GAC 010 Test 3 Sample Questions 2.docx
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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
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