GMAT（QUANTITATIVE）数学练习试卷4（共92题）

GMAT（QUANTITATIVE）数学练习试卷4（共4套）（共92题）GMAT（QUANTITATIVE）数学练习试卷第1套一、单项选择题(本题共25题，每题1.0分，共25分。)1、Acertaintrucktravelingat55milesperhourgets4．5milespergallonofdieselfuelconsumed．Travelingat60milesperhour，thetruckgetsonly3．5milespergallon．Ona500-miletrip，ifthetruckusedatotalof120gallonsofdieselfuelandtraveledpartofthetripat55milesperhourandtherestat60milesperhour，howmanymilesdidittravelat55milesperhour?A、140B、200C、250D、300E、360标准答案：E知识点解析：Letmbethenumberofmilesthetrucktraveledat55milesperhour.Itfollowsthat500-misthenumberofmilesthetrucktraveledat60milesperhour.Then,m/4.5isthenumberofgallonsthetruckusedwhiletravelingat55milesperhourandisthenumberofgallonsthetruckusedwhiletravelingat60milesperhour,so=120.Solvingthisequationgivesm=360.ThecorrectanswerisE.2、Amerchantpaid$300forashipmentofxidenticalcalculators．Themerchantusedtwoofthecalculatorsasdemonstratorsandsoldeachoftheothersfor＄5morethantheaverage(arithmeticmean)costofthexcalculators．Ifthetotalrevenuefromthesaleofthecalculatorswas＄120morethanthecostoftheshipment，howmanycalculatorswereintheshipment?A、24B、25C、26D、28E、30标准答案：E知识点解析：Themerchantpaid$300forashipmentofxcalculators,sotheaveragecost,indollars,percalculatorwas300/x.Themerchantsold(x-2)ofthematthepriceof5+300/xdollarseach,foratotalrevenueof(x-2)(5+300/x)dollars,whichwas120+300=420.Manipulatingtheequation(x-2)(5+300/x)=420givesx2-26x-120=0or(x-30)(x+4)=0,whichcanbesolvedbyfactoring.Itfollowsthattherewere30calculatorsintheshipment.ThecorrectanswerisE.3、Acartraveled462milespertankfulofgasolineonthehighwayand336milespertankfulofgasolineinthecity．Ifthecartraveled6fewermilespergalloninthecitythanonthehighway，howmanymilespergallondidthecartravelinthecity?A、14B、16C、21D、22E、27标准答案：B知识点解析：Letgbethenumberofgallonsofgasolinein1tankful.Thenthenumberofmilespergallonwhiletravelingonthehighwayis462/gandthisnumberis6morethan336/g，whichisthenumberofmilespergallonwhiletravelinginthecity.Solving462/g=6+336/ggivesg=21.Therefore,thenumberofmilespergallonwhilethecarwastravelinginthecitywas336/21=16.ThecorrectanswerisB.4、MachinesXandYrunatdifferentconstantrates，andmachineXcancompleteacertainjobin9hours．MachineXworkedonthejobaloneforthefirst3hoursandthetwomachines，workingtogether，thencompletedthejobin4morehours．HowmanyhourswouldithavetakenmachineY,workingalone，tocompletetheentirejob?A、18B、C、D、E、标准答案：A知识点解析：MachineXcancompletethejobin9hours,soitcompleted1/3ofthejobbeforemachineYstartedworking.SincemachineXworked4morehours,itcompleted7/9ofthejob,whichleft2/9ofthejobformachineYtododuringthe4hoursitworkedtogetherwithmachineXtocompletethejob.Atthisrate,machineYwouldtake(2/9)(4)=18hourstodotheentirejobworkingalone.Alternatively,machineXcancompletethejobin9hours,soitcompleted—ofthejobbeforemachineYstartedworking.Thisleaves2/3ofthejob,whichtookmachinesXandY,workingtogether,4hourstocomplete.So,ifworkingtogetherthemachinestook4hourstocomplete2/3ofthejob,itwouldtakethem6hourstodothewholejobworkingtogether.Furthermore,ratesareadditive,soweshouldconvertworkingtimestorates.MachineX’srateis—jobsperhour.Bothmachinestogetherhavearateof1/6jobsperhour.WemustdeterminemachineY’sindividualrate—forthewholejob.First,addmachineX’sratetomachineY’sratetogetthecombinedrate.Then,4/9+1/Y=1/61/Y=1/6-1/91/Y=2/36Y=18ThecorrectanswerisA.5、Ifs/t=2，thentheValueofwhichofthefollowingcanbedetermined?A、ⅠonlyB、ⅢonlyC、ⅠandⅡonlyD、ⅡandⅢonlyE、Ⅰ，Ⅱ，andⅢ标准答案：C知识点解析：Substitute2tforsintheexpressionsgiveninI,II,andIII.√I.Valuecanbedetermined:2t/s=2t/2t=1√II.Valuecanbedetermined:×Ⅲ.Valuecannotbedetermined:equals0ift=1,andequals1/3ift=2ThecorrectanswerisC.6、Ifk≠0，andk-=x/k，thenx=A、-3-k2B、k2-3C、3k2-3D、k-3-2k2E、k-3+2k2标准答案：C知识点解析：Multiplyingbothsidesoftheequationbykgivesk2-(3-2k2)=xorx=3k2-3.ThecorrectanswerisC.7、ThesumoftheagesofDorisandFredisyyears．IfDorisis12yearsolderthanFred，howmanyyearsoldwillFredbeyyearsfromnow，intermsofy?A、y-6B、2y-6C、(y/2)-6D、(3y-2)-6E、(5y/2)-6标准答案：D知识点解析：LetDandFrepresentDoris’sandFred’scurrentages,respectively.ItisgiventhatD+F=yandD=F+12.Itfollowsthat(F+12)+F=yand.Therefore,Fred’sageyyea’sfromnowwillbeThecorrectanswerisD.8、Theshadedregioninthefigureaboverepresentsarectangularframewithlength18inchesandwidth15inches．Theframeenclosesarectangularpicturethathasthesameareaastheframeitself．Ifthelengthandwidthofthepicturehavethesameratioasthelengthandwidthoftheframe，whatisthelengthofthepicture，ininches?A、

B、

C、

D、

E、

标准答案：A知识点解析：Letkbetheproportionalityconstantforthefractionaldecreasefromthedimensionsoftheframetothedimensionsofthepicture.Thatis,let18kbethelengthofthepictureandlet15kbethewidthofthepicture.Wearegiventhat(18)(15)=2(18k)(15k).Hence,k2=1/2andk=Therefore,thelengthofthepictureis18k=ThecorrectanswerisA.9、Thesurfacedistancebetween2pointsonthesurfaceofacubeisthelengthoftheshortestpathonthesurfaceofthecubethatjoinsthe2points．Ifacubehasedgesoflength4centimeters，whatisthesurfacedistance，incentimeters。betweenthelowerleftvertexonitsfrontfaceandtheupperrightvertexonitsbackface?A、

B、

C、

D、

E、

标准答案：B知识点解析：Theleftfigurebelowshowsacubewithedgelength4,wherePisthelowerleftvertexonitsfrontfaceandQistheupperrightvertexonitsbackface.ItisclearthattheshortestpathonthesurfacebetweenPandQconsistsofapathonthefrontfacejoinedtoapathonthetopface,orapathonthefrontfacejoinedtoapathontherightface.Infact,eachofthesetwoapproachescanbeusedinessentiallythesamewaytogiveapathfromPtoQ,whoselengthisthesurfacedistancebetweenPandQ.Tosimplifytheexposition,wewillconsiderthecasewherethesurfacedistanceisthelengthofacertainpathPFonthefrontfaceplusthelengthofacertainpathPTonthetopface.Themiddlefigurebelowshowsthetopfaceofthecubeliftedabout45degrees,andtherightfigurebelowshowsonlythefrontandtopfacesofthecubeafterthetopfacehasbeenlifted90degrees.Sincerotationsofthetopfaceaboutthis"hinge"donotchangethelengthofanypathonthefrontfaceoronthetopface,thisrotationby90degreeswillnotchangethelengthofPporthelengthofPTandhencethisrotationwillnotchangethesumofthelengthsofPFandPT.Intherightfigurebelow,theshortestpathfromPtoQisthedashedsegmentshowninthefigure.Moreover,thelengthofthisdashedsegmentisthesurfacedistancebetweenPandQbecauseifPFandPTdidnotcorrespondtoportionsofthisdashedsegmentonthefrontandtopfaces,respectively,thenthesumofthelengthsofPFandPTwouldbegreaterthanthelengthofthedashedsegment,sincetheshortestdistancebetweenPandQintherightfigurebelowisthelengthofthelinesegmentwithendpointsPandQ.Fromthediscussionabove,itfollowsthatthesurfacedistancebetweenPandQisthelengthofthedashedsegmentintherightfigure,whichiseasilyfoundbyusingthePythagoreantheorem:ThecorrectanswerisB.10、Thefigureaboveshows2circles．ThelargercirclehascenterA，radiusRcm．andisinscribedinasquare．ThesmallercirclehascenterC，radiusrcm，andistangenttothelargercircleatpointBandtothesquareatpointsDandF，IfpointsA，B，C，andEarecollinear，whichofthefollowingisequaltoR/r?A、

B、

C、

D、

E、

标准答案：D知识点解析：BecauseisadiagonalofsquareCDEF,whichhassidelengthr,itfollowsfromthePythagoreanantheoremthatr2+r2=(CE)2,andhenceCE=.Therefore,BE=andAE=R+.Since2(AE)isthediagonallengthofthelargesquare,whichhassidelength2R，itfollowsfromtheabovetipthat2(AE)=(2R),orAE=R.Alternatively,anappropriateapplicationofthePythagoreantheoremgivesR2+R2=(AE)2,orAE=R.NowsubstituteforAEandsolveforR/r.FromthelastequationwegetThecorrectanswerisD.11、ThefigureaboveshowsthedimensionsofarectangularboardthatistobecutintofouridenticalpiecesbymakingcutsatpointsA，B，andC，asindicated．IfX=45．whatisthelengthAB?(1foot=12inches)A、5ft6inB、C、5ft3inD、5ftE、4ft9in标准答案：C知识点解析：ThefigureaboveshowstheleftsideoftherectangularboardwithpointsE,F,G,andHaddedandsegmentadded.WearetodeterminethevalueofAB,whichequalsthevalueofFH.Since△AFGisanisoscelestriangle,itfollowsthatFG=FA=0.5ft.Moreover,EF=GHbecausethefourpiecesoftherectangularboardhavethesamedimensions.Therefore,sinceEHishalfthelengthofthe20ftboard,EH=10ftandwehaveEF+FG+GH=10ft,or2(GH)+0.5ft=10ft,orGH=4.75ft.Hence,AB=FH=FG+GH=0.5ft+4.75ft=5.25ft,whichequals5ft3in.ThecorrectanswerisC.12、Inthefigureabove，theareaoftheshadedregionisA、

B、

C、

D、

E、

标准答案：D知识点解析：First,thediagonalofthesquarewithsidesoflength4is4.Fromthisitfollowsthattheareaoftheshadedregionconsistsoftheareaofanequilateraltrianglewithsides4minus—theareaofasquarewithsidesoflength4.Thus,theareaoftheshadedregionis.ThecorrectanswerisD.13、Ifeachsideof△ACDabovehaslength3andifABhaslength1，whatistheareaofregionBCDE?A、

B、

C、

D、

E、

标准答案：B知识点解析：TheareaofregionBCDEistheareaof△ACDminustheareaof△ABE.SinceAACDisequilateral,itsareais.△ABEisa30-60-90trianglewithsidelengths1,,and2andarea.Thus,theareaofregionBCDEisThecorrectanswerisB.14、Inthefigureabove，threesquaresandatrianglehaveareasofA，B，C，andXasshown．IfA=144，B=81，andC=225，thenX=A、150B、144C、80D、54E、36标准答案：D知识点解析：Thesidelengthsofthesquareswithareas144,81,and225are12,9,and15,respectively,sothetrianglewithareaXhassides12,9,and15.Because122+92=152,thetrianglewithareaXisarighttrianglewithlegsoflength12and9.Thus,X==54.ThecorrectanswerisD.15、Inthefigureabove，PQisadiameterofcircleO，PR=SQ，and△RSTisequilateral．IfthelengthofPQis2．whatisthelengthofRT?A、

B、

C、

D、

E、

标准答案：D知识点解析：SincePR=SQitfollowsthatRO=OS,soOisthemidpointofRS.Since△RSTisequilateraland0isthemidpointof,△ROTisa30°-60°-90°triangle,andsinceisaradiusofthecirclewithdiameter2,OT=1.Usingtheratiosofthesidesofa30°-60°-90°triangle,itfollowsthatRT=.ThecorrectanswerisD.16、ThefigureaboveshowssomeofthedimensionsofatriangularplazawithanL-shapedwalkalongtwoofitsedges．Ifthewidthofthewalkis4feetandthetotalareaoftheplazaandwalktogetheris10，800squarefeet，whatisthevalueofx?A、200B、204C、212D、216E、225标准答案：A知识点解析：Theareaofthetriangularplazaisgivenby(100-4)=48x,andtheareaofthewalkwayisgivenby(x+4)(4)+(96)(4).Sincethetotalareaoftheplazaandwalkwayis10,800,itfollowsthat52x+400=10,800andx=200.ThecorrectanswerisA.17、Acircularrim28inchesindiameterrotatesthesamenumberofinchespersecondasacircularrim35inchesindiameter．IfthesmallerrimmakesXrevolutionspersecond，howmanyrevolutionsperminutedoesthelargerrimmakeintermsofX?A、48π/XB、75XC、48XD、24XE、X/75标准答案：C知识点解析：Sincethesmallerrimhasdiameter28inchesandrotatesxrevolutionspersecond,itrotates2Snxinchespersecond.Ifyrepresentsthenumberofrevolutionsthelargerrimrotatespersecond,thenthelargerrimrotates35πyinchespersecond.Sincetherimsrotatethesamenumberofinchespersecond,itfollowsthat28πx=35πy.Theny=28x/35=4x/5inchespersecondor4x/5(60)=48xinchesperminute.ThecorrectanswerisC.18、Theannualstockholders’reportforCorporationXstatedthatprofitswereup10percentoverthepreviousyear，althoughprofitsasapercentofsalesweredown10percent．Totalsalesforthatyearwereapproximatelywhatpercentofsalesforthepreviousyear?A、78％B、90％C、110％D、122％E、190％标准答案：D知识点解析：LetP1andS1betheprofitandsalesforthepreviousyear,andletP2andS2betheprofitandsalesforthefollowingyear.ItisgiventhatP2=1.1P1andP2/S2=0.9(P1/S1).SubstitutingthefirstequationintothesecondequationgivesTherefore,S2isapproximately122%ofS1.ThecorrectanswerisD.19、Acertainbrandofhousepaintmustbepurchasedeitherinquartsat$12eachoringallonsat$18each．Apainterneedsa3-gallonmixtureofthepaintconsistingof3partsblueand2partswhite．Whatistheleastamountofmoneyneededtopurchasesufficientquantitiesofthetwocolorstomakethemixture?(4quarts=1gallon)A、$54B、$60C、$66D、$90E、$144标准答案：C知识点解析：Tomake3gallonsofthemixturerequires12quarts,andtheleastamountofmoneyisachievedbypurchasingthegreatestnumberofgallonsandleastnumberofquarts.LettingBbethenumberofquartsofbluepaintneededandWbethenumberofquartsofwhitepaintneeded,itfollowsthatB+W=12,whereB=3W/2sincethemixturehasabluetowhiteratioof3to2.ThisgivesB=andW=.Sincethepaintcanbepurchasedinwholequartsonly,thepaintermustpurchase8quartsor2gallonsofblue,and5quartsor1gallonplus1quartofwhiteforatotalof(2+l)($18)+$12=$66.ThecorrectanswerisC.20、ThetableaboveshowsthepercentofchangefromthepreviousmonthinCompanyX’ssalesforFebruarythroughJuneoflastyear．ApositivepercentindicatesthatCompanyX’ssalesforthatmonthincreasedfromthesalesforthepreviousmonth．andanegativepercentindicatesthatCompanyX’ssalesforthatmonthdecreasedfromthesalesforthepreviousmonth．ForwhichmonthwerethesalesclosesttothesalesinJanuary?A、FebruaryB、MarchC、AprilD、MayE、June标准答案：D知识点解析：Explicitcalculationincorporatingafewnumericalshortcutsgivesthefollowing,whereJistheJanuarysalesamount.Februarysales:1.1JMarchsales:0.85(1.1J)=0.85J+0.85(0.1J)=0.85J+0.085J=0.935JAprilsales:1.2(0.935J)=0.935J+0.2(0.935J)=0.935J+0.187J=1.122JMaysales:0.9(1.122J)=1.0098JJunesales:1.05(1.0098J)(Mayisclearlycloser)Alternatively,fromitfollowsthatapercentchangeofx%followedbyapercentchangeofy%isequaltoapercentchangeof(x+y)%plusx%ofypercentagepoints(equivalently,plusy%ofxpercentagepoints).PercentchangefromJanuarythroughFebruary:Thepercentchangeisgivenas+10%.PercentchangefromJanuarythroughMarch:Usingtheruleaboveforx=+10andy=-15gives(+10-15)%+(+0.10)(-15%),or-5%-1.5%=-6.5%.PercentchangefromJanuarythroughApril:Thisisequivalenttoa-6.5%change(percentchangefromJanuarythroughMarch)followedbya+20%change,andhenceusingtheruleaboveforx=-6.5andy=+20gives(-6.5+20)%+(+0.20)(-6.5%),or+13.5%-1.3%=+12.2%.PercentchangefromJanuarythroughMay:Thisisequivalenttoa+12.2%change(percentchangefromJanuarythroughApril)followedbya-10%change,andhenceusingtheruleaboveforx=+12.2andy=-10gives(+12.2-10)%+(-0.10)(+12.2%),or+2.2%-1.22%=+0.98%.PercentchangefromJanuarythroughJune:Thisisclearlygreaterthan+0.98%+5%,andhencegreaterinmagnitudethantheresultforMay.Fromtheresultsabove,theleastchangeinthemagnitudeofthepercentchangefromJanuaryoccurredforMay.ThecorrectanswerisD.21、Astorebought5dozenlampsat＄30perdozenandsoldthemallat＄15perlamp．Theprofitoneachlampwaswhatpercentofitssellingprice?A、20％B、50％C、D、100％E、500％标准答案：C知识点解析：Forthisproblemitisespeciallyimportanttokeepyourfocusonwhatisaskedandtoignoreextraneousdetails.Thecostperlampis$30/dozen=$30/12andthesellingpriceperlampis$15,sotheprofitperlampis$(15-30/12).Therefore,foreachlamptheprofitasapercentofthesellingpriceisThecorrectanswerisC.22、StoreNgivesa50percentdiscountonthelistpriceofallitsitemsandStoreWgivesa60percentdiscountonthelistpriceofallitsitems．Ifthelistpriceofthesameitemis20percenthigherinStoreW，whatpercent(moreorless)ofthesellingpriceinStoreNisthesellingpriceoftheiteminStoreW?A、10％lessB、4％lessC、2％lessD、10％moreE、12％more标准答案：B知识点解析：Let$PbethelistpriceoftheiteminStoreN.Thetableshowsthelistandsellingpricesoftheiteminthetwostores.Forexample,thesellingpriceoftheiteminStoreWis60percentlessthantheitem’slistpriceof$1.2P,or(0.4)($1.2P)=$0.48P.Theamount,indollars,bywhichthesellingpriceoftheiteminStoreWislessthanthesellingpriceoftheiteminStoreNis0.5P-0.48P=0.02P,whichis0.02P/0.5P=4%lessasapercentofthesellingpriceinStoreN.Alternatively,wecanassignaspecificvaluetoPandcarryoutthecomputationsusingthisvalue.Inpercentproblems,thecomputationsareusuallysimplerwhen100isused,soletP=100.(If,say,ofPhadbeeninvolved,thenP=300mightbeabetterchoice.)Therefore,thelistpriceoftheiteminStoreNis$100,thediscountpriceoftheiteminStoreNis(50%)($100)=$50,thelistpriceoftheiteminStoreWis(120%)($100)=$120,andthediscountpriceoftheiteminStoreWis(40°/o)($120)=$48.Thus,thequestionbecomesthefollowing.Whatpercent(moreorless)of50is48?Asimplecomputationshowsthat48islessthan50.ThecorrectanswerisB.23、Amerchantpurchasedajacketfor＄60andthendeterminedasellingpricethatequaledthepurchasepriceofthejacketplusamarkupthatwas25percentofthesellingprice．Duringasale，themerchantdiscountedthesellingpriceby20percentandsoldthejacket．Whatwasthemerchant’sgrossprofitonthissale?A、＄0B、＄3C、＄4D、＄12E、＄15标准答案：C知识点解析：Thepurchaseprice,indollars,ofthejacketwas60.IfSrepresentsthesellingprice,indollars,thenS=60+0.25S,fromwhichS=80.Duringthesale,thediscountedsellingprice,indollars,was0.8(80)=64,sothemerchant’sgrossprofit,indollars,was64-60=4.ThecorrectanswerisC.24、Whenacertainstretchofhighwaywasrebuiltandstraightened，thedistancealongthestretchwasdecreasedby20percentandthespeedlimitwasincreasedby25percent．Bywhatpercentwasthedrivingtimealongthisstretchreducedforapersonwhoalwaysdrivesatthespeedlimit?A、

B、

C、

D、

E、

标准答案：B知识点解析：LetDandrbethedistanceandspeedlimit,respectively,alongthestretchofhighwaybeforeitwasrebuilt.Then,thedistanceandspeedlimitalongthestretchofhighwayafteritwasrebuiltaregivenby0.8Dand1.25r.ItfollowsthatthepercentreductionintimeisAlternatively,ithelpstouseactualnumberswhencalculatingpercentchange,with100beingthemostmathematicallyconvenientnumbertouse.Forthisproblem,setthespeedlimitat100milesperhourandthedistanceat100miles.A20percentdecreaseindistancemakesthenewdistance80miles.A25percentincreaseinspeedlimitmakesthenewspeedlimit125milesperhour,which,ofcourse,isunrealistic,butveryconvenienttoworkwith.Then,thenewtimedistance/rateis80/125hours,ThepercentchangeintimeisThecorrectanswerisB.25、AfactoryassemblesProductXfromthreecomponents，A，B，andC．One0feachcomponentisneededforeachProductXandallthreecomponentsmustbeavailablewhenassemblyofeachProductXstarts．IttakestwodaystoassembleoneProductX．AssemblyofeachProductXstartsatthebeginningofonedayandisfinishedattheend0fthenextday．ThefactorycanworkonatmostfiveProductXsatonce．Ifcomponentsareavailableeachdayasshowninthetableabove，whatisthelargestnumber0fProductXsthatcanbeassembledduringthethreedayscoveredbythetable?A、3B、5C、6D、9E、10标准答案：B知识点解析：WewilldeterminethelargestnumberofProductXsthatcanbeassembledduringallthreedaysbyconsideringseparatelythelargestnumberthatcanbeassembledif0,1,2,or3ProductX’sbeginassemblyonMonday.0ProductX’sbeginassemblyonMonday:Inthiscase,atmostthreeProductXscanbeginassemblyonTuesday(becauseonlythreeunitsofComponentAareavailableonWednesday),andhenceatmost0+3=3ProductXscouldbeassembledduringthethreedays.1ProductXbeginsassemblyonMonday:Inthiscase,atmostthreeProductXscanbeginassemblyonTuesday(becauseonlythreeunitsofComponentAareavailableonWednesday),andhenceatmost1+3=4ProductXscouldbeassembledduringthethreedays.2ProductX’sbeginassemblyonMonday:Inthiscase,atmostthreeProductX’scanbeginassemblyonTuesday(becauseonlythreeunitsofComponentAareavailableonWednesday),andhenceatmost2+3=5ProductXscouldbeassembledduringthethreedays.3ProductX’sbeginassemblyonMonday:Inthiscase,atmosttwoProductXscanbeginassemblyonTuesday(becausethefactorycanworkonatmostfiveProductXsatonce),andhenceatmost3+2=5ProductXscouldbeassembledduringthethreedays.Therefore,thelargestnumberofProductXsthatcanbeassembledduringthethreedaysis5.ThecorrectanswerisB.GMAT（QUANTITATIVE）数学练习试卷第2套一、单项选择题(本题共30题，每题1.0分，共30分。)1、Workingataconstantrate,acopymachinemakes20copiesofaonedocumentperminute.Ifthemachineworksatthisconstantrate,howmanyhoursdoesittaketomake4,800copiesofaonedocument?A、4B、5C、6D、7E、8标准答案：A知识点解析：Thecopymachineproduces20copiesoftheonedocumenteachminute.Becausethereare60minutesinanhour,theconstantrateof20copiesperminuteisequalto60x20=1,200copiesperhour.Withthemachineworkingatthisrate,theamountoftimethatittakestoproduce4,800copiesofthedocumentis=4hours.ThecorrectanswerisA.2、Ifx+y=2andx2+y2=2,whatisthevalueofxy?A、-2B、-1C、0D、1E、2标准答案：D知识点解析：x+y=2giveny=2-xsubtractxfrombothsidesx2+(2-x)2=2substitutey=2-xintox2+y2=22x2-4x+4=2expandandcombineliketerms2x2-4x+2=0subtract2frombothsidesx2-2x+1=0dividebothsidesby2(x-1)(x-1)=0factorx=1seteachfactorequalto0y=1usex=1andy=2-xxy=1multiply1and1Alternatively,thevalueofxycanbefoundbyfirstsquaringbothsidesoftheequationx+y=2.x+y=2given(x+y)2=4squarebothsidesx2+2xy+y2=4expandandcombineliketerms2+2xy=4replacex2+y2with22xy=2subtract2frombothsidesxy=1dividebothsidesby2ThecorrectanswerisD.3、ThesumSofthefirstnconsecutivepositiveevenintegersisgivenbyS=n(n+1).Forwhatvalueofnisthissumequalto110?A、10B、11C、12D、13E、14标准答案：A知识点解析：Giventhatthesumofthefirstnevennumbersisn(n+1),thesumisequalto110when110=n(n+l).Tofindthevalueofninthiscase,weneedtofindthetwoconsecutiveintegerswhoseproductis110.Theseintegersare10and11;10x11=110.Thesmallerofthesenumbersisn.ThecorrectanswerisA.4、Acertainharborhasdockingstationsalongitswestandsouthdocks,asshowninthefigure;anytwoadjacentdockingstationsareseparatedbyauniformdistanced.Acertainboatleftthewestdockfromdockingstation#2andmovedinastraightlinediagonallyuntilitreachedthesouthdock.Iftheboatwasatonetimedirectlyeastofdockingstation#4anddirectlynorthofdockingstation#7,atwhichdockingstationonthesouthdockdidtheboatarrive?A、#7B、#8C、#9D、#10E、#11标准答案：B知识点解析：Theboattraveledinastraightlinefromdockingstation#2onthewestdocktooneofthedockingstationsonthesouthdock,passingthroughasinglepointthatisbothdueeastofdockingstation#4andduenorthofdockingstation#7.CallthispointP.HavingtraveledtoP,theboatwasboth2dsouthofitsstartingpointand2deastofitsstartingpoint.Therefore,travelinginastraightline,theboattraveledoneunitsouthforeveryoneunittraveledeast.AndbecauseatpointPtheboatwasadistancednorthofthesouthdock,theboatmusthavereachedthesouthdockatapointwhichisadistanceofdeastofdockingstation#7(whichisduesouthofpointP).Thispointisthepositionofdockingstation#8.Theboatthereforearrivedatdockingstation#8.ThecorrectanswerisB.5、6(87.30+0.65)-5(87.30)=A、3.90B、39.00C、90.90D、91.20E、91.85标准答案：D知识点解析：Thisquestionismostefficientlyansweredbydistributingthe6over87.30and0.65,andthencombiningthetermsthatcontainafactorof87.30,asfollows:6(87.30+0.65)-5(87.30)=6(87.30)+6(0.65)-5(87.30)=(6-5)87.30+6(0.65)-87.30+3.90=91.20ThecorrectanswerisD.6、PointsA,B,C,andD,inthatorder,lieonaline.IfAB=3cm,AC=4cm,and8D=6cm,whatisCD,incentimeters?A、1B、2C、3D、4E、5标准答案：E知识点解析：ThefigureshowspointsA,B,C,andDaswellasthegivenmeasurements.SinceAC=AB+BC,itfollowsthat4=3+BC,andsoBC=1.Then,sinceBD=BC+CD,itfollowsthat6=1+CD,andsoCD=5.Alternately,AD=AB+BD=3+6=9.Also,AD=AC+CD,so9=4+CDandCD=5.ThecorrectanswerisE.7、Whatisthevalueofx2yz-xyz2,ifx=-2,y=1,andz=3?A、20B、24C、30D、32E、48标准答案：C知识点解析：Giventhatx=-2,y=1,andz=3,itfollowsbysubstitutionthatx2yz-xyz2=(-2)2(1)(3)-(-2)(1)(32)=(4)(l)(3)-(-2)(l)(9)=12-(-18)=12+18=30ThecorrectanswerisC.8、Asouvenirvendorpurchased1,000shirtsforaspecialeventatapriceof$5each.Thevendorsold600oftheshirtsonthedayoftheeventfor$12eachand300oftheshirtsintheweekfollowingtheeventfor$4each.Thevendorwasunabletoselltheremainingshirts.Whatwasthevendor’sgrossprofitonthesaleoftheseshirts?A、$1,000B、$2,200C、$2,700D、$3,000E、$3,400标准答案：E知识点解析：Thevendor’sgrossprofitonthesaleoftheshirtsisequaltothetotalrevenuefromtheshirtsthatweresoldminusthetotalcostforalloftheshirts.Thetotalcostforalloftheshirtsisequaltothenumberofshirtsthevendorpurchasedmultipliedbythepricepaidb