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1、2020PascalSolution滑铁卢竞赛题答案1.Incents,the?vegivenchoicesare50,90,95,101,and115cents.Thedi?erencesbetweeneachoftheseand$1.00(or100cents),incents,are100?50=50100?90=10100?95=5101?100=1115?100=15Thedi?erencebetween$1.01and$1.00isthesmallest(1cent),so$1.01isclosestto$1.00.Answer:(D)doczj/doc/448ec407e8710
2、1f69e3195c9.ingthecorrectorderofoperations,(20?16)(12+8)4=4204=804=20Answer:(C)3.Wedividethe750mLof?ourintoportionsof250mL.Wedothisbycalculating750250=3.Therefore,750mListhreeportionsof250mL.Since50mLofmilkisrequiredforeach250mLof?our,then350=150mLofmilkisrequiredintotal.Answer:(C)4.Thereare8?guresi
3、ntotal.Ofthese,3aretriangles.Therefore,theprobabilityis3.Answer:(A)5.Wesimplifytheleftsideandexpressitasafractionwithnumerator1:19+118=218+118=318=16Therefore,thenumberthatreplacestheis6.Answer:(C)6.Thereare16horizontalsegmentsontheperimeter.Eachhaslength1,sothehorizontalsegmentscontribute16totheper
4、imeter.Thereare10verticalsegmentsontheperimeter.Eachhaslength1,sotheverticalsegmentscontribute10totheperimeter.Therefore,theperimeteris10+16=26.(Wecouldarriveatthistotalinsteadbystartingata?xedpointandtravellingaroundtheoutsideofthe?gurecountingthenumberofsegments.)Answer:(E)7.Since33=333=39=27,then
5、33+33+33=27+27+27=81=9Answer:(B)8.Thedi?erencebetweenthetwogivennumbersis7.62?7.46=0.16.Thislengthofthenumberlineisdividedinto8equalsegments.Thelengthofeachofthesesegmentsisthus0.168=0.02.PointPisthreeofthesesegmentstotherightof7.46.Thus,thenumberrepresentedis7.46+3(0.02)=7.46+0.06=7.52.Answer:(E)9.
6、A12by12gridofsquareswillhave11interiorverticallinesand11interiorhorizontallines.(Inthegiven4by4exle,thereare3interiorverticallinesand3interiorhorizontallines.)Eachofthe11interiorverticallinesintersectseachofthe11interiorhorizontallinesandcreatesaninteriorintersectionpoint.Thus,eachinteriorverticalli
7、neaccountsfor11intersectionpoints.Therefore,thenumberofinteriorintersectionpointsis1111=121.Answer:(B)10.Becausethecentralanglefortheinteriorsector“Lessthan1houris90?,thenthefractionofthestudentswhodolessthan1hourofhomeworkperdayis90?360?=14.Inotherwords,25%ofthestudentsdolessthan1hourofhomeworkperd
8、ay.Therefore,100%?25%=75%ofthestudentsdoatleast1hourofhomeworkperday.Answer:(E)11.Solution1Sincethereismorethan1four-leggedtable,thenthereareatleast2four-leggedtables.Sincethereare23legsintotal,thentheremustbefewerthan6four-leggedtables,since6four-leggedtableswouldhave64=24legs.Thus,therearebetween2
9、and5four-leggedtables.Ifthereare2four-leggedtables,thenthesetablesaccountfor24=8legs,leaving23?8=15legsforthethree-leggedtables.Since15isdivisibleby3,thenthismustbethesolution,sothereare153=5three-leggedtables.(Wecancheckthatifthereare3or4four-leggedtables,thenthenumberofremaininglegsisnotdivisibleb
10、y3,andifthereare5four-leggedtables,thenthereisonly1three-leggedtable,whichisnotallowed.)Solution2Sincethereismorethan1tableofeachtype,thenthereareatleast2three-leggedtablesand2four-leggedtables.Thesetablesaccountfor2(3)+2(4)=14legs.Thereare23?14=9morelegsthatneedtobeaccountedfor.Thesemustcomefromaco
11、mbinationofthree-leggedandfour-leggedtables.Theonlywaytomake9from3sand4sistousethree3s.Therefore,thereare2+3=5three-leggedtablesand2four-leggedtables.Answer:(E)12.Solution1Thetotalareaoftherectangleis34=12.Thetotalareaoftheshadedregionsequalsthetotalareaoftherectangle(12)minustheareaoftheunshadedreg
12、ion.Theunshadedregionisatrianglewithbaseoflength1andheight4;theareaofthisregionis12(1)(4)=2.Therefore,thetotalareaoftheshadedregionsis12?2=10.Solution2Theshadedtriangleonthelefthasbaseoflength2andheightoflength4,sohasanareaof12(2)(4)=4.Theshadedtriangleontherighthasbaseoflength3(atthetop)andheightof
13、length4,sohasanareaof12(3)(4)=6.Therefore,thetotalareaoftheshadedregionsis4+6=10.Answer:(C)13.SincetheratioofboystogirlsatCayleyH.S.is3:2,then33+2=35ofthestudentsatCayleyH.S.areboys.Thus,thereare35(400)=12005=240boysatCayleyH.S.SincetheratioofboystogirlsatFermatC.I.is2:3,then22+3=25ofthestudentsatFe
14、rmatC.I.areboys.Thus,thereare25(600)=12005=240boysatFermatC.I.Thereare400+600=1000studentsintotalatthetwoschools.Ofthese,240+240=480areboys,andsotheremaining1000?480=520studentsaregirls.Therefore,theoverallratioofboystogirlsis480:520=48:52=12:13.Answer:(B)14.Whenthegivennetisfolded,thefacenumbered5w
15、illbeoppositethefacenumbered1.Therefore,theremainingfourfacesshareanedgewiththefacenumbered1,sotheproductofthenumbersis2346=144.Answer:(B)15.Thepercentage10%isequivalenttothefraction110.Therefore,t=110s,ors=10t.Answer:(D)16.Sincethebaseofthefoldedboxmeasures5cmby4cm,thentheareaofthebaseoftheboxis5(4
16、)=20cm2.Sincethevolumeoftheboxis60cm3andtheareaofthebaseis20cm2,thentheheightoftheboxis60=3cm.Therefore,eachofthefouridenticalsquareshassidelength3cm,becausetheedgesofthesesquaresformtheverticaledgesofthebox.Therefore,therectangularsheetmeasures3+5+3=11cmby3+4+3=10cm,andsohasarea11(10)=110cm2.Answer
17、:(B)17.Solution1SinceSURisastraightline,thenRUV=180?SUV=180?120?=60?.SincePWandQXareparallel,thenRVW=VTX=112?.SinceUVWisastraightline,thenRVU=180?RVW=180?112?=68?.Sincethemeasuresoftheanglesinatriangleaddto180?,thenURV=180?RUV?RVU=180?60?68?=52?Solution2SinceSURisastraightline,thenRUV=180?SUV=180?12
18、0?=60?.SincePWandQXareparallel,thenRST=RUV=60?.SinceSTXisastraightline,thenRTS=180?VTX=180?112?=68?.Sincethemeasuresoftheanglesinatriangleaddto180?,thenURV=SRT=180?RST?RTS=180?60?68?=52?Answer:(A)18.Solution1WhenCatherineadds30litresofgasoline,thetankgoesfrom18fullto34full.Since34?18=68?18=58,then58
19、ofthecapacityofthetankis30litres.Thus,18ofthecapacityofthetankis305=6litres.Also,thefullcapacityofthetankis86=48litres.To?lltheremaining14ofthetank,Catherinemustaddanadditional1448=12litresofgas.Becauseeachlitrecosts$1.38,itwillcost12$1.38=$16.56to?lltherestofthetank.Solution2Supposethatthecapacityo
20、fthegastankisxlitres.Startingwith1ofatank,30litresofgasmakesthetank3full,so1x+30=3xor5x=30orx=48.Theremainingcapacityofthetankis14x=14(48)=12litres.At$1.38perlitre,itwillcostCatherine12$1.38=$16.56to?lltherestofthetank.Answer:(C)19.Theareaofasemi-circlewithradiusris1r2sotheareaofasemi-circlewithdiam
21、eterdis12(12d)2=18d2.ThesemicircleswithdiametersUV,VW,WX,XY,andYZeachhaveequaldiameterandthusequalarea.Theareaofeachofthesesemicirclesis18(52)=258.ThelargesemicirclehasdiameterUZ=5(5)=25,sohasarea18(252)=6258.Theshadedareaequalstheareaofthelargesemicircle,minustheareaoftwosmallsemicircles,plustheare
22、aofthreesmallsemicircles,whichequalstheareaofthelargesemicircleplustheareaofonesmallsemicircle.Therefore,theshadedareaequals6258+258=6508=3254.Answer:(A)20.Thesumoftheoddnumbersfrom5to21is5+7+9+11+13+15+17+19+21=117Therefore,thesumofthenumbersinanyrowisone-thirdofthistotal,or39.Thismeansaswellthatth
23、esumofthenumbersinanycolumnordiagonalisalso39.Sincethenumbersinthemiddlerowaddto39,thenthenumberinthecentresquareis39?9?17=13.Sincethenumbersinthemiddlecolumnaddto39,thenthenumberinthemiddlesquareinthebottomrowis39?5?13=21.591317x21Sincethenumbersinthebottomrowaddto39,thenthenumberinthebottomrightsq
24、uareis39?21?x=18?x.Sincethenumbersinthebottomlefttotoprightdiagonaladdto39,thenthenumberinthetoprightsquareis39?13?x=26?x.Sincethenumbersintherightmostcolumnaddto39,then(26?x)+17+(18?x)=39or61?2x=39or2x=22,andsox=11.Wecancompletethemagicsquareasfollows:195159131711217Answer:(B)21.Welabelthenumbersin
25、theemptyboxesasyandz,sothenumbersintheboxesarethus8,y,z,26,x.Sincetheaverageofzandxis26,thenx+z=2(26)=52orz=52?x.Werewritethelistas8,y,52?x,26,x.Sincetheaverageof26andyis52?x,then26+y=2(52?x)ory=104?26?2x=78?2x.Werewritethelistas8,78?2x,52?x,26,x.Sincetheaverageof8and52?xis78?2x,then8+(52?x)=2(78?2x
26、)60?x=156?4x3x=96x=32Therefore,x=32.Answer:(D)22.SinceJKLMisarectangle,thentheanglesatJandKareeach90?,soeachofSJPandQKPisright-angled.BythePythagoreanTheoreminSJP,wehaveSP2=JS2+JP2=522+392=2704+1521=4225SinceSP0,thenSP=4225=65.SincePQRSisarhombus,thenPQ=PS=65.BythePythagoreanTheoreminQKP,wehaveKP2=P
27、Q2?KQ2=652?252=4225?625=3600SinceKP0,thenKP=3600=60.(InsteadofusingthePythagoreanTheorem,wecouldnoteinsteadthatSJPisascaled-upversionofa3-4-5right-angledtriangleandthatQKPisascaled-upversionofa5-12-13right-angledtriangle.Thiswouldallowustousetheknownratiosofsidelengthstocalculatethemissingsidelength
28、.)SinceKQandPZareparallelandPKandWQareparallel,thenPKQWisarectangle,andsoPW=KQ=25.Similarly,JPZSisarectangleandsoPZ=JS=52.Thus,WZ=PZ?PW=52?25=27.Also,SYRMisarectangle.SinceJMandKLareparallel(JKLMisarectangle),JKandMLareparallel,andPQandSRareparallel(PQRSisarhombus),thenMSR=KQPandSRM=QPK.SinceSMRandQ
29、KPhavetwoequalangles,thentheirthirdanglesmustbeequaltoo.Thus,thetriangleshavethesameproportions.Sincethehypotenusesofthetrianglesareequal,thenthetrianglesmustinfactbeexactlythesamesize;thatis,thelengthsofthecorrespondingsidesmustbeequal.(WesaythatSMRiscongruenttoQKPby“angle-side-angle.)Inparticular,
30、MR=KP=60.Thus,ZY=SY?SZ=MR?JP=60?39=21.Therefore,theperimeterofrectangleWXYZis2(21)+2(27)=96.Answer:(D)23.First,wenotethat2020=10(201)=2(5)(3)(67)andso20202=223252672.ConsiderNconsecutivefour-digitpositiveintegers.FortheproductoftheseNintegerstobedivisibleby20202,itmustbethecasethattwodi?erentinteger
31、saredivisibleby67(whichwouldmeanthatthereareatleast68integersinthelist)oroneoftheintegersisdivisibleby672.SincewewanttominimizeN(andindeedbecausenoneoftheanswerchoicesisatleast68),welookforalistofintegersinwhichoneisdivisibleby672=4489.Sincetheintegersmustallbefour-digitintegers,thentheonlymultiples
32、of4489thewemustconsiderare4489and8978.First,weconsideralistofNconsecutiveintegersincluding4489.Sincetheproductoftheseintegersmusthave2factorsof5andnosingleintegerwithin10of4489hasafactorof25,thenthelistmustincludetwointegersthataremultiplesof5.Tominimizethenumberofintegersinthelist,wetrytoinclude448
33、5and4490.Thusourcandidatelistis4485,4486,4487,4488,4489,4490.Theproductoftheseintegersincludes2factorsof67(in4489),2factorsof5(in4485and4490),2factorsof2(in4486and4488),and2factorsof3(sinceeachof4485and4488isdivisibleby3).Thus,theproductofthese6integersisdivisibleby20202.Therefore,theshortestpossibl
34、elistincluding4489haslength6.Next,weconsideralistofNconsecutiveintegersincluding8978.Here,thereisanearbyintegercontaining2factorsof5,namely8975.Sowestartwiththelist8975,8976,8977,8978andchecktoseeifithastherequiredproperty.Theproductoftheseintegersincludes2factorsof67(in8978),2factorsof5(in8975),and
35、2factorsof2(in8976).However,theonlyintegerinthislistdivisibleby3is8976,whichhasonly1factorof3.Toincludeasecondfactorof3,wemustincludeasecondmultipleof3inthelist.Thus,weextendthelistbyonenumberto8979.Therefore,theproductofthenumbersinthelist8975,8976,8977,8978,8979isamultipleof20202.Thelengthofthisli
36、stis5.Thus,thesmallestpossiblevalueofNis5.(Notethataquickwaytotestifanintegerisdivisibleby3istoadditsdigitandseeifthistotalisdivisibleby3.Forexle,thesumofthedigitsof8979is33;since33isamultipleof3,then8979isamultipleof3.)Answer:(A)24.Welabelthetermsx1,x2,x3,.,x2020,x2020.SupposethatSisthesumoftheodd-
37、numberedtermsinthesequence;thatis,S=x1+x3+x5+x2007+x2020Weknowthatthesumofallofthetermsis5307;thatis,x1+x2+x3+x2020+x2020=5307Next,wepairuptheterms:eachodd-numberedtermwiththefollowingeven-numberedterm.Thatis,wepairthe?rsttermwiththesecond,thethirdtermwiththefourth,andsoon,untilwepairthe2020thtermwi
38、ththe2020thterm.Thereare1005suchpairs.Ineachpair,theeven-numberedtermisonebiggerthantheodd-numberedterm.Thatis,x2?x1=1,x4?x3=1,andsoon.Therefore,thesumoftheeven-numberedtermsis1005greaterthanthesumoftheodd-numberedterms.Thus,thesumoftheeven-numberedtermsisS+1005.Sincethesumofallofthetermsequalsthesu
39、moftheodd-numberedtermsplusthesumoftheeven-numberedterms,thenS+(S+1005)=5307or2S=4302orS=2151.Thus,therequiredsumis2151.Answer:(C)25.Beforeweanswerthegivenquestion,wedeterminethenumberofwaysofchoosing3objectsfrom5objectsandthenumberofwaysofchoosing2objectsfrom5objects.Consider5objectslabelledB,C,D,E
40、,F.Thepossiblepairsare:BC,BD,BE,BF,CD,CE,CF,DE,DF,EF.Thereare10suchpairs.Thepossibletriplesare:DEF,CEF,CDF,CDE,BEF,BDF,BDE,BCF,BCE,BCD.Thereare10suchtriples.(Canyouseewhytherearethesamenumberofpairsandtriples?)LabelthesixteamsA,B,C,D,E,F.WestartbyconsideringteamA.TeamAplays3games,sowemustchoose3ofth
41、eremaining5teamsforAtoplay.Aswesawabove,thereare10waystodothis.Withoutlossofgenerality,wepickoneofthesesetsof3teamsforAtoplay,sayAplaysB,CandD.Wekeeptrackofeverythingbydrawingdiagrams,joiningtheteamsthatplayeachotherwithaline.Thusfar,wehaveABCDTherearetwopossiblecasesnoweithernoneofB,CandDplayeachot
42、her,oratleastonepairofB,C,Dplayseachother.Case1:NoneoftheteamsthatplayAplayeachotherInthecon?gurationabove,eachofB,CandDplaytwomoregames.TheyalreadyplayAandcannotplayeachother,sotheymusteachplayEandF.ThisgivesAEFBCDNofurtherchoicesarepossible.Thereare10possibleschedulesinthistypeofcon?guration.These
43、10combinationscomefromchoosingthe3teamsthatplayA.Case2:SomeoftheteamsthatplayAplayeachotherHere,atleastonepairoftheteamsthatplayAplayeachother.GiventheteamsB,CandDplayingA,thereare3possiblepairs(BC,BD,CD).Wepickoneofthesepairs,sayBC.(Thisgives103=30con?gurationssofar.)ABCDItisnownotpossibleforBorCto
44、alsoplayD.IfitwasthecasethatC,say,playedD,thenwewouldhavethecon?gurationABCDEFHere,AandChaveeachplayed3gamesandBandDhaveeachplayed2games.TeamsEandFareunaccountedforthusfar.Theycannotbothplay3gamesinthiscon?gurationasthepossibleopponentsforEareB,DandF,andthepossibleopponentsforFareB,DandE,withthe“Ban
45、d“Dpossibilitiesonlytobeusedonce.AsimilarargumentshowsthatBcannotplayD.Thus,BorCcannotalsoplayD.Sowehavethecon?gurationABCDHere,Ahasplayed3games,BandChaveeachplayed2games,andDhasplayed1game.BandCmustplay1moregameandcannotplayDorA.TheymustplayEandFinsomeorder.Thereare2possiblewaystoassignthesegames(BEandCF,orBFandCE.)Thisgives302=60con?gurationssofar.SupposethatBplaysEandCplaysF.ABCDEFSofar,A,BandCeachplay3gamesandE,FandDeachplay1game.Theonlywaytocompletethecon?gurationistojoinD,EandF.ABCDEFTherefore,thereare60possibleschedulesinthiscase.Intotal,thereare10+60=70possibleschedules.Answer:(E)