GAC 010 Test 3 Sample Questions 2.docx

GAC 010 Test 3 Sample Questions 2.docx

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GAC010Test3SampleQuestions2

GAC010Test3-ExamSampleQuestions

Question1[11marks]

a.）Thefollowingdatashowtheages,tothenearestyear,of60

employeeswhentheystartedworkingatacertainfactory.

Ages

NoOfEmployees

10-19.9

3

20-29.9

16

30-39.9

18

40-49.9

10

50-59.9

8

60-69.9

5

Total

60

i.）Graphthefrequencyhistogram[3marks]

ii.）Howmanywereunder40whentheystartedworkingthere?

[1mark]

iii.）Howmanywereaged30andoverwhentheyfirststarted

workingthere?

[1mark]

b.）Usethetablebelowtodetermine:

i.）Taxableincomeon$57678when

Allowabledeductionsfortheyearamountto$1745

ii.）Howmuchannualtaxapersonwouldpaywhenhis

Grossincomeis$57678?

Taxableincome

Taxonthisincome

$1-$6000

Nil

$6001-$20000

17centsforeach$1over$6000

$20001-$50000

$2380+30centsforeach$1over$20000

$50001-$60000

$11380+42centsforeach$1over$50000

$60001andover

$15580+47centsforeach$1over$60000

Source:

2001ATOTaxGuide

Solutions

1.a.）i.）

a.）ii.）3+16+18=37

a.）iii.）18+10+8+5=41

b.）i.）TaxableIncome=gross-deductions

=$57678-$1745

=$55933

b.）ii.）TaxPayableon$57678is

$57678-$1745=$55933

on$50000=$11380

on$55933-$50000=$5933*.42

=$2491.86

TaxPayableon$55933=$11380+$2491.86

=$13871.86

Question2[16marks]

Thestemandleafdiagramdisplaysthescoresobtainedby40studentsintheirfinalMathematicsexam.

Stem

Leaf

4

57

5

3578

6

245689

7

1112344556666889

8

3557889

9

33588

a.）Calculatethemean,medianandmode.[5marks]

b.）Calculatetherange,andthelowerandupper

quartiles.[5marks]

c.）Constructaboxandwhiskerdiagram.[3marks]

d.）Hence,commentonthesymmetryorotherwiseof

thedata.[1mark]

e.）Findthevarianceandstandarddeviation.[2marks]

Solutions

a.）Mean=Sx

n

Median=X（n+1）=X41/2=X20.5

2

Mode=76

b.）Range=HighestValue-Lowestvalue

=98-45

=53

Q1=X（n+1）

4

Q1=X（41）

4

=X10.25

Q3=X3（n+1）

4

Q3=X3（41）

4

=X30.75

c.）BoxandWhisker

diagram

______________________________________________________

30405060708090

45Q1MdQ398

d.）Thedataisskewedtotheleft（negativelyskewed）

duetothegreaternumberofsmallervalues.

e.）

VarianceisS^2=S（x-x̅）^2*f

n-1

Standarddeviationiss=√s^2

Question3[15marks]

a.）HouseholdsinacertaintownweresurveyedtodeterminewhethertheywouldsubscribetoaPayTVChannel.Thehouseholdswereclassifiedaccordingto“high”,“medium”,and“low”incomelevels.Theresultsofthesurveyweresummarizedinthefollowingtable.

IncomeLevel

WillSubscribe

Willnotsubscribe

Total

High

1650

1650

3300

Medium

1600

6340

7940

Low

520

1900

2420

Total

3770

9890

13660

Ahouseholdischosenatrandom.Findtheprobabilitythatthehousehold:

i.）willsubscribe.[1mark]

ii.）hasalowincomelevel.[1mark]

iii.）willsubscribeandahasalow-incomelevel.[2marks]

iv.）willsubscribeorhasalowincomelevel.[2marks]

v.）willsubscribegiventhatthehouseholdhasalowincome

level.[2marks]

b.）i.）12peoplearetobedividedintotwogroupswith8inone

and4intheother.

Inhowmanywayscanthisbedone?

[2marks]

ii.）Atelevisionfirmhasinstock9televisionsetswith32

inchtubesand6setswith28inchtubes.Foursetsare

removedatrandom.Whatistheprobabilitythatthey

areallofthesamesize?

[3marks]

iii.）AparticularfirmhasintypeofTVtubeisalmostcertain

tofailwithin3500hours.Ifthelifeofatubefollowsa

normaldistributionwithastandarddeviationof150

hours,findthemeanlifeofthesetubes.[2marks]

Solutions

3.a）

i.）（willsubscribe）=3770=0.27

13660

ii.）P（lowlevelincome）=2420=0.17

13660

iii.）P（willsubscribeandhaslowlevelincome）=520=0.03

13660

iv.）P（willsubscribeorhaslowlevelincome）=3770+1900

13660

=0.41

v.）P（willsubscribegivenhouseholdhaslowlevelincome）

=520=0.21

2420

b.）i.）Numberofarrangements

nCr=n!

r!

（n-r）!

12C8*4C4=12!

*4!

8!

（12-8）!

4!

（4-4）!

=12!

*4!

8!

（4!

）4!

（0）!

=12*11*10*9*8!

*1

8!

（4）!

=12*11*10*9

4*3*2*1

=11880=495

24

ii.）P（samesize）=P（LLLL）+P（SSSS）

=（9*8*7*6）+（6*5*4*3）

1514131215141312

=0.10

iii.）Almostcertaintofailis3standarddeviationsabovethe

mean.

therefore,meanis

μ+3σ=3500

μ=3500-3σ

=3500-3*150

=3500-450

=3050

or,

3500-150=3350

3350-150=3200

3200-150=3050

3050320033503500

Question4[13marks]

Mostmovietrailersarenormallydistributedwithameanof2minutesandastandarddeviationof30seconds.

a.）Findtheprobabilitythatamovietrailerselectedatrandom

willbe

i.）lessthan160seconds

ii.）lessthan80seconds

iii.）over180seconds

iv.）over100secondsbutlessthan140seconds.

b.）Ifyouwatch100ofthesemovietrailers,howmanywillbe

under140seconds?

Solutions

4.a.）i.）

LetX=timeinseconds

μ=120,σ=30X:

N（120,30^2）

z=X-μ=X-120

σ30

P（X<160）=P（Z<160-120）=P（Z<140）

3030

=P（Z<4.66）

=.5+P（0≤Z≤1.66）

=.5+.4515

=.9515

120160

________________________________________

04.66Z

a.）ii.）

P（X<80）=P（Z<80-120）=P（Z<-40）

3030

=P（Z<-1.33）

=.5-P（-1.33

=.5-P（0

=.5-.4082

=.0918

80120

_____________________________________________

-1.330Z

a.）iii）

P（X>180）=P（Z>180-120）=P（Z>60）

3030

=P（Z>2）

=.5-P（0

=.5-.4772

=.0228

120180

__________________________________________

02Z

a.）iv）

P（100

3030

=P（-20

3030

=P（-.66

=P（-.66

=P（0

=.2454+.3413

=.5867

100120150

_____________________________________________

-.6601Z

b.）

P（X<140）=P（Z<140-130）

30

=P（Z<.33）

=.5+P（0

=.5+.1293

=.6293

So,for100trailers:

100*.6293=62.93=>about63trailerswillbeunder140seconds.

120140

______________________________________________

0.33

Question5[14marks]

a.）Frankrecentlyboughtasmallboat.Heput$8000asdownpaymentandborrowedtherestofthemoney,withpaymentsof$400attheendofevery4monthsfor4years.Iftheinterestis8%p.a.compoundedthreetimesayear,whatisthecurrentvalueoftheboat?

[4marks]

Unit7F,Ex7.7,p.235

b.）MybrotherTony,wholivesinCanada,justbought2houses.thefirstcosted$60400andthesecond$80600.Lateron,hesoldthefirstoneataprofitof15%andthesecondatalossof12%.Howmuchprofitorlossdidtheagentgainorloseintotal?

[4marks]

Unit6D,Ex6.3,p.207

c.）Anewlyopenedoffice’spurchasedequipment-computers,printers,faxmachines,etc.costs$25000.Johnputsdowna$3000depositandagreestopaythebalancein3yearsat7%interestonthebalance.

i.）IfMisthemonthlyrepayment,showthattheamountowedattheendofthefourthmonthis

22000*1.0058^4-M（1.0058^3+1.0058^2+1.0058+1）

[2marks]

Unit7G

ii.）Findthemonthlyrepaymenttorepaytheloanin36months.[4marks]

Unit7G,Ex7.11,p.238（ApplicationofGeometricSeries）

Solutions

5a.）

OrdinaryAnnuities

PresentValueofanOrdinaryAnnuity

Thepresentvalue（A）ofanordinaryannuityisgivenby:

A=R[1-（1+i）^-n]

i

Where:

Rperiodicpaymentoftheannuity

i=interestrateperperiod（asadecimal）

n=thenumberofperiodicpaymentstobemade

Theterm1-（1+i）^-ncanbeabbreviatedtoɑn⅂i

i

thereforethepresentvalueofanordinaryannuityisA=Rɑn⅂i,where

ɑn⅂i=1-（1+i）^-n

i

================

Deposit=$8000

R=$400

n=4*3=12

i=8%/3=.026

Presentvalue（A）=$8000+A

=$8000+400[1-（1.026）^-12]

.026

=$8000+400[10.195931]

=$8000+4078.37

=$12078.37

b.）

costofhouse1=60400

costofhouse2=80600

costofbothhouses=141000

Mark-uponhouse1=.15*60400=9060

Salepriceofhouse1=60400+9060=69460

Lossonhouse2=.12*80600=9672

Salepriceonhouse2=80600-9672=70928

Totalsalespriceofhouses1and2=69460+70928=140388

Totalloss=141000-140388=612

$612is（612/141000）*100=4.340

=about4.3%ofTotalCostPrice

c.）

Thebalancelefttopayis25000-3000=22000

Thecalculationmustfirstbebasedonmonths

Thereforemonthlyinterestrate=7%=.58%=.0058

12

so1+r=1.0058

100

Numberofmonths=3*12=36

i.）

LetAibetheamountstillowingafterimonthsandMthemonthlyrepayments.

A1=22000*1.0058-M

A2=A1*1.0058-M

=（22000*1.0058-M）*1.0058-M

=22000*1.0058^2-1.0058M-M

=22000*1.0058^2-M（1.0058+1）

A3=A2*1.0058-M

=（220001.0058^2-M（1.0058^1+1）*1.0058-M

=22000*1.0058^3-（1.0058^1M-M）*1.0058-M

=22000*1.0058^3-1.0058^2-1.0058M-M

=22000*1.0058^3-M（1.0058^2+1.0058+1）

A4=A3*1.0058-M

=....

=22000*1.0058^4-M（1.0058^3+1.0058^2+1.0058+1）

ii.）

A36=22000*1.0058^36-M（1.0058^35+1.0058^34+...+1）

（usingthegeometricseriesformulaforSn）

TheSumofnTermsofaGeometricProgressio