Math For Junior High Schools

By Baffour Asamoah

In this book, each chapter is broken down into short, manageable sections intended to completely alleviate or reduce to a large extent, the canker that is termed in Ghanaian context as “Math's Phobia” by providing in its content, concise notes and commentary with well illustrated diagrams on each topic and subtopic of the Basic School Mathematics Syllabus.
The book reflects the authors experience that students work better from work examples than abstract discussion of principles, hence the provision of numerous and comprehensive range of worked examples. A numbered, step-by-step approach to problem solving, trial test and challenge problems to cater for all levels pupils are greatly featured in this book. Not left out in this book are “exercises” on each subtopic which contains an extensive selection of Multiple- choice, Fill-ins, True or False, Essay- type questions similar in standard to exam style questions. Tackling these exercises is no doubt, an excellent form of revision. Answers to all exercises are also provided to help students assess their level of progress.
Taking into accounts, full analysis of the pattern and level of difficulty of examination questions, the last part of the book is “some solved past questions”, and “Likely Examination Questions” (Objectives and Theory) with answers within its contents.

• Language: English
• Publisher: Lulu.com
• Release date: Nov 26, 2020
• ISBN: 9781716388941

Book preview

PREFACE

Baffour – Ba Series, is a series of Mathematics, covering the Basic Education Certificate Examination Syllabus.

The author of this book has been teaching Mathematics since completing college. Several years of his teaching experience have revealed the causes and short falls that contribute to the poor performance of students of Mathematics in BECE, which is gradually becoming a thorn in the flesh of many students and a national canker.

In this book, each chapter is broken down into short, manageable sections intended to completely alleviate or reduce to a large extent, the canker that is termed in Ghanaian context as Math‟sPhobia by providing in its content, concise notes and commentary with well illustrated diagrams on each topic and subtopic of the Basic School Mathematics Syllabus.

The book reflects the authors experience that students work better from work examples than abstract discussion of principles, hence the provision of numerous and comprehensive range of worked examples. A numbered, step-by-step approach to problem solving, trial test and challenge problems to cater for gifted pupils are greatly featured in this book. Not left out in this book are exercises oneachsubtopic which contains an extensive selection of Multiple-choice, Fill-ins, True or False, Essay- type questions similar in standard to the BECE

questions. Tackling these exercises is no doubt, an excellent form of revision. Answers to all exercises are also provided to help students assess their level of progress.

Taking into accounts, full analysis of the pattern and level of difficulty of examination questions,

the last part of the book is some solved past questions, and "Likely

Examination Questions" ( ObjectivesandTheory)with answers within its contents .

Baffour – Ba Series,Mathematics for J.H.S.

Page i

ACKNOWLEDGEMENTS

The success story of this book cannot be told without appreciating the immense contributions made by some individuals. The writer is of the conviction that one‟s effort needs to be complemented by others in order to achieve success as proclaim in the Akan adage "obaako nsa nso nyameani kata," meaning "the palm of one person cannot cover the entire face of God".

A myriad of services including spiritual, moral, inspirational, motivational and financial support, typing, editing, printing, designing and what have you, was deeply rendered by friends, family members, students, staff members, some Co- Tutors of J.H.S, S.H.S and

Colleges of Education to make this work a success. To mention a few: Mrs. Joyce Mensah, Mr. Asamoah Koduah Thick, Mr. Fredrick Adjei (Man – Exhorter), Mr. D. K. Boateng (Headteacher, Asuofua D.A. J.H.S. A), Mr. Adjei B. Kwabena (PBC,Nkawie), Mr. Opoku Ware Ghabby, Mr. Issah Peter, Mr. Jerry Dadzie, Mr. Prince Owusu, Mr. Justice Agyapong, Mr. Baafi Eric ( Ericus), Mr. Kofi Boakye Ansah, Mr. Robert Sarpong, Mrs. Rita Otchere, Mrs. Matilda Boakye, Mrs. Winifred Addai, Maybel Owusu-Ansah, Lawrencia Blay, Mr.

Ocloo Denis, Zambobia Zeita, Mr. Boadi Augustine, Mr. Kojo Boakye, Benjamin Osei –

Mensah (Ireland), Mr. Isaac Owusu Mensah, and Mr. K.D. Bosiako, I really appreciate your contributions. And to the entire staff of Asuofua D.A. J.H.S. and Agona S. D. A S. H. S, I say God richly bless you all for making my dreams a reality.

I also wish to acknowledge the fact that the book is not absolutely free from errors of typing, grammar and inaccuracy. These occurred as a result of oversight but not ignorance and incompetence on the part of the author and editors. This should therefore, not undermine the credibility of the book, for they say to err is human. However, your comments, corrections, suggestions and criticism are warmly welcomed for consideration and rectification in the next edition.

Baffour – Asamoah

S. D. A. S. H. S. – Agona

Editors

Mr. Opoku John (Patase M.A. J. H. S)

Mr. Brown Michael ( Asuofua D/A J.H.S)

Mr. Asamoah Koduah Thick (Afoako D/A J. H. S)

Mr. George Osei Mensah (Dublin Institute of Technology, Ireland) Baffour – Ba Series,Mathematics for J.H.S.

Page ii

TABLE OF CONTENTS

Chapter

Page

Chapter

page

1. Numbersand Numerals - - - - - - 1-28

2. Setsand Operations on Sets- - - 29-35

A Number and a Numeral

Definition of a Set

Counting and Writing Numerals up

Describing a Set

to Hundred Million (100,000,000)

Representation of Sets

The Place Value of a Base Ten

Types of Set

Numeral

Relationship between Sets (Subsets,

The Place Value Chart

Equal sets and Equivalent sets)

The Value of a Digit in a Base Ten

Listing the subsets of a set

Numeral

Equal Sets

Writing Numerals in Words

Equivalent Sets

Numerals for Number Words

Intersection of Sets

Comparing and Ordering Numerals

Union of Sets

Using < and >

Identifying the Missing Number on

3. CommonFractions - - - - - - - 36 - 54

a Number Line

Definition of Fractions

Rounding Numbers (To the nearest

Types of Fractions

10, 100, 1000 and millions)

Proper

Fractions

as

Decimal

Even Numbers and Odd Numbers

Fractions

Factors

Mixed

Fractions

as

Improper

Prime and Composite Numbers

Fractions

The Sieve of Eratosthenes

Terminating

and

Recurring

Prime Factorization of Numbers

Decimals

Highest or Greatest Common Factor

Equivalent Fractions

(H. C. F or G. C. F)

Comparing and Ordering Fractions

Using Prime Factorization to Find

Addition

and

Subtraction

of

H.C.F

Fractions

Adding Two or More Fractions with

Least Common Multiples (L.C.M)

Like Denominators

Using Prime Factorization to Find

Addition

and

Subtraction

of

L.C.M.

Fractions with Unlike Denominators

Operations on Whole Numbers

Multiplication

and

Division

of

Fractions

Properties of Operations on Whole

Operations

Involving

Complex

Numbers

Fractions ( BODMAS)

Baffour – Ba Series,Mathematics for J.H.S.

Pageiii

Spending in Fractions (Application Prime Factorization

of Addition

and Subtraction of

Fractions)

7. Relations - - - - - - - - - - 73 - 83

Finding the Total Quantity Given

Definition of Relations

the

Fraction(s)

Spent

and

the

Types of Relations

Remaining Quantity

Finding the Domain or Co-Domain,

Given The Rule

4. Shapeand Space- - - - - - - - - - 55 - 58

The Range of a Relation

Meaning of a Solid

Relations as Ordered Pair

Shape of Some Common Solids :

Cylinder,

Cube, Cuboid,

Cone,

8. Algebraic ExpressionsI - - - - - 84 - 90

Prism and Sphere

Definition

Net of Some Common Solids

Like Terms and Unlike Terms

Relationship between Edges, Faces

Simplifying Algebraic Expressions

and Vertices of Common Solids

Addition

and

Subtraction

of

Algebraic Expressions

5. Lengthand Area - - - - - - - - - - 59 - 64

Grouping Algebraic Expressions

Meaning of Perimeter

Multiplication

and

Division

of

Perimeter of a Triangle

Algebraic Expressions

Perimeter (P) of a Square

Expansion of Algebraic Expressions

Perimeter of a Rectangle

Factorization

of

Algebraic

Perimeter of Other Polygons

Expressions

Perimeter

of

a

Circle

(Circumference)

9. Capacity, Mass,Time &Money--91-120

Capacity

6. Powersof Numbers- - - - - - - - 65 - 72

Addition

and

Subtraction

of

The idea of Powers of Numbers

Capacity

(Indices)

Mass

Multiplication of Powers of Natural

Changing from Kilograms to Grams

Numbers

and Vice – Versa

Division of Powers of Numbers

Addition and Subtraction of Masses

Zero Power of a Natural Number

of Objects

Complex Exponents

Word Problems Involving Capacity

Expressing

a

Number

as

an

and Mass

Exponent

Time

Exponential Equations

The Clock or Watch

Negative Exponents

Telling the Time on the Analog

Rational Exponents

Clock

Baffour – Ba Series,Mathematics for J.H.S.

Page ii

Units of Time

Fractions (With Power of 10) as

Conversions Between Units of Time

Decimal Fraction

(Minutes to Seconds,

Seconds to

Decimal Places

Minutes, Seconds to Hours, Days to

Common

Fraction

as

Decimal

Hours and Vice – Versa )

Fractions

Addition and Subtraction of Time

Decimal

Fractions

as

Common

Word Problem Involving Time

Fractions

Complex Word Problems Involving

Ordering Decimal Fractions

Time

Addition

and

Subtraction

of

Money

Decimal Fractions

Types of Currency

Multiplication of Decimals

Addition and Subtarction of Money

Division of Decimals

Value of a Given Number of

Decimal Places

Denominations of Money

Standard Forms

The Coin and Note Denominations

Numerals for Standard Forms

Word Problems Involving Addition

Additon and Subtraction of standard

and Subtraction of Money

Forms

Change of Denominations (From

Other

Application

of

Standard

Coins to Notes and Notes To Coins)

Forms

Significant Figures

10. Integers- - - - - - - - ---- - - - -121 – 130

The Idea of Integers

12. Percentages- - - - - - - - - 144 – 155

Basic Concept of Integers

Definition

Importance of Positive and Negative

Percentages as Common Fractions

Signs

Fractions

and

Decimal

as

Integers on the Number Line

Percentages

Comparing Integers

Ordering Fractions, Decimals and

Ordering Integers

Percentages

Addition and Subtraction of Integers

Percentage of a Given Quantity

on a Number Line

Expressing

One

Quantity

as

a

Multiplication

and Division of

Percentage of Another

Integers

Increasing a Quantity by a given

Expressions

Involving

Brackets,

Percentage

Multiplication, Division, Addition

Decreasing a Quantity by a given

and Subtraction

Percentage

11. Decimal Fractions - - - - - 131 - 143

Depreciation

Definition

Baffour – Ba Series,Mathematics for J.H.S.

Pageiii

13. Handling Data - - - - - - - - 156 – 169

Extracting

Information

from

a

Idea of Statistics

Given Bar Chart

Types of Data

Stem and Leaf Plot

Sources and Collection of Data

Measures of Central Tendencies

16: Equations and Inequalities – 203- 220

(Mode, Median and Mean of a Raw

Definition of Equations

Data)

Variables

Frequency and Frequency Diagrams

Type 1 Involving Multiplication

Mode, Median and Mean on the

Type 2 Involving Division/Fraction

Frequency Table

Type 3 Involving All Procedure(s)

Relative Frequency

Equations Involving Two Variables

Equations Involving Brackets

14. Probability - - - - - - - - - - - - 170 – 179

Fractional Equations

Definition

General Rules for Solving Linear

Random Experiment

Equations

Outcome of an Experiment

Word Problem Involving Equations

Sample Space of an Experiment

Hints on Forming Word Problems

Event

Steps for Solving Word Problems

Probability of a Letter in a Word

Consecutive Numbers

Equally Likely Outcomes

Inequalities

Finding the Number of an Event,

Solving Inequalities

n(E), Given the Probability, P and

Inequalities on a Number Line

Number of Sample Space, n(S)

Word Problems Involving Linear

Probability of an Event Given the

Inequalities

Probability of another Event in the

Same Experiment

17. Angles- - - - - - - - - - - - - - 221 – 240

Lines and Planes

15. Representationof Data - - - 180 – 202

The Circle

Forms of Data Representation

Measurement of Angles

Pie – Chart

Types of Angles

Calculating The Angles of a Pie

Properties of Angles

Chart

Angles of Parallel Lines

Drawing a Pie-Chart

Special Alternate Angles

Given the Pie Chart to Determine

Types of Triangles

Other Values

Properties of a Triangle

The Bar Chart / Graph

18. Rational Numbers - - - - 241 – 246

Drawing a Bar Chart

Idea of Rational Numbers

Baffour – Ba Series,Mathematics for J.H.S.

Page iv

Rational Numbers on the Number

21. The Number Plane - - - 272 - 295

Line

Explaining a Number Plane

Terminating and Non-terminating

Co-ordinate Axes

Decimals

Quadrants

Irrational Numbers

Naming the Axes

Repeating Decimals as Fractions

Numbering the Axes

Subsets of Rational Numbers

The scale of a Graph

Properties of Operations on Rational

Plotting Points on the Number Plane

Numbers

Identifying the Coordinates of a

Point on the Number Plane

19. Geometrical Construction - - 247– 162

Linear Graphs

Construction of a Line Segment

Drawing Linear Graphs

Bisecting a Line

The Gradient of a Line

Construction of a Perpendicular at a

Gradient of a Line from Two Points

Given Point on a Straight Line (AB)

Gradient of a Line from a Given

Construction of a Perpendicular at

Equation/Relation

the End of a Given Line (AB)

Gradient of a Line from a Linear

Construction of a Perpendicular

Graph

from a Given Point (K) to a Given

Drawing a Line at a Given Point on

Line AB

the x and y – axes

Constructing a Line Perpendicular

Length of a Line Joining Two

to a Given Line AB

Points

Bisecting an Angle

Equation of a Straight Line

Construction of Angles (900, 450,

1200, 600, 300, 150, 750, 1050) at a

22. Quadrilaterals - - - - - - 296 – 298

Given Point

Meaning of Quadrilaterals

The Idea of Locus

Properties of Quadrilaterals

Constructing Triangles

Finding the Unknown Angle of a

The Circum-circle

Quadrilateral

The Inscribed Circle

Drawing Other Figures

23. Ratio and Proportion - - - -299 - 312

Meaning of Ratio

20. Agebraic Expressions II - - 263 - 271

Proportion

Change of subject

Proportion with a Variable

Substitution

Direct Proportion

Factorization by Grouping

Indirect Proportion

Binomial Expansion

Sharing According to a Given Ratio

Baffour – Ba Series,Mathematics for J.H.S.

Page v

Finding

the

Total

Amount

Volume of a Cylinder

(Quantity) Shared Given the Ratio

Volume of a Pyramid

and the Share of One Person

Volume of a Cone

Volume of a Sphere

24: Mappings- - - - - - - - - - -313 – 323

Volume of a Combined Solids

Idea of Mapping

Important Formulas

Types of Mappings

Rule of a Mappings

27. Surface Area of Solids - - -364 – 372

Finding the Image under a Given

Description of a surface and Surface

Mapping

Area

Finding the Image under a Given

Surface Area of a Rectangular Solid

Rule

Surface Area of a Cube

Inverse Mapping

Surface Area of a Cone

Table of Values

Curved Surface Area of a Cone

Surface Area of a Pyramid

25. Areas- - - - - - - - - - - - - - 324 – 347

Surface Area of a Cylinder

Areas of Planes

Surface Area of a Sphere

Units of Area

Practical Problems

Area of a Triangle

Area of Isosceles and Equilateral

28. Rates- - - - - - - - - - - - 373 - 399

Triangles

Simple Interest (I)

Area of a Trapezium

Calculating for Principal (P), Rate

Relationship

between

Radius,

(R) and Time (T)

Diameter and Circumference of a

Commission

Circle

Finding the Total Amount Given the

Area of a Circle

Rate and Commission

Area of a Rectangle

Finding the Rate, given the Total

Area of a Square

Sales and the Commission

Area of a Parallelogram

Discount

Area of Complex Shapes

Finding

the

Original

Price

or

Word Problems Involving Area of

Marked Price given the Discount

Complex Shapes

Rate and the New Price.

Profit and Loss

26. Volume of Solids - - - - - - 348 – 363

Profit and Loss Percentage

Meaning of Volume

Finding the Cost Price or Selling

Volume of a Cuboid /Rectangular

Price given the Percentage Profit or

Solid

Loss

Volume of a Cube

Meaning of Rate

Baffour – Ba Series,Mathematics for J.H.S.

Page vi

Average Speed (S)

Illustration of Diagrams

Scale Drawing (Scaleof a Map)

Two Set Problems

Foreign Exchange Conversion

32. Rigid Motion - - - -430- 449

29. Bearings- - - - - - - - - 400 – 410

Meaning of Transformation

Meaning of Bearings

Translation

Using the North Pole and Measuring

Enlargement

in the Clockwise Direction

Reflection

Using the North and South Poles

Rotation

and Measuring Towards the East or

Drawing the Graph

West

Distance Bearing

33. Enlargement & Similarities 450- 459

Back

Bearings

or

Opposite

Scale Factor

Bearings.

Properties of Enlargement

Similar Figures

30. Vectors - - - - - - - - - - 411 – 418

Symmetrical Objects and Lines of

Defining a Vector

Symmetry

Representation of Vectors

Line of Symmetry of Some Figures:

Position Vectors

Rectangle,

Square,

Equilateral

Column Vectors

triangle, kite

Inverse or Negative Vector

Rotational Symmetry

Equal Vectors

Zero Vector

34. Money and Taxes - - - - - 460 - 476

Magnitude or Length of a Vector

Meaning of Wages and Salaries

Scalar Multiplication

Calculations Involving Wages

Addition of Vectors

Calculations Involving Salaries

Subtraction of Vectors

Banking

Relating Free Vectors and Position

Transactions and Services at the

Vectors

Banks

The Resultant Vector

Insurance

Calculations

Involving

Equal

Income Tax

Vectors

Value Added Tax (V. A. T)

Cartesian Form of a Vector

National Health Insurance Levy

(NHIL)

31. Setsand Diagrams -- - - -419 - 429

Custom Duties

Venn Diagrams

Complement of a Set

35. Quadratic Expressions- - - 477 – 483

Universal Set

Describing a Quadratic Expression

Baffour – Ba Series,Mathematics for J.H.S.

Page vii

Factors of Quadratic Expressions of Abundant Numbers:

the Forms: x 2 + bx + c , x2 – bx + c, Perfect Numbers

x2 + bx – c, x2 – bx – c

Amicable Numbers

Factors of Perfect Squares

Relating Even and Prime Numbers

Factors of Expressions of the Form:

Number Pattern (Sequence)

ax2 + bx + c, where x

1 or x

1

Sequence

of

Even

and

Odd

Factors of Expressions of the Form:

Numbers

a2 – b2

Fibonacci Sequence

Triangular Numbers

36. Quadratic Equations484- 487

Pascal‟s Triangle

Describing a Quadratic Equation

The Magic Square (

Factors of Quadratic Equations

Factors

of

Equations

with

39. Number Bases - - - - - - 517 - 528

Difference of Two Squares (a2 − b2

Base Ten Numerals (Decimals)

= 0)

Basis of Base Ten Numerals

Solving Equations of the Form: a2 −

Base Five Numerals

b2= 0

Writing Base Five Numerals

Word Problems Involving Quadratic

Place Value of Base Five Numerals

Equations

Addition in Base Five

Subtraction in Base Five

37. Polygons- - - - - - - - - - 488 – 508

Base Two Numerals (Binary)

Meaning of a polygon

Writing Base Two Numerals

Sum of Interior Angles of a Polygon

Addition in Base Two

Interior and Exterior Angles of a

Subtraction in Base Two

Polygon

Changing to Base Ten

A Regular Polygon

Changing from Base Ten to Other

Sum of

Exterior

Angles of

a

Bases

Polygon

Conversion between Non Decimal

The Right – angled Triangle

Bases

The Pythagoras Theorem

Equations Involving Number Bases

Pythagorean Triples

Application of Pythagoras Theorem

40. SimultaneousEquations - - 529 - 536

Trigonometry (SOH, CAH, TOA)

Introduction

Angles of Elevation and Depression

Solving Simultaneous Equations by

Elimination,

Substitution

and

38. Number Investigations - - - 509 – 516

Graphical Methods

A number and Sum of its factors

Word

Problems

Involving

Deficient Numbers

Simultaneous Equations

Baffour – Ba Series,Mathematics for J.H.S.

Pageviii

41. The Circle - - - - - - 537 – 539

Length of an Arc and Area of Sector

The idea of a circle

Perimeter of Sector

Parts of a Circle

Likely Examination Questions(Objectivesand Theory) - - -

- - - - 540 - 568

Answersto Exercises- - -- - -- -- -- -- - - - -

--

-- - -

- --

- - - - 569 - 591

Answersto likely Examination Questions - - -- - - -- - - - - -- - - - - -- - -- 592

Bibliograhy -- -- -- -- -- -- -- -- -- -- -- -- --- --- ---- -- -- --

-- - 593

Copyright© 2014

Allrigths reserved under copyright law

No part of this publication may be published, produced or stored on a print or any retrievable media without approval or authorization

For suggestions & views / copies;

Please call any of the following numbers:

0208211966 / 0244282977

ISBN : 978 9988 3 0743 1

Baffour – Ba Series,Mathematics for J.H.S.

Page ix

1

NUMBERS AND NUMERALS

Baffour– Ba Series

A Number and a Numeral

The Place Value of a BaseTen Numeral

A number is an idea and as such cannot be

The position of a digit in a number

seen or felt. In order to portray the meaning

determines the value of that digit. For

of a number, it is expressed as a word or a

instance, in the number 222, although each

symbol. Symbols that represent numbers are

digit is 2, their value differ with respect to

called numerals. For example,

6 is a

their

positions(or

Place)

or

order

of

symbol of six and as such a numeral.

arrangement. Thus;

2

2

2

Counting and Writing Numerals up to

Hundred Million (100,000,000)

Ones

Tens

The numerals used in counting and writing

Hundreds

are called Natural numbers. These are 1,2,and 3…,the dots indicating that they

The values of the digits in a number

continue without end. Natural numbers are

therefore increase from right to left and

also called Counting numbers.When zero decreases

from

left

to

right.

The

is put together with the natural numbers, a

understanding of place value is enhanced by

new

group

of numbers

called Whole

the use of instrument called Abacus shown

numbers isobtained. That is 0, 1,2,3…

below;

Counting large quantities of objects one

after the other is extremely difficult. This is

made easy by counting in groups of tens,

hundreds or thousands. Counting in groups

of ten is called the base ten or decimal

Tth

Th

H

T

system.Thebase ten or decimal system uses ten digits namely; 0,1,2,3,4,5,6,7,8,9. These

From the right, the wires of the abacus

digits are combined with each other in an

indicate Ones, Tens, Hundreds, Thousands,

orderly manner to write as many base ten

Ten thousand, Hundred thousand, One

numerals as required.

million and so on. Thus, the number 3504

can be represented on the abacus by using

Baffour – Ba Series,Mathematics for J.H.S.

Page 1

beads to represent the digits as shown 4 = Ten (10),5 = Ones (1)

below;

2. On the Abacus, sketch and identify the

place value of each digit of 683.

Solution

Therefore the place value of each digit is

6 = Hundreds, 8 = Ten and 3 = ones

Tth

Th

H

T

This indicates that the place value of each

digit in 3504 is 3 = Thousands, 5 =

Hundreds, 0 = Tens, 4 = Ones . The number

3504 represented on the abacus can be

explained as:

3 thousand + 5 hundred + 0 tens + 4 ones

H

T

O

= (3 ×103) + (5 ×102) + (0 × 10) + (4 × 1)

= 3000 + 500 + 0 + 4 = 3504

3. Identify the place value of each digit of

67124 on the abacus

Worked Examples

1. Sketch an abacus showing 2,345

Solution

Solution

Tth

Th

H

T

O

Th

H

T

O

Therefore the place value of each digit is

6= Ten thousands (10,000), 7 = Thousands

Therefore the place value of each digit is

(1,000), 1= Hundreds (100), 2 = Tens (10)

2 = Thousands (1,000), 3 = Hundreds (100),

and 4 = Ones

Baffour – Ba Series,Mathematics for J.H.S.

Page 2

The PlaceValue Chart

GROUP

MILLIONS

THOUSANDS

HUNDREDS

NAME

PERIODS

PERIODS

PERIODS

H

T

O

H

T

O

H

T

O

n

dn

PLACE

a

d

s

n

VALUE

illio

n

u

a

sd

s

M

n

o

s

d

u

n

s

s

d

illio

h

th

T

o

as

re

n

e

d

M

n

illio

h

d

e

n

d

T

u

n

T

e

re

n

M

o

u

O

T

d

e

re

n

h

H

n

T

d

e

T

u

nu

T

H

H

POWER

OF TEN

Reading from right to left, each place is 10 times that of the preceding one.

The digits in the numeral 356 represent 3 Hundreds, 5 Tens and 6 Ones. Similarly, in 241.376, the place value of each digit is displayed below: 2

4

1

.

3

7

6

Thousandth

Hundredth

Tenth

Ones

Tens

Hundreds

It is noteworthy that the place value of numbers less than one (digits at the right of the decimal point) is read from the left as Tenth, Hundredth, Thousandth, Ten-thousandth etc. In 241.376, the place value of each digit is;

2 = Hundreds, 4 = Tens, 1= Ones, 0.3 = Tenth, 0.07 = Hundredth, 0.006 = Thousandth Baffour – Ba Series,Mathematics for J.H.S.

Page 3

3. Find in base ten, the value of 4 in 1435 to base ten

Solution

(1 × 52) + (4 × 51) + (3 × 50)

The value of a Digit in a BaseTen

= 25 + 20 + 3

Numeral

The value of 4 in 1435 = 20

To determine the value of a digit of a base

ten numeral, multiply the digit by its

4.What is the value of „4‟ in the number

respective place value. For example, given

2,043,507?

ABC to determine the value of each, go

through the following process:

Solution

I. Identify the place value of each of the

digits as shown below;

2

0

4

3

5

0

7

A

B

C

Ones

The value of 4 in the number 2,043,507

Tens

Hundreds

= 4 × 104 = 4 × 10,000 = 40,000

II. Multiply each digit by its place value to

5. What is the value of the digit „9‟ in the

determine its value.

number 624.93?

A × 100 = 100A, B × 10 = 10B, C × 1 = C

Solution

Worked Examples

6 2 4 .

9 3

1. Find the value of 3 in 43014

Solution

Tenth

In 43014, the place value of the digit 3 is

The place value of 9 is tenth

Value of 9

thousands.The value of 3 = 3 1000 = 3,000

= 9

=

= 0.9

2. Determine the value of 4 in 142063

6. What is the value of „5‟ in 620.153?

Solution

The place value of the digit 4 = Ten

Solution

thousands.

620.153

The value of 4 = 4× 10,000 = 40,000

Hundredth

Baffour – Ba Series,Mathematics for J.H.S.

Page 4

3) 2121

7) 25.84

The place value of 5 = hundredth (

)

4) 684

8) 1408

Value of 5 = 5 ×

=

=

B. 1. What is the value of „4‟ in the numeral

872, 043

Exercises 1.1

2. What is the value of „6‟ in the numeral

A. Sketch an abacusshowing the

17, 365, 921?

following;

3. What is the value of the digit 2 in

1) 407

5)105268

829,214,764?

2) 2011

6) 20435

4. What is the value of „8‟ in 25.84

Writing Numeralsin Words

To write a given numerals in words:

Represent the numerals on the place value chart to identify the place value of each digit starting from right to left.

Then write and read each numeral and its respective place value from the left to right.

Worked Examples

1. Write 342 in words.

Solution

H

T

O

3

4

2

Therefore, 342 is written in words as Three hundred and forty two.

That is:3 × 100 + 4 × 10 + 2 × 1= 300 + 40 + 2= 342 read as Three hundred and forty- two

2. Write 16,201,920 in words

Solution

Place

Hm

Tm

M

Hth

Ttm

Th

H

T

O

Value

Baffour – Ba Series,Mathematics for J.H.S.

Page 5

Powers

108

107

106

105

104

103

102

101

100

of ten

Numerals

1

6

2

0

1

9

2

0

Add millions to get sixteen million i. e 1 × 107 + 6 × 106, add thousands to get two hundred and one thousand, i. e 2 × 105 + 0 × 104 + 1 × 103 = 201,000, and then add hundreds, tens and ones to get nine hundred and twenty. i. e. 9 × 102 + 2 × 101 + 0 × 100 = 920. By combination, 16,201,920 is written in words as Sixteen million, two-hundred and one thousand, nine hundred and twenty.

3. Write the words that represent 30,820,452

Solution

Place

Hm

Tm

M

Hth

Ttm

Th

H

T

O

value

Powers

108

107

106

105

104

10

102

101

100

of Ten

3

Numerals

3

0

8

2

0

4

5

2

Add millions to get Thirty million ie 3 × 107 + 0 × 106 =30,000,000, add thousands to get eight hundred and twenty thousand, i.e 8 × 105 + 2 × 104 + 0 × 103 = 820,000, and then add hundreds, tens and ones to get four hundred and fifty- two. i. e. 4 × 102 + 5× 101 + 2 × 100 =

452. By combination, 30,820,452 is written in words as Thirty million, eight hundred and twenty thousand, four hundred and fifty- two.

4. Write 2,196 in words.

Solution

Place

Hm

Tm

M

Hth

Ttm

Th

H

T

O

value

Baffour – Ba Series,Mathematics for J.H.S.

Page 6

Powers

108

107

106

105

104

103

102

101

100

of Ten

Numerals

2

1

9

6

Add thousands to get two thousands. i.e 2 × 103 = 2000, and then add hundreds and ones to get one hundred and ninety-six. That is;1 × 102 + 9 × 101 + 6 × 100 = 196. By combination, 2,196 is written in words as Two thousand, one hundred and ninety –six.

5. Write 10,201 in words

Solution

Place

Hm

Tm

M

Hth

Ttm

Th

H

T

O

value

Powers

108

107

106

105

104

103

102

101

100

of Ten

Numerals

1

0

2

0

1

Add thousands to get ten thousand. i. e 1 × 104 + 0 ×103 = 10,000, and then add hundreds, tens and ones. i. e. 2 × 102 + 0 × 101 + 1 × 100 = 201.Therefore 10,201 is written in words as

Ten thousand two hundred and one.

Exercises 1.2

Write the following numerals in words:

1)

520

7) 110, 067

2)

501, 010

8) 10,004

3)

7,578,750

9) 22,135

4)

2, 610,763

10) 12,546

5) 75,210

11) 1,500

6) 80,001

12) 1,001

To write the numerals for a number word,

I. Write each numeral with its place value

Numeralsfor Number Word

attached at the right.

Baffour – Ba Series,Mathematics for J.H.S.

Page 7

Millions → Thousands (three digits) →

= 43,000,000 + 905,000 + 00 + 10 + 2

Hundreds (two digits) → Tens → Ones

= 43, 905, 012

II. Multiply each numeral and its place

Therefore the numeral is 43,905,012

value to determine its actual value

III. Sum up the values of all the digits

3. Write in numerals, two hundred and ten

IV. Counting from the right, place a comma

thousand, eight hundred and six.

after every three digits

Solution

Write the numerals as 210 thousands, 8

Note: The order is from left to right .Thus, hundreds, 0 tens and 6 ones

nineteen million, four hundred and sixty

= (210 × 1000) + (8 ×100) + (0 × 10) +

five thousand, five hundred and twenty-one

(6 × 1)

is written as 19 million, 465 thousands, 5

= 210,000 + 800 + 0 + 6

hundreds, 2 tens and 1 unit. Therefore the

= 210,806

numeral is 19,465,521.

Therefore the number is written in numeral

as 210,806.

Worked Examples

1. Write sixteen million, and twenty-three

Exercises 1.3

thousand, eight hundred and ninety four in

A. Write the numerals for each of the

numerals.

following:

Solution

1. Three million, five hundred and twenty

Write the numeral as 16 million, 023

eight thousand, two hundred and twenty –

thousand, 08 hundred, 9 tens and 4 ones.

five

= (16 × 106) + (23 × 1000) + (8 × 100) +

2.

Fifty

three

million,

nine

hundred

(9 × 10) + (4 × 1)

thousand, four hundred and twenty-two

= 16,000,000 + 23,000 + 800 + 90 + 4

3. Eight hundred and thirty-three thousand,

= 16,023,894.

one hundred and seventy

Therefore the numeral is 16,023,894

4. Twelve thousand and twelve

5. Twelve million, five hundred and thirty-

2. Write the numerals for Forty-three

four

million, nine hundred and five thousand and

6. Seven hundred and nineteen

twelve.

7. Eighteen thousand, seven hundred and

five

Solution

8. One thousand one hundred and one

Write the numeral as 43 million, 905

thousand, 0 hundreds, 1 tens and 2 ones

B. 1. In 2008, it was announced that there

= (43 × 106 ) + (905 × 1000) + (0 × 100) +

were nine million, four hundred thousand

(1 × 10) + ( 2 × 1)

Baffour – Ba Series,Mathematics for J.H.S.

Page 8

eligible

voters.

Write

this

number

in

The number line is used to compare and

numerals.

order whole numbers.

2. In 1996/97 academic year, there were two

Consider the number line below;

million,

seven

hundred thousand,

six-

hundred and fifty one children enrolled in

basic schools in Ghana. Write this in

50

60

70

90

80

numerals.

When this is extended to the left, we have

3.

In

2008

elections,

at

Odotobri

numbers less than 50 and when it is

constituency, NPP had twenty six thousand,

extended to the right, we have numbers

four hundred and twelve votes, NDC had

five thousand five hundred and six votes

greater than 90. On this number line, the

and CPP had three thousand and nine votes.

following observations can be made:

Write the numeral for the votes of each

1) 50 < 70

2) 70 < 90

political party.

3) 50 > 30

4) 30 > 10

4. An articulator truck is loaded with eight

hundred and thirteen bags of cement. What

Exercises 1.4

numeral represents this number?

List any 4 observations on the above

number line.

5. A Boeing 749 air craft has a capacity of

Identifying the Missing Number on a

one thousand one hundred and twenty-one

Number Line

passengers. Express its capacity in numerals

The number line is equally spaced so it

increases and decreases by a fixed value.

6. Write two hundred and fifty-seven in

figures.

To find the value of a missing number on a

number line:

Comparing and Ordering Numerals

I. Find the interval between any two

UsingLessthan (<) or Greater than (>)

consecutive numbers on the number line

To compare whole numbers is to put in >

II. Then add the value of the intervals to the

(greater than) or < (less than) between two

whole

number

preceeding the

missing

whole numbers.

number

III. The sum obtained is the missing

To order whole numbers is to arrange two

number.

or more whole numbers in either ascending

or descending order by putting in < or >

Worked Examples

between them.

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Page 9

What is the missing number on the number Interval = 200 – 100 = 100

line below?

x = 200 + 100 = 300

Therefore the missing number is 300.

Exercises 1.5

30

x

35

40

50

A. Put in lessthan (<) or greater than (>) million

million

million

to make the following statementstrue.

million million

1) 30

50

4) 80

90

Solution

2) 21

11

5) 88

84

1st number = 30 million

3) 110

101

6) 86

90

2nd number = 35 million

Interval = (35 – 30) million= 5 million

x = 40 million + 5 million = 45million

B. Put in > or < in the following

Therefore the missing number is 45 million

statement

1)

381925

381921

2. On the number line below, identify the

2)

381920

381922

missing number.

3)

381920

381925

4)

80000

90000

5)

88000

84000

700

x

900

1000

1100

C. Represent the following setof

numbers on the number line.

Solution

1) 4, 8, 12, 16, 20

Let the 1st number = 1100 and the 2nd

2) 15, 30, 45, 60, 75

number = 1000

3) 60, 120, 180, 240

Interval = 1100 – 1000 = 100

4) 26, 28, 30, 32

x = 700 + 100 = 800

5) 15, 20, 25

Therefore the missing number is 800

6) 2600, 2800, 3000, 3200

7) 15000, 18000, 20000, 23000

3. Determine the missing number in the

8) 25000, 50000,

number line below;

9) 8200, 8400, 8600, 8800, 9000

10) 4000, 12000, 20000, 28000

100

200

x

400

500

D. Find the missing numbers on the

following number lines:

Solution

the 1st number = 100

1. 1000

let the 2nd number =200

2000

4000

Baffour – Ba Series,Mathematics for J.H.S.

Page 10

funeral, Ghanaians that watched Ghana-2.

Brazil U-20 finals on T.V in 2009 etc. In

100

125

200

these instances, it is impossible to give the

A

B

exact figures or number of people but

3.

possible to use a rounded number as an

54

E

F

G

62

estimate.

4.

Rounding a number to a given place, that is:

nearest 10, 100, 1000 and millions depends

Q

6000

6500

R

S

on the digit to the right of the place. If that

digit is 5 or more than 5, round up and if

5.

that digit is less than 5, round down. It is

820

N

800

M

880

therefore necessary to first identify the digit

that represents the place to which you are

6. The number line below shows the

rounding and the next immediate digit on

distance from Cape coast in kilometers:

the right of the said digit:

a. How far is Salt pond from cape coast?

b. How far is Winneba from Accra and from

Guidelines

Cape coast?

Cape

I. Identify the digit that represents the place

to which you are rounding

Accra

Saltpond

Winneba

Coast

II. Consider the immediate digit on the right

of the digit identified.

1500

III. If that digit is 5 or more, round up by

c. which of the following towns is nearest to

adding 1 to the place digit, and the digit(s)

winneba, Accra or Cape coast?

after the place become zero (s).

d. how is these towns shown on the number

IV. If the next immediate digit on the right

line?

is less than 5, round down by adding

nothing to the place digit and the digit(s)

Rounding Numbers (To the nearest Tens,

after the place becomes zero(s).

Hundreds , Thousandsand millions)

Numbers are rounded when they cannot be

Worked Examples

stated with exactness or precision. When

Round the following to the nearest thousand

numbers are rounded, the intention is to

1) 9, 410

2) 98,653

give an estimate of the number. For

instance,

the

number

of

people

who

Solution

welcomed President Obama to Ghana in

1)

9

4

1

0

2009, the number of people at a political

Thousands

party‟s rally, the number of people at a

Baffour – Ba Series,Mathematics for J.H.S.

Page 11

The digit representing the place value 1000

is 9. The digit at the right of 9 is 4 so we

round down (because 4 < 5), by adding 0 to

9 whilst the rest of the digits after 9 become

932 = 930 (to the nearest ten)

zeros.

5. Round 6540133 to the nearest million.

+0

9410

Solution

+1

Thousands

6540133

9410 = 9000 (To the nearest thousand)

Millions

2.

9 8 6 5 3

6,540,133 = 7,000,000 (to the nearest

million)

Thousands

The digit representing the place value

Exercises 1.6

thousand is 8. The digit at the right of 8 is 6.

A.1.Round 826 to the nearest ten.

Since 6 > 4, round up by adding 1 to 8 and

2. Round 17,553 to the nearest hundred.

the rest of the digits after 8 becomes zero.

3. Round 5,679 to the nearest hundred.

That is 98653 = 99000

4.Round 3,228 to the nearest thousand.

3. Round 24, 953 to the nearest thousand

5.Round 67,483 to the nearest thousand.

6. Round 43,572 to the nearest ten.

Solution

7.Round 7,759,321 to the nearest million.

24953 to the nearest thousand

8. Round 8,256,000 to the nearest million.

9. Round 12,532,200 to the nearest million.

+

10. Round 9,825 to the nearest

2 49 5 3

a. Ten

b. Thousand

B. Round 51,624,362 to the nearest:

Thousands

1) Ten

24953 = 25000 to the nearest thousand.

2) Hundred

3) Thousand

4. Round 932 to the nearest ten.

4) Hundred thousand

5) Million

Solution

9 3 2

Baffour – Ba Se

T r

eie

n s

s ,Mathematics for J.H.S.

Page 12

C. 1. A school received a yearly capitation not exactly divisible by two. Examples of

grant of GH¢1250.00. How much is this to

odd number are 1, 3, 5, 7, 9, 11, 13, 15, 17,

the nearest hundred?

19... These constitute the first ten odd

numbers

2. The daily consumption of water in a

certain district is 2,567,348 liters. What is

In two or more digit numbers, odd numbers

the daily consumption to the nearest ten

are determined by verifying the last digit to

thousand liters?

see if it is odd. If it is so, then the entire

number is odd. For example in 4247, the

3. Round 342.532 to the nearest

last digit, 7, is odd and therefore, the entire

a. Tenth

b. Tens

c. Ones.

number, 4247 is an odd number.

D. Identify the place to which each of the

Exercises 1.7

following is rounded:

A.1. Differentiate between an even number 1) 452 = 450

4) 9,770 = 10,000

and an odd number.

2) 24,361 = 24,000

5) 28,944 = 29,000

2. List the first five even numbers.

3) 29,904 = 30,000

6) 51,142 = 50,000

3. The first even number is…..

4. The first odd number is……

Even Numbers and Odd Numbers

5. List the first five odd numbers

Even numbers are numbers that are exactly

divisible by two. This means that when two

B. Determine which of the following

divides a number without a reminder, that

numbers is evenor odd;

number is said to be an even number. For

1) 77

6) 19,151,123

example, 2, 4, 6, 8, 10, 12, 14, 16, 18,

2) 623

7) 7,662,107

20…These constitutes the first ten even

3) 12519

8) 6,755

numbers.

4) 383576

9) 5,022

5) 20,006,500

10) 1,211

To determine whether a two or more digit

number is an even number, check and see if

C.1. Find the sum of the even number

the last digit is an even number. If it is so,

between 70 and 80

then the entire number is an even number.

2. Find the sum of even numbers from 70 to

For example, in 3114, since the last digit, 4

80

is an even number, 3114 is also an even

3. List the odd numbers between 60 and 70

number.

4. Find the difference between the sum of

odd numbers between 20 and 30 and the

Odd numbers are numbers that give a

sum of odd numbers between 80 and 90.

reminder when divided by two. In order

5. Round your answer in question 4 to the

words, odd numbers are numbers that are

nearest hundred.

Baffour – Ba Series,Mathematics for J.H.S.

Page 13

1) 4

2) 18

3) 33

4)13

5) 27

Factors

6)25 7) 15

8) 49

9) 56

10) 20

A factor is any number that divides another

number completely without any remainder.

B. UseTrue or False to answerthe

That is; if A B = C, then B and C

are

following questions;

factors of A. For example, if 12 3 = 4, then

1) 2 is a factor of 29

3 and 4 are factors of 12. Also 12

6 = 2,

2) 3 is a factor of 40

so 6 and 2 are factors of 12, 12 ÷ 1 = 12, so

3) 5 is a factor of 50

1 and 12 are factors of 12. All the factors of

4) 1 is a factor of any number.

12 can be listed as 1, 2, 3, 4, 6, and 12.

5) Every number is a factor of itself.

6) Zero is a factor of every number.

When listing the factors of a number, it is

7) If A B = C, then A, B and C are factors

advisable to list them in ascending order

of A.

8) If a number divides another number

Worked Examples

leaving a remainder, then the divisor is a

1. What are the factors of 10?

factor of the dividend.

Solution

Multiples

Factors of 10

The multiples of a number, A, is all the

10

1 = 10, 1 and 10 are factors of 10.

numbers that are divisible by A. For

10

, 2 and 5 are factors of 10

example, 3, 6, 9,12,15,18 are multiples of 3.

The factors of 10 are 1, 2,5,10 (in

The multiples of a number, A, is written by

increasing order)

adding A, successively to the sum of itself.

List all the factors of 9.

i. e multiples of A = A, A+A, A+A+A,…

= A, 2A,

3A,…

Solution

Thus, multiples of 5 = 5, 5+ 5, 5 + 5 + 5…

9

1 = 9, 1 and 9 are factors of 9.

= 5, 2(5),

3(5) …

9

= 3, 3 and 3 are factors of 9

= 5,

10,

15…

. Factors of 9 = 1, 3, 9

Exercises 1.9

3. List all the factors of 7.

A. Write the first 5 multiples of the

following;

Solution

1) 7

2) 20

3) 9

4) 17 5) 10 6) 4

7

1 = 7, 1 and 7 are factors of 7

The factors of 7 = 1 and 7

B. Indicate whether the following

statementis true or false, if A

B = C,

Exercises 1.8

1. C is a multiple of A

A. List all the factorsof the following:

Baffour – Ba Series,Mathematics for J.H.S.

Page 14

2. C is a multiple of B

3. B is a multiple of A

It is noteworthy that not all even numbers

4. A is a multiple of B

are composite numbers and vice- versa.

5. B is a multiple of C

6. C is not a multiple of C

Worked Examples

Prime and CompositeNumbers

Determine whether the following are prime

Study the factors of the numbers below;

or composite numbers.

A

B

1)

21

2)

15

3) 17

4)29

2 = 1,2

4 = 1,2,4

3 = 1,3

6 = 1,2,3,6

Solution

5 = 1,5

9 = 1,3,9

1. Factors of 21 = 1, 3,7,21

7 = 1,7

10 = 1,2,5,10

21 is a composite number

11 = 1,11

12 = 1,2,3,4,6,12

2. Factors of 15 = 1,3,5,9

The numbers in group A have exactly two

15 is a composite number

factors namely, one and the number itself.

Such numbers are called prime numbers.

3. Factors of 17 = 1, 17

Examples

are;

2,3,5,7,11,13,17,19,23,29.

17 is a prime number.

These

constitute

the

first

ten

prime

numbers. Note that one is not a prime

4. Factors of 29 = 1, 29

number and not all odd numbers are prime

29 is a prime number.

numbers and vice-versa.

Exercises 1.10

On the other hand, all the numbers in group

Determine

whether

the

following

B have more than two factors. Such

numbers are prime or composite

numbers

are called

composite numbers.

1) 13

2) 11

3) 4

4) 40

5) 83

Thus composite numbers have more than

6) 6

7) 19

8) 30

9) 31 10) 100

two factors. Examples are; 4, 6, 8, 9, 10, 12,

14, 15, 16, 18, 20. These constitute the first

ten composite numbers.

The Sieve of Eratosthenes

The sieve of Eratosthenes was introduced by Eratosthenes a Greek mathematician to determine the prime numbers between 1 and 100. The steps involved in using the sieve of Eratosthenes are:

Step1. Write all the numbers from 1 to 100

Baffour – Ba Series,Mathematics for J.H.S.

Page 15

Step 2: Circle 2 and cross out all multiples of 2.

Step 3: Circle 3 and cross out all multiples of 3 (those that have not been crossed out) Step 4 : Circle the next number that has not been crossed, i.e. 5 and cross out all multiples of 5

which have not been crossed out.

Step 5: Circle the next number that has not been crossed, i.e. 7 and cross out all the multiples of 7

that have not been crossed out.

Step 6: Circle all the remaining numbers that have not been crossed.

Conclusion: All the circled or ringed numbers are prime numbers Prime Factorization of Numbers

By disintegration,

Any composite number can be written as a

20 = 4 × 5

product of prime numbers. For example:

= 2 × 2 × 5

12 can be written as 2 × 2 × 3. That is:

= 22 × 5

12 = 2 × 2 × 3 = 22 × 3. The product 22 × 3

is called the Primefactorization of 12.

2. Express 12 as a product of prime factors.

Worked Examples

Solution

1. What is the prime factorization of 20

Method 1

Using the factor tree

Solution

Method 1

12

using the factor tree

20

2

6

2

10

2

3

12 = 2 × 2 × 3 = 22 × 3

2

5

= 2 × 2 × 5= 22 × 5

Method 2

12 = 3 × 4

Method 2

= 3 × 2 × 2

Baffour – Ba Series,Mathematics for J.H.S.

Page 16

= 3 × 22

315 = 3 × 3 × 5 × 7 = 32 × 5 × 7

3. Find the prime factorization of 294

5. Write 24 as a product of prime factors.

Solution

Solution

Method 1

Method 1

294

24

2

1

2

12

49

3

2

6

7

7

294 = 2 × 3 × 7 × 7 = 2 × 3 × 72

2

3

24 = 2 × 2 × 2 × 3 = 23 × 3

Method 2

294 = 7 × 42

Method 2

= 7 × 7 × 6

24 = 2 × 12

= 7 × 7 × 2 × 3

= 2 × 2 × 6

= 2 × 3 × 7 × 7

= 2 × 2 × 2× 3

= 2 × 3 × 72

= 23 × 3

4. Write the prime factorization of 315.

Exercises1.11

A. Find the prime factorization of the

Solution

following:

Method 1

1) 75

2) 35 3) 2700 4) 56

5) 34

315 = 5 × 63 = 5 × 7 × 9 = 5 × 7 × 3 × 3

= 3 × 3 × 5 × 7= 32 × 5 × 7

6) 150 7) 40

8) 42

9) 108 10) 500

Method 2

B. Express the following

as prime

315

factorization;

1. 22 × 62

2. 22 × 42

3

105

Highest or Greatest Common Factor

35

(H. C. F or G. C. F)

3

To find the highest common factor of two or

more numbers:

5

7

Baffour – Ba Series,Mathematics for J.H.S.

Page 17

I. List all the factors of each number II. Choose the factors common to all

Solution

III. Choose the biggest number among the

36 = 6 × 6 = 2 × 3 × 2 × 3= 22 × 32

common factors as the H.C.F

45 = 5 × 9 = 5 × 3 × 3= 3 × 3 × 5 = 32 × 5

Common factors = 32 and 32

Worked Examples

1. Find the highest common factor of 28 and

The least factor is 32

70.

H.C.F. = 32 = 3 × 3 = 9

Solution

2. What is the H.C.F. of 24 and 50?

factors of 28 = 1,2,3,4,7,14,28

factors of 70 = 1,2,5,7,10,14,35,70

Solution

common factors = 1,2,7,14

24 = 4 × 6 = 2 × 2 × 2 × 3 = 23 × 3

50 = 5 × 10 = 5 × 5 × 2 = 2 × 52

H.C.F = 14

Common factors = 2 and 23

The least factors is 2

2. Find the HCF of 12 and 18.

H.C.F. of 24 and 50 is 2

Solution

3. What is the H.C.F. of 23 × 32 and 23 × 34

Factors of 12 = 1, 2, 3, 4, 6, 12

Factors of 18 = 1, 2, 3, 6, 9, 18

Solution

Common factors = 1, 2, 3, 6

23 × 32 and 23 × 34

H.C.F = 6

The common factors are 2 and 3. The least

UsingPrime Factorization to Find H.C.F

power of 2 is 23 and the least power of 3 is

The simplest way to find the H.C.F. of two

32

or more numbers is by using the method of

H.C.F = 23 × 32 = 2 × 2 × 2 × 3 × 3

prime factorization. This is done by going

= 8 × 9 = 72.

through the following steps:

I. Write each of the numbers into its prime

4. What is the HCF of 18, 42 and 90?

factors

II. Cross out all prime factors that is not

Solution

common to all.

18 = 3 × 6 = 3 × 3 × 2 = 2 × 32

III. Out of the remaining common factors,

42 = 6 × 7 = 2 × 3 × 7

select those with the least power and

90 = 2 × 45 = 2 × 5 × 9 = 2 × 5 × 3 × 3

multiply to get the H.C.F.

= 2 × 32 × 5

Common factors of 18, 42 and 90 are 2 and

Worked Examples

3. Least factors are 2 and 3.

1. Find the H.C.F. of 36 and 45

H.C.F. of 18, 42 and 90 = 2 × 3 = 6

Baffour – Ba Series,Mathematics for J.H.S.

Page 18

Worked Examples

5. Find the greatest common factor of 35

1. What is the L.C.M. of 6 and 8?

and 70.

Solution

Solution

Multiples of 6 = 6, 12,18,24,30,36,42,48…

35 = 5 × 7

Multiples of 8 = 8, 16,24,32,40,48,56,64…

70 = 5 × 14 = 5 × 2 × 7= 2 × 5×7

Common multiples = 24, 48…

Common factors are 5 and 7

L.C.M. = 24

H.C.F. of 35 and 70 = 5 × 7 = 35

2. Find the L.C.M. of 5 and 7

Exercises 1.12

A. Find the H. C.F. of the following

Solution

numbers.

Multiples of 5 =

1) 30 and 65

2) 12 and 10

5,10,15,20,25,30,35,40,45,50,55,60,65,60…

3) 28 and 63

4) 18 and 40

Multiples of 7 =

5) 32 and 33

6) 40 and 56

7,14,21,28,35,42,49,56,63,70…

7) 36 and 28

8) 18 and 24

Common multiples = 35, 70 …

9) 54 and 27

10) 30 and 21

L.C.M. = 35

B. Find the H.C.F.

of the following

3. What is the L.C.M. of 6 and 9?

numbers by using prime factorization

method.

Solution

1) 24 and 18

6) 16, 24 and 64

Multiples of 6 = 6, 12, 18, 24,30,36,42, 48

2) 81 and 18

7) 28, 35 and 70

54, 60, 66…

3) 90 and 105

8) 60, 96 and 120

Multiples of 9 = 9, 18, 27, 36, 45, 54, 63…

4) 72 and 48

9) 72, 126 and 162

Common multiples = 36, 54…

5) 11, 13 and 17

10) 9, 15and 27

L.C.M. = 36

LeastCommonMultiples (L.C.M) The

4. Find the L.C.M of 2, 4 and 6.

L.C.M. of two or more natural numbers is

found by:

Solution

I. Listing the multiples of each number (as

Multiples of 2= 2, 4, 6, 8, 10, 12, 14, 16…

many as you can)

Multiples of 4 = 4,8,12,16,20,24,28,32,36…

II. Choosing the common multiples among

Multiples of 6 = 6, 12,18,24,30,36,42,48…

the numbers

Common multiples = 12, 24, 36…

III. Choosing the least number among the

L.C.M. = 12

common multiples

Baffour – Ba Series,Mathematics for J.H.S.

Page 19

Using Prime Factorization to Find the Step I

L.C.M.

of Two

or More

Numbers

18 = 2 × 9 = 2 × 3 × 3 = 2 × 32

The prime factorization method of finding

42 = 6 × 7= 2 × 3 × 7

the L.C.M of two or more numbers is

90 = 9 × 10 = 3 × 3 × 2 × 5 = 2 × 32 × 5

employed when the numbers involved are

StepII

big. The steps involved are:

factors are 2, 3, 5 and 7.

I. Write each of the numbers into its prime

StepIII

factorization

Highest number of times that:

II. Select all the numbers that can be found

2 occurs = 2,3 occurs = 32 , 5 occurs = 5

in all the prime factorization of

all the

and 7 occurs = 7

given numbers

L.C.M. of 18, 42 and 90 = 2 × 32 × 5× 7

III. Find the highest or the greatest number

= 2 × 9 × 5 × 7 = 630

of times that each factor occurs in each

given number.

3.

Find

the

least

common

multiples

IV. The product of the highest or greatest

(L.C.M.) of 12 and 20.

number of times all the factors occur is the

L.C.M.

Solution

12 = 2 × 6 = 2 × 2 × 3 = 22 × 3

Worked Examples

20 = 2 ×10 = 2 × 2 ×5 = 22 × 5

1. Find the L.C.M. of 4, 6, and 12.

The factors are 2, 3 and 5.

The highest number of times that:

Solution

2 occurs = 22, 3 occurs = 3, 5 occurs = 5

StepI.

L.C.M. of 12 and 20 = 22 × 3 × 5

4 = 2 × 2 = 22

= 2 × 2 × 3 × 5= 60

6 = 2 × 3

12 = 2 × 6 = 2 × 2 × 3 = 22 × 3

StepII.

4. Determine the least common multiple of

16, 30 and 36.

Factors of the three given numbers are 2

and 3

StepIII.

Solution

16 = 2 × 2 × 2 × 2 = 24

Highest number of times

30 = 2 × 15= 2 × 3 × 5

2 occurs = 22 ,

3 occurs = 3.

36 = 6 × 6 = 2 × 3 ×2 × 3 = 22 × 32

L.C.M. of 4, 6 and 12 = 22 × 3

All