisible by 3
by 3
4
if the last
2 digits are
divisible by 4
224
last 2 digits:
24 which is
divisible by 4
5
if the last
digit is 0 or 5
5, 10, 15, 120,
1 225,…
6
if it is divisible
by both 2
and 3
336
336 is
divisible by
both 2 and 3
7
none
8
if its last 3
digits is
divisible by 8
3 432
432 is
divisible by 8
9
if the sum of
its digits is
divisible by 9
828
8+2+8=
18 which is
divisible by 9
10
if it ends in 0
110, 2 890,…
27
6. Find the missing number and/or factor.
4
Number
Factors
_____
1, 2, 4, 8
_____
___, 2, 3, 6, 9, 18
12
1, 2, 3, ___, 6, 12
_____
1, 2, 3, 5, 6, 10, 15, ___
_____
___, 3, 5, 9, 15, ____
_____
1, 2, 3, 4, 6, 8, 12, 16, ___, 48
Example
2
3. Find all the possible factors of each of the
following numbers:
(a) 16
Rule
Chapter 1 - Numbers, Factors and Multiples
Prime Factors
Prime factors are factors of a number that are prime numbers. For example, the factors of 6 are
1, 2, 3 and 6 but the prime factors of 6 are only 2 and 3.
Prime Factorisation
Prime factorisation is the process of writing a number as a product of its prime factors.
Example
Express 24 as a product of its prime factors.
Method 1: Factor Tree
Method 2: Division Method
2
2
2
3
24
4
x
Note: You can choose
any two factors.
6
2 x 2 x 2 x
3
Therefore, 24 = 2 × 2 × 2 × 3
= 23 × 3 24
12
6
3
1
Divide by prime
numbers, starting
from the smallest
until the division
ends with 1.
Expanded form
Index form
GeoGebra - Finding factors in index form on GeoGebra
Step 1: Open Geogebra
RECALL
Step 2: Click on View – CAS
Step 3: Type the number(s) for which you need to find the prime factors.
For example (i) 12 (ii) 36 (iii) 120
Step 4: Select “factors” function :
Index (Power) form
In Grade 6, you learnt about
index (power) notation.
A number can be written in
index form as follows:
Index / Power
8=2×2×2=23
Base
81 = 3 × 3 × 3 × 3 = 34
EXERCISE 1.3
1. Express each of the numbers as a product of its prime factors in (i) expanded form, (ii) index form:
(a) 60(b) 42(c) 88(d) 32(e) 120
(f) 36(g) 75(h) 100(i) 64(j) 208
5
Chapter 1 - Numbers, Factors and Multiples
2. Express each of the following numbers as a product of its prime factors :
(a) 20
(b) 96
(c) 112
(d) 240
(e) 450
STOP AND THINK
3. Find the smallest number that has three different prime factors.
Highest Common Factor (H.C.F.)
The Highest Common Factor (H.C.F.) of two or more
numbers is the largest factor that is common to
these numbers. H.C.F. is also known as the Greatest
Common Factor (G.C.F.) or the Greatest Common
Divisor (G.C.D.).
H.C.F. of two and three numbers
Method 1: Listing the factors
Where do we use H.C.F. in Mathematics?
NOTE TO TEACHER
Prompt students to think of:
1. when they want to equally
distribute 2 or more items into
their largest grouping.
2. when they reduce or simplify
fractions.
Example
Find the H.C.F. of 18 and 24.
Solution
Factors of 18 = 1, 2, 3, 6, 9, 18
Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24
Common factors of 18 and 24 = 1, 2, 3, 6
Highest Common Factor (H.C.F.) of 18 and 24 = 6
Note: This method is not
practical when we have to find
the H.C.F. of large numbers.
Method 2: Using Prime Factorisation (in expanded form and index form)
Example 1
Find the H.C.F. of 18 and 24.
Solution
2
3
3
18
9 3 1 2
2
2
3
In Expanded Form:
24
18 = 2 ×
3×3
12
24 = 2 × 2 × 2 × 3
6
3 H.C.F. of 18 and 24 = 2 × 3 = 6
1
In Index Form:
18 = 2 × 32
24 = 23 × 3
H.C.F. of 18 and 24 = 2 × 3 = 6
6
Note: We select the
common factors of the
2 numbers and find
their product.
Note: We select the
common factors of the
2 numbers which have
the smallest power and
find their product.
Chapter 1 - Numbers, Factors and Multiples
Example 2
Find the H.C.F. of 45, 60 and 90.
Solution
3
3
5
45
2 60
15
2 30
5
3 15
1
5 5 1
2
45 =
3 ×5
2
60 = 2 × 3 × 5
90 = 2 × 32 × 5
H.C.F. of 45, 60 and 90 = 3 × 5 = 15
2
3
3
5
Note: We select the
common factors of the 3
numbers which have the
smallest power and find
their product.
90
45
15
5
1
Method 3: Division by common prime factor(s)
Example
Find the H.C.F. of 72, 84 and 96.
Solution
Step 1: We divide the 3 numbers by a prime factor that is exactly divisible by all 3 numbers at
the same time.
Step 2: Repeat this process until all 3 numbers cannot be divided by the same prime factor at
the same time.
2
72 , 84 , 96
2
36 , 42 , 48
18, 21 and 24 cannot be divided
by 2 at the same time, so we
3
18 , 21 , 24
divide them by 3.
6
7
8
6, 7 and 8 cannot be divided anymore by a
common prime factor at the same time.
So, the H.C.F. of 72, 84 and 96 = 22 × 3 = 12
EXERCISE 1.4
1. Find the H.C.F. of the following numbers:
(a) 10 and 15 (b) 8 and 20
(e) 18 and 32
(f) 44 and 110
(c) 14 and 35
(g) 84 and 108
(d) 16 and 24
(h) 150 and 210
2. Find the H.C.F. of the following numbers:
(a) 9, 12 and 24
(b) 14, 21 and 49
(e) 12, 36 and 48
(f) 25, 45 and 70
(c) 72, 84 and 108
(g) 11, 33 and 44
(d) 32, 48 and 64
(h) 14, 35 and 42
3. Find the largest number which is a factor of both 48 and 72.
4. Find the greatest number which exactly divides both 100 and 120.
7
Chapter 1 - Numbers, Factors and Multiples
Multiples
Multiples of a number are obtained by multiplying the
number by the whole numbers such as 1, 2, 3, 4 and so on.
Consider the multiples of 3.
Multiples of 3 are obtained by multiplying 3 by 1, 2, 3, 4, 5,
6 and so on.
So the multiples of 3 are 3, 6, 9, 12, 15, 18, ……
The multiples of a number are infinite.
Example
Write down all the multiples of 9 between 70 and 120.
Note:
A number is a multiple of
its factors.
For example, 2 × 3 = 6
We say that 6 is a multiple of
2 and 3.
Also, 2 and 3 are factors of 6.
FUN ACTIVITY:
FACTORS AND MULTIPLES
SEARCH
Locate and circle the following
factors and multiples in the
table below. This can be done
horizontally, vertically or
diagonally.
Solution
1
5
3
6
9 12 9
Let us consider 9 x 6 = 54 and 9 x 7 = 63.
Both 54 and 63 are multiples of 9 but are less than 70.
2
1
2
3
6
7 22 1
4
1 16 27
We now consider 9 x 8, 9 x 9 and so on till we obtain all
the multiples of 9 between 70 and 120.
Multiples of 9 between 70 and 120 = 72, 81, 90, 99, 108, 117.
EXERCISE 1.5
1. In the list below, circle the numbers which are multiples of 6 :
36, 1, 12, 3, 16, 24, 9, 32
2. The following are multiples of a number.
Write down the number in each case.
(a) 10, 45, 55, 20
(b) 28, 35, 70, 14
(c) 48, 20, 16, 28
3 18
14 11 3 25 6 12 36
1
2
5 10 9
1
1 15 1
3
8
7
4 10
1. Factors of 12
2. First four multiples of 3
3. Factors of 10
4. First four multiples of 9
5. Factors of 22
6. Factors of 15
7. First five multiples of 2
8. Factors of 6
9. First four multiples of 4
10. Factors of 25
11. Factors of 14
12. Factors of 9
There are 5 numbers which are left.
What type of numbers are these?
3. Write down all the multiples of 8 between 40 and 100.
4. I am a multiple of 9. I have 2 digits. I am odd and I am also a multiple of 5. Which number am I?
5. I am a multiple of 15. I have 3 digits and I am the smallest of all the 3-digit multiples of 15.
Which number am I?
6. I am a multiple of 3. I am a 1-digit number. I am a square number. Which number am I?
8
Chapter 1 - Numbers, Factors and Multiples
The Least Common Multiple (L.C.M.)
The Least Common Multiple (L.C.M.) of two or more numbers
is the smallest (least) multiple of the common multiples.
Method 1: Listing multiples
STOP AND THINK
Where do we use L.C.M. in Mathematics?
NOTE TO TEACHER
Example
Find the L.C.M. of 8 and 12.
Solution
Multiples of 8 = 8, 16, 24, 32, 40, 48, 56, 64, 72,….
Prompt students to think of:
1. comparison of fractions.
2. addition and subtraction
of fractions with different
denominators.
Multiples of 12 = 12, 24, 36, 48, 60, 72, 84, ….
The common multiples are 24, 48, 72,…..
So, the L.C.M. of 8 and 12 is 24 (we choose the least common multiple).
Method 2: Using Prime Factorisation (in Expanded Form and Index Form)
Example 1
Find the L.C.M. of 24 and 60.
Solution
2
24
2
60
2
12
2
30
2
6
3
15
3
3
5
5
1
1
In Expanded Form
24 = 2 × 2 × 2 × 3
60 = 2 × 2 ×
3×5
L.C.M. of 24 and 60 = 2 × 2 × 2 × 3 × 5 = 120
In Index Form
24 = 23 × 3
60 = 22 × 3 × 5
L.C.M. of 24 and 60 = 23 × 3 × 5 = 120
To find the L.C.M., we find all the prime factors of the
given numbers.
Note: The common prime factors are taken only once.
We select all the prime factors. For the common factors,
we select the one with the highest power. If the power is
the same, we select the prime factor only once.
9
Chapter 1 - Numbers, Factors and Multiples
Example 2
CHECK THIS LINK
Find the L.C.M. of 18, 40 and 45.
https://www.geogebra.org/m/
Ciaukiwm
Solution
In Expanded Form
18 = 2 ×
40 = 2 × 2 × 2
45 =
http://www.projectmaths.ie/
geogebra/hcf-and-lcm/
3 × 3
×5
3×3 ×5
L.C.M. of 18, 40 and 45 = 2 × 2 × 2 × 3 × 3 × 5 = 360
In Index Form
18 = 2 × 32
40 = 23 ×
5
45 =
32 × 5
L.C.M. of 18, 40 and 45 = 23 × 32 × 5 = 360
Method 3: Repeated division by prime factors
Example
Find the L.C.M. of 12, 18 and 27.
Solution
2
2
3
3
3
12
6
3
1
1
1
,
,
,
,
,
18
9
9
3
1
1
, 27
, 27
, 27
, 9
, 3
1
Since 27 is not exactly divisible by 2, we rewrite 27 in the next two lines.
Since 9 is not exactly divisible by 2, we rewrite 9 in the next line.
Division is complete as we have obtained 1 for all the numbers.
L.C.M. of 12, 18 and 27 = 22 × 33 = 108
EXERCISE 1.6
1. Find the L.C.M. of the following numbers:
(a) 8 and 10
(b) 6 and 14
(c) 15 and 25
(e) 32 and 40
(f) 18 and 48
(g) 72 and 108
2. Find the L.C.M. of the following numbers:
(a) 6, 8 and 12
(b) 10, 15 and 25
(d) 20, 32 and 40
(e) 24, 48 and 60
10
(d) 10 and 35
(h) 64 and 96
(c) 14, 20 and 35
(f) 30, 36 and 45
Chapter 1 - Numbers, Factors and Multiples
Word problems involving H.C.F. and L.C.M.
Example 1
Alice has to make bouquets with 48 red roses and 20 yellow roses.
She needs to use all the roses to make similar bouquets with the
same combination of red and yellow roses in each bouquet. Find
the greatest number of red and yellow roses in each bouquet.
Solution
Alice has to make the greatest number of similar bouquets. She has to equally distribute the
red roses and yellow roses in each bouquet.
As she has to distribute the 2 items into their largest grouping to find the greatest number of
similar bouquets, she must calculate the highest common factor (H.C.F.) of 48 and 20.
48 = 24 × 3
20 = 22 × 5
H.C.F. of 48 and 20 = 22 = 4
Since the H.C.F. is 4, the greatest number of similar bouquets that Alice can make is 4.
That is,
Number of red roses in each bouquet = (48 ÷ 4) = 12
Number of yellow roses in each bouquet = (20 ÷ 4) = 5
Example 2
Two ice cream vans regularly come to sell ice cream in a children’s park. One comes every 4
days and the other comes every 6 days. If the two ice cream vans came together last Saturday,
when will they next come together again?
Solution
Here, we have two events: 1st event is when the 1st ice cream van comes to sell ice cream every
4 days and the 2nd event is when the 2nd van comes to sell ice cream every 6 days.
We should find when these two events will occur at the same time again in the future (common
multiples).
Since we should find when the soonest possible that the 2 vans will come together, we calculate
the least common multiple (L.C.M.) of 4 and 6.
4 = 22 6=2×3
L.C.M. of 4 and 6 = 22 × 3 = 12
So, the 2 ice cream vans will next come together in 12 days, on a Thursday.
11
Chapter 1 - Numbers, Factors and Multiples
EXERCISE 1.7
1. Sheena has two pieces of ribbon of lengths 18 cm and 24 cm respectively. She wants to cut
both pieces into smaller pieces of equal length that are as long as possible. What would be
the length of each piece cut?
2. Roy has 18 red marbles and 12 white marbles. If he wants to place them in identical groups
without any marbles left over, what is the greatest number of groups Roy can make?
3. A cinema runs its movies in two different halls everyday, 24 hours a day. One movie runs for
80 minutes and the second one runs for 120 minutes. Both movies start at 1.00 p.m. When
will the movies begin again at the same time?
4. Two boats leave the jetty at the same time. One of them returns to the
jetty every 25 minutes and the second one returns every 30 minutes.
When will the boats next be at the jetty at the same time?
5. Abdul baked 300 cookies, 480 muffins and 240 cupcakes. He wants to pack them into
containers to sell at a fair. Each container should be similar in terms of the number of cookies,
muffins and cupcakes.
(a) What is the greatest number of containers he can pack?
(b) How many cookies, muffins and cupcakes will there be in each container?
Summary
•
•
•
•
•
•
•
•
•
•
12
Even numbers are numbers which are divisible by 2 and end with 0, 2, 4, 6 or 8.
Odd numbers leave a remainder of 1 when divided by 2. Examples include 5, 21,
65 and so on.
Prime numbers are numbers having only 2 factors (1 and the number itself).
Composite numbers are numbers having more than 2 factors such as 14, 27, 48.
Square numbers are obtained when a whole number is multiplied by itself.
Examples: 1, 16, 81, 144.
Examples of triangular numbers are 1, 3, 6, 10, 15.
The factors of a number are the numbers that exactly divide the number, that is,
without leaving any remainder. Example: Factors of 16 are 1, 2, 4, 8 and 16.
The multiples of a number are obtained by multiplying that number by whole
numbers. Example: Multiples of 16 are 16, 32, 48, 64 ...
The Highest Common Factor (H.C.F.) of two or more numbers is the largest
factor that is common to these numbers.
The Least Common Multiple (L.C.M.) of two or more numbers is the smallest
multiple that is common to these numbers.
2
INTEGERS
Chapter 2 - Integers
Learning Objectives
By the end of this chapter, you should be able to:
• distinguish between positive and negative integers.
• represent positive and negative integers on a number line.
• compare and order integers.
• perform arithmetic operations on integers.
• find the square root of square numbers.
• perform arithmetic operations mentally.
In our everyday life, we find different types of numbers such
as whole numbers, fractions and decimals. Numbers play an
important role in the daily activities that we are involved in,
such as cooking, elections and sports.
Whole numbers are also called integers. However, fractions
and decimals are not integers.
CHECK THAT YOU CAN:
•
Perform the four
arithmetical operations
with numbers.
Identify square numbers.
•
KEY TERMS
Integers in real life
•
•
•
•
•
•
•
Lift
Thermometer
-5 c
Whole numbers
Integers
Positive Integers
Negative Integers
Number Line
Square roots
Square numbers
Note: We notice that numbers
can be positive or negative. In
this chapter, we will learn about
positive and negative integers.
In later chapters, you will come
across negative fractions and
negative decimals.
-2 c
0 c
40 c
30 c
38 c
Weather forecast
Observe the above pictures. What do you notice about the integers on each of the pictures
above?
Can you find other examples where you can see or use integers in real life?
13
Chapter 2 - Integers
Integers: positive, negative and zero
Positive integers are the whole numbers that are greater than
0 such as 15, 60 and 4 078.
Negative integers are the whole numbers that are less than 0,
for e.g. – 5 , – 10 , – 40. We read these numbers as ‘negative five’,
‘negative ten’ and ‘negative forty’ respectively. We represent
negative integers by placing a negative or minus ( – ) sign in
front of them.
Negative integers can be used to represent floors in buildings
(e.g. –1 and –2 are used to denote the basements), altitudes
below sea levels, temperatures (e.g. negative temperatures in
some European countries) or financial situations (e.g. a loss of
Rs 500 in business may be represented by – 500).
NOTE TO TEACHER
Prompt students to notice and discuss
the use of negative numbers. Provide
other examples where integers are
used in real life such as to define time
(AD and BC).
Note: When there is no negative
sign in front of an integer, then it
is understood that it is a positive
integer. It is conventional to omit
the positive or plus (+) sign in
front of positive numbers.
Zero is neutral and it is therefore neither positive nor negative.
Note: denotes the set of integers,
the set of negative integers.
+
denotes the set of positive integers and
–
denotes
Example 1
DID YOU KNOW
Here is a list of numbers:
6 , 3 , –89 , 4 , 9.6 , 0 , –653 , 902 , 0.78 , –7 , 40 , –300 000
5
From the above list, write down all the
(a) integers (b) positive integers (c) negative integers.
Solution
History of negative numbers
The Chinese were one of the earliest
known people who used negative
numbers in their calculations. They
used rods for counting. Black rods
were used for negative numbers
while red rods were used for
positive numbers.
(a) 6 , 3 , –89 , 0 , –653 , 902 , –7 , 40 , –300 000
Note: Fractions and decimals are not integers:
4
, 9.6, 0.78
5
(b) Positive integers : 6 , 3 , 902 , 40
Note: All these integers do not have a negative sign and are
therefore positive. Zero is neither positive nor negative.
(c) Negative integers: –89 , –653 , –7 , –300 000
Note: All these integers have a negative sign in front and
zero is neither positive nor negative.
14
The Indian
mathematician
Brahmagupta
later set out the
rules for dealing
with positive
and negative numbers and zero. He
used the term ‘fortune’ to represent
positive numbers and ‘debt’ to
represent negative numbers.
Chapter 2 - Integers
Example 2
Write each of the following as an integer (positive or
negative):
(a) An altitude of 20 000 metres above sea level
(b) 2 floors below the ground floor
Solution
(a) As the altitude is above sea level, we denote it by a
positive integer, that is 20 000.
(b) As the floors are below the ground floor, we denote
them by negative integers. Therefore, 2 floors below the
ground floor is –2.
DID YOU KNOW
The lowest point on land is the
shores of the Dead Sea (a salt lake
bordered by Jordan, Israel and
Palestine). It is 420 metres below sea
level or at an altitude of –420. The
highest point above sea level is the
peak of the Mount Everest which is
8 850 metres above sea level or at an
altitude of + 8 850 .
Example 3
Write down the temperature shown on each of these thermometers.
(a)
(b)
Solution
(a) –25⁰ C
(b) 26⁰ C
EXERCISE 2.1
1. Here is a list of numbers:
1 , 0.7 , –63 , 5 , 6 , –10 , –50 , 78 , 34 , 90 , –100 , 596 000 , –9 308 , 85 930 , 6.7
2
From the above numbers, list down all
(a) integers
(b) positive integers (c) negative integers.
2. Write each of the following as an integer (positive or negative):
(a) 50 C below zero
(b) A profit of Rs 1 000
(c) 350 metres below sea level
(d) A debt of Rs 5 000
(e) A loss of Rs 650
(f) A temperature of 300 C above zero
15
Chapter 2 - Integers
3. Write down the temperature shown on each of these thermometers.
(a)
(b)
(c)
(d)
(e)
(f)
Representing integers on a number line
A number line can be used to represent numbers. In earlier Grades, you learnt that the
numbers 0, 1, 2, 3 and so on can be represented on a number line as follows:
We now extend our number line to the left hand of zero to include negative numbers as follows:
Negative integers
Neutral
Positive integers
The number zero separates the positive integers from the negative integers.
The arrows on both sides of the number line indicate that the number line can be extended on
both sides as the numbers are infinite.
Drawing a number line to represent integers
Example 1
Draw a number line to represent all the integers
(a) from –5 to 3
(b) between –7 and 2.
Solution
(a)
We include both the integers –5 and 3 as we have to represent all integers from –5 to 3.
16
Chapter 2 - Integers
(b)
NOTE TO TEACHER
Consolidate the notion of 'between',
'from' and 'to'.
We do not include –7 and 2 as we have to represent all
integers between –7 and 2.
Caution:
When drawing a number line,
ensure that the markings are
equally spaced to represent the
numbers on the number line.
Example 2
Fill in the missing integers on the number line below:
(a)
-2 -1 0 1
(b)
3
(c)
-4 -3
-1 0 1 2
(d)
-3 -2
0 1
3
-10 -5
5 10
20
Solution
(a) Between the integers 1 and 3, we have the integer 2.
-2 -1
(b) Between the integers –3 and –1, we have the integer –2.
0
1
2
3
-4 -3 -2 -1 0 1
2
(c) Between the integers 1 and 3, we have the integer 2. Similarly, between 0 and –2 ,
we have the integer –1.
-3 -2 -1 0 1
2
3
(d) Here, there is a difference of 5 between any two markings.
The integer between 10 and 20 is 15 and the integer between 5 and –5 is 0.
-10 -5 0 5 10 15 20
EXERCISE 2.2
1. Using separate number lines, represent the following:
(a) integers from –3 to 3
(b) integers between –4 and 1.
2. Copy and complete the number lines below :
(a)
(b)
(c)
(d)
-25
25
17
Chapter 2 - Integers
Order of Integers
STOP AND THINK
Consider the number line below:
The positive numbers are greater than the negative numbers
and 0. As we move to the right of the number line, the numbers
get bigger and as we move to the left of the number line, the
numbers get smaller.
For example, 9 is greater than 2 or 2 is less than 9.
We use the symbol > to denote ‘greater than’ and the symbol
< to denote ‘less than’ or ‘smaller than’.
Have you ever heard on the
weather news that there will be
a decrease in temperature or that
temperatures in European countries
have dropped to below 00 C or that
the temperature is –100 C ?
Compare a temperature of 250 C
and a temperature of –100 C. Which
temperature is colder and why?
Therefore, which of the numbers 25
and -10 is smaller?
So, 9 > 2 or 2 < 9.
Example
RECALL
Rewrite the following integers in ascending order:
Ascending order:
Start with the smallest to the
greatest (increasing order).
–3 , –1 , 6 , 8 , –11 , –6
Solution
We use a number line to help us to determine the smallest and
largest integer.
Hence, the numbers in ascending order (starting with the least) are:
–11, –6 , –3, –1, 6, 8
EXERCISE 2.3
1. Express the following statements using the symbol “<”, or “ >”.
(a) Five is less than ten.
(b) Four is greater than negative one.
(c) Negative twelve is less than negative nine.
(d) One hundred and two is greater than negative two hundred.
2. Fill in each box with “<”, or “ >”.
(a) –3
18
7
(b) 5
–5
(c) –6
–2
Descending order:
Start with the greatest to the
smallest (decreasing order).
Chapter 2 - Integers
(d) 0
–2
(g) –12
–26
(e) –24
–25
(f) 8
–9
(h) –15
–11
(i) 2
0
3. For each of the pairs of temperatures, write down the lower
temperature.
(a) 60 C or 40 C(b) –50 C or 10 C
(c) 00 C or 30 C (d) –60 C or –40 C
4. Rewrite the following integers in ascending order.
(a) –7 , 13 , 7 , –5 , 4 , –24 , 22
STOP AND THINK
Siya, Feroz, Nathan and Aisha were
all waiting for their respective
buses to arrive after school. Siya’s
bus leaves 10 minutes before 3
p.m, Feroz’s bus leaves 15 minutes
after 3 p.m, Nathan’s bus leaves 5
minutes before Feroz’s bus, and
Aisha’s bus leaves 3 minutes after
Siya’s bus.
Can you find out the times at which
each of the buses will leave?
(b) –250 , 500 , –612 , –435 , 110
(c) 6 , –5 , 4 , –4 , 0 , –2 , 3
Write down these times in
ascending order.
5. Arrange the following integers in descending order.
(a) 52 , –8 , 11 , –16 , –33 , –19 , 26
(b) 4 , –7 , –15 , –1 , 1 , 5 , –8
(c) –235 , –421 , –158 , 125 , 239 , –127
4
Addition and Subtraction of Integers
3
2
One way of carrying out addition and subtraction of integers is
to make use of a number line.
We can consider a vertical number line as a ladder. As we climb
up the ladder, the numbers increase and as we go down the
ladder, the numbers decrease.
+
1
0
–1
–
–2
–3
Example
(a) Find the value of 1 + 3.
Solution
is at 1
He climbs up the ladder by
3 steps and reaches step 4.
Hence, 1 + 3 = 4
(b) Find the value of 4 – 5.
Solution
5
4
3
is at –1
He moves down the ladder by
2 steps and reaches step –3.
Hence, –1 – 2 = –3
2
–5
1
He moves down the ladder by 0
–1
5 steps and reaches step –1.
So, 4 – 5 = –1
1
0
(c) Find the value of –1 – 2.
Solution
3
is at 4
+3
2
4
(d) Evaluate – 3 + 5.
Solution
2
1
is at –3 and moves up 5
1
0
–1
–2
–3
2
–2
steps to end up at 2.
Hence, –3 + 5 = 2
Alternatively, it can be
represented as
–4 –3 –2 –1
0
+5
–1
–2
–3
0
1
2
3
19
Chapter 2 - Integers
EXERCISE 2.4
Evaluate each of the following.
(a) 4 – 1
(b) 10 – 12 (d) 6 – 8
(e) –4 + 4
(h) 0 – 6
(g) –5 + 9
(j) –3 – 7
(k) –8 – 3 STOP AND THINK
(c) –7 + 4
(f) –11 + 6
(i) –2 – 5
(l) –10 + 10
How do you use the number line to
calculate –34 + 45?
What will be the value of –245 + 130 ?
Rules for addition and subtraction of Integers
Consider the following.
(a) (–1) – (2) = –3
-4 -3 -2 -1 0 1
(b) (–2) – 3 = –5
Observation:
(a) (–1) – (2) = –(1 + 2) = –3
In general,
(b) (–2) – 3 = –(2 + 3) = –5
(–a) – (b) = –(a + b).
Other rules involving integers are:
a + (–b) = a – b
a – (+ b) = a – b
a – (–b) = a + b
-6 -5 -4 -3 -2 -1
Note: To distinguish
between the operators –
and + and a positve or a
negative integer, we may
use brackets.
(–a) + (–b) = –(a + b)
Example 1
Evaluate
(a) 1 + ( –3)
(b) 6 – (+7).
Solution
Solution
1 + (–3) = 1 – 3
At 1, we move 3 steps to the left.
So, we are at –2.
6 – (+7) = 6 – 7
At 6, we move 7 steps to the left.
Hence, 1 + (–3) = –2
-5 -4 -3 -2 -1 0 1 2 3 4 5
So, 6 – (+7) = –1.
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6
Example 2
Evaluate
(a) 7 + (–6)
(b) –1 + (–4)
(c) 2 + (–4)
(d) –3 – (–2)
Solution
(a) 7 + (–6)
=7–6
=1
(b) –1 + (–4)
(c) 2 + (–4)
= –1 – 4
=2–4
= – (1 + 4)
= –2
= –5
Note: If a > b, then a – b > 0 and if a < b, then a – b < 0.
20
(d) –3 – (–2)
= –3 + 2
= –1
Chapter 2 - Integers
EXERCISE 2.5
1. Find the value of :
(a) 2 + 3
(b) 10 + 12
(c) 9 + 21
(d) –2 – 3
(e) –7 – 5
2. Find the value of:
(a) 6 – (+4)
(b) 8 – (+5)
(c) 3 – (+4)
(d) 7 + (–5)
(e) 5 + (–7)
3. Find the value of:
(a) 2 – (–5)
(b) 3 – (–7)
(c) –8 – (–4)
4. Find the value of :
(a) 3 – (–2)
(b) 3 – (+8)
(g) (–3) + (–2)
(h) 4 – (–3)
(c) 0 – (–10)
(i) 11 – (–5)
5. Evaluate
(a) 8 – (+2) – (–3)
(d) – 3 – (–3) – (+ 6)
(d) 10 – (–6)
(d) –4 + 6
(j) 5 – (+5)
(b) 7 + (–10) – (–5)
(e) 9 – (+4) – (+5)
(f) –6 – 3
(e) 25 – (–15)
(e) –5 – (+7)
(k) 0 – (+6)
(f) –12 + 6
(l) –9 + (–3)
(c) (–2) + (–5) + 3
(f) 4 – (–5) + (–6)
Example
(a) Evaluate 15 – 25
Method 1
25
We can make use of a number line as follows:
10 + 15
-10
0
15
We can carry out this operation in two parts.
We first subtract 15 from 15 to obtain 0 (moving to the left of the number line).
We again subtract 10 from 0 to obtain –10 (moving to the left of the number line).
So, 15 – 25 = –10.
A number line may not however be convenient for operating on large numbers. We
therefore need other methods for addition and subtraction of integers.
Method 2
15 is placed in the positive column whereas 25 is
placed in the negative column.
We write 25 as (10 + 15).
15 is common to both columns, so we cancel it,
i.e., (– 15) + 15 = 0. 10 is left in the negative column.
So, 15 – 25 = – 10.
_
25
+
15
10 + 15
21
Chapter 2 - Integers
NOTE TO TEACHER
(b) Evaluate –120 + 250
Prompt students to decompose the bigger
number into a number common to both
columns and discuss how the numbers
cancel out.
250 is placed in the positive column and 120
in the negative column.
We write 250 as (120 + 130).
130 is left after cancelling the common number 120.
130 is in the positive column.
_
+
120
250
So, the answer is 130.
120 + 130
_
(c) Evaluate (–42) + (–38)
+
42
+
38
(–42) + (–38) = –42 – 38
80
Both numbers are negative and are thus placed under the same negative column.
We add both numbers 42 + 38.
We obtain 80 and since it is found in the negative column, the answer is –80.
(d) Evaluate –36 – (–21 ) + ( –15)
–36 – (–21 ) + (–15) = –36 + 21 – 15
21 is placed in the positive column.
36 and 15 are placed under the same negative column
and are added. We obtain 51.
51 can be written as (21 + 30). We cancel 21.
30 is left in the negative column.
_
+
36 +
15
51
21
21 + 30
So, the answer is –30.
EXERCISE 2.6
1. Evaluate:
(a) –16 + 30
(e) –35 + (–17)
(b) –12 + (–17)
(f) 110 – 55 + 15
(c) 53 + (–23) (g) –540 – (–220) – (+ 50)
2. Fill in each box with the correct answer.
(a) –8 +
= –6(b) 8 –
(c) 12 +
– 3 = 17
22
= –7
(d) – 14 – 12 –
= –32
(d) 62 – (+85)
(h) 225 + (–450) – (–375)
Chapter 2 - Integers
3. The number line below shows the temperature, in
degrees celsius, of different cities around the world on a
particular day.
New York
London Paris
–4⁰ C
0⁰ C
Port-Louis
30⁰ C
3⁰ C
(a) What is the difference in temperature between the
hottest and the coldest places shown above?
(b) What is the difference in temperature between New
York and Paris?
4. Rose participates in a quiz competition. She earns positive
marks for each correct answer she gives and earns
negative marks for each incorrect answer she gives. In
four successive rounds, her scores were 20, –5, 25 and –10.
What is her total score after the four rounds?
5. Complete the magic square
given so that the sum of the
numbers in each row, column
or diagonal is –15.
–4
–2
–3
–1
–6
6. Complete the magic star so that the sum of the numbers
along any line is 6.
–4
2
–1
1
3
5
6
7
Multiplication of Integers
We can think of multiplication of integers as repeated
addition.
Consider 5 × 3.
5 × 3 = 5 + 5 + 5 = 15 (There are 3 groups of 5 in 15).
In the same way, we can multiply a negative integer by a
positive integer.
Consider (–5) × 3.
(–5) × 3 = (–5) + (–5) + (–5) = –15 (There are 3 groups of –5 in –15).
Activity
Assume that at the beginning
of this activity, you have a bank
account balance of Rs 1 800.
Directions:
(i) Reproduce the 7 cards below on
bristol paper.
(ii) Cut out the cards.
(iii) Place them in ascending order.
(iv) Draw one card at a time and do
the necessary calculation to
know the balance each time.
(v) Note: Your bank will allow you
to withdraw more money
than is in your account. If you
do take more money out of
your account than you have in
it, it will result in a negative
balance.
(vi) Calculate your final bank
balance.
(vii) Repeat the process one more
time but this time, mix them
up and do the calculations
again.
(viii) What do you observe?
1 You take Rs 700 out of
your account to buy
some clothes.
2 You receive Rs 250 as
gift from your aunt
and deposit it into your
account.
3 You take Rs 1 400 out
of your account to buy
books.
4 You take Rs 50 out of
your account to pay a
library fine.
5 You take Rs 225 out of
your account to buy a
birthday gift for your
best friend.
6 You win a maths
competition and receive
Rs 550 as reward. You
deposit the money into your
account.
7 You take Rs 150 out of
your account to pay for
your class outing.
23
Chapter 2 - Integers
Column 1
Column 2
Column 3
5 × 3 = 15 –5
3 × 5 = 15 –5
–5 × 3 = –15 +5
5 × 2 = 10 –5
5 × 1 = 5 –5
2 × 5 = 10 –5
1 × 5 = 5 –5
–5 × 2 = –10 +5
–5 × 1 = – 5 +5
5×0=0
0×5=0
–5 × 0 = 0
5 × (–1) = ?
–1 × 5 = ?
–5 × -1 = ?
5 × (–2) = ?
–2 × 5 = ?
–5 × -2 = ?
5 × (–3) = ?
–3 × 5 = ?
–5 × -3 = ?
1. Consider column 1. What do you notice?
Note: a x (–b) = a x –b
As we move down the column, the numbers decrease by 5.
We conclude that
5 × (–1) = –5 as 0 – 5 = –5.
Similarly, 5 × (–2) = –10 (as –5 – 5 = –10) and 5 × (–3) = –15 (as –10 – 5 = –15).
2. Now, consider column 2. What do you notice?
As we move down the column, the numbers decrease by 5.
We conclude that
–1 x 5 = –5, –2 x 5 = –10 and –3 x 5 = –15.
3. We now consider column 3. What do you notice?
As we move down the column, we observe that the numbers increase by 5.
We conclude that
(–5) × (–1) = 5 (as 0 + 5 = 5)
Similarly (–5) × (–2) = 10 (as 5 + 5 = 10) and (–5) × (–3) = 15 (as 10 + 5 = 15).
What do you notice about multiplication of:
(i) two positive integers?
(ii) two negative integers?
(iii) one positive integer and one negative integer?
Note: We can also observe
that any number (whether
positive or negative) when
multiplied by 0 gives 0.
Example 1
Evaluate
(a) 4 x 5
(c) –4 x 5
Sign of 1st
number
(b) 4 x –5 (d) –4 x –5
+
Solution
(a) 4 x 5 = 20
(c) –4 x 5 = –20
24
(b) 4 x –5 = –20
(d) –4 x –5 = 20
Note: We multiply the
2 numbers and then
consider the signs.
-+
-
Sign of 2nd
number
+
-+
Sign of
product
of the two
numbers
+
+
-
Chapter 2 - Integers
Example 2
Evaluate
(a) (–6) × 5 × (–4)
POINTS TO REMEMBER
The product of two
numbers with same
signs will result in
a positive number
whereas the product
of two numbers with
different signs will
result in a negative
number.
(b) (–8) × (–4) × (–2)
Solution
(a) (–6) × 5 × (–4)
(–6) × 5 = –30
Then, –30 × (–4) = 120
(b) (–8) × (–4) × (–2)
(–8) × (–4) = 32
Then, 32 × (–2) = –64
EXERCISE 2.7
1. Find the value of
(a) 5 × 6
(b) 14 × 5
(c) (–3) × (–8)
2. Find the value of
(a) (–8) × 7 (b) 9 × (–5)
(e) 0 × (–9)
(f) (–11) × 11
3. Find the value of
(a) 8 × 3 × (–3)
(e) (–32) × 0 × 45
(i) (–5) × (–1) × (–6)
(d) (–10) × (–6)
(c) (–12) × 4
(g) (–12) × 10
(e) (–40) × (–20)
(d) 11 × (–3)
(h) 15 × (–6)
(b) (–4) × 3 × (–7) (c) 18 × 0 × (–2)
(f) 4 × (–3) × (–4)
(g) (–11) × 5 × (–4)
(j) (–4) × (–4) × (–5) × (–10)
(d) 9 × (–2) × 2
(h) (–13) × (–5) × (–5)
4. Fill in the blank spaces with the correct integer value.
(a) (–9) × _____ = –36
(b) _____ × 11 = –66
(c) (–22) × _____ = 110
(d) 2 × _____ × –7 = –70
5. The temperature of a room is 30⁰ C and needs to be lowered for a certain freezing process.
If the temperature is decreased by 4⁰ C every hour, what will be the room temperature 12
hours after the process begins?
6. Bryan plays a computer game. He is awarded 3 points each time he hits a moving target.
However, each time he misses the target, he loses 2 points. At the end of the game, he
makes 12 hits and 8 misses. What is his score?
7. Complete the magic square below in which the magic product along any row, column
or diagonal is 1 000.
–5
–8
10
25
Chapter 2 - Integers
Division of Integers
Division is normally considered as the inverse operation of multiplication. Let’s have a look at
the following table:
Multiplication
Division
Sign of 1st
number
2 × 5 = 10
10 ÷ 5 = 2
10 ÷ 2 = 5
–2 × 5 = –10
–10 ÷ 5 = –2
–10 ÷ –2 = 5
–2 × –5 = 10
10 ÷ –5 = –2
10 ÷ –2 = –5
3 × 5 = 15
15 ÷ 5 = 3
15 ÷ 3 = 5
–3 × 5 = –15
–15 ÷ 5 = –3
–15 ÷ –3 = 5
–3 × –5 = 15
15 ÷ –5 = –3
15 ÷ –3 = –5
Sign of 2nd
number
+
+
-+
-
Sign of
quotient
of the two
numbers
+
+
-+
-
What do you observe?
Division involving Zero
STOP AND THINK
Zero divided by any integer value gives zero.
E.g. 0 ÷ 5 = 0
0 ÷ –5 = 0
What is 5 ÷ 0?
Example
Evaluate
(a) 24 ÷ 6
(b) 24 ÷ –6
(c) –24 ÷ 6
(d) –24 ÷ –6
Note: a ÷ (–b) = a ÷ –b
Solution
(a) 24 ÷ 6 = 4
(c) –24 ÷ 6 = –4
(b) 24 ÷ –6 = –4
(d) –24 ÷ –6 = 4
Note: We divide the 2 numbers
and then consider the signs.
EXERCISE 2.8
Find the value of:
(a) 42 ÷ 7
(e) 54 ÷ (–9)
(i) 64 ÷ (–4)
(b) (–28) ÷ 7 (f) (–121) ÷ (–11)
(j) (–55) ÷ 5
(c) 16 ÷ (–4)
(g) 0 ÷ (–45)
(k) 625 ÷ (–5)
(d) (–8) ÷ (8)
(h) (–48) ÷ (–4)
(l) (–72) ÷ 8
Square Numbers and Square Roots
Consider the square number 16.
It can be written as 16 = 4 × 4.
We say that the square of 4 is 16 or
‘4 squared’ is 16 and the square root of 16 is 4.
RECALL
squared is
Square Roots
4
16
is the square root of
We use the symbol √ to denote square root of a number, i.e., √16 = 4.
26
A number multiplied
by itself is called a
square number.
Examples include 1,
4, 16 and so on.
Chapter 2 - Integers
Example
Find
(a) √25
(b)
√81
(c)
Note: The square of (–4) or (–4)
squared is also 16 as –4 squared is
–4 × –4 which is 16. Thus,16 has
two square roots –4 and 4.
However, the symbol √ means
the principal square root or the
positive square root.
For example: The square roots of 16
are 4 and −4.
√169
Solution
We know that
(a) 5 × 5 = 25 (b) 9 × 9 = 81
√25 = 5
√81 = 9
(c) 13 × 13 = 169
√169 = 13
But √16 = 4 (NOT –4).
Square roots of large numbers
We use the division method to find square roots of large numbers.
Example
Find (a) √196
(b) √900
Solution
(a) We first use the division method to write
196 as a product of its prime factors.
So, 196 = 2 x 2 x 7 x 7
√196 = √2 x 2 x 7 x 7
=
=
2 x
14
7
2
2
7
7
196
98
49
7
1
(b) We first use the division method to write
900 as a product of its prime factors.
√900 = √2 x 2 x 3 x 3 x 5 x 5
=
=
2 x
30
3
x 5
2
2
3
3
5
5
900
450
225
75
25
5
1
Note: For each pair of similar factors we take only one. We then multiply the selected
factor from the pairs to find the square root.
EXERCISE 2.9
FUN FACT:
1. List the square numbers between :
(a) 20 and 70
(b) 95 and 150
2. Find the positive square root of
(a) 36
(b) 100
(c) 144
What do you notice about the
date 4/4/16?
(d) 225
3. Find the value of
(a) √49
(b) √121
(c) √256
(d) √576
(e) 400
We can say that the 4th April
2016 is a square root day.
Which date can be called a
square root day before and
after 4/4/16 ?
27
Chapter 2 - Integers
Mental Arithmetic
Mental arithmetic can be useful in our lives
and we can solve many problems without
performing written calculations. Mental
arithmetic can help us make decisions
faster and can provide a second check on
written calculations.
STOP AND THINK
How would you do the following
calculations mentally:
(i) 199 +199?
(ii) 350 - 49?
There are different strategies that can be used for mental
arithmetic and we illustrate some of these strategies in the
following section.
Addition and Subtraction
Strategy 1: Partitioning
In this strategy, we mentally break up the numbers into units,
tens, hundreds and so on. We then perform the addition or
subtraction by working with the largest numbers first (e.g.
hundreds followed by tens and then units).
Note: It is NOT important for you to
know the names of the strategies
but it is important to understand the
strategies so that you can use them
when doing calculations.
Example 1
Calculate 124 + 238 mentally.
STOP AND THINK
Solution
124 + 238
Step 1: We first add the hundreds: 100 + 200 = 300
Step 2: We then add the tens: 20 + 30 = 50
Step 3: Add the units: 4 + 8 = 12
Finally we add all the numbers
300 + 50 + 12 = 362.
You will need to decide which
strategy is best for you to use
depending on the numbers that
are involved in the calculation.
E.g. It may be more difficult to use
the partitioning strategy for the
following subtraction: 748 – 259.
Why do you think this is so?
Example 2
Calculate 459 – 325 mentally.
Solution
Method 1
Step 1: 459 –300 = 159
Step 2: 159 – 20 = 139
Finally, 139 – 5 = 134.
28
Method 2
Step 1: We first subtract the hundreds: 400 – 300 = 100.
Step 2: We then subtract the tens: 50 – 20 = 30.
Step 3: We then subtract the units: 9 – 5 = 4.
Finally, 459 – 325 = 100 + 30 + 4 = 134.
Chapter 2 - Integers
Strategy 2: Grouping using multiples of 10.
In this strategy, we group numbers in order to obtain multiples of 10 so that it is easier to perform
the calculation.
Example 1
Example 2
Calculate 96 + 25 mentally.
Calculate 353 – 55 – 45 mentally.
Solution
Solution
96 + 25 = (96 + 4) + 21
= 100 + 21
= 121
353 – 55 – 45 = 353 – (55 + 45)
= 353 – 100
= 253
Step 1: We first break up number 25 into
4 and 21.
Step 2: We regroup 96 and 4 to obtain 100.
Step 3: We add 21 to 100 to obtain 121.
Step 1: We first combine 55 and 45 to get
100.
Step 2: We then subtract 100 from 353 to
obtain 253.
Strategy 3: Compensation
In this strategy, we modify one of the numbers ‘to make’ a ten, hundred or thousand and so on
and then adjust the calculation by adding or subtracting the difference between the original
number and the modified number.
Example 1
Example 2
Calculate 673 + 99.
Calculate 82 – 36.
Solution
Solution
We first add 673 to 100 since 99 is close
to 100: 673 + 100 = 773
We first perform 82 – 40 = 42 since 40
is close to 36.
But we have added 1 in excess.
Thus, we subtract 1 from the answer:
But we have subtracted 4 in excess.
Thus, we add 4 to the answer.
773 – 1 = 772
42 + 4 = 46
So,
So,
673 + 99 = 772.
EXERCISE 2.10
NOTE TO TEACHER
1. Calculate mentally
(a) 18 + 24
(b) 28 + 33
(d) 94 + 58 (e) 59 + 59 (g) 143 + 259
(h) 46 + 96 + 123
2. Calculate mentally
(a) 57 – 9
(b) 72 – 7
(g) 54 – 37
(h) 288 – 46
82 – 36 = 46.
(c) 124 – 8
(i) 94 – 66
(c) 28 + 67
(f) 99 + 299
(d) 45 – 12
(j) 75 – 48
When doing mental arithmetic, allow
students to use jottings if necessary
to help them to keep track of long
calculations.
(e) 68 – 25
(k) 137 – 19
(f) 155 – 21
(l) 254 – 27
29
Chapter 2 - Integers
Multiplication
Strategy 1: Multiplying and dividing by 10, 100 etc. and then adjusting
Example
Find 5 × 24.
Solution
We first multiply 24 by 10 instead of 5, that is 24 × 10 = 240.
We then adjust by dividing by 2 as we know that 5 is half of 10,
that is, 240 ÷ 2 = 120.
So,
5 × 24 = 120.
Note: We can use this
strategy when multiplying
by 50 (multiply by 100
and then divide by 2).
Multiplying by 25 is the
same as multiplying by 100
and dividing by 4.
Strategy 2: Splitting one of the numbers and then performing the calculation
Example
Find
(a) 7 × 32 (b) 13 × 39
Solution
(b) 13 × 39
(a) 7 × 32
7 × 32 = 7 × (30 + 2)
= (7 × 30) + (7 × 2)
= 210 + 14
= 224
As 39 is close to 40, we first perform the
13 × 39 = 13 × (40 – 1) calculation 13 × 40 (40 groups of 13).
However, we need only 39 groups of 13.
We now subtract (1 × 13) or 13 from our
520 – 13 = 507
answer.
So, 13 × 39 = 507
EXERCISE 2.11
Calculate mentally
(a) 4 × 26
(b) 6 × 49
(c) 7 × 108
(d) 209 × 9
(e) 220 × 40
(f) 14 × 25
(g) 18 × 50
(h) 22 × 13
(i) 60 × 15
(j) 33 × 21
Division
Strategy: Splitting the number into two or more parts before performing
division
In this strategy, we split the number into two or more parts so that we have numbers that are
exactly divisible by the other number to make calculations easier.
30
Chapter 2 - Integers
Example 1
Example 2
Find 48 ÷ 3.
Find 295 ÷ 5.
Solution
Solution
We break up 48 into two parts. We
take numbers that are divisible by 3,
that is, 30 and 18.
We break up 295 into two parts.
We take numbers divisible by 5,
that is, 250 + 45.
48 ÷ 3 = 48 = 30 + 18 = 30 + 18
3
3
3
3
= 10 + 6 = 16
295 ÷ 5 = 295 = 250 + 45 = 250 + 45
5
5
5
5
= 50 + 9 = 59
EXERCISE 2.12
Calculate mentally
(a) 42 ÷ 3
(b) 72 ÷ 4
(g) 108 ÷ 4
(h) 288 ÷ 9
(c) 545 ÷ 5
(i) 264 ÷ 12
(d) 90 ÷ 6
(j) 240 ÷ 16
(e) 98 ÷ 7
(k) 345 ÷ 15
(f) 128 ÷ 8
(l) 576 ÷ 18
Summary
•
Integers are both positive, negative and include zero. All positive integers
are to the right of zero and all negative integers are to the left of zero.
Negative integers
•
Positive integers
Addition and Subtraction of Integers
In general, (–a) – (b) = –(a + b)
a – (–b) = a + b
•
Neutral
a + (–b) = a – b
(–a) + (–b) = –(a + b)
Multiplication and Division of Integers
Sign of 1st
number
+
-
+
a – (+ b) = a – b
Sign of 2nd Sign of product/quotient of
number
the two numbers
+
-
+
+
+
-
•
Square numbers and Square Roots:
Square numbers are numbers obtained when a whole number is multiplied by
itself. Examples are 1, 4, 9, 16,.....
Square root: √1 = 1, √4 = 2, √16 = 4, ....
•
Mental arithmetic:
Mental arithmetic is a useful way for us to estimate or check our calculations.
There are different mental calculation strategies that can be used for addition,
subtraction, multiplication and division.
31
3
ORDER OF OPERATIONS
Chapter 3 - Order of Operations
Learning Objectives
By the end of this chapter, you should be able to:
• perform operations according to the BODMAS convention.
• use the commutative, associative and distributive properties of operations.
Investigate:
Case 1
CHECK THAT YOU CAN:
8+7–6=?
8+7–6=?
• Add, subtract, multiply and
divide integers.
KEY TERMS
Lina
Sam
Lina does addition first,
followed by subtraction,
i.e., 8 + 7 = 15 15 – 6 = 9
•
•
•
•
Sam does subtraction
first, followed by
addition,
i.e., 7 – 6 = 1
8+1=9
BODMAS
Commutative
Associative
Distributive
We can observe that for addition and subtraction, the order in which the operations are carried
out does not matter.
Case 2
6×4÷2=?
Lina
Lina does multiplication
first, followed by division,
i.e., 6 × 4 = 24
24 ÷ 2 = 12 6×4÷2=?
Sam
Sam does division first,
followed by multiplication,
i.e., 4 ÷ 2 = 2
6 × 2 =12
Here also, the order in which the operations multiplication and division are carried out does not
matter.
32
Chapter 3 - Order of Operations
Case 3
8 + 7 – 6 × 4 = –9
8 + 7 – 6 × 4 = 12
NOTE TO TEACHER
Prompt students to think of the
result based on the order of the
operations.
Lina
Sam
Can you find out why Lina and Sam got different answers for the same problem?
Lina may have done the
following calculation:
8 + 7 = 15
6 x 4 = 24
15 – 24 = –9
Sam may have done the
following calculation:
7–6=1
1x4=4
8 + 4 = 12
What is the correct solution?
–9
STOP AND THINK
Does it make a difference if
1. you put your right shoe first and
then your left shoe or your left
shoe first and then your right
shoe?
2. I say goodbye first on the phone
and then hang up or I hang up first
and then say goodbye?
Using a calculator, we can see that Lina is right. The correct answer is indeed –9. Why is Lina right?
In real life, there is an order of doing things.
For example, can you wear your shoes and then wear your socks?
Similarly, in Mathematics, there are certain rules that are used to perform
arithmetic calculations. The four operations of addition, subtraction,
multiplication and division are carried out in a specific order to obtain the
correct solution.
Order of operations: BODMAS Convention
We make use of the mathematical convention, “BODMAS”, to perform the operations in the
correct order.
BODMAS stands for :
Bracket (calculation inside brackets come first)
Orders (numbers involving powers or square roots)
Division
(Division and Multiplication rank equally)
Multiplication
Addition
(Addition and Subtraction rank equally)
Subtraction
}
}
33
Chapter 3 - Order of Operations
Rules:
STOP AND THINK
1. We first perform operations within Brackets, starting with
the innermost brackets and moving outwards.
Example: [4 + (5 – 2) x 1] = [4 + 3 x 1] = 4 + 3 = 7
2. We perform:
• Orders (Numbers involving powers or square roots)
• Division or Multiplication whichever comes first,
6
6
6
6
Using +,– , x and ÷ only once
to fill the 3 boxes above, how
many different values can be
obtained?
Two examples are given below.
Example 1:
6 x 6 + 6 ÷ 6 = 37
Example 2:
6 ÷ 6 x 6 + 6 = 12
from left to right
• Addition or Subtraction whichever comes first,
from left to right
NOTE TO TEACHER
Case 3 is therefore correctly solved as follows:
8+7–6×4
= 8 + 7 – 24
Multiplication comes first and is performed.
= 15 – 24
Addition is performed.
= –9
Subtraction is performed.
Encourage students to explain
their solutions based on the order
in which they carried out the
operations.
Example 1
Calculate
(a) 15 ÷ 3 × 4 ÷ 2
(b) 8 – 2 + 5 × 3 (c) 16 + 22 × 5 – 56 ÷ 7
Solution
As Multiplication and Division rank equally, we perform the operations from left to right:
15 ÷ 3 × 4 ÷ 2 = 5 × 4 ÷ 2
Division is performed.
= 20 ÷ 2
Multiplication is performed.
= 10
Division is performed.
(b) 8 – 2 + 5 × 3 = 8 – 2 + 15 Multiplication is performed first.
= 6 + 15 Subtraction comes first on the line.
= 21Then Addition is performed.
(c) 16 + 22 × 5 – 56 ÷ 7 = 16 + 4 × 5 – 56 ÷ 7
Square 2
= 16 + 20 – 8
Multiplication and Division are performed.
= 36 – 8
Addition is performed.
= 28Subtraction is then performed.
34
Chapter 3 - Order of Operations
Example 2
Evaluate (a) 2 × (5 – 7) – 4 + 9
(b) [(4 + 21) ÷ 5 + 3] × 2
Solution
(a) 2 × (5 – 7) – 4 + 9 = 2 × (–2) – 4 + 9
We perform the operation within brackets first.
Multiplication is performed.
= –4 – 4 + 9
= –8 + 9
Subtraction comes first.
= 1Then Addition is performed.
(b) [(4 + 21) ÷ 5 + 3] × 2 = [(25) ÷ 5 + 3] × 2
Inner Brackets, i.e. , (4 + 21), is performed.
= [5 + 3] × 2
Division within brackets is performed.
= 8 × 2
Operations within Brackets, is done next.
= 16 Multiplication is carried out last.
Example 3
Ishita brought 20 cookies at school to share with her friends for her birthday. She has two best
friends to whom she gave 3 cookies each. She then gave 2 cookies to each of her five friends
who bought her a gift. Write an expression for the number of cookies left and evaluate it.
20 cookies
2 best friends:
3 cookies each
5 friends:
2 cookies each
Left = ?
Solution
Number of cookies left = 20 – [(2 × 3) + (5 × 2)]
= 20 – [6 + 10]
= 20 – 16
=4
This means Ishita has 4 cookies left.
Activity
Your school is organising a “Fun Day” where your class will be responsible for the game
“Treasure Hunt”. Treasures will be hidden in 8 classrooms and the classrooms are named as
follows: A1, A2, A3, B4, B5, B6, C7 and C8.
You need to prepare some clues to get to the treasures that you hide. You decide to use
BODMAS as clues.
For example, a student has to solve the following question to know that one of the treasures
is hidden in classroom A1:
A [2 × 3 – 20 ÷ 4] = A1
Now, you have to formulate 7 questions as clues for the 7 other classrooms.
35
Chapter 3 - Order of Operations
EXERCISE 3.1
1. Evaluate
(a) 12 + 5 – 7
(b) 2 × 4 – 6
(c) 8 + 6 ÷ 3 × 4
(d) 11 × 5 ÷ 5 × 3
(e) 16 – 4 + 8 ÷ 2
(f) 24 ÷ 6 × 12 ÷ 4
(g) 18 + 3 × 0 – 5
(h) 25 ÷ 5 + 12 – 7 (i) (–9) × 2 ÷ 6 – 5
(j) 13 – 6 + 3 × 11
(k) 13 + 1 – 9 × 4 ÷ 6
(l) 8 × 8 + 8 – 8 ÷ 8
(m) (–3) + 7 – 4 × (–10) ÷ 2
(n) 22 ÷ 2 × 0 – 8 + 12
(o) 24 – 36 ÷ 3 – 4 × 2
(p) 15 – 8 ÷ 4 + 16
(q) 50 ÷ 52 × 3 + 10 – 6
(r) 2 × 62 ÷ 9 – 12 + 28
2. Find the value of
(a) 7 + (9 – 5 ) × 3
(b) 8 × (10 × 2) ÷ 4
(c) 12 ÷ 3 × (4 + 7)
(d) 6 + [14 – (8 × 2)] × 3
(e) [(1 + 17) ÷ 6] × 5
(f) (12 × 4) ÷ 6 + 9
(g) 66 ÷ 6 × (4 + 5) + 12
(h) (5 × 8 ) ÷ 4 × (2 + 3)
(i) [(–6) × 3] ÷ (–9) × 7 – 4
(j) 4 + (10 – 8 × 2 ÷ 2) × 7
(k) 24 – [4 – (0 – 8)] ÷ 6
(l) [56 ÷ 8 × (8 – 10) ] + 15
3. Circle the correct answer.
STOP AND THINK
(a)1
2
4 = 6. What do the boxes represent?
3
A. + , – , –
B. – , + , +
C. – , + , –
D. + , – , +
(b) 3 – 5 + 2 = -------A. 2B. 0C. 4D. –4
Fill in the boxes with
appropriate operations
(+ , – , x , ÷ ).
2
3
5 = 30
2
3
5 = 11
2
3
5 = –13
2
3
5 = 1.2
(c) 3 x 2 + 4 x 5 = ------A. 90
B. 26
C. 50
D. 66
(d) –3 + 5 x 2 = -----A. 7
B. 4
C. 13
D. 16
(e) 32 + 5 x – 6 = -----A. –84B. 39C. –21
STOP AND THINK
Put brackets where necessary for
each of the statements to hold true:
(a) 2 x 4 + 5 – 9 x 3 = –14
(b) 28 ÷ 3 + 4 x 6 + 3 = 27
(c) –2 + 6 x 4 – 9 ÷ 3 = 13
D. –24
4. Irfaan says that 36 ÷ 3 + 6 = 4. Do you agree with the answer? Justify your answer.
5. Sonam says that 5 + 4 x 23 = 37. Do you agree with the answer? Justify your answer.
36
Chapter 3 - Order of Operations
6. 30 students and 2 teachers went on an educational tour at Casela. The bus fare for each
student was Rs 50. Each teacher paid Rs 10 more than the students for the bus fare. The Casela
entry fee was Rs 100 and each student paid Rs 20 less than the normal price. What is the total
amount paid by the students and teachers for the tour?
7. Isha is always stressed on the eve of her Mathematics examination. Her father left her a coded
message. She needs your help to read it.
The Message
4 21 25
5 11 16 16
11 15 19
12 21
26 6 21 25 12
2 6 3 15 1
21 13
4 21 25 !
Work out the answers to help her decode the message.
W = 17 – 3 × 4
D = 21 + 3 – 12
E = 15 – 2 – 10
I=7×2–3
Y = 2 + (10 – 8)
G = 7 + (2 – 3) × 5
O=6+5×3
M = 15 + 8 ÷ 2
A = 2 x (7 – 3) + 7
U = 10 ÷ 2 × 5
T = 4 – (6 x 4) ÷ 8
R = 18 – 22 x 3
P=6+5x4
F=3×5–2
V = 10 – 5 + 2
L = (5 – 3) × ( 6 + 2)
After having read the message, Isha felt much better and coded this message for her dad.
FIND OUT
11
16 21 7 3
4 21 25
12 15 12 !
Do you know what cryptology
means?
What was her message to her dad?
8. Use the four operations and brackets where necessary to make the following sentences true.
One example has been done for you.
Example: 1 2 3 4 5 = 1
Solution: [(1 + 2) ÷ 3 + 4] ÷ 5 = 1
(a) 1
(b) 1
(c) 1
(d) 1
2
2
2
2
3
3
3
3
4
4
4
4
5
5
5
5
6=1
6 7=1
6 7 8=1
6 7 8 9=1
[Note: For this question there are different possible solutions.]
37
Chapter 3 - Order of Operations
EXERCISE 3.2
Tick the appropriate box.
True
False
(a) 2 + 3 = 3 + 2
(b) 4 × 5 = 5 × 4
(c) 6 – 3 = 3 – 6
(d) 16 ÷ 2 = 2 ÷ 16
(e) (5 + 3) + 4 = 5 + (3 + 4)
(f) (8 – 2) – 3 = 8 – (2 – 3)
CHECK THIS LINK
(g) (2 × 4) × 5 = 2 × (4 × 5)
http://www.math-play.com/Orderof-Operations-Millionaire/order-ofoperations-millionaire.html
(h) 5 × (2 + 6) = (5 × 2) + (5 × 6)
(i) 7 × (6 – 5) = (7 × 6) – (7 × 5)
(j) (16 ÷ 2) ÷ 4 = 16 ÷ (2 ÷ 4)
Observations:
1. Even if the order of the numbers is changed, we obtain the same result in the case of addition
and multiplication, i.e., a + b = b + a and a × b = b × a.
2. We say that addition and multiplication are commutative.
3. If the order of the numbers is changed for subtraction and division, we do not obtain the
4.
5.
6.
7.
same result, i.e., a – b ≠ b – a and a ÷ b ≠ b ÷ a.
We say that subtraction and division are not commutative.
Even if the place of the brackets changes for addition and multiplication, we obtain the same
result, i.e., (a + b) + c = a + (b + c) and (a × b) × c = a × (b × c).
We say that addition and multiplication are associative.
If the place of the brackets changes for subtraction and division, the results are different, i.e.,
(a – b) – c ≠ a – (b – c) and (a ÷ b)÷ c ≠ a ÷(b ÷ c).
8. We say that subtraction and division are not associative.
9. For a × (b + c) = (a × b) + (a × c) and a × (b – c) = (a × b) – (a × c), we say that multiplication is
distributive over addition and subtraction.
EXERCISE 3.3
1. Fill in the boxes with “=” or “≠”.
(a) 12 + 8
(d) 18 – 10
38
8 + 12
10 – 18
(b) (–7) × (–4)
(e) 60 ÷ 12
(–4) × (–7)
12 ÷ 60
(c) 25 × 40
40 × 25
Chapter 3 - Order of Operations
2. Fill in the blanks with the following: associative, distributive, commutative.
(a) When two numbers are added, the sum is the
same regardless of the order of the numbers, for
example, a + b = b + a. This is the -------- property.
(b) When three numbers are added, the sum is the
same regardless of the grouping of the numbers,
for example, (a + b) + c = a + (b + c). This is the
-------- property.
(c) When two numbers are multiplied together, the
product is the same regardless of the order of the
numbers, for example, a × b = b × a. This is the
-------- property.
(d) The sum of two numbers multiplied by a third
number is equal to the sum of each number
multiplied by the third number, for example,
a × (b + c) = (a × b) + (a × c). This is the -------- property.
(e) When three numbers are multiplied, the product
is the same regardless of the order of the numbers,
for example, (a × b) × c = a × (b × c). This is the
-------- property.
FUN ACTIVITY:
MATHEMATICS GAMES
3
4 (25 – 5)
+ 4(11 – 3)
6 (3 + 4) +
(2 x 3 – 1)
2 (3 + 1) +
8 (7 – 2)
3 (7 + 5) +
2
(92 – 3(8 + 3))
–1
2 (8 + 7)
3 (4 x 3) +
2 (10 – 4)
52 + 4(7 – 2) 32 + 2(42) +
1 (18)
+3
3
Toby and Ellie played tic-tac-toe. They followed the
order of operations to solve the problems that are
shown on the grid. Toby went first. He played X,
and Ellie played O. All of Toby's answers came out
to be 48 and all of Ellie’s answers came out to be 47.
At the end of the game, which boxes had X's and
which boxes had O's?
Who was the winner?
Summary
• When we have a problem to solve involving different operations, we proceed as follows:
• We first perform operations within Brackets, starting with the innermost brackets and
moving outwards. Example: [9 + (3 x 2) – 5] x 3
= [9 + 6 – 5] x 3 = [15 – 5] x 3 = 10 x 3 = 30
• We perform:
- Orders (Numbers involving powers or square roots)
Example: 23 + 5 = 8 + 5= 13
- Division or Multiplication whichever comes first, from left to right
Example: 12 x 9 ÷ 3 = 108 ÷ 3 = 36
- Addition and Subtraction whichever comes first, from left to right
Example: 6 + 5 – 4 = 11 – 4 = 7
• Addition and multiplication are commutative while subtraction and division are not
commutative. For example, 6 x 4 = 4 x 6 or 5 + 3 = 3 + 5
• Addition and multiplication are associative while subtraction and division are not
associative. For example, (2 x 4) x 5 = 2 x (4 x 5)
• Multiplication is distributive over addition and subtraction, i.e., 7 x (2 + 3) = (7 x 2) + (7 x 3)
39
4
FRACTIONS AND DECIMALS
Chapter 4 - Fractions and Decimals
Learning Objectives
By the end of this chapter, you should be able to:
• demonstrate an understanding of the concept of fractions as part of a whole, as a measure,
as an operator, as a quotient and as a ratio.
• work with different types of fractions (proper, improper and mixed).
• compare and order fractions.
• add, subtract, multiply and divide fractions.
• solve word problems involving fractions.
• demonstrate conceptual understanding of decimals, including place value.
• compare and order decimals.
• add, subtract, multiply and divide decimals.
• convert fractions to decimals and vice versa.
• solve word problems involving decimals.
Fractions in real life
CHECK THAT YOU CAN:
Fractions are used in different situations everyday. For example,
we use fractions to find the amount of fuel, to read and write
time, to share food and so on.
•
Perform arithmetic
operations.
Find H.C.F. and L.C.M. (up to
3 numbers).
•
KEY TERMS
Fuel indicator
Half past three
Quarter of an apple
Representing Fractions
In earlier grades, you learnt that fractions can be represented
in the following ways:
1. Fractions are used to express parts of whole. For example,
the cake below represents a whole divided into 10 equal
parts.
1
1 part is represented as .
10
•
•
•
•
•
•
Fractions
Numerator, Denominator
Proper and Improper fractions
Mixed numbers
Equivalent fractions
Decimals
DID YOU KNOW
The word ‘fraction’ comes from
the Latin word ‘frangere’ which
means ‘to break into pieces’.
RECALL
a
.
b
“a” is called the numerator.
Consider a fraction
“b” is called the denominator.
40
Chapter 4 - Fractions and Decimals
2. Fractions are used to represent parts of a group. The
fraction 4 represents 4 blue marbles out of 9 marbles.
9
3. Fractions can also be represented on a number line. For
example, the number line below represents fractions
1 , 1 and 3 .
4
4 2
1
4
0
1
2
3
4
1
Types of Fractions
DID YOU KNOW
History of Fractions
Fractions didn't exist in Europe until
the 17th century. Around 1800 BC, the
Egyptians wrote fractions where they
put a mouth picture (which meant
part) above a number. In Ancient
Rome, fractions were only written
in words whereas the Babylonians
wrote their fractions in sixtieths
which made reading numbers very
confusing. Through Arab traders,
Indian numerals were spread to
Arabia where they were then used
to write fractions as we now use
them, that is, the numerator above
the denominator separated by a
horizontal or slant line.
CHECK THIS LINK
1. Proper fractions: A fraction is said to be proper if its
numerator is less than its denominator.
Examples are 1 , 3 , 5 , 11
2 5
9 12
2. Improper fractions: A fraction is said to be improper if its
numerator is greater than or equal to (≥) its denominator.
Examples are 3 , 5 , 17 , 8 , 12
2
4 9
8 12
3. Mixed Numbers: Mixed numbers consist of a whole
number and a proper fraction together.
1
1
Examples are 1
, 3 1 , 17
2
6
4
https://www.geogebra.org/m/
P4ybW7Re
DID YOU KNOW
Improper Fractions and Mixed
numbers in real life
We use mixed numbers more in real
life than improper fractions. We
often say the following:
• "I will have dinner in 1 and a half
1
hours (1 h)" but we never say:
2
“I will have dinner in 3 hours.”
EXERCISE 4.1
2
3
1.
in words is
8
1
A. three quarters
B. three eighths
C. one third
D. one and three eighths
• “I will be on vacation in 3
2
months” but we never say:
7
"I will be on vacation in 2
months."
41
Chapter 4 - Fractions and Decimals
2. Seven twelfths in figures is:
A.
17
12
B.
7
10
C.
7
12
D.
12
7
3. Which fraction is indicated by the arrow?
0
A.
1
1
5
B.
3
5
C.
2
2
5
D.
2
4
4. What fraction is represented by the blue-coloured petals?
A.
2
8
B.
3
8
C.
3
7
D.
8
3
5. Ashley bought a large sized pizza with 8 equal slices. He ate 5 slices of this pizza. What
fraction represents the portion he has eaten?
3
A. 7
B.
3
8
C.
5
8
D.
8
5
6. Study the number line below and write down the fractions represented by the letters A, B,
C, D, E, F, G, H and I.
A
B
-2
C D
-1
E
0
F
G
H
1
I
2
3
7. Classify the following fractions as proper fractions, improper fractions and mixed numbers:
1
3
8
2 5
6 9 12
1 9
,
,3 ,
,
,3 ,
, 1 125 , 12 5 , 5
3 4
7 9 17
2 8
Converting an Improper Fraction into a Mixed Number
A fraction
a
can also be expressed as a division: “a” divided by “b” (a÷ b).
b
Example
Convert 62 into mixed numbers.
7
Solution
62
is the same as 62 ÷ 7
7
62 = 8 6
Therefore,
7
7
42
8
7 62
- 56
6
Activity
Whole number
Numerator of proper fraction
Using the digits 4, 5 and 6
only once, three different
mixed numbers can be
formed.
Write down the 3 different
possibilities.
Chapter 4 - Fractions and Decimals
Converting a Mixed Number into an Improper Fraction
Example
Convert the following mixed numbers into improper fractions:
1
17
8
3
(a) (b)
4
20
Solution
(a)
+
(b)
1
(3 x 4) + 1
=
3
4
x 4
=
8
17 = (8 x 20) + 17
20
20
13
4
=
177
20
EXERCISE 4.2
1. Convert the following improper fractions into mixed numbers:
7
2
10
(b) 3
5
(c) 4
11
(d) 8
(e)
14
11
16
(f) 5
13
(g) 12
21
(h) 10
43
(i) 25
(j)
83
8
(a) 2. Convert the following mixed numbers into improper fractions:
1
3
2
3
(a) 1(b) 3 (c) 5(d) 2
2
4
5
7
5
(e) 8
9
8
(f) 7
11
3
(g) 23 10
(h) 3
17
25
1
(i) 24
3
7
(j) 10 8
1
(k) 4 6
(l) 6
3
8
Reducing a fraction to its lowest terms
Example
Reduce the following to its lowest term (a) 16
Solution
(a) 16
8 ÷2
4 ÷2
2
16 ÷ 2
=
=
=
=
24 24 ÷ 2
12 ÷ 2
6 ÷2
3
(b) 12
24
48
(b)
6
3
1
12
6
3
1
12
=
=
=
=
48
48
24
12
4
24
12
4
43
Chapter 4 - Fractions and Decimals
Equivalent Fractions
CHECK THIS LINK
Study the figures below.
1
2
1
2
Fig. 1
1
4
1
4
1
4
1
4
1
8
1
8
1
8
1
8
1 1
8 8
Fig. 2
1
8
1
8
Note: Equivalent fractions are
equal in value but with different
numerators and denominators.
Fig. 3
The shaded areas above represent the following fractions:
Fig. 1,
http://www.abcya.com/
equivalent_fractions_
bingo.htm
4
2
in Fig. 2, and 8 in Fig. 3.
4
1
in
2
The shaded part of each circle represents the same fraction of the circle.
Therefore, we can say that
1
2
4
= =
.
2
4
8
1 2
4
,
and
are called equivalent fractions.
2 4
8
Note: To obtain equivalent
fractions of higher order, we
multiply both numerator and
denominator by the same whole
number.
Consider 1 , 2 , 3 , 4 ,…..
4 8 12 16
1x2
2
=
4x2
8
multiply both
numerator and denominator by 2
1x3
3
=
4x3
12
multiply both
numerator and denominator by 3
4
4 ÷4
1
=
=
16 16 ÷ 4
4
divide both
numerator and denominator by 4
Note: To obtain equivalent
fractions of lower order, we divide
both numerator and denominator
by the same whole number.
Example
Complete the following to find the equivalent fractions:
(a)
2
=
5
15
Solution
(a)
x3
2
6
=
5
15
x3
44
(b)
12
=
32
8
(b)
(c)
9
÷4
12
3
=
32
8
÷4
=
3
= 6
27
÷3
(c)
1
9
=
÷3
x2
6
3
=
54
27
x2
Chapter 4 - Fractions and Decimals
EXERCISE 4.3
1. Reduce each of the following fractions to its lowest terms:
8
(a)
12
4
(b) 20
3
(c) 42
18
(d)
30
24
42
(e)
32
22
132
250
125
(f)(g) (h)(i) (j)
108
33
144
1000
500
2. Express each of the following mixed numbers in its simplest form:
2
(a) 1
16
4
(b) 1
10
5
(c) 3
15
3
(d) 2
12
(e) 4
15
18
24
(f) 8 26
12
(g) 9
40
42
(h) 6
60
63
(i) 7
84
(j) 10
7
56
3. Copy and complete the following equivalent fractions:
(a)
4
= 5 25
(b)
2
8
= 3
(c)
5
=
6 36
(d)
16
=
18 9
(e)
3
21
= 28
(f)
45
=
100 20
(g)
3
=
10 40
60
= (h)
7
=
= 8
16
32
(i)
15 45
=
=
108
12
7
11
equivalent to
? Explain your reasoning.
28
44
4. (a) Is
(b) Is
16
4
equivalent to
? Explain your reasoning.
25
5
5. Show that the following pairs of fractions are equivalent.
(a)
10
4
and
15
6
(d) 33 and 55
60
36
78
(b) 39 and 90
45
(e) 2
1
30
and
2
12
54
(c) 18 and 63
21
(f) 1
4
12
and 1
5
15
6. Write down 3 equivalent fractions for each of the following fractions:
(a)
1
7
2
(b) 3
5
(c) 8
2
(d) 9
(e)
3
5
45
Chapter 4 - Fractions and Decimals
Comparing and Ordering Fractions
We can easily compare fractions with same denominators
8
5
(for example,
and
) by comparing the numerators.
9
9
Consider 5
9
Since 8 > 5,
8
5 .
>
9
9
and 8
9
STOP AND THINK
Ziad and Taj, both made a pizza at home.
3
Ziad gave
of his pizza to his younger
8
4
brother and Taj gave of his pizza to his
7
sister. Who shared the bigger part of his
pizza, Ziad or Taj?
To compare fractions with different denominators, we first convert them into fractions with the
same denominators. We then compare the numerators.
Example 1
Which fraction is greater: 5 or 11 ?
12
16
Solution
L.C.M. of 12 and 16 is 48
We express each fraction with denominator 48.
Caution:
We cannot compare just the
numerators. We need to ensure
that the fractions have the same
denominators before comparing
the numerators.
5
5 x 4 20
=
=
12
12 x 4 48
11 11 x 3 33
=
=
16 16 x 3 48
We compare the numerators 33 and 20; 33 > 20
11
5
Therefore,
>
.
16 12
Example 2
Arrange
5 7
2
,
and
in ascending order.
6 8
3
Solution
L.C.M. of 3, 6 and 8 is 24.
5
5 x 4 20
=
=
6
6 x 4 24
7 x 3 21
7
=
=
8 x 3 24
8
We compare the numerators: 16 < 20 < 21
Hence, in ascending order we have
46
2 , 5 , 7 .
3 6 8
2 2 x 8 = 16
=
3 3 x 8 24
Chapter 4 - Fractions and Decimals
EXERCISE 4.4
1. Arrange the following fractions in ascending order:
(a)
1 1
1
, and 3 2
5
(b)
5 3
1
,
and 12 8
16
(c)
3 7
5
,
and
10 15
6
2. Arrange the following fractions in descending order:
(a)
3 1
6
,
and 4 6
7
(b)
2
5
7
,
and 5
8
10
(c)
17 10
5
,
and
24 16
6
3. For each of the given pairs, determine which fraction is greater:
(a)
5
3
or 6
4
(b)
3
4
or 5
7
(c)
2
2
or 7
9
(d)
3
4
or
11
5
(e)
5
11
or
8
16
(f)
15
21
or
16 24
4. Fill in the blanks with “<” or “>”:
(a)
5
4
(b) 8 ____ 9
3 ____ 7
8
4
(c) 13 ____ 15
16
24
(d)
5 ____ 6
6
7
Addition and Subtraction of Fractions
Two or more fractions can be added or subtracted by simply adding or subtracting the
numerators if their denominators are the same.
Fractions with different denominators must first be converted to fractions with the same
denominator before carrying out addition or subtraction.
One way of adding or subtracting fractions is by using diagrams.
1
1
Example: Evaluate
+
.
3
4
1
3
+
1
4
=
4
12
+
3
12
=
7
12
47
Chapter 4 - Fractions and Decimals
Example 1
Evaluate
5
1
+
7
7
(a)
1
1
+
(b) 4
3
(c)
5 11
–
6 15
(d)
2
1
3
–
+
3
8
4
Solution
5
1
6
+
=
7
7
7
(a)
1
1
1x4
1x3
=
+
+
3
4
3x4
4x3
4
3
= +
12 12
7
=
12
(b)
5 11 5 x 5 11 x 2
=
–
–
6 15 6 x 5 15 x 2
25 22
–
=
30 30
3
=
30
1
=
10
(c)
3
2
1
–
+
4
3
8
(d)
For fractions with common
denominators, we simply add
the numerators.
The L.C.M. of 3 and 4 is 12.
The L.C.M. of 6 and 15 is 30.
Caution:
Fractions should be simplified
to their lowest terms.
3x6 2x8 1x3
–
+
4x6 3x8 8x3
18 16
3
–
+
=
24 24 24
18 – 16 + 3
=
24
5
=
24
=
The L.C.M. of 4, 3 and 8 is 24.
Simplify the numerator from left
to right
EXERCISE 4.5
1. Evaluate the following, giving your answer in its simplest form:
(a)
3 2
+
7 7
(b)
10 4
–
21 21
(c)
13 11
7
–
+ 18 18 18
(d)
5 11 7
+
–
36 36 36
2. Evaluate the following giving your answer in its simplest form:
1 2
(a) 4 + 7
2 3
(b) 5 + 8
7
3
(c) 8 – 5
11 5
(d) 12 – 8
(e) 1 + 1 + 1
3 4 5
(f) 5 – 1 – 7
6
8 12
(g) 5 + 2 – 1
6 5 3
(h) 3 + 1 – 5 4
2 12
48
Chapter 4 - Fractions and Decimals
Addition and Subtraction of Mixed Numbers
Example
Calculate:
Evaluate:
4
1
5
1
1
–1
(a) 5 9 –1
(b) 2 (b) + 3
6
18
3
4
(c)
Solution:
Solution
(a)
5
1
4
–1
6
9
Method 1
Subtract the whole numbers first then
subtract the fractions
5
Method 2
Convert the mixed numbers into improper
fractions first, then subtract. Then convert
back into mixed numbers.
4
1
1
4
–1
= (5–1)+(
– )
9
6
6
9
5
49 7
1
4
–1
=
–
9
6
9
6
8 3
– )
18 18
5
= 4 +( )
18
98 21
–
18 18
77
=
18
= (4)+(
= 4
(b)
2
=
5
18
= 4
5
18
5
1
1
–1
+3
18
4
3
Method 1
Add and/or subtract the whole numbers
first then add and/or subtract the fractions
5
1
1
–1
+3
18
4
3
1 1 – 5
= (2 + 3 – 1) + ( +
)
3 4 18
2
12 + 9 – 10
36
11
=4 +
36
11
=4
36
= 4+
Method 2
Convert the mixed numbers into improper
fractions first, then add or subtract. Then
convert back into mixed numbers.
5
1
1
2
+3
–1
18
3
4
=
7 13 – 23
+
3 4 18
84 + 117 – 46
36
155
=
36
=
= 4
11
36
49
Chapter 4 - Fractions and Decimals
EXERCISE 4.6
Evaluate the following, giving your answer in its simplest form:
(a) 2
1
3
+ 1 5
5
(b) 5
5
1
– 3 8
6
(c) 2
4
1
+ 7
4
(d) 3
1
5
5
+1 –2
6
9
12
(e) 2
3
1
3
– 1 + 3 4
6
8
(f) 5
4
1
3
–3 +
5
4 10
Multiplication of Fractions
Multiplication of a fraction by a whole number
Example
Evaluate 3 x
1
2
Solution
Method 1
3x
1
1
means taking three times, that is,
2
2
3
1
1 1 1
+ + =
=1
2
2
2 2 2
Method 2
3x
3x
1
1
means taking of three wholes,
2
2
3
1
1
=
x
1
2
2
3x1
=
1x2
3
=
2
Multiplication of a fraction by another fraction
Example
Caution:
Cancelling can be done only vertically
and diagonally. We can never cancel
the numbers horizontally.
Evaluate:
(a)
2 15
x
5 16
(b)
Solution
(a)
1
4
x 9 x 7
15 14 12
3
2 15
2 15
x
=
x
5 16
5 16 8
1
=
50
1x3 = 3
8
1x8
(b)
4
x 9 x 7
15 14 12
1
2
1
3
1
=
4
x 9 x 7
155 14 2 12 4
=
1x1x1
1
=
5x1x2
10
1
2
Multiplication involving mixed numbers
Example
Chapter 4 - Fractions and Decimals
Evaluate:
(a) 1
1
2
1
(b) 1 x 2 x 1
6
5
10
3 x 15
5 17
Solution
1
3
(a)
3 15 = 8 x 15
1 x
17
5 17 1 5
(b)
8x3
1 x 17
24
=
17
7
=1
17
=
=
11 x 1 x 7
5 x 5 x1
77
25
2
= 3
25
=
Reciprocal of a Fraction
RECALL
Fraction
Reciprocal
7
2
1
5
3 =3
1
2
7
5=
6
11 12 7
1
2
1
=
X
X
1 x2 x1
10 5 5
61
10
5
6
5
1
1
3
Division of Fractions
Example
To obtain the reciprocal of a fraction,
we simply invert the numerator and
denominator (that is, we turn the
fraction upside down).
NOTE TO TEACHER
Explain to students the meaning
of division of fractions. Example
1 ÷ 1 means how many quarters
2 4
are there in one half?
Evaluate:
9
3
(a) ÷
5 20
6
(b) ÷ 3
11
(c) 2
2
4
÷1
15
5
Solution
(a)
1
4
Step 1: Replace “÷” by “×” and invert the 2nd fraction.
9
3 20
3
÷
x
=
93
5 20 1 5
=
1x4
=
1x3
Step 2: Multiply the 1st fraction by the reciprocal of the 2nd
fraction.
4
3
= 1
Step 3: Convert the improper fraction into mixed number.
1
3
51
Chapter 4 - Fractions and Decimals
2
1
6
6
x
÷3 =
31
11
11
(b)
=
=
(c)
2
17
4
14
÷1
÷
=
15
15
5
5
2
3
14 x 15
=
5
17
2x1
11 x 1
1
14 x 3
1 x 17
42
=
17
8
=2
17
=
2
11
EXERCISE 4.7
1. Evaluate:
(a)
1
3
× 3
4
(b)
4 5
× 9 8
(c)
3 15
× 4 16
(d)
2
3
× 3
8
(e)
10 39
×
13 40
(f)
5
9
×
6 10
(g)
4 15
× 5 16
(h)
11 3
× 12 5
(i)
5 48
x
12 65
2. Evaluate:
(a)
1
× 6
3
1
(b) 7 × 2
(d)
1
3
14
×
×
3
7
15
(e)
2
6
8
×
× 2 3
13
9
(f)
19 15
5
×
×
24 38
6
(c) 1
1
15
× 8
27
3. Evaluate:
(a)
1
5
÷ 2
2
(b)
5
3
÷ 8
4
(c)
1
3
÷
2
4
(d)
4
3
÷ 11 22
(e)
3
16
÷ 7
35
(f)
49
63
÷
160 140
1
(b) 5 ÷ 2
(c)
7
3
÷ 1
11
4
4. Evaluate:
(a)
5
÷ 6 6
(d) 4
52
1
3
÷ 5 10
(e) 2
3
5
÷ 1 4
12
(f) 3
1
1
÷5
2
4
Chapter 4 - Fractions and Decimals
Arithmetical Operations on Fractions
Example
Evaluate the following:
(a)
(
1
1
+
4
2
)÷
RECALL
3
8
(b)
9
8
1
1
1
x2 –3 ÷
4
5 15
3
In Chapter 3, you learned about
the BODMAS rule. This rule can
also be applied to fractions or
decimals.
Solution
(a)
(
1
1
+
4
2
) ÷ 38
=
(
)
1
2+1
3
÷
4
8
2
(b)
9
7
3
2
3
28 x 9 – 16 x 15
=
41
51
31
81
(Brackets come first
followed by division)
(Do multiplication and division before
subtraction)
3 x 8
41
31
=
8
1
1
1
x2 –3 ÷
4
5 15
3
= 21 – 6
= 15
=2
EXERCISE 4.8
1. Evaluate the following:
(a)
1
1
2
× ( + )
4
2
3
(b) (
3 1
– ) × 15
5 2
(c) ( 2
5 2
3
+ ) ÷ 6 3
4
(e) 4
1
1
1
×(5 –2 )
5
6
3
(f) 6
(d) (
1
3
2
–1 )÷
6
4
3
2
1
1
÷(3 –2 )
3
2
4
2. Evaluate the following giving your answer in its simplest form:
(a)
1
1 2
8
× ( + ) ÷ 4
2 3
3
3
2 1 5
×( + –
)
5
3 2 12
5
1
3 1
+ ) ÷ ( × )
8
6
4 6
(d) 2
1
5
2
1
3
×(
+ ) × ( 4 – )
5
9
3
2
4
3 1
1
2
× –
÷ 4 6 12 3
(f) 6
1
1
1
1
×3 –2 ×4
8
7
4
6
(c) (
(e)
(b)
53
Chapter 4 - Fractions and Decimals
Word problems involving Fractions
Example
1
In a school,
of the students will participate in the sports day activities. If there are 750
3
students in the school, how many students will not participate in the sports day activities?
Solution
Method 1
Method 2
Number of students who will participate in the
sports day activities
Fraction of students who will not participate in
the sports day activities
=
250
1
x 750 = 250
3
1
=1–
1
2
=
3
3
Number of students who will not participate in
the school sports activities
Number of students who will not participate in
the school sports activities
= 750 – 250
= 500
2
= x 750 = 500
3
250
1
EXERCISE 4.9
1
1. Simran is given Rs 75 to go to school every day. On a particular day, she spent 3 of her
1
money to buy a burger,
on water and 2 on ice-cream. Calculate
5
5
(a) the amount spent on ice-cream,
(b) the total amount spent,
(c) how much money is left.
1
2. Kavi, Rodney, Yanish and Mathieu rent a car and went on a road trip. Kavi drove for
of
2
1
the trip, and Rodney drove for 4 of the trip. Yanish and Mathieu divided the rest of the
driving evenly between them. If the entire trip was 110 km, how many km did Mathieu
drive?
1
3. The difference between two fractions is
. The larger one is 5 . Find the smaller fraction
36
12
in its lowest term.
2
4. After spending
of his money on cakes, Mahesh had Rs 45 left. How much money did
5
he have at first?
2
5. Minta bought a box of 225 oranges. She threw away 20 rotten oranges and gave
of the
5
remainder to Justin. How many oranges are there left?
54
Chapter 4 - Fractions and Decimals
6. A fruit seller had 300 apples, 200 oranges and 150 pears in his stock on Sunday. He
1
1
3
managed to sell
of the apples,
of the oranges and
of the pears on Monday. On
4
3
5
1
Tuesday he sold 40 apples, 100 oranges and
of the remaining pears.
2
(a) Calculate the number of fruits he sold on Monday.
(b) Express the number of pears he sold on Tuesday as a fraction of the number of pears he
had on Sunday.
(c) Find the number of fruits left on Wednesday.
7. Mehrine and Zaara brought an equal amount of money to the shopping mall. Mehrine
spent Rs 500 on a pair of shoes and Zaara bought a dress for Rs 2 000. After their shopping,
2
Zaara had
of what Mehrine had left. How much money did Mehrine bring for
3
shopping?
1
8. Joan and Luciana collect stickers. Luciana has 340 stickers. of the number of stickers that
5
2
Luciana has is equal to
of the number of stickers that Joan has. How many stickers does
17
Joan have?
Decimals
We often use decimal numbers in our everyday lives, such as when dealing with money and in
measurements of length, mass, time and so on.
Decimals in real life
Electronic balance
Supermarket receipt
Decimal system
DID YOU KNOW
We use the Hindu-Arabic or decimal number system
to represent numbers. The position of a digit in a
number represents its place value.
Consider the number 3.567.
3.567
Units
Fuel reading
Thousandths
Hundredths
Tenths
Decimal fraction notation
Decimal numbers have been used by Ancient
Chinese and in medieval Arabia and was widely
accepted by mathematicians by the year 1500
but the notation used for decimals today, i.e.,
the whole number with the decimal point and
the fractional part, was introduced less than
500 years ago. John Napier made the decimal
point common in mathematics.
55
Chapter 4 - Fractions and Decimals
Definition
STOP AND THINK
A decimal number consists of a whole number part, a decimal
point and a fractional part.
Compare the following:
7 and 7. 0.
What can you say about the
numbers?
Decimal Point
Whole Number part
2.84
Fractional Part
The decimal point separates the whole number part, which is
found to the left of the decimal point from the fractional part
of the number which is found to the right of the decimal point.
For example:
3.75
STOP AND THINK
Where will you place 2.75 and
2.57 on a number line?
EXERCISE 4.10
1. What is the value of 7 in each of the following?
(a) 0.007
(b) 70.32
(c) 0.708
(d) 17.59
(e) 0.478
2. What is the value of each digit in the following?
(a) 0.58
(b) 59.48
(c) 0.008
(d) 407.638
(e) 40.7854
3. Answer True or False.
(a) 0.5 means one half (b)
(c) The value of 2 in 200.45 is two hundredths
(d)
7
70
35
0.35 is same as
10
0.7 equals to
Common fractions and their corresponding decimal numbers
56
1
2
1
4
0.5
0.25
3
4
0.75
1
5
0.2
1
8
0.125
Chapter 4 - Fractions and Decimals
Converting fractions to decimals
Fractions whose denominators are powers of 10
Example
Convert the following into decimals.
9
(a) 1000
(b) 191
100
Solution
(a)
(b)
9 = 0 . 0 0 9.
1000
191 = 1. 9 1.
100
Note: There are two zeros in the denominator, so
move the decimal point two places to the left of
the numerator.
Note: There are three zeros in the denominator, so
move the decimal point three places to the left of
the numerator.
Fractions whose denominators can be expressed as powers of 10.
Example 1
RECALL
Convert the following to a decimal.
3
4
(a)(b)
5
8
To convert a fraction to a decimal,
we first express the fraction into an
equivalent fraction with denominator
10, 100, 1000 and so on.
Solution
(a)
3 x 25
= 75
4 x 25
100
Therefore,
Method 2
(b) Method 1
Perform division
5
Note:
means 5 divided by 8.
8
5 x 125
= 625
8 x 125
1000
3
= 0.75
4
Therefore,
5
= 0.625
8
5 2 4
8 5.000
0.625
We divide the numerator by
the denominator.
5
= 0.625
8
Example 2
Express the following mixed numbers as decimals.
(a)
2
3
10
(b) 1
1
4
Solution
(a)
2
3
= 2.3
10
( 103 = 0.3 in decimal (
(b) 1
1
= 1.25
4
25
= 0.25 (
( 14 = 14 xx 2525 = 100
57
Chapter 4 - Fractions and Decimals
Converting decimals to fractions
Example
Convert each of the following into a fraction in its lowest term:
(a) 0.18
(b) 2.38
Solution
(a) 0.18 =
18
9
=
100 50
(b) 2.38 = 2
=2
38
100
19
50
EXERCISE 4.11
1. Match the following.
Fraction
Decimal
2
10
5
100
59
100
87
10
2
1000
204
100
56
10000
0.59
0.0056
0.2
2.04
0.05
0.002
8.7
2. Convert the following decimals into fractions, giving your answers in the lowest term.
(a) 0.39
(b) 0.75
(c) 0.4
(d) 0.36
(e) 0.412
(f ) 0.02
(g) 0.04
(h) 0.032
(i) 0.65
(j) 0.0925
3. Express each of the following decimals as a mixed number in its lowest term.
(a) 2.4
(b) 7.05
(c) 6.03
(d) 11.40
(e) 5.55
4. Express the following as decimals.
8
(a) 10
47
(b) 100
215
(c) 10
25
(d) 1 000
(e)
22
(f ) 5
7
(g) 25
(h) 2
7
10
1
(i) 4 5
7
(j) 9 8
58
17
200
Chapter 4 - Fractions and Decimals
Comparing and ordering decimals
Example
Caution:
It is only when adding zeros at the end of
a decimal number that the value of the
number does not change.
E.g 4.65 is the same as 4.65000
BUT 4.65 is NOT the same as 4.00065
Arrange in descending order:
3.27, 7.35, 7.53, 3.72, 3.709, 7.027
Solution
Note: We can also make use of a number line.
We can make use of a table:
Step 1: We place each number according to its place value.
Step 2: Fill in the empty boxes with zeros as the value of the decimal number does not
change.
Step 3: Compare the digits in the first column (here it is the Units column) and choose the
largest one.
Step 4: If the digits are equal, then we move to the next column and compare (here the
Tenth column). We do the same if the digits are equal in the Tenth column.
As it is descending order, we start with the largest number.
Decimal
number
3.27
7.35
7.53
3.72
3.709
7.027
Units
3
7
7
3
3
7
Decimal
point
.
.
.
.
.
.
Tenth
Hundredth
Thousandth
2
3
5
7
7
0
7
5
3
2
0
2
0
0
0
0
9
7
Therefore, solution is 7.53, 7.35, 7.027, 3.72, 3.709, 3.27
EXERCISE 4.12
1. Arrange each of the following set of numbers in ascending order:
(a) 0.3, 0.8, 0.1, 0.7, 0.5
(b) 0.08, 0.8, 0.18, 1.8, 0.108
(c) 56.23, 5.623, 5.236, 0.5623, 52.63
(d) 10.023, 10.203, 10.2003, 10.0023, 10.302
2. Arrange each of the following set of numbers in descending order:
(a) 10.02, 1.002, 12.02, 2.001, 20.02
(b) 7.095, 7.905, 7.995, 7.059, 7.509
(c) 65.84, 65.084, 65.804, 65.884, 65.88
(d) 15.047, 1.5047, 15.0407, 15.747, 15.477
59
Chapter 4 - Fractions and Decimals
3. Compare each of the following decimals by using > , = and <.
(a) 2.3 ___ 3.02
(b) 0.23 _____ 23
(c) 0.508 _____ 5.08
(d) 7.75 _____ 7 +
100
7
5
+
10
100
4. Answer the following questions:
(a) Margaret has 1 489 cents and Paul has Rs 14.80. Who has less money?
(b) Kiara is 137 cm and Disha is 1.43 m tall. Who is taller?
(c) Dilshad weighs 53.85 kg and Annelise weighs 50 300 g. Who is heavier?
Addition and Subtraction of Decimals
When we add or subtract two or more decimal numbers, we need to place the numbers in the
correct place value column. The decimal points need to be aligned one under the other.
Example
Evaluate (a) 12.43 + 4.687
Solution
(a)
(b) 45.043 – 28.21
1 1
12 . 4 3 0
+ 4.687
17.117
(b)
1
3 4 1
45.0 43
– 2 8 . 210
16 . 8 3 3
EXERCISE 4.13
1. Find the value of
(a) 3.6 + 4.2
(b) 2.53 + 1.24
(c) 55.75 + 56.4
(d) 31.7 + 21.005
(g) 0.0053 + 1.049
(h) 125.01 + 15.5 + 2.7
(b) 4.89 – 2.25
(c) 12.64 – 8.3
(d) 11.47 – 9.521
(e) 8.004 – 6.563 (f) 0.432 – 0.041
(g) 1.65 – 0.391
(h) 103.4 – 64.27
(e) 205.5 + 41.47 (f) 13.12 + 27.6
2. Evaluate
(a) 6.4 – 2.2
3. Evaluate
(a) 5.55 + 4.2 – 6.34
(b) 12.45 + 0.321 – 2.516
(c) 8.214 – 3.028 + 5.141
(d) 13.2 – 7.11 + 8.221
(e) 2.397 – 4.215 + 7.233
(f) 21.07 – 11.18 + 42.02
60
Chapter 4 - Fractions and Decimals
Multiplication of Decimal Numbers
Multiplication of a decimal number by powers of 10
RECALL
Example
In Grade 6, you learnt that
when multiplying a decimal
number by
(i) 10, the decimal point
moves 1 digit to the right.
(ii) 100, the decimal point
moves 2 digits to the right
(iii) 1000, the decimal point
moves 3 digit to the right.
Calculate:
(a) 1.56 × 10
(b) 0.00327 × 1 000
Solution
(a) 1.56 × 10 = 1 . 5 6
(b) 0.00327 × 1 000 = 0 . 0 0 3 2 7
= 3.27
= 15.6
Multiplication of a decimal number by a whole number
Example
Calculate: 0.56 × 50
Solution
0.56 × 50
= 0.56 × 10 × 5
= 5.6 × 5
= 28
Multiplication of a decimal number by another decimal number
Example
Calculate: 4.2 × 5.3
Solution
Method 1
4.2 = 4 + 0.2
5.3 = 5 + 0.3
4.2 × 5.3
= 20 + 1.2 + 1.0 +0.06
= 22.26
4
0.2
5
5 × 4 = 20
5 x 0.2 = 1.0
0.3
0.3 × 4 = 1.2
0.3 × 0.2 = 0.06
Method 2
A faster way of multiplying decimals is as follows:
Multiply 4.2 by 10 to get 42.
Multipy 5.3 by 10 to get 53.
Multiply 42 by 53.
1
42
x 53
126
+ 2100
2226
We divide the answer by 100 (10 x 10) as we have multiplied by 10 twice.
4.2 x 5.3 = 2 226 ÷ 100 = 22.26
61
Chapter 4 - Fractions and Decimals
EXERCISE 4.14
1. Calculate:
(a) 0.3 × 10
(b) 1.21 × 10
(c) 4.06 × 1000 (d) 0.7 × 100
(e) 3.452 × 1 000
(f) 30.7 × 100 (g) 0.143 × 100 (h) 21.7 × 100
2. Calculate:
(a) 0.4 × 8
(b) 0.27 × 3
(c) 2.4 × 12
(d) 0.07 × 25
(e) 0.42 × 30
(f) 0.7 × 50
(g) 6.72 × 16
(h) 1.5 × 2 000
3. Calculate:
(a) 0.2 × 0.6
(b) 0.32 × 0.9
(c) 2.25 × 1.3
(d) 1.72 × 2.1
(e) 52.3 × 0.2
(f) 2.31 × 2.9
(g) 0. 042 × 7.3
(h) 0.86 × 2.52
Division of decimal numbers
Division of a decimal number by powers of 10
Example
Calculate: (a) 283.31 ÷ 100
RECALL
(b) 7350.2 ÷ 1 000
Solution
(a) 283.31 ÷ 100
=283.31
(b) 7350.2 ÷ 1 000
=7350.2
= 2.8331
= 7.3502
Division of a decimal number by a whole number
Example
Calculate: 1.2 ÷ 4
Solution
Method 1
1
4 1. 2
0.3
1.2 ÷ 4 = 0.3
Method 2
1.2 ÷ 4 = 12 ÷ 4
10
3
12
1
=
x
=
10
4 1 10
3
= 0.3
62
In Grade 6, you learnt that when
dividing a decimal number by
(i) 10, the decimal point moves
1 digit to the left.
(ii) 100, the decimal point
moves 2 digits to the left.
(iii) 1000, the decimal point
moves 3 digit to the left. On
page 57 you also learnt how to
convert fractions into decimals.
Chapter 4 - Fractions and Decimals
Division of a decimal number by another decimal number
Example
Calculate: 4.8 ÷ 0.2
Caution:
The decimal point should
be moved to the right by
the same number of digits
in the numerator and in
the denominator
Solution
4.8 ÷ 0.2 =
4.8
0.2
Move decimal point 1 digit to the right in
the numerator and denominator
= 48
2
= 24
EXERCISE 4.15
1. Calculate:
(a) 2.9 ÷ 10
(b) 5.2 ÷ 100
(c) 0.347 ÷ 10
(d) 5.9 ÷ 1 000
(e) 12.53 ÷ 100
(f) 99.7 ÷ 10
(g) 646.1 ÷ 100
(h) 0.0025 ÷ 100
2. Calculate:
(a) 3.6 ÷ 9
(b)
0.42 ÷ 7
(c)
16.5 ÷ 5
(d) 248.2 ÷ 2
(e)
33.9 ÷ 3
(f)
256.8 ÷ 32
(a) 7.2 ÷ 60
(b)
242.8 ÷ 20
(c)
55.22 ÷ 11
(d) 0.0345 ÷ 50
(e)
0.216 ÷ 16
(f)
42.12 ÷ 12
3. Calculate:
4. Calculate:
(a) 3.8 ÷ 0.2
(b) 0.234 ÷ 0.3
(c) 0.0012 ÷ 0.6
(d) 156.3 ÷ 0.6
(e) 745.5 ÷ 1.5
(f) 0.07 ÷ 0.0028
Word problems involving decimals
Example
Ali bought 6 copybooks for Rs 97.20.
(a) Find the cost of 1 copybook.
(b) A discount of Rs 1 was given per copybook when buying a pack of 12 copybooks.
Find the price of a pack of 12 copybooks.
63
Chapter 4 - Fractions and Decimals
Solution
(a) Cost of 6 copybooks = Rs 97.20
Cost of 1 copybook = Rs (
97.20
)
6
(b) Discounted price of 1 copybook
= Rs 16.20 – Rs 1 = Rs 15.20
Price of 12 copybooks
= Rs 15.20 x 12 = Rs 182.40
= Rs 16.20
Therefore, price of 1 pack = Rs 182.40
EXERCISE 4.16
1. Ashwinee bought 15 party hats at Rs 12.50 each and 15 balloons at Rs 7.50 each. How much
money did she spend?
2. Ali is 0.85 m taller than Anish while Anju is 0.5 m shorter than Anish. If the total height of
the three children is 3.05 m, find the height of each child.
3. A pen and a copybook cost Rs 27.50 together. Deven bought 8 copybooks and 5 pens for
Rs 182.50. Find the price Sushila paid for 2 copybooks and 3 pens.
4. Sheila won a Rs 100 gift voucher at her local bakery.
(a) If she buys a slice of pizza for Rs 57.50 and uses the rest of the gift voucher to buy
chocolate chip cookies that cost Rs 2.50 each, how many cookies can Sheila buy?
(b) If instead she buys a loaf of bread for Rs 28.75 and uses the rest to buy muffins, how
many muffins will Sheila get if one muffin costs 1.5 times as much as one chocolate chip
cookie?
5. Study the following table and answer the questions given below:
Items
Cost
Cinema ticket : 1 Adult
1 Child
Rs 175
Rs 125
Ice cream
Rs 25.75 per unit
Popcorn
Rs 32.50 per packet
Fruit juice
Rs 18.25 per juice
Mr and Mrs Samy went to the cinema with their three children.
(a) How much will they pay for the cinema tickets?
(b) Mr Samy bought a fruit juice and an ice cream for each person, including himself. How
much did he pay for the fruit juice and ice cream?
(c) Mrs Samy bought a packet of popcorn while one of the children ate one more ice cream.
How much money did the family spend in all?
64
Chapter 4 - Fractions and Decimals
Summary
a
: a is the numerator and b is the denominator.
b
•
Fraction
•
Proper fraction: the numerator is less than the denominator. Examples are 2 , 6 , 115 .
•
Improper fraction: the numerator is greater than or equal to the denominator.
3
11
132
Examples are 5 , 17 , 5 , 13 .
4
•
•
•
•
•
•
•
•
•
•
•
•
5
13
A mixed number consists of a whole number and a proper fraction. Examples are
1
•
15
4
2
2 , 6 , 45
4
5
3
Equivalent fractions are equal in value with different numerators and denominators.
Examples are 2 and 12 .
3
18
To add or subtract two or more fractions, they must have the same denominator and
we only add or subtract the numerators.
To add or subtract fractions with different denominators, we first find the L.C.M. of
the denominators and then add or subtract the numerators.
When multiplying fractions, we can cross out the numbers vertically or diagonally
but never horizontally.
To obtain the reciprocal of a fraction, we simply invert the fraction.
For division of fractions, we always invert the second fraction and change the division
sign to the multiplication sign.
A decimal number consists of a whole number part, a decimal point and a fractional
part.
The value of the digits before the decimal point have values in units, tens, hundreds
and so on.
The value of the digits after the decimal point have values tenths, hundredths,
thousandths and so on.
To convert fractions into decimals, we first convert the fraction into an equivalent
fraction with denominators 10,100 and so on. Another way is to perform the division,
that is, we divide the numerator by the denominator.
To add and subtract decimals, we place the numbers according to their place value
and align the decimal point one under the other before proceeding with the addition
and subtraction.
To multiply and divide decimal numbers by 10, 100, 1000, we move the decimal point
to the right or to the left by 1 digit, 2 digits and 3 digits respectively.
65
5
ANGLES
Chapter 5 - Angles
Learning Objectives
By the end of this chapter, you should be able to:
• recognise and use fundamental geometrical terms.
• distinguish among different types of angles.
• measure and construct angles using geometrical instrument.
• identify parallel lines and transversals.
• identify complementary, supplementary, vertically opposite, corresponding, alternate
and co-interior angles.
• find unknown angles using notions of complementary, supplementary, corresponding,
alternate and co-interior angles.
Angles in real life
CHECK THAT YOU CAN:
•
•
Draw straight lines.
Perform the four arithmetic
operations.
KEY TERMS
•
•
•
•
•
•
•
•
•
•
•
•
Points and Lines
Rays and Line Segments
Angles: Acute Angle, Obtuse
Angle, Right Angle, Straight
Angle, Reflex Angle, Angle
at a point
Parallel Lines
Perpendicular Lines
Transversal
Complementary Angle
Supplementary Angle
Vertically Opposite Angles
Corresponding Angles
Co-interior Angles
Alternate Angles
DID YOU KNOW
History of Angles
Measurement of angles, using the number 360, dates back to the Babylonians, who used a number system based on units of 60
rather than units of 10. Today, angles are measured using the unit degree, where 360 degrees describes one full turn. In ancient
Egypt, for instance, angles were measured using the sun’s shadows and markings made on stone tables.
66
Chapter 5 - Angles
Introduction
Points, lines, line segments, rays and planes
A point has no dimensions (i.e, length, width or depth). It is usually denoted by a dot and is
used to specify a location. A capital letter is used to indicate a point. E.g. Point Y
Y
.
Lines, line segments and rays have one dimension.
A line has no end points. Thus, we represent a line by including arrows at both ends to indicate
line
that the line continues endlessly in both directions.
A line segment starts with a point and ends with another point. We can draw and measure line
segments as they have definite lengths. A line segment is part of a line.
A
e.g. line segment AB
line segment
B
A ray is a half-line and is also part of a line. It starts at a specific point, say A and continues
endlessly in one direction only.
A
ray
A
A
or
or
A plane has length and width but no height. It is a flat surface with two dimensions. Example of
a plane is the surface of a table.
EXERCISE 5.1
1. Match the following:
Point
Ray
Line segment
Line
viewed from top
2. Represent the following on a sheet of paper:
(a)
(b)
(c)
(d)
(e)
A point P
A line segment XY
A ray starting at Z
A plane
A line
Plane
Angles
The word angles come from the Latin word angulus, which means
‘a corner’. Angles can be found everywhere in our daily life. One of
the common angles you may come across is the right angle (90⁰).
An angle is formed between two lines or rays or line segments
when they intersect. The point of intersection is called the vertex
of the angle. The two lines or rays or line segments are also referred
to as the arms of the angle.
Arm of the angle
Vertex
Angle
67
Chapter 5 - Angles
RECALL
Types of angle
An angle of 180⁰
is called a straight
angle.
Activity:
180⁰
A right angle is an
angle of 90⁰.
Identify all the different
angles you can find in the
bicycle below. Indicate
which type of angles they
are and estimate the size
of each angle.
90⁰
An acute angle is
an angle less than
90⁰.
An obtuse angle is
an angle between
90⁰ and 180⁰.
Angles of more
than 180⁰ but
less than 360⁰
are called reflex
angles.
A complete turn
makes an angle of
360⁰.
NOTE TO TEACHER
Students may be asked to find examples of
each type of angle in their own classroom
or school. They can also estimate the size of
each angle, before checking their estimations
by measuring if possible.
360⁰
EXERCISE 5.2
1. For each of these angles, identify whether it is acute, obtuse, reflex or right-angled.
(a)
(b)
(c)
(d)
(e)
(f)
2. Represent the following in your copybook.
(a) An acute angle
(b) An obtuse angle
3. State the type of angle for each of the following angles:
(a) 160⁰ (b) 12⁰ (c) 321⁰ (d) 90⁰ (e) 180⁰ (f) 256⁰
68
(c) A reflex angle
(g) 190⁰
(h) 360⁰
Chapter 5 - Angles
Types of Lines
STOP AND THINK
Parallel Lines
Consider the 2 pairs of lines below
(Pair A and Pair B).
Two lines which are at the same distance apart and that
never meet are parallel to each other (Fig. 1).
Note: Arrows are used to show that two lines are parallel.
Are the two lines in pair A parallel? Why?
Are the two lines in pair B parallel? Why?
A:
Fig. 1
B:
Perpendicular lines
Two lines that meet or cross at 90⁰ are perpendicular to
each other (Fig. 2).
How many right angles are there in Fig. 2?
Fig. 2
Transversals
Draw a pair of parallel lines using the two sides of your
ruler (Fig. 3).
Fig. 3
Now, draw a line at any angle in order to intersect both
lines as shown in Fig 4. What do you observe? How many
angles are formed between the ‘new’ line and the pair of
parallel lines? The 'new' line is called a transversal.
NOTE TO TEACHER
Explain to students that a transversal
can also cut more than 2 lines and
that the lines need not necessarily be
parallel.
transversal
Fig. 4
EXERCISE 5.3
1. Draw on different diagrams
(a) a pair of parallel lines,
(b) two lines which are perpendicular to each other,
(c) a pair of parallel lines and a transversal.
2. ABCD is a rectangle.
(a) Which side is parallel to AB?
(b) Which side(s) is/are perpendicular to AD?
A
B
D
C
3. LMNO is a trapezium.
(a) Which side is parallel to LM?
(b) Which side(s) is/are perpendicular to LO?
L
O
M
N
69
Chapter 5 - Angles
Measuring Angles
Anti-clockwise
scale
Clockwise
scale
The instrument used to measure angles is called
a protractor. Some protractors are full circles and
can be used to measure angles up to 360⁰.
However, most protractors are semi-circular in
shape and can be used to measure angles up to
180⁰ as shown in Fig. 5.
Outer scale
Consider the angle POQ in Fig. 6. To denote the
angle shown in the figure, we use the following
^
notations:
POQ
POQ or
P
O
Inner scale
Fig. 5
The protractor has two sets of numbers,
ranging from 0⁰ to 180⁰, one starting from
the left and the other from the right.
Note: It is a good practice to estimate the
size of the angle before measuring it.
Q
Fig. 6
Baseline
Centre
Notations to write angles
Using the protractor to measure angles
Example 1
Consider the figure below.
Z
CHECK THIS LINK
https://www.mathplayground.com/
measuringangles.html
X
Y
^
(i) What type of angle is XYZ?
^
(ii) Using the clockwise scale, measure XYZ.
Solution
^
(i) XYZ is obtuse.
(ii) Step 1: Place your protractor so that its centre lies on
the vertex of the angle and the base line is placed along
one of the arms of the angle
Z
(here it is XY).
X
Step 2:
Y
• Start reading the scale at zero.
• We use the clockwise scale.
• Go round this scale until you reach the other arm of the
angle.
• Read the size of the angle (120⁰) .
^
So, XYZ = 120⁰.
70
Caution:
A common error in measuring angles
is not to start at zero.
Ensure that you start reading at zero.
This is why it is useful to make an
estimate of the angle first.
STOP AND THINK
Can you think of another way to
measure this angle using your
protractor?
Chapter 5 - Angles
Example 2
Using the anti-clockwise scale, measure
ABC.
STOP AND THINK
C
Solution
How will you measure a reflex angle?
A
B
We first make an estimate of the angle.
ABC is an acute angle.
Step 1: Place your protractor so that its
centre lies on the vertex of the angle and
the base line is placed along one of the
arms of the angle, say, BA.
Step 2: Measure the angle ABC using the
anti-clockwise scale.
Hence,
C
35⁰
A
ABC = 35⁰.
B
EXERCISE 5.4
1. State the size of the angles shown on the protractor below :
(a)
BOA
(b)
BOC
(c)
BOD
(d)
BOE
C
A
B
B
O
O
D
B
B
O
E
O
71
Chapter 5 - Angles
2. Measure each of the following angles.
(a) (b)(c)
(d) (e)(f)
FUN ACTIVITY:
Using masking tape, make several
straight intersecting lines on your desk.
Measure as many angles as you can
and say whether each of these angles is
acute, obtuse, reflex, right or straight.
Constructing Angles
To construct an angle of a given size, you will need to use a
ruler and a protractor.
CHECK THIS LINK
http://www.mathplayground.com/
alienangles.html
Example
^
Construct ABC = 50⁰
Step 1: Draw a line segment AB as shown
below.
A
Step 2: Place the centre of the
protractor at B with the base line of the
protractor aligned to line segment AB
as shown in the figure below.
B
A
Step 3: Use the clockwise scale to read 50⁰.
Mark this position 'C' with a pencil.
C
B
Step 4: Remove the protractor and
join the points B and C to form line
^
segment BC. ABC = 50⁰
C
500
A
72
B
A
B
Chapter 5 - Angles
EXERCISE 5.5
1. Construct a right angle.
2. Use a ruler and protractor to construct the following angles:
(a) 40⁰
(b) 135⁰
(g) 95⁰
(h) 200⁰
(c) 170⁰
(d) 300
(e) 240⁰
(f) 300⁰
(i) 18⁰
(j) 210⁰
(k) 190⁰
(l) 320⁰
3. Find the following in degrees, stating the type of angle
formed:
(a)
3
of a straight angle
4
2
of a straight angle
5
1
(d) 3 of a right angle
2
(c) 0.6 of a straight angle
1
(e) 18 of a complete turn
(g) 19 of a complete turn
20
(h)
A right angle makes an angle of
90⁰.
2.
A straight angle makes an angle
of 180⁰.
3.
A complete turn makes an angle
of 360⁰.
1
of a complete turn
8
(a) 44⁰ as a fraction of a straight angle.
(b) 35⁰ as a fraction of a right angle.
(c) 130⁰ as a fraction of a right angle.
(d 207⁰ as a fraction of a straight angle.
(e) 75⁰ as a fraction of a complete turn.
(f) 333⁰ as a fraction of a complete turn.
5. Consider the angle formed from the hour hand to the minute
hand on a clock in a clockwise direction. Match the type of
angle formed when the clock shows:
03 20
08 10
16 50
07 25
1.
(f) 0.4 of a straight angle
4. Write in its simplest form:
21 00
RECALL
(b)
FUN ACTIVITY:
Measure as many angles as you can on the
stickman below.
Reflex angle
Right angle
Obtuse angle
Acute angle
Straight angle
6. Find the angle that represents 3 of a complete turn.
5
73
Chapter 5 - Angles
Complementary Angles
Complementary angles form a right angle and have a sum of 90⁰.
60⁰
For example, 30⁰ and 60⁰ are complementary angles.
30⁰
Can you give some more examples of complementary angles?
Supplementary Angles
140⁰
Supplementary angles form a straight angle and have a sum of 180⁰.
40⁰
For example, 140⁰ and 40⁰ are supplementary angles. Can you give some more examples of
supplementary angles?
Angles at a point
The sum of the angles at a point is 360⁰.
a
e
a + b + c + d + e = 360⁰.
DID YOU KNOW
c
b
d
Angle at a point is also known as a
full turn or complete turn or a
perigon.
Example
Work out the size of the unknown angles. Justify each answer.
(a)(b)(c)
30⁰
a
40⁰
b
65⁰
110⁰
a
55⁰
Solution
(a)(b)(c)
55⁰ + a = 90⁰
40⁰ + 65⁰ + b = 180⁰
30⁰ + 90⁰ + 110⁰ + a = 360⁰
(right angle)
a = 90⁰ – 55⁰
a = 35⁰
(straight angle)
(angles at a point)
105⁰ + b = 180⁰
b = 180⁰ – 105⁰
b = 75⁰
230⁰ + a = 360⁰
a = 360⁰ – 230⁰
a = 130⁰
EXERCISE 5.6
1. Write down the complementary angle of
(a) 40⁰
74
(b) 70⁰
(c) 14⁰
(d) 48⁰
(e) 65⁰
(f) 85⁰
Chapter 5 - Angles
2. Write down the supplementary angle of
(a) 25⁰
(b) 70⁰
(c) 45⁰
(d) 115⁰
(e) 165⁰
(f)140⁰
3. Calculate the angle marked x in each of the following diagrams.
(a)(b)(c)
x
50⁰
4.
x
40⁰
x x
x
30⁰
x
Calculate angle y in each of the following cases.
(a)(b)(c)
y
120⁰
y
120⁰
50⁰
y
y
5. Calculate angle z in each of the following cases.
(a)
(b)
z
z
z
130⁰
120⁰
120⁰
100⁰
6. (a) Given angle b is twice angle a, find a and b.
(b) Find angle c.
145⁰
a
3c
b
65⁰
4c
5c
2c
c
7. If there are 4 angles at a point and they are in the ratio 1 : 2 : 3 : 6, calculate the size of the
largest angle.
8. There are seven angles at a point. One of them is 600 and the other six angles are all equal.
What is the size of one of the remaining angles?
75
Chapter 5 - Angles
Angles formed by straight lines
Vertically Opposite Angles
Activity
1. Draw two lines that cross each other, at any angle, as shown in the figure.
a
2. Label the point of intersection 0.
b
0
d
c
What can you say about the sum of the angles
a, b, c and d?
NOTE TO TEACHER
3. Now, using your protractor, measure angles a , b , c and d.
What do you notice?
What can you say about angles a and c?
What can you say about angles b and d?
Discuss students’ solutions in the
class and encourage them to
observe that the angles vertically
opposite each other are equal,
irrespective of the size of the
angles. This can be pointed out
by using the different examples
obtained from different students
in the class.
Compare your measurements with that of your friends.
Vertically opposite angles are angles which are opposite
each other and which are equal.
In this case, we have a = c and b = d.
EXERCISE 5.7
1. Calculate the unknown angles in each of the following cases.
(a)(b)(c)
110⁰
x
65⁰
60⁰
x
x
(d)(e)(f)
50⁰
68⁰
x
y
x
75⁰
x
35⁰ 35⁰
10⁰
76
70⁰
Chapter 5 - Angles
Angles formed by Parallel Lines
When a transversal crosses a pair of parallel lines, different angles are formed.
NOTE TO TEACHER
Corresponding Angles
a
d
You can ask your students to
draw a pair of non-parallel lines
and a transversal and check if
corresponding angles are equal in
this case.
b
c
e
h
f
g
Step 1: Using the two sides of your ruler, draw a pair of parallel lines.
Step 2: Draw a transversal to cross the pair of parallel lines.
Step 3: Label each of the angles formed as shown in the above diagram.
Step 4: Using a protractor, measure each of the angles.
Angle a = ?
Angle e = ?
What do you notice? What can you conclude?
Similarly, measure angle c and angle g. What do you notice?
We say that angles a and e are corresponding angles and are equal as they are in the same
position between the parallel lines and the transversal. Also, c and g are corresponding angles.
Can you find other pairs of corresponding angles in your diagram?
Example
Find the unknown angles in each of the following.
a
(a)(b)
67⁰
75⁰
b
Solution
(a)
a = 75⁰
Since they are
corresponding angles
(b)
b = 67⁰
Since they are
corresponding angles
77
Chapter 5 - Angles
Co-Interior Angles
The angles between parallel lines on the same side of the
transversal are called co-interior angles. They add up to 180⁰. c
and f are co-interior angles,
Note: “Co” means together
and “interior” means inside.
c
i.e., c + f = 180⁰
Activity
f
Measure angles c and f.
c=?
f= ?
c
Similarly, d and e are also co-interior angles and are supplementary,
i.e., d + e = 180⁰
d
f
e
What do you notice
about their sum?
Example
Find the unknown angles in each of the following.
(a)(b)
b
a
110⁰
48⁰
Solution
(a)
a = 180⁰ – 110⁰ = 70⁰
since they are co-interior
angles.
(b)
b = 180⁰ – 48⁰ = 132⁰
since they are co-interior
angles.
Alternate Angles
Angles d and f are contained in a 'z-shape' figure.
They are called alternate angles and they are equal,
i.e,
d
d=f
f
Angles c and e are contained in a ‘reverse’ z figure.
Activity
Measure angles d and f
in each figure.
d=?
f= ?
d
They are called alternate angles and they are equal.
c
c=e
78
e
d
f
f
What do you notice?
Chapter 5 - Angles
Example 1
FUN ACTIVITY:
MATHEMATICS GAMES
Find the unknown angles in each of the following.
(a)(b)
a
Can you identify alternate,
corresponding and co-interior
angles in the alphabets of
your name?
118⁰
b
54⁰
Solution
(a) a = 54⁰
(they are alternate angles)
(b) b = 118⁰
(they are alternate angles)
Example 2
Find the unknown angles a and b in the diagram.
53⁰
b
a
62⁰
Solution
We first draw a line parallel to the 2 given
parallel lines as shown in the diagram
(in blue).
a = 53⁰ + 62⁰ = 115⁰
53⁰
b
53⁰
a
62⁰
(we have 2 alternate angles)
b = 360⁰ – 115⁰ = 245⁰
62⁰
(sum of angles at a point = 360⁰)
EXERCISE 5.8
1.
Calculate angle x in each of the following cases.
(a)
(b)
(c)
x
50⁰
x
2.
140⁰
110⁰
x
Calculate angle x in each of the following cases.
(a)
(b)
(c)
125⁰
x
35⁰
x
x
65⁰
79
Chapter 5 - Angles
3.
Calculate angle x in each of the following cases.
(a)
(b)
(c)
x
x
4.
68⁰
130⁰
40⁰
x
Find the unknown angles in each of the following cases.
(a)
(b)
115⁰
(c)
125⁰
140⁰
x
y
x
x
z
y
y
z
z
(d)
(e)
135⁰
50⁰
60⁰
x
y
(f)
x
y
x
z
45⁰
35⁰
z
(g)
(h)
35⁰
x
y
(i)
58⁰
125⁰
x
y
75⁰
x
w
z
y
z
125⁰
5.
Find the unknown angles given that a =
z
4
b.
5
a
c
d
b
6.
Find the unknown angles in the diagram
given that angle b is thrice angle a.
a
b
80
c
Chapter 5 - Angles
7.
Find the unknown angles in the diagram
given that angle b is twice angle a.
a
b
63⁰
8.
Find the unknown angles in the diagram
given that angle a is twice angle b.
a
b
c
d
9. Find the unknown angles in the diagram
given that a = 2 b and d = 2c.
3
b
a
c
10.
d
Find the unknown angles:
(b)
(a)
(c)
68⁰
114⁰
a
b
a
313⁰
121⁰
50⁰
(d)
(e)
(f)
d
12⁰
118⁰
a
25⁰
130⁰
a
46⁰
54⁰
b
a
c
a
(g)
(h)
56⁰
(i)
a
a
b
31⁰
b d
c
a
55⁰
30⁰
81
Chapter 5 - Angles
Summary
•
An angle is formed between two lines or rays or line segments when they
intersect. The point of intersection is called the vertex of the angle.
vertex
•
angle
Angles
An acute angle is less than 90⁰.
An obtuse angle is an angle between
90⁰ and 180⁰.
A right angle is exactly 90⁰.
180⁰
An angle of 180⁰ is called a straight
angle
Angles of more than 180⁰ but less
than 360⁰ are called reflex angles.
Angle at a point is equal to 360⁰.
82
360⁰
Chapter 5 - Angles
Summary
•
Lines
•
Parallel lines are lines which are at the
same distance apart and which never
meet each other.
•
Perpendicular lines meet or cross at 90⁰
to each other.
•
A line which cuts a pair of parallel lines is
called a transversal.
•
Angles are measured or constructed using a protractor.
•
Complementary angles: sum = 90⁰
Supplementary angles: sum = 1800
a + b = 90⁰
a
x
b
•
Angles at a point.
e
a + b + c + d + e = 360⁰
d
When two lines cross, vertically opposite angles are equal.
b
a
b
•
x + y = 180⁰
c
b
a
•
y
a
Angles formed by parallel lines:
(a) Corresponding angles:
a=b
a
b
(b) Co-interior angle:
c + d = 180⁰
c
d
(c) Alternate angles form a ‘Z’:
e=f
e
f
83
6
POLYGONS
Chapter 6 - Polygons
Learning Objectives
By the end of this chapter, you should be able to:
• identify and name polygons (up to decagon), including regular polygons.
• differentiate among scalene, isosceles and equilateral triangles in terms of lengths and
angles.
• find unknown angles in triangles.
• differentiate among different types of quadrilaterals ( rectangle, square, parallelogram,
rhombus, kite, trapezium, arrow head).
• find unknown angles in quadrilaterals.
Polygons in real life
CHECK THAT YOU CAN:
Polygons can be found all around us in our everyday life.
For instance, the buildings that surround us often involve
polygons. Architects, artists and designers often use
different types of polygons when designing buildings,
houses and decorations.
•
•
•
Identify polygons.
Identify a vertex or vertices.
Determine why a polygon is
said to be regular.
KEY TERMS
Picnic table
Kite
House
Honeycomb
Tile
Stop Sign
Introduction
A polygon is a closed (bounded) flat surface (shape) with
many sides. A polygon can be either regular or irregular.
Regular polygons are those where all the sides are equal
and all the angles are equal. Irregular polygons are
polygons whose sides are not all of equal length and
whose angles are not all of equal size.
Examples of polygons:
Categorise the above polygons into regular and irregular polygons.
84
•
•
•
•
•
•
•
•
•
•
Polygon
Regular polygon
Angle
Interior angles
Triangle
Quadrilateral
Line Segments
Diagonals
Vertex/vertices
Sides
DID YOU KNOW
The word polygon has Greek
origins. “Poly” means many and
“gons” comes from the Greek word
“gonus” meaning angles.
NOTE TO TEACHER
• All the sides of the polygon must
be straight lines.
• The end point and the starting
point must be the same.
Chapter 6 - Polygons
Activity 1
STOP AND THINK
Look around your classroom and school premises to find
different polygons. Try to identify the polygons that you
have found and name them.
Are the figures below polygons?
Justify your answer.
Compare your results with that of your classmates.
Polygons are named by the number of sides they have.
Name of
Polygon
Shape of
Polygon
Number of
sides
Number of
Angles
Triangle
‘Tri’ means “three”
e.g. Tri-cycle
having three wheels
3
3
Quadrilateral
‘Quad”
means four
e.g. quadricolour or
four colours
4
4
Pentagon
“penta”
means five
5
5
Hexagon
“hexa”
means six
6
6
Heptagon
“Hepta”
means seven
7
7
Octagon
“Octa”
means eight
e.g. octopus
8
8
Nonagon
“nona”
means nine
9
9
Decagon
“deca”
means ten
10
10
85
Chapter 6 - Polygons
Activity 2
DID YOU KNOW
The Pentagon which is found in
Washington, US, is the headquarters
of the U.S. Department of Defense.
As its name suggests, the building is
in the shape of a pentagon, that is, a
five-sided structure.
For each of the polygons in the previous table,
draw a polygon that is irregular.
EXERCISE 6.1
The design for this building was
originally made as the construction
was to be made on a plot of land
with borders on five sides. However,
the building was later constructed
on another plot of land, where the
shape of the pentagon was not
necessary. Nevertheless, the original
design remained as there was no
time to make a new design.
Write down the name of each of the following polygons:
(a)
(b)
(c)
(d)
Triangle
Vertex
A triangle is a polygon with three sides (edges),
three angles and three vertices. Angles a, b and
c are called interior angles since they are found
inside the polygon (triangle).
c
Angle
b
Side or edge
a
Angle Properties of a Triangle
Activity 3
A
Step 1: Using your ruler, draw a triangle of any size.
Step 2: Label the triangle ABC and the angles a, b and c
a
respectively (as shown in Fig. 1).
Step 3: Using your protractor, measure each of the angles
and note down the size of each angle.
Angle a =
Angle b =
b
Angle c =
c
B
C
Fig. 1
Step 4: Find the sum of angles a, b and c. What do you observe?
Compare these results with that of your classmates.
Step 5: Cut out the triangle ABC. Tear off the three corners (as shown in Fig. 2)
and place them without overlapping as shown in Fig. 3.
a
What do you notice?
a+
b+
c = 180⁰
(A straight angle is equal to 180⁰)
Hence, the sum of the interior angles of a triangle is equal to 180⁰.
86
c
b
Fig. 2
c
a
Fig. 3
b
Chapter 6 - Polygons
Finding unknown angles in a triangle
Example
Solution
Find the value of a.
65⁰
60⁰
65⁰ + 60⁰ + a = 180⁰ (Sum of interior angles in a triangle is 180⁰)
125⁰ + a = 180⁰
a = 180⁰ - 125⁰
a = 55⁰
a
EXERCISE 6.2
1. Find the size of each of the marked angles.
(i)
(ii)
40⁰
e
b
55⁰
a
74⁰
59⁰
(iv)
(iii)
42⁰
(v)
53⁰
125⁰
80⁰
(vi)
40⁰
46⁰
g
p
11⁰
f
64⁰
Types of Triangles
We can classify triangles based on the length of their sides or based on the size of their angles.
1. Scalene Triangle
Note: ∆ is used to
represent a triangle.
A scalene triangle is a triangle in which all the sides are of
different lengths and all its angles are of different sizes.
Example:
AB ≠ BC ≠ AC and
Angle BAC ≠ angle ABC ≠ angle BCA
So, ∆ ABC is a scalene triangle.
2. Isosceles Triangle
A
65⁰
105⁰
10⁰
B
C
An isosceles triangle is a triangle in which two of its sides and two of its angles are equal.
P
Example:
∆ PQR is an isosceles triangle because it has two sides
3 cm
3 cm
of equal length, namely PQ = PR and angle PQR = PRQ.
equal
Note: The angles opposite the two equal sides of an isosceles
triangle are also equal, that is, angle PQR = angle PRQ.
Q
5 cm
R
87
Chapter 6 - Polygons
3. Equilateral Triangle
FIND OUT
An equilateral triangle has all its sides of the same length.
It is a regular polygon and all its angles are of the same size as well.
X
Example:
∆ XYZ is an equilateral triangle.
2 cm
2 cm
XY = XZ = YZ
^
^
^
and YXZ = XYZ = YZX = 60⁰
Z
Y
How do you call ∆ ABC ?
A
B
C
2 cm
4. Right-Angled Triangle
A
A right-angled triangle is a triangle with one angle equal to 90⁰.
Example:
^
In ∆ ABC, ABC = 90⁰ .
Stands for 90⁰
So, ∆ ABC is called a right-angled triangle.
C
B
5. Acute-Angled Triangle
L
An acute-angled triangle is a triangle with all its angles less than 90⁰.
Example:
In ∆ LMN, the 3 angles are all less than 90⁰.
So, ∆ LMN is called an acute-angled triangle.
6. Obtuse-Angled Triangle
40⁰
80⁰
60⁰
N
M
An obtuse-angled triangle is a triangle with one angle obtuse, i.e., less than 180⁰ but
greater than 90⁰.
X
Example:
Y = 140⁰ which is an obtuse angle.
18⁰
So, ∆ XYZ is called an obtuse-angled triangle.
140⁰
22⁰
Z
Y
EXERCISE 6.3
1. Classify the following triangles according to (a) their lengths, (b) their angles.
P
(i)
X
(ii)
G
(iii)
Y
Q
R
H
Z
88
I
Chapter 6 - Polygons
(iv)
(v)
B
(vi)
L
C
M
N
A
D
F
E
Example
Find the unknown angles.
(a) (b)
(c)
p
35⁰ y
38⁰
s
t
x
42⁰
a
b
50⁰
Solution
(a) s + 38⁰ + 90⁰ = 180⁰
s = 180⁰ – 90⁰ – 38⁰
s = 52⁰
[Sum of angles in a triangle = 180⁰]
(b) t = 42⁰
t + 42⁰ + p = 1800
p = 180⁰ – 2 (42⁰)
p = 180⁰ – 84⁰
p = 96⁰
(c) x = 50⁰
y = 180⁰ – (2 x 50⁰) – 35⁰ y = 45⁰
[Since the triangle is an isosceles triangle, it has 2 equal angles]
[Sum of angles in a triangle = 180⁰]
[Since t = 42⁰]
[Since the triangle is an isosceles triangle: it has 2 equal angles]
[Sum of angles in a triangle = 180⁰]
a = 180⁰ – 50⁰ – 35⁰ a = 950
[Sum of angles in a triangle = 180⁰]
b = 180⁰ – (50⁰ + 45⁰) [Sum of angles in a triangle = 180⁰] or b = 180⁰ – 95⁰ [Straight angle]
b = 85⁰
b = 85⁰
EXERCISE 6.4
1. Find the unknown angles:
q
(i)
55⁰
(ii)
(iii)
48⁰
t
s
75⁰
89
Chapter 6 - Polygons
(iv)
(v)
a
(vi)
55⁰
x
d
80⁰
c
b
2. Find the unknown angles:
(i)
(ii)
(iii)
30⁰
e
c
25⁰
d
b
a
p
24⁰
(iv)
n
n
q
55⁰ q
40⁰
m
(v)
r
s
p
(vi)
p
q
b
r
t
x
30⁰
s
a
65⁰
z
y
50⁰
20⁰
c
Quadrilaterals
A quadrilateral is a simple closed figure formed by joining 4 line segments. It has 4 sides, 4
vertices, 4 angles and 2 diagonals. 'Quad' means four and 'lateral' means sides.
B
vertex
A
RECALL
diagonals
Diagonals:
angle
D
C
The lines AC and BD that
join the opposite vertices
are called diagonals.
side
The diagonal AC divides the quadrilateral ABCD into 2 triangles (∆ ADC and ∆ ABC). Similarly,
diagonal BD divides quadrilateral ABCD into 2 triangles (∆ BAD and ∆ BCD).
We know that the sum of the interior angles of any triangle is 1800.
Therefore, sum of the 4 interior angles of a quadrilateral = 2 x 180⁰ = 360⁰
^
^
^
^
i.e, A + B + C + D = 360⁰
90
Chapter 6 - Polygons
Types of Quadrilaterals and their properties
Name
Square
Rectangle
Parallelogram
Properties
Shape
All 4 sides are equal.
Opposite sides are parallel.
All angles are right angles.
The diagonals which are of equal
length bisect each other at 90⁰.
Opposite sides are parallel and equal.
All angles are right angles.
The diagonals of a rectangle are equal
in length.
The diagonals bisect each other but
not at 90⁰.
a
b
Opposite sides are parallel and equal.
Opposite angles are equal.
The diagonals bisect each other.
b
a
B
Rhombus
Opposite sides are parallel.
All sides are equal.
Opposite angles are equal.
The diagonals bisect each other at 900 .
C
A
D
Trapezium
M
One pair of opposite sides is parallel.
Angles and lengths can be of different
sizes.
L
Kite
One pair of opposite angles is equal.
Two pairs of adjacent sides are equal.
The longer diagonal bisects the other
diagonal at 90⁰.
Arrowhead
It is also called a concave quadrilateral.
It has at least one pair of adjacent sides
which are equal.
It has one reflex angle.
N
O
91
Chapter 6 - Polygons
Cultural Connection: Tangram
FIND OUT
The tangram is a dissection puzzle consisting
of seven flat shapes. It originates from China
and dates back to centuries ago. The seven
pieces of the puzzle can be assembled
together in different ways to form different
shapes and polygons.
Do you know what is meant
by tessellation?
How is tessellation related
to polygons?
Activity 4: Create your own tangram
Check this link: https://www.tangram-channel.com/draw-your-own-tangram/
to create your own tangram.
Now that you have your tangram drawn, cut out the different
pieces and rearrange them to obtain different shapes.
Explore how many different polygons you can obtain with all
seven pieces of your tangram. Two examples are given below.
FUN ACTIVITY:
1. I am a quadrilateral with
both pairs of opposite
sides equal. Which shape
can I be?
2. I am a quadrilateral with
just one pair of opposite
angles equal. Which
shape am I?
EXERCISE 6.5
Name each of the following quadrilaterals.
Finding unknown angles in quadrilaterals
Example 1
Find the unknown angle x.
A
Solution
70⁰
x + 98⁰ + 124⁰ + 70⁰ = 360⁰
(sum of angles in a quadrilateral = 360⁰)
x = 360⁰ - 292⁰
x = 68⁰
92
D
98⁰
x
124⁰
C
B
Chapter 6 - Polygons
Example 2
Find the unknown angle x.
R
P
48⁰
Solution
x = 90⁰- 48⁰ = 42⁰
(Triangle PTS is right angled)
x
S
T
Example 3
Find the unknown angle x.
N
M
Solution
x = 180⁰ - 78⁰ = 102⁰
(Co-interior angles are supplementary)
P
x
78⁰
O
Example 4
Find the unknown angles x and y.
S
Solution
x = 28⁰
(Alternate angles as ST and VU are parallel)
y = 1800 – ( 49⁰+ 28⁰) = 103⁰
T
x
49⁰
y
28⁰
U
V
Example 5
Find the unknown angles x and y.
630
x
Solution
x = 180⁰ – 63⁰ = 117⁰ (co-interior angles)
y = 63⁰ (opposite angles are equal)
y
Example 6
Find the unknown angles x and y.
A
D
Solution
x=
1800 – 1420
= 19⁰
2
y = 1800 - (59⁰ x 2) = 620
(∆ ABD is an
isoceles triangle)
1420
x
590
B
y
C
93
Chapter 6 - Polygons
Example 7
Find the unknown angle x.
X
W
x
Solution
^
XWY = 290 ( alternate angles)
x = 1800 – (290 x 2) = 1220
290
(∆ WXY is an isoceles triangle)
Y
Z
EXERCISE 6.6
1. For each of the given quadrilaterals, calculate the unknown angles.
(a)
(b)
(c)
x
450
31
0
230
1040
y
x
x
(d)
(e)
(f)
y
y
x
5x
420
x
x
(g)
y
980
(h)
370
(i)
y
x
500
290
560
1060
860
x
y
320
x
2. ABCD is a rectangle with diagonals AC and BD as
shown in the diagram. Angle AEB = 1440.
Find the value of a.
A
B
1440
a
D
94
240
E
C
Chapter 6 - Polygons
B
A
3. In the diagram, ABCD is a trapezium.
Angle c is twice angle b. Find the
value of the unknown angles.
c
1400
a
b
D
C
Summary
1. The sum of the interior angles of a triangle is 1800.
2. Types of Triangles
X
A
P
650
equal
1050
B
10
0
Scalene triangle
R
Q
C
Z
Y
Isosceles triangle
Equilateral triangle
400
800
Right-angled triangle
180
1400
600
Acute-angled triangle
220
Obtuse-angled triangle
3. The sum of the interior angles of a quadrilateral is 3600.
4. Types of Quadrilaterals
B
a
b
A
b
a
C
D
Parallelogram
Rectangle
Square
Rhombus
N
M
L
O
Trapezium
Kite
Arrow Head
95
LENGTH, PERIMETER AND AREA
Chapter 7 - Length, Perimeter and Area
7
Learning Objectives
By the end of this chapter, you should be able to:
• distinguish among different units of length and units of area.
• convert length and area from one unit to another.
• perform arithmetic operations involving length.
• find the perimeter and area of 2-D figures.
• solve word problems involving length, perimeter and area.
Length in real life
CHECK THAT YOU CAN:
24 cm
5 mm
Grain of rice
4 mm
Ant
Copybook
• Perform arithmetic operations
involving numbers.
• Find the perimeter and area of
a square and a rectangle.
Cap Malheureux
KEY TERMS
75 cm
Tennis racket
Souillac
Cap Malheureux to Souillac: 76.7 km
Length refers to the distance between two points. The
measurement of length dates back to early civilisations such
as the Ancient Egyptians and Roman civilisations, where
body parts were common units for measurement of length.
The cubit, foot and handspan (Fig. 1) are just a few of the
earliest known units of lengths. These units were adapted
over the years by the Greeks, French and English amongst
others. The metric system was later introduced in the 17th
century and was widely used by the 19th century. It is still
used in almost all countries in the world.
•
•
•
•
•
•
•
•
•
kilometre
metre
centimetre
millimetre
perimeter
area
hectare; arpent
perche; toise
mm2, cm2, m2, km2
FIND OUT
Can you find out some measuring
instruments that can be used to
measure lengths and distances?
DID YOU KNOW
Fig. 1
Today, we measure length using the unit metre (m). The
millimetre (mm) and centimetre (cm) are used to measure
shorter lengths and the kilometre (km) is used to measure
longer lengths or distances.
96
The original definition of the metre
has changed over time and today
the metre is defined as the distance
travelled by light in vacuum during
1
a time interval of
of a
299792458
second.
Check this link to find out more
about the metre:
http://www.bipm.org/metrology/
length/units.html
Chapter 7 - Length, Perimeter and Area
EXERCISE 7.1
STOP AND THINK
Estimate the following, using an appropriate unit.
(a) the distance between your house and your school
(b) the distance between your table and the whiteboard/
blackboard in your class
(c) the thickness of your Mathematics textbook
(d) the thickness of one sheet of paper in your copybook
Conversion of length from one unit to another
How will you measure the length
of the following:
NOTE TO TEACHER
Prompt students to use a string
and place it along the wavy
line. They can then use a ruler
to measure the length of string
afterwards.
× 1 000
× 1 000
km
× 100
m
÷ 1 000
× 10
cm
÷100
mm
÷10
÷ 1 000
Note: Use a ruler to observe that 1 cm = 10 mm.
centimetre
Example 1
Convert the following:
(a) 3.5 km to m
(b) 2 575 mm to m
(c) 2 1 m to cm
2
Solution
(a) 1 km = 1 000 m
To convert km to m, we multiply by 1 000.
Therefore, 3.5 km = (3.5 × 1 000) m = 3 500 m.
FIND OUT
You may have heard of the units
‘inch’, ‘yard’, ‘foot’ and ‘mile’ for
measurement of lengths or
distances. These units form part of
the British Imperial system of units
and are sometimes still used in
certain parts of the world.
You may find out more about the
Imperial system of units (e.g.
where are these still used in the
world today and the conversion to
metres).
(b) 1 m = 1 000 mm
To convert mm to m, we divide by 1 000.
2 575
Therefore, 2 575 mm = (
) m = 2.575 m.
1 000
(c) 1 m = 100 cm
To convert m to cm, we multiply by 100.
Therefore, 2 1 m = ( 5 × 100) cm = 250 cm.
2
2
NOTE TO TEACHER
Encourage students to use a ruler
where one side of the scale is in
metric unit and the other side is
in inches.
Prompt them to determine the
equivalent of one inch in cm.
97
Chapter 7 - Length, Perimeter and Area
Example 2
Fill in the blanks using = (equal), < (less than), > ( greater than).
(a) 0.1 m_____ 10 cm
(b) 0.25 cm_____ 25 mm
Solution
(a) 1 m = 100 cm
So 0.1 m = (0.1 × 100) cm = 10 cm
Therefore 0.1 m = 10 cm
(b) 1 cm = 10 mm
So 0.25 cm = (0.25 × 10) mm = 2.5 mm
Therefore 0.25 cm < 25 mm
EXERCISE 7.2
1. Convert the following measurements to the units indicated:
(a) 7 cm to mm
(b) 3 km to mm
(c) 8 m to cm
(e) 9 km to m
(f) 48.7 m to cm
(g) 8.26 m to mm
(i) 4.75 km to mm
(j) 8 000 mm to m
(k) 35 mm to cm
(m) 250 cm to m
(n) 6 750 m to km
(o) 35.6 cm to m
2. Fill in the blanks using = , < , > .
(a) 0.2 m_____ 20 cm
(b) 1.78 mm____ 178 cm
(d) 3.5 km____ 3 500 m
(e) 0.03 km_____ 300 cm
(g) 2 mm_______ 0.2 cm
(h) 24 km____ 24 mm
(d) 62.8 cm to mm
(h) 4 km to cm
(l) 450 cm to m
(p) 4 605 mm to m
(c) 0.25 cm_____ 25 mm
(f) 79 cm____ 79 m
(i) 1 mm____0.1 cm
3. Circle the greater length for each pair.
(a) 4 cm, 20 mm
(b) 3 km, 300 m
(e) 5.5 m, 555 cm
(f) 6 m, 4 mm
(c) 14 km, 1 410 m
(g) 800 cm, 8.1 m
4. Arrange in ascending order:
(a) 95 mm, 23 km, 105 cm, 2 m
(c) 1 000 cm, 560 mm, 5 km, 3 m
(e) 14 km, 1 015 mm, 254 cm, 2.054 m
(b) 18 km, 1 cm, 12 mm, 87 m
(d) 3.5 m, 2 000 mm, 49 cm, 7 km
(f) 16 km, 1 550 m, 10 000 cm, 0.9 km
(d) 50 mm, 5 m
(h) 7 km, 700 m
Arithmetic operations involving length
Example 1
Evaluate 8 km + 300 m – 800 m, giving your answer in km.
Solution
We first convert 300 m and 800 m into km.
300
800
) km = 0.3 km
800 m = (
) km = 0.8 km
1 000
1 000
We then perform the operations starting from left to right, that is, addition followed by subtraction.
8 km + 300 m - 800 m = 8 km + 0.3 km - 0.8 km = 7.5 km
300 m = (
98
Chapter 7 - Length, Perimeter and Area
Example 2
Evaluate 0.5 km – 300 m + 70 cm, giving your answer in m.
Solution
We first convert 0.5 km into m and 70 cm into m.
We then perform the operations starting from left to right, that is, subtraction
followed by addition.
70
0.5 km = (0.5 x 1 000) m = 500 m and 70 cm = ( 100 ) m = 0.7 m
0.5 km - 300 m + 70 cm = 500 m – 300 m + 0.7 m
= 200 m + 0.7 m
= 200.7 m
EXERCISE 7.3
DID YOU KNOW
Evaluate:
(a) 5 cm 7 mm + 13 cm 4 mm = ______cm ______ mm
(b) 15 cm 4 mm – 8 cm 5 mm = ______cm ______ mm
(c) 4 m 30 cm + 18 m 4 cm = ______m ______ cm
(d) 6 m 75 cm – 3 m 97 cm = ______ m ______ cm
(e) 13 km 300 m + 18 km 400 m = ______ km ______ m
(f) 24 km 400 m – 5 km 555 m = ______ km ______ m
The distance from the Earth to Pluto
is around 7.5 billion kilometres.
Word problems involving length
Example
Hemisha walks a distance of 670 m from home to reach the bus stop. She then takes a bus
and travels 10 km 450 m. If the market is 15 km away from her house, find the distance left to
reach the market. Give your answer in km.
Home
Bus Stop
670 m
10 km 450 m
?
Market
15 km
Solution
We first convert 670 m and 10 km 450 m into km.
670 m = (670 ÷ 1 000) km = 0.67 km
10 km 450 m = 10 km + (450 ÷ 1 000) km = (10 + 0.45) km = 10.45 km
So, the total distance covered by Hemisha by bus and on foot = (10.45 + 0.67) km = 11.12 km
Therefore, distance left to cover to reach the market = 15 km – 11.12 km = 3.88 km
99
Chapter 7 - Length, Perimeter and Area
EXERCISE 7.4
1. Circle the correct answer.
(a) Convert 4.55 km to m.
A. 455 m
B. 4 555 m
C. 4 550 m
D. 455 m
(b) 5 cm 7 mm + 8 cm 4 mm =
A. 13 cm 74 mm
B. 14 cm 01 mm
C. 13 m 11 mm
D. 14 m 10 mm
(c) John's bedroom is 12 m long. Yash’s bedroom is 4 m longer than John's while Akhil’s bedroom
is 5 m shorter than Yash’s bedroom. What is the length of Akhil’s bedroom?
A. 16 m
B. 17 m
C. 13 m
D. 11 m
(d) A lorry is 12 m 65 cm long and when a trailer is attached to it, the total length is 20 m. The
length of the trailer is
A. 32 m 65 cm B. 7 m 35 cm
C. 8 m 35 cm
D. 7 m 65 cm
(e) Carina bought 23 m 75 cm of ribbon. She cuts 10 pieces each of 2 m 30 cm from it. Find the
length of ribbon left.
A. 20 m 30 cm
B. 3 m 75 cm C. 3 m 45 cm
D. 75 cm
2. Reema bought 9 m 75 cm of metallic fencing. She used 6 m 95 cm from it to make an enclosure.
Find the length of metallic fencing left in metres.
3. Kiren’s house is 7 km 300 m away from school and Anne's house is 11 km 432 m away from
school. Whose house is further from the school and by how much?
4. The total length of three sticks is 18.27 m. If the lengths of two sticks are 4 m 28 cm and 7 m 46 cm,
calculate the length of the third stick.
5. Kartik’s pencil box is 16 cm long. Hanshika’s pencil box is 4 cm shorter than Kartik's. Sumayya's
pencil box is 2 cm longer than Kartik's. What is the difference between the longest and
shortest pencil box?
Perimeter
Perimeter is the distance all around a two-dimensional
(2-D) shape or figure. The word perimeter comes from
the Greek word 'peri,' meaning around, and 'metron,'
which means measure. In Mathematics, the perimeter
refers to the total length of the sides of a figure.
Fencing around a garden
We often use perimeter in our daily life. For example,
we use perimeter in order to find the length of
fencing needed to surround a garden, the length of
wood needed to frame a picture or the length of wire
needed for electrical installations in a room.
Photo frame
100
Chapter 7 - Length, Perimeter and Area
Finding the perimeter of 2-D figures
The perimeter of a figure is obtained by adding the length of all its sides.
Example
Find the perimeter of the figure.
12 cm
B
C
5 cm
5 cm
A
D
7 cm
Solution
Perimeter = 5 cm + 12 cm + 5 cm + 7 cm = 29 cm.
EXERCISE 7.5
1. Find the perimeter of the following figures:
(a)
6 cm
5 cm
(b)
6 cm
4 cm
2 cm
3 cm
9m
(c)(d)
10 m
2m
6m
8m
9m
5m
3m
6m
(e)
8m
3m
8m
(f)
2m
5m
6m
4m
STOP AND THINK
The perimeter of a square is 16 cm. Find the length of one side.
Find the dimensions of a rectangle which has the same perimeter as the square. Is there more than one possibility?
101
Chapter 7 - Length, Perimeter and Area
2. Consider the following diagrams:
3m
9m
9m
4m
5m
B
C
A
8m
7m
6m
(a) Which shape has the greatest perimeter?
(b) If 5 identical shapes as C are placed next to each other forming a larger rectangle, find its perimeter.
(c) The three shapes are placed as follows: B, A, C so that the shapes touch one another (as
shown below). Find the perimeter of the resulting figure.
STOP AND THINK
How can you find the perimeter of
a circle?
B
A
C
NOTE TO TEACHER
Encourage students to think of
a piece of thread to measure the
perimeter of a circle.
Will the perimeter be the same if the figures are arranged as A, B, C?
Word problems involving perimeter
EXERCISE 7.6
1. The perimeter of a square tile is 40 cm. What is the length of each side of the tile?
2. The perimeter of a rectangular card is 18 cm. The card is 50 mm long. How wide is it? Give
your answer in mm.
3. A walking path goes around the edges of a park as shown below. Calculate the distance
covered in km by a jogger if he completes
400 m
550 m
(a) one complete track
(b) three complete tracks .
270 m
300 m
400 m
200 m
4. (a) The perimeter of a rectangular vegetable garden is 18 m. The length of the vegetable
garden is 3 m longer than its width. Calculate its length.
(b) A square has length 7.2 cm. Find its perimeter.
(c) An equilateral triangle has length 8 mm. Find its perimeter.
5. A wire is bent into a rectangular shape with
length 1.25 m and width 75 cm. If the same
wire is now bent into a square shape, what
will be the length of the square?
102
75 cm
1.25 m
Chapter 7 - Length, Perimeter and Area
6. Nabeehah has 25 m of wire. She needs to cut the wire and form
(a) a rectangle of length 4 m and width 1.5 m,
(b) an equilateral triangle of side 2.3 m ,
(c) a circle and
(d) a square, as shown below.
If the length around the circle is 3 m, what is the length of the
square?
4m
STOP AND THINK
The following isosceles trapezium
is composed of 7 match sticks.
Modify the position of three match
sticks only in order to obtain two
equilateral triangles.
1.5 m
2.3 m
7. Mr Khan wants to fence his rectangular yard, which is 24 m long
and 20 m wide.
(a) Calculate the length of fencing he needs.
(b) Fences are sold in rolls of 10 m at Rs 540 each.
How many rolls should he buy and how much will it cost?
Area
If one match stick is 5 cm long,
will there be any difference in
perimeter between the original and
final figure?
DID YOU KNOW
The term 'area' is used to refer to the amount of space inside
the boundary of a 2-D shape or figure. Builders, architects,
farmers or engineers need to calculate areas as part of their
daily job. Calculating area is an important and useful skill,
which we can use in our everyday tasks, such as finding the
area of a room to determine the number of tiles needed or
the size of a new carpet to be fitted in the room.
We measure or calculate small areas using the units mm2 or
cm2 and larger areas using the units m2, hectare or km2.
The unit hectare (Ha) is mostly used in the measurements of
land.
Before the French Revolution,
different units of measurement
of length and area were used
such as the toise, perche and
arpent. As a former French
colony, Mauritius has inherited
measurements of land in perche,
arpent and toise. These units are
still commonly used in Mauritius
when it comes to measurements
of land.
Can you find out more about
these units of measurements?
The units ‘arpent’, ‘perche’ and ‘toise’ are also used in
Mauritius to determine the area of the land.
An architect uses area
in designing the layout
of a building
Area of solar
panel
A gardener uses area
to make flower beds
620 hectares of tea cultivated
in Mauritius in 2016
1 arpent
4221 m2
1 perche
42.21 m2
1 toise
3.80 m2
1 hectare
10 000 m2
Area of stamps are
measured in mm2
Sale of land
103
Chapter 7 - Length, Perimeter and Area
FIND OUT
EXERCISE 7.7
Can you find the area of the
smallest continent?
State a most convenient unit to measure the area of:
(a) a classroom.
(b) the surface of a table for 12 persons.
(c) a page in your mathematics book.
(d) Mauritius.
(e) the Indian Ocean.
Conversion from one unit of area to another
Fig. 1 shows a square of area 1 cm2 and Fig. 2 shows a square of area 1 mm2.
1 cm
1 mm
1 cm
Note: The 2 squares of
length 1 cm and 1 mm
are not drawn to scale.
1 mm
Fig. 1
Fig. 2
How many squares of area 1 mm2 will fit onto Fig. 1?
There will be 100 squares of area 1 mm2.
1 cm
= 10 mm
1 cm x 1 cm = 10 mm x 10 mm
1 cm2
= 100 mm2
1
or 1 mm2 =
cm2
100
Therefore 1 cm2 = 100 mm2 or
Similarly,
1 m2
10 000 cm2
1 km2
1 000 000 m2
1 km2
100 Ha
× 1 000 000
km2
÷ 1 000 000
Note: 1 Hectare (Ha) = 10 000 m2
104
× 10 000
m2
× 100
cm2
÷10 000
mm2
÷100
Caution:
1m2 ≠ 100 cm2
Chapter 7 - Length, Perimeter and Area
Example
Convert the following:
(a) 7 m2 to cm2 (d) 68 000 m2 to Ha (b) 5.5 mm2 to cm2 (e) 7 750 Ha to km2
(c) 0.0097 km2 to m2
Solution
(a) 1 m2 = 10 000 cm2
7 m2 = (7 × 10 000) cm2 = 70 000 cm2
(b) 1 mm2 = 1 cm2
100
1
5.5 mm2 = (5.5 x
) cm2 = 0.055 cm2
100
(c) 1 km2 = 1 000 000 m2
0.0097 km2 = (0.0097 × 1 000 000) m2 = 9700 m2
1
Ha
10 000
1
68 000 m2 = (
x 68 000) Ha = 6.8 Ha
10 000
(e) 1 Ha = 1 km2
100
1
7 750 Ha = (7 750 x
) km2 = 77.5 km2
100
(d) 1 m2 =
EXERCISE 7.8
1. Convert the following to mm2 :
(a) 3 cm2
(b) 2.5 cm2
(c) 0.45 cm2
3
(d) 5 cm2
(e) 0.004 cm2
(c) 0.045 m2
(d) 800 mm2
(e) 56 mm2
2. Convert the following to cm2 :
(a) 5 m2
(b) 3.6 m2
3. Convert the following to m2 :
(a) 2 km2
(b) 0.04 km2
(c) 2 hectares
(d) 12 500 cm2
(e) 40.5 hectares
4. Convert the following to km2 :
(a) 6 hectares
(b) 46 805 000 m2
(c) 9.34 hectares
(d) 56 000 m2
105
Chapter 7 - Length, Perimeter and Area
Finding area of plane figures
RECALL
1. Square
Example
Area of square = length × width
Find the area of the following:
However since a square has all sides
equal, area is often calculated as
12 cm
5 cm
Area = length x length or (length)2
STOP AND THINK
Solution
Area = 5 cm × 5 cm
= 25 cm2
Area = 12 cm × 12 cm
= 144 cm2
2. Rectangle
Example
Consider a square of any length.
Calculate its area.
What will happen to the area if the
length of the square doubles?
RECALL
Find the area of the following:
Area of rectangle = length × width
15 m
4 cm
STOP AND THINK
5 cm
Solution
Area = 5 cm × 4 cm
= 20 cm2
3m
Area = 3 m × 15 m
= 45 m2
EXERCISE 7.9
1. Complete the following tables.
(a)
Length of square
Area of square
6 cm
20 mm
8 km
81 km2
10 000 m2
106
1. A square and a rectangle each
have an area of 64 cm2.What
are the possible dimensions of
the rectangle?
2. A square and a rectangle each
have a perimeter of 36 cm.
What are the possible areas of
the rectangle?
Chapter 7 - Length, Perimeter and Area
(b)
Length of a rectangle
Width of a rectangle
3 cm
6 cm
8m
15 m
9 mm
Area of a rectangle
36 mm2
10 km
15 km2
2. If the length of a rectangle is 36 cm and the area is 720 cm2, what is the width of the rectangle?
3. What is the area of a square with a perimeter of 44 m?
4. A square having length 6 m has the same perimeter as a rectangle. Find the area of the
rectangle if its length is 4 cm longer than its width.
5. The length of a rectangle is twice its width. If its area is 288 cm2, find the length of the rectangle.
6. The figure shows a hole cut in the middle of a cardboard. Calculate the shaded area.
10 cm
4 cm
4 cm 10 cm
7. PQRS is a square of length 12 cm. A square LMNS of area 25 cm2 is removed from it. What will
be the length of LP?
P
Q
M
L
S
N
R
8. The area of a square is twice the area of a rectangle with dimensions 45 cm and 10 cm. Find
the length of the square.
9. The width of a rectangle is 4 cm less than its length and its perimeter is 48 cm. Find its area.
107
Chapter 7 - Length, Perimeter and Area
10. A block of apartments is in the shape of a rectangle
as shown in the diagram. It has a perimeter of
110 m. It is divided into 5 identical rectangular
apartments as shown in the diagram. Find the
area of each apartment.
20 m
11. A man decides to sell his plot of land measuring 16 m by 32 m at the price of Rs 2 400 per
metre square. Find the total cost of this plot of land.
12. Tina has to get the flooring of her room covered with carpet. The dimensions of her room
are 3 m by 2.5 m. If 1 m2 of carpet costs Rs 125, find the amount of money Tina will have to
spend to buy carpet for her room.
13. The area of a room is 9 m2. How many square tiles each of length 30 cm would you require to
completely fill the floor of the room?
14. A real estate agent bought 3 hectares of land. He divided the land into 50 equal plots and
sold it to 50 clients at the cost of Rs 2 000 per square metre. How much does one plot of
land cost?
3. Area of triangles
Finding the area of triangle ABC
A
D
A
w
B
l
B
C
Fig. 1
C
Fig. 2
Step 1: Draw rectangle ABCD on a piece of paper (as shown in Fig. 1).
Step 2: Area of rectangle ABCD = length × width = l × w.
Step 3: Cut out rectangle ABCD as in Fig.2.
We can observe that area of triangle ABC = half area of the rectangle = 1 × l × w
2
Now, in a triangle we refer to the ‘length’ as base and the ‘width’ as the height. In this case, this
is also the perpendicular height.
So, area of triangle ABC =
1
× b × h, where b is the base of the triangle and h is the perpendicular height.
2
Similarly, for any type of triangle:
Area of triangle = 1 × base × height
2
h
h
b
108
h
b
b
Chapter 7 - Length, Perimeter and Area
Example
Find the area of the following triangles.
(a)
(b)
6 cm
STOP AND THINK
What is the area of each of the triangles?
What do you notice? Explain why.
5 cm
10 cm
4 cm
5 cm
Solution
9 cm
1
×b×h
2
= ( 1 × 4 × 5) cm2
2
= 10 cm2
Area =
1
×b×h
2
= ( 1 × 10 × 6) cm2
2
= 30 cm2
Area =
5 cm
9 cm
EXERCISE 7.10
1. Calculate the area of the following triangles.
(a)
6 cm
(b)
(c)
8 mm
4m
12 mm
8 cm
(d)
3m
7 cm
11 cm
he
igh
t
2. Given the area of the triangle is 15 cm2, find the height.
6 cm
3. Given the area of the triangle is 0.48 cm2, find the base.
1.2 cm
base
4. A gardener wants to divide his triangular plot of
land to plant lettuces and pumpkins as shown in the
diagram. Find the ratio of the area planted with lettuces
to the area planted with pumpkins.
15 m
lettuces
10 m
pumpkins
12 m
109
Chapter 7 - Length, Perimeter and Area
4. Area of parallelogram
Consider a parallelogram.
h
h
b
b
We can cut a right angled triangle (blue) and send it on the other side to form a rectangle.
Area of rectangle = length × width
NOTE TO TEACHER
=b×h
Area of parallelogram = base × perpendicular height
Encourage students to try out this
cutting activity to derive the area
of a parallelogram.
EXERCISE 7.11
1. Find the area of the following parallelograms.
(a)
(b)
6 cm
7 cm
8 cm
11 cm
2. Given the area of the parallelogram is 28 mm2, find the height.
height
7 mm
3. Given the area of the parallelogram is 0.6 m2, find the base.
3m
base
4. If the height of a parallelogram is 11 m and the base is 6 m, what is the area of the
parallelogram?
110
Chapter 7 - Length, Perimeter and Area
5. Kites
A kite has 2 pairs of equal adjacent sides and its diagonals meet each other at right angles.
Line of symmetry
Diagonals
Height
Height
Base
Base
Since we have two pairs of equal sides in a kite, the vertical diagonal is a line of symmetry. So
we have two identical triangles on both sides of the vertical diagonal.
Now, the area of triangle is 1
2 × base × height.
The base of the triangle is the vertical diagonal.
The height of the triangle is 1
2 of the horizontal diagonal.
Therefore, area of one of the triangles
NOTE TO TEACHER
1
= 2 × base × height
Encourage students to do the
= 1 × vertical diagonal × 1 horizontal diagonal
cutting to see how a kite consists
2
2
of 2 identical triangles.
= 1 × vertical diagonal × horizontal diagonal
4
Thus, the area of the kite = 2 × area of triangle
1
= 2 × 4 × horizontal diagonal × vertical diagonal
=1
2 × horizontal diagonal × vertical diagonal
Area of a kite = 1
2 × product of the diagonals
Example
1. Find the area of the kite.
2. Find the area of the
given figure.
5 cm
5 cm
12 cm
8 cm
Solution
Solution
Area = 1 x product of the diagonals
2
= ( 1 x 12 x 5) cm2
2
= 30 cm2
Area = 1
2 x product of the diagonals
= ( 1 x 5 x 8) cm2
2
= 20 cm2
111
Chapter 7 - Length, Perimeter and Area
EXERCISE 7.12
Calculate the area of the following kites.
1.
2.
3.
4m
6 cm
2 cm
5m
19mm
10 cm
5m
4m
6 cm
38mm
6. Rhombus
The diagonals of a rhombus bisect each other at right angles. A rhombus is a special type of kite
where the 4 sides are equal. We can therefore apply the formula for area of kite to find the area
of a rhombus.
d2
d1
Area of rhombus = 1 x product of the diagonals
2
= 1 x d1 x d2 (provided diagonals are known)
2
Alternatively, since a rhombus is also a special kind of parallelogram, its area can also be
calculated as base x height.
Area of rhombus = base × height
Note: The height should be
the perpendicular height.
d2
height
d1
base
EXERCISE 7.13
1. Find the area of the following figures.
(a)
(b)
4m
6 cm
5m
5 cm
112
Chapter 7 - Length, Perimeter and Area
2 (a) Given the area of the rhombus is 35 cm2, find the height.
h
7 cm
(b) Given that the area is 7.5 mm2, find the length of its base.
3 mm
base
3. Given the area of a rhombus is 54 cm2, and one of its diagonals is 12 cm, find the length of
the other diagonal.
7. Trapezium
A trapezium is a quadrilateral having one pair of parallel sides.
a
h
b
The parallel sides are a and b while h is called the perpendicular height.
To find the area of a trapezium, we cut it into 2 parts and then join the two parts to form a
parallelogram.
a
h
b
h
2
b
a
The new figure formed consists of 2 trapezia joined together to now form a parallelogram.
Area of parallelogram = base × height
So,
= (a + b) × h
2
Area of trapezium = 1
2 × (a + b) × h
= 1 × (sum of parallel sides) × h
2
NOTE TO TEACHER
Encourage students to do the
cutting so as to obtain the
parallelogram.
113
Chapter 7 - Length, Perimeter and Area
Example
Calculate the area of the following figures.
1.6 m
(a)
2.9 cm
(b)
0.9 cm
2m
Solution
1.3 cm
4m
Solution
Area = 1
2 (sum of parallel sides) × h
= 1 (1.6 + 4) × 2 m2
2
= 5.6 m2
1
Area = 2 (sum of parallel sides) × h
2
= 1
2 (2.9 + 1.3) × 0.9 cm
= 1.89 cm2
EXERCISE 7.14
1. Calculate the area of the following figures.
6 cm
(a)
(b)
2.5 cm
3 cm
4 cm
5.5 cm
8 cm
12 m
(c)(d)
12 cm
4 cm
5m
8m
9 cm
2. Given that a trapezium with area 124 cm2 has parallel sides of lengths 18 cm and 13 cm
respectively, find the height of the trapezium.
3. A trapezium of height 10 cm has an area of 65 cm2. Given that one of the parallel sides is of
length 8 cm, find the length of the other side.
114
Chapter 7 - Length, Perimeter and Area
8. Area of composite figures
Example
Find the area of the given figure.
6 cm
STOP AND THINK
4 cm
Is there another method to calculate
the area of this irregular shape?
2 cm
3 cm
Solution
4 cm
Divide the figure into two rectangles and find all missing lengths.
Area of rectangle A = 4 cm x 7 cm = 28 cm2
Area of rectangle B = 4 cm x 2 cm = 8 cm2
B
7 cm
A
Total area = 28 cm2 + 8 cm2 = 36 cm2
4 cm
2 cm
3 cm
EXERCISE 7.15
1. Calculate the area of the following figures:
(a)
3 cm
7 cm
(b)
10 mm
5 cm
5 mm
(c)
3 mm
8 cm
3 mm
6 cm
4 cm
10 cm
6 cm
0.4 m
4m
2m
2m
2m
0.6 m
4m
5m
6m
0.4 m
1.6 m
6m
2.4 m
0.4 m
4m
(f)
1.0 m
7m
(e)
10 m
(d)
0.4 m
2. Circle the correct answer.
(a) Convert 60.8 cm2 into mm2.
A. 0.0608 mm2
B. 0.608 mm2
C. 6 080 mm2
D. 608 mm2
C. 0.65 m2
D. 65 000 m2
(b) Convert 6.5 hectares into metre square.
A. 650 m2
B. 6 500 m2
115
Chapter 7 - Length, Perimeter and Area
(c) If the height of a trapezium is 8 cm and the sum of its parallel sides is 16 cm, then the area of
the trapezium is
A.
85 cm2
B. 54 cm2
C. 64 cm2
(d) The area of the shaded region in the figure is
A. 27 cm2
C. 40.5 cm2
D. 32 cm2
A
15 cm
D
E
B
B. 135 cm2
D. 108 cm2
9 cm
C
(e) ABCD is a square. Its area is 100 cm². The area of the shaded square AQRP is 16 cm².
The length of QB is
A
A. 84 cm
C. 10 cm
Q
B
B. 6 cm
P
D. 4 cm
Summary
R
D
C
× 1 000
× 1 000 000
× 1 000
km
× 100
m
÷ 1 000
× 10 000
× 100
× 10
cm
÷100
km2
mm
÷ 1 000 000
÷10
m2
cm2
÷10 000
mm2
÷100
÷ 1 000
• To find the perimeter of a figure, we find the total length of the sides.
• Area refers to the amount of space inside a 2-D shape.
Name of figure
Square
Rectangle
Parallelogram
Rhombus
Kite
Trapezium
116
Area
(Length)2
Length × Width
Base × Height
Conversion of units
Base × Height or 1
2 × product of diagonals
1 arpent = 4 221 m 2
1
2 × product of diagonals
1 (sum of parallel sides) × height
2
1 perche = 42.21m 2
1 toise = 3.80m 2
1 Ha =10 000 m 2
8
PERCENTAGE
Chapter 8 - Percentage
Learning
Objectives
By the end of this chapter, you should be able to:
• recognise the use of percentages in real life situations.
• convert a percentage to a fraction and/or decimal and vice versa.
• solve word problems involving percentages.
Percentages in real life
CHECK THAT YOU CAN:
We come across percentages in many situations in our
everyday life. For example, during festive seasons such as
Christmas and New Year or on special occasions such as
Mother’s Day we come across brochures, boards and signs
showing for example 10 % discount or Sale 25 % off.
• Add, subtract, multiply and
divide fractions and decimals.
• Convert fractions to decimals
and vice versa.
KEY TERMS
•
•
•
•
Percent
Percentage
Fractions
Decimals
Interest on home loan
Sales
DID YOU KNOW
%
Food labels
Cloth Label
FURNITURE LTD
Royal City Tel: 57101010
“Per cent” or “percent” comes from
the Latin word “per centum” which
means “one out of hundred”. The
percent sign used to consist of two
circles separated by a horizontal
line until the recent symbol %
made its appearance.
Cash Sale Receipt
Name: Yan Paul
Address: 35, Ville Road, Curepipe
Quantity
1
Description
Table
15% VAT
Total
Salesman’s signature
Date: 10/8/2017
Tel: 59998888
Price (Rs)
8 000
1 200
9 200
Customer’s signature
Part of payment
receipts showing
15% VAT
Introduction
RECALL
A percent is equivalent
to a fraction in which the
denominator is 100.
Percentages are just another way of writing fractions, with
a denominator of 100. The advantage of using percentages
is that it is easier to compare percentages than fractions.
For instance, in newspapers, we can very often read about
percentage of inflation in our country, the percentage
increase or decrease in the rate of interest, the percentage
number of accidents occurring every year and so on.
117
Chapter 8 - Percentage
Representation of percent
Example
In the 100-square grid, 30 squares are shaded.
Fraction shaded is 30 = 30 % where the symbol % denotes 1 .
100
100
Converting percentages into fractions
To convert a percentage into fraction, we change the symbol % to a denominator of 100.
Example
Convert the following percentages to fractions reduced to their lowest terms:
(a) 12 %
(b) 55 %
(c) 130 %
(d) 4
2
%
3
Solution
12
3
(a) 12 % = 100 = 25
130 13
(c) 130 % =
= 100 10
55
=
100
2
14
(d) 4 % = % =
3
3
(b) 55 % =
11
20
1
14
7
14
x
=
=
100 300 150
3
EXERCISE 8.1
1. Convert the following percentages into fractions reduced to their lowest term.
(a) 35 % (b) 85 %
(c) 62 %
(d) 76 %
(e) 16 %
(f) 42 %
(g) 23 %
(h) 95 %
2. Express the following percentages as fractions reduced to their lowest term.
(a) 120 %
3
(e) 2 %
4
(b) 140 %
1
(f) 7 % 2
(c) 245 %
1
(g) 12 %
5
(d) 630 %
1
(h) 39 %
3
Converting a fraction into percentage
So far, we have seen how percentages can be converted into fractions. We now look at the
conversion of fractions into percentages.
3
For example, consider the fraction
.
25
118
Chapter 8 - Percentage
3
We first need to convert 25 into an equivalent fraction with denominator 100.
x4
3
12
=
25
100
x4
We multiply both numerator and denominator by 4.
3
12
So
is same as
which is 12 parts out of 100 parts, hence 12 %.
25
100
Similarly, 3 = 3 x 125 = 375 = 37.5 = 37.5 %
8
8 x 125 1 000 100
POINTS TO REMEMBER:
Common percentages and their equivalent fractions:
Fraction
1
2
Percentage
50 %
1
3
33
1
%
3
1
4
1
5
25 %
20 %
1
8
12
1
%
2
2
3
3
4
2
66 3 %
75 %
EXERCISE 8.2
Convert the following fractions into percentages.
4
7
11
2
5
(a) (b)
(c)
(d)
(e)
5
10
20
25
8
2
4
3
8
21
(f) 5
(g) 1
(h) 5(i)(j)
25
50
15
35
Expressing a given amount as a percentage
Example
Express the following into percentages.
(a) 10 out of 50
(b) 20 cm out of 800 cm
(c) 40 g out of 10 kg
Solution
(a) 10 out of 50 =
Caution:
Both quantities should
be in the same unit.
10
× 100 % = 20 %
50
(b) 20 cm out of 800 cm =
(c) 40 g out of 10 kg =
20
× 100 % = 2.5 %
800
40
× 100 % = 4 = 0.4 %
10 x 1 000
10
Convert 10 kg into g
i.e. 10 kg = (10 x 1 000) g
119
Chapter 8 - Percentage
EXERCISE 8.3
Express the following as percentages:
(a) 35 out of 70
(b) 80 L out of 240 L
(d) Rs 55 out of Rs 2 200
(e) 4 kg out of 8 000 g
(c) 120 km out of 1 000 km
(f) 30 cm out of 6 m
POINTS TO REMEMBER:
• In short, when we convert a fraction into a
percentage we multiply by 100 %.
• When we convert a percentage into a fraction, we divide by 100 as shown in
the diagram.
÷ 100
Percentage
Fraction
x 100 %
Converting a decimal number into percentage
0.25 means
So
25
which is equivalent to 25 shaded squares in the 100-square grid.
100
25
= 25%
100
We can convert the decimal number to a fraction and then
convert the fraction to a percentage.
or
A decimal number can be converted to a percentage by
multiplying the decimal value by 100 %.
Example
Convert the following decimal numbers into percentages:
(a) 0.1
(b) 1.2
(c) 0.125
Solution
(a) 0.1 = 0.1 × 100 % = 10 %
(b) 1.2 = 1.2 × 100 % = 120 %
(c) 0.125 = 0.125 × 100 % = 12.5 %
Note: A fraction can be expressed as a decimal or a percentage.
Example:
50%
1
2
120
0.5
Chapter 8 - Percentage
EXERCISE 8.4
CHECK THIS LINK
Convert the following decimals into percentages:
(a) 0.5
(b) 1.02
(c) 0.01
(d) 0.043
(e) 0.0802
(f) 2.921
(g) 1.035
(h) 0.4006
(i) 12.014
(j) 25.769
Decimal to percentage game:
http://www.xpmath.com/forums/
arcade.php?do=play&gameid=32
Converting percentage into a decimal number
Consider 62 %. We can represent this on a 100-square grid as follows:
62
62 % =
= 0.62
100
What do you observe?
÷ 100
Percentage
To convert a percentage into
a decimal number we can
divide the number by 100 as %
1
represents
.
100
Decimal
x 100 %
Example
Convert the following percentages into decimal numbers:
1
(a) 30 %
(b) 2 %
(c) 135 % (d) 16.5 %
2
Solution
(a) 30% =
(c) 135% =
30
= 0.3
100
1 35
= 1.35
100
(e) 25
3
%
4
2.5
(b) 2 1 % =
= 0.025
2
100
16.5
(d) 16.5% =
= 0.165
100
(e) 25 3 % = 25.75 = 0.2575
4
100
EXERCISE 8.5
Convert the following percentages into decimals:
(a) 12 %
(b) 55 %
(c) 99 %
(g) 22 %
(h) 1050 %
(i) 178.4 %
(d) 20.7 %
1
(j) 5 %
2
(e) 26.2 %
2
(k) 32 %
5
(f) 8.5 %
1
(l) 7 %
4
121
Chapter 8 - Percentage
Word Problems involving percentages
Example 1
Shreya's monthly budget for her house expenses is Rs 36 000. She spends 9% of the
budget on cleaning products. How much does she spend on cleaning products?
Solution
Amount of money spent on house expenses = Rs 36 000
Percentage spent on cleaning products = 9
100 %
Rs 36 000
100 %
Rs 36 000
36 000
×9
9 %
Rs
9%
100
9
Amount spent on cleaning products =
× Rs 36 000 = Rs 3 240
100
(
)
Example 2
In a school, there are 1 200 students, of which 44 % are boys. Find the number of girls in the school.
100 % = 1 200 students
Solution
Method 1
44%
56%
Boys
Girls
56
Number of girls =
× 1 200 = 672
100
Method 2
44
× 1 200 = 528
100
Number of girls = 1 200 – 528 = 672
Number of boys =
EXERCISE 8.6
1. Calculate
(a) 5 % of Rs 200
(b) 40 % of 8 000 km
(c) 24 % of 120 L
(d) 110 % of 250 mm.
2. During a fancy fair organised at school, the total money collected was Rs 54 000.
The following table represents the amount of money collected from the different items:
Items
Amount of money in Rs
Food and drinks
27 000
Entertainment
6 000
Games
12 000
Donations
9 000
What percentage of the money collected does each item represent?
3. Mary-Lou made a fruit salad using red and green apples. 46 % of the apples are red. If she
used a total of 50 apples, how many green apples did she use?
122
Chapter 8 - Percentage
4. Rakesh hired a new worker to lay bricks in his construction business. If the worker laid 150
bricks and this represents 30 % of the work, how many bricks would be needed to complete
the work?
1
5. In a cake mix of 1.5 kg, 5 represents cocoa, 25 % is flour and the rest consists of a secret
ingredient. What is the mass of the secret ingredient?
6. Kiren is a philanthropist. She donates money to Miracle Association which promotes
animals’ welfare. She usually donates Rs 5 000 every month which represents 6.25 % of her
monthly salary. What is Kiren's monthly salary?
7. In a village of 35 000 inhabitants, 35 % are men, 40 % are women and the rest are children.
(a) Calculate the percentage of inhabitants who are children.
(b) How many more women are there than children?
8. A fruit seller sells only oranges. On one day, after having sold 30 % of the oranges, he had 140
left. How many oranges did he have at the beginning?
Summary
•
•
Percentage Notation: 10 % means 10 out of 100. It can be written as 10 or 1 .
100
10
Converting percentage into fraction and vice versa:
÷ 100
Percentage
Fraction
x 100 %
•
Converting percentage into decimal and vice versa:
÷ 100
Percentage
Decimal
x 100 %
10
( 100
× 500 ) = Rs 50
•
Percentage Calculation: 10 % of Rs 500 = Rs
•
Percentage expressed as fraction and decimal:
Example
50%
1
2
0.5
123
9
Chapter 9 - Ratio and Proportion
RATIO AND PROPORTION
Learning Objectives
By the end of this chapter, you should be able to:
• demonstrate understanding of ratio and direct proportion.
• compare two quantities multiplicatively in terms of a ratio and proportion.
• solve word problems involving ratio and direct proportion.
Ratio and Proportion in real life
CHECK THAT YOU CAN:
Ratios and proportions appear everywhere in our daily
life. Examples include cooking, construction or scale
map drawings.
• Multiply and divide fractions.
• Convert mixed numbers to
improper fractions.
• Find H.C.F. and L.C.M. of
numbers.
Belle Mare
KEY TERMS
Flic en Flac
3 cups of water
Ratio of
water : juice
3:1
Oats Cookies
What you need
• Ratio
• Proportion
• Direct proportion
1 cup of juice
Serving:
40 biscuits
2 cups of sugar
2 cups of flour
2 cups of oats
1 cup coconut powder
Mix ingredients together
Bake at 180 ο C
The actual distance from Belle Mare
to Flic en Flac is approximately 60
km but on a map, the distance can
be represented by 3 cm.
In the recipe, the ratio of coconut
powder to flour is 1 : 2. This serves
40 biscuits. For 80 biscuits, we will
use proportion to calculate how
much coconut powder, sugar, oats
and flour will be used.
Introduction
Ratios and proportions were used in the design of
ancient sacred and ceremonial sites in all traditional
civilisations. The golden ratio was used to design
Cathedrals, Hindu temples, Pagodas, Mosques and
monuments like the Eiffel Tower and the Taj Mahal.
The golden ratio is also found in nature.
124
RECALL
Ratio is written in the
form a: b, for example,
1:2.
One example of direct
proportion is: If a
pencil costs Rs 5, then
40 pencils cost Rs 200.
DID YOU KNOW
History of Ratio and Proportion
Ratio and proportion originally
emerged from human activities. For
example, ancient Egyptian builders
used ratio and proportion in the
construction of the pyramids.
The golden ratio relates to the length
of a person's forearm divided by the
length of the same person's hand.
Find the value of the golden ratio.
Chapter 9 - Ratio and Proportion
Ratios
Ratios are used to compare two quantities of the same kind or quantities measured in the same unit.
Compare the number of pink to green spinners.
We say that the ratio of pink spinners to that of green spinners is 2 : 3, i.e., pink : green
2:3
The ratio 2 : 3 can also be written in fractional form as follows:
number of pink spinners = 2
number of green spinners 3
Example
Orange juice is prepared by mixing one cup of concentrated orange juice to 3 cups of water.
(a) State the ratio of concentrated orange juice to water.
(b) Write down this ratio in fractional form.
(c) If you use 2 cups of concentrated orange juice, how many cups of water will you need?
(d) If you use 4 cups of concentrated orange juice, how many cups of water will you need?
(e) If you use 15 cups of water, how many cups of concentrated orange juice will you need?
Solution
NOTE TO TEACHER
(a) Ratio of concentrated orange juice to water = 1 : 3
(b) Number of cups of concentrated orange juice = 1
3
Number of cups of water
(c) Concentrated orange juice : water = 1 : 3
1 : 3
x2
x2
2 : 6
Number of cups of water needed = 6
(d)
x4
1 : 3
x4
4 : 12
Number of cups of water needed = 12
(e) ÷5
1 : 3
÷5
5 : 15
Number of cups of concentrated orange juice = 5
Elaborate more on the concept
of equivalent ratios.
Observation:
Ratio
Fraction
(a)
1:3
1
3
(c)
2:6
2= 1
6 3
(d)
4 : 12
4 = 1
12 3
(e)
5 : 15
5 = 1
15 3
These ratios and the corresponding
fractions are all equivalent. We
call these ratios equivalent ratios,
that is, 1 : 3, 2 : 6, 4 : 12, 5 : 15 are
all equivalent.
125
Chapter 9 - Ratio and Proportion
Simplifying ratios
Just like fractions, ratios are usually written in their simplest form.
Example 1
Simplify (a) 3 : 9
Solution
(b) 36 : 24
(a) The H.C.F. of 3 and 9 is 3.
So we divide both numbers by 3. (b) The H.C.F. of 36 and 24 is 12.
So we divide both numbers by 12.
3 : 936 : 24
÷3
÷3
÷ 12
1 : 3
÷ 12
3:2
Example 2
Simplify the ratio 2 2 : 5 1
6
3
Solution
8 : 31
2 2 : 5 1 = 3
6
3 6
= 8 x 6 : 31 x 6
3
6
= 16 : 31
Convert to improper fraction
Multiply by 6, by the L.C.M. of 3 and 6
Example 3
Simplify 1.2 : 2.4
Solution
1.2 : 2.4
12 : 24 1 : 2 (We multiply by 10 to obtain whole numbers)
(We divide by the H.C.F. which is 12)
EXERCISE 9.1
1. (a) Express the following ratios in their simplest form:
(i) Balls to Rubik Cubes,
(ii) Balls to Skateboard,
(iii) Skateboards to Rubik Cubes,
(iv) Rubik Cubes to total number of toys.
(b) Express each of the ratios as fractions.
126
Note: You can also
divide by 2, 3, 4, ...
untill you reach the
simplest form.
Chapter 9 - Ratio and Proportion
2. Copy and complete the following table.
Ratio of
Ratio form
Fractional form
4 pens to 5 pencils
7 cars to 3 buses
10 tables to 32 chairs
3. For every 5 packets of sweet popcorn sold, 3 packets of snacks were sold on a particular day
at a cinema. Write down the ratio of the number of packets of sweet popcorn to snacks sold.
4. Simplify the following ratios:
(a) 2 : 6
(b) 5 : 20
(c) 8 : 24
(d) 10 : 30
(e) 13 : 39
(f) 27 : 18
(g) 22 : 99
(h) 80 : 120
(i) 144 : 60
(j) 250 : 750 (k) 1500 : 500
(l) 2800 : 800
5. Arshad prepares pancakes for his son’s birthday party. He uses 6 cups of flour, 1 cup of sugar
2
and 2 cups of water. Write down, in its lowest terms, the ratio of
(a) sugar to water
(b) flour to sugar.
6. Simplify the following ratios:
2: 5
2: 2
1: 3
1: 2
(a) (b)
(c)
(d)
7 7
5 5
4 8
3 3
3: 9
(e) 2 (f) 0.5 : 2
(g) 2.4 : 3.6
(h) 3 3 : 2 1
5 10
8
4
7. In a class of 35 pupils, 20 are girls and the rest are boys. Write down the ratio of girls to boys.
8. A bouquet contains 18 roses and 12 gerberas. What is the ratio of roses to the total number of
flowers in the bouquet?
CHECK THIS LINK
9. Write down an equivalent ratio for the following ratios:
(a) 2 : 5 (b) 1 : 1
(c) 3 : 8
(d) 12 : 17
http://studymaths.co.uk/workout.
php?workoutID=73
http://www.worksheetmath.com/
Number/Ratio/EquivalentRatios
Word problems involving ratio
Example 1
Haider and Hannah share Rs 8 000 in the ratio 9 : 11. How much does Hannah get?
9 parts
Solution
11 parts
Rs 8 000
9 + 11 = 20 parts The ratio 9 : 11 means there will be a total of 20 parts
8 000 ÷ 20 = Rs 400
Divide the total amount by the number of parts to find one ‘part’
11 × 400 = Rs 4 400
Multiply 400 by 11 to find the amount Hannah receives
Hence, Hannah’s share = Rs 4 400
127
Chapter 9 - Ratio and Proportion
Example 2
The ratio of boys to girls participating in a Mathematics competition was 5 : 4. If there were
45 boys, how many students participated in the competition altogether?
Solution
Boys : Girls
x9
5: 4
45 : ?
x9
Number of girls = 4 × 9 = 36
Total number students = 45 + 36 = 81
Example 3
In a box of 24 pens, the ratio of the number of red pens to blue pens is 1 : 3. If 2 red pens
are removed from the box, and replaced by 6 blue pens, find the new ratio of the number
of red pens to the number of blue pens.
Solution
Ratio of red to blue pens = 1 : 3
Total number of parts = 1 + 3 = 4
Initially,
1
Number of red pens = 4 × 24 = 6
Number of blue pens = 3 × 24 = 18
4
Now, there are : 6 – 2 = 4 red pens
and 18 + 6 = 24 blue pens
New ratio of red pens : blue pens = 4 : 24
=1:6
Example 4
In an animal park, the ratio of deer to tortoises was 3 : 5. There were 8 more tortoises than
deer. Some tortoises disappeared and the ratio of deer to tortoises became 2 : 3. How many
tortoises disappeared?
Solution
Ratio of deer to tortoises = 3 : 5
Difference in ratio = 5 – 3 = 2
2 parts
8 animals
1 part
4 animals
3 parts
12 deer
5 parts
20 tortoises
New ratio of deer : tortoises
2 : 3
x6
x6
12 : ?
There are still 12 deer since only tortoises have disappeared.
Therefore, there are now 18 tortoises and so 2 have disappeared.
128
Chapter 9 - Ratio and Proportion
EXERCISE 9.2
1. Chocolate cookies and oats cookies are placed on a tray in the ratio 3 : 5. If there are 15
chocolate cookies, how many oats cookies are there on the tray?
2. Vimla donates a sum of money every year to an orphanage and to a disabled centre in the
ratio 9 : 5. If she donates Rs 1 350 to the orphanage, calculate
(a) the sum of money Vimla donates to the disabled centre,
(b) the total amount of money she spent as donation.
3. In a pattern of 65 triangles, the ratio of red triangles to green triangles is 6 : 7. Calculate the
number of green triangles.
4. The cost of a book and a pen is in the ratio 10 : 1. If the difference in the price of the book and
the pen is Rs 180, find the total cost of the book and the pen.
2 : 1
5. A sum of Rs 550 is shared between Anisha and Valeisha in the ratio
. Find each one’s
3 4
share.
6. The difference between two numbers is 24 and their ratio is 5 : 2. Find the two numbers.
7. In 2017, the ratio of Jenny's age to Hans was 1 : 5 and the sum of their ages was 36. How old
were they in 2014?
8. In a school library, the ratio of French books to Mathematics books is the same as the ratio
of Mathematics books to English books. If there are 120 French books and 180 Mathematics
books, find the number of English books.
9. For Christmas, grandmother gave Rs 1 000 to her two grandchildren Ashley and Priya. The
money was shared between Ashley and Priya in the ratio 2 : 3. Grandfather then gave Rs 200
to each one of them. Find the new ratio of Ashley’s money to Priya’s money.
10. In a school of 600 pupils, the ratio of boys to girls was 2 : 1. Some girls left the school and the
new ratio of boys to girls is 4 : 1. How many girls left the school?
11. The ratio of Sayan's age to Pari’s age is 2 : 5. If Sayan is 6 years younger than Pari, find the sum
of their ages.
12. For the Sports Day at school, for every 6 girls who participated, there were 10 boys. If there
were 36 more boys than girls, find the total number of participants.
Direct Proportion
Sarah saves some money over a period of 5 months. Each month she adds some money in her
savings box. The table below shows the amount of money Sarah has in her savings box at the
end of each month.
Number of months
1
2
3
4
5
Amount of money (in Rs)
50
150
240
300
420
1 : 50
1 : 50
2 : 150
1 : 75
3 : 240
1 : 80
4 : 300
1 : 75
5 : 420
1 : 84
Ratio
In simplified form We observe that the ratio is not constant. Thus, the amount of money and the number of months
are not in direct proportion.
129
Chapter 9 - Ratio and Proportion
The table below shows the relationship between the number of mugs and the cost of mugs.
Number of mugs
1
2
3
4
5
6
Cost of mugs (Rs)
60
120
180
240
300
360
1 : 60
2 : 120
3 : 180
4 : 240
5 : 300
6 : 360
1 : 60
1 : 60
1 : 60
1 : 60
1 : 60
1 : 60
Ratio
In simplified form
The ratio is equal in each column. We say that the cost of mugs is in direct proportion to the
number of mugs.
NOTE TO TEACHER
Discuss with students how the cost of the mugs has been calculated. Encourage them to find the relationship between
number of mugs and cost, that is, as number of mugs increases, cost of mugs also increases in proportion.
Example 1
A recipe for cupcakes requires 3 cups of flour to make 10 cupcakes.
(a) How many cupcakes can be made from 12 cups of flour?
(b) How many cups of flour are needed to make 50 cupcakes?
Solution
Method 1
Method 2
(a) 3 cups of flour
10 cupcakes
10
1 cup of flour
3 cupcakes
x4
10
12 cups of flour
× 12 = 40 cupcakes
3
Number of cupcakes made from 12 cups of flour = 40
(b) 10 cupcakes
3 cups of flour
3
1 cupcake
x5
10 cup of flour
3
50 cupcakes
× 50 = 15 cups of flour
10
Number of cups of flour to make 50 cupcakes = 15
3 cups of flour
10 cupcakes
12 cups of flour
40 cupcakes
10 cupcakes
3 cups of flour
50 cupcakes 15 cups of flour
Example 2
If 5 milkshakes cost Rs 375, find the cost of 7 milkshakes.
Solution
5 milkshakes cost Rs 375
375
1 milkshake costs Rs (
) = Rs 75
5
7 milkshakes cost Rs (75 x 7) = Rs 525
130
Hint: We need to find the cost of 1 milkshake.
x4
x5
Chapter 9 - Ratio and Proportion
EXERCISE 9.3
1. If 1 sweet costs Rs 1.50, find the cost of 4 sweets.
2. If 14 pens cost Rs 154, find the cost of 5 pens.
3. If a copybook costs Rs 12, how many such copybooks can be bought for Rs 108?
4. If 8 kg of tomatoes cost Rs 336, how many kg of tomatoes can be bought for Rs 462?
5. A university student worked as a trainee worker in a company during 3 months and earned
a total sum of Rs 4 200. How much did she earn for the first 2 months if she got the same
amount every month?
6. Raksha decides to invite 12 friends for her birthday party. Her parents plan to give each of
her friends a small gift to bring back home. If 50 cm of ribbon is needed to wrap one gift,
calculate the length of ribbon needed to wrap all gifts.
7. Vanisha got her graduation photograph of length 15 cm and width 10 cm enlarged in
proportion. If the width of the enlarged photograph is 26 cm, what will be the length of the
enlarged photograph?
8. Vinesh’s telephone bill for the month of May was as follows:
Line rental Rs 90
Call charges (Rs 2 per minute)
Rs 198
Total chargesRs 288
(a) Calculate the number of minutes Vinesh spent on the phone for the month of May.
(b) If Vinesh’s bill for the month of June was Rs 272, find the amount paid for call charges.
9. A salesgirl’s basic salary is Rs 9 500. She also earns Rs 60 for every 2 perfumes sold.
(a) If she sells 30 perfumes during a particular month, calculate her total salary.
(b) If her total salary for the following month was Rs 10 880, calculate the number of perfumes
she sold during that particular month.
Summary
Ratios
CHECK THIS LINK
https://www.ixl.com/math/grade-7/
solve-proportions-word-problems
•
Ratios are used to compare two quantities of the same kind or quantities measured in
the same unit.
a
• Ratios can be written in the form a : b or .
b
Proportion
•
Two quantities are in direct proportion if an increase in one quantity results in an
increase in the other quantity in the same ratio. Similarly, if a decrease in one quantity
results in a decrease in the other quantity in the same ratio, we say that the two quantities
are in direct proportion.
131
10
INDICES
Chapter 10 - Indices
Learning Objectives
By the end of this chapter, you should be able to:
• identify and use laws of indices involving positive exponents (multiplication law, division
law, power law and zero index).
Introduction
CHECK THAT YOU CAN:
Indices are a convenient way to express numbers.
In Grade 6, you learnt about indices as powers. For
instance, 25 is read as 2 to the power of 5.
• Perform arithmetic operations
with numbers.
KEY TERMS
2 is called the base and 5 is called the power or index
or exponent.
The index or power of a number indicates the number
of times the number is multiplied by itself.
25 = 2 × 2 × 2 × 2 × 2
Index form
•
•
•
•
•
•
powers
base
exponent
indices
index form
expanded form
Expanded form
DID YOU KNOW
Similarly , 53 = 5 × 5 × 5
and
24 = 2 × 2 × 2 × 2.
History
EXERCISE 10.1
1. Write in expanded form:
(b) 92(c) 35
(a) 73
(e) 104
(f ) 46
(d) 57
2. Write in index form:
(a) 4 × 4 × 4 × 4
(c) 6 × 6 × 6 × 6 × 6 × 6
(e) 7 × 7
(b) 2 × 2 × 2
(d) 10 × 10 × 10 × 10 × 10 × 10
(f ) 9 × 9 × 9 × 9
Ancient Greek Mathematician,
Euclid, used the term “power” to
represent the number of times a
number is multiplied by itself.
In the 17th century, the use of
raised numbers to denote powers
or exponents was introduced.
It is in 1637 that Rene Descartes
used positive indices in the same
way as we write them today.
Laws of Indices
Note: Index is the singular of indices.
One way of working with powers is through the expanded
form. A more practical and convenient way to work with
powers is to use the laws of indices.
132
Chapter 10 - Indices
Multiplication law
Consider the following:
𝑎
𝑏
𝑎 × 𝑏
51
52
51 × 52
51
52
54
53
53
52
51× 53
52 × 53
53
54
55
51
55
54
𝑎 × 𝑏 (expanded form)
𝑎 × 𝑏 (index form)
5×5×5
53
5×5×5×5
55
Copy and complete the above table.
Observe the third and fifth columns. What do you notice?
STOP AND THINK
1. Is 23 × 42 = 85? Why?
2. Is 24 × 25 = 220? Why?
Multiplication Law
𝑎𝑥 × 𝑎𝑦 = 𝑎𝑥+𝑦
Caution:
1. This law is valid only when the bases
are the same.
2. We do not multiply the powers but
add them.
Example
Simplify the following:
(a) 45 × 47 (b) a2 × a5
(c) 5p2 × 3p4 (d) p3 × q4 × q5 × p3
Solution
(a) 45 × 47= 45+7 = 412 (b) a2 × a5= a2+5 = a7
(c) 5p2 × 3p4 = 5 x p2 x 3 x p4 = (5 × 3)p2+4 = 15p6
(d) p3 × q4 × q5 × p3 = p3+3q4+5 = p6q9
EXERCISE 10.2
1. Simplify
(a) 92 × 94
(f ) 63 × 69
(b) 46 × 43 (g) 76 × 72 × 73
2. Simplify
(a) a9 × a7
(b) q3 × q4
(f ) e6 × e2 × e (g) 𝑦3 × 𝑦4 × 𝑦2
(c) 36 × 35 (h) 156 × 1511
(c) v7 × v2
(h) m8 × m7
(d) 55 × 5 (i) 26 × 25 × 23
(d) h10 × h13 (i) t2 × t9 × t
(e) 108 × 107
(j) 86 × 82 × 8
(e) m2 × m9
(j) z2 × z3 × z5
133
Chapter 10 - Indices
3. Simplify
(a) a6 × 2a7
(b) 3p7 × 2p4
(f ) 3t6 × 4t2 × 5t (g) 3𝑥3 × 4𝑥4
4.
(a)
(d)
(g)
Simplify
a2 × a7 × b5 × b3
d 5 × e2 × d3 × e4 3p2 × q3 × 3p9 × q5
(c) 5q6 × 2q3
(h) 8m2 × 4m6
(d) 8q6 × 2q
(i) t6 × 4t7 × 5t
(b) p2 × q4 × q4 × p3 (e) 3s2 × t3 × 3s9
(h) 2t6 × t2 × 2m2 × 3m5
(e) 2m6 × 4m2
(j) 7k5 × k × 3k3
(c) m3 × m2 × n3 × n5
(f ) 5t6 × 3t2 × 2y2 × y5
(i) 3𝑥3 × 4y2 × 5𝑥2 × 7y6
Division Law
Consider the following:
𝑎
3
2
𝑏
2
2 ÷ 2
2⁴
2
2 ÷ 2
24
23
24 ÷ 23
25
25
26
27
22
23
23
24
2
𝑎 ÷ 𝑏
3
1
2
1
2
4
2
𝑎 ÷ 𝑏
(expanded form)
2 ×12 × 2
1 2 ×12
1
1
𝑎 ÷ 𝑏
(index form)
21
1
2×2×2×2
1 2 ×12
21
Copy and complete the above table.
Observe the third and fifth columns. What do you notice?
Division Law
𝒂𝒙
𝒂𝒙 ÷ 𝒂𝒚 = 𝒚 = 𝒂𝒙−𝒚 𝒂 = 0
𝒂
Example
Simplify the following:
(a) 410 ÷ 46 (b) a8 ÷ a7 4𝑥7
𝑎4 × 𝑎 5
(c)
(d) 2
𝑎 × 𝑎3
2𝑥
134
STOP AND THINK
1. Is 36 ÷ 32 = 33? Why?
2. Is 67 ÷ 23 = 34? Why?
Caution:
1. This law is valid only when the bases
are the same.
2. We do not divide the powers but
subtract them.
Chapter 10 - Indices
Solution
(a) 410 ÷ 46 = 410 − 6 = 44
2
4𝑥7
4 x 𝑥7
(c)
=
2𝑥
12 x 𝑥
= 2𝑥7 − 1
(b) a8 ÷ a7 = a8 − 7 = a1 = a
(d)
= 2𝑥6
𝑎4 × 𝑎 5 𝑎4 + 5
=
𝑎2 × 𝑎3 𝑎2 + 3
𝑎9
= 𝑎5 = 𝑎9 − 5
= 𝑎4
Note: For part (d), we use
both multiplication and
division laws of indices.
EXERCISE 10.3
1. Simplify
(a) 89 ÷ 87
(f ) 76 ÷ 7
(b) 99 ÷ 94
(g) 57 ÷ 52
(c) 36 ÷ 35
(h) 109 ÷ 107
2. Simplify
(a) a9 ÷ a7
(f ) z3 ÷ z
(b) q3 ÷ q
(g) q5 ÷ q2
(c) t7 ÷ t2
(h) m12 ÷ m9
3. Simplify
(a) 4𝑥7
2𝑥2
(b) 12𝑦7
3𝑦5
(c) 21𝑝8
3𝑝3
(d) 10𝑡9
𝑡4
(b) 𝑎7 × 𝑎5
𝑎 4 × 𝑎3
(c) 𝑝4 × 𝑝6
𝑝3 × 𝑝3
(d) 𝑚4 × 𝑚5
𝑚2 × 𝑚 4. Simplify
(a) 34 × 35
32 × 32
(f ) 𝑡4 × 𝑡6 × 𝑡2
𝑡2 × 𝑡3 × 𝑡3
(g) 𝑝4 × 𝑝9
𝑝2 × 𝑝3
(h) 𝑓4 × 𝑓5
𝑓2 × 𝑓
(d) 1010 ÷ 108
(i) 156 ÷ 153
(d) 𝑚8 ÷ 𝑚3
(i) t6 ÷ t
(e) 26 ÷ 24
(j) 67÷ 62 ÷ 6
(e) t9 ÷ t6
(j) 𝑥7 ÷ 𝑥6
(e) 30𝑥7
6𝑥2
(f ) 24𝑒7
3𝑒5
(e) 𝑎4 × 𝑎5 × 𝑎6
𝑎 2 × 𝑎3 × 𝑎2
(i) 𝑎4 × 𝑎5 × 𝑎6 (j) l 4 × l 5 × l 6
𝑎 2 × 𝑎3
l2 × l3
Power Law
Consider the following:
(53)2 = 53 × 53 = 5 × 5 × 5 × 5 × 5 × 5 = 56
(22)3 = 22 × 22 × 22 = 2 × 2 × 2 × 2 × 2 × 2 = 26
We can observe that (53)2 = 53 × 2 = 56, that is, we multiply the powers.
The same applies for (22)3 = 22 × 3 = 26
Power Law
(𝒂𝒙)𝒚 = 𝒂𝒙𝒚
135
Chapter 10 - Indices
Example
Simplify each of the following:
(a) (52)4 (b) (34)3 (c) (52)4 × (54)3 (d) (𝑝2)4 × (𝑝5)3
(e) (72)2 × (73)2 (f ) (𝑎2)6 × (𝑎3)2
(72)4
(𝑎2)2
Solution
(a) (52)4 = 52 × 4 =58(b) (34)3 = 34 × 3 = 312
(c) (52)4 × (54)3 = 58 × 512 = 58 + 12 = 520
(d) (𝑝2)4 × (𝑝5)3 = 𝑝8 × 𝑝15 = 𝑝8 + 15 = 𝑝23
(e) (72)2 × (73)2 74 × 76 710
(f ) (𝑎2)6 × (𝑎3)2 𝑎12 × 𝑎6 𝑎18
=
=
= 𝑎18 − 4 = 𝑎14
= 710 − 8 = 72
=
=
2 4
8
8
2 2
4
4
7
7
(𝑎 )
𝑎
𝑎
(7 )
EXERCISE 10.4
1. Simplify
(a) (92)4
(f ) (𝑡3)5
(b) (53)4
(g) (𝑝2)8
2. Simplify
(a) (42)4 × (44)3
(f ) (𝑡2)5 × (𝑡7)3
3. Simplify
(a) (𝑝2)4 × (𝑝5)4
(𝑝3)4
(c) (92)5
(h) (𝑡6)7
(b) (35)2 × (34)2
(g) (𝑡3)5 × (𝑡4)3
(b) (s5)4 × (s2)4
(s3)4 × (s4)2
(d) (𝑥2)8
(i) (102)5
(c) (72)2 × (74)3
(h) (z2)2 × (z8)3
(e) (𝑝9)4
(j) (y2)4
(d) (𝑥5)3 × (𝑥6)2
(e) (𝑝5)4 × (𝑝7)3
(i) (𝑡5)3 × (𝑡6)3 × (𝑡2)4
(c) (𝑡2)4 × (𝑡3)4
(𝑡3)2 × (𝑡2)2
(d) (𝑥6)4 × (𝑥4)3
(𝑥3)4 × (𝑥2)4
Zero power or zero index
𝑎
𝑏
23
23
32
32
43
43
𝑎 ÷ 𝑏 (expanded form)
1
1
2×2×2=1
1 2 ×12 ×12
1
3 ×13 = 1
1 3 ×13
1
1
1
4 × 4 × 4 =1
1 4 ×14 ×1 4
What do you observe?
𝒂0 = 1 , 𝒂 = 0
Any number (apart from 0) to the power of 0 is 1.
136
1
(e) (𝑏2)2 × (𝑏3)3
(𝑏2)4 × 𝑏3
𝑎 ÷ 𝑏 (index form)
23 − 3 = 20
32 − 2 = 30
43 − 3 = 40
Chapter 10 - Indices
Example 1
Evaluate:
(a) 90
Solution
(b) 𝑝0
(a) 90 = 1
(c) –(20)
(b) 𝑝0 = 1
8
(e) 𝑝0
(d) 4𝑎0
STOP AND THINK
What is the difference
between 𝑎0 and −(𝑎0) ?
(c) –(20) = –1 x (20) = –1 x 1 = –1
8
8
(e) 0 =
=8
𝑝
1
(d) 4𝑎0 = 4 x 𝑎0 = 4 x 1 = 4
Example 2
Evaluate:
(a) 32 + 30
(b) (62 × 60) + 6
(c) 82 + 8 – 80
Solution
(a) 32 + 30 = 9 + 1 = 10
(b) (62 × 60) + 6 = (36 × 1) + 6 = 36 + 6 = 42
(c) 82 + 8 – 80 = 64 + 8 – 1 = 71
EXERCISE 10.5
1. Write down the value of
(a) 10
(b) 70
(c) 1000
(d) –(80)
(e) 𝑚0
2. Evaluate
(f ) –(𝑝0)
(a) 40 + 4
(b) 30 + 32 + 33
(c) 53 – 52 + 50
(e) (72)3 x 70
74
(f ) 32 x 30 + 3
(g) 13 + 20 – 32
(g) 5𝑏0
(h) –3𝑥0
(d) 20 + 22 x 23
Summary
𝑎3 = 𝑎 × 𝑎 × 𝑎
index form
expanded form
where 𝑎 is the base and 3 is the index or power or exponent.
Laws of Indices:
Multiplication law :
Division Law
:
Power Law
:
Zero power
:
𝑎 𝑥 x 𝑎 y = 𝑎 𝑥 + y
𝑎 𝑥 ÷ 𝑎 y = 𝑎 𝑥 − y , 𝑎 = 0
(𝑎 𝑥) y = 𝑎 𝑥 x y = 𝑎 𝑥 y
𝑎 0 = 1, 𝑎 = 0
137
MONEY
Chapter 11 - Money
11
Learning Objectives
By the end of this chapter, you should be able to:
• decompose notes and coins.
• convert rupees into cents and vice versa.
• recognise the different currencies ($, £, €).
• convert from one currency to another (Rs, $, £, €).
• solve problems involving money.
Introduction
We use money to buy different commodities such as food
and clothes but also to pay for services such as doctor’s fees,
water bills and electricity bills. Payments can be made using
notes and coins but also using plastic cards such as debit
and credit cards. This method is convenient for people who
do not want to keep large amounts of cash with them.
Mauritian Currency
In Mauritius, we use the Mauritian Rupee (Rs) as currency
and the different coins and notes are shown below.
CHECK THAT YOU CAN:
• Perform the four mathematical
operations.
• Work with decimals.
FIND OUT
a. What is the meaning of
currency?
b. How do we compare the
currency used in different
countries?
c. What is the difference between
Rs 119.99 and Rs 120?
d. How is money processed
electronically?
DID YOU KNOW
Before the development of money,
people used the barter system to
obtain the goods and services that
they needed.
Use this link to learn more about the
history of money:
http://www.telegraph.co.uk/finance/
businessclub/money/11174013/
The-history-of-money-from-bartertobitcoin.html
NOTE TO TEACHER
• Discuss the various features on
particular notes.
• Students may be asked to work
on a project on the development
of the currency system.
138
Chapter 11 - Money
Example
Mrs Tina went to Best Bourbon Supermarket and her bill amounted to Rs 975.50. Below is her
receipt where some numbers have faded.
Find the amount she spent on biscuits and the amount of change she received.
Solution
1 1 1
Cost of Fish : Rs 359.75
Cost of Milk : Rs 270.50 +
Cost of Fruits : Rs 127.30
Amount spent on
fish, milk and fruits : Rs 757.55
Total amount of money spent
n
ourbo
Best B arket
m
Super
1 1
6 4 4 1
= Rs 97 5.50
Amount spent on fish, milk and fruits = Rs 757.55 –
Amount spent on biscuits = Rs 21 7.95
0 9 9 9 1
Amount given = Rs 10 00.00
Total amount spent Amount of change = Rs 97 5.50 –
= Rs 24 .50
Activity 1
FastPhone mobile company proposes the following
options for internet packages:
Option 1: Daily 40 MB mobile internet, validity: 24 hours,
Price: Rs 15
Option 2: Weekly 250 MB mobile internet, validity: 7 days,
Price: Rs 59
Option 3: Monthly 1 GB mobile internet, validity: 1 month,
Price: Rs 279
(a) Sana needs mobile internet during weekdays only.
Which one of the options 1 or 2 will be the best for her?
(b) Robin has to complete his school project during the
month of August. He needs to take mobile internet
to help him in his research work. Which one of the
options 1, 2 or 3 will be the best deal for him?
9.75
Rs 35 0
0.5
Rs 27
7.30
2
1
Rs
........
..
..........
h
1> Fis
ilk
2> M s
uit
3> Fr uits
sc
4> Bi
5.50
Rs 97
Total
00
Rs 1 0
nt
Amou
n
e
v
i
g
nt
Amou
ed
n
r
u
t
e
r
.........
..........
DID YOU KNOW
Some countries rarely use coins or
do not use coins at all; they only
use paper money or notes. Some of
these countries are: Laos, Belarus
and Myanmar.
CHECK THIS LINK
Who wants to be a hundredaire?
http://w w w.math-play.com/
millionaire-money-game/
millionaire-style-money-game.
html
FIND OUT
What do MB and GB mean?
139
Chapter 11 - Money
Example
Sohail goes to the bakery every morning to buy loaves of bread costing Rs 2.60 each. From
Monday to Friday, he buys 6 loaves of bread daily. His mother gives him a Rs 50 note, a Rs 25
note, a twenty rupee coin, a five rupee coin and some 20 cent coins every Monday morning
to buy bread for the week.
(a) How much money has he spent on loaves of bread by Friday night?
(b) How many loaves of bread can he buy on Saturday so that there will be just enough
money to buy 5 loaves of bread on Sunday?
(c) Find out how many 20 cent coins Sohail’s mother gave him on Monday morning.
Solution
Amount of money Sohail has at the beginning of the week to buy bread (excluding the 20
cent coins) = Rs (50 + 25 + 20 + 5) = Rs 100
Days of the week
Number
of loaves of bread
Cost
Amount left
(excluding 20 cent coins)
Monday
6
Rs 15.60
Rs 84.40
Tuesday
6
Rs 15.60
Rs 68.80
Wednesday
6
Rs 15.60
Rs 53.20
Thursday
6
Rs 15.60
Rs 37.60
Friday
6
Rs 15.60
Rs 22
Saturday
?
Sunday
5
?
Rs 13
Note: We make
use of a table
as our strategy
to solve this
problem.
?
Rs 9
We now use the table to answer the questions.
(a) Amount of money spent by Friday night = Rs (15.60 x 5) = Rs 78
(b) Amount spent from Monday to Friday and Sunday = Rs (78 + 13) = Rs 91
Amount left, excluding the 20 cent coins = Rs 9
To find the number of loaves of bread that can be bought with Rs 9, we divide Rs 9 by the
9
cost of one loaf of bread (Rs 2.60), i.e.,
= 3.46
2.60
Why 4?
Therefore the number of loaves of bread that can be bought with Rs 9 = 4
This is because Sohail still has some 20 cent coins that he can add to buy 4 loaves of bread.
(c) Cost of 4 loaves of bread = Rs 2.60 x 4 = Rs 10.40
Number of 20 cent coins = [Rs 10.40 – Rs 9] = 1.40 = 7
0.20
0.20
140
Chapter 11 - Money
Investigate:
On 1st January 2017, you were given Rs 10 as pocket money. This amount doubled every day. On
which date will you first have more than Rs 1 000, assuming you saved all your pocket money?
EXERCISE 11.1
1. In his money box, Jeff has three 25-rupee notes, ten 5-rupee coins, fifteen 1-rupee coins and
four 20-cent coins. How much money does Jeff have?
2. Mother buys a bottle of milk at Rs 32.90, a packet of sugar at Rs 48.50 and a bunch of bananas
at Rs 25. Find how much money mother spends in all.
3. Jimmy buys 5 packets of biscuits at Rs 25.50 each.
(a) Find the total cost of the 5 packets of biscuits.
(b) If he pays with a Rs 200 note, how much change will he receive?
4. The cost of 5 bars of chocolate is Rs 60.
(a) What is the cost of 20 bars of chocolate?
(b) How many bars of chocolate can be purchased with Rs 150? How much money will be left?
5. Mr Goodman has Rs 10 520 as savings in the bank. At the end of the month, he receives Rs 25 065
as net salary. His expenses for the month amount to Rs 18 750. Calculate the amount of money
Mr Goodman now has in his bank account, assuming his salary was paid directly into the bank
by his employer.
6. Two sisters, Pari and Ruchi, decide to save their pocket money for Mother’s Day. On the
eve of Mother’s Day, Pari has Rs 1 250 and Ruchi has Rs 875. If they each decide to buy a gift
costing Rs 925, calculate
(a) how much money Pari must give to Ruchi in order for Ruchi to have enough money to buy her gift,
(b) how much money does Pari have left after Mother’s Day?
7. Two friends, Kavi and Rishi, decide to go out for a meal after work on Tuesday. There are two
restaurants which have special offers on Tuesday.
At Restaurant Chef, if you buy a meal at Rs 125, you get a second meal half price.
At Restaurant Savanna, a meal costs Rs 80 from Monday to Thursday.
In which restaurant will it be cheaper for Kavi and Rishi to eat? Explain your answer.
8. Last Sunday Mr. and Mrs. Akbar decided to visit the Lakeside Zoo accompanied by their
daughter aged 16 and their two sons aged 9 and 5.
Cost of ticket
Weekdays (Rs)
Cost of ticket
Weekends (Rs)
Children ticket (7 to 12 years)
Rs 50
Rs 75
Adult ticket
Rs 85
Rs 125
Family tickets (2 adults and 2 children)
Rs 225
Rs 300
Type of ticket
Free Tickets for Children under 7 years old.
What is the best price deal for the Akbar family? Explain your answer.
141
Chapter 11 - Money
9. Circle the correct answer.
(a) At different market stalls, the prices of potatoes are as follows:
• 500 g at Rs 13
• 1 kg at Rs 26
• 2 kg at Rs 50
• 4 kg at Rs 98
Which one is the best bargain?
A. 500 g at Rs 13
B. 1 kg at Rs 26
C. 2 kg at Rs 50
D. 4 kg at Rs 98
(b) 4 painters and a supervisor earn a total of Rs 2 800 daily. If a painter earns Rs 300 less than
the supervisor, how much do 9 painters and 3 supervisors earn in 5 days?
A. Rs 14 000
(c)
B. Rs 12 320
C. Rs 34 500
D. Rs 43 500
Miksha pays a total of Rs 600 for the following vegetables:
• 5 kg of tomatoes at Rs 90 per kg
• 3
1
kg of potatoes at Rs 17 per kg
2
• 1.25 kg of onions at Rs 18 per kg and 2 kg of pumpkin.
What is the cost of 1 kg of pumpkin?
A. Rs 34
B. Rs 68
C. Rs 237.50
D. Rs 532
(d) Mr Harris has Rs 5 000 in the bank. He needs to pay a loan of Rs 7 000, house rent at
Rs 5 525 and insurance at Rs 2 580. What is the least amount of money that Mr Harris
has to deposit in the bank in order to pay for all his expenses?
A. Rs 10 105 B. Rs 10 100
C. Rs 15 100 D. Rs15 105
10. There is a sale on fruit juice at Windy Supermarket:
Deal 1: Buy 1.5 L of fruit juice at Rs 39.
Deal 2: Buy 1 L of fruit juice at Rs 38 and get 0.5 L of fruit juice free.
Which deal is the better value for money? Explain why.
STOP AND THINK
CHECK THIS LINK
Which deal is the best? Explain why.
(a) 4 ‘pains au chocolat’ for Rs 92
or
6 ‘pains au chocolat’ for Rs 132
(b) 2 large pizzas for Rs 399
or
Buy 2 large and get 1 large pizza free
given that one large pizza costs Rs 250
(c) A box of 1 kg of cereals at Rs 129
or
a box of 500 g of the same cereal at Rs 70
(d) 1 kg of milk powder at Rs 170
or
Buy 1 kg of milk powder at Rs 259, get 500 g of milk powder free.
142
www.scootle.edu.au/ec/viewing/L1929/
index.html
NOTE TO TEACHER
• Encourage students to work out how they can
use the online shop (in the above link) to buy
the items on their shopping lists at the cheapest
prices.
• Prompt students to think of different solutions
and to explain their reasoning based on the
number of items required.
• Encourage students to use different supermarket
promotion leaflets and to compare the prices of
specific food items to determine the best deals.
Chapter 11 - Money
Currencies and Exchange Rates
All countries do not use the same currency. For
example, in Mauritius we use the Mauritian Rupee
(Rs) while in Reunion Island, the Euro (€) is used.
As we travel to other countries and trade with them,
it is important to recognise the different foreign
currencies and convert money from one currency to
another.
Banks and Foreign Exchange dealers allow us to
change currencies. The exchange rate is not always
fixed and changes all the time. For instance, $1 may
be equivalent to Rs 35 today but Rs 38 in 6 months.
The table on the right shows the currency used in
different countries and its corresponding symbol.
Investigate:
Can you list the different coins and notes that
are used in the countries mentioned in the
"Recall" table?
Which note and coin have the lowest value?
Example
Country
Currency
Symbol
Mauritius
Mauritian
Rupee
Rs
United States
US Dollar
$
United
Kingdom
Pound
Sterling
£
Countries in
Europe, for
example,
Germany,
Italy, France,
Spain, Greece,
Portugal
Euro
€
DID YOU KNOW
Somebody who collects coins is
called a numismatist.
DID YOU KNOW
Given that US $1 = Rs 35 and € 1 = Rs 38, convert
(a) $100 to rupees (Rs),
(b) Rs 70 000 to US dollars ($),
(c) € 250 to rupees (Rs).
Interesting fact:
Here are some conversions of the weakest foreign
currencies in the world still in use today:
Foreign
Country
Currencies
Solution
(a) $1 = Rs 35
$100 = Rs (100 × 35) = Rs 3 500
(b) Rs 35 = $1
1
Rs 1 = $
35
Rs 70 000 = $(
RECALL
To Mauritian
Rupee (Rs)
Iranian Rial
Iran
Rs 1 = 934.34 Rial
Vietnamese
Dong (VND)
Vietnam
Rs 1 = 653.53 VND
Indonesian
Indonesia Rs 1 = 382.17 IDR
Rupiah (IDR)
1
× 70 000) = $2 000
35
(c) € 1 = Rs 38
€ 250 = Rs (250 × 38) = Rs 9 500
Laos
(LAK- LAO KIP)
Laos
Rs 1 = 233.18 LAK
Note: These figures are based on the exchange
rate in July 2017.
143
Chapter 11 - Money
Fun activity: Currencies Around the world
Complete the cross word below with the currency of the
different countries.
1
2
Across
1. Mauritius
3. Former currency of Estonia
6. USA
3
4
7. Japan
8. Bangladesh
9. Iran
10. England
DID YOU KNOW
Have you ever seen a 100 trillion dollar
note?
This note was used before the year 2009 in
Zimbabwe but was equivalent to just 40
US cents, that is, around Rs 13.
People needed a bundle of these bank
notes to buy just a few household
items.
5
6
Down
2. Argentina
4. Russia
5. Thailand
6. Kuwait
7. China
7
8
DID YOU KNOW
9
The rupee is used in different
countries. In Mauritius, we have the
Mauritian Rupee (MUR). We use the
Indian Rupee (INR) in India and the
Pakistani Rupee (PKR) in Pakistan.
10
Investigate:
Ruchina is planning her travel to Europe and is trying
to find out the best exchange rate available to change
Rs 100 000 into Euros. Can you help Ruchina find the
best foreign exchange deal?
Hint: You may find the exchange rates on the Bank of
Mauritius website https://www.bom.mu
Similarly, the dollar is used in various
countries. We have the US Dollar, the
Singaporean Dollar, the Australian
Dollar and the Canadian Dollar
amongst others.
Activity 2
Kiren wishes to buy a new dress. She finds the dress in a shop in Mauritius at Rs 3 450.
She decides to check the price of the dress online and finds different options.
Option 1: Mauritius
Price: Rs 3 450
Bus fare to go to the shop
and come back: Rs 50
Option 2: US
Price: US $ 35
Delivery charge to
Mauritius: free
Option 3: UK
Price: £ 30
Delivery charge to
Mauritius: £ 10
Option 4: France
Price: € 28
Delivery charge to
Mauritius: Free
Can you help Kiren decide where she should buy the dress and explain why?
Hint: You need to find the exchange rates for the different foreign currencies.
144
Chapter 11 - Money
EXERCISE 11.2
1. Study the following conversion table and answer the questions below.
(a) Convert the following into Mauritian Rupees
(i) US $ 60, (ii) € 25, (iii) £ 30.
US $ 1 = Rs 35
(b) Convert
(i) Rs 1 155 into US dollars,
£ 1 = Rs 45.25
(ii) Rs 1 810 into pound sterling,
€ 1 = Rs 38.50
(iii) Rs 1 386 into Euros.
RECALL
£ 1 = 100 pence
€ 1 = 100 cents
Rs 1= 100 cents
$ 1 (US Dollar) = 100 cents
2. The rate of exchange between pound sterling (£) and US dollars ($) was £ 1 = $ 2.80.
Calculate (a) the number of dollars received in exchange for £ 120,
(b) the number of pound sterling received in exchange for $ 224.
3. Last year, Samantha went to Italy.
(a) She changed Rs 130 560 into Euros at the rate of € 1 = Rs 40.80.
How many Euros did she get?
(b) She bought a gift for 85 Euros. How much was the gift worth in Mauritian Rupees?
4. Sanjana goes to the bank to change $ 200 into Mauritian Rupees (Rs). The bank charges $ 5
and then changes the rest of the money at the rate of $ 1 = Rs 40. How much money does
Sanjana receive in Rupees?
5. Nillen is travelling to Spain to attend a conference. He changes Rs 140 000 into Euros (€) at
a rate of € 1 = Rs 35.
(a) Find how many Euros he receives.
(b) He spent € 3 750 in Spain and changes the rest to Mauritian Rupees. Find the amount
of money that he has left, in Rupees.
6. Riya changed $ 600 into pound sterling (£) when the exchange rate was £ 1 = $ 2.40. She
later changed all the pound sterling back into dollars when the exchange rate was £ 1 = $ 2.60.
How many dollars did she receive?
7. Circle the correct answer.
(a) Vinesh is going to America. The exchange rate is £ 1 = $ 1.22. However, the smallest
note that the exchange bureau has in stock is $ 50. How many dollars does he need
to exchange in order to have at least £ 1 000?
A. $ 1 200
B. $ 1 225
C. $ 1 250
D. $ 1 220
(b) The exchange rate between dollars and Euros is $1 = € 0.80. Ben changes $275 into
Euros. Calculate the number of Euros Ben receives.
A. 20
B. 22
C. 220
D. 2 200
Summary
•
•
•
Money is used to buy commodities such as food and clothing as well as pay for
services such as telephone bills.
Different countries use different currencies.
The rate of exchange from one currency to another changes everyday.
145
12
TIME
Chapter 12 - Time
Learning
Objectives
By the end of this chapter, you should be able to:
• express times in terms of the 12-hour and 24-hour clock.
• convert times from one unit to another (hour, minute, second).
• use GMT in practical situations.
• solve real life problems involving time.
Time in real life
CHECK THAT YOU CAN:
• Perform arithmetic
operations with numbers.
KEY TERMS
Time on
a clockface
Departure times
flight schedule
Stopwatch
Time on
a mobile phone
Time on
a watch
•
•
•
•
•
•
12-hour clock
24-hour clock
a.m., p.m.
GMT, Meridian
Hour, Minute,
Noon, Midnight
Dates on
a calendar
DID YOU KNOW
Introduction
Time is used to measure the duration of an event or the
gap between different events.
For example, Tony watched a movie for two hours and
Sonam walked for one hour from home to school.
The Standard Unit of time is the second (s). Time is
measured in other units as well: minutes (min), hours (h),
days, weeks, months, years and centuries.
HISTORY
In the past, the sun, the moon as well
as other objects such as candles were
used to measure time and the passage
of time.
Before the invention of the Clock, time
was measured using Sundials, the
Sand clock or hourglass and the water
clock.
RECALL
1 day = 24 hours
1 week = 7 days
1 month = 30 or 31 days
(except February = 28 or 29 days)
1 year = 52 weeks = 12 months
146
1 hour = 60 minutes
1 minute = 60 seconds
1 hour = 3 600 seconds
1 year = 365 or 366 days
Sand clock
Sundial
Hourglass
Water clock
Chapter 12 - Time
EXERCISE 12.1
Reading time on a clock
Write down the time shown on each clock face below: (a) In figures
(b) In words.
(i) (ii)(iii)
Convert time from one unit to another (Hour, Minute, Second)
× 3 600
× 60
h
RECALL
× 60
min
÷ 60
1 h = 60 min
1 min = 1 h
60
1 h = 3 600 s
1s= 1 h
3 600
1 min = 60 s
1 s = 1 min
60
s
÷60
÷ 3 600
Example
Convert
(a) 8 h into min
(b) 700 s into min
Solution
(a) 1 h = 60 min
8 h = (60 x 8) min
= 480 min
(b)
1
1 s = 60 min
700 s = ( 1 × 700) min
60
= 35 min
3
= 11 2 min
3
EXERCISE 12.2
1. Express each of the following in minutes.
(a) 2 h
(b) 5 h
(c) 600 s
(d) 7 200 s
2. Express each of the following in seconds.
(a) 15 min
(b) 1 1 h
(c) 4 1 min
(d) 4 h
4
2
1
(e) 1 h
3
(e) 3 min
147
Chapter 12 - Time
3. Express each of the following in hours.
(a) 180 min (b) 1 800 min (c) 7200 s
(d) 150 s
(e) 500 min
4. Convert
(a) 320 seconds into minutes and seconds.
(b) 2 hours 23 minutes into minutes
(c) 210 minutes into hours and minutes
5. Write the following times in ascending order.
(a) 1.3 h , 45 min, 1 h 30 min, 85 min
(b) 3 000 s, 1 h 10 min, 40 min, 2 h, 60 min
Express times in terms of the 12-hour and 24-hour clock
Analogue Clock (12-hour clock/ a.m./ p.m.)
(using the minute and the hour hands)
Digital Clock (24-hour clock)
(using numbers)
There are two main ways to show the time, one using the 12-hour clock (a.m. /p.m.) and the
other using the 24-hour clock.
Example
Copy and complete the table below.
Time in words
Three o’clock in the morning
12 hour
3 a.m.
Half past two in the afternoon
14 30
3.45 p.m.
Twenty minutes past six in the morning
24 hour
15 45
6.20 a.m.
Twenty five minutes past nine in the evening
21 25
Solution
Time in words
12 hour
24 hour
3 a.m.
03 00
Half past two in the afternoon
2.30 p.m.
14 30
Quarter to four in the afternoon
3.45 p.m.
15 45
Twenty minutes past six in the morning
6.20 a.m.
06 20
Twenty five minutes past nine in the evening
9.25 p.m.
21 25
Three o’clock in the morning
148
Chapter 12 - Time
EXERCISE 12.3
1. Match the time on the analogue clock to that on the digital one.
2. Write the following in figures using
(i) 12-hour clock system, (ii) 24-hour clock system
DID YOU KNOW
(a) Quarter past two in the morning.
(b) Ten minutes past eight in the evening.
(c) Half past six in the afternoon.
(d) Five o’clock in the morning.
3. Write the following times in words.
(a) 1.25 a.m.
(b) 02 35
(c) 11 45
(d) 13 25
(e) 22 20
(f ) 7.40 p.m.
•
•
a.m. / p.m.
a.m. stands for "ante
meridiem" that is the
time from midnight to
noon.
p.m. stands for "post
meridiem" that is the
time from noon to
midnight.
149
Chapter 12 - Time
Addition and Subtraction of time
Example
Find (a) 2 h 35 min + 6 h 42 min
(b) 10 h 15 min – 4 h 27 min
Solution
(a) We first add the minutes and hours separately.
2h+6h=8h
35 min + 42 min = 77 min
But 77 min = 60 min + 17 min
= 1 h + 17 min
Therefore, 2 h 35 min + 6 h 42 min
= 8 h + 1 h + 17 min
= 9 h + 17 min = 9 h 17 min
9
60
(b) 10 h 15 min
– 4 h 27 min
5 h 48 min
Note: Since 27 min cannot be subtracted
from 15 min, we borrow 1 h from the 10 h
and send it as 60 min to the minute column.
EXERCISE 12.4
1. (a)(b)(c)(d)
3 h 30 min
5 h 15 min
7 h 55 min
23 h 25 min
+ 9 h 28 min
– 2 h 55 min
+ 6 h 10 min
– 18 h 45 min
2. (a) 7 min 35 s + 14 min 55 s
(c) 16 min 45 s – 14 min 57 s
(b) 34 min 20 s – 28 min 12 s
(d) 55 min 35 s + 12 min 10 s
Word problems involving Time
Example
A car reached its destination at 21 40 after travelling for 55 minutes. At what time did the car
leave for its destination, assuming it did not stop on the way?
Solution
Travelling time: 55 min
Destination
?
Departure time
20
60
Arrival time
= 21 40 –
Travelling time =
55
Departure time = 20 45
The car left at 20 45 for its destination.
150
21 40
Arrival time
Chapter 12 - Time
EXERCISE 12.5
1. Sadna watched a movie of duration 2 hours 35 minutes. The movie started at 3.15 p.m.
At what time did it end?
2. The football match started at 2.30 p.m. Anil was 25 minutes late. At what time did he reach
the playground?
3. The time on the clockface is 08 45. What time was it 35 minutes ago?
4. My watch is 4 minutes slow and it shows 15 33. What is the exact time?
5. Suzy left her home at 07 45 and reached her office at 08 55. How long did she take to travel
from her home to her office?
6. Kavita left her office at 4 o’clock in the afternoon. She took 15 minutes to walk to the bus
stop and waited for 10 minutes for the bus. She travelled 55 minutes by bus to reach home.
At what time did she reach home?
7. A plane left Perth at 10.30 p.m. (Mauritian time) on Saturday. It took 7 hours 15 minutes
to reach SSR International Airport. At what time did it land at SSR International Airport in
Mauritius?
8. A meeting is due to start at 2.15 p.m. However, it started 25 minutes late and its duration was
one and a half hour. At what time did the meeting end?
9. Ashley covered a journey in three parts. He rode the first part of his journey on a bicycle
for 35 minutes, then walked the second part for three quarter of an hour and finally
travelled by car for another 720 seconds. How many minutes did he take in all to cover the
whole journey?
Greenwich Mean Time (GMT)
The Greenwich Mean Time or the Greenwich Meridian Time or
simply GMT is a standard for setting the time zones. The GMT is the
time measured from the Greenwich Meridian Line at 00 longitude
situated at Greenwich in England. Countries found on the East of the
Meridian Line are ahead of the GMT whereas countries on the West
of that line are behind the GMT.
There are 24 time zones, each of 15 degrees longitude as shown
on the World Standard Time Zones Map. All places found in the
same time zone have the same local time. The local time is the
actual time in your country. For example, Mauritius and Dubai
have the same time zone.
FIND OUT
What are the different time
zones which exist?
A few examples are given:
Eastern Time, Pacific Time,
Australia Central Time.
Find out their meanings.
DID YOU KNOW
In the year 1884, at the International
Meridian Conference in Washington,
the GMT was recommended as the
centre of world time and the basis for
a global system of time zones.
151
Chapter 12 - Time
Mauritius
For example, when it is 10 30 in London (GMT), it is 14 30 (GMT + 4) in Mauritius and it is
02 30 (GMT –8) in Los Angeles.
This is illustrated in the figure above.
Example 1
Mauritius is 4 hours ahead of GMT and Moscow is 3 hours ahead of GMT. If it is 16 00 in
Mauritius, what is the (a) GMT, (b) local time in Moscow?
4 hours
Solution
(a) Local time in Mauritius is 16 00 = GMT + 4 hours
GMT = 16 00 – 4 hours = 12 00
(b) Method 1
Local time of Moscow is GMT + 3 hours
Local time of Moscow = 12 00 + 3 hours = 15 00
GMT
Moscow
Mauritius
16 00
3 hours
Method 2
Since there is 1 hour difference
between Moscow and Mauritius,
local time in Moscow
= 16 00 – 1 h = 15 00
Example 2
An aeroplane leaves England for Mauritius at 17 30 local time on Saturday. Given that the
time in Mauritius is 4 hours ahead of the time in England, what is the time in Mauritius
when the plane lands in Mauritius if the trip is 12 hours long?
Solution
Plane left at
17 30 +
17 30
England
Duration of flight
12
hours
Time in England when
29 30
plane lands in Mauritius = 05 30 on Sunday morning (29 30 – 24 00)
Arrival time in Mauritius = (05 30 + 4 h) = 09 30 on Sunday morning.
152
4 hours
?
Mauritius
Chapter 12 - Time
EXERCISE 12.6
1. Copy and complete the table given below:
Country
Difference in time with GMT
Local Time
Mauritius
GMT + 4 hours
13 15
India
1
GMT + 5 hours
2
Sydney
Japan
19 15
GMT + 9 hours
Egypt
11 15
Greenland
06 15
Mexico
GMT – 7 hours
2. Chicago is 6 hours behind GMT and Abu Dhabi is 3 hours ahead of GMT.
(a) What is the time in Chicago if it is 11.40 a.m. in Abu Dhabi?
(b) What is the time in Abu Dhabi if the time is 21 15 in Chicago?
3. The local time in Mauritius is 9 hours ahead of Ottawa (located in Canada).
(a) Look at the two clocks below. One shows the local time in Mauritius.
Show the local time in Ottawa on the other clock.
MauritiusOttawa
(b) Using the 24-hour clock, write down the local time in Mauritius if it is 14 15 in Ottawa.
4. A plane leaves London at 03 35 and reaches Jakarta in 12 hours 20 minutes.
(a) What time will it be in London when the plane reaches Jakarta?
(b) If Jakarta is 7 hours ahead of London, what is the time when the plane lands in Jakarta?
5. The table below shows the information of the time zones for Mauritius and Perth.
Country
Hours ahead of or behind GMT
Mauritius
+4
Perth
+8
Sandrine travelled from Mauritius to Perth on Sunday. Her flight left Mauritius at 10 55 and it
took 7 hours 46 minutes. Find the day and time when she reached Perth.
153
Chapter 12 - Time
Calendar
Example
Observe the calendar in Fig. 1 and answer the
following questions.
(a) Describe the pattern along
(i) each row
(ii) each column.
(b) From this calendar, can you find out on which day
was (i) the 26th of June 2017,
(ii) the 14th of August 2017?
Solution
Fig. 1
(a) (i) The numbers along each row follow a pattern where we add 1 each time to get the
next term.
(ii) The numbers along each column follow a pattern where we add 7 to get the next term.
(b) (i) Since the 1st of July was a Saturday, we count backward.
Therefore, 26th June 2017 was a Monday.
(ii) Similarly, since 31st July was a Monday, we count forward and thus 14th August 2017
was a Monday.
EXERCISE 12.7
1. The 24th April 2017 was a Monday. On which day was the
12th April 2017?
2. The 25th May 2007 was a Friday. On which day was the 10th
June 2007?
FIND OUT
What is a leap year?
How do we know a year is leap or
common?
When did the last leap year occur?
When will the next leap year occur?
3. From the list below, find the leap years and common years:
1997, 1988, 2000, 1978, 2019, 2032, 2105
Summary
•
•
•
•
•
•
154
12-hour clock: time is written using a dot and is followed by a.m. or p.m.
For example, 7.15 a.m. and 1.35 p.m.
a.m. means ‘ante meridiem’ that is ‘before noon’.
p.m. means ‘post meridiem’ that is ‘after noon.
The 24-hour clock does not use a.m. and p.m. For example, quarter past seven in
the morning is written as 07 15 and quarter past seven in the afternoon is written
as 19 15.
GMT stands for Greenwich Mean Time. The local time in Mauritius is 4 hours
ahead of the GMT.
Countries to the East of the Meridian Line are ahead of the GMT (GMT +) whereas
countries to the West of that line are behind the GMT (GMT –).
13
SPEED
Chapter 13 - Speed
Learning Objectives
By the end of this chapter, you should be able to:
• demonstrate an understanding of the terms speed and average speed.
• convert speed from one unit to another.
• solve real life problems involving speed (including average speed).
Real life examples of Speed
The Cheetah is known to be the
fastest animal on land. It can go
up to 113 km/h.
The Three-toed sloth, native
to America, is known to be the
slowest animal in the world,
moving up to a maximum speed
of 0.001 m/s.
Usain Bolt is one of the fastest
men in the world. He reached a
top speed of 12.27 m/s.
The speed of light is around 300
million m/s.
CHECK THAT YOU CAN:
• Perform the four
arithmetic operations.
• Work with units of length
(km, m, cm).
• Work with units of time
(h, min, s).
KEY TERMS
•
•
•
•
•
Speed
Instantaneous Speed
Average Speed
Distance
Time
DID YOU KNOW
Galileo Galilei, an
Italian physicist,
is believed to
have been the
first to measure
speed by taking into account the
distance covered and the time taken.
The Bugatti Veyron 16.4 Super
Sport is one of the fastest cars in
the world capable of going up to
431.072 km/h.
STOP AND THINK
Mr Farid was driving on the motorway at 110 km/h. When he saw this sign post, he applied his brakes gently.
Why did he do that?
155
Chapter 13 - Speed
Introduction
In Grade 6, you learnt that the speed of an object is the distance travelled by this object per unit
of time. Speed measures how fast an object is moving.
Consider the following two cases.
Case 1: Bradley and Tina run a distance of 200 m in 1 min and 50 s respectively. Who is faster?
Case 2: Two students Anna and Elsa participate in a race. Anna covers a distance of 150 m
in 25 s while Elsa covers a distance of 400 m in 40 s. Who is faster?
In Case 1, we have the same distance but different times. We can therefore easily compare and
say that Tina is faster as she takes less time than Bradley to complete the same distance.
In Case 2, we have two different times and two different distances. To find out who is faster, we
can compare the rate at which they are running.
Anna: 25 s
1s
150 m
Elsa:
150 m = 6 m
25
40 s
1s
400 m
400 m = 10 m
40
Therefore, Elsa is faster as she covers 10 m in 1 s while Anna covers 6 m in 1 s.
We have calculated the distance travelled per unit of time, also called speed. So, speed can be
calculated as follows
Speed = Distance
Time
Speed is commonly measured in
• metres per second (m/s or ms-1) or
• kilometres per hour (km/h or kmh-1).
STOP AND THINK
Consider a car.
(i) What is its speed before the driver starts the car?
(ii) What might be its speed after a short while?
(iii) What will be its speed just before reaching the traffic light showing red?
What can you observe?
DID YOU KNOW
Instantaneous Speed
A speedometer measures and shows
the instantaneous speed of a vehicle,
that is, the speed measured at a
particular or specific time.
Example: If the speedometer of a
car displays
80 km/h, this
means that the
instantaneous
speed of the car
is 80 km/h.
Average Speed
Consider a car that covers a distance of 180 km in 3 h. The speed of the car varies during the journey.
Thus, we use the term average speed to represent the overall speed of the car throughout the journey.
Average speed is different from instantaneous speed. It is defined as the total distance covered
divided by the time interval and does not take into account the speed variations that may have
occurred during short time intervals.
156
Chapter 13 - Speed
Example 1
A cyclist travelled a distance of 80 km in 2 hours. Find his average speed.
Solution
80 km
80
1 hour
km = 40 km
2
Average speed of cyclist = 40 km/h
2 hours
Average speed is given by:
Average Speed = Total Distance travelled
Total Time taken
Note: The terms speed and average speed are often used interchangeably.
Example 2
Ashvin travels a distance of 80 km in 6 hours.
Find his average speed.
Note: Constant speed is
when the speed does not
change, that is, it neither
increases nor decreases.
Average Speed =
Distance
Time
RECALL
Average Speed =
80 km
6h
In Grade 6, you learnt
1
Average Speed = 13 3 km/h
S
EXERCISE 13.1
S= D
T
D
T
where D stands for distance,
S stands for speed and T
stands for time.
1. Calculate the average speed in each of the following cases:
(a) Distance travelled = 100 km, Time taken = 5 hours
(b) Distance travelled = 280 m, Time taken = 40 seconds
1
(c) Distance travelled = 375 km, Time taken = 1 hours
4
(d) Distance travelled = 960 m, Time taken = 1 200 seconds
2. Jack covers a distance of 2.5 km in 2 hours. Find his average speed in km/h.
3. A school bus covers a distance of 40 km in 50 minutes. Find its average speed in km/h.
4. Alexia drives her car through a distance of 425 km in 300 minutes. What is the average
speed in kilometres per hour?
5. A baseball is thrown at a distance of 140 m. What is its speed in m/s if it takes 0.8 seconds to
cover the distance?
6. Rajiv rides his bike at a constant speed of 18 km/h for 81 km and another 72 km at a
constant speed of 24 km/h. Find his average speed for the whole trip.
157
Chapter 13 - Speed
Finding distance given time and speed
RECALL
Example
Dilshad walks from her parents’ house to the park at an average
speed of 6 km/h. The journey took 2 hours. What is the distance
between the house and the park?
In Grade 6, you learnt
D=SxT
D
Solution
S
T
1 hour
6 km
2 hours
(6 × 2) km = 12 km
Distance between the house and the park = 12 km
EXERCISE 13.2
1. Calculate the distance covered in each of the following:
(a) Speed = 24 km/h, Time = 4 hours
(b) Speed= 40 m/s, Time = 2 minutes
(c) Speed = 120 km/h, Time = 70 minutes
(d) Speed = 25 m/s, Time = 9 seconds
Caution:
Check the unit of time
and speed before
proceeding with this
question.
2. A car travelled at an average speed of 60 km/h for 45 minutes. What is the distance covered?
3. Rekha walks at an average speed of 2.5 m/s. What distance is covered in 1 hour?
4. A train is travelling at an average speed of 40 km/h. How far can it travel in 240 minutes?
5. Khalid drives from Town A to Town B at an average speed of 17 m/s. The drive took him
80 minutes. Find the distance from Town A to Town B in km.
6. Juan cycles from his home to school at an average speed of 25 km/h. Given that the journey
takes 30 minutes, find the distance between his home and school.
7. How far will you travel if you ride your motorbike for 15 minutes at 50 km/h?
1
8. Emilie walked 10 km in 2 hours and then cycled at an average speed of 15 km/h for 2 2 hours.
Find the total distance travelled by Emilie.
DID YOU KNOW
Pace of Life
•
•
•
158
The pace of life in different countries can be compared
by measuring the average walking speed of people in
those countries.
Singapore has the fastest pace of life (people take 10.55
seconds to cover 18.3 metres).
Malawi has the slowest pace of life.
FIND OUT
Can you find the speed
at which the moon
orbits the Earth and the
speed at which the Earth
revolves around the sun?
Chapter 13 - Speed
Finding time given speed and distance
Example 1
RECALL
Lizzie hires a helicopter to travel a distance of 720 km at an
average speed of 160 km/h. How long did the journey take?
In Grade 6, you learnt
Solution
1h
1
160 h
( 1 x 720) h = 4.5 h
160
160 km
1 km
720 km
T= D
S
D
S
T
Example 2
A car covers a distance of 110 km at 40 km/h and a further 60 km in 75 minutes.
Find the total time taken in hours.
Solution
40 km
1 km
110 km
1h
1
40 h
( 1 x 110) h = 11 h = 2 3 h
40
4
4
(40 km/h)
110 km
60 km
Time taken = ?
75 min
3
Total time = 2 4 h + 75 min
3
1
=2 4 h+14 h=4h
EXERCISE 13.3
1. Calculate the time taken to travel
(a) 90 km at a speed of 40 km/h
(c) 380 m at a speed of 19 m/s
(b) 1450 km at a speed of 80 km/h
(d) 108 m at a speed of 3.6 m/s.
2. Aisha runs the 100 m race at an average speed of 5 m/s. Find the time taken for Aisha to
finish the race.
3. How long will it take you to travel 150 km in a hot air balloon at an average speed of 20 km/h?
4. An ant covers a distance of 150 cm to reach its colony. If it crawls at a speed of 8 cm/s, how
long does it take to complete the journey?
5. Elena rides her bicycle at a constant speed of 10 m/s. How long will she take to travel a
distance of 1 875 m? Give your answer in minutes.
6. Rakesh travels a distance of 6 400 km in an aeroplane. For one quarter of the distance, the
aeroplane flies at an average speed of 640 km/h and for the rest of the distance, it flies at
an average speed of 768 km/h. How long does the trip take?
159
Chapter 13 - Speed
Investigate:
A cheetah spots a gazelle 500 m and sprints towards it at a
speed of 110 km/h. At the same time, the gazelle runs away
from the cheetah at a speed of 80 km/h. How many seconds
does it take for the cheetah to catch the gazelle?
Cheetah
Harder problems
Gazelle
Example 1
A car travelled at a uniform speed of 72 km/h for 2 hours and then travelled at a speed of
65 km/h for another 1.5 hours.
Find the average speed of the car for the whole trip.
Solution
1h
2h
72 km
(72 x 2) km = 144 km
1h
1.5 h
72 km/h
65 km/h
2 hours
1.5 hours
65 km
(65 x 1.5) km = 97.5 km
Total distance covered = (144 + 97.5) km = 241.5 km
Total time taken = (2 + 1.5) h = 3.5 h
Average speed =
Total distance covered
Total time time taken
= 241.5 km
3.5 h
= 69 km/h
FIND OUT
Average speed ≠
Why?
(72 + 65)
km/h
2
Example 2
Town X and Town Y are 600 km apart. At 08 15, a taxi leaves town X, travelling towards
Town Y at an average speed of 90 km/h. At the same time, a lorry leaves Town Y and travels
towards Town X. The taxi and the lorry meet at 12 15. Find the average speed of the lorry.
Solution
x
90 km/h
600 km
Number of hours travelled before taxi and lorry meet = 12 15 – 08 15 = 4 hours
Distance travelled by car after 4 hours = Average speed of car x time taken
= (90 x 4) km
= 360 km
160
Y
Chapter 13 - Speed
(600 – 360) km
x
Y
Y
360 km
Distance travelled by lorry after 4 hours = 600 km – 360 km = 240 km
Average speed of lorry = Distance travelled
Time taken
= 240 km = 60 km/h
4h
EXERCISE 13.4
1. Terry drove for 4 hours at an average speed of 65 km/h and at 50 km/h for another 3 1 hours.
2
What is his average speed for the whole journey?
2. A man makes a journey of 66 km. He cycles for the first 57 km at an average speed of 12 km/h.
He walks the rest of the distance and takes a total of 7 hours for the whole journey. Find the
average walking speed of the man.
3. A car travels a distance of 210 km. It travels the first 112 km at an average speed of 32 km/h.
The rest of the distance is completed at an average speed of 42 km/h.
(a) Find the total time taken to cover the distance of 210 km.
(b) Find the average speed for the whole journey.
4. At exactly the same time, two motorcycles leave a supermarket and travel in opposite
directions. One motorcycle travels at an average speed of 55 km/h and the other travels at an
average speed of 63 km/h. After 5 hours, how far apart will they be?
5. Raj left home at 08 00 and walked to school at an average speed of 5 km/h. His sister, Shalini
left home 10 minutes later and cycled to school. They both reached school at 08 30. Calculate
Shalini’s average speed.
1
6. A car travelling at 80 km/h takes 1 2 h to travel a certain distance. The car returns by the same
route and decreases its speed by 20 km/h. How much more time will the car take for the
return journey?
7. The distance between Alpha Town and Beta Town is 312 km. John leaves Alpha Town at the
same time that Peter leaves Beta Town. They meet after 3 hours. If John drives 4 km/h faster
than Peter, find their respective speeds.
8. Humeira drives from Snowville to Summerhills. After 2 hours, she found out that she has
covered a distance of 60 km but that if she continued at the same speed, she would be
30 minutes late. Thus she increased her speed by 10 km/h. She reached Summerhills
10 minutes earlier than the due time. Find the distance between Snowville and Summerhills.
9. A car and a double-decker bus leave Townsville and travel in opposite directions along a
straight road. After having travelled for 4 hours, they were 500 km apart. The average speed
of the bus was 45 km/h. Find the average speed of the car.
161
Chapter 13 - Speed
Conversion of speed from one unit to
another (compound units)
It is important that the units are consistent when performing
calculations. If the distance is in kilometres and the time in
hours, then the speed will be in kilometres per hour, that is, km/h.
If the distance is in metres and the time in seconds, then the
speed will be in metres per second, that is, m/s.
STOP AND THINK
Pavel Kulizhnikov, a Russian, became
the fastest speed skater in the world
on 20th November 2015 in Salt Lake City.
His average speed was 16.45 m/s. The
wolf can run at an average speed of 55
km/h. Who is faster?
Sometimes we need to convert km/h to m/s or vice versa.
Converting m/s to km/h
Example
RECALL
Convert 10 m/s into km/h.
1 h = 3 600 s
1 km = 1 000 m
Solution
10 m
(10 × 3 600) m
( 10 × 3 600 ) km = 36 km
1 000
Therefore, 10 m/s = 36 km/h
In 1 s
In 3 600 s
In 1 h
STOP AND THINK
A car is driving along a road where the speed
limit is 60 km/h. The traffic police uses a
radar gun and measures the car to be 250 m
away. After 1.2 s, it measures the car to be
227.8 m away.
(i) Find the average speed of the car in m/s.
(ii) Is the car travelling above the speed
limit? Explain your answer.
Converting km/h to m/s
Example
Convert 72 km/h into m/s.
Solution
In 1 h
In 3 600 s
In 3 600 s
In 1 s
72 km
72 km
(72 × 1 000) m
( 72 × 1 000 ) m = 20 m
3 600
Therefore, 72 km/h = 20 m/s
Hint: Find the speed of the car in km/h.
EXERCISE 13.5
1. Convert the following into m/s.
(a) 3 km/h
(b) 120 km/h
(e) 10.8 km/h
(f) 7.2 km/h
162
(c) 5.76 km/h
1
(g) 13 km/h
2
(d) 54 km/h
(h) 198 km/h
Chapter 13 - Speed
2. Convert the following into km/h.
(a) 3 m/s
(e) 9 m/s
(b) 25 m/s
1
(f) 15 m/s
4
(c) 80 m/s
(d) 0.5 m/s
(g) 42 m/s
(h) 14.8 m/s
Example
Peter walks at an average speed of 6 km/h. What is his average speed in metres per second?
Solution
6 km
6 km
(6 × 1 000) m
5
( 6 × 1 000 ) m = m
3
3 600
5
Therefore, average speed = m/s
3
In 1 h
In 3 600 s
In 3 600 s
In 1 s
EXERCISE 13.6
1. A hedgehog runs at the top speed of 5 m/s. What is its speed in kilometres per hour?
2. An eagle flies at 54 km/h. Express this speed in m/s.
1
3. Ben can cover 18 2 km in 5 hours. Find his speed in (a) km/h (b) m/s.
1
4. A car covers a distance of 60 km in 1 2 hours whereas a bus covers 24 000 m in 45 minutes.
Find the ratio of their average speeds.
5. Hillside and Riverland are 375 km apart. Sarita leaves Hillside at 07 00 and travels at an average
speed of 80 km/h to Riverland. At the same time, Iqbal leaves Riverland for Hillside. They meet at 09 30.
(a) Find the distance travelled by Sarita when she meets Iqbal.
(b) Find the average speed in m/s of Iqbal before he meets with Sarita.
Summary
•
The speed of an object is the distance travelled by this object per unit of time.
•
We may calculate speed, distance and time as follows:
(i) Speed = Distance
Time
(ii) Distance = Speed × Time
(iii) Time = Distance
Speed
163
14
MASS
Chapter 14 - Mass
Learning
Objectives
By the end of this chapter, you should be able to:
• distinguish among different units of mass: mg, g, kg and tonnes (t).
• convert mass from one unit to another.
• perform arithmetic operations involving mass.
• solve word problems involving mass.
Mass of different objects in real life
CHECK THAT YOU CAN:
•
The mass of a 10 cm pigeon tail
feather is approximately 50 milligrams (mg).
Perform arithmetic operations
involving integers, fractions
and decimals.
A paperclip weighs
about 1 gram (g).
KEY TERMS
milligrams, grams,
kilograms, tonnes
A dictionary has a mass of
about 1 kilogram (kg).
Some cars weigh
about 2 tonnes (t).
DID YOU KNOW
Introduction
The measurement of mass of objects dates back to the
early times where measurements were necessary for trade.
During that time, the mass of an object was measured
using stones or seeds. As civilisations evolved, there was a
need for more accurate measurements of mass.
An object having a larger mass is measured in kilograms
(kg) and metric tonnes (t) whereas an object that has a
smaller mass is measured in milligrams (mg).
EXERCISE 14.1
Give an appropriate unit to measure the mass of the
following.
(a)
(b)
(c)
(d)
A full suitcase
164
The amount of matter in any given
object is referred to as its mass.
Each object has its own mass based
on the amount of matter it contains.
The weight of an object is the force
that gravity exerts on it.
The mass of an object remains the
same no matter where it is situated
whereas the weight of an object
varies with different locations
depending on the gravitational pull.
Yet, in everyday language, ‘weight’
is often used to refer to the mass of
an object and ‘to weigh an object’
usually means to determine its
mass.
Chapter 14 - Mass
Conversion of mass from one unit to another
1 kg
1 kg = 1 000 g
1g=
1000
1 t = 1 000 kg 1 kg = 1 t
1000
1 g = 1 000 mg
1 mg = 1 g
1000
× 1 000
t
× 1 000
kg
÷1 000
× 1 000
g
÷1 000
mg
÷1 000
The kilogram which is the base
unit of mass is equal to the mass of
the international prototype of the
kilogram.
This prototype is kept at the
‘Bureau International des Poids
et Mesures in Sèvres, France'. It is
made of the metals platinum and
iridium and is in the form of a small
cylinder. Measurements of mass
follow this prototype.
You can learn more about the
kilogram on the following
website:
http://www.bipm.org/en/bipm/
mass/ipk/
Example
Convert the following:
(a) 5 g to mg
(b) 7.7 kg to g
DID YOU KNOW
(c) 8750 kg to t
FIND OUT
Solution:
(a) 1 g = 1 000 mg
To convert g to mg, we multiply by 1 000
Therefore, 5 g = (5 × 1 000) mg = 5 000 mg
(b) 1 kg = 1 000 g
To convert kg to g, we multiply by 1 000
Therefore, 7.7 kg = (7.7 × 1 000) g = 7 700 g
Can you find out about the
different devices that can be used
to measure the mass of objects?
DID YOU KNOW
The mass of an ant is between 1 mg
and 5 mg.
(c) 1 t = 1 000 kg
To convert kg to t, we divide by 1 000
Therefore, 8 750 kg = (8 750 ÷ 1 000) t = 8.75 t
EXERCISE 14.2
1. Convert the following to milligrams.
(a) 25 g
(b) 51 g
(e) 120 kg
(f )
3
g
4
5
kg
8
1
(g) 3 g
2
(c)
(d) 4.75 g
(h) 12.345 kg
165
Chapter 14 - Mass
2. Convert the following to grams.
(a) 5 kg
(b) 10 mg
(f ) 0.06 kg
(g) 5 mg
3. Convert the following to kilograms.
(a) 6 t
(b) 4 750 g
1
(e) 0.005 t
(f ) t
4
4. Convert the following to tonnes.
(a) 5 000 kg
(b) 6 500 kg
(c) 3.29 kg
3
(h) kg
4
1
kg 8
(i) 1.25 mg
(d)
3
(c) 2 8 t
(g) 725 g
(d) 40 000 mg
(h) 3 500 000 mg
(c) 3 000 000 g
(d) 22 500 g
(e) 4 t
(j) 3.4 t
Arithmetic Operations involving mass
Example 1
Evaluate the following, giving your answer in g.
4 kg + 300 g
Solution
To add 4 kg and 300 g, we first convert the 4 kg into g.
4 kg = (4 x 1 000) g = 4 000 g
STOP AND THINK
Consider an astronaut in space.
Will he/she have a weight?
What about his/her mass?
Will it change in space?
4 kg + 300 g = 4 000 g + 300 g = 4 300 g
Example 2
Evaluate the following, giving your answer in kg.
5 t + 40 kg + 3 500 g
Solution
We first convert 5 t into kg and 3 500 g into kg
5t
= (5 × 1 000) kg = 5 000 kg
3 500 g = (3 500 ÷ 1 000) kg = 3.5 kg
5 t + 40 kg + 3 500 g = 5 000 kg + 40 kg + 3.5 kg
= 5 043.5 kg
166
NOTE TO TEACHER
If we take the example of an
astronaut in space, he/she does
not have a weight as there is no
gravitational force in space. However,
this does not mean that he/she does
not have a mass. The astronaut still
has the same mass even though he/
she is floating in space.
Chapter 14 - Mass
Example 3
Evaluate the following, giving your answer in g.
5 kg + 500 mg – 80 g
Solution
We first convert 5 kg and 500 mg into g. We then perform the operations.
5 kg = (5 × 1 000) g = 5 000 g
500 mg = (500 ÷ 1 000) g = 0.5 g
Perform operations starting from left to right, i.e, addition followed by subtraction.
5 kg + 500 mg – 80 g = 5 000 g + 0.5 g – 80 g
= 4 920.5 g
Example 4
Evaluate the following, giving your answer in mg.
0.875 kg + 300 g - 800 mg
Solution
We first convert 0.875 kg and 300 g into mg. We then perform the operations.
0.875 kg = (0.875 × 1 000 000) mg = 875 000 mg
300 g = (300 × 1 000 ) mg = 300 000 mg
Perform operations starting from left to right, i.e, addition followed by subtraction.
0.875 kg + 300 g – 800 mg = 875 000 mg + 300 000 mg – 800 mg
= 1 174 200 mg
EXERCISE 14.3
1. Evaluate the following:
(a) 12 g + 8 kg = ____ kg
(c) 22 kg + 25 g = ____ g
(e) 175 g + 0.5 kg = ____ g
(g) 530 g – 4 000 mg = ____ g
(b) 378 kg – 2 100 g = ____ kg
(d) 3 kg 350 g + 5 kg 230 g = ___ g
(f ) 3 t + 350 kg – 2.5 t = ____ kg
(h) 4.75 t – 2 300 kg = ____ t
2. Evaluate the following:
(a) 5 g 300 mg + 8 g 400 mg = __ g __ mg
(c) 15 g 200 mg – 8 g 550 mg = __ g __ mg
(b) 9 g 275 mg + 3 g 65 mg = __ g __ mg
(d) 4 g 40 mg – 2 g 240 mg = __ g __ mg
3. Evaluate the following:
(a) 4 kg 300 g + 18 kg 400 g = ____ kg _____ g
(c) 15 kg 900 g – 8 kg 550 g = ____ kg _____ g
(b) 78 kg 5 g + 5 kg 23 g = ____ kg _____ g
(d) 3 kg 60 g – 1 kg 780 g = ____ kg _____ g
167
Chapter 14 - Mass
Word problems involving mass
Example
The total mass of a basket containing 10 apples, each having the same mass, is 3 kg 200 g.
If the mass of the empty basket is 300 g, find the mass of one apple.
Solution
3 200 g
Mass of 10 apples + Mass of empty basket
= 3 kg 200 g
= (3 × 1 000) g + 200 g = 3 200 g
?
300 g
Mass of 10 apples
Mass of
empty
basket
Mass of empty basket = 300 g
Therefore, mass of 10 apples = 3 200 g – 300 g
= 2 900 g
Mass of 1 apple = (2 900 g ÷ 10) = 290 g
EXERCISE 14.4
1. Circle the correct answer:
(a) The mass of a container is 3 t 450 kg. The mass of 10 such containers is
A. 30 t 450 kg
B. 3 t 45 kg
C. 34 t 450 kg
D. 34 t 500 kg
(b) The mass of a box containing 5 identical books is 4 kg 850 g. What is the mass of each
book assuming that the mass of the box is negligible?
A. 0.097 kg
B. 9 700 mg
C. 970 g
D. 9700 g
(c) A box of chocolate bars has a mass of 1.6 kg. If one bar of chocolate has a mass of 25 g and
the box contains 60 such bars, what is the mass of the box when it is empty?
A. 1.5 kg
B. 3.1 kg
C. 1 kg
D. 0.1 kg
(d) The mass of Anu is 37.5 kg while that of Jack and Laila are 34.56 kg and 35.65 kg respectively.
What is their combined mass?
A. 107 710 g
B. 107 710 kg
C. 107.710 t
D. 1 070 000 mg
(e) To bake a 200 g cake, you need to use 70 g of butter. What is the mass of butter you would
need to bake a cake of mass 1 kg?
A. 14 000 g
B. 350 g
C. 70 g
D. 270 g
2. Robin’s eraser has a mass of 30 grams. Find the mass of 250 such erasers in kilograms.
168
Chapter 14 - Mass
3. The mass of a car is 2.5 t. If the mass of the driver is 85 kg and there are 3 passengers with a
combined mass of 155 kg in the car, calculate the mass of the car when it is full with the driver
and the 3 passengers.
1
4. A bag full of fruits contains 5 kg of apples, 4 kg of oranges and some grapes. Find the
2
mass of grapes if the total mass of fruits is 10 kg 200 g, assuming that the mass of the bag is
negligible.
5. A bag of rice has a mass of 2.5 kg. 50 such bags of rice are placed inside a big box. If the empty
box has a mass of 3 500 g, find the total mass of the box with the bags of rice.
6. Priya wants to buy a pumpkin of mass 1.8 kg at the market.
(a) What is the mass of the pumpkin in g?
(b) If 1 kg of pumpkin costs Rs 25, how much would she pay if she finally decides to buy only
half of the pumpkin?
7. A vegetable planter harvested 3 t of potatoes in his field. He then sold all the potatoes to 20
vegetable sellers at Rs 15 per kg. If each vegetable seller bought an equal amount of potatoes,
calculate the mass of potatoes in kg that each one bought and the amount of money each one
paid.
8. A lorry has a mass of 4 t when empty. It carries concrete blocks for a construction company.
The total mass of the lorry and the blocks is 8.5 t. If 1 concrete block has a mass of 5 kg, find
the number of concrete blocks on the lorry.
9. A man bought 5 kg of tomatoes, 6 kg of onions and 4.5 kg of potatoes. He placed these
vegetables inside a big bag which has a mass of 400 g when empty. He bought the bag at Rs 25,
one kg of potato at Rs 16 and one kg of onions at Rs 18.
(a) Calculate the total mass of the bag with the vegetables.
(b) If the man spent Rs 300 in all, find the cost of one kg of tomatoes.
Summary
•
•
We have different units of mass such as tonnes (t), kilograms (kg),
grams (g) and milligrams (mg) amongst others.
The mass of large objects are measured in t and kg whereas the mass
of small objects are measured in g and mg.
× 1 000
t
× 1 000
kg
÷1 000
× 1 000
g
÷1 000
mg
÷1 000
169
15
ALGEBRAIC EXPRESSIONS &
ALGEBRAIC EQUATIONS
Chapter 15 - Algebraic Expressions & Algebraic Equations
Learning Objectives
By the end of this chapter, you should be able to:
• use letters to represent unknown quantities.
• recognise algebraic terms, coefficients and expressions.
• perform addition and subtraction of algebraic expressions.
• expand and simplify algebraic expressions involving brackets of the form m(x + y) where m is a
whole number, fraction or decimal.
2xy2
• perform multiplication and division of simple algebraic expressions (e.g.
).
4xy
• substitute numbers in algebraic expressions.
• perform addition and subtraction on algebraic fractions with numerical denominators.
• distinguish between an algebraic expression and an algebraic equation.
• demonstrate an understanding of additive and multiplicative inverses.
• solve simple linear equations involving additive and multiplicative inverses (ax + b = c,
where a, b and c are whole numbers, fractions or decimals).
Algebra is a branch of mathematics dealing with symbols
and rules for manipulating operations (+, –, ×, ÷). Symbols,
such as alphabets, represent quantities with unknown
values, also known as variables.
CHECK THAT YOU CAN:
• Use laws of indices.
• Perform the four operations
(+, –, ×, ÷).
Consider the following:
KEY TERMS
I think of a number and subtract 2 from it. What is the answer?
If the number that I think of is represented by
statement can be written as – 2.
, then this
Similarly, if Anakin has a number of CDs and Leila gives him
5 CDs, then the total number of CDs that Anakin has is
+5
where
represents the number of CDs Anakin had at the
beginning.
EXERCISE 15.1
•
•
•
•
•
•
•
•
Symbol
Variable
Unknown quantity
Expression
Equation
Inverse operation
Additive inverse
Multiplicative inverse
Note: You can choose any symbol to
represent the unknown quantity.
Use an appropriate symbol to rewrite the following:
(a) I think of a number. What is the resulting number if I add 6 to it?
(b) I think of a number and multiply it by 3. What will be the resulting number?
(c) Luke has a number of kiwis in his lunch bag. He ate 4 kiwis. How many kiwis are left?
(d) Ken has a number of songs saved on his smartphone. He downloads 7 more songs.
How many songs are there in all on his phone?
170
Chapter 15 - Algebraic Expressions & Algebraic Equations
Using letters to denote unknown quantities
Instead of using
can use letters.
or
DID YOU KNOW
to represent the unknown quantity, we
History
The choice of the letter is not significant, that is, any letter can be
chosen but very often we use letters which help us to remember
the variables easily. For example, h is used to denote ‘height’, l to
represent ‘length’, a to represent ‘number of apples’ and so on.
Algebra comes from the
Arabic word, ‘Al Jabr’meaning
‘reduction’. It was first used in
the year 1551.
Let us now consider the statement:
I think of a number and divide it by 4.
What will be the resulting number?
The origins of algebra can be
traced back to the Babylonians,
ancient Egyptians and ancient
Greeks amongst other
civilisations.
We can use any letter to denote the unknown number, say s.
Therefore, the resulting number will be s ÷ 4 or
s
.
4
Can you find out more about the
origins of algebra?
Similarly, if Jade had 10 pencils and then lost some of them, then she has (10 – x)
pencils left where x represents the number of pencils lost.
Mathematical Expressions
When numbers or letters are written down together with other symbols used for operations
(+, –, ×, ÷), we have a mathematical expression.
Examples of mathematical expressions : 3 + 4 – 5
a × b ÷ c.
3, 4 and 5 are the terms of the expression '3 + 4 – 5'
+ and – are the symbols used as operations.
Similarly, a, b and c are the terms of the expression 'a × b ÷ c'
x and ÷ are the symbols used as operations.
Conventions when using letters
1. When a letter is multiplied by an unknown quantity, we omit the multiplication sign as far as
possible, for example, we write 2 × a as 2a. We usually write the number before the variable,
that is 2a and not a2.
2. When two or more variables are used in an algebraic product, we write them in alphabetical
order, for example, m × l = lm .
3. We usually omit writing 1 before a variable, for example 1 × c = c × 1 = c and not 1c .
171
Chapter 15 - Algebraic Expressions & Algebraic Equations
EXERCISE 15.2
1. Rewrite the following statements using letters:
(a) I think of a number and multiply it by 4. What is the resulting number?
(b) I think of a number and square the number. What is the answer?
(c) I think of a number and then add it to 7. What is the answer?
(d) I have 6 pencils and give a few to my sister. How many pencils do I have left?
(e) Mother gave me some grapes to eat. After eating 10, what will be the number of grapes left?
2. Write down a mathematical expression for each of the following:
(a) Sum of a and 18
(b) 5 more than b
(c) Subtract c from 14
(d) Subtract 12 from d
(e) Product of e and f
(f) g times 10
(g) Multiply h by 2, then add 8
(h) The sum of j and k is divided by 6
Algebraic expressions
An algebraic expression contains numbers, variables (represented by letters) and the symbols
used for operations (–, +, ×, ÷) that are put together to represent the value of something.
Examples of algebraic expressions are: 2x – 1, 3y, 4a – 5b + 8, x2.
Terms, Coefficients and Constant Term
NOTE TO TEACHER
A term consists of a coefficient and a variable.The + and
– signs in an algebraic expression separate it into terms.
Encourage students to write the
x (multiplication sign) and the x
(variable) differently.
Example: 4a – 3x consists of 2 terms: 4a and 3x.
The number that is placed in front of a term is called the coefficient of that variable.
Example: In 3x, 3 is the coefficient and x is the variable.
A term containing no variable is called a constant term.
Example: In 8a + 5, 5 is the constant term.
Can you write an algebraic expression involving 3 terms?
CHECK THIS LINK
https://www.mathgames.
com/skill/6.6-identify-termscoefficients-and-monomials
EXERCISE 14.3
For each of the following expressions, complete the sentence found on the right:
(a) 9m
The coefficient of m is _______ .
(b) 7x
The coefficient of x is _______ .
(c) –2z
The coefficient of z is _______.
2
(d) 4x The coefficient of x2 is _______.
172
Chapter 15 - Algebraic Expressions & Algebraic Equations
(e) –5ab
The coefficient of ab is is _______.
(f) 6c – 2d
The coefficient of _______ is 6 and the coefficient of d is _______.
(g) 9m + 7n – 3y
The coefficient of m is _______, the coefficient of n is _______ and the
coefficient of y is _______ .
(h) 6e + 3f – 4
The coefficient of f is _______, the coefficient of _______ is 6 and the
constant term is _______.
(i) 3x2 – 2x + 7
The coefficient of x is _______, the coefficient of x2 is _______ and the
_______ is 7.
Example
Chen buys 3 apples and 4 oranges. If the cost of one apple is Rs x and the cost of one orange
is Rs y, write down an expression for the total amount he will pay.
Solution
Cost of 3 apples = 3 × (cost of 1 apple) = Rs (3 × x) = Rs 3x
Cost of 4 oranges = 4 × (cost of 1 orange) = Rs (4 × y) = Rs 4y
Total amount to be paid = Rs (3x + 4y)
EXERCISE 15.4
1. Write down an expression for each of the following:
(a) Vidoushi bought 10 balls from Super Deal supermarket at Rs x each. How much did she pay
for the balls?
(b) Ishika has Rs 500 less than Mayumi. If Mayumi has Rs z, find the amount of money Ishika has.
(c) Ouma is twice as old as Katherine. If Katherine is x years old, what is Ouma’s age in terms
of x ?
(d) Find the average of x, y and z.
(e) The length of a rectangle is l cm and its width is w cm. Find the perimeter of the rectangle.
(f) Akshay went to buy vegetables. He bought 5 kg of potatoes at Rs p per kg, 2.5 kg of
tomatoes at Rs q per kg and 1 kg of onions at Rs r per kg. Write down an expression for the
total amount of money he spent.
Like and unlike terms
We call 3a, 5a and 2a like terms as they have exactly the same variable. Similarly, 3a2b and 5a2b
are like terms as they have exactly the same variable combination (a2b) and a is raised to the
same power in both terms.
However, 4x3y and 8xy3 are unlike terms as in the first term
(4x3y), x is to the power of 3 and y is to the power of 1 while
in the second term (8xy3), x is to the power of 1 and y is to
the power of 3.
STOP AND THINK
Are 5ab and 3ba like terms? Explain.
173
Chapter 15 - Algebraic Expressions & Algebraic Equations
Example
Note: In terms like xy, x, y or x2, the
coefficient of each term is 1.
In terms like –x, –z, –y3 the coefficient
of each term is -1.
Identify the like terms in the following expression:
3x + 5y + 3xy – 7y + 8x – 10xy
Solution
3x + 5y + 3xy – 7y + 8x – 10xy
Caution:
2x and x2 are unlike terms:
2x = 2 times x and x2 = x times x.
2
x y and xy2 are unlike terms:
x2y = x times x times y and
xy2 = x times y times y.
(terms in x)
(terms in y)
(terms in xy)
The like terms are (i) 3x and 8x
(ii) 5y and –7y
(iii) 3xy and –10xy
EXERCISE 15.5
Identify the like terms in each of the given list of terms:
(a) 2x, 4y, –2z, –5x(b) 3x, 5xy, x2, –7xy, x2y, 9xy, xy2
(c) 3pqr, 6pr, –4pq, 10pqr, –6pqs
(d) 4x, 7y, –3, 9xy, –12yz, –8, 2xz, 19
(e) 1 ab, 3 ac, 1 ab, – 2 bc, 4ab
2
3
5
4
(f) x2y, 3xy2, 4xz2, 5x2y
Simplification of expressions containing like terms of one type
Expressions involving like terms can be simplified, that is, like terms can be added or subtracted
by collecting all the like terms together.
Example
Simplify the following: (a) 2x + 3x
(b) 4m + 6m – 3m
(c) 1 xy + 2 xy – 1 xy
4
3
2
Solution
If an expression contains like terms, it can be simplified as follows:
(a) 2x + 3x = 5x[Collect all the coefficients of x together]
[2 + 3 = 5]
(b) 4m + 6m – 3m = 7m[Collect all the coefficients of m together]
[4 + 6 – 3 = 7]
(c) 1 xy + 2 xy – 1 xy = ( 1 + 2 – 1 )xy
4
3
2
4
3
2
174
= 5 xy
12
[Collect all the coefficients of xy together]
[
1 + 2 – 1 = 5 ]
4
3 2
12
Chapter 15 - Algebraic Expressions & Algebraic Equations
EXERCISE 15.6
Simplify the following:
(a) 3x + 5x
(b) 2xy + 7xy
(c) 9x2 + 11x2
(d) 4xyz + 6xyz
2
1
3
(g) 2f – f
(h) x + x
3
2
4
(k) ab2 + 9ab2 + 21ab2
(e) 13a – 5a
(f) 14ab – 7ab
3
1
(i)
e – e
(j) 19x – 12x – 6x
5
6
(l) 5 abc + 10 abc + 6 abc(m) –5ab + 8ab – 3ab
(n) – 3xy – 5xy + 11xy
(o) 13x2 – 19x2 + 31x2 – 8x2
Simplification of expressions containing like terms of more than
one type
To simplify the expression 2a + 3b + 4c + 5a + 2c + 4b + a which contains 3 types of like
terms involving a, b and c, we group like terms of the same type together.
Example
Simplify (a) 2a + 3b + 5a + 4b + a
(c) 3 a + 1 c – 1 a – 1 c
4
3
2
5
(b) 4a + 3b – 5c + 2b – 3a + 3c
Solution
(a) 2a + 3b + 5a + 4b + a = 2a + 5a + a + 3b + 4b
= 8a + 7b
Note: We add all the terms
involving a together, then terms
involving b together.
(b) 4a + 3b – 5c + 2b – 3a + 3c = 4a – 3a + 3b + 2b – 5c + 3c
= a + 5b – 2c
(c) 3 a + 1 c – 1 a – 1 c = 3 a – 1 a + 1 c – 1 c
4
3
2
5
2
3
5
4
= 1 a+ 2 c
4
15
EXERCISE 15.7
Simplify.
(a) 12x + 3y + 8x + 9
(b) 16ab + 13 + ab + 21
(c) 8d – 3e + 5d – 7e(d) 39x2y + 21xy2 + 5xy2 – 17x2y
3
1
1
(f) s –
(e) 3abc + 5ac – 7abc – 8ac
t+s+ t
7
8
2
(g) 23st + 8uv – 41st – 14uv(h) 28xy – 36yz – 10xy + 9yz
(j) 4.9x – 2.5y – 1.6x - 3.7x
(i) 45x – 34 – 27x + 16
175
Chapter 15 - Algebraic Expressions & Algebraic Equations
Simplifi ation of expressions involving multiplication
We already know that:
5 x 5 = 52
5 x 5 x 5 = 53
5 x 5 x 5 x 5 = 54
Similarly,
a x a = a2
a x a x a = a3
a x a x a x a = a4 , and so on.
Example 1
Simplify (a) 4 x a x a
(b) 3 x 5 x b x b x b
Solution
(a) 4 x a x a = 4 x a2 = 4a2
(b) 3 x 5 x b x b x b = 15 x b3 = 15b3
a2
b3
Example 2
Simplify the following:
(a) 3u × 5u
(b) 6x2 × 4x3 (c) 2x2y × –8xy2
Solution
(a) 3u × 5u
Method 1
3u × 5u = 3 × u × 5 × u [We write the expression in expanded form first]
= (3 × 5) × (u × u)
[We rearrange the coefficients together and
2
the variables together]
= 15 × u = 15 u2[We simplify]
Method 2
3u × 5u = (3 × 5)u1+1 = 15u2
(b) 6x2 × 4x3
Method 1
6x2 × 4x3 = 6 × x × x × 4 × x × x × x
= (6 × 4) × (x × x × x × x × x)
= 24x5
Method 2
6x2 × 4x3 = (6 × 4)x2+3 = 24x5 [Using multiplication law of indices]
[We proceed as in part (a)]
[Using multiplication law of indices]
(c) 2x2y × –8xy2
Method 1
2x2y × –8xy2 = 2 × x × x × y × –8 × x × y × y
= (2 × –8) × (x × x × x) × (y × y × y)
= –16x3y3
Method 2
2x2y × –8xy2 = (2 × –8) x2+1y1+2 = –16x3y3
[Using multiplication law of indices]
176
Chapter 15 - Algebraic Expressions & Algebraic Equations
EXERCISE 15.8
1. Simplify the following:
(a) 12x × 3
(b) 4a × 5a
(c) 6a2 × 2a
(e) –5s3 × 4s2
(f) –17t8 × -9t10
(g) xy2 × 8x5y7
(d) 3z9 × 12z6
1
2
(h) x4y × xy5
2
5
(l) 4x9y2 × 3x3y5 × x10
(i) 7x3y × 8x2y7 × 2y15 (j) 13a2b4 c × 12a5b3c (k) 15u8v6 × 6u4v12
(o) 6m2 n3p × –4 mn2p3
(m) 3a8b3 × 4 a12b4 (n) 10a4b3c9 × 12a2b12c6
9
Simplifi ation of expressions involving division
To divide like and unlike terms, we rewrite the expression in expanded form and cross out the
common factors as in the examples below.
Example
Simplify the following:
(a) 12m ÷ 3
(b) 2c
8c
(c) 28x2y ÷ 7x
(d)
4m3n2
12m2n
Solution
4
(a) 12m ÷ 3 = 12 x m = 4 × m = 4m
31
Method 1: Expanded form
1
1
1
2c
2xc
(b) 8c =4 8 x c1 = 4 4
1
(c) 28x2y ÷ 7x = 28 × x × x × y = 4xy
1 7 × x1
1
1
1
1
(d) 4m3n2 = 4 × m × m × m × n × n = m × n = mn
12m2n
12 × m × m × n
3
3
3
1
1
1
Method 2: Division law of indices
2c = 2 1 - 1 = 1 0 1
c
c =
(Note: c0 = 1)
8c
8
4
4
28x2y ÷ 7x = 28 x2 - 1y = 4xy
7
1
4m3n2
4 m3-2n2-1 mn
=
=
12m2n 123
3
EXERCISE 15.9
Simplify the following:
(a) 2x10 ÷ x4(b) 9y7 ÷ 3y5
(c) s6t8 ÷ s5t2
(e) 24a12b4 c7 ÷ 21a11b2c5
(f) 32x2y8
(g) 3 u15v9
x5y4
9 u6v9
(i) 48x13y5z9 × 2x3y2 (j) 112a8b2c7 × a3b9 c5
28x11y4z6
100a9b10 c12
(d) 4v9w ÷ 12v3w
(h) a10b4
3 a 6 b3
177
Chapter 15 - Algebraic Expressions & Algebraic Equations
Simplifi ation of expressions involving brackets
It can be checked that
(a) 3(2 + 5) = (3 x 2) + (3 x 5)
(b) 4(5 – 3) = (4 x 5) – (4 x 3)
(c) –2(7 + 4) = (–2 x 7) + (–2 x 4)
More generally, a (b + c) = ab + ac and a (b – c) = ab – ac.
We use the distributive law to simplify expressions involving brackets, as illustrated in the
following examples.
Example
Expand and simplify the following:
(a) 9 (x + y)
(b) 5 (x + 6)
(c) 2 (s – 12)
(d) 3 (t – 5)
Solution
RECALL
(a) 9 (x + y) = (9 × x) + (9 × y)(b) 5 (x + 6) = (5 × x) + (5 × 6)
= 5x + 30
= 9x + 9y
(c) 2 (s – 12) = (2 × s) – (2 × 12)
(d) 3 (t – 5) = (3 × t) – (3 × 5)
= 3t – 15
= 2s – 24
A positive number
multiplied by a
negative number gives
a negative number.
EXERCISE 15.10
Expand and simplify.
(a) 3 (2x + 10y)
(b) 4 (2x + 8)
(c) 20 (7x + 3z)
(d) 9 (3x + 11)
(e) 7 (u – 23)
(f) 14 (2x – 3y)
(g) 20 (3a – 7b)
(h) 8 (2m – 3n)
Example
Expand and simplify.
(a) –4 (a + 8)
(b) –5 (8x – 2y)
Solution
(a) –4 (a + 8) = (–4 × a) + (–4 × 8)
= –4a – 32
(d) –5 (8x – 2y) = (–5 × 8x) – (–5 × 2y)
= –40x + 10y
RECALL
1. A negative number
multiplied by a positive
number gives a negative
number.
2. A negative number
multiplied by a negative
number gives a positive
number.
EXERCISE 15.11
Expand and simplify.
(a) –7 (d + 9)
(b) –2 (3x + 2y)
(e) –5 (10e + 9f )
(f) –17 (3u + 6v)
(i) –18 (12a – 6b)
(j) –5 (2ab – 8cd)
178
(c) –3 (2ab + 4cd) (d) –6 (16x + y)
(g) –12 (8uv – 10uw) (h) –(7x – 8y)
(k) –9 (4xy – 7yz) (l) –2 (4ab2 – 11a2b)
Chapter 15 - Algebraic Expressions & Algebraic Equations
Example
Expand and simplify the following:
(a) 3u (4u – 7)
(b) –2x (3x – 4y)
Solution
(a) 3u(4u – 7) = (3u × 4u) – (3u × 7)
= 12u2 – 21u
(b) –2x (3x – 4y) = (–2x × 3x) – (–2x × 4y)
= –6x2 + 8xy
EXERCISE 15.12
Expand and simplify.
(a) x (x + 9)
(b) a (2a + b)
(e) –5a (6b – 9c)
(f) –7xy (4x + 8y)
(c) 2xy (3x + 6y)
(g) –2m (3m – 4n)
(d) –v (2v – 12)
(h) –6d (–2e – 3f )
Example
Simplify
(a) 3 (2a + 4b) + 4 (3a + 2b)
(b) 3 (3a – 5b) – 4 (2a – 3b)
(c) –3 (2a – 3b) + 4 (–a + 2b)
Solution
(a) 3 (2a + 4b) + 4 (3a + 2b) = 6a + 12b + 12a + 8b
= 6a + 12a + 12b + 8b
= 18a + 20b
(b) 3 (3a – 5b) – 4 (2a – 3b) = 9a – 15b – 8a + 12b
= 9a – 8a – 15b + 12b
= a – 3b
Caution:
Be careful when removing brackets
after a negative sign.
(c) –3 (2a – 3b) + 4 (–a + 2b) = –6a + 9b – 4a + 8b
= –6a – 4a + 9b + 8b
= –10a + 17b
EXERCISE 15.13
Expand and simplify.
(a) 6 (x + y) + 4 (x – y)
(d) 10 (s – 3t) + 12 (2s + 5t)
(g) 12 (x – 7) – 18 (2x – 9)
(b) 2 (u + v) + 9 (2u + v)
(e) 4 (a + 9) + 6 (a – 10)
(h) 31 (c + 4d) – 24 (2c – 5d) (c) 10 (s + 3t) + 12 (2s – 5t)
(f) 17 (2a + 3b) – 20 (4a + b)
(i) 15 (y – 6) – 2 (y – 14)
179
Chapter 15 - Algebraic Expressions & Algebraic Equations
Substituting values in expressions
Example
Given that a = 3, b = –1 and c = 7, find the value of:
(a) abc
(b) 2a – 3b
(c) ab + 4c
(d) b2 + 5ac
(e) 7bc – 2ab
Solution
(a) abc = (3) (–1) (7) (b) 2a – 3b = 2(3) – 3(–1)
= –21
= 6 – (–3)
=6+3
=9
(c) ab + 4c = (3)(–1) + 4(7)
= –3 + 28
= 25
(d) b2 + 5ac = (–1)2 + 5(3)(7) (e) 7bc – 2ab = 7(–1)(7) – 2(3)(–1)
= 1 + 105 = –49 + 6
= 106
= –43
EXERCISE 15.14
If a = 2 , b = 3 and c = –1, find the numerical value of each of the following expressions:
(b) 4c – b
(c) 2a + 3b – 5c
(d) ac2
(a) 3a + b
(f ) –(ab2)
(g) 3a2bc
(h) 3a(bc)2
(e) (ac)2
(i) abc3(j) a(bc)3 (k) a2 + b2 (l) abc + c3
(m) 3ab – 4c
(n) 2ab + 3bc2
(o) 4ab2 – 6abc
(p) 2a(3b – 4c)
Equations
An equation is a mathematical sentence/statement which indicates that both sides of the '=' sign
have the same value.
Examples of equations are
(a) 3 + 7 – 2 = 5 + 3
(b) 5 × 3 + 4 = 7 + 6 × 2
An equation has two sides, the left-hand side (L.H.S.) and the right hand side (R.H.S.).
For example
3×5 = 7+8
L.H.S
Equals
R.H.S
We can think of an equation as being a scale which should balance on both sides (L.H.S. = R.H.S.).
An equation may have any number of variables. In Grade 7, we will learn about equations
involving one variable only.
180
Chapter 15 - Algebraic Expressions & Algebraic Equations
Fig. 1 and Fig. 2 represent equations 1 = 1 and 8 = 3 + 5.
1 kg
1 kg 1 kg
+
1 kg
1 kg 1 kg 1 kg
1 kg
1 kg
1 kg 1 kg
1 kg 1 kg 1 kg 1 kg
Fig. 1
1 kg 1 kg 1 kg
Fig. 2
x
3
Fig. 3
In the same way, we can have an equation involving a variable as shown in Fig. 3. This is called an
algebraic equation (x = 3).
Examples of algebraic equations: 2a = 8, x – 3 = 5,
y
= 7.
2
What is the difference between an expression and an equation?
An expression can be simplified and evaluated whereas an equation can be solved.
When we are asked to solve an equation, it means that we need to find the value of the variable/
unknown so that the equation becomes a true sentence.
Suppose that we are asked to solve 2a = 8. This means that we need to find the value of a so that
when we multiply this value by 2, the answer is 8. This value of a (i.e., a = 4) which will make the
statement correct is called the solution of the equation.
Forming an equation
Equations are helpful in solving word problems. Consider this problem:
I think of a number, subtract 6 from it and the result is equal to 10.
To form an equation, we first denote our unknown number by a.
The problem can then be written mathematically as: a – 6 = 10.
Note: You can use any
other letter of your
choice.
Similarly, if I have the following problem:
When I add 10 to a number, the result is the same as twice the number. Find the number.
How do we form this equation?
Let the number be x.
The problem can be written mathematically as x + 10 = 2x.
x, the unknown, is to be found in this equation.
181
Chapter 15 - Algebraic Expressions & Algebraic Equations
EXERCISE 15.15
1. State whether each of the following is an equation or an expression:
(b) 3(4 + 2x)
(c) x(x + 5)
(d) 4 × a + a – 18
(a) x + 10 = 12
(f) 5 – 3(2x + 1)
(g) –2m + 4 = 8
(h) 4u + 5 = 2u – 7
(e) 14 – y = 8 + y
2. For each of the statements below, write an equation in x. The first one has been done for you.
x + 7 = 15
(a) A number added to 7 is equal to 15 (b) 18 subtracted from a number is equal to 21
(c) Twice a number is equal to 16
(d) A number divided by 5 is equal to 60
(e) Thrice a number added to 8 is equal to 29
Solving algebraic equations
Consider the equation
x + 6 = 11
What value of x will make the equation true?
In other words, we need to find the value of x to make this statement true.
On one side, we have x + 6 and on the other side we have 11.
+ 6 has to be eliminated from the L.H.S. To do so, we have to remove (subtract) 6 on both sides
for the scale to be balanced as illustrated in the diagram.
Cancelled out
x + 6 = 11
+6– 6=0
x + 6 – 6 = 11 – 6
x = 5
x+6
11
x+6–6
subtract 6
11– 6
x
5
subtract 6
The value of x that will make the equation true is 5.
We say that the solution to the equation is x = 5.
Additive inverse
The additive inverse is the number that needs to be added so that the sum of the two numbers is zero.
For example, the additive inverse of 2 is –2 since 2 + (–2) = 2 – 2 = 0 .
and the additive inverse of –1.5 is 1.5 since (–1.5) + 1.5 = 0 .
Similarly, if we have 3x = 12, we need to find the value of x to balance the equation on both sides.
If we multiply by 1 on both sides, we have
3
Note: An important step after obtaining the value of x is to
1 × 3x = 12 × 1
check if this value is correct. This is done by substituting the
3
3
value obtained into the original equation.
x=4
For example, we replace
x = 4 into 3x = 12, that is, 3 × 4 = 12 , that is, L.H.S. = R.H.S.
182
Chapter 15 - Algebraic Expressions & Algebraic Equations
Multiplicative inverse
The multiplicative inverse is the number that needs to be multiplied so that the product of the
two numbers is 1.
1
1
For example, the multiplicative inverse of 3 is 3 since 3 x 3 = 1 and
the multiplicative inverse of – 1 is –6 since – 1 x –6 = 1.
6
6
EXERCISE 15.16
1. Write down the additive inverse of each of the following:
3
3
7
(a) 4
(b)
(c) 1
(d) –3
(e) –
7
11
4
(f) –2.1 (g) 0.75
(f) 6.5 (g) –8.3
2. Write down the multiplicative inverse of each of the following:
1
3
(c)
(d) –
(e) – 4
4
13
9
3. Is 1 the additive inverse of –1 1 ? Explain your answer.
3
3
3
5
4. Is 1 the multiplicative inverse of
? Explain your answer.
5
3
(a) 5
(b) –9
Solving equations involving the additive inverse
Example
Solve each of the following equations:
(a) x + 5 = 1
(b) y – 8 = 17
(c) a + 3 = –2
(d) m – 6 = –9
Solution
(a) x + 5 = 1
x + 5 + (– 5) = 1 + (– 5)
x+5–5=1–5
x = –4
(b) y – 8 = 17
y – 8 + 8 = 17 + 8
y = 25
(c) a + 3 = –2
a + 3 + (–3) = –2 + (-3)
a + 3 – 3 = –2 – 3
a = –5
(d) m – 6 = –9
m – 6 + 6 = –9 + 6 m = –3
(Add the additive inverse, i.e., ‘- 5’ to both sides to eliminate ‘5’)
Check:
-4 + 5 = 1
(Add the additive inverse, i.e., ‘8’ to both sides to eliminate ‘-8’)
Check:
25 - 8 = 17
(Add the additive inverse, i.e., ‘- 3’ to both sides to eliminate ‘3’)
Check:
-5 + 3 = -2
(Add the additive inverse, i.e., ‘6’ to both sides to eliminate ‘-6’)
Check:
-3 - 6 = -9
183
Chapter 15 - Algebraic Expressions & Algebraic Equations
EXERCISE 15.17
1. Solve the following equations.
(a) x + 5 = 9
(b) z + 2 = 18
(e) z + 20 = 20
(f) 10 + t = 32
1
(i) c + = 4
(j) s + 0.9 = 1.6
2
2. Solve the following equations.
(a) x – 6 = 15
(b) z – 24 = 19
(e) w – 1 = 0
(f) t – 10 = –4
(c) y + 7 = –3
(g) a + 12 = –19
(k) e + 1 = 7
4
(d) m + 13 = –14
(h) b + 2.8 = 5
(l) v + 2 = 5 1
3
3
(c) m – 14 = 18
(g) a – 1 = 3
8
(d) u – 9 = –7
(h) b – 2.3 = –3.9
Solving equations involving the multiplicative inverse
Example
Solve the following equations:
(a) 4a = –16
(b) b = 12
5
(c) –2x = 8
(d) – m = 3
4
Solution
(a) 4a = –16
4 × a = –16
1 × 4 × a = 1 × –16
4
4
1 × a = –4
a = –4
b
(b) 5 = 12
b × 5 = 12 × 5
5
b = 60
(c) –2x = 8
– 1 × –2x = – 1 × 8
2
2
x = –4
m
(d) – = 3
4
m
–4 × = –4 × 3 4
m = –12
(multiply both sides by the multiplicative inverse, i.e., ‘ 1 ' to eliminate ‘4’)
4
Check:
4 × -4 = -16
(multiply both sides by the multiplicative inverse, i.e., ‘5’ to eliminate ‘ 1 ’)
5
Check:
60 = 12
5
(multiply both sides by the multiplicative inverse, i.e., ‘- 1 ’ to eliminate ‘-2’)
2
Check:
-2 × -4 = 8
(multiply both sides by the multiplicative inverse, i.e., ‘-4’ to eliminate ‘- 1 ’)
4
Check:
-12 × - 1 = 3
4
EXERCISE 15.18
1. Solve the following equations.
(a) 3x = 24
(b) 5m = 100
(e) –8a = 16
(f) –9r = –36
184
(c) 19y = 38
(g) 0.5b = 2.05
(d) 2z = –5
(h) –0.2u = 1.8
Chapter 15 - Algebraic Expressions & Algebraic Equations
2. Solve the following equations.
x
(a) 3 = 12
(b) – z = 9
6
m
a
(e) – = 10
(f) –
= –10
2
10
y
= –5
4
(g) – a = –8
7
(c)
y
(d) 13 = –1
(h) b = –4
0.3
Solving equations involving more than one operation
Example
Solve the following equations:
x
(a) 2x – 6 = 14
(b)
+ 5 = –7(c) –4s – 9 = 11
3
Solution
(a) 2x – 6 = 14
2x – 6 + 6 = 14 + 6 (add ‘6’ to both sides)
2x = 20
1
1
2x ×
= 20 × (multiply both sides by ’ 1 ’ )
2
2
2
x = 10
x
(b) 3 + 5 = –7
x + 5 – 5 = –7 – 5 (add ‘–5’ to both sides)
3
x = –12
3
x × 3 = –12 × 3 (multiply both sides by ‘3’)
3
x = –36
(c) –4s – 9 = 11
–4s – 9 + 9 = 11 + 9
(add ‘9’ to both sides)
–4s = 20
1
– × –4s = – 1 × 20 (multiply both sides by ‘– 1 ’)
4
4
4
s = –5
EXERCISE 15.19
1. Solve the following equations.
(a) 3x + 6 = 13
(b) –2x + 7 = 1
(c) 4y – 15 = 10
(d) 4 – a = 11
(e) –7y – 24 = –3
z
(g) + 9 = –2
4
(i) 6z + 7 = –42
(f) –3z – 12 = –17
a
(h) – 1 = 18
3
b
(j) 7 –
= –3
5
(l) –4b – 8 = –2
(k) –7a – 12 = –21
Check:
2 × 10 – 6 = 20 - 6 = 14
Check:
-36 = -12
3
-12 + 5 = -7
Check:
-4 × -5 – 9 = 20 - 9 = 11
GeoGebra
Solve the equation 3x – 5 = 10 using Geogebra.
1. Click on View and select CAS.
2. Enter the equation 3x – 5 = 10 in the bar.
3. Click on the icon
to solve the equation and obtain the answer x = 5.
185
Chapter 15 - Algebraic Expressions & Algebraic Equations
Solving equations involving brackets (in the form m(x + a) = b)
Example
Solve the following equations:
(a) 5(a + 7) = 15
(b) –3(m – 2) = 9
Solution
(a) 5(a + 7) = 15
Method 1
(a) 5(a + 7) = 15
5a + 35 = 15 [Using distributive law to
5a + 35 – 35 = 15 – 35 expand the brackets]
5a = –20
1 × 5a = 1 × –20
5
5
a = –4
(b) –3(m – 2) = 9
Method 1
–3m + 6 = 9
–3m + 6 – 6 = 9 – 6
–3m = 3
1
– × –3m = – 1 × 3
3
3
m = –1
Method 2
5 (a + 7) = 15
1 × 5(a + 7) = 1 × 15
5
5
(a + 7) = 3
a+7–7=3–7
a = –4
Check:
5(-4 + 7) = 5 x 3 = 15
Method 2
– 1 × –3(m – 2) = – 1 × 9
3
3
m – 2 = –3
m – 2 + 2 = –3 + 2
m = –1
Check:
-3(-1 - 2) = -3 x -3 = 9
EXERCISE 15.20
Solve the following equations:
(a) 3(x + 2) = 12
(b) 7(x – 5) = 35
(c) 3(z + 7) = –3
(d) 2(x – 2) = 10
(e) –5(x – 1) = 5
(f) 2(–x + 3) = –12
(g) 12 = 2(x + 3)
(h) 3(t – 2) = 0.42
Summary
•
•
•
•
•
•
186
We use letters to represent unknown quantities (e.g. I think of a number and add 4 can
be written as a + 4).
An algebraic expression consists of terms. E.g. 4x + 2y consists of two terms.
The coefficient is the number placed in front of a variable, e.g., 6 is the coefficient of z in
6z + 3 and 3 is the constant term.
To add or subtract algebraic terms, the terms should be like terms. E.g. 2a + 4a = 6a and
3a – 5 ≠ –2a.
We use the distributive law to expand m (a + b) = ma + mb.
An expression can be simplified while an equation can be solved.
16
PATTERNS AND SEQUENCES
Chapter 16 - Patterns and Sequences
Learning
Objectives
By the end of this chapter, you should be able to:
• identify and complete number patterns.
Patterns in real life
CHECK THAT YOU CAN:
Pencil Case
Basket
• Identify the different types of
numbers: odd, even, prime, square
and triangular.
• Work with integers.
• Work with fractions and decimals.
Beanie hat
KEY TERMS
Porcelain Set
Mat
Curtain
Introduction
• Number Pattern
• Sequence
Mathematics is the study of patterns and relationships. In our everyday life, we come across
patterns which may consist of a repeated arrangement of shapes, colours and/or numbers.
When solving mathematical problems, one strategy that can be used is to ‘look for a pattern’. A
pattern in mathematics involves any predictable regularity.
We can find different number patterns in our everyday life such as in calendars and in number charts.
Number patterns can occur in different forms. For instance, we can have repeating patterns,
growing patterns and shrinking number patterns.
In a repeating number pattern, the numbers repeat themselves in a regular way.
Examples of repeating number patterns
(a) 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, …
(b) –1, 5, 6, –1, 5, 6, –1, 5, 6, ….
Can you give another example of a repeating number pattern?
187
Chapter 16 - Patterns and Sequences
In a growing number pattern, the numbers increase in an orderly and predictable manner.
Examples of growing number patterns
+5
+5
ACTIVITY 1
+5 +5
(a) 5, 10, 15, 20, 25, … pattern of multiples of 5
62
72
82
92
(b) 36, 49, 64, 81, … pattern of square numbers
+1
3
(c)
+1
3
+1
3
1 2
1
, , 1, 1 , … (number patterns involving fractions)
3 3
3
Can you give another example of a growing number pattern?
In a shrinking number pattern, the numbers decrease in
an orderly and predictable manner.
Examples of shrinking number patterns
-1
-1
-1
-1
(a) -5, -6, -7, -8, -9, …
-0.2 -0.2 -0.2
-0.2
A group of students decide to play
a game. They cut pieces of paper of
the same size and write A, B, C, …
on them. They write A on one piece
of paper, B on two pieces of paper, C
on three pieces and so on.
A
BB
C C
C
(i) On how many pieces of paper will
they write the letter F?
(ii) They wrote a letter on 20 pieces
of paper. What is this letter?
(iii) How many pieces of paper have
they already used by the time
they start to write the letter H?
(b) 5.0, 4.8, 4.6, 4.4, 4.2, ….
Can you give another example of a shrinking number pattern?
EXERCISE 16.1
1. Complete the following:
(a) 2, 5, 8, _____ , 14, _____ , _____ .
(c) 64, _____ , 72, 76, 80, _____ .
(e) _____ , _____ , 225, 250, 275, 300 .
(g) 39 000, 40 000, _____ , _____, 43 000 .
(i)10, 20, 40, 70, _____, _____ .
(b) 3, 6, 9, _____ , _____ , 18 .
(d) 49, 64, 81, 100, _____ , _____ .
(f ) 100, 200, 300, _____ , _____, 600 .
(h) 2 000, _____ , 3 000, 3 500, 4 000, _____ .
(j) –1, –2, –3, –4, _____, _____ .
2. Find the next two terms in each of the following:
(a) 6, 12, 18, 24, _____ , _____ .
(b) 60, 72, 84, 96, _____ , _____ .
(c) 5, 25, 125, 625, _____ , _____ .
(d) 9, 27, 81, 243, _____ , _____ .
3. Find the two missing terms in each of the following:
(a) 1 500, 1 400, 1 300, 1 200, _____ , _____ .
(b) _____ , 37, _____ , 29, 25, 21.
(c) 20 500, 20 000, 19 500, 19 000, _____ , _____ .
(d) 160, 148, 138, 130, _____ , _____ .
(e) 2 300, 3 300, 4 300, 5 300, _____ , _____ .
(f) 4 800, 4 700, 4 600, 4 500, _____ , _____ .
(g) –2, –4, –6, –8, _____ , _____ .
(h) –5, –10, –15, –20 , _____ , _____ .
188
Chapter 16 - Patterns and Sequences
4. Find the next three terms in each of the following:
3 6 9 12
,
,
,
, _____ , _____ , _____ .
20 20 20 20
7
7
7
7
(c)
,
,
, , _____ , _____ , _____ .
15 21 27 33
1
2
3
4
(e) – , – , – , – , _____ , _____ , _____ .
5
5
5
5
(a)
1
1
(b) 1 2 , 3 , 4 , 6, _____ , _____ , _____ .
2
1
2 1
(d) 5 , 5, 4 , 4 , _____ , _____ , _____ .
3
3 3
(f ) – 7 , – 5 , – 3 , – 1 , _____ , _____ , _____ .
8
8 8 8
5. Write down the missing terms in each of the following:
(a) 2.0, 2.2, 2.4, 2.6, _____ , _____ .
(b) 6.5, 7.0, _____ , 8.0, _____ .
(c) 12.1, 12.3, _____ , 12.7, 12.9, _____ .
(d) 1.0, 2.1, 3.2, 4.3, _____ , _____ .
(e) 4.8, 6.1, 7.4, 8.7, _____ , _____ .
(f ) 1.3, 1.7, _____ , 2.5, _____ .
(g) –3.3, –5.3, _____ , –9.3, _____ , –13.3 (h) –6.4, –3.2, _____ , –0.8, _____ , –0.2
6. Circle the correct answer:
(a) 1, 4, 9, _____ , 25
A. 15
B. 16
C. 20
D. 18
C. –2
D. –4
C. 10
D. 1
C. 25
D. 26
C. 10
D. 12
C. 28
D. 29
C. 0.00002
D. 0.002
B. 124
C. 216
D. 81
1
3
, 2 , 4, _____ .
2
4
1
3
A. 5 B. 6 2
4
1
C. 5
4
D. 6
(b) –1, 1, –2, 2, –3, _____ .
A. 3
B. 4
(c) 1 000 000, 100 000, 10 000, _____ .
A. 100
B. 1 000
(d) 10, 17, _____ , 31, 38, 45
A. 24
B. 23
(e) _____ , 15, 19, 23, 27
A. 13
B. 11
(f ) 3, 11, 19, _____ , 35
A. 27
B. 26
(g) 0.2, 0.02, 0.002, _____ , 0.00002
A. 0.02
B. 0.0002
(h) 1, 8, 27, 64, _____ .
A. 125
(i) 1
1
4
189
Chapter 16 - Patterns and Sequences
Number patterns from patterns of figu es
Example
Consider the figures below and answer the questions.
Fig. 1
Fig. 2
Fig. 3
(a) Draw Figure 4.
(b) Complete the table below. What do you observe?
Figure
Number of dots
1
3
2
6
3
10
4
5
Solution
(b)
(a)
Figure
No. of dots
1
3
2
6
3
10
4 5
15 21
The number of dots represents the
triangular numbers.
Fig. 4
EXERCISE 16.2
1. Consider the following patterns of counters:
Fig. 1
(a) Draw Figures 4 and 5.
(b) Describe the pattern.
2. Complete each of the following:
(a)
(b)
190
Fig. 2
Fig. 3
Chapter 16 - Patterns and Sequences
(c)
FUN ACTIVITY:
LOOKING FOR A
PATTERN
(d)
In a group of 3 pupils, each one
would gift the other a flower on
Friendship Day.
How many flowers will be
exchanged on that day?
3. The figures below have been formed using matchsticks.
If instead of 3 pupils, there are 4,
5 or 6 pupils, how many flowers
will be exchanged on that day?
Now, in a group of 10 pupils, how
many flowers will be exchanged?
Fig. 1
Fig. 2
Fig. 3
(a) Draw Figures 4 and 5.
(b) How many matchsticks will be used in Figure 10?
(c) In which Figure will 64 matchsticks be used?
4.
Diagram 1
Diagram 2
Diagram 3
Diagram 4
The table below shows the number of small squares in each of the above diagrams.
Diagram number
1
2
3
4
Number of unshaded squares
3
6
9
12
Total number of squares
4
9
16
25
Number of shaded squares
1
3
7
5
6
Copy and complete the table.
Sequence
A sequence is a specific type of pattern that involves a mathematical rule.
For example: 1, 4, 7, 10, …
To obtain the next term, we add 3 to the previous term.
191
Chapter 16 - Patterns and Sequences
Activity 2
Describe each of the following number sequences.
NOTE TO TEACHER
(a) 2, 4, 6, 8, 10, …
Prompt students to think
of the types of numbers to
describe each of the different
sequences.
(b) 1, 4, 9, 16, 25, …
(c) 1, 8, 27, 64, …
(d) –1, –2, –3, –4, –5, …
(e) 1, 3, 6, 10, 15, …
(f ) 1 , 1 , 1 , 1 , …
2 3 4 5
Can you find out what the next term will be in each of the different parts?
EXERCISE 16.3
Describe each of the following number sequences and write down the next two terms.
(a) 4, 9, 14, 19, ___ , ___
(b) 12, 10, 8, 6, ___ , ___
(c) 3, 9, 27, 81, ___ , ___
(d) 0, 11, 22, 33, ___ , ___
(e) –5, –8, –11, –14, –17, ___ , ___
(f ) –0.8, –0.6, –0.4, –0.2, ___ , ___
Activity 3
We can find different number patterns in a
calendar.
(a) The 1st of May 2017 is on a Monday. What
can you observe about the dates of the
other Mondays in the month of May?
(b) Consider the boxed area for the 1st , 2nd,
8th and 9th of May 2017.
Add the two numbers that are diagonally
opposite one another. What do you
notice?
Now, try multiplying the two numbers
diagonally opposite each other.
(c) Repeat the above steps by considering
another 4 dates in the calendar, for
instance, 17, 18, 24, 25.
What do you notice? Is there a pattern?
Can you describe the pattern?
(d) Can you find other number patterns in
the calendar? Use different months.
192
NOTE TO TEACHER
The sums of the numbers in diagonally opposite
corners are always equal in a calendar.
The products of the numbers in diagonally opposite
corners always differ by 7 in a calendar.
Example of other patterns in calendar: consider the
column 3, 10, 17, 24, 31.
3 + 31 = 34
10 + 24 = 34
Double 17 is 34
Consider row 14, 15, 16, 17, 18, 19, 20.
14 + 20 = 34
15 + 19 = 34
16 + 18 = 34
Double 17 is 34
Chapter 16 - Patterns and Sequences
EXERCISE 16.4
1. If the 29th September 2017 was a Friday, find on which date was the first Friday of September 2017?
2. If the 25th May 2008 was a Sunday, on which day was 1st May 2008?
3. If the 3rd of March was a Sunday, on which day was the 31st of March?
4. If the 6th February was a Tuesday, what day was it 20 days later?
5. If the 24th November was a Sunday, what day was it 15 days before?
Investigate:
(a) (1 x 7) + 1 = 8
(12 x 7) + 2 = 86
(123 x 7) + 3 = 864
(b) 1 = 110
11 = 111
121 = 112
(c) (0 x 9) + 8 = 8
(9 x 9) + 7 = 88
(98 x 9) + 6 = 888
What do you notice about each of the patterns?
Add three more lines to the different patterns.
Summary
•
•
A number pattern is defined as a set of numbers which are arranged in
such a way that each successive term/number follows the preceding
one according to a specific rule.
A sequence is a specific type of pattern that involves a mathematical
rule.
193
17
COORDINATES
Chapter 17 - Coordinates
Learning Objectives
By the end of this chapter, you should be able to:
• identify the axes in a Cartesian plane (x - y plane).
• locate and plot points in the Cartesian plane.
• determine the equation of lines parallel to the x and y axes.
• draw lines in the form x = h and y = k, where h and k are constants.
• find point of intersection of horizontal and vertical lines.
CHECK THAT YOU CAN:
• Locate a point on a given graph.
• Identify the coordinates of given
points on a graph.
Coordinates in real life
Have you ever noticed the coordinates on the weather
chart tracking a cyclone or have you ever thought about
how an aeroplane keeps track of the positions of other
aeroplanes to avoid collision in the sky?
KEY TERMS
• Cartesian plane
• Points; Coordinates
• Axis; Axes
• x-axis; y-axis
• Ordered pair
• Origin
• Parallel lines
• Point of intersection
We use coordinates and the coordinate system in both
of the above situations. Coordinates are used to describe
the positions of objects or places on maps. The coordinate
system is also used by air traffic controllers and pilots to
determine the location of the aeroplane. Landscapers
also make use of the coordinate system to determine
where to place buildings and other amenities.
_
00S
DID YOU KNOW
History of Coordinates
0 _
10 S
200S_ _
_
500E
600E
700E
800E
Tracking and predicting the path of cyclones using coordinates
194
_
_
400E
_
_
300E
_
_
300S_
900E
The coordinate system
we currently use is called
the Cartesian system.
It is named after the
Rene Descartes
French mathematician
Rene Descartes (1596 - 1650) who
developed the concept of a plane to
locate a point accurately. He thought
of this as he was lying on his bed
one day and decided to find a way to
describe the exact position of a fly,
which was on the ceiling. He decided
that by drawing two perpendicular
lines and then labelling them with
numbers, he would be able to describe
the position of the fly accurately.
In his honour, a graph showing the
coordinates of a point is known as the
Cartesian plane (or the x- y plane).
Chapter 17 - Coordinates
Activity 1
Learning to read and identify coordinates on a map
y-axis
9
XYZ Shopping Centre
8
LMN Shopping Centre
7
6
Petrol Pump
5
Jewel Shop
Sally Shop
4
3
Curepipe Hotel
Curepipe Post Office
Chinese Restaurant
2
Coffee Bar
1
0
-1
-2
District Court
1
2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20 21 22 23
Bank
Sun College Curepipe
Bees Shop
Municipality of Curepipe
-3
-4
Sam is a retired professor on vacation in Mauritius and he is staying at Curepipe Hotel.
The following is his itinerary for the day:
8:30 – He starts his day by refuelling his car.
9:00 – He sends a card to his mother for Mothers’ day.
9:30 – He has to exchange money at the bank.
10:30 – He gives an inspirational talk on time management to
Grade 11 students of Sun College Curepipe.
12:00 – He has to meet a friend for lunch at Chinese Restaurant.
13:30 – He goes to the District Court to pay a speeding fine.
DID YOU KNOW
Latitude and Longitude coordinates
We can find the latitude and longitude
of any point on Google Maps. You can
click on the map to get the address and
its GPS Coordinates. Find the latitude
and longitude of Port Louis using this
link:
https://www.gps-coordinates.net/
14:00 – He goes shopping.
17:00 – He has a coffee break.
18:00 – He goes for a walk around the Municipality of Curepipe.
19:00 – He has dinner at Curepipe Hotel.
195
x-axis
Chapter 17 - Coordinates
(a) Write down the coordinates of:
(i) the Petrol Pump where he refuels his car
(iii) the bank where he might exchange his money
(v) the Chinese Restaurant
(ii) the Post Office of Curepipe
(iv) Sun College Curepipe
(vi) the District Court.
(b) Sam went shopping at different locations given by the following coordinates:
(5, 9), (19, 8), (3, 5), (13, –2) and (13, 4).
Name each of these places.
(c) Where can Sam go for a cup of coffee? Write down the coordinates of the place you have chosen.
(d) The distance between Municipality of Curepipe and Curepipe Hotel is 180 m. He walks at
a constant speed of 0.5 m/s. If he leaves the Municipality at 18 58 and dinner is at 19 00 sharp,
will he be on time?
Introduction to the Cartesian plane
In Grade 6, you learnt about positive coordinates. We now extend the x-axis and y-axis to
include negative values of x and negative values of y as shown in Fig. 1.
y-axis
6
y is positive
5
POINTS TO REMEMBER:
4
3
2
1
-6 -5 -4 -3 -2 -1
y is negative
x is negative
0
-1
-2
-3
-4
-5
1
2
3
4
5
x is positive
6
x-axis
1. Points on the Cartesian plane are
denoted by capital letters.
E.g P, A, B, D…
2. The plural of axis is axes.
3. The x-coordinate is written first and
then the y-coordinate, i.e. ,
(x-coordinate, y-coordinate).
-6
Fig. 1
We call Fig. 1 the number plane or the coordinate plane or the Cartesian plane. In the Cartesian
plane, the x-axis and y-axis intersect each other at right angles. This point of intersection is
called the origin and has coordinates (0, 0). It is denoted by O.
Coordinates of a point on the Cartesian plane
Ordered pair
Coordinates are written in the form (x-coordinate, y-coordinate) and this is called an ordered
pair. For instance, (4, 9) and (9, 4) do not indicate the same position.
For the point (4, 9), the x-coordinate is 4 and the y-coordinate is 9 whereas for (9, 4), the
x-coordinate is 9 and the y-coordinate is 4.
196
Chapter 17 - Coordinates
Example 1
Note: An easy way to remember
which of the x-coordinate or
y-coordinate is written or read
first is to think of the letters in their
alphabetical order. As x is before
y in the alphabet list, we always
write or read the x coordinate first
followed by the y coordinate.
Plot each of the following points on a separate Cartesian plane.
(a) A (4, 7)
(b) B (–2, 6)
(c) C (–5, –3)
y-axis
(a)
(d) D (3, –1)
8
8
A (4 , 7)
7
B (-2 , 6)
6
-3 -2 -1
7
6
5
5
4
4
3
3
2
2
1
1
0
-1
1
2
3
4
5
x-axis
-3 -2 -1
-2
y-axis
(c)
-5 -4 -3 -2 -1
0
-1
1
2
3
4
5
x-axis
-2
Consider the point A. Both the
x-coordinate and the y-coordinate are
positive. We start at the origin and move
4 units to the right and 7 units up.
C (-5 , -3)
y-axis
(b)
For the point B, the x-coordinate is negative.
We start at the origin and move 2 units to
the left and 6 units up as the y-coordinate
is positive.
y-axis
(d)
6
6
5
5
4
4
3
3
2
2
1
1
0
-1
1
2
3
x-axis
-5 -4 -3 -2 -1
0
-1
-2
-2
-3
-3
-4
-4
We start at the origin for point C and move
5 units to the left as the x-coordinate
is negative and 3 units down as the
y-coordinate is negative.
1
2
3
x-axis
D (3 , -1)
For the point D, the x-coordinate is positive
and we therefore move 3 units to the right,
starting at the origin and 1 unit down as
the y-coordinate is negative.
197
Chapter 17 - Coordinates
Example 2
y-axis
Write down the coordinates of P, Q, R and S.
P
8
Step 1: To start, place your finger at the
origin.
7
R
6
Step 2: Then move your finger to the right
along the x-axis until your finger is lined up
under point P.
5
4
3
Units moved to the right = 3
2
x-coordinate = 3
1
Step 3: Now, move your finger up from the
x-axis until your finger reaches point P.
-6 -5 -4 -3 -2 -1
Units moved up = 8
1
-1
2
3 x-axis
S
-2
Q
y-coordinate = 8
0
-3
Therefore coordinates of P = (3, 8)
We now proceed in a similar way to read the coordinates of Q, R and S.
Q = (–6, –3)
(6 units to the left followed by 3 units down)
R = (–5, 6)
(5 units to the left followed by 6 units up)
S = (2, –2)
(2 units to the right followed by 2 units down)
EXERCISE 17.1
y-axis
1. Plot the following coordinates on the
Cartesian plane using graph paper.
H
5
A
P (2 , 5), Q (–3 , 4), R (–2 , –6), S (4 , –3),
T (2 , –4), U (–1 , –2), V (5 , 5),
2. Write down the coordinates of the points in
Fig. 2.
4
G
B
3
J
-6 -5 -4 -3 -2 -1
I
1
C
0
1
-1
2
-3
L
-5
-6
-7
3
D
-2
-4
K
-8
Fig. 2
198
F
2
W (4 , 6),
X (–4 , 3), Y (–1 , –6), Z (2 , –6)
E
6
4
5
6 x-axis
Chapter 17 - Coordinates
3. Below is the map of Zoo Park. Write down the coordinates representing the location of each
of these animals.
y-axis
6
4
3
Lions
Elephants
2
(
(
(
(
(a) Lions
(b) Giraffes (c) Elephants (d) Crocodiles
Giraffes
5
,
,
,
,
)
)
)
)
Crocodiles 1
-6 -5 -4 -3 -2 -1
0
-1
1
2
3
4
5
x-axis
6
-2
4. Write down the coordinates of the following places on the map below.
y-axis
6
5
Grand-Bois
4
South College
South Sugar Estate
3
2
1
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1
-1
0
1
2
3
4
5
6
7
8
9 10 11 12 13
x-axis
-2
-3
(a) South College
(c) Grand-Bois
Gris Gris Public Beach
( , )
( , )
(b) Gris Gris Public Beach
(d) South Sugar Estate
( , )
( , )
5. Write down
(a) the coordinates of four points having the same x-coordinate,
(b) the coordinates of four points having the same y-coordinate,
(c) the coordinates of four points having the same x and y coordinates.
6. Draw the x-axis from –4 to 4 and the y-axis from –1 to 5 on a graph paper.
(a) Plot and label the points A (–3, 2), B (2, 2), C (4, 5) and D (–1, 5).
(b) Join the points to obtain the shape ABCD.
(c) What is the special name given to this shape?
199
Chapter 17 - Coordinates
7. PQRS is a rectangle. The coordinates of P, Q and R are (6, 9), (10, 9), and (10, 3) respectively. By
plotting the points on graph paper, write down the coordinates of S.
8. The following three points A (–2, –1), B (4, –1), C (4, 5) are the three vertices of a square ABCD.
Plot these points on graph paper, then determine the coordinates of the fourth point D. Plot
that point and label it.
9. Two points have been plotted on the diagram below. Write down the third point to make an
y-axis
isosceles triangle.
3
2
1
-3 -2 -1
0
-1
1
2
3
x-axis
-2
-3
10. The diagram shows three plotted points (2,1), (0, 3) and (–3,0). Find the coordinates of the
y-axis
fourth point to make a rectangle.
3
2
1
-3 -2 -1
0
-1
1
2
3
x-axis
-2
-3
11. WXYZ is a parallelogram. The diagonals XZ and YW meet at T. Given that X, Y and T have
coordinates (7, 2), (9, 6) and (5, 4) respectively, write down the coordinates of the points Z and W,
by plotting these points on graph paper. Hint: The diagonals of a parallelogram bisect each other.
12. Fill in the blanks with: x-axis, y-axis
(a) (0, 4) is a point on the ...........................
(b) (5, 0) is a point on the ...........................
13. (a) When a point lies on the y-axis, the x-coordinate is always
A. 1
B. –1
C. y
D. 0
(b) When a point lies on the x-axis, the y-coordinate is always
A. –1
200
B. 0
C. 1
D. x
Chapter 17 - Coordinates
Activity 2
a) Plot the following coordinates on the same graph:
A (2, 0)
B ( 5, 0)
C ( –3, 0)
D ( –6, 0)
What do you observe?
b) Plot the following coordinates on the same graph:
W (0, –8)
X (0, 4)
Y (0, –1)
Z (0, 7)
What do you observe?
All points that lie on the x-axis have their y-coordinate as 0. They are in the form (x, 0).
All points that lie on the y-axis have their x-coordinate as 0. They are in the form (0, y).
Now join the points that you have plotted in (a).
What do you observe?
Similarly, join the points that you have plotted in (b).
What do you observe?
Equations of lines parallel to the x-axis and y-axis
All the coordinates of every point on a straight line have a common relationship. This relationship
can be expressed algebraically as an equation in terms of x and/or y which is called the equation
of line.
Lines parallel to the y-axis (vertical lines)
Consider the graph below.
Look at the vertical line on the graph. This vertical
line is parallel to the y-axis. The coordinates of
the points on the line are (2, 0), (2,1), (2, –1), (2, –2).
y-axis
3
These points can be written in the table showing
the x-coordinate and the y-coordinate.
2
x(2 , 1)
1
-2
0
-1
1
x (2 , 0)
2
3
-1
x(2 , -1)
-2
x(2 , -2)
-3
4
x-axis
x
2
2
2
2
y
–2
–1
0
1
What do you notice?
Since the x-coordinate is same throughout, i.e., 2, the equation of the straight line is x = 2.
201
Chapter 17 - Coordinates
Note: The equation x = 2 is a vertical line parallel to the y-axis. Any vertical line which
is parallel to the y-axis will pass through points having the same x-coordinate (x = h), and
therefore we say that the line has equation x = h.
Example 1
Write down the equation of the line which passes through the following points.
(a) ( 4, –3), ( 4, 0), ( 4, 1), ( 4, 2), ( 4, 5), ( 4, 6),
(b) ( –2, 1), ( –2, –2), ( –2, –1), ( –2, 0), ( –2, 1), ( –2, 2), ( –2, 3).
Solution:
(a) ( 4, –3), ( 4, 0), ( 4, 1), ( 4, 2), ( 4, 5), ( 4, 6),
Since the x-coordinate of all these points have the same value, that is, 4, therefore, the
equation of the straight line is x = 4.
(b) ( –2, 1), ( –2, –2), ( –2, –1), ( –2, 0), ( –2, 1), ( –2, 2), ( –2, 3),
Since the x-coordinate of all these points have the same value, that is, –2, therefore, the
equation of the straight line is x = –2.
Lines Parallel to the x-axis (Horizontal lines)
Consider the horizontal line on the graph below.
y-axis
(-4 , 3)
x
(0, 3)
(-2 , 3)
x
3x
(2 , 3)
(4 , 3)
x
x
2
1
-4 -3 -2
-1
0
-1
1
2
3
4
This horizontal line is parallel to
the x-axis. The coordinates of the
points on the line are (-4, 3), (-2, 3),
(0, 3), (2, 3), (4, 3). These points can
be written in the table showing the
x-coordinate and the y-coordinate.
x-axis
What do you notice?
Since the y-coordinate is 3 throughout,
the equation of the straight line is y = 3.
x
y
–4
–2
3
3
3
3
3
0
2
4
Note: The equation y = 3 is a horizontal line parallel to the x-axis.
Any horizontal line which is parallel to the x-axis will pass through points having the same
y-coordinate (y = k), and therefore has equation y = k.
202
Chapter 17 - Coordinates
Example
Write down the equation of the line which passes through the following points.
(a) ( –3, 5), ( –2, 5), ( –1, 5), ( 0, 5), ( 1, 5), ( 2, 5),
(b) ( –1, –4), ( 0, –4), ( 1, –4), ( 2, –4), ( 3, –4), ( 4, –4), ( 5, –4).
Solution:
(a) ( –3, 5), ( –2, 5), ( –1, 5), ( 0, 5), ( 1, 5), ( 2, 5),
Since the y-coordinate of all these points have the same value, that is, 5, therefore, the
equation of the straight line is y = 5.
(b) ( –1, –4), ( 0, –4), ( 1, –4), ( 2, –4), ( 3, –4), ( 4, –4), ( 5, –4)
Since the y-coordinate of all these points have the same value, that is, –4, therefore, the
equation of the straight line is y = –4.
STOP AND THINK
What is the equation of the x-axis?
What is the equation of the y-axis?
EXERCISE 17.2
1. In each of the following, identify the coordinates of some of the points on the straight line
and use these points to find the equation of the straight line.
y-axis
(a)
-4 -3 -2
-1
3
3
2
2
1
1
0
-1
1
2
3
-4 -3 -2
-1
4 x-axis
-4 -3 -2
-1
0
-1
-2
-2
-3
-3
y-axis
(c)
y-axis
(b)
3
2
2
1
1
-1
1
2
3
4 x-axis
2
3
4 x-axis
1
2
3
4 x-axis
y-axis
(d)
3
0
1
-4 -3 -2
-1
0
-1
-2
-2
-3
-3
203
Chapter 17 - Coordinates
2. Write down the equation of the line which passes through the following points:
(a) (–5, 1), (–4, 1), (–1, 1), ( 0, 1), ( 4, 1)
(b) ( –4, –5), ( –2, –5), ( –0, –5), ( 3, –5), ( 5, –5)
(c) ( 4, –3), ( 4, –2), ( 4, 0), ( 4, 4), ( 4, 5)
(d) (–5, –4), (–5, –1), ( –5, 3), ( –5, 4), ( –5, 5)
3. Write down the coordinates of four points on each of the following lines.
(a) x = 6
(d) y = 2 (b) x = 4
(c) x = –3
(e) y = –5 (f ) y = 7
4.Write down the coordinates of five points on the x-axis. Hence, state the equation of the x-axis.
5.Write down the coordinates of five points on the y-axis. Hence, state the equation of the y-axis.
6.(a) Circle the points that do not lie on the line x = –6.
(–6, 0), ( 4, –4), (–6, 2), (–6, 3), ( 3, 1), (–6, 4)
(b) Circle the points that lie on the line y = 10.
( 5, 4), (–2, 10), ( –1, 10), ( 6, 3), ( 8, 10), ( 1, 10)
7. Write down the equation of the line which is parallel to the x-axis and passes through the point:
(a) (2 , –3)
(b) (–1, 4)
(c) (3, 5)
(d) (–3, –6)
8. Write down the equation of the line which is parallel to the y-axis and passes through the point:
(a) (1, 1)
(b) (–2, –3)
(c) (–6, 4)
(d) (3, 9)
9. (a) Draw the following straight lines on the same Cartesian plane.
(a) y = 4
(b) y = 3
(c) y = –2
(d) y = –5
(b) On another graph, draw the following straight lines.
(a) x = 5
(b) x = 3
(c) x = –4
(d) x = –1
10. (a) (2, 1), (2, –4), (2, –3), (2, 6) are points on the line
A. x = 0
B. x = 2
C. y = 2
D. y = 1
(b) (3, –1), (–2, –1), (–6, –1), (5, –1), are points on the line
A. y = 3
B. y = –1
C. x = 1
D. x = –1
(c) The line x = 3 is parallel to the
A. Line x = –1 B. y - axis C. Line y = 3
D. x - axis
(d) The line through the point (3, -6) and parallel to the y-axis has equation
204
A. x = –3
B. y = –6
C. x = 3
D. y = –6
Chapter 17 - Coordinates
Finding point of intersection of horizontal and vertical lines
The point of intersection of a horizontal line and a vertical line is the point where the two
lines meet.
Example
y-axis
Q
3
y=3 P
2
-3
-2
R
0
-1
These two lines intersect
at the point P. The point
P has x-coordinate 2 and
y-coordinate 3.
x =2
1
-4
The diagram shows the lines
x = 2 and y = 3.
1
-1
2
3
4
x-axis
The coordinates of P is (2, 3).
We say that the point of
intersection of the two lines
is P (2, 3).
In the above diagram, the line y = 3 intersects the y-axis (x = 0) at the point Q. Therefore, Q
has coordinates (0, 3)
Similarly, in the above diagram, the line x =2 intersects the x-axis (y = 0) at the point R.
Therefore, the coordinates of R is (2, 0)
EXERCISE 17.3
1. (a) Draw and label the following lines on the same Cartesian Plane with axes from –5 to 5.
(i) x = –2
(ii) x = 5
(iii) x = 2
(iv) x = –4
(v) y = 1
(vi) y = –2
(vii) y = 4
(viii) y = –3
(b) Use the graph to write the point of intersection of the following pairs of lines.
(i) x = 2 and y = –2
(v) x = –4 and y = 1
(ii) x = 5 and y = 1
(vi) x = 5 and y = –3
(iii) x = 2 and y = 4
(vii) x = –2 and y = 4
(iv) x = –4 and y = 4
(viii) x = –2 and y = 1
2. The line y = 5 intersects the y-axis at the point D. Find the coordinates of D.
3. The line x = –3 intersects the x-axis at the point F. Find the coordinates of F.
4. Write the coordinates of the points where the following pairs of lines meet
(a) x = 4 and y = 9
(b) x = –6 and y = –4
(c) y = 5 and x = 7
5. Write the coordinates of the point of intersection of the x-axis and y-axis. What is the name
given to this point?
6. The point of intersection of a horizontal line and a vertical line is (–1, 3). Find the equations of
the two straight lines.
205
Chapter 17 - Coordinates
7. Circle the correct answer.
(a) The axes intersect at the point
A. (1, 0)
B. (0, –1)
C. (0, 1)
D. (0, 0)
(b) The line x = 5 intersects the line y = –2 at the point
A. (–2, 5)
B. (0, 5)
C. (–2, 0)
D. (5, –2)
Summary
•
•
•
•
•
•
•
A point is represented by an ordered pair (x, y), that is, (x-coordinate, y-coordinate).
The axes intersect at the point (0, 0) called the origin.
The equation of the x-axis is y = 0.
The equation of the y-axis is x = 0.
Lines parallel to the y-axis have equation of the form x = h and are vertical lines.
Lines parallel to the x-axis have equation of the form y = k and are horizontal lines.
The vertical line x = h intersects the horizontal line y = k at the point (h, k).
GeoGebra
1. Locating points A(2, 4), B (–2, 3), C (–4, –1) and D (–3, –2).
Step:
Click on the icon
to plot the points A (2, 4),
B (–2, 3), C (–4, –1) and D (–3, –2) on the graph.
2. Drawing lines x = 4 and y = 3 and finding their
point of intersection.
Step 1:
Click on the “Input” bar at the bottom of the
GeoGebra interface and type the equation
x = 4. Press enter. The line x = 4 will appear.
Step 2:
Click on the “Input” bar again and type
the equation y = 3. Press enter.
The line y = 3 will appear.
Step 3:
Click on the icon
and place the cursor on
the point of intersection of the two lines to
draw the point of intersection, which is A (4, 3).
• Click on Name and Value on the box A
• A (4, 3) is the point of intersection.
206
SYMMETRY
18
Chapter 18 - Symmetry
Learning Objectives
By the end of this chapter, you should be able to:
• determine the number of lines of symmetry in plane shapes.
• locate and draw the line(s) of symmetry in a given shape.
• complete plane figures given line(s) of symmetry (horizontal, vertical and slant).
Symmetry in real life
Symmetry forms part of our daily life and it can be observed all around us: for instance, in nature,
art, architecture and design as well as in different cultures all over the world.
Symmetry is fascinating because of its artistic and
decorative aspects. In architecture, symmetry is important
to ensure that buildings and structures are stable. Human
beings and animals also have a symmetric body to
support forward movement.
CHECK THAT YOU CAN:
• Identify polygons.
• Determine why a polygon is
said to be regular.
KEY TERMS
• Line of Symmetry
• Mirror Line
• Axis of Symmetry
Symmetry in plants
DID YOU KNOW
History of Symmetry
Symmetry in nature
Symmetry of the Eiffel
tower in Paris
Symmetry of the Taj Mahal
in India
Symmetry in animals
Symmetry is used in almost every field. Architects, fashion
designers, jewellers, car manufacturers, artists and many
others make use of symmetry in their respective professions.
The concept
of symmetry
emerged in
Italy around
the start
of the 15th
Century. It
was believed
that nature’s
Top of a church in Rose-Hill
forms were
symmetric and so nothing could be termed
beautiful unless its shape was symmetric.
It became necessary that all new buildings
be symmetric and that existing public
buildings as well as churches have
symmetric facades.
207
Chapter 18 - Symmetry
Line of Symmetry
When a figure is folded such that the two halves are
mirror images of each other, that is, they can fit exactly
on top of each other, the figure is said to have refle tive
symmetry. This line is called the line of symmetry or the
axis of symmetry or the mirror line.
Line of Symmetry
Cultural connections
Paper cutting is a traditional popular folk art in China. Papers
are folded, cut and unfolded to create varied, complex patterns
and symmetrical designs. Paper cut outs are usually placed
at the entrance of houses and are considered to bring good
luck. Today, Chinese paper cutting is often used for decorative
purposes.
Chinese paper cuttings
Islamic art and designs that feature in the decoration of
mosques and buildings have as distinctive features symmetrical
arrangements of geometrical forms.
CHECK THIS LINK
http://www.scootle.edu.au/ec/
viewing/L7801/index.html
RECALL
Islamic Art and Designs
Activity 1
Lines of symmetry can be horizontal,
vertical or inclined/slant provided the
halves fit exactly onto each other.
Finding line(s) of symmetry in given shapes
Trace the following figures on a sheet of paper.
Cut the different figures and fold them in halves such that they fit exactly onto each other.
What can you say about the line(s) of symmetry of each figure?
STOP AND THINK
Can you find out how many lines of
symmetry a circle has?
Note: Figures that have line(s) of symmetry are called symmetric figures.
208
Chapter 18 - Symmetry
Example
FUN ACTIVITY:
Draw the line(s) of symmetry (if any) of the following figures.
(a)
(b)
(c)
Materials needed:
1 sheet of paper and water colour
of 4 different colours (red, blue,
yellow, green).
Procedure:
Fig. 1
Fig. 2
Fig. 3
Solution
To locate the line(s) of symmetry of each figure, we need to
decide how to fold the figure (horizontally, vertically or inclined)
into two so that the two parts fit exactly onto each other.
Step 1: Fold your sheet of paper in
half and then open it.
Step 2: Now, put a small amount of
the different colours directly from
the tubes of paint onto half of the
paper as illutrated below:
(a) Fig. 1 can be folded into halves in the
following ways:
Step 3: Fold the paper back into
two equal parts, as in step 1.
Step 4: Press on the paper, moving
your hand on the surface of the
paper gently so that the paint is
spread out.
Combining the different possibilities, we obtain:
Step 5: Now, carefully, open up the
sheet to reveal your painting.
What can you say about the design
on each half of the paper?
Can you locate the line(s) of
symmetry, if any?
(b) Fig.2 is not symmetrical because when we fold it into
halves, they do not fit exactly onto each other.
Can you find out w y?
NOTE TO TEACHER
Prompt students to think about the
lengths and dimensions of Fig. 2 to
ensure that the inclined line is not a
line of symmetry.
(c) Fig. 3 can be folded into halves in six different ways and
the dotted lines drawn are the lines of symmetry.
209
Chapter 18 - Symmetry
EXERCISE 18.1
1. Draw the line(s) of symmetry, if any,
of the following triangles.
2. Locate and draw the line(s) of symmetry,
if any, of the following quadrilaterals.
(a)
(a)
6 cm
6 cm
(b)
(b)
8 cm
8 cm
(c)
500
(c)
1000
(d)
(d)
(e)
(e)
210
(f )
Chapter 18 - Symmetry
3. Draw the line(s) of symmetry, if any,
of the following.
(a)
4. Circle the correct answer.
(a) How many lines of symmetry does a
square have?
A. 1
C. 3
B. 2
D. 4
(b) The figure has ------ line(s) of symmetry.
(b)
A. 0
C. 2
B. 1
D. 3
(c) A circle has ------- line(s) of symmetry.
A. 0
C. 10
B. 4
D. infinite
(d) How many line(s) of symmetry does a
regular decagon have?
(c)
A. 0
C. 9
B. 5
D. 10
(e) The figure has ------- line(s) of symmetry.
A. 0 C. 2
(d)
B. 1
D. 3
(f ) A rhombus has ------- line(s) of symmetry.
A. 0
C. 2
B. 1
D. 4
(g) Which of the following has no line of
symmetry?
(e)
A. rectangle
C. parallelogram B. rhombus
D. circle
(h) How many line(s) of symmetry does the
figure have?
A. 0
C. 6
B. 3
D. infinite
211
Chapter 18 - Symmetry
5. Draw a polygon with (a) 1 line of symmetry
(c) 3 lines of symmetry.
(b) 2 lines of symmetry
6. Draw a pentagon with (a) no line of symmetry
(b) 1 line of symmetry.
7. Draw a hexagon with (a) no line of symmetry
(b) 2 lines of symmetry.
8. Copy and complete the following table.
RECALL
A regular polygon is a polygon having
all its sides and angles equal.
Regular Polygons
212
Number
of Sides
Name of polygon
Number of Lines
of Symmetry
Chapter 18 - Symmetry
Activity 1
For this activity, you will need 3-4 square pieces of paper (10 cm by 10 cm).
Step 1: Take one piece of paper and fold it in half.
Step 2: Draw a design and cut out the design (Ensure that you do not cut away the fold line).
Step 3: Open up the piece of paper.
What do you observe? How many lines of symmetry does the figure have? What type of
lines of symmetry (vertical, horizontal, slant) does it have?
With another piece of paper, fold the paper horizontally in half. Then fold the paper vertically
in half. Then fold the paper diagonally so that you obtain a triangle as shown in Fig. 4.
Repeat steps 2 and 3.
Fig. 4
What do you observe?
Investigate:
Can you find out how such paper designs were made?
How many lines of symmetry does each of them have?
Try to make such a paper design by yourself.
Hint: Use the paperfolding and cutting technique.
CHECK THIS LINK
Activity 2
wild.maths.org/fold-and-cut
A piece of paper is folded in four (Fig. 5a) and a
small triangle is cut out as shown in Fig. 5b. Can
you draw the result you think you would obtain
when the paper is opened out?
Fig. 5a
Fig. 5b
You can check your answer by doing the paper
folding and cutting.
NOTE TO TEACHER
(1) Prompt students to think about
how the way they have folded the
paper is connected to the line of
symmetry, i.e. whether it will be
vertical, horizontal or inclined as
well as the number of times they
have folded the paper.
(2) Encourage students to think about
the properties of equidistance and
perpendicularity.
213
Chapter 18 - Symmetry
EXERCISE 18.2
1. State the number of lines of symmetry
(if any) of each of the following.
2. Draw the lines of symmetry (if any) of
each of the following:
(a)
(a)
(b)
(b)
(c)
(c)
(d)
(d)
(e)
(e)
(f )
FUN ACTIVITY:
1. I am a quadrilateral and I have
no line of symmetry. Who am I?
(g)
214
2. I am an alphabet and I have 2
lines of symmetry. I am also
found in the word MATHS.
Who am I?
Chapter 18 - Symmetry
3. Draw the line(s) of symmetry, if any, of
the following plane figures.
(a)
4. Determine whether the dotted line(s)
are lines of symmetry.
(a)
(b)
(b)
(c)
(c)
5. Two students, Siya and Rohan, have been
given a piece of square paper folded in
halves where a design has been cut out
as follows:
(d)
(e)
They have been asked to provide the
solution to the following question: What
will be the design obtained when the
paper will be opened?
(f )
Siya’s answer
Rohan’s answer
Who has the correct solution?
215
Chapter 18 - Symmetry
6. Copy and complete the following figures given that the dotted lines are lines of symmetry.
(a)(b)
(c)(d)
Completing figu es given inclined lines of symmetry
In Grade 7, we will learn how to complete a figure given a slant line of symmetry. Consider the
following symmetric figure. MR (the slant or inclined line of symmetry) divides the figure into halves.
• Using a divider, measure the perpendicular
distance from B to the line of symmetry
and the perpendicular distance from Bl to
the line of symmetry. Do the same for C
and Cl and D and Dl.
• What do you notice?
Observation:
• A and Al are at the same point on the line
of symmetry.
• Bl, Cl and Dl are at equal (perpendicular)
distance from the line of symmetry as B,
C and D respectively.
• BBl, CCl, DDl are at right angles
(perpendicular) to the line of symmetry.
NOTE TO TEACHER
Explain to students how to use the divider.
216
Chapter 18 - Symmetry
In any symmetric figure, we notice that:
• Any point on the line of symmetry (e.g. A) does not move.
• Each pair of corresponding points (e.g. B and Bl, C and Cl , D and Dl) are at equal distances from
the line of symmetry on each side.
• Each pair of corresponding points forms a perpendicular line to the line of symmetry (BBl is
perpendicular to the line of symmetry).
Example
Copy and complete the following figure such that it is symmetrical about the given dotted line.
Solution
• We consider one point at a time.
• We draw a line perpendicular to the line of symmetry for each point.
• There are two ways of proceeding.
Method 1
Method 2
• Count the number of squares diagonally
from one point to the line of symmetry.
For example, as illustrated in the above
1
diagram, there are 1 small squares from
2
the line of symmetry to the point.
• We now reproduce the count perpendicular
to the line of symmetry, but on the other
side, i.e., 1 1 small squares as illustrated in
2
the diagram.
• We use the divider to ensure that
the two points are at equal distances
from the line of symmetry.
217
Chapter 18 - Symmetry
EXERCISE 18.3
1. Copy and complete the other half of each symmetric shape given below.
(a)(b)
(c)(d)
(e)(f )
218
Chapter 18 - Symmetry
2. Copy and complete the following figures such that they are symmetrical about the given
dotted lines.
(a)(b)
(c)(d)
(e) (f )
219
Chapter 18 - Symmetry
3. Shade one more small square so that the diagram has
exactly one line of symmetry.
4. Shade two more small squares such that the diagram has
one line of symmetry.
5. Shade two small triangles such that the diagram has exactly
one line of symmetry.
6. Shade two more small squares such that the figure has
exactly 2 lines of symmetry.
7. Complete the diagram so that it has exactly one line of
symmetry.
Summary
• If a plane figure is folded into two so that the two halves fit exactly onto each
other, then the figure is said to have a line of symmetry.
• Plane figures can also have infinite lines of symmetry or no line of symmetry.
220
GEOMETRICAL CONSTRUCTIONS
19
Chapter 19 - Geometrical Constructions
Learning Objectives
By the end of this chapter, you should be able to:
• perform simple geometrical constructions using ruler, set squares, protractor, a pair of
compasses and dividers as well as digital tools.
• construct a line parallel to a given line.
• construct a line parallel to a given line passing through a given point.
• construct the perpendicular bisector of a line segment.
• construct the bisector of an angle.
Geometrical Constructions
CHECK THAT YOU CAN:
Geometry is present in our daily life. It is used by a large
number of people especially in fields such as engineering or
architecture.
• Draw a straight line.
• Measure the length of a line.
KEY TERMS
• Line segment
• Parallel line
• Perpendicular bisector/ Mediator
• Angle bisector
Geometry in architecture
Geometry in art
DID YOU KNOW
Geometry in nature
Geometrical Instruments
Geometry can also be used
in music.The information in
music can be represented by
a series of shapes. Geometry
in music was further
developed in the 18th century
by the mathematician
Leonhard Euler who
developed geometric tools to
analyse music.
The table below shows some geometrical instruments that you will use in geometry.
Name
Set Squares
Description
There are 2 types of set
squares used to draw parallel
and perpendicular lines.
Name
The Pair of
Compasses
Description
The pair of compasses is an
instrument to draw circles or
arcs.
221
Chapter 19 - Geometrical Constructions
Construction of a line segment
Suppose we want to draw a line segment of length 10.4 cm. We can either use a pencil and a
ruler or use a pair of compasses.
Method 1: Pencil and Ruler
Step 1: Mark a point A on paper.
Step 2: Measure a distance of 10.4 cm with the ruler and mark the end point with letter B.
Step 3: Join the two points.
Method 2:
The second method involves the use of a pair of compasses in the following way:
Step 1: Draw a straight line and mark a point A Step 2: Place the pointer of the pair of
at the beginning of the line as shown in Fig. 1. compasses on the ‘0’ mark of the ruler and
measure 10.4 cm as shown in Fig. 2.
A
15
16
140
150
14
17
130
160
13
18
120
12
19
110
180
170
11
20
100
190
10
21
90
200
9
22
80
210
8
23
70
220
7
24
60
230
6
25
50
240
5
26
40
250
4
27
30
260
3
28
20
270
2
29
30
10
280
1
0
290
0
300
Fig. 2
Fig. 1
Step 3: With the same measurement as Step 4: Mark with the letter ‘B’, the ‘cut’ on the
in step 2, place the pointer of the pair of line. You have a line segment, AB, of length
compasses on the point ‘A’ and draw an arc 10.4 cm as shown in Fig. 4.
to ‘cut’ the line as shown in Fig. 3.
A 10.4 cm B
A
Fig. 3
EXERCISE 19.1
Fig. 4
NOTE TO TEACHER
• Remind students about the difference
1. Using a pair of compasses, construct a line segment of length
(a) 4.3 cm,
(b) 10.6 cm.
2. Construct a line segment AB of length 80 mm using a pair of
compasses.
(a) Mark the point C where AC = 3.5 cm.
(b) Measure the length of BC.
between a line and a line segment.
• Explain to students what is meant by
an arc.
Note: Ensure that your pencil has a fine
tip in order to draw thin lines for accurate
diagrams.
3. (a) Using a pair of compasses, construct a line segment,
XY, of length 5.3 cm.
(b) Given that XZ is twice XY, use your compasses to construct XZ. Hence, measure the length of XZ.
222
Chapter 19 - Geometrical Constructions
Construction of a line parallel to a given line
We construct a line DC, parallel to a given line AB in the following way:
250
5
240
6
230
7
220
8
210
9
200
10
190
11
180
12
170
13
160
14
150
15
140
16
130
17
120
18
110
19
100
20
90
21
80
22
70
23
60
24
50
25
40
26
30
300
260
4
27
20
290
270
3
28
10
2
29
0
1
30
Fig. 5
0
B
A
280
Step 1: Place the set square on the line AB
and the ruler perpendicular to the line AB as
shown in Fig. 6.
Consider the line AB as shown in Fig. 5.
NOTE TO TEACHER
• Prompt students to ‘realise’
that the parallel line can be
either below or above the
given line, depending on how
we place the set square.
A
Fig. 6
B
300
290
280
270
260
250
240
230
220
210
200
190
180
170
160
150
140
130
120
110
100
90
80
70
60
50
40
30
20
10
0
300
290
280
270
260
250
240
230
220
210
200
190
180
170
160
150
140
130
120
110
100
90
80
70
60
50
40
30
20
10
30
30
29
29
28
28
27
27
26
26
25
25
24
24
23
23
22
22
21
21
20
20
19
19
18
18
17
17
16
16
15
15
14
14
13
13
12
12
11
11
10
10
9
9
8
8
7
7
6
6
5
5
4
4
3
3
B
2
2
Fig. 7
1
1
A
0
0
0
Step 2: Move the set square along the ruler Step 3: Draw the line DC parallel to AB as
to the position where the parallel line is to be shown in Fig. 8.
drawn as shown in Fig. 7.
C
D
A
Fig. 8
B
We thus have a line DC which is parallel to AB.
C
D
B
A
Fig. 9
GeoGebra - Constructing a line parallel to a given line
Construction of parallel lines can also be done
using the software GeoGebra. Here are the steps:
1. Choose the line tool
B to draw the line.
. Click on the point
2. Click on the tool
, select parallel line.
Click on the line AB, drag a line to the right
or left of the line AB. A line parallel to AB will
automatically appear.
223
Chapter 19 - Geometrical Constructions
EXERCISE 19.2
Copy the lines below in your copybook. Using your set square and ruler, construct a line parallel
to the given lines.
(a)
(b)
(c)
(d)
(e)
(f )
(g)
(h)
(i)
(j)
Construction of a line parallel to a given line, passing through a
given point
To construct a line parallel to a given line AB and passing through a point C, we proceed in the
following way:
Consider the line AB and the point C as shown Step 1: Place the pointer of your pair of
compasses on the point C and draw an arc
in Fig. 10.
such that it intersects the line AB. Label the
A
point of intersection D as shown in Fig. 11.
C
A
C
D
B
Fig. 10
Fig. 11
Step 2: Now place the pointer of your pair of
compasses (with the same measurement as in
step 1) on the point D and draw another arc
that passes through the point C and 'cuts' the
line AB as in Fig. 12. Label this point E.
Step 3: With your pair of compasses, measure
the distance between C and E. Using this
measurement and D as the centre, cut the first
arc drawn to obtain the point F. Join C and F.
This is the parallel line (as shown in Fig. 13).
B
F
A
A
C
C
D
D
E
E
Fig. 12
224
B
Fig. 13
B
Chapter 19 - Geometrical Constructions
EXERCISE 19.3
Copy the lines and points below in your copybook. Using your pair of compasses and ruler,
construct a line parallel to the given line shown and passing through the given point.
(a)
(b)
(c)
(d)
(e)
(f )
(g)
(h)
GeoGebra - Constructing a line parallel to a given line passing through a given point
1. Click on the line tool
to draw a line.
2. Click on any point on the grid, e.g C in the
diagram shown.
3. Click on the tool
and select parallel
line. Click on the point C, then on the line AB. A
line parallel to AB, passing through the point
C will automatically appear.
Construction of a perpendicular bisector of the line
To construct the perpendicular bisector of line AB, the following steps are used:
Consider line AB as shown in Fig. 14.
A
Fig. 14
Step 1: Set the compass open greater than
half of the length of line AB. Using A as centre,
with the pair of compasses draw two arcs (one
above the line AB and one below the line AB)
as shown in Fig. 15.
B
Step 2: Without changing the measurement
set on the pair of compasses and using B as
centre, draw two other arcs such that they
now cut the first two arcs formed. Mark their
points of intersection with the letters C and D
as shown in Fig. 16.
C
A
B
B
A
D
Fig. 15
Fig. 16
225
Chapter 19 - Geometrical Constructions
Step 3: Use a ruler to join the points C and D Step 4: Mark with the letter X, the intersection
as shown in Fig. 17.
of the two lines. Now, measure angle CXB and
angle CXA. What do you notice? Measure the
C
length of AX and XB. What do you notice? The
line CD is called the perpendicular bisector of
C
the line AB.
Fig. 18
A
B
A
X
B
D
D
NOTE TO TEACHER
Fig. 17
• Use students’ answers to prompt
GeoGebra - Constructing the perpendicular
bisector of a line segment
1. Choose the line tool
them to define X as the midpoint of
line AB and explain the concept of
perpendicular bisector.
to draw a line.
2. Click on the tool
and select perpendicular
bisector. Then click on the point A. A perpendicular
line will appear. Finally, click on the point B to fix
the perpendicular bisector.
EXERCISE 19.4
1. Copy the following lines in your copybook. Using a pair of compasses, draw the perpendicular
bisector of each of these lines.
(a)
(b)
(c)
(d)
(e)
(f )
(g)
(h)
2. Draw a line AB of length 4 cm. Construct the perpendicular bisector line of AB.
3. Draw a line PQ of length 5 cm. Construct the perpendicular bisector line of PQ.
226
Chapter 19 - Geometrical Constructions
Construction of a bisector of an angle
To draw the bisector of an angle, we proceed as explained below.
^
Construct ABC = 600 and draw its angle bisector.
Step 1: Use the protractor to Step 2: Place the pointer of
draw angle ABC (600)
your pair of compasses on the
point B and draw an arc on AB
and another arc on BC. Label
them X and Y respectively.
Step 3: Now place the pointer
of your pair of compasses on X
(using the same measurement
as in Step 2) and draw another
arc.
A
A
A
X
600
B
B
C
X
600
C
B
Y
600
C
Y
Step 4: Using the same Step 5: Join the point Z to Measure angle ABZ and angle
CBZ.
measurement as in Steps 2 point B.
What do you notice?
and 3, repeat the process with
The line BZ divides the angle
the pointer on Y and cut the
ABC into 2 equal parts, i.e., BZ
‘new’ arc. Label this point of
intersection Z.
bisects angle ABC.
A
A
X
B
Z
600
Y
A
Z
X
B
C
600
Y
X
C
B
Z
600
Y
C
EXERCISE 19.5
1. Draw the angle bisector of each of the following angles.
(a)
(b)
(c)
2. Construct the following angles and draw their angle bisector.
(a) Angle ABC = 800
(b) Angle XYZ = 1200 (c) Angle PQR = 420
227
Chapter 19 - Geometrical Constructions
Summary
• We use different geometrical instruments for constructions such as ruler, set
squares, protractor and a pair of compasses.
• To construct a line, we use a ruler and a pair of compasses:
300
0
290
1
280
2
270
3
260
4
250
5
240
6
230
7
220
8
210
9
200
10
190
11
180
12
170
13
160
14
150
15
140
16
130
17
120
18
110
19
100
20
90
21
80
22
70
23
60
24
50
25
40
26
30
27
20
28
10
29
30
• To construct a line parallel to a given line,
we use a ruler and set squares:
0
A 10.4 cm B
A
D
C
A
B
• To construct a line parallel to a given line and passing through a point, we use the
pair compasses and a ruler:
F
A
A
C
C
D
D
E
E
B
B
• To construct the perpendicular bisector of a line, we use a pair of compasses and a ruler:
C
C
A
X
A
B
B
D
D
• To construct the angle bisector of a given angle, we use a pair of compasses and a ruler:
A
A
X
B
600
C
B
600
Y
228
C
20
REFLECTION
Chapter 20 - Reflection
Learning Objectives
By the end of this chapter, you should be able to:
• demonstrate an understanding of the notion of reflection.
• reflect points, line segments and polygons in lines of reflection.
• reflect figures and shapes in lines of reflection.
• locate the line of reflection given the object and the image (through construction).
Refle tion in real life
CHECK THAT YOU CAN:
We come across reflection almost everyday and
everywhere: for example, reflection in mirrors, glass and
lakes. The side-view mirror in a vehicle is an example of
one of the important uses of reflection in our daily life.
Reflection in a lake
Cat looking at its reflection in
a mirror
Side-view mirror
Reflection in a lake
Refle tion
When you stand in front of the mirror, you see your
reflection in the mirror.
L
A
A1
M
(Mirror)
KEY TERMS
• Reflection
• Line of Reflection
• Mirror Line
• Axis of Reflection
• Object
• Image
• Invariant
DID YOU KNOW
The ancient Greek
mathematician Euclid
(sometimes called
Euclid of Alexandria)
stated the law of
reflection around
300 BC.
X
Fig. 1
• Identify the axes in a Cartesian Plane.
• Locate and plot points on the
Cartesian Plane.
• Draw horizontal lines (y = k) and
vertical lines (x = h) where h and k
are constants.
• Use ruler, set squares and a pair of
compasses.
• Construct the perpendicular bisector/
mediator of a line segment.
• Complete plane figures given lines of
symmetry (vertical, horizontal and
slant).
Fig. 2
Fig. 1 represents the person standing in front of the mirror
and Fig. 2 represents his reflection in the mirror.
Euclid of
Alexandria
The law of reflection states that light
travels in a straight line and reflects
from a smooth surface at the same
angle at which it hits it.
229
Chapter 20 - Reflection
What do you notice in the two figures?
Fig. 1 is symmetrical about the line LM.
In chapter 18, we learnt about symmetry. Symmetry is an important property of reflection. If a
shape is reflected in a mirror line, then this mirror line will be the line of symmetry. LM is called
the line of refle tion. Fig. 1 is called the object and Fig. 2 is called the image.
Consider a point A in Fig. 1. Observe the distance between the point A and the mirror line and
the distance between the point A1 and the mirror line.
We can see that the perpendicular distance of the object from the mirror line is equal to the
perpendicular distance of the image from the mirror line.
Now, if you consider the point X which is found on the mirror line LM, we can see that it does
not move/change when it is reflected in the line LM.
We say that X is invariant, that is, points which are found on the mirror line are invariant (do
not change).
Activity 1
y-axis
Refle ting the point A (4, 2) in the y-axis.
Materials needed: Graph paper, ruler, sharp
pencil, pair of compasses.
Make a copy of the graph. The given point
A is called the object. When reflected in the
y-axis, we obtain the image of point A, say A1.
6
5
4
3
A
2
1
-6 -5 -4 -3 -2 -1 0
-1
1
2
3
4
6 x-axis
5
Fig. 3
Method 1: Finding the image A1 of the point A.
y-axis
Step 1: From the point A (4, 2), draw a
dotted line which cuts the y-axis at right
angles, say at the point M, and extend
this dotted line beyond the mirror line as
shown in Fig. 4.
Line of reflection
6
5
4
3
2
M
A
1
Step 2: Using a pair of compasses,
measure the distance AM (Fig. 4).
Step 3: Now, keeping the same
measurement as in step 2, place the pointer
of the pair of compasses on M and draw an
arc to intersect the dotted line as shown in
Fig. 5. Where the arc cuts the dotted line,
we have the point A1.
-6 -5 -4 -3 -2 -1 0
-1
1
2
3
4
6 x-axis
5
Fig. 4
y-axis
Line of reflection
6
5
4
A1
3
2
A
M
1
-6 -5 -4 -3 -2 -1 0
-1
230
Fig. 5
1
2
3
4
5
6 x-axis
Chapter 20 - Reflection
Method 2: Finding the image A1 of the point A, by
counting the number of squares.
y-axis
Line of reflection
6
Step 1: From the point A (4, 2), draw a dotted line
which cuts the y-axis at right angle, say, at the
point M, and extend this dotted line beyond the
mirror line as shown in Fig. 6.
5
4
A1
3
2
M
A
1
Step 2: We count the number of squares from
A to M, that is, there are 4 squares, and then
count 4 squares from M on the extended line.
Label this point A1.
-6 -5 -4 -3 -2 -1 0
-1
Can you try to do the same for a reflection in the x-axis?
1
2
3
4
5
6 x-axis
Fig. 6
FIND OUT
What is the difference between
a line of reflection and a line of
symmetry?
EXERCISE 20.1
Use graph paper to answer the following questions.
What are the coordinates of the image of the point:
(a) P (1, 2) when it is reflected in the y-axis,
A. (1, −2)
B. (−1, 2)
C. (−2, −1)
D. (2, 1)
STOP AND THINK
Find the coordinates of the image of
A (100, -120) when reflected in the
(i) x axis (ii) y-axis.
(b) Q (−2, 3) when it is reflected in the x-axis,
A. (−2, −3)
B. (−3, 2)
C. (2, 3)
D. (3, −2)
(c) R (4, 1) when it is reflected in the x-axis,
A. (−1, −4)
B. (−4, 1)
C. (4, −1)
D. (1, 4)
(d) S (−2, −4) when it is reflected in the y-axis?
A. (−2, 4)
B. (2, −4)
C. (4, 2)
D. (−4, −2)
GeoGebra - Refl cting the point A(4, 2) in the y-axis
Step 1: Plot the point A(4, 2) using the icon
Step 2: Click on the icon
Step 3: Click on the point A(4, 2), then click on the y-axis.
Point A' (-4, 2) will appear on the opposite side of the y-axis.
231
Chapter 20 - Reflection
Activity 2
A
Refle ting the line segment AB in the
x-axis given A (-1, 4) and B (5, 1).
Materials needed: Graph paper, ruler,
sharp pencil, pair of compasses.
Make a copy of Fig. 7.
y-axis
4
3
2
-2 -1 0
-1
Line of reflection
B
1
1
2
3
4
5
6
x-axis
-2
-3
-4
Fig. 7
Step 1: We draw a line from A,
perpendicular to the x-axis as shown
in Fig. 8.
Step 2: We find the image of A by
reflecting A in the x-axis and label it A1.
We do the same for B.
Step 3: Join points A1 and B1 and this is
the image of the line segment AB.
A
y-axis
4
3
2
0
-2
-1
Line of reflection
B
1
1
2
3
4 5
6
x-axis
B1
-2
-3
A1
-4
Fig. 8
GeoGebra - Refl cting the line segment AB in the x-axis given A(-1, 3) and B(5, 1)
Step 1: Plot the points A(-1, 3) and B(5,1) using the icon
Step 2: Draw the line AB by clicking the icon
then click on the
point A and the point B. The line segment AB will appear.
Step 3: Click on the icon
.
Step 4: Click on the line segment AB, then on the x-axis. The line A1B1
will automatically appear on the opposite side of the x-axis.
STOP AND THINK
Observe the picture of the ambulance below.
Can you explain why the word AMBULANCE is
written in this way?
232
Chapter 20 - Reflection
Example
On graph paper, draw the images
of the square ABCD under a reflection in the
(a) y-axis
(b) line y = −1
y-axis
6
5
4
D
C
A
B
3
2
1
-6 -5 -4 -3 -2 -1 0
-1
1
2
3
4
5
6 x-axis
Solution
Fig. 9
(a)
y-axis
Line of reflection
6
5
C1
D1
4
D
C
3
2
1
A1
B1
A
-6 -5 -4 -3 -2 -1 0
-1
B
1
2
3
4
5
6 x-axis
-2
-3
-4
-5
-6
Fig. 10
y-axis
(b)
6
5
4
D
C
3
2
1
A
-6 -5 -4 -3 -2 -1 0
-1
B
1
2
3
4
5
6 x-axis
y = -1
We reflect each point one by one in the
given line of reflection and then join the
points to obtain the image of ABCD.
-2
Line of reflection
-3 A2
B2
-4
-5
-6
D2
C2
Fig. 11
GeoGebra - Refl cting the rectangle ABCD in the slanting line given
A(1, 5), B(5, 5), C(5, 7) and D(1, 7).
233
Chapter 20 - Reflection
EXERCISE 20.2
Answer the whole of this question on graph paper.
(a) Reflect ABCD in the y-axis
(b) Reflect PQRSTU in the x-axis
y-axis
y-axis
5
4
C
5
D
4
3
U
3
B
2
1
2
A
-3 -2 -1 0
-1
1
R
1
2
3
4
5
x-axis
6
(c) Reflect KLMN in the line x-axis
-3 -2 -1 0
-1
P
Q
1
2
3
S
T
4
5
6
x-axis
(d) Reflect ABCDEF in the line x = –4
y-axis
y-axis
6
6
N
5
F
4
3
2 K
1
-3 -2 -1 0
-1
D
M
L
1
2
3
A
4
5
6
7
x-axis
(e) Reflect PQR in the line x = 2
y-axis
C
B
3
2
1
y-axis
P
4
Q
K
3
2
L
1
R
1
2
3
4
1 x-axis
(f ) Reflect KLMN in the line y = –1
N
5
234
4
-5 -4 -3 -2 -1 0
-1
6
-3 -2 -1 0
-1
5
E
5
6
7
8
x-axis
6
5
M
4
3
2
1
-8 -7 -6 -5 -4 -3 -2 -1 0
-1
1
x-axis
Chapter 20 - Reflection
Activity 3
Refle ting the rectangle ABCD in the slanting line, l, given A (1, 5), B (5, 5), C (5, 7) and D (1, 7).
Materials needed: Graph paper, ruler, sharp pencil, pair of compasses, set square.
1. Make a copy of the graph found in Fig. 12.
y-axis
8
7
y-axis
l
D
8
C
7
6
6
5
5
4
A
B
4
3
3
2
2
1
1
0
1
2
3
4
5
7 8 x-axis
6
l
D
C
C1
B
A
0
B1
A1
1
2
3
4
Fig. 12
D1
5
7 8 x-axis
6
Fig. 13
2. Using the same technique as in Activity 1 and 2, locate the images of the points A, B, C and
D respectively and finally draw the image A1B1C1D1 of the rectangle ABCD as shown in Fig. 13.
You can use a set square to make sure your dotted lines are perpendicular to the mirror line.
EXERCISE 20.3
1. Make a copy of the following graphs. Reflect each of the given objects in the given line l.
(a)
y-axis
R
S
-6 -5 -4 -3
T
(b)
l
l
y-axis
6
6
5
5
4
4
3
3
2
2
1
1
-2 -1 0
-1
1
2
3
4
5
6 x-axis
-6 -5 -4 -3
-2 -1 0
-1
-2
-2
-3
-3
-4
-4
-5
-5
-6
-6
D
C
A
1
B
2
3
4
5
6 x-axis
235
Chapter 20 - Reflection
2. Make a copy of the following graphs. Reflect each of the given objects in the given line l.
(a)
(b)
l
y-axis
S
P
l
y-axis
6
6
R5
Q
5
4
4
3
3
2
2
1
1
-6 -5 -4 -3 -2 -1 0
-1
1
2
3
4
5
6 x-axis
F E
-6 -5 -4 -3 -2 -1 0
-1
-2
-2
-3
-3
-4
-4
-5
-5
-6
-6
1
2
3
4
D
5
6 x-axis
C
B
A
Locating the mirror line given object and image
The line of reflection or the mirror line can be located by drawing the perpendicular bisector or
mediator of the line joining a given point (provided it is not invariant) and its reflected image.
Example 1
y-axis
Fig. 14 shows the triangle ABC and
its reflected image A1B1C1. Draw the
mirror line.
4
C1
C
3
A1
B1
2
A
1
B
-6 -5 -4 -3 -2 -1 0
-1
1
2
3
4
5
6 x-axis
5
6
Fig. 14
Solution
• Join A to A1 .
• Construct the mediator of the line
AA1.
• This mediator represents the line of
reflection.
• We could have also considered the
line BB1 or the line CC1.
The y-axis (x = 0) is the line of reflection.
236
Line of reflection
y-axis
4
C1
A1
B1
C
3
2
A
1
-6 -5 -4 -3 -2 -1 0
-1
B
1
Fig. 15
2
3
4
x-axis
Chapter 20 - Reflection
Example 2
The trapezium ABCD has been
mapped onto the trapezium A1B1C1D1
by a reflection as shown in Fig. 16.
Write down the equation of the mirror
line.
y-axis
D
4
C
3
2
1
-3 -2 -1 0
-1
A
A1
B
B1
1
-2
2
3
4
6
x-axis
C1
D1
-3
5
Fig. 16
y-axis
Solution
We join A to A1 and draw the
perpendicular bisector as shown in
Fig.17.
4
The mirror line is the line y =1.
1
D
C
3
2
Note: We could have constructed the
-3 -2 -1 0
-1
perpendicular bisector of BB1 or CC1
-2
or DD1.
-3
A
A1
1
B
B1
2
3
y=1
5
6
x-axis
2
3
4
4
C1
D1
Fig. 17
EXERCISE 20.4
y-axis
1. The triangle ABC has been mapped onto
triangle A1B1C1 by a reflection. Which of the
following is the equation of the mirror line?
A. y = −1
C. x = −2
B. x = −1
D. y = −2
B1
1
-5 -4 -3 -2 -1 0
-1
A1
C1
A
B
1
5 x-axis
-2
-3
C
2. The point R (−1, 5) has been mapped onto the point R1 (−1, −5) by a reflection. What is the
equation of the line of reflection?
A. y = −1
B. x = −1
C. y = 0
D. y = 5
237
Chapter 20 - Reflection
3. The kite ABCD has been mapped onto the kite
A1B1C1D1 by a reflection. What is the equation
of the mirror line?
A. x = 0
B. x = 6
C. y = 6
D. y = 0
y-axis
2
A
1
0
-1
D
1
2
B
-3
4
5
B1
6
7
A1
8
D1
9 10 x-axis
-2
-3
C
-4
C1
y-axis
4. The trapezium ABCD is reflected onto
trapezium A1B1C1D1. What is the equation of
the mirror line?
A. y = −5
C. x = 1
B. x = −5
2
1
-5 -4 -3 -2 -1 0
-1
A
1
2
3
5 x-axis
B
-2
D. y = 0
4
-3
-4
D -5
D1
C
C1
-6
-7
-8
A1
B1
-9
y-axis
5. The rectangle ABCD is mapped onto
rectangle A1B1C1D1 by a reflection. Which of
the following is the equation of the mirror
line?
A. x = 1
B. x = 0
C. y = 0
D. y = 1
6
5
A
B
D
D1
C
C1
4
3
2
1
0
-1
1
2
-3
4
5
6
7
8
x-axis
-2
-3
-4
A1
B1
6. The point Q(−2, 1) has been mapped onto the point Q1(2, 1) by a reflection. What is the equation of
the line of reflection?
A. y = −1
238
B. x = 0
C. y = 1
D. x = −2
Chapter 20 - Reflection
8. The kite PQRS has been mapped onto
the kite P1Q1R1S1 by a reflection. Make a
copy of the following graph and construct
the axis of reflection.
7. The parallelogram KLMN has been
mapped onto the parallelogram K1L1M1N1
by a reflection. Make a copy of the following
graph and draw the mirror line.
y-axis
K
y-axis
L
2
1
-5 -4 -3 -2 -1 0
-1
N
N1
1
2
3
4
S
5 x-axis
M
-2
M1
-3
-7
2
1
-5 -4 -3 -2 -1 0
-1
-5
-6
3
R1
P1
R
1
2
3
4
5 x-axis
-2
Q1
L1
Q
4
S1
-4
K1
6
5 P
-3
-8
-4
-9
-5
9. The triangle UVW has been mapped onto the triangle U1V1W1 by a reflection. Copy the graph
given below and construct the line of reflection.
y-axis
7
W1
6
W
5
4
3
U1
V
2
1
0
-1
V1
U
1
2
-3
4
5
6
7
8
x-axis
Summary
• The perpendicular distance of the object from the mirror line is equal to the
perpendicular distance of the image from the mirror line.
• The image of any point on the mirror line is the point itself, that is, any point on the
mirror line is invariant (does not move).
• The line of reflection or the mirror line can be located by drawing the perpendicular
bisector or mediator of the line joining a given point (provided it is not invariant)
and its reflected image.
239
21
SETS
Chapter 21 - Sets
Learning Objectives
By the end of this chapter, you should be able to:
• demonstrate an understanding of the concept of sets.
• distinguish among different types of sets.
• identify and use set notations.
• find the cardinal number of a given set.
• find union and intersection of 2 sets.
• represent 2 sets in a Venn Diagram.
• shade required region(s) in a Venn Diagram.
Introduction
Humans tend to classify things to
make better sense of the world
around them. Take a fridge for
instance. We place frozen food in
the freezer, butter in the butter
compartment, soft drinks in the
bottle holder and fruits and vegetables in the crisper,
that is, to better preserve our different food items. Thus
we separate them into specific groups. We collect the
objects based on their common characteristics and put
them together, thus forming sets.
Similarly, in mathematics, numbers, letters, geometrical
shapes and many other things are usually classified in
collections known as sets.
Sets in Mathematics
Consider the following objects.
240
CHECK THAT YOU CAN:
• Identify
- prime/composite numbers,
- odd/even numbers,
- integers, whole numbers,
- triangular numbers,
- square numbers.
• Work with factors and multiples.
KEY TERMS
•
•
•
•
•
•
•
•
•
•
•
•
•
•
Element/Member
Set Notations
Cardinal Number of a Set
Null/Empty Set
Finite and Infinite Sets
Equivalent Sets
Equal Sets
Disjoint Sets
Universal Set
Complement of a Set
Subsets
Union of 2 sets
Intersection of 2 Sets
Venn Diagrams
Chapter 21 - Sets
Can we separate them into groups?
How many groups can we have?
Explain how you have grouped them.
We observe that each group contains objects which share the same characteristics and each
object is unique. We call each of these groups a set.
Definitio
A set is a well-defined collection of objects. The objects in a set are called its elements or
members.
We describe a set by:
(i) listing all its elements, separated by commas, within braces
or curly brackets.
e.g. {a, e, i, o, u}
(ii) a phrase or sentence within braces.
e.g. {Vowels of the alphabet}
Note: The use of brackets is
necessary to define a set.
Note: Since each member of
a set is unique, no elements
can be repeated. The order in
which the elements of a set are
listed does not matter.
Note: Capital letters are used to name sets.
e.g. A = {2, 3, 5, 7, 11},
B = {even numbers less than 10}.
We say that P = {students of Grade 7 Yellow in St Martin Secondary School} is a well-defined set
whereas Q = {students} is not a well-defined set since we do not know which students,
which grade, which school and so on.
Thus, it is very important to describe a set precisely.
Example 1
List the elements of the following sets.
(a) A = {odd numbers between 4 and 20}
(b) B = {factors of 24}
(c) K = {Days of the week starting with the letter S}
(d) L = {Letters of the word MATHEMATICS}
(e) N = {Multiples of 3 from 2 to 15}
Solution
(a) A = {5, 7, 9, 11, 13, 15, 17, 19}
(b) B = {1, 2, 3, 4, 6, 8, 12, 24}
(c) K = {Saturday, Sunday}
(d) L = {M, A, T, H, E, I, C, S}
(e) N = {3, 6, 9, 12, 15}
Caution:
There must be no
repetition of objects in a
set. So for part (d), we do
not put M twice or T twice
or A twice in the set even
if they appear twice in the
word MATHEMATICS.
241
Chapter 21 - Sets
Example 2
Describe the following sets in words.
(a) P = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}
(b) Q = {8, 12, 16, 20, 24, 28}
(c) R = {Red, Blue, Yellow, Green}
(d) X = {1, 4, 9, 16, 25, 36, 49}
Solution
(a) P ={prime numbers less than 30}
(b) Q = {multiples of 4 from 8 to 28}
(c) R = {colours of the Mauritian flag}
(d) X = {square numbers less than 50}
EXERCISE 21.1
1. Which of the following collections is not well-defined?
A.
C.
{Days starting with the letter T}
{Intelligent persons}
B.
D.
{Students in your class}
{Countries of the world}
2. Which of the following are sets? Justify your answer.
(a) The collection of all the months beginning with the letter ‘A’.
(b) The collection of all difficult chapters in your Physics textbook.
(c) The collection of all beautiful flowers in the Pamplemousses Botanical Garden.
(d) The collection of all mountain ranges in Mauritius.
(e) The collection of all integers less than 5.
3. List the elements of the following sets.
(a) A = {Factors of 32}
(b) B = {Even numbers from 2 to 12}
(c) C = {Roman numerals greater than 5 but less than 15}
(d) M = {Multiples of 6 between 10 and 40}
(e) N = {The first 15 prime numbers}
(f ) V = {Integers greater than −3 but less than 3}
4. Describe the following sets in words.
(a) {1, 2, 3, 4, 5, 6, 7, 8, 9}
(b) {Pamplemousses, Plaines-Wilhems, Port-Louis}
(c) {−3, −2, −1, 0, 1, 2, 3}
(d) {1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120}
(e) {Antarctic, Indian, Pacific, Arctic, Atlantic}
(f ) {Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune}
242
Chapter 21 - Sets
The Symbols ∈ and ∉
Consider the set A = {2, 3, 5, 7, 11}.
We say that 2 is an element of the set A as 2 is found in the set A whereas 12 is not an element
of A as 12 is not found in the set A.
A simpler way to write these statements is to make use of the symbols ∈ and ∉ where ∈ means
‘is an element of’ while ∉ means ‘is not an element of’. Thus, 2 ∈ A and 12 ∉ A.
Example
Fill in the blanks with the correct symbol ∈ or ∉.
(a) 𝑎 _____ {c, a, t}
(c) 31 _____ {composite numbers}
(b) 117 _____ {multiples of 9}
(d) Kite _____ {Triangles}
Solution
(a) 𝑎 ∈ {c, a, t}(c) 31 ∉ {composite numbers} (31 is a prime number)
(b) 117 ∈ {multiples of 9}
(d) Kite ∉ {Triangles} (a kite has 4 sides)
EXERCISE 21.2
1. A = {2, 3, 5, 7, 11, 13}
B = {b, a, s, k, e, t}
C = {7, 14, 21, 28, 35, 42} D = {c, h, a, m, b, e, r}
Tick True or False for each of the following:
TrueFalse
(a) 7 ∈ A
(b) c ∉ B
(c)
28 ∉ C
(f)
k∉B
(d) b ∈ D
(e) s ∈ A
(g) 49 ∉ C
(h) z ∈ D
2. Fill in the blanks with the correct symbol ∈ or ∉.
(a) m
(c) IV
(e) July
(g) 36
(i) 14
_____ {e, x, a, m}
_____ {VI, VII, VIII, IX, X}
_____ {Months having 31 days}
_____ {Multiples of 8}
_____ {Factors of 72}
(b) 4
(d) bat
(f ) Red
(h) 2
(j) 55
_____ {1, 3, 5, 7, 9}
_____ {b, a, t}
_____ {Colours of the Mauritian flag}
_____ {Prime numbers}
_____ {Triangular numbers}
243
Chapter 21 - Sets
Cardinal Number of a Set
The cardinal number of a set is the number of elements in the set.
The cardinal number of a given set A is denoted by n (A).
Caution:
The cardinal
number should not
be written within
brackets or braces.
e.g. Given A = {2, 3, 6, 7, 11}, n (A) = 5.
Example
Find the cardinal number of each of the following sets.
(a) X = {0, 2, 4, 6, 8, 10, 12}
(b) Y = {Months of the year}
(c) Z = {Factors of 12}
Solution
(a) n(X) = 7
(b) n(Y) = 12 as Y = {January, February, .... , December}
(c) Since Z = {1, 2, 3, 4, 6, 12}, n(Z) = 6
EXERCISE 21.3
1. Find the cardinal number of each of the following sets.
(a) A = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}
(b) B = {Days of the week}
(c) C = {Factors of 64}
(d) Y = {Multiples of 6 between 15 and 60}
(e) Z = {Prime numbers from 31 to 61}
2. Consider the following sets.
A = {Days of the week starting with the letter G},
B = {Multiples of 15 between 16 and 26} and
C = {Months with 32 days}.
What is the cardinal number of each of these sets?
Null or Empty Set ,
or
A null or empty set is a set having no elements. It is denoted by { } or the Greek letter
e.g. A = {Factors of 36 from 20 to 30}
The factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, 36
We can see that there are no factors of 36 from 20 to 30.
So A = { } or A = .
Also, the cardinal number of a null or empty set is 0.
So, in this case, n (A) = 0
244
(phi).
Chapter 21 - Sets
Example
Caution:
{0} is not an empty
set, it contains the
element 0 and so its
cardinal number is 1.
Determine whether the following sets are null/empty.
(a) A = {Whole numbers greater than 10 but less than 11}
(b) B = {Right-angled isosceles triangles}
(c) C = {Prime numbers divisible by 4}
Solution
(a) A = { } since there is no whole number between 10 and 11.
(b) B is not an empty set as we have an isosceles triangle which
can also be right-angled.
(c) C = { } as a prime number is divisble by 1 and itself only.
EXERCISE 21.4
1. What is the cardinal number of the null set?
A.
B. 0
C. {0}
D. 1
2. Which of the following is an empty set?
A. {1}
B.
C. 0
D. {0}
3. Determine whether the following sets are null/empty.
(a) A = {Integers greater than −2 but less than −1}
(b) B = {0}
(c) C = {Odd numbers divisible by 2}
(d) D = {Multiples of 17 between 20 and 30}
(e) L = {Even numbers divisible by 3}
Finite and Infini e Sets
A fini e set is a set containing a fixed number (or a finite number) of elements.
e.g. Given A = {Days of the week}, since n (A) = 7, A is a finite set.
An infini e set is a set containing an infini e (uncountable) number of elements.
e.g. The set of all whole numbers {1, 2, 3, 4, 5, ...} is an infinite set indicated by trailing dots.
Example
Classify the following as finite or infinite sets.
(a) K = {Multiples of 10}
(b) I = {Integers}
(c) L = {Letters of the alphabet}
Solution
(a) K is an infinite set as K = {10, 20, 30, ...}
(b) The set of all integers is an infinite set
as I = {..., −3, −2 ,−1, 0, 1, 2, 3, ...}
(c) L is a finite set as n (L) = 26
STOP AND THINK
Consider the sets:
A = {1, 2, 3, ... 1 000}
B = {...., -3, -2, -1, 0}
C= {-1 000 000, ..., -2,-1, 0}
Which of these sets are infinite? Why?
245
Chapter 21 - Sets
EXERCISE 21.5
1. Which of the following is a finite set?
A. {prime numbers}
C. {factors of 144}
B. {odd numbers}
D. {multiples of 10}
2. Classify the following as finite or infinite sets.
(a) X = {Factors of 100}
(c) Z = {the set of all positive integers}
(e) Q = {Multiples of 13}
(b) Y = {Vowels of the alphabet}
(d) P = {Even numbers between 0 and 100}
(f ) R = {Triangular numbers}
Equivalent Sets
Two sets are equivalent if they contain the same number of elements but the elements can be different.
e.g. Given A = {1, 2, 3, 4} and B = {w, x, y, z}, then A and B are equivalent sets since n (A) = n (B) = 4.
Example
Determine whether the following pairs of sets are equivalent.
(a) A = {1, 4, 9, 16, 25} and B = {c, h, a, r, m}
(b) C = {Multiples of 5 from 10 to 25} and D = {Consonants in the word ‘pickle’}
(c) L = {Months starting with the letter D} and M = {Factors of 9}
Solution
(a) A and B are equivalent. [n(A) = n(B) = 5]
(b) C and D are equivalent. [C = {10, 15, 20, 25}, D = {p, c, k, l}, n(C ) = n(D) = 4]
(c) L and M are not equivalent. [L = {December}, M = {1, 3, 9}, n(L) = 1, n(M) = 3]
Equal Sets
Two sets are equal if and only if they contain exactly the same elements which may or may
not be listed in the same order.
e.g. Given A = {s, e, a, m} and B = {s, a, m, e}. Then A and B are equal sets, also written as A = B,
even if the elements are not in the same order.
Note: A and B are also equivalent sets since n(A) = n(B) = 4.
Example
Determine whether the following pairs of sets are equal.
(a) A = {1, 2, 3, 4, 5} and B = {4, 5, 2, 1, 3}
(b) C = {2, 3, 5, 7} and D = {Prime numbers less than 10}
(c) L = {Months starting with the letter J} and M = {April, August}
Solution
Caution:
Two equal sets are
equivalent but two
equivalent sets are
not necessarily equal.
(a) A = B as both sets contain exactly the same elements.
(b) C = D.
[D = {2, 3, 5, 7}]
(c) L and M are not equal. [L = {January, June, July}, L = M since April ∉ L and August ∉ L]
246
Chapter 21 - Sets
EXERCISE 21.6
1. Determine whether the following pairs of sets are equivalent.
(a) A = {1, 3, 6, 10, 15, 21, 28, 36, 45} and B = {v, e, h, i, c, u, l, a, r}
(b) L = {Factors of 12} and M = {Factors of 32}
(c) P = {t, a, d, p, o, l, e} and Q = {p, r, i, n, c, e}
(d) X = {Multiples of 5} and Y = {Prime factors of 90}
2. Determine whether the following pairs of sets are equal.
(a) A = {a, c, r, e} and B = {r, a, c, e}
(b) C = {4, 6, 8, 9} and D = {Composite numbers less than 10}
(c) L = {Days starting with the letter T} and M = {Tuesday, Thursday}
(d) J = {1, 2, 8, 16} and K = {Factors of 16}
(e) V = {t, e, n, s} and W = {s, e, n, t}
Disjoint Sets
Two sets are disjoint if they have no elements in common.
e.g. Given A = {2, 3, 5, 7} and B = {4, 6, 8, 9, 10}, then A and B are disjoint sets since there are
no common elements in sets A and B.
Example
Determine whether the following pairs of sets are disjoint.
(a) A = {1, 4, 9, 16, 25} and B = {2, 3, 5, 7, 11}
(b) C = {c, h, i, p, s} and D = {Consonants in the word ‘examination’}
(c) L = {Factors of 12} and M = {Multiples of 4 from 9 to 19}
Solution
(a) A and B are disjoint as there are no common elements in sets A and B.
(b) C and D are disjoint. [D = {m, n, t, x}]
(c) L and M are not disjoint. [L = {1, 2, 3, 4, 6, 12}, M = {12, 16}, 12 ∈ L and 12 ∈ M]
EXERCISE 21.7
1. Determine whether the following pairs of sets are disjoint.
(a) A = {1, 8, 27, 64, 125} and B = {11, 13, 17, 19, 23}
(b) C = {c, h, a, i, r} and D = {d, e, s, k}
(c) L = {Prime factors of 12} and M = {Prime factors of 35}
(d) X = {Even numbers between 0 and 10} and Y = {Prime numbers less than 10}
(e) T = {Multiples of 5 from 11 to 31} and U = {Factors of 40}
247
Chapter 21 - Sets
Subset
If every element of a given set A is also an element of a set B, then we say that A is a subset of B.
e.g. If A = {6, 9} and B = {4, 6, 8, 9, 10}, then A is a subset of B.
Note: (i) All the elements of the set A are in the set B, thus A is a subset of B.
(ii) The null or empty set is a subset of every set.
(iii) Any set is a subset of itself.
The Symbols ⊂ and ⊄
Consider the sets P = {g, a, p}, Q = {p, a, r, t} and R = {g, r, a, p, e}. P is a subset of the set R as all
the elements in set P are also found in set R whereas Q is not a subset of R as t ∈ Q but t ∉ R.
These statements can be expressed more easily by making use of the symbols ⊂ and ⊄
where ⊂ means ‘is a subset of’ while ⊄ means ‘is not a subset of’.
Thus, P ⊂ R and Q ⊄ R.
Example
Fill in the blanks with the correct symbol: ⊂ or ⊄.
(a) {a} _____ {m, a, t}
(b) {2, 3} _____ {Triangular numbers}
(c) { } _____ {k, i, n, g}
(d) {Acute-angled triangles} _____ {Triangles}
Solution
(a) {a} ⊂ {m, a, t}
(b) {2, 3} ⊄ {Triangular numbers} as 2 is not a triangular number
(c) { } ⊂ {k, i, n, g}
(d) {Acute-angled triangles} ⊂ {Triangles}
EXERCISE 21.8
1. How many subsets does the null set have?
A.
0 B.
1 C.
2 D.
3
2. Which of the following is a subset of the set {2, 4, 6}?
A.
{1, 2}
B.
{2, 3}
C.
{2, 4}
D.
{3, 4}
D.
{1, 2, 3}
3. Which of the following is not a subset of the set {1, 3, 5, 7}?
A.
{1}
B.
C.
{3, 7}
4. Which of the following is not a subset of the set of integers?
A. {fractions}
B. {even numbers}
C. {square numbers}
D. {positive integers}
5. How many subsets does the set {6, 12} have?
A.
248
0 B.
1 C.
2 D.
4
Chapter 21 - Sets
6. Fill in the blanks with the correct symbol: ∈, ∉, ⊂ or ⊄.
(a) x _____ {b, r, e, x, i, t}
(b) {2, 3, 5} _____ {1, 3, 5, 7, 9}
(c) {I, III} _____ {I, II, III, IV, V}
(d) {tap} _____ {t, a, p}
(e) {June} _____ {Months having 30 days}
(f ) Pink _____ {Colours of the Mauritian flag}
(g) 63 _____ {Multiples of 9}
(h) {2, 4, 6} _____ {Composite numbers}
(i) 18 _____ {Factors of 36}
(j) 27 _____
(k) {A, B, C} _____ {c, a, b, i, n}
(l) T _____ {f, i, e, s, t, a}
{Square numbers}
(m) {b, u, t} _____ {t, u, b}
7. (a) Copy and complete the following table.
Set
Number of Elements
Subsets
Number of Subsets
Pattern
{}
0
{}
1
20
{a}
1
{ }, {a}
2
21
{a, b}
{a, b, c}
{a, b, c, d}
(b) How many subsets does a set with 7 elements have?
(c) If a set has 256 subsets, how many elements does it contain?
(d) A set has 64 subsets. How many elements does it have?
(e) Given that a set has n elements, how many subsets does it have?
Universal Set,
Consider set A = {even numbers less than 10} and B = {odd numbers less than 10}.
Then A = {2, 4, 6, 8} and B = {1, 3, 5, 7, 9}.
The set of all elements found in sets A and B is {1, 2, 3, 4, 5, 6, 7, 8, 9}.
This is called the Universal set ( ) which is {1, 2, 3, 4, 5, 6, 7, 8, 9}.
The Universal set contains the elements of all the sets under consideration. It is denoted by
the Greek letter : [xi, pronounced ksi].
249
Chapter 21 - Sets
Complement of a Set
The complement of a set A is the set of all elements which are not found in set A but are
found in the universal set. It is denoted by A'.
Note: A and A' are
e.g. If = {1, 2, 3, 4, 5, 6} and A = {2, 4, 6}, then A' = {1, 3, 5}.
disjoint sets and
n(A) + n(A') = n( ).
EXERCISE 21.9
1. Let = {1, 2, 3, ..., 10} , P = {multiples of 4} and Q = {factors of 9}. Which of the following
statements are correct?
(a) P' = {4, 8} (b) P' = {4}
(c) Q' = {1, 2, 3, 4, 5, 7, 9, 10}
(d) P' = {1, 2, 3, 5, 6, 7, 9, 10}
(e) Q' = {9} (f ) Q' = {2, 4, 5, 6, 7, 8, 10}
(g) Q' = {1, 3, 6, 9} (h) P' = {2, 4, 5, 6, 7, 8, 9}
2.
= {a, b, h, i, n, p, r, s, t, u}
A = {p, a, i, n, t}
B = {r, u, s, h}
List the elements of:
(a) A'
(b) B'
3. Given = {m, e, t, a, l, w, o, r, k} and A = {m, a, r, k, e, t}, the complement of the set A is
A. {m, e, t, a, l}
B. { }
C. {w, o, r, k}
D. {l, o, w}
Intersection of Sets, ⋂
The intersection of two given sets A and B is the set of elements common to both A and B.
It is denoted by A ⋂ B.
e.g. If A = {2, 3, 5, 7}, B = {3, 5, 7, 9}, then A ⋂ B = {3, 5, 7} as 3, 5 and 7 are common to both sets.
EXERCISE 21.10
1. Circle the correct answer.
Given P = {t, w, i, g} and Q = {w, a, g}, find P ⋂ Q .
A. {w, i, g}
B. {w, g}
C. {w, a, g}
2. Given X = {5, 10, 15, 20} and Y = {10, 20, 30}, find X ⋂ Y .
Union of Sets, ⋃
The union of two given sets A and B is the set of all elements,
which belong to either A or B or both. It is denoted by A ⋃ B.
e.g. If A = {h, e, a, t}, B = {t, a, m, e}, then A ⋃ B = {h, e, a, t, m}.
250
D. {t, w, i, g}
Caution:
Since no elements in
a set can be repeated,
though a, e and t are
found in both sets A
and B, they appear only
once in the set A ⋃ B.
Chapter 21 - Sets
EXERCISE 21.11
1. Circle the correct answer.
(a) Given P = {r, i, p, e} and Q = {p, e, a, r}, find P ⋃ Q .
A. {a, e, i, p, r}
B.
C. {p, e, r}
(b) Given P = {2, 3, 5, 7} and Q = {1, 3, 5, 7}, find P ⋃ Q .
A. {1, 2, 3, 5, 7}
B. {3, 5, 7}
C. { }
D. {1, 2, 3}
D. {2}
Example
= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
A = {1, 3, 6, 10}
B = {4, 6, 8, 9, 10}
(i) List the elements of:
(a) A ⋂ B
(b) A ⋃ B
(c) A'
(ii) Write down the value of:
(a) n(A ⋂ B)
Solution
(d) A' ⋃ B
(b) n(A ⋃ B)
(i) (a) A ⋂ B = {6, 10}
(c) A' = {2, 4, 5, 7, 8, 9}
(ii) (a) n(A ⋂ B) = 2
(c) n(A') = 6
(c) n(A')
(d) n(A' ⋃ B)
(b) A ⋃ B = {1, 3, 4, 6, 8, 9, 10}
(d) A' ⋃ B = {2, 4, 5, 6, 7, 8, 9, 10}
(b) n(A ⋃ B) = 7
(d) n(A' ⋃ B) = 8
EXERCISE 21.12
= {d, e, r, m, a, t, o, g, l, y, p, h, i, c, s}
P = {d, r, e, a, m}
Q = {d, e, m, o, g, r, a, p, h, y}
(i) List the elements of:
(a) P ⋂ Q
(b) P ⋃ Q
(c) Q'
(d) P ⋃ Q'
(ii) Write down the value of:
(a) n(P ⋂ Q)
(b) n(P ⋃ Q)
(c) n(P')
(d) n(P ⋃ Q')
(e) n(P ⋂ Q)'
Note:
(i) (A ⋃ B)’ means the
complement of (A ⋃ B).
(ii) A ⋃ B’ means the
union of set A and the
complement of set B.
(iii) A’ ⋃ B’ means the
complement of the set A
union the complement of
the set B.
(iv) (A ⋂ B)’ means the
complement of the set
(A ⋂ B), i.e., those elements
which are not found in the
set A ⋂ B.
251
Chapter 21 - Sets
Venn diagrams
DID YOU KNOW
Venn diagrams are commonly used in Mathematics to
represent and solve problems involving sets.
History of Venn Diagrams
The British
mathematician
John Venn (1834
- 1923) created
graphics which
were pictorial
representations
of the relationships between sets.
These eventually became known as Venn
diagrams. Nowadays, Venn diagrams
have become an essential tool in the
teaching of Mathematics and are also
widely used in media, by scientific
researchers, geologists, engineers,
political advisors, psychologists and
many others. Google designed a doodle
illustrating Venn diagrams for the 180th
birthday of John Venn.
The Universal set is commonly represented by a rectangle
in a Venn diagram and any subset of the Universal set is
represented by a circle. An example is shown below.
A
Representing sets on a Venn diagram
Apart from the universal set, we can also represent
different sets on a Venn diagram as shown below. The sets
are defined by the shaded regions in the Venn diagrams.
(a)
(b)
A
A
(d)
A
https://www.google.com/doodles/johnvenns-180th-birthday
(c)
A
A
A'
B
(e)
A
A⋃B
(f )
B
A
B
A
252
A⊂B
(b)
A
A⋂B
(A ⋃ B)'
(A ⋂ B)'
We also have different types of sets as illustrated below.
(a)
B
B
A and B are disjoint sets
B
Chapter 21 - Sets
Example 1
= {t, r, o, u, b, l, e, m, a, k, i, n, g, s}
(a) X = {t, r, o, u, b, l, e}
Y = {m, a, k, i, n, g, s}
Illustrate the given sets on a well-labelled Venn diagram.
Caution:
We do not use commas to
seperate the elements in
the Venn Diagram.
(b) A = {t, r, l, m, k, e}
B = {m, a , g, s}
Illustrate the given sets on a well-labelled Venn diagram.
Solution
t r o
u b l
e
(a) There are no elements common to both sets.
X
X and Y are called disjoint sets and n(X ⋂ Y) = 0
(b) A ⋂ B = {m}
Since there is one element common to both
sets A and B, the 2 sets overlap.
u
A
m a k
i n g
s
o
Y
n
a
t l
g
m
r k
s
e
i
b
B
Example 2
= {b, l, u, e, p, r, i, n, t, s}
P = {p, i, n, s}
Q = {s, p, r, i, n, t}
(i) List the elements of:
(a) P ⋂ Q (b) P ⋃ Q (c) P' ⋂ Q
(ii) Illustrate the given sets on a well-labelled Venn diagram.
(iii) Hence or otherwise, write down the value of
(a) n(P ⋂ Q)
(b) n(P ⋃ Q)
(c) n(P' ⋂ Q)
Solution
= {b, l, u, e, p, r, i, n, t, s}
(i) (a) P ⋂ Q = {p, i, n, s}
(b) P ⋃ Q = {s, p, r, i, n, t}
(c) Since P' = {b, l, u, e, r, t}, P' ⋂ Q = {r, t}
(ii)
b
l
r
u
p
n
P
Q
i s
t
e
(iii) (a) n(P ⋂ Q) = 4
(b) n(P ⋃ Q) = 6
(c) n(P' ⋂ Q) = 2
253
Chapter 21 - Sets
EXERCISE 21.13
1. Study the Venn diagram given.
(a) List the elements of:
(i) A
(ii) B
(iv) A ⋃ B (v) A'
1
(iii) A ⋂ B
(b) Find the value of:
(i) n(A)
(ii) n( )
(iv) n(A ⋃ B)
(v) n(A ⋃ B)'
4.
5
7
8
6
B
9
l
i
P
n
= {1, 2, 3, 4, 5, 6, 7, 8}
A = {Factors of 8}
B = {Multiples of 4}
(a) List the elements of:
(i) A ⋂ B
(ii) A ⋃ B
(iii) A′
(b) Illustrate the given sets on a Venn diagram.
(c) Write down the value of
(i) n(A ⋂ B)
(ii) n(A ⋃ B) (iii) n(A′ ⋃ B)
= {g, y, m, n, a, s, t, i, c}
X = {g, y, m, s}
Y = {c, a, n}
(a) List the elements of:
(i) X ⋃ Y
(ii) X′
(b) Illustrate the given sets on a Venn diagram.
(c) Write down the value of
(i) n(X ⋂ Y)
(ii) n(X ⋃ Y)
(iii) n(X′ ⋂ Y)
Shading Venn Diagrams
Example
For the two intersecting sets A and B, shade on separate diagrams,
the regions representing:
(a) B
A
B
(b) A'
(c) A' ⋂ B
(d) A' ⋃ B
254
2
(iii) n(A ⋂ B)
2. Study the Venn diagram given.
(a) List the elements of:
(i) P
(ii) Q
(iii) P ⋂ Q
(iv) P ⋃ Q
(v) Q′
(vi) P′⋂ Q
(b) Find the value of:
(iii) n(P ⋂ Q)
(i) n(Q)
(ii) n( )
(iv) n(P ⋃ Q)
(v) n(P ⋃ Q)′
3.
A
4
3
t o
u c
h
g
Q
y
Chapter 21 - Sets
Solution
(a) To represent the set B, the whole of the set B is shaded in the Venn diagram.
A
B
(b) To represent the set A', the whole region of the universal set, , outside the set A is shaded
in the Venn diagram.
A
B
(c) The sets A' and B are shaded using different types of shadings. The overlap of both
shadings gives the set A' ⋂ B. Only the required region is shaded in the final answer.
A
A
B
Working
B
A' ⋂ B
(d) The sets A' and B are shaded using different types of shadings. The whole shaded areas
give the set A' ⋃ B. Only the required region is shaded in the final answer.
A
B
Working
A
B
A' ⋃ B
255
Chapter 21 - Sets
EXERCISE 21.14
1. For the two intersecting sets P and Q, shade on
separate diagrams, the regions representing:
(a) P
(b) Q'
(c) (P ⋂ Q)'
(d) (P ⋃ Q)'
2. For the two disjoint sets L and M, shade on separate
diagrams, the regions representing:
(a) L ⋃ M (b) L'
(c) L ⋂ M' (d) L' ⋃ M (e) (L ⋃ M)'
3. Given the two sets X and Y such that X ⊂ Y, shade on
separate diagrams, the regions representing:
(a) X
(b) X' (c) X' ⋂ Y
(d) (X ⋃ Y)'
P
Q
L
M
Y
X
Summary
•
•
•
•
•
•
•
•
•
•
•
•
•
•
256
A set is a well-defined collection of objects, e.g. {a, e, i, o, u} or {vowels}.
Capital letters are used to name sets.
∈ means ‘is an element of’ while ∉ means ‘is not an element of’.
The cardinal number of a given set A is the number of elements in the set A. It is
denoted by n(A).
A null or empty set is a set having no elements. It is denoted by { } or .
Two sets are equivalent if they contain the same number of elements.
Two sets are equal if they contain exactly the same elements which may or may
not be in the same order.
Two sets are disjoint if they have no elements in common.
If every element of a given set A is also an element of a set B, then A is a subset of B.
⊂ means ‘is a subset of’ while ⊄ means ‘is not a subset of’.
The complement of a set A is the set of all elements found in but which are not
found in set A. It is denoted by A'.
The intersection of two given sets A and B is the set of elements common to both
A and B. It is denoted by A ⋂ B.
The union of two given sets A and B is the set of all elements, which belong to
either A or B or both. It is denoted by A ⋃ B.
Venn diagrams are used to represent sets.
22
STATISTICS
Chapter 22 - Statistics
Learning Objectives
By the end of this chapter, you should be able to:
• collect, classify and tabulate statistical data.
• construct and use frequency tables, pictograms and bar charts.
• interpret data in pictograms and bar charts.
Introduction
Today statistics is an important tool used by many
people in their jobs or businesses. Every day, we come
across information that is presented in a number of
different ways and in different forms.
We can find these information in newspapers, magazines,
social media, news programmes or different official
websites.
CHECK THAT YOU CAN:
• Use tally marks.
• Perform arithmetic operations.
KEY TERMS
•
•
•
•
•
•
Data Collection
Tally Marks
Frequency
Frequency table
Pictogram
Bar chart
DID YOU KNOW
History of statistics
The word statistics is derived from
the latin word ‘status’or Italian word
‘statista’ or german word ‘statistik’
each of which means ‘state’.
The information is generally presented in the form
of tables and charts. Graphs are used to interpret the
information collected in a more useful and meaningful
way. We commonly refer to these information as data.
World Statistics Day was officially
designated by the United Nations
in 2010 and is celebrated every 5
years. The first ever World Statistics
Day was celebrated on October 20,
2010.
Statistics involves the study of data collected and its interpretation.
257
Chapter 22 - Statistics
Data collection and tabulation
Activity 1
The mathematics teacher of a Grade 7 class asked each of his 25 pupils how they spent
their afternoon after school. The responses were as follows:
Watch TV
Watch TV
Play Games
Homework Play Games
Play Games
Homework
Watch TV Play Games
Read Books
Watch TV
Play Games
Read Books
Watch TV
Play Games
Play Games
Watch TV
Homework
Read Books
Homework
Watch TV
Read Books
Read Books
Watch TV
Watch TV
We say that the teacher conducted a survey. The information gathered is called data. The
process is known as data collection.
The data can then be organised in a table consisting of three columns as shown below.
Copy and complete the table.
Activities
Tally
Frequency
Play Games
IIII II
7
Read Books
This table is called a Frequency table. A Frequency table shows the number of times
(frequencies) a particular observation occurs. Such tables are used to construct several
types of charts in statistics.
From the table above, write down
(a) how many students watch TV after school,
(b) the most common activity after school,
(c) the least common activity after school.
258
Chapter 22 - Statistics
EXERCISE 22.1
1. A survey was carried out to find students’ preference regarding the types of books they like to read.
Part of the information is represented in the frequency table below. Copy and complete the table.
Type of book
Tally
Action
IIII IIII II
IIII IIII IIII I
Drama
Frequency
Fiction
8
Horror
5
IIII IIII IIII
Romance
2. The data below shows the number of cups of coffee that 24 employees working in a company
have per day:
2
1
5
1
3
0
1
2
0
4
2
2
2
4
2
1
1
5
2
2
4
3
4
5
(a) Copy and complete the frequency table below.
Number of cups of coffee
Tally
Frequency
0
1
2
3
4
5
(b) How many cups of coffee do most employees drink per day?
3. In a Grade 7 class, 4 students: Nilesh, David, Kavi and Dev submitted their names to be elected
as class captain. A vote was carried out by the teacher and the results were as follows:
Nilesh
KaviDevNileshDevNileshDev
KaviDavid
DevDevKaviDevDev
KaviDev DavidDevDevKaviDavid
DevNileshKaviDavidDavidDevKavi
(a) Complete the frequency table below:
Name
Nilesh
David
Kavi
Dev
Tally
Frequency
(b) Who was elected as class captain?
(c) If the class were to select a vice class captain, who would it be? Explain why.
259
Chapter 22 - Statistics
Pictograms
DID YOU KNOW
A pictogram is a chart which consists of pictures or
symbols to represent data. To create a pictogram, we
construct a table with a key. The key may be a symbol
or a picture which represents the data.
Example of a key can be
or
to represent 4 flowers
to represent 1 student.
History of Pictograms
In earlier times, pictograms
also known as pictographs
were drawings carved or
painted on the surface of
rocks. They consisted of
pictures of the objects they
represented and were used mainly to record the
events in the lives of the prehistoric humans.
Example 1
The pictogram below shows the number of pizzas ordered by Grade 7 students of St William
School over a period of 5 months.
Pizzas ordered over a period of 5 months
Months
Number of Pizzas ordered
February
March
April
May
June
Key:
represents 10 pizzas
(a) How many pizzas were ordered in February?
(b) How many more pizzas were ordered in March than in April?
(c) How many pizzas were ordered over the 5 months?
Solution
(a) Number of pizzas ordered in February= (2 × 10) = 20
(b) Number of pizzas ordered in March = (4 × 10) = 40
1
Number of pizzas ordered in April = (1 × 10) + ( × 10) = 15 or (1.5 × 10) = 15
2
Therefore there were (40 – 15) = 25 more pizzas ordered in March than in April.
1
2
(c) Total number of keys = 2 + 4 + 1 + 1 + 4
1
= 13
2
Number of pizzas ordered over the 5 months = (13 × 10) = 130
260
Chapter 22 - Statistics
Example 2
Manisha recorded the number of drinks sold from a drinks machine at the school canteen
over 1 week. The information is given in the table below.
Day
Monday
Tuesday
Wednesday
Thursday
Friday
Number of drinks
50
45
30
20
55
(a) On which day were the most number of drinks sold ?
(b) How many more drinks were sold on Friday than on Tuesday ?
(c) Draw a pictogram to illustrate the information given in the above table. Use
to
represent 10 drinks.
(d Express the number of drinks sold on Monday to that sold on Thursday as a ratio in its
simplest form.
(e) Express the number of drinks sold on Wednesday as a percentage of the total number of
drinks sold over the week.
Solution
(a) The most number of drinks were sold on Friday.
(b) Number of drinks sold on Friday = 55
Number of drinks sold on Tuesday = 45
Therefore there were (55 – 45) = 10 more drinks sold on Friday than in Tuesday.
(c)
Number of drinks sold from a drinks machine over a week
Day
Number of drinks sold over 1 week
Monday
Tuesday
Wednesday
Thursday
Caution:
When you need to use
only part of the key
for e.g. half of the key,
ensure that you divide
the key in such a way
that you obtain two
equal parts. Explain
why these keys are not
appropriate:
Top half
Friday
Bottom half
.
(d) Ratio = 50 : 20 = 5 : 2
(e) Total number of drinks sold = 50 + 45 + 30 + 20 + 55 = 200
Percentage = 30 x 100 % = 15 %
200
POINTS TO REMEMBER:
• A key is essential in a pictogram as it shows the meaning of the symbol.
• Each symbol should be of the same size.
• A title should be given to the pictogram.
261
Chapter 22 - Statistics
EXERCISE 22.2
1. The following pictogram shows the number of handbags sold at a shop over the last six
months.
Number of handbags sold at a shop over the last six months
Months
Number of handbags sold
January
February
March
April
May
June
Key:
represents 10 handbags
(a) During which month was the highest number of handbags sold?
(b) During which two months were the same number of handbags sold?
(c) During which month were 55 handbags sold?
(d) How many more handbags were sold in March than in May?
(e) Express the number of handbags sold in June to the number sold in February as a ratio in its simplest form.
(f) Express the number of handbags sold in the months of March and June as a percentage of
the total number of handbags sold over the last six months.
2. Navina, Samantha and Pazani saved their pocket money for the month of April so that they
could buy presents for Mothers’ Day. The table below shows their savings.
Navina
Samantha
Pazani
Rs 280
Rs 160
Rs 420
(a) Use the above data to complete the pictogram below using an appropriate key and answer
the questions which follow:
Navina
Samantha
Pazani
(b) Provide an appropriate title for the pictogram.
(c) How much more money did Navina save than Samantha?
(d) Navina wants to buy roses for her mother. One rose costs Rs 20. How many roses will she get?
(e) Pazani buys a dress at Rs 350 for her mother and a card with the rest of the money.
How much does the card cost?
262
Chapter 22 - Statistics
3. Liberty Ltd exports flowers. The following pictogram shows the number of flowers
shipped during 6 weeks.
Number of flowers shipped during 6 weeks
Weeks
Number of flowers
Week 1
Week 2
Week 3
Week 4
Week 5
Week 6
(a) Given that the number of flowers shipped in Week 1 was 5 500, find the value represented
by the key
.
(b) Find the number of flowers shipped in Week 3.
(c) If the number of flowers shipped in Week 6 is twice the number of flowers shipped in
Week 3, complete the pictogram for Week 6.
(d) How many flowers were shipped altogether during the 6 weeks?
4. Eric cultivates dragon fruits for the local market. The following pictogram shows the
number of dragon fruits that he harvested from 2012 to 2016.
Each
represents 2 000 dragon fruits.
Number of dragon fruits harvested from 2012 to 2016
Months
Number of dragon fruits harvested
2012
2013
2014
2015
2016
(a) How many dragon fruits were harvested in 2014?
(b) How many dragon fruits were harvested from 2012 to 2016 inclusive?
(c) What percentage of the total number of dragon fruits was harvested in 2013?
263
Chapter 22 - Statistics
Bar Charts
Number of students
A bar chart is a representation of data obtained by drawing rectangular bars of equal width. The
length of the bars is proportional to the value that they represent. Equal spaces are left between
two consecutive bars. The bars are represented vertically or horizontally. Vertical bar charts
are also called column bar charts and horizontal bar charts are also called row bar charts.
An example is illustrated below.
Points to remember when drawing
Means of transport used by students
a bar chart:
• Title of the chart
250
(e.g. Means of transport used by
students)
200
• Label axes
(e.g. Number of students, Means of
150
transport)
• Suitable scale
100
(e.g. 0, 50, 100, ...)
• Draw bars of equal width
50
• Leave equal spaces in between bars
0
• Length of each bar represents
Bus
Car
Bicycle
On Foot
its respective frequency.
Means of transport
Example
The results of a survey of favourite pets in a class of 30 students were as follows:
Favourite Pets
Dogs
Cats
Rabbits
Birds
Fish
No. of students
10
8
3
4
5
Study the above frequency table and answer the following questions:
(a) On graph paper, draw a vertical bar chart using a scale of 1 cm to represent 1 pet.
(b) What is the most favourite pet?
(c) Calculate the percentage of students who prefer rabbits.
(d) What fraction of the class chose fish as their favourite pet?
(e) What is the ratio of students who prefer birds to cats?
Solution
(a) 10
Favourite pets in a class of 30 students
9
(b) The most favourite pet is "Dogs".
= 10 %
(d) Fraction of the class who chose fish
= 5 = 1
30
6
(e) Ratio of birds : cats
4 : 8
1 : 2
264
7
Number of students
(c) Number of students who prefer rabbits = 3
3
% of students who prefer rabbits = x 100 %
30
8
6
5
4
3
2
1
0
Dogs
Cats
Rabbits Birds
Favourite Pets
Fish
Chapter 22 - Statistics
Activity
Harry, Wesley, James, Alvina and Sophie sat for a selection test for the National Mathematics
Olympiad.
Wesley obtained 10 marks less than James. The ratio of Harry's marks to James’ marks is 3 : 2.
Alvina obtained 15 marks less than Harry. Sophie obtained 30 marks more than Wesley.
Complete the following bar chart and find out who was selected for the competition, given
that the student with the highest mark was selected.
National Mathematics Olympiad
90
80
70
Marks obtained
60
50
40
30
20
10
0
Harry
Wesley James
Alvina
Name of Participants
Sophie
GeoGebra
The bar chart can also be drawn using the mathematical softwares Microsoft Excel and Geogebra.
Below are snapshots from Geogebra and Excel on “Favourite fruits of students in a class” given
in the following table:
Name of fruit
Number
of students
Page
13
Banana
Apple
Mango
Orange
10
15
8
7
16
14
No of students
12
10
8
6
4
2
0
Banana
Apple
Mango
Orange
Name of fruits
Chart drawn using Microsoft Excel
Page 14 ex 3
9
Chart drawn using GeoGebra software
265
Chapter 22 - Statistics
EXERCISE 22.3
1. The table below shows the number of different types of vehicles sold in the year 2016 in Country A.
Type of vehicles
Lorry
Car
Van
Bus
Number of vehicles
400
1 100
700
150
Using a scale of 1 cm to represent 100 vehicles, illustrate the above information on a vertical
bar chart on graph paper.
2. Twenty five students of Grade 7A were asked to state their favourite subjects and the results
were recorded in the following table.
Subjects
Maths
English
French
Science
Number of students
12
4
2
7
On graph paper, draw a horizontal bar chart to represent the above data using an appropriate scale.
3. The bar chart below shows how the students of Grade 7B travel to Ashford Secondary School
daily.
8
Number of students
7
6
5
4
3
2
1
0
Walk
Van
Bicycle Bus
Mode of Transport
(a) Use the information to complete the following table.
Transport
Number of students
Car
Car
Bicycle
Bus
Walk
1
(b) How many students are there in Grade 7B?
(c) How many students go to school on foot?
(d) What is the most frequent mode of transport of the students of Grade 7B?
(e) How many more students come to school by bus than by car?
(f) Provide a suitable title for the above bar chart.
266
Van
Chapter 22 - Statistics
4. Use the bar chart to answer the following questions.
Favourite fruit of students in a class
8
7
Number of students
6
5
4
3
2
1
0
Apples Mangoes Longans Grapes
Favourite Fruits
Litchis
Oranges
(a) How many students are there in the class?
A. 3
B. 8
C. 30
D. 35
(b) How many more of them prefer litchis to mangoes?
A. 6
B. 9
C. 2
D. 14
(c) What is the ratio of students who prefer mangoes to apples?
A. 3 : 6
B. 8 : 3
C. 2 : 1
D. 1 : 2
(d) What fraction of students prefer oranges?
3
A. B. 35
C. 1
3
5
35
(e) What percentage of students prefer grapes?
A. 7% B. 20%
C. 35%
D. 3
8
D. 30%
5. Lakshita from class A and Lakshana from class B decide to do a survey on the favourite
subject of their respective classmates. Each class has a population of 40 students.
Lakshita represents her results on a pictogram and Lakshana represents hers on a bar chart.
Favourite subject of Lakshita's classmates
Subject
Number of students
English
French
Maths
Science
Social Studies
represents 2 students
267
Chapter 22 - Statistics
Favorite subject of Lakshana's classmates
CHECK THIS LINK
14
http://www.mathsisfun.com/
data/data-graph.php
Number of students
12
10
8
6
4
2
0
English French
Maths Science
Social Studies
(a) What is the ratio of students who prefer English in Lakshana's class to that of students who
prefer English in Lakshita's class?
A. 14 : 9
B. 9 : 14 C. 7 : 9
D. 9 : 7
(b) What percentage of students prefer Science in (i) Lakshana's class, (ii) Lakshita's class?
(c) Draw a bar chart to represent the information given in Lakshita's pictogram.
(Use an appropriate scale).
6. The pictogram below shows the total savings of a group of people in 2017.
Key: Rs represents Rs 15 000.
Total savings of a group of people in 2017
Day
Total savings of a group of people
Liam
Rs
Rs
Rs
Rs
Sayan
Rs
Rs
Rs
Rs
Aditi
Rs
Rs
Rs
Devraj
Rs
Rs
Rs
Emma
Rs
Rs
Sanya
Rs
Rs
Rs
Rs
Rs
Rs
Rs
Study the pictogram above and answer the following questions.
(a) How much money did Sanya save? (b) Who saved the greatest amount of money? How much did he/she save?
(c) What is the ratio of Emma’s savings to that of Devraj?
Express your answer in simplest form.
268
Chapter 22 - Statistics
(d) Express Aditi’s savings as a percentage of Emma’s savings.
(e) The amount of money saved by Sayan in 2017 represents 25% of his total income for the
year 2016. Calculate Sayan's income in 2016.
(f) Given that $1 = Rs 36, how many dollars was Liam’s savings worth in 2017?
(g) Aditi’s brother, Vihaan, saved twice as much as his sister.
On a sheet of graph paper and using an appropriate scale, draw a bar chart to illustrate
the savings of the seven persons.
7. Pablo recorded the number of text messages that he received per day over a period of 3 weeks.
The results are shown below.
15
15
10
9
10
11
10
14
15
12
14
13
9
13
10
13
15
15
11
12
14
(a) Draw a frequency table to show Pablo's results.
(b) On a sheet of graph paper and using a scale of 1 cm to represent 1 day, illustrate the
results of the frequency table on a bar chart.
(c) What is the most frequent number of text messages?
(d) Express the number of days where he received 9 text messages as a percentage of the
number of days where he received 15 text messages. (e) Use a pictogram with an appropriate key to illustrate the information of the frequency
table above.
(f) Which of the bar chart or pictogram was easier for you? Give your reasons.
Investigate:
You need to prepare a talk on the eating habits of people at breakfast. To do so, you will need
information about what people eat in the morning (for example: bread, cereal, fruits, eggs, etc)
to be able to deduce a common eating habit.
• Ask around 20–25 people in your family, friends or neighbours about what they have as
breakfast.
• Draw a frequency table to show the information collected. Use the following categories
(1) Bread, (2) Cereal, (3) Fruits, (4) Other and (5) None.
• Use a mathematical software to draw either a pictogram or a bar chart to represent the
information that you have collected.
• Comment on the chart drawn. (What is the most common food that most people take for
breakfast? Describe the eating habit of the group of people in your survey,etc.)
Summary
• A frequency table shows the number of times a feature is observed.
• A pictogram is a chart which consists of pictures or symbols to represent data.
• A key is essential when drawing a pictogram.
• A bar chart is a representation of data. We use equally spaced rectangular bars
of equal width in a bar chart.
• A title is important when drawing either a pictogram or a bar chart.
269