2010PascalSolution滑铁卢竞赛题答案16
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2010PascalSolution滑铁卢竞赛题答案16
第1/10页2010PascalContestS；1.Incents,the?vegivencho；Thedi?erencesbetweeneach；100?50=00?95；Thedi?erencebetween$1.01；Answer:(D)；2.Usingthecorrectorderof；(20?16)×(12+8)4×208
第1/10页 2010PascalContestSolutions1.Incents,the?vegivenchoicesare50,90,95,101,and115cents.Thedi?erencesbetweeneachoftheseand$1.00(or100cents),incents,are100?50=00?95=5?100=15Page2Thedi?erencebetween$1.01and$1.00isthesmallest(1cent),so$1.01isclosestto$1.00.Answer:(D)2.Usingthecorrectorderofoperations,(20?16)×(12+8)4×2080===20444Answer:(C)3.Wedividethe750mLof?ourintoportionsof250mL.Wedothisbycalculating750÷250=3.Therefore,750mListhreeportionsof250mL.Since50mLofmilkisrequiredforeach250mLof?our,then3×50=150mLofmilkisrequiredintotal.Answer:(C)4.Thereare8?guresintotal.Ofthese,3aretriangles..Therefore,theprobabilityis35.Wesimplifytheleftsideandexpressitasafractionwithnumerator1:121311+=+==Therefore,thenumberthatreplacesthe??is6.Answer:(C)6.Thereare16horizontalsegmentsontheperimeter.Eachhaslength1,sothehorizontalsegmentscontribute16totheperimeter.Thereare10verticalsegmentsontheperimeter.Eachhaslength1,sotheverticalsegmentscontribute10totheperimeter.Therefore,theperimeteris10+16=26.(Wecouldarriveatthistotalinsteadbystartingata?xedpointandtravellingaroundtheoutsideofthe?gurecountingthenumberofsegments.)Answer:(E)7.Since33=3×3×3=3×9=27,then√√√===9Answer:(B)8.Thedi?erencebetweenthetwogivennumbersis7.62?7.46=0.16.Thislengthofthenumberlineisdividedinto8equalsegments.Thelengthofeachofthesesegmentsisthus0.16÷8=0.02.PointPisthreeofthesesegmentstotherightof7.46.Thus,thenumberrepresentedis7.46+3(0.02)=7.46+0.06=7.52.Answer:(E)Answer:(A)9.A12by12gridofsquareswillhave11interiorverticallinesand11interiorhorizontallines.(Inthegiven4by4example,thereare3interiorverticallinesand3interiorhorizontallines.)Eachofthe11interiorverticallinesintersectseachofthe11interiorhorizontallinesandcreatesaninteriorintersectionpoint.Thus,eachinteriorverticallineaccountsfor11intersectionpoints.Therefore,thenumberofinteriorintersectionpointsis11×11=121.Answer:(B)10.Becausethecentralanglefortheinteriorsector“Lessthan1hour”is90?,thenthefractionof90?1.thestudentswhodolessthan1hourofhomeworkperdayis=360?4Inotherwords,25%ofthestudentsdolessthan1hourofhomeworkperday.Therefore,100%?25%=75%ofthestudentsdoatleast1hourofhomeworkperday.Answer:(E)11.Solution1Sincethereismorethan1four-leggedtable,thenthereareatleast2four-leggedtables.Sincethereare23legsintotal,thentheremustbefewerthan6four-leggedtables,since6four-leggedtableswouldhave6×4=24legs.Thus,therearebetween2and5four-leggedtables.Ifthereare2four-leggedtables,thenthesetablesaccountfor2×4=8legs,leaving23?8=15legsforthethree-leggedtables.Since15isdivisibleby3,thenthismustbethesolution,sothereare15÷3=5three-leggedtables.(Wecancheckthatifthereare3or4four-leggedtables,thenthenumberofremaininglegsisnotdivisibleby3,andifthereare5four-leggedtables,thenthereisonly1three-leggedtable,whichisnotallowed.)Solution2Sincethereismorethan1tableofeachtype,thenthereareatleast2three-leggedtablesand2four-leggedtables.Thesetablesaccountfor2(3)+2(4)=14legs.Thereare23?14=9morelegsthatneedtobeaccountedfor.Thesemustcomefromacombinationofthree-leggedandfour-leggedtables.Theonlywaytomake9from3sand4sistousethree3s.Therefore,thereare2+3=5three-leggedtablesand2four-leggedtables.Answer:(E)12.Solution1Thetotalareaoftherectangleis3×4=12.Thetotalareaoftheshadedregionsequalsthetotalareaoftherectangle(12)minustheareaoftheunshadedregion.Theunshadedregionisatrianglewithbaseoflength1andheight4;theareaofthisregionis1(1)(4)=2.Therefore,thetotalareaoftheshadedregionsis12?2=10.Solution2Theshadedtriangleonthelefthasbaseoflength2andheightoflength4,sohasanareaof1(2)(4)=4.2Theshadedtriangleontherighthasbaseoflength3(atthetop)andheightoflength4,so1hasanareaof(3)(4)=6.Therefore,thetotalareaoftheshadedregionsis4+6=10.Answer:(C)3=3ofthestudentsatCayley13.SincetheratioofboystogirlsatCayleyH.S.is3:2,thenH.S.areboys.3Thus,thereare(400)=boysatCayleyH.S.2SincetheratioofboystogirlsatFermatC.I.is2:3,then=2ofthestudentsatFermatC.I.areboys.1200Thus,thereare2(600)==240boysatFermatC.I.Thereare400+600=1000studentsintotalatthetwoschools.Ofthese,240+240=480areboys,andsotheremainingstudentsaregirls.Therefore,theoverallratioofboystogirlsis480:520=48:52=12:13.Answer:(B)14.Whenthegivennetisfolded,thefacenumbered5willbeoppositethefacenumbered1.Therefore,theremainingfourfacesshareanedgewiththefacenumbered1,sotheproductofthenumbersis2×3×4×6=144.Answer:(B)15.Thepercentage10%isequivalenttothefraction1Therefore,t=10s,ors=10t.1.Answer:(D)16.Sincethebaseofthefoldedboxmeasures5cmby4cm,thentheareaofthebaseoftheboxis5(4)=20cm2.Sincethevolumeoftheboxis60cm3andtheareaofthebaseis20cm2,thentheheightof60=3cm.theboxisTherefore,eachofthefouridenticalsquareshassidelength3cm,becausetheedgesofthesesquaresformtheverticaledgesofthebox.Therefore,therectangularsheetmeasures3+5+3=11cmby3+4+3=10cm,andsohasarea11(10)=110cm2.Answer:(B)17.Solution1SinceSURisastraightline,then∠RUV=180??∠SUV=180??120?=60?.SincePWandQXareparallel,then∠RVW=∠VTX=112?.SinceUVWisastraightline,then∠RVU=180??∠RVW=180??112?=68?.Sincethemeasuresoftheanglesinatriangleaddto180?,then∠URV=180??∠RUV?∠RVU=180??60??68?=52?2010PascalContestSolutionsSolution2SinceSURisastraightline,then∠RUV=180??∠SUV=180??120?=60?.SincePWandQXareparallel,then∠RST=∠RUV=60?.SinceSTXisastraightline,then∠RTS=180??∠VTX=180??112?=68?.Sincethemeasuresoftheanglesinatriangleaddto180?,thenPage5∠URV=∠SRT=180??∠RST?∠RTS=180??60??68?=52?Answer:(A)18.Solution11fullto3full.WhenCatherineadds30litresofgasoline,thetankgoesfrom8415Since3?1=6?=5,thenofthecapacityofthetankis30litres.Thus,1ofthecapacityofthetankis30÷5=6litres.Also,thefullcapacityofthetankis8×6=48litres.1ofthetank,Catherinemustaddanadditional4×48=12litresofgas.To?lltheremaining14Becauseeachlitrecosts$1.38,itwillcost12×$1.38=$16.56to?lltherestofthetank.Solution2Supposethatthecapacityofthegastankisxlitres.1Startingwithofatank,30litresofgasmakesthetank3full,so1x+30=3xor5x=30orx=48.1x=1(48)=12litres.TheremainingcapacityofthetankisAt$1.38perlitre,itwillcostCatherine12×$1.38=$16.56to?lltherestofthetank.Answer:(C)119.Theareaofasemi-circlewithradiusrisπr2sotheareaofasemi-circlewithdiameterdis11π(1d)2=8πd2.22ThesemicircleswithdiametersUV,VW,WX,XY,andYZeachhaveequaldiameterand1thusequalarea.Theareaofeachofthesesemicirclesisπ(52)=25π.1ThelargesemicirclehasdiameterUZ=5(5)=25,sohasarea8π(252)=625π.8Theshadedareaequalstheareaofthelargesemicircle,minustheareaoftwosmallsemicircles,plustheareaofthreesmallsemicircles,whichequalstheareaofthelargesemicircleplustheareaofonesmallsemicircle.π+25π=650π=325π.Therefore,theshadedareaequals625Answer:(A)20.Thesumoftheoddnumbersfrom5to21is5+7+9+11+13+15+17+19+21=117Therefore,thesumofthenumbersinanyrowisone-thirdofthistotal,or39.Thismeansaswellthatthesumofthenumbersinanycolumnordiagonalisalso39.Sincethenumbersinthemiddlerowaddto39,thenthenumberinthecentresquareis39?9?17=13.Sincethenumbersinthemiddlecolumnaddto39,thenthenumberinthemiddlesquareinthebottomrowis39?5?13=21.59x131721包含各类专业文献、高等教育、幼儿教育、小学教育、外语学习资料、中学教育、行业资料、2010PascalSolution滑铁卢竞赛题答案16等内容。　
　Pascal 或 Java 中的一种编写程序解决 8 或 10 ...每道试题用 时将从竞赛开始到试题解答被判定为正确...滑铁卢大学 加拿大 美国亚特兰大 Charles University ...pascal 竞赛题求解【题目描述】 在异乡打拼的小李同志迷上了一款叫诺斯克的台球游戏，而且随着练习的深入，他总是能在某些神奇的时刻开启外挂模式，此时小李将指哪打哪，直至无球可_百度作业帮
pascal 竞赛题求解【题目描述】 在异乡打拼的小李同志迷上了一款叫诺斯克的台球游戏，而且随着练习的深入，他总是能在某些神奇的时刻开启外挂模式，此时小李将指哪打哪，直至无球可
pascal 竞赛题求解【题目描述】 在异乡打拼的小李同志迷上了一款叫诺斯克的台球游戏，而且随着练习的深入，他总是能在某些神奇的时刻开启外挂模式，此时小李将指哪打哪，直至无球可打。现在小李想让你帮他计算下当他开启外挂模式的时候最多可以取得多少分数。注意：台面上的球数经常会异于传统斯诺克。 斯诺克比赛的基本规则如下：一、彩球共分8种颜色，红（1分）、黄（2分）、绿（3分）、棕（4分）、蓝（5分）、粉（6分）、黑（7分）、白（主球，控制白球来打其余球）。二、当台面上有红球的时候你必须先击打一个红球，然后能且只能击打一个彩球（不包括红球），此时落袋的彩球将会被放回桌面，一直重复该过程。三、当打完规则二的彩球（不包括红球）发现已经没有红球时，按照彩球的分值从低到高将其依次击入袋中。【输入】输入仅有一行，共7个用空格隔开的整数，分别为当前台面上红、黄、绿、棕、蓝、粉、黑球的数目。 【输出】输出仅有一行，共1个整数，表示小李可以得到的最高分。【样例输入】2 0 1 0 3 0 2 【样例输出】48 【样例说明】台面上共有红球2个、绿球1个、蓝球3个、黑球2个，获得最高分的打法是红-黑-红-黑-绿-蓝-蓝-蓝-黑-黑，共可以获得48分。【数据规模】保证最后得分不会超过2的31次方-1
var\x09i,ans,max:\x09color:array[1..7]&of&begin\x09for&i:=1&to&7&do\x09begin\x09\x09read(color[i]);\x09\x09ans:=ans+color[i]*i;\x09\x09if&color[i]&0&then&max:=i;\x09\x09if&max&1&then&ans:=ans+color[1]*\x09write(ans);end.2009PascalContest滑铁卢竞赛题_百度文库
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