Revisi Cepat PT3 Mathematics KSSM Form 1, 2 & 3 - PDF Flipbook

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PENERBIT
ILMU
BAKTI SDN. BHD.

Contents

PENERBITFORM 1 Chapter 5 Algebraic Expressions
ILMU
BAKTI SDN. BHD.Chapter 1 Rational Numbers 1 55

1.1 Integers 2 5.1 Variables and Algebraic

1.2 Basic Arithmetic Operations Expressions 56

Involving Integers 5 5.2 Algebraic Expressions Involving

1.3 Positive and Negative Basic Arithmetic Operations 59

Fractions 9 Summative Assessment 5 62

1.4 Positive and Negative Chapter 6 Linear Equations 64

Decimals 13 6.1 Linear Equations in One

1.5 Rational Numbers 16 Variable 65

Summative Assessment 1 18 6.2 Linear Equations in Two

Chapter 2 Factors and Multiples Variables 68

21 6.3 Simultaneous Linear Equations in

2.1 Factors, Prime Factors and Two Variables 71

Highest Common Factor (HCF) 22 Summative Assessment 6 77

2.2 Multiples, Common Multiples and Chapter 7 Linear Inequalities 80

the Lowest Common 24 7.1 Inequalities 81
Multiple (LCM)
7.2 Linear Inequalities in One
Summative Assessment 2 27
Variable 82

Chapter 3 Squares, Square Summative Assessment 7 85
Roots, Cubes and
Cube Roots 29 Chapter 8 Lines and Angles 87

3.1 Squares and Square Roots 30 8.1 Lines and Angles 89
3.2 Cubes and Cube Roots 34
Summative Assessment 3 39 8.2 Angles Related to Intersecting

Lines 99

8.3 Angles Related to Parallel Lines

Chapter 4 Ratios, Rates and and Transversals 102

Proportions 41 Summative Assessment 8 106

4.1 Ratios 42 Chapter 9 Basic Polygons 110

4.2 Rates 44

4.3 Proportions 45 9.1 Polygons 111

4.4 Ratios, Rates and Proportions 46 9.2 Properties of Triangles and the

4.5 Relationship Between Ratios, Interior and Exterior Angles of

Rates and Proportions, with Triangles 112

Percentages, Fractions and 9.3 Properties of Quadrilaterals
Decimals 51
and their Interior and Exterior

Summative Assessment 4 53 Angles 115

Summative Assessment 9 119

ii

Chapter 10 Perimeter and Area Chapter 3 Algebraic Formulae
122 199

10.1 Perimeter 123 3.1 Algebraic Formulae 200
10.2 Area of Triangles, Parallelograms, Summative Assessment 3 205

Kites and Trapeziums 125 Chapter 4 Polygon 207

10.3 Relationship Between Perimeter

and Area 128 4.1 Regular Polygon 208

Summative Assessment 10 132 4.2 Interior Angles and Exterior
PENERBIT
ILMU Angles of Polygons 213
BAKTI SDN. BHD.Chapter 11 Introduction of Set Summative Assessment 4 218

136

11.1 Set 137 Chapter 5 Circles 221

11.2 Venn Diagrams, Universal Sets, 5.1 Properties of Circles 222
5.2 Symmetry and Chords 225
Complement of a Set and Subsets 5.3 Circumference and Area
228
139 of a Circle 232
Summative Assessment 5
Summative Assessment 11 141

Chapter 12 Data Handling 143

12.1 Data Collection, Organisation Chapter 6 Three-Dimensional
Geometrical Shapes
and Representation Process, 234

and Interpretation of Data

Representation 144 6.1 Geometric Properties of Three-

Summative Assessment 12 159 Dimensional Shapes 235

Chapter 13 The Pythagorasʹ 6.2 Nets of Three-Dimensional

Theorem 163 Shapes 238

13.1 The Pythagorasʹ Theorem 164 6.3 Surface Area of Three-

13.2 The Converse of the Dimensional Shapes 239

Pythagorasʹ Theorem 166 6.4 Volume of Three-Dimensional

Summative Assessment 13 168 Shapes 243

Summative Assessment 6 249

FORM 2 Chapter 7 Coordinates 252

Chapter 1 Patterns and 7.1 Distance in a Cartesian
Sequences 170
Coordinate System 253

1.1 Patterns 171 7.2 Midpoint in the Cartesian
1.2 Sequences 173
1.3 Patterns and Sequences 175 Coordinate System 255
Summative Assessment 1
181 7.3 The Cartesian Coordinate System

258

Chapter 2 Factorisation and Summative Assessment 7 259
Algebraic Fractions
184 Chapter 8 Graphs of Functions
262

2.1 Expansion 185 8.1 Functions 264
8.2 Graphs of Functions 268
2.2 Factorisation 189 Summative Assessment 8 277

2.3 Algebraic Expressions and Basic

Arithmetic Operations 193

Summative Assessment 2 197

iii

Chapter 9 Speed and 281 Chapter 3 Consumer
Acceleration Mathematics: Savings
282 And Investments,
9.1 Speed 286 Credit and Debt 377
9.2 Acceleration 289
Summative Assessment 9 3.1 Savings and Investments 379

Chapter 10 Gradient of a Straight 3.2 Credit and Debt Management

Line 292 397

10.1 Gradient 293 Summative Assessment 3 406
Summative Assessment 10 299
PENERBIT Chapter 4 Scale Drawings 408
Chapter 11 IsometricILMU 302
TransformationsBAKTI SDN. BHD.4.1 Scale Drawings 409
303 Summative Assessment 4 415
11.1 Transformations 306
11.2 Translation 309 Chapter 5 Trigonometric Ratios
11.3 Reflection 313 418
11.4 Rotation
11.5 Translation, Reflection and 317 5.1 Sine, Cosine and Tangent of Acute
319
Rotation as an Isometry 322 Angles in Right-Angled Triangles
11.6 Rotational Symmetry
Summative Assessment 11 419

Summative Assessment 5 431

Chapter 6 Angles and Tangents

Chapter 12 Measures of Central of Circles 434

Tendencies 326 6.1 Angle at the Circumference and

12.1 Measures of Central Central Angle Subtended by an
Tendencies
327 Arc 436
Summative Assessment 12 339
6.2 Cyclic Quadrilaterals 440

6.3 Tangents to Circles 443

Chapter 13 Simple Probability344 6.4 Angles and Tangents of

13.1 Experimental Probability 345 Circles 448
13.2 Probability Theory Involving
347 Summative Assessment 6 450
Equally Likely Outcomes
13.3 Complement of an Event 350 Chapter 7 Plans and Elevations
351
Probability 353 453
13.4 Simple Probability
Summative Assessment 13 7.1 Orthogonal Projections 454
7.2 Plans And Elevations 458
Summative Assessment 7 464

FORM 3 356 Chapter 8 Loci in Two 468
Dimensions
Chapter 1 Indices 357 469
358 8.1 Loci 470
1.1 Index Notation 364 8.2 Loci in Two Dimensions 475
1.2 Law of Indices Summative Assessment 8
Summative Assessment 1

Chapter 2 Standard Form 366 Chapter 9 Straight Lines 479

2.1 Significant Figures 367 9.1 Straight Lines 480
2.2 Standard Form 370 Summative Assessment 9 491
Summative Assessment 2 375
SCAN THE QR CODE TO
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• Answers

iv

Chapter 1 Rational Numbers

Diagrammatic Notes

1.1 Integers 1.3 Positive and Negative Chapter
Fractions
1

Integer is a whole number that can 1.4 Positive and Negative Form 1
be positive, negative or zero. Decimals
PENERBIT
ILMUNegative number –1.2 –0.4 0.8 1.6
BAKTI SDN. BHD.
3 negatif 3 1 4 2 2 4 1 3 positive
2 5 5 5 5 5 5 5 5
Positive 1 Positive fraction –1 –1 – – 0 1 1 fraction
number number

–3 –2 –1– 1O 1 2 3 –1.6 –0.8 0.4 1.2
–2
–3 Combined Operations

Negative number 2.5 + 0.8 × 2 Perform
= 2.5 + 1.6 multiplication first
= 4.1
1.2 Basic Arithmetic
Operations Involving 400 – 20.5  5 Perform division
Integers = 400 – 4.1 first
= 395.9
a + (+b) = a + b (+) × (+) = +
a + (–b) = a – b (–) × (–) = + 12.5 – 1 + 3.5 If the operation involves
a – (+b) = a – b (+) ÷ (+) = + (+ and –) or (× and ÷),
a – (–b) = a + b (–) ÷ (–) = + = 11.5 + 3.5 calculate from the left to
= 15 the right

Arithmetic Operation Laws 16.4 × 10  (4 + 96)
= 164  100
The Commutative Law = 1.64 Perform in
a+b=b+a bracket first
a×b=b×a

The Distributive Law 1.5 Rational Numbers
a × (b + c) = a × b + a × c
a × (b – c) = a × b – a × c Rational numbers ➡ numbers that
p
can be written as the fraction q where

The Associative Law p and q are integers and where q ≠ 0.

(a + b) + c = a + (b + c) Example:
(a × b) × c = a × (b × c) 0.2

Identity Law –5 2 4
5
a + 0 = a 1 0.2 = 2 –5 = –5
a + (– a) = 0 a× a =1 10 1 4 14
a × 0 = 0 2 5 = 5
a×1=a = 1
5

1

Chapter Chapter Learning Area: Numbers and Operations

1 1 Rational Numbers

PENERBIT Form11.1 Integers (b) Temperatures at two different
ILMU places
BAKTI SDN. BHD.Positive and Negative
Numbers Malaysia: North Pole:
31°C –22°C
 1 Positive numbers are numbers
that are greater than zero. Example 1
Positive numbers are written with
a ‘+’ sign or without any sign. Write down each value described in the
Example: + 15, +50, +226, 31, 8.95, statements below as a positive value or
14—27 a negative value.
(a) Dividend of 12%
 2 Negative numbers are numbers (b) Increase of 4 000 litres in the
that are smaller than zero.
Negative numbers are written volume of water
with a ‘–’ sign. (c) A debt in the amount of RM2 500
Example: –29, –54.03, –33—59 (d) A drop in the velocity of a car by

Negative numbers Positive numbers 45 km h–1

(–) 0 (+) Solution
(a) +12
 3 Positive and negative numbers (b) +4 000
exist in our daily lives.
(a) Position above sea level

2 000 2 000 m (c) –2 500
above (d) –45
sea level

0 Sea level Integers
-155 155 m below
sea level  1 Integers include the positive
whole numbers {1, 2, 3, …}, zero {0}
and the negative whole numbers
{–1, –2, –3, …}.

Key Terms • Integer – Integer

• Zero – Sifar 2

Tree Map  3 For a number that is smaller
than 0, its value decreases as its
Integers position is further left on the
number line. These numbers are
Negative integers Zero Positive integers negative integers. Chapter

–1, –2, –3, –4, ... 0 1, 2, 3, 4, ... 2 is smaller than 9. 1
0 is greater than –7.
Diagram 1.1 Components of integers Form 1
PENERBIT
ILMU 2 Fractions and decimals are not
BAKTI SDN. BHD.integers.

Decreases in value Increases in value

Example 1 –10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10
Identify and circle all the integers.
Scan the QR Code or visit PAK-21
Solution https://www.youtube.com/
watch?v=5oHJcmYbHvA
68 0.45 36 –8.99 to view a video on the
0 definition and other examples
–3.5 –9 involving integers.

2 1 508 – 5 Info Gallery
2 6
3
–92 7 73 • There are two types of 0
number lines:

(a) Horizontal number line 0

 3 Integers can be written in (b) Vertical number line
numerals or in words.
• The arrow heads at both ends of
Example: –5 (in numeral) →
the number line mean that the
negative five (in words)
numbers extend to infinity in both

directions.

Represent Integers on a Example 1

Number Line Represent the following integers on a
number line.
  1 Integers can be represented in (a) 60, 18, 12, 0, 48
different ways. Representation on
a number line gives us an idea of 0 12 18 48 60
the value of a number.
(b) –5, –7, 1, 6, 4, –2
  2 For a number that is greater than
zero, its value increases as its –7 –5 –2 1 4 6
position is further right on the
number line. All these numbers
are positive integers.

Key Terms • Decimal – Perpuluhan • Number line – Garis nombor

• Fraction – Pecahan 3

Example 2 Integers Ascending Descending
order order
Determine the smaller integer based
Chapter on its position on the number line. (a) –11, 9, –1, –11, –4, 9, 2, 0, –1,
Give your reason. 0, 2, –4 –1, 0, 2, 9 –4, –11
1
Integers Reason (b) 35, –6, –50, –32, 56, 35, –6,
–32, –25, –25, –6, –25, –32,
–50, 56 35, 56 –50
PENERBIT Form1(a) 5, 18 5 is smaller as it is on the
ILMU left of 18
BAKTI SDN. BHD.
(b) 5, –3 –3 is smaller as it is on Quick Review 1.1
the left of 5
 1 Mr Paker and his wife, Serena, are
Compare and Arrange in the lobby of a hotel. Mrs Serena
Integers in Order takes the lift up to their room on
level 7 while Mr Paker goes two
 1 We can compare the values of floors down from the lobby to the
integers using the number line. car park. State the position of Mr
Paker and that of Mrs Serena from
Ascending order the hotel lobby.

–10 –9 –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 10  2 The peak of Mount Everest is
8 850 m above sea level while the
Descending order lowest point, Challenger Deep, is
Diagram 1.2 Position of integers on the at a depth of 10 911 m below sea
level. Represent the position of
number line the peak of Everest and that of the
Challenger Deep from the sea level.
Example 1
 3 Identify and circle all the integers
Compare the numbers and write ‘is in the list of numbers below.
greater than’ or ‘is less than’.
(a) 9 is greater than 4 23 – —21 –41 24.79

9 is located –8.8 –0.067 200 1–12—41
to the right
4 9 of 4.  4 Rearrange the following list of
integers in the correct order.

(b) –47 is less than 23 Integers Ascending Descending
order order

–47 23 –47 is located (a) 7, 11, 0,
to the left –1, 10, –3
of 23.

Example 2 (b) –15, –65,
–23, –17,
Rearrange the following list of integers 17, 31
in the correct order.

Key Terms

• Order – Tertib

4

Basic Arithmetic Example 2
1.2 Operations (a) 7 – 9 = –2

Involving Integers –9 Chapter

Add and Subtract Integers 7 1
–2 –1 0 1 2 3 4 5 6 7
  1 We can use the number line to
add and subtract integers. (b) –1 – 5 = –6 Form 1

 2 Addition can be carried out –5
by moving to the right on the
number line.
PENERBIT –1
ILMU
BAKTI SDN. BHD. –6 –5 –4 –3 –2 –1 0

Example 1 +4  4 Subtraction is the inverse
(a) 3 + 4 = 7 operation of addition. Therefore,
subtracting a negative number is
3 equivalent to adding a positive
number.
0 1 2 3 4 5 6 7
Example 3
(b) –5 + 4 = –1 –4 – (–7) = –4 + 7 = 3

–5 +7

+4 –4
–6 –5 –4 –3 –2 –1 0 1 –4 –3 –2 –1 0 1 2 3

Alternative Method Multiply and Divide Integers

Coloured chips can be used to solve  1 The product of two integers that
problems involving the addition and have the same sign is positive.
subtraction of integers.
Let + represent a positive integer  2 The product of two integers that
and – represent a negative integer. have different signs is negative.

–5 – – – – –  3 Multiplication of two integers:
+4 + + + + Balance –1
(+) × (+) = +
 3 Subtraction can be carried out by (–) × (–) = +
moving to the left on the number (+) × (–) = –
line. (–) × (+) = –

 4 The product of any integer with
zero is zero.
Example: 8 × 0 = 0
(–13) × 0 = 0

5

Chapter Example 1 (c) –10 × (9) = Brackets Divide/ Flow Map
(d) –12 × (–4) = () multiply Add/
1 Calculate: (left to
(a) –6 × –6 = subtract
(b) 5 × (–8) = right) (left to
right)
Solution

(a) –6 × –6 = 36
(b) 5 × (–8) = –40
(c) –10 × (9) = –90
(d) –12 × (–4) = 48
PENERBIT Form1 Example 1
ILMU 5 When you divide two integers
BAKTI SDN. BHD.with the same sign, the quotientSolve and compare the values obtained.
is positive. Does each of the following pairs of
operations give the same or different
 6 When you divide two integers values?
with diferent signs, the quotient (a) 505 244 + (–234 619 – 6 785) and
is negative.
505 244 – (234 619 – 6 785)
 7 Division of two integers: (b) 280 × (–70 ÷ 50) and [280 × (–70)]

(+) ÷ (+) = + ÷ 50
(–) ÷ (–) = + (c) [71 760 ÷ (–3)] × 40 – 5 and
(+) ÷ (–) = –
(–) ÷ (+) = – 71 760 ÷ (–3 × 40) – 5

Example 2 Solution
(a) 505 244 + (–234 619 – 6 785)
Calculate: (c) 72 ÷ (–8) = = 505 244 + (–241 404)
(a) 24 ÷ 3 = (d) –54 ÷ (–9) = = 263 840
(b) –24 ÷ 4 =
505 244 – (234 619 – 6 785)
Solution = 505 244 – (227 834)
= 277 410
(a) 24 ÷ 3 = 8
(b) –24 ÷ 4 = –6 The values obtained are different.
(c) 72 ÷ (–8) = –9 (b) 280 × (–70 ÷ 50)
(d) –54 ÷ (–9) = 6 = 280 × (–1.4)
= –392
Perform Computations
Involving Combined Basic [280 × (–70)] ÷ 50
Arithmetic Operations of = (–19 600) ÷ 50
Integers = –392

 1 Calculations involving brackets The values obtained are the same.
and combined operations of (c) [71 760 ÷ (–3)] × 40 – 5
integers follow the order of = (–23 920) × 40 – 5
operations for whole numbers = (–956 800) – 5
(BODMAS). = –956 805

71 760 ÷ (–3 × 40) – 5
= 71 760 ÷ (–120) – 5
= –598 – 5
= –603

The values obtained are different.

6

Describe the Laws of  3 Associative Law
(a) For the integers a, b and c,
Arithmetic Operations (a + b) + c = a + (b + c)
(a × b) × c = a × (b × c)
  1 Identity Law Example: Chapter
(6 + 7) + 8 = 6 + (7 + 8)
(a) For any integer a, 2 × (3 × 4) = (2 × 3) × 4 1
(b) The operations of subtraction
a + 0 = a a + (–a) = 0 and division are not associative. Form 1
a × 0 = 0 a × —a1 = 1
  4 Distributive Law
a×1=a For the integers a, b and c, the
PENERBIT product of a with the sum of b
ILMU(b) The additive identity is 0. If and c is equal to the total of the
BAKTI SDN. BHD. products ab and ac. Subtraction
0 is added to any number, the is also distributive.

number remains unchanged. a × (b + c) = a × b + a × c
a × (b – c) = a × b – a × c
Example: 9 + 0 = 9
Example:
0 + 8 = 8 (a) 8(7 + 6) = (8 × 7) + (8 × 6)
(b) 4(9 – 6) = (4 × 9) – (4 × 6)
(c) The multiplicative identity

is 1. If 1 is multiplied by any

number, the number remains

unchanged.

Example: 1 × 35 = 35

–6 × 1 = –6

  2 Commutative Law

For the integers a and b,

a + b = b + a and a × b = b × a

Example: –11 + (–8) = (–8) + (–11) Example 1
–11 × (–8) = (–8) × (–11)
Determine the laws involved in the
following operations.

Info Gallery Operation Law

(a) Subtraction is not commutative. (a) 9(6 + 13) Distributive law
a–b≠b–a = 9(6) + 9(13)
Example: 18 – 4 = 14
4 – 18 = –14
Therefore, 18 – 4 ≠ 4 – 18 (b) 20 + (9 + 5) Associative law
= (20 + 9) + 5
(b) Division is also not commutative.
a÷b≠b÷a
Example: –24 ÷ (–8) = 3 (c) 65 × 0 = 0 Identity law

–8 ÷ (–24) = —13 (d) 17 × 4 = 14 × 7 Commutative law

Therefore, –24 ÷ (–8) ≠ (–8) ÷ (–24) (e) 32 × 1 = 32 Identity law

Key Terms • Associative law – Hukum kalis sekutuan
• Distributive law – Hukum kalis agihan
• Identity law – Hukum identiti
• Commutative law – Hukum kalis tukar tertib

7

Chapter Perform Computations Example 2
Efficiently
1 The diagram below shows the
Using the laws of basic arithmetic arrangement of some numbers in a
operations can help make calculations grid.
faster and more efficient.
–7 5 –8
x
1
PENERBIT Form1Example 1
ILMU
BAKTI SDN. BHD.Calculate each of the following

without using a calculator.

(a) (13 + 45) + 7 Commutative When each cell in the grid above is
= (45 + 13) + 7 law
filled with a number, the sum of the
= 45 + (13 + 7) Associative
= 45 + 20 law numbers in each row, column and

diagonal are the same. Find the value

= 65 of x. Evaluating

(b) 1 700 × 17 Distributive law Solution

= (1 000 + 700) × 17 –7 + 5 + (–8) = –10
–7 + x + 1 = –10
= 17 × 1 000 + 17 × 700 –6 + x = –10
x = –4
= 17 000 + 11 900

= 28 900

Solve Problems Worked Example

Example 1 The temperature in a town at 8
p.m. is –19°C. Six hours later, the
The diagram below shows a number temperature drops by 5°C before
line. rising again by 8°C in the next five
hours. Calculate the temperature in
–14 –9 r 1 6 s 16 the town at 7 a.m.

Find the value r – s. Solution
Gathering information:
Solution 8 p.m. → –19°C
Number interval: 7 a.m. → ?°C
6 – 1 = 5
–9 – (–14) = –9 + 14 Solving the problem:
= 5 8 p.m. + 6 hours = 2 a.m.
r = –9 + 5 –19°C – 5°C = –24°C
= –4 2 a.m. + 5 hours = 7 a.m.
s = 6 + 5 –24°C + 8°C = –16°C
= 11
r – s = –4 – 11 Answer: –16°C
= –15

8

Quick Review 1.2   3 Negative fractions are fractions Chapter
that are smaller than zero.
 1 Using the number line, solve the Negative fractions are indicated 1
following. by the (–) sign in front.
(a) –7 + 3 Form 1
(b) –9 + 0 Negative fractions: – —27 , – -—1235–, – -—3239–
(c) 8 + 13 + (–15)
(d) –12 + (–16) + 5 Example 1
PENERBIT
 2 Calculate each of the followingILMU Identify the positive fractions and
using the laws learnt without theBAKTI SDN. BHD.
aid of a calculator. negative fractions in the following list:
(a) 7 × 208
(b) 69 × 67 + 69 × 33 0.9, – 1 —95 , –11—21 , –2.3, –4.1, – —32 , 14,
(c) 734 × 13 + 266 × 13 –0.7, –19—4
(d) 63 × 8 – 13 × 8 14 is equal to —114–.
(e) 16 × 6 + 24 × 6 + 51 × 6 + 9 × 6 Solution
(f ) (54 + 89) + 46 Positive fractions: –11—12 , –1—94 , 14
Negative fractions: – —23 , – 1—95
 3 In a treasure hunt competition,
Nina is at Station 1 which is Represent Positive and
located at step 59 from the base Negative Fractions on a
of a tower. She climbs down Number Line
24 steps to collect a token and
climbs up 68 steps to the second  1 Like integers, the values of
station. Find the location of the fractions can be represented on
second station from the base of a number line.
the tower.
 2 Positive fractions are on the right
 4 The melting point of mercury is of zero while negative fractions
–39°C while the freezing point are on the left of zero.
of alcohol is –114°C. What is the
difference between the melting
point of mercury and the freezing
point of alcohol?

1.3 Positive and –1 – —32 – —13 0 —31 —32 1
Negative Fractions Diagram 1.3  Position of positive fractions

 1 Fractions consist of positive and and negative fractions on a
negative fractions. number line

  2 Positive fractions are fractions Example 1
that are greater than zero (0).
Positive fractions are indicated Mark the given fractions on the
by the (+) sign in front or by the number line below.
absence of any sign. (a) 2—21 , —34 , 2, 1—41

Positive fractions: —12 , —35 , -—1191– 0 1 3

9

(b) –1—32 , –2—23 , –1, –2—31  2 Equivalent fractions are
determined by multiplying or
Chapter –3 –2 dividing the numerator and
denominator by the same whole
1 Solution number.
(a) Interval: —1––4–——0 = —41–
Bridge Map
PENERBIT Form1
ILMU0 —43 1 1—14 2 2—21 3 Fractions —12 × 2 as –12—14 ÷÷ 7 as –11—52 ÷÷ 3
BAKTI SDN. BHD. Equivalent × 2 7 3
(b) Interval: —––2–—–—6(—–—3—) = —61–
to —24 —32 —45

Diagram 1.4 Examples of equivalent fractions

–3 –2—32 –2—31 –2 –1—31 –1 Example 1
–2—64 –2—62 –1—62
Determine which fraction is greater
in value.

(a) —1252– or —2235– Same denominator;
compare the
Compare and Arrange
numerators.
Positive and Negative Thus, —2235– is greater
than —1225–.
Fractions in Order
(b) —1175– or —1295– Same numerator;
 1 We can compare the values of compare the
fractions in the following ways.
(a) If the fractions have the denominators.
same denominator, compare Thus, —1157– is greater
the numerators. The larger than —1295–.
numerator indicates the larger
fraction. (c) —25 or —75 —52 ××–—77 = —3145– ;
(b) If the fractions have the —75 ××–—55 = —3255–
same numerator, compare Compare the
the denominators. The larger numerators; 25 is
denominator indicates the greater than 14,
smaller fraction. Thus, —75 is greater
(c) If the fractions have than —52 .
different numerators or
denominators, change them
to equivalent fractions with
the same denominator. Then,
compare.

Key Terms

• Equivalent fraction – Pecahan setara

10

Example 2 Leave the denominator unchanged. Chapter
(a) Compare and arrange the Simplify the answer if necessary.
Example: 1
following fractions in ascending
order. —18–1– – —16–1– = —12–1–

—67–, —14–, —54–, —32– ➞ —14–, —32–, —54–, —67–  4 To subtract fractions with Form 1

(b) Compare and arrange the different denominators, convert
following fractions in descending
order.

– —61–, —23–, – —65–, 1—13– ➞ 1—13–, —32–, – —61–, – —65–
PENERBIT thefractionstoequivalentfractions
ILMU
BAKTI SDN. BHD. with the same denominator first.

Example:

—97–×× 4– —15–2– × 3 The LCM of 9
× 3 and 12 is 36.
4

Perform Computations = —32–68– – —13–56– = —13–63–

Involving Combined Basic  5 Multiplication of fractions is
performed by multiplying the
Arithmetic Operations numerators and the denominators
separately.
of Positive and Negative
Example:
Fractions —14–3– × —43– = —15–22– = —13–3– or

 1 To add fractions with the same —14–3–1 × —43–1 = —13–3–
denominator, add the numerators
of the fractions involved and leave   6 Division of fractions can be
the denominator unchanged. performed according to the
Simplify the answer if necessary. following steps:
(a) Change the ÷ sign to ×
Example: (b) Invert the second fraction.
—15–2– + —13–2– = —18–2– = —23– (c) Multiply the numerators and
the denominators separately.
  2 To add fractions with different Example:
denominators, convert the
fractions to equivalent fractions —19–1– ÷ —13–1–
with the same denominator first. = —19–1–31 × —13–1–11 = 3
Example:
 7 Combined operations on
—31–×× 4 + —17–2– positive and negative fractions
4 are performed according to the
BODMAS principle.
= —4–1–+—2—7 = —11–12–

 3 To subtract fractions with the
same denominator, find the
difference between the numerators
of the fractions involved.

11

Example 1 Difference: 2—41– – —61–

Chapter Solve each of the following. = —49– – —16–
= —2–—71—2–—2––
1 1 2 1 2(a) – —89– ÷ —11–8– – 4 × –1—38– = 2—11–2–
1 2 = – —89– × —11–8– – 4 × – —18–1–

= –16 + —12–1–
= –10—12–
PENERBIT Form1 Answer: 2—11–2– spoons of butter
ILMU
BAKTI SDN. BHD.1 2(b) —85– ÷ 3 + – —17–2– ÷ —32–Worked Example
1 2 = —85– × —13– + – —17–2– × —23–
The cost price of Zakaria's house is
= —25–4– – —78–×× 3 RM475 000. He then sells his house for
3 RM—34 million. What is his percentage
profit? Round off the answer to two
= —5–—2–—42—1– decimal places.

= – —21–64–÷÷ 8 Solution
8 Gathering information:
S elling price: R=M—34—43×mRMill1io0n00 000
= – —23– = RM750 000
Devising a plan:
Solve Problems Subtraction, division, multiplication
Carrying out the plan:
Example 1 RM750 000 – RM475 000 = RM275 000
Laila’s cake recipe requires 2—41– spoons
of butter while Miza’s recipe requires —RR—MM—2477—55 –00–00–—00 × 100% = 57.89%
—61– spoon of butter. How much more Answer: 57.89%
butter does Laila need for her cake
Quick Review 1.3
recipe compared to Miza?
 1 Represent the following numbers
Solution
Amount of butter Laila needs: on the given number line.
2—14– spoons of butter
Amount of butter Miza needs: – —12 , —78 , –1—58 , —43 , –1—41
—61– spoon of butter


–2 –1 0 1

12

 2 Solve. Example 1 Chapter
Represent the following decimals on
1 2(a) –1–—45 ÷ – —54 + –1—37 × 0 the number line. 1
(a) 0.6, 1.2, 1.8, 3.2
1 2(b) – –1—41 × (–2) + (–16) ÷ —49
0 2.6
 3 a—52reoffisthh.e—13anoimf tahles in an aquarium Form 1
fish are gourami (b) –1.5, –2.9, –2.2, –2.7
PENERBIT
ILMUfish. What fraction of the animals –2 –1
BAKTI SDN. BHD.
in the aquarium are gourami fish? Solution
(a) Interval ⇒ 2.6 ÷ 13 = 0.2
Applying
0 0.6 1.2 1.8 2.6 3.2
 4 Siti uses —41 bowl of coconut milk
fsohreeuascehdd—21ishbsohwelcoofockosc. oOnnuet day, (b) Interval ⇒ 1 ÷ 10 = 0.1
milk,
–2.9 –2.7 –2.2 –2 –1.5 –1
and on the following day, she
H1—o41wmmoarenybdoiwshlseos fdcidocSoitniut
used Compare and Arrange
milk.
Positive and Negative
cook in the two days?
Decimals in Order
1.4 Positive and
Negative Decimals  1 To arrange decimal numbers in
order, take note of the whole
Represent Positive and numbers first.
Negative Decimals on a
Number Line  2 If the whole numbers are the
same, identify the value of the
 1 Decimals are numbers that digit after the decimal point at
represent fractions with the tenths, hundredths, followed
denominators 10, 100, 1 000 and by the thousandths place value,
so on. and so on.

 2 Decimals can also be represented  3 For negative numbers, the larger
on a number line. the digit, the smaller its value.

Hundredths

18.6 7 5

Positive decimals Whole Thousandths

number Tenths

–4 –3.5 –3 –2.5 –2 –1.5 –1 –0.5 0 0.5 1 1.5 2 Example 1
Negative decimals
(a) Arrange the numbers below in
Diagram 1.5 Position of positive and ascending order.
negative decimals on the
number line 5.09, 4.9, 5.19, 4.19, 4.09

13

(b) Arrange the numbers below in Example 3
descending order.
13.271 + 16.27 – 9.902 =
–0.206, –2.06, –0.026, –0.26, –2.26
Chapter 13.21 71 29.541
Solution —— +—— ——21——96——..––52––——47——01–– —— –—— ——1——99——..––96––——30——92––
1 (a) 4.09, 4.19, 4.9, 5.09, 5.19
(b) –0.026, –0.206, –0.26, –2.06, –2.26
PENERBIT Form1
ILMUPerform Computations  3 To multiply a decimal number by
BAKTI SDN. BHD. 10, 100 or 1 000, move the decimal
Involving Combined Basic point to the right according to the
number of zeros in the multiplier.
Arithmetic Operations
 4 The number of decimal places in
of Positive and Negative the product is the same as the
total number of decimal places
Decimals in the numbers being multiplied.

 1 Addition and subtraction of Example 4 2 decimal places
decimals is performed using the 1 decimal place
following steps: 6.54 × 7.3 =
(a) Arrange the decimal numbers 6.5 4 2 + 1 = 3 decimal
according to place value. Make places
sure that the decimal points × 7.3
are aligned vertically. 11 19 6 2
(b) Add zeros, if necessary, to
ensure that the number of + 45 7 8
decimal places in each number 47.7 4 2
are the same.
(c) Add up or subtract the digits  5 To divide a decimal number by 10,
according to place value from 100 or 1 000, move the decimal
the right. point to the left according to the
number of zeros in the divisor.
Example 1
 6 Steps to divide a decimal number
5.62 + 9.801 = 15.421 by another decimal number:
(a) Change the divisor to a
51 .620 0 is added whole number by shifting the
—— +—— ——1——59——..––48––——20——11–– to 5.62 decimal point to the right.
(b) Move the decimal point in
Example 2 the dividend to the right
according to the number of
19.54 – 5.719 = 13.821 places in (a). Add zeros if the
decimal point moves outside
18 15 3 10 0 is added the original number.
to 19.54 (c) Perform division.
19.540

——– —— ——1——35——..––87––——21——19––
 2 Combined operations involving

addition and subtraction are

performed from left to right.

14

Example 5 (c) –4.2 + (3.9– 2.73) × (–1.2)
(a) 18.36 ÷ 2.7 = (b) 0.9 ÷ 0.12 =
=–4.2 + [1.17 ×(–1.2)]

6.8 7.5 =–4.2 –1.404 Calculation in the Chapter
27)183.6 12)90.0 =–5.604 brackets is performed
1
first, followed by
multiplication
– 162 – 84 Division is
21 6 6 0
–21 6 –6 0 performed first Form 1

PENERBIT 00 (d) –3.09 – 6.63 – 2.15
ILMU –1.7
BAKTI SDN. BHD. 8 Steps to divide a decimal by a = –3.09 + 3.9 – 2.15

fraction: = –1.34

(a) Change the operation of

division to a multiplication. Solve Problems

(b) Invert the fraction.

(c) Multiply the decimal by the Worked Example

numerator of the fraction

and divide the product by The table below shows the list of
prices of the items that Sham bought.
the denominator, or divide the

decimal by the denominator Item Price

and multiply the quotient by

the numerator. Cinema ticket RM12.00

Example 6 Popcorn RM3.50/cup

36.9 ÷ —67– = 36.9 × —67– or =36.9× —67– Canned drink RM2.70/can
= —2—5—68—.3–– = 6.15 × 7
Chocolate RM4.20/bar

= 43.05 = 43.05 (a) Find the total price of two cinema

 9 C o mp u t a t i o n s i nvo l v i n g tickets, three cups of popcorn, two
combined operations of decimals
must follow the BODMAS canned drinks and four bars of
principle.
chocolate.

(b) Sham has RM50. Has he got

enough money to buy the items in

(a)? Give the reason. Analysing

Example 7 Solution
Gathering information:
(a) (40 – 1.065) ÷ 1.25 × 5 (a) 2 cinema tickets: 2(RM12.00)
3 cups of popcorn: 3(RM3.50)
= 38.935 ÷ 1.25 × 5 2 cans of drink: 2(RM2.70)
4 chocolate bars: 4(RM4.20)
= 31.148 × 5 Calculation in (b) Sham’s money: RM50

= 155.74 the brackets is Devising a plan:
Division is performed first performed first (a) Multiplication, addition
(b) Subtraction
(b) 14.8 – (–1.95) ÷ (–0.3) + 2.08
= 14.8 – 6.5 + 2.08
= 10.38

15

Chapter Carrying out the plan:  5 Find the total length of seven
(a) 2(RM12.00) + 3(RM3.50) + bamboo sticks, each 6.72 m long
1 and four wooden sticks, each
2(RM2.70) + 4(RM4.20) 2.98 m long.
= RM24.00 + RM10.50 + RM5.40 +
1.5 Rational Numbers
RM16.80
= RM56.70 Recognise and Describe
(b) RM50 – RM56.70 = –RM6.70 Rational Numbers
His money is not enough to buy
 1 Rational numbers are numbers
the items because he is still short
of RM6.70 that can be written as the fraction
PENERBIT Form1 —qp– where p and q are integers, with
ILMUQuick Review 1.4 q  0.
BAKTI SDN. BHD.  2 Integers, fractions, finite decimals
 1 Represent the following numbers and recurring decimals are rational
on the given number line.
numbers. Examples include 17, 0,
–2.25, –2.75 , –0.5, 1.75 , –1.5 –29, —75–, 4.56, 3.333333..., 0.83

–3 –2 –1 0 1 2 Info Gallery
 2 Find the value of each of the
following. 0.83 means that the digits 83 repeat
(a) 0.97 + 2.358 to infinity.
(b) 45.76 – 23.4 – 11 0.83 = 0.83838383…
(c) 0.0109 × 1 000 Another example: 0.067
(d) 37.5 × 2.77 = 0.067067067…
(e) 0.364 ÷ 100
(f ) 15 ÷ 0.6 Example 1
(g) 5.6 ÷ —52
 3 Calculate the value of each of the Mark ‘ 3 ’ for a rational number and
following.
(a) 93.7 + 21.2 – 87.14 ‘ 7 ’ for an irrational number.
(b) 15.6 ÷ 1—15 × 4.2
(c) –0.5(–1) – (–5)(0.4) – (–5)(8.4) (a) 3—25– 3
(b) – —76– 3
(d) 5.86 + 2.9 × 0.2 – 1.92 (c) —15–2– 3

(e) –2.92 + (8.6 + 2.99) × 0.3 (d) p 7
 4 Encik Hasri bought 96 apples for
(e) 4 3
RM95.20. He packed the apples in
plastic containers with 6 apples in (f) –9 3
each plastic container. How much
is each plastic container of apples
he sold if he incurred a loss of
RM32.80 after all the apples were
sold?

Key Terms

• Rational number – Nombor nisbah

16

Perform Computations Devising a plan: Chapter
Involving Combined Basic Multiplication, addition
Arithmetic Operations of 1
Rational Numbers Carrying out the plan:
Rina’s savings: Form 1
Computations involving combined —43 × RM50.80 = RM38.10
operations of rational numbers are Nik’s savings:
similar to computations involving 3 × RM38.10 = RM114.30
integers. Total savings of all three:
RM50.80 + RM38.10 + RM114.30
= RM203.20

Answer: RM203.20
PENERBITExample 1
ILMU
BAKTI SDN. BHD.Solve.
(a) —51– + (–2) – 7—43– = —4–—–—4—02—0–——1—5—5
= – —12–9–0–1– Quick Review 1.5

 1 Mark (3) for a rational number
÷ —14– + = –9—21–01– and mark (7) for an irrational
(b) 2—83– 0.62 × 5 number.

= —18–9–2 × —41–1+ 3.1 (a) 1.212212212...

(b) –13.6

= —12–9–×× 5 + —13–01– (c) 0
5 (d) –1—61
= —91–05– + —31–10– (e) 23

= —11–2–0–6–  2 Calculate.
= 12.6 (a) – —32 × 71.88 ÷ 2

Solve Problems 1 2 1 2(b) —150—40 + —160—90 × —130—20 – 0.17

Worked Example (c) –6–2—12 + 4.12

Bibah has saved RM50.80. Rina’s  3 Sally has 300 cm ribbon. She uses
savings is —43 of Bibah’s savings. Nik
has saved three times of Rina’s savings. 1.6 m of the ribbon to tie a gift.
How much is the total savings of all
three of them? Her daughter buys her another
Solution: m1—45oremsiomf itlhare
Gathering information: ribbon. How many
Bibah’s savings: RM50.80 gifts can be tied

using the available ribbon?

17

1Chapter Summative Assessment
1

PENERBIT Form1Section A
ILMU
BAKTI SDN. BHD.Instructions: Each question is followed by four options, A, B, C and D. Answer all
questions and choose the best option.

 1 Arrange the following  4 –8—21– + 0.8 ÷ 5 =
temperatures in descending A −8.34 C −1.54
order. B −7—14–
D 6.34
4°C, −5°C, 10°C, −7°C, 0°C
A −5°C, 4°C, −7°C, 10°C, 0°C  5 Seha, Aziah and Mei Ling were
B −5°C, −7°C , 0°C, 4°C, 10°C assigned to complete an art
C −7°C, −5°C, , 0°C, 4°C, 10°C, project. Seha took 40 minutes
D 10°C, 4°C, 0°C , −5°C, −7°C to finish her project. Aziah took
—54– of the time Seha completed
 2 Which of the following is an her project. Mei Ling finished
hers 23 minutes later after
integer? Aziah completed hers. Find the
time, in minutes, taken by Mei
A 11.2 C −12 Ling to complete her task.
B 3—91– A 36
D −0.37 B 42
C 50
 3 Which of the following is a D 55

rational number?
A —05–
C 0.251251

B  3 D π

Section B and C

Instructions: Answer all questions.

 1 (a) Circle the integer. (b) In the answer space, place

[1 mark] the digits 2, 3, 4, 5, 6 and

Answer: 7 in the circles so that the

—21– –5 –—16– 5.7 sum along each side of the

triangle is 14. Evaluating

[3 marks]

18

Answer: Answer:

(i) –2.8, , –1, –0.1, Chapter
(ii) – —54– , , , – —51– , 0
1
 4 (a) Choose suitable numbers
from the numbered cards Form 1
below to complete the
arrangement of integers
in ascending order in the
answer space.
 2 (a) Mark (3) for a truePENERBIT
statement and (7) for a falseILMU 0 –11 3 –6
statement.BAKTI SDN. BHD. [2 marks]
[2 marks]
Answer:
Answer:
(i) 5—29– is greater –9 , , –4 , 1 , , 6
than—59–2–.
(b) (i) Arrange the following
(ii) –7—16– is smaller numbers in descending
than –1—76–. order.

(b) Circle all the rational 3—56–, –1, 2—13–, –3.5, –2—32–

numbers. (ii) Find the difference
[2 marks] between the largest
integer and the smallest
 7 1—51– –8  5 integer.
[2 marks]
0.38 p —73– 1—02–
Answer:
(b) (i) Descending order:

 3 (a) In the answer space, fill in (ii) Difference:
the blanks with the symbol
‘=‘ or ‘’.  5 The fund collected for flood
[2 marks] victims amounts to RM3 580.
—14– of the fund is used to buy
Answer: food and —53– of the balance
is spent on management of
(i) 18–(34+25) (18–34)+25 expenses.

(ii) 5×(2.8÷0.7) (5×2.8)÷0.7

(b) Complete each number
pattern in the answer space.
[2 marks]

19

Chapter (a) Complete the table to show (b) Represent 2—18– and 1.75
the amount of money on the number line in the
1 spent on food and on answer space.
management of expenses. [2 marks]
[3 marks]
Answer:
Evaluating

Answer:
PENERBIT Form1 1 2 3
ILMU Amount spent
BAKTI SDN. BHD.
Food Management (c) Calculate
of expenses 10.4 ÷ 2—35– × 3.4

[3 marks]

Answer:



(b) Calculate the amount of the

fund left?

[1 mark]

Answer:

(d) The maximum capacity of
a water tanker is 950 litres.
Determine the number of
trips that 7 drivers must
make to supply 13 300 litres
of water to a housing area.

 6 (a) In the answer space, fill Evaluating
the boxes with the correct
symbol or number. [3 marks]
[2 marks]
Answer:
Answer:
−13 + (−26) – 32

= −13 26 −32

= −32

= –71

20