mst121 prerequisite revision

7/28/2019 MST121 Prerequisite Revision

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Using Mathematics

Starting points

MST121

Revision Pack


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ContentsHowtousethispack 2DiagnosticQuiz 3SolutionstoDiagnosticQuiz 13Module1 Numbers 27

1.1 Numbersystemandnotation 271.2 Calculating 301.3 Multiples,factorsandprimenumbers 341.4 Fractions,decimalsandpercentages 371.5 Calculatingwithsignednumbers 451.6 Workingwithpowers,indicesand logarithms 461.7 Workingwithroots 521.8 Ratioandproportion 55

Module2 Measures 592.1 Units 592.2 Inequalities 602.3 Formulas 612.4 Measuringandclassifyingangles 622.5 Statisticalmeasures 66

Module3 Somebasicfigures 713.1 Triangles 713.2 Rectangles 773.3 Circles 793.4 Areas 80

Module4 Coordinatesandlines 834.1 PointsandCartesiancoordinates 834.2 Lines,gradientsand intercepts 84

Module5 Algebra 895.1 Introduction 895.2 Expressions 895.3 Equations 975.4 Inequalities 106

Module6 Trigonometry 1116.1 Trigonometricratios 1116.2 Areaofageneraltriangle 118

Module7 Graphsandfunctions 1217.1 Graphs 1217.2 Functions 125

Module8 Geometry 1418.1 Propertiesofplanefigures 1418.2 Areasandvolumesofsolids 144

SolutionstoExercises 149Index 163


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How to use this packThispackhasbeendesignedtohelpyoutomakethemostefficientuseofwhatevertimeyouhaveforrevision. It is intwomainparts: anauditsectionandarevisionsection. Thereisalsoan index.TheGuide to Preparationcontainsdetailedadviceaboutyourgeneralpreparation forthecourse(s)andforyourmathematicalpreparation inparticular.Youshouldworkthrough theGuide to Preparation,anduse thispackasdirected there.Audit Section

ThissectioncontainstheDiagnosticQuizand itsSolutions.Revision Section

Thissectioncomprisesrevisionnotes,workedexamplesandpracticeexerciseswithsolutions.The index istheretohelpyoufindtopics,butcanalsobeachecklistofmathematicalwordswhichyoushouldunderstand. Ifyoucomeacrosswordsorsymbolswhichareunfamiliar,thenperhapsyoushouldstartyourownmathematicaldictionary.

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Diagnostic Quiz

Thequestionsthatfollowaredesignedtogiveyoua lookatanumberof TheGuide to Preparationmathematicaltechniquesthatwillberequired inMST121(andMS221). explainshowtomakebest

useoftheRevision PackandDont

worry.

This

quiz

is

not

atest

or

an

examination.

Only

you

will

inparticularthisquiz.knowhowwellorhowbadlyyoudid.

Donthurry. Youcantakeaslongasyoulike,anddothequestions inanyorder. Youmaywellfindthatinanyonequestionyoucandoonepartandthengetstuckorslipuponanother.Asyouworkoutyouranswers,keepanoteabouthowconfidentyoufeelaboutthem. Whenyouhavefinishedthequestions,checkyouranswerswiththesolutions,whichstartonpage13. ThesolutionscontainreferencestotheappropriatesectionsoftheRevisionsection,soyoucanthenspendsometimeworkingonthetopicswithwhichyouhaddifficulty.Youmayfindithelpfultoreturntosomequestionsafteryouhaverevisedatopictocheckthatyouarethenabletoanswerthemcorrectly. ThequestionscannotcovereverythingincludedintheRevisionsection,soevenifyougetthemallright,youmaystillfind ithelpfultoreadthroughthatsection.Good luckenjoythechallenge!

1 Numbers

Question 1

Theuseofpowersisverycommon inmathematics,and itis importanttoknowwhattheymean. Seehowmanyofthefollowingyoucando. Ifyougetstuckonone,trythenextone.(a) Evaluateeachofthefollowing,withoutusingyourcalculator.

32, 83, 41, (12

)2, (8)2, (2)3, 32, (3)2, 32.(b) Useyourcalculatortofindeachofthefollowing.

(i) 76 (ii) 3.24 (iii) (3.2)4 (iv) (3.2)4(v) (3.2)4 correcttotwosignificantfigures

Question 2

(a) Aftercarryingoutaseriesofnumericaloperations,acalculatorgivesthefollowinganswer.

1.917150743E5Explainwhatthismeans.

(b) Howwouldyouexpectthesamecalculatortodisplaythefollowingnumber?

32190818670

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REVISION PACK

Question 3

It isveryeasytomakeamistakewhenyouuseacalculator,so it isimportanttocheckthatyouranswerisreasonable. Youcandothisbyestimatingtheanswer.(a) Giveroughestimatesforeachofthefollowingnumbers.

(i) 413 (ii) 2782 (iii) 12.4 (iv) 0.1253 (v) 189025(b) Useyourestimatestoobtainanapproximateanswertoeachofthe

calculationsbelow.12.4

(i) 4132782 (ii) 413189025 (iii)0.1253

27820.1253(iv)

12.4Checktheaccuracyofyouranswersbyusingacalculator.

Question 4

Knowingthecorrectorder inwhichtodealwiththearithmeticoperationsinanexpressionisimportant.Calculateeachofthefollowing.(a) 32+25 + 6 (b) (32 + 2) 5 + 6 (c) 32 + 2(5 + 6) (d) (32+2)52 +6 (e) (32+2)(52 + 6) Question 5

Byfindingtheprimefactorsofeachofthenumbers12,20and45,findthesmallestwholenumberthatcanbedividedexactlybyallthreenumbers.Question 6

Sometimesnumbers lookverydifferent,butareactuallythesame.(a) Useacalculatortowriteeachofthefollowing fractionsasdecimals,

correcttothreedecimalplaces.3 15 13 87

(i) (ii) (iii) 1 (iv) 2140 62 84 93

(b) Withoutusingyourcalculator,evaluateeachofthefollowing.(i) 2

3+27 (ii)

56

23 (iii)

415+

720 (iv) 3

252

78

(v) 381

511 (vi) 1

344

23

Question 7

Inthisquestion,trytouseyourcalculatoronlywhenreallynecessary,andthenwithasfewkeystrokesaspossible.Findthereciprocalofeachofthefollowingnumbers,givinganswersthatarenotexactcorrecttothreesignificantfigures.(a) 10 (b) 5 (c) 101 (d) 72.5 (e) 0.00356

4


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DIAGNOSTIC QUIZ

Question 8

Percentagesoccur inmanywalksoflife, includingMST121!(a) Increaseeachofthefollowingnumbersbythepercentageshown.

(i) 14by10% (ii) 142by115%(b) Bywhatpercentagehas300beenincreasedordecreasedtoarriveat

thefollowinganswers?(i) 317.5 (ii) 60

Question 9

(a) Evaluateeachofthefollowingwithoutusingyourcalculator. Thenseeifyoucanobtainthesameanswerbyusingyourcalculator.(i) 3+(4)6 (ii) 34 + (6) (iii) 3+(4 + 6) (iv) 3(4 + 6) (v) (6)(5) (vi) 6(35)

(b) Calculateeachofthefollowing,withoutusingyourcalculator. 3(i) 14400 (ii) 0.81 (iii) 12 (iv) 24

1(v)10000

Question 10

Withoutusingyourcalculator,writeeachofthefollowingnumbersinlog10form. Forexample,log10 100=2.

1(a) 102 (b) (c) 0.001

100Question 11

(a) Useyourcalculatortofindeachofthefollowing,givingyouranswerscorrecttofourdecimalplaces.(i) ln3.142 (ii) ln14.16 (iii) ln14658 (iv) ln0.0324

(b) Useyourcalculatortofindthenumbers,correcttothreesignificantfigures,whosenatural logarithmsareasfollows.(i) 0.812 (ii) 1.623 (iii) 2.976 (iv) 0.0356

Question 12

Threepeopledecidetosplitafoodbillbetweenthem,takingintoaccountthenumberofmealseatenathome. Theyagreethat itshouldbesplit intheratio2to3to5. Thebill is70.How much does each person pay? 2 Measures

Question 13

(a) Markthefollowingnumbersonanumberline,andhencearrangetheminascendingorder.

2.9, 4.3, 0.2, 3, 3.5, 1.5.(b) Ineachofthefollowingparts, inserttheappropriate>or


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REVISION PACK

Question 14

Makexthesubjectoftheformula ineachofthefollowingequations.(a) 2x=y3 (b) p=x2 + 3 (c) (x+ 3)2 =t(d) y= 1

2

x (e) m= 3px2

Question 15

Inmathematicsweoftenuseradians,ratherthandegrees,tomeasureangles.(a) Howmanyradiansarethereinonecompleteturn?(b) Sketchanangleof60. Whatisthemeasureofthisangle inradians?

3(c) Sketchanangleof radians. Whatisthemeasureofthisangle in

2degrees?

Question 16

Theword averagehasmorethanonemeaning. Somematchboxesarelabelled Averagecontents50matches.(a) Thecontentsoffiveboxesofmatcheswerecheckedandfoundtobe:

50, 47, 48, 51, 51.Findthemean,medianandmodeofthecontentsofthefiveboxes.

(b) Thecontentsofafurtherfiveboxeswerecheckedandfoundtobe:50, 48, 48, 51, 48.

Combinethesefigureswiththosefrompart(a),andcalculatethemean,medianandmodeforthetenboxes.

3 Some basicfigures

Question 17

Itisimportanttobeabletoobtainanswersfrominformationinadiagram.Intherectangle inFigure0.1,thesideAB isof length6mandthesideBC isof length8m. Youmayfind ithelpfultoinsertthesefigures inthediagram.

A D

B C

Figure 0.1

6


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DIAGNOSTIC QUIZ

(a) UsePythagorasTheoremtofindthe lengthofsideAC.(b) CalculatetheperimeterofthetriangleABC.(c) Findtheareaoftherectangle,andhencetheareaofthetriangleABC.(d) Confirmthattheareaofthetriangleyouhaveobtainedinpart(c)is

correct,byfindingtheareadirectly.(e) Apathofwidthonemetre isconstructedwhose inneredge isthe

rectangleABCDandwhoseouteredge isa largerrectangle. What istheareaofthepath?

Question 18

InFigure0.2,O isthecentreofacircle,andA,B,C andD lieonthatcircle.The linesegmentsAC andAB arenotequalin length,andAOD isadiameter.

A

B

CO

D

Figure 0.2

(a) ExplainwhyangleACD is90.(b) Identify

(i) threeisoscelestriangles(therearemorethanthree);(ii) twoscalenetriangles(therearemorethantwo);(iii) tworight-angledtriangles.

(c) IfangleDAC is40andangleCBO is15,showthatangleCOD is80andfindtheotherangles inthefigure.

(d) Iftheradiusofthecircle is10cm,find,correcttothenearestwholenumber:

(i) theareaofthetriangleOCD;(ii) the lengthofthearcCD.

(e) What istheareaofthesectorOC Dtothenearestwholenumber?

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REVISION PACK

4 Coordinates and lines

Question 19

Coordinatesareashorthandwayofdescribingthepositionsofpointsrelativetoafixedpoint(theorigin)andapairofaxes. ThepointsA,BandC areplottedonFigure0.3.

y

3A

2

1

2 1 0 1 2 3x

1

C2

B

Figure 0.3

(a) WritedownthecoordinatesofA,B andC.(b) PlotthepointsD(1,3),E(1,2.5)andF(0.5,2).(c) DrawthestraightlinewhichpassesthroughAandF.

(i) What isthegradientofthis line?(ii)

Where

does

it

cross

the

x-axis?

(iii) Wheredoesitcrossthey-axis?

5 Algebra

Question 20

Ifx=8,findthevalueofeachofthefollowingexpressions.(a) (i) 5x (ii) 5x (iii) 3x2 (iv) 3(x+ 2) (b) (i) x2 + 4 (ii) (x+ 4)2 (iii) (x+24)12 (iv) x2 3x4Question 21

Simplifyeachofthefollowingexpressionsasfaraspossible.(a) 2a+ 3b4a (b) 2a+ 3b4c(a5b)(c) 2a+ 3b4c+ 2(a5b) (d) 2a+ 3b4c2(a5b)(e) 2a+ 3b4c+c(25b)Question 22

Multiplyoutthebracketsineachofthefollowingexpressions,andsimplifytheresultasfaraspossible.(a) (b+ 1)(b+ 2) (b) (c2)(c+ 5) (c) (df)2(d) (3x+4)(2x7) (e) (3x4y)(5y+ 6x)

8


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DIAGNOSTIC QUIZ

Question 23

Factorisingistheoppositeofmultiplyingout. Youstartwiththesumofanumberoftermsandaimtoexpressitasaproduct. Factoriseeachofthefollowingexpressions.

2(a) ab+a (b) ab+ac (c) ab+ 2ac+a (d) a2 b2(e) a2 + 2ab+b2 (f) 6a2 +a

1

Question 24

Solveeachofthefollowingequations.(a) (i) 2x5=15 (ii) 2(x5)=15 (iii) 2(x5)2 = 32

(iv) 82x=x+ 7 25

(b) (i) p2 + 2p4=0 (ii) 36=v2

(iii) 5t = 26 Inparts(i)and(iii),giveyouranswerscorrecttotwodecimalplaces.

Question 25

Solvethefollowingpairofsimultaneousequations.x+ 2y= 4

2x3y=6Question 26

Solveeachofthefollowing inequalities,andillustrateeachansweronanumberline.(a) 2x35 (b) 2x4 5

6 TrigonometryQuestion 27

Inthetriangle inFigure0.4,angleA is90, angle B is70andthehypotenusehas length15cm.Usingtrigonometricratios,orotherwise,findtheunknownsidesandangleofthetriangle. Givelengthscorrecttothreesignificantfigures.

A

B C

Figure 0.4

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REVISION PACK

7 Graphs and functions

Question 28

Thegraphonthe leftinFigure0.5showsajourney inwhichChriswalksfromhometothenewsagent,buysanewspaper,andwalksback,stoppingtotalktoafriendontheway. Thespeedofwalking isconstantineachsectionofthejourney. Usethegraphtodeduce:(a) (i) thedistance, inkilometres,fromChrisshometothenewsagent;

(ii) thespeedofwalking ineachsectionofthejourney, inkilometresperhour;(iii) thedistance,inkilometres,fromthenewsagenttothepointatwhichtheconversationwiththefriendtakesplace.

(b) ExplainwhythegraphontherightofFigure0.5cannotrepresentajourney.

distance in kilometres distance from start

time in minutes0

0.2

0.4

0.6

0.8

1.0

10 20 30 40 50

(a) time in minutes (b)

Figure 0.5

Question 29

Inabuildingdevelopmentallplotsarerectangular. Inonesection(A)allplotswillhaveanareaof500m2. Inanothersection(B)theareasofplotswillvary,butthedimensionsofeachplotwillbesuchthatthe lengthisofthebreadth.Letlmetresandbmetresrepresentthe lengthandbreadth,respectively,ofaplot.(a) Forplots insectionA,describe inwordstherelationshipbetweenthe

lengthandbreadth,andrepresentthisrelationship insymbols.(b) Forplots insectionB,representtherelationshipbetweenthe length

andbreadth insymbols.(c) (i) WhichofgraphsI,IIandIIIinFigure0.6representsthe

relationship inpart(a),andwhichgraphrepresentstherelationship inpart(b)?(ii) Describetherelationshiprepresentedinthegraphyoudidnotchoose.

8

5

10


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DIAGNOSTIC QUIZ

l

bgraph I graph II graph III

b

l

b

Area

Figure 0.6

Question 30

Sketchthegraphofeachofthefollowingequations. (Accurateplotsarenotrequired. Nocalculatorsallowed!)

3(a) y=x2 (b) y=x2 2 (c) y= (x2)2 (d) y=xQuestion 31Sketchthegraphofeachofthefollowingfunctions.(a) f(x) = sin xforxbetween2and2(b) g(x) = 1 + cos xforxbetween2and2

x(c) h(x) = e

8 Geometry

Question 32

Tworectangularpiecesofpapereachmeasure31cmby20cm. Oneisusedtomakeanopen-endedtubeofcircularcross-sectionand length20cm.Theotherisusedtoformanopen-endedprism,alsoof length20cm,whosecross-sectionisanequilateraltriangle.Foreachshape,1cm(ofthe31cm)istakenupbytheneedtosecuretheedges.

20cm

31cm

1cm

20cm

20cm

Figure 0.7

(a) Find:(i) theradiusofthecircularcross-sectionandhencetheareaofthecircularholeatthebaseofthecylinder;(ii) theareaofthetriangularholeattheendoftheprism;(iii) thesurfaceareaofeachshape;(iv) theratioofthevolumeofthecylindertothatoftheprismformedbyclosingofftheends.

(b) Canyouthinkofanyeverydayexamplesofsuchcylindersandprisms?11


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REVISION PACK

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Solutions to Diagnostic Quiz

Solution 11 1(a) 32 = 9; 83 =512; 41 =4

; ( 12

)2 =4

; (8)2 =64; (2)3 =8; SeeSection1.1.1

32

= 9 ; (3)2 = 9; 32 =9.(b) (i) 117649 (ii) 104.8576 (iii) 104.8576

(iv) 104.8576 (v) 0.0095(to2s.f.)Solution 2

This isthecalculatorswayofshowingverylargeandverysmallnumbers. SeeSection1.1.(a) 1.917150743E5 means 1.917150743105,which isequalto

0.00001917150743.(b) 32190818670=3.21908186701010,whichthecalculatorwillshow

as3.219081867E10.Solution 3

Youmayhaveroundedthefigurestoothervalues,soyourestimated SeeSection1.2.answerscoulddifferslightlyfromours. The importantthingistomakesurethatyoucanestimateananswer inyourhead.(a) (i) 400 (ii) 3000 (iii) 10 (iv) 1/10(0.1) (v) 200000(b) (i) 4003000=1200000;

calculatedvalue=1148966.(ii) 400200000=80000000;calculatedvalue=78067325.

1

(iii) 10 =100;10calculatedvalue=98.9625(to4d.p.).

1(iv) 3000 10=30;

10calculatedvalue=28.1117(to4d.p.).

Solution 4

(a) 32+10+6=48 (b) 345+6=170+6=176 SeeSection1.2.(c) 32+22=54 (d) 3425+6=850+6=856(e) 34(25+6)=3431=1054Solution 5

12=223,20=225,45=335;sothesmallestnumber SeeSection1.3.divisibleby12,20and45 is22335=180.

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REVISION PACK

Solution 6

(a) Eachofthefollowinganswersisgivencorrecttothreedecimalplaces.3 15

eeSection1.4. (i) = 0.075 (ii) = 0.24240 62

13 13(iii) = 0.155,so1 = 1.155.

84 8487 87(iv) = 0.935,so21 = 21.935.93 93

14 6 20 5 4 1 16 21 37(b) (i) + = (ii) = (iii) + =

21 21 21 6 6 6 60 60 6017 23 136 115 21 3 16 6

(iv) = = (v) =5 8 40 40 40 8 11 11

7 14 7 3 3(vi) = =

4 3 4 14 8Solution 7

Didyouusethereciprocalbutton(marked1/xorx1)onyourcalculatorforparts(d)and(e)?

eeSection1.4. (a) 0.1 (b) 0.2 (c) 10 (d) 0.0138(to3s.f.)(e) 281(to3s.f.)Solution 8

eeSection1.4. (a) (i) 10%of14is1.4,soa10% increasegives14+1.4 = 15.4.(ii) 15%of142 is21.3,soa115% increaseis142+21.3=163.3. Thetotalistherefore142+163.3=305.3.

(b) (i) Theactualincreaseis17.50;thepercentage increaseis17.5/300%=5.83%(to2d.p.).(ii) Theactualdecreaseis240;thepercentagedecreaseis240/300%=80%.

Solution 9

eeSection1.5. (a) (i) 3+(4)6 = 3 46 = 7(ii) 34 + (6)=346 = 7(iii) 3+(4 + 6) = 3 + 2 = 5 (iv) 3(4 + 6) = 3 2 = 6 (v) (6)(5)=(6)/(5)=6/5 = 1.2

3 5(vi) 6(35)=6 = 6 = 10 5 3Yourcalculatorprobablygavedifferentanswersforsomeofthecalculations.

eeSection1.7. (b) (i) 14400= 122 102 = 12 10=120 81 92 9

(ii) 0.81= = = = 0.9100 102 10

(iii) 12=22 3,so 12= 22 3 = 2 3. 3 3(iv) 24=23 3,so 24= 23 3 = 2 3 3.

1

1 1

(v)

10000

=

100

2

, so

=

= = 0.01.10000 1002 100(Theanswerstoparts(iii)and(iv)havebeen leftin surd form,butcouldbeexpressedindecimalformwiththeaidofacalculator.)

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SOLUTIONS TO DIAGNOSTIC QUIZ

Solution 101

(a) log10 102 =2 (b) log10 =2 (c) log10 0.001=3 SeeSection1.6.100Solution 11

Ifyouhavedifficulty inobtainingtheanswershown,checkyourrounding. SeeSection1.6.(a) (i) 1.1449 (ii) 2.6504 (iii) 9.5927 (iv)

3.4296

(b) (i) 2.25 (ii) 5.07 (iii) 19.6 (iv) 1.04Solution 12

Thetotaloftheratios is2+3+5=10. Onetenthof70is7,sothe SeeSection1.8.firstpersonpays27 = 14,thesecondpays37 = 21andthethirdpays57 = 35.Solution 13

(a)3 0.2 1.5 2.9 4.3

4 532101234 6

3.5

Figure 0.8

So,arranged inascendingorder,thenumbersare SeeSection1.1.3.5, 3, 0.2, 1.5, 2.9, 4.3.

(b) (i) 2.93 SeeSection2.2.Solution 14

(a)(b)

x= y32 orx=

12(y3)

x2 =p3,sox=p3 (provided p3) SeeSection2.3.

(c) x+ 3 = t, so x=3 t(providedt0)(d) 2y=x, so x= (2y)2 = 4y2(e) Multiplythroughbyx2 togive

mx2 = 3p,fromwhich

x2 = 3p.m

Sox=

3pm (providedp/m0).

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REVISION PACK

Solution 15

eeSection2.4. (a) Onecompleteturn=2radians(=360).(b)

60

Figure 0.9

60 60= 2radians= radians

360 3(c)

3p radians2

Figure 0.10

3 3 360radians= =270

2 2 2Solution 16

eeSection2.5. (a) (50+47+48+51+51)5 = 49.4,sothemean is49.4matches.Themedianisthemiddlevaluewhenthevalueshavebeenarrangedinorderofmagnitude,so is50matches. Themode isthevaluethatoccursmostoftenthat is,51matches.

(b) Themean is49.2matches. Therearenowtenvalues,sothereisnosinglemiddlevalue. Themedian isthemeanof48and50,thetwomiddlevaluesthat is,49matches. Themode is48matches.

Solution 17 eeSections3.1,3.2,3.4 (a) AC= (62 + 82) m = 36+64m= 100m=10m.nd6.2. (b) PerimeteroftriangleABC= 6 m + 8 m + 10 m = 24m.

(c) Areaofrectangle=6m8 m = 48 m2,soareaoftriangleABC is24m2.

1(d) Areaoftriangle= baseheight,so

21

areaoftriangleABC= 8 m 6 m = 24 m2.2

16


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(e)

A D

B C

A

B C

D8 m

10 m

8 m6 m

Figure 0.11

Thediagramshowsthemeasurementsafterthepath,ofwidth lm,hasbeenadded.NewrectangleABCD hasarea10m

8 m = 80 m2.

OriginalrectangleABCDhasarea48m2, from above. Soareaofpath=80m2 48m2 = 32 m2.

Solution 18

(a) AD isadiameterandtheangle inasemicirclesubtendedbya SeeSections3.1,3.3,3.4diameterisalways90. and6.2.

(b) (i) Isoscelestriangleshavetwosidesequal,solookfortrianglesformedbyradiiofthecircle. ExamplesincludetrianglesOBC,OAB,OBD,OCD.(ii) Scalenetriangleshavenosidesequal. ExamplesincludetrianglesABC,BC D,ACD,ABD.(iii) Sincetheangle inasemicircle is90,ACDandABDareright-angledtriangles.

(c) TriangleAOC is isosceles(AO=OC= radius), so angleACO= angle DAC= 40.

HenceangleAOC=180(40+ 40)=100.

AOD isastraight line,soangleCOD =180

100

= 80

.

Theotheranglesareshown inFigure0.12,overleaf.

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(c) SeeFigure0.13.(i) Thegradientofthe linethroughAandF is 3(2) = 5 = 2.

2(0.5) 2.5(ii) Thelinecrossesthex-axisat(1

2,0).

(iii) Thelinecrossesthey-axisat(0,1).Solution 20Ifyouranswersarewrong inthissection,donotbedisheartened;justtry SeeSection5.2.toidentifyyourmistakesfromtheworkingshownhere.(a) (i) 5(8)=40

(ii) (5)(8)=40(iii) 3(8)2 = 242 = 26(iv) 3(8 + 2) = 3 (6)=18

(b) (i) (8)(8)+4=64+4=68(ii) (8 + 4)2 = (4)2 = 16

1 1(iii) (8+24) = 16 = 16=+42 2(iv) (8)2 3(8)4 = 64 + 24 4 = 84

Solution 21

(a) 2a4a=2a, so 2a+ 3b4a= 3b2a. SeeSection5.2.(b) 2a+ 3b4c(a5b) = 2a+ 3b4ca+ 5b=a+ 8b4c(c) 2a+ 3b4c+ 2(a5b) = 2a+ 3b4c+ 2a10b= 4a7b4c(d) 2a+ 3b4c2(a5b) = 2a+ 3b4c2a+ 10b= 13b4c(e) 2a+ 3b4c+c(25b) = 2a+ 3b4c+ 2c5bc

= 2a+ 3b

2c5bcSolution 22

(a) b(b+ 2) + 1(b+ 2) = b2 + 2b+b+ 2 = b2 + 3b+ 2 SeeSection5.2.(b) c(c+ 5) 2(c+ 5) = c2 + 5c2c10=c2 + 3c10(c) d(df)f(df) = d2 df f d +f2 =d2 2df +f2(d) 3x(2x7)+4(2x7)=6x2 21x+ 8x28=6x2 13x28(e) 3x(5y+ 6x)4y(5y+ 6x) = 15xy+ 18x2 20y2 24xy

2= 18x2 9xy20ySolution 23(a) a(b+ 1) SeeSection5.2.(b) a(b+c)(c) a(b+ 2c+a)(d) (ab)(a+b)(e) (a+b)2(f) (3a1)(2a+ 1)

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REVISION PACK

Solution 24

eeSection5.3. (a) (i) 2x= 15 + 5, so 2x= 20 and x=10.(ii) 2x10=15,so2x= 25 and x= 12.5.(iii) 2(x5)2 =32,so(x5)2 =16. Takingsquarerootsgivesx5 = 4,fromwhichx= 5 + 4 = 9 or x= 5 4 = 1. Thesolutionsarex= 1 and x= 9. Alternatively,it ispossibletomultiplyoutandfactorisetheresult,asfollows.

(x5)2 = 16 becomes

x2 10x+ 25 = 16 sothat

x2 10x+ 9 = 0,fromwhich

(x9)(x1)=0.Sox= 9 or x=1. Thefirstmethodiseasier,however.(iv) 8=x+ 7 + 2x, so 8 = 3x+ 7 giving 8 7 = 3x. Hence3x= 1

1andsox= .3

(b) (i) Usingtheformulaforthesolutionofaquadraticequationgives 2 22 4(4)= 2 4 + 16 p=

2 2 2 20 22 5=

2=

2 .

Sop=1 5,thatis,p= 1.24orp=3.24(to2d.p.).

25(ii) Multiplyingbothsidesof36=

v2byv2 gives36v2 =25,from

which25

2 =v36.

Takingsquarerootsgivesv=5 .6

(iii) Becausetheunknowntis intheindexposition, it isnecessarytotakelogsofbothsides,whichgivestlog5=log26. So

log26t= = 2.02 (to2d.p.).

log5(Itdoesntmatterwhetheryouuse logstobase10or ln,providedyouareconsistent: thesameanswerwillbeobtained.)

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Solution 25

There ismorethanonewayofsolvingapairofsimultaneousequations.First,welabeltheequations:

x + 2y = 4, (0.1)2x 3y =6. (0.2)

FromEquation(0.1),x = 4 2y. Thensubstitutingforx inEquation(0.2)gives2(42y)3y =6,

sothat84y 3y =6,

fromwhich7y = 14.

Hencey =2. Substitutingthisvalue inEquation(0.1)givesx + 4

=

4,

sothatx = 0. Thesolution isx = 0, y = 2. Solution 26

(a) 2x 5 + 3, so 2x 8,givingx 4.

x > 4

0 1 2 3 4 5 6

Figure 0.14

(b) 2x < 1 + 4, so 2x < 5,givingx < 2.5.

x < 2.5

2 1 0 1 2 3 4

Figure 0.15

(c) Careisneededwiththisone!3> 5 + x, so 3 5> x,giving2> x, that is, x < 2.

x < 2

5 4 3 2 1 0

Figure 0.16

SeeSection5.4.

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REVISION PACK

eeSection6.1. Solution 27Thesumoftheanglesofthetriangle is180,sotheremainingangle is

1809070= 20.Inthetriangle,BC isthehypotenuse,so

AB=BCcos70= 15 cm cos70= 5.13cmand

AC=BCsin70= 15 cm sin70= 14.1 cm.(Youcanobtainthesameanswersbyusingdifferentratiosoftheangle20.Havingobtainedthelengthofoneofthesides,youcouldhaveusedPythagorasTheoremtocalculatethe lengthofthethirdside.)Solution 28

eeSection7.1. (a) (i) ThedistancefromChrisshometothenewsagentis0.9km.(ii) Hisspeedinthefirstpartofjourneyis0.9kmin10minutes,whichis5.4km/hour.Attheshop,thespeed is0km/hour.Hisspeedinthefirstpartofjourneyhome is0.5kmin5minutes,whichis6km/hour.Duringtheconversation,thespeed is0km/hour.Hisspeedinthefinalpartofjourneyis0.4kmin5minutes,whichis4.8km/hour.(iii) Conversationtakesplace0.5kmfromthenewsagent.

(b) Thisgraphsuggestsaperiodofnomovement,followedbyaninstantaneouschange indistance,followedbynomovement. Allmovementstakesometimetocomplete,sothemiddlesectionisimpossible.

Solution 29

eeSection7.1. (a) Foreachplot insectionA,thelengthtimesthebreadthequals500m2.Insymbols,

lb=500.(b) Foreachplot insectionB,

l= 85

b.(c) (i) GraphIcorrespondstol= 8

5

b. GraphIIIcorrespondstolb=500.(ii) GraphIIcorrespondstoarelationship inwhichthearea increasesindirectproportiontothebreadth,that is,

area=kb,wherek isaconstant. (Eachplothasconstantlength.)

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Solution 30

Theappropriategraphsaregivenbelow. SeeSection7.2.

(a) y (b) y

25 25

20 20

15 15

10 10

5 5

4 2 2 4x

4 2 2 4x

2

5

y y(c) (d)

25

20

15

10

512 21

6

4

8

2

2

4

6

8

x

4 2 2 4 6x

Figure 0.17

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REVISION PACK

Solution 31

eeSection7.2. Theappropriategraphsaregivenbelow.(a) f(x) = sin(x)

y

x

1

1

2p 3p/2 p p/2 p/2 p 3p/2 2p

Figure 0.18

(b) g(x) = 1 + cos(x)

y

2

1

2p p p 2px

Figure 0.19

x(c) h(x) = e

y

1

4

2x

21

2

6

8

10

Figure 0.20

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Solution 32

(a) (i) Letrbetheradiusofthecircularhole. Thecircumferenceofthe SeeSections8.1and8.2.circle(=2r) is 30 cm, so

r= 30/2= 4.77cm.(Thisanswerhasbeenroundedtotwodecimalplacesforconvenience,butthecalculatorgivesmoreplaces. Itistheunroundedversionthatshouldbeusedtofindthearea. Alsomakesurethatyouhaveusedthekeyonyourcalculator,andnotanapproximatevaluefor.)Areaofholeisr2 = 71.62cm2 correcttotwodecimalplaces. (Usingtheroundedversionfortheareagives71.48!)(ii) Lengthofeachsideofthetriangle is30cm/3 = 10 cm.

Areaoftriangle= 1 10cm10cmsin602 3

= 50 cm2 = 25 3 cm2.2

Soarea is43.30cm2 (to2d.p.).(iii) Sincethere isnotoporbottomtoeithershape,thesurfaceareaofeachisjusttheareaofthepaperused,minustheareaoftheone-centimetreseam. Sosurfaceareaofeachis

20cm30cm=600cm2.(iv) Volumeofcylinder is

areaofcirclelength=r2 length.The lengthis20cmandr= 30/2frompart(i). Sovolumeofcylinder is

2

30 cm2

20cm=1432.39cm3 (to2d.p.).2

Volumeofprism is

areaoftriangularbaselength=25 3 cm2 20cm=866.03cm3 (to2d.p.).

Soratioofthevolumeofthecylindertothatoftheprism isapproximately

1432.39= 1.65.

866.03(Theratio isanapproximateonesincethenumbersquotedareroundedfigures,notthefullcalculatoraccuracy.)

(b) Therearemanytubes incommonusage;forexample,theinsideoftoiletandkitchenrolls,thetubularpartofaSmartiestube.Theprismhasmore limiteduse,althoughToblerone issold inprism-shapedboxes.

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REVISION PACK

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Module 1 Number s

1.1 Number system and notation

Naming number s

Overthecenturies,ashumanactivitybecamemorecomplex,therewasaneedtodeveloptheconceptofnumberfordifferentpurposes. Initiallytheonlyneedwasforthecountingofnumbers:

one,two,three,. . . . Thesepositivewholenumbersarenowcommonlyreferredtoasthenaturalnumbers. Ascommercial,scientificandmathematicalactivitydeveloped,sodidfractions,negativenumbers,theconceptofzero,andthevarioussymbols,systemsandnotationsforexpressingthem.Thenegativeandpositivewholenumbersand0(zero)arereferredtoastheintegers,andcanbevisualisedas lyingonanumber line.

negative integers zero positive integers

4 3 2 1 0 1 2 3 4

Figure 1.1

Fractions,alsoreferredtoasrationalnumbers,areobtainedbydividingtwointegersandcanbewrittenas barfractions(oftencalledcommonorvulgarfractions),suchas 12 . Rationalnumbersincludetheintegersbecauseeachintegercanberewrittenasabarfraction;forexample,2= 2 . All Also0= 0

1, forexample.

1

rationalnumberscanbewrittenasdecimalswhicheitherterminateforexample, 5

4= 1.25and4

10=0.4 or recurforexample,

1 = 0.333333333 . . . and811

=0.727272727 . . .. (Theshorthandfora3

recurringdecimal isadotabovetherecurringdigitorateachendofaset 0,1,2,3,4,5,6,7,8and9= 0.3, = 0.7 2,and 41 arethedigits.= 0.12 3.)ofrecurringdigits;forexample, 1

3

8

11 333

Figure1.2 illustratesthecompositionofthesetofrationalnumbers.

All natural numbersare integers.

All integersare rationals.

naturalnumbers

integers

rationals

Figure 1.2

Therearealsoinfinitedecimalsthatneitherterminatenorrecur,andthesearecalledirrationalnumbers. Theyincludenumbers like

2 = 1.4142. . ., 5 = 2.2360. . .,= 3.1415. . . ande= 2.7182. . .. istheGreekletterpi. eisthebasefornaturalThesetofnumbersthatincludestherationalsandthe irrationals iscalled logarithms.

thesetofrealnumbers. (Squarerootscanbewrittenaseither 3or 3. Inthistextweuseonlythelatter.)

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4

REVISION PACK

Eachrealrationalor irrationalhas itsplaceonthenumber line. SomeexamplesareshowninFigure1.3. Thegapsbetweentheintegersonthenumberlinearecompletelyfilledbyrationalandirrationalnumbers.

negative integers zero positive integers

5 4/10 1/2 2 e p

4 3 2 1 0 1 2 3

Figure 1.3

Aparticularrealnumbermaybewritteninavarietyofforms;forexample,two(word),2(rationalnumber),+2(positive integer),2.0(decimal), 2 or4Powersandrootsare2

(fractions),21 (power)and 4(positivesquareroot)representthesamexplainedinmoredetaillater number. Differentformshavebeendevelopedtosuitparticularpurposes.nthissection.

Experimenttoensurethatyoucaninputthedifferentformsofnumbersonyourparticularcalculator. Forexample,doyouneedtopressfunctionkeysbeforeorafteryouenterdigits?Place value

TheIndianArabic(orHinduArabic)systemmostgenerallyusedinthewestern world has ten numeral symbols (the digits 0, 1, 2, 3, 4, 5, 6, 7, 8,9)whicharecombinedinaplace-valuesystemstructure. Theplace-valuesystemisbasedonpowersoftenandextendstodecimals.Forexample,thenumeral23789.56(twenty-threethousandsevenhundredand

eighty-nine

point

five

six)

has

seven

digits,

the

value

of

each

of

which

is indicated inthefollowingtable.

Ten Thousands Hundreds Tens Units tenths hundredthsthousands

integer/decimal 10000 1000 100 10 1 0.1 0.01powerform 104 103 102 101 100 101 102example 2 3 7 8 9 5 6

1

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MODULE 1 NUMBERS

Indices

Indexnotation isaconcisewayofsymbolisingtherepeatedmultiplicationofanumberby itself. Forexample,multiplying10stogethersuccessivelygives:

1010=100, 101010=1000, 10101010=10000.Usingtheshorthandnotation,wewrite:

102 =100, 103 =1000, 104 =10000, . . . . Inthisexample,10isthebasenumberandthesuperscriptnumberatthetopright indicateshowmanyofthesebasenumbershavebeenmultipliedtogether. Thissuperscriptnumber isvariouslycalledthepower,indexorexponent. Thisshorthand isextendedasfollows:

1101 = 10, 100 = 1, 101 =

10 (or0.1),1 1

102 =1010 = 100 (or0.01), . . . .

Thenotationalsoextendstofractionalpowers. Forexample:10

11/2 = 10, 101/2 = .10

Ajustificationforassigningthevalue 10tothesymbol101/2 (or100.5) is giveninSection1.6.

Example 1.1

Evaluateeachofthefollowingpowers,firstbyhandandthenusingyourcalculator.(a) 53 (b) 106 (c) 23Solution

(a) 53 = 5 55=1255 [yx]3[=]125(scientificcalculator) Calculatorkeysareindicatedor5 []3[ENTER]125(graphicscalculator) bysquarebrackets []. The

actualkeysusedvaryfrom(b) 106 = 10 1010101010=1000000

onemakeofcalculatorto10 [yx]6[=]1000000 another.or10 []6[ENTER]1000000

1 1(c) 23 =2

2

2

=8

(or0.125)2 [yx] 3 [ +/][=]0.125or2 [] [()]3 [ENTER] .125

Exercise 1.1

(a) Evaluateeachofthefollowing byhand.(i) 104 (ii) 105 (iii) 34

(b) Useyourcalculatortofindeachofthefollowing.(i) 65 (ii) 4.34 (iii) (7.1)3 (iv) (8)7 (v) 124

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cientificcalculatorsfollowhisconventionbutmost

non-scientificonesdonot.

REVISION PACK

Scientific notation

Anyrealnumbercanbeexpressedintheform(numberbetween1and10,butnot including10)(poweroften).

For example:253 = 2.53100 = 2.53102,

25.3 = 2.53

10 = 2.53

101,2.53 = 2.531 = 2.53100,0.253 = 2.530.1 = 2.53101,0.0253 = 2.530.01 = 2.53102.

Thiswayofwritinganumberindicatesitsorderofmagnitudeastheexponent,and isusefulwhencalculatingusingverylargeorverysmallnumbers,andparticularlysowhendealingwithamixtureof largeandsmallnumbers. Itiscommonlyreferredtoasscientificnotationbut isalsoknownasstandard form. Itisthisnotationwhichscientificcalculatorsusetoshowverylargeandverysmallnumberssometimesthepoweris indicatedbytheletterE(forexponent). Forexample,2.53E1means0.253.Exercise 1.2

(a) Expresseachofthefollowingnumbersinscientificnotation.(i) 1427000000 (ii) 8075 (iii) 0.00327 (iv) 0.5672(v) 0.0000004007

(b) Expresseachofthefollowingnumbers infull.(i) 3.298105 (ii) 7.654101 (iii) 1.098103(iv) 3.41010

1.2 Calculating

Order of operations

Imaginethatsomeoneasksyoutodothefollowingcalculations(readthemout loudwith . . . . . . . . . . . . beingadramaticpause).

What is2add3 . . . . . . . . . . . . times4?What is2. . . . . . . . . . . . add3times4?

Thefirst is likelytohaveproducedtheanswer20,thesecond14.Sincetherearenodramaticpauses inwrittencalculations(oroncalculatorsorcomputers),there isauniversallyacceptedconventiontoovercomethisandsimilarpossiblemisunderstandings.

Incalculations,operationsareperformed inthefollowingorder. Brackets Indices DivisionandMultiplication(theorderdoesnotmatter) AdditionandSubtraction(theorderdoesnotmatter)

Onewayofrememberingthis isthemnemonicBIDMAS.

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MODULE 1 NUMBERS

Exercise 1.3

(a) Insertbracketsineachofthefollowingcalculationswherenecessarytoemphasisetheorderinwhich itmustbeperformedaccordingtotheaboveconvention. Thendothecalculationswithoutusingacalculator.

15+5(i) 3+52 (ii) 103 3 (iii) (iv) 64 + 2

3 + 7 (v)

2

2

+ 3

10

2

(b) Whathappenswhenyouuseyourcalculatorwithoutthebrackets?Process

Foranymathematicalproblemrequiringanextensivecalculationtherearefivestages.1 Establishthecalculationtobedone.2 Make an estimate(particularlyadvisablewhenusingacalculator).3 Dothefullcalculation,usingacalculatororcomputerifappropriate.4 Verifythesolution;firstcompareyourexactansweragainstthe

estimate,toascertainwhetherit isabouttherightsize;thendosomekindofcheck(forexample,puttingthesolutionbackintotheoriginalproblem,orredoingthecalculationadifferentway).

5 Checkthattheanswermakessenseinthecontextoftheoriginalproblem. Roundasnecessary.

Example 1.2

Abuscompanyownsanumberof42-seatercoachesandhas185passengerswishingtogotoLondon. Howmanycoachesareneeded?Solution

Stage1 18542Stage2 Approximateto easynumberstoobtainanestimate:

20040=5Stage3 18542=4.4047. . . (usingacalculator)Stage4 Calculatoranswerisnottoofarfromestimate,and

calculatoranswer42=185(sotherewerenoerrorskeyingin)

Stage5 Fractionsofacoacharenot sensible,sorounduptonearestwholenumber. Answer: 5coaches.

Estimating

Anestimate isaroughanswerproducedbyusingapproximatenumbers.Estimatinganswerstonumericalcalculations is importantas itprovidesawayofcheckingthatthefinalanswer isofthecorrectorderofmagnitude,whetherit isproducedby handorusingacalculator. Theprocessinvolvessubstitutingeasynumbersthatcanbeworkedonmentallyorquicklyusingpaperandpencil.

Roundingtothenearestintegerisnotappropriateinthiscase,sincetheresultwouldleavesomepassengersstranded.

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REVISION PACK

Thesymbolisreadas ispproximatelyequalto.

Example 1.3

Estimatetheanswerstothefollowingcalculations;thenfindtheexactanswersusingacalculator.(a) 44281+3729 (b) 7002633 (c) 379.420.23(d) 0.0461.5Solution

(a) 4428144000and37294000;soanapproximateanswer is48000.Usingacalculatorgivestheexactanswer: 44281+3729=48010.

(b) 70027000and633600;soanapproximateansweris6400. Usingacalculatorgives7002633=6369.

(c) 379.42400and0.230.2;soanapproximateanswer is80. (Youmighthavesaidthat0.230.25,givinganapproximateanswerof100.) Usingacalculatorgives379.420.23=87.2666.

(d) 0.0461.5 = 0.46150.4515=0.03. Usingacalculatorwhichdisplays10digitsgives0.0461.5 = 0.0306666667.

Exercise 1.4

Carryouteachofthefollowingcalculationsusingyourcalculator,havingfirstmadeanestimateoftheansweryouexpect.(a) 441.75.2 (b) 53.470.922.2 (c) 217.5 + 60.317.7

Notetheuseofbracketsin (d) (1285329)0.023part(d)toindicatethattheubtractionistobeperformed

beforethemultiplication. RoundingIfthefinalanswertoanumericalcalculation involvesanumberwithalargenumberofdigits, it isoftenappropriatetogivearoundedanswer.

nExample1.2,4.4047. . . was Unlessthenatureofaparticularproblemdetermineswhetherasensibleoundedupto5,notdownto answer isobtainedbyroundingdownorup,aconventiononhowtoround. Useoftheconventionis numbers isused.

llustratedinExample1.4.Toroundtoagivennumberofdecimalplaces: lookatthedigitwhich isonemoreplacetotherightofthe

numberof

places;

roundupifthisdigit is5ormore,anddownotherwise.Toroundtoagivennumberofsignificantfigures: lookatthedigitwhichisthatnumberofplacestotherightofthe

firstnon-zerodigit; roundupifthisdigit is5ormore,anddownotherwise.

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MODULE 1 NUMBERS

Example 1.4

(a) Round0.0306666667tofourdecimalplaces.(b) Round47.1372536tothefollowingnumbersofdecimalplaces.

(i) two(ii) three(iii) four

(c) Roundeachofthefollowingtotwosignificantfigures.(i) 2295673 (ii) 0.027294 (iii) 0.0000405108

Solution

(a) Thedigit inthefifthdecimalplacein0.0306666667is6,which ismorethan5,sothenumber isroundedup,giving0.0306666667=0.0307(to4d.p.). Forconveniencethewords

(b) (i) Thedigit inthethirddecimalplace in47.1372536 is7,so decimalplacesand47.1372536=47.14(to2d.p.). significantfiguresareoftenabbreviatedtod.p. ands.f.(ii) Thedigit inthefourthdecimalplacein47.1372536is2,so respectively.47.1372536=47.137(to3d.p.).(iii) Thedigit inthefifthdecimalplace in47.1372536 is 5, so 47.1372536=47.1373(to4d.p.).

(c) (i) Thedigittwoplacestotherightofthefirstnon-zerodigit,fromtheleft, in2295673 is9,so2295673=2300000(to2s.f.).(ii) Thedigittwoplacestotherightofthefirstnon-zerodigit,fromtheleft, in0.027294 is2,so0.027294=0.027(to2s.f.).(iii) Thedigittwoplacestotherightofthefirstnon-zerodigit,fromtheleft,in0.0000405108is5,so0.0000405108=0.000041(to2s.f.).

Exercise 1.5

(a) Round2.1415996tothefollowingnumbersofdecimalplaces.(i) one (ii) three (iii) four (iv) five

(b) (i) Round63056totwosignificantfigures.(ii) Round0.038toonesignificantfigure.(iii) Round0.04006tothreesignificantfigures.

(c) (i) Round23.009toonedecimalplace.(ii) Round9999tothreesignificantfigures.(iii) Round6080totwosignificantfigures.(iv) Round16.99toonedecimalplace.

Youneedtobecarefulwhenusingroundednumbersincalculations. Ifyouroundanumberthatistheresultofacalculation,andthenusethisroundednumberaspartofafurthercalculation,thentheanswerthatyouobtainforthe latercalculationmaybe inaccurate. Toavoidsuchroundingerrors,itsbesttousethefull-calculator-accuracyversionsofnumbers incalculations. Storingthesenumbers inyourcalculatorsmemorywillhelpyoutodothis.

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Thefirstsixprimenumbersre 2, 3, 5, 7, 11 and 13.

REVISION PACK

1.3 Multiples, factor s and prime number s

Withtheadventofcalculators,theuseofmultiplesandfactorsindealingwithwholenumbersisdeclining,butanunderstandingandfacilitywithmultiplesandfactorsremainsvery important foralgebra.Multiples

Themultiplesofapositivewholenumberarethosepositivewholenumbersintowhichitdividesexactly: forexample,themultiplesof6are6,12,18,24, . . .;thatis,themultiplesofanumberarethatnumbermultipliedby1,2,3,4, . . .. However,findingthecommonmultiplesoftwoormorenumbers ismoredifficult,andtofindthelowestcommonmultiple(LCM)needsaknowledgeoffactors.Factors

Theprocessoffindingallthewholenumbersthatdivideexactlyintoagivenwholenumberiscalled findingthefactors. Factorisingmeanswritinganumberastheproductoftwoormorewholenumbers. Forexample,48=124 and 48 = 8 32.Aprimenumber isapositivewholenumber,otherthan1,that isdivisibleonlybyitselfand1. Factorisinganumbersothateachfactorisaprimenumberisoftenveryusefulthisprocessiscalledfindingtheprimefactors. Forexample,thefactorsof48whichareprimeare2and3;theprimefactorsof48are

48=22223 (= 24 3).Theprocedureforfindingtheprime factorsofanumberaareasfollows.It is illustrated inExample1.5(b)fora=60.To find the prime factor s ofa

Divideaby2(thefirstprimenumber)repeatedlyuntiltheresult isnotdivisibleby2. Ifthenumberofsuchdivisions isN, then 2N isafactorofa.Iftheresultisaprimenumber,theprocessiscomplete. Ifnot,repeattheprocessforeachsuccessiveprimenumber,replacing2by3,5,7,. . . ,asnecessary.

Example 1.5

(a) Findthefactorsof60.(b) Findtheprimefactorsof60.

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MODULE 1 NUMBERS

Solution

(a) Thequestionisaskingwhatpairsofwholenumberswhenmultipliedtogethermake60. Theyare

160, 230, 320, 415, 512, 610.Thusthefactorsof60are1,2,3,4,5,6,10,12,15,20,30and60.

(b) Divide60by2: 60=2

30. 30 isdivisibleby2.Divide30by2: 30=215. 15 isnotdivisibleby2.Divide15by3: 15=35. 5 isprime.

60 divide by 2

2 30 divide by 2

2 15 divide by 3

3 5prime,so stop

Figure 1.4

Sotheprime factorsof60are2,2,3,5. Thus60=2235 = 22 35.

Exercise 1.6(a) Findthefactorsof36.(b) Findtheprimefactorsof36.Highest common factor

Thehighestcommon factor(HCF)ofanytwopositivewholenumbers isthe largestnumberwhichisafactorofboth.Orputanotherway,theHCF istheproductofthelowestpowerofeachprimefactorcommontoboth.

Thefollowingexample illustrateshowtofindtheHCF.

Example 1.6

With small numbers, you may beablejusttowritedowntheprimefactors,withoutusingtheprocedure.

Thisdefinitionextendseasilytothreeormorenumbers.

What isthehighestcommonfactorofthenumbers18and30?

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Again,thisdefinitionextendsothreeormorenumbers.

REVISION PACK

Solution

First,writeeachnumberintermsofitsprimefactors. Wehave18=232 and 30=235.

Takingthe lowestpowerofthecommonprimefactorsgivesHCFof18and30=23 = 6.

Exercise 1.7

(a) Findtheprimefactorsof:(i) 8 (ii) 16 (iii) 32.

(b) Whatisthehighestcommonfactorof8,16and32?Lowest common multiple

Thelowestcommonmultiple (LCM)ofanytwopositivewholenumbersisthesmallestpositivewholenumberwhichisamultipleofboth.Orputanotherway,theLCM istheproductofthehighestpowerofeachprimefactoroccurring.

Example 1.7

What isthelowestcommonmultipleofeachofthefollowingsetsofnumbers?(a) 10,25. (b) 8,24,60.Solution

(a) Intermsofprimefactors10canbewrittenas25and25as55 = 52. TheLCMof10and25istheproductofthehighestpowerofeachprimefactor,thatis

252 = 2 25=50.SotheLCMof10and25is50.

(b) 8=23.24=2223 = 23 3.60=2235 = 22 35.SotheLCMof8,24,and60 is23 35=120.

Exercise 1.8

(a) WhatistheLCMof28and36?(b) WhatistheLCMof7,10and14?

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MODULE 1 NUMBERS

1.4 Fractions, decimals and percentages

Somerealnumberscanbeexpressedasexactfractions,that is,asonewholenumberdividedbyanotherwholenumber,suchas:

3 1 = 9 25

(=3 5); 244

(=9 4);7

(=2 7, orequally2 7.)Thenumberwrittenabovethe lineorbar inafraction iscalledthenumerator,andthenumberwrittenbelowthelineiscalledthedenominator.Converting fractions into decimals

Fractionsareconvertedintodecimalsbydividingthenumeratorbythedenominator.

Example 1.8

Convert 3 , 21 and2 intodecimals.5 4 7

Solution

(a) 35

= 3 5 = 0.6(b) 21 = 9 = 9 4 = 2.25

44

(c) 27

=(2 7)=0.285714285714. . .

In(a)and(b)above,thedivisionprocessterminates.In(c)above,thedivisionprocessdoesnotterminatebutgoes intoarepeatingloopthatgeneratesthedigits285714overandoveragain. The

issaidtoberecurring,andcanbewrittenas0.28571 4,decimalfor 27

wherethetwodotsabovethedigits2and4 indicatetheblockofdigitsthatisrepeated.Iftheresultofacalculation isafinalanswertoaquestion, itcanberoundedbymakingastatementlike: 2

7=0.2857tofourdecimalplaces.

Numberswhicharegoingtobeused insubsequentcalculationsshouldnotberounded.Exercise 1.9Converteachofthefollowingfractions intodecimals.(a) 1

5(b) 1

3(c) 33

4(d) 11

8(e) 33

7

Equivalent fractions

Whenthenumeratoranddenominatorofafractionarebothmultipliedordividedbythesamewholenumber(otherthan0),thenewfractionobtained isequaltotheoriginalone,andthetwofractionsarecalledequivalent fractions. Theyarerepresentedbythesamepointonthenumberline.

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REVISION PACK

Example 1.9

Writedownthefractionsequivalentto 35

whichareobtainedbymultiplyingitsnumeratoranddenominatorbyeachofthefollowingnumbers.(a) 2 (b) 3 (c) 10 (d) 20 (e)

1

Solution

(a) 3 25 2 =

610 (b)

3 35 3 =

915 (c)

3 105 10 =

3050

(d) 3 205 20 =

60100 (e)

3 15 1 =

35

Inacompletesetofequivalentfractions,thefractionwiththesmallestpossiblepositivedenominator issaidtobethefraction inits lowest terms. Thisisfoundfromanyoneoftheequivalentfractionsbycancelling,thatis,bysuccessivedivisionofthenumeratoranddenominatorbyeachoftheircommonprime factors.(ThisisequivalenttodivisionbytheHCFofthenumeratoranddenominator.)

Anywholenumberthatdividesintoeachoftwonumbersiscalledacommonactorofthosenumbers.

Example 1.10

28Express(a)42

and(b) 360 asfractions intheirlowestterms.240

Solution

(a) We have 28 = 2 2 7 and 42 = 2 3 7,sothecommonprimefactorsare2and7. Hence

28 =42

4

6

2

3

(dividingby7)(dividingby2).=

360 = 36 (dividingby10=2 5)(b)240 24

= 32

(dividingby12=2 2 3)Inpractice,spottinglargecommonfactors,ashere,isanefficientwaytoproceed.

Calculations involving fractions

Adding and subtracting fractions

Twofractionsareadded(orsubtracted)byconvertingeachofthemintoequivalentfractionswiththesamedenominator,adding(orsubtracting)theseequivalentfractionsand,whereappropriate,convertingtheresulttoafraction in its lowestterms.

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MODULE 1 NUMBERS

Example 1.11

Evaluateeachofthefollowingandexpresstheresult in itslowestterms.

Inthissolutiontheequivalentfractiondenominatorused isthelowestcommonmultipleofthedenominatorsoftheoriginal fractions.However,thesamefinalresultwouldbeobtainedbyusingtheproductofthedenominators:

Thereasonforchoosingtouselowestcommondenominators istokeepthenumbers involvedincalculatingtheequivalentfractionssmallenoughtobedonementallyquicklyandeasily.

Multiplying fractions

Theproductoftwofractions isobtainedbymultiplyingtheirnumeratorsandmultiplyingtheirdenominators,togiveanewnumeratoranddenominatorrespectively,then,whereappropriate,convertingtheresulttoafraction in its lowestterms.

Example 1.12

Evaluateeachofthefollowingandexpresstheresult in itslowestterms.(a) Solution

(c) +

Thenextexampleshowshowtodealwithdivision involvingfractions.

Example 1.13

Noticethatinthiscasethedenominatoroftheequivalentfractions istheproductofthedenominatorsoftheoriginal fractions.

.=

= =

10 +15 = 2550 50

(a) +Solution

+

+

(b)

(b)

(a)

(c)

(a)

3

10

5

12=3

128

12=12

43

1

2

5

10

3

10+2

10=3

10

1

5

1= 250310+15

11

54

5=22

3(b)11

32

+3=11632

5

8

66

34

1=2=213632

15=533284

(b)2132

Evaluateeachofthefollowingandexpresstheresult in itslowestterms.(a) 4 (c) (d) (b)1

2

1

2

1

4

2

3

1

3

4

5

2

3

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REVISION PACK

Solution

(a) 412

Thisquestionisthesameas howmanyhalvesarethere in4?. Theansweris8asthereare2halvesin1,sotherewillbe4times2halvesin4.

.

Analternativewayofsimplifyingaproductoffractionsistoemploycrosscancellation. Forexample,

(here2hasbeendivided into4and2)

Dividing fractions

Todivideanumberbyafraction,multiplythenumberbythefractionturnedupsidedown. Anothernameforthis upsidedownnumber is reciprocal,atermwhichisexplainedmorefully inthenextsubsection.

Mixed number s

Amixednumber ismadeupofawholenumberandafraction. Forexample,thenumber2 isamixednumber. Itcanbeconvertedintotheequivalent improperortop-heavyfraction(wherethenumerator is largerthanthedenominator)asfollows:

2 .

is the same as 4

= 8.

.

That is,4

TheanswerinExample1.13(d)wasintheformofanimproperfraction:

Inwords,thisquestion isaskinghowmanyquartersinonehalf?.Theansweris2.

= 2.

An improperfractioncanbeconvertedintoamixednumberbydividingthetopbythebottomtofindthenumberofwholeones;theremaindergivesthenumberoffractionparts leftover.For example, (75=1,remainder2).Themethodsforcalculatingwithmixednumbersareexactlythesameasthoseforthemethodsgiven inthepreviousexamplesforproperfractions(inwhichthenumerator issmallerthanthedenominator)providedthatyouchangethemixedfractiontoan improperfractionfirst.

(b)

(c)(d)

65

21

12

5=122

4=4

6=12=35102

21

= 2.

1

2

6=331

3

23

1

+

2

5

4=122

.65

=53

1=124

2=4

=2

= 1

5

7

5

12=12

2

11

42

333

4=24535

45

= 2 + 12

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Exercise 1.10

(a) Evaluateeachofthefollowingandexpresstheresultin its lowestterms.

2 1 3 3 18 2 2 3(i) (ii) 1 (iii) (iv) 4 1+ + 3 6 4 8 25 5 7 5

(b) Evaluateeachofthefollowingandexpresstheresult in its lowestterms.

4 2 4 2 (iii) 1225 389 (iv) 42 2(i) (ii) 1 5 7 5 7 7 32 114

(v) 1 13

Reciprocals

Thewordreciprocal isthemathematicaltermforhowmanyofa Thetermappliestorealnumbers,butinitiallyithelps1

2istherefore2asthereare2particularnumberarein1. Thereciprocalof

tothinkintermsofpositivehalvesin1.wholenumbers.

Thereciprocalofanumberis1dividedbythatnumber. Forexample,thereciprocalof10 is 1

10. Ineffectthenumberis inverted,because10when

10

1.writtenasarationalnumberinlowestterms is

1

10or inpowerSothereciprocalof10canbewritten infractionformas

formas101. Asadecimal, itsvalue is0.1.There isakeyforfindingreciprocalsonyourcalculatorthoughtheanswerisgivenasadecimalratherthanafraction. Trythisonyourcalculatornowforthenumber10byputtingin10andthenpressingthereciprocalkey,which ismarked [1/x] or [x1]. Withsomecalculatorstheanswer0.1willbedisplayedimmediately,withothersyoumayneedtopress[ENTER].Leavethe0.1 inthedisplayonyourcalculatorandthenpressthereciprocalkeyagain. Youwillseethatthereciprocalof0.1(1 ) is10.

10

The inverseoperationof findingareciprocal isthesameoperation, i.e.findingareciprocal;this isanunusualresult. Itmeansthatanumbermultipliedby itsreciprocalalwaysequals1.

1

10= 1. Inthecaseof10: 10

Inthecaseof 110

: 110

10=1.1

2= 1. Inthecaseof2: 2

1Ingeneral: n = 1.

nExample 1.14

Writedownthereciprocalofeachofthefollowingnumbers,givingtheanswerinbothfractionanddecimalform.

1

4(a) 5 (b) (c) 100 (d) 3Solution

3(a) 15

= 0.2 (b) 41

= 4 (c) 1100 = 1100 =0.01 (d) 1 = 0.3

Exercise 1.11

Writedownthereciprocalofeachofthefollowingnumbers,givingtheanswerinbothfractionanddecimalform.(a) 8 (b) 10 (c) 1

5(d) 25

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REVISION PACK

Percentages

Insomecontexts,andespeciallywhenmakingcomparisons, it isconventionaltoexpressafractionasapercentage. This is a way of expressingfractionswithdenominator100;forexample, 3

5= 60 iswritten

100

%isreadpercentwhich as60%.meansperhundred.

Afraction

is

converted

into

percentage

form

by

multiplying

it

by

100

andwriting%aftertheresult.

Example 1.15

2 27 5Expresseachofthefollowingfractionsaspercentages: ,20

, .5 6

100%= 20027 100%= 2700%=135%20 20

5

2

5%=40%

5

6 100%= 500% = 83.33%(to2d.p.)6

Exercise 1.12

Expresseachofthefollowingfractionsasapercentage.1 1 3 1 3(a)

Tofindagivenpercentageofanumber,changethepercentagetoafractionoradecimalanduseittomultiplythenumber.

(b) (c) 3 (d) 1 (e) 35 3 4 8 7

Example 1.16

Find6%of340.Solution

6% is 6100

or0.06.So6%of340 is0.06340=20.40.

Exercise 1.13

Findeachofthefollowing.isthesymbolforgram. (a) 40%of250g (b) 4%of250g (c) 104%of250g

2

1(d) 17Tofindonequantityasapercentageofanother,dividethefirstquantitybythesecond(toformafraction)andthenexpressthefractionasapercentage.

% of 150 (e) 200%of50

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MODULE 1 NUMBERS

Example 1.17

Express25gasapercentageof125g.Solution

25

(25

125)

100%

=

125

100%

=

20%.

Exercise 1.14

(a) Express32.50asapercentageof50.00(b) Express23asapercentageof115.(c) Express55asapercentageof50.Percentage increase

Findinganewpricewhentheoriginalpricehasgoneupby10%canbedoneintwostages: byfirstfindingthe10% increaseandthenadding ittotheoriginalprice. Sometimesit isusefultofindthenewpriceusingjustonestage.

Example 1.18

Thepriceofan80CDplayergoesupby10%. What isthenewprice?Solution

Firstmethod: 1010%of80 is 80=0.180=8,sothe increaseis8.100

Thenewprice is80+8 = 88.Secondmethod:Theoriginalcostof80canberepresentedby100%,so increasingthisby10%to110%isthesameasworkingout110%of80.

11080=1.180=88.100

Theadvantageofthisone-stagemethod isthatitcanbeusedtoworkbackwardstofindtheoriginalprice ifthenewpriceandthepercentageincreaseareknown.

Example 1.19

Thenewpriceofatelevisionis225andthis is25%morethantheoriginalprice. Whatwastheoriginalprice?

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REVISION PACK

Solution

An increasefrom100%to125%needsamultiplierof125/100=1.25. Sooriginalprice1.25=newprice.

Taking25%offthenewprice Sothenewpricedividedby1.25givestheoriginalprice:doesnotgivethecorrect

nswer. (Trythisifyouare 2251.25=180.notconvinced.)

Exercise 1.15

Fill inthegapsinthetablebelow.Originalprice() Percentageincrease Multiplier Newprice()100 30% 1.3(130/100) 130

45 3% 1.03(103/100) 46.35230 15%120 150%

20% 1.2 43.2076 17.5% 1.175

17.5% 30

Percentage decrease

Theone-stagemethodcanalsobeusedforfindingreducedprices.

Example 1.20

Oneofthe80CDplayersisshop-soiled,soitisreducedby15%. Whatisthenewprice?Solution

Theoriginalcostof80canberepresentedby100%,sodecreasingthisby15%isthesameasworkingout(10015)%or85%of80. Thenewpriceis

85 80=0.8580=68.100

Exercise 1.16

Fill inthegapsinthetablebelow.Originalprice() Percentagedecrease Multiplier Newprice()100 20% 0.8 80200 25%

45 5%50% 9030% 154

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MODULE 1 NUMBERS

1.5 Calculating with signed number s

Oftenacalculation involvesamixtureofpositiveandnegativenumbers.Belowisasummaryoftheruleswhichgovernthearithmeticofsigned(positiveornegative)numbers.

Addinganegativenumberisequivalenttosubtractingthecorrespondingpositivenumber,e.g.4+(2)=42.Subtractinganegativenumberisequivalenttoaddingthe Thisisoftendescribedascorrespondingpositivenumber,e.g.4(2)=4+2. twominusesmakeaplus.Multiplyingordividingtwonumberswiththesamesigngivesapositiveanswer,e.g.(4)(2)=42,(4)(2)=42.Multiplyingordividingtwonumberswithoppositesignsgivesanegativeanswer,e.g.(+4)(2)=(42)=8,(+4)(2)=(42)=4

2

=2.

Example 1.21

Evaluateeachofthefollowing.(a) 5+(8)(b) (5)+(8)(c) 5(8)(d) (5)(8)(e) (7)(3)(f) (7)3(g) (12)4(h) (12)(4)Solution

(a) 5+(8)=58 = 3(b) (5)+(8)=(5)8 = 13(c) 5(8) = 5 + 8 = 13 (d) (5)(8)=(5)+8=3(e) (

7)

(

3)=+(7

3)=7

3 = 21 (f) (7)3 = (73)=21(g) (12)4 = (124)=3(h) (12)(4)=+(124)=3.

Exercise 1.17

Evaluateeachofthefollowing.(a) (i) (2)+(7) (ii) (5)+8 (iii) 3+(5)(b) (i) 4

(

2) (ii) (

3)

(

5) (iii) (

3)

(

3)(c) (i) 4(3) (ii) (2)(7) (iii) 3(9)(d) (i) 24(6) (ii) (40)(8) (iii) (45)15

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1.6 Working with powers, indices and logarithms

Inanumberoftheforman,aiscalledthebaseandnthepower(indexorexponent).Calculations involvingdifferentbasesraisedtothesamepowercanbemadesimplerbyfactorisingfirst.

Example 1.22

Rewriteeachofthefollowing infactorisedformandhencesimplify itmentally.(a) 22 (b) 513

42 173

Solution 22 12 = 1(a) 22 =

4=

42 2 4

2

2

2 1(Infulldetail: 22 = 22 =

42 = = 2

2

=1

2

=4

.)42 44 4 4 22 2513(b) 513 =

17= 33 = 27

173

Exercise 1.18

Factoriseandsimplifyeachofthefollowingmentally. Thencheckyourresultusingacalculator.(a) 1002 (b) 44 (c) 93

252 24 33

Thereareseveralrulesforcombiningpowers.Combiningpowers,andnegativepowersTomultiplytwopowerswiththesamebase,addtheindices:

m m+n 32 = 34+2a an =a e.g.34 = 36.Todividetwopowerswiththesamebase,subtractthe indices:

ma an =amn e.g.34 32 = 342 = 32.Tofindthepowerofapower,multiplytheindices:

m)n(a =amn e.g.(34)2 = 342 = 38.Anumberraisedtoanegativepower isthereciprocalofthenumberraisedtothecorrespondingpositivepower.

ma = 1/am e.g.32 = 1/32 and32 = 1/32.

Fractional powers

Theaboverulesapplytofractionalpowersaswellas integerpowers. Forexample,

101/2 101/2 = 101/2+1/2 = 101 = 10.

3 3

10issaidasthecuberoot Thus(101/2)2 =10,andso101/2 = 10. Similarly101/3 = 10.f10.

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MODULE 1 NUMBERS

Example 1.23

Evaluateeachofthefollowing,firstasapower,thenasanumber.(a) (i) 72 73

(ii) 712

712(iii) 75 73(iv) 75 75(v) 73 75

Solution

(a) (i) 72 73 = 72+3 = 75 =16807Check: (77)(777)=(77777).

Thisconfirmsthat712 = 7.(ii) 712 712 = 7 1 1 = 71 = 7+2 2

7

(iii) 75 73 = 753 = 72 = 49 5 77777Check: =

73 7777

(iv) 75 75 = 755 = 70 = 1 5 77777

Check: =75 77777

1(v) 73 75 = 735 = 72 =

72

73 777Check: =

75 77777

77= .1

1= = 1.

1= 0.0204(to4d.p.)

1= .

77

Infact,a0 = 1 for any real numbera.

Exercise 1.19

(a) Evaluateeachofthefollowing,firstasapower,thenasanumber.(i) 102 103(ii) 103 101(iii) (10)3 (10)2(iv) 1013 1013 1013(v) 102 101

(b) Evaluateeachofthefollowing,firstasapower,thenasanumber.(i) 107 104(ii) 104 107(iii) (10)7 (10)4(iv) 1032 1012(v) 102 101

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ndex,powerandexponentllmeanthesame.

00=102.

Logarithmswereoriginallydevised in the seventeenth

enturytosimplifyomplicatedcalculationshroughtheuseoftables.

Theywerealsothebasisofliderules,whichhavebeeneplacedbyelectronicalculators.

REVISION PACK

Logarithms: common and natural

Beforestartingthissectionmakesureyouarefamiliarwithpower(index)notationandcanconfidentlymanipulatenumberswritteninthisform.Thenumber100canbewritteninpowernotationas:

102,where

the

base

is

10

and

the

power

is

2.

In

this

example

we

say

that

the

logarithmtothebase10ofthenumber100is2,andwrite

log10 100=2.Logarithmswhichusethebase10arecalledcommon logarithmsandarewrittenas log10 orsometimessimplyas log.Anypositivenumbersuchas2,3or5canbeusedasabaseforalogarithm.Natural logarithmsusethebasee, where eisan irrationalnumberwithavalueofapproximately2.71828. Therearetwokindsofshorthandfornatural logarithms: loge and ln. Inthistextweuseln.Youneedtobeabletochangefrompowerformtothecorrespondinglogarithmic formandbackagainwithconfidence,byhandforsimplecasesandusingacalculatorforawkwardnumbers.

Example 1.24

(a) Writethenumber16asapowerof4(powerform)andthenwritedownthecorrespondinglogarithmic form.

(b) Changelog5 625=4topowerformandthenwritethepowerformasasimplenumber.

Solution(a) Powernotation: 16=42.

Logarithmicform: log4 16=2. The latterisreadas logtothebase4 of 16 equals 2. Inboththeseforms,16 isthenumber,4isthebaseand2 isthe indexorlogarithm.

(b) Thebase is5andtheindex is4,sothepowerform is54 andthesimplenumber is625.

Exercise 1.20

Fill inthegapsinthetablebelow.Power form81=341 = 60

0.2 = 515 = 5

1

x= 24

Logarithmic formlog2 4 = 2 log5 125=3

logx8 = 2

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MODULE 1 NUMBERS

Herearesomemoreexamplesphrased inadifferentway.

Example 1.25

Findthelogarithmof8tothebase2.SolutionWrite8asapowerof2: 8=23; so log2 8 = 3.

Exercise 1.21

(a) Findthe logarithmtothebase2ofeachofthefollowing.(i) 16(ii) 2(iii) 0.5

(b) Withoutusingacalculator,findthelogarithmtothebase10ofeachofthefollowing.(i) 1000(ii) 10 Youwillhavenoticedthat(iii) 0.1 log10 10andlog2 2 have the

value1. Thisresultisgenerallytrue,i.e.logaa= 1.

The inverse of a logarithm

Ifyouknowthe logarithmofanumberusingaknownbase,howdoyoufindthenumberitself? Forexample,ifthe logarithmtobase10ofsomenumber

x, say, is

4,

what is

x? We

have

log10 x= 4,

andthecorrespondingpowerformgivesx:x= 104 =10000.

10000 issaidtobetheantilogarithmof4tobase10.

Example 1.26

Findthe

antilogarithm

of

5to

the

base

2.

Solution

Here,ifxistheantilogarithm,5 = log2 x,

sothatx= 25 = 32.

Sotheantilogarithmof5tothebase2 is32.

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REVISION PACK

Exercise 1.22

(a) Withoutusingacalculatorworkouttheantilogarithmtothebase10ofeachofthefollowing.(i) 5 (ii) 0 (iii) 2

(b) Findtheantilogarithmtothebase3ofeachofthefollowing.(i) 2 (ii) 1 (iii)

1

Thelogarithmtoagivenbaseofanumberisthepowertowhichthebase israisedtocalculatethenumber.

Mostoftheindicesencounteredsofarhavebeenintegersbutthesystemofpowernotationextendstofractional indices. Thenumber1canbewrittenas100 andthenumber10as101. Thenumbersbetween1and10canalsobewritteninpowernotationusingthebase10andthecorrespondingindices liebetween0and1. Oneofthese,theindex0.5,hasalreadybeenusedasanalternativenotationforasquareroot: 10=100.5 = 10 12 .Thevalueofthesquarerootof10 is3.162(to3d.p.),that is

10=100.5 = 3.162(to3d.p.).

Sothelogarithmtothebase10of 10=3.162(to3d.p.) is0.5:

log10 3.162log10 10=0.5.Notethat,althoughthe index0.5isexactlyhalfwaybetween0and1theactualnumber(3.162) is lessthanhalfwaybetween1and10.

Calculatorsmayhave Logarithmicvaluesofnumberscanbefoundaslists in logtablesbutnowupersededlogarithmsasa theyaremoreusuallyobtainedbypressingtheappropriatecalculatoralculatingtoolbuttwowell keys. Scientificandgraphicscalculatorshavethefacilitytodealwithbothnownmeasurementscales commonandnatural logarithms. Forcommon logarithms,tothebase10,

uselogarithms. Oneisthe thekeytouseisgenerallymarked[LOG]andtheinverse(orRichterscaleforearthquakes antilogarithm) isthesecondfunctionkeymarked [10x].

ndtheotheristhebelequivalentto10decibels) Fornaturallogarithms,tothebasee,thekeytouseisgenerallymarkedystemformeasuringthe [LN]andthe inverse(orantilogarithm) isthesecondfunctionkeymarkedntensityofsound. [ex].

Findthesekeysonyourcalculatorandestablishhowtheywork: formostscientificcalculators,youenterthenumberthenkey(s);formostgraphicscalculators,youpressthekey(s)thenthenumberand [ENTER].Exercise 1.23(a) Useyourcalculatortofindeachofthefollowing,givingyouranswerto

fourdecimalplaces.(i) log10 1.5489(ii) log10 15.1149(iii) log10 734.256

(b) Useyourcalculatortofindeachofthefollowing,givingyouranswertofourdecimalplaces.(i) ln1.5489(ii) ln15.1149(iii) ln734.256

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Exercise 1.24

(a) Useyourcalculatortofindthenumberswhose logarithmstothebase10arethefollowing,givingyouranswerstofourdecimalplaces.(i) 0.5267 (ii) 0.0023 (iii) 2.4593

(b) Useyourcalculatortofindthenumberswhosenatural logarithmsarethefollowing.(i) 0.5267 (ii) 0.0023 (iii) 2.4593Combining logarithmsSince logarithmsareindices,therulesformultiplyinganddividingpowerscanbeadaptedforcalculationswith logarithms.Rule 1

Thelogofaproduct isthesumofthe logs. Forexample,forthenumbersaandb:

log10(a

b) = log10 a+ log10 b,ln(ab) = ln a+ ln b.

Rule 2

Thelogofaquotientisthedifferencebetweenthe logs. Forexample,fortwonumbersaandb:

alog10 b = log10 alog10 b,aln = ln alnb.b

Rule 3

Thelogofanumberraisedtoapoweristheproductofthepowerandthe logofthenumber. Forexample,forthenumberaandpowern:

n nlog10 a =nlog10 a and lna =nlna.

This lastresultwillbeusedlaterforsolvingexponentialequations(seeModule5).

Example 1.27

Simplifyeachofthefollowing.(a) log10 6 + log10 5 (b) 3 ln 2 ln4Solution

(a) log10 6 + log10 5 = log10(65) (byRule1)= log10 30

(b) 3 ln 2 ln4=ln23 ln4 (byRule3)= ln 8 ln4

8= ln (byRule2)

4= ln 2

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2

REVISION PACK

Exercise 1.25

Simplifyeachofthefollowing.(a) (i) log10 12+log10 3 (ii) log10 12 log10 6 (iii) 3log10 5(b) (i) ln15+ln7 (ii) ln35 ln14 (iii) 4ln3

1.7 Working with roots Rootsareoftenbestconsideredasfractionalpowers;forexample,

3writteninpowerformis21/2, 27is271/3. However,therearesomepointsworthunderstandingwhenusingtherootform. Eachpositivenumberhastwosquareroots,onepositiveandone

ometimes iswritten in negative;forexample,thetwosquarerootsof9are3(3 3=9)andplaceof, and adenotes 3 (3 3=9). Thesymbol denotesthepositivesquareroot,so bothsquarerootsofa. thetwosquarerootsof9are 9 = 3 and 9 = 3. Calculatorsgive

onlypositivesquareroots,soyoumustrememberaboutthenegativeones.

Anumberdividedbyoneof itssquarerootsgivesthatsquareroot;forexample,3 = 3 because 3 3 = 3.

3 Thepositivesquarerootofapositivenumbercanberewrittenasa

productofthepositivesquarerootsoftheprime factorsofthat number;forexample, 6 = 2 3. (Thenumbers 2 = 1.414. . . and

3 = 1.732. . . are irrationalnumberswritteninsurdform. It isconventiontoremoveasurdinthedenominatorofafractionbymultiplyingthenumeratoranddenominatorbythatsurd.)

Sometimesit ismoreappropriatetosimplifyanexpressioninvolvingsquarerootsusingfactorsratherthancalculateanumericalsolutionusingacalculatorparticularly iftheresult istobeusedinfurthercalculations.TosimplifyapositivesquarerootRewritethesquarerootastheproductofthepositivesquarerootsofthenumbersprimefactors.

e.g. 175= 5 5 7 = 5 5 7.Multiplytogetherpairsof identicalsquareroots.

e.g. 175= 5 5 7 = 5 7.Theaboveprocesscanbeshortenedifyoucanspothowtowriteasquare

Theperfectsquaresare1(12),(22),9(32), . . . . rootastheproductofaperfectsquareandanotherfactor. Forexample,

108=9 12=912 (9 isaperfectsquare) = 3 3 4 = 3 3 2 = 6 3,

or

108= 36

3 = 36

3 (36isaperfectsquare)= 6 3.

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MODULE 1 NUMBERS

Example 1.28

(a) Evaluate 2 8withoutusingacalculator.(b) Simplify 60.Solution (a) 2 8 = 28 = 16=4(b) Write60asaproductof itsprimefactors: 2235. Then

60= 2235 = 2 2 3 5 = 2 3 5 = 2 15.

Exercise 1.26

(a) Evaluateeachofthefollowingwithoutacalculator. 9 18(i) 100 (ii)

(iii)

9 2(b) Simplifyeachofthefollowingbyfactorisingfirst. 15

(i) 200 (ii) 112 (iii) 256 (iv) 3

Exercise 1.27

(a) Asquarehasedgelengtha. What isthelengthofitsdiagonal?(b) The longestedgeofanisoscelesright-angledtriangle isa. What isthe

lengthoftheshorteredges? A-sizepaperhasedgesof lengths1and 2.(c) Whatisthe lengthofthediagonalofA-sizepaper?(d) ShowthatcuttingA-sizepaper inhalf,paralleltotheshorteredge,

createstworectanglesthataresimilartotheoriginalrectangle.(e) ThelargestpossiblesquareiscutfromanA-sizerectangle. Whatare

thedimensionsofthewastepiece?(f) Twoequalsquares(as largeaspossible),sidebyside,arecutfrom

A-sizepaper,leavingarectangle. Writedownthedimensionandtheareaofeachpiece. Checkyouransweragainsttheareaoftheoriginalrectangle.

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REVISION PACK

Surd form

In some of the exercises above we have left an answer in surd form(e.g 2and 5)wheretherewasnoexactnumericalformforthenumber. Ifyouinput 2intoyourcalculator itwilldisplaysomething like1.414213562.This is,ofcourse,averygoodapproximationto 2,moreaccuratethanweshouldneedformostpurposes,butit isanapproximation. In fact, there isnowaytowritedowntheexactvalueof 2exceptbyusingthissymbol. So,inmanycircumstances,ratherthanwritedownsomeapproximation like1.4or1.4142weleavetheresult insurdformas

2. Thisensuresthatouranswer isabsolutelycorrect,andenablesustoensurethatsubsequentcalculationsareaccurate. Forexample,suppose thattheexactanswertoaquestion is 2,andwewrite itdownas 2ratherthan1.4or1.4142. If,later,weneedtosquarethisanswerthenit willbecalculatedas 2 2=2ratherthan1.41.4 = 1.96or1.41421.4142=1.99996164.YoucanusePythagorasTheoremtocalculatetheheightofanequilateraltriangle. Forexample,forthetriangleshowninthefigure,

h2

+ ( 1

2)2

= 12

= 1.

1 1h

1

212

3So, ifwehaveh2 =4

, then h= 3/4 = 3/2. Frequentlywewritethisas

3 . Here 3 isasurd,standingfortheexactpositivesquarerootof3. To2

threedecimalplaces, 3 = 1.732,but(1.732/2)2 + ( 1

2)2 = (0.866)2 + 0.25=0.99956,while

3( 3/2)2 + ( 14

+12

)2 =4

= 1 exactly. Surdscanbemanipulatedexactlyasyouwouldmanipulateothernumbers.Someexamplesaregivenbelow.

Example 1.29

Withoutusingacalculator,simplifythefollowing. (a) 2 6 (b) 4/ 2 (c) 18Solution (a) 2 6 = 2 2 3 = 2 3(Usuallywrittenmoresimplyas

2 3.) (b) 4/ 2 = 2 2/ 2 = 2 2 2/ 2 = 2 2

(c) 18= 92 = 9 2 = 3 2

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Example 1.30

Twopeoplecontributetoabudgetintheratio10to30. Putthisratioinits lowesttermsandwriteeachcontributionasafractionofthewholebudget.Solution

Divideeachquantity intheratio10:30bytheirHCF,inthiscase10.Theratiobecomes1:3.Thereare4partsaltogether,soonepersoncontributes 1

4ofthewhole

budgetandtheothercontributes 34

.

Exercise 1.29

Completethetable.Ratio

(comparingpartwithpart) Ratio(in lowestterms) Asfractions(partsofthewhole)4 : 16 10:56 : 9 1 : 7 10:20:3030:25

Example 1.31

600istobesharedbetween2people intheratio2:3. Howmuchwilleachpersonget?Solution

Thereare5parts,soonepersonwillget 25

of600,whichis240. Theotherpersonwillget 3

5of600,which is360.

Exercise 1.30

Shareeachofthefollowingamountsofmoney intheratioshown.(a) 400between2peopleintheratio3:5(b) 250between3peopleintheratio1:2:2(c) 1000between2people intheratio3:2Map scales

Mapsandplansaredrawntoscaletorepresentsomething inthephysicalworld. Inascaledrawing,all lengthsaremultipliedbythesamescalefactoroftendenotedbythe letterk.

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MODULE 1 NUMBERS

Example 1.32

Amaphasascaleof1:25000. Thismeansthatthescalefactoris25000.(a) Afootpathalongsideacanalmeasures12.5centimetresonthemap.

Howfar, inkilometres,would itbetowalkalongthispath?(b) Thedistancebetweentwoplacesbyroad is8km. What lengthwould

thisberepresentedbyonthemap?Solution

(a) 12.5cmonthemaprepresentsadistanceof12.525000cm.12.525000cm=312500cm.Thereare100cm inametre,sothelengthofthepath is

312500cm=3125mor3.125km.

100(b) 8 km = 8 1000m

= 8 1000100cm=800000cm

800 000Onthemapthislengthwouldberepresentedby25 000

cm=32cm.

Exercise 1.31

Amaphasascale1:50000.(a) What istheactual lengthofaroadshownas16.5cm?(b) Whatwouldbethelengthonthemapofapathof length15km?Sometimesratiosaregivenassinglenumbers;forexample,theratioofthecircumferenceofacircleto itsdiameter isafixednumbercalledpi, the symbolforwhich is. = 3.14159. . ..Enlargement

Intheenlargementofaphotograph,thephotohastostay inproportioninordertoavoiddistortion. Thatis,theproportionate increase inheightofthephoto isthesameastheproportionate increase inwidth.

Example 1.33

Aphotohasaheightof15cmandawidth12cm. Whatwillbethenewwidthiftheheight is increasedto20cm?Solution

20The lengthmultiplier is15

. Thewidthwillgoup inthesameproportion,sothesamemultiplier isused. Thenewwidthis

2012cm=16cm.15

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Exercise 1.32

Completethetableshowingthedifferentenlargementorreductionsizesforthephotographabove.Height 15 20 30 10Width 12 6 40

Proportion

AlanandBethdecidethattheyneedtoputmoremoney intotheirbudgetbuttheyagreetokeeptheamountsinthesameproportion, i.e.BethwillstillpaytwiceasmuchasAlan. IfAlancontributes110,thenBethwillput in220. Theratio is220:110=2:1,i.e. itstaysthesame.Exercise 1.33

Intheabovesituation,ifBethpays450,howmuchwillAlancontribute?Direct and inverse proportion

Iftwoquantitiesaredirectlyproportionaltoeachother,multiplyingonequantitybyanumbermeansthattheotherquantity ismultipliedbythesamenumber. Anexampleofdirectproportion is: thegreateryourrunningspeed,thegreaterthedistancetravelled inafixedtime. Wesaythatspeed isdirectlyproportionaltodistancetravelled.Iftwoquantitiesareinverselyproportional,multiplyingonequantitybyanumbermeansthattheotherquantityisdividedbythesamenumber(ormultipliedby itsreciprocal). Anexampleofinverseproportion is: thefasterthespeedofajourney,theshorterthetimetaken. Inthiscase,speed

is

inversely

proportional

to

time

taken.

Weshallreturntotheseideas inModule7.

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Module 2 MeasuresMeasuressuchasacountof10000carsandaspeedof10kmperhourillustratetheneedforquantitiestobe labelledandforunitstobespecified,sothattheycanbecomparedandusedtoestablishrelationships.

2.1 Units

Systeme Inter national dUnit es (The inter national system of units)

This internationallyagreedsystem,referredtoasSIunits, isbasedonthemetricsystem. Itcomprisesa limitednumberofbaseunitswhichhaveassociatedsymbols. Examplesaremetre(m)for length,second(s)fortime, litre(l)forcapacity. Forlargeorsmallquantitiesthesebaseunitsarecombinedwithprefixesto indicatemagnitude inpowersoften;forexample,

kilometre

(km)

and

millisecond

(ms).

The

table

shows

some

of

theseprefixes.Table 2.1

Prefix Symbol Figures Words Powersof10mega M 1000000 million 106kilo k 1000 thousand 103

1 one 100centi c 0.01 hundredth 102milli m 0.001 thousandth 103micro 0.000001 millionth 106

Baseunitscanbecombinedtoproducederivedunits. Here are some examples.

area m2 (squaremetres)volume m3 (cubicmetres)velocity ms1 (metrespersecond)acceleration ms2 (metrespersecondpersecond

ormetrespersecondsquared)Notethatinaderivedunit,afinespaceisplacedbetweenbaseunits.

(mu) isaGreekletter,pronouncedmew. ThefullGreekalphabetisgivenintheGuide to Preparation.

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Thepictorialrepresentationfinequalitiessuchas

P > 150000arelookedatinModule5.

REVISION PACK

2.2 Inequalities

The=signindicatesthattwothingshaveexactlythesamevalue. Ineverydaylifeyouare likelytomeetsituationswhichinvolvearangeofvaluesandusephrasessuchasmorethanorlessthanortheirequivalentinexpressionslike allpricesare lessthan1and thepopulationofMiltonKeynesisgreaterthan150000.Theseexpressionsareexamplesofinequalities. Themathematicalsymbolsusedtoexpressrelationshipsof inequalityareasfollows.

Symbol Meaning isgreaterthan is greater than or equal to

Remember: ifyouthinkofan inequalitysymbolasanarrow itpointstothesmallervalue,asFigure2.1shows.

is greater than is less than

larger > smaller smaller < larger

Figure 2.1

Examplesoftruestatementsinvolvingthe inequalitysymbols< and> are5< 7, 7> 5, 1< 0, 5> 7.

Truestatements,involvingandare11 and 1 1.

Thesesymbols

can

also

be

used

with

unknowns.

For

example,

if

P is

the

populationofMiltonKeynesthenthesecondintroductoryexamplemaybewritten

P > 150000.

Exercise 2.1

(a) Writeeachofthefollowingstatementsinwords.(i) 16> 12(ii) 11< 9

(b) Writeeachofthefollowingstatementsinsymbols.(i) 9 isgreaterthan11.(ii) Thenumberofstudentsintheclassroom isat least20.

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MODULE 2 MEASURES

2.3 Formulas

Anexpressionoftherelationshipbetweenquantitiesasanequality iscalledaformula;forexample:

circumferenceofacircle=2timesradius.Formulasareoftenexpressedinsymbols;forexample:

C= 2r,whereC isthecircumferenceandr istheradius.Anotherexampleis:

areaofatriangle = 12

baseheightor,usingtheobvioussymbolsbforbaseandhforheight(seeFigure2.2),

A= 12

bh.

base

height

Figure 2.2

Thetwoformulasabove involvesymbolswhichincludeunits. Forexample,ifr=6cmthen

C= 2

6 cm = 12cm.Sometimesthesymbolsinaformularepresentjustnumericalqualities. Anexample is as follows. ToconvertfromdegreesFahrenheittodegreesCelsius,taketheFahrenheitmeasurementandsubtract32;multiplytheresultby 5

9. Usingtheobvious

symbolsforthenumbersinvolvedgivestheformulaC= 5

9(F32).

Formulasare ineffectarecipefordoingacalculationwhenallbutoneofthequantities isknown. Substitutinginformulas isstraightforwardprovidedconsiderationisgiventotheunits,whereappropriate,andtheformula is therightwayround. Sometimesaformulaneedstoberearrangedor undone. Onewayofdoingthisis illustratedbelow. Ifyouareconfidentinan

alternativemethod,usethat.

Example 2.1

UndotheformulaC= 59

(F32)sothatitreads F =something.

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Solution

WerepresenttheformulaC= 59

(F32)asadiagramtoshowtheorderoftheoperations,from lefttoright.

F 32 F32 59 59

(F32)(=C) (2.1)Toundotheformulatoobtain

F = something,weundoeachoftheaboveoperations inreverseorder, i.e.startingattheright,atC.

C 59 CC+ 32 +32(F =)9 95 5

ThisgivestheformulaF = 9

5C+ 32.

The importantfeaturetonotice inproducingthesediagrams isthatyouneedtostart(Equation2.1)withthesymbol(F) that you wish to make the subjectoftheformula.

Exercise 2.2

Rearrangeeachofthefollowingformulasasindicated. Youshouldusethedoingandundoingmethod inpart(a).(a) Theareaofatriangleformula,A= 1

2bh, to give a formula for h, the

height,intermsofAandb.distance

(b) speed= togiveaformulafortime

(i) distance, (ii) time.(c) Thecircumferenceofacircleformula,C= 2r, to give a formula for r,

theradius.

2.4 Measuring and classifying angles

Anangle isanamountofrotationorturning,normallytakenaspositivewhenmeasured intheanticlockwisesenseanglesmeasuredclockwisearetakentobenegative;seeFigure2.3.

B B

Oanticlockwise rotation

A O Aclockwise rotation

Figure 2.3

Anglescanbemeasuredinvariousunitsforexample,degrees,radiansandgradients,butonlythefirsttwoareusuallyused inmathematics.

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MODULE 2 MEASURES

TheBabyloniansdividedthecircumferenceofacircleinto12equalsectionsanddividedeachofthosesectionsinto30equalparts. Theresulting partcorrespondstothepresent-daydegree,becausethereare360degrees inafullrevolution. Figure2.4showshowtheBabylonianandthepresent-daysystemsarerelated.

Figure 2.4

Alessarbitrarywayofdividingthecircumferenceofacircle(2r) is to divide itbytheradius(r). Thisgives 2r = 2partsforacomplete

rrevolution,foranyradius.Itisthisobservationwhich leadstothedefinitionofaradian.

Theanglesubtendedatthecentreofacirclebyanarcequalinlength Anarcofacircleisanypartofthecirclejoiningtwototheradius isdefinedtobeoneradian;seeFigure2.5. Therearepointsonthecircle.2radians inafullrevolution.

1 radian

r

r

Figure 2.5

Exercise 2.3

Drawroughdiagramstoindicateeachofthefollowingangles. Labeltheanglesinbothdegreesandradians. (Thesymbolforthedegree is; for example,360degrees iswritten360.)(a)(b)

1

4

1

3

revolutionrevolution

(c) 16

revolution

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Converting angle measures

Since2radians=360,360 2

1 radian = and1= radians.radian=57.30(to2d.p.); 2 360

= 0.017radians(to3d.p.). Theseequivalencesleadtotherulesbelowforconversionfromradianstodegrees,andviceversa.

360 2=2 360xradians= x and y y radians.

Example 2.2

Convert113.5toradians.Solution

2113.5= 113.5 radians360

= 1.98radians(to2d.p.)

Exercise 2.4

Unlessdegreesarespecified, (a) Converteachofthefollowingtoradians,givingyouranswertotwooushouldassumeanangle decimalplaces.

measurementisgivenin(i) 54 (ii) 125 (iii) 667.18

adians.(b) Converteachofthefollowingtodegrees,givingyouranswerstoone

decimalplace.(i) 2/7radians (ii) 1radian (iii) 0.5 radians

Calculatorscanbesettowork indegreesorradians(andsometimesgradientsaswell). Makesureyouknowhowtowork indegreesandradiansonyourcalculator.Somecalculatorscanbeusedtoconvertbetweendegreesandradiansautomatically. Ifthisfunctionisavailableonyourcalculator,repeattheaboveexerciseusing it.

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MODULE 2 MEASURES

Classifying angles

Severaldifferenttypesof(positive)anglearenamed,asfollows.Anacuteangle is lessthanaquarterrevolutionthat is, lessthan90, or /2 radians.

acute angle

Figure 2.6

Arightangle isaquarterrevolutionthatis,exactly90, or /2 radians.

Notetheuseofthemarktodenotearightangle.

Figure 2.7

Anobtuseangle isbetween90and180(butnotequaltoeither)that is,between/2 and radians.

obtuse angle

Figure 2.8

Areflexangle isbetween180and360that is,betweenand2radians.

reflex angles

Figure 2.9

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2.5 Statistical measur es

Statistics involvescollecting,analysing, interpretingandcommunicatingnumericaldata. Oneaspectofanalysingasetofdataistocompare itwithanotherset,sothereneedstobeagreementaboutformsofmeasure.Onewayofprovidingasummaryofabatchofdata isanaverage,andtherearethreetypesofaverage:mean,modeandmedian. Which is mostappropriatetousedependsonwhattheaverage istobeusedfor; insomecasesmorethanonemaybeneeded.Mean

Themean iswhatmostpeopleandthemediausewhentalkingaboutanaverage. Itisfoundbydividingthesumofallthe itemvalues inabatchbythenumberof items(thebatchsize). Thisgivestheformula

sumofvaluesmean= .

batchsizeThemathematicalsymbolfor sumofistheGreeklettercapitalsigma.Ifthevaluesaredenotedbyx,thebatchsizebyn, and the mean by

x(readasxbar),thentheformulaforthemeanmaybewritten insymbolsas

xx= (readasxbarequalssigmaxovern).

nSomescientificcalculatorshavestatisticalfunctionswhichusethesesymbolstodenotethekeystouse. Thevaluesareenteredbyusingaspecialkey;onsomecalculatorsthisismarked [+].

Example 2.3

Theheightsinmetres(measuredtothenearestcentimetre)ofagroupofsevenpeoplearegivenbelow.

1.52 1.72 1.66 1.81 1.69 1.59 1.77What isthemeanheightofthegroup?Solution

Thenumberofvaluesinthebatch(thebatchsize)is7. Hence1.52+1.72+1.66+1.81+1.69+1.59+1.77

meanheight= m711.76

= m7

= 1.68m.

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,

MODULE 2 MEASURES

Exercise 2.5

Findthemeanofeachofthefollowingbatchesofdata.(a) 23 21 26 26 24 (b) 101 107 98 92 115 102Dataareusually listeddifferentlywhenseveralvaluesoccurmorethanonceandcareneedstobetakentoincludeallthedata. Inthesecases,itismoreappropriatetouseanadaptedformulaforthemean,asshown inthefollowingexample.

Example 2.4

Anumberofpeoplewereaskedwhatsizeofhouseholdtheylived inthatis,howmanypeoplelived intheirhousehold. TheygavetheresponsesshowninTable2.2. What isthemeannumberofpeopleperhousehold?

Table 2.2Sizeofhousehold Numberofresponses

1 32 23 14 35 1

Solution

Letthevaluesofsizeofhouseholdbedenotedbyx,andthenumberofeachsuchhouseholdbyf (forfrequency). Theformulaforthemeanmaythenbewrittenas

xff

where means sum of. The calculation is as follows. Totalnumberofpeople inthehouseholds=(13)+(22)+(31)

+ (4 3)+(51)=27(this is xf).

Total number ofhouseholds = 3 + 2 + 1 + 3 + 1 =10(thisis f).

Meannumberofpeopleperhousehold =

xff =

2710 = 2.7.

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Exercise 2.6

Table2.3belowgives informationonthenumberofbrothersandsistersthatchildrenhaveinaparticularschoolclass. Findthemeannumberofsiblingsofthechildren,correcttoonedecimalplace.

Table 2.3

Numberofsiblings Frequency0 71 18 2 53 24 05 1

Limitations!

One limitationofmeans isthatextremevaluescancausedistortionsandmake

the

summary

meaningless.

(For

example,

the

mean

of

101,

1,

1,

1

and1 is21.) Tomakeameanamore informativesummary,theran

mst121 prerequisite revision

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