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GMAT(QUANTITATIVE)數(shù)學(xué)練習(xí)試卷4(共4套)(共92題)GMAT(QUANTITATIVE)數(shù)學(xué)練習(xí)試卷第1套一、單項(xiàng)選擇題(本題共25題,每題1.0分,共25分。)1、Acertaintrucktravelingat55milesperhourgets4.5milespergallonofdieselfuelconsumed.Travelingat60milesperhour,thetruckgetsonly3.5milespergallon.Ona500-miletrip,ifthetruckusedatotalof120gallonsofdieselfuelandtraveledpartofthetripat55milesperhourandtherestat60milesperhour,howmanymilesdidittravelat55milesperhour?A、140B、200C、250D、300E、360標(biāo)準(zhǔn)答案:E知識(shí)點(diǎn)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標(biāo)準(zhǔn)答案:E知識(shí)點(diǎn)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標(biāo)準(zhǔn)答案:B知識(shí)點(diǎn)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、標(biāo)準(zhǔn)答案:A知識(shí)點(diǎn)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標(biāo)準(zhǔn)答案:C知識(shí)點(diǎn)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標(biāo)準(zhǔn)答案:C知識(shí)點(diǎn)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標(biāo)準(zhǔn)答案:D知識(shí)點(diǎn)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、
標(biāo)準(zhǔn)答案:A知識(shí)點(diǎn)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、
標(biāo)準(zhǔn)答案:B知識(shí)點(diǎn)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.a(chǎn)ndisinscribedinasquare.ThesmallercirclehascenterC,radiusrcm,andistangenttothelargercircleatpointBandtothesquareatpointsDandF,IfpointsA,B,C,andEarecollinear,whichofthefollowingisequaltoR/r?A、
B、
C、
D、
E、
標(biāo)準(zhǔn)答案:D知識(shí)點(diǎn)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標(biāo)準(zhǔn)答案:C知識(shí)點(diǎn)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、
標(biāo)準(zhǔn)答案:D知識(shí)點(diǎn)解析:First,thediagonalofthesquarewithsidesoflength4is4.Fromthisitfollowsthattheareaoftheshadedregionconsistsoftheareaofanequilateraltrianglewithsides4minus—theareaofasquarewithsidesoflength4.Thus,theareaoftheshadedregionis.ThecorrectanswerisD.13、Ifeachsideof△ACDabovehaslength3andifABhaslength1,whatistheareaofregionBCDE?A、
B、
C、
D、
E、
標(biāo)準(zhǔn)答案:B知識(shí)點(diǎn)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標(biāo)準(zhǔn)答案:D知識(shí)點(diǎn)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、
標(biāo)準(zhǔn)答案:D知識(shí)點(diǎn)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標(biāo)準(zhǔn)答案:A知識(shí)點(diǎn)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標(biāo)準(zhǔn)答案:C知識(shí)點(diǎn)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%標(biāo)準(zhǔn)答案:D知識(shí)點(diǎn)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標(biāo)準(zhǔn)答案:C知識(shí)點(diǎn)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.a(chǎn)ndanegativepercentindicatesthatCompanyX’ssalesforthatmonthdecreasedfromthesalesforthepreviousmonth.ForwhichmonthwerethesalesclosesttothesalesinJanuary?A、FebruaryB、MarchC、AprilD、MayE、June標(biāo)準(zhǔn)答案:D知識(shí)點(diǎn)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%標(biāo)準(zhǔn)答案:C知識(shí)點(diǎn)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標(biāo)準(zhǔn)答案:B知識(shí)點(diǎn)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標(biāo)準(zhǔn)答案:C知識(shí)點(diǎn)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、
標(biāo)準(zhǔn)答案:B知識(shí)點(diǎn)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標(biāo)準(zhǔn)答案:B知識(shí)點(diǎn)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)數(shù)學(xué)練習(xí)試卷第2套一、單項(xiàng)選擇題(本題共30題,每題1.0分,共30分。)1、Workingataconstantrate,acopymachinemakes20copiesofaonedocumentperminute.Ifthemachineworksatthisconstantrate,howmanyhoursdoesittaketomake4,800copiesofaonedocument?A、4B、5C、6D、7E、8標(biāo)準(zhǔn)答案:A知識(shí)點(diǎn)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標(biāo)準(zhǔn)答案:D知識(shí)點(diǎn)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標(biāo)準(zhǔn)答案:A知識(shí)點(diǎn)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標(biāo)準(zhǔn)答案:B知識(shí)點(diǎn)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標(biāo)準(zhǔn)答案:D知識(shí)點(diǎn)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標(biāo)準(zhǔn)答案:E知識(shí)點(diǎn)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標(biāo)準(zhǔn)答案:C知識(shí)點(diǎn)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標(biāo)準(zhǔn)答案:E知識(shí)點(diǎn)解析:Thevendor’sgrossprofitonthesaleoftheshirtsisequaltothetotalrevenuefromtheshirtsthatweresoldminusthetotalcostforalloftheshirts.Thetotalcostforalloftheshirtsisequaltothenumberofshirtsthevendorpurchasedmultipliedbythepricepaidb
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