GMAT（QUANTITATIVE）數(shù)學(xué)練習(xí)試卷1（共231題）

GMAT（QUANTITATIVE）數(shù)學(xué)練習(xí)試卷1（共9套）（共231題）GMAT（QUANTITATIVE）數(shù)學(xué)練習(xí)試卷第1套一、單項選擇題(本題共30題，每題1.0分，共30分。)1、Ifthesmallerof2consecutiveoddintegersisamultipleof5,whichofthefollowingcouldNOTbethesumofthese2integers?A、-8B、12C、22D、52E、252標準答案：C知識點解析：Sincethesmallerofthe2consecutiveoddintegersisamultipleof5,letitberepresentedbySnforsomeintegern.Thentheotheroddintegercanberepresentedby5n+2.Thesumofthesetwointegersis10n+2.Thesumis-8whenn=-1and5n=(5)(-l)isodd.Thesumis12whenn=1and5n=(5)(1)isodd.Thesumis22whenn=2,but5n=(5)(2)isnotodd.ThereisnoneedtochecktheDandEbecauseithasbeendeterminedthat22cannotbethesumofthe2consecutiveoddintegers.Forcompleteness,thesumis52whenn=Sand5n=(5)(5)isodd.Thesumis252whenn=25and5n=(5)(25)isodd.ThecorrectanswerisC.2、Inthefigureabove,iftrianglesABC,ACD,andADEareisoscelesrighttrianglesandtheareaofAABCis6,thentheareaofAADEisA、18B、24C、36D、E、36標準答案：B知識點解析：Because△ABCisanisoscelesrighttrianglewithlegs,itfollowsthatAB=BCandtheareaof△ABCis(AB)(BC)/2=(1/2)(AB)2.Itisgiventhattheareaof△ABCis6,andthus(1/2)(AB)2=6,or(AB)2=12,orAB=.ApplyingthePythagoreantheoremto△ABCgives(AC)2=(AB)2+(BC)2,andhence(AC)2==12+12=24,orAC=.Because△ACDisanisoscelesrighttrianglewithlegs,andAC=,itfollowsthatAC=CD=.ApplyingthePythagoreantheoremto△ACDgives(AD)2=(AC)2+(CD)2,andhence(AD)2==24+24=48,orAD=.Because△ADEisanisoscelesrighttrianglewithlegs,andAD=,itfollowsthatAD=DE=,andhencetheareaof△ADEis(1/2)(AD)(DE)==24.ThecorrectanswerisB.3、Eachof27white1-centimetercubeswillhaveexactlyonefacepaintedred.Ifthese27cubesarejoinedtogethertoformonelargecube,asshownabove,whatisthegreatestpossiblefractionofthesurfaceareathatcouldbered?A、11/27B、13/27C、1/2D、5/9E、19/27標準答案：B知識點解析：Allbutoneofthe1-centimetercubes,namelythe1-centimetercubeinthecenter,hasatleastonefacelyingonthesurfaceofthelargecube.Thegreatestpossiblefractionofthesurfaceofthelargecubewillberedwheneachofthese27-1=26non-center1-centimetercubesisorientedsothatitsredfaceliesonthesurfaceofthelargecube.Sincethesurfaceofthelargecubehas6faces,eachconsistingof9facesfromthe1-centimetercubes,thesurfaceofthelargecubeconsistsofatotalof(6)(9)=54facesfromthe1-centimetercubes.Therefore,thegreatestpossiblefractionofthesurfaceareathatcouldberedis26/54=13/27.ThecorrectanswerisB.4、Thefigureaboveisconstructedbyseparatingacircularregioninto6equalpartsandrearrangingthepartsasshown.Ifthediameterofthecircleisd,whatistheperimeterofthefigureabove?A、πdB、2πdC、πd+2D、πd+dE、2πd+d標準答案：D知識點解析：Theperimeterconsistsof6arcsand2segments.Thetotallengthofthe6arcsisthecircumferenceofthecircle,whichisnd.Eachsegmentisaradiusofthecirclewithlengthd/2.Thereforetheperimeterofthefigureisπd+2(d/2)=πd+d.ThecorrectanswerisD.5、Onascaledrawingofatriangularpieceofland,thesidesofthetrianglehavelengths5,12,and13centimeters.If1centimeteronthedrawingrepresents3meters,whatisthearea,insquaremeters,ofthepieceofland?A、90B、180C、240D、270E、540標準答案：D知識點解析：Since52+122=132,thepieceoflandisintheshapeofarighttriangle.Thelengthsofthelegsofthepieceofland,inmeters,are(3)(5)=15and(3)(12)=36since1centimeteronthedrawingrepresents3metersonthepieceofland.Therefore,thearea,insquaremeters,ofthepieceoflandis(1/2)(15)(36)=270squaremeters.ThecorrectanswerisD.6、Whichofthefollowinggivesallpossiblevaluesofxinthefigureabove?A、1<x<4B、1<x<7C、3<x<5D、4E、5標準答案：B知識點解析：Becausethesumofthelengthsoftwosidesofatrianglemustbegreaterthanthelengthofthethirdside,3+4=7>xand3+x>4orx>l.Combining7>xandx>1gives7>x>1or17、Inthefigureabove,linesaredrawnattheverticesofthequadrilateralasshown.Whatisthesumofthedegreemeasuresofthemarkedangles?A、450B、360C、270D、240E、180標準答案：B知識點解析：Thesumofthemarkedanglesisthesumof4straightanglesminusthesumoftheinterioranglesofthequadrilateralor(4)(180)-360=360.ThecorrectanswerisB.8、Thedimensionsofareamofpaperareinchesby11inchesbyinches.Theinsidedimensionsofacartonthatwillholdexactly12reamsofpapercouldbeA、inby11inby12inB、17inby11inby15inC、17inby22inby3inD、51inby66inby15inE、102inby132inby30in標準答案：B知識點解析：Incubicinches,thetotalvolumeof12reamsofpaperis12(8.5×11×2.5)=2(8.5)×11×6(2.5)=17×11×15,whichisexactlythetotalvolumerepresentedbyB.ThecorrectanswerisB.9、Intheracetrackshownabove,regionsⅠandⅢaresemicircularwithradiusr.IfregionⅡisrectangularanditslengthistwiceitswidth,whatistheperimeterofthetrackintermsofr?A、2r(π+2)B、2r(π+4)C、2r(π+8)D、4r(π+2)E、4r(π+4)標準答案：B知識點解析：Thefigureshowsthatthe(vertical)widthoftheracetrackisequaltothediameterofthesemicircles,whichis2r,andthusthe(horizontal)lengthoftheracetrackis2(2r)=4r.Therefore,theperimeteroftheracetrack,whichconsistsoftwosemicirculararcsandtwolengthsoftherectangle,is2(πr)+2(4r)=2r(π+4).ThecorrectanswerisB.10、Thefigureabove,whichisdividedinto6sectorsofequalarea,containsanarrowrepresentingaspinner.Ifthespinnerisrotated3,840degreesinaclockwisedirectionfromthepositionshown,whichofthefollowingindicatesthesectortowhichthearrowonthespinnerwillpoint?A、

B、

C、

D、

E、

標準答案：B知識點解析：Thespinnerrotates3840/360=32/3=revolutions,whichcanalsobeseenbyobservingthat3,840degrees-3,600degrees=240degrees=2/3ofarevolution.Aftermaking10revolutions,thespinnerisbacktothestartingpointandstillneedstorotate2/3ofthewayaroundthecircle.Sincerotating2/3ofthewayaroundthecircleisequivalenttorotatingthrough4ofthe6sectorsofthecircle,thespinnerwillpointtothesectorwiththesquare.ThecorrectanswerisB.11、Whenarectangularvatthatis3feetdeepisfilledto2/3ofitscapacity,itcontains60gallonsofwater.Ifgallonsofwateroccupies1cubicfootofspace,whatisthearea,insquarefeet,ofthebaseofthevat?A、4B、8C、12D、150E、225標準答案：A知識點解析：2Whenfilledto2/3capacity,thevatcontains60gallonsofwater(i.e.,(2/3)(capacity)=60),andsowhenitisfilledtocapacityitcontains(2/3)(60)=90gallonsofwater,whichoccupies90/7.5=12cubicfeet.IfArepresentstheareaofthebaseofthevat,insquarefeet,thenthevolumeofthevatis3A=12cubicfeet,andhenceA=4.ThecorrectanswerisA.12、ThefigureaboverepresentsanantennatowerwithtwoguywiresthatextendfrompointQ,40feetabovetheground,topointsPandRasshown.Ifthetwowireshaveequallength,approximatelywhatisthetotallength,infeet,ofthetwowires?A、60B、80C、100D、120E、180標準答案：C知識點解析：BecausePQ=QR,itfollowsthat△PQRisisoscelesandtheperpendicularfromQtobisectsTherefore,eachoftherighttrianglesinthefigurehaslegsthatare30ftand40ft.BythePythagoreantheorem,PQ==50andthetotallengthofthetwoguywiresis2(50ft)=100ft.ThecorrectanswerisC.13、Thefiguresaboveshowasealedcontainerthatisarightcircularcylinderfilledwithliquidto1/2itscapacity.Ifthecontainerisplacedonitsbase,thedepthoftheliquidinthecontaineris10centimetersandifthecontainerisplacedonitsside,thedepthoftheliquidis20centimeters.Howmanycubiccentimetersofliquidareinthecontainer?A、4,000πB、2,000πC、1,000πD、400πE、200π標準答案：A知識點解析：Thefigureontheright,whichshowsthecylinderfilledto1/2itscapacity,indicatesthattheradiusofthecylinderis20cm.Thefigureontheleftindicatesthattheheightoftheliquidis10cmUsingV=πr2h,whereristheradiusandhistheheightoftheliquidinthecylinder,thevolumeofliquidinthecylindricalcontainerisπ(20)2(10)=4,000πcubiccentimeters.ThecorrectanswerisA.14、Thefigureaboveshowsadrop-leaftable.Withallfourleavesdownthetabletopisasquare,andwithallfourleavesupthetabletopisacircle.Whatistheradius,inmeters,ofthetabletopwhenallfourleavesareup?A、1/2B、C、1D、E、2標準答案：B知識點解析：Thefigureaboveshowsthecirculartabletopviewedfromabove.Thediameterofthecircleisthediagonalofthesquare,whichis1meteronaside.BythePythagoreantheorem,thediagonalismetersandtheradiusismeters.ThecorrectanswerisB.15、Ifthediameterofacircularskatingrinkis60meters,theareaoftherinkisapproximatelyhowmanysquaremeters?A、90B、180C、900D、2,800E、10,800標準答案：D知識點解析：Ifthediameterofthecircularskatingrinkis60meters,thentheradiusis30metersandtheareaisπ(30)2=900πsquaremeters.Assumingthevalueofπtobeslightlygreaterthan3,therinkisapproximately2,800squaremeters.ThecorrectanswerisD.16、Whatisthegreatestnumberofblocks8centimetersby6centimetersby9centimetersthatwillfitintoastoragespacethatis60centimetersby72centimetersby96centimeters?A、60B、840C、896D、960E、1,080標準答案：D知識點解析：Thevolumeofthestoragespaceis60×72×96andeachblockis6×8×9,sothenumberofblocksthatwillfitis=(60×72×96)/(6×8×9)=60/6×72/9×96/8=10×8×12=960.ThecorrectanswerisD.17、ClosingPricesofStockXDuringaCertainWeek(indollars)Acertainfinancialanalystdefinesthe"volatility"ofastockduringagivenweektobetheresultofthefollowingprocedure:findtheabsolutevalueofthedifferenceinthestock’sclosingpriceforeachpairofconsecutivedaysintheweekandthenfindtheaverage(arithmeticmean)ofthese4values.WhatisthevolatilityofStockXduringtheweekshowninthetable?A、2B、3C、4D、6E、7標準答案：D知識點解析：TheeasiestwaytodothisproblemmightbebyjustcountingeverythirdbulbgoingclockwisearoundthecirclestartingatBulb1,whichflashes,skipping2bulbsandgettingtoBulb4,whichflashes,skipping2bulbsandgettingtoBulb7,whichflashes,skipping2bulbsandgettingtoBulb2,whichflashes,skipping2bulbsandgettingtoBulb5,whichflashes,skipping2bulbsandgettingtoBulb8,whichflashes,skipping2bulbsandgettingtoBulb3,whichflashes,andfinallyskipping2bulbsandgettingtoBulb6,whichflashes.Now,all8bulbshaveflashedonceandthelastonetoflashwasBulb6.ThecorrectanswerisD.18、Acertainfinancialanalystdefinesthe"volatility"ofastockduringagivenweektobetheresultofthefollowingprocedure:findtheabsolutevalueofthedifferenceinthestock’sclosingpriceforeachpairofconsecutivedaysintheweekandthenfindtheaverage(arithmeticmean)ofthese4values.WhatisthevolatilityofStockXduringtheweekshowninthetable?A、0.50B、1.80C、2.00D、2.25E、2.50標準答案：D知識點解析：ThevolatilityofStockXduringtheweekistheaverageofthe4valuesassociatedwiththe4pairsofconsecutivedaysduringtheweek.ThecorrectanswerisD.19、Ify=，forwhatvalueofxwillthevalueofybegreatest?A、-5B、-3/5C、0D、3/5E、5/3標準答案：E知識點解析：Sincetheabsolutevalueofanyrealnumberisgreaterthanorequaltozero,itfollowsthat|3x-5|≥0.Also,foranyrealnumberxwehavex2≥0,andhence-x2<0.Subtracting3frombothsidesofthelastinequalitygives-x2-3≤-3.Therefore,thenumeratoroftheexpressionforyisgreaterthanorequaltozeroandthedenominatoroftheexpressionforyisnegative.Itfollowsthatthevalueofycannotbegreaterthan0.However,thevalueofyisequalto0when|3x-5|=0,or3x-5=0,orx=5/3.Therefore,thevalueofxforwhichthevalueofyisgreatest(i.e.,wheny=0)isx=5/3.ThecorrectanswerisE.20、Whatvaluesofxhaveacorrespondingvalueofythatsatisfiesbothxy>0andxy=x+y?A、x≤-1B、-1C、0<x≤1D、x>1E、Allrealnumbers標準答案：D知識點解析：First,usexy=x+ytosolveforyintermsofx.xy=x+ygivenxy-y=xsubtractyfrombothsidesy(x-1)=xfactory=x/(x-1)dividebothsidesbyx-1Notethatthedivisionbyx-1requiresx≠1，andthusx=1needstobeconsideredseparately.However,ifx=1,thenxy=x+ybecomesy=1+y,whichisnottrueforanyvalueofy.Usingy=x/(x-1),itfollowsthattheinequalityxy>0isequivalenttox2/(x-1)>0.Sincex2>0foreachvalueofx,thequotientx2/(x-1)canonlybepositivewhenx≠0andx-1ispositive,orwhenx>1.Alternatively,thecorrectanswercanbefoundbyeliminatingtheincorrectanswers,whichcanbeaccomplishedbyconsideringtheendpointsoftheintervalsgivenintheanswerchoices.Case1:Ifx=-1,thenxy=x+ybecomes-y=-1+y,ory=1/2.However,inthiscasexy=(-1)(1/2)isnegative,andthusxy>0isnottrue.Therefore,theanswercannotbeAorE.Case2:Ifx=0,thenxy=0,andthusxy>0isnottrue.Therefore,theanswercannotbeBorE.Case3:Ifx=1,thenxy=x+vbecomesy=1+y,whichisnottrueforanyvalueofy.Therefore,theanswercannotbeCorE.SincetheanswercannotbeA,B,C,orE,itfollowsthattheanswerisD.ThecorrectanswerisD.21、EmployeeX’sannualsalaryis$12,000morethanhalfofEmployeeY’sannualsalary.EmployeeZ’sannualsalaryis$15,000morethanhalfofEmployeeX’sannualsalary.IfEmployeeX’sannualsalaryis$27,500,whichofthefollowingliststhesethreepeopleinorderofincreasingannualsalary?A、Y,Z,XB、Y,X,ZC、Z,X,YD、X,Y,ZE、X,Z,Y標準答案：E知識點解析：Lettingx,y,andzrepresenttheannualsalary,indollars,ofEmployeeX,EmployeeY,andEmployeeZ,respectively,thefollowinginformationisgiven:(1)x=12,000+y/2(2)z=15,000+x/2(3)x=27,500From(1)and(3),itfollowsthat27,500=12,000+y/2ory=2(27,500-12,000)=31,000.From(2)and(3),itfollowsthatz=15,000+27500/2=28,750.Therefore,x<z<y.ThecorrectanswerisE.22、TheformulaabovegivesthecontributionC,indollars,toacertainprofit-sharingplanforaparticipantwithasalaryofsdollars.Howmanymoredollarsisthecontributionforaparticipantwithasalaryof$70,000thanforaparticipantwithasalaryof$50,000?A、$800B、$1,400C、$2,000D、$2,400E、$2,800標準答案：D知識點解析：Foraparticipantwithasalaryof$70,000,C=0.1(S70,000)+0.04($70,000-$60,000)=$7,000+$400=$7,400.Foraparticipantwithasalaryof$50,000,C=0.1($50,000)=$5,000.Thedifferenceis$7,400-$5,000=$2,400.ThecorrectanswerisD.23、Nextmonth,RonandCathywilleachbeginworkingpart-timeat3/5oftheirrespectivecurrentsalaries.IfthesumoftheirreducedsalarieswillbeequaltoCathy’scurrentsalary,thenRon’scurrentsalaryiswhatfractionofCathy’scurrentsalary?A、1/3B、2/5C、1/2D、3/5E、2/3標準答案：E知識點解析：LettingRandC,respectively,representRon’sandCathy’scurrentsalaries,itisgiventhat3R/5+3C/5=C.Itfollowsthat3R/5=2C/5andR=ThecorrectanswerisE.24、DavidandRonareorderingfoodforabusinesslunch.Davidthinksthatthereshouldbetwiceasmanysandwichesastherearepastries,butRonthinksthenumberofpastriesshouldbe12morethanone-fourthofthenumberofsandwiches.HowmanysandwichesshouldbeorderedsothatDavidandRoncanagreeonthenumberofpastriestoorder?A、12B、16C、20D、24E、48標準答案：E知識點解析：LetSbethenumberofsandwichesthatshoulcbeorderedandletPbethenumberofpastriesthatshouldbeordered.ThenDaviddesiresS=2PandRondesiresP=12+S/4.S=2PgivenS=2(12+S/4)P=12+S/4S=24+S/2distributivelawS/2=24subtractS/2frombothsidesS=48multiplybothsidesby2ThecorrectanswerisE.25、Thecostofpurchasingeachboxofcandyfromacertainmailordercatalogisvdollarsperpoundofcandy,plusashippingchargeofhdollars.Howmanydollarsdoesitcosttopurchase2boxesofcandy,onecontainingspoundsofcandyandtheothercontainingtpoundsofcandy,fromthiscatalog?A、h+stvB、2h+srvC、2hstvD、2h+s+t+vE、2h+v(s+t)標準答案：E知識點解析：Thecost,indollars,topurchasethe2boxesofcandyisthesumof2shippingchargesandthecostofs+tpoundsofcandy.cost=(2shippingcharges)+(v)(s+/)cost=2(h)+(v)(s+t)cost=2h+v(s+i)ThecorrectanswerisE.26、Ifx≠1/2，thenA、3x2+3x/2-8B、3x2+3x/2-4C、3x2-4D、3x-4E、3x+4標準答案：C知識點解析：Alternatively,sometimesitiseasierorquickertotestone-variableexpressionsforequalitybysubstitutingaconvenientvalueforthevariableandeliminatinganswerchoicesforwhichthevalueoftheexpressioninthatanswerchoicedoesnotequalthevalueofthegivenexpression.Forexample,choosex=0,sincecalculationsforx=0areminimal.Then,asshowninthetablebelow,but3x2+3x/2-8=-8and3x+4=4,neitherofwhichequals-4,soanswerchoicesAandEcanbeeliminated.Anotherconvenientvaluetochooseforxis1.ThereisnoneedtoevaluateanswerchoicesAandEat1sincetheyhavealreadybeeneliminated.Asshown,whenx=1,but3x2+3x/2-4=1/2≠-1，soanswerchoiceBcanbeeliminated.Athirdconvenientvalueforxis-1.ThereisnoneedtoevaluateanswerchoicesA,B,andEat-1sincetheyhavealreadybeeneliminated.Asshown,whenx=-1,but3x-4=-7≠-1,soanswerchoiceDcanbeeliminated.Notethat,ifx=-1hadbeenchoseninitially,A,B,D,andEwouldhavebeeneliminatedimmediatelysince3x2+3x/2-8=≠-1，3x2+3x/2-4=≠-1，3x-4=-7≠-1,and3x+4=1≠-1.ThecorrectanswerisC.27、Ifx2+bx+5=(x+c)2forallnumbersx,wherebandcarepositiveconstants,whatisthevalueofb?A、

B、

C、

D、

E、

標準答案：C知識點解析：Giventhatx2+bx+5=(x+c)2,since(x+c)2=x2+2cx+c2,itfollowsthat5=c2andb=2c.Thepossiblevaluesofcarebutsincecispositive,c=andb=2c=.ThecorrectanswerisC.28、LastyearShannonlistenedtoacertainpublicradiostation10hoursperweekandcontributed$35tothestation.Ofthefollowing,whichisclosesttoShannon’scontributionperminuteoflisteningtimelastyear?A、$0.001B、$0.010C、$0.025D、$0.058E、$0.067標準答案：A知識點解析：Sincethereare52weeksin1yearand60minutesin1hour,10hoursperweekisequivalentto(10)(52)(60)=31,200minutesperyear.Shannon’s35$35contributionisthen35/31200dollarsperminute,whichisclosestto$0,001perminute.ThecorrectanswerisA.29、Eachofthe20employeesatCompanyJistoreceiveanend-of-yearbonusthisyear.Agneswillreceivealargerbonusthananyotheremployee,butonly$500morethanCherylwillreceive.NoneoftheemployeeswillreceiveasmallerbonusthanCheryl.IftheamountofmoneytobedistributedinbonusesatCompanyJthisyeartotals$60,000,whatisthelargestbonusAgnescanreceive?A、$3,250B、$3,325C、$3,400D、$3,475E、$3,500標準答案：D知識點解析：Sincethetotalamountofthebonusesisfixed,thelargestpossiblebonusthatAgnescanreceivewilloccurwhenthetotalamountreceivedbythe19employeesotherthanAgnesisthesmallestpossible.LetAbethebonus,indollars,thatAgnesreceives.Then,indollars,Cherylwillreceive(A-500),andeachoftheremaining18employeeswillreceivebetween(A-500)andA.Therefore,thetotalamountreceivedbythe19employeesotherthanAgnesissmallestwheneachofthese19employeesreceives(A-500)dollars.19(A-500)+A=60,000totalof20bonusesis$60,00019A-9,500+A=60,000distributivelaw20A-9,500=60,000combineliketerms20A=69,500add9,500tobothsidesA=3,475dividebothsidesby20ThecorrectanswerisD.30、Beth,Naomi,andJuanraisedatotalof$55forcharity.Naomiraised$5lessthanJuan,andJuanraisedtwiceasmuchasBeth.HowmuchdidBethraise?A、$9B、$10C、$12D、$13E、$15標準答案：C知識點解析：LetB,N,andJbetheamountsraised,respectivelyandindollars,byBeth,Naomi,andJuan.B+N+J=55givenN=J-5givenJ=2BgivenJ=55-B-NsubtractB+NfrombothsidesoffirstequationJ=55-B-(J-5)N=J-52B=55-B-(2B-5)J=2BB=12solveforBThecorrectanswerisC.GMAT（QUANTITATIVE）數(shù)學(xué)練習(xí)試卷第2套一、單項選擇題(本題共30題，每題1.0分，共30分。)1、Howmanyhourswerebudgetedforacertainjob?(1)Thenumberofhoursbudgetedforthejobplustheactualnumberofhoursusedforthejobwas140.(2)Theactualnumberofhoursusedforthejobminusthenumberofhourbudgetedforthejobwas8.A、Statement(1)ALONEissufficient,butstatement(4)aloneisnotsufficienttoanswerthequestionasked.B、Statement(2)ALONEissufficient,butstatement(3)aloneisnotsufficienttoanswerthequestionasked.C、BOTHstatement(1)and(4)TOGETHERaresufficienttoanswerthequestionasked,butNEITHERstatementALONEissufficient.D、EACHstatementALONEissufficienttoanswerthequestionasked.E、Statement(1)and(4)TOGETHERareNOTsufficienttoanswerthequestionasked,andadditionaldataspecifictotheproblemareneeded.標準答案：C知識點解析：假設(shè)Thenumberofhoursbudgetedforthejob為x，theactualnumberofhoursusedforthejob為y，那么得到：單獨(1)x+y=140，一個方程兩個未知數(shù)有無數(shù)解；單獨(2)y-x=8，一個方程兩個未知數(shù)有無數(shù)解；把(1)和(2)結(jié)合起來考慮，兩個未知數(shù)，兩個方程，可以求解，所以答案為C。2、Isthepositiveintegermequaltothecubeofaninteger?(1)Foreveryprimenumberq,ifqisafactorofm,thenq3isalsoadivisorofm.(2)isaninteger.A、Statement(1)ALONEissufficient,butstatement(12)aloneisnotsufficienttoanswerthequestionasked.B、Statement(2)ALONEissufficient,butstatement(11)aloneisnotsufficienttoanswerthequestionasked.C、BOTHstatement(1)and(12)TOGETHERaresufficienttoanswerthequestionasked,butNEITHERstatementALONEissufficient.D、EACHstatementALONEissufficienttoanswerthequestionasked.E、Statement(1)and(12)TOGETHERareNOTsufficienttoanswerthequestionasked,andadditionaldataspecifictotheproblemareneeded.標準答案：B知識點解析：單獨(1)，q，q3是m的因子，m不一定是完全立方數(shù)，比如m=q4同樣能滿足q，q3是m的因子。單獨(2)，是整數(shù)，滿足完全立方數(shù)的定義，所以選擇B。3、Whatistheunitsdigitofpositiveintegerx?(1)xdividedby9hasaremainderof5.(2)xdividedby3hasaremainderof2.A、Statement(1)ALONEissufficient,butstatement(20)aloneisnotsufficienttoanswerthequestionasked.B、Statement(2)ALONEissufficient,butstatement(19)aloneisnotsufficienttoanswerthequestionasked.C、BOTHstatement(1)and(20)TOGETHERaresufficienttoanswerthequestionasked,butNEITHERstatementALONEissufficient.D、EACHstatementALONEissufficienttoanswerthequestionasked.E、Statement(1)and(20)TOGETHERareNOTsufficienttoanswerthequestionasked,andadditionaldataspecifictotheproblemareneeded.標準答案：E知識點解析：Statement(1)中x=9n+5，隨著n的不同，x的個位數(shù)也不同，所以單獨(1)無法回答問題。Statement(2)中x=3m+2，隨著m的不同，x的個位數(shù)也不同，單獨(2)也無法回答問題。把(1)和(2)結(jié)合起來，x=9n+5，除以3余2，所以結(jié)合起來的性質(zhì)仍然等同于(1)，無法回答問題，所以E為正確答案。4、Theaverage(arithmeticmean)ofMconsecutiveoddintegersis10.Whatisthesmallestnumberinthesequence?(1)TherangeoftheMconsecutiveoddintegersis14.(2)ThegreatestnumberoftheMconsecutiveoddintegersis15.A、Statement(1)ALONEissufficient,butstatement(28)aloneisnotsufficienttoanswerthequestionasked.B、Statement(2)ALONEissufficient,butstatement(27)aloneisnotsufficienttoanswerthequestionasked.C、BOTHstatement(1)and(28)TOGETHERaresufficienttoanswerthequestionasked,butNEITHERstatementALONEissufficient.D、EACHstatementALONEissufficienttoanswerthequestionasked.E、Statement(1)and(28)TOGETHERareNOTsufficienttoanswerthequestionasked,andadditionaldataspecifictotheproblemareneeded.標準答案：D知識點解析：設(shè)最小的奇數(shù)為a1，M個連續(xù)奇數(shù)是一個等差數(shù)列，公差為2，它們的平均數(shù)為10，那么它們的和為10M，在Statement(1)中，am-a1=14，那么M=7，知道了和、項數(shù)、公差，首項當然可以求得，所以單獨(1)可以回答問題。在Statement(2)中，設(shè)am為原數(shù)列中最大的奇數(shù)，則am=15，那么S=10M=(a1+15)/215=a1+(M-1)×2從方程組里面能夠求得a1，也就是數(shù)列中的最小值，所以單獨(2)也可以回答問題。因此D為正確答案。5、Ifrs=s,whatisthevalueofr+s?(1)r=4(2)5=0A、Statement(1)ALONEissufficient,butstatement(36)aloneisnotsufficienttoanswerthequestionasked.B、Statement(2)ALONEissufficient,butstatement(35)aloneisnotsufficienttoanswerthequestionasked.C、BOTHstatement(1)and(36)TOGETHERaresufficienttoanswerthequestionasked,butNEITHERstatementALONEissufficient.D、EACHstatementALONEissufficienttoanswerthequestionasked.E、Statement(1)and(36)TOGETHERareNOTsufficienttoanswerthequestionasked,andadditionaldataspecifictotheproblemareneeded.標準答案：A知識點解析：對于Statement(1)，r=4，要滿足4s=S，s必然為0，可以求得r+s。對于Statement(2)，s=0，那么r可以為任意值，所以r+s不確定，因此答案為A。6、DoesthelineLpassquadrantIV？(1)TheslopeofLisnegative.(2)y-interceptoflineLis-5.A、Statement(1)ALONEissufficient,butstatement(44)aloneisnotsufficienttoanswerthequestionasked.B、Statement(2)ALONEissufficient,butstatement(43)aloneisnotsufficienttoanswerthequestionasked.C、BOTHstatement(1)and(44)TOGETHERaresufficienttoanswerthequestionasked,butNEITHERstatementALONEissufficient.D、EACHstatementALONEissufficienttoanswerthequestionasked.E、Statement(1)and(44)TOGETHERareNOTsufficienttoanswerthequestionasked,andadditionaldataspecifictotheproblemareneeded.標準答案：D知識點解析：對于Statement(1)，只要斜率是負數(shù)，那么L一定過第四象限。對于Statement(2)，直線L在y軸上的截距為負數(shù)，那么無論斜率是多少，直線總是過第四象限。因此答案為D。7、Ifd=123.4546,andd1isthenumberobtainedbyroundingdtothenearesthundredths,thend1=A、123.45B、123.46C、123.455D、123.5E、123.454標準答案：A知識點解析：答案為A，四舍五入到百分位只能看千分位如何，千分位大于等于5就進1，千分位小于5就舍掉，千分位后面如何一概不管。8、Whichofthefollowingcannotbeexpressedasthesumoftwoprimenumbers?A、21B、14C、18D、28E、23標準答案：E知識點解析：這道題目用排除法當然可以迅速解出來，比如：21=19+2，14=7+7，18=13+5，28=23+5，只剩下23，當然選擇E。但是如果利用上面的定理，則速度更快。五個選項中B、C、D都為大于2的偶數(shù)，根據(jù)定理必然可以表達為兩個質(zhì)數(shù)的和；剩下選項A、E均為奇數(shù)，若兩數(shù)相加為奇數(shù)，則這兩數(shù)必然為一奇一偶。而這個數(shù)又要能夠表達為兩個質(zhì)數(shù)的和，那么必須這一奇一偶都為質(zhì)數(shù)，我們知道在所有質(zhì)數(shù)中2是唯一的一個偶數(shù)，因此若A或E能夠表達為兩質(zhì)數(shù)的和，其中一個一定為2，選項A：21-2=19，E：23-2=21，19為質(zhì)數(shù)，21=3×7不為質(zhì)數(shù)，因此E是正確答案。9、如果某一自然數(shù)除了1之外只有2個因子，則這個自然數(shù)必為：A、奇數(shù)B、偶數(shù)C、4的倍數(shù)D、某一質(zhì)數(shù)的平方E、質(zhì)數(shù)標準答案：D知識點解析：既然這個數(shù)除了1之外只有2個因子，則這個數(shù)一共有3個因子(把因子1加上)，而3是一個大于2的質(zhì)數(shù)，根據(jù)性質(zhì)4得到它必然是某一質(zhì)數(shù)的3-1=2次方，所以D為正確答案。10、如果n是正整數(shù)，并且n2的個位數(shù)是4，(n+1)22的個位數(shù)是1，那么(n+5)5的個位數(shù)是多少?A、2B、7C、6D、1E、9標準答案：B知識點解析：n2的個位數(shù)是4，那么n的個位數(shù)可能為2或者8；(n+1)22的個位數(shù)是1，22除以4余數(shù)為2，那么(n+1)2的個位數(shù)也為1；如果n的個位數(shù)為2，那么n+1的個位數(shù)是3，(n+1)2的個位數(shù)就是9；如果n的個位數(shù)為8，那么n+1的個位數(shù)是9，(n+1)2的個位數(shù)恰好是1，所以n的個位數(shù)為8，那么n+5的個位數(shù)是3，(n+5)3的個位數(shù)必然為7。B為正確答案。11、WhichofthefollowingnumbersisNOTthesumofthreeconsecutiveoddintegers?A、15B、75C、123D、297E、313標準答案：E知識點解析：三個連續(xù)奇數(shù)的平均數(shù)必然是中間那個數(shù)，因此三個連續(xù)奇數(shù)的和必然能夠被3整除，利用一個數(shù)能夠被3整除的特性，只有E：313的各位數(shù)之和不能夠被3整除，所以選擇E。12、Inacertainbox,thereare3t