新加坡入学考试英语数学

Thisquestionpaperconsistsof15printedpages(excludingthispage)Name:___________________________________Identityno:_________________________Date:__________________SeatNo:_________BCAACADEMYSCHOOLOFBUILDING&DEVELOPMENTSINGAPOREMATHEMATICSSCREENINGTESTSetA1.5HOURSInstructionstocandidates1.Donotturnoverthispageuntilyouaretoldtodoso.2.Checkthatyouhavethecorrectexampaper,numberofpagesandquestions.3.ThispaperconsistsofTEN(10)questions(100marks).AnswerALLquestions4.WriteyourName,ICNO.andSeatNo.onthiscoverpage.5.AllanswersaretobewritteninTHISbooklet.6.DoNOTtearoutanypage.ThisbookletisthepropertyofBCAAcademyandmustnotberemovedfromthetestcentre.7.Mobilephonesaretobeswitchedoffandelectronicequipmentsarenotallowedtobeused.8.Candidatesaretobringtheirownnon-programmablescientificcalculator.Unlessotherwisestated,leaveyouranswersin3significantfigures.Unlessthequestionsrequiretheanswersintermof,thecalculatorvalueforshouldbeused.Ifworkingisneededforanyquestion,itmustbeshownwiththeanswer.Omissionofessentialworkingwillresultinlossofmarks.ForOfficialUse:TestCentre:TestDate:Marks(/100):Marker:Checker:MathematicsScreeningTestBCAAcademyPage11.Simplifythefollowingexpressionsusingfractionsonly.Showyourworkingclearly.(a)01715143(5marks)(b)21)16.0((12.96)21(5marks)(a)01715143=43(b)21)16.0((12.96)21=91MathematicsScreeningTestBCAAcademyPage2MathematicsScreeningTestBCAAcademyPage32.(a)Givenif27yx=81and3yx=27,find922yx.(5marks)(b)()(),findz.(5marks)(a)922yx=6561(b)z=35MathematicsScreeningTestBCAAcademyPage43.(a)Thefollowingfigureshowsaright-angledtriangleanditsdimensions.Findx.(4marks)()+()()X=4737(x+3)cm(2x-1)cm(3x-2)cmMathematicsScreeningTestBCAAcademyPage53.(b)Apencilcasecontains2green,6redand12orangepens.Apenispickedoutatrandomandreplaced.Asecondpenisthenpickedoutatrandomandnotreplaced.Athirdpenisfinallypickedoutatrandom.Findtheprobabilitythatallthreepenspickedareofdifferentcolours.(6marks)P=47554MathematicsScreeningTestBCAAcademyPage64.(a)Thediagramshowsfourpoints,A,B,C,D,onlevelground.ItisgiventhatDisdueNorthofC,ABC=52°,ACD=120°andACB=90°.Find:(i)BCD,and(ii)BearingofBfromA(5marks)(i)BCD=150(ii)eastbynorth18CNorthDAB12052MathematicsScreeningTestBCAAcademyPage74(b)GiventhatAD=BE=12m.BEC=55°andADC=68°.FindlengthAB.(5marks)AB=)55sin68(sin1268°55°ABDECMathematicsScreeningTestBCAAcademyPage85.Inthediagram,PSTisastraightline,QR=21mandRS=62m.TheangleofdepressionofRfromQis23.TheangleofdepressionofTfromRis66.H=(23sin21)mRST=67L=(66sin47sin62)m(a)HowmuchhigherisQthanR?(b)CalculateRST.(c)CalculatethelengthofST.(10marks)21m62m66°23°SRQPTMathematicsScreeningTestBCAAcademyPage96.OABandORSaretwosectorswithcommoncentreOandradiiOAandORrespectively.PQRSarectangleinwhichPQ=18cmandQR=rcm.RSistangenttothearcANBatN,PO=OQandMN=4cm.(10marks)ROS=-arcsin94099360L=m)8)94099360arcsin-(481(S=)94099360arcsin(281Calculate(a)ROSinradians,(b)theperimeteroftheshadedregion,and(c)theshadedarea.Hint:Lengthofarc,S=rθAreaofsector=where,r=radiusθ=angleofthearc(inradian)MathematicsScreeningTestBCAAcademyPage10MathematicsScreeningTestBCAAcademyPage117.Inthediagram,thepointsA,B,CandDlieonacircle.Oisthecentreofthecircle.AOCisadiameterandEDisatangentofthecircle.ADO=34andBAC=28CalculateCDE=34ABD=56BDO=28BCD=118(a)CDE(b)ABD(c)BDO(d)BCD(10marks)ADECBBMathematicsScreeningTestBCAAcademyPage128.(a)Findthevaluesofx,yandzequationofthefollowing.(5marks)=X=25Y=23Z=1831127322zyx5114327xMathematicsScreeningTestBCAAcademyPage138.(b)Giventhat2132nm=55,findthevaluesofmandn.(5marks)M=-5N=5MathematicsScreeningTestBCAAcademyPage149.(a)CB,(b)CE,(c)CF,(d)AF,(e)DE.(10marks)CB=abCE=31(ab)CF=43(ab)AF=ab4347DE=ab3197ABCFEDabInthediagram,AB=a,AC=bandAD=95AC.EandFarepointsonCBsuchthatCE:EB=1:2andCF:FB=3:1.Express,assimplyaspossible,intermsofaandb,MathematicsScreeningTestBCAAcademyPage1510.(a)Simplifylog3125•log5121•log113.(5marks)log3125•log5121•log113=6(b)Giventhatsin40°=a,expressthefollowingintermsofa.(i)tan40°(ii)cos40°(5marks)tan40°=221a1aacos40°=21aMathematicsScreeningTestBCAAcademyPage16(THISPAGEISINTENTIONALLYLEFTBLANK)MathematicsScreeningTestBCAAcademyPage17(THISPAGEISINTENTIONALLYLEFTBLANK)ENDOFPAPER