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1、2007PascalSolution滑铁卢竞赛题答案1.Calculating,3(7?5)?5=32?5=6?5=1.Answer:(B)2.Sincexislessthan?1andgreaterthan?2,thenthebestestimateofthegivenchoicesis?1.3.Answer:(B)3.Theshadedsquarehassidelength1sohasarea12=1.Therectanglehasdimensions3by5sohasarea35=15.Thus,thefractionoftherectanglethatisshadedis115.Ans
2、wer:(A)4.Calculating,25?52=32?25=7.Answer:(E)5.In3hours,Leonaearns$24.75,soshemakes$24.753=$8.25perhour.Therefore,ina5hourshift,Leonaearns5$8.25=$41.25.Answer:(E)6.Calculating,64+3664+36=8+6100=1410=75.Answer:(A)7.Solution1IntotalMeganandDaninherit$1010000.Sinceeachdonates10%,thenthetotaldonatedis10
3、%of$1010000,or$101000.Solution2Megandonates10%of$1000000,or$100000.Dandonates10%of$10000,or$1000.Intotal,theydonate$100000+$1000=$101000.Answer:(A)8.WethinkofBCasthebaseofABC.Itslengthis12.Sincethey-coordinateofAis9,thentheheightofABCfrombaseBCis9.Therefore,theareaofABCis12(12)(9)=54.Answer:(B)9.Cal
4、culatingthegivendi?erenceusingacommondenominator,weobtain58?116=916.Since916islargerthaneachof12=816and716,thenneither(D)nor(E)iscorrect.Since916islessthaneachof34=1216,then(A)isnotcorrect.Asadecimal,9=0.5625.Since35=0.6and59=0.5,then91659,so(C)isthecorrectanswer.Answer:(C)10.SinceM=20073,thenM=669.
5、SinceN=M3,thenN=6693=223.SinceX=M?N,thenX=669?223=446.Answer:(E)11.Themeanof6,9and18is6+9+183=333=11.Thusthemeanof12andyis11,sothesumof12andyis2(11)=22,soy=10.Answer:(C)12.InPQR,sincePR=RQ,thenRPQ=PQR=48?.SinceMPNandRPQareoppositeangles,thenMPN=RPQ=48?.InPMN,PM=PN,soPMN=PNM.Therefore,PMN=12(180?MPN)
6、=12(180?48?)=12(132?)=66?.Answer:(D)13.Theprimenumberssmallerthan10are2,3,5,and7.Thetwoofthesenumberswhicharedi?erentandaddto10are3and7.Theproductof3and7is37=21.Answer:(B)14.Sincetherewere21maleswritingandtheratioofmalestofemaleswritingis3:7,thenthereare7321=49femaleswriting.Therefore,thetotalnumber
7、ofstudentswritingis49+21=70.Answer:(D)15.Solution1The?rststackismadeupof1+2+3+4+5=15blocks.Thesecondstackismadeupof1+2+3+4+5+6=21blocks.Thereare36blocksintotal.Westartbuildingthenewstackfromthetop.Sincetherearemorethan21blocks,weneedatleast6rows.For7rows,1+2+3+4+5+6+7=28blocksareneeded.For8rows,1+2+
8、3+4+5+6+7+8=36blocksareneeded.Therefore,Claracanbuildastackwith0blocksleftover.Solution2Sincethenewstackwillbelargerthanthesecondstackshown,letusthinkaboutaddingnewrowstothissecondstackusingtheblocksfromthe?rststack.The?rststackcontains1+2+3+4+5=15blocksintotal.The?rsttworowsthatwewouldaddtothebotto
9、mofthesecondstackwouldhave7and8blocksinthem,foratotalof15blocks.Thisusesalloftheblocksfromthe?rststack,withnoneleftover,andcreatesasimilarstack.Therefore,thereare0blocksleftover.Answer:(A)16.Solution1Thesumofthenumbersinthesecondrowis10+16+22=48,sothesumofthenumbersinanyrow,columnordiagonalis48.Inth
10、e?rstrow,P+4+Q=48soP+Q=44.Inthethirdrow,R+28+S=48soR+S=20.Therefore,P+Q+R+S=44+20=64.Solution2Thesumofthenumbersinthesecondrowis10+16+22=48,sothesumofthenumbersinanyrow,columnordiagonalis48.Fromthe?rstrow,P+4+Q=48soP+Q=44.Fromthe?rstcolumn,P+10+R=48soP+R=38.Subtractingthesetwoequationsgives(P+Q)?(P+
11、R)=44?38orQ?R=6.Fromoneofthediagonals,R+16+Q=48orQ+R=32.Addingtheselasttwoequations,2Q=38orQ=19,soR=32?Q=13.Also,P=44?Q=25.Fromthelastrow,13+28+S=48,orS=7.Thus,P+Q+R+S=25+19+13+7=64.Solution3ThesumofallofthenumbersinthegridisP+Q+R+S+10+16+22+28+4=P+Q+R+S+80Butthesumofthethreenumbersinthesecondcolumn
12、is4+16+28=48,sothesumofthethreenumbersineachcolumnis48.Thus,thetotaloftheninenumbersinthegridis3(48)=144,soP+Q+R+S+80=144orP+Q+R+S=64.Answer:(C)17.Atpresent,thesumofNorinesageandthenumberofyearsthatshehasworkedis50+19=69.Thistotalmustincreaseby85?69=16beforeshecanretire.Aseveryyearpasses,thistotalin
13、creasesby2(asherageincreasesby1andthenumberofyearsthatshehasworkedincreasesby1).Thus,ittakes8yearsforhertotaltoincreasefrom69to85,soshewillbe50+8=58whenshecanretire.Answer:(C)18.BythePythagoreanTheoreminPQR,PQ2=PR2?QR2=132?52=144,soPQ=144=12.BythePythagoreanTheoreminPQS,QS2=PS2?PQ2=372?122=1225,soQS
14、=1225=35.Therefore,theperimeterofPQSis12+35+37=84.Answer:(D)19.Sincethereciprocalof310is1x+1,then1x+1=1031x=73x=37sox=37.Answer:(C)20.DrawalinefromFtoBC,paralleltoAB,meetingBCatP.ADCBEFPSinceEBisparalleltoFPandFEB=90?,thenEBPFisarectangle.SinceEB=40,thenFP=40;sinceEF=30,thenBP=30.SinceAD=80,thenBC=8
15、0,soPC=80?30=50.Therefore,theareaofEBCFissumoftheareasofrectangleEBPF(whichis3040=1200)andFPC(whichis12(40)(50)=1000),or1200+1000=2200.SincetheareasofAEFCDandEBCFareequal,theneachis2200,sothetotalareaofrectangleABCDis4400.SinceAD=80,thenAB=440080=55.Therefore,AE=AB?EB=55?40=15.Answer:(D)21.Letus?rst
16、considerthepossibilitiesforeachintegerseparately:?Thetwo-digitprimenumbersare11,13,17,19.Theonlyonewhosedigitsadduptoaprimenumberis11.Therefore,P=11.?SinceQisamultipleof5between2and19,thenthepossiblevaluesofQare5,10,15.?Theoddnumbersbetween2and19thatarenotprimeare9and15,sothepossiblevaluesofRare9and
17、15.?Thesquaresbetween2and19are4,9and16.Only4and9aresquaresofprimenumbers,sothepossiblevaluesofSare4and9.?SinceP=11,thepossiblevalueofQare5,10and15,andTistheaverageofPandQ,thenTcouldbe8,10.5or13.SinceTisalsoaprimenumber,thenTmustbe13,soQ=15.WenowknowthatP=11,Q=15andT=13.Sincethe?venumbersarealldi?ere
18、nt,thenRcannotbe15,soR=9.SinceR=9,Scannotbe9,soS=4.Therefore,thelargestofthe?veintegersisQ=15.Answer:(B)22.BythePythagoreanTheorem,PR=QR2+PQ2=152+82=289=17km.Asafarunsatotaldistanceof8+15+7=30kmat21km/hinthesametimethatFlorencerunsatotaldistanceof17+7=24km.Therefore,Asafasspeedis30=5ofFlorencesspeed
19、,soFlorencesspeedis421=84km/h.Asafarunsthelast7kmin721=13hour,or20minutes.Florencerunsthelast7kmin7845=3584=512hour,or25minutes.SinceAsafaandFlorencearriveatStogether,thenFlorencearrivedatR5minutesbeforeAsafa.Answer:(E)23.Thetotalareaofthelargercircleis(22)=4,sothetotalareaoftheshadedregionsmustbe51
20、2(4)=53.SupposethatADC=x?.Theareaoftheunshadedportionoftheinnercircleisthusx360ofthetotalareaoftheinnercircle,orx360(12)=x360(sinceADCisx360ofthelargestpossiblecentralangle(360?).Theareaoftheshadedportionoftheinnercircleisthus?x360=360?x360.Thetotalareaoftheouterringisthedi?erenceoftheareasoftheoute
21、randinnercircles,or(22)?(12)=3.Theshadedareaintheouterringwillbex360ofthistotalarea,sinceADCisx360ofthelargestpossiblecentralangle(360?).Sotheshadedareaintheouterringisx360(3)=3x360.Sothetotalshadedarea(whichmustequal53)is,intermofx,3x360+360?x360=360+2x360.Therefore,360+2x360=53=600360,so360+2x=600
22、orx=120.Thus,ADC=120?.Answer:(B)24.First,wecompletethenextseveralspacesinthespiraltotrytogetabettersenseofthepattern:1716151413185431219612112078910212223242526Wenoticefromthisextendedspiralthattheoddperfectsquareslieonadiagonalextendingdownandtotherightfromthe1,since1,9,25andsoonwillcompleteasquare
23、ofnumberswhentheyarewritten.(Tryblockingoutthenumberslargerthaneachofthesetoseethis.)Thispatterndoescontinuesincewheneachoftheseoddperfectsquaresisreached,thenumberofspacesuptothatpointinthesequenceactuallydoesformasquare.The?rstoddperfectsquarelargerthan2007is452=2025.2025willlie18spacestotheleftof
24、2007inthisrow.(Therowwith2025willactuallybelongenoughtobeabletomove18spacestotheleftfrom2025.)Theoddperfectsquarebefore2025is432=1849,so1850willbethenumberdirectlyabove2025,astherowcontaining1849willcontinueonemorespacetotherightbeforeturningup.Since1850isdirectlyabove2025,then1832isdirectlyabove200
25、7.Theoddperfectsquareafter2025is472=2209,so2208willbethenumberdirectlybelow2025,since2209willbeonespacetotherightandonedown.Since2208isdirectlybelow2025,then2190isdirectlybelow2007.Therefore,thesumofthenumbersdirectlyaboveandbelow2007is1832+2190=4022.Answer:(E)25.Forxand3xtoeachhaveevendigitsonly,xm
26、ustbeinoneofthefollowingforms.(Here,a,b,crepresentdigitsthatcaneachbe0,2or8,andnisadigitthatcanonlybe2or8.)?nabc(2333=54possibilities)?na68(23=6possibilities)?n68a(23=6possibilities)?68ab(33=9possibilities)?n668(2possibilities)?668a(3possibilities)?6668(1possibility)?6868(1possibility)Intotal,therea
27、re82possibilitiesforx.Ingeneralterms,thesearetheonlyformsthatwork,sincedigitsof0,2and8inxpro-duceevendigitswithaneven“carry(0or2)thuskeepingalldigitsin3xeven,whilea6maybeused,butmustbefollowedby8or68or668inordertogiveacarryof2.Moreprecisely,whydotheseformswork,andwhyaretheytheonlyformsthatwork?First
28、,wenotethat30=0,32=6,34=12,36=18and38=24.Thus,eachevendigitofxwillproduceanevendigitinthecorrespondingpositionof3x,butmaya?ectthenextdigittotheleftin3xthroughits“carry.Notethatadigitof0or2inxproducesnocarry,whileadigitof8inxproducesanevencarry.Therefore,noneofthesethreedigitscanpossiblycreateanodddi
29、gitin3x(eitherdirectlyorthroughcarrying),astheyeachcreateanevendigitinthecorrespondingpositionof3xanddonota?ectwhetherthenextdigittotheleftisevenorodd.(Weshouldnotethatthecarryintoanydigitin3xcanneverbemorethan2,sowedonothavetoworryaboutcreatingacarryof1fromadigitinxof0or2,oracarryof3fromadigitof8in
30、xthroughmultiplecarries.)Soadigitof2or8canappearinanypositionofxandadigitof0canappearinanypositionofxexceptforthe?rstposition.Adigitof4canneverappearinx,asitwillalwaysproduceacarryof1,andsowillal-wayscreateanodddigitin3x.Adigitof6canappearinx,aslongasthecarryfromthepreviousdigitis2tomakethecarryforw
31、ardfromthe6equalto2.(Thecarryintothe6cannotbelargerthan2.)Whenthishappens,wehave36+Carry=20,andsoa2iscarriedforward,whichdoesnota?ectwhethernextdigitisevenorodd.Acarryof2canoccurifthedigitbeforethe6isan8,orifthedigitbeforethe6isa6whichisprecededby8orby68.Combiningthepossibleusesof0,2,6,and8givesusthelistofpossibleformsabove,andhence82possiblevaluesforx.Answer:(A)