GRE（QUANTITATIVE）模擬試卷1（共257題）

GRE（QUANTITATIVE）模擬試卷1（共9套）（共257題）GRE（QUANTITATIVE）模擬試卷第1套一、數(shù)值比較題(本題共10題，每題1.0分，共10分。)1、Whenx+2isdividedby5theremainderis3ColumnAColumnBTheremainderwhen2xisdividedby5A、ifthequantityinColumnAisgreaterB、ifthequantityinColumnBisgreaterC、ifthetwoquantityareequalD、iftherelationshipcannotbedeterminedfromtheinformationgiven標(biāo)準(zhǔn)答案：B知識(shí)點(diǎn)解析：當(dāng)x+2被5除時(shí)，余數(shù)(remainder)是3。解：本題的正確答案是(B)。設(shè)n為整數(shù)，則有：x+2=5n+3；x=5n+1。所以x被5除時(shí)余數(shù)是1。2、Whenintegernisdividedby9,theremainderis2.ColumnAColumnBTheremainderwhen2nisdividedby3A、ifthequantityinColumnAisgreaterB、ifthequantityinColumnBisgreaterC、ifthetwoquantityareequalD、iftherelationshipcannotbedeterminedfromtheinformationgiven標(biāo)準(zhǔn)答案：C知識(shí)點(diǎn)解析：整數(shù)n被9除時(shí)，余數(shù)(remainder)為2。n被3除的余數(shù)2。解：本題的正確答案為(C)。根據(jù)題意整數(shù)，n可以表示為：n=9m+2，其中m是整數(shù)，顯然9m可以被3整除，所以n被3除時(shí)余數(shù)仍為2。3、ColumnAColumnBTheproductoftwointegersis6.Thesumofthetwointegers3A、ifthequantityinColumnAisgreaterB、ifthequantityinColumnBisgreaterC、ifthetwoquantityareequalD、iftherelationshipcannotbedeterminedfromtheinformationgiven標(biāo)準(zhǔn)答案：D知識(shí)點(diǎn)解析：本題的正確答案為(D)。兩整數(shù)乘積為6，則這兩個(gè)整數(shù)可同時(shí)取正值，也可同時(shí)取負(fù)值。當(dāng)這兩個(gè)數(shù)同時(shí)取正值時(shí)，它們的和大于3，而當(dāng)它們同時(shí)取負(fù)值時(shí)，則它們的和將顯然小于3，所以本題為無法判斷。4、xisthesumofthefirst25positiveevenintegers,yisthesumofthefirst25positiveoddintegers.ColumnAColumnBxy+25A、ifthequantityinColumnAisgreaterB、ifthequantityinColumnBisgreaterC、ifthetwoquantityareequalD、iftherelationshipcannotbedeterminedfromtheinformationgiven標(biāo)準(zhǔn)答案：C知識(shí)點(diǎn)解析：x是前25個(gè)正偶數(shù)(positiveeveninteger)之和，y是前25個(gè)正奇數(shù)(positiveoddinteger)之和。解：本題的正確答案為(C)。由第n個(gè)偶數(shù)總比第n個(gè)奇數(shù)多1(n為正整數(shù))，可知前25個(gè)正偶數(shù)之和比前25個(gè)正奇數(shù)之和大25。5、Ms.Rogersboughtanelectricrangeontheinstallmentplan.Thecashpriceoftherangewas$400.Theamountshepaidwas$120downand12monthlypaymentsof$28each.ColumnAColumnBTheamountshepaidfortheelectricrangeinexcessofthecashprice$56A、ifthequantityinColumnAisgreaterB、ifthequantityinColumnBisgreaterC、ifthetwoquantityareequalD、iftherelationshipcannotbedeterminedfromtheinformationgiven標(biāo)準(zhǔn)答案：C知識(shí)點(diǎn)解析：Rogers太太用分期付款的方法購(gòu)買了一個(gè)電氣灶，這個(gè)灶的現(xiàn)金價(jià)格為400美元，她首期付120美元，以后每月付28美元，分12個(gè)月付完。解：本題的正確答案是(C)。設(shè)她超出現(xiàn)金價(jià)格的付款為x，則有400+x=120+28×12so,x=56；注：down=downpayment首期付款6、ColumnAColumnB(RS)2+(ST)2(RT)2A、ifthequantityinColumnAisgreaterB、ifthequantityinColumnBisgreaterC、ifthetwoquantityareequalD、iftherelationshipcannotbedeterminedfromtheinformationgiven標(biāo)準(zhǔn)答案：D知識(shí)點(diǎn)解析：解：本題的正確答案是(D)。從圖中看，好像∠RST是鈍角，(RS)2+(ST)2>(RT)2，但是該題的圖中并沒有表明Figuredrawntoscale，所以考生只能讀圖，不能度量，從而也就無法判斷∠RST是否大于90o，所以本題為無法判斷。7、ColumnAColumnBThelengthofarcABCThelengthofarcADCA、ifthequantityinColumnAisgreaterB、ifthequantityinColumnBisgreaterC、ifthetwoquantityareequalD、iftherelationshipcannotbedeterminedfromtheinformationgiven標(biāo)準(zhǔn)答案：D知識(shí)點(diǎn)解析：本題的正確答案為(D)。由BC=AD=x，只能推出它們的對(duì)應(yīng)弧長(zhǎng)BC，AD相等，但是AB弧與CD弧的長(zhǎng)度卻不能確定，因此本題的為無法判斷。只有當(dāng)BC∥AD時(shí)才能得到ABC弧的長(zhǎng)度與BCD弧的長(zhǎng)度相等。(BC=AD保證了與圓心等距)。8、Allfacesofthesolidabovearerectangular.ColumnAColumnBThelengthofdiagonalMN15A、ifthequantityinColumnAisgreaterB、ifthequantityinColumnBisgreaterC、ifthetwoquantityareequalD、iftherelationshipcannotbedeterminedfromtheinformationgiven標(biāo)準(zhǔn)答案：A知識(shí)點(diǎn)解析：圖中物體的所有的面都是矩形。解：本題的正確答案為(A)。由圖中所給的條件可知，該物體是一個(gè)長(zhǎng)方體，三條相交的邊相互垂直。所以對(duì)角絨(diagonal)MN等于任意三條相交的邊的平方和，即：MN2=(62+82)+122=244>225=1529、TriangulargardenABCisredesignedbyincreasingthelengthofACby20percenttopointCanddecreasingthelengthofABby20percenttopointB.ColumnAColumnBTheareaoftheoriginalgardenABCTheareaoftheredesignedgardenAB’C’A、ifthequantityinColumnAisgreaterB、ifthequantityinColumnBisgreaterC、ifthetwoquantityareequalD、iftherelationshipcannotbedeterminedfromtheinformationgiven標(biāo)準(zhǔn)答案：A知識(shí)點(diǎn)解析：重新設(shè)計(jì)三角形花園ABC，使其AC邊長(zhǎng)增加20％到C’點(diǎn)；AB邊長(zhǎng)減少20％到B’點(diǎn)。解：本題的正確答案為(A)。從表面上看，好像是變化前后花園的面積大小不變，但是若考生根據(jù)題意列出方程就可發(fā)現(xiàn)，變化后花園的面積是減小了：S(△AB’C’)=0.96S(△ABC)10、TheboundariesofregionsXandYareparallelograms.ColumnAColumnBTheareaofXifitssideshavelengths6and10TheareaofYifitssideshavelengths5and12A、ifthequantityinColumnAisgreaterB、ifthequantityinColumnBisgreaterC、ifthetwoquantityareequalD、iftherelationshipcannotbedeterminedfromtheinformationgiven標(biāo)準(zhǔn)答案：D知識(shí)點(diǎn)解析：X和Y是平行四邊形。解：本題的正確答案為(D)。比較的左項(xiàng)為邊長(zhǎng)分別是6和10的平行四邊形X的面積；比較的右項(xiàng)為邊長(zhǎng)分別是5和12的平行四邊形Y的面積。平行四邊形公式為：S=a×bsinθ，θ為a和b的夾角，因而僅僅a×b乘積相等并不能表明他們的面積相等，面積的大小還受a和b的夾角大小的制約。二、計(jì)算題(本題共14題，每題1.0分，共14分。)11、Howmanypositivewholenumberslessthan81areNOTequalsquaresofwholenumbers?A、9B、70C、71D、72E、73標(biāo)準(zhǔn)答案：D知識(shí)點(diǎn)解析：小于81且不等于整數(shù)的平方(square)的正整數(shù)(positivewholenumbers)有多少個(gè)?解：本題的正確答案是(D)。該題也就是問小于81的非完全平方數(shù)有多少個(gè)。小于81的完全平方數(shù)有1，4，9，16，25，36，49，64共8個(gè)，而小于81的正整數(shù)有80個(gè)，所以小于81的非完全平方數(shù)有80-8=72個(gè)。12、Ofthefollowing,whichistheclosestapproximationto?A、1B、5C、10D、20E、100標(biāo)準(zhǔn)答案：C知識(shí)點(diǎn)解析：本題的正確答案為(C)。五個(gè)選項(xiàng)中的數(shù)值之間差距較大，因此可以在一定的范圍內(nèi)做近似計(jì)算，并且可以保證結(jié)果的可靠性。觀察后發(fā)現(xiàn)0.51的平方的值與0.261相近，而97.942的值又與100相近，所以根據(jù)題意可得：13、Inhowmanyoftheintegersbetween1and100doesthedigit5occurexactlyonce?A、10B、18C、19D、20E、21標(biāo)準(zhǔn)答案：B知識(shí)點(diǎn)解析：在1至100之間的數(shù)字中，5僅在它們中出現(xiàn)1次的整數(shù)有多少?解：本題的正確答案是(B)。解答本題的關(guān)鍵是考生一定要記住5只能在這些數(shù)字中出現(xiàn)1次，5在其中出現(xiàn)兩次的數(shù)字應(yīng)排除掉，即5要么在個(gè)位，要么在十位，個(gè)位與十位都出現(xiàn)5的數(shù)字(55)應(yīng)排除掉。個(gè)位上有5的數(shù)字是：5，15，25，35，45，65，75，85和95；十位上有5的數(shù)字是：50，51，52，53，54，56，57，58和59，一共是18個(gè)。14、ThenumberofconnectionsCthatcanbemadethroughaswitchboardtowhichTtelephonesareconnectedisgivenbytheformulaC=T(T-1)/2.Howmanymoreconnectionsarepossiblewith30telephonesthanwith20telephones?A、435B、245C、190D、45E、10標(biāo)準(zhǔn)答案：B知識(shí)點(diǎn)解析：連接有T門電話的交換機(jī)所能形成的連絡(luò)數(shù)目C由公式C=T(t-1)/2給定。30門電話能比20門電話多出多少連絡(luò)?解：本題的正確答案是(B)。30門電話的連絡(luò)數(shù)目：C(30)=30×(30-1)/2=435；20門電話的連絡(luò)數(shù)目：C(20)=20×(20-1)=190；C(30)-C(20)=435-190=24515、Thedailyrateforahotelroomthatsleeps4peopleis$39foronepersonandxdollarsforeachadditionalperson.If3peopletaketheroomforonedayandeachpays$21fortheroom,whatisthevalueofx?A、6B、8C、12D、13E、24標(biāo)準(zhǔn)答案：C知識(shí)點(diǎn)解析：某旅館的4人間的每天收費(fèi)是住1個(gè)人39美元，并且每多1人加收x美元。若3個(gè)人住該屋1天，并且每人付21美元，問x的值是多少?解：本題的正確答案是(C)。解答本題的關(guān)鍵是對(duì)“andxdollarsforeachadditionalperson”的理解。若考生能把這句話理解到位，那么就很容易把這道題做對(duì)。住一個(gè)人收39美元，住兩個(gè)人收39+x美元，住3個(gè)人收39+2x美元，由題意可得：39+2x=21×3→x=1216、Thecost,indollars,foranappliancerepairatacertaincompanyis1.2p+20h,wherepisthewholesalepriceoftheparts,indollars,andhisthenumberofhoursittakestorepairtheappliance.Whatisthecostofrepairinganapplianceifthewholesalepriceofthepartsis$15andittakes2hourstorepairit?A、$12B、$18C、$20D、$40E、$58標(biāo)準(zhǔn)答案：E知識(shí)點(diǎn)解析：某公司一設(shè)備的修理成本，以美元計(jì)，為1.2p+20h，其中p是零件的批發(fā)價(jià)格(以美元計(jì))，h是修理設(shè)備所用的小時(shí)數(shù)。如果零件的批發(fā)價(jià)是15美元，并且花了2小時(shí)修理，那么修理這個(gè)設(shè)備的成本是多少?解：本題的正確答案為(E)。解答該題時(shí)，只要把數(shù)據(jù)代入題中所給的公式即可：1.2×15+2×20=5817、Thescoresreportedforacertainmultiple-choicetestwerederivedbysubtracting1/3ofthenumberofwronganswersfromthenumberofrightanswers.Ona40-questiontest,ifnoneofthequestionswasomittedandthescorereportedwas20,howmanywronganswerswerethere?A、5B、10C、15D、25E、30標(biāo)準(zhǔn)答案：C知識(shí)點(diǎn)解析：某一多項(xiàng)選擇考試的分?jǐn)?shù)的計(jì)算是用答對(duì)的題數(shù)減去答錯(cuò)題數(shù)的1/3。在一個(gè)由40道題組成的測(cè)驗(yàn)中，若無題遺漏，且考試分?jǐn)?shù)為20，問答錯(cuò)的題數(shù)有多少個(gè)?解：本題的正確答案為(C)。設(shè)答錯(cuò)題數(shù)為x個(gè)，則由題意可得：(40-x)-1/3x=20→x=1518、Art,Bob,andCarmenshareaprizeof$400.IfArtreceivestwiceasmuchasBob,andifBobreceives1/2asmuchasCarmen,howmuchdoesCarmenreceive?A、$20B、$40C、$80D、$120E、$160標(biāo)準(zhǔn)答案：E知識(shí)點(diǎn)解析：Art，Bob，Carmen分享400美元的獎(jiǎng)金，若Art接受的獎(jiǎng)金數(shù)是Bob的兩倍，并且Bob接受的獎(jiǎng)金數(shù)是Carmen的1/2，問Carmen的獎(jiǎng)金是多少?解：本題的正確答案是(E)。設(shè)Carmen的獎(jiǎng)金為x，Bob的為1/2x，則Art是Bob的兩倍也為x，由題意可得：x+1/2x+x=400→x=16019、ToobtainanFHAmortgagefor$50,000ormore,thehomebuyermusthaveadownpaymentequalto4percentofthefirst$25,000ofthemortgageamountand5percentoftheportioninexcessof$25,000.Atsettlementthebuyerpaysamortgageinsurancepremiumequalto3percentofthemortgageamount.WhatisthemaximumFHAmortgage,ifany,abuyercanobtainifthebuyerhasonly$6,000availableforthedownpaymentandinsurancepremium?A、$62500B、$71875C、$78125D、$125000E、ThehomebuyercannotobtainanFHAmortgage.標(biāo)準(zhǔn)答案：C知識(shí)點(diǎn)解析：為了獲得50000美元或更多的FHA的住房抵押貸款，購(gòu)房者的首期付款必須相當(dāng)于第一期住房抵押貸款25000美元的4%和超過25000美元以上部分的5％。在入住時(shí)，購(gòu)房者付一個(gè)相當(dāng)于住房抵押貸款量3％的住房貸款保險(xiǎn)費(fèi)。若一個(gè)購(gòu)房者僅有6000美元付首期付款和保險(xiǎn)費(fèi)，問該購(gòu)房者能從FHA獲得的最大的住房貸款額是多少?解：本題的正確答案是(C)。題目中出現(xiàn)了許多金融業(yè)的詞語(yǔ)，例如mortgage指住房抵押貸款，downpayment指首期付款，insurancepremium指保險(xiǎn)費(fèi)，另外Settlement指入住或安家，住房抵押貸款指付了首期的downpayment后就可以搬入新房了。設(shè)最大住房抵押貸款額為x，則由題意可得：25000×4％+(x-25000)×5％+x×3%=6000→x=7812520、Inaclassof120students,60percentcanspeakFrenchandtherestcanspeakonlyEnglish.If25percentofthoseintheclasswhocanspeakFrenchcanalsospeakEnglish,howmanyofthestudentsintheclasscanspeakEnglish?A、54B、60C、66D、84E、90標(biāo)準(zhǔn)答案：C知識(shí)點(diǎn)解析：某個(gè)班級(jí)有120名學(xué)生，其中有60%的學(xué)生講法語(yǔ)，其余的學(xué)生人只會(huì)講英語(yǔ)。若講法語(yǔ)的學(xué)生中有25％的學(xué)生也可以講英語(yǔ)，問該班級(jí)中有多少人會(huì)講英語(yǔ)?解：本題的正確答案是(C)。班級(jí)里講法語(yǔ)的學(xué)生中有25％的學(xué)生會(huì)講英語(yǔ)，則會(huì)講英語(yǔ)的學(xué)生總數(shù)為：120×60％×25％+120×40％=6621、If9treesareoriginallyplantedinacircularpatternasshownabove,whatistheleastnumberoftreesthatmustbetransplantedsothatthe9treeswillbein2straightrows?A、4B、5C、6D、7E、8標(biāo)準(zhǔn)答案：B知識(shí)點(diǎn)解析：若9棵樹最初以如左圖所示的圓形種植，要使9棵樹排在兩條直線上，問最少需移植幾棵樹?解：本題的正確答案為(B)。圓上的任意三點(diǎn)都不在同一條直線上，又根據(jù)在同一平面內(nèi)的兩點(diǎn)確定一條直線可知，要使這9棵樹排在兩條直線上，則至少得有9-2×2=5棵樹被移植。22、Ifasolidpyramidhas4verticesand4faces,howmanyedgesdoesthepyramidhave?A、2B、3C、4D、6E、8標(biāo)準(zhǔn)答案：D知識(shí)點(diǎn)解析：假如一個(gè)立體四角錐有4個(gè)頂點(diǎn)和4個(gè)面，該四角錐有多少條邊?本題的正確的答案是(D)。Solid，n．固體，立體(幾何)；Pyramidn．錐體(四角錐)；Vertices，n．頂點(diǎn)(pl.)。23、Acertainrectanglehasperimeter54.Iftheratioofthelengthoftherectangletothewidthis5to4,whatisthelengthoftherectangle?A、30B、27C、24D、18E、15標(biāo)準(zhǔn)答案：E知識(shí)點(diǎn)解析：一長(zhǎng)方形的周長(zhǎng)等于54，其長(zhǎng)寬比為5：4，問該長(zhǎng)方形的長(zhǎng)為多少?解：本題的正確答案為(E)。設(shè)該長(zhǎng)方形的長(zhǎng)和寬分別是5x，4x，由題意可得：(5x+4x)×2=54→x=3，5x=1524、Inthefigureabove,ifx,y,andzareintegerssuchthatxA、59and91B、59and135C、91and178D、120and135E、120and178標(biāo)準(zhǔn)答案：C知識(shí)點(diǎn)解析：若x，y和z都是整數(shù)，且滿足x三、圖表題(本題共2題，每題1.0分，共2分。)25、Ifthepercentincreaseinpopulationfrom1910to1920hadbeenapproximatelythesameasthepercentincreasefrom1900to1910,the10-yearincrease,inmillions,from1910to1920,wouldhavebeenapproximatelyA、3B、6C、16D、19E、29標(biāo)準(zhǔn)答案：D知識(shí)點(diǎn)解析：標(biāo)題：美國(guó)的人口(1890-1980年官方的統(tǒng)計(jì)結(jié)果)若1910年至1920年人口增長(zhǎng)的百分?jǐn)?shù)與1900至1910年人口增長(zhǎng)的百分?jǐn)?shù)相同，那么以百萬(wàn)為單位，從1910至1920年的十年期間人口大約會(huì)增長(zhǎng)多少?解：本題的正確答案為(D)。從人口數(shù)據(jù)表中可知，1900至1910年人口增長(zhǎng)的百分比為：16/76×100％，若按照這個(gè)比率計(jì)算，1910至1920年人口的增長(zhǎng)量為：92×16/76=19.3726、Accordingtothegraphabove,atapproximatelywhatspeed,inmilesperhour(mph),istheenergyperdistanceusedinrunningapproximatelytwicetheenergyperdistanceusedinwalking?A、Between6.5and7mphB、Between5.5and6mphC、Between4.5and5mphD、Between3.5and4mphE、Between2.5and3mph標(biāo)準(zhǔn)答案：E知識(shí)點(diǎn)解析：標(biāo)題：能量與速度根據(jù)上圖，大約在多大速度(以mph為單位)時(shí)，跑每單位距離消耗的能量約是走單位距離所消耗能量的2倍?解：本題的正確答案為(E)。題目中圖形的橫軸代表速度(mph)，縱軸代表單位距離所消耗的能量(焦耳/米)；圖中的實(shí)線是跑步時(shí)速度與能量之間的關(guān)系，虛線代表走路時(shí)速度與能量之間的關(guān)系。根據(jù)以上分析再結(jié)合圖形可知，在速度為2.5--3mph時(shí)，跑每單位距離的路程所消耗的能量約是走單位距離的路程所消耗能量的2倍。GRE（QUANTITATIVE）模擬試卷第2套一、數(shù)值比較題(本題共9題，每題1.0分，共9分。)CompareQuantityAandQuantityB,usingadditionalinformationcenteredabovethetwoquantitiesifsuchinformationisgiven,andselectoneofthefollowingfouranswerchoices:(A)QuantityAisgreater.(B)QuantityBisgreater.(C)Thetwoquantitiesareequal.(D)Therelationshipcannotbedeterminedfromtheinformationgiven.Asymbolthatappearsmorethanonceinaquestionhasthesamemeaningthroughoutthequestion.1、x<0QuantityAQuantityBx5x4標(biāo)準(zhǔn)答案：B知識(shí)點(diǎn)解析：Ifx<0,thenx5<0andx4>0.Thusx5<x4,andthecorrectanswerisChoiceB.Thisexplanationusesthefollowingstrategy.Strategy8:SearchforaMathematicalRelationship2、QuantityAQuantityB(x+4)(y+3)(x+3)(y+4)標(biāo)準(zhǔn)答案：D知識(shí)點(diǎn)解析：Tocompare(x+4)(y+3)and(x+3)(y+4),tryplugginginafewvaluesforxandy.Case1:x=0andy=0.Inthiscase,QuantityAisequalto(4)(3),or12,andQuantityBisequalto(3)(4),or12.SoQuantityAisequaltoQuantityB.Case2:x=0andy=1.Inthiscase,QuantityAisequalto(4)(4),or16,andQuantityBisequalto(3)(5),or15.SoQuantityAisgreaterthanQuantityB.Inonecase,QuantityAisequaltoQuantityB,andintheothercase,QuantityAisgreaterthanQuantityB.ThereforethecorrectanswerisChoiceD.Thisexplanationusesthefollowingstrategies.Strategy10:TrialandErrorStrategy13:DetermineWhetheraConclusionFollowsfromtheInformationGiven3、representsthedecimalinwhichthedigitbisrepeatedwithoutend.QuantityAQuantityB1.0標(biāo)準(zhǔn)答案：A知識(shí)點(diǎn)解析：Bythedefinitiongiveninthequestion,representsthedecimalinwhichthedigit3isrepeatedwithoutend;thatis,=0.333…Itfollowsthatisgreaterthan0.3.Similarly,isgreaterthan0.7.Therefore+isgreaterthan1;thatis,QuantityAisgreaterthanQuantityB.ThecorrectanswerisChoiceA.Thisexplanationusesthefollowingstrategy.Strategy1:TranslatefromWordstoanArithmeticorAlgebraicRepresentation4、Acompanyplanstomanufacturetwotypesofhammers,typeRandtypeS.ThecostofmanufacturingeachhammeroftypeSis$0.05lessthantwicethecostofmanufacturingeachhammeroftypeR.標(biāo)準(zhǔn)答案：A知識(shí)點(diǎn)解析：NotethatQuantityAandQuantityBbothincludethecostofmanufacturing1,000hammersoftypeS.Ifyouremovethatcostfrombothquantities,theproblemisreducedtocomparingthecostofmanufacturing1,000hammersoftypeRwiththecostofmanufacturing500hammersoftypeS.SincethecostofmanufacturingeachhammeroftypeSis$0.05lessthantwicethecostofmanufacturingeachhammeroftypeR,itfollowsthatthecostofmanufacturing1,000hammersoftypeRisgreaterthanthecostofmanufacturing500hammersoftypeS.ThusthecorrectanswerisChoiceA.Thisexplanationusesthefollowingstrategies.Strategy8:SearchforaMathematicalRelationshipStrategy9:Estimate5、標(biāo)準(zhǔn)答案：C知識(shí)點(diǎn)解析：Since=43,youcansimplifyQuantityAasfollows.=1+4+42=21SinceQuantityBisalso21,thecorrectanswerisChoiceC.Thisexplanationusesthefollowingstrategy.Strategy5:SimplifyanArithmeticorAlgebraicRepresentation6、ListX:2,5,s,tListY:2,5,tTheaverage(arithmeticmean)ofthenumbersinlistXisequaltotheaverageofthenumbersinlistY.QuantityAQuantityBs0標(biāo)準(zhǔn)答案：D知識(shí)點(diǎn)解析：Sinceyouaregiventhattheaverageofthe4numbersinlistXisequaltotheaverageofthe3numbersinlist7,itfollowsthat.TomakeiteasiertocompareQuantityAandQuantityB,youcansimplifythisequationasfollows.3(7+s+t)=4(7+t)21+3s+3t=28+4t3s=7+tFromtheequation3s=7+t,ift=-7,thens=0,butift=0,thens>0.Inonecase,thequantitiesareequal,andintheothercase,QuantityAisgreater.ThereforethecorrectanswerisChoiceD.Thisexplanationusesthefollowingstrategies.Strategy1:TranslatefromWordstoanArithmeticorAlgebraicRepresentationStrategy10:TrialandErrorStrategy13:DetermineWhetheraConclusionFollowsfromtheInformationGiven7、TheradiusofcircleAis12greaterthantheradiusofcircleB.標(biāo)準(zhǔn)答案：A知識(shí)點(diǎn)解析：SincetheradiusofcircleAis12greaterthantheradiusofcircleB,iftheradiusofcircleBisr,thentheradiusofcircleAisr+12.ThecircumferenceofcircleAminusthecircumferenceofcircleBis2π(r+12)-2πr,whichsimplifiesto24π.Sinceπ>3,itfollowsthat24π>24(3);thatis,24π>72.SinceQuantityBis72,thecorrectanswerisChoiceA.Thisexplanationusesthefollowingstrategies.Strategy8:SearchforaMathematicalRelationshipStrategy9:Estimate8、QuantityAQuantityBx120標(biāo)準(zhǔn)答案：B知識(shí)點(diǎn)解析：Notethattherearefouranglesinthefigure:thethreeinterioranglesoftriangleABCandtheexteriorangleatvertexA.SincethemeasureoftheexteriorangleatvertexAisxdegrees,itfollowsthatthemeasureofinteriorangleAis(180-x)°.Also,sinceAB=BC,itfollowsthattriangleABCisisoscelesandthemeasuresofinterioranglesAandCareequal.ThereforethemeasureofinteriorangleCisalso(180-x)°.Sincethesumofthemeasuresoftheinterioranglesofatriangleis180°andyouaregiventhatthemeasureofinteriorangleBis40°,itfollowsthat40+(180-x)+(180-x)=180.Solvingtheequationforxgivesx=110.SinceQuantityBis120,thecorrectanswerisChoiceB.Thisexplanationusesthefollowingstrategies.Strategy4:TranslatefromaFiguretoanArithmeticorAlgebraicRepresentationStrategy8:SearchforaMathematicalRelationship9、S={1,2,3,4,5,6,7,8}標(biāo)準(zhǔn)答案：A知識(shí)點(diǎn)解析：Recallthatthenumberofr-membersubsetsofasetwithnmembersisequalto.SoQuantityAisequalto=70.Similarly,QuantityBisequalto=56.ThusthecorrectanswerisChoiceA.Thisexplanationusesthefollowingstrategy.Strategy1:TranslatefromWordstoanArithmeticorAlgebraicRepresentation二、計(jì)算題(本題共15題，每題1.0分，共15分。)10、InrighttriangleABC,theratioofthelengthsofthetwolegsis2to5.IftheareaoftriangleABCis20,whatisthelengthofthehypotenuse?A、7B、10C、D、E、標(biāo)準(zhǔn)答案：E知識(shí)點(diǎn)解析：TheratioofthelengthsofthelegsofrighttriangleABCis2to5,soyoucanrepresentthelengthsas2xand5x,respectively,wherex>0.Sincetheareaofthetriangleis20,itfollowsthat=20.Solvingthisequationforxgivesx=2.Sothelengthsofthetwolegsare2(2)and2(5),or4and10,respectively.Therefore,bythePythagoreantheorem,thelengthofthehypotenuseis.Thissquarerootcanbesimplifiedasfollows.ThecorrectanswerisChoiceE.Thisexplanationusesthefollowingstrategies.Strategy5:SimplifyanArithmeticorAlgebraicRepresentationStrategy8:SearchforaMathematicalRelationship11、Accordingtosurveysatacompany,20percentoftheemployeesownedcellphonesin1994,and60percentoftheemployeesownedcellphonesin1998.From1994to1998,whatwasthepercentincreaseinthefractionofemployeeswhoownedcellphones?A、3%B、20%C、30%D、200%E、300%標(biāo)準(zhǔn)答案：D知識(shí)點(diǎn)解析：From1994to1998,thepercentofemployeeswhoownedcellphonesincreasedfrom20%to60%.Thusthepercentincreaseinthefractionofemployeeswhoownedcellphoneswas(100%),or200%.ThecorrectanswerisChoiceD.Thisexplanationusesthefollowingstrategy.Strategy1:TranslatefromWordstoanArithmeticorAlgebraicRepresentation12、Iftisanintegerand8m=16t,whichofthefollowingexpressesmintermsoft?A、2tB、2t-3C、23(t-3)D、24t-3E、24(t-3)標(biāo)準(zhǔn)答案：D知識(shí)點(diǎn)解析：Notethatallofthechoicesareexpressionsoftheform2raisedtoapower.Expressing8and16aspowersof2,youcanrewritethegivenequation8m=16tas23m=(24)t,whichisthesameas23m=24t.Solvingformandusingtherulesofexponents,yougetm==24t-3.ThecorrectanswerisChoiceD.Thisexplanationusesthefollowingstrategies.Strategy5:SimplifyanArithmeticorAlgebraicRepresentationStrategy8:SearchforaMathematicalRelationship13、Threepumps,P,R,andT,workingsimultaneouslyattheirrespectiveconstantrates,canfillatankin5hours.PumpsPandR,workingsimultaneouslyattheirrespectiveconstantrates,canfillthetankin7hours.HowmanyhourswillittakepumpT,workingaloneatitsconstantrate,tofillthetank?A、1.7B、10.0C、15.0D、17.5E、30.0標(biāo)準(zhǔn)答案：D知識(shí)點(diǎn)解析：Workingsimultaneously,pumpsPandRfill1/7ofthetankin1hour.Workingsimultaneously,thethreepumpsfill1/5ofthetankin1hour.Therefore,workingalone,pumpTfills,or2/35,ofthetankin1hour.Thus,workingalone,pumpTtakes35/2hours,or17.5hours,tofillthetank.ThecorrectanswerisChoiceD.Thisexplanationusesthefollowingstrategies.Strategy1:TranslatefromWordstoanArithmeticorAlgebraicRepresentationStrategy8:SearchforaMathematicalRelationship14、Theperimeterofaflatrectangularlawnis42meters.Thewidthofthelawnis75percentofitslength.Whatistheareaofthelawn,insquaremeters?A、40.5B、96C、108D、192E、432標(biāo)準(zhǔn)答案：C知識(shí)點(diǎn)解析：Letsandtbethewidthandlength,inmeters,ofthelawn,respectively.Thentheperimeteris2s+2tmeters,sothat2s+2t=42.Also,therelationshipbetweenthewidthandthelengthcanbetranslatedass=0.75t.Substituting0.75tfor5intheequationfortheperimeteryields42=2(0.75t)+2t=3.5t.Sot==12ands=(0.75)(12)=9.Sincest=108,theareaofthelawnis108squaremeters.ThecorrectanswerisChoiceC.Thisexplanationusesthefollowingstrategies.Strategy1:TranslatefromWordstoanArithmeticorAlgebraicRepresentationStrategy8:SearchforaMathematicalRelationship15、Thegreatestofthe21positiveintegersinacertainlistis16.Themedianofthe21integersis10.Whatistheleastpossibleaverage(arithmeticmean)ofthe21integers?A、4B、5C、6D、7E、8標(biāo)準(zhǔn)答案：C知識(shí)點(diǎn)解析：Youaregiventhatthemedianofthe21positiveintegersis10andthegreatestofthe21integersis16.Thismeansthatwhenthe21integersarelistedinorderfromleasttogreatest,the1stthrough10thintegersarebetween1and10,inclusive;the11thintegeris10;the12ththrough20thintegersarebetween10and16,inclusive;andthe21stintegeris16.Theleastpossibleaverageofthe21integerswouldbeachievedbyusingtheleastpossiblevalueofeachintegerasdescribedaboveinthereorderedlist:the1stthrough10thintegerswouldeachbe1;the11thintegerwouldbe10;the12ththrough20thintegerswouldeachbe10;andthe21stintegerwouldbe16.Fortheleastpossibleintegers,thesumis(10)(1)+10+(9)(10)+16,or126.Thereforetheleastpossibleaverageis126/21,or6.ThecorrectanswerisChoiceC.Thisexplanationusesthefollowingstrategies.Strategy1:TranslatefromWordstoanArithmeticorAlgebraicRepresentationStrategy8:SearchforaMathematicalRelationship16、Itxandyareintegersandx=,whichorthefollowingcouldbethevalueofy?A、15B、28C、38D、64E、143標(biāo)準(zhǔn)答案：B知識(shí)點(diǎn)解析：Tosimplifytheequationx=,dividethenumeratoranddenominatorofthefractionby39togetx=.Fromthesimplifiedequation,youcanseethatxisanintegerifandonlyifyisafactorof(2)(4)(5)(7)(11).Toanswerthequestion,youneedtocheckeachoftheanswerchoicesuntilyoufindtheonethatisafactorof(2)(4)(5)(7)(11).ChoiceA:15.Since15=(3)(5)and3isnotafactorof(2)(4)(5)(7)(11),neitheris15.ChoiceB:28.Since28=(4)(7)andboth4and7arefactorsof(2)(4)(5)(7)(11),sois28.Youcanchecktheotherchoicestoconfirmthatnoneofthemisafactorof(2)(4)(5)(7)(11).ThecorrectanswerisChoiceB.Thisexplanationusesthefollowingstrategies.Strategy5:SimplifyanArithmeticorAlgebraicRepresentationStrategy8:SearchforaMathematicalRelationship17、Ofthe40specimensofbacteriainadish,3specimenshaveacertaintrait.If5specimensaretobeselectedfromthedishatrandomandwithoutreplacement,whichofthefollowingrepresentstheprobabilitythatonly1ofthe5specimensselectedwillhavethetrait?A、

B、

C、

D、

E、

標(biāo)準(zhǔn)答案：E知識(shí)點(diǎn)解析：Inthecontextofthisproblem,representsthenumberofwaysrspecimenscanbeselectedwithoutreplacementfromnspecimens.Theprobabilitythatonly1ofthe5specimensselectedfromthe40specimenswillhavethetraitisequaltonumberofwaystoselect5specimens,only1ofwhichhasthetraitnumberofwaystoselect5specimensThenumberofways5specimenscanbeselectedfromthe40specimensis.Toselect5specimens,only1ofwhichhasthetrait,youhavetoselect1ofthe3specimensthathavethetraitandselect4ofthe37specimensthatdonothavethetrait.Thenumberofsuchselectionsistheproduct.Sotheprobabilitythatonly1ofthe5specimensselectedwillhavethetraitisrepresentedby.ThecorrectanswerisChoiceE.Thisexplanationusesthefollowingstrategies.Strategy1:TranslatefromWordstoanArithmeticorAlgebraicRepresentationStrategy8:SearchforaMathematicalRelationship18、Twodifferentpositiveintegersxandyareselectedfromtheoddintegersthatarelessthan10.Ifz=x+yandzislessthan10,whichofthefollowingintegerscouldbethesumofx,y,andz?Indicateallsuchintegers.A8B9C10D12E14F15G16H18標(biāo)準(zhǔn)答案：A，D，G知識(shí)點(diǎn)解析：Theonlypairsofpositiveoddintegersxandythatarelessthan10andsatisfytheconditionx+y<10arethepair1and3,thepair1and5,thepair1and7,andthepair3and5.Sincez=x+y,itfollowsthatthesumofx,y>andzisequalto2z.Thesumforeachofthefourpossiblepairsisfoundasfollows.1and3:z=4,andthesumofx,y,andzis2z,or8.1and5:z=6,andthesumofx,y,andzis12.1and7:z-8,andthesumofx,y,andzis16.3and5:z=8,andthesumofx,y,andzis16.Thustheonlypossiblevaluesofthesumofx,y,andzare8,12,and16.ThecorrectanswerconsistsofChoicesA,D,andG.Thisexplanationusesthefollowingstrategies.Strategy1:TranslatefromWordstoanArithmeticorAlgebraicRepresentationStrategy8:SearchforaMathematicalRelationshipStrategy11:DivideintoCases19、Foracertainprobabilityexperiment,theprobabilitythateventAwilloccuris1/2andtheprobabilitythateventBwilloccuris1/3.WhichofthefollowingvaluescouldbetheprobabilitythattheeventAB(thatis,theeventAorB,orboth)willoccur?Indicateallsuchvalues.A、

B、

C、

標(biāo)準(zhǔn)答案：B，C知識(shí)點(diǎn)解析：SinceyouknowthattheprobabilityofeventAis1/2andtheprobabilityofeventBis1/3butyouarenotgivenanyinformationabouttherelationshipbetweeneventsAandB,youcancomputeonlytheminimumpossiblevalueandthemaximumpossiblevalueoftheprobabilityoftheeventAB.TheprobabilityofABisleastifBisasubsetofA;inthatcase,theprobabilityofABisjusttheprobabilityofA,or1/2.TheprobabilityofABisgreatestifAandBdonotintersectatall;inthatcase,theprobabilityofABisthesumoftheprobabilitiesofAandB,or.WithnofurtherinformationaboutAandB,theprobabilitythatAorB,orboth,willoccurcouldbeanynumberfrom1/2to5/6.Oftheanswerchoicesgiven,only1/2and3/4areinthisinterval.ThecorrectanswerconsistsofChoicesBandC.Thisexplanationusesthefollowingstrategies.Strategy8:SearchforaMathematicalRelationshipStrategy11:DivideintoCases20、Ifaandbarethetwosolutionsoftheequationx2-5x+4=0,whatisthevalueof?Giveyouranswerasafraction.標(biāo)準(zhǔn)答案：5/2知識(shí)點(diǎn)解析：Factoringthequadraticequationx2-5x+4=0,youget(x-1)(x-4)=0,sothetwosolutionsareeithera=1andfr=4ora=4andb=1.Notethataandbareinterchangeableintheexpressionsothevalueoftheexpressionwillbethesameregardlessofthechoicesofaandb.Thusandthecorrectansweris5/2.Thisexplanationusesthefollowingstrategies.Strategy5:SimplifyanArithmeticorAlgebraicRepresentationStrategy8:SearchforaMathematicalRelationship21、From2011to2012,Jack’sannualsalaryincreasedby10percentandArnie’sann