Year 9 Teacher Pack 3

TEACHER’S PACK 3YEAR 9

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Introduction v

CD-ROM Manual x

Algebra 1 & 2

1.1 Sequences 21.2 Pattern spotting 41.3 Functions 61.4 Graphs 81.5 Limits of sequences 10

Number 1

2.1 The four rules governing fractions 122.2 Percentages and compound

interest 142.3 Reverse percentages and

percentage change 162.4 Direct and inverse proportion 182.5 Ratio in area and volume 202.6 Numbers between 0 and 1 222.7 Reciprocal of a number 242.8 Rounding and estimation 26

Algebra 3

3.1 Simultaneous equations 283.2 Solving by substitution 303.3 Find the nth term for a quadratic

sequence 323.4 Equations involving fractions 343.5 Inequalities 363.6 Graphs showing direct

proportion 383.7 Solving simultaneous equations

by graphs 40

Shape, Space and Measures 1

4.1 Pythagoras’ theorem 424.2 Solving problems using

Pythagoras’ theorem 444.3 Loci 464.4 Congruent triangles 484.5 Circle theorems 504.6 Tessellations and regular

polygons 524.7 Practical Pythagoras 54

Handling Data 1

5.1 Statistical investigations 565.2 Scatter graphs and correlation 585.3 Scatter graphs and lines of

best fit 605.4 Time series graphs 625.5 Two-way tables 645.6 Cumulative frequency diagrams 665.7 Estimation of a mean from

grouped data 68


6.1 Similar triangles 706.2 Metric units for area and volume 726.3 Length of an arc and area

of a sector 746.4 Volume of a cylinder 766.5 Rates of change 78

Number 2

7.1 Standard form 807.2 Multiplying with numbers in

standard form 827.3 Dividing with numbers in

standard form 847.4 Upper and lower bounds 1 867.5 Upper and lower bounds 2 887.6 Recurring decimals 907.7 Efficient use of a calculator 92

Algebra 4

8.1 Index notation with algebra 948.2 Square roots, cube roots and

other fractional powers 968.3 Quadratic graphs 988.4 Cubic graphs 100

Handling Data 2

9.1 Probability statements 1029.2 Mutually exclusive events and

exhaustive events 1049.3 Combining probabilities and

tree diagrams 1069.4 Estimates of probability 108

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Contents

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10.1 Fractional enlargements 11010.2 Trigonometry: The tangent of

an angle 11210.3 Trigonometry: The sine and

cosine of an angle 11410.4 Solving problems using

trigonometry 116

Algebra 5

11.1 Expansion 11811.2 Factorisation 12011.3 Quadratic expansion 12211.4 Quadratic factorisation 12411.5 Change of subject 126

Solving Problems and Revision

12.1 Fractions, percentages and decimals 128

12.2 The four rules; ratio; standard form 130

12.3 Rules of algebra and linearequations 132

12.4 Graphs 13412.5 Shape, Space and Measures 13612.6 Handling Data 138

SATs Paper 140

Handling Data 3

13.1 Revision of statistical techniques 160

13.2 A handling data project 162


14.1 Shape and space revision 16414.2, 14.3 Shape and space investigation 166

14.4 Symmetry revision 16814.5, 14.6 Symmetry investigation 170

Handling Data 4

15.1 Revision of probability 17215.2 A probability investigation 174

GCSE Preparation

16.1 Reinforcement of Number 17616.2 Reinforcement of Number 17816.3 Reinforcement of Number 18016.4 Reinforcement of Number 18216.5 Reinforcement of Number 18416.6 Reinforcement of Number 18616.7 Reinforcement of Number 188

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14

13

12

11

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© HarperCollinsPublishers Ltd 2003 v

This is the higher-level teaching text for Year 9. It accompanies Maths Frameworking Year 9 Pupil Book 3 andcaters for students working at Levels 6–8. Students who are working at Levels 3–5 are catered for by Pupil Book 1and Teacher’s Pack 1. Those working at Levels 5–7 are catered for by Pupil Book 2 and Teacher’s Pack 2.

Maths Frameworking has been based totally on the finalised National Numeracy Strategy document. The detailedlesson plans deliver core material from the Framework’s medium-term plans. We have reduced the teaching timefrom the 105 hours recommended in the NNS to around 85 one-hour lessons. This should enable teachers someflexibility to include tests, extended activities and revision classes in their teaching programme, and allows for thenormal events that disrupt teaching time.

The lesson plans have the following features:

� Framework objectives to identify the key learning outcomes from the Framework� Engaging Oral and mental starter activities to involve the whole class� Main lesson activities to help you lead students into exercise questions� Plenary guidance to round off the NNS three-part lesson� Key words which highlight when to introduce Framework Vocabulary terms� Extra Homework questions to consolidate and extend learning� Answers for all pupil book exercises, homework and SAT-style questions

PLUS

Chapter numbers and titles in Maths Frameworking follow the NNS medium-term plans. Due to the break causedby KS3 National Tests, the following specific approach has been taken to tackling the framework objectives inlater chapters:

� Chapter 12 is devoted to revision and is followed by a mock SATs paper, for practice prior to KS3 National Tests.Full answers and a tutorial section for the test, suitable for individual student use, are provided on the CD-ROM.

� Chapters 13 to 15 allow students to consolidate and extend knowledge of Handling Data and Shape, Spaceand Measures through a range of investigation tasks.

� Chapter 16 consolidates Number and Algebra and prepares students for KS4 work.

The Oral and mental starters are designed to work with minimal specialised equipment – a blackboard and apiece of chalk would suffice – but resources such as OHPs, A3-sized target boards, counting sticks, numbersquares, student white boards and number fans make the activities easier to present and more accessible tostudents. A selection of inexpensive or free numeracy resources are available from Collins. See the accompanyingwebsite www.mathsframeworking.com for details.

The authors recognise that ICT provision in schools is varied and we have tried not to commit teachers to anactivity that they could not carry out. However, suggestions for activities using ICT are included throughout thelesson plans and Pupil Book exercises. Some lessons also address cross-curricular issues such as Literacy andCitizenship.

For further information and resources for ICT integration, visit www.mathsframeworking.com

The NNS is intended to improve standards. This can only be done by the good work of teachers in the classroom.The authors appreciate the good work teachers do and hope that Maths Frameworking proves a suitable resourceto help them.

Kevin Evans, Keith Gordon, Trevor Senior and Brian Speed

The free CD-ROM that comes with each teacher pack allows you to extract text and graphics from thelesson plans, to help produce customised lessons for individualised teaching programmes. Diagrams canalso be reproduced for use on overhead projectors or electronic whiteboards. Full details of how to usethis resource are given on pages x–xii.

Introduction

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Objectives

Using and applying mathematics to solving problems� Solve increasingly demanding problems and evaluate solutions; explore

connections in mathematics across a range of contexts: number, algebra,shape, space, and measures, and handling data.

� Represent problems and synthesise information in algebraic, geometric orgraphical form; move from one form to another to gain a differentperspective on the problem.

� Solve substantial problems by breaking them into simpler tasks, using arange of efficient techniques, methods and resources, including ICT.

� Present a concise, reasoned argument, using symbols, diagrams, graphsand related explanatory text; give solutions to problems an appropriatedegree of accuracy.

� Suggest extensions to problems, conjecture and generalise; identifyexceptional cases or counter-examples, explaining why.

Numbers and the number systemPlace value, ordering and rounding� Use rounding to make estimates.

� Write numbers in standard form.

� Understand upper and lower bounds; round numbers to three decimalplaces and a given number of significant figures.

Fractions, decimals, percentages, ratio and proportion� Use efficient methods to add, subtract, multiply and divide fractions,

interpreting division as a multiplicative inverse; cancel common factorsbefore multiplying or dividing.

� Recognise when fractions or percentages are needed to compareproportions; solve problems involving percentage changes.

� Use proportional reasoning to solve a problem, choosing the correctnumbers to take as 100%, or as a whole; compare two ratios; interpretand use ratio in a range of contexts, including solving word problems.

� Understand the implications of enlargement for area and volume.

� Recognise and use reciprocals.

� Know that a recurring decimal is an exact fraction.

� Use algebraic methods to convert a recurring decimal to a fraction insimple cases.

Chapter title

Chapter titleSolving Problems and Revision


Handling Data 3



Number 1

Number 2

Number 2

Number 1

Handling Data 2

Number 1


Number 1


Number 1

Number 1

Number 2

Number 2

Lesson number and title

Lesson number and title12.1 Fractions, percentages and

decimals12.2 Long multiplication and

division; ratio; directed numbers

12.3 Rules of algebra and linearequations

12.4 Graphs12.5 Shape, Space and Measures

13.2 A handling data project

14.2, 14.3 Shape and spaceinvestigation

14.5, 14.6 Symmetry investigation

14.2, 14.3 Shape and spaceinvestigation

14.5, 14.6 Symmetry investigation

2.8 Rounding and estimation

7.1 Standard form

7.4 Upper and lower bounds 17.5 Upper and lower bounds 2

2.1 The four rules governingfractions

9.2 Mutually exclusive events andexhaustive events

9.3 Combining probabilities andtree diagrams

2.2 Percentages and compoundinterest

12.1 Fractions, percentages anddecimals

2.3 Reverse percentages andpercentage change

2.4 Direct and inverse proportion12.1 Fractions, percentages and

decimals12.2 Long multiplication and

division; ratio; directednumbers

2.5 Ratio in area and volume

2.7 Reciprocal of a number

7.6 Recurring decimals

7.6 Recurring decimals

Framework Objectives Matching Chart This chart matches the National Strategy Framework Objectives (Year 9, core and extension) to specific lesson plans inMaths Frameworking Year 9 Teacher’s Pack 3.

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Objectives

Calculations� Understand the effects of multiplying and dividing by numbers between 0

and 1; use the laws of arithmetic and inverse operations.

� Estimate calculations by rounding numbers to one significant figure andmultiplying or dividing mentally.

� Use known facts to derive unknown facts; extend mental methods ofcalculation, working with decimals, fractions, percentages, factors,powers and roots; solve word problems mentally.

� Check results using appropriate methods.

Calculator methods� Use the reciprocal key of a calculator.

� Enter numbers in standard form into a calculator and interpret the display.

� Use a calculator efficiently and appropriately to perform complexcalculations with numbers of any size, knowing not to round duringintermediate steps of a calculation; use the constant, ( and sign changekeys, function keys for powers, roots and fractions, brackets and thememory.

� Enter numbers and interpret the display in context (negative numbers,fractions, decimals, percentages, money, metric measures, time).

AlgebraEquations, formulae and identities� Solve a pair of simultaneous linear equations by eliminating one

variable.

� Construct and solve linear equations with integer coefficients (with andwithout brackets, negative signs anywhere in the equation, positive ornegative solution), using an appropriate method.

� Solve linear inequalities in one variable, and represent the solution set ona number line; begin to solve inequalities in two variables.

� Solve problems involving direct proportion using algebraic methods,relating algebraic solutions to graphical representations of the equations;use ICT as appropriate.

� Link a graphical representation of an equation or a pair of equations tothe algebraic solution.

� Simplify or transform algebraic expressions by taking out single-termcommon factors; add simple algebraic fractions.

� Square a linear expression, expand the product of two linear expressionsof the form x ± n and simplify the corresponding quadratic expression;establish identities such as a2 – b2 = (a + b)(a – b).

� Derive and use more complex formulae, and change the subject of aformula.

Sequences, functions and graphs� Generate terms of a sequence using term-to-term and position-to-term

definitions of the sequence, on paper and using ICT.

� Generate sequences from practical contexts and write an expression todescribe the nth term of an arithmetic sequence.

� Find the inverse of a linear function.

� Construct functions arising from real-life problems and plot theircorresponding graphs.

� Find the next term and the nth term of quadratic sequences and functionsand explore their properties.

� Generate points and plot graphs of linear functions (y given implicitly interms of x), e.g. ay + bx =0, y + bx + c =0, on paper and using ICT; givenvalues for m and c, find the gradient of lines given by equations of theform y = mx + c.

� Plot graphs of simple quadratic and cubic functions, e.g. y = x2, y = 3x2 + 4, y = x3.

Chapter title

Number 1

Number 1

Number 2

Number 2

Number 2


Algebra 3

Algebra 3

Algebra 3

Algebra 3

Algebra 3

Algebra 5

Algebra 5

Algebra 5

Algebra 1 & 2

Algebra 1 & 2

Algebra 1 & 2

Algebra 1 & 2

Algebra 3

Algebra 3Algebra 5Solving Problems and Revision

Algebra 4


2.6 Numbers between 0 and 1

2.8 Rounding and approximation

Oral and mental starter activitiesthroughout

Throughout

2.7 Reciprocal of a number

7.2 Multiplying with numbers instandard form

7.3 Dividing with numbers instandard form

7.7 Efficient use of a calculator

12.1 Fractions, percentages anddecimals

12.2 The four rules, ratios anddirected numbers

3.1 Simultaneous equations3.2 Solving by substitution

3.1 Simultaneous equations3.2 Solving by substitution

3.5 Inequalities

3.6 Graphs showing directproportion

3.7 Solving simultaneous equationsby graphs

11.1 Expansion11.2 Factorisation

11.2 Factorisation11.3 Quadratic factorisation

11.5 Change of subject

1.1 Sequences1.2 Pattern spotting1.5 Limits of sequences

1.2 Pattern spotting

1.3 Functions

1.4 Graphs

3.3 Find the nth term for aquadratic sequence

3.6 Graphs showing directproportion

3.7 Solving simultaneous equationsby graphs

8.3 Quadratic graphs8.4 Cubic graphs

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Objectives

Integers, powers and roots� Know and use the index laws (including in generalised form) for

multiplication and division of positive integer powers; begin to extendunderstanding of index notation to negative and fractional powers,recognising that the index laws can be applied to these as well.

� Use ICT to estimate square roots and cube roots.

Shape, space and measuresGeometrical reasoning: lines, angles and shapes� Understand and apply Pythagoras’ theorem.

� Understand congruence.

� Apply the conditions SSS, SAS, ASA or RHS to establish the congruenceof triangles.

� Know that the tangent at any point on a circle is perpendicular to theradius at that point; explain why the perpendicular for the centre to thechord bisects the chord.

� Distinguish between practical demonstration and proof.

� Explain how to find, calculate and use the interior and exterior angles ofregular polygons.

� Know that if two 2-D shapes are similar, corresponding angles are equaland corresponding sides are in the same ratio.

� Solve problems using properties of angles, of parallel and intersectinglines, and of triangles and other polygons, justifying inferences andexplaining reasoning with diagrams and text.

� Visualise and use 2-D representations of 3-D objects.

Transformations� Enlarge 2-D shapes given a fractional scale factor; recognise the similarity

of the resulting shapes; understand the implications of enlargement forarea and volume.

� Transform 2-D shapes by combinations of translations, rotations andreflections, on paper and using ICT; know that translations, rotations andreflections preserve length and angle and map objects on to congruentimages; identify reflection symmetry in 3-D shapes.

Construction and loci� Find the locus of a point that moves according to a more complex rule,

involving loci and simple constructions.

� Know from experience of constructing them that triangles given SSS,SAS, ASA or RHS are unique, but that triangles given SSA or AAA are not.

Coordinates� Find points that divide a line in a given ratio, using the properties of

similar triangles.

Measures and mensuration� Use units of measurement to calculate, estimate, measure and solve

problems in a variety of contexts; convert between area measures (mm2 to cm2, cm2 to m2, and vice versa) and between volume measures(mm3 to cm3, cm3 to m3, and vice versa).

� Know and use the formulae for length of arcs and area of sectors ofcircles.

� Calculate lengths, areas and volumes in right prisms, including cylinders.

� Understand and use measures of speed (and other compound measuressuch as density or pressure) to solve problems; solve problems involvingconstant or average rates of change.

Chapter title

Algebra 4

Algebra 4





















8.1 Index notation with algebra:negative powers

8.2 Square roots, cubes roots andother fractional powers

8.3 Square roots, cube roots andother fractional powers

4.1 Pythagoras’ theorem4.2 Solving problems using

Pythagoras’ theorem4.7 Practical Pythagoras

4.4 Congruent triangles


4.5 Circle theorems

4.5 Circle theorems4.7 Practical Pythagoras

4.6 Tessellations and regularpolygons

10.1 Fractional enlargements

12.5 Shape, Space and Measures

14.4 Symmetry revision

10.1 Fractional enlargements

14.4 Symmetry revision

4.3 Loci


6.1 Similar triangles

6.2 Metric units for area andvolume

14.1 Shape and space revision

6.3 Length of an arc and area of asector

6.4 Volume of a cylinder

6.5 Rates of change

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Objectives

� Begin to use sine, cosine and tangent in right-angled triangles to solveproblems in two dimensions.

� Know and use the formulae for the circumference and area of a circle.

� Calculate the surface area and volume of right prisms.

Handling dataSpecifying a problem, planning and collecting data� Suggest a problem to explore using statistical methods, frame questions

and raise conjectures.

� Design a survey or experiment to capture the necessary data from oneor more sources; determine the sample size and degree of accuracyneeded; design, trial and if necessary refine data collection sheets.

� Identify possible sources of bias and plan how to minimise it.

� Discuss how data relate to a problem; identify possible sources, includingprimary and secondary sources.

Design and use two-way tables.

Processing and representing data, using ICT as appropriate� Select, construct and modify, on paper and using ICT, suitable graphical

representation to progress an enquiry, including:– line graphs for time series; – scatter graphs to develop further understanding of correlation;identify key features present in the data.

� Select, construct and modify, on paper and using ICT, suitable graphicalrepresentation to progress including lines of best fit by eye, understandingwhat they represent.

� Find the median and quartiles for large data sets; estimate the mean,median and interquartile range of a large set of grouped data.

� Find summary values that represent the raw data, and select the statisticsmost appropriate to the problem.

Interpreting and discussing results� Analyse data to find patterns and exceptions, look for cause and effect

and try to explain anomalies.

� Interpret graphs and diagrams and draw inferences to support or castdoubt on initial conjectures; have a basic understanding of correlation.

� Examine critically the results of a statistical enquiry, and justify choice ofstatistical representation in written presentations, recognising thelimitations of any assumptions and their effect on conclusions drawn.

� Communicate interpretations and results of a statistical enquiry usingselected tables, graphs and diagrams in support, using ICT as appropriate.

� Identify what extra information may be required to pursue a further line ofenquiry.

Probability� Use the vocabulary of probability in interpreting results involving

uncertainty and prediction.

� Identify all the mutually exclusive outcomes of an experiment; know thatthe sum of probabilities of all mutually exclusive outcomes is 1 and usethis when solving problems.

� Understand relative frequency as an estimate of probability and use thisto compare outcomes of experiments.

� Estimate probabilities from experimental data.

� Compare experimental and theoretical probabilities in a range ofcontexts; appreciate the difference between mathematical explanationand experimental evidence.

Chapter title




Handling Data 1

Handling Data 1

Handling Data 1Handling Data 3


Handling Data 1

Handling Data 1

Handling Data 3

Handling Data 1

Handling Data 1

Handling Data 3

Solving Problems and RevisionHandling Data 3

Solving Problems and RevisionHandling Data 3

Handling Data 3

Handling Data 3

Handling Data 3


Handling Data 2

Handling Data 4



Handling Data 4


10.2 Trigonometry: The tangent of anangle

10.3 Trigonometry: The sine andcosine of an angle

10.4 Solving problems usingtrigonometry



5.1 Statistical investigations

5.1 Statistical investigations

5.1 Statistical investigations13.2 A handling data project

5.1 Statistical investigations13.1 Revision of statistical

techniques

5.5 Two-way tables

5.2 Scatter graphs and correlation5.3 Scatter graphs and lines of

best fit5.4 Time series graphs


5.3 Scatter graphs and lines of bestfit

5.6 Cumulative frequency diagrams5.7 Estimation of mean from

grouped data

13.1 Revision of statisticaltechniques

12.6 Handling data13.2 A handling data project



12.6 Handling data13.2 A handling data project


9.1 Probability statements15.1 Revision of probability

9.2 Mutually exclusive events andexhaustive events

9.3 Combining probabilities andtree diagrams

15.1 Revision of probability

9.4 Estimates of probability15.1 Revision or probability15.2 A probability investigation

9.4 Estimates of probability15.1 Revision of probability

15.1 Revision of probability15.2 A probability investigation

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This free Maths Frameworking Teacher Pack CD-ROM provides all the pages of this pack in PDFformat. These can be read by Adobe AcrobatReader. If your computer does not already havethe Acrobat Reader software it can be installeddirectly from the CD-ROM (please refer to theinstallation instructions below).If your computer already has Acrobat Readerinstalled, follow these steps to view the MathsFrameworking Teacher Pack CD-ROM:

Macintosh� Insert the Maths Frameworking Teacher Pack

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PCFor Windows:� Click the ‘Start’ button and select ‘Run’.� Type ‘D:\MF.pdf’. If you are not using the D

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Using the MathsFrameworking TeacherPack CD-ROMThese pages contains brief guidance to help youto move around the CD, to enlarge and printpages and to adapt any of the activities to suityour own requirements. For further, extensive help in using Acrobat Readerwith the CD-ROM, select ‘Reader Online Guide’from the ‘Help’ menu within Acrobat Reader.

Navigating the CD-ROMUse the black, triangular direction buttons at thetop of the screen to move forwards or backwardsbetween pages of text.You can also navigate your way around byclicking on the ‘bookmarks’ to each lesson, thatappear on the left hand side of the screen. If a plusor minus sign appears to the left of a bookmarkthen you can click on this to show or hidesubordinate bookmarks.

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OS) key as you drag across the document. To select all the text on the page, choose Edit >Select All.

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Attributions

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Restrictions on use

All rights reserved. Maths Frameworking Year 9Teacher Pack 3 CD-ROM must not be sold,rented, leased, sub-licensed, lent, assigned ortransferred, in whole or in part, to third parties. Nopart of this CD-ROM may be reformatted,adapted, varied or modified by the user other thanspecifically for teaching purposes whereenlargements and/or minor adaptations may benecessary. This CD-ROM may not be reproducedor transmitted in any form or by any means,without the permission of the publishers otherthan the form of printed copies for single use only.

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© HarperCollinsPublishers Ltd 2003

First published 2003

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For use with Maths Frameworking Year 9 Pupil Book 3

Maths Frameworking Year 9Lesson Plans

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Oral and mental starter� Put on the board 102 and ask: ‘What does this represent?’� You want the response ‘Ten squared’, as well as ‘A hundred’ and ‘Ten times ten’.� Then put on the board 103 and ask: ‘What does this represent?’� You want the response ‘Ten cubed’, as well as ‘A thousand’ and ‘Ten times ten

times ten’. You may even get ‘One hundred times ten’.� Now put on the board 104 and ask: ‘What does this represent?’� The response is now ‘Ten to the power four’ and ‘Ten thousand’.� Now jump to 106 and ask: ‘What does this represent?’� You are looking for ‘Ten to the power six’ and ‘One million’.� Next, put on the board 109 and ask: ‘What does this represent?’ ‘One billion’ is

the response that you want.� Then put on the board 1012: ‘What does this represent?’ ‘One trillion’ is the

response that you want.� Talk about the use made of powers to represent large numbers.� Then ask them about 10100: ‘What does this represent?’ This is a googol, but you

are unlikely to get a response.

Main lesson activity� Put on the board T(n) = 2n + 7 and explain that this is a rule describing a

sequence. It gives the nth term in the sequence, from which every term can befound by substituting the integers 1, 2, 3, 4, … .

� Get the class to generate this sequence: 9, 11, 13, 15, … .� Notice how the sequence goes up in 2s and that the first term is 2 + 7.� Go through with the class the building up of the sequence with T(n) = 3n + 2.

This generates 5, 8, 11, 14, … .� Notice that the sequence goes up in multiples of 3, the number in front of the n,

plus 2.� Show the class that this rule will also work for negative numbers. � Go through the sequence with T(n) = 4n – 3. This sequence is 1, 5, 9, 13, … .� Now put on the board 9, 13, 17, 21, … and ask the class what the nth term is for

this sequence.� Ask the students for the reasons why they have suggested various rules. Lead

them to the fact that, since 4 is added on every time to obtain the next term, thenth term will start with 4n. Then see what has to be added to 4 to get the firstterm of 9. This will be 5. So the nth term will be 4n + 5.

� Show the class the sequence 3, 8, 13, 18, … and ask what the nth term is for thissequence.

� They should be able to identify the first part as 5n since 5 is added on each time.� Ask what must be added to 5 to get the first term of 3. The answer is –2. So the

nth term is given by 5n – 2.� Put on the board the sequence 1, 4, 9, 16, 25, 36, … . Then ask the class what

they notice about the differences between the consecutive terms.

2 © HarperCollinsPublishers Ltd 2003

Algebra 1 & 2CHAPTER

1

LESSON

1.1

Framework objectives – Sequences

Generate terms of a sequence using position-to-term definitions of the sequence,on paper and using ICT.

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� After some discussion, put on the board the first and second differences:1 4 9 16 25 36

First differences 3 5 7 9 11Second differences 2 2 2 2

Explain that when a sequence has the same second differences, it is a quadraticsequence. That is, a sequence whose nth term contains n2. In the exampleabove, T(n) = n2.

� Finally, run through the sequence T(n) = n(n + 2), and show that the seconddifferences are constant.

� The class can now do Exercise 1A from Pupil Book 3.

Plenary� Discuss with the class the results of the investigation in Exercise 1A, especially if

any student has completed Question 7.


Exercise 1A Answers

1 a 17, 21, 25 b 23, 28, 33 c 30, 37, 44 d 34, 44, 55 e 50, 66, 84f 29, 40, 53

2 a 5, 7, 9, 11 b 1, 4, 7, 10 c 9, 13, 17, 21 d 2, 7, 12, 17 e 1, 4, 9, 16f 2, 5, 10, 17 g 2, 6, 12, 20 h 8, 14, 22, 32

3 a 4n + 2 b 7n + 1 c 3 – 4n d 3n – 18 e 0.2n + 2.2 f 0.3n + 1.4n (2n + 1)

g ———— h ————(3n – 1) (5n – 1)

4 a 2n b 1005 a 6, 12, 20, 30, 42 b 0, 0, 2, 6, 12 c 1, 2, 5, 10, 17 d 2, 4, 6, 14, 34

e 3, 6, 9, 12, 396 b Both are 2 c 27 b Both are 4 c 6 d 2A

Ho

me

wo

rk 1 Write down the first four terms of each sequence whose nth term is given below.

a 3n + 1 b 4n – 2 c n2 + 7 d n(n + 3) e (n + 3)(n – 1)

2 Find the nth term of each of the following sequences.

a 5, 7, 9, 11, … b 2, 5, 8, 11, … c 1, 4, 9, 16, … d 3, 6, 11, 18, …

3 Find the nth term of each of the following sequences of fractions.

a 1–2, 2–3, 3–4, 4–5, … b 1–3, 2–5, 3–7, 4–9, …

4 Find the nth term of each of the following sequences.

a 3.5, 5, 6.5, 8, 9.5, … b 5.1, 7.2, 9.3, 11.4, … c 3.6, 6.1, 8.6, 11.1, …

Answers1 a 4, 7, 10, 13 b 2, 6, 10, 14 c 8, 11, 16, 23 d 4, 10, 18, 28 e 0, 5, 12, 212 a 2n + 3 b 3n – 1 c n2 d n2 + 2

n n3 a ——— b ——–—

(n + 1) (2n + 1)4 a 1.5n + 2 b 2.1n + 3 c 2.5n + 1.1

� sequence� nth term� quadratic

sequence� first difference� second

difference

Key Words

Extension Answers

There are many different solutions to each part. The following includes one example ofeach:a 1, 1.5, 2, 2.5, 3 b 0.25, 0.5, 0.75, 1, 1.25 c 2, 4, 6, 8, 10d 5, 10, 15, 20, 25 e 2.5, 5, 7.5, 10, 12.5

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Oral and mental starter� Draw on the board three measuring jugs labelled 1 gallon, 3 pints and 5 pints.� Tell the class that the gallon jug is full.� Ask if anyone knows how many pints there are in a gallon. There are 8 pints.� Working in pairs, ask which pair can work out, using just the three jugs, how to

divide the liquid into two equal measures of half a gallon (4 pints).� The solution is as follows:

�� From the full gallon jug, fill the 5 pint jug, leaving a measure of 3 pints in thegallon jug.

�� From the 5 pint jug, fill the 3 pint jug leaving 3 pints, 3 pints and 2 pints.�� Pour all of the 3 pint jug into the gallon jug, giving 6 pints and 2 pints.�� Pour the 2 pints from the 5 pint jug into the 3 pint jug, leaving 6 pints and

2 pints.�� From the gallon jug, fill the 5 pint jug, leaving 1 pint, 2 pints and 5 pints.�� From the 5 pint jug, fill up the 3 pint jug, leaving 1 pint, 3 pints and 4 pints. �� Now just pour the 3 pint jug into the gallon jug to leave 4 pints in the gallon

jug and 4 pints in the 5 pint jug.

Main lesson activity� Draw on the board a circle with a chord (any straight line from one part of the

circumference to another).� Ask how many regions there are in the circle. (There are two.)� Now draw another chord in the circle, intersecting the first one, and ask: ‘How

many regions are in the circle now?’ (There are four.)� Build up a table of results while doing this, showing the number of lines and the

number of regions.� Ask the class if they can tell you the possible maximum number of regions when

another chord is drawn.� Many might suggest six due to the sequence starting 2, 4, … . Put the suggestions

on the board.� Now draw in the chord to intersect both chords already in the circle. Count with

the class the number of regions. There are seven regions. Put this in the table.� Now ask again: ‘What is the maximum number of regions if I draw another

chord in the circle?’� Some may spot the pattern, which gives 11. If so, then get them to explain the

pattern to the rest of the class. Show that this is true. The explanation is that 2 isadded, then 3, then 4 and so on.

� Now ask if anyone can describe the term-to-term rule. Encourage classdiscussion here, and clarity of explanation. There may be a few goodsuggestions, all different from each other but still correct.

� The rule is:T(n) = Term + n

where the build-up is from the term immediately before.� Explain this terminology: T(n) is the nth term and Term is the term immediately

before it.

� The class can now do Exercise 1B from Pupil Book 3.


LESSON

1.2

Framework objectives – Pattern spotting

Generate terms of a sequence using term-to-term and position-to-term definitions ofthe sequence, on paper and using ICT.

Generate sequences from practical contexts and write an expression to describe thenth term of an arithmetic sequence.

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Plenary� Discuss with the class how good their predictions have been and whether they

got better as the lesson went on.� Explain that it is good to try to make a prediction, as this means they are actually

looking at the pattern, which should lead to a refinement in the rule beinglooked for.

� To get results from complicated situations, it is very often useful to create simplediagrams in order to look at the pattern.


Exercise 1B Answers

1 a 10 c 15, 21 e T(n) = Term + (n – 1)2 a You probably predicted 10 b Probably not, for there are only nine lines

c 12, 15 e T(1) = 0 , T(2) = 1, T(3) = 3, after this T(n) = Term + 33 a 20 c 27, 35 e T(n) = Term + n4 Two-hundred-and-ten pin bowling

Ho

me

wo

rk Look at the following diagrams.

a Before drawing a diagram, can you predict, from the table, the number of crosses which are inDiagram 4?

b Draw Diagram 4, and count the number of crosses there are. Were you right?

c Now predict the number of crosses for Diagrams 5 and 6.

d Check your results for part c by Drawing diagrams 5 and 6.

e Write down the term-to-term rule for the sequence of crosses. (Hint 4 = 22, 8 = 23)

Answersa You will get the following results.

e T(n) = Term + 2n

Diagram 1 2 3 4 5 6Crosses 1 5 13 29 61 125

Diagram 1 2 3 4 5 6Crosses 1 5 13

321

� prediction

Key Words

Extension Answers

1–2(n2 + 3n + 6)

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Oral and mental starter � Ask the question: ‘Is 1234 × 5678 = 7 006 652?’� Is there a quick way of checking without doing the whole multiplication?� One check is to look at the product of the last digit of each number: 4 × 8 = 32.

The end digit of this number must be the end digit of the original sum. Here theyare both 2, so it might be correct.

� Another check is to do the digit sum scan. That is, add the digits of 1234, 5678and the predicted answer, 7 006 652, which gives 10, 26 and 26.

� Add the digits of any of these three numbers greater than 9, here giving 1, 8 and 8.

� Now check that the product of the first two values is the same as the third value.� Here they are the same – both 8 – so the answer might be correct.� Try this out: ‘Is 314 × 783 = 245 762?’� The digit sums come to 8, 9 and 8.� The product, 8 × 9 = 72, has digit total 9. This is different from the last digit sum,

so the prediction is not correct.� Let the class try this out with a calculator to convince them that this procedure

always works.

Main lesson activity � Put on the board x → 4x and ask the class whether they remember what this is.

They should tell you that it is the function x maps to 4x. Ensure that the class usethe correct terminology.

� Put in a column underneath the numbers x → 1, 2, 3, 4 and ask if anyone cantell you what each number will map to. You should get the response:

1 → 4, 2 → 8, 3 → 12, 4 → 16� Now continue the diagram by making each term map back to itself. That is:

1 → 4 → 1, 2 → 8 → 2, 3 → 12 → 3, 4 → 16 → 4 Ask the class if they can tell you what function will map each value back towhere it started. You may need to hide the original column of numbers in orderto focus only on the other two columns.

� You want to get the response x → 1–4 x.� Tell them that this is the inverse function of x → 4x. Explain that the term

‘inverse’ means here doing the opposite process to return to the original values.� Ask the class: ‘What is the inverse function of x → 5x?’ You may need to go

through this in the same way which you did for x → 4x. Ask them to work outthat the inverse function of x → 5x is x → 1–5 x.

� Discuss what is happening with the above inversion. That is, division is theinverse of multiplication and vice versa. Ask: ‘When we see a pair of inversefunctions, are they both inverses of each other?’ You may need to refer to both x → 4x and x → 5x in order to show that this is true.

� Now put on the board the function x → x + 3, and ask if anyone can proposewhat its inverse will be. Discuss each proposal and ask for a reason why theproposal was made.

� You need to lead the students to the inverse being x → x – 3.� Discuss with them the inverse of x → x + 7, which is x → x – 7.� Now put on the board the function x → 4x + 3 and ask for its inverse.� This is not easy to see, so you will need to break down the function into its two

steps of × 4 and + 3. Show this in flow diagram notation as:

x 4x + 3+ 3× 4


LESSON

1.3

Framework objectives – Functions

Find the inverse of a linear function.

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� Given that the inverse is the opposite process which returns the mapped values to their original values, the flowdiagram needs to be viewed in reverse. Reverse the arrows and start with x at the right-hand side, to give:

(x – 3)———

4x

Remind the class that the inverse of × is ÷, and of + is –. This gives the inverse function:(x – 3)

x → ———4

� Show that this is the inverse function of x → 4x + 3 by choosing a starting set of numbers, say, 1, 2, 3 and 4.

� The class can now do Exercise 1C from Pupil Book 3.

Plenary� Ask: ‘What is the inverse of multiplication?’ (Division.)� Ask: ‘What is the inverse of addition?’ (Subtraction.)� Ask: ‘What is the inverse of squaring?’(Taking the square root.)� Discuss the problems that this last inverse has. For example, the square of a

negative number is the same as the square of its positive value. So what aboutthe inverse?

– 3÷ 4


Exercise 1C Answers

1 a x → 1–2x b x → 1–5x c x → x – 6 d x → x – 1 e x → x + 3 f x → 5x(x – 3) (x – 1) (x + 3) (x + 2) (x – 7)

2 a x → ——— b x → ——— c x → ——— d x → ——— e x → ———2 3 4 5 4

(x + 5)f x → ———

63 Two different types of example are:

12i x → 10 – x 1 → 9 → 1 ii x → —– 1 → 12 → 1

x2 → 8 → 2 2 → 6 → 23 → 7 → 3 3 → 4 → 3

(x – 6) (x + 12)4 a x → ——— b x → ———— c x → 4x – 3 d x → 5x + 2 e x → 2(x – 3)

2 3f x → 2(x + 7)

5 a {2, 4, 6, 8, 10} b 2, 4, 6, 8, 10 c Yes9 The lines are symmetrical about the line y = x

Ho

me

wo

rk 1 Write down the inverse of each of the following functions.

xa x → 3x b x → x + 8 c x → 6 + x d x → –– e x → 2x + 1 f x → 4x + 3 g x → 3x – 52

2 Write down two different types of inverse function and show that they are self inverse functions.

3 Write down the inverse of each of the following functions.

(6 + x)a x → 3(x + 5) b x → 1–2(x + 5) c x → ———4

4 a On a pair of axes, draw the graph of the function x → 2x + 3.

b On the same pair of axes, draw the graph of the inverse of x → 2x + 3.

c Comment on the symmetries of the graphs.

Answers(x – 1) (x – 3) (x + 5)

1 a x → 1–3 x b x → x – 8 c x → x – 6 d x → 2x e x → ——— f x → ——— g ———2 4 3

2 There will be a variety of different correct answers3 a x → 1–3x – 5 b x → 2x – 5 c x → 4x – 64 c The graphs are reflections of each other in the line y = x

� inverse

Key Words

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Oral and mental starter� Ask: ‘Who knows how many miles are equivalent to 8 km?’ (5 miles)� Use this fact to ask quick-fire questions about equivalences of the following:

� Discuss how they need to think: ‘How many 8s? Then multiply that by 5.’� Now ask for approximations, such as 10 kilometres and 60 kilometres. Again, look for approximations of

eights. So, 10 km is just over eight, which will make it just over 5 miles. For an approximation, call it 6 miles. For 60 kilometres, divide by 8, giving 71–2, which gives 5 × 71–2 = 35 + 21–2 = 371–2. As this is anapproximation, round to 38 miles.

� Finish off with the class trying mentally to convert approximately each of the following:

� As the concern is to find approximations, any answer which is close to the correct mileage will be acceptable.� The main intention is to practise mental division by 8 and multiplication by 5.

Main lesson activity� Draw on the board a pair of axes with the horizontal axis labelled ‘Time’ and the vertical labelled ‘Distance’. � Ask: ‘What might the graph look like if it were representing a car being driven at a steady speed?’(Straight line.)

You may want to discuss that a steeper line represents a faster speed but you will need to create values for thegraph in order to show this.

� Draw another pair of axes on the board and ask: ‘What shape would the graph have if the car were slowingdown?’

� It would be a curve whose steepness becomes less and less until the curve is horizontal.� Now draw a third pair of axes on the board and ask what shape the graph would be for a car accelerating from

standstill to a steady speed.� This will show a different type of curve with a gradually increasing gradient which eventually becomes a

straight line (but not horizontal).� Discuss with the class the fact that graphs illustrate typical speeds. In real life, however, there would be many

changes in the speed of a car, resulting in a range of different graphs from the ones drawn.� Tell the class how useful graphs can be and that they can hold a lot of information.� Look at the graph in Pupil Book 3, page 10, which represents a race between three boys. Ask the class to tell

you the story behind the graph.

� The class can now do Exercise 1D from Pupil Book 3.

Kilometres 20 35 50 70 90 100 200Miles 13 22 31 44 56 63 125

Kilometres 16 24 32 40 64 80 96Miles 10 15 20 25 40 50 60


LESSON

1.4

Framework objectives – Graphs

Construct functions arising from real-life problems and plot their correspondinggraphs.

Extension Answers

a b 59 million

0

1801

1821

1841

1861

1881

1901

1921

1941

1961

1981

2001

10203040506070

Year

Popu

latio

n (m

illio

ns)

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Plenary� Draw on the board a pair of axes labelled ‘Time’ on the horizontal axis and

‘Distance’ on the vertical axis.� Ask a student to draw on the axes a graph representing his/her journey to school

that day.� Discuss this graph and whether it actually shows the variations in speed, the

stationary times, etc.� If time permits, ask a student to sketch a graph of an aircraft journey from

London to Amsterdam.


Ho

me

wo

rk 1 Sketch graphs to show how the depth of water varies with time when water drips steadily into thefollowing containers.

2 Sketch distance–time graphs to illustrate each of the following situations.

a A car accelerating away from traffic lights.

b A train slowing down to a standstill in a railway station.

c A car travelling at a steady speed and then having to accelerate to overtake another vehiclebefore slowing down to travel at the same steady speed again.

3 Sketch a graph to show the depth of water in a bath where it is filled initially with just hot water,then the cold water is also turned on. After 2 minutes, a child gets into the bath, splashes about for5 minutes before getting out, and pulling out the plug. It takes 6 minutes for the water to drainaway.

Answers1 2

3 The graph may look something like this:

t

d

a b c

t

d

t

d

t

d

a b c

a b c

� axes� gradient� acceleration

Key Words

Exercise 1D Answers

1 a 300 m b 75 m c 42 seconds d D: 40 seconds, E: 60 secondse Rocket D travelled a greater distance vertically. The graph is not calibrated for horizontal distance

2 a The cold tap was turned on, increasing the rate of flow into the bath b After 4 minutesc The more water in the bath, the quicker it will flow out, gradually slowing down d 3 minutes

3 a b c d

4 a C b A c B 5 a C b A c B 6 a b c d e

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Oral and mental starter� Ask: ‘Who can multiply 15 by 13 mentally?’ (195)� Should anyone be able to do this, ask them to explain how they arrived at the

answer. If a different method from what will be given in the lesson is used, takeit as an alternative method.

� If no one is able to do this mentally, or as an alternative from the explanationoffered, explain that you can split the sum into two parts.

� As 5 × 3 = 15, a multiplication can be split when multiplying by 15.� For example, to find 15 × 13, multiply 13 by 3 to get 39 and then multiply 39

by 5.� There are various ways to multiply 39 by 5. One way is to multiply 39 by 10 and

then halve the result. The number needed is half of 390. Halve this in two parts:half of 300 + half of 90 = 150 + 45 = 195.

� Ask the class to practise multiplying 16, 21, 34, 42 and 57 by 15. The tablebelow can be used to check the results.

Main lesson activity� Put on the board: ‘Divide by 5 and add 4.’� Tell the class that this is a rule for creating a sequence and that this is the term-

to-term rule.� Ask someone to give you a number between 0 and 100. Use this to start the

sequence. If you wish to have more control over the numbers, then choose yourown starting number, say 1.

� Using the term-to-term rule, this will generate:1, 4.2, 4.84, 4.968, 4.9936, 4.99 872, 4.999 744, 4.999 949

� Ask the class: ‘Do you notice anything about the numbers?’ They should spotthat the terms are getting closer and closer to 5.

� The class will need calculators to do this, or they could use a spreadsheet ifavailable. If using a spreadsheet, they will need to know how to set up a formulaand be able to copy it from one cell to another.

� The class can now do Exercise 1E from Pupil Book 3.

Number 16 21 34 42 57Number × 15 240 315 510 630 855


LESSON

1.5

Framework objectives – Limits of sequences

Generate terms of a sequence using term-to-term and position-to-term definitions ofthe sequence.

Exercise 1E Answers

1 a 1, 3.5, 4.75, 5.375, 5.6875, 5.843 75, 5.921 875, 5.960 937 5, 5.980 468 75,5.990 234 375, 5.995 117 188, 5.997 558 594

b 6 c Sequence always gets closer and closer to 62 a 1, 4.5, 6.25, 7.125, 7.5625, 7.78 125, …

b 8 c Sequence always gets closer and closer to 83 a 1, 5.5, 7.75, 8.875, 9.4375, 9.71 875, 9.859 375, …

b 10 c Sequence always gets closer and closer to 104 a 125 Gets closer to 4.56 Gets closer to 6 7 Gets closer to 7.5 8 Gets closer to 9

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Plenary� Put on the board the term-to-term rule ‘Divide by 2 and add 10’.� Ask: ‘Can anyone suggest what value this sequence will get closer and closer to?

Why do you think that?’� Students should be using the results from Questions 1 to 4, Exercise 1E, to help

them to see a simple link, and that this sequence will get closer to 20.� Now change the term-to-term rule to ‘Divide by 3 and add 10’, and ask the

same question.� Using their results from Questions 5 to 8, the class may need help to see the link

to multiplying the two numbers and halving the result to get the answer. Here,this gives the result of 15.


Ho

me

wo

rk 1 A sequence starting at 1 has the term-to-term rule Add 3 and divide by 2.

a Find the first 10 terms generated by this sequence.

b To what value does this sequence get closer and closer?

c Use the same term-to-term rule with different starting numbers. What do you notice?

2 Repeat Question 1, but change the term-to-term rule to Add 4 and divide by 2.

3 What would you expect the sequence to do if you used the term-to-term rule Add 7 and divide by 2?

4 What will the sequence get closer to using the term-to-term rule Add A and divide by 2?

5 Investigate the term-to-term rule Add A and divide by 3.

Answers1 b 3 c Always gets closer to 32 b 4 c Always gets closer to 43 Always gets closer to 74 A

A5 The terms in the sequence will get closer to –– .

2

� sequence� term-to-term

Key Words

Extension Answers

ABSequence always gets closer to ———– .

(A – 1)

SATs Answers

1 a

b

c 3n + 1 d 5n + 4n

2 a ———— b ii c 2–5, 3––10, 4––17 d i(2n + 1)

Pattern number Expression for Expression for number of grey tiles number of white tiles

n n + 1 2n

Pattern number Number of grey tiles Number of white tiles5 6 10

16 17 32

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Oral and mental starter� Use a target board such as the one shown on the right.� Start by asking the students, as a group or individually, to give the first five

multiples of various numbers.� Once the idea of the multiple is well established, ask for the lowest common multiple

(LCM) of a pairs of numbers. (It may be necessary to remind students of the definition.)� Continue for as long as necessary.

Main lesson activity� This is essentially a lesson on the four rules governing fractions.� The students will have met already the addition and subtraction of fractions, so they just need reminding of the

basic rules.� Ask the class to work out 25–9 + 3 7––15.� Outline the two methods for solving this addition. First, change to improper (top-heavy) fractions and proceed

as follows:25–9 + 3 7––15 = 23––9 + 52––15

= 115–––45 + 156–––45 = 271–––45 = 6 1––45

� Second, separate the whole numbers from the fractions:25–9 + 3 7––15 = 2 + 3 + 5–9 + 7––15

= 5 + 25––45 + 21––45

= 5 + 46––45

= 5 + 1 1––45 = 6 1––45

� Discuss the comparative advantages and disadvantages of each method. For example, the first method involveslarger numbers.

� Now ask the students how to work out 25–6 – 12–5.� Show them both methods.

25–6 – 12–5 = 17––6 – 7–5= 85––30 – 42––30 = 43––30 = 113––30

25–6 – 12–5 = 2 – 1 + 5–6 – 2–5= 1 + 25––30 – 12––30

= 1 + 13––30 = 113––30

� Ensure that the students understand the fact that the whole numbers are subtracted, the result of which is addedto the outcome of subtracting the fractions.

� This is the first time that the students will have met multiplication and division of fractions. � One way to introduce this topic is to use calculators to investigate the answers to products such as 1–5 × 3–5, 5–7 × 1–4

and 3–4 × 3–8. The students will see the rule very quickly.� Now repeat with 1–2 × 4–5, 3–4 × 8–9 and 4––15 × 3–8. Ask why the rule doesn’t appear to work. Discuss cancelling in the initial

product and in the answer. � Explain that it is better to cancel the initial fractions, as this makes the calculations easier and means that the

answer does not need to be cancelled down. Demonstrate with:4 1 25 5 3 1 5–– × –— × –— = –—9 3 28 7 10 2 42

� Now ask for the answer to 21–5 × 17–8. Explain how this is done, namely: 11 153 33

21–5 × 17–8 = –— × –— = –— = 41–851 8 8


Number 1CHAPTER

2

LESSON

2.1

Framework objectives – The four rules governing fractions

Use efficient methods to add, subtract, multiply and divide fractions. Cancelcommon factors before multiplying and dividing.

2 5 7 8

24 15 18 3

6 9 27 14

12 20 10 25

9B3LP_02.qxd 18/09/2003 15:18 Page 12

� Make sure students know that the mixed numbers cannot be separated as in addition or subtraction. They mustbe converted to improper (top-heavy) fractions.

� Division is a little harder to see. If calculators are used to investigate problems such as 2–7 ÷ 1–3, 3–4 ÷ 4–5 and 2–3 ÷ 8–9,some students may see the method.

� Most are unlikely to see that it requires turning the dividing fraction upside down and multiplying by it.Demonstrate this with the above examples.

� This is an easy method to use but not an easy one to understand. It can best be explained by examples such ashow many halves in 7? The answer is, of course, the same as multiplying by 2.

� Now ask for the answer to 22–3 ÷ 15–9. Explain how this is done, namely:8 14 8 4 9 3 12

22–3 ÷ 15–9 = –– ÷ –— = –– × –— = –— = 15–73 9 31 14 7 7

� Do more examples if necessary.


Plenary� Write the following problem on the board: 41–5 ÷ 101–2 × 3–8.� Work through it, cancelling whenever possible. The answer is 3––20.� Repeat for 11–8 × 41–3 × 2––13 (answer: 3–4) and (3–8 + 5–6) ÷ (3–4 + 1––18) (answer: 11–2).


Exercise 2A Answers

1 a 211––12 b 417––30 c 311––15 d 35–6 e 519––20 f 71–3 g 817––18 h 11 1––242 a 2 1––12 b 3 7––30 c 1 1––15 d 21–6 e 113––20 f 32–3 g 417––18 h 223––243 a 111––28 b 26––63 c 13––20 d 9––28 e 17––96 f 22––754 a 2––15 b 9––16 c 5––28 d 1–6 e 3––10 f 1–2 g 3––11 h 2––15 i 2–3 j 1–45 a 32–5 b 84–5 c 4 d 21–4 e 1 1––21 f 4–5 g 14–5 h 62–5 i 42–5 j 11 7––106 a 4 b 11–9 c 11–4 d 7––18 e 22–3 f 2–3 g 16––49 h 1 1––12 i 5–6 j 2–37 a 2 b 2 3––16 c 2 d 1 1––24 e 115––19 f 12––19 g 71–2 h 5 1––16 i 7––22 j 42–38 2–9 cm2 9 24 9––32 cm2 10 45–6 = 4 lengths 11 22–5 m

Extension Answers

Magic number is 1

Ho

me

wo

rk 1 Convert each of the following pairs of fractions to equivalent fractions with a common denominator.Then work out each answer, cancelling down and/or writing as a mixed number if appropriate.

a 22–5 + 21–4 b 22–3 + 11–8 c 25–8 – 15––12 d 3 5––12 – 13–4

2 Work out each of the following. Cancel before multiplying when possible.

a 1–6 × 3–8 b 2–3 × 3–4 c 2–9 × 3––16 d 41–5 × 13–7 e 23–8 × 13–5

3 Work out each of the following. Cancel at the multiplication stage when possible.

a 1–4 ÷ 1–3 b 3––16 ÷ 9––14 c 1–6 ÷ 1–3 d 25–8 ÷ 7––16 e 23–5 ÷ 3––10

Answers1 a 413––20 b 319––24 c 1 5––24 d 12–32 a 1––16 b 1–2 c 1––24 d 6 e 34–53 a 3–4 b 7––24 c 1–2 d 6 e 82–3

� convert� equivalent

fraction� mixed number� top-heavy

fraction� improper

fraction� lowest common

multiple� cancelling

Key Words

2––153–5

4––15

7––151–3

1–5

2–51––15

8––15

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Oral and mental starter� Use a target board such as the one shown on the right.� Ask the students for the equivalent multiplier when there is a percentage

increase or decrease. For example, 1.05 for the multiplier for a 5% increase and0.95 for a 5% decrease.

� Go round the class picking students at random and asking for the appropriatemultipliers for an increase and/or a decrease for percentages on the board.

Main lesson activity� Start by asking the students what they know about interest and how it works.

Then move on to compound interest. � Emphasise the basis of compound interest, namely: an amount of money (the

principal) is invested at an annual percentage rate(R%), and over a period ofyears the value increases by R% each year.

� Make sure that the students understand that the interest for each year is added tothe principal of the previous year to give the new principal for the followingyear.

� Work through an example. First, use the method of working out the yearlyinterest and adding it on for each year; then use a multiplier. For example,calculate how much £300 will earn when invested for 3 years at 4% interest perannum.

First method Increase and additionAfter first year: 4% of £300 = £12. So, at end of first year, you have

£300 + £12 = £312.After second year: 4% of £312 = £12.48. So, at end of second year, you

have £312 + £12.48 = £324.48.After third year: 4% of £324.48 = £12.98. So, at end of third year, you

have £324.48 + £12.98 = £337.46.

Second method Use a multiplierAfter first year: £300 × 1.04 = £312After second year: £312 × 1.04 = £324.48After third year: £324.48 × 1.04 = £337.46

� Point out that the last value is rounded to the nearest penny.� Most students are likely to prefer the second method. However, some will prefer

the structured nature of the first method.� Demonstrate the use of powers on a calculator. For example, the above result is

identical to 300 × (1.04)3. This gives 337.4592, which has to be rounded to thenearest penny.

� Discuss the advantage (quick) and the disadvantage (any keying errors mean noworking from which to gain partial credit).

� Repeat with an example that decreases each year. For example, an ant colonyhas 30 000 ants. They start to die off at the rate of 22% per day. How many antswill be left after 7 days?

� The first method is too lengthy. The second method (using the constantmultiplier of 0.78) gives:

30 000, 23 400, 18 252, 14 236.56, 11 104.52, 8661.52, 6755.99,5269.67

Hence, there are 5270 ants left at the end of 7 days.

� The class can now do Exercise 2B From Pupil Book 3.


LESSON

2.2

Framework objectives – Percentages and compound interest

Recognise when fractions or percentages are needed to compare proportions.

Solve problems involving percentage changes.

5% 10% 22% 13%

16% 25% 14% 35%

20% 17% 6% 15%

8% 12% 71–2% 2%

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Plenary� Discuss the advantages/disadvantages of using a multiplier and powers

compared with other methods.� If the class is able enough, establish the formula for compound interest:

Percentage rateTotal at end of period = Initial amount × (1 ± ————————)

Time period

100So, take as an example 45 000 units decreasing at 6% each day for 3 days,which gives:

6Total after 3 days = 45 000 × (1 – ——)

3= 37 376.28 units

100


Exercise 2B Answers

1 a 1.12 b 0.95 c 0.92 d 1.07 e 0.96 f 1.02 g 1.032 h 1.025i 0.85 j 1.06 k 0.974 l 1.005 m 0.76 n 0.93 o 1.175

2 a £216.49 b £3740.06 c £214.90 d £19 348.42 e £80.77 3 a £2348.27 b £219.15 c £1334.494 a 33 662 b 18 8375 6 days6 12 days

Ho

me

wo

rk 1 How much would you have in the bank if you invest as follows?

a £450 at 3% interest per annum for 4 years.

b £6000 at 4.5% interest per annum for 7 years.

2 Stocks and shares can decrease in value as well as increase. How much would your stocks andshares be worth if you had invested as follows?

a £1000, which lost 14% each year for 3 years.

b £750, which lost 5.2% each year for 5 years.

Answers1 a £506.48 b £8165.172 a £636.06 b £574.25

� compoundinterest

� multiplier

Key Words

Extension Answers

a £547.50 b £47.54 c £47.30d £47.07, £46.83, £46.60, £46.37, £46.14, £45.91, £45.68, £45.45, £45.23, £45e £555.12

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Oral and mental starter� Use a target board such as the one shown on the right.� Ask the students for the equivalent percentage increase or decrease for each

multiplier. For example, 1.05 represents a 5% increase. � Go round the class picking students at random and asking for the appropriate

percentage increase and/or decrease for the multipliers on the board.

Main lesson activity� Reverse percentages and how to choose the appropriate quantity to take as

100% are now going to be covered.� Ask the students for the original amount when the new value, after a 30%

increase, is £195.� Many may suggest that it is the same as a 30% decrease of £195, which is

£136.50. However, when the latter amount is increased by 30%, the answer is £177.45.

� Explain that £195 represents not 100% but 130%.� Solve the problem using first the unitary method, and then a multiplier.

Method 1 Unitary methodThis involves finding a single unit value, which in this case is the value of 1%.£195 represents 130%£1.50 represents 1% (Dividing both sides by 130.)£150 represents 100% (Multiplying both sides by 100.)

Method 2 Use a multiplierA 30% increase is represented by the multiplier 1.30.Hence, divide £195 by 1.3 to find the original amount. This gives:195 ÷ 1.3 = £150

� Discuss the disadvantages/advantages of each method. The students willprobably prefer to use the multiplier, as it is easier to work out and has fewersteps.

� Repeat with other examples if necessary.� Now ask the students: ‘What is the percentage increase from £550 to £704?’� Do the calculation on the board:

Actual increase = £704 – £550 = £154£154

Percentage increase = ——– × 100 = 28%£550

� Emphasise that £550 is the original amount.



LESSON

2.3

Framework objectives – Reverse percentages and percentagechange

Use proportional reasoning to solve a problem, choosing the correct numbers totake as 100%, or as a whole.

1.05 0.90 1.22 0.87 0.6

1.16 0.75 1.14 1.35 1.3

0.8 0.81 1.06 0.65 0.02

0.92 1.12 1.175 0.98 1.88

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Plenary� Introduce the ‘magic road sign’, which is a mnemonic to help to recall the

different combinations used in the calculation of percentage change and reversepercentages.

That is:New value = Original value × MultiplierMultiplier = New value ÷ Original valueOriginal value = New value ÷ Multiplier


Exercise 2C Answers

1 1200 g2 £2603 Camera £190, heater £60, printer £70, washer £250, sofa £450, computer £18004 £1285 £506 41.4%7 4.2%8 946 1549 £15

10 South-east England 24%, Scotland 16%, Yorkshire 15%, East Anglia 36%

Ho

me

wo

rk 1 A packet of biscuits claims to be 24% bigger! It now contains 26 biscuits. How many did it havebefore the increase?

2 After a 10% price decrease, a hi-fi system now costs £288. How much was it before the decrease?

3 This table shows the cost of some items after 171–2% VAT has been added. Work out the cost of eachitem before VAT.

4 A pair of designer jeans is on sale at £96, which is 60% of its original price. What was the originalprice?

5 A pair of boots, originally priced at £60, were reduced to £36 in a sale. What was the percentagereduction in the price of the boots?

Answers1 21 biscuits 2 £3203 Radio £96, table £112, cooker £280, bed £3204 £1605 40%

Item Cost inc VAT Item Cost inc VAT

Radio £112.80 Cooker £329

Table £131.60 Bed £376

� reversepercentage

� unitary method� multiplier� power

Key Words

Extension Answers

a 19.6% b 11.4% c 12.7% d 79.6% e 2.2%

Newvalue

Originalvalue Multiplier

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Oral and mental starter� Give the class a ratio, such as 2 : 5, and ask them to find the missing value in

the equivalent ratio … : 15 (6 : 15). Repeat with, for example, 10 : … (10 : 25),… : 35 (14 : 35).

� Students could use number fans or mini white boards on which to write theiranswers.

� Repeat with other ratios.

Main lesson activity� Ask the class to solve the following problem:

If five exercise books have a total of 175 pages, how many pages will fourexercise books have?

� Students may have an intuitive idea of the answer (140 pages) but outline theunitary method. Five books have 175 pages so, one book has 35 pages.Therefore, four books have 4 × 35 = 140 pages.

� Alternatively this can be considered as a ratio, giving 5 : 175 = 1 : 35 = 4 : 140.(Note: this relates to the starter activity.)

� Repeat with the following examples.

�� Nine canteen tables can seat 72 people. How many people can sit at eightcanteen tables?If nine tables seat 72 people, one table would seat eight people. Therefore,eight tables will seat 8 × 8 = 64 people.Alternatively, 9 : 72 =1 : 8 = 8 : 64.

�� A bus with 20 passengers on board takes 10 minutes to travel from Silkstoneto Barnsley. How long will the same bus with 10 people on board take to dothe same journey?The number of passengers does not affect the speed of the bus. So, the timetaken will be exactly the same, 10 minutes.

� The following examples use inverse proportion. Students find this concept quitedifficult.

� Ask the class how to do this problem.

Five men build a wall in 9 days. How long would it take six men?The wall would take one man 5 × 9 = 45 days. Therefore, six men take 45 ÷ 6 = 7.5 days.

� Repeat with the following examples.

Three taps can fill a sink in 15 minutes. How long would five taps take tofill the sink?One tap would fill the sink in 3 × 15 = 45 minutes. Now, 5 × 9 = 45, so itwould take five taps 9 minutes.When I travel at 60 mph, my car does 10 miles to a litre of petrol. Howmany miles will I get per litre if I travel at 30 mph?This is impossible to answer as the petrol consumption of a car is not justdependent on speed.



LESSON

2.4

Framework objectives – Direct and inverse proportion

Reduce a ratio to its simplest form. Use the unitary method to solve simple wordproblems involving ratio, direct proportion and inverse proportion.

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Plenary� Put the following problem on the board: xy = 24.� Ask the class what values x and y could take. These could be written on the

board.� Now ask what would happen to y if x doubles? (y halves)� Similarly, what happens to x if y trebles? (x assumes one third of its first value)� Ask what happens when x is divided by 4? ( y is multiplied by 4)� Repeat with xy = 36 and other fractions. � If time allows, discuss the connection with the inverse proportion questions.


Exercise 2D Answers

1 £39.20 2 1 h 40 min 3 2 days 4 3 h 20 min 5 £2.86 6 14 min7 6 min 8 3 days 9 240 miles 10 40 min 11 12 days

12 a 22.4 cm b 4.375 km 13 a 144 miles b 4.58 14 48 min 15 216 min16 a 12 s b 6 s 17 360 min 18 9 days 19 5 min 20 15 min 21 21 tins22 2 h 24 min 23 2.5 cm 24 20 miles 25 36 h 26 9 days 27 22–3 tubes28 180 miles 29 7.5 h 30 253 exam papers

Ho

me

wo

rk 1 In 4 hours a man earns £45. How much does he earn in 5 hours?

2 A man walking one dog takes 20 minutes to walk one mile. How long will it take him to cover onemile if he walks two dogs?

3 In a week, grass grows 21 mm. How much does it grow in 4 days?

4 Fifty litres of petrol costs £35. How much will 20 litres of petrol cost?

5 Eight men dig a ditch in 9 days. How long would six men take?

6 A camping party of three has enough food to last them 4 days. If another person joins the party, howlong will the food last?

7 At £6 an hour, Jack takes 16 hours to earn enough for a guitar. If he had earned £8 an hour, howlong would it have taken him to earn the money?

8 Three bell ringers ring a tune on 6 bells in 5 minutes. How long would four bell ringers take to ringthe same tune?

Answers1 £56.25 2 20 min 3 12 mm 4 £14 5 12 days 6 3 days 7 12 h 8 5 min

� ratio� unit� direct proportion� inverse

proportion

Key Words

Extension Answers

a 2d b 3d c 1–2d d 3d e 8d f 1–8d

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Oral and mental starter� Give the class practice on squares and cubes. They should know the squares up

to 15 × 15 and the cubes up to 5 × 5 × 5.� Do this as a mental test or ask students individually to give you answers to

questions such as: four cubed; the cube root of eight; nine squared, and so on.

Main lesson activity� This section of Pupil Book 3 opens with an investigation which is reproduced

below.� Let students work through the investigation.� Use the plenary (see next page) to bring the students together to check their

answers.� Investigation

These three blocks are similar. This means that the ratio height : length : width isthe same for all three blocks.a Work out the area of the front face of each block.b Work out the volume of each block.

Work out each of the following ratios and write it in the form 1 : n.c i Length of block A to length of block B.

ii Area of the front face of block A to area of the front face of block B.iii Volume of block A to volume of block B.

d i Length of block A to length of block C.ii Area of the front face of block A to area of the front face of block C.iii Volume of block A to volume of block C.

e i Length of block B to length of block C.ii Area of the front face of block B to area of the front face of block C.iii Volume of block B to volume of block C.

Look at your answers to parts c, d and e. What do you notice?Explain the connection between the ratio of the lengths, areas and volumes ofsimilar shapes.

� After the investigation, the class can now do Exercise 2E from Pupil Book 3.


LESSON

2.5

Framework objectives – Ratio in area and volume

Understand the implications of enlargement for area and volume.

AB

C

2 cm 4 cm 12 cm

2 cm

4 cm

12 cm

18 cm

3 cm

6 cm

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Plenary� The plenary can be done at any appropriate time and is used to ensure that

every student has grasped the concept of linear, area and volume scale factorsand the connection between them.

� Stop students working and discuss the results of the investigation.� Make sure they understand that a linear increase by a factor of 2, say, means an

area increase by a factor of 4 and a volume increase by a factor of 8.� Generalise this as linear 1 : a, area 1 : a2, volume 1 : a3.� Students have difficulty with the concept of a scale factor. Ensure they

understand that it is the original length, area or volume which is multiplied bythe scale factor to get the new length, area or volume.


Exercise 2E Answers

1 b i 1 : 3 ii 1 : 9 iii 9 × 5 = 45 cm2 c i 1 : 5 ii 1 : 25iii 25 × 5 = 125 cm2

2 b i 1 : 4 ii 1 : 64 iii 64 × 6 = 384 cm3 c i 1 : 5 ii 1 : 125iii 125 × 6 = 750 cm3

3 90 cm2

4 4 cm2

5 64 cm3

6 2 cm3

7 a 1–4 b 1–88 a 1 : 100 b 1 : 10 000 c 1 : 1 000 000

Ho

me

wo

rk 1 Two similar, plane shapes, A and B, have lengths in the ratio 1 : 4. The area of shape A is 10 cm2.

What is the area of shape B?

2 Two similar, plane shapes, P and Q, have lengths in the ratio 1 : 2. The area of shape Q is 100 cm2.

What is the area of shape P?

3 Two similar solids, C and D, have lengths in the ratio 1 : 3. The volume of solid C is 15 cm3.

What is the volume of solid D?

4 Two similar solids, R and S, have lengths in the ratio 1 : 2. The volume of solid S is 72 cm3.

What is the volume of solid R?

Answers1 160 cm2 2 25 cm2 3 405 cm3 4 9 cm3

� linear scalefactor

� area scale factor� volume scale

factor

Key Words

Extension Answers

a Height : radius in same ratio b i 12.57 cm2, 28.27 cm2, 78.54 cm2

ii 50.27 cm3, 169.65 cm3, 785.40 cm3 c i 1 : 2.25 ii 1 : 6.25 iii 1 : 3.375iv 1 : 15.625 d i 1 : 9 ii 1 : 27

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Oral and mental starter� This brief starter could be missed out.� Quickly recall the rules for multiplying and dividing with directed numbers. Do

this with a few examples, such as:–2 × +3 = –6 –4 ÷ –1 = +4 –5 ÷ 0.5 = –10

Main lesson activity� This lesson is essentially an investigation. The class can work in groups or

individually. Keep the introduction as brief as possible.� The introduction and Example 2.15 from Pupil Book 3 are reproduced here.

The special numbers –1, 0 and 1 divide the number line into four sets ofnumbers: A, B, C and D.A contains all the numbers less than –1. B contains all the numbers between –1and 0. C contains all the numbers between 0 and 1 and D contains all thenumbers greater than 1.

Example 2.15 a What happens when a number from set A is multiplied bya number from set D?

b What happens when a number from set B is divided by 1?

a Choose any number from set A, say –2. Choose anynumber from set D, say +3. Multiply them together:

–2 × +3 = –6.

The answer belongs to set A.Try other combinations of numbers from set A and set D.For example:

–4 × +4 = –16 –1.5 × 5 = –7.5 –5 × 1.5 = –7.5

They all belong to set A. So, a number from set Amultiplied by a number from set D always gives a numberin set A.

b Pick numbers from set B and divide each one by 1. Forexample:

–0.4 ÷ 1 = –0.4 – 2–3 ÷ 1 = – 2–3 –0.03 ÷ 1 = –0.03

The answers are the same as the values from set B. So,they all give numbers in set B.

� This example, or similar examples, could be worked through to start theinvestigation.

� The class can now do Exercise 2F from Pupil Book 3.

0–1–2–3 1

0–1 1

2 3

A B C D


Framework objectives – Numbers between 0 and 1

Use the laws of arithmetic and inverse operations.

Understand the effects of multiplying and dividing by numbers between 0 and 1.

LESSON

2.6

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Plenary� The answers to the two tables (Exercise 2F, Questions 1 and 2) should be

discussed with the whole class. Highlight particularly the significance ofmultiplying and dividing by numbers between 0 and 1.

� Familiarise them with the term counter-example.


Exercise 2F Answers

1 2

3 a ii b iii c iii d ii4 For example: a –8 ÷ –1 = 8 b 7 × –1.2 = –8.4 c 9 ÷ –0.2 = –45

d –6 × –0.5 = +3

Ho

me

wo

rk Say which of these statements is true. If it is not true, give a counter-example.

a The square of a number between 0 and 1 is also between 0 and 1.

b The square of a number between 0 and –1 is also between 0 and –1.

c Dividing any number by a number between 0 and 1 always gives a bigger answer.

d Dividing any positive number by a number between 0 and 1 always gives a bigger answer.

Answersa True b False, for example: (–0.4)2 = 0.16 c False, for example –7 ÷ 0.5 = –14, which is smaller d True

� counter-example� generalise� conclude� deduce

Key Words

× A –1 B 0 C 1 D

A D D C/D 0 A/B A A

–1 D 1 C 0 B –1 A

B C/D C C 0 B B A/B

0 0 0 0 0 0 0 0

C A/B B B 0 C C C/D

1 A –1 B 0 C 1 D

D A A A/B 0 C/D D D

Extension Answers

No firm conclusions can be reached. The results depend on the numbers chosen.

÷ A –1 B 0 C 1 D

A C/D D D A A A/B

–1 C 1 D A –1 B

B C C C/D A/B B B

0 0 0 0 0 0 0

C B B A/B C/D C C

1 A –1 B C 1 D

D A/B A A D D C/D

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Oral and mental starter� As preparation for the mental test in the SATs papers, this starter concentrates on

SATs style questions using powers. Ask each question twice and allow about 10seconds for the students to answer.1 What is nine squared?2 What is the square root of 144?3 The diagram shows a cube made from smaller cubes. [Draw this on the

board.] What is the volume of the big cube?4 The length of one side of a square is 4 cm. What is the area of the square?5 The area of a square is 64 cm2. What is the perimeter of the square?6 x squared is 4. Which two values can x have?7 Look at the inequality. [Write x2 <17 on the board.] What is the largest

integer value that x could be?8 What number is five cubed?9 The nth term of a sequence is n plus one all squared. [Write (n + 1)2 on

board.] What is the fifth term of the sequence?10 What is the square root of nine-sixteenths?

� It would be worthwhile discussing the techniques involved when giving theanswers.

� It is useful to repeat this test within a few days to see whether scores improve.

Answers 1 81 2 12 3 8 (cm3) 4 16 cm2 5 32 cm 6 2 and –27 4 8 125 9 36 10 Three-quarters

Main lesson activity� Every student will need a calculator.� First, define the reciprocal of x as 1 ÷ x.� Give some examples, such as the reciprocal of 2 is 1 ÷ 2 = 0.5, the reciprocal of

10 is 1 ÷ 10 = 0.1.� Now ask the class to find the reciprocal key on their calculators.� This will be marked in various ways. For example, it may look like one of the

following:

� Do some examples and get the class to practise the use of this key.� Some students may find it just as easy to divide the number into 1, rather than

use the reciprocal key.� Next, ask them to find the reciprocal of each of these numbers: 40 (1 ÷ 40 =

0.025) and 0.8 (1 ÷ 0.8 = 1.25).� Do more examples, if necessary.� In Exercise 2G there are two questions (3 and 4) which are investigations. These

could be done as a class activity.

� The class can now do Exercise 2G from Pupil Book 3.


LESSON

2.22.7

Framework objectives – Reciprocal of a number

Recognise and use reciprocals.

Know how to use the reciprocal key on a calculator.

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Plenary� Invite the class to give you the value of each power from 10–3. Then write them

on the board in the form of a table.

10–3 10–2 10–1 100 101 102 103

0.001 0.01 0.1 1 10 100 1000

� Now ask them to find the reciprocals of these numbers.� Establish the relationship between positive powers, negative powers and

reciprocals. For example:10–4 = 1 ÷ 104 10–7 = 1 ÷ 107 10–a = 1 ÷ 10a


Exercise 2G Answers

1 a 1, 0.5, 0.33 … , 0.25, 0.2, 0.166 … , 0.142 857 14 … , 0.125, 0.111 … , 0.1, 0.0909 … , 0.0833 … , 0.076 923 0769 … , 0.0714 285 71 … , 0.0666 … ,0.0625, 0.0588 … , 0.055 … , 0.0526 … , 0.05

b Reciprocals of 1, 2, 4, 5, 8, 10, 16, 202 a 0.0333 … b 200 c 0.0125 d 800 e 0.0005 f 500 g 0.01

h 0.000 001 (10–6)a b

3 Reciprocal of — is —b a

4 a They are perpendicular to each otherb A, –2, 1–2; B, 1, –1; C, 3–2, – 2–3; D, 1–4, –4; E, 3, –1–3c When a pair of lines is perpendicular, their gradients are the negative reciprocalsof each other

5 a 2 b –2 Always true6 No, you are not allowed to divide by zero.

Ho

me

wo

rk 1 a Find, as decimals, the reciprocals of all the integers from 21 to 25.

b Which of the reciprocals are recurring decimals?

2 Find the reciprocals of each of the following numbers. Round your answers if necessary.

a 50 b 0.004 c 60 d 0.625

Answers1 a 0.047 619 … , 0.045 45 … , 0.043 47 … , 0.041 66 … , 0.04b All except the reciprocal of 25 which terminates, and the reciprocal of 23 which neither recurs or terminates2 a 0.02 b 250 c 0.016 66 … d 1.6

� reciprocal� divide into

Key Words

Extension Answers

For any power n, the reciprocal of 2n is 2–n

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Oral and mental starter� Have a set of Follow-me cards dealing with approximations. A set of 20 are

given below. These should be given to pairs of students to allow discussion. It issuggested that each pair be given a few moments to come up with an estimate(the teacher can write each one on the board) and that jottings are allowed.

1 START. You are 0.52 × 0.62 2 I am approx. 0.3. You are 892 × 0.48.3 I am approx. 450. You are 0.29 × 0.31 4 I am approx. 0.09. You are 58 × 72.5 I am approx. 4200. You are 0.32 × 61 6 I am approx. 18. You are 312 × 0.32.7 I am approx. 90. You are 217 × 53 8 I am approx. 10 000. You are 0.092 × 0.129 I am approx. 0.009. You are 6.2 × 0.72 10 I am approx. 4.2. You are 187 × 0.68.

11 I am approx. 140. You are 3.2 × 0.33 12 I am approx. 0.9. You are 39 × 51.13 I am approx. 2000. You are (72)2 14 I am approx. 4900. You are 0.32 × 0.1115 I am approx. 0.03. You are 504 × 189 16 I am approx. 100 000. You are 96 × 0.11.17 I am approx. 10. You are (0.68)2 18 I am approx. 0.49. You are 92 × 89.19 I am approx. 8100. You are 0.092 × 321 20 I am approx. 27. END.

Main lesson activity� Ask the class to round the following numbers as indicated: 368 to the nearest

100, 23.9 to the nearest 10, 0.0713 to one decimal place (dp).� Ask them whether the answers have anything in common.� They may notice – or may need to be prompted to state – that the answers all

have one digit apart from zero.� These are examples of rounding to one significant figure (1 sf).� This is a concept that many students find hard to grasp. Emphasise that there will

be only one digit apart from zeros.� Do more examples of rounding to one significant figure, such as:

3789 ≈ 4000 0.265 ≈ 0.3 0.25 ≈ 0.3 0.198 ≈ 0.20.636 ≈ 0.6 0.0621 ≈ 0.06 0.0764 ≈ 0.08

� Now give some examples of how rounding to one significant figure can be usedto estimate answers to calculations. For example:

320 × 398 ≈ 300 × 400 = 120 0000.092 × 476 ≈ 0.09 × 500

= 0.9 × 50 = 9 × 5 = 45

(29 + 88) ÷ (2.3 × 0.053) ≈ (30 + 90) ÷ (2 × 0.05) = 120 ÷ 0.1 = 1200 ÷ 1 = 1200

17 ÷ 0.42 ≈ 20 ÷ 0.4 = 200 ÷ 4 = 50

� Do more examples, if necessary.

� The class can now do Exercise 2H from Pupil Book 3.


LESSON

2.22.8

Framework objectives – Rounding and estimation

Estimate calculations by rounding numbers to one significant figure and multiplyingor dividing mentally.

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Plenary� Put the following on the board:

(14.7 × 3.9) ÷ (0.96 + 0.59)� Ask a student to round to 1 sf and approximate the answer. That is:

(10 × 4) ÷ (1 + 0.6) = 40 ÷ 1.6� At this stage, the problem becomes too difficult to do mentally.� Ask whether it is possible to approximate differently. For example:

(15 × 4) ÷ (1 + 0.5) = 60 ÷ 1.5 = 40� Point out that approximations do not have to be made to 1 sf when a more

sensible approximation is possible.� Try this out with 23.6 × 7.8 ÷ 49.2. Rounding to 1 sf gives 20 × 8 ÷ 50, but

rounding to 25 × 8 ÷ 50 = 4 is easier to do mentally.


Exercise 2H Answers

1 a 600 b 0.3 c 0.07 d 100 e 0.3 f 0.09 g 300 h 0.4 i 0.7j 0.6 k 900 l 1

2 a 80 000 b 1 500 000 c 4200 d 160 000 e 8100 f 0.18 g 0.063h 0.04 i 100 j 140 k 240 l 420

3 a 6 b 3000 c 5 d 400 e 40 f 50 g 500 h 600 i 400 j 50k 400 l 0.3

4 a 0.7 × 600 = 420 b 300 ÷ 0.5 = 600 c 3000 ÷ 0.6 = 5000d 0.06 × 0.2 = 0.012 e (20 × 0.05) ÷ 0.4 = 2.5f (0.04 + 0.06) × (0.07 + 0.08) = 0.015 g (200 × 0.1) × (800 ÷ 0.02) = 800 000h 0.5 × (30 ÷ 0.4) = 40 i 52 × 8 ÷ 0.22 = 25 × 8 ÷ 0.04 = 200 ÷ 0.04 = 5000j (20 × 0.4) ÷ (2 + 0.5) = 3.2

5 a 4000 b 10 000 c 250 000 d 0.001 e 20 f 1900

Ho

me

wo

rk 1 By rounding each value to one significant figure, estimate the answer to each of the following.

a 0.83 × 793 b 618 ÷ 0.32 c 812 ÷ 0.38

d 0.78 × 0.049 e (38 × 3.2) ÷ 0.487 f (2.7 + 6.3) × (0.52 – 0.17)

Homework answers1 a 0.8 × 800 = 640 b 600 ÷ 0.3 = 2000 c 800 ÷ 0.4 = 2000 d 0.8 × 0.05 = 0.04

e (40 × 3) ÷ 0.5 = 240 f (3 + 6) × (0.5 – 0.2) = 2.7

� most significantdigit

� least significantdigit

� approximate� significant figure

Key Words

Extension Answers

a 350 b 4220 c 4200 d 0.619 e 0.62 f 300 g 4700 h 4700i 0.079 j 978.3 k 978 l 980

SATs Answers

1 a 14.5% b 17 255 c 39.3%d Not enough information: total number of police officers not given for both years

2 a 48% b 1 : 5.8 c 15––343 a 70 × 1.09; 70 × 0.9 is 90% of 70, 70 × 1.9 is 70 increased by 90%, 70 × 0.09 is

9% of 70 b 0.861 1 2 2

4 For example: Let t = 1, w = 1. Then – + – = 1 + 1 = 2 but –––– = – = 1, and 2 ≠ 11 1 1+ 1 2

5 a 13 b 28.2%c In 1961, birthrate in Scotland was lower than birthrate in Northern Ireland

6 £45 7 t = 4 r = 7.58 2 : 3 9 21%

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Oral and mental starter� Ask the class for the largest multiple of 9 less than 100. (It is 99.) � Repeat this for the largest multiples less than 100 of these numbers:

7 (98) 8 (96) 13 (91) 3 (99)� Discuss how students found the answers. They would probably have added on from known facts or used trial

and improvement. Working from known facts is a good way to find an answer.� Now ask for the largest multiple of 7 under 300. Demonstrate the following method for finding it.

Start from 300 ÷ 7, which can be seen to be about 40.Calculate 7 × 40 = 280, add on sevens to get 287, then 294, which is the largest multiple (as it is only 6 less than 300).

� Now ask individual students to find each of the following using a similar method.Largest multiple of 6 less than 500 (498).Largest multiple of 7 less than 400 (399).Largest multiple of 8 less than 300 (296).Largest multiple of 11 less than 700 (693).Largest multiple of 15 less than 800 (795).

Main lesson activity � Put on the board the equation x + y = 8. Ask the class to give you some solutions to the equation. Put them all

on the board. Examples are: x = 5, y = 3; x = 1, y = 7; x = 0, y = 8; x = –1, y = 9.� Now put on the board the equation 3x – y = 12. Ask the class to give you some solutions of this equation.

Again, put them all on the board. Examples are: x = 4, y = 0; x = 6, y = 6; x = 2, y = –6; x = 5, y = 3.� Each equation has many solutions, but how many solutions will fit both equations simultaneously?� There is only one such solution to these two equations: x = 5, y = 3.� Explain to the class that solving simultaneous equations is an important part of mathematics and there are three

main ways of finding the unique solution.� Today, you are going to show them the method of elimination.� Put the two equations on the board again in the traditional simultaneous mode:

x + y = 8 (i)3x – y = 12 (ii)

Add the two equations in columns, to get:4x = 20 (Notice that, +y + –y = 0.)

x = 5Now substitute x = 5 into equation (i) to give:

5 + y = 8y = 3

� Show the class that this is the same solution which they had before.� Explain that, since there are two identical terms apart from their signs, the two equations are added together to

eliminate this variable.� Next, put on the board another pair of simultaneous equations:

5x + 2y = 20 (i)3x + 2y = 16 (ii)

Since there are two identical terms (2y), 2y can be eliminated by subtracting one equation from the other. Thiscan be either top from bottom or bottom from top. It really depends on avoiding a negative quantity in theresulting equation.


Algebra 3CHAPTER

3

LESSON

3.1

Framework objectives – Simultaneous equations

Solve a pair of simultaneous equations by eliminating one variable.

9B3LP_03.qxd 18/09/2003 15:18 Page 28

Hence, subtract equation (ii) from equation (i). This gives:2x = 4 (Notice that 2y – 2y = 0.)

x = 2Substitute x = 2 into equation (ii). (Chosen as it has the smallest numbers.)

6 + 2y = 162y = 10

y = 5� Show here, how to check the solution by substituting both x and y values into equation (i). That gives:

5 × 2 + 2 × 5 = 20, which is correct for equation (i).� If the class require further examples to be worked through, use those on page 42 in Pupil Book 9.3.


Plenary� Ask the class: ‘When do we add and when do we subtract a pair of simultaneous

equations?’� Use this as a discussion starter, but do ensure that every member of the class

understands that subtraction is undertaken when there are two identical termswith the same sign, and that addition is undertaken when terms are identical butthe signs are not.


Exercise 3A Answers

1 x = 4, y = 1 2 x = 1, y = 4 3 x = 3, y = 1 4 x = 5, y = 2 5 x = 7, y = 16 x = 5, y = 3 7 x = 4, y = 2 8 x = 2, y = 4 9 x = 3, y = 5 10 x = 2, y = 3

11 x = 4, y = 3 12 x = 5, y = 4

Extension Answers

1 a 29p b 19p2 £1.10

Ho

me

wo

rk 1 4x + y = 14 2 6x + 3y = 33

2x + y = 8 2x + 3y = 21

3 3x + y = 10 4 5x + 2y = 22

8x – y = 1 7x – 2y = 2

5 5x – 4y = 36 6 5x + 3y = 50

2x – 4y = 6 9x – 3y = 48

Answers1 x = 3, y = 2 2 x = 3, y = 5 3 x = 1, y = 7 4 x = 2, y = 6 5 x = 10, y = 3.5 6 x = 7, y = 5

� simultaneousequation

� elimination� substitution

Key Words

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Oral and mental starter� Tell the class they are going to play a quick game of 24.� Give them four numbers: 1, 2, 4 and 6. Ask who can make a sum out of these

numbers whose answer is 24. Once one sum has been made, make another.� Two possible answers to this game are:

(4 × 6) × (2 – 1) (4 – 1) × (2 + 6)Discuss the possible solutions and see who has made the most.

� Next give them 1, 2, 4 and 9. Ask who can make a sum out of these numberswhose answer is 24.

� One solution is 42 + 9 – 1.� This game can be extended to more sets of four numbers, if you wish.

Main lesson activity� Remind the class that the last lesson was devoted to solving simultaneous

equations by adding or subtracting in order to eliminate one of the twovariables.

� Now, they are going to solve equations where it is possible to eliminate one ofthe variables by simple substitution.

� Put on the board this pair of simultaneous equations:x – 2y = 1 (i)2x + y = 12 (ii)

From equation (i) make x the subject:x = 1 + 2y

Next, substitute this value for x in equation (ii):2(1 + 2y) + y = 12

2 + 4y + y = 122 + 5y = 12

5y = 12 – 2 = 10y = 2 (Divide both sides by 5.)

Now substitute y = 2 in equation (i), which gives:x = 1 + 2yx = 1 + 2 × 2 = 1 + 4x = 5

� Go through another example, such as:2x + y = 7 (i)

3x – 2y = 7 (ii)From equation (i), make y the subject:

y = 7 – 2xSubstitute this in equation (ii) to give:

3x – 2(7 – 2x) = 73x – 14 + 4x = 7

7x = 21x = 3 (Divide both sides by 7.)



LESSON

3.2

Framework objectives – Solving by substitution

Solve a pair of simultaneous linear equations by eliminating one variable.

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Plenary� Tell the class that there will be times when they have to make a choice about

which method to use – elimination or substitution. Take, for example:3x + 4y = 31

x – 4y = 5� Work through the solution using both methods. The solution is x = 9 and y = 1.� You will find that most students see that the elimination method (adding the two

equations) will be the quicker method here.


Exercise 3B Answers

1 x = 4, y = 1 2 x = 2, y = 3 3 x = 4, y = 3 4 x = 7, y = 3 5 x = 2, y = 56 x = 4, y = 7 7 x = 2, y = 1 8 x = 3, y = 5 9 x = 6, y = 3 10 x = 8, y = 2

11 x = 1, y = 5 12 x = 4, y = 5

Ho

me

wo

rk 1 3x + y = 8 2 6x + 4y = 36

2x + 5y = 27 2x + y = 11

3 5x + 2y = 47 4 3x + y = 24

3x – y = 26 5x + 2y = 41

5 7x – 4y = 16 6 8x – 4y = 36

x – y = 1 x + 3y = 8

Answers1 x = 1, y = 5 2 x = 4, y = 3 3 x = 9, y = 1 4 x = 7, y = 3 5 x = 4, y = 3 6 x = 5, y = 1

� substitution� variable

Key Words

Extension Answers

1 £2.25 2 1 h 20 min

9B3LP_03.qxd 18/09/2003 15:18 Page 31

Oral and mental starter� Tell the class that last night in a pub quiz, you got correct answers to 28 questions out of 37. Ask them:

‘Approximately what percentage is this?’� Jot down the answers on the board. (76% is the correct estimate.)� Ask the students closest to explain how they estimated the answer – no calculators allowed.� One way is to try to get to a fraction out of 100. This can be done by multiplying both numbers by 3. This

would give 84 out of 117. Take from the 84 a third of the extra 17 from the denominator, which is about 6. Thisgives 78% – not a bad estimate.

� Do encourage class discussion and a sharing of techniques here, as some students will have other quiteinteresting ways to explore.

� Once various strategies have been considered, ask for a few more to see whether they apply.

Main lesson activity� Ask the class what the nth term is of the sequence 3, 5, 7, 9, … .� You want the answer T(n) = 2n + 1.� Remind the class that any linear sequence can be expressed by an nth term such as T(n) = An + B.� Today, however, you are going to be looking at quadratic sequences. Remind the class that a quadratic

sequence is one whose second differences are the same throughout the sequence.� The nth term of any quadratic sequence is given by T(n) = An2 + Bn + C.

You may want to show that this is true by substituting some values for A, B and C, generating the sequence, andseeing what the second differences are.

� Having established that this is the way to express a quadratic nth term, put on the board the sequence 10, 24,44, 70, 102.

� Then show the differences:10 24 44 70 102

First differences 14 20 26 32Second differences 6 6 6

� In a previous investigation it was established that half the second difference is A. Hence, A = 3, which gives thenth term as:

T(n) = 3n2 + Bn + CWhen n = 1, T(1) = 10. Hence, 10 = 3 + B + C ⇒ B + C = 7When n = 2, T(2) = 24. Hence, 24 = 12 + 2B + C ⇒ 2B + C = 12This gives a pair of simultaneous equations to solve:

B + C = 7 (i)2B + C = 12 (ii)

Subtract equation (i) from equation (ii), to obtain B = 5.Substitute B = 5 into equation (i) to give C = 2.Hence, the nth term is T(n) = 3n2 + 5n + 2.

� You should check this with the class on the next few terms.� Explain that you now have a routine for finding the nth term of a quadratic sequence T(n) = An2 + Bn + C.

Step 1 Find the second difference. Halve it to give A. Step 2 Find T(1) and T(2), or any other pair, to give a pair of simultaneous equations to solve for B and C. Step 3 Solve the simultaneous equations in B and C, given by Step 2.

� You may need to go through another example with the class. Use one of the examples in Pupil Book 3.

� The class can now do Exercise 3C from Book 3.

43 out of 77 62 out of 81 34 out of 92 51 out of 123

Find a third (to get to Find a quarter (to get to Find a tenth (92 needs about a Find a fifth (to get to 25) and multiply by 4 20) and multiply by 5 tenth added on to get to 100) 25) and multiply by 4

of 34 and add it on

56% 75% 37% 40%


LESSON

3.3

Framework objectives – Find the nth term for a quadraticsequence

Find the next term and the nth term of quadratic sequences.

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Plenary� Put on the board the triangular numbers together with their associated triangles

of circles. Ask if anyone can tell you what the nth triangular number is? (Answer: 1–2n(n + 1))

� Work through this with the class as a means of bringing together the strategy forfinding the nth term of a quadratic sequence.


Exercise 3C Answers

1 a No, second differences are 0 b Yes, second differences are 2c Yes, second differences are 1 d Yes, second differences are 2e Yes, second differences are 6 f No, second differences are all different

2 a 14 b 24 c 293 a 9, 19, 33, 51, 73 b 5, 18, 37, 62, 93

n (n2 + 4)4 a n2 + 7n b 3n2 + n + 5 c 3n2 + 2n – 3 d ———– e ——————

(n2 + 3) (n2 + 4n + 4)

Ho

me

wo

rk 1 Find the first five terms of each of the following sequences given by:

a T(n) = n2 + 7n – 3 b T(n) = 5n2 + 3n + 1 c T(n) = 6n2 – 5n

2 Find the nth term for each of the following quadratic sequences.

a 13, 25, 41, 61, 85 b 12, 18, 26, 36, 48 c 7, 14, 27, 46, 71

d 1–3, 2–7, 3––13, 4––21, 5––31 e 1–9, 4––18, 9––31, 16––48, 25––69 f 12––21, 25––46, 44––83, 69—–132, 100—–193

Answers1 a 5, 15, 27, 41, 57 b 9, 27, 55, 93, 141 c 1, 14, 39, 76, 1252 a 2n2 + 6n + 5 b n2 + 3n + 8 c 3n2 – 2n + 6 d n/(n2 + n + 1) e n2/(2n2 + 3n + 4)

f (3n2 + 4n + 5)/( 6n2 + 7n + 8)

� quadraticsequence

� seconddifference

Key Words

Extension Answers

1–2 (n + 3)(n + 4) – n

9B3LP_03.qxd 18/09/2003 15:18 Page 33

Oral and mental starter� Ask for some equivalent fractions to 1–2. After a few correct suggestions, such as

2–4, 5––10, … , ask for an equivalent fraction to 1–2 that uses 34.� The two possible answers are 34––68 and 17––34.� Now ask for two fractions equivalent to 1–3 that use the number 12 (12––36 and 4––12).� Repeat this with the following examples:

Equivalent to 1–5 using the number 45. ( 45–––225 and 9––45)Equivalent to 2–3 using the number 18. (18––27 and 12––18)Equivalent to 1–4 using the number 28. ( 28–––112 and 7––28)Equivalent to 3–4 using the number 36. (36––48 and 27––36)

Main lesson activityx

� Put on the board the equation –– = 5 and ask the class how to solve it.3

� You should be given the response ‘multiply both sides by 3’, which gives x = 15.(4x + 5)

� Then, put on the board the equation ———— = 7 and ask ‘How do we solve this?’3

� Again it is a matter of simplifying, step-by-step, the side which has the variable, x.� Start by multiplying both sides by 3 to give 4x + 5 = 21. Then subtract 5 from

both sides to give 4x = 16. Finally, divide both sides by 4 to give x = 4.� Next, write the following example on the board:

x – 1 2x + 8——– = ————

2 6� Explain that when there is a fraction on both sides, the first step is to find the

product of the denominators and multiply both sides by it. Here, this gives: 2 × 6 = 12,

12 × (x – 1) 12 × (2x + 8)—————— = ——————

2 6� Now remind the students about cancelling fractions. Hence, the denominator on

each can be eliminated to give: 6(x – 1) = 2(2x + 8)

� Expand each side to give 6x – 6 = 4x + 16. Show how to add 6 to each side,then subtract 4x from each side to give 2x = 22 and x = 11.

� When the class understand this process, you may want to work through a moredifficult problem, such as:

4(2x + 1) 2(2x – 4)———— = ————

5 3� Again, multiply both sides by the product of the denominators and then cancel

down, to obtain:15 × 4(2x + 1) 15 × 2(2x – 4)——————— = ———————

5 3which cancels down to 12(2x + 1) = 10(2x – 4).Expanding and simplifying gives the solution x = –13.



LESSON

3.4

Framework objectives – Equations involving fractions

Solve linear equations involving fractions.

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Plenary12

� Put on the board the equation —– = 2 and ask the class if anyone can solve it.x

� Some students should see that the solution is x = 6, but discuss with the class themethods that they might have used in order to get that solution.

245� Now give them —— = 14 and ask how this could be solved.

x� Multiply both sides by x to give 245 = 14x. Then divide both sides by 14 to

obtain x = 17.5.


Exercise 3D Answers

1 a x = 15 b t = 15 c m = 24 d x = 12 e w = 82 a 22 b 2.17 c 2.753 a 0.8 b 3.2 c 16 d –24 a 3 b 17 c 19 d –21.5 e 4 f –5.5

Ho

me

wo

rk 1 Solve each of the following equations.

3x 3t 6m 2x 2wa —– = 12 b —– = 6 c —– = 18 d —– = 8 e —– = 65 5 8 5 7

2 Solve each of the following equations.

x + 1 x + 5 2x + 4 3x + 1a ——– = 5 b ——– = 8 c ——— = 6 d ——— = 23 4 5 8


x – 1 x + 1 2x + 3 x – 2 3x – 2 x + 4a ——– = ——– b ——— = ——– c ——— = ——–3 4 3 2 5 2


5 3 4 5 7 5a ——– = ——– b ——— = ——— c ——— = ———x – 1 x + 1 3x – 2 2x + 1 5x – 2 3x + 5

Answers1 a x = 20 b t = 10 c m = 24 d x = 20 e w = 212 a x = 14 b x = 27 c x = 13 d x = 53 a x = 7 b x = –12 c x = 244 a x = –4 b x = 2 c x = 11.25

� denominator� cancelling

Key Words

Extension Answers

1 a 1––11 b 7––13 c – 1–3 d 47––32 e 17––5 f 26––52 a 5––14 b – 8–3 c 1–5 d 73––70 e 0.57 f 0.6

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Oral and mental starter� Ask the class if anyone can tell you the solution to x2 = 45.� It should quickly become clear that since 45 is not a square number, only an approximate answer can be given.� We are looking for the two whole numbers between which the solution lies. These will be 6 and 7, as 62 = 36

and 72 = 49. The solution looks closer to 7 than 6, so we might estimate this as 6.7 (and –6.7).� Repeat this process for the following numbers:

x2 = 58, between 7 and 8, estimate 7.6x2 = 158, between 12 and 13, estimate 12.6x2 = 179, between 13 and 14, estimate 13.4

Main lesson activity� Tell the class that if your salary was tripled and then they took £100 away you would still earn less than two

thousand pounds a week. Then ask them how much you might be earning? (You could use a similar problem,with the intention being to get them thinking about a solution that involves an inequality.)

� If you were to set the problem out as an equation it would look like3x – 100 < 2000

Go through the solution of this inequality, explaining that the rules are exactly the same as for normalequations.This will lead to 3x < 2100 x < 700So, you are saying that you earn less than £700 per week.

� Show how this can be illustrated on a number line as:

� Explain the use of an empty circle to show a strict inequality and a solid circle to show the added equality used.� Now put up on the board the inequation 4x + 5 ≥ 19. Show how this reduces to x ≥ 3.5 and show this on a

number line.


–1 0 1 2 3 4 5 6

0 700


LESSON

3.5

Framework objectives – Inequalities

Solve linear inequalities in one variable and represent the solution set on a numberline; begin to solve inequalities in two variables.

Exercise 3E Answers

1

2 a 1, 2, 3, 4, 5, 6, 7, 8 b 1, 3, 5 c 2, 4, 6, 8 d 2, 3, 5 e 1, 4, 9, 16 f 2, 3, 5, 7, 11 g 3, 6, 9 3 a x ≤ 5 b x > 16 c x < 3 d x ≥ 104

2 3 4 5 6 7

a

4 5 6 7 8

c

1 2 3 4 5

e

0 1

d

1 2 3 4

f–3 –2 –1 0 1 2 3 4

b

0 1 2

a

8 9 10

b

5 6 7

c

1 2 3

d

2 3 4

e

0 1 2

f

–2 –1 0

g

–3 –2 –1

h

3 4 5

i

–1 0 1

j

–3 –2 –1

k

5 6 7

l

11 12 13

m

3 4 5

n

2 3 4

o

9B3LP_03.qxd 18/09/2003 15:18 Page 36

Plenary� Put on the board a sketch of the graph y = x and discuss with the class where the

region y > x actually is. Discuss the line itself and the fact that as this is y = x theregion does not include the line.

� Draw on the board a sketch of the graph y = x2 and repeat the discussion for theregion y > x2.


Ho

me

wo

rk 1 Solve the following inequalities and illustrate their solutions on number lines.

a 5x + 7 ≥ 22 b 2x – 3 ≤ 10 c 4x + 3 < 11

d 2(x + 4) > 20 e 4(3t + 7) ≤ 16 f 2(5x – 4) ≥ 17

2 Write down the values of x that satisfy the conditions given.

a 2(4x + 3) < 50, where x is a positive, prime number.

b 2(3x – 1) ≤ 60, where x is a positive, square number.

c 4(5x – 3) ≤ 100, where x is positive but not a prime number.

3 Solve the following inequalities and illustrate their solutions on number lines.

a 5x – 4 < 11 b 3(2x + 5) ≤ 9x > –1 x > –4

Answers1

2 a 2, 3, 5 b 1, 4, 9 c 1, 43

–2 –1 0 1 2 3 4 –5 –4 –3 –2 –1 0

a b

2 3 4

5 6 7

6 6.5 7

–2 –1 0

1 2 3

2 3

a b c

d e f

Extension Answers

1

2 x ≤ 4, y ≥ 1, y ≤ x

x

ya

6

y ≥ 6

x

ybx ≤ 4

x

yc

2

5

1

y ≤ 3x + 2

4

� inequality� inequation

Key Words

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Oral and mental starter� Tell the class that you were looking at a DVD yesterday and noted that the

running time was 135 minutes. Ask the class how many hours and minutes thisis (2 hours 15 minutes).

� Discuss with the class how to calculate this, using multiples of 60 to find thenumber of hours with the remainder being minutes.

� Now ask: ‘What fraction of an hour is 15 minutes?’ Ask for both a fraction and adecimal answer (1–4 and 0.25).

� Next ask: ‘What fraction of an hour is 10 minutes?’ This is 1–6 as a fraction, but thedecimal is more awkward. Can the class determine what the decimal answerwould be without using a calculator?

� Starting from known facts, 1–3 = 0.333 3333 and 1–6 is half of this. Hence, 1–6 = 0.166 6667. Alternatively, 3–6 = 1–2 = 0.5, and 4–6 = 2–3 = 0.666 6667, so 1–6 = 4–6 – 3–6 = 0.666 6667 – 0.5= 0.166 6667.

� Then ask: ‘What fraction of an hour is 5 minutes?’. This is 1––12, which is half of 1–6 or0.083 333.

� Talk about the potential confusion of using decimal notation in time. That is,1.50 could mean 11–2 hours or 1 hour 50 minutes. The students should be surethey know which unit is being used.

Main lesson activity� Tell the class that you took a taxi journey late at night and watched the taxi fare

change on the display. The table shows the fares at various times after gettinginto the taxi.

� Draw a pair of axes with time on the x-axis and the fare on the y-axis. Labelthem. Mark the x-axis with one hour as the principal unit, subdivided into 5-minute sections. Mark the y-axis in pounds up to £10. Ensure there is enoughroom to extend the y-axis down to at least –£4. Now plot the points and draw asuitable straight line through them.

� Explain to the class that as the line is straight, the two variables have a linearrelationship. Furthermore, the fare increase in a time interval is directlyproportional to the length of time. This means that the fare always increases bythe same amount for a given increase in time. Show that this works out as £1.70every 5 minutes.

� Ask the students what the equation of the line will be in terms of time, t, inhours, and fare, f, in pounds. Remind them of the general equation of a straightline: y = mx + c, where m is the gradient and c is the y–axis intercept.

� In one hour, the fare would have increased by 12 × £1.70, which is £20.40.Hence, the gradient is 20.4. Extend the vertical axis down to –£4. Thendemonstrate that the line intercepts it at f = –£3.8. Therefore, the equation of theline is f = 20.4t – 3.8.

� Explain that the extra point used to find the equation of the line is impossible inreal-life situations, since a negative fare is meaningless.

� Ask what the fare would be at 00:50 if the same rate continued.� The graph can be used, or the equation. But if the equation is used, 50 minutes

must be changed into hours (50––60 = 0.833 hours), giving £13.20.


Time 00:15 00:20 00:30 00:35Fare (£) 1.30 3.00 6.40 8.10


LESSON

3.6

Framework objectives – Graphs showing direct proportion

Solve problems involving direct proportion, relating algebraic solutions to graphicalrepresentation of the equations.

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Plenary� Ask the class what is meant by ‘directly proportional’.� They need to understand that this means that one variable always increases by

the same amount for a given increase in the other variable. The graph of twovariables which are directly proportional will always be linear (that is, a straightline).

� Discuss the fact that some students did not find it easy to obtain the equations ofthe lines in Pupil Book 3, Exercise 3F. Using the idea of y = mx + c will alwayshelp them in this respect.


Exercise 3F Answers

1 b Yes c C = 1.5t – 8 d 16°C2 b Yes c f = 48 000t – 58 000 d Between 12 and 13 minutes past one

w3 b Yes c L = —— + 10 d 1700 g

1004 b Yes c S = 52t – 156 d 3385 b Approximately, yes c B = Approximately 0.66H d Approximately 760 cm

Ho

me

wo

rk A baby squid is weighed from birth at midday for its first 5 days. The results are shown in the tablebelow.

a Plot the points on a graph and join them with a suitable line.

b Is the increase in weight during a time interval directly proportional to the length of the interval?

c Write down the equation of the line showing the relationship between the weight (W ) and the age(D) of the squid.

d If the relationship held, at what age would the squid first weigh over 15 kg?

Answersb Yes c W = 1.4D + 0.3 d Day 11

Day 1 2 3 4 5Weight (kg) 1.7 3.1 4.5 5.9 7.3

� directlyproportional

� linearrelationship

Key Words

Extension Answers

a Approximately 10.40 AM b 5 times

9B3LP_03.qxd 18/09/2003 15:18 Page 39

Oral and mental starter� Ask whether any student can estimate the answer to 37 × 8.� You are hoping for ‘About 300’.� Discuss with the class the various methods they used, such as rounding 37 up to

40, then 8 × 40 = 320. But since 37 was rounded up, round down the answer to300.

� Ask for the approximation of 28 × 43.� You are hoping for ‘About 1200’ – maybe from 30 × 40.� Discuss the different strategies the class have adopted.� Ask the following, ensuring that everybody is taking part and knows what they

are doing.

� Should the class find these too simple, challenge them with the three- and two-digit product estimations:

Main lesson activity� Put a pair of axes on the board and draw on the line with the equation

y = 2x + 1 from (–1, –1) to (3, 7).� Tell the class this is the graph of y = 2x + 1. Ask them what the graph actually

represents.� What you want as a response is: ‘It’s a collection of solutions to the equation’.

Every point on the graph represents a different possible solution. For example,(0, 1) represents x = 0, y =1, (1, 3) represents x = 1, y =3, and so on.

� Ask the class for some solutions to the equation and write them down, ascoordinates.

� Now, using the same axes, put on the board the graph with the equation y = 4x – 2, from (–1, –6) to (3, 10).

� Again, ask for some solutions for this equation. For example, (0, –2) representingx = 0, y = –2 and (2, 6) representing x = 2, y = 6.

� Ask the question: ‘What is so special about the point where the two lines cross?’� One of the students should comment that the coordinates are on each graph. So,

at that point the solution of each equation is the same. In other words, this is thesolution of the two simultaneous equations.

� Show that the point where these two lines cross is (11–2 , 4) and that this point doesindeed satisfy both equations.

� Explain to the class that an alternative way to find the solution of simultaneousequations is to draw their graphs.


237 × 76 319 × 88 423 × 579 792 × 617

200 × 80 = 16 000 300 × 90 = 27 000 400 × 600 = 240 000 800 × 600 = 480 000

33 × 58 67 × 72 38 × 59 19 × 98 25 × 36

30 × 60 = 1800 70 × 70 = 4900 40 × 60 = 2400 20 × 100 = 2000 100 × 36 ÷ 4 = 900You know it will be You know it will be Did anyone spot this less. less. way?


LESSON

3.7

Framework objectives – Solving simultaneous equations bygraphs

Link a graphical representation of a pair of equations to the algebraic solution.

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Plenary� Draw on the board a pair of axes. Then draw on the lines whose equations are

x + y = 4 and x + y = 7.� Ask whether these two simultaneous equations have a solution.� This should prompt some useful class discussion about solutions and

intersections. The fact that these two lines are parallel to each other indicatesthat there can be no solution to them as a pair.


Exercise 3G Answers

1 x = 1.5, y = 3.5 2 x = 2.5, y = 1.5 3 x = 4.5, y = 3.5 4 x = 1.5, y = 3.55 x = 1.3, y = 2.7 6 x = 2.2, y = 1.6 7 x = 1.2, y = 3.3 8 x = 2.1, y = 1.9

Ho

me

wo

rk 1 a On the same pair of axes, draw the graphs of the equations y = 2x + 1 and y = 2x + 3.

b Explain why there is no solution to this pair of simultaneous equations.

2 a Does every pair of linear simultaneous equations have a solution?

b Explain your answer to part a.

3 a Does every pair of simultaneous equations which do have a solution, have a unique solution?

b Explain your answer to part a.

4 Sketch a pair of graphs, one quadratic and one linear, which represent a pair of simultaneousequations that will have only one solution.

Answers1 b Because the two lines are parallel, they do not intersect2 a No b When the pair represent a pair of parallel lines3 a No

b When one equation can be simplified to the other equation, there are an infinite number of solutions4 U-shaped curve and a straight line, which is a tangent to the curve, will have only one point in common

� intersection

Key Words

SATs Answers

1 y = 7.37

2 a 2 b 1 c –– d 23

3 a (1 × 1) + (2 × n) + (3 × 5) + (4 × 6) + (5 × 3) = 55 + 2n b 15 + n c 104 x = –1, y = –1

Extension Answers

1 Two solutions: x = 1.9, y = 2.6 and x = –3.2, y = 8.92 Two solutions: x = 0.9, y = 1.7 and x = –4.5, y = 15.8

9B3LP_03.qxd 18/09/2003 15:18 Page 41

Oral and mental starter� Ask the class to find the square and the square root keys on their calculators.� Ask individual students to explain how to use the two keys by giving examples.� Next, ask the class to copy the table below and then complete it as quickly as possible. They must only use

their calculators for the square numbers that they do not know.

� Now repeat the activity for square roots. Ask the class to give their answers to one decimal place.

� Give the class the answers.

Main lesson activity� Tell the class that they are going to use Pythagoras’ theorem, which was discovered over two thousand years

ago. Explain that it is used to calculate the length of sides in right-angled triangles.� Draw the diagram on the right on the board or OHP. Remind the class that in a right-

angled triangle, the longest side, which is always opposite the right angle, is called thehypotenuse.

� Ask the class to do the activity in Pupil Book 3, page 59.� Pythagoras’ theorem

Pythagoras was a Greek philosopher and mathematician who was born in about 581 BC on the island ofSamos, just off the coast of Turkey.The following famous theorem about right-angled triangles is attributed to him.

In any right-angled triangle, the square of the hypotenuse is equal to the sum ofthe squares of the other two sides.Pythagoras’ theorem is usually written as:

c2 = a2 + b2

� Show the class how to find the length of a hypotenuse by doing the following example. Calculate the length x in the triangle shown on the right.Using Pythagoras’ theorem:

x2 = 82 + 72

= 64 + 49= 113

So, x = √——113 = 10.6 cm (1 dp).

� Then show them how to work this out on a scientific calculator.

This may not work on some makes of calculator, and you may need to show some students how to do it ontheir calculator.

8 x2 + 7 x2 = x =

x 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

x2 1 4 9 16 25 36 49 64 81 100 121 144 169 196 225 256 289 324 361 400

√––x 1 1.4 1.7 2 2.2 2.4 2.6 2.8 3 3.2 3.3 3.5 3.6 3.7 3.9 4 4.1 4.2 4.4 4.5

x 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

√––x

x 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

x2


Shape, Space and Measures 1CHAPTER

4

LESSON

4.1

Framework objectives – Pythagoras’ theorem

Understand and apply Pythagoras’ theorem.

Hypotenuse

x

7 cm

8 cm

c

b

a

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� Next, show the class how to find the length of a shorter side by doing the following example. Calculate the length x of the triangle on the right.Using Pythagoras’ theorem:

x2 + 72 = 102

x2 = 102 – 72

= 100 – 49= 51

So, x = √—–51 = 7.1 cm (1 dp).

� Then, show them how to work this out on a scientific calculator.


Plenary� Ask the class to explain Pythagoras’ theorem.� Tell them that in the next lesson they will apply Pythagoras to solve problems.

10 x2 – 7 x2 = =x

Extension Answers

When a2 + b2 = c2, ∠C is a right angle When a2 + b2 < c2, ∠C is an acute angleWhen a2 + b2 > c2, ∠C is an obtuse angle

a2 b2 c2 a2 + b2 Is a2 + b2 = c2? Is ∠C right-angled, Is a2 + b2 > c2? acute or obtuse?Is a2 + b2 < c2?Write =, > or <

9 16 25 25 = Right-angled16 25 49 41 < Acute25 36 49 61 > Obtuse25 144 169 169 = Right-angled16 64 100 80 < Acute49 64 81 113 > Obtuse


Exercise 4A Answers

1 a 5.8 cm b 9.2 cm c 12.0 cm d 13.4 cm e 5.7 cm f 3.6 m g 4.1 m h 10.9 m2 a 6.2 cm b 10.2 cm c 6.6 cm d 8.7 cm e 4.5 m f 4.0 m g 4.4 m h 4.4 m3 x = 1.4 cm, y = 1.7 cm, z = 2.0 cm 4 10.8 cm 5 7.1 cm

Ho

me

wo

rk 1 Calculate the length of the hypotenuse in each of the following right-angled triangles. Give your answers to one decimal place.

2 Calculate the length of the unknown side in each of the following right-angled triangles. Give youranswers to one decimal place.

3 a Calculate x in the right-angled triangle shown on the right.

b Calculate the area of the triangle.

Answers1 a 3.6 cm b 10.8 cm c 18.9 cm 2 a 4.9 cm b 7.2 cm c 6.6 cm 3 a 7 cm b 84 cm2

5 cm

14 cm

9.8 cm7.2 cm7 cm

12 cma

b

ca b c

2 cm 6 cm

16 cm

10 cm3 cm9 cm

a b

c

a b c

x

10 cm7 cm

24 cm

25 cmx

� hypotenuse� Pythagoras’

theorem� square� square root

Key Words

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Oral and mental starter� The class will need calculators for this starter.� Ask the class to copy and complete the following table.

� Ask the class whether they can see the rule for squaring a number that ends in 5.Answer:

The last two digits of x2 are 25 and the digit(s) before the 25 is (are) the productof the number in the tens column of x and this number plus 1.

Main lesson activity� Explain to the class that Pythagoras’ theorem can be used to solve various

practical problems. When solving a problem, it is helpful to proceed as follows.�� Draw a diagram for the problem, clearly showing the right angle.�� Decide whether the hypotenuse or one of the shorter sides needs to be found.�� Label the unknown side x.�� Use Pythagoras’ theorem to calculate x.�� Round your answer to a suitable degree of accuracy.

� Show the class how to do the following problem.

A ship sails 25 km due east. Then it sails for a further 45 km due south. Calculatethe distance the ship would have travelled if it had sailed the direct route.

First, draw a diagram to show the distances sailed by the ship. Then label thedirect distance x.

Now use Pythagoras’ theorem:

x2 = 252 + 452

= 625 + 2025= 2650

So, x = √———2650 = 51.5 km (1 dp).


x 15 25 35 45 55 65 75

x2 225 625 1225 2025 3025 4225 5625

x 15 25 35 45 55 65 75

x2


LESSON

4.2

Framework objectives – Solving problems using Pythagoras’theorem


45

25

x

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Plenary� Ask the class to give a summary of how to go about solving a practical problem

using Pythagoras’ theorem.


Exercise 4B Answers

1 108 km 2 10.8 m 3 7.7 m 4 2.56 m 5 1.78 m6 a 36.1 cm b 22.4 cm7 a 7.4 cm b 22.2 cm2

8 4.5 km9 a 4 b 5

10 a 4.5 b 4.8 c 4.2 d 9.8

Extension Answers

a Some further Pythagorean triples with a odd are given below.

b For a odd and greater than 1, b = 1–2 (a2 – 1) and c = 1–2 (a2 + 1)c Yes

a b c

3 4 55 12 137 24 259 40 4111 60 6113 84 8515 112 113

Ho

me

wo

rk 1 A plane flies due east for 120 km from airport A to airport B. It then flies due north for 280 km toairport C. Finally, it flies directly back to airport A. Calculate the direct distance from airport C toairport A. Give your answer to the nearest kilometre.

2 The length of a football pitch is 100 m and the width of the pitch is 80 m. Calculate the length of adiagonal of the pitch. Give your answer to the nearest metre.

3 The regulations for the safe use of ladders states: For a 6 m ladder, the foot of the ladder must beplaced between 1.5 m and 2.2 m from the building.

a What is the minimum height the ladder can safely reach up the side of a building?

b What is the maximum height the ladder can safely reach up the side of a building?

4 Calculate the area of an equilateral triangle whose side length is 10 cm. Give your answer to onedecimal place.

Answers1 305 km2 128 m3 a 5.6 m b 5.8 m4 43.3 cm2

� Pythagoras’theorem

Key Words

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Oral and mental starter� Ask the class to imagine a stick standing upright in the ground. Then tell them to

imagine a fly moving around the stick, so that it is always exactly 5 cm from it. � Next, invite the class to describe the shape which shows all the different

positions at which the fly could be. Allow them to work in pairs or groups, andgive them about 5 minutes to discuss their answers. (Answer: surface of acylinder with a hemisphere on top.)

Main lesson activity� Tell the class that they are going to learn how to find loci for more complicated

situations than those which they met in Year 8.� Remind them that a locus is the movement of a point according to a given set of

conditions or a rule.� Go over the two important constructions of Example 4.4 in Pupil Book 3, which

can now be stated to be loci.

�� The locus of a point which is always �� The locus of a point which is equidistant from each of two fixed equidistant from two fixed lines points, A and B, is the perpendicular AB and BC, which meet at B, is bisector of the line joining the two the bisector of the angle ABC.points.

� Explain that a locus can sometimes be a region, as shown in the three examplesto the right.�� A point which moves so that it is always 5 cm from a fixed point X has a locus

which is a circle of radius 5 cm, with its centre at X.

�� The locus of a set of points which are 5 cm or less from a fixed point X is theregion inside a circle of radius 5 cm, with its centre at X.

Note that the region usually is shaded.

�� The locus of a set of points that are less than 5 cm from a fixed point is theregion inside a circle of radius 5 cm, with its centre at X.

Note that the boundary usually is drawn as a dashed line to show that thepoints which are exactly 5 cm from X are not to be included.


CB

A

A B


LESSON

4.3

Framework objectives – Loci

Find the locus of a point that moves according to a more complex rule, involvingloci and simple constructions.

X

X

X

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Plenary� Ask the class to give the definition of a locus.� Then ask them to make up some examples of their own.


Exercise 4C Answers

3 a b c

4 5 6

7 8

9 All points on the surface of a sphere with radius 10 cm 10 All points on the surface of a half cylinder with half of a hemisphere at each end, both with a radius of 5 cm

X X X

A B X

D

C

B

A

YX A B

X

RQ

P

� angle bisector� perpendicular

bisector� equidistant� locus� loci� region

Key Words

Ho

me

wo

rk 1 Using a ruler and compasses, construct the locus which isequidistant from the points A and B.

2 Using a ruler and compasses, construct the locus which is equidistantfrom the perpendicular lines AB and BC.

3 Draw a diagram to show the locus of a set of points which are 4 cm orless from a fixed point X.

4 Two alarm sensors, 6 m apart, are fitted to the side of a house, as shownbelow. The sensors can detect movement to a maximum distance of 5 m.

Draw a scale drawing to show the region that canbe detected by both sensors. Use a scale of 1 cmto 1 m.

Answers1 Perpendicular bisector of AB 42 Angle bisector of angle ABC3 Shaded region inside the circle of radius 4 cm

Extension Answers

1 Locus is a cycloid (see diagram)

2 Locus is an arc of the circle whose radius is 3 m4 Point is where perpendicular bisector of AB meets road

A B5 cm

A

BC

5 cm

5 cm

6 m

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Oral and mental starter� Ask the class how to recognise congruent shapes. Their answer should be: ‘Two shapes are congruent when

they are exactly the same shape and size’.� Draw on the board the following triangles, or use a prepared OHT.

� Ask the class which pairs of triangles are congruent, inviting individual students to explain their answers.� The following questions could also be asked:

�� Do you need dimensions to show that the triangles are congruent?�� Do you have to draw the triangles exactly to prove that they are congruent?�� Do you need to be given any angles to show that the triangles are congruent?

� Ask them whether they can draw other pairs of triangles which are congruent but include angles.

Main lesson activity� Tell the class that they will now learn how to prove that two triangles are congruent when information is given

on both triangles. � Remind the class that they already know how to construct triangles from given information. Then summarise

the following on the board or using an OHT.

� Now apply these conditions to show that the two triangles given on the right are congruent.Invite individual students to state which angles and which sides are equal. Write their responses on the board, which should be as follows:

∠B = ∠X ∠C = ∠Y BC = XYThen ask the class whether this proves that ∆ABC is congruent to ∆XYZ, to which they should respond affirmatively. Get them to state the condition of congruence (ASA).

� Explain to the class that it is a convention to show congruence by using the symbol ≡. Hence, the congruence of these two triangles may be written as:

∆ABC ≡ ∆XYZ� The class can now do Exercise 4D from Pupil Book 3.

SideSide

Hypotenuse

AngleSideSide

SideSideSide

Angle Angle

Three sides (SSS) Two sides and theincluded angle (SAS)

Two angles and theincluded side (ASA)

Right angle, hypotenuseand side (RHS)

8 cm

6 cm 6 cm

8 cm 8 cm 8 cm8 cm 8 cm

6 cm

10 cm6 cm

6 cm6 cm

6 cm

10 cm

8 cm7 cm

9 cm

A

E F G H

B

7 cm

8 cm

8 cm

8 cm

6 cm9 cmC D


LESSON

4.4

Framework objectives – Congruent triangles

Know from experience of constructing them that triangles given SSS, SAS, ASA orRHS are unique, but that triangles given SSA or AAA are not.

Apply the conditions SSS, SAS, ASA or RHS to establish the congruence of triangles.

Explain how to find, calculate and use the interior and exterior angles of regularpolygons.

Understand congruence.

A Z X

Y

B

5 cm

5 cm

C

70°

56°

56°

70°

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Plenary� Ask the class to write in their books the four conditions which show that two

triangles are congruent.


Exercise 4D Answers

1 a ∠C = ∠D, BC = DE, AC = DF (SAS) b GH = KL, GI = JL, HI = JK (SSS)c ∠N = ∠R, ∠O = ∠P, NO = PR (ASA) d ∠T = ∠V = 90°, SU = WX (hypotenuse), TU = VX (RHS)

2 ∠X = 40° (angles in a triangle). So, ∠B = ∠X, ∠C = ∠Y, BC = XY (ASA)3 a Angles are the same but no sides are given. So, the two triangles could be drawn with different side lengths. AAA is

not a condition for congruence4 ∠ADB = ∠ADC = 90°, AB = AC (hypotenuse), AD is a common side. So, ∆ABD ≡ ∆ACD (RHS)5 ∆ACD ≡ ∆ABC ≡ ∆ABD ≡ ∆BCD, ∆AXB ≡ ∆CXD, ∆AXD ≡ ∆BXC

Ho

me

wo

rk 1 Show that each of the following pairs of triangles are congruent. Give reasons for your answers andstate which condition of congruence you are using.

a b

c d

2 ABCD is a rectangle and E is the mid-point Answersof AB. 1 a ∠C = ∠D, BC = DE, AC = DF (SAS)

b GH = JK, GI = JL, HI = KL (SSS)c ∠N = ∠R, ∠O = ∠Q, NO = QR (ASA)d ∠T = ∠W = 90°, SU = VX (hypotenuse),

TU = WX (RHS)2 AE = EB, AD = BC, ∠A = ∠B (SAS)

Explain why ∆AED is congruent to ∆BEC.

� congruent� congruence

Key Words

A F

B C D E6 cm 6 cm

5 cm 5 cm

40° 40°

G J K

LH I8 cm

9 cm 9 cm 8 cm

7 cm

7 cm

M Q

O P

R

10 cm

10 cm

N55° 75°

75°

55°

T U

S

V W

X

15 cm

15 cm

9 cm

9 cm

D C

A BE

Extension Answers

1 Draw, in turn, each diagonal and show that the two triangles formed are congruent (SSS). Hence, opposite angles areequal

2 a AB = AD, BC = CD, AC a common side. So, ∆ABC ≡ ∆ADC (SSS). b ∆ABE ≡ ∆ADE, ∆DCE ≡ ∆ΒCE3 Four different triangles are possible, ignoring reflections or rotations4

Side 2

Side 2Side 3

Angle

Side 1

Arc for the radius of side 2 cuts side 3 in two places. So, two triangles can be drawn.

Arc for the radius of side 2 touches side 3, forming a right angle. This is condition RHS.

Arc for the radius of side 2 does not meet side 3. So, it is not possible to draw the triangle.

Side 2 Side 3

Angle

Side 1

Side 2Side 3

Angle

Side 1

Case 1 Case 2 Case 3

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Oral and mental starter� Draw a circle and a vertical line on the board or OHP, as in the diagram.� Tell the class to imagine the line getting closer to the circle, passing through the

circle and then moving to the other side of the circle. � Ask the class to draw on their white boards or on the board, the different

situations which can occur.Answer:

Main lesson activity� Remind the class of the following terms for parts of a circle that they met in Year 8.

A circle is a set of points equidistant from a centre, O.Circumference The distance around a circle.

Radius The distance from the centre of a circle to itscircumference.

Diameter The distance from one side of a circle to the other,passing through the centre.

Chord A line which cuts a circle into two parts.

Tangent A line that touches a circle at a single point on thecircumference.

� Alert the class to the fact that they are going to meet two important circle theorems in Exercise 4E:


OO

The radius at the point of contact of a tangent to a circle is perpendicular to the tangent.

The perpendicular bisector of a chord passes through the centre of a circle.


LESSON

4.5

Framework objectives – Circle theorems

Distinguish between practical demonstration and proof.

Know that the tangent at any point on a circle is perpendicular to the radius at thatpoint. Explain why the perpendicular from the centre to the chord bisects the chord.

O

O

Circumference, C

Diameter, d

Radius, r

Chord

Tangent

d

r

Touches ata point

Cuts the circletwice

Passes throughthe centre

Touches ata point again

Does not touchagain

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Plenary� Invite individual students to explain on the board or OHP the two circle

theorems they have met in the lesson.


Exercise 4E Answers

3 a 28° b 65° c 50° d 120° e 51° f 29° g 123° h 48° i 50°4 a 8.5 cm b 22.9 cm c 6.4 cm d 10.9 cm5 Centre of the circle is the intersection of the perpendicular bisector of EF and the perpendicular drawn from either E or F6 Centre of the circle is the intersection of the perpendicular bisector of XY and the perpendicular bisector of XZ

Extension Answers

1 a 33° b 42° c 70° d 52° e e = f = 67°2 In ∆ABO, ∠ABO = 90° (angle in a semicircle). OB is a radius of the small circle and since ∠ABO = 90°, AB is a

tangent (radius is perpendicular to tangent)3 Join OA, OB and OP. In ∆AOP and ∆BOP, OA = OB (radii), ∠OAP = ∠OBP = 90° (radii are perpendicular to tangents),

OP is a common side. So, ∆AOP ≡ ∆BOP (RHS), hence AP = BP

Ho

me

wo

rk 1 Calculate the size of the lettered angle in each of the following diagrams.

2 Use Pythagoras’ theorem to calculate the length x in each of the following diagrams. Give youranswers to one decimal place.

3 A circle passes through the three points A, B and C. On a copy of the diagram, construct the circle, using a ruler and compasses.

Answers1 a 47° b 52° c 55° d 58° e 34° f 61° 2 a 19.3 cm b 7.1 cm c 4.2 cm d 10.5 cm3 Centre of the circle is intersection of perpendicular bisector of AB and perpendicular bisector of BC (and

perpendicular bisector of AC)

a b

7 cm

18 cm

20 cm

14 cm

xx

O•O

O

O

O•O

OO

43°

122°

56°

61°

38°

110°a

d

e f

b

c

a b c

d e f

� chord� radius� tangent� Pythagoras’

theorem

Key Words

A

B

C

c

3 cm 3 cm

3 cmx

Od

8.5 cm10 cm

x •O

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Oral and mental starter� Show the class two transparencies, each displaying a set of parallel lines 4 cm

apart, as below.

� Ask them what shapes are formed when one transparency is placed on top of theother and one of them is rotated.

� Then show them that only squares and rhombuses can be formed.� Now ask the class what shapes can be formed when one of the transparencies

has parallel lines 2 cm apart. (Answer: rectangles and parallelograms)

Main lesson activity� Remind the class about tessellations, which they met in Year 7. A tessellation is a

pattern made on a plane (flat) surface with identical shapes which fit togetherexactly, leaving no gaps. Explain that it is usual to draw up to about ten of theshapes to show the tessellating pattern.

� Show the class how equilateral triangles and squares tessellate.� Ask the class whether any other regular polygons will tessellate. Explain that

they will be doing a practical activity to discover which of the regular polygonstessellate and the reason why.

� Remind the class how to find the size of the interior angle of a regular polygonby showing them an example for a regular pentagon.

x is an exterior angle of the regular pentagon.

The sum of the exterior angles for any polygon is 360°. Since all the exteriorangles are equal, it follow that:

x = 360° ÷ 5 = 72°y is an interior angle of the regular pentagon. Hence:

y = 180° – 72° = 108°� For Exercise 4F, the class will require squared paper, isometric paper, card for

making regular polygon templates and scissors. Sets of commercially produced,regular polygons can also be useful, so that the students can easily visualise theshapes, particularly if they have difficulty in making their own templates.



LESSON

4.6

Framework objectives – Tessellations and regular polygons

Explain how to find, calculate and use the interior and exterior angles of regularpolygons.

xy

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Plenary� Invite a student to the board and give her/him a regular polygon, asking the

student to show whether it tessellates.� Continue to invite individual students to the board, giving each a different

polygon.


Exercise 4F Answers

2 b There are gaps 3 b There are gaps (squares)4 a

b Size of the interior angle divides exactly into 360°c No, each interior angle of a regular nonagon is 140°, which does not divide

exactly into 360°

Regular polygon Size of each interior Does polygon angle tessellate?

Equilateral triangle 60° YesSquare 90° YesRegular pentagon 108° NoRegular hexagon 120° YesRegular octagon 135° No

Ho

me

wo

rk 1 Work out, by making templates or by drawing diagrams, which of the following regular polygonstessellate, and which do not. In each case, write down a reason for your answer.

a Equilateral triangle b Square c Regular pentagon d Regular hexagon e Regular octagon

2 Draw a diagram to show how squares and equilateral triangles together form a tessellating pattern.

Answers1 a Yes, interior angle is 60°, which divides exactly into 360°

b Yes, interior angle is 90°, which divides exactly into 360°c No, interior angle is 108°, which does not divide exactly into 360°d Yes, interior angle is 120°, which divides exactly into 360°e No, interior angle is 135°, which does not divide exactly into 360°

2 For example:

� exterior angle� interior angle� regular polygon� tessellate

Key Words

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Oral and mental starter� Tell the students to imagine a square.� Now tell them to imagine cutting the square along one of its diagonals.� Next, ask the students to describe the two shapes that are left. Answer: two isosceles right-angled triangles.� Then tell them to imagine cutting the square again, but this time along a line which is parallel to the diagonal. � Finally, get the students to describe the two shapes that are left. Answer: an isosceles right-angled triangle and a

pentagon.

Main lesson activity� Explain to the class that the aim of the lesson is to show the difference between a practical demonstration and a

proof. A practical demonstration shows how a rule or theorem works by using a specific example, whereas aproof shows how the rule or theorem works for all cases. A theorem is usually proved by using algebra.

� The activity in Pupil Book 3, page 75 gives a practical demonstration to show Pythagoras’ theorem. Eachstudent will need a sheet of thin card and a pair of scissors. For completeness, the activity is given below.�� In your book, draw the right-angled triangle X, as below.

�� On the card, draw eight more triangles identical to X. Cut them out and place them to one side.�� On your original triangle, X, draw squares on each of the three sides of the triangle. Label them A, B and C,

as below.

�� On the card, draw another diagram identical to this. Cut out the squares A, B and C.�� Arrange the cut-outs of the eight triangles and three squares as in the two diagrams below.

�� What can you say about the total area of Diagram 1 and of Diagram 2?�� Now remove the four triangles from each diagram.�� What can you say about the areas of squares A, B and C?�� Show how this demonstrates Pythagoras’ theorem.Answer Diagram 1 and Diagram 2 have the same total area.

Area of A + Area of B = Area of CSo, 16 + 9 = 25 cm2, which is 42 + 32 = 52 cm2.

� More able students can do the Extension Work, which gives a proof of Pythagoras’theorem. Proof of Pythagoras’ theoremA right-angled triangle has sides a, b and c.

XX

X

X

Diagram 2

A

B

C

X X

X X

Diagram 1

3 cm

4 cm

5 cmX


LESSON

4.7

Framework objectives – Practical Pythagoras

Distinguish between a practical demonstration and a proof.


a

cb

3 cm

4 cm

5 cm

C

A

BX

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The diagram on the right can be drawn using four of these triangles.Area of each triangle = 1–2ab.So, area of the four triangles = 2ab.Area of the large square = (a + b)2.Area of the large square can also be written as c2 + 2ab. Hence:

(a + b)2 = c2 + 2aba2 + 2ab + b2 = c2 + 2ab

a2 + b2 = c2

which is Pythagoras’ theorem.

Plenary� Ask the class to explain the difference between a practical demonstration and a

proof.


SATs Answers

1 a Sum of interior angles of a triangle = 180°, so 2 × 180° = 360° b 540° c 900°2 a 20.8 cm b 9.8 cm 3 A and C, three equal sides (SSS)4

5 a AD = 33.8 cm, AC = 31.2 cm and CD = 13 cm, so perimeter = 78 cmb AC2 + CD2 = 1142.44 and AD2 = 1142.44, so by Pythagoras’ theorem ∆ACD is right-angled

6 a ∠ABO = x (∆ABO is isosceles, since radii are equal), ∠CBO = y (∆CBO is isosceles, since radii are equal)b x + x + y + y = 180°, 2x + 2y = 180°. So, x + y = 90° = ∠ABC

2 m

� proof� prove� Pythagoras’

theorem

Key Words

a

a

a

a

c

cc

c

b

b

b

b

Ho

me

wo

rk Practical demonstration with a differenceCut out an 8 cm by 8 cm square and then cut it up Now rearrange the four pieces to make a into two right-angled triangles and two trapezia, rectangle, as in the diagram below.as in the diagram below.

What is the area of the square and of the rectangle?

Can you explain why this practical demonstration does not work?

AnswersThe area of the square is 64 cm2 and the area of the rectangle is 65 cm2. The apparent increase of 1 cm2 in thearea of the rectangle is due to the fact that the diagonal of the rectangle is not a straight line. This can be shownby applying Pythagoras’ theorem.The length of the diagonal in the rectangle should be √

——194 = 13.928 388 28. This diagonal is the length of the

hypotenuse of the triangle plus the length of the sloping side of the trapezium, which is √—–73 + √

—–29 =

13.929 168 55.This shows that the diagonal of the rectangle cannot be a straight line.

5 cm3 cm

3 cm5 cm

5 cm

3 cm

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Oral and mental starter� The students can work in small groups for this activity.� Write the following sources of information on a sheet: Questionnaire, Printed

tables in books, Internet, Computer database and Observation sheet.� Give the students cards, or a list, containing the following topics.

Primary Data Secondary DataNumber of left-handed students Long jump performances by

in the class international athletesTV viewing habits of students Car engine sizesReaction times of students Populations of various countriesWhether students are better at Football results in Europe

catching with their left or Prices of different makes of right hand second-hand cars

The amounts of pocket moneyreceived by males and femalesin school

� They should discuss each topic and decide how best to investigate each one,using the sources given above.

Main lesson activity� Explain to the class that the aim of the lesson is to look at how to plan a

statistical investigation. � Point out that sometimes the most difficult part is to decide a topic to investigate.

(Steps 1 and 2 in the table below.)� Explain that to help them they will be given a planning sheet so that they may

work systematically through their problem.� Take an example of your choice or use the example given below.


Handling Data 1CHAPTER

5

LESSON

5.1

Framework objectives – Statistical investigations

Suggest a problem to explore using statistical methods, frame questions and raiseconjectures.

Discuss how data relate to a problem. Identify possible sources, including primaryand secondary sources.

Design a survey or experiment to capture the necessary data from one or moresources. Determine the sample size and degree of accuracy needed. Design, trialand if necessary refine data collection sheets.

Identify possible sources of bias and plan how to minimise it.

Step Example1 Decide which general topic to The cost of housing in different parts of

study the UK

2 Specify in more detail Comparing the costs in Wales andEngland

3 Consider questions which you Is the average price higher in Wales?could investigate Is there a bigger difference in the prices

in England than in Wales?

4 State your hypotheses (Your The price is higher in Walesguesses at what could happen) There is more variation in price in

England

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� You could now show the students the three examples in Pupil Book 3.


Plenary� Use a group’s planning sheet to discuss the points on it.� Ask other groups to contribute points that can be added to the planning sheet.� Explain that the homework is to produce an individual plan for a different topic.

The students could use ideas already used by other groups.


Exercise 5A Answers

Answers will vary but should be similar in style to the examples.

Ho

me

wo

rk Take a different topic to those already studied and prepare a new planning sheet.

AnswersAnswers will vary but should be similar in style to the examples.

� questionnaire� printed table� database� survey� statistic� bias� census

Key Words

5 Sources of information required Internet. Estate agents. Building societiesand banks mortgage reports. Governmentdata: for example, Office for NationalStatistics http://www.statistics.gov.uk/

6 Relevant data Average house prices in differentcounties of Wales and England

7 Possible problems Counties of different sizes may affect theaverage unfairly, causing your data to bebiased

8 Data collection Make sure that your sample size is bigenough to draw valid conclusions

9 Decide on level of accuracy If results are within, say, £1000 pounds required of each other, you may decide that the

results do not support the hypothesis thatthe prices are different

10 Determine sample size Make sure that you collect enough datafrom both countries

11 Construct tables for large sets of raw data in order to make work manageable

12 Decide which statistics are most If there are a few extreme values, you suitable may choose to ignore the mean, as this

will distort the results

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Oral and mental starter� Write a simple, open, multiplication table on the board or OHP, as shown.� Prompt the students to tell you how to complete it. � Now add on an extra column, with zero.� Complete the extra column, with the students’ help.� Prompt them to extend the table, into negative numbers.� Use patterns (for example, subtracting 3s in the first row) to

complete the columns.� Now prompt the students to fully extend the table downwards,

as shown. � Show the class that, using patterns, they

have just proved the rules formultiplying positive and negativenumbers: for example, + × – = –.

Main lesson activity� Introduce the class to the table below, which gives the rules for combining two

scatter graphs, which have a common axis, to obtain the resulting correlation.

As can be seen from the table, the resulting graph can have its axes in eitherorder, as this does not affect the correlation.An easy way to remember these rules is by comparing them with the rules formultiplying together positive and negative numbers, as shown below.

� Combining two graphs showing no correlation can be misleading, as the answercould be a graph with either positive, negative or no correlation. Hence youcannot tell just by using the rules.Take, for example, the case of the fish caught off Rhyl.

In this example, the two left-hand graphs show no correlation but combininghours of sunshine with temperature gives a positive correlation.

Multiply (×) + 0 –+ + 0 –0 0 The exception 0– – 0 +

Positive Negative correlation No correlation correlation

Positive correlation Positive No correlation NegativeNo correlation No correlation Cannot tell No correlationNegative correlation Negative No correlation Positive


LESSON

5.2

Framework objectives – Scatter graphs and correlation

Select, construct and modify, on paper and using ICT, suitable graphicalrepresentation to progress an enquiry, including scatter graphs to develop furtherunderstanding of correlation. Identify key features in the data.

× +3 +2 +1+3+2+1

× + 0 –+ + 0 –0 0 0 0– – 0 +

× +3 +2 +1 0 –1 –2 –3+3 9 6 3 0 –3 –6 –9+2 6 4 2 0 –2 –4 –6+1 3 2 1 0 –1 –2 –30 0 0 0 0 0 0 0

–1 –3 –2 –1 0 1 2 3–2 –6 –4 –2 0 2 4 6–3 –9 –6 –3 0 3 6 9

Hou

rs o

f sun

shin

e

Number of fish caught daily off Rhyl

Tem

pera

ture

(°C

)

Number of fish caught daily off Rhyl

Hou

rs o

f sun

shin

e

Temperature (°C)

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� Tell the class that 12 students’ marks were collected from their tests inthree different subjects, and two scatter graphs created. Then put thesegraphs on the board.

� Ask the class to tell you what the subject of each test could be.� Then tell the class that they are actually Mathematics (Test A), Science

(Test B) and Art (Test C).� Invite them to discuss the correlation between the Mathematics and

Science scores, and between the Mathematics and Art scores.� Now prompt the class to tell you the correlation between the Science and

the Art scores. In this case, there would be negative correlation.� Now ask them to look at the rules for combining two correlation graphs

in Pupil Book 3. They could copy into their books the table forcorrelations and the table for multiplying together positive and negativenumbers.

� Point out how similar the rules are, but emphasise the exception to therule, namely, two graphs, each showing no correlation, do not necessarilymean that the derived graph would have no correlation.


Plenary� Finish the lesson with a short test of multiplications of positive and negative

whole numbers, to reinforce the rules for combining two scatter graphs.1 +8 × –3 2 –6 × –5 3 +11 × 0 4 –13 × –13 5 +8 × –66 –9 × –4 7 0 × +14 8 +15 × –15 9 +14 × +14 10 –7 × –10

Answers1 –24 2 30 3 0 4 169 5 –486 36 7 0 8 –225 9 196 10 70


Test

B

Test A

Test

C

Test AExercise 5B Answers

1

2 a Positive correlation b No correlation c No correlation3 a Negative correlation b Negative correlation c Positive correlation

Correlation between Q and Ra Negative d Cannot tell g Negativeb No correlation e Positive h No correlationc Positive f No correlation i No correlation

� scatter graph� correlation� positive

correlation� negative

correlation� no correlation

Key Words

Ho

me

wo

rk 1 The test results of ten students are recorded for four different subjects. Here are the results.

a Plot the data for French and Spanish on a scatter graph.b Describe the relationship between French and Spanish.c Plot the data for English and Music on a scatter graph.d Describe the relationship between English and Music.e Plot the data for Spanish and English on a scatter graph.f Describe the relationship between Spanish and English.g Use your answers to parts d and f to state the correlation between Music and Spanish.

Answersb Positive correlation d Negative correlation f Positive correlation g Negative correlation

Student French Spanish English MusicA 45 52 63 35B 64 60 56 45C 22 30 46 58D 75 80 70 30E 47 60 55 42F 15 24 40 50G 80 74 68 42H 55 65 53 48I 85 77 75 41J 33 47 51 50

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Oral and mental starter� Using students’ Show me white boards or sheets of paper, ask the class to draw

and show you a sketch illustrating positive correlation. � Ask a student to give an example of two variables having positive correlation. � Ask those who think that their sketches show strong positive correlation, to keep

showing their sketches. � Discuss different strengths of positive correlations.� Repeat for negative correlation.� Briefly talk about no correlation.

Main lesson activity� Continuing from the Oral and mental starter, ask a student to draw a strong

positive correlation on the board or OHP.� Introduce the idea that it could represent the test results for Paper 1 and Paper 2

of the SATs. Label the axes with graduation marks and ‘Paper 1’ and ‘Paper 2’.� Tell them that a student was absent for the second paper but, although they

could not have an official level, the school wanted to know how the studentmight have performed.

� Ask the class to describe the trend. Students who do well on Paper 1 usually dowell on Paper 2.

� Explain that in order to make an accurate guess at the missing student’s Paper 2score, you are going to put a trend line (called a line of best fit) on the scattergraph. Discuss where the line should go. The Pupil Book suggests using a rulerto draw a line between the plotted points, passing as close as possible to all ofthem, but you could also show them the idea of covering the points with afinger. In this case the line of best fit will run down the middle of the finger.

� Draw in a line of best fit. It is important to stress the need to use a ruler and alsothat different students would have slightly different lines of best fit.

� Now ask a student to come out and read off the Paper 2 score.



LESSON

5.3

Framework objectives – Scatter graphs and lines of best fit

Select, construct and modify, on paper and using ICT, suitable representation toprogress an enquiry, including scatter graphs to develop further understanding ofcorrelation, lines of best fit by eye, understanding what they represent.

9B3LP_05.qxd 18/09/2003 15:19 Page 60

Plenary� Look at an example and discuss the problem of trying to extrapolate from a line

of best fit. You could use the questions posed in the extension. � Stress that students will need to be able to explain, for specific examples,

reasons why it might not be valid to extrapolate data outside the given range.


Exercise 5C Answers

1 c Darren, as the point representing him is not close to the line of best fit.2 c The older they are, the closer they live to the shops.3 c Answers will vary according to line of best fit drawn but should be approximately

£12 – £14.

Extension Answers

Explanations which imply that it is not sensible to extrapolate lines of best fit outside therange of the original data.

In Question 2, the distance which a child lives from the shops will depend on the parent.So, the age of the child is not a factor in the question as he/she is not the homeowner. InQuestion 3, they could imply that a 20-year-old may be working and have a muchgreater income.

Ho

me

wo

rk 1 The table shows the scores of some students in a music exam and in a maths exam.

a Plot the data on a scatter graph. Use the x-axis for the music exam scores, from 0 to 100, and they-axis for the maths exam scores, from 0 to 100.

b Draw a line of best fit.

c One person did not do quite as well as expected on the maths test. Who do you think it was?Give a reason.

2 A survey is carried out to compare the ages of people with the reaction time in a test.

a Plot the data on a scatter graph. Use the x-axis for the range of ages, from 0 to 90 years, and they-axis for reaction times, from 0 to 1 seconds.

b Draw a line of best fit.

c Use your line of best fit to estimate the reaction time of a 30-year-old.

d Explain why it would not be sensible to use the line of best fit to predict the reaction time ofsomeone aged 100.

Answers1 c Student E, as their point is not as close to the best fit line as the other students’ points2 c Answers will vary according to line of best fit drawn but should be approximately 0.2

d Line of best fit is for range 24–83 years old. 100 years old is outside this range, so correlation might notcontinue to be linear

Age (years) 45 62 83 24 76 63 44 42 37 50

Reaction time (seconds) 0.15 0.31 0.58 0.20 0.62 0.43 0.21 0.25 0.18 0.49

Student A B C D E F G H I JMusic 35 48 72 23 76 51 45 60 88 17Maths 42 57 80 32 65 69 50 71 94 25

� scatter graph� correlation� line of best fit� positive

correlation� negative

correlation� interpret

Key Words

9B3LP_05.qxd 18/09/2003 15:19 Page 61


Oral and mental starter� Ask individual students to give different units of time. You may need to give an

example, such as months.� Write any correct answers on the board or OHP.� For each example, draw a horizontal scale and put a few labels on it, as shown

on the right. � Other examples which they may offer could include year, season, hour and days

of the week.

Main lesson activity� Tell the class that the aim is to look at different types of graph involving time.� Explain that for all the types of graph which they are going to look at or produce,

the time axis is always the horizontal axis.� Invite the class to look at Graph 1 in Pupil Book 3 (mean temperature difference

from normal for the UK in 2002).� Ask them to give some facts from the graph. For example, in 10 out of 12

months, the mean temperature was above normal; February and March showedthe greatest differences. You could prompt them by asking what they can tell youabout February and March.

� Then ask the students to look at the winter months and the summer months andcompare the weather. They should observe that there are bigger temperaturechanges in the winter months.

� Now tell the class that you want them to look at the other graphs in Pupil Book 3. This activity could be done individually or in small groups. Suggest thatthey write down any key features of the graphs.


LESSON

5.4

Framework objectives – Time series graphs

Select, construct and modify, on paper and using ICT, suitable graphicalrepresentation to progress an enquiry, including line graphs for time series. Identifykey features present in the data.

January February March

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Plenary� Choose a graph and ask a group to list some key features.� Ask the other students to add to it.� Explain that when there are two similar graphs, such as two rainfall graphs, it is

important to compare them, looking for both similarities and differences.

Exercise 5D Answers

1 a Time becomes shorter b 3–4 c After third bounced In theory, the ball never comes to rest. In practice, it would eventually stop

bouncing2 a Mean temperature was exceeded on 10 months out of 12. Extreme changes in

temperature were greater above normal (about 3.3°C) than below normal (about –1.2°C)

b As data is for only one year, this could be exceptional. Data would need to berecorded over at least 10 years for valid conclusions to be drawn.

3 a June b Septemberc Different pattern for average rainfall each month and different pattern for number

of days of rain each monthd Perth (approx values) 3 + 3 + 4 + 7 + 13 + 17 + 18 + 16 + 14 + 9 + 7 + 4 = 115.

Brisbane (approx values) 13 + 14 + 14 + 11 + 10 + 7 + 7 + 6 + 7 + 10 + 10 + 11= 120. So, Brisbane has 5 more days of rainfall.

Ho

me

wo

rk

Write a brief report on the similarities and differences between the visits from the UK for NorthAmerica and Western Europe. Make at least three statements. Try to give reasons for your answers.

AnswersMore UK people visit Western Europe than North America (three to four times more), probably because WesternEurope is nearer, so the costs are less. There is greater variation in the number of visitors for Western Europe frommonth to month (April being the most popular month). July and August are the most popular months to visit NorthAmerica

Nov Dec Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov

2001 2002

North American Western Europe

0

500

1000

1500

Vis

itors

(×10

00)

� time seriesgraphs

� raw data� key features� line graphs

Key Words

9B3LP_05.qxd 18/09/2003 15:19 Page 63

Oral and mental starter� Put the class into small groups of four or five students. � Give the groups a mental test of ten questions. Appoint one person from each

group as team captain to record a team’s answer to each question.� After the answers have been given, ask the team captains to record their answers

in a two-way table on the board or OHP, using ticks for correct answers, as shown.

Test1 £4.99 × 4 2 25% of 60 3 1–2 of a 1–24 600 × 4000 5 72 ÷ 0.2 6 What is the HCF of 36 and 48?7 Write down one answer to x2 + x = 0 8 Increase £132 by 20%9 What is the square root of 196? 10 Give both solutions to (5 + x)2 = 81

Answers1 £19.96 2 15 3 1–4 4 2 400 0005 360 6 12 7 x = 0 or x = –18 £158.40 9 14 or –14 10 x = 4 and x = –14

Main lesson activity� Keeping the students in their groups, explain that their task is to collect data from

the whole class and record it.� Use the data collection sheets as shown below, or the students can design their own.� One student from each group is the ‘collector’, the rest of the group are the

‘informers’. Each ‘collector’ goes from group to group collecting their data.

� Now use other combinations to form different two-way tables. For example:

Favourite colourBlue Red Yellow Green

Sport

Favourite Computerhobby Music

Group Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q101 � � � � � �2 � � � � � �3 � � � � �


LESSON

5.5

Framework objectives – Two-way tables

Design and use two-way tables.

Favourite subjectBoys Girls

EnglishScienceArtMaths

Favourite colourBoys Girls

BlueRedYellowGreen

Favourite TV programmeBoys Girls

SoapDramaCartoonNewsOther

Favourite musicBoys Girls

RockPopDanceR n B

Favourite foodBoys Girls

ChipsSaladPizzaBurger

Favourite hobbyBoys Girls

SportComputerMusic

9B3LP_05.qxd 18/09/2003 15:19 Page 64

� Having collected their data, the students can record it in their books. In each case, ask the students to pick outa key feature. A key feature could be that the data appears random (no relationship between the two variables).


Plenary� Ask the class to select a table where they saw a relationship.� Look at, for example, boys’ favourite colour and boys’ favourite music. Are their

responses different from girls?� Write any relationships on the board. Ask the class what they could do to test

whether the results were representative of the school.


Exercise 5E Answers

1 a b Boxed toys are worth more than unboxed toysbut the percentage difference in value reducesas the condition deteriorates.

2 a For the age range 10 to 12, a larger percentage of boys have mobile phones. b As the boys and girls get older, both For the age range 13 to 15, a larger percentage of girls have them. percentages increase.

3 In June, July and August, 252 birthdays but in November, December and January, 228 birthdays. This would supportthe claim. Answers may vary depending on how the data is analysed but the conclusion should be the same.

4 160 cm and above: 20 boys but only 16 girls. This would support the claim. Answers may vary depending on how thedata is analysed but the conclusion should be the same.

Condition Difference between boxed and not boxedExcellent 100% – 60% = 40%Very good 80% – 50% = 30%Good 60% – 40% = 20%Average 40% – 25% = 15%Poor 20% – 10% = 10%

Ho

me

wo

rk 1 Two fair spinners are spun and the scores areadded together to get a total score. This isrecorded in the two-way table, shown below.

a Complete the table of total scores.b List all the total scores which are prime

numbers.c State the most likely total scores.d Write down the probability of getting a total

score of 7. Give your answer as a fraction inits simplest form.

e Write down the probability of getting a totalscore of 5. Give your answer as a fraction inits simplest form.

2 A year group recorded the days of the week onwhich they were born. Here are the results.

a Write a comment on the births of boys andgirls.

b Write a comment about the number of birthson different days of the week.

Answers1 a

b 2, 3, 5 and 7 c 4 and 5 d 1––12 e 1–42 a Each day, number of births of boys is close to that

of girlsb Fewer births on Saturdays and Sundays

� two-way table� relationship� data� tally� frequency

Key Words

Second spinner+ 1 2 31 2 3First2 3spinner34

Day Boys GirlsMonday 23 19Tuesday 19 25Wednesday 27 28Thursday 31 26Friday 35 41Saturday 14 17Sunday 12 11Total 161 167

Second spinner+ 1 2 3

First 1 2 3 4

spinner 2 3 4 53 4 5 64 5 6 7

4

3

2

1

3 2

1

Extension Answers

a 9––40 b 31––80 c 1––10 d 67––80 e 11––16

9B3LP_05.qxd 18/09/2003 15:19 Page 65

Oral and mental starter� Write a blank table on the board or OHP. � Ask the class to indicate if their birthday is in January. Record the number in the

table.� Repeat for February and so on for the whole year.� Now ask them to look at the table and tell you how many students have a

birthday in the first month of the year, the first two months of the year, the firstthree months and so on. Add extra columns to your table and record the data asshown.

� Prompt the students to explain how they worked out each answer. � Complete the cumulative frequency column in the table. For example:

Main lesson activity� Continuing from the Oral and mental starter, explain to the class they are about

to learn how to estimate medians and quartiles for large sets of grouped data.� Draw the axes for a cumulative frequency graph on the board. Write ‘Month’

along the horizontal axis and ‘Cumulative frequency’ along the vertical axis. � Plot the origin and say this is the beginning of January so there are no birthdays.

Discuss with the class where they think the next point should be plotted. Use thedata obtained in the starter.

� Stress that in order to include, for example, all five students, you must plot at theend of January and so on – the upper class boundaries.

� Complete the graph and ask the class to copy the tables and the graph ontograph paper, choosing suitable scales so that it uses most of the page.

� Now ask the class whether they think it would be better to join the points usingstraight lines or a curve. Explain that they can use either unless a questionspecifies ‘polygon’ or ‘curve’.

� If some students are waiting for others, they can start to draw the graphs fromExercise 5F.

n n 3n� Explain that because there is a large set of data, you can use ––, ––, —–, where n

2 4 4is the total frequency, to read off the values of the median, lower quartile andupper quartile. Show the class how to do this on the board and then ask them toread off their values.

� Write on the board ‘Median = …’, ‘Lower quartile = …’, ‘Upper quartile = …’and ask them to copy and fill in their answers.

� Now ask the class to compare their answers. There may be slight differences.Explain that the slight differences in the graphs and the fact that they are usinggrouped data are why their results are estimates.

� Finish off by giving the class the formula for finding the interquartile range:Interquartile range = Upper quartile – Lower quartile

� Write ‘Interquartile range = …’ And ask them to work out their value and write itdown.


Month Number of students Birthdays Cumulative frequencyJanuary 5 First month 5February 7 First two months 12March 2 First three months 14… … … …


LESSON

5.6

Framework objectives – Cumulative frequency diagrams

Find the median and quartiles for large data sets. Estimate the median andinterquartile range of a large set of grouped data.

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Plenary� Look back at one of the questions – for example, the birthday data – and ask the

students what the median tells them. Ask those with a birthday before the mediandate to put up their hands. Hopefully, this will be approximately half the class.

� Repeat this for the quartiles and for students with a birthday between the lower and upper quartiles.


Exercise 5F Answers

1 a

c Median = 13.2 °C, IQR = 17 – 10.2 = 6.8 °C2 a

c Median = 11.7 min, IQR = 17.9 – 7.1 = 10.8 min3 a

c Median = 14 kg, IQR = 18.2 – 10.8 = 7.4 kg

Mass, M (kg) Cumulative frequencyM ≤ 5 0M ≤ 10 15M ≤ 15 46M ≤ 20 68M ≤ 25 80

Time, t (min) Cumulative frequencyt ≤ 5 5t ≤ 10 17t ≤ 15 26t ≤ 20 33t ≤ 25 37t ≤ 30 40

Temperature, T (°C) Cumulative frequencyT ≤ 5 8T ≤ 10 23T ≤ 15 65T ≤ 20 90T ≤ 25 100

Ho

me

wo

rk For each table of data:

a Copy and complete the cumulative frequency table.

b Draw the cumulative frequency graph.

c Use your graph to estimate the median and the interquartile range.

1 The height of 100 plants.

2 The time that the school bus is late on 40 days.

Answers1 a 2 a

c Median = 7.7 min, IQR = 12.5 – 4.2 = 8.3 minc Median = 27.4 cm, IQR = 36 – 17.9 = 18.1 cm

Time, t (min) Cumulative frequencyt ≤ 5 12t ≤ 10 27t ≤ 15 33t ≤ 20 40

Height, h (cm) Cumulative frequencyh ≤ 10 6h ≤ 20 30h ≤ 30 57h ≤ 40 87h ≤ 50 100

Time, t (min) Number of days Time, t (min) Cumulative frequency0 < t ≤ 5 12 t ≤ 55 < t ≤ 10 15 t ≤ 10

10 < t ≤ 15 6 t ≤ 1515 < t ≤ 20 7 t ≤ 20

Height, h (cm) Number of plants Height, h (cm) Cumulative frequency0 < h ≤ 10 6 h ≤ 10

10 < h ≤ 20 24 h ≤ 2020 < h ≤ 30 27 h ≤ 3030 < h ≤ 40 30 h ≤ 4040 < h ≤ 50 13 h ≤ 50

� cumulativefrequency

� upper classboundary

� quartile� lower quartile� upper quartile� interquartile

range� polygon� curve

Key Words

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Oral and mental starter� Introduce a counting stick, as shown below.

� Tell the class that, for example, one end is 12 and the other end is 16. Then ask them the value of the mid-point.

� Repeat this, increasing the level of difficulty each time. For example, using fractions and decimals.� Now give the class the end values. Tell them that you want, for example, the product of eight and the mid-

values. So, using 12 and 16, the answer would be 8 × 14 = 112.� Repeat.

Main lesson activity� Tell the class that they are going to be estimating the means of large sets of grouped data. � Remind them of the key words which they have met before in this context, such as mid-value, frequency,

mean, estimate and so on. You could ask the students for definitions.� Put the numbers 10, 12, 12, 12 on the board and ask the class to tell you the mean. Hopefully, they will say

11.5. � Now write the same data in a grouped frequency table and ask the same question.

� Clearly, you should get the same answer, but students may suggest the answer 11 (as it is half way between 10 and 12).

� Point out that every piece of data has to be counted, so there are three 12s (36) and a 10 (total 46). This is thendivided by 4 (total frequency) to give the answer 11.5.

� Now refer them to the example in Pupil Book 3, page 96. Point out that, as in the last section on cumulativefrequency, any results will be estimates because you do not know the original (individual) data as we aredealing with grouped data.

� Explain that in order to estimate the mean, we have to use mid-values as these offer the best guess of the valueof the average of the data in a class interval.

� Work through the example to show the class how to estimate the mean using a table with four columns, asshown below.

� Show the students how they can now calculate an estimate of the mean.1541

� The estimate of the mean time = ——– = 15.41 seconds.100


Time, t (seconds) Frequency, f Mid value, x, of time f × x(seconds) (seconds)

13 < T ≤ 14 12 13.5 16214 < T ≤ 15 21 14.5 304.515 < T ≤ 16 39 15.5 604.516 < T ≤ 17 20 16.5 33017 < T ≤ 18 8 17.5 140

Total = 100 Total = 1541

Frequency10 112 3


LESSON

5.7

Framework objectives – Estimation of a mean from groupeddata

Estimate the mean of a large set of grouped data.

Exercise 5G Answers

1 c i 4.5 kg ii 27.35 cm2 19.8 °C3 8.4 min (8 min 24 s)

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Plenary� Discuss the effect of extreme values on a mean.� Ask the class to give you a definition of an extreme value. Point out that

sometimes these are called rogue values and sometimes they are called outliers.� Remind the class that when tables contain people’s ages, for example, 0–2 years,

would have a mid-value of 1.5 years. (They will need this information to enablethem to do the homework.)

� Remind the class how to find which class interval the median lies in and referthem to the Extension Work. They may wish to do this as an oral exercise.


Ho

me

wo

rk Copy and complete each table of values given below.

a Complete each table including the totals.

b Calculate an estimate of each mean.

1

2

Answers1 a 2 a

b Estimate of mean = 15.13 years b Estimate of mean = 4.5 hours

Time, t, Frequency, f Mid value, f × x(hours) x, of time (years)

(hours)0 < t ≤ 2 2 1 22 < t ≤ 4 7 3 214 < t ≤ 6 10 5 506 < t ≤ 8 5 7 35


Age, A, Frequency, f Mid-value, f × x(years) x, of age (years)

(years)11–12 5 12 6013–14 8 14 11215–16 12 16 19217–18 5 18 90


Time, t, (hours) Frequency, f Mid-value, x, of time (hours) f × x (hours)0 < t ≤ 2 2 1 22 < t ≤ 4 74 < t ≤ 6 106 < t ≤ 8 5

Total = Total =

Age, A (years) Frequency, f Mid-value, x, of age (years) f × x (years)11–12 5 12 6013–14 815–16 1217–18 5

Total = Total =

� grouped data� estimate� mean� median� mid-class value� frequency table

Key Words

SATs Answers

1 a False (1988 – 725, 1998 – 500) b Cannot be certain, trend may not continue2 a 65 b 30 and 50 c Game A and Game B positive relationship, Game A and Game C no relationship

d Game B and Game C no relationship3 a Positive correlation (wider the diameter, higher the tree)

b Point if plotted on scatter graph is not near to line of best fitc Approximately 5.2 m d False, false, false, false

4 a 4.85 g b 0.1 g c Point at 12.30 pm. Bird is under the mass to survive, as furthest away from line of best fit.5 a A b Approximately 350 h6 a Old 7 ≤ x < 8, new 6 ≤ x < 7

b For example, the new version has shorter sentences: that is, there are more words per sentence in old version.c 20% – 30%

7 a £30.50 – £32 b 83% c £9.50 – £11.50 e A – true, B and C – false8 a 28 years b 16 to 18 years c Younger people, on average, went to the theme park

Extension Answers

1 i 4 < M ≤ 6 ii 24 < L ≤ 28 2 20 < T ≤ 30 3 7 < t ≤ 9

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Oral and mental starter� Invite the class to imagine two 10p coins, one of which is held still while the other coin is rolled around it so

that the coins are always in contact.� Now, ask the class how many revolutions the moving coin will make before it returns to its starting position.

(Answer: Two)

Main lesson activity� Remind the class about the properties of an enlargement by showing them how a triangle is enlarged by a scale

factor of 2.

Triangle ABC has been mapped onto triangle A�B�C� by an enlargement of scale factor 2.Under an enlargement all of the angles are the same size and corresponding sides are in the same ratio.So, AB : A�B� = AC : A�C� = BC : B�C� = 1: 2.

A�B� A�C� B�C�This can also be written as –––– = –––– = –––– = 2.

AB AC BC� Explain to the class that the two triangles are said to be similar.

Two triangles are similar if their angles are the same size or their correspondingsides are in the same ratio.Explain that only one of these conditions is required to show that the trianglesare similar.

� Show the class how to use similar triangles by completing the following threeexamples. Some revision on parallel lines may be required.

� Example 1Show that the two triangles on the right are similar.In triangle ABC, ∠C = 80° (the sum of the angles in atriangle = 180°) and in triangle XYZ, ∠X = 48° (the sum ofthe angles in a triangle = 180°).Since the angles in both triangles are the same, triangleABC is similar to triangle XYZ.Example 2Triangle ABC is similar to triangle DEF. Calculate the length of the side DF.Let the side DF = x.Since the triangles are similar, corresponding sides are in the same ratio.

DE EF DFSo, ––– = ––– = ––– = 3.

AB BC ACx 15

Therefore, –– = — = 3. So, x = 18 cm.6 5



6

LESSON

6.1

Framework objectives – Similar triangles

Find points that divide a line in a given ratio, using the properties of similartriangles.

A

B

B�

C

A�

C�

X

Y Z80° 52°

B C

A

48°

52°

B C5 cm

4 cm6 cm

AD

E F15 cm

12 cm x

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Example 3In the triangle, EB is parallel to DC. Calculate the length of the side DC.∠AEB = ∠ADC (corresponding angles in parallel lines) and ∠ABE = ∠ACD(corresponding angles in parallel lines). So, triangle AEB is similar to triangle ADC (since ∠A is common to both triangles).Let the side DC = x. Since triangle AEB is similar to triangle ADC, the corresponding sides are in the sameratio.

DC ACSo, ––– = ––– = 2.

EB ABx 8

Therefore, –– = –– = 2. So, x = 6 cm.3 4


Plenary� Ask the class to explain the two conditions needed to show that two triangles are

similar.� Check that the students understand the difference between similar triangles and

congruent triangles.


Exercise 6A Answers

1 a Yes b No c Yes d No 2 a 3 equal angles b 8 cm 3 a 3 equal angles b 12 cm4 a ∠PST = ∠PQR (corresponding angles), ∠PTS = ∠PRQ (corresponding angles), ∠P common b 6 cm5 a ∠BAE = ∠AED (alternate angles), ∠ABD = ∠BDE (alternate angles), ∠ACB = ∠DCE (vertically opposite angles)

b 6 cm 6 12 cm 7 40 m 8 90 cm

Ho

me

wo

rk 1 State whether each of the pairs of triangles below are similar.

2 a Explain why triangle ABC is similar to triangle PQR.

b Find the length of the side QR.

3 In the triangle below DE is parallel to BC. Find the length of BC.

Answers1 a Yes b No c Yes d No2 a 3 equal angles b 8 cm3 15 cm

D C

A

E B3 cm

4 cm

4 cm

x

Extension Answers

1 a Triangle ABC and triangle ADE, 9 cm b Triangle PQR and triangle STR, 6 cmc Triangle XYZ and triangle VWY, 5 cm d Triangle JKL and triangle JMN, 5 cm

� similar� similarity

Key Words

42°

5 cm

4 cm

2 cm 4 cm 8 cm

10 cm

51°

87°

42°30°46°

30°

114°

a

c d

b

2cm

6 cm6 cm 4 cm

9 cm

9 cm

88°35°A

C

B9 cm

12 cm R

P

Q57°88°

6 cm

A

B C

ED4 cm

6 cm6 cm

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Oral and mental starter� Multiplying and dividing by 10, 100, 1000 and 10 000 will be revised.� On the board, draw the grid on the right, or use a prepared OHT.� Tell the class that the starting number is 3.5.� Ask individual students to point to the number that is:

3.5 × 10 3.5 × 100 3.5 × 1000 3.5 × 10 000� Then ask the class to explain the rules for multiplying by powers of 10.� Next, ask individual students to point to the number that is:

3.5 ÷ 10 3.5 ÷ 100 3.5 ÷ 1000 3.5 ÷ 10 000� Finally, ask the class to explain the rules for dividing by powers of 10.

Main lesson activity� Tell the class that the lesson is about converting the metric units of area, volume and capacity.� Remind them that the metric units for area are: the square millimetre (mm2),

the square centimetre (cm2) and the square metre (m2).� Draw two squares on the board and explain why

1 cm2 = 100 mm2.� Draw another two squares on the board and explain to the

class why 1 m2 = 10 000 cm2.� The students can now copy the following into their books:

Metric units of area100 mm2 = 1 cm2

10 000 cm2 = 1 m2

10 000 m2 = 1 hectare (ha)1 hectare = 100 ares

Note that, for measuring the area of fields, the m2 is too small,while the km2 is too large. Hence, a more conveniently sizedunit is used – the are, which is 100 m2. Land area is usuallygiven in units of 100 ares, where 100 ares = 1 hectare.

� Remind the class that the metric units for volume are: the cubic millimetre (mm3), the cubic centimetre (cm3) and the cubic metre (m3).

� Draw two cubes on the board and explain why 1 cm3 = 1000 mm3.

� Draw another two cubes on the board and explain why 1 m3 = 1 000 000 cm3.

� The students can now copy the following into their books:Metric units of volume1000 mm3 = 1 cm3

1 000 000 cm3 = 1 m3

� Remind the class that the metric units for capacityare: the litre (l), the centilitre (cl) and the millilitre (ml). They can now copy the following intotheir books:

Metric units of capacity1 m3 = 1000 litres1000 cm3 = 1 litre1 cm3 = 1 millilitre

� Stress the following:�� To convert from a large unit to a smaller unit, always multiply by the conversion factor.�� To convert from a small unit to a larger unit, always divide by the conversion factor.

LESSON

6.2

Framework objectives – Metric units for area and volume

Use units of measurement to calculate, estimate, measure and solve problems in avariety of contexts; convert between area measures (mm2 to cm2, cm2 to m2, andvice versa) and between volume measures (mm3 to cm3, cm3 to m3, and viceversa).

35 0.0035 3500

0.000 35 3.5 0.35

350 0.035 35 000

1 cm

1 cm21 cm

10 mm

100 mm210 mm

1 m

1 m21 m

100 cm

10 000 cm2100 cm

1 cm31 cm 1000 mm310 mm

1 cm 10 mm1 cm 10 mm

1 m31 m 1 000 000 cm3100 cm

1 m 100 cm1 m 100 cm

9B3LP_06.qxd 18/09/2003 15:19 Page 72


� Show the class the following examples of conversion.35 000 cm2 to m2 0.35 cm3 to mm3 3500 cm3 to litres

�� 35 000 cm2 = 35 000 ÷ 10 000 = 3.5 m2

�� 0.35 cm3 = 0.35 × 1000 = 350 mm3

�� 3500 cm3 = 3500 ÷ 1000 = 3.5 litres


Plenary� Write the following on the board and ask the students to fill in the blanks.

1 _____ mm2 = 1 cm2 4 _____ mm3 = 1 cm3 6 _____ litres = 1 m3

2 _____ cm2 = 1 m2 5 _____ cm3 = 1 m3 7 _____ cm3 = 1 litre3 _____ m2 = 1 hectare (ha)

Answers 1 100 2 10 000 3 10 000 4 1000 5 1 000 000 6 10007 1000

Exercise 6B Answers

1 a 40 000 cm2 b 70 000 cm2 c 200 000 cm2 d 35 000 cm2 e 8000 cm2

2 a 200 mm2 b 500 mm2 c 850 mm2 d 3600 mm2 e 40 mm2

3 a 8 cm2 b 25 cm2 c 78.3 cm2 d 5.4 cm2 e 0.6 cm2

4 a 2 m2 b 8.5 m2 c 27 m2 d 1.86 m2 e 0.348 m2

5 a 3000 mm3 b 10 000 mm3 c 6800 mm3 d 300 mm3 e 480 mm3

6 a 5 m3 b 7.5 m3 c 12 m3 d 0.65 m3 e 0.002 m3

7 a 8 litres b 17 litres c 0.5 litre d 3000 litres e 7200 litres 8 a 8.5 cl b 120 cl c 84 ml d 4.5 litres e 2400 ml9 160 10 a 10 800 m2 b 1.08 hectares 11 150 litres 12 6 days 13 500

Ho

me

wo

rk 1 Express each of the following in mm2.

a 3 cm2 b 8 cm2 c 4.5 cm2 d 0.8 cm2

2 Express each of the following in m2.

a 40 000 cm2 b 70 000 cm2 c 32 000 cm2 d 5000 cm2

3 Express each of the following in cm3.

a 2 m3 b 9 m3 c 3.7 m3 d 0.3 m3

4 Express each of the following in litres.

a 8000 cm3 b 12 000 cm3 c 23 500 cm3 d 250 cm3

5 A rectangular park is 620 m long and 340 m wide. Find the area of thepark in hectares.

6 Calculate the volume of the box on the right. Give your answer in litres.

Homework Answers1 a 300 mm2 b 800 mm2 c 450 mm2 d 80 mm2 2 a 4 m2 b 7 m2 c 3.2 m2 d 0.5 m2

3 a 2 000 000 cm3 b 9 000 000 cm3 c 3 700 000 cm3 d 300 000 cm3

4 a 8 litres b 12 litres c 23.5 litres d 0.25 litre 5 21.08 hectares 6 10 litres

� square millimetre� square

centimetre� square metre� hectare� cubic millimetre� cubic centimetre� cubic metre� litre

Key WordsExtension Answers

1 250 2 a 1296 square inches b 46 656 cubic inches3 4840 square yards ≈ 0.405 hectares

40 cm10 cm

25 cm

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Oral and mental starter� Revise the names and spelling of the various parts of a circle.� First, ask the students to sketch a circle on their whiteboards or in their books.

Then ask them to draw and label all the different parts of a circle which they canremember.

� Ask individual students to show the answers on their whiteboards, or to writeseparate answers on the board. Check their spelling.

� The following terms should be covered: centre, radius, diameter, arc,circumference, chord, sector, segment, and tangent.

Main lesson activity� Remind the class how to calculate the circumference and the area of a circle

using the formulae:C = πd = 2πrA = πr2

Remind them that π = 3.142 or they can use the key on their calculator.� Draw the diagram below on the board or on a prepared OHT.

Explain that the arc, AB, is part of the circumference,and that the sector, AOB, is a slice of the circleenclosed by the arc AB and the radii OA and OB.∠AOB is the angle of the sector and is usually denotedby the Greek letter θ (pronounced theta).

θ� Explain that the length of the arc AB as a fraction of the circumference is –––––.

360θ

So, the length of the arc AB = ––––– × πd.360

θSimilarly, the area of the sector AOB = ––––– × πr2

360� Complete the following example, making sure that the students obtain the

correct answers using their calculators.Calculate: a the length of the arc AB.

b the area of the sector AOB.Give the answers to three significant figures.

30a Length of the arc AB = ––––– × π × 16 = 4.19 cm (3 s.f.).360

30b Area of the sector AOB = ––––– × π × 82 = 16.8 cm2 (3 s.f.).360


π


LESSON

6.3

Framework objectives – Length of an arc and area of a sector

Know and use the formulae for length of arcs and area of sectors of circles.

A Bsector

arc

Oθ

AB

O

30°8 cm

r

d

O

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Plenary� Draw this diagram on the board.� Ask the class to write down the formulae for the length

of the arc AB and the area of the sector AOB.� Check the students’ answers.


Exercise 6C Answers

1 i a 10.5 cm b 52.4 cm2 ii a 5.50 cm b 19.2 cm2 iii a 4.19 cmb 25.1 cm2 iv a 5.24 m b 10.5 m2 v a 5.24 m b 6.54 m2

vi a 18.8 m b 67.9 m2

2 a 23.5 m b 34.4 m2

3 109 mm4 29 cm2

Ho

me

wo

rk In this exercise take π = 3.142 or use the key on your calculator.

1 Calculate (i) the length of the arc and (ii) the area of the sector for each of the following circles.

Give your answers correct to three significant figures.

2 Calculate the total perimeter of the sector on the right. Give your answer correct to three significant figures.

3 Calculate the area of the sector below. Give your answer correct to three significant figures.

Answers1 a i 2.62 cm ii 6.55 cm2 b i 5.59 cm ii 22.3 cm2 c i 23.6 cm ii 118 cm2

2 33.4 cm 3 16.0 cm2

3.5 cm 3.5 cm150°

π

Extension Answers

a i 2π cm ii 6π cm2 b i 4π cm ii 32π cm2 c i 2π cm ii 12π cm2

d i 6π cm ii 27π cm2 e i 4π cm ii 6π cm2

� arc� circumference� radius� sector

Key Words

A B

O

θ

8 cm

40°5 cm30° 10 cm

135°

a b c

12 cm 12 cm

45°

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Oral and mental starter� Show the class various 3-D shapes such as: cube, cuboid, square-based

pyramid, tetrahedron, triangular prism, cylinder, sphere.� Ask them to identify each shape and the spelling of each name. Write all the

names on the board.

Main lesson activity� Remind the class how to calculate the volume of a prism. The volume, V, of a

prism is found by multiplying the area A of its cross-section by its length l:V = Al

� Explain to the class that the cross-section of a cylinder is a circle with radius r.So, the area of the cross-section is A = πr2.If the height of the cylinder is h, then the volume, V, for the cylinder is given bythe formula:

V = πr2 × h = πr2h

� Carefully go through the following example on the board, making sure that thestudents obtain the correct answer using their calculators.Calculate the volume of the cylinder, giving the answer correct to threesignificant figures.

V = π × 42 × 7 = 352 cm3 (3 s.f.)



LESSON

6.4

Framework objectives – Volume of a cylinder

Calculate lengths, areas and volumes in right prisms, including cylinders.

r

h

l

A

4 cm

7 cm

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Plenary� Draw a cylinder on the board.� Ask the class to explain how the formula for the volume of a cylinder is derived.


Exercise 6D Answers

1 a 1130 cm3 b 226 cm3 c 75.4 m3 d 37.7 m3 e 9.82 cm3

2 1062 mm3

3 cylindrical tin (a 2352 cm3 b 2300 cm3 c 2356 cm3)4 a 35.3 litres b 1415 a 1.51 m3 b 1131 litres6 a 10 000 cm3 b 250 cm2 c 8.9 cm

Ho

me

wo

rk

In this exercise take π = 3.142 or use the key on your calculator.

1 Calculate the volume of each of the following cylinders. Give your answers correct to threesignificant figures.

2 The diagram below shows a metal pipe of length 1 m. It has an internal diameter of 2.8 cm, and anexternal diameter of 3.2 cm. Calculate the volume of metal in the pipe. Give your answer correct tothe nearest cubic centimetre.

3 A cylindrical can holds 2 litres of oil. If the height of the can is 25 cm, calculate the radius of thebase of the can. Give your answer correct to one decimal place.

Answers1 a 339 cm3 b 157 cm3 c 101 m3

2 188 cm3 3 5.0 cm

3.2 cm

2.8 cm

1 m

3 cm

12 cm

5 cm

2 cm4 m

8 ma b c

π

Extension Answers

1 a 207 cm2 b 37.7 m2

2 r = 5.4 cm, h = 10.8 cm

� cross-section� cylinder� radius� volume

Key Words

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Oral and mental starter� Explain to the class that the starter is about converting minutes into fractions and decimals of an hour.� Copy the table below on the board or use a prepared OHT.

The first example has already been completed.

Time in minutes 30 minutes 15 minutes 45 minutes 20 minutes 6 minutes 10 minutesTime as a fraction 1–2 hourof an hourTime as a decimal 0.5 hourof an hour

� Ask the students to copy and complete the table in their books or on their whiteboards.Answers (Explain that sometimes the answers may have to be given to three decimal places.)

Time in minutes 30 minutes 15 minutes 45 minutes 20 minutes 6 minutes 10 minutesTime as a fraction 1–2 hour 1–4 hour 3–4 hour 1–3 hour 1––10 hour 1–6 hourof an hourTime as a decimal 0.5 hour 0.25 hour 0.75 hour 0.333 hour 0.1 hour 0.167 hourof an hour

Main lesson activity� Explain to the class that a rate of change is the way of comparing how one quantity changes with another. Rates

of change can always be recognised because their units contain ‘per’ or ‘p’ or ’/’ which means ‘for every’.� Write the following examples of compound measures on the board:

Speed – with units miles per hour (mph), kilometres per hour (km/h) or metres per second (m/s).Density – with units grams per cubic centimetre (g/cm3).Fuel consumption – with units miles per gallon (mpg) or kilometres per litre (km/l ).

� Explain that speed is the distance travelled per unit of time and that the relationship between speed, distanceand time can be expressed by the following three formulae:

Distance DistanceSpeed = ––––––––– Distance = Speed × Time Time = –––––––––

Time SpeedWhen we refer to speed, we usually mean average speed, as it is unusual to maintain the same, exact speed inone journey.

� Draw the diagram below and explain that the relationships between distance D, time T and speed S can beremembered by using this triangle:

Covering up the quantity you want to find leads to the three formulae:

D DS = — D = ST T = —

T S

� Show the students how to use the triangle by doing the following examples.� Example 1

A train travels at an average speed of 80 mph. Find the distance travelled by the train in 21–2 hours.Using the formula D = ST, the distance travelled = 80 × 21–2 = 200 miles. Example 2A car travels 120 km on a motorway at an average speed of 80 km/h. Find the time taken for the journey.

D 120Using the formula T = —, the time taken = –––– = 1.5 hours = 11–2 hours or 1 hour 30 minutes.

S 80� Explain that density is the mass of a substance per unit of volume and that the formula for density is:

MassDensity = –––––––

Volume


LESSON

6.5

Framework objectives – Rates of change

Understand and use measures of speed (and other compound measures such asdensity or pressure) to solve problems; solve problems involving constant oraverage rates of change.

D

S T

9B3LP_06.qxd 18/09/2003 15:19 Page 78

� Draw the diagram below and explain that the relationships between density D, mass M and volume V can beremembered by using this triangle:

Covering up the quantity you want to find leads to the three formulae:

M MD = — M = DV V = —

V D

� Show the students how to use the triangle by doing the following examples.Example 1The volume of a wooden block is 20 cm3 and its mass is 18 g. Find the density of the wood.

M 18Using the formula D = —, the density of the wood is ––– = 0.9 g/cm3.

V 20Example 2Find the mass of a plastic model, which has a volume of 30 cm3 and a density of 1.6 g/cm3.Using the formula M = DV, the mass of the model = 1.6 × 30 = 48 g.


Plenary� Summarise the lesson by asking the class to write down the relationships between

distance D, time T and speed S by using the triangle at the top of the page.� Repeat for density.


Exercise 6E Answers

1 a 240 miles b 180 miles c 30 miles d 40 miles 2 a 52.5 mph b 42 mph c 60 mph d 63 mph3 a 3 hours b 2 hours 30 minutes c 3 hours 20 minutes d 1 hour 45 minutes4 a 75 mph b 120 km/h c 125 miles d 225 km e 12.5 seconds f 3 hours 20 minutes5 216 metres 6 600 mph 7 a 10 m/s b 25 m/s c 331–3 m/s8 156.8 g 9 5 litres 10 1.6 g/cm3

Ho

me

wo

rk 1 Find the distance travelled by a hiker who walks for 3 hours at an average speed of 2.5 mph.

2 Find the time taken to drive a car 125 km at an average speed of 75 km/h.

3 A runner runs a 1000 m race in 3 minutes 20 seconds. Find his average speed in m/s.

4 Find the density of a gold ingot that has a mass of 4825 g and a volume of 250 cm3.

5 The density of sea water is 1.05 g/cm3. If a bucket with a capacity of 5 litres is filled with seawater,find the mass of the water in the bucket. Give your answer in kilograms.

6 The density of cork is 0.25 g/cm3. Find the volume of a block of cork that has a mass of 120 g.

Answers1 7.5 miles 2 1 hour 40 minutes 3 5 m/s 4 19.3 g/cm3 5 5.25 kg 6 480 cm3

M

D V

Extension Answers

1 42 mph 2 48 mph 3 17.1 km/h 4 450 g 5 a 90 miles b 20 gallons

SATs Answers

1 a 71–2 hours b 465 mph c 60 miles 2 a 60 mph b 30 mph c 40 mph3 a 15 cm b 12 cm c Yes, the angles are the same 4 a B b A c 2.83 cm 5 0.72 cm

� density� pressure� speed

Key Words

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Oral and mental starter� Use a target board, such as the one shown on the right, to do this exercise in

multiplying and dividing powers.� Point to two powers and ask a student to multiply or divide them.� Ensure that the class know which way round to do the division.� Recall the rules for multiplying and dividing powers.

Main lesson activity� Students may have met the idea of standard form through extension activities.� Explain that a way is needed to express and use very large and very small

numbers without writing out many zeros. For example 3 400 000 000 can bewritten as 3.4 × 109, and 0.000 072 can be written as 7.2 × 10–5.

� Ask students if they can see the connection with the original number and thepower of 10.

� Make sure they know and understand the definition of a standard form number:A × 10n, where 1 ≤ A < 10 and n is an integer.

� Demonstrate how to convert ordinary numbers to standard form. For example:540 000 = 5.4 × 105 0.000 005 89 = 5.89 × 10–6

89 630 000 = 8.963 × 107

� The class may realise that this is a matter of moving the digits and counting thenumber of places that they move.

� Repeat with examples such as23 × 106 = 2.3 × 10 × 106 = 2.3 × 107

and0.56 × 10–4 = 5.6 × 10–1 × 10–4 = 5.6 × 10–5

� Repeat with more examples as necessary.� Now reverse the process and ask the class to convert numbers in standard form

to ordinary numbers. For example:3.4 × 105 = 340 000 3.6 × 10–7 = 0.000 000 36

� They should be able to relate this to work done previously with powers of 10.� Repeat with more examples as necessary.



Number 2CHAPTER

7

LESSON

7.1

Framework objectives – Standard form

Write numbers in standard form.

Use algebraic methods to convert a recurring decimal to a fraction in simple cases.

x3 x5 x–4 x–2

x8 x–7 x7 x–6

x–3 x6 x–8 x4

x2 x10 x9 x–5

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Plenary� Make sure that the class is aware of the difference between a calculator

display showing standard form as 4.39 and the actual standard form number 4.3× 109. This is a common error in SATs and GCSE exams.

� Put some calculator displays on the board. Then ask individual students toconvert them: first, to a standard form number and next to an ordinary number.


Exercise 7A Answers

1 a 5.69 × 103 b 1.2 × 106 c 9.38 × 105 d 7.78 × 104 e 3.965 × 108

f 5.61 × 102 g 7.3 × 101 h 4.3 × 109

2 a 3.4 × 10–3 b 5.6 × 10–2 c 3.71 × 10–5 d 9.2 × 10–6 e 7.6 × 10–1

f 5 × 10–4 g 7.2 × 10–6 h 4 × 10–4

3 a 8.9 × 106 b 5.3 × 10–3 c 1.8 × 104 d 3.33 × 107 e 6.7 × 10–6

f 8.923 × 103 g 7.35 × 10–1 h 9 × 10–5

4 a 2 300 000 b 456 c 670 000 d 3590 e 9 000 000 f 2 010 000g 34 780 h 87 300 000

5 a 0.000 067 b 0.0385 c 0.000 78 d 0.005 39 e 0.000 008 f 0.167g 0.003 21 h 0.000 000 66

6 a 4 600 b 0.057 66 c 930 d 0.001 22 e 50 000 f 0.305 g 4 820 000h 0.0543

7 a 4.3 × 106 b 5.68 × 103 c 7.8 × 103 d 5.8 × 102 e 9.4 × 10–4

f 2.01 × 10–4 g 8 × 10–4 h 8 × 10–2 I 2.5 × 10–3 j 5.6 × 10–3

k 6.7 × 104 l 3.59 × 104

8 a 880 b 53 200 c 0.003 14 d 0.903 e 0.001 82 f 7 950 000g 504 000 000 h 0.000 684 2

Extension Answers

a 1.5 × 108 km b 2.5 × 10–3 cm c 2.21 × 109 s d 9 × 10–28 gme 1.868 × 107 g

Ho

me

wo

rk 1 Write each of the following numbers in standard form.

a 63 000 000 b 0.000 74 c 322 000 d 83 300

e 0.000 000 71 f 92 321 g 0.009 35 h 0.000 0005

2 Write each of the following standard form numbers as an ordinary number.

a 4.9 × 104 b 4.36 × 10–3 c 8.4 × 103 d 5.68 × 10–2

e 8 × 109 f 4.82 × 10–4 g 9.2 × 106 h 6.03 × 10–1

3 Write each of the following numbers in standard form

a 68 × 103 b 37.8 × 10–5 c 0.87 × 10–3 d 58 × 10–4

Answers1 a 6.3 × 107 b 7.4 × 10–4 c 3.22 × 105 d 8.33 × 104 e 7.1 × 10–7 f 9.2321 × 104 g 9.35 × 10–3

h 5 × 10–7

2 a 49 000 b 0.004 36 c 8400 d 0.0568 e 8 000 000 000 f 0.000 482 g 9 200 000h 0.603

3 a 6.8 × 104 b 3.78 × 10–4 c 8.7 × 10–4 d 5.8 × 10–3

� standard form� integer� less than� less than or

equal to

Key Words

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Oral and mental starter� Use a target board like the one shown on the right.� Ask individual students to give each number in standard form.

(The numbers are straightforward enough to allow this to be donementally.)

Main lesson activity� Tell the class that they are going to learn how to multiply together two numbers

in standard form.� Give them: (3 × 106) × (3 × 102) and ask them what they think the answer is.

Some students may respond intuitively that it is 9 × 108.� Explain the process of separating the numbers and the powers:

3 × 3 × 106 × 102.� Now give the class: (6 × 106) × (3 × 102). Ask them for the answer. Some will

probably respond 18 × 108. � Ask them what is wrong with this. Explain that it is not in standard form.� Demonstrate how to get it in standard form. That is, 1.8 × 10 × 108 = 1.8 × 109.� Repeat with: (2.5 × 10–2) × (8 × 10–5). This gives 20 × 10–7 = 2 × 10 × 10–7 =

2 × 10–6.� Do other examples as necessary.� Next give the class: (4.56 × 103) × (2.13 × 10–7). They will realise that this is

difficult without the use of a calculator.� Explain how to use a calculator to enter numbers in standard form.

4.56 × 103 is entered as:

2.13 × 10–7 is entered as:

� Note that on some makes of calculator, the appropriate key may be marked EEor with some other notation, and that the sign change key may operatedifferently.

� Explain the calculator notation which relates to the previous plenary.� Do the above calculation. The display should say 0.00 097 128 or 9.7128–04.� Convert this to the standard form number 9.7128 × 10–4 or round to 9.71 × 10–4

(3 sf).� Make sure that no students enter 4.56 × 103 as:

This will give a value of 4.56 × 104.


3EXP01×65.4

+/–7EXP31.2

3EXP65.4


LESSON

7.2

Framework objectives – Multiplying with numbers in standardform

Enter numbers in standard form into a calculator and interpret the display.

134 5477 0.023 0.25 2000

35 0.067 0.007 0.039 270

50 6000 0.134 0.03 0.001

0.256 256 0.002 1300 100

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Plenary� Give the class the following problem: 0.012 × 0.006.� They should be able to do this as 12 × 6 = 72, with six decimal places in the

answer: 0.000 072.� Ask them to ‘translate’ the calculation to a standard form problem which would

be: (1.2 × 10–2) × (6 × 10–3) = 7.2 × 10–5.� Repeat with:

0.085 × 0.04 = 0.0034(8.5 × 10–2) × (4 × 10–2) = 34 × 10–4 = 3.4 × 10–3

� Discuss the similarities and the advantages of each method.


Exercise 7B Answers

1 a 8 × 105 b 1.2 × 108 c 8 × 107 d 9 × 10–5 e 3.2 × 10–7 f 4.2 × 10–2

g 2.1 × 10–1 h 2.6 × 103 i 5.6 × 103 j 2.25 × 105 k 1.12 × 103

l 3.6 × 107

2 a 9.46 × 109 b 1.152 × 108 c 1.288 × 1010 d 5.51 × 10–5 e 4.672 × 105

f 1.674 × 10–1 g 2.99 × 104 h 1.311 × 103 i 4.6656 × 107

3 a 9.82 × 1010 b 7.28 × 107 c 7.27 × 109 d 2.35 × 10–6 e 4.05 × 10–3

f 5.84 × 10–2 g 2.95 × 106 h 1.56 × 10–6 i 6.13 × 1013 j 1.63 × 10–5

4 1.25 × 109 bytes5 1.2 m

Extension Answers

The American mathematician Edward Kasner asked his nine-year-old nephew toinvent a name for the number, ten to the power of one hundred (10100). His nephewcalled it a googol. Another mathematician came up with googolplex, and defined itto be ten to the power of googol. There is no limit to the largest number in practical use, but remember that the totalnumber of elementary particles in the known universe is only about 1080.

Ho

me

wo

rk 1 Do not use a calculator for this question. Work out each of the following and give your answer instandard form.

a (4 × 102) × (2 × 106) b (5 × 103) × (4 × 102) c (6 × 10–3) × (2 × 10–4)

d (9 × 10–2) × (3 × 108) e (5 × 10–5) × (8 × 10–3) f (7 × 103) × (7 × 103)

2 You may use a calculator for this question. Work out each of the following and give your answer instandard form. Do not round off your answers.

a (2.1 × 105) × (3.4 × 103) b (3.2 × 103) × (1.5 × 104) c (3.6 × 103) × (2.8 × 10–8)

d (1.5 × 10–2) × (2.5 × 10–4) e (3.8 × 10–4) × (2.8 × 104) f (8.6 × 104) × (1.5 × 10–7)

Answers1 a 8 × 108 b 2 × 106 c 1.2 × 10–6 d 2.7 × 107 e 4 × 10–7 f 4.9 × 107

2 a 7.14 × 108 b 4.8 × 107 c 1.008 × 10–4 d 3.75 × 10–6 e 1.064 × 101 f 1.29 × 10–2

� standard form� addition of

powers

Key Words

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Oral and mental starter� Use a target board such as the one shown on the right.� Ask individual students to give the ordinary numbers

shown by the numbers in standard form. (The numbers aresimple enough for the students to do this mentally.)

Main lesson activity� This lesson focuses on dividing one number in standard form by another number

in standard form. � Give the class: (9 × 106) ÷ (3 × 102) and ask them what they think the answer is.

Some students may intuitively respond that it is 3 × 104.� Explain the process of separating the numbers and the powers:

(9 ÷ 3) × (106 ÷ 102). Make sure they are happy with the position of themultiplication sign.

� Now give the class: (4 × 106) ÷ (8 × 102). Ask them for the answer. Some willprobably respond 0.5 × 104.

� Ask them what is wrong with this. Explain that it is not in standard form.� Demonstrate how to get it in standard form. That is, 5 × 10–1 × 104 = 5 × 103.� Repeat with: (2 × 10–2) ÷ (8 × 10–5). This gives 0.25 × 103 = 2.5 × 10–1 × 103 =

2.5 × 102.� Do with other examples as necessary.� Next give the class: (4.56 × 103) ÷ (2.13 × 10–7). They will realise that this is

extremely difficult without the use of a calculator.� Make sure that they can enter this into their calculators using the EXP (or

equivalent) key and the sign change key.� The display should say 2.140 845 0710

� Round off and convert this to the standard form number 2.14 × 1010 (3 sf).


2 × 103 3.2 × 10–2 5.4 × 102 4.3 × 10–1

7 × 10–4 6.3 × 10–3 6.3 × 103 5.1 × 102

3.6 × 103 8.1 × 102 2.5 × 103 3 × 10–3

4.1 × 101 6.8 × 10–1 4 × 105 8 × 10–2

6 × 10–5 5.2 × 10–3 1.7 × 102 5.6 × 10–1


LESSON

7.3

Framework objectives – Dividing with numbers in standard form

Divide numbers in standard form.

Enter numbers in standard form into a calculator and interpret the display.

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Plenary� Give the class this problem: 0.012 ÷ 0.006.� They should be able to do this as:

0.012 ÷ 0.006 = 0.12 ÷ 0.06 = 1.2 ÷ 0.6 = 12 ÷ 6 = 2

� Ask them to ‘translate’ the calculation to a standard form problem, which wouldbe:

(1.2 × 10–2) ÷ (6 × 10–3)= 0.2 × 101 = 2� Repeat with:

0.055 ÷ 0.05 = 1.1(5.5 × 10–2) ÷ (5 × 10–2) = 1.1 × 100 = 1.1

� Discuss the similarities and the advantages of each method.


Exercise 7C Answers

1 a 3 × 104 b 7.5 × 102 c 2 × 10–1 d 3 × 105 e 5 × 105 f 1.2 × 109

g 9 × 10–7 h 4 × 108 i 1.6 × 105 j 5 × 102 k 7 × 107 l 3 × 103

2 a 4.5 × 104 b 4.05 × 102 c 2.8 × 10–5 d 3.4 × 102 e 8 × 10–9 f 6 × 109

g 7 × 104 h 8 × 105 i 2 × 102

3 a 5.61 × 104 b 1.37 × 103 c 1.32 × 10–4 d 2.92 × 102 e 5.98 × 1011

f 2.88 × 109 g 6.23 × 10–12 h 4.48 × 10–4 i 2.5 × 103 j 1.2 × 103

4 6.366 × 103 km 5 33.3·

Ho

me

wo

rk 1 Do not use a calculator for this question. Work out each of the following and give your answer instandard form.

a (8 × 105) ÷ (2 × 103) b (4 × 105) ÷ (5 × 107) c (6 × 103) ÷ (2 × 10–4)

d (1.2 × 10–3) ÷ (3 × 10–2) e (6 × 106) ÷ (8 × 10–1) f (5 × 102) ÷ (8 × 10–3)

2 You may use a calculator for this question. Work out each of the following and give your answer instandard form. Do not round off your answers.

a (6.15 × 105) ÷ (1.5 × 102) b (3.15 × 106) ÷ (1.4 × 10–1)

c (3.19 × 103) ÷ (1.45 × 10–2) d (2.32 × 10–3) ÷ (2.9 × 10–5)

e (5.85 × 10–3) ÷ (6.5 × 103) f (1.495 × 106) ÷ (4.6 × 10–2)

Answers1 a 4 × 102 b 8 × 10–3 c 3 × 107 d 4 × 10–2 e 7.5 × 106 f 6.25 × 104

2 a 4.1 × 103 b 2.25 × 107 c 2.2 × 105 d 8 × 101 e 9 × 10–7 f 3.25 × 107

� standard form� division� subtraction of

powers

Key Words

Extension Answers

Milli = one thousandth (10–3); micro = one millionth (10–6); nano = one billionth(10–9); pico = one trillionth (10–12); femto = one quadrillionth (10–15); atto = onequintillionth (10–18); zepto = one sextillionth (10–21); yocto = one septiillionth (10–24)

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Oral and mental starter� Use a target board such as the one shown on the right.� Ask individual students to give the limits of each

number chosen by you to a given accuracy. Forexample, 100 marbles (to nearest 10), 95 to 104; 80 cm (to nearest cm), 79.5 to 80.5.

� The students may need to have some of the rules ofupper and lower bounds explained to them,particularly the difference between discrete andcontinuous data.

Main lesson activity� Ensure that the class can work out the limits, called the upper and lower

bounds, of numbers given to degrees of accuracy.� Make sure they are happy with the difference between limits of accuracy for

discrete quantities, such as a number of people, and continuous quantities, suchas the weight of people.

� Some students may be happier to have a ‘rule’ for continuous data:Lower bound = Value – Half the degree of accuracyUpper bound = Value + Half the degree of accuracy

For example, 500 given to the nearest 10 has a lower bound of 500 – half of 10 = 495 and an upper bound of 500 + half of 10 = 50523 cm to the nearest centimetre has upper and lower bounds of 23 ± half acentimetre: 22.5 ≤ 23 < 23.5

� Emphasise that the upper value is a limit and can be given as a value and neednot be truncated to 23.4 or 23.4999.

� This is a concept students find hard to grasp. Justify it by using the notation:22.5 ≤ Value < 23.5.

In other words, the value can equal the lower bound but cannot equal the upperbound.

� Give students the following example:One piece of wood is 25 cm long and another is 30 cm long, bothmeasurements to the nearest centimetre. If the pieces are joined end to end,what are the upper and lower bounds of the combined length?Upper and lower bounds of first piece: 24.5 ≤ Length < 25.5Upper and lower bounds of second piece: 29.5 ≤ Length < 30.5Lowest possible combined length: 24.5 + 29.5 = 54 cmLargest possible combined length: 25.5 + 30.5 = 56 cmUpper and lower bounds of combined length: 54 ≤ Combined length < 56 cm

� Next, do the following example:One bag of marbles contains 200 marbles, measured to the nearest 10. Anotherbag of marbles contains 120 marbles, measured to the nearest 10. If both bagsare combined, what are the upper and lower bounds for the number of marblesin the bag.Upper and lower bounds for first bag: 195 ≤ Marbles ≤ 204Upper and lower bounds for second bag: 115 ≤ Marbles ≤ 124Emphasise why these limits are precise.Upper and lower bounds for combined bags: 310 ≤ Marbles ≤ 328



LESSON

7.4

Framework objectives – Upper and lower bounds 1

Understand upper and lower bounds for discrete data and continuous data.

100 marbles 25 cm 500 eggs 300 g

5 kg 25 m 400 people 200 ml

5 litres 80 cm £ 2000 400 sweets

50 g 100 cars 150 mm 220 ducks

600 mph 600 people 40 km 120 bees

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Plenary� Give the class the following problem.

a = 5 to nearest unit, b = 16 to nearest unit.What are the upper and lower bounds of a + b (20, 22), a × b (69.75,90.75), a – b (10, 12), a ÷ b (0.2727 … , 0.3548 …)?

Discuss the required combinations, particularly the subtraction and division.


Exercise 7D Answers

1 a 25 ≤ Toffees ≤ 34 b 15 ≤ Rice < 25 g c 65 ≤ Speed < 75 mphd 69.5 ≤ Speed < 70.5 mph e 15 ≤ String < 25 cm f 19.5 ≤ String < 20.5 cmg 495 ≤ Cake < 505 g h 49.5 ≤ Donut < 50.5 g i 999.5 ≤ Jug < 1000.5 cm3

j 39.5 ≤ Hard drive < 40.5 Gb2 a 395 ≤ Length < 405 cm b 295 ≤ Width < 305 cm

c 13.8 ≤ Perimeter < 14.2 m3 a 9.5 × 9.5 cm b 10.5 × 10.5 cm c No, could be guaranteed to cover 95 cm

d No, if all tiles are 9.5 cm, he will need 32 × 22 = 7044 a 6250 b 67455 a 195 ≤ Number ≤ 204 b 195 ≤ Mass < 205 g

c Number is discrete data, mass is continuous6 a 23 500 b 244 990 ≈ 245 0007 a 225 b 2348 No, if each slab is only 39.5 cm, he will only have enough for 19.75 m9 Yes, 9 mugs would be 205 × 9 = 1845. Smallest jug is 1850 ml

10 No, Melanie’s lowest possible total is 43 + 48 + 62 + 38 + 52 = 243, which is notenough

Ho

me

wo

rk Do not use a calculator for Questions 1 and 2.

1 Find the upper and lower bounds between which the following quantities lie.

a In a hive there are 2000 bees to the nearest 100.

b The amount of honey in a jar is 200 ml to the nearest 10 ml.

c The width of a field is 70 m to the nearest metre.

d The mass of a loaf is 0.6 kg to the nearest 100 grams.

2 A poster is 2.5 metres by 1.5 metres, each measurement accurate to the nearest 10 cm.

a What are the upper and lower bounds for the length of the poster?

b What are the upper and lower bounds for the width of the poster?

c What are the upper and lower bounds for the perimeter of the poster?

3 A bottle of water holds 1 litre to the nearest centilitre.

a What is the smallest possible amount in the bottle?

b What is the greatest possible amount that 10 bottles could hold?

Answers1 a 1950 ≤ Bees < 2049 b 195 ≤ Honey < 205 ml c 69.5 ≤ Width < 70.5 m d 0.55 ≤ Mass < 0.65 kg2 a 2.45 ≤ Length < 2.55 m b 1.45 ≤ Width < 1.55 m c 7.8 ≤ Perimeter < 8.2 m3 a 99.5 cl b 10.05 litres

� upper bound� lower bound� strict inequality� greater than/less

than

Key Words

Extension Answers

a For largest length left, take the shortest length of cut-off piece from the longest lengthof original: 120.5 – 84.5 = 36 cm. For shortest length left, take the longest length ofcut-off piece from the shortest length of original: 119.5 – 85.5 = 34 cm.

b For largest possible length, divide the smallest possible area of the rectangle by thesmallest possible width: 42.5 ÷ 1.5 = 28.33 cm. For smallest possible length, dividethe largest possible area by the largest possible width: 47.5 ÷ 2.5 = 19 cm.

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Oral and mental starter� Recall the methods of multiplying integers and decimals

expressed to one significant figure. For example, 0.7 × 6 = 4.2, 0.03 × 8 = 0.24.

� Use this to ask for answers to a variety of similar questions. As these can be hard to verbalise and for students to conceptualise, agrid, as shown on the right, could be used.

� Particular squares could be pointed at and individual studentsasked, for the product.

� Reverse the process by asking for the missing number in, for example, 0.4 × ? = 0.32 (0.8), or the answer to 0.048 ÷ 0.6 = ? (0.08).

� Repeat with examples such as 0.04 ÷ ? = 0.08 (1–2), 0.5 × ? = 0.035 (0.07).

Main lesson activity� This lesson is concerned with making calculations with limits.� Following the previous plenary, the students will have some idea of how the

maximum and minimum values are combined. To help them the following tablecould be copied into their books.For two numbers a and b with upper and lower bounds amax ≤ a < amin and bmax ≤ b < bmin

Operation Maximum Minimuma + b amax + bmax amin + bmina – b amax – bmin amin – bmaxa × b amax × bmax amin × bmina ÷ b amax ÷ bmin amin ÷ bmax

� Work through the following examples.A jar contains 850 ml of oil (accurate to the nearest 10 ml). A cup containing250 ml (to the nearest 10 ml) is poured out. What are the upper and lowerbounds of the amount of oil remaining in the jar?

Least amount left: 845 – 255 = 590.Greatest amount left: 855 – 245 = 610Upper and lower bounds of amount left: 590 ≤ Amount left < 610 ml

A runner sprints 400 metres in 55 seconds, each measurement to the nearestunit. What are the upper and lower bounds of the runner’s speed?

Lowest speed: 399.5 ÷ 55.5 = 7.20 m/sGreatest speed: 400.5 ÷ 54.5 = 7.35 m/s (3 sf)

The same runner runs for 40 minutes at a speed of 8 mph, both values to twosignificant figures. What are the upper and lower bounds of the distance run?

Least distance: 39.5 × 7.5 = 296.25 ÷ 60 = 4.94 milesGreatest distance: 40.5 × 8.5 = 344.25 ÷ 60 = 5.74 miles (3 sf)



LESSON

7.5

Framework objectives – Upper and lower bounds 2

Understand upper and lower bounds for discrete data and continuous data.

× 0.8 0.3 9 0.6

0.7

5

0.02

0.01

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Plenary� Ask the class to consider which combination of maximum and minimum value

of a, b, c and d would give the maximum and minimum values of an expressionsuch as:

a – b a2–––––– or –––––––– × dc – d (b – c)

� Ensure that the students understand the need to combine upper and lowerbounds in such a way that they maximise and minimise the value of suchcombined expressions.


Exercise 7E Answers

1 a 13.75 ≤ Area < 22.75 cm2 b 11.25 ≤ Area < 21.25 cm2

c 97.75 ≤ Area < 118.75 cm2

2 a 11.5 ≤ a < 12.5, 17.5 ≤ b < 18.5, 23.5 ≤ c < 24.5 b i 201.25 ≤ a × b < 231.25ii 1.4 ≤ a ÷ a < 1.61 iii 681.5 ≤ (a + b)c < 759.5 iv 552.25 ≤ c2 < 600.25

3 a 82.4 mph (1.37 miles per min) b 77.7 mph (1.29 miles per min)4 a 1137.75 miles b 1062.75 miles5 a 2.903 b 3.103 g6 9.75 ≤ Side < 10.25 cm (3 sf)

a + b7 a 94.62 ≤ a × b × c < 101.3 (4sf) b 1.109 ≤ –––––– < 1.156 (4sf)c

c 0.75 ≤ a + b + c < 1.05 d 1.44 ≤ (b – a)2 < 1.968 9.61 ≤ Width < 10.4 cm (3 sf)9 1.65 ≤ Mass coat < 2.75 kg

10 No, rod of 4.05 cm would be required to fit into hole which is 3.5 cm

Ho

me

wo

rk 1 a = 10, b = 20 and c = 30. All values to the nearest whole number.

a Write down the upper and lower bounds of a, b and c.

b Work out the upper and lower bounds of each of the following.

i a × b ii c ÷ a iii (a × b) + c iv c2

2 A rectangle has an area of 120 cm2, measured to the nearest 10 cm2. The length is 15 cm, measuredto the nearest cm.

a What is the greatest possible width of the rectangle?

b What is the least possible width of the rectangle?

Answers1 a 9.5 ≤ a < 10.5, 19.5 ≤ b < 20.5, 29.5 ≤ c < 30.5 b i 185.25 ≤ a × b < 215.25

ii 2.81 ≤ c ÷ a < 3.21 (3 sf) iii 214.75 ≤ (a × b) + c < 245. 75 iv 870.25 ≤ c2 < 930.252 a 8.62 cm (3 sf) b 7.42 cm (3 sf)

� upper bound� lower bound� strict inequality� greater than/

less than

Key Words

Extension Answers

1 352.3 ≤ Surface area < 402.8 cm2 (4 sf)2 214.3 ≤ Surface area < 219.1 cm2 (4 sf)

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Oral and mental starter� The class will need calculators.� Using a target board, like that shown on the right, ask the class whether they can

predict which of the fractions, when converted to a decimal, will terminate andwhich will recur.

� They have met these ideas before but may need to be reminded of them.� After they have made their predictions, ask them to work out the decimals on

their calculators.� These can then be written on the board and the recurring notation explained to

the class. For example: 4––15 = 0.26·, 32––33 = 0.9

·6·.

� Ask if there is any relationship between the denominators which giveterminating decimals? They are all multiples of powers of 2 (2, 4, 8, 16, …),powers of 5 (5, 25, 125, …), powers of 10 (10, 100, 1000, …), or products ofthese, such as 40 (4 × 10), 50 (5 × 10) … .

Main lesson activity� This follows on from the Oral and mental starter.� Ask: ‘How would you write a recurring decimal as a fraction: for example, 0.454

545 … = 0.4·5·?’

� Give them the chance to see whether they can find the answer by trial andimprovement. They may find the answer of 5––11.

� Outline the following method.Let F = 0.454 545 45 (1)Then 100F = 45.454 545 (2)Subtract equation (1) from (2): 99F = 45

45 5Divide through by 99: F = ––– = ––– (Cancel by 9).

99 11� Ask: ‘Why multiply by 100?’� Establish that we have to multiply by the power of 10 equivalent to the number

of recurring digits.� Now repeat with 0.7

·( 7–9 ), 0.2

·34

·( 26–––111).

� Ask: Can anyone see a connection or a short cut?� Establish that when there is just one recurring digit, the denominator will be 9;

when there are two recurring digits, the denominator will be 99; and when thereare three recurring digits, the denominator will be 999.

� Be careful! This rule works only when the recurring digits are the only digits afterthe decimal point. It will not work with a number such as 0.377 777 7 … = 0.37

·.

� Give some examples:6 2 39 13 675 25

0.6·

= – = – 0.3·9·

= ––– = ––– 0.6·75

·= ––––– = –––

9 3 99 33 999 37� Put this recurring decimal on the board: 0.277 777 77 … = 0.27

·.

� Give the students a few moments to see whether they can find its fraction by trialand improvement.

� Now work through the procedure.Let F = 0.277 777 77 … (1)Multiply by through10: 10F = 2.777 777 77… (2)Subtract equation (1) from (2): 9F = 2.5

2.5 5Divide through by 9: F = –––– = –––

9 18� Repeat with 0.166 66 … (1–6) and 0.416 666 … ( 5––12)



LESSON

7.6

Framework objectives – Recurring decimals

Know that a recurring decimal is an exact fraction. Use algebraic methods toconvert a recurring decimal to a fraction in simple cases.

1–23–7

7––202–3

13––18

11––307––12

43––503–8

7–9

3––113–4

1––2812––25

5–6

9––257––60

31––404–5

3––13

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Plenary� Ask the class to write down from their calculators the values of π, √

–2 and √

–5.

� Ask them if they can spot the difference between these decimals and those withwhich they have been working.

� The values of π, √–2 and √

–5 do not terminate nor do they recur.

� Define these as irrational numbers. That is, decimals which do not terminate orrecur.

� The other numbers are known as rational numbers. That is, numbers which canbe expressed as a terminating decimal or a recurring decimal.


Exercise 7F Answers

1 a 0.5·71 428· b 0.7·524· c 0.6·9· d 0.3· e 0.2·

2 1–9 = 0.1· , 2–9 = 0.2· , 3–9 = 0.3· , 4–9 = 0.4· , 5–9 = 0.5· , 6–9 = 0.6· , 7–9 = 0.7· , 8–9 = 0.8· , 9–9 = 0.9·

Recurring digits are the same as original numerators3 1––11 = 0.0·9· , 2––11 = 0.1·8· , 3––11 = 0.2·7· , 4––11 = 0.3·6· , 5––11 = 0.4·5· , 6––11 = 0.5·4· , 7––11 = 0.6·3·

8––11 = 0.7·2· , 9––11 = 0.8·1· , 10––11 = 0.9·0·

Recurring digits form the nine times table up to 904 1–7 = 0.1·42 857· , 2–7 = 0.2·85 714· , 3–7 = 0.4·28 571· , 4––11 = 0.5·71 428· , 5–7 = 0.7·14 285· ,

6–7 = 0.8·57 142·

First recurring digits are in numerical order5 a 5––11 b 107–––333 c 8–9 d 27––37 e 4––33 f 9––11 g 123–––1111 h 26––33 i 89–––111 j 9–9 = 16 a 4––90 b 28—–495 c 578—–999 d 352——4995 e 23—–6667 a 2389—–990 b 11––18 c 5 39—–185 d 22021——4950

Ho

me

wo

rk 1 Write each of the following fractions as a recurring decimal.

a 4–7 b 85–––101 c 17––33

2 Write each of the following recurring decimals as a fraction in its simplest form.

a 0.5·4· b 0.2

·46

· c 0.2· d 0.1

·2· e 0.37

·

Answers1 a 0.5·71 428· b 0.8·415· c 0.5·1·

2 a 6–11 b 82–––333 c 2–9 d 4––33 c 17––45

� recurringdecimal

� terminatingdecimal

Key Words

Extension Answers

1––13 = 0.0·76 923· , 2––13 = 0.1·53 846· , 3––13 = 0.2·30 769· , 4––13 = 0.3·07 692· , 5––13 = 0.3·84 615· , 6––13 = 0.4·61 538· , 7––13 = 0.5·38 461· , 8––13 = 0.6·15 384· , 9––13 = 0.6·92 307· , 10––13 = 0.7·69 230· , 11––13 = 0.8·46 153· , 12––13 = 0.9·23 076· , 13––13 = 1

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Oral and mental starter� Due to the variety of calculators in use and the problems this may cause, there is

no Oral and mental starter. However, some of the main keys listed in theFramework objectives could be discussed and found for the different makes ofcalculator. Some students may need help to locate the inverse/shift keys, whichare needed on many calculators in order to do powers.

Main lesson activity� This is a review lesson on the effective use of calculators.� The class should have met most of the keys before.� Do a variety of problems using the appropriate keys. Make sure the students can

find these keys on their calculators.� Work out, for example, √

———π ÷ 5.32

—–. This requires the square root key, the π key

and the square key. The answer is 0.334 425 254 9 ≈ 0.33.� Discuss the need to round answers to a sensible degree of accuracy.� Now take 4.55. This requires the power key. The answer is 1845.281 25 ≈ 1845.� As the next example, work out:

(7–8 – 5––12) ÷ (11––15 + 7––12)This requires the fraction key and the brackets keys. The answer is 55–––158.

� Finally, work out:3.56(43.2 – 17.48)—————————(4.53 – 1.2) 0.92

This requires the brackets keys. The answer is 29.887 452 67 ≈ 29.89 ≈ 30.



LESSON

7.7

Framework objectives – Efficient use of a calculator

Use a calculator efficiently and appropriately to perform complex calculations withnumbers of any size, knowing not to round during intermediate steps of a calculation.Use the constant, π and sign change keys, the function keys for powers, roots andfractions, brackets and the memory.

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Plenary� This is the last time that calculators will be the focus of the lesson, so each

student needs to be sure that he/she can use his/her particular model. Ensure that they can identify the following keys: brackets, memory (M in, M+, M–), π,square root, square, power, fraction.

� This leaves only three principal keys that will be needed in future: sine, cosineand tangent.


Exercise 7G Answers

1 a 153.9 b 0.627 c 8202 a 13.69 b 32 768 c 1.95 d 0.000 002 23 a 3 b 8 c 11 d 1.54 a 2 b 4 c 5 d 0.65 a 15.8 b 70.4 c 512 d 104.22 e 7.876 a 138––99 b 1–––210 c 413––20

� fraction� brackets� square root� square� power� pi� memory keys

Key Words

Extension Answers

0.618

Ho

me

wo

rk 1 Use a calculator to evaluate each of these.

63.4 × 21.02 19 7a [2.42 + (6.7 – 1.04)]2 b ——————— c —– – —–2.9(4.5 – 1.72) 21 18

2 Use the power key to evaluate each of these.

a 2.75 b 42.8752–3

Answers1 a 130.4164 b 165.3 c 65–––1262 a 143.5 b 12.25

SATs Answers

1 Using film of 24 photos costs £56.10. Using film of 36 photos costs £61.40. For 360photos, 24 photo films are £5.30 cheaper

2 a 0.636 62 b 0.528 683 a = 1500, b = 204 13 403.076 92 gallons, 13 000 gallons5 a 1.01 × 105 N/m2 b 16.66:1 c 5.79 × 1010 km3

6 a 4 × 10–4 b 4 × 10–5 c 4.4 × 10–4

7 a ii Time, in hours, taken by spaceship to travel to Proxima Centauriiii time in years, taken by spaceship to travel to Proxima Centauri

b 11 years

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Oral and mental starter� Ask the class approximately how many litres there are to a gallon. The correct answer is about four and a half.� Then ask how many litres there are in 8 gallons. The answer is 36.� Ask them how they arrived at this answer and discuss the various methods used.� One way is to multiply the 4 by 8, getting 32, then to find the result of 8 halves, which is 4. Next, add the two

together, giving 32 + 4 = 36.� Now ask how many litres there are in: 12 gallons (54 litres); 22 gallons (99 litres); 17 gallons (761–2 litres).� Now ask approximately how many gallons are in 60 litres.� Discuss the strategy. For an approximation, you would not try to divide by 4.5. Rather, you would divide by

4 and also by 5 and go for a midway answer. Here this gives 60 ÷ 4 = 15 and 60 ÷ 5 = 12. Approximatelyhalfway between 15 and 12 is 13.5, which is rounded up to 14.

� Ask the class how many gallons there are in: 100 litres (22 gallons); 70 litres (16 gallons); 40 litres (9 gallons);134 litres (30 gallons).

Main lesson activity� Write on the board x2 + x3. Ask the class if they can simplify this expression. The expression cannot be

simplified because it contains two unlike terms.� Now, write on the board x2 × x3. Ask the class if they can simplify this expression. If someone comes up with

the correct answer of x5, ask him/her to explain how. � Have a discussion with the class about multiplying expressions containing powers and adding the powers. Ask

if anyone can generalise the rule. You are looking for something like:xa × xb = xa + b

� Ask if anyone can prove this rule. You want something like:k2 × k3 = (k × k) × (k × k × k) = k × k × k × k × k = k5 = k2 + 3

� Now, write on the board 4m2 × 3m. Ask the class to simplify this expression. If someone comes up with thecorrect answer of 12m3, ask him/her to explain how, i.e., that they multiplied the coefficients and added thepowers. If not, remind the class of this and that m is the same as m1 and work through some more examples.

� Next, move on to dividing powers. Write on the board the expression 8m5 ÷ 2m3 and ask if anyone can tell youwhat the answer should be. Again, if someone comes up with the correct answer of 4m3, ask him/her to explainhow, i.e., that they divided the coefficients and subtracted the powers. If not, work through this with the classand introduce some further examples.

� You should explore the proof of this with an example such as:k × k × k × k × k × k

k6 ÷ k2 = –––––––––––––––––– = k × k × k × k = k4 = k6 – 2

k × k� Now write on the board the expression m3 ÷ m5 and ask the students to simplify this expression. You should get

the response m–2. Make sure that the students understand that numbers and variables with negative indicesfollow the same rules as numbers and variables with positive indices.

� Ask the students what they think m–2 represents. If the class covered standard form recently, they may bring thisknowledge to this discussion.

� Introduce the investigation on page 139 of Pupil Book 9.3. Go through the first two parts with the class, then letthem complete the investigation individually or in pairs. Discuss the results with the whole class.

1� Finally, ensure that all of the students understand that n–n = — .

xn



Algebra 4CHAPTER

8

LESSON

8.1

Framework objectives – Index notation with algebra: negative powers

Know and use the index laws (including in generalised form) for multiplication anddivision of positive integer powers; begin to extend understanding of index notationto negative and fractional powers, recognising that the index laws can be appliedto these as well.

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Plenary� Discuss with the class why powers are used – they shorten very large numbers

and also make very small numbers accessible.� Remind the class of the large number 10100, called the googol. Tell them that this

was the answer to the million pound question on the television programme WhoWants to be a Millionaire, when Major Charles Ingram was suspected ofcheating.


Exercise 8A Answers

1 a 4x5 b 7m4 c n2 d x5 e 4m4 f 6x4 g 15t4 h 5m3 i g9 j m4

k 4t2 l 15m8 m 48q6 n abm5 o cdy4

1 1 1 1 5 4 8 a2 a — b — c — d — e — f — g — h –m k2 x3 n4 m2 y x3 b

3 a 5–1 b 4–1 c 3x–1 d x–2 e m–6 f 9–1 g 5x–4 h Am–3

4 a 4x–2 b 7m–2 c n–3 d 4x–3 e 4m–5 f 4x–2 g 15t–2 h 5m–2 i g–1

5 a x7 b m7 c n2 d 2x8 e 3m–3 f 3x3 g 12t–7 h 6m5 i 6g–5

6 a 3x2 b 30m c 6n d 12x10 e 6m–2 f 24x–2 g 60t3 h 8m–1 i 9gj 4m–1 k 30t7 l 24m7 m 6q12 n kpm6 p dey8

Extension Answers

1 a Any value larger than 1 or smaller than 0 b Values between 0 and 1c x = 0 and x = 1

2 a Any value larger than 5 or less than 0 b Values between 0 and 5c x = 0 and x = 5

Ho

me

wo

rk 1 Expand the following, and find their value (use a calculator if necessary).

a 26 b 35 c 64 d 45 e 172 f 143 g 272 h 114

2 Write down the following in index form:

a t × t × t × t b t × t × t × t × t c m × m d q × q × q

3 a Write m + m + m + m + m + m as briefly as possible.

b Write t × t × t × t × t × t as briefly as possible.

c Show the difference between 6m and m6.

d Show the difference between t4 and 4t.

4 Simplify each of the following:

a 2x3 × 4x7 b 12t6 ÷ 3t c 20m5 ÷ 5m3 d 3y × 2y5 e x–2 ÷ x–3

5 Simplify each of the following, leaving your answer in fraction form:

a x3 ÷ x5 b 4m2 ÷ m5 c 8x–4 ÷ 2x d 2x5 ÷ 3x8 e Ax × Bx–5 f Ax ÷ Bx–5

Answers1 a 64 b 243 c 1296 d 1024 e 289 f 2744 g 729 h 146412 a t4 b t5 c m2 d q3

3 a 6m b t6 c 6m = m + m + m + m + m + m, m6 = m × m × m × m × m × md t4 = t × t × t × t, 4t = t + t + t + t

4 a 8x10 b 4t5 c 4m2 d 6y6 e x1 4 4 2 AB Ax6

5 a — b —– c —– d —— e —– f ——x2 m3 x5 3x3 x4 B

� power� index

Key Words

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Oral and mental starter� Tell the class that approximation is a valuable skill. It enables a quick check to

be made on calculations – maybe even to ensure that they have used theircalculators correctly.

� For example, ask the class if they can estimate the answer to 41 672 divided by 8. Ask for quick approximations and write them up on the board. After a fewhave been given, discuss with the class any strategy which they used to estimatedivision by 8.

� One way is to halve the dividend, then to halve the result, and finally to halvethe second result to complete the division. Doing this with 41 672 gives theapproximate sequence:

41 672 → 20 000 → 10 000 → 5000� Finally, ask the class to estimate the following (all divisions by 8):

629 537 → 300 000 → 150 000 → 80 000173 492 → 90 000 → 45 000 → 22 00023 774 → 12 000 → 6000 → 3000117 008 → 60 000 → 30 000 → 15 000

Main lesson activity� Ask the class for the square root of 25.� If only 5 is given, remind the class that –5 is also a square root of 25.� Now ask the class to estimate the square root of 20. Put their suggestions on the

board.� Discuss with them any answers that are obviously wrong, and how they know

they are wrong. For example, the square root of 20 must be larger than 4 as 4 × 4 = 16, which is less than 20. Similarly, the square root of 20 must be lessthan 5.

� Hence, the square root of 20 must lie between 4 and 5. It looks halfway, so agood estimate would be 4.5.

� Ask the class to work out the correct answer using their calculators. You can usethis as an opportunity to show or remind them how to find square roots on theircalculators.

� The calculator will show the square root of 20 to be 4.472 135 955. This roundsto 4.5 – so, the approximation was correct. Ensure that the class know thecorrect symbol, √

–, for square root.

� Ask the class for the cube root of 8. You will need to explain that the cube root isthe number that gives 8 when multiplied by itself three times. It is written as3√

––8 . In this example, the answer is 2.

� Ask the class why they think that there are no negative answers to cube roots. � Write on the board a few of the better-known cube roots: the cube root of 27 is

3; the cube root of 125 is 5, and the cube root of 1000 is 10. Discuss why thenumbers for which the roots are given have grown so quickly.

� Introduce the investigation on page 141 of Pupil Book 3. Go through the firstexample with the class, then let them complete the investigation.

� Finally, discuss with the class the generation xn = n√–x.



LESSON

8.2

Framework objectives – Square roots, cube roots and otherfractional powers

Estimate square roots and cube roots.

Begin to extend understanding of index notation to negative and fractional powers,recognising that the index laws can be applied to these as well.

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Plenary� Ask the class which is bigger, the square root of ten thousand or the cube root of

a million. The answer is they are both the same, 100.� Go through the better-known square roots and cube roots as a quick

competition, seeing which set of students (row, column, table) remembers themost. Use the list of roots in Pupil Book 3, page 141, but ensure that all studentshave their textbooks shut.


Exercise 8B Answers

1 a x = 3, x = –3 b x = 6, x = –6 c x = 7, x = –7 d x = 11, x = –112 a x2 = 50 ÷ 2 = 25, hence x = 5 and x = –5 b x2 = 36 ÷ 4 = 9, hence x = 3 and x = –33 a 4 b 5 c 9 d 10 e 2 f 10 g –4 h –5 i 5 j 3 k 6 l 2 m 3 n 5 p 2 q 94 a 5.1 b 7.4 c 9.7 d 10.4 e 16.6 5 a 2 b 1 c 5 d 3 e 10 f –4 g –1 h –10 i 0.1 j 0.26 b i 4.4 ii 4.6 iii 3.6 iv 6.4 v 12.6 7 a 3√

—–50 b √

—–30 c √

—–20 d √

—–35 e √

—–15 f √

—–40

8 b i 3.6 ii 1.7 iii 7.9 iv 6.3 v 9.1 c i 4.0 ii 1.9 iii 7.4 iv 6.8 v 9.59 a 33 b 19 c 42 d 34 e 44

Ho

me

wo

rk 1 Estimate the square root of each of the following. Then use a calculator find the result to onedecimal place and see how close you were.

a √—–46 b √

—–31 c √

—–74 d √

——129 e √

——215

2 Without a calculator, state the cube roots of each of the following numbers.

a 64 b 343 c 216 d 729 e 512

3 a Estimate the integer closest to the cube root of each of the following.

i 96 ii 110 iii 55 iv 297 v 3000

b Use a calculator to find the accurate value of the above. Give your answers to one decimalplace.

4 State which, in each pair of numbers, is the larger.

a √—–20, 3√

—–55 b √

—–28, 3√

——149 c √

—–18, 3√

—–79

5 Estimate the cube root of each of these numbers without a calculator.

a 15 b 61 c 400 d 150 e 850

6 Try to estimate the cube root of each of these numbers without using a calculator.

a 25 000 b 8000 c 57 000 d 41 000 e 83 000

7 Write down the value of each of the following without using an index.

a 491–2 b 512

1–3 c 161–4 d 1024

1–5 e (–343)1–3

Answers1 a 6.8 b 5.6 c 8.6 d 11.4 e 14.7 2 a 4 b 7 c 6 d 9 e 83 i 4.6 ii 4.8 iii 3.8 iv 6.7 v 14.4 4 a √

—–20 b 3√

——149 c 3√

—–79

5 a 2.5 b 3.9 c 7.4 d 5.3 e 9.5 6 a 29 b 20 c 38 d 34 e 447 a 7 b 8 c 2 d 4 e –7

Extension Answers

a Only for A = B = 0b Always truec Never trued Always true

� square root� cube root

Key Words

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Oral and mental starter� Ask if anyone has been to France or Germany recently. If someone has, ask how

many euros he/she got for a pound.� There are approximately 1.50 euros to the pound. Hence, an easy way to make

a comparison while shopping in Europe is to divide the price in euros by 3, thendouble the result to get the approximate price in pounds.

� Go through a few examples, such as:� 25 euros: Dividing by 3 gives about 8, doubling this gives 16. So 25 euros

is about £16.37 euros: Dividing by 3 gives about 12, doubling this gives 24. So 37 euros

is about £24.� Now ask the students to try this themselves, to convert the following euro prices

to the approximate equivalent in pounds: 9 euros (£6), 15 euros (£10), 22 euros(£14), 29 euros (£20), 49 euros (£32).

Main lesson activity� Put on the board y = mx + c, and ask the class what type of equation this is.� You want the answer ‘linear’. Then ask them what is special about the graph

drawn from an equation like this.� You want the answer: ‘It’s a straight line’. If they do not also mention that m is

the gradient and c is the y-axis intercept, try to tease this from them.� Then put on the board the equation y = ax2 + bx + c, and ask if anyone knows

what type of equation this is.� You want the answer ‘quadratic’. This may not be known, so explain the term,

and that it is used because the highest power is a square.� Tell the class that, today, they are going to look at the graphs of equations like

this. Put the equation y = x2 – 4x + 3 on the board. Then tell them that you aregoing to go through the drawing of this graph with them.

� Start by putting a table on the board, as below.

x –1 0 1 2 3 4

x2 1 0 1 4 9 16

–4x 4 0 –4 –8 –12 –16

3 3 3 3 3 3 3

y = x2 + 2x + 3 8 3 0 –1 0 3

� Complete this table with help from the class. Plot the coordinates given by thetable on a pair of axes drawn on the board. Plot the coordinates to show asmooth curve.

� Explain that the U-shaped curve is typical for a graph from a quadratic equation,but do tell them that when the x2 part is negative, the graph will be an invertedU-shape.

� Ensure the class realise that quadratic graphs will always be curved, with nostraight lines and no kinks.



LESSON

8.3

Framework objectives – Quadratic graphs

Plot graphs of simple quadratic functions.

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Plenary� Talk about the characteristic U-shape of a quadratic graph, and inverted U-shape

when the x2 coefficient is negative.� Compare the difference between a quadratic graph and a linear graph. In

particular, the fact that in both cases, for each value of x, there is only one valueof y. However, when given a value of y, there is only one possible value for x ina linear graph, but in a quadratic graph there could be no value for x, one valuefor x or two values for x. You will need to point out where this happens on agraph.

� Point out to the class that this will be useful to them when they come to lookingat solutions of quadratic equations.


Exercise 8C Answers

1 a 9, 4, 1, 0, 1, 4, 9 b 13, 8, 5, 4, 5, 8, 13 c 12, 6, 2, 0, 0, 2, 6d 3, –1, –3, –3, –1, 3, 9

2 a

b

c

d

3 b All the curves go through the origin, but the larger the value of a, (coefficient ofx2), the wider the graph becomes

4 b They are all parallel curves, with c being the y-axis intercept

x –3 –2 –1 0 1 2 3

y 4 0 –2 –2 0 4 10

x –4 –3 –2 –1 0 1 2

y 5 0 –3 –4 –3 0 5

x –4 –3 –2 –1 0 1 2

y 8 3 0 –1 0 3 8

x –3 –2 –1 0 1 2 3

y 18 8 2 0 2 8 18

Ho

me

wo

rk 1 A sledge sliding down a slope has travelled a distance, d metres, in time, t seconds, where d = 5t + t2.

a Draw a graph to show the distances covered up to 6 seconds.

b Find the distance travelled after 3.8 seconds.

c Find the time taken to travel 50 metres.

2 The cost, C pence, for plating knives of length L cm is given by the formula C = 50L + 7L2.

a Draw a graph to show the cost of plating knives up to 10 cm long.

b What would be the cost of plating a knife 8.7 cm long?

c What would be the length of a knife costing £4 to plate?

Answers1 b 33.4 m c 5 s 2 b £9.65 c 4.8 cm

� quadratic� intercept� gradient

Key Words

Extension Answers

1 b The graph is an inverted U shape2 b y = –x2

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Oral and mental starter� Ask the class whether anyone knows the square root of 400.� After some guesses, someone may come up with the correct answer of 20. Ask

him/her how they worked it out.� The quick way to get the solution is to recognise that 400 = 4 × 100. The square

root of each of these numbers is known, which gives the answer 2 × 10 = 20.� Ask for the square root of 900. The correct answer of 30 should now be

forthcoming, but may still create a discussion before everyone sees thereasoning.

� Continue in this way, asking for square roots of: 1600 → 40; 4900 → 70; 12 100 → 110; 810 000 → 900; 90 000 → 300.

� Now ask if anyone can estimate the square root of 377.� Let the question provoke discussion in the class with different suggestions being

put on the board. If the correct answer of 19 is offered, ask how the answer wasworked out.

� Explain that the number can be broken down into 3.77 × 100. The square root of100 is 10, and the square root of 3.77 can be estimated at between 1 and 2, butquite close to 2. The approximation 1.9 is quite reasonable, giving the root as1.9 × 10 = 19.

� Repeat with other examples, such as: 805 → 28; 548 → 23; 971 → 31; 457 → 21.

Main lesson activity� Put on the board y = mx + c. Ask the class what type of equation this is and what

is special about its graph. They should remember that it is linear and has astraight line graph.

� Then put on the board the equation y = ax2 + bx + c. Ask if anyone knows whattype of equation this is and what is special about its graph.

� You want the answer ‘quadratic’, and that the graph has a U-shape or aninverted U-shape.

� Then put on the board the equation y = x3 + 4x2, and ask the class if they knowwhat type of equation this is.

� Explain that it is called a cubic equation because the highest power in theequation is a cube.

� Tell the class you are going to go through the drawing of this graph with them.� Start by putting a table on the board, as below.

x –5 –4 –3 –2 –1 0 1 2

x3 –125 –64 –27 –8 –1 0 1 8

4x2 100 64 36 16 4 0 4 16

y = x3 + 4x2 –25 0 9 8 3 0 5 24� Complete this table with help from the class. Plot the coordinates given by the

table on a pair of axes drawn on the board. Join up the coordinates to show asmooth curve with two turning points.

� Explain that this is a typical shape for a graph from a cubic equation. However,sometimes the two turning points coincide, giving the graph a single twist. Oneexample of such a graph is given by y = x3 + 3x2 + 3x +1 (see Pupil Book page146). Also tell them that if the x3 part is negative, the graph will start in thesecond quadrant (top left-hand corner) instead of in the third quadrant (bottomleft-hand corner.)

� Ensure the class realise that this type of graph will always be curved, havingeither two bends or a single twist.



LESSON

8.4

Framework objectives – Cubic graphs

Plot graphs of simple cubic functions.

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Plenary� Talk about the shape of a cubic graph always having two turning points even

though they may coincide and act as only one.� Start a discussion about how many turning points they might expect in the graph

of y = Ax4 + Bx3 + Cx.� Following the pattern linear, no turning points; quadratic, one turning point;

cubic, two turning points; an equation whose highest power was four wouldprobably have three turning points.

� Someone might like to find out!


Ho

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rk 1 By drawing suitable graphs, solve this pair of simultaneous equations:

2x + y = 5 y = x3 – 1There is only one solution.

2 The distance, d metres, a rocket is above the ground is given by

d = 2t + t3

where t is the time in seconds.

Draw the distance–time graph for the first 3 seconds.

Answers1 x = 1.45, y = 2.12 b t 0 1 2 3

d 0 3 12 33

SATs Answers

21 k = 3, m = 6 2 a 2n b — c n2n

3 a A wider U-shape. Goes through (0,0) and the y-axis is a line of symmetry.b y = –x2 c y = x2 + 1 d y > x2 and y < 2

4 a –0.04 b 1.6 m

Extension Answers

Exercise 8D Answers

1 Values of y: –8, –1, 0, 1, 8 2 Values of y: 0, 10, 8, 0, –8, –10, 03 Values of y: –3, 3, 3.375, 3, 2.625, 3, 9 4 Values of y: –32, –10, –4, –8, –1, 225 x = 1.2 y = 2.7 6 t 0 1 2 3

v 0 2 9 28

� cubic� turning points

Key Words

a b cy yy

xx x

y = 1–x y = 1––x2 y = 1––x3

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Oral and mental starter� Use a counting stick as a probability scale.

� Explain to the class that the stick is a probability scale. Tell them that one end is zero. Ask them for the value ofthe other end.

� Now point at the centre of the stick and ask them to tell you an event which has a probability of 0.5. (For example, obtaining a Head on a fair coin.)

� Now say to the class that you will point out a probability of something that happens and they will have to showyou the probability that it does not happen. For example, you point to 0.1 on the stick, the class would point to0.9. Repeat this quickly for different values.

� Finish by asking why ‘Event happens’ and ‘Event does not happen’ is not always at 0.5 (as there are twochoices). Ask the class to give you examples to explain the reason, such as dice P(6) = 1–6, P(not 6) = 5–6.

Main lesson activity� Remind the class that they should be able to find the probability of a range of situations, all of which they have

already met. Draw their attention to the tabular summaries in Pupil Book 3.� Move on to a brief discussion of independent events, giving them this definition:

Two events are said to be independent when the outcome of one of them does not affect the outcome ofthe other event.

� Tell the class that most of the combined events they have dealt with so far have been independent. Give as anexample a dice being rolled at the same time as a coin is tossed. Ask them whether they think that the score onthe dice may affect how the coin lands – Heads or Tails.

� Two other aims of this lesson are to give students an understanding of when statements can be misleading eventhough they have apparently been deduced logically, and to increase their familiarity with the vocabulary ofprobability.

� Write on the board a = b and b = c. Ask the class to give you an equation connecting a and c. They shouldrespond a = c.

� Now replace the statements with the following sentence. Amy likes Bob and Bob likes chips.

Ask the class to give you a statement connecting Amy and chips! Hopefully, the class will realise that the logichere does not necessarily work.

� Explain that sometimes statements may seem logical but are not necessarily true.� Tell the class that you have five names on five separate pieces of paper, four boys and one girl. Ask them which

statement is true when one piece of paper is picked at random.�� There is an even chance of picking the girl because there are both boys and girls (meaning there are two

choices).�� There is no chance of picking the girl because she is outnumbered.�� The probability of picking the girl is one fifth because there is a one in five chance.

� Explain to the class that these are the sort of statements they will have to consider. Their answers will require anexplanation, not just responses such as ‘It is incorrect’. Their explanations could include what the truestatements should be.


0 1



9

LESSON

9.1

Framework objectives – Probability statements

Use the vocabulary of probability in interpreting results involving uncertainty andprediction.

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Plenary� Ask the class to think of different sentences which have some logic to them but

which are incorrect. (You could give them a hint by asking for a sentence about,for example, eye colour or shoe size.)

� Prompt them with an example of your own, such as: ‘There is a 50% chance that Jon is wearing size 10 shoes because either heis size 10 or he isn’t.’

� Ask a student to explain why the statement is incorrect.


Exercise 9A Answers

1 a Probability of rolling 6 on a fair dice is 1–6. The fact that Ashad thinks he is unlucky does not affect his chance ofstarting the game

b If this were true, it would rain foreverc Probability of snow depends which part of the world you are in and at what time of yeard Only true if there were equal numbers of mint, chocolate and plain sweets

2 a Incorrect, it could happen todayb Correct, assuming it is a fair coinc Incorrect, it is possible that the bus could be late tomorrowd Incorrect, there is an equal chance of picking out red or blue

3 a Not possible to know, as number of winning squares not given for each caseb Grid 1c Proportion of winning squares is Grid 1 1–4, Grid 2 1–3, Grid 3 1–2. So, greatest chance of winning is using Grid 3

4 a Not independent because when Jonathan writes computer programs he is unlikely to be watching TV at the sametime

b Independent c Independent

Ho

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rk 1 Write down a reason why each of these statements is incorrect. a A bag contains black and white cubes, so there is a 50% chance of picking a black cube.b A bag contains black and white cubes. Last time I picked out a black cube, so this time I will pick

out a white cube.c A bag contains one black cube and many white cubes. So, I have no chance of picking out the

black cube.

2 Here are three different bags of cubes.A There are four black cubes and four white cubes in the bag.B There are two black cubes and five white cubes in the bag.C There are seven black cubes and five white cubes in the bag.

Here are three statements about the bags of cubes.X There is a probability of 2–5 that I will pick a black cube.Y There is an even chance that I will pick a black cube.Z There is a probability of 5––12 that I will pick a white cube.

For each bag, say whether the statements are correct or incorrect.

Answers1 a This would only be correct if there were an equal number of black and white cubes

b Provided there are still some black cubes in the bag, there is a chance that black might be picked outc As in part b, there is a chance. It would only be impossible if the black cube had been taken from the bag

2 I Incorrect, C Correct

� certain� even chance� probability� probable

Key Words

Extension Answers

5 by 5 grid with 7 winning squares has a probability of 7––25 (0.28) to produce a winning square6 by 6 grid with 10 winning squares has a probability of 10––36 (0.27· ) to produce a winning square10 by 10 grid with 27 winning squares has a probability of 27–––100 (0.27) to produce a winning squareGreatest chance of finding a winning square is on 5 by 5 grid

A B CX I I IY C I IZ I I C

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Oral and mental starter� Use a target board as shown on the right.� Ask the class to pick out fractions and decimals which add up to one.

Encourage them to give additions which involve more than twofractions/decimals: for example, 0.5 + 0.125 + 3–8.

� Ask them how they worked it out when they mixed fractions withdecimals.

� Now give them one value and ask them to make it add up to 1, with two otherfractions or decimals. They may use either the target board values or their own.For example: you say 0.2 and they give 0.5 + 0.3.

Main lesson activity� Give the class the example of a bag which contains three blue, two yellow and

five green balls, from which only one ball is allowed to be picked at random.The probabilities for picking each colour are:

P(blue) = 3––10

P(yellow) = 2––10 = 1–5P(green) = 5––10 = 1–2

The events ‘picking a blue ball’, ‘picking a yellow ball’ and ‘picking a green ball’can never happen at the same time, given that only one ball is allowed to betaken out. Such events are call mutually exclusive, because they do not overlap.

� Using the same example, deal with the probability of an event which will nothappen. Thus, the probability of not picking out a blue ball is give by:

P(not blue) = 7––10

because there are seven outcomes which are not blue balls. Then note thatP(blue) + P(not blue) = 3––10 + 7––10 = 1

So, knowing P(Event happening), then:P(Event not happening) = 1 – P(Event happening)

� Now ask the class to sum the probabilities of picking the three coloured balls.Hopefully, they will get the correct answer of 1. Then tell them that becausethere are no other possibilities, they are called exhaustive events.

� Tell them that the events are also mutually exclusive and that when events areboth exhaustive and mutually exclusive, their probabilities always add up to 1.



LESSON

9.2

Framework objectives – Mutually exclusive events andexhaustive events

Identify all the mutually exclusive outcomes of an experiment. Know that the sumof probabilities of all mutually exclusive outcomes is 1 and use this when solvingproblems.

Use efficient methods to add, subtract, multiply and divide fractions.

1–2 0.25 1–4 0.85 1–8

0.15 3–4 0.125 5–8 0.6257–8 0.1 3–8 0.5 0.75

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Plenary� Tell the class that in a box you have a set of mathematical instruments and other equipment: for example, pens,

pencils, pairs of compasses, protractors, and small and large rulers.� You say that in one hand you have, for example, pencils, and in the other hand you have rulers. The class have

to tell you whether the content of your hands is exhaustive, mutually exclusive or neither. � Repeat this with different combinations.


Exercise 9B Answers

1 a 0.4 b 0.3 c 0.952 a 2–5 b 16––25 c 9––25 d 3–53 a 0.55 b 0.45 c 0.75 d 0.254 a Mutually exclusive b Mutually exclusive c Both d Neither

Ho

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rk 1 Ten pictures are shown, which are all face down. A picture is picked at random.

a What is the probability of choosing a picture of a guitar?

b What is the probability of choosing a picture of a guitar or a boat?

c What is the probability of choosing a picture of a horse or a doll?

d What is the probability of choosing a picture which is not of a boat?

2 A bag contains a large number of discs, each labelled either A, B, C or D. The probabilities that a disc picked at random will have a given letter are shown below.

P(A) = 0.2 P(B) = 0.4 P(C) = 0.15 P(D) = ?

a What is the probability of choosing a disc with a letter D on it?

b What is the probability of choosing a disc with a letter A or B on it?

c What is the probability of choosing a disc which does not have the letter C on it?

Answers1 a 3––10 b 7––10 c 3––10 d 3–52 a 0.25 b 0.6 c 0.85

� mutuallyexclusive

� exhaustive� probability� expectation

Key Words

Extension Answers

5, 20, 40, 24, 60, 90

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Oral and mental starter� Write some fractions on the board: for example 2–6, 5––10 and 6–8.� Ask the class to tell you each fraction in its simplest form, or ask them to cancel

each fraction down to its simplest form.� Then ask them how they did it.� Now ask them to give you equivalent fractions to, for example, 1–2 , 1–5 and 2–7.� Next, invite the class to give you the answer to 1–2 × 2–3. You could remind them that

this is the same as 1–2 of 2–3. Ask them how they worked it out. Hopefully, theiranswers will cover different methods, such as: knowing that 1–2 of 2–3 is 1–3;multiplying the numerators together and then the denominators to obtain 2–6 which is cancelled down to 1–3; and cancelling the 2s first to obtain 1–1 × 1–3 andthen 1–3.

� Now write 2–9 × 3–4 on the board. Tell the class that the answer they give must be insimplest form.

� This activity can be repeated for different fractions. Hopefully, the class willrecognise that it is easier and more efficient to simplify the fractions before doingthe multiplication.

Main lesson activity� Explain that the aim of this lesson is to look at probability questions: in

particular, to solve problems which involve two or more events, using differentmethods. The class will also be working with fractions and may wish to refer tochapter 2 to remind themselves of the various methods for simplifying fractions.

� Start by explaining the meaning of the word ‘independent’. You could ask theclass to write down a definition. For example: ‘Two events are independent if theoutcome of one is not affected by the outcome of the other’.

� Draw a blank sample space diagram on the board for two events, such as rollinga fair dice and throwing a fair coin.

Dice1 2 3 4 5 6

CoinHeadTail

� Point out that there are twelve different outcomes so that the probability of, forexample, a Head and a 6 is 1––12.

� Now draw a tree diagram on the board, as shown on the right. Coin Dice� Explain that it is important for the diagram to have all the labels,

Coin, Dice, Head, Tail, etc. and all the probabilities, completed. The class should help to complete the details.

� Now explain that when working across the branches the probability of getting a Head and a 6 is 1–2 × 1–6 = 1––12, as before.

� Ask the class to tell you the other combinations. Then explain that this information could also be put into a table, as shown below.

DiceP(6) = 1–6 P(Not a 6) = 5–6

CoinP(Head) = 1–2

1–2 × 1–6 = 1––121–2 × 5–6 = 5––12

P(Tail) = 1–21–2 × 1–6 = 1––12

1–2 × 5–6 = 5––12

� Explain to the class that they are all valid methods. Tell them that unless they aretold what to do, they may use whichever method they prefer.



LESSON

9.3

Framework objectives – Combining probabilities and treediagrams

Identify all mutually exclusive outcomes of an experiment.

Use efficient methods to add, subtract, multiply and divide fractions. Cancelcommon factors before multiplying or dividing.

Head

Tail

1–2

1–2

6

Not a 6

1–6

6

Not a 65–6

1–6

5–6

9B3LP_09.qxd 18/09/2003 15:20 Page 106

Plenary� Explain to the class that they have been looking in Exercise 9C at two events. Then

show them the tree diagram in the extension question which takes in three events.� Ask the class how many different combinations there are in this example. Point

out that 23 = 8. Then invite them to tell you how many combinations they wouldexpect for four events (16).

� Explain that they can either work out the probabilities for every possiblecombination or just the combinations required to answer the question.


Exercise 9C Answers

1 First dice Second dice 2 a 3–4 b 1–3 c 1–6

1–3 × 1–2 = 1–6

5 or 6

Not 5 or 6

1–3

2–3

Even

Not even

1–2

Even

Not even1–2

1–2

1–2

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rk 1 A builder is working on a patio. The probability that the weather is fine is 0.6, and the probabilitythat he has all the materials is 0.9. To complete the job in a day, he needs the weather to be fineand to have all the materials.a Draw a tree diagram to show all the possibilities.b Calculate the probability that he completes the job in a day.c Calculate the probability that it is not fine and he does not have all the materials.

2 A game is played three times. The probability of winning each time is 1–2.a Show that the probability of winning all three games is 1–8.b What is the probability of winning exactly one game?

Answers1 a Weather Materials

b 0.54 c 0.04 2 a 1–2 × 1–2 × 1–2 = 1–8 b 3–8

Fine0.6

0.9

0.1

0.9

0.1

0.4 Not fine

Has all materialsDoes not have all materials

Has all materialsDoes not have all materials

Extension Answers

a First game Second game Third game b WWL, WLW, LWW, WWW c 4––27 + 4––27 + 4––27 + 8––27 = 20––27

Win

Lose1–3

2–3

Win

Lose1–3

2–3

Win

Lose1–3

2–3

Win

Lose1–3

2–3

Win

Lose1–3

2–3

Win

Lose1–3

2–3

Win

Lose1–3

2–3

� tree diagram� mutually

exclusive� independent� two-way table

Key Words

3 aSecond bag

Red 1–5 Blue 4–5First bag Red 1–2 1––10

2–5Blue 1–2 1––10

2–5

b 1––10c 1––10 + 2–5 = 1–2

9B3LP_09.qxd 18/09/2003 15:20 Page 107


LESSON

2.2

Oral and mental starter� Tell the class that in a test someone scored 7 out of 10. Ask a student to give you the score as a fraction of 10.

Ask another student to give you this score as a decimal. � Repeat this for other simple scores, leading to, for example, 9––10, 4––10, 24–––100 and 32––50.� Gradually increase the level of difficulty up to scores which will need rounding: for example, 20 out of 30.

Encourage the students to round to two decimal places. � Finally, ask them to work out mentally simple fractions of quantities: for example, 7––10 of 100, 2–5 of 20.

Main lesson activity� The aim is to recall the work on experimental probability, to introduce the term relative frequency and to move

on to expectation.� Ask the class to consider a coin which is tossed ten times and lands on Heads nine times. Ask them whether the

coin is biased. They will probably say that it is. � Now suggest that when the coin is tossed a further ten times, it landed on Heads only once. Ask the class

whether they still think that it is biased. They will probably say that you cannot tell. � Ask the class how they could improve the experiment. Hopefully, they will tell you to carry out more trials.� Tell them that relative frequency is about carrying out repeated trials and obtaining estimates of probability

from experimental data.� Write down on the board the formula:

Number of successful trialsRelative frequency = —————————————

Total number of trials� Emphasise that the greater the number of trials, the closer the estimates of probability get to the theoretical

probability.� Draw a graph showing relative frequency against number of trials on the board and show the class how to

record the results. Use the coin example, plotting at 10 trials, 20 trials and so on. Point out to the class that arelative frequency graph will usually approach a value and that this value is the best estimate of the probability.

� Ask the class why it is not valid to use the graph to read off at say, 15 trials. Hopefully they will realise that thatdata between 10 and 20 trials may not follow the steady trend line.

� Now say to the class that an experiment has been carried out many times and you believe that the estimate isreliable. For example, the coin landed on Heads 70 times out of 100, so the relative frequency is 70–––100. Ask themhow many times they would expect this coin to land on Heads were it tossed 200 times. Stress that you areusing the estimate of probability and not the theoretical probability.

� Then tell them that the expected number of successes can be calculated using the formula:Expected number of successes = Relative frequency × Number of trials


9.4

Framework objectives – Estimates of probability

Estimate probabilities from experimental data.

Understand relative frequency as an estimate of probability and use this to compareoutcomes of experiments.

Exercise 9D Answers

1 a 32–––100 b Yes, it is probably biased: more 4s than other scores c 160 (32 × 5)2 a b 0.64 c 0.64 × 200 = 128

3 a b 0.28 c 0.28 × 75 = 21

4 a Relative frequency of Heads 7––10 = 0.7 12––20 = 0.6 18––30 = 0.6 22––40 = 0.55 28––50 = 0.56

Number of trials 10 25 50 100Number of times blue cube chosen 3 8 15 28Relative frequency 0.3 0.32 0.3 0.28

Number on throws 10 20 30 40 50Number of times it lands point up 6 13 20 24 32Relative frequency of landing point up 0.6 0.65 0.67 0.6 0.64

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Plenary� Refer to the question in the extension work, which involves taking relative

frequency from a graph.� Explain to the class that relative frequency questions can be asked using

experimental data either collected in a table or presented as a graph. Point outthat the relative frequency plots can be read from a graph and that the number ofsuccessful trials can be calculated by working backwards.

� Discuss briefly why it is not possible to read values at intermediate points on agraph.


Ho

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rk A spinner has different coloured sectors. It is spun 100 times and the number of times it lands on blueis recorded at regular intervals. The results are shown in the table.

a Copy and complete the table.b What is the best estimate of the probability of landing on blue?c How many times would you expect the spinner to land on blue in 2000 spins?d If there are two sectors of the spinner coloured blue, how many sectors do you think there are

altogether? Explain your answer.

Answersa

b 0.26 c 0.26 × 2000 = 520d Given spinner is fair, 0.26 ≈ 1–4, so a quarter of sectors are blue. So, altogether eight sectors.

Relative frequency 0.3 0.25 0.25 0.275 0.26

Number of spins 20 40 60 80 100

Number of times lands on blue 6 10 15 22 26

Relative frequency 0.3

� relativefrequency

� estimate� probability� expectation� limit

Key Words

Extension Answers

a 6 b 0.65 c 65, assuming player performs as on graphd Relative frequency not plotted for 15 throws, so cannot find number of hits

SATs Answers

1 a 7––25 b 336 (accept 332 to 340)2 a This would give 2.5 blue counters

b 1 blue, 6 red, 9 green, 4 yellow3 a Sue, most throws b 171–––300 c 167, 125, 8

d Experiment, therefore subject to random variation,and theoretical results do not give whole-numberanswers in this case e 1––––1296

4 a 1–2 × 1–2 × 1–2 = 1–8 b WWL or WLW or LWW = 1–8 + 1–8 + 1–8 = 3–85 a Win, Lose, Lose b 45% c 0.6

d 0.6

0.4

Exercise 9D (Cont’d)

b

c 0.56 d 112e Probably, as best estimate is not 0.5

5 a

b Only know one out of first two games was won, notwhich one c 0.70

d Number of wins 1 3 4 6 7

00

0.2

0.4

0.6

0.8

1

2 4 6 8 10Number of games

Rel

ativ

e fr

eque

ncy

00

0.2

0.4

0.6

0.8

1

10 20 30Numbers of throws

40 50

Rel

ativ

e fr

eque

ncy

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Oral and mental starter� Imagine a cube which has an edge length of 2 cm. What is the volume of the

cube? (Answer: 8 cm3) � Now imagine the cube is twice as big. What is its volume now?

(Answer: 64 cm3) � Make sure that the students appreciate that ‘twice as big’ means that the edges of

the cube are multiplied by two.� This starter can be extended by asking the class to make the original cube three

times or four times larger and allowing them to work out their answers on paper.

Main lesson activity� Remind the class about enlargement, which was covered in Year 8.� It may be necessary to revise enlargement, using the ray

method, by drawing on the board the diagram shownon the right, or show a prepared OHT.

� State that the object triangle ABC has been enlarged bya scale factor of 2 about the centre of enlargement, O,to give the image triangle A�B�C�. The dashed lines arecalled the rays or guidelines for the enlargement. Theobject and image are on the same side of O. The scale factor is positive. This is called positive enlargement.

� Next, explain to the class that there is fractional enlargement.� Draw on the board the diagram shown on the right,

or show a prepared OHT.� Explain that each side of ∆A�B�C� is half the

length of the corresponding side of ∆ABC. Also that OA� = 1–2 of OA, OB� = 1–2 of OB and OC� = 1–2 of OC.

� That is, ∆ABC has been enlarged by a scale factor of 1–2 about the centre ofenlargement, O, to give the image ∆A�B�C�. The object and the image are on thesame side of O, with the image smaller than the object. The scale factor is afraction. This is called fractional enlargement.

� Point out that, under any enlargement, corresponding angles on the object andimage remain the same.

� Now demonstrate fractional enlargement on a grid. Explain that the grid may ormay not have coordinate axes x and y.

� Show the class how the image points can be easily found by counting grid unitsin the vertical and horizontal directions. This can be an alternative to drawingrays.

� When there are coordinate axes, the centre of enlargement is sometimes givenas the origin (0, 0). The coordinates of the image shape are then the coordinatesof the object shape multiplied by the fractional scale factor.




10

LESSON

10.1

Framework objectives – Fractional enlargements

Know that if two 2-D shapes are similar, corresponding angles are equal andcorresponding sides are in the same ratio.

Enlarge 2-D shapes, given a fractional scale factor; recognise the similarity of theresulting shapes; understand the implications of enlargement for area and volume.

B

C

A

O

B�

C�

A�

ImageObject

C

A

BO

C�

A�

B�Image Object

9B3LP_10.qxd 18/09/2003 15:21 Page 110

Plenary� Ask what the class knows about each of these when making enlargements.

�� The relationship between lengths on the image and lengths on the object.�� The relationship between angles on the image and angles on the object.�� Fractional scale factor (makes a shape smaller on the same side of the centre

of enlargement).


Exercise 10A Answers

3 a Vertices at (3, 5), (5, 3), (1, 3) b Vertices at (1, 3), (3, 3), (3, 2), (1, 2)c Vertices at (1, 3), (3, 2), (1, 2)

4 a 1–2 b (2, 2) c 16 cm2 and 4 cm2, 4:1 d 1–4

Ho

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rk 1 Draw copies of (or trace) each of the following 2 Copy the diagram below onto a coordinate shapes. Enlarge each one by the given scale grid and enlarge the triangle by scale factor 11–2factor about the centre of enlargement O. about the origin (0, 0).

a Scale factor 1–3

b Scale factor 1–2

Answers2 Vertices at (3, 3), (3, 6), (7.5, 3)

O

O

� object� image� centre of

enlargement� scale factor� fractional

enlargement� similar

Key Words

00

1 2 3 4 5 6 7 8

1

2

3

4

5

6

7

8

Extension Answers

2 a i 22 cm2 ii 6 cm3 b i 88 cm2, 4 ii 48 cm3, 8c i 198 cm2, 9 ii 162 cm3, 27 d i k2 ii k3

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Oral and mental starter� On the board draw the table below, or use a prepared OHT.

� Ask individual students to cross out the pairs of ratios that are identical.� The last student asked should then simplify the remaining ratio.

Main lesson activity� NB You may wish to use more than one lesson to cover the material in this lesson plan and the associated

pupil book.� Explain to the class that the next few lessons will be about an important branch of Mathematics called

Trigonometry which is used to calculate the lengths of sides and the size of angles in right-angled triangles. Itsmain use is in areas of engineering, navigation and surveying.

� Draw the right-angled triangle on the right on the board.� Explain the following terms for the right-angled triangle ABC:

�� The side opposite the right angle is always the longest side and is known as the hypotenuse (H)

�� The side opposite the angle in question (labelled �) is called theopposite side (O)

�� The side next to the angle in question is called the adjacent side (A)� Draw another triangle on the board with ∠A labelled �. Ask the students to point out the opposite and adjacent

sides.� Introduce the investigation on page 171 of Pupil Book 3. It will be useful if students can work in pairs or small

groups, as some discussion is helpful.� Summarise what the investigation has shown by explaining that this value is called the tangent of angle A, and

is shortened to tanA. � Tell the students that the value of the tangent of any angle is stored in their calculators. � Show the students how to find the tangent of angles using their calculators. Some students may need individual

help as different models have different key sequences. The pupil book shows some of the more commonsequences. Work through some examples with the students to ensure that they have all mastered using theircalculators for this.

Opposite� Using a diagram similar to the one shown here, explain that the formula tan� = ————–

Adjacentcan be used in conjunction with the ‘tan’ key to find the lengths of sides and the size ofangles in right-angled triangles.

� Explain that once they have found tan�, they will need to find the value of � and to do this they will need touse the inverse tan function on their calculators. Ensure that all of the students know how to do this and workthrough some examples, such as: find the value of � when tan� = 0.6.

� Show the class how to calculate � in the diagram below and explain that angles are usually given to 1 decimalplace.

OppositeUsing the formula: tan� = ————–

Adjacent4

= –– = 0.667 6

So, � = 33.7° (1dp)Make sure that the students know how to use their calculators to do this.

1:10 1:8 1:12 2:80 1:48:200 1:2 20:720 1:5 8:8001:20 10:300 5:50 1:36 5:251:40 20:80 1:30 100:800 10:500

40:80 4:48 1:100 10:200 1:25


LESSON

10.2

Framework objectives – Trigonometry: The tangent of an angle

Begin to use (sine, cosine and) tangent in right-angled triangles to solve problems intwo dimensions.

6 cm

4 cm

�

Hypotenuse (H)Opposite side (O)

Adjacent side (A)

�

A

HO

�

9B3LP_10.qxd 18/09/2003 15:21 Page 112

� Now show the class how to calculate x in the diagram below and explain thatlengths are usually given to three significant figures.

OppositeUsing the formula: tan� = ————–

Adjacentx

tan60° = ––4

Multiplying both sides by 4, gives: 4tan60° = xSo, x = 6.93 cm (3sf)Make sure that the students know how to use their calculators to do this.


Plenary� Explain that tan� is the ratio between the opposite side and the adjacent side in a

right-angled triangle.� Tell the students that next lesson they will be looking at two other trigonometric

ratios that will involve using the hypotenuse.


Exercise 10B Answers

1 a 0.532 b 1.732 c 1 d 0.158 e 0.759 f 11.6642 a 26.6° b 13.0° c 40.2° d 56.3° e 66.8° f 84.3°3 a 1.51 b 3.73 c 10.0 d 39.9 e 6.09 f 86.64 a 16.7° b 26.6° c 24.4° d 74.1° e 45.0° f 66.8°5 a 4.25 cm b 3.25 cm c 14.0 cm d 32.0 cm e 3.29 cm f 40.5 cm6 31.0° 7 10.7 cm

Extension Answers

1 a 6.93 cm b 17.0 cm c 2.77 cm d 28.4 cm2 a 56.2 m b 78.7°

Ho

me

wo

rk 1 Find the value of � for each of the following. Give your answers to 1 decimal place.

a tan� = 0.12 b tan� = 0.956 c tan� = 3.45

2 Find the value of each of the following. Give your answers to 3 significant figures.

a 5tan31° b 14tan51° c 23tan58.9°

3 Calculate the angle marked � in each of the following triangles. Give your answers to 1 decimal place.

4 Calculate the length of the side marked x in each of the following triangles. Give your answers to 3 significant figures.

Answers1 a 6.8° b 43.7° c 73.8° 2 a 3.00 b 17.3 c 38.13 a 21.8° b 29.1° c 67.4° 4 a 3.26 cm b 2.55 cm c 26.7 cm

7 cm25°

15°62°

9.5 cm

xx x

a b

14.2 cm

c

10 cm

4 cm�

18 cm

10 cm

�

a b

5 cm

12 cm

�

c

� adjacent� angle of

depression� angle of

elevation� hypotenuse� opposite� tangent (tan)

Key Words

4 cm60°

x

9B3LP_10.qxd 18/09/2003 15:21 Page 113

Oral and mental starter� Draw the triangle on the right on the board or on a prepared OHT.� Let the class work in pairs or small groups and ask them to work out how the sin

and cos keys on their calculators can be used to find lengths and angles. � Allow a five-minute discussion.

Main lesson activity� NB You may wish to use more than one lesson to cover the material in this

lesson plan and the associated pupil book.O

� Remind the class that the tangent of the angle � = tan� = –– and that this is the A

ratio of two sides of a right-angled triangle.Point out to the students that the tangent does not include the hypotenuse, H, and tell them that there are two trigonometric ratios, which do include H.

� Using the same diagram, tell the students that:O

�� the ratio –– is called the sine of the angle � and is written as sin�HA

�� the ratio –– is called the cosine of the angle � and is written as cos�H

� Explain that the sine and cosine are used in the same way as the tangent to findthe lengths of sides and size of angles in right-angled triangles when the lengthof the hypotenuse is involved.

� Explain to the class that the and keys on their calculators are used in

exactly the same way as the key.

� The three trigonometric ratios should now be summarised as follows:O A O

sin� = –– cos� = –– tan� = ––H H A

� Show the class how to do the following two examples:Example 1 Calculate the angle marked � in the diagram on the right. Give

your answer to 1 decimal place.The adjacent side and the hypotenuse are given, so use cosine:

Adjacent 8cos� = —————– = —– = 0.8

Hypotenuse 10So, � = 36.9° (1dp)

Example 2 Calculate the length of the side marked x on the diagram on theright, giving your answer to 3 significant figures.The angle and the hypotenuse are given, and the opposite side isrequired, so use sine:

Oppositesin� = —————–Hypotenuse x

sin 38° = ––5

Multiply both sides by 5 to give: 5sin38° = xSo, x = 3.08 cm (3sf)


A

HO

�

tan

cossin


LESSON

10.3

Framework objectives – Trigonometry: The sine and cosine of anangle

Begin to use sine, cosine (and tangent) in right-angled triangles to solve problems intwo dimensions.

A

HO

�

A

HO

�

8 cm

10 cm

�

x

38°

5 cm

9B3LP_10.qxd 18/09/2003 15:21 Page 114

Plenary� Ask the class to give a summary of the three trigonometric ratios.


Exercise 10C Answers

1 a 0.309 b 0.500 c 0.912 d 0.875 e 0.500 f 0.0802 a 5.7° b 32.0° c 58.8° d 66.4° e 47.2° f 16.7°3 a 1.59 b 3.54 c 12.2 d 1.98 e 3.57 f 9.004 a 17.5° b 34.8° c 28.0° d 75.5° e 40.9° f 69.7°5 a 3.80 cm b 4.62 cm c 14.8 cm d 1.97 cm e 8.23 cm f 12.3 cm6 19.0 cm7 49.3°

Extension Answers

1 a 12.8 cm b 15.9 cm c 9.23 cm d 49.0 cm2 a

b sin� = cos(90° – �)

� 0° 10° 20° 30° 40° 50° 60° 70° 80° 90°sin� 0.00 0.17 0.34 0.50 0.64 0.77 0.87 0.94 0.98 1.00cos� 1.00 0.98 0.94 0.87 0.77 0.64 0.50 0.34 0.17 0.00

Ho

me

wo

rk 1 Find the value of � for each of the following. Give your answers to 1 decimal place.

a sin� = 0.25 b sin� = 0.854 c cos� = 0.752 d cos� = 0.235

2 Find the value of each of the following. Give your answers to 3 significant figures.

a 5sin62° b 12sin52.6° c 21cos86° d 3.7cos42.3°

3 Calculate the angle marked � in each of the following triangles. Give your answers to 1 decimalplace.

4 Calculate the length of the side marked x in each of the following. Give your answers to 3significant figures.

Answers1 a 14.5° b 58.6° c 41.2° d 76.4°2 a 4.41 b 9.53 c 1.46 d 2.743 a 19.5° b 29.9° c 26.8°4 a 5.67 cm b 1.40 cm c 24.1 cm

11 cm

31° 17°55°

4.8 cmx x

x

a b

42 cm

c

9 cm3 cm

�

28 cm

25 cm�

13 cm

15 cm

�

a b c

� adjacent� hypotenuse� opposite� cosine (cos)� sine (sin)

Key Words

9B3LP_10.qxd 18/09/2003 15:21 Page 115

Oral and mental starter� Draw the triangle on the right on the board or on a prepared OHT.� Ask the class to write down in their books or on their whiteboards the three

trigonometric ratios for the angle �.� Check their answers.

Main lesson activity� Explain to the class that the lesson is about solving problems using trigonometry. � Go through the following stages which should be taken when solving a problem

involving trigonometry: 1. Draw a sketch of the right-angled triangle in the problem. Tell the students

that even when a diagram or picture accompanies the problem, it is a goodidea to redraw the triangle.

2. Mark on the sketch all the known sides and angles, including the units.3. Identify the unknown side or angle by labelling it x or �.4. Decide and write down which ratio you need to use to solve the problem. At

this point you may wish to introduce some mnemonics which the studentscan use to help them remember the three ratios.

5. Solve the problem and give your answer to a suitable degree of accuracy.This is usually three significant figures for lengths and one decimal place forangles.

� Demonstrate how to follow these stages by working through an example, suchas the one given below. Example A window cleaner has a ladder that is 6 m long. He leans it against

a wall so that the foot of the ladder is 2 m from the wall. Calculatethe angle the ladder makes with the wall. 1/2 Draw a sketch for the problem and write on all the known

sides and angles:

3 Identify the angle required by labelling it �:

4 Identify the trigonometric ratio to be used:The opposite and hypotenuse are known, so sine should be

Oused to solve the problem. The ratio required is sin� = —–.

H5 Solve the problem:

2sin� = –– = 0.333.

6So, � = 19.5 (1dp).


2 m (O)

6 m (H)�

2 m

6 m


LESSON

10.4

Framework objectives – Solving problems using trigonometry

Begin to use sine, cosine and tangent in right-angled triangles to solve problems intwo dimensions.

A

HO

�

9B3LP_10.qxd 18/09/2003 15:21 Page 116

Plenary� Write on the board ‘SOHCAHTOA’ or a similar mnemonic.� Ask individual students to explain how the mnemonic can be used to remember

the three trigonometric ratios.


Exercise 10D Answers

1 7.22 m 2 69.7 m 3 40.0° 4 058° 5 24.4 m 6 58.1°7 a 7.19 cm b 7.01 cm c 25.2 cm2

Extension Answers

1 a 238 km b 26.5 km c 239 km2 a 3.44 cm b 8.60 cm2 c 43.0 cm2

SATs Answers

1 a 4 cm b 40° c 12 cm2 a 10:14 = 5:7 and 8:12 = 2:3 so corresponding sides are not in the same ratio

b 11.2 cm3 a 9.98 m b 2.9°4 a 056° b 1.40 km5 a AC2 = 62 + 82 = 100, so AC = 10 cm b 11.7 cm c 5.9°

Ho

me

wo

rk 1 The stays on a flagpole are 10 m long and make an angle of 65°with the horizontal ground. Calculate the height of the flagpole.

2 The diagram on the right shows a ramp for wheelchairs. Calculate the angle the ramp makes with the ground.

3 A helicopter takes off from an army base on a bearing of 075° and flies for 52 km.

a How far east has the helicopter flown?

b How far north has the helicopter flown?

4 A plane takes off from an airport, climbing at a constant angle. When the plane has flown for 3.2 km, it reaches an altitude of 1000 m. Calculate the angle at which the plane is climbing.

5 The diagram on the right shows a wooden truss of a roof. Calculate the height, h, of the roof.

Answers1 9.06 m 2 10.9° 3 a 50.2 km b 13.5 km 4 18.2° 5 2.28 m

� cosine (cos)� sine (sin)� tangent (tan)

Key Words

10 m

65°

1.3 m

25 cm

9.8 m

h25° 25°

9B3LP_10.qxd 18/09/2003 15:21 Page 117

Oral and mental starter � Ask the class how many pounds equal one kilogram in weight. Accept the answer 2 as a start, and work

towards (or point out) the correct answer, which is 2.2.� Now ask what the equivalent of 6 kg is in pounds. Ask them how they worked this out, probably 6 × 2 = 12

added to 6 × 0.2 = 1.2, giving 13.2.� Using a target board such as the one shown below, work your way around the class asking individual students

for pound equivalents of kilogram weights.

Main lesson activity� Draw a rectangle on the board, showing the dimensions 4 cm and 5 cm. Ask the class to give you the area of

the rectangle. When someone gives you the correct answer (20 cm2), ask them how they calculated it (4 cm × 5 cm).

� Draw a second rectangle on the board, this time showing the dimensions 4x and 3 cm. Again, ask the class forthe area of the rectangle. This should provoke some discussion leading to the correct answer of 12x. (The unitsare not an important issue for this lesson, so ignore them.)

� Now draw on the board a rectangle with the dimensions 3x + 2 and 5. Again, ask the class for the area of therectangle. The correct answer (15x + 10) will be seen by some students more easily than others.

� Show how this rectangle can be split up, using the 3x and the 2 to give two rectangles – one with thedimensions 3x and 5, and the other with the dimensions 2 and 5. This should help all the students see that theyneed to find the area of each smaller rectangle and add them together to give the total area (15x + 10).

� Show that the last example is the same as 5(3x + 2) and link this to the finding of the area using two rectangles.Show that the brackets are expanded by multiplying each term by 5 to arrive at 15x + 10.

� Now draw on the board a rectangle with the dimensions 4x + 3 and 2x. Again, ask the class for the area of therectangle. Lead the students through the splitting of the rectangle to give two rectangles – one with thedimensions 4x and 2x and the other with the dimensions 3 and 2x. These can then be multiplied to find thearea of each rectangle and added together to give the total area 8x2 + 6x. Again, show the link to the expansionof the bracket 2x(4x + 3).

� Now put on the board the expression 3(4x + 2) and ask the class what this might mean. You want the response:‘The area of a rectangle with dimensions 3 and 4x + 2’. You also want them to refer to the expansion of thebracket to give 12x + 6.

� Repeat the process for x(3x + 2) to give 3x2 + 2x.� Put on the board the expression 3(4x – 1) and ask the class what this might represent. One answer is a rectangle

with sides 4x – 1 and 3. Show how, if this is split into two rectangles, the larger one has the dimensions 4x by 3with the smaller one, the dimensions 1 and 3, taken from it.

� Show that the expansion of 3(4x – 1) can be done in the same way as before to give 12x – 3.


8 5 15 4 73 12 9 11 20

10 13 100 2 3


Algebra 5CHAPTER

11

LESSON

11.1

Framework objectives – Expansion

Simplify or transform algebraic expressions by taking out single-term commonfactors.

Square a linear expression, expand the product of two linear expressions of theform x ± n and simplify the corresponding quadratic expression. Establish identitiessuch as a2 – b2 = (a + b)(a – b).

9B3LP_11.qxd 18/09/2003 15:21 Page 118

Plenary� Put on the board a rectangle with the dimensions 3x + 7 and 4x. Ask the class for

the area of the rectangle.� Now put on the board a rectangle with the dimensions x + 7 and x + 3. Again,

ask the class for the area of the rectangle.� This should provoke discussion whereby the rectangle with dimensions x + 7

and x + 3 is split into four rectangles with a total area of x2 + 7x + 3x + 21.



1 a 3x + 6 b 5t + 20 c 4m + 12 d 2y + 14 e 12 + 4m f 6 + 3k g 5 + 5t h 14 + 7x2 a 2x – 6 b 4t – 12 c 3m – 12 d 6y – 30 e 20 – 5m f 6 – 2k g 8 – 4t h 15 – 3x3 a 8x + 8 b 18t – 24 c 10m – 15 d 9y + 21 e 9 – 9m f 8 + 16k g 6 – 12t h 4 + 6x4 a 12t + 3 b 15x + 10 c 10x – 2 d 24x – 8 e 28t – 145 a x2 + 3x b t2 + 5t c m2 + 4m d y2 + 8y e 2m + m2 f 3k + k2 g 2t + t2 h 5x + x2

6 a x2 – 2x b t2 – 4t c m2 – 3m d y2 – 6y e 5m – m2 f 2k – k2 g 3t – t2 h 6x – x2

7 a 4x2 + 3x b 2t2 – 3t c 3m2 – 2m d 4y2 + 5y e 4m – 5m2 f 3k + 2k2 g 4t – 3t2 h x + 4x2

8 a 2x2 + 3x b 5t – 3t2 c 4m + 5m2 d 7k2 – 2k9 a 9x + 14 b 10t + 27 c 18 – 8m d 2k + 26 e 4x – 12 f 9x – 7 g 6 – x h 8 – 2x i 13m + 2 j 16m – 3

k 4x – 7 l 6x – 36 m 9x – 14 n 14x – 1910 a AB = y – 5 b AB = 3y c AB = y + 1 d AB = 4y + 1

CD = 4x – 1 CD = x + 3 CD = 3x – 3 CD = 3x + 1

Ho

me

wo

rk 1 Expand each of the following.

a x(3x + 4) b t(3t – 1) c m(4m – 3) d y(5y + 3)

e m(5 – 4m) f k(1 + 6k) g t(3 – 4t) h x(2 + 5x)

2 Expand and simplify each of the following.

a 3(m + 2) + 2(1 – 3m) b 4(2k + 3) + 2(1 – 3k)

c 5(3x – 2) + 3(2 – 4x) d 4(5x + 2) + 5(1 – 5x)

3 Write down the missing lengths in each of the following rectangles.

a b

Answers1 a 3x2 + 4x b 3t2 – t c 4m2 – 3m d 5y2 + 3y e 5m – 4m2 f k + 6k2 g 3t – 4t2 h 2x + 5x2

2 a 8 – 3m b 2k + 14 c 3x – 4 d 13 – 5x3 a AB = 4x + 5, CD = 4y – 3 b AB = 5x + 3, CD = y + 5

8x + 4

3y + 5

?A3x + 1 B

2y

D

C

?

?A3x

4y

7x + 5

B

3

C

D?

� dimension� expansion

Key Words

Extension Answers

1 One possible method for each is shown.a 1–

a + 1–b = b––ab + a––

ab = (b + a)——ab = (a + b)——

ab

b a–b + c–d = ad––

bd + bc––bd = (ad +–––——bc)

bd

2 b It is always the same as the starting number.c x [multiply by 3] = 3x [add 15] = 3x + 15 → [divide by 3] = x + 5 [subtract 5] = x

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Oral and mental starter� A formula for the approximate conversion of temperatures from degrees

Fahrenheit to degrees Celsius is C = 1–2(F – 32), where C is the temperature indegrees Celsius and F is the temperature in degrees Fahrenheit.

� Ask the class to use this formula to estimate the equivalent of 100° Fahrenheit indegrees Celsius (Answer 34°). Discuss with the class the strategy they used to dothis mentally.

� Work through a couple more examples, such as:

66 36 34 17

90 60 58 29

� Using a target board such as the one shown below, work your way around theclass asking the students to convert temperatures in degrees Fahrenheit to theirapproximate equivalents in degrees Celsius.

Main lesson activity� Put on the board a rectangle with 7 cm2 written inside the outline. Ask the class

what dimensions the rectangle could have. The simplest answer is 1 cm by 7 cm. Explain that coming up with these numbers involves finding a pair offactors.

� Now put on the board a rectangle with the area 12 cm2 written inside. Again,ask the class what dimensions the rectangle could have. There is more than onesimple choice here, and any factor pair will do (for example, 2 cm by 6 cm, 3 cm by 4 cm).

� Now put on the board a rectangle with the expression 3x + 3 written inside. Tellthe class that this is the area of the rectangle and ask what dimensions therectangle could have. Lead the discussion so that the class reaches the correctanswer of 3 and x + 1.

� In order to reach the answer above, a pair of factors has to be found which,when multiplied together, give 3x + 3. Explain that this is called factorisationand is the opposite process of expansion, which they covered in the previouslesson.

� Next, put on the board the expression 6x + 9. Invite them to imagine that this isthe area of a rectangle. Ask them what the dimensions of the rectangle could be.They need to factorise the expression to create a bracket with two factors insideand a term outside. For this example, factorisation gives 3(2x + 3). Show that thisexpands to give 6x + 9.

� Now put on the board the expression x2 + 5x and again ask what dimensions arectangle of this area could have. Help the class to see that this factorisation willbe x(x + 5), again showing that the expansion will give the original expression.


34 109 38 40 7355 42 76 89 5061 32 57 71 8899 93 103 67 72

÷ 2– 2– 30

÷ 2– 2– 30


LESSON

11.2

Framework objectives – Factorisation

Simplify or transform algebraic expressions by taking out single-term commonfactors; add simple algebraic expressions.

Square a linear expression, expand the product of two linear expressions of theform x ± n and simplify the corresponding quadratic expression. Establish identitiessuch as a2 – b2 = (a + b)(a – b).

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Plenary� Ask the class what is meant by the term ‘factorisation’. You want responses

which show their understanding of how to break down an expression into twoterms which will multiply together to give the original expression.

� Show them again the two stages which they have gone through today with someexamples such as:

6 + 9x = 3(2 + 3x)5x2 – 3x = x(5x – 3)



1 a 3(x + 2) b 2(2t + 3) c 4(m + 2) d 5(y + 2) e 2(4 + m) f 3(1 + 2k)g 5(1 + 3t) h 3(4 + x)

2 a 2(x – 2) b 4( t – 3) c 3(m – 3) d 3(2y – 3) e 7(2 – m) f 3(7 – k)g 4(3 – 2t) h 3(5 – x)

3 a 3(4x + 1) b 2(3t – 2) c 3(3m – 1) d 3( y + 2) e 3(5 – m) f 4(3 + k)g 2(3 – t) h 3(9 + x)

4 a 3x + 4 b 5 + 3t c 2m – 3 d 4 – 2t5 a x(x + 3) b t( t + 4) c m(m + 5) d y( y + 7) e m(3 + m) f k(4 + k)

g t(3 + t) h x(1 + x)6 a x(x – 3) b t(3t – 5) c m(m – 2) d y(4y – 5) e m(2 – m) f k(4 – 3k)

g t(5 – t) h x(7 – 4x)7 a x(3x + 4) b t(5t – 3) c m(3m – 2) d y(4y + 5) e m(4 – 3m) f k(2 + 5k)

g t(4 – 3t) h x(2 + 7x)8 a 3x + 4 b 2 + 3m c 3 – 2t d 4x – 19 a n, n + 1, n + 2 b 3n + 3 c 3(n + 1)

d Sum of 3 integers = 3n + 3 = 3(n + 1). (n + 1) is an integer, hence 3(n + 1) isalways a multiple of 3.

Ho

me

wo

rk 1 Factorise each of the following.

a 3x + 9 b 4t + 12 c 2m + 8 d 5y + 15

e 10 + 2m f 4 + 6k g 10 + 15t h 12 + 9x

i 6x – 4 j 8t – 12 k 6m – 9 l 20y – 8

m 21 – 7m n 18 – 3k p 12 – 10t q 15 – 5x

2 Factorise each of the following.

a x2 + 5x b t2 + 3t c m2 + 4m d y2 + 8y

e 6m + m2 f 2k + k2 g 7t + t2 h x + x2

i x2 – 4x j 2t2 – 3t k m2 – 5m l 5y2 – 4y

m 3m – m2 n 6k – 5k2 p 6t – t2 q 8x – 5x2

Answers1 a 3(x + 3) b 4(t + 3) c 2(m + 4) d 5( y + 3) e 2(5 + m) f 2(2 + 3k) g 5(2 + 3t) h 3(4 + 3x)

i 2(3x – 2) j 4(2t – 3) k 3(2m – 3) l 4(5y – 2) m 7(3 – m) n 3(6 – k) p 2(6 – 5t) q 5(3 – x)2 a x(x + 5) b t( t + 3) c m(m + 4) d y( y + 8) e m(6 + m) f k(2 + k) g t(7 + t) h x(1 + x)

i x(x – 4) j t(2t – 3) k m(m – 5) l y(5y – 4) m m(3 – m) n k(6 – 5k) p t(6 – t) q x(8 – 5x)

Extension Answers

a 2 and x2 + 2x, x and 2x + 4, 2x and x + 2b For x = 1, all pairs of values multiply to give an area of 6. For x = 2, all pairs of

values multiply to give an area of 16. For x = 3, all pairs of values multiply to givean area of 30.

c 2 and 6x2 + 9x, 3 and 4x2 + 6x, 6 and 2x2 + 3x, x and 12x + 18, 2x and 6x + 9, 3x and 4x + 6, 6x and 2x + 3

� factorisation

Key Words

9B3LP_11.qxd 18/09/2003 15:21 Page 121

Oral and mental starter� This starter is concerned with calculating percentages.� Set percentages in context by talking about getting a 20% reduction on the price

of a certain item you purchased recently.� Ask the class how they would calculate 20% of a value? They might use the

strategy of finding the value of 10% then doubling it, or they might divide theoriginal value by 5.

� Give them an example, such as: ‘What is 20% of £34’. Work through theexample, using the first method which most students will regard morestraightforward. That is, 10% of £34 is £3.40, doubling it gives 20% as £6.80.

� Using a target board such as the one shown below, work your way around theclass, asking individual students to work out 20% of the given value.

Main lesson activity� Put on the board a rectangle with sides labelled as (x + 2) and (x + 4). Ask the

class if they know what the area of the rectangle is.� This should lead to a discussion on splitting the rectangle up into 4 parts, with

areas of x2, 4x, 2x and 8. Add these up to get x2 + 6x + 8, which is the area of theoriginal rectangle.

� Next, put on the board the pair of brackets (x + 5)(x + 3). Ask the class if anyonecan expand, or multiply out this pair of brackets. Let the discussion flow fromthe geometric application of finding the area of a rectangle to a possible way ofmultiplying out the brackets term by term.

� Show both methods. Tell the students they must use whichever method they findsimpler for them. Ultimately, they will be expected to multiply out brackets withno props at all.

� Then, put on the board a pair of brackets with a negative sign involved: forexample (x + 6)(x – 2). Show the class that the easiest way to do this is tomultiply term by term, to get x2 – 2x + 6x – 12, which simplifies to x2 + 4x – 12.

� Finally, put on the board (x – 4)2. Ask if anyone can square this bracket. Look outfor the response of x2 + 16, which is the most common incorrect answer.

� Demonstrate this expansion, which is (x – 4)(x – 4), giving x2 – 8x + 16.


£45 35 kg 50 minutes £82 29 kg65 minutes £67 18 m 2 hours £39130 kg 180 minutes £89 75 kg 49 m£234 83 m 24 hours £26 130 kg


LESSON

11.3

Framework objectives – Quadratic expansion

Square a linear expression, expand the product of two linear expressions of theform (x ± n) and simplify the corresponding quadratic expression.

9B3LP_11.qxd 18/09/2003 15:21 Page 122

Plenary� Put on the board (x + a)(x + b) and discuss with the class what its expansion will

give: x2 + ax + bx + ab.� Next, put on the board (x + a)2 and discuss with the class what its expansion will

give: x2 + 2ax + a2.� If you feel that the class is ready for the next step, then write on the board

(ax + b)(cx + d ). Discuss with the class what its expansion will yield: acx2 + adx + bcx + bd.



1 x2 + 7x + 12 2 x2 + 6x + 5 3 x2 + 9x + 14 4 x2 – 2x – 8 5 x2 + x – 126 x2 – 4x – 5 7 x2 – x – 6 8 x2 + 5x – 6 9 x2 – x – 12 10 x2 – 3x + 2

11 x2 – 9x + 18 12 x2 – 9x + 20 13 4 – 3x – x2 14 10 – 3x – x2 15 18 + 3x – x2

16 x2 + 10x + 25 17 x2 – 6x + 9 18 4 – 4x + x2

19 f (x + y)(x – y) = x2 – xy + xy – y2 = x2 – y2

20 a 800 b 400 c 280 d 75 e 400 f 35 g 62 h 58 i 1997

Ho

me

wo

rk 1 Expand and simplify each of the following expressions.

a (x + 4)(x + 7) b (x + 3)(x – 6) c (x – 5)(x + 7)

d (x + 3)(x – 5) e (x – 4)(x – 3) f (x – 8)(x + 5)

g (x + 3)2 h (x – 5)2 i (x + 4)(x – 4)

2 Without using a calculator, find the result of each of the following calculations.

a 752 – 252 b 9.72 – 0.32 c 18.72 – 1.32

Answers1 a x2 + 11x + 28 b x2 – 3x – 18 c x2 + 2x – 35 d x2 – 2x – 15 e x2 – 7x + 12 f x2 – 3x – 40

g x2 + 6x + 9 h x2 – 10x + 25 i x2 – 162 a 5000 b 94 c 348

� expand� quadratic

Key Words

Extension Answers

1 8x2 + 14x + 3 2 12x2 + 23x + 10 3 8x2 + 2x – 3 4 10x2 + 26x – 125 8x2 – 22x + 12 6 25x2 – 30x + 9

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Oral and mental starter� Ask the class: ‘Who sends text messages to their friends?’. Ask several of the

students who do, how much they cost.� Put on the board the different rates for text messages. At the time of printing,

these are generally 8p, 10p or 12p. But do use those which your students offer,unless they only offer 10p!

� Ask how many text messages each student sends per day. Using the price permessage that the student quoted earlier, work out how much students spend ontext messaging per day.

� Ask other students how many messages they send per week. Again, using theprice per message, work out how much various students spend on textmessaging per week.

� Talk about top up cards costing £10, £15 or £20. How many text messages canthe students send from one top up at the different values?

� This is a rich source of mental work that goes best when real live data is beingused: that is, current charges. Try not to be exclusive. If there are some studentsnot able to text, ensure that they are included in the discussions.

� This can be expanded into estimating how many text messages each studentmight make in a year and therefore how much this will cost.

Main lesson activity� Put on the board the expression x2 + 8x + 12. Ask the class if they can simplify

this expression. Lead the discussion to factorising with a pair of brackets.� The x2 indicates that each bracket must start with two single xs, and the pair of

brackets will be in the form (x + a)(x + b).� Talk about the last plus sign in the expression, which indicates that the signs in

the brackets will be the same, and the same as the first sign in the expression.Hence, the two signs in the pair of brackets will be pluses.

� Ask what information the 12 gives. Help the class to see that this represents theproduct of a and b. Hence, a factor pair of 12 is required: 1, 12 or 2, 6 or 3, 4.

� Ask what information the 8x gives. Help the class to see that this shows that a + b must be 8. So, a factor pair which adds up to 8 is required. This leads tothe conclusion that a and b are 2 and 6 respectively. (They can be either wayround.) Hence, the factorisation is: (x + 2)(x + 6).

� Go briefly through the expansion to show that this does indeed give x2 + 8x + 12.

� Now put on the board the expression x2 – 8x + 16 and ask the class to factorise it.� Discuss the fact that because the second sign of the expression is a plus, the

signs will be the same, and the same as the first sign in the expression, which isa minus. Hence, the brackets will be of the form (x – a)(x – b).

� This means that a factor pair of 16 is required which adds up to 8, leading to (x – 4)(x – 4) or (x – 4)2.

� Next, put on the board the expression x2 – 3x – 10, and lead the class into adiscussion on the factorisation of this expression.

� As the second sign of the expression is a minus, the two signs in the bracketsmust be different. The first sign of the expression is negative, which indicatesthat the larger of the factor pair must also be negative.

� This means that as the signs are different, the factor pair of 10 must have adifference of 3, which leads to 5 and 2. Hence, the factorisation is (x + 2)(x – 5).



LESSON

11.4

Framework objectives – Quadratic factorisation

Simplify or transform algebraic expressions by taking out common factors.

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Plenary� Put on the board x2 + Ax + B.� Ask what clues should be looked for in the factorisation of these expressions.� Discuss the signs. When the last sign in the expression is a plus, both bracket

signs will be the same as the first sign in the expression. When the last sign inthe expression is a minus, the bracket signs will be different.

� Discuss the factor pair needed for B. When the last sign in the expression is aplus, the factor pair must add up to A. When the last sign in the expression is aminus, the factor pair must have a difference of A.



1 (x + 4)(x + 6) 2 (x + 2)(x + 12) 3 (x + 3)(x + 6) 4 (x – 2)(x – 9)5 (x – 3)(x – 4) 6 (x – 2)(x – 6) 7 (x + 6)(x – 4) 8 (x + 11)(x – 4) 9 (x – 2)(x + 6)

10 (x + 4)(x – 11) 11 (x – 9)(x + 7) 12 (x – 10)(x + 9) 13 (x + 5)2 14 (x – 6)215 (x – 1)2 16 (x + 2)(x – 2) 17 (x + 5)(x – 5) 18 (x + 10)(x – 10)

Ho

me

wo

rk Factorise each of the following.

1 x2 + 6x + 8 2 x2 – 9x + 20 3 x2 + 3x – 4

4 x2 – 4x – 12 5 x2 + 4x + 4 6 x2 – 14x + 49

7 x2 – 16 8 x2 – 1 9 x2 – 4x – 21

Answers1 (x + 4)(x + 2) 2 (x – 5)(x – 4) 3 (x + 4)(x – 1) 4 (x + 2)(x – 6) 5 (x + 2)2 6 (x – 7)2 7 (x + 4)(x – 4)8 (x + 1)(x – 1) 9 (x – 7)(x + 3)

� factorisation

Key Words

Extension Answers

1 (3x + 1)(x + 1) 2 (3x + 1)(x – 2) 3 (3x + 2)2 4 (2x – 1)(x – 5) 5 (4x + 5)(x – 5)6 (3x + 4)(2x – 5)

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Oral and mental starter� Ask the students to put up their hands if they can multiply by 11. Many hands will go up.� Tell them to keep their hands up if they can work out 11 × 28 in their heads. Discuss the different strategies the

students have used to get their various answers.� One strategy is to do it in two parts and add the results together:

11 × 28 = (10 × 28) + (1 × 28) = 280 + 28 = 308.� Go through another example using this method with them:

11 × 34 = (10 × 34) + (1 × 34) = 340 + 34 = 374.� Using a target board such as the one shown below, work your way around the class asking individual students

to multiply the number at which you point by 11.

Main lesson activity � This lesson is about changing the subject of a formula. Put on the board the formula C = 250 + 5W. Explain that

this formula is used to calculate the cost of advertisements in a certain newspaper, where C is the cost in penceof the advertisement and W is the number of words in the advertisement. Explain that C is the subject of theformula because it is the variable (letter) in the formula which stands on its own, usually on the left-hand sideof the equals sign.

� Ask how much it would cost to place an advertisement with 20 words in the newspaper. Use this example toverify that all the students can substitute W = 20 into the formula to get C = 250 + 5 × 20 = 250 + 100 = 350,giving the cost as £3.50.

� Now tell the class that you want to place an advertisement in the newspaper, on which you want to spend £10.Ask them how many words you would use in the advertisement.

� In order to work this out, they need to rearrange the formula so that W is the subject. Remind the class that thesame rules apply which they have used previously with equations. That is, the same thing is done to both sidesof the formula.

� Work through the example. To get W on its own, start by simplifying the right-hand side of the formula:

C = 250 + 5WSubtracting 250 from both sides gives: C – 250 = 5W

C – 250 Dividing both sides by 5 gives: ———— = W

5C – 250

Turning the expression round gives: W = ————5

1000 – 250So, if C = 1000, then: W = —————— = 150.

5


17 25 51 67 9242 78 19 34 8394 85 36 48 23


LESSON

11.5

Framework objectives – Change of subject

Derive and use more complex formulae and change the subject of a formula.

SATs Answers

1 a i 100 cm2, 80 cm2, 32 cm2 ii 252 cm2 iii 252 b i n2 cm2, 2n cm2, 6 cm2 ii n2 + 5n + 62 a 5(2y + 4) and 2(5y + 10) b 12( y + 24) c 7( y + 2) d 2y2(3y – 1)

3d3 a a = 1500; b = 200 b —–

54 a (y + 3)2 = (y + 3)(y + 3) = y2 + 6y + 9 ≠ y2 + 9 b i y2 + 7y + 10 ii y2 – 12y + 36 iii 6y2 – y – 40

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Plenary� Discuss with the class the similarities between the process of changing the

subject of a formula and that of solving an equation.� To close the lesson, work with the class to make T the subject of the following

formula:3K + 5T2

W = —————4


Ho

me

wo

rk 1 Change the subject of each of the following formulae as indicated.

a Make I the subject of the formula W = IPT.

b i Make P the subject of F = P + MK. ii Make M the subject of F = P + MK.

c i Make m the subject of T = 3m + 2n. ii Make n the subject of T = 3m + 2n.

abhd Make b the subject of V = ——.

3

19R2 The formula C = —— + 40 is used to calculate the cost in pounds of making a boiler of radius 8

R (cm).

a Make R the subject of the formula.

b Use this formula to find the radius of a boiler that cost £150 to make.

3 Draw a graph of each of the following equations on the same pair of axes.

a y – 2x – 1 = 0 b y – 2x – 3 = 0 c y – 2x + 1 = 0 d y – 2x + 3 = 0

Comment on the similarities and differences between the graphs.

AnswersW (F – P) (T – 2n) (T – 3m) 3V

1 a I = —– b i P = F – MK ii M = ——— c i m = ———— ii n = ———— d b = —–PT K 3 2 ah8(C – 40)

2 a R = ————– b 46.3 cm19

3 All graphs should be parallel to each other ( y = 2x) and cut the y-axis at the negative value of the constant inthe equation.

� subject of aformula

Key Words

Extension Answers

They are all parallel to each other (y = –6x) and they cut the y-axis at the negative valueof the product of the constant in the equation multiplied by the denominator of the y coefficient.

Exercise 11E Answers

V V (S – U) (S – U)1 a i I = –– ii R = –– b i U = S – FT ii F = ———– iii T = ———–

R I T F(P – 2w) (P – 2b) 2A 2A

c i b = ———— ii w = ———— d i b = —– ii h = —–2 2 h b

5(F – 32)2 a C = ———— b i –53.9 °C ii 19.2 °C iii 13.6 °C

93 a 24.5 cm2 b 11.8 cm2 4 a £1.60 b 0.75 cm 5 a 125.7 cm2 b 9.6 cm (9.5 cm if π key on calculator used) c 1.5 cm6 a N = 5, R = 3, A = 6, so N + R – A = 5 + 3 – 6 = 2, so N + R – A = 2.

c N + R – A = 2, so N + R = 2 + A, so N + R – 2 = A, i.e. A = N + R – 2. N = 10, R = 9, so A = 177 a y = 2x – 3 b y = 3x – 5 c y = 9 – 2x d y = 14 – 3x e y = 4x + 3 f y = 5x + 18 a i y: –1, 1, 3, 5, 7 ii y: –4, –1, 2, 5, 8 iii y: 7, 5, 3, 1, –19 They are all parallel to each other ( y = x) and cut the y-axis at the negative value of the constant in the equation.

10 They are all parallel to each other ( y = –x) and cut both the x-axis and the y-axis at the negative value of the constantin the equation.

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Oral and mental starter� The following is a 10-question, SATs-style, mental test on the theme of fractions.

Repeat each question twice and allow 10 seconds to answer.1 What fraction of one metre is thirty five centimetres? [Write ‘1 m’ and

‘35 cm’ on the board.]2 Look at these numbers. Which one of them is the decimal equivalent of

three-eighths? [Write the decimals on the board.]0.38 3.8 0.375 3.125 3.08

3 What is half of 4.7?4 What is the sum of three-eighths and one-quarter?5 Add one point six to one quarter.

3 26 What is three-quarters squared? [Write ( ––) on the board.]4

7 What is a quarter of 6.2?8 What decimal is equivalent to the fraction five-eights?9 Four-ninths of a number is thirty six. What is the number?

10 What is the square root of 16––49?Answers 1 35—–100 or equivalent 7––20 2 0.375 3 2.35 4 5–8 or equivalent (0.625)

5 1.85 6 9––16 7 1.55 8 0.625 9 81 10 4–7

Main lesson activity� This is a revision lesson on Number, principally covering fractions, decimals and

percentages.The questions in the Pupil Book exercise are graded from Level 5 to Level 8 asfollows:

Q. 1, 6, 7 Level 5Q. 2, 8, 9 Level 6Q. 3, 4, 10 Level 7Q. 5, 11, 12, 13 Level 8

� Before letting students start the questions you can go through key points (assuggested below) or discuss some specific questions with the class to remindthem of the methods used.

General�� Equivalence of fractions, percentages and decimals

Fractions�� Equivalent fractions�� Cancelling�� Converting mixed numbers to top heavy fractions and vice versa�� Adding and subtracting fractions


Solving Problems and RevisionCHAPTER

12

LESSON

12.1

Framework objectives – Fractions, percentages and decimals

Revision of Number:

Solve increasingly demanding problems and evaluate solutions.

Solve problems involving percentage changes.

Use proportional reasoning to solve a problem, choosing the correct numbers totake as 100%, or as a whole.

Recognise when fractions or percentages are needed to compare proportions; solveproblems involving percentage changes.

Enter numbers and interpret the display in context (negative numbers, fractions,decimals, percentages, money, metric measures, time).

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Decimals�� Ordering�� Changing between metric measurement units

Percentages�� Finding a percentage of a quantity�� Finding one quantity as a percentage of another�� Calculating percentage increase and decrease�� Compound interest�� Reverse percentage


Plenary� Go through the answers to the exercise. Discuss and clarify those with which

students had difficulty.

HomeworkIt is assumed that during the revision period students will be given a past SATspaper to work through at home. Students will have seen some questions before inthe Maths Frameworking Pupil Books, so a mock SATs paper is provided in thisTeacher’s Pack, after the Chapter 12 lesson plans. The mock paper consists of SATs-style questions which students will not have encountered before.

Additional homework questions are provided below, for further practice on thetopics covered in this lesson.



1 a 14––15 b 7––18 c 4 3––202 134.43 a i 0.08 ii 2000 b 400–4504 a 45 × 0.85 × 0.9 b 1.135 a Always even b Always an integer6 74.5 kg7 £63.058 £36.219 a 200 000 b 425 000 c 3.9%

10 a 31.4% b 36% c Unleaded 76.2p per litre. Lead replacement 77p per litre11 a 800 b 12.2%12 a £836 b £499 c £295.53 d 6 days13 a 16.521 174 86 b 16.5

Ho

me

wo

rk 1 In a sale, a hi-fi is reduced by 15%. The sale price of the hi-fi is £459, what was the original price?

2 For each part of the question, where n is always an integer, write down the answer which is true andexplain your choice.

n2 – 2a When n is even, ——— is:2

Always odd Always even Sometimes odd, sometimes even

n2 – 2b When n is odd, ——— is:2

Always an integer Always a fraction Sometimes an integer, sometimes a fraction

Answers1 £540 2 a Always odd b Always a fraction

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Oral and mental starter� The following is a 10-question, SATs-style, mental test on the theme of

percentages. Repeat each question twice and allow 10 seconds to answer.1 What is ten percent of thirty five pounds?2 What is twenty percent of two hundred pounds?3 Thirty percent of a number is nine. What is the number?4 A CD costing thirteen pounds is reduced in a sale by ten percent. What is the

new price of the CD?5 There are 20 chocolates in a box, of which 12 have soft centres. What

percentage of the chocolates have soft centres?6 Fifteen percent of a number is twelve. What is the number?7 After a 10% reduction, a cooker was priced at £180. What was the original

price?8 What percentage is equivalent to the fraction seven-eighths?9 Six percent of a number is nine. What is the number?

10 What percentage of forty five is twenty seven?Answers 1 £3.50 2 £40 3 30 4 £11.70 5 60% 6 80 7 £200

8 87.5% 9 150 10 60%

Main lesson activity� This is a revision lesson on Number, principally covering the four rules, ratios

and standard form.� The questions in the Pupil Book exercise are graded from Level 5 to Level 8 as

follows:Q. 1, 7, 8 Level 5Q. 2, 3, 9, 10 Level 6Q. 4, 5, 11 Level 7Q. 6, 12, 13 Level 8


General�� Basic knowledge of tables up to 10 × 10

Four rules�� Setting out in columns for addition and subtraction�� Using box method or column methods for long multiplication�� Using chunking for long division

Directed numbers�� Using a number line�� Combining signs when adding and subtracting: ++, –+, etc

Ratio�� Adding ratios�� Dividing up an amount into a given ratio�� Multiplying by a ratio to get individual amounts

Standard form�� Writing numbers in standard form�� Calculating with numbers in standard form



LESSON

12.2

Framework objectives – The four rules; ratio; standard form

Revision of Number:

Solve increasingly demanding problems and evaluate solutions.

Interpret and use ratio in a range of contexts, including solving word problems.

Enter numbers and interpret the display in context (negative numbers, fractions,decimals, percentages, money, metric measures, time).

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1 17 bins with £7 left over2 £183 a 14.4 b 90 c 0.064 a = 4, b = 8, c = 35 a 59 b –26 and 82 c 44 d (–1)2, (–2)6, 82 and 59

6 a iii 32 × 103 is larger than 33 × 102 b 3.6 × 103 and 36 × 102 c 54 × 10–3

7 a 15 b £56 175 c £80.258 a 52 mph b 4 hours and 40 minutes 9 a 400 kg b 8 bags

10 347 tickets11 1 : 412 a 1 × 10–6 b 5 × 104

13 a 3.32 × 107 b 8.46 × 108 c 1.93 × 107

Ho

me

wo

rk 1 p = 1.2 × 107, q = 2.5 × 108, r = 6.3 × 10–3

Work out each of the following, giving your answer in standard form.

a p + q b p × q c r2

2 Light green paint is made by mixing yellow paint and blue paint in the ratio 2 : 3.

Dark green paint is made by mixing yellow paint and blue paint in the ratio 1 : 3.

One litre of light green paint and one litre of dark green paint are poured into a large bucket.

How much more yellow paint needs to be added to the bucket to produce light green paint?

Answers1 a 2.62 × 108 b 3 × 1015 c 3.969 × 10–5

2 250 ml

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Oral and mental starter� The following is a 20-question, SATs-style, mental test on the theme of the four rules. Repeat each question

twice and allow 10 seconds to answer.1 Four boxes of pencils cost six pounds. How much will seven boxes of pencils cost?2 Multiply nought point seven by ten.3 A chocolate bar costs one pound forty pence. I buy four bars. How much change will I get from a ten

pound note?4 I am thinking of two numbers. When I add them together I get nine. When I multiply them together I get

twenty. What are the numbers ?5 Multiply nought point six by nought point five.6 Double sixty six.7 What is the total cost of five video tapes at four pounds ninety five pence each?8 How many seconds are there in fifteen minutes?9 Work out the value of this. [Write 23 × 32 on the board.]

10 What is three minus nought point two?11 How much must be added to this number to make one hundred. [Write 63.5 on the board.]12 Multiply together the first three prime numbers.13 I have saved thirty seven pounds in twenty pence coins. How many coins is that?14 Divide thirty by nought point three.15 What number is nought point nought one less than five point three? [Write 0.01 and 5.3 on the board.]16 Multiply together nought point nought four and nought point two. [Write 0.04 and 0.2 on the board.]17 Divide forty point two by nought point one. [Write 40.2 and 0.1 on the board.]18 What is the value of this. [Write √

——225 × 23 on the board.]

19 Work out nought point nought three squared. [Write 0.03 on the board.]20 Divide eight by fifty [Write 8 and 50 on the board.]

Answers 1 £10.50 2 7 3 £4.40 4 4 and 5 5 0.3 6 132 7 £24.75 8 900 9 7210 2.8 11 36.5 12 30 13 185 14 100 15 5.29 16 0.008 17 402 18 12019 0.0009 20 0.16

Main lesson activity� This is a revision lesson on Algebra, principally covering the basic rules and solving linear equations.� The questions in the Pupil Book exercise are graded from Level 5 to Level 8 as follows:

Q. 1, 2, 3 Level 5Q. 4, 5, 6, 7 Level 6Q. 8, 9, 10 Level 7Q. 11, 12, 13 Level 8

� Before letting students start the questions you can go through key points (as suggested below) or discuss somespecific questions with the class to remind them of the methods used.

Basic algebra�� Using letters to represent variables�� The difference between a term, an expression and an equation

Manipulative algebra�� Substituting numbers into expressions�� Expanding brackets�� Factorising�� Collecting like terms�� Cancelling


LESSON

12.3

Framework objectives – Rules of algebra and linear equations

Revision of Algebra:

Represent problems and synthesise information in algebraic form.

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Linear equations�� Rearranging – collecting together variables and numbers on the LHS and RHS

respectively�� Inverse operations (change sides, change signs)�� Checking answers by substituting into original equation






1 a n2, 4n, 16 b n2 + 8n + 16, (n + 4)22 a 4x – 20 b 11x + 3 c 5x + 2 d 17x + 16 e 5x + 223 a i 21 ii 10 iii 50 b i z = 3 ii z = 22 iii z = –14 a 6x + 3 = 12, x = 1.5 b 3y – 6 = y + 7, y = 6.55 a 2n + 4 b n + 6 c 2 – n d 3n – 16 a i 45 ii 26 iii –11 b i 3(x + 2y) ii x(x + 1) iii 2a(2b + 3)7 a x = 1.5 b x = 2 c x = –0.58 a 3(4x – 6) and 6(2x – 3) b 6(y – 2) c 3y(3y – 2)

2n 4 9 169 a ——— b ––, —–, —– c 1

2n + 3 5 10 172

10 a i 3 ii 4 b –– xy2 c i x – 24 ii 9y + 2x3

11 a 7 b –2.5 c 0 or 412 a x2 – 8x + 16 b x2 + x –20 c 12x2 – 5x – 213 2

Ho

me

wo

rk 1 a Explain why (x – 4)(x – 4) ≠ x2 – 16

b Expand and simplify each of the following.

i 2(x – 3) + 3(2x – 1) ii (x + 4)(x – 7)


2x + 3a 6 + 2x = 8 + 4x b ——— = 5 c 5(1 + x) = 3(x + 2)3

Answers1 a (x – 4)(x – 4) = x2 – 4x – 4x + 16 = x2 – 8x + 16

b i 8x – 9 ii x2 – 3x – 282 a –1 b 6 c 0.5

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Oral and mental starter� The following is a 10-question, SATs-style, mental test on reading diagrams.

Each student will need a sheet with the diagrams below to refer to. Teachers mayfind the Teacher’s Pack CD useful to prepare these.

� Repeat each question twice and allow 10 seconds to answer.1 The bar chart shows the number of children in some families. How many

families are represented?2 This is a centimetre grid. What is the area of the shaded square?3 Which diagram shows strong positive correlation?4 By looking at the timetable, work out how long the journey from Barnsley to

High Green takes.5 What number is the arrow pointing to?6 Add one more square to the grid so that it has rotational symmetry of order 2.7 Which diagram shows the graph x + y = 5?8 What is the exterior angle of a regular hexagon?9 The pie chart shows the proportion of men and women in a

sports club. Work out the angle which represents the women.10 Look at the octagon. Which diagram shows the octagon after

a ninety degree rotation in a clockwise direction?Answers 1 21 2 17 cm2 3 c 4 44 minutes 5 –1.4 6 7 c

8 60° 9 72° 10


LESSON

12.4

Framework objectives – Graphs

Revision of Algebra:

Represent problems and synthesise information in graphical form.

1

2

8 9 10

3a b c d

5

6 7 a

c

b

d

6

4

2

01 2 3 4 5

Num

ber

of fa

mili

es

Number of children 4 BarnsleyBirdwellHigh GreenSheffield

09 3209 5510 1610 36 –2 –1 0

5

5 55

Women (20%)

Men (80%)

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Main lesson activity� This is a revision lesson on Algebra, principally covering graphs.� The questions in the Pupil Book exercise are graded from Level 5 to Level 8 as follows:

Q. 1, 2 Level 5Q. 3, 4, 5, 6 Level 6Q. 7, 8, 9 Level 7Q. 10, 11 Level 8

� Before letting students start the questions you can go through key points (as suggested below) or discuss somespecific questions with the class to remind them of the methods used.

Basic graphs�� y = mx + c as the formula of a straight line�� Significance of m and c for gradients, parallel lines and intercept with y–axis�� Quadratic graphs

Real life graphs�� Travel graphs or distance–time graphs�� Gradient of line represents speed�� Horizontal line represents no motion






1 a 7:15 b 7:45 c 8 : 40 d 9:30 2 a (6, 0) and (–1, 7)

b

3 a true b cannot be sure, graph is levelling off c cannot be sure – there may not be a causal link

4 a b c (3, 5)

5 a y = –2 b y = –2x c x = –0.5 and y = 1 6 a b 13.5 square units

7 a A and E b C and D c A and F d D and F8 a 9:30 b 160 kph9 a Test (8, 0) and (0, 4) in the equation b x + y = 6

10 Draw line at y = 2, x = ±2.511 a y ≤ x, y ≥ 1, x ≤ 2 b R

001234

1 2 3 4

6

4

2

0

–2

–4

–6

–6 –4 –2 0 2 4 6

x = 2y = 3x

y = –3

6

4

2

0

–2

–4

–6

–6 –4 –2 0 2 4 6

y = 2x – 1y = 2x + 2

y = x + 2

6

4

20

2

4

6

–6 –4 –2 0 2 4 6

Ho

me

wo

rk 1 By drawing the graphs y = 2x, y = –2 and x = 3, work out the area of the triangle enclosed by all three lines.

2 Give the four inequalities which describe the shaded region.

Answers1 16 square units2 y ≤ 4, y ≥ x, x ≥ 1, x ≤ 3

001234

y

x1 2 3 4

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Oral and mental starter� The following is a 10-question, SATs-style, mental test on the theme of shape,

space and measures. Repeat each question twice and allow 10 seconds toanswer.1 How many lines of symmetry does a parallelogram have?2 I face south-east and turn anticlockwise through 270 degrees. What direction

am I now facing?3 I drive six and a half kilometres in 10 minutes. What is my average speed in

kilometres per hour?4 What is the value of angle a? [Draw the diagram shown on the

board.]5 What is the area of this triangle? [Draw the triangle shown on the board.]6 Draw the shape that you get after rotating this T shape by 90° clockwise.

[Draw the shape shown on the board.]7 What is the approximate area of a circle with a radius of 2 centimetres?8 The distance from home to school is about 12 kilometres. Approximately

how many miles is that?9 What is the area of this triangle? [Draw the triangle shown on the board.]

10 This picture shows a shape cut in half along a line of symmetry. [Draw the picture shown on the board.] What is the name of the original shape?Answers 1 0 2 south-west 3 39 km/h 4 50° 5 12 cm2

6 7 12–14 cm2 8 7–8 miles 9 6 cm2 10 Kite

Main lesson activity� This is a revision lesson on Shape, Space and Measures.� The questions in the Pupil Book exercise are graded from Level 5 to Level 8 as

follows:Q. 1, 5, 6 Level 5Q. 7, 8, 9, 10 Level 6Q. 2, 3, 11 Level 7Q. 4, 12, 13 Level 8


Volume and area�� Recall of formulae for area of rectangle, triangle, parallelogram, circle�� Recall of formulae for volume of cubes, cuboids and prisms

Symmetry�� Line symmetry�� Rotational symmetry

Angles�� Definition of acute, obtuse and reflex�� Measuring angles�� Angles at a point and on a straight line�� Alternate and corresponding angles


LESSON

12.5

Framework objectives – Shape, Space and Measures

Revision of Shape, Space and Measures:

Represent problems and synthesise information in geometric form.

Solve problems using properties of angles, of parallel and intersecting lines, and oftriangles and other polygons, justifying inferences and explaining reasoning withdiagrams and text.

130°

a°

5 cm

6 cm

4 cm

3 cm5 cm

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Enlargements�� Scale factor and centre of enlargement�� Similar triangles

Right-angled triangles�� Pythagoras�� Trigonometry






1 a = 54°, b = 82°, c = 152°2 27.5 cm2

3 a 300 cm3 b 25 cm2

4 x = 18 cm, y = 10 cm5 a 288 cm2 b 4 c 16 : 16 a 80 km b no, 100 kph ≈ 62 mph

c 93 miles, 50 km is about 31 and 3 × 31 = 937 a check sides measure 5 cm, 8 cm and 7 cm b 82°8 a a = 39°, b = 39°, c = 43° b angle ADB = angle DBE so AD is parallel to BE9 40 cm

10

11 a 6.32 cm b 5.66 cm12 a Area = 33.5 cm2, Perimeter = 24.4 cm b 4.09 cm13 x = 16.6 cm y = 36.7°

Ho

me

wo

rk 1 Find the length x in the triangle shown on the right.

2 Find the length x and the angle y in each of the triangles shown below.

a b

Answers1 15.2 cm 2 a 10.8 cm b 54.3°

20 cm13 cm

x cm

12 cm

7 cm

y°

7 cm

x cm

57°

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Oral and mental starter� The following is a 10-question, SATs-style mental test on the theme of Handling

Data. Repeat each question twice and allow 10 seconds to answer.1 What is the mean of these numbers? [Write 10, 10 and 25 on the board.]2 What is the range of these numbers? [Write on the board: 4, 8, 2, 9, 7,

12, 1, 3, 7, 8, 2 and 3.]3 A dance class contains both boys and girls. The probability that a member of

the class, picked at random, is a girl is 0.7. What is the probability that amember picked at random is a boy?

4 A bag contains only red, blue and green balls. The table shows theprobability of choosing each colour, when a ball is picked from the bag atrandom. [Draw the table shown.]What is the probability that the ball picked is blue or green?

5 An ordinary, fair, six-sided dice is rolled. What is the probability that the diceshows a score of 7?

6 Two ordinary, fair, six-sided dice are rolled. What is the probability that thetotal score is 3?

7 What is the median of these numbers? [Write on the board: 7, 8, 10, 13, 15and 20.]

8 The table shows the number of pets owned by ten students.How many pets are owned altogether?

9 The mean mass of thirty text books is nought point seven kilograms.[Write 0.7 on the board.] What is the total mass of the textbooks?

10 A referendum returned one hundred yes votes and five hundred novotes. Which pie chart best represents these data?

Answers 1 15 2 11 3 0.3 4 0.65 5 0 6 2––36 = 1––18 7 11.5 8 219 21 kg 10 c

Main lesson activity� This is a revision lesson on Handling Data.� The questions in the Pupil Book exercise are graded from Level 5 to Level 8 as

follows:Q. 1, 2, 3 Level 5Q. 4, 5, 8, 9 Level 6Q. 6, 10 Level 7Q. 7, 11 Level 8


Probability�� Language and definition of probability�� Writing probabilities as fractions, decimals or percentages�� Sample space diagrams

a b c d


LESSON

12.6

Framework objectives – Handling Data

Revision of Handling Data:

Interpret graphs and diagrams and draw inferences to support or cast doubt oninitial conjectures; have a basic understanding of correlation.

Analyse data to find patterns and exceptions, look for cause and effect and try toexplain anomalies.

Red Blue Green0.35 0.2 0.45

Number of pets Frequency1 22 53 3

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Averages�� The three averages used for discrete data�� Range of a set of data�� Mean of a table of discrete data�� Mean of a table of grouped data

Surveys�� Methods of sampling�� Unbiased questions with unambiguous response boxes

Scatter diagrams�� Types of correlation�� Using line of best fit for predicting values





Exercise 12F Answers

1 Any combination where the number of red to blue is in the ratio 2:1, e.g. 20 red 10 blue2 a Q b R c P and R 3 a 6 b 5 c i False, there is no mode to start with ii False, both old and new marks were above median

iii True, total will be 2 more4 0.35 a Not representative b Overlap of responses 6 a 3x b 3x + 17 a 56 b 39 to 71 = 32 c 33%8 The percentage value of a car decreases as the mileage increases or there is a negative correlation

b 50% c 28 000 miles9 a 30 b 4.9 10 3.75 min or 3 min 45 s 11 a 0.36 b 0.48 c 80

Ho

me

wo

rk 11 When two dice are rolled the probability of a double one is —–.36

a When two dice are rolled what is the probability of a double 2?

b Which answer shows the probability of a treble six when three dice are rolled.

1 1 3 1— —— —— —18 216 216 42

2 The bar chart shows the distances that 50 students threw a discus.

a What is the probability that a pupil chosen at random will have thrown the discus more than 30 metres?

b What is the probability that a pupil chosen at random will have thrown the discus more than 45 metres?

c Work out the mean length of throw for the 50 pupils.

Answers1 a 1––36 b 1–––2162 a 0.4 b 0.09 c 26.4 m

00

5

10

15

10 20Distance (m)

Num

ber

of p

upils

30 40 50

6

10

14

11 9

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Level 5

A 3 × 3 × 3 cube is made from 27 different coloured small cubes.

The small cubes with 3 faces showing are coloured grey.

The small cubes with 2 faces showing are coloured white.

The small cubes with 1 face showing are dotted.

The small cubes with 0 faces showing are striped.

(a) Complete the table below to show the number of small cubes of each colour thatare used.

3 marks

(b) A 4 × 4 × 4 cube is made in the same way. Complete the table below to show thenumber of small cubes of each type it has and the total number of cubes used.

2 marks

Grey cubes ..........................

White cubes ..........................

Dotted cubes ..........................

Striped cubes 8

Total: ..........................

Grey cubes ..........................

White cubes ..........................

Dotted cubes ..........................

Striped cubes ..........................

Total: 27

1


PRACTICE PAPER THIS IS A NON-CALCULATOR PAPER

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This is how Helen works out 25% of 240 in her head.

(a) Show how Helen can work out 25% of 180 in her head.

2 marks

(b) This is how Jim works out 25% of 240 in his head.

Show how Jim can work out 25% of 460 in his head.

2 marks

(a) A school needs 250 pencils.

The pencils come in packs of 12.

How many packs must the school order?

Show your working.

packs2 marks

3

“50% of 240 is 120 so 25% of 240 is 60”

“10% of 240 is 2420% of 240 is 485% of 240 is 12

so 25% of 240 is 48 + 12 = 60”

2


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(b) Rulers cost 22p each.

How much do 250 rulers cost?

Show your working. Give your answer in pounds.

£ 3 marks

(a) There are n cubes in this tower.

Linda adds another 7 cubes to the tower.

Write an expression to show the total number of cubes inLinda’s tower.

1 mark

(b) Qayser builds another tower.

This tower is m cubes high.

Write an expression to show the total number of cubes in Qayser’s tower.

1 mark

Tower M

m

Tower N

n

4


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(c)

The number of cubes in each tower is the same.

Which expression below is true?

Put a tick ( � ) by the correct expression.

1 mark

(d) Brian builds two more towers like Tower N and Tower M. Brian’s towers have anequal number of cubes in them, and their heights are related by this expression: n = m + 6.

Work out the value of m.

m = 1 mark

n = m × 2

n = m × 4

n = m ÷ 2

n = m ÷ 4

Tower M

m

Tower N

n


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The graphs below show the number of goals scored per match by Team A and Team Bover 50 games.

(a) How many goals did Team A score altogether in the 50 games?

Goals: 2 marks

(b) Which team scored three or more goals in 20% of their games?

Explain your answer.

2 marks

(c) Eli says that the graphs show that Team B is more successful than Team A.

Give a reason why this may not be true.

1 mark

Team B

0

5

10

15

20

0 1 2

Number of goals

Num

ber

of g

ames

3 4 50

5

10

15

20

0 1 2

Number of goals

Team A

Num

ber

of g

ames

3 4 5

5


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Level 6

Three people A, B and C run a race.

Below are five different distance–time graphs.

Fill in the gaps below to show which runner matches up with which graph.

Runner A sets off quickly, slows down and then speeds up

Graph

Runner B runs at a steady speed

Graph

Runner C sets off quickly and then slows down

Graph

3 marks

Time

Distance

Graph 1

Time

Distance

Graph 2

Time

Distance

Graph 3

Time

Distance

Graph 4

Time

Distance

Graph 5

6


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Use the graphs to match each line with its equation.

The first one has been done for you.

3 marks

Each shape in this question has an area of 20 cm2.

No diagram is drawn to scale.

(a) Calculate the length of the base of the parallelogram.

area = 20 cm2 base = ................... cm

1 mark

5 cm

base

8

105

5

10

0D

C

A

B

x

y

LINE A

LINE B

LINE C

LINE D

y = x

x = 5

y = 5

x + y = 5

y = 5x

7


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(b) Calculate the height of the triangle.

area = 20 cm2 height = ................ cm

1 mark

(c) What is the value of h in this trapezium?

area = 20 cm2 a + b = 4 cm h = ......................... cm

1 mark

A different trapezium has an area of 20 cm2 and a height of 5 cm. What is the valueof a + b in the trapezium?

area = 20 cm2 h = 5 cm a + b = .................... cm

1 mark

b

h

a

height

10 cm


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(d) Look at this rectangle:

Calculate the value of x and use it to find the perimeter of the rectangle.

Show your working.

perimeter = ...................... cm

2 marks

This is a series of patterns with black and white squares.

(a) Complete this table:

2 marks

pattern number number of black squares number of white squares

5

12

Pattern Number 1 Pattern Number 2 Pattern Number 3

9

3x + 1

area = 20 cm2

5x – 5


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(b) Complete this table by writing expressions:

2 marks

(c) Write an expression to show the total number of tiles in pattern number n.

Simplify your expression.

1 mark

(d) A different series of patterns is made with squares.

For this series of patterns, write an expression to show the total number of tiles inpattern number n. Show your working and simplify your expression.

2 marks

Pattern Number 1 Pattern Number 2 Pattern Number 3

pattern number expression for the expression for the number of black squares number of white squares

n


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(a) Each of these calculations has the same answer, 75.

Fill in each gap with a number.

2 marks

(b) Solve these equations to find the values of a, b and c.

a = b = c =

3 marks

4a + 3

2b – 25

c2 – 6

= 75

2.5 × 30

0.25 × ...................

75 ÷ 1

= 75

7.5 ÷ ...................

10


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Level 7

In the scale drawing, the shaded area represents a rectangular flowerbed.

There is a path all around the flowerbed.

The shortest distance from the flowerbed to the edge of the path is always 2 m.

On the diagram, draw accurately the position of the edge of the path.

2 marks

This is what a student wrote:

Show that the student was wrong.

2 marks

a

2

b

3

a + b

5+ =

12

Scale: 1 cm to 1 m

11


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Passengers have been complaining to a bus company about how long they have towait for a bus.

An inspector records the waiting times of 100 passengers on one of the company’s busroutes on one day.

Results

(a) Use the graph to estimate the probability that a passenger chosen at random willwait for 15 minutes or longer.

1 mark

(b) Use the graph to estimate the probability that a customer chosen at random willwait for 7.5 minutes or less.

1 mark

00 5 10 15

Waiting Time (minutes)

Num

ber

of

Pas

seng

ers

20 25 30

5

10

15

20

25

30

35

40

13


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(c) Calculate an estimate of the mean waiting time per passenger.

Show your working.

You may complete the table below to help you with the calculation.

minutes

2 marks

(d) The inspector wants to improve the survey.

She records the waiting times of more customers.

Give a different way the inspector could improve the survey.

1 mark

Waiting time Mid-point of bar Number of f x(minutes) ( x ) passengers ( f )

0– 2.5 22

5–

10–

15–

20–

25–30


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(a) Find the values of a and b when p = 5.

2p3

a = ——5

a = 1 mark

p2( p + 1)b = —————

3p

b = 1 mark

(b) Simplify this expression as fully as possible:

15cd——–

3d

1 mark

(c) Multiply out and simplify these expressions:

3(x + 4) – 2(3 – 2x)

1 mark

(x + 3)(x + 5)

1 mark

(x + 1)(x – 2)

1 mark

(x – 4)2

1 mark

14


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Level 8

(a) The diagram shows the graph with equation y = x3.

On the same axes sketch the graph of y = x3 + 2.

1 mark

(b) Curve A is a reflection in the x-axis of y = x3.

What is the equation of curve A?

1 mark

(c) Curve A can also be obtained from the graph of y = x3 using a differenttransformation.

Describe this transformation.

1 mark

x

yy = x3A

A

x

yy = x3

15


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(d) The shaded region is bounded by the curve y = x3 and the line y = 2x.

Circle two inequalities which together fully describe the shaded region.

y < x3 x < 0 y < 0 y < 2x

y > x3 x > 0 y > 0 y > 2x

2 marks

(a) Which number is the greater?

3 × 104 4 × 103

Explain your answer.

1 mark

(b) Circle the number that has the same value as 5 × 102.

52 52 500 0.5 × 103 –5 × 10–2

2 marks

16

x

y

y = x3

y = 2x


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(c) (3 × 102) × (2 × 102) can be written more simply as 6 × 104.

Write these values as simply as possible.

(4 × 105) × (2 × 10–1)

1 mark

8 × 1010————4 × 102

1 mark

Here are 10 shapes.

(a) Complete the table.

1 mark

Number of shapes with Number of shapes with Number of shapes with straight edges only curved edges only both straight edges and

curved edges

17


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(b) A shape is chosen at random.

What is the probability that it does not have both straight and curved edges?

1 mark

(c) What is the probability that a shape is chosen that has straight edges if you alreadyknow that it has curved edges?

1 mark

(d) Two of the shapes are chosen at random.

Which calculation shows the probability that both shapes have curved edges?

6 6 6 5 6 6—– × —– —– × —– —– + —–10 10 10 10 10 10

6 5 6 5 6 5—– + —– —– × –– —– + ––10 10 10 9 10 9

1 mark


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This shape is made using two circles.

The radii of the circles are 3a and 2a.

(a) Find the area of each circle, in terms of a and π, and show that the total area of theshape is 5πa2.

3 marks

(b) The shaded area, 5πa2, of the shape is 20 cm3.

Write an equation in the form a = … … , leaving your answer in terms of π. Show your working to simplify your equation.

2 marks

3a 2a

18


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Oral and mental starter� The class should work in small groups for this activity.� Explain to the class that the lesson will be used to revise handling data topics that they have already covered, in preparation

for an investigation next lesson. Tell the class that this activity is called ‘The A–to–Z of handling data’.� Ask each group to think of as many handling data topics and words beginning with different letters of the alphabet as they

can, excluding those to do with probability. You may wish to ask for a few suggestions and write them on the board, such asaverage, bias, data, graph, etc.

� Ask the groups to write down an example for each word they come up with. For example: the average of 2 and 4 is 3.� Give the groups a few minutes to discuss their answers, then ask the groups to give you the topics to write on the board,

covering the meaning of each one.

Main lesson activity� Continuing from the oral and mental starter, ask the students to give you responses to the following types of questions.

1 What sort of things should you think about when 7 What is a frequency table?planning a question and response section for a 8 Name as many different types of frequency diagrams questionnaire? as you can.

2 What is a census? 9 What is a stem–and–leaf diagram?3 Name some ways of recording data. 10 How do you work out a mode, a median, a mean and a 4 What is the difference between primary and range for a set of data?

secondary data? 11 How do you estimate the median and interquartile range 5 How can you choose a random sample from a group from a cumulative frequency graph (for large sets of

of people? data).6 What is a two-way table? 12 How do you estimate the mean for a table of grouped data.

� You should look for answers which refer to the following.1 Avoid: leading questions; missing or overlapping responses; any form of bias.2 A census is a survey of a whole population.3 Tally chart, data collection sheet or observation sheet.4 Primary data is data obtained directly by the person carrying out the research whereas secondary data has already been

collected by someone else.5 A random sample of a group could be obtained by putting everyone’s name in a bag, then drawing a portion of the

names out.6 A two-way table records two sets of related information within one table, for example make of car and colour of car.7 A frequency table is a table showing the number of times (frequency) that each particular value or item is recorded in a

survey or experiment.8 Pictogram, bar chart, pie chart, line graph, frequency polygon, histogram.9 A stem–and–leaf diagram is an ordered set of numerical data, grouped to show how the data is distributed. It is in effect

a bar chart using a list of numerical data. 10 The mode is the most common value; the median is the middle value of an ordered set of data; the mean is the total of

all the values divided by the number of items of data; the range is the difference between the largest and smallest values.n n 3n11 The median, lower and upper quartiles are obtained by reading off at ––, ––, —–, where n is the total 2 4 4

frequency. The interquartile range is the difference between the upper and lower quartile values.12 The estimate of the mean is calculated by multiplying the mid-class values by the frequency for that class, summing up

these answers and dividing by the total frequency.� Ask the class to look through the table of vocabulary in the Pupil Book, to make sure they are familiar with the terms there.




13

LESSON

13.1

Framework objectives – Revision of statistical techniques

Discuss how data relate to a problem; identify possible sources, including primaryand secondary sources.

Find summary values that represent the raw data, and select the statistics mostappropriate to the problem.

Extension Answers

The vertical axis (population) starts at 56 500 000, making population appear to more than doublein ten years. In fact it has only risen by about 2.4% as stated.

9B3LP_13.qxd 18/09/2003 15:22 Page 160

Plenary� Explain to the class that it is important to be able to draw appropriate statistical diagrams

and calculate statistics such as averages, but that it is equally important to be able tointerpret them, commenting on how useful they are.

� Ask the class if it is appropriate to use a pie chart to represent 20 categories of data [withso many categories, a bar chart might be easier to interpret], or ask the class if it issensible to use a scatter graph for 5 pairs of data [there are probably not enough data toproduce a meaningful graph].

� Briefly refer to the mode, median and mean and explain the effect of an extreme value onthe mean. Use the phrase, ‘it would be inappropriate to use the mean as one value hasdistorted the data’. For example, the data 50, 50, 50, 50, 100 give a mean of 50 when the100 is excluded, but a mean of 60 when the 100 is included.



1 a Other categories not given, e.g. ‘cycle’ or ‘other’ b The categories overlap c 8.15 AM is in two categories2 a b 45% c 18 ≤ T < 20 d 14 ≤ T < 16

3 a 0 5 6 8 9 b 14 c 15 d 81 0 1 1 2 2 4 4 5 6 6 7 8 9 92 0 Key 0 | 6 means 6 students

4 a There are longer bars for females over 55 years old.b Females may live longer generally in France; more men than women were killed in WWII (and WWI).

5 a mode = 1, median = 4, mean = 11.7 b mode = 5, median = 5, mean = 5.4 c mode = £4.50, median = £3.50, mean = £3.41d mode = 18, median = 20, mean = 20.8

6 a Check that pie chart shows the following data:b For this to be true the classes would have to

have equal numbers of students in.7 a 8 Mean = 132.35 cm

c Median = 132.7 cm, IQR = 138.7 – 125.6 = 13.1 cm

Height, h (cm) Cumulative frequencyh ≤ 100 0h ≤ 120 7h ≤ 130 39h ≤ 140 80h ≤ 150 97h ≤ 160 100

Class A B C DAngle (°) 72 54 144 90

Boys Girls12 ≤ T < 14 2 214 ≤ T < 16 4 116 ≤ T < 18 3 218 ≤ T < 20 0 420 ≤ T < 22 1 1

Ho

me

wo

rk 1 The weights (in kg) of 24 men are given below. a Use the data to copy and complete the frequency table.

b In which class is the median weight?c Complete a table of cumulative frequencies,

draw the cumulative frequency graph and use it to calculate the median and interquartile range.

d Explain why these weights are not representative of the whole adult population.2 These tables show the average monthly temperatures for Paris and Madrid over the course of one year.

Paris

Madrid

a Draw suitable graphs to represent both sets of data.b Comment on the differences between the average monthly temperatures in Paris and Madrid.

Answers1 a c

b 60 ≤ W < 70d Only males in results. Median = 62.9 kg, IQR = 72.0 – 51.6 = 20.4 kg.

2 a Histograms or line graphs drawn. b Average temperatures consistently higher in Madrid.

Weight (kg) Cumulative frequencyW ≤ 40 0W ≤ 50 4W ≤ 60 10W ≤ 70 17W ≤ 80 22W ≤ 90 24

Weight, W (kg) Tally Frequency40 ≤ W < 50 |||| 450 ≤ W < 60 |||| | 660 ≤ W < 70 |||| || 770 ≤ W < 80 |||| 580 ≤ W < 90 || 2

Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec°C 5.3 6.7 9.7 12.0 16.1 20.8 24.6 23.9 20.5 14.7 9.3 6.0

Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec°C 3.7 3.7 7.3 9.7 13.7 16.5 19.0 18.7 16.1 12.5 7.3 5.2

62 48 55 67 81 40 45 59 58 6272 65 70 82 66 48 59 68 71 6554 57 76 74

Weight, W (kg) Tally Frequency40 ≤ W < 5050 ≤ W < 6060 ≤ W < 7070 ≤ W < 8080 ≤ W < 90

� data collection� raw data� primary source� secondary source� frequency table� frequency

diagram� population

pyramid� scatter graph� cumulative

frequency� lower quartile� upper quartile� interquartile

range

Key Words

9B3LP_13.qxd 18/09/2003 15:22 Page 161

Oral and mental starter� Write on the board, ‘A solo pop singer is more likely to be female than male, but a singer in a band is more

likely to be male than female.’� Ask the class how they could investigate this.� Answers which they might say or be prompted to say could include:

Make a list of known singers and bandsDo a survey using the pop charts

� You could discuss the fact that the chart data will only look at the most successful singers and therefore maygive a biased result. You could then suggest that the initial hypothesis needs to say that it refers to the mostsuccessful singers.

Main lesson activity� The activities given in this section could easily take two lessons, depending on the amount of detail asked for

when carrying out the investigation. You may wish to ask the students to collect certain data prior to the lesson,or you may decide to provide the students with secondary data.

� Continuing from the oral and mental starter, show the class the handling data cycle and the related checklist forcompleting a handling data investigation, on p. 218 of Pupil Book 3. The list is also reproduced opposite.

� Go through and consider how each point applies to the pop singers example from the oral and mental starter:�� [Statement of topic] Compare number of male singers and female singers in the charts.�� [Hypothesis] ‘In the charts, a solo singer is more likely to be female than male, but a singer in a band is

more likely to be male than female.’�� [Sample size] Look at, say, the top 50.�� [Foreseen problems] Charts will change from week to week – may need to use charts over several weeks.�� [Identify any sources of bias and plan how to minimise them] A chart may only look at one type of music,

so consider different types of chart.�� [How to obtain data] Use pop charts from different sources (this information could be obtained from the

Internet).�� [Identify what extra information may be required] Ask yourself the question, “How can I extend this

problem, using more complex techniques which will provide more reliable results? �� [Data collection sheet] Record the number of bands and the number of solo artists, record the number of

males and females in each case.�� [Analysis] Calculate average numbers per week. Represent in percentage bars or pie charts.�� [Factors affecting results] Songs stay in the charts for several weeks, so the weeks looked at should really be

far apart in the year, rather than consecutive.�� [Limitations of any assumptions made] The mean number of weeks may be distorted by an extreme value

(one singer may be in the charts for a relatively long time).�� [Conclusion] State whether you agree with initial hypothesis based on your results.

� The class can now do one of the investigations in Exercise 13B from Pupil Book 3, working in small groups.


LESSON

13.2

Framework objectives – A handling data project

Select, construct and modify, on paper and using ICT, suitable graphicalrepresentation to progress an enquiry.

Interpret graphs and diagrams and draw inferences to support or cast doubt oninitial conjectures; have a basic understanding of correlation.

Communicate interpretations and results of a statistical enquiry using selectedtables, graphs and diagrams in support, using ICT as appropriate.

Solve substantial problems by breaking them into simpler tasks, using a range ofefficient techniques, methods and resources, including ICT.

Identify possible sources of bias and plan how to minimise it.

Identify what extra information may be required to pursue a further line of enquiry.

Analyse data to find patterns and exceptions, look for cause and effect and try toexplain anomalies.

Examine critically the results of a statistical enquiry, and justify choice of statisticalrepresentation in written presentations, recognising the limitations of anyassumptions and their effect on conclusions drawn.

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Plenary� Having observed the students working on a choice of investigations, you may

wish to give one or more groups the opportunity to present their findings so farto the rest of the class.

� It is important that the students are encouraged to be critical of their own workand that they can recognise any limitations.

� Discuss how the problem can be extended using more complex data, forexample, analysing large sets of continuous data and carrying out more complexcalculations, for example, using the interquartile range rather than the range inorder to overcome problems with extreme values.

Checklist for completing a handling data investigation� Specify the problem and plan

�� statement of problem or topic to investigate�� hypothesis stating what you think the investigation will show�� how you will choose your sample and size�� any practical problems you foresee�� identify any sources of bias and plan how to minimise them�� how you will obtain your data�� identify what extra information may be required to extend the project

� Collect data from a variety of sources�� follow initial plan and use a suitable data-collection sheet

� Process and represent data�� analysis of your results using appropriate statistical calculations and diagrams

� Interpret and discuss data�� comparison of results with your original hypothesis�� list of any factors which might have affected your results and how you could

overcome these in future�� consider the limitations of any assumptions made�� a final conclusion


Ho

me

wo

rk Choose one of the following tasks.

1 Complete the investigation started in the lesson by writing up the report.

2 Collect data in order to investigate the pop singers example.

3 Carry out and write up a detailed investigation of your own choice.

� conjecture� hypothesis� sample� bias� investigation� extension� enquiry� limitations

Key Words

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Oral and mental starter� The class should work in pairs or small groups for this activity.� Tell them that they are going to revise perimeter, area and volume, which they

have already covered.� Ask each pair or group to write down the formulae for the perimeter and the

area of any 2-D shapes. � Now ask each pair or group to write down the formulae for the surface area and

the volume of any 3-D shapes. � Give the class a few minutes to discuss their answers.

Main lesson activity� Once the class has finished writing down the formulae, ask individual students

to draw a shape on the board or OHT and below it give the formula for itsperimeter, area or volume.

� Continue this activity until all the shapes are covered.




14

LESSON

14.1

Framework objectives – Shape and space revision

Use units of measurement to calculate and solve problems in a variety of contexts.

Know and use the formulae for the circumference and area of a circle.

Calculate the surface area and volume of right prisms.

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Plenary� Tell the class that in the next lesson they will be working on a shape and space

investigation, and that they may need to use some of the formulae which theymet during the present lesson.



1 a i 12 cm ii 9 cm2 b i 28 cm ii 45 cm2 c i 16 cm ii 12 cm2

d i 52 cm ii 128 cm2 e i 30 cm ii 30 cm2

2 a i 18.8 cm ii 28.3 cm2 b i 28.3 cm ii 63.6 cm2

c i 31.4 cm ii 78.5 cm2 d i 39.6 cm ii 124.7 cm2

3 a i 150 cm2 ii 125 cm3 b i 160 cm2 ii 100 cm3 c i 108 cm2 ii 48 cm3

d i 736 cm2 ii 960 cm3

4 a 226 cm3 b 295 cm3 c 7.39 m3

5 a 6.4 cm b 31.8 cm2

6 7.1 cm

Extension Answers

366 m, 6060 m2

Ho

me

wo

rk 1 Find the area of each of the following shapes.

a b c d

2 Calculate i the circumference and ii the area of each of the following circles. Take π = 3.14 or usethe key on your calculator. Give your answers to one decimal place.

a b

3 Calculate the volume of this prism.

Answers1 a 45 cm2 b 24 cm2 c 180 cm2 d 60 cm2

2 a i 50.3 cm ii 201.1 cm2 b i 62.8 cm ii 314.2 cm2

3 96 m3

π

� area� circumference� surface area� volume

Key Words

9 cm

5 cm 8 cm

6 cm

12 cm

15 cm

15 cm

6 cm

5 cm

8 cm 20 cm

2 m

5 m

3 m

12 m

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Oral and mental starter� Ask individual students to give the formulae for the perimeter, the area or the volume of various 2-D and 3-D

shapes.

Main lesson activity� Each investigation given in this section will take up to two lessons to complete.� The class can work in pairs or small groups. The whole class may undertake the same investigation, or pairs or

groups may choose their own investigation from the four given.� Before the investigations are started, briefly go through the methods of doing an investigation:

�� Draw some easy examples first, making all diagrams clear with all measurements shown.�� Put your results in a table with suitable headings.�� Look for any patterns among the entries in the table.�� Describe and explain any patterns which you spot.�� Try to find a rule or formula to explain each pattern.�� Try another example to see whether your rule or formula does work.�� Summarise your results with a conclusion.�� If possible, extend the investigation by introducing different questions.

� Note: the students may need centimetre-squared paper.� The class can now do Exercise 14B from Pupil Book 3.


LESSONS

14.2

14.3

Framework objectives – Shape and space investigation

Present a concise, reasoned argument, using symbols, diagrams and relatedexplanatory text.

Suggest extensions to problems, conjecture and generalise: identify exceptionalcases or counter-examples, explaining why.


The layout below for each investigation is minimal. The students should always explain their choice of presentation andlink all their diagrams with the text. They may use different approaches to the investigations and these should be noted. Theuse of algebra would also enhance the students’ work and should be encouraged wherever possible. All the investigationsare helpful to the preparation for the GCSE coursework assessment.

1 It is expected that the students will draw a sequence ofsquares and complete a table similar to the one on the right. Note that units are not necessary for this investigation.From the table, the students will notice that the 4 × 4 squarehas the same value for its perimeter and area. They should seethat this is the only solution, by noticing that, after the 4 × 4 square, the area is always greater than the perimeter. Theyshould then check this using another example. (It could also be shown by drawing a graph.)Some students may be able to show that this is the only solution by using an algebraic approach.

For a square of side l, P = 4l and A = l2.For the square to have the same value for P and A, 4l = l2.This is true only for l = 0 or l = 4. Clearly, l = 0 has no meaning, so l = 4 is the only solution.

To extend the investigation, the students could consider rectangles.By drawing a number of rectangles, they should find that the value of the perimeter equals thevalue of the area when l = 6 and w = 3.They may think that this is the only solution, but intuition may tell them that other solutionsmay exist. For completeness, this can be shown by an algebraic approach: For a rectangle of length l and width w, P = 2l + 2w and A = lw.For the rectangle to have the same value for P and A, 2l + 2w = lw, which gives:

lw – 2w = 2lw( l – 2) = 2l

2lw = ——

l – 2

This shows that there are an infinite number of solutions. For example, when l = 10, w = 2.5. Other solutions can nowbe found. Considering circles would also extend the investigation.

l

l

w

l

Size of square 1 × 1 2 × 2 3 × 3 4 × 4 5 × 5Perimeter (P) 4 8 12 16 20Area (A) 1 4 9 16 25

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Plenary� Having observed the students working on the investigations, you may wish to

discuss one of their choices.


Exercise 14B (cont’d)

2 The students should start by completing a table of results toshow the length (l), perimeter (P) and area (A) of the squares.They can then find the ratios l : P and l : A.From the table, the students can see that the ratio l : P isalways 1 : 4 and the ratio l : A is 1 : n, where n is the length ofthe side of the square. They should now test their rules on new data.They may be able to generalise: for a square of size n × n, l : P = 1: 4 and l : A = 1: n.The investigation could be extended by looking at cubes and considering the ratio of the length of a side to the surfacearea and the ratio of the length of a side to the volume.

3 The students should start by considering squares of different sizes. They will realise that it will be easier to use evennumbers for the side of the square. They should then present their results in a table, as shown below, giving their finalanswers to a suitable degree of accuracy. (One decimal place is suggested.)

The students should now see that the percentage waste is always 21.5% for squares of any size. A further examplewould show this. An algebraic approach could also be used. For a square of side 2r, the area of the square is 4r2 and the area of the coinis π r2.The area of the coin as a percentage of the area of the square is:

π r2 π—— × 100 = –– × 100 = 78.54r2 4

So, the percentage waste is 21.5%.The investigation could be extended to four coins stamped from a square or by considering rectangular sheets of metal.

4 a For a 6 × 2 table, there are two bounces. b For a 6 × 2 table, the ball goes down pocket C.c After drawing different sized tables, including square tables, the students

should complete a table to show their results. The table on the right showssome results for 17 different tables, with the ball always starting frompocket A. From the table, the students should be able to spot various patterns. Forexample: a square table has no bounces and the ball ends up in pocket C;when the length and width of the table are reversed, the number ofbounces is the same but the ball may not end up in the same pocket. The following is a summary.• For a table of length l and width w, write it as the ratio l : w.• When l : w cannot be simplified, the number of bounces, N, is given by

the formula:N = l + w – 2

• When l : w can be simplified to give the ratio a : b, the number ofbounces, N, is given by the formula:

N = a + b – 2 • When l and w (or a and b if simplified) are both odd, the ball ends up in

pocket C.• When only l is odd (or a if simplified), the ball ends up in pocket D.• If only l is even (or a if simplified), the ball ends up in pocket B.The students may write down these conditions, but in all cases theyshould test their rules or formulae on further examples.

Size of square Area of square Area of coin Area of coin as % % wasteof area of square

2 cm × 2 cm 4 cm2 3.14 cm2 78.5% 21.5%4 cm × 4 cm 16 cm2 12.57 cm2 78.5% 21.5%6 cm × 6 cm 36 cm2 28.27 cm2 78.5% 21.5%

Size of table Number of Pocketbounces

1 × 1 0 C2 × 2 0 C3 × 3 0 C

2 × 1 1 B3 × 1 2 C4 × 1 3 B

3 × 2 3 D4 × 2 1 B5 × 2 5 D6 × 2 2 C

4 × 3 5 B5 × 3 6 C6 × 3 1 B

2 × 3 3 B3 × 4 5 D3 × 5 6 C3 × 6 1 D

Ho

me

wo

rk Complete the investigation you started in the lesson.

Size of square l P A l:P l:A1 × 1 1 4 1 1:4 1:12 × 2 2 8 4 1:4 1:23 × 3 3 12 9 1:4 1:3

� investigate� generalise

Key Words

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Oral and mental starter� The class should work in pairs or small groups for this activity.� Tell them that they are going to revise reflection and rotational symmetry, which

they have already covered.� Ask each pair or group to draw and write down the names of any 2-D shapes,

and then to draw on them all the lines of symmetry. � Now ask them to repeat the activity but to write down the order of rotational

symmetry below each one. Give the class a few minutes to discuss theiranswers.

� Then ask them to write down how to find the number of planes of symmetry fora 3-D shape.

Main lesson activity� Once the class have finished writing down their answers, ask individual students

to draw a shape on the board or OHT and to insert its lines of symmetry or togive its order of rotational symmetry.

� Continue this activity until it is clear that the class understand the concepts.� Briefly remind them how to find the number of planes of symmetry for a 3-D

shape by showing them that a cuboid has three planes of symmetry.



LESSON

14.4

Framework objectives – Symmetry revision

Visualise and use 2-D representations of 3-D objects.

Use rotations and reflections on paper. Identify reflection symmetry in 3-D shapes.

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Plenary� Tell the class that in the next lesson they will be working on a symmetry

investigation and that they may need to use some of the concepts which theymet during the present lesson.



1 a 2 b 2 c 6 d 4 e 52 a 2 b 1 c 1 d 43 a 2 b 2 c 5 d 4 e 24 a 4 b 3 c 4 d 25 a 2 b 3 c 2 d 36 Cuboid with two square faces

Ho

me

wo

rk Design a logo for a badge for your school, which has both reflection and rotational symmetry.

� reflection symmetry

� rotationalsymmetry

� plane ofsymmetry

Key Words

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Oral and mental starter� Ask individual students to explain reflection symmetry, rotational symmetry and planes of symmetry.

Main lesson activity� Each investigation given in this section will probably take two lessons.� The class can work in pairs or groups and the class can all work on the same investigation or be allowed to

make their own choice from the four investigations given.� Before the class start the investigation, briefly go through the methods of doing an investigation:

�� Draw some easy examples first, showing any lines of symmetry and/or stating the order of rotationalsymmetry on the diagrams.

�� Explain anything you notice from the diagrams.�� Describe and explain any patterns which you spot.�� Summarise your results with a conclusion.�� If possible, extend the investigation by introducing different questions.�� Note: the students may need tracing paper, mirrors, centimetre-squared paper and a selection of 3-D solids.



LESSONS

14.5

14.6

Framework objectives – Symmetry investigation

Present a concise, reasoned argument, using symbols, diagrams and relatedexplanatory text.

Suggest extensions to problems, conjecture and generalise: identify exceptionalcases or counter-examples, explaining why.


The layout below for each investigation is minimal. The students should always explain their choice of presentation andlink all their diagrams with the text. They may use different approaches to the investigations and these should be noted. Theuse of algebra would also enhance the students’ work and should be encouraged wherever possible. All the investigationsare helpful to the preparation for the GCSE coursework assessment.

1 It is expected that the students will draw diagrams to show the number of lines of symmetry for a tile with differentnumbers of shaded squares. Examples are shown below. (Reflections and rotations are omitted.)

The students should now notice that for five shaded squares, the diagrams are the same as for four squares but with theshading reversed. Similarly for six, seven and eight squares. This could now be usefully summarised in a table.The investigation could be extended by using a different size of tile.

1 line of symmetry

One square

Two squares

Three squares

Four squares

1 line of symmetry 2 lines of symmetry

4 lines of symmetry

1 line of symmetry

1 line of symmetry

2 lines of symmetry

4 lines of symmetry

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Plenary� Having observed the students working on the investigations, you may wish to

discuss one of their choices.


Exercise 14D (cont’d)

2 There are 12 different pentominoes (excluding mirror images). The students should be encouraged to show their resultsin a table.

For students who extend the investigation, there are 35 different hexominoes.

3 Allowing rotations, the T-shape will fit in the 3 × 3 grid four times.The students should now draw square grids of different sizes and find the number of ways in which the T-shape fitsinside each grid. The table below shows the number of ways for three different square grids.

The students should now be able to spot the rule: the numbers of ways give the even square numbers. They should testanother example to confirm that the rule works. The students may also be able to use algebra in this investigation and arrive at a formula.For an n × n grid, the total number of ways, T, the T-shape will fit inside the grid is given by:

T = 4(n – 2)2 for n > 2To extend the investigation, the students could consider rectangular grids or use a different shape.

4 For this investigation, it is useful for the students to have access to a collection of different solids. The table below isnot exhaustive.

The students may notice that the symmetry number for any prism is twice the order of rotational symmetry for thecross-section of the prism.

3-D solid Outline Symmetry numberCube Square 24Cuboid Rectangle 4Regular tetrahedron Equilateral triangle 12Square-based pyramid Square 4Regular octahedron Square 4Regular triangular prism Equilateral triangle 6Regular pentagonal prism Regular pentagon 10Regular hexagonal prism Regular hexagon 12Cylinder Circle ∞Sphere Circle ∞

Size of grid 3 × 3 4 × 4 5 × 5Number of ways 4 16 36

Pentomino

Number of lines of symmetry 4 2 1 1 1 1Order of rotational symmetry 4 2 1 1 1 1

Pentomino

Number of lines of symmetry 0 0 0 0 0 0Order of rotational symmetry 2 1 1 1 1 1

� investigate� generalise

Key Words

Ho

me

wo

rk Complete the investigation you started in the lesson.

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Oral and mental starter� Explain to the class that the lesson will be used to revise the probability topics they have already covered, in preparation for

an investigation next lesson.� Ask the class a few simple probabilities. Ask the students to explain how they have worked each answer out. Some

examples are given below.�� rolling a 6 on a fair, six-sided dice �� picking a blue cube when there are 3 blue, 4 red �� rolling a 3 or a 4 on a fair, six-sided dice and 3 green cubes in a bag�� rolling an odd number on a fair, six-sided dice �� not picking a blue cube in the above example�� throwing a head on a fair coin �� it not raining, if the probability of rain is 3–4

Main lesson activity� Continuing from the oral and mental starter, ask the students to give you responses to the following types of questions.

1 A box contains 60 sweets, of which 10 are strawberry.a What is the probability of picking a strawberry sweet?b What is the probability of picking a sweet that is not strawberry?c A second box of sweets of the same make contains 300 sweets, how many would you expect to be strawberry?

2 A six-sided dice is rolled and lands on 6 four times out of 24 rolls. Do you think that the dice is fair? Give a reason.3 Two coins are thrown. How many different outcomes are there?4 What is the difference between theoretical and experimental probability?5 What could you do to test if a spinner is biased? 8 What are exhaustive events?6 What are independent events? 9 When do you use a tree diagram?7 What are mutually exclusive events? 10 What is relative frequency?

� You should look for answers which refer to the following.1 a 10––60 or 1–6 b 5–6 c 502 It appears biased as 6 out of 24 is 1–4. If fair you would expect it to be close to 1–6.3 Four, HH, HT, TH, TT4 Theoretical probability looks at equally likely outcomes whereas experimental probability is based on the results of an

experiment or number of trials.5 Roll it many times and compare the results to see if the experimental probabilities are close to the theoretical

probabilities based on the assumption that the spinner is fair.6 Independent events are events where the outcome of one event is not affected by the outcome of the other event and

vice versa.7 Mutually exclusive events are events that cannot happen at the same time.8 Exhaustive events are events that cover every possible outcome.9 A tree diagram is used to help calculate combined probabilities for more than one event.10 Relative frequency is an estimate of probability based on experimental data.

Number of successful trialsRelative frequency = ————————————— .

Total number of trials� Ask the class to look through the table of vocabulary in the Pupil Book, to make sure they are familiar with the terms there.




15

LESSON

15.1

Framework objectives – Revision of probability

Use the vocabulary of probability in interpreting results involving uncertainty andprediction.Identify all the mutually exclusive outcomes of an experiment; know that the sumof probabilities of all mutually exclusive outcomes is 1 and use this when solvingproblems.Estimate probabilities from experimental data.Compare experimental and theoretical probabilities in a range of contexts;appreciate the difference between mathematical explanation and experimentalevidence.


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Plenary� Explain to the class that it is important to be able to calculate probabilities from

experimental and theoretical situations in order to make and test hypotheses.� Ask the class to explain the meaning of bias. Point out that bias can be tested either by

looking at raw data or by comparing experimental and theoretical probabilities.� As an example, show the class the table below, which gives the results of spinning a

four-sided spinner 40 times. Ask the class if they think the spinner is biased, just bylooking at these experimental frequencies.

� Now ask them to work out the theoretical probabilities and compare with theexperimental probabilities.

� From inspection of the data, the spinner appears to be biased against landing on 2,because this occurs much less than the other numbers. Comparison of the experimentalwith the theoretical probability verifies this conclusion (experimental = 5––40 = 0.125;theoretical = 1–4 = 0.25).

� Finally, point out that minor differences between expected frequencies and actualfrequencies do not necessarily mean that there is bias.

Number on spinner 1 2 3 4Frequency 12 5 10 13



1 a 8; HHH, HHT, HTH, THH, TTH, THT, HTT, TTT b 1–8 c 3–8 d 7–82 a 0.4 b 0.6 3 a 7––50 b 12––50 = 6––25 c 31––504 a 6––50 or 0.12 b 1–5 or 0.2 c Probably not fair as the probabilities are quite different. d 30 5 a 1–5 b 3–5 c 3––206 a b You only know that 4 out of the first 5 games were won.

c 0.76d Number of games 5 10 15 20 25

Relative frequency of wins 0.8 0.7 0.67 0.75 0.76Number of wins 4 7 10 15 19

00 5 10 15

Number of games

Rel

ativ

e fr

eque

ncy

20 25

0.2

0.4

0.60.8

1

Ho

me

wo

rk Two four-sided spinners are each spun 80 times. The results are shown below.

For each spinner state whether you think it is biased by comparing i the individual frequencies ii the experimentaland theoretical probabilities.

1st spinner

2nd spinner

AnswersThe 1st spinner is probably not biased: i as the frequencies are all close to 20ii as the experimental probabilities are 20––80 = 0.25, 21––80 = 0.2625, 19––80 = 0.2375, 20––80 = 0.25, which are all close to the theoretical

probability of 0.25.The 2nd spinner is possibly biased: i as the frequencies are not close to 20ii as the experimental probabilities are 25––80 = 0.3125, 17––80 = 0.2125, 16––80 = 0.2, 22––80 = 0.275, which are not very close to the theoretical

probability of 0.25. It could still be argued here that these are sufficiently close to 0.25 to suggest the spinner is fair.

Number on spinner 1 2 3 4

Frequency 25 17 16 22

Number on spinner 1 2 3 4

Frequency 20 21 19 20

� event� outcome� random� probability scale� experimental

probability� theoretical

probability� relative

frequency� expectation � bias� fair� sample� sample space � exhaustive� independent� mutually

exclusive

Key Words

Extension Answers

1 a Independent as 2nd roll is not affected by outcome of 1st roll.b Not independent – the chances of winning with 2nd ticket are increased as there is 1 ticket

fewer to choose from.c Not independent.

2 a Not mutually exclusive as 2 is both even and prime.b Mutually exclusive as the outcomes do not overlap.c Not mutually exclusive as ‘at least one tail’ includes ‘two tails’.

3 a Exhaustive as all possible outcomes are included.b Not exhaustive as the outcome landing on the number 3 is not included.c Exhaustive as all possible outcomes are included.

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Oral and mental starter� Write on the board, ‘Teenagers are better at probability than adults’.� Ask the class how they could investigate this.� Students might suggest writing a set of probability questions to be given to both

teenagers and adults. They could then record the results for their samples andcompare the experimental probabilities of answering particular questionscorrectly for teenagers and adults.

� You could discuss how they would decide which people to use in the sample.

Main lesson activity� The activities given in this section could easily take two lessons, depending on

the amount of detail asked for when carrying out the investigation. You maywish to ask the students to collect certain data prior to the lesson or you maydecide to provide the students with secondary data.

� Continuing from the oral and mental starter, show the class the handling datacycle and the related checklist for completing a probability investigation, on pp. 232–233 of Pupil Book 3. The list is also reproduced opposite.

� Go through and consider how each point applies to the example from the oraland mental starter:�� [Statement of topic] Compare the abilities of teenagers and adults at

probability.�� [Hypothesis] ‘Teenagers are better at working out theoretical probabilities

than adults’.�� [Sample size] Look at 30 teenagers and 30 adults. It would not be sensible to

use students in your class who have just revised probability. You may evenwant to test a different area of mathematics if using your fellow students as asample.

�� [Foreseen problems] Adults may be reluctant to answer the questions given.Choosing the sample may be difficult.

�� [Identify any sources of bias and plan how to minimise them] Avoid usingall one age group for adults.

�� [Data collection] Record the number of correct and incorrect answers, thenumber of people who declined to do the questions and any other factorswhich may affect your results.

�� [Identify what extra information may be required] Ask yourself the question,“How can I extend this problem, using more complex techniques which willprovide more reliable results?”

�� [Analysis] Produce statistical diagrams to compare the success rates ofteenagers and adults, and calculate the experimental probability of eachanswering a probability question correctly.

�� [Limitations of any assumptions made] Children may have had a morerecent experience than adults of work on probability, whereas adults mayquickly understand the topic, if they were given a short briefing.

�� [Conclusion] State whether you agree with the initial hypothesis based onyour results.

� The class can now do Exercise 15B from Pupil Book 3, working in small groups.


LESSON

15.2

Framework objectives – A probability investigation

Compare experimental evidence in a range of contexts; appreciate the differencebetween mathematical explanation and experimental evidence.


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Plenary� Having observed the students working on a choice of investigations, you may

wish to give one or more groups the opportunity to present their findings so farto the rest of the class.

� It is important that the students are encouraged to be critical of their own workand that they can recognise any limitations.

� Discuss how the problem can be extended using more complex analysis. � Discuss the effect of a small number of trials on the reliability of any relative

frequencies as estimates of theoretical probability.

Checklist for completing a probability investigation� Specify the problem and plan

�� statement of problem or topic to investigate�� hypothesis stating what you think the investigation will show�� how you will choose your sample and size�� any practical problems you foresee�� identify any sources of bias and plan how to minimise them�� how you will obtain your data�� identify what extra information may be required to extend the project

� Collect data from a variety of sources�� follow initial plan and use a suitable data-collection sheet

� Process and represent data�� analysis of your results using appropriate statistical calculations and diagrams

� Interpret and discuss data�� comparison of results with your original hypothesis�� list of any factors which might have affected your results and how you could

overcome these in future�� consider the limitations of any assumptions made�� a final conclusion


Ho

me

wo

rk Choose one of the following tasks.

1 Complete the investigation started in the lesson by writing up the report.

2 Collect data in order to investigate the ability of teenagers and adults at working out theoreticalprobabilities.

3 If you have completed the report of your first investigation, then carry out and write up anotherdetailed investigation of your own choice.

� hypothesis� experimental

data� mathematical

explanation� statistical report� experimental

probability� theoretical

probability� Relative

frequency

Key Words

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Oral and mental starter� Ask students to mentally solve x – 1 = 0. They should find that x = 1 quickly

(answers can be written on mini white boards).� Repeat with x + 4 = 0 (–4), x – 3 = 0 (3), x + 3 = 0 (–3).� Now ask students to give two values for a and b that solve a × b = 0. Ask

students to write their answers down on mini white boards as a = ? and b = ?.� Check answers. Some could be written on the board. Discuss the common

characteristics, which should be that either a or b should be zero in each case.� Now ask students to write down a value for x that will make (x – 2)(x + 4) = 0.� Once again check answers. Hopefully they will be either x = 2 or x = –4.

Main lesson activity� This is a lesson on solving quadratics that factorise.� Ask students if they can find a value for x that solves the quadratic equation

x2 + 6x – 7 = 0.� They may spot the answer x = 1, but are unlikely to spot the answer of x = –7.� Outline the method. First, factorise and then solve each bracket equal to zero.

i.e. x2 + 6x – 7 = 0 ⇒ (x – 1)(x + 7) = 0. So, either x – 1 = 0 ⇒ x = 1, or x + 7 = 0 ⇒ x = –7.

� Repeat with other examples such as:x2 – x – 6 = 0 (x = 3 or –2), x2 + 4x + 3 = 0 (x = –1 or –3), x2 – 8x + 15 = 0 (x = 3 or 5).


� The class can now do Exercise 16A from Pupil Book 3 or Exercise 10K (page 256) from the Higher Mathematics for GCSE textbook.


GCSE PreparationCHAPTER

16

LESSON

16.1

Framework objectives – Reinforcement of Number

Solving quadratic equations.

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Plenary� Ask the students if they can solve 2x2 – 3x + 1 = 0.� They may spot x = 1 as a solution. Give a clue that the other answer is a fraction

between 0 and 1.� They should find 1–2 fairly quickly.� Discuss ways that this could be solved mathematically.� The students may establish that the factorisation is (x – 1)(2x – 1) = 0 and that

these brackets solve to 1 and 1–2.� Do more examples with a non-unit coefficient of x2, if time allows, such as

2x2 – 5x + 3 = 0.



1 a x = –1 or 1 b x = 2 or –5 c x = 3 or –6 d x = –4 or –3 e x = –2 or –7f x = 3 or 8 g x = 8 or –1 h x = –3 i x = 4

2 a x = –1 or –2 b x = –5 or –6 c x = –2 or –4 d x = 3 or 2 e x = 5 or 2f x = 1 or 4 g x = –5 h x = 4 i x = 5 or –3 j x = 3 or –5 k x = 6 or –4l x = 3 or –2 m x = 9 or 1 n x = 6 or –3 p x = –1

Ho

me

wo

rk 1 Solve these equations.

a (x + 3)(x – 4) = 0 b (x – 1)(x + 6) = 0 c (x – 7)(x + 6) = 0

d (x + 5)(x + 2) = 0 e (x – 3)(x + 6) = 0 f (x – 9)(x – 3) = 0

2 First factorise, then solve these equations.

a x2 + 8x + 15 = 0 b x2 + 13x + 30 = 0 c x2 + 4x – 5 = 0

d x2 – 9x + 14 = 0 e x2 + 4x – 21 = 0 f x2 – 4x + 4 = 0

Answers1 a x = –3 or 4 b x = 1 or –6 c x = 7 or –6 d x = –5 or –2 e x = 3 or –6 f x = 9 or 32 a x = –5 or –3 b x = –3 or –10 c x = 1 or –5 d x = 2 or 7 e x = 3 or –7 f x = 2

� quadraticequations

� unitarycoefficient

Key Words

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Oral and mental starter� Give students the factors of 12 in pairs and the factors of 15 on each side of a

vertical line, i.e.1 2 3 1 312 6 4 15 5

� Ask the students to find a combination of products using one pair of factors fromeach side that give a total of 28. This first exemplar may need to bedemonstrated i.e.

2 3 2 × 5 + 6 × 3 = 28.6 5

� Repeat with totals of 29(3 × 3 + 4 × 5), 27(3 × 5 + 4 × 3 or 1 × 15 + 12 × 1), 41(1 × 5 + 12 × 3).

� Now ask for the total 8. Suggest to the students that negative values could beused. The answer is

2 3 2 × –5 + 6 × 3 = 8.6 –5

� Repeat with totals of 24(2 × 15 + 6 × –1), –3(3 × –5 + 4 × 3).� Do more examples if time available.

Main lesson activity� This is a lesson on factorising quadratics with a non-unit coefficient for x2.� Students have met the idea in previous plenaries, but a method for solution has

not been outlined.� Ask the students to factorise 2x2 + 9x + 4.� They may have an intuitive idea of the answer but they also need to have a

method outlined.� There are several methods, two of which are outlined in the Pupil Book.� Example: Find factors of the x2 coefficient and of the constant term.

For 2x2 + 9x + 4, the factors of 2 are 1 × 2 and the factors of 4 are 1 × 4.Now find a combination that gives the coefficient of x, e.g.

1 1 1 × 1 + 2 × 4 = 92 4

The brackets are then the ‘opposite’ to the pairs i.e. (x + 4)(2x + 1).� Example: Factorise 6x2 – 17x + 12.

The brackets must start (3x …)(2x …) or (6x …)(x …) and the constant term hasfactors 3 × 4, 2 × 6, 1 × 12.By trial and improvement we can find that the combination (3x – 4)(2x – 3) works.


� The class can now do Exercise 16B from Pupil Book 3 or Exercise 10L (page 257) from the Higher Mathematics for GCSE textbook.


LESSON

16.2


Factorisation of quadratics of the form ax2 + bx + c, where a > 1.

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Plenary� Ask the students if they can solve the quadratic equation 2x2 – 3x – 2 = 0.� They should be able to put all previous ideas together to explain the process of

factorisation and solving each bracket.e.g. (2x + 1)(x – 2) = 0 giving x = – 1–2 or x = 2

� Repeat with other examples, if time allows, such as 3x2 – 5x – 2 = 0.



1 a 2x2 + 11x + 5 b 3x2 + 9x – 12 c 4x2 – 18x + 20 d 6x2 – 15x – 21e 4x2 + 24x + 36 f 9x2 – 24x + 16 g 6x2 – 22x – 8 h 8x2 + 10x – 3 i 4x2 – 1

2 a (2x + 1)(x + 3) b (x + 2)(2x + 5) c (x + 4)(3x + 1) d (x – 1)(2x + 1)e (2x + 1)(3x + 2) f (x – 2)(2x + 3) g (x + 3)(2x – 3) h (2x + 1)2i (4x – 1)(x + 2) j (5x + 1)(x + 2) k (3x – 1)(x + 1) l (4x + 1)(2x + 1)m (x – 2)(3x + 1) n (2x + 1)(3x – 1) p (4x + 1)(x – 3) q (2x – 3)(2x + 5)r (x – 7)(2x + 5) s (x – 5)(2x + 5) t (3x – 1)(x + 5) u (3x + 1)2v (2x + 3)(5x – 1)

Ho

me

wo

rk 1 Expand these brackets into quadratic expressions.

a (3x + 1)(x – 4) b (3x – 1)(x + 5) c (2x – 1)(2x + 3)

d (3x – 2)(3x + 2) e (3x – 1)2 f (2x + 5)2

2 Factorise the following quadratic expressions.

a 2x2 – 7x – 4 b 2x2 + 13x + 15 c 3x2 + 5x – 2

d 4x2 + 23x – 6 e 6x2 – 5x + 1 f 6x2 + 11x + 3

g 5x2 – 26x + 5 h 6x2 – 5x – 6 i 4x2 – 16x + 15

Answers1 a 3x2 – 11x – 4 b 3x2 + 14x – 5 c 4x2 + 4x – 3 d 9x2 – 4 e 9x2 – 6x + 1 f 4x2 + 20x + 252 a (2x + 1)(x – 4) b (2x + 3)(x + 5) c (3x – 1)(x + 2) d (4x – 1)(x + 6) e (3x – 1)(2x – 1)

f (3x + 1)(2x + 3) g (x – 5)(5x – 1) h (3x + 2)(2x – 3) i (2x – 3)(2x – 5)


� non-unitarycoefficients

Key Words

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Oral and mental starter� Ask students to mentally solve 2x – 1 = 0. They should find that x = 1–2. Answers

can be written on mini white boards.� Repeat with 3x + 4 = 0 (–11–3), 4x – 3 = 0 ( 3–4 ), 2x + 3 = 0 (–11–2), 5x – 1 = 0 ( 1–5),

3x – 2 = 0 ( 2–3).� Do more examples if necessary.� Now ask students to write down a value for x that will make (3x – 2)(4x + 1) = 0.� Once again check answers. They should be x = 2–3 or x = – 1–4.

Main lesson activity� This is a lesson on solving quadratic equations with non-unitary coefficients.� Ask students if they can find a value for x that solves the quadratic equation

2x2 – x – 1 = 0.� They may spot the answer x = 1, but are unlikely to spot the answer of – 1–2.� Outline the method. First factorise and then solve each bracket equal to zero.� i.e. 2x2 – x – 1 = 0 ⇒ (2x + 1)(x – 1) = 0, so either x – 1 = 0 ⇒ x = 1 or

2x + 1 = 0 ⇒ x = – 1–2.� Repeat with other examples such as 3x2 + 2x – 1 = 0 (x = 1–3 or –1),

12x2 + 5x – 2 = 0 (x = – 2–3 or 1–4), 8x2 – 14x – 15 = 0 (x = – 3–4 or 21–2).� Do more examples if necessary.� Ask students to solve the quadratic equation x2 + x = 12.� This is not in the correct form to factorise and solve. Encourage students to

always rearrange quadratic equations into the form ax2 + bx + c = 0.� The equation above then becomes x2 + x – 12 = 0, which can be factorised and

solved to give x =3 or –4.� Repeat with 2x(x – 7) = 6 – 3x. This must be expanded and then collected into

the correct form.2x2 – 14x = 6 – 3x ⇒ 2x2 – 11x – 6 = 0 ⇒ (2x + 1)(x – 6) = 0 ⇒ x = – 1–2 or 6.


� The class can now do Exercise 16C from Pupil Book 3 or Exercise 10M (page 258) from the Higher Mathematics for GCSE textbook.


LESSON

16.3


Solving quadratic equations of the form ax2 + bx + c = 0, where a > 1.

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Plenary� Write the following equation on the board x4 – 5x2 + 4 = 0.� Explain that the equation has 4 solutions since the highest power of x is 4.� Ask students if they can spot any of the solutions. They may spot 1 and 2 but

may not spot –1 and –2.� Outline the method of solution (x2 – 1)(x2 – 4) = 0.

So, x2 – 1 = 0 ⇒ x2 = 1 ⇒ x = ±1 or x2 – 4 = 0 ⇒ x2 = 4 ⇒ x = ±2.� Repeat with 4x4 – 37x2 + 9 = 0.

(4x2 – 1)(x2 – 9) = 0 ⇒ x = ± 1–2 or ±3.



1 a x = 1–2 or 3 b x = 2 or – 1–3 c x = –4 or 1–2 d x = –11–2 or 1–2 e x = –6 or 1–2f x = –2 or 1–4 g x = –21–2 or – 1–3 h x = 1–3 or – 1–2 i x = –11–2 j x = 1–4 or 1–5k x = 21–2 or 2–3 l x = –1–3 m x = 1–2 n x = – 5–6 or 1 p x = –12–3 or 2

2 a x = –2 or 1 b x = 1 or –1 c x = – 1–2 or 1 d x = – 2–3 or 1 e x = 1–4 or 2f x = 1–2 g x = 5 or –5 h x = – 1–2 or 11–3 i x = – 1–4 or 1–5

Ho

me

wo


a 2x2 – 15x + 7 = 0 b 3x2 – 5x + 2 = 0 c 2x2 – 9x – 5 = 0

d 24x2 + 14x – 5 = 0 e 6x2 + 23x + 20 = 0 f 6x2 – 23x + 7 = 0

2 Solve these equations

a x2 + x = 6 b 2x(x + 4) = 3(x – 1) c 8x2 – 3x + 4 = 2x2 + 2x + 3

Answers1 a 7 or 1–2 b 2–3 or 1 c – 1–2 or 5 d 1–4 or – 5–6 e –21–2 or –11–3 f 31–2 or 1–32 a 2 or –3 b –3 or 1–2 c 1–3 or 1–2



� roots

Key Words

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Oral and mental starter–4 + √

—–41 –4 + 6.5 2.5

� Ask students to estimate the value of ————– (≈ ———— = —— = 1.25).2 2 2

–3 – √—–52 –5 + √

—–31 –1 – √

—–13

� Repeat with ————– (≈ –2.5), ————– (≈ 0.25), ————– (≈ –1.1), 4 2 4

–4 + √—–21

————– (≈ 0.25).2

–5 ± √—–60

� Repeat with (two answers required) ————– (≈ 1.25 or –6.25), 2

–1 ± √—–17

————– (≈ 0.75 or –1.25).4

Main lesson activity� This is a lesson that introduces the quadratic formula.� Introduce the formula:

–b ± √———b2 – 4ac

—x = ————————

2aand explain that a, b and c are the coefficients of x2, x and the constant term inthe general quadratic equation ax2 + bx + c = 0.

� Work through some examples but encourage students to follow some basicprinciples:

Substitute values in brackets before attempting to work anything out.Take particular care with: The change of sign of b.

Squaring b inside the root, particularly if it is negativeWorking out 4 × a × c correctly particularly if c is negative.Dividing the whole top line by 2a, not just the square root.

� Example: Solve 2x2 + 3x – 4 = 0.First, identify a, b and c (a = 2, b = 3, c = –4.

–(3) ± √———(3)2 – 4(2)

——(–4)

——Substitute into the formula x = ———————————–

2(2)–(3) ± √

—–41 –3 ± 6.403

Now evaluate the square root x = —————– = —————–2(2) 4

–3 + 6.403 +3.403So, the solutions are x = —————– = ———– = 0.8508 or

4 4–3 – 6.403 –9.403—————– = ———– = –2.3508

4 4Answers are normally given to 2 decimal places, so x = 0.85 or –2.35.

� Repeat with 3x2 – 4x – 2 = 0 (1.72, –0.39).� Repeat with x2 – 4x – 1 = 0, but this time leave the answer in surd form.

4 ± √—–20

The answers are ————– or 2 ± √––5 .

2

� The class can now do Exercise 16D from Pupil Book 3 or Exercise 10N (page 259) from the Higher Mathematics for GCSE textbook.


LESSON

16.4


The quadratic formula.

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Plenary� Ask students to solve 2x2 + 3x + 5 = 0 using the quadratic formula.� This example could be worked through on the board by a student.� The formula gives

–3 ± √—–––31

x = ————– which cannot be solved due to the square root.4

� Explain that when a quadratic equation is solved, the roots are where the graphcrosses the x-axis.

� This gives three situations

� This will be useful for the investigation that ends this chapter.

2 roots 1 (repeated) root No roots



1 a x = 4 or –6 b x = 1 or –7 c x = –2 or 1–2 d x = 21–2 or –4 e x = 0 or –5f x = 6 or –6

2 a x = 1.14 or –2.64 b x = 1.59 or –1.26 c x = 1.32 or –5.32d x = 2.22 or –0.22 e x = 0.16 or –6.16 f x = 0.19 or –2.69g x = 1.82 or 0.18 h x = 2.19 or –0.69 i x = 4.56 or 0.44

(3 ± √––5) –4 ± √

–—20 –4 ± √

–—12

3 a ———— b ————– or –2 ± √––5 c ————– or –2 ± √

––3

2 2 2–6 ± √

–—32 –10 ± √

–—92 –2 ± √

––8d ————– or –3 ± √

––8 e ————— or –5 ± √

–—23 f ———— or –1 ± √

––2

2 2 2

Ho

me

wo

rk 1 Solve these equations using the quadratic formula. All answers are whole numbers or fractions.

a x2 + 4x – 5 = 0 b 2x2 + 5x – 3 = 0 c 6x2 – 19x + 10 = 0

2 Solve these equations, giving your answers to 2 decimal places.

a x2 + 7x – 10 = 0 b 2x2 – x – 4 = 0 c 4x2 + x – 7 = 0

3 Solve these equations, giving your answer in surd form.

a x2 – 4x – 2 = 0 b x2 + 6x – 1 = 0 c x2 + 5x – 2 = 0

Answers1 a 1 or –5 b 1–2 or –3 c 2–3 or 21–22 a 1.22 or –8.22 b 1.69 or –1.19 c 1.20 or –1.45

4 ± √–—24 –6 ± √

–—40 –5 ± √

–—33

3 a ———— or 2 ± √––6 b ———–— or –3 ± √

–—10 c ———–—

2 2 2

� quadraticformula

� coefficients

Key Words

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Oral and mental starter� Ask students to fill in the missing number in problems such as –22 – 9 = ?

(Answer –13).� Repeat with other examples such as –32 + 5 (answer –4), –42 – 1 (answer –17).� Make sure that students understand that –32 = –(3 squared) and not (–3) squared.� Repeat with more examples if necessary.

Main lesson activity� This is a lesson on completing the square.� Ask students to expand (x – a)2 and (x + a)2.� Hopefully they will get the answer x2 – 2ax + a2 and x2 + 2ax + a2.� Now ask if they can fill in the missing numbers in x2 + 4x – 3 = (x + …)2 – … .� Eventually they will probably come up with the answer (x + 2)2 – 7.� Discuss how this relates to the identities established at the start of the lesson.� Work through the following examples on completing the square.� Example: Write x2 + 6x in the form (x + a)2 – b.

The value, a, inside the bracket is half the coefficient of x. The value b is thesquare of this, So, x2 + 6x = (x + 3)2 – 9.

� Repeat with x2 + 12x ((x + 6)2 – 36), x2 – 4x ((x – 2)2 – 4) and x2 + 30x ((x + 15)2 – 225).

� Example: Write x2 + 6x – 7 in the form (x + a)2 – b.The value of a is half the coefficient of x. The value b is the sum of the existingconstant term and the square as before.So, x2 + 6x – 7 = (x + 3)2 – 9 – 7 = (x + 3)2 – 16.

� Repeat with x2 + 12x + 15 ((x + 6)2 – 21), x2 – 4x – 3 ((x – 2)2 – 7) and x2 + 30x + 100 ((x + 15)2 – 125).

� Demonstrate how to solve equations using completing the square.� Example: Solve x2 + 6x – 4 = 0.

x2 + 6x – 4 = 0 ⇒ (x + 3)2 – 13 = 0 ⇒ (x + 3)2 = 13 ⇒ x + 3 = ±√—–13 ⇒

x = –3 ± √—–13.

� Repeat with x2 – 6x – 2 = 0 (x = 3 ±√—–11) and x2 + 10x + 15 = 0 (x = –5 ± √

—–10 ).

� Complete more examples if necessary.

� The class can now do Exercise 16E from Pupil Book 3 or Exercise 10P(page 260) from the Higher Mathematics for GCSE textbook.


LESSON

16.5


Solving quadratics by completing the square.

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Plenary� Ask students to solve x2 + 3x – 2 = 0 using the completing the square method.� A student could be asked to come to the board to give the solution.� The odd coefficient of x will cause problems, particularly when taking a square

rooti.e. x2 + 3x – 2 = 0 ⇒ (x + 11–2)2 – 41–4 = 0 ⇒ (x + 11–2)2 = 41–4 ⇒ x + 11–2 = ±√

—–41–4

⇒ x = –11–2 ± √—–41–4 .

� Now ask students to solve 2x2 + 4x – 3 = 0 using the completing the squaremethod.

� 2x2 + 4x – 3 = 0 ⇒ 2(x2 + 2x) – 3 = 0 ⇒ 2[(x + 1)2 – 1] – 3 = 0 ⇒2(x + 1)2 – 2 – 3 = 02(x + 1)2 = 5, ⇒ (x + 1)2 = 21–2 ⇒ x + 1 = ±√

—–21–2 ⇒ x = –1 ± √

—–21–2 .



1 a (x + 4)2 – 16 b (x – 1)2 – 1 c (x – 6)2 – 36 d (x – 7)2 – 49 e (x + 2)2 – 4f (x + 1)2 – 1

2 a (x + 4)2 – 17 b (x – 1)2 + 2 c (x – 6)2 – 31 d (x – 7)2 – 42 e (x + 2)2 – 7f (x + 1)2 – 6 g (x + 3)2 – 11 h (x + 5)2 – 34 i (x – 3)2 – 6

3 a x = –4 ± √–—17 b x = 3 or –1 c x = 6 ± √

–—31 d x = 7 ± √

–—42 e x = –5 or 1

f x = 3 ± √–—11 g 5 ± √

–—24 h 3 ± √

––5 i 4 ±√

–—11 j x = 1 ± √

––2 k x = –1 ± √

––6

l x = –6 ± √–—43

Ho

me

wo

rk 1 Complete the square for the following.

a x2 + 12x b x2 – 6x c x2 – 20x

2 Rewrite the following quadratic expressions by completing the square.

a x2 + 12x – 9 b x2 – 6x + 3 c x2 – 20x + 100

3 Solve the following quadratic equations using the completing the square method.

a x2 + 12x – 9 = 0 b x2 – 6x + 3 = 0 c x2 – 20x + 100 = 0

d x2 – 10x + 5 = 0 e x2 + 4x – 7 = 0 f x2 – 8x – 5 = 0

Answers1 a (x + 6)2 – 36 b (x – 3)2 – 9 c (x – 10)2 – 1002 a (x + 6)2 – 45 b (x – 3)2 – 6 c (x – 10)23 a x = –6 ± √

–—45 b x = 3 ± √

––6 c x = 10 d x = 5 ± √

–—20 e x = –2 ± √

–—11 f x = 4 ± √

–—21

� completing thesquare

� coefficient

Key Words

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Oral and mental starter� Ask students to find the square root of 81.� They will obviously give the answer of 9, (answers can be written on mini white

boards) but may not give –9. Remind them that both answers are possible.� Now ask students for the square root of x2. They will, hopefully, give ±x as the

answer.� Now ask for the square root of 4x2. Some will give the answer ±4x. Make sure

that they understand the answer is ±2x.� Repeat with the square roots of 9y2, 16z2, 25x2, 81x2, 100y2 etc.

Main lesson activity� This is a lesson on factorising the quadratic the difference of two squares.� Start by asking students to expand expressions of the type (x + 2)(x – 2) = (x2 – 4)

e.g. (x – 4)(x + 4) = (x2 – 16), (2x – 1)(2x + 1) = (4x2 – 1), (3x – 6)(3x + 6) = (9x2 – 36) etc.� Eventually ask for the expansion of (a – b)(a + b).� Discuss with students the similarities between the results.� Establish the result (a – b)(a + b) = a2 – b2.� Explain that this is known as the difference of two squares.� Now ask if students can reverse the result. i.e. can they factorise x2 – 36?� Explain that this can be written down as x2 – 62 = (x – 6)(x + 6).� Emphasise that it is important to identify both squares since failure to do this can

lead to errors such as (x – 36)(x + 36).� Repeat with x2 – 100 ((x –10)(x + 10)), 4x2 – 9 ((2x – 3)(2x + 3)), 9x2 – 16y2

((3x – 4y)(3x + 4y)).� Do more examples if necessary.

� The class can now do Exercise 16F from Pupil Book 3 or Exercise 10J (page 255) from the Higher Mathematics for GCSE textbook.


LESSON

16.6


Difference of two squares.

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Plenary� Ask the students to simplify (x + 2)2 – (x – 4)2.� If possible, ask a student who wants to expand brackets to work through the

problem on the board, i.e. (x + 2)(x + 2) – (x – 4)(x – 4) = (x2 + 4x + 4) – (x2 – 8x + 16) = x2 + 4x + 4 – x2 + 8x – 16 = 12x – 12.

� Now, ask if the original expression could be written in any other way. Somestudents may recognise this as the difference of two squares.

� Make sure that the students understand that the difference of two squares is anidentity, i.e. any values or expressions can be used for a and b in a2 – b2 = (a – b)(a + b).

� Using brackets (because of the minus signs), write the above as:((x + 2) – (x – 4))((x + 2) + (x – 4))

= (x + 2 – x + 4)(x + 2 + x – 4) = (6)(2x – 2)= 12x – 12.

� Discuss the advantages of both methods.



1 a x2 – 1 b x2 – 25 c x2 – y2 d 4x2 – 1 e x2 – 4y2 f 4x2 – 9y2

2 a (x – 10)(x + 10) b (x – 2)(x + 2) c (x – 6)(x + 6) d (x – 9)(x + 9)e (x – 8)(x + 8) f (x – 11)(x + 11) g (x – z)(x + z) h (2x – 5)(2x + 5)i (x – 3y)(x + 3y) j (4x – 3)(4x + 3) k (2x – 5y)(2x + 5y) l (5x – 8)(5x + 8)m (3 – x)(3 + x) n (2x – 6)(2x + 6) p (6x – 1)(6x + 1)

Ho

me

wo


a (x + 11)(x – 11) b (2x – 3)(2x + 3) c (5x – 2y)(5x + 2y)


a x2 – 144 b x2 – 225 c 4x2 – 36

d 81x2 – 64 e x2 – 4y2 f 16x2 – 121

g x2 – 9z2 h 4x2 – 25y2 i 81x2 – 16y2

Answers1 a x2 – 121 b 4x2 – 9 c 25x2 – 4y2

2 a (x – 12)(x + 12) b (x – 15)(x + 15) c (2x – 6)(2x + 6) d (9x – 8)(9x + 8) e (x – 2y)(x + 2y)f (4x – 11)(4x + 11) g (x – 3z)(x + 3z) h (2x – 5y)(2x + 5y) i (9x – 4y)(9x + 4y)

� difference of twosquares

Key Words

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Oral and mental starterThere is no starter as the investigation will need to be introduced via a class discussion.

Main lesson activityThe investigation in the Pupil Book is reproduced here. Students will need access to graphical calculators orcomputers with a graph-drawing application package. The investigation will take more than one lesson.

Introduce the investigation and ask the students to suggest ideas for getting started.

For example, they could start by looking at graphs of y = x2, y = 2x2, y = 3x2 etc.Then look at graphs of y = x2 + 1, y = x2 – 2, y = x2 + 3 etc.Then look at graphs of y = x2 + 2x, x2 – 3x, x2 + 5x etc.Once they have an idea of the effect of a, b and c individually they could then look at more general equations.Students should be encouraged to look at the roots from the quadratic formula or the completing the squaremethod. The vertex also could be investigated via the completing the square method.

The graph of a quadratic equation has a characteristicshape called a parabola.

Investigate the relationship between the values a, b andc in the graph y = ax2 + bx + c and the points P, Q(where the graph crosses the x-axis), the point R (wherethe graph crosses the y-axis) and the point S (the ‘vertex’or turning point of the graph).

You should use a graph plotting program or a graphicalcalculator to help you.You will not gain any credit for spending a lot of timedrawing graphs accurately.You will gain credit for a systematic investigation intothe effect of a, b and c on the graph.


LESSON

16.7


An investigation into y = ax2 + bx + c.

x

y

Q

SR

P

y = ax2 + bx + c

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PlenaryThere is no plenary. Students could be stopped working at various points and ageneral discussion on progress could be held. This can be helpful for students whoare having difficulty but avoid the ‘bush fire’ effect of one student giving the rest ofthe group the answer.


Ho

me

wo

rk Students could be asked to continue the work at home if they have computer facilities or asked to lookon the Internet for information on the quadratic equation.

� roots� factors� vertex� intersect� coefficient

Key Words

GCSE Answers

1 0 and 52 a (x – 4)(x – 2) b x = 4, x = 23 a i 3(pq – 2r) ii (c – 4)(c – 5)

b x = –7 or x = 2p – 5

4 a (3p + 1)(p + 5) b ———3p + 1

5 –5.74, 1.746 a 9x2y6 b 3.14, –0.647 1.28, –0.788 a = 9 b = 2 c = –59 a = –5 b = –7

10 p = 2 q = –4

9B3LP_16.qxd 18/09/2003 15:22 Page 189


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ISBN 0 00 713881 4

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PRACTICE PAPER ANSWERS AND TUTORIAL

Maths Frameworking Year 9 Teacher’s Pack 3 Practice Paper Answers ISBN 0 00 713881 4

Part Mark Answer Tutorial

(a) 2 For full marks you have • This question is testing whether you can understand a to show something like: rule and carry it out, so your answer has to follow the same 10% of 180 = 18 method as the example.20% of 180 = 36 • You will be given some credit for using a different correct 5% of 180 = 9 method such as the method shown in part (b):

25% of 180 is 50% of 180 is 9036 + 9 = 45 25% of 180 = 45

(b) 2 For full marks you have • Again this question is testing whether you can to show something like: understand a rule and carry it out, so your answer has to 50% of 460 = 230 follow the same method as the example.25% of 460 = 115 • You will be given some credit for using a different correct

method such as the method shown in part (a):10% of 460 = 4620% of 460 = 925% of 460 = 23

25% of 460 is 92 + 23 = 115

Total 4

Question 2


(a) 3 8 grey cubes • You can count cubes of each type that are showing.12 white cubes • You can still gain 1 mark if you get three of the four 6 dotted cubes numbers right, or get the right numbers in the wrong order.1 striped cube

(b) 2 8 grey cubes • You might need to make a drawing of a 4 × 4 × 4 cube. 24 white cubes You have to realise that there are 64 cubes altogether.24 dotted cubes • You can still gain 1 mark if you get three of the four 8 striped cubes numbers right, or get the right numbers in the wrong order.Total 64 cubes

Total 5

Question 1



(a) 2 21 packs • This is long division. There are a couple of ways of doinglong division. The first is the traditional method:

2 0 r. 1012)2 5 0

2 4

1 00

1 0

• As the remainder is fairly easy to work out you may havedone this as a short division.

2 0 r. 1012)2 51 0

• Another method is the method of repeated subtraction:

2 5 010 × 12 = 1 2 0

1 3 0

10 × 12 = 1 2 0

1 0

The number of 12s in 250 is 20 remainder 10, which meansthat the school will need 21 packs.• An answer of 21 gains 2 marks. If you make one mistake,or say 20 packs, you will only gain 1 mark.

(b) 2 5500 p • This is long multiplication.1 £55 • There are several ways of doing this. One way is the grid

method.

or 250× 22

50050005500

• An answer of 5500 gains 2 marks. If you make onemistake you can still gain 1 mark.• To gain the last mark you need to convert your answerfrom pence to pounds. This means dividing by 100. You canstill gain this mark for dividing by 100, even if yourcalculation is wrong.

Total 5

20

2

200

4000

400

50

1000

100

5000

500

5500

Question 3




(a) 1 n + 7 • There are lots of rules about what you can and cannot dowith algebra. Another acceptable answer is 7 + n, but x + 7is wrong. You cannot change letters. Neither can yousimplify expressions wrongly. So n + 7 = 7n would bewrong but c = n + 7 would be allowed as it could be readas ‘number of cubes = n + 7’. It is safer to write down theexpression without using equals signs and not to try tosimplify the answer unless the question says ‘write youranswer as simply as possible’ or ‘simplify your answer’.

(b) 1 4m • You can write multiplications such as m × 4 and 4 × meither way around and they would be acceptable but do notwrite m4 as this is not correct.There are four cubes in each layer and m layers so theanswer is 4 × m or 4m.You can check your answer using numbers. Suppose thenumber of cubes is 20 then m would be 5 as 5 × 4 = 20.

(c) 1 You should have this • Each tower has the same number of cubes, so because box ticked: there are 4 cubes in each layer in the second tower it only n = m × 4 needs to be 1–4 of the height.

(d) 1 2 • You now know that n = 4m and n = m + 6.You can combine these by writing an equation m + 6 = 4m.Subtracting m from each side gives 3m = 6 and dividing by3 gives m = 2.• You could also answer this question by trying differenttower sizes. If m = 2 there will be 8 cubes in tower M, sothe other tower is 8 cubes high, which is 2 + 6 as required.

Total 4

Question 4


(a) 2 59 1 goal was scored in 14 games = 14 goals2 goals were scored in 8 games = 16 goals3 goals were scored in 7 games = 21 goals4 goals were scored in 2 games = 8 goals14 + 16 + 21 + 8 = 59 goals altogether

(b) 2 Team B 20% of 50 games is 10 games.You can work this out by saying 10% of 50 = 5 and doubling it.Team B scored 3 or more goals in 10 matches (5 + 2 + 3).

(c) 1 Any valid reason: Many different reasons could be acceptable such as; ‘It does not show how many games each team won’;‘It does not show how many goals were conceded’;‘Successful might mean different things to different people’.

Total 5

Question 5




3 A = Graph 4 • Graph 1 shows a runner going at a steady speed as the B = Graph 1 graph is a straight line. C = Graph 2 Graph 2 covers distance very quickly at the beginning and

then less quickly at the end so this runner is getting slower.Graph 3 covers distance very slowly at the beginning andthen more quickly at the end so this runner is getting faster.Graph 4 covers distance very quickly at the beginning, thenless quickly in the middle and then more quickly at the end. Graph 5 covers distance very slowly at the beginning, thenmore quickly in the middle and then less quickly at the end.• You would gain 1 mark for each correct answer.

Total 3

Question 6


3 LINE B y = 5 • Lines that are drawn horizontally, parallel to the x-axis, LINE C x = 5 are of the form y = a, where a is the value where the line LINE D x + y = 5 crosses the y-axis.

• Lines that are drawn vertically, parallel to the y-axis, areof the form x = b, where b is the value where the linecrosses the x-axis.• Line B passes through the point 5 on the y-axis and sohas the equation y = 5. You would gain no marks if youwrote ‘y is 5’.• Line C passes through the point 5 on the x-axis and sohas the equation x = 5. You would gain no marks if youwrote ‘x is 5’.• Line D passes through points where the coordinates addup to 5 (x + y = 5) such as (0, 5), (1, 4), (2, 3), (5, 0) so theequation is x + y = 5.There is one mark for each correct answer.

Total 3

Question 7


(a) 1 4 • Some formulae for the area of 2-D shapes that you needto know will be given at the front of the test papers. Othersyou need to learn. The formula for the area of aparallelogram is A = bh.If A = 20 and h = 5, then 20 = b × 5 and so b = 4.

(b) 1 4 • The formula for the area of a triangle is:

bhA = —–

2

10 × hIf A = 20 and b = 10, then 20 = ——— and so h = 4.2

Question 8



(c) 1 h = 10 • The formula for the area of a trapezium is:

(a + b)hA = ————

2

If A = 20, then (a + b) × h = 40. You need to find a numberwhich when multiplied by 4 gives 40, so h = 10.

1 a + b = 8 If A = 20, then (a + b) × h = 40. So (a + b) × 5 = 40, so a + b = 8.

(d) 2 x = 3 • Find x first by noticing that the two expressions for the Length = 10 cm length of the rectangle must be equal. This gives the Width = 2 cm equation:Perimeter = 24 cm 3x + 1 = 5x – 5 (Take 3x from both sides)

1 = 2x – 5 (Add 5 to both sides)6 = 2x (Divide both sides by 2)x = 3

• The length of the rectangle is 3x + 1 = 10 or 5x – 5 = 10.The formula for the area of a rectangle is A = lw. If A = 20and l = 10, then 20 = 10w. This gives w = 2.You would gain 1 mark if you showed that x = 3 or that l = 10.So the perimeter is 10 + 2 + 10 + 2 = 24 cm.

Total 6


(a) 2 • From the three patterns given, notice that the number of black squares is always 2 times the pattern number, and the number of white squares is always the patternnumber squared.• You would gain 1 mark if you gave two or three correctvalues or if you got the colours the wrong way round.

(b) 2 • The question asks for expressions for the number ofblack and white squares, so you must use n in youranswers, as shown in the table. Do not use equations suchas n = n × 2, as you would lose marks. You could write‘number of black squares = 2n’, but try to avoid usingequations when questions ask you to write expressions.For 2n you could also write 2 × n or n + n, and for n2 youcould also write n × n.

(c) 1 2n + n2 • The total number of squares in pattern number n is 2nfor the black squares + n2 for the white squares which is written 2n + n2.

(d) 2 3n +n2 • The number of black squares in the pattern follows thesequence 3, 6, 9 … increasing by 3 each time.• The total number of squares in pattern number n is 3nfor the black squares + n2 for the white squares which is written 3n + n2.

Total 7

n 2n n2

5 10 2512 24 144

Question 9




(a) 1 300 • If 2.5 × 30 = 75, then 0.25 × 300 = 75. Notice that to findtwo multiplications that have the same answer, you candivide one of the numbers by an amount (10 here), andmultiply the other number by the same amount.• You can, of course, do a division sum without acalculator:

75 750 7500—— = —— = ——– = 3000.25 2.5 25

1 0.1 or 1––10 • If 75 ÷ 1 = 75, then 7.5 ÷ 0.1 = 75. Notice that to find divisions that have the same answer, you can divide one ofthe numbers by an amount (10 here), and divide the othernumber by the same amount.

(b) 1 a = 18 • 4a + 3 = 75 (Subtract 3 from both sides)4a = 72 (Divide both sides by 4)

a = 18

1 b = 50 • 2b – 25 = 75 (Add 25 to both sides)2b = 100 (Divide both sides by 2)

b = 50

1 c = 9 or • c2 – 6 = 75 (Add 6 to both sides)c = –9, or both c2 = 81 (Take the square root of both sides)

c = 9

Since √—–81 is also –9, this answer would be accepted.

Total 5

Question 10



2 The completed diagram • The line on the diagram that shows the position of the should look like this: path must be drawn 1 square from the edge of the flower

bed on the sides.The line at the corners is a quarter-circle, radius 2 cm,centred at the corner, drawn using compasses.• Some allowance is made for diagrams that are slightlyinaccurate, but try to avoid drawing lines freehand. You areexpected to show that you can use a ruler and a pair ofcompasses in the tests.• You would gain 1 mark for an otherwise correct diagramwith the circle arcs wrongly drawn or omitted.

Total 2

Question 11



2 Any correct method to • The easiest way to show that the statement is wrong is show that the statement by substituting numbers for a and b into both sides of the is wrong. equation and then showing that the answers obtained are

not the same.For example, let a = 2 and b = 3.

2 3Then the left-hand side becomes –– + –– = 1 + 1 = 22 3(2+3) 5and the right-hand side becomes ——— = –– = 1.

5 52 is not equal to 1, so the values on both sides are notequal, and we have found an example to prove the studentsis wrong.

Total 2

Question 12


(a) 1 • The number of passengers waiting 15 minutes or longeris found by adding the last three frequencies of the barchart: 8+ 7 + 5 = 20.So P (passenger waits for 15 minutes or longer) = 20 1—— = ––.

100 5• Probability is written as a fraction or a decimal. Alwayscancel fractions if it is possible, although you would not losemarks if you did not. You would get no marks, though, forwriting ‘20 out of 100’ or ‘20 in 100’.

(b) 1 • There were 40 passengers waiting from 5 to 10 minutes,so approximately 20 passengers were waiting between 5and 7.5 minutes. Therefore the number of passengerswaiting 7.5 minutes or less is approximately 22 + 20 = 42.So P (passenger waits for 7.5 minutes or less) = 42 21—— = –— or 0.42.

100 50

(c) 2 10.15 • The last column of the table is given below:fx55

300225140157.5137.5

1015total waiting time for all passengers

The mean = —————————————————total number of passengers

1015= ——–100

= 10.15• You would gain 1 mark if your total in the fx column waswrong but you correctly divided this by 100.

42 21—— or –— or 0.42100 50

20 1—— or –– or 0.2100 5

Question 13




(d) 1 Possible answers are: • Make sure that you read the question carefully. You are decrease the time already told one way of improving the survey, so you need intervals for recording; to give a different way. Do not give answers that are do it on different days; irrelevant, such as ‘draw a pie chart’ or ‘use the mode’.do it on different routes.

Total 5


(a) 1 50 • When p = 5,

2 × 53a = ———

52 × 5 × 5 × 5= —————— = 2 × 25 = 50

5

1 10 • When p = 5,

5 × 5 × 6b = ————

3 × 55 × 6= ——–

330= —–3

= 10

(b) 1 5c 15cNotice how the d cancels, leaving —— = 5c. 3

(c) 1 7x + 6 This question involves multiplying out brackets.You need to know that a(b + c) = ab + ac and –a(b + c) =–ab – ac and –a(b – c) = –ab + ac. Multiplying out the brackets gives 3x + 12 – 6 + 4x. Notice the change in sign when multiplying out the secondbracket.Collecting like terms together gives 7x + 6.

1 x2 + 8x + 15 • The next three parts involve expanding two brackets, so you need to know that:(a + b) (c + d) = a(c + d) + b(c + d)

= ac + ad + bc + bd(x + 3) (x + 5) = x2 + 5x + 3x + 15

= x2 + 8x + 15

1 x2 – x – 2 • (x + 1) (x – 2) = x2 – 2x + x – 2= x2 – x – 2

1 x2 – 8x + 16 • (x – 4)2 = (x – 4)(x – 4)= x2 – 4x – 4x + 16= x2 – 8x + 16

Total 7

Question 14




(a) 1 Your graph should • y = x3 + 2, so the y-value is 2 more than at each point look like this: on y = x3. This makes the curve translate 2 units in the

positive y direction.

(b) 1 y = –x3 • Curve A is a reflection of y = x3 in the x-axis, so all the yvalues will now become negative.

(c) 1 Reflection in the y axis. • All the points on the graph can also re-plotted with thesign of the x values negative so the graph has been reflectedin the y-axis.

(d) 2 y < 2x and • The shaded region is below the line y = 2x, so y < 2x is y > x3 one inequality that describes the region. The shaded region

is also above the curve y = x3, so y > x3 is the otherinequality that describes the region.

• You will gain 1 mark for each correct inequality. Youwould get no marks if you ringed y < 2x and y > 2x or y < x3 and y > x3, as these are conflicting inequalities.

Total 5

x

y

y = x3 + 2

Question 15


(a) 1 3 × 104 is greater than • This part is about using numbers written in standard 4 × 103. form. 3 × 104 is a shorter way of writing 3 × 10000 = 30000 3 × 104 = 30 000 and 4 × 103 is a shorter way of writing 4 × 1000 = 4000.4 × 104 = 4000 • For example 4 000 000 can be written as 4 × 1 000 000 or,

in standard form 4 × 106.

(b) 2 500 and 0.5 × 103 • 5 × 102 = 5 × 100 = 500.This is the same as 0.5 × 103 or 0.5 × 1000.

(c) 1 80000 or 8 × 104 • This part uses the fact 10a × 10b = 10a+b

(4 × 105) × (2 × 10–1) = 8 × 105–1 = 8 × 104

= 8 × 10 000 = 80 000

1 200 000 000 or 2 × 108 10a

• This part uses the fact —– = 10a–b

10b

8 × 1010———— = 2 × 108 or 200 000 0004 × 102

Total 5

Question 16




(a) 1 • There are 4 shapes with only straight edges.

There are 3 shapes with only curved edges.

The are 3 shapes with both straight and curved edges.

(b) 1 7––10 • There are 7 shapes which do not have both straight andcurved edges (4 + 3) or (10 – 3) as there are 3 shapes withboth straight and curved edges.Therefore P (a shape does not have both straight andcurved edges) = 7––10.

(c) 1 1––2 • There are 6 shapes altogether which have curved edgesand 3 of them also have straight edges.Therefore P(shape has straight edges given that it hascurved edges) = 3––6 = 1––2.

(d) 1 6––10 × 5––9 • For this question you need to know that for two events Aand B:P(A and B) = P(A) × P(B)provided A and B are independent events (i.e. an outcomefrom one event does not determine an outcome of the otherevent).P(first shape has a curved edge) = 6––10

• If another shape is chosen this leaves 9 shapes to choosefrom, of which 5 have curved edges.So, P(second shape has curved edges) = 5––9.P(both shapes have curved edges) = 6––10 × 5––9.

Total 3

Question 17

Number Number Number of of of

shapes shapes shapes with with with

straight curved both edges edges straight only only edges

and curved edges

4 3 3




(a) 3 9πa2 – 4πa2 = 5πa2 • This is an example of an unstructured question. Thismeans that you have to do more than one step to get theanswer. It usually involves showing a fair amount ofworking. The formula for the area of a circle is A = πr2.Now use this formula to calculate the area of each separatecircle.For the large circle, r = 3a.So, A = πr2 = π × (3a)2 = π × 3a × 3a = 9πa2

For the small circle r = 2a.So A = πr2 = π × (2a)2 = π × 2a × 2a = 4πa2

The shaded area is given by 9πa2 – 4πa2 = 5πa2

• Remember to take away the area of the small circle.• You would gain 2 marks if you found the area of bothcircles but did not obtain the final answer.• You would gain one mark if you only found the area ofone circle.

(b) 2 • 5πa2 = 20 (Divide both sides by 5π)20 4

a2 = —– = –– (take the square root of both sides)5π π—–4 2

a = –– = ——√ π √––π

• You would only gain 1 mark if you tried to substitute3.142 for π, since the question asks for the answer in termsof π.

Total 5

2a = ——

√––π

Question 18


Oral and mental starter� Ask students to mentally solve x – 1 = 0. They should find that x = 1 quickly

(answers can be written on mini white boards).� Repeat with x + 4 = 0 (–4), x – 3 = 0 (3), x + 3 = 0 (–3).� Now ask students to give two values for a and b that solve a × b = 0. Ask

students to write their answers down on mini white boards as a = ? and b = ?.� Check answers. Some could be written on the board. Discuss the common

characteristics, which should be that either a or b should be zero in each case.� Now ask students to write down a value for x that will make (x – 2)(x + 4) = 0.� Once again check answers. Hopefully they will be either x = 2 or x = –4.

Main lesson activity� This is a lesson on solving quadratics that factorise.� Ask students if they can find a value for x that solves the quadratic equation

x2 + 6x – 7 = 0.� They may spot the answer x = 1, but are unlikely to spot the answer of x = –7.� Outline the method. First, factorise and then solve each bracket equal to zero.

i.e. x2 + 6x – 7 = 0 ⇒ (x – 1)(x + 7) = 0. So, either x – 1 = 0 ⇒ x = 1, or x + 7 = 0 ⇒ x = –7.

� Repeat with other examples such as:x2 – x – 6 = 0 (x = 3 or –2), x2 + 4x + 3 = 0 (x = –1 or –3), x2 – 8x + 15 = 0 (x = 3 or 5).


� The class can now do Exercise 16A from Pupil Book 3 or Exercise 10K (page 256) from the Higher Mathematics for GCSE textbook.


GCSE PreparationCHAPTER

16

LESSON

16.1


Solving quadratic equations.

Plenary� Ask the students if they can solve 2x2 – 3x + 1 = 0.� They may spot x = 1 as a solution. Give a clue that the other answer is a fraction

between 0 and 1.� They should find 1–2 fairly quickly.� Discuss ways that this could be solved mathematically.� The students may establish that the factorisation is (x – 1)(2x – 1) = 0 and that

these brackets solve to 1 and 1–2.� Do more examples with a non-unit coefficient of x2, if time allows, such as

2x2 – 5x + 3 = 0.



1 a x = –1 or 1 b x = 2 or –5 c x = 3 or –6 d x = –4 or –3 e x = –2 or –7f x = 3 or 8 g x = 8 or –1 h x = –3 i x = 4

2 a x = –1 or –2 b x = –5 or –6 c x = –2 or –4 d x = 3 or 2 e x = 5 or 2f x = 1 or 4 g x = –5 h x = 4 i x = 5 or –3 j x = 3 or –5 k x = 6 or –4l x = 3 or –2 m x = 9 or 1 n x = 6 or –3 p x = –1

Ho

me

wo


a (x + 3)(x – 4) = 0 b (x – 1)(x + 6) = 0 c (x – 7)(x + 6) = 0

d (x + 5)(x + 2) = 0 e (x – 3)(x + 6) = 0 f (x – 9)(x – 3) = 0

2 First factorise, then solve these equations.

a x2 + 8x + 15 = 0 b x2 + 13x + 30 = 0 c x2 + 4x – 5 = 0

d x2 – 9x + 14 = 0 e x2 + 4x – 21 = 0 f x2 – 4x + 4 = 0

Answers1 a x = –3 or 4 b x = 1 or –6 c x = 7 or –6 d x = –5 or –2 e x = 3 or –6 f x = 9 or 32 a x = –5 or –3 b x = –3 or –10 c x = 1 or –5 d x = 2 or 7 e x = 3 or –7 f x = 2


� unitarycoefficient

Key Words

Oral and mental starter� Give students the factors of 12 in pairs and the factors of 15 on each side of a

vertical line, i.e.1 2 3 1 312 6 4 15 5

� Ask the students to find a combination of products using one pair of factors fromeach side that give a total of 28. This first exemplar may need to bedemonstrated i.e.

2 3 2 × 5 + 6 × 3 = 28.6 5

� Repeat with totals of 29(3 × 3 + 4 × 5), 27(3 × 5 + 4 × 3 or 1 × 15 + 12 × 1), 41(1 × 5 + 12 × 3).

� Now ask for the total 8. Suggest to the students that negative values could beused. The answer is

2 3 2 × –5 + 6 × 3 = 8.6 –5

� Repeat with totals of 24(2 × 15 + 6 × –1), –3(3 × –5 + 4 × 3).� Do more examples if time available.

Main lesson activity� This is a lesson on factorising quadratics with a non-unit coefficient for x2.� Students have met the idea in previous plenaries, but a method for solution has

not been outlined.� Ask the students to factorise 2x2 + 9x + 4.� They may have an intuitive idea of the answer but they also need to have a

method outlined.� There are several methods, two of which are outlined in the Pupil Book.� Example: Find factors of the x2 coefficient and of the constant term.

For 2x2 + 9x + 4, the factors of 2 are 1 × 2 and the factors of 4 are 1 × 4.Now find a combination that gives the coefficient of x, e.g.

1 1 1 × 1 + 2 × 4 = 92 4

The brackets are then the ‘opposite’ to the pairs i.e. (x + 4)(2x + 1).� Example: Factorise 6x2 – 17x + 12.

The brackets must start (3x …)(2x …) or (6x …)(x …) and the constant term hasfactors 3 × 4, 2 × 6, 1 × 12.By trial and improvement we can find that the combination (3x – 4)(2x – 3) works.


� The class can now do Exercise 16B from Pupil Book 3 or Exercise 10L (page 257) from the Higher Mathematics for GCSE textbook.


LESSON

16.2


Factorisation of quadratics of the form ax2 + bx + c, where a > 1.

Plenary� Ask the students if they can solve the quadratic equation 2x2 – 3x – 2 = 0.� They should be able to put all previous ideas together to explain the process of

factorisation and solving each bracket.e.g. (2x + 1)(x – 2) = 0 giving x = – 1–2 or x = 2

� Repeat with other examples, if time allows, such as 3x2 – 5x – 2 = 0.



1 a 2x2 + 11x + 5 b 3x2 + 9x – 12 c 4x2 – 18x + 20 d 6x2 – 15x – 21e 4x2 + 24x + 36 f 9x2 – 24x + 16 g 6x2 – 22x – 8 h 8x2 + 10x – 3 i 4x2 – 1

2 a (2x + 1)(x + 3) b (x + 2)(2x + 5) c (x + 4)(3x + 1) d (x – 1)(2x + 1)e (2x + 1)(3x + 2) f (x – 2)(2x + 3) g (x + 3)(2x – 3) h (2x + 1)2i (4x – 1)(x + 2) j (5x + 1)(x + 2) k (3x – 1)(x + 1) l (4x + 1)(2x + 1)m (x – 2)(3x + 1) n (2x + 1)(3x – 1) p (4x + 1)(x – 3) q (2x – 3)(2x + 5)r (x – 7)(2x + 5) s (x – 5)(2x + 5) t (3x – 1)(x + 5) u (3x + 1)2v (2x + 3)(5x – 1)

Ho

me

wo


a (3x + 1)(x – 4) b (3x – 1)(x + 5) c (2x – 1)(2x + 3)

d (3x – 2)(3x + 2) e (3x – 1)2 f (2x + 5)2


a 2x2 – 7x – 4 b 2x2 + 13x + 15 c 3x2 + 5x – 2

d 4x2 + 23x – 6 e 6x2 – 5x + 1 f 6x2 + 11x + 3

g 5x2 – 26x + 5 h 6x2 – 5x – 6 i 4x2 – 16x + 15

Answers1 a 3x2 – 11x – 4 b 3x2 + 14x – 5 c 4x2 + 4x – 3 d 9x2 – 4 e 9x2 – 6x + 1 f 4x2 + 20x + 252 a (2x + 1)(x – 4) b (2x + 3)(x + 5) c (3x – 1)(x + 2) d (4x – 1)(x + 6) e (3x – 1)(2x – 1)

f (3x + 1)(2x + 3) g (x – 5)(5x – 1) h (3x + 2)(2x – 3) i (2x – 3)(2x – 5)



Key Words

Oral and mental starter� Ask students to mentally solve 2x – 1 = 0. They should find that x = 1–2. Answers

can be written on mini white boards.� Repeat with 3x + 4 = 0 (–11–3), 4x – 3 = 0 ( 3–4 ), 2x + 3 = 0 (–11–2), 5x – 1 = 0 ( 1–5),

3x – 2 = 0 ( 2–3).� Do more examples if necessary.� Now ask students to write down a value for x that will make (3x – 2)(4x + 1) = 0.� Once again check answers. They should be x = 2–3 or x = – 1–4.

Main lesson activity� This is a lesson on solving quadratic equations with non-unitary coefficients.� Ask students if they can find a value for x that solves the quadratic equation

2x2 – x – 1 = 0.� They may spot the answer x = 1, but are unlikely to spot the answer of – 1–2.� Outline the method. First factorise and then solve each bracket equal to zero.� i.e. 2x2 – x – 1 = 0 ⇒ (2x + 1)(x – 1) = 0, so either x – 1 = 0 ⇒ x = 1 or

2x + 1 = 0 ⇒ x = – 1–2.� Repeat with other examples such as 3x2 + 2x – 1 = 0 (x = 1–3 or –1),

12x2 + 5x – 2 = 0 (x = – 2–3 or 1–4), 8x2 – 14x – 15 = 0 (x = – 3–4 or 21–2).� Do more examples if necessary.� Ask students to solve the quadratic equation x2 + x = 12.� This is not in the correct form to factorise and solve. Encourage students to

always rearrange quadratic equations into the form ax2 + bx + c = 0.� The equation above then becomes x2 + x – 12 = 0, which can be factorised and

solved to give x =3 or –4.� Repeat with 2x(x – 7) = 6 – 3x. This must be expanded and then collected into

the correct form.2x2 – 14x = 6 – 3x ⇒ 2x2 – 11x – 6 = 0 ⇒ (2x + 1)(x – 6) = 0 ⇒ x = – 1–2 or 6.


� The class can now do Exercise 16C from Pupil Book 3 or Exercise 10M (page 258) from the Higher Mathematics for GCSE textbook.


LESSON

16.3


Solving quadratic equations of the form ax2 + bx + c = 0, where a > 1.

Plenary� Write the following equation on the board x4 – 5x2 + 4 = 0.� Explain that the equation has 4 solutions since the highest power of x is 4.� Ask students if they can spot any of the solutions. They may spot 1 and 2 but

may not spot –1 and –2.� Outline the method of solution (x2 – 1)(x2 – 4) = 0.

So, x2 – 1 = 0 ⇒ x2 = 1 ⇒ x = ±1 or x2 – 4 = 0 ⇒ x2 = 4 ⇒ x = ±2.� Repeat with 4x4 – 37x2 + 9 = 0.

(4x2 – 1)(x2 – 9) = 0 ⇒ x = ± 1–2 or ±3.



1 a x = 1–2 or 3 b x = 2 or – 1–3 c x = –4 or 1–2 d x = –11–2 or 1–2 e x = –6 or 1–2f x = –2 or 1–4 g x = –21–2 or – 1–3 h x = 1–3 or – 1–2 i x = –11–2 j x = 1–4 or 1–5k x = 21–2 or 2–3 l x = –1–3 m x = 1–2 n x = – 5–6 or 1 p x = –12–3 or 2

2 a x = –2 or 1 b x = 1 or –1 c x = – 1–2 or 1 d x = – 2–3 or 1 e x = 1–4 or 2f x = 1–2 g x = 5 or –5 h x = – 1–2 or 11–3 i x = – 1–4 or 1–5

Ho

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a 2x2 – 15x + 7 = 0 b 3x2 – 5x + 2 = 0 c 2x2 – 9x – 5 = 0

d 24x2 + 14x – 5 = 0 e 6x2 + 23x + 20 = 0 f 6x2 – 23x + 7 = 0

2 Solve these equations

a x2 + x = 6 b 2x(x + 4) = 3(x – 1) c 8x2 – 3x + 4 = 2x2 + 2x + 3

Answers1 a 7 or 1–2 b 2–3 or 1 c – 1–2 or 5 d 1–4 or – 5–6 e –21–2 or –11–3 f 31–2 or 1–32 a 2 or –3 b –3 or 1–2 c 1–3 or 1–2



� roots

Key Words

Oral and mental starter–4 + √

—–41 –4 + 6.5 2.5

� Ask students to estimate the value of ————– (≈ ———— = —— = 1.25).2 2 2

–3 – √—–52 –5 + √

—–31 –1 – √

—–13

� Repeat with ————– (≈ –2.5), ————– (≈ 0.25), ————– (≈ –1.1), 4 2 4

–4 + √—–21

————– (≈ 0.25).2

–5 ± √—–60

� Repeat with (two answers required) ————– (≈ 1.25 or –6.25), 2

–1 ± √—–17

————– (≈ 0.75 or –1.25).4

Main lesson activity� This is a lesson that introduces the quadratic formula.� Introduce the formula:

–b ± √———b2 – 4ac

—x = ————————

2aand explain that a, b and c are the coefficients of x2, x and the constant term inthe general quadratic equation ax2 + bx + c = 0.

� Work through some examples but encourage students to follow some basicprinciples:

Substitute values in brackets before attempting to work anything out.Take particular care with: The change of sign of b.

Squaring b inside the root, particularly if it is negativeWorking out 4 × a × c correctly particularly if c is negative.Dividing the whole top line by 2a, not just the square root.

� Example: Solve 2x2 + 3x – 4 = 0.First, identify a, b and c (a = 2, b = 3, c = –4.

–(3) ± √———(3)2 – 4(2)

——(–4)

——Substitute into the formula x = ———————————–

2(2)–(3) ± √

—–41 –3 ± 6.403

Now evaluate the square root x = —————– = —————–2(2) 4

–3 + 6.403 +3.403So, the solutions are x = —————– = ———– = 0.8508 or

4 4–3 – 6.403 –9.403—————– = ———– = –2.3508

4 4Answers are normally given to 2 decimal places, so x = 0.85 or –2.35.

� Repeat with 3x2 – 4x – 2 = 0 (1.72, –0.39).� Repeat with x2 – 4x – 1 = 0, but this time leave the answer in surd form.

4 ± √—–20

The answers are ————– or 2 ± √––5 .

2

� The class can now do Exercise 16D from Pupil Book 3 or Exercise 10N (page 259) from the Higher Mathematics for GCSE textbook.


LESSON

16.4


The quadratic formula.

Plenary� Ask students to solve 2x2 + 3x + 5 = 0 using the quadratic formula.� This example could be worked through on the board by a student.� The formula gives

–3 ± √—–––31

x = ————– which cannot be solved due to the square root.4

� Explain that when a quadratic equation is solved, the roots are where the graphcrosses the x-axis.

� This gives three situations

� This will be useful for the investigation that ends this chapter.

2 roots 1 (repeated) root No roots



1 a x = 4 or –6 b x = 1 or –7 c x = –2 or 1–2 d x = 21–2 or –4 e x = 0 or –5f x = 6 or –6

2 a x = 1.14 or –2.64 b x = 1.59 or –1.26 c x = 1.32 or –5.32d x = 2.22 or –0.22 e x = 0.16 or –6.16 f x = 0.19 or –2.69g x = 1.82 or 0.18 h x = 2.19 or –0.69 i x = 4.56 or 0.44

(3 ± √––5) –4 ± √

–—20 –4 ± √

–—12

3 a ———— b ————– or –2 ± √––5 c ————– or –2 ± √

––3

2 2 2–6 ± √

–—32 –10 ± √

–—92 –2 ± √

––8d ————– or –3 ± √

––8 e ————— or –5 ± √

–—23 f ———— or –1 ± √

––2

2 2 2

Ho

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rk 1 Solve these equations using the quadratic formula. All answers are whole numbers or fractions.

a x2 + 4x – 5 = 0 b 2x2 + 5x – 3 = 0 c 6x2 – 19x + 10 = 0

2 Solve these equations, giving your answers to 2 decimal places.

a x2 + 7x – 10 = 0 b 2x2 – x – 4 = 0 c 4x2 + x – 7 = 0

3 Solve these equations, giving your answer in surd form.

a x2 – 4x – 2 = 0 b x2 + 6x – 1 = 0 c x2 + 5x – 2 = 0

Answers1 a 1 or –5 b 1–2 or –3 c 2–3 or 21–22 a 1.22 or –8.22 b 1.69 or –1.19 c 1.20 or –1.45

4 ± √–—24 –6 ± √

–—40 –5 ± √

–—33

3 a ———— or 2 ± √––6 b ———–— or –3 ± √

–—10 c ———–—

2 2 2

� quadraticformula

� coefficients

Key Words

Oral and mental starter� Ask students to fill in the missing number in problems such as –22 – 9 = ?

(Answer –13).� Repeat with other examples such as –32 + 5 (answer –4), –42 – 1 (answer –17).� Make sure that students understand that –32 = –(3 squared) and not (–3) squared.� Repeat with more examples if necessary.

Main lesson activity� This is a lesson on completing the square.� Ask students to expand (x – a)2 and (x + a)2.� Hopefully they will get the answer x2 – 2ax + a2 and x2 + 2ax + a2.� Now ask if they can fill in the missing numbers in x2 + 4x – 3 = (x + …)2 – … .� Eventually they will probably come up with the answer (x + 2)2 – 7.� Discuss how this relates to the identities established at the start of the lesson.� Work through the following examples on completing the square.� Example: Write x2 + 6x in the form (x + a)2 – b.

The value, a, inside the bracket is half the coefficient of x. The value b is thesquare of this, So, x2 + 6x = (x + 3)2 – 9.

� Repeat with x2 + 12x ((x + 6)2 – 36), x2 – 4x ((x – 2)2 – 4) and x2 + 30x ((x + 15)2 – 225).

� Example: Write x2 + 6x – 7 in the form (x + a)2 – b.The value of a is half the coefficient of x. The value b is the sum of the existingconstant term and the square as before.So, x2 + 6x – 7 = (x + 3)2 – 9 – 7 = (x + 3)2 – 16.

� Repeat with x2 + 12x + 15 ((x + 6)2 – 21), x2 – 4x – 3 ((x – 2)2 – 7) and x2 + 30x + 100 ((x + 15)2 – 125).

� Demonstrate how to solve equations using completing the square.� Example: Solve x2 + 6x – 4 = 0.

x2 + 6x – 4 = 0 ⇒ (x + 3)2 – 13 = 0 ⇒ (x + 3)2 = 13 ⇒ x + 3 = ±√—–13 ⇒

x = –3 ± √—–13.

� Repeat with x2 – 6x – 2 = 0 (x = 3 ±√—–11) and x2 + 10x + 15 = 0 (x = –5 ± √

—–10 ).

� Complete more examples if necessary.

� The class can now do Exercise 16E from Pupil Book 3 or Exercise 10P(page 260) from the Higher Mathematics for GCSE textbook.


LESSON

16.5


Solving quadratics by completing the square.

Plenary� Ask students to solve x2 + 3x – 2 = 0 using the completing the square method.� A student could be asked to come to the board to give the solution.� The odd coefficient of x will cause problems, particularly when taking a square

rooti.e. x2 + 3x – 2 = 0 ⇒ (x + 11–2)2 – 41–4 = 0 ⇒ (x + 11–2)2 = 41–4 ⇒ x + 11–2 = ±√

—–41–4

⇒ x = –11–2 ± √—–41–4 .

� Now ask students to solve 2x2 + 4x – 3 = 0 using the completing the squaremethod.

� 2x2 + 4x – 3 = 0 ⇒ 2(x2 + 2x) – 3 = 0 ⇒ 2[(x + 1)2 – 1] – 3 = 0 ⇒2(x + 1)2 – 2 – 3 = 02(x + 1)2 = 5, ⇒ (x + 1)2 = 21–2 ⇒ x + 1 = ±√

—–21–2 ⇒ x = –1 ± √

—–21–2 .



1 a (x + 4)2 – 16 b (x – 1)2 – 1 c (x – 6)2 – 36 d (x – 7)2 – 49 e (x + 2)2 – 4f (x + 1)2 – 1

2 a (x + 4)2 – 17 b (x – 1)2 + 2 c (x – 6)2 – 31 d (x – 7)2 – 42 e (x + 2)2 – 7f (x + 1)2 – 6 g (x + 3)2 – 11 h (x + 5)2 – 34 i (x – 3)2 – 6

3 a x = –4 ± √–—17 b x = 3 or –1 c x = 6 ± √

–—31 d x = 7 ± √

–—42 e x = –5 or 1

f x = 3 ± √–—11 g 5 ± √

–—24 h 3 ± √

––5 i 4 ±√

–—11 j x = 1 ± √

––2 k x = –1 ± √

––6

l x = –6 ± √–—43

Ho

me

wo

rk 1 Complete the square for the following.

a x2 + 12x b x2 – 6x c x2 – 20x

2 Rewrite the following quadratic expressions by completing the square.

a x2 + 12x – 9 b x2 – 6x + 3 c x2 – 20x + 100

3 Solve the following quadratic equations using the completing the square method.

a x2 + 12x – 9 = 0 b x2 – 6x + 3 = 0 c x2 – 20x + 100 = 0

d x2 – 10x + 5 = 0 e x2 + 4x – 7 = 0 f x2 – 8x – 5 = 0

Answers1 a (x + 6)2 – 36 b (x – 3)2 – 9 c (x – 10)2 – 1002 a (x + 6)2 – 45 b (x – 3)2 – 6 c (x – 10)23 a x = –6 ± √

–—45 b x = 3 ± √

––6 c x = 10 d x = 5 ± √

–—20 e x = –2 ± √

–—11 f x = 4 ± √

–—21

� completing thesquare

� coefficient

Key Words

Oral and mental starter� Ask students to find the square root of 81.� They will obviously give the answer of 9, (answers can be written on mini white

boards) but may not give –9. Remind them that both answers are possible.� Now ask students for the square root of x2. They will, hopefully, give ±x as the

answer.� Now ask for the square root of 4x2. Some will give the answer ±4x. Make sure

that they understand the answer is ±2x.� Repeat with the square roots of 9y2, 16z2, 25x2, 81x2, 100y2 etc.

Main lesson activity� This is a lesson on factorising the quadratic the difference of two squares.� Start by asking students to expand expressions of the type (x + 2)(x – 2) = (x2 – 4)

e.g. (x – 4)(x + 4) = (x2 – 16), (2x – 1)(2x + 1) = (4x2 – 1), (3x – 6)(3x + 6) = (9x2 – 36) etc.� Eventually ask for the expansion of (a – b)(a + b).� Discuss with students the similarities between the results.� Establish the result (a – b)(a + b) = a2 – b2.� Explain that this is known as the difference of two squares.� Now ask if students can reverse the result. i.e. can they factorise x2 – 36?� Explain that this can be written down as x2 – 62 = (x – 6)(x + 6).� Emphasise that it is important to identify both squares since failure to do this can

lead to errors such as (x – 36)(x + 36).� Repeat with x2 – 100 ((x –10)(x + 10)), 4x2 – 9 ((2x – 3)(2x + 3)), 9x2 – 16y2

((3x – 4y)(3x + 4y)).� Do more examples if necessary.

� The class can now do Exercise 16F from Pupil Book 3 or Exercise 10J (page 255) from the Higher Mathematics for GCSE textbook.


LESSON

16.6


Difference of two squares.

Plenary� Ask the students to simplify (x + 2)2 – (x – 4)2.� If possible, ask a student who wants to expand brackets to work through the

problem on the board, i.e. (x + 2)(x + 2) – (x – 4)(x – 4) = (x2 + 4x + 4) – (x2 – 8x + 16) = x2 + 4x + 4 – x2 + 8x – 16 = 12x – 12.

� Now, ask if the original expression could be written in any other way. Somestudents may recognise this as the difference of two squares.

� Make sure that the students understand that the difference of two squares is anidentity, i.e. any values or expressions can be used for a and b in a2 – b2 = (a – b)(a + b).

� Using brackets (because of the minus signs), write the above as:((x + 2) – (x – 4))((x + 2) + (x – 4))

= (x + 2 – x + 4)(x + 2 + x – 4) = (6)(2x – 2)= 12x – 12.

� Discuss the advantages of both methods.



1 a x2 – 1 b x2 – 25 c x2 – y2 d 4x2 – 1 e x2 – 4y2 f 4x2 – 9y2

2 a (x – 10)(x + 10) b (x – 2)(x + 2) c (x – 6)(x + 6) d (x – 9)(x + 9)e (x – 8)(x + 8) f (x – 11)(x + 11) g (x – z)(x + z) h (2x – 5)(2x + 5)i (x – 3y)(x + 3y) j (4x – 3)(4x + 3) k (2x – 5y)(2x + 5y) l (5x – 8)(5x + 8)m (3 – x)(3 + x) n (2x – 6)(2x + 6) p (6x – 1)(6x + 1)

Ho

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a (x + 11)(x – 11) b (2x – 3)(2x + 3) c (5x – 2y)(5x + 2y)


a x2 – 144 b x2 – 225 c 4x2 – 36

d 81x2 – 64 e x2 – 4y2 f 16x2 – 121

g x2 – 9z2 h 4x2 – 25y2 i 81x2 – 16y2

Answers1 a x2 – 121 b 4x2 – 9 c 25x2 – 4y2

2 a (x – 12)(x + 12) b (x – 15)(x + 15) c (2x – 6)(2x + 6) d (9x – 8)(9x + 8) e (x – 2y)(x + 2y)f (4x – 11)(4x + 11) g (x – 3z)(x + 3z) h (2x – 5y)(2x + 5y) i (9x – 4y)(9x + 4y)

� difference of twosquares

Key Words

Oral and mental starterThere is no starter as the investigation will need to be introduced via a class discussion.

Main lesson activityThe investigation in the Pupil Book is reproduced here. Students will need access to graphical calculators orcomputers with a graph-drawing application package. The investigation will take more than one lesson.

Introduce the investigation and ask the students to suggest ideas for getting started.

For example, they could start by looking at graphs of y = x2, y = 2x2, y = 3x2 etc.Then look at graphs of y = x2 + 1, y = x2 – 2, y = x2 + 3 etc.Then look at graphs of y = x2 + 2x, x2 – 3x, x2 + 5x etc.Once they have an idea of the effect of a, b and c individually they could then look at more general equations.Students should be encouraged to look at the roots from the quadratic formula or the completing the squaremethod. The vertex also could be investigated via the completing the square method.

The graph of a quadratic equation has a characteristicshape called a parabola.

Investigate the relationship between the values a, b andc in the graph y = ax2 + bx + c and the points P, Q(where the graph crosses the x-axis), the point R (wherethe graph crosses the y-axis) and the point S (the ‘vertex’or turning point of the graph).

You should use a graph plotting program or a graphicalcalculator to help you.You will not gain any credit for spending a lot of timedrawing graphs accurately.You will gain credit for a systematic investigation intothe effect of a, b and c on the graph.


LESSON

16.7


An investigation into y = ax2 + bx + c.

x

y

Q

SR

P

y = ax2 + bx + c

PlenaryThere is no plenary. Students could be stopped working at various points and ageneral discussion on progress could be held. This can be helpful for students whoare having difficulty but avoid the ‘bush fire’ effect of one student giving the rest ofthe group the answer.


Ho

me

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rk Students could be asked to continue the work at home if they have computer facilities or asked to lookon the Internet for information on the quadratic equation.

� roots� factors� vertex� intersect� coefficient

Key Words

GCSE Answers

1 0 and 52 a (x – 4)(x – 2) b x = 4, x = 23 a i 3(pq – 2r) ii (c – 4)(c – 5)

b x = –7 or x = 2p – 5

4 a (3p + 1)(p + 5) b ———3p + 1

5 –5.74, 1.746 a 9x2y6 b 3.14, –0.647 1.28, –0.788 a = 9 b = 2 c = –59 a = –5 b = –7

10 p = 2 q = –4


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Year 9 Teacher Pack 3

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Transcript of Year 9 Teacher Pack 3