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SAMPLEQUESTIONPAPER08

Class-X(2017–18)

Mathematics

Timeallowed:3HoursMax.Marks:80

GeneralInstructions:

(i)Allquestionsarecompulsory.

(ii)Thequestionpaperconsistsof30questionsdividedintofoursectionsA,B,CandD.

(iii)SectionAcontains6questionsof1markeach.SectionBcontains6questionsof2marks

each.SectionCcontains10questionsof3markseach.SectionDcontains8questionsof4

markseach.

(iv)Thereisnooverallchoice.However,aninternalchoicehasbeenprovidedinfour

questionsof3markseachandthreequestionsof4markseach.Youhavetoattemptonlyone

ofthealternativesinallsuchquestions.

(v)Useofcalculatorsisnotpermitted.

SECTION–A

1.Thesumandproductofzerosofaquadraticpolynomialare and–7respectively.

Writethepolynomial?

2.CantwopositiveintegershavetheirH.C.FandL.C.Mas12and512respectively?Justify.

3.If ,thendeterminethevalueof .

4.WritetherelationbetweenMean,ModeandMedian.

5.IfthestraightlinejoiningtwopointsP(5,8)andQ(8,k)isparalleltox-axis,thenwritethe

valueofk.

6.AtangentPQatapointPofacircleofradius5cmmeetsalinethroughthecentreOata

pointQsothatOQ=12cm.WritethelengthofPQ.

SECTION–B

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7.The7thtermofanA.P.is–4andits13thtermis–16.Findthesumofitsfirst19terms.

8.Ifthepoints(4,3)and(x,5)lieonthecircumferenceofthecirclewhosecentreis(2,3),

thenfindthevalueofx.

9.Showthat isirrational.

10.InFig-1,ifEF||BCandFG||CD,provethat, .

11.AquadrilateralABCDisdrawntocircumscribeacircle(fig-2).

Provethat,AB+CD=AD+BC.

12.Fromasolidcylinderwhoseheightis2.4cmanddiameter1.4cm,aconicalcavityofthe

sameheightandsamediameterishollowedout(fig-3).Findthetotalsurfaceareaofthe

remainingsolid.

SECTION–C

13.Findtherootsoftheequation3x2–7x–2=0bythemethodofcompletingthesquare.

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14.Solvethepairoflinearequations8x+5y=9and3x+2y=4bycross-.multiplication

method.

15.Povedthatifintwotriangles,sidesofonetriangleareinthesameratioofthesidesofthe

othertriangle,thentheircorrespondinganglesareequal.

16.ProvethatthepointsA(–5,4),B(–1,–2)andC(5,2)aretheverticesofanisoscelesright-

angledtriangle.

Or

TheverticesofatriangleareA(-1,3),B(1,-1)andC(5,1).Findthelengthofthemedian

throughthevertexC.

17.Cardsmarkedwithnumbers3,4,5,…,50areplacedinaboxandmixedthoroughly.One

cardisdrawnatrandomfromthebox.Findtheprobabilitythatnumberonthedrawncard

isatwodigitnumberwhichisaperfectsquare.

Or

Adieisthrownonce.Findtheprobabilityofgetting(i)anevennumber(ii)anumbergreater

than3(iii)acomposiitenumber

18.Adieisthrownonce.Findtheprobabilityofgetting(i)aprimenumber;(ii)anodd

number.

19.Solveforx:

Or

Iftherootsoftheequation(b–c)x2+(c–a)x+(a–b)=0areequal,

thenprovethat2b=a+c.

20.If ,thenprovethat,

21.TwotangentsPAandPBaredrawntoacirclewithcentreOfromanexternalpointP.

Provethat∠APB=2∠OAB.

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Or

AcirclewithcentreO,diameterABandachordADisdrawn.Anothercircleisdrawnwith

AOasdiametertocutADatC.ProvethatBD=2OC.

22.Statethe‘FundamentalTheoremofArithmetic’.UseEuclid’sdivisionalgorithmtofind

theHCFof196and38220.HencefindtheLCMofthesenumbers.

SECTION–D

23.Drawapairoftangentstoacircleofradius5cmwhichareinclinedtoeachotheratan

angleof60°.

Or

DrawalinesegmentABoflength8cm.TakingAascentre,drawacircleofradius4cm.and

constantthepairoftangentsofthecirclefrompointBandmeasuretheirlengths.

24.Checkgraphicallywhetherthepairofequationsx+y=8andx–2y=2isconsistent.Ifso,

solvethemgraphically.Alsofindthecoordinatesofthepointswherethetwolinesmeetthe

y-axis.

25.If and showthat,

Or

If provethat

26.Ifthemedianofthedistributiongivenbelowis28.5,findthevaluesofxandy.Thesumof

allfrequencyis60

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Class-interval 0-10 10-20 20-30 30-40 40-50 50-60

Frequencry 5 x 20 15 y 5

27.Findtheareaoftheshadedregioninfig-5,whereABCDisasquareofside20cm.

28.Theangleofelevationofacloudfromapoint60mabovealakeis30oandtheangleof

depressionofthereflectionofthecloudinthelakeis60o.Findtheheightofthecloud.

29.Ametallicrightcircularcone20cmhighandwhoseverticalangleis60oiscutintotwo

partsatthemiddleofitsheightbyaplaneparalleltoitsbase.Findthevolumeofthefrustum

soobtained.

30.The termandthesumoffirstntermsofanA.Parerespectivelyare and and

.Provethat, .

Or

Findthesumoffirst40positiveintegersdivisibleby6.Alsofindthesumoffirst20positive

integersdivisibleby5or7.

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CBSESAMPLEPAPER08

CLASSX-Mathematics

Solutions

SECTION–A

1.Sumofzeroes=

Productofzeroes=

Coefficientofx2=5andcoefficientofx=1andthetermfreefromx=–35

Thepolynomialis

2.No.Therecannotexisttwonumberssatisfyingthegivencondition,becausehereL.C.M(=

512)isnotdivisiblebyH.C.F(=12).

3. i.e.,

4.3Median=Mode+2Mean

5.k=8

6.

SECTION–B

7.7thterm, .Byquestion,a+6d=–4…(i)

13thterm, .Byquestion,a+12d=–16…(ii)

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Nowsumoffirst19terms

=

[Adding(i)and(ii),weget,2a+18d=–16]

8.Lengthofradius=

i.e.,

i.e.,(x–2)2=0

i.e.,x=2

Hence,valueofx=2.

9.Ifpossible,letusassumethat isrationalandequalsto

i.e., ,whereaandbarepositiveintegersprimetoeachotherandb>1

i.e., …..(i)

From(i),weseethat, isnotaninteger,asaandbareprimetoeachother,so

arealsoprimetoeachother,but3bisaninteger

i.ein(i),afractionequalstoaninteger,whichcontradictsourinitialassumption.

Hence, isirrational.1

10.Since,EF||BC,

Since,FG||CD,

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By(i)and(ii),

11.Since,thelengthsoftangentsdrawnfromanexternalpointtoacircleareequal.

AP=AS…(i)BP=BQ…(ii)

CQ=CR…(iii)DR=DS…(iv)

Now,AB+CD

=AP+PB+CR+RD

=AS+BQ+CQ+DS

=(AS+DS)+(BQ+CQ)

=AD+BC

Henceproved.

12.Heightofthesolidcylinder(h)=2.4cm

Diameterofitsbase(2r)=1.4cm.

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Thereforeitsbaseradius(r)=0.7cm

Heightanddiameteroftheconicalcavityareequaltothoseofthecylinder.

Remainingsurfacearea

=(Curvedsurfacearea(outside)ofcylinder)+(surfaceareaofitsbottom)+(curvedsurface

areaoftheconicalcavity)

= 1

SECTION–C

13.

i.e., [multiplyingbothsidesby3]

i.e.,

i.e.,

i.e.,

i.e.,

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i.e.,

i.e.,

Therefore,therootsare and

14.

Solvingequations(i)and(ii)bycross-multiplicationmethod,weget,

i.e.,

i.e.,

i.e.,

Hencethesolutionsarex=–2,y=5.

15.Let ABCand DEFbetwotrianglessuchthat,

.

Toprovethat,

FromABandACcuttingAP=DE,AQ=DFletusjoinPandQ

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so,

PQ||BC[Ifalinedividesanytwosidesofatriangleinthesameratio,thenthelineis

paralleltothethirdside.]

So, (commonangle)

Therefore,DABCandDDEFareequiangularandsotheircorrespondingsidesareinthesame

ratio.

Hence, i.e.,

So

i.e.,PQ=EF

DEF APQ(S-S-S)

So

Hencethecorrespondinganglesofthetrianglesareequal.

16.Lengthof units,

Lengthof units,

Lengthof units,

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Hereweget,

Hencethetriangleisanisoscelestriangle.

Also,weobservethat

Therefore,byPythagorastheorem,DABCisaright-angledtriangle(rightangleatB)

Hencethetriangleisanisoscelesright-angledtriangle.

17.Accordingtothequestion,cardsaremixedthoroughlyandonecardisdrawnatrandom

fromthebox,sotheeventofdrawingacardisequallyandlikely.

Sincecardsaremarkedwithnumbers3,4,5,…,50,

Sothereare48cards

Herethetotalnumberofpossibleoutcomes=48.

LettheeventofdrawingacardatrandombearingtwodigitperfectsquarenumberbeE.

ThenthenumberofoutcomesfavourabletotheeventE=4

(heretwodigitperfectnumbersare16,25,36,49)

Therefore,

18.Whenadieisthrownonce,thenthenumberoftotaloutcomes=6

LettheEbetheeventofgettingoneprimenumber.

Hereprimenumbersare2,3,5

ThenthenumberoutcomesfavourabletoE=3

Hence,

AgainletFbetheeventofgettinganoddnumber.

Hereoddnumbersare1,3,5

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Hence,

19. (x 1,2,3)

i.e., (givenx 1,2,3)

i.e.,

i.e.,

i.e.,

i.e.,x=0orx=4

Hencethesolutionsarex=0andx=4.

20.Given,

i.e.,

i.e., [multiplyingbothsidesby ]

i.e.,

i.e.,

i.e.,

Henceproved.

21.Accordingtothequestion,fromanoutsidepointPtwotangentsPAandPBaredrawntoa

circlewithcentreO(fig-4).

Toprovethat,∠APB=2∠OAB.

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Sincethelengthsoftangentsdrawnfromanexternalpointtoacircleareequal.

So,PA=PB.i.e., PABisisosceles.

[since,radiusthroughpointofcontactisperpendiculartothetangentatthepointofcontact]

Henceproved.

22.FundamentalTheoremofArithmetic:

Everycompositenumbercanbeexpressed(factorised)asaproductofprimes,andthis

factorisationisunique,apartfromtheorderinwhichtheprimefactorsoccur.

Since867>255,weapplythedivisionlemmato867and255,toget

867=255 3+102

255=102 2+51

102=51 2+0

Theremainderis0(zero)anddivisoris51.

ThereforetheHCFof867and255is51.

HenceLCMof867and255

=

SECTION–D

23.Constructacircleofradius5cm.

LetitscentrebeO.

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Nowconstructoneradius(sayOP)ofthecircle.

AtO,drawanangleof60°,

anddrawaperpendiculartoOPatP,90°.

ExtendtheselinestomeetatT(say).

Now,takingTascentreandaradiusequaltoTP

drawanarcwhichcutsthecircleatQ.JoinT,Q.

HenceTPandTQarethetwotangentstothegivencircleofradius5cm

whichareinclinedtoeachotheratanangleof60°

[Forjustificationoftheconstruction:

i.e., ]

24.

x 0 4 8

y=8–x 8 4 0

Threesolutionsforequation(i)aregiveninthetable:

Threesolutionsforequation(ii)aregiveninthetable:

x 0 2 8

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–1 0 3

DrawingLineAC

DrawingLinePR

PlottingpointsA(0,8),B(4,4)andC(8,0)ongraphpaperthestraightlineACisobtainedas

graphoftheequation

(i)PlottingpointsP(0,–1),Q(2,0)andR(8,3)ongraphpaperthestraightlinePRisobtained

asgraphoftheequation

(ii)Fromthegraph,itisclearthatapointM(6,2)commontoboththelinesACandPR.

Sothepairofequationsisconsistentandthesolutionsoftheequationsarex=6andy=2.

FromthegraphitisseenthatthecoordinatesofthepointswherethelinesACandPRmeets

they-axisare(0,8)and(0,–1)respectively.

25.Given, …(i)

and …(ii)

adding(i)and(ii),weget,

andsubtracting(ii)from(i),weget,

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.Henceproved.

26.

Classinterval Frequency CumulativeFrequency

0-10 5 5

10-20 x 5+x

20-30 20 25+x

30-40 15 40+x

40-50 y 40+x+y

50-60 5 45+x+y

Total 60

Itisgiventhat,n=60

i.e.,45+x+y=60i.e.,x+y=15

Themedianis28.5,whichliesintheclass20-30

So,l=20,f=20,cf=5+x,h=10

Weknow,

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Here,

Thereforey=15–8=7.

27.LetthesquarebeABCDofside20cm.

AreaofthesquareABCD= .

Diameterofeachcircle(infig-5)=

Thereforeradiusofeachcircle=5cm.

Soareaofeachcircle=

Totalareaoffoursquares=

Henceareaoftheshadedregioninthefig-5

= 1

28.LetintheadjacentfigureECbethesurfaceofwaterinthelake.

Aisthepositionoftheobserver.AE=60m.

AlsoletBisthepositionofcloudandDbeitsimageforfig.

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InthelakeandBF=hmetre

So,BC=CD=(h+60)m(seefig)

FC=60m.

Byquestion,

In ABF,

In AFD,

[sinceDF=DC+CF]

By(i)and(ii),

Henceheightofthecloudfromthewatersurfaceofthelake

=BC=60+60m=120m

29.LetADHbeametallicrightcircularcone,whoseheightis20cm.iscutintotwopartsat

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themiddleofitsheightbyaplaneparalleltoitsbase.ThefrustumisEBFHCDE

Giventhat, ,

AC=20cm1

Accordingtoquestion,AB=BC=10cm.

In ABF,

AgaininIn ACH,

ThereforevolumeofthefrustumEBFHCDE=

= [hereH=AC=20cmAB=10cm]

=

Hencetherequiredvolume=

30.LetfirsttermandthecommondifferenceoftheAPbeaanddrespectively.

Then

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Accordingtoquestion,

Thisisanidentity.

Nowputting,n=2n–1andm=2m–1,weget,

Henceproved.