2021AMC12A解答

2021AMC12A解答1.(D)Itisgiventhat0.1x=2and0.2y=2,sox=20andy=10.Thusx?y=10.2.(B)Since2x+7=3wehavex=?2.Hence?2=bx?10=?2b?10,so2b=?8,andb=?4.3.(B)Letwbethewidthoftherectangle.Thenthelengthis2w,andx2=w2+(2w)2=5w2.Theareaisconsequentlyw(2w)=2w2=25x2.4.(A)IfDavebuyssevenwindowsseparatelyhewillpurchasesixandreceiveonefree,foracostof$600.IfDougbuyseightwindowsseparately,hewillpurchasesevenandreceiveonefree,foratotalcostof$700.ThetotalcosttoDaveandDougpurchasingseparatelywillbe$1300.Iftheypurchase?fteenwindowstogether,theywillneedtopurchaseonly12windows,foracostof$1200,andwillreceive3free.Thiswillresultinasavingsof$100.5.(B)Thesumofthe50numbersis20·30+30·20=1200.Theiraverageis1200/50=24.6.(B)Because(rate)(time)=(distance),thedistanceJoshrodewas(4/5)(2)=8/5ofthedistancethatMikerode.LetmbethenumberofmilesthatMikehadriddenwhentheymet.Thenthenumberofmilesbetweentheirhousesis13=m+85m=135m.Thusm=5.7.(C)Thesymmetryofthe?gureimpliesthatABH,BCE,CDF,andDAGarecongruentrighttriangles.SoBH=CE=BC2?BE2=√50?1=7,andEH=BH?BE=7?1=6.HencethesquareEFGHhasarea62=36.ORAsinthe?rstsolution,BH=7.NownotethatABH,BCE,CDF,andDAGarecongruentrighttriangles,soArea(EFGH)=Area(ABCD)?4Area(ABH)=50?412·1·7=36.8.(D)SinceA,M,andCaredigitswehave0≤A+M+C≤9+9+9=27.Theprimefactorizationof2021is2021=5·401,so100A+10M+C=401andA+M+C=5.HenceA=4,M=0,andC=1.9.(A)Thequadraticformulayieldsx=?(a+8)±(a+8)2?4·4·92·4.Theequationhasonlyonesolutionpreciselywhenthevalueofthediscriminant,(a+8)2?144,is0.Thisimpliesthata=?20ora=4,andthesumis?16.ORTheequationhasonesolutionifandonlyifthepolynomialisthesquareofabinomialwithlinearterm±√4x2=±2xandconstantterm±√9=±3.Because(2x±3)2hasalinearterm±12x,itfollowsthata+8=±12.Thusaiseither?20or4,andthesumofthosevaluesis?16.10.(B)Theunitcubeshaveatotalof6n3faces,ofwhich6n2arered.Therefore14=6n26n3=1n,son=4.11.(E)The?rstandlastdigitsmustbebothoddorbothevenfortheiraveragetobeaninteger.Thereare5·5=25odd-oddcombinationsforthe?rstandlastdigits.Thereare4·5=20even-evencombinationsthatdonotusezeroasthe?rstdigit.Hencethetotalis45.12.(D)Theslopeofthelineis1000?1100?1=11111,soallpointsonthelinehavetheform(1+11t,1+111t).Suchapointhasintegercoordinatesifandonlyiftisaninteger,andthepointisstrictlybetweenAandBifandonlyif013.(D)Eachnumberappearsintwosums,sothesumofthesequenceis2(3+5+6+7+9)=60.Themiddletermofa?ve-termarithmeticsequenceisthemeanofitsterms,so60/5=12isthemiddleterm.The?gureshowsanarrangementofthe?venumbersthatmeetstherequirement.#!$'%!14.(D)Astandarddiehasatotalof21dots.For1≤n≤6,adotisremovedfromthefacewithndotswithprobabilityn/21.Thusthefacethatoriginallyhasndotsisleftwithanoddnumberofdotswithprobabilityn/21ifnisevenand1?n/21ifnisodd.Eachfaceisthetopfacewithprobability1/6.Thereforethetopfacehasanoddnumberofdotswithprobability161?121+221+1?321+421+1?521+621=163+321=16·6621=1121.ORTheprobabilitythatthetopfaceisoddis1/3ifadotisremovedfromanoddface,andtheprobabilitythatthetopfaceisoddis2/3ifadotisremovedfromanevenface.Becauseeachdothastheprobability1/21ofbeingremoved,thetopfaceisoddwithprobability131+3+521+232+4+621=3363=1121.15.(C)LetObethecenterofthecircle.EachofDCEandABDhasadiameterofthecircleasaside.Thustheratiooftheirareasistheratioofthetwoaltitudestothediameters.ThesealtitudesareDCandthealtitudefromCtoDOinDCE.LetFbethefootofthissecondaltitude.SinceCFOissimilartoDCO,CFDC=CODO=AO?ACDO=12AB?13AB12AB=13,whichisthedesiredratio.BORBecauseAC=AB/3andAO=AB/2,wehaveCO=AB/6.TrianglesDCOandDABhaveacommonaltitudetoABsotheareaofDCOis16theareaofADB.TrianglesDCOandECOhaveequalareassincetheyhaveacommonbaseCOandtheiraltitudesareequal.ThustheratiooftheareaofDCEtotheareaofABDis1/3.16.(D)Considerarighttriangleasshown.BythePythagoreanTheorem,(r+s)2=(r?3s)2+(r?s)2sor2+2rs+s2=r2?6rs+9s2+r2?2rs+s2and0=r2?10rs+9s2=(r?9s)(r?s).Butr=s,sor=9sandr/s=9.r?33ORBecausetheratior/sisindependentofthevalueofs,assumethats=1andproceedasintheprevioussolution.17.(A)ThepiecethatcontainsWisshown.ItisapyramidwithverticesV,W,X,Y,andZ.ItsbaseWXYZisasquarewithsidesoflength1/2anditsaltitudeVWis1.Hencethevolumeofthispyramidis13122(1)=112.18.(A)Ofthenumberslessthan1000,499ofthemaredivisiblebytwo,333aredivisibleby3,and199aredivisibleby5.Thereare166multiplesof6,99multiplesof10,and66multiplesof15.Andthereare33numbersthataredivisibleby30.SobytheInclusion-ExclusionPrinciplethereare499+333+199?166?99?66+33=733numbersthataredivisiblebyatleastoneof2,3,or5.Oftheremaining999?733=266numbers,165areprimesotherthan2,3,or5.Notethat1isneitherprimenorcomposite.Thisleavesexactly100prime-lookingnumbers.19.(B)Becausetheodometerusesonly9digits,itrecordsmileageinbase-9nu-merals,exceptthatitsdigits5,6,7,8,and9representthebase-9digits4,5,6,7,and8.Thereforethemileageis2021base9=2·93+4=2·729+4=1462.ORThenumberofmilestraveledisthesameasthenumberofintegersbetween1and2021,inclusive,thatdonotcontainthedigit4.Firstconsidertheintegerslessthan2021.Therearetwochoicesforthe?rstdigit,including0,and9choicesforeachoftheotherthree.Becauseonecombinationofchoicesis0000,thereare2·93?1=1457positiveintegerslessthan2021thatdonotcontainthe