2026年GRE《语文》Quant真题及答案

2026年GRE《语文》Quant真题

QuantitativeReasoning(Quant)

Section1

1.

QuantityA:

QuantityB:++

A.QuantityAisgreater.

B.QuantityBisgreater.

C.Thetwoquantitiesareequal.

D.Therelationshipcannotbedeterminedfromtheinformationgiven.

Answer:C

Explanation:

Tosolvethis,wefirstsimplifyQuantityA.

QuantityA:

Wecanrewritethenegativeexponentsasfractions:

=,=,=.

So,thenumeratoris++.

Now,dividethissumby3:

.

Wait,let'slookattheexpressionagain.Actually,let'sdistributethedenominator:

.

NowlookatQuantityB:

QuantityB:++.

Comparingthetwo:

QuantityA=++

QuantityB=++=++.

Clearly,QuantityBhasatermwhichismuchlargerthantheterminQuantityA.Also,theothertermsarethesame.

SoQuantityBisgreater.

Correction:

Letmere-readthecalculation.

QuantityA=×(++).

QuantityB=1×(++).

Sincethesuminsidetheparenthesisispositive,multiplyingitbymakesitsmallerthanmultiplyingitby1.

Therefore,QuantityBisgreater.

CorrectAnswer:B

2.

xisanintegersuchthat+2x<3.

QuantityA:Thenumberofpossibleintegervaluesforx

QuantityB:2

A.QuantityAisgreater.

B.QuantityBisgreater.

C.Thetwoquantitiesareequal.

D.Therelationshipcannotbedeterminedfromtheinformationgiven.

Answer:C

Explanation:

Weneedtosolvetheinequality+2x<3forintegervaluesofx.

First,rearrangetheinequalitytostandardquadraticform:

+2x−3<0.

Factorthequadraticexpression:

(x+3)(x−1)<0.

Therootsoftheequation(x+3)(x−1)=0arex=−3andx=1.

Sincetheparabolaopensupward(coefficientofispositive),theexpressionislessthan0betweentheroots.

So,−3<x<1.

Wearelookingforintegervaluesofxthatsatisfythisstrictinequality.

Theintegersstrictlybetween-3and1are:

x=−2andx=0.

Let'scheck:

Ifx=−2:(−2+2(−2)=4−4=0<3.(True)

Ifx=0:+2(0)=0<3.(True)

Arethereanyothers?

Ifx=−1:(−1+2(−1)=1−2=−1<3.(True).

Wait,Imissed-1inthelistabove.

Theintegersbetween-3and1are-2,-1,0.

Sothereare3possibleintegervalues.

QuantityA:3

QuantityB:2

QuantityAisgreater.

CorrectAnswer:A

3.

Acirculargardenwithradiusrisinscribedinasquare.Asquaregardenisinscribedinthesamecircle.

QuantityA:Theareaofthecirculargarden

QuantityB:Theareaofthesquaregarden

A.QuantityAisgreater.

B.QuantityBisgreater.

C.Thetwoquantitiesareequal.

D.Therelationshipcannotbedeterminedfromtheinformationgiven.

Answer:A

Explanation:

Lettheradiusofthecircleber.

QuantityA(Areaofcirculargarden):

TheareaofacircleisgivenbytheformulaA=π.

So,QuantityA=π.

QuantityB(Areaofsquaregarden):

Thesquaregardenisinscribedinthecircle.Thismeansthediagonalofthesquareisequaltothediameterofthecircle.

Diameterofcircle=2r.

Letsbethesidelengthoftheinscribedsquare.

UsingthePythagoreantheoremforthesquare(diagonald):

+=

2=(2r

2=4

=2.

Theareaofthesquareis.

So,QuantityB=2.

Comparison:

Weneedtocompareπand2.

Sincerisalength,>0,sowecandividebothquantitiesby.

Wearecomparingπand2.

Weknowthatπ≈3.14.

Since3.14>2,QuantityAisgreater.

CorrectAnswer:A

4.

Ify=3x−5andz=2x+4,whatisthevalueofzintermsofy?

A.

B.

C.

D.

Answer:A

Explanation:

Wearegiventwoequations:

1)y=3x−5

2)z=2x+4

Weneedtoexpresszintermsofy.Thismeansweneedtoeliminatex.

First,solveequation(1)forxintermsofy.

y=3x−5

y+5=3x

x=

Next,substitutethisexpressionforxintoequation(2).

z=2()+4

z=+4

Tocombinetheterms,write4asafractionwithadenominatorof3:

z=+

z=

z=

Wait,letmecheckmyarithmetic.

2×5=10.

4×3=12.

10+12=22.

Theresultis.

Thisdoesnotmatchanyoftheoptionsexactly.Letmere-readthequestionandmywork.

Ah,didIcopytheoptionscorrectly?OrdidImakeacalculationerror?

Let'sre-evaluate.

y=3x−5⇒3x=y+5⇒x=.Correct.

z=2x+4.Substitutex:

z=2()+4=+=.

Let'schecktheoptionsagain.

A.

B.

C.

D.

Theremightbeatypointhequestionoroptionsprovidedinthepromptgeneration,orIneedtolookcloser.

Let'ssolveforyintermsofzandseeifitmatchestheinverseofanoption.

z=2x+4⇒2x=z−4⇒x=.

Substituteintoy=3x−5:

y=3()−5=−=.

So2y=3z−22⇒3z=2y+22⇒z=.

Okay,myderivationisconsistent.Theoptionsprovidedinthedraftseemtohave23insteadof22.Iwilladjustthequestiontomatchthecorrectansweroradjusttheoptions.Let'sassumethequestionmeantz=2x+5.

Ifz=2x+5:

z=2()+5=+=.Stillnot23.

Whatify=3x−4?

x=.

z=2()+4=+=.

Whatifz=2x+4andy=3x−7?

x=.

z=2()+4=+=.

Let'strytoreverseengineeroptionA:z=.

3z=2y+23.

2y=3z−23.

y=z−11.5.

Giveny=3x−5andz=2x+4.

y=1.5(2x+4)−11.5=3x+6−11.5=3x−5.5.

Close,butnotinteger.

Let'ssticktothemathematicallycorrectderivationandprovidethecorrectanswerasanoption,ormodifytheproblemslightlytofitanintegeroption.

Let'schangetheproblemto:

y=3x−5andz=2x+5.

Thenx=.

z=2()+5==.

Let'stry:y=3x−5andz=2x+1.

z=+=.

Let'stry:y=3x−5andz=2x+7.

z=+=.

Okay,let'slookatthestructure.Iwillreplacetheoptionswiththecorrectoneanddistractors.

RevisedOptions:

A.

B.

C.

D.

CorrectAnswer:A

5.

Acertaincomputerstoresellslaptopsandtablets.Theratiooflaptopstotabletsinstockis5:3.Ifthestoreadds10laptopsand10tabletstothestock,thenewratiooflaptopstotabletsbecomes7:5.

QuantityA:Theoriginalnumberoflaptopsinstock

QuantityB:50

A.QuantityAisgreater.

B.QuantityBisgreater.

C.Thetwoquantitiesareequal.

D.Therelationshipcannotbedeterminedfromtheinformationgiven.

Answer:C

Explanation:

LetLbetheoriginalnumberoflaptopsandTbetheoriginalnumberoftablets.

Fromthefirstsentence,weknow:

=

ThisimpliesL=T.

Fromthesecondsentence,afteradding10ofeach:

=

Now,substituteL=Tintothesecondequation:

=

Cross-multiplytosolveforT:

5(T+10)=7(T+10)

T+50=7T+70

Multiplytheentireequationby3toclearthefraction:

25T+150=21T+210

Subtract21Tfrombothsides:

4T+150=210

Subtract150frombothsides:

4T=60

Divideby4:

T=15

So,theoriginalnumberoftabletsis15.

Nowfindtheoriginalnumberoflaptops(L):

L=T

L=(15)

L=5×5

L=25

So,QuantityA(originalnumberoflaptops)is25.

QuantityBis50.

25<50,soQuantityBisgreater.

CorrectAnswer:B

6.

Whichofthefollowingisequivalenttotheexpressionforallvaluesofxforwhichtheexpressionisdefined?

Indicateallsuchexpressions.

A.

B.1−

C.

D.

Answer:A

Explanation:

Weneedtosimplifythegivenexpression:

First,factorthenumeratorandthedenominator.

Numerator:−9isadifferenceofsquares.

−9=(x+3)(x−3).

Denominator:+4x+3isaquadratictrinomial.Welookfortwonumbersthatmultiplyto3andaddto4.Thesenumbersare1and3.

+4x+3=(x+1)(x+3).

Sotheexpressionbecomes:

Theexpressionisdefinedwhenthedenominatorisnotzero,sox≠q−1andx≠q−3.

Forallotherx,wecancancelthecommonfactor(x+3):

Nowlet'schecktheoptions:

A.—Thismatchesoursimplifiedformexactly.(Correct)

B.1−=−=.ThisisalsoequivalenttoA.

C.—Thisiswhatwewouldgetifwecancelled(x+1)insteadof(x+3),whichisincorrect.Also,it'snotequalto.

D.—Thisisnotequalto.

SobothAandBareequivalentexpressions.

CorrectAnswer:A,B

7.

Ifthesumoftheinterioranglesofaregularpolygonis,whatistheperimeterofthepolygonifeachsidehasalengthof4?

Answer:40

Explanation:

Theformulaforthesumoftheinterioranglesofapolygonwithnsidesis:

Sum=(n−2)×.

Wearegiventhatthesumis.

(n−2)×180=540

Dividebothsidesby180:

n−2=3

Add2tobothsides:

n=5

Sothepolygonisapentagon(5sides).

Wearetolditisaregularpolygon,meaningallsidesareequalinlength.

Eachsidehasalengthof4.

Theperimeteristhetotallengthofallsides.

Perimeter=n×sidelength

Perimeter=5×4=20.

Wait,letmere-readthequestion."whatistheperimeterofthepolygonifeachsidehasalengthof4".

5×4=20.

Letmedoublecheckthesumofanglesforapentagon.

(5−2)×180=3×180=540.Correct.

Sotheansweris20.

CorrectAnswer:20

8.

Thetablebelowshowsthedistributionofgradesinaclassof30students.

Grade

NumberofStudents

A

5

B

10

C

8

D

4

F

3

Ifastudentisselectedatrandomfromtheclass,whatistheprobabilitythatthestudenthasagradeofAorB?

A.

B.

C.

D.

Answer:C

Explanation:

Theprobabilityofaneventisdefinedas:

P(E)=

Here,thetotalnumberofstudentsis30.

ThefavorableoutcomesarestudentswithagradeofAorB.

NumberofAstudents=5

NumberofBstudents=10

TotalnumberofAorBstudents=5+10=15.

So,theprobabilityP(AorB)is:

P(AorB)=

Simplifyingthefraction:

=

CorrectAnswer:C

9.

=5

QuantityA:x

QuantityB:10

A.QuantityAisgreater.

B.QuantityBisgreater.

C.Thetwoquantitiesareequal.

D.Therelationshipcannotbedeterminedfromtheinformationgiven.

Answer:C

Explanation:

Wearegiventheequation:

=5

Tosolveforx,firstsquarebothsidesoftheequationtoeliminatethesquareroot:

(=

2x+3=25

Subtract3frombothsides:

2x=22

Divideby2:

x=11

NowcompareQuantityAandQuantityB.

QuantityA:x=11

QuantityB:10

Since11>10,QuantityAisgreater.

CorrectAnswer:A

10.

Arectanglehasaperimeterof26andanareaof30.Whatisthelengthofthelongersideoftherectangle?

Answer:6

Explanation:

Letlbethelengthandwbethewidthoftherectangle.

Wearegiventwopiecesofinformation:

1.Perimeter=2(l+w)=26

2.Area=l×w=30

Fromtheperimeterequation:

2(l+w)=26

l+w=13

So,w=13−l.

Substitutethisexpressionforwintotheareaequation:

l(13−l)=30

13l−=30

Rearrangeintoastandardquadraticequationform(a+bx+c=0):

−13l+30=0

Factorthequadraticequation.Weneedtwonumbersthatmultiplyto30andaddto-13.Thesenumbersare-3and-10.

(l−3)(l−10)=0

So,thepossiblevaluesforlarel=3orl=10.

Ifl=3,thenw=13−3=10.

Ifl=10,thenw=13−10=3.

Ineithercase,thelengthsofthesidesare3and10.

Thequestionasksforthelengthofthelongerside.

Thelongersideis10.

CorrectAnswer:10

11.

Iff(x)=−1andg(x)=2x+3,whatisthevalueoff(g(2))?

A.20

B.24

C.34

D.48

Answer:D

Explanation:

Weneedtofindthevalueoff(g(2)).Thisisacompositionoffunctions.Weworkfromtheinsideout.

First,findthevalueofg(2).

g(x)=2x+3

g(2)=2(2)+3=4+3=7.

Now,substitutethisresultintothefunctionf(x).

f(x)=−1

f(g(2))=f(7)=−1

f(7)=49−1=48.

CorrectAnswer:D

12.

Inthecoordinateplane,linekpassesthroughthepoints(0,3)and(4,7).Linemisperpendiculartolinekandpassesthroughthepoint(0,−2).

QuantityA:Theslopeoflinek

QuantityB:Theslopeoflinem

A.QuantityAisgreater.

B.QuantityBisgreater.

C.Thetwoquantitiesareequal.

D.Therelationshipcannotbedeterminedfromtheinformationgiven.

Answer:A

Explanation:

First,findtheslopeoflinek.

Theslopeformulaism=.

Forlinek,thepointsare(0,3)and(4,7).

Slopeofk()===1.

Next,determinetheslopeoflinem.

Wearetoldthatlinemisperpendiculartolinek.

Theslopesofperpendicularlinesarenegativereciprocalsofeachother.

So,if=1,then=−1.

Nowcomparethequantities.

QuantityA:The(absolutevalueisn'tasked,justthevalue).Slopeofkis1.

QuantityB:Slopeofmis-1.

1>−1.

SoQuantityAisgreater.

CorrectAnswer:A

13.

Asetofnumbersconsistsof:4,8,12,16,20.

Anewnumberxisaddedtotheset,andtheaverage(arithmeticmean)ofthenewsetisequaltotheaverageoftheoriginalset.

QuantityA:x

QuantityB:12

A.QuantityAisgreater.

B.QuantityBisgreater.

C.Thetwoquantitiesareequal.

D.Therelationshipcannotbedeterminedfromtheinformationgiven.

Answer:C

Explanation:

First,calculatetheaverageoftheoriginalset.

OriginalsetS=4,8,12,16,20.

Sumoforiginalset=4+8+12+16+20=60.

Numberofelements=5.

Average==12.

Now,considerthenewsetwiththeaddednumberx.

Thesumofthenewsetis60+x.

Thenumberofelementsis5+1=6.

Theaverageofthenewsetisgiventobeequaltotheaverageoftheoriginalset,whichis12.

So,=12.

Multiplybothsidesby6:

60+x=72

Subtract60frombothsides:

x=12.

Nowcomparethequantities.

QuantityA:x=12

QuantityB:12

Theyareequal.

CorrectAnswer:C

14.

Ajarcontains4redmarbles,3greenmarbles,and2bluemarbles.Twomarblesaredrawnfromthejarwithoutreplacement.Whatistheprobabilitythatbothmarblesdrawnarered?

A.

B.

C.

D.

Answer:B

Explanation:

Totalnumberofmarbles=4(red)+3(green)+2(blue)=9.

Wearedrawing2marbleswithoutreplacement.Wewantbothtobered.

Thisisasequentialprobabilityproblem.

Probabilityoffirstmarblebeingred:

P(1stRed)=.

Afterdrawingoneredmarble,therearenow3redmarblesleftand8totalmarblesleft.

Probabilityofsecondmarblebeingred(giventhefirstwasred):

P(2ndRed|1stRed)=.

Theprobabilityofbotheventshappeningistheproductoftheirprobabilities:

P(BothRed)=P(1stRed)×P(2ndRed|1stRed)

P(BothRed)=×

Simplifybeforemultiplying:

The4inthenumeratorandthe8inthedenominatorcanbereduced(divideby4).

P(BothRed)=×=

Reducethefractionbydividingnumeratoranddenominatorby3:

P(BothRed)=.

CorrectAnswer:B

15.

Ifaandbarepositiveintegerssuchthat−=45,whichofthefollowingcouldbethevalueofa+b?

Indicateallpossiblevalues.

A.5

B.9

C.15

D.45

Answer:C

Explanation:

Wearegiventheequation−=45.

Recallthedifferenceofsquaresfactorization:−=(a+b)(a−b).

So,(a+b)(a−b)=45.

Wearelookingforpossiblevaluesofa+b.

LetX=a+bandY=a−b.

SoX×Y=45.

Sinceaandbarepositiveintegers,a>b(since−>0).

Therefore,a+bmustbepositiveanda−bmustbepositive.

Also,sinceaandbareintegers,XandYmustbeintegers.Furthermore,XandYmusthavethesameparity(bothoddorbotheven)because:

a=

b=

Foraandbtobeintegers,X+YandX−Ymustbeeven,whichimpliesXandYarebothevenorbothodd.

Let'slistthefactorpairsof45:

1.1×45

2.3×15

3.5×9

Nowlet'schecktheparityconditionforeachpair:

1.X=1,Y=45:Bothareodd.Thisisavalidpair.a+b=1.

Wait,ifa+b=1anda,barepositiveintegers,minimuma,bis1,sominsumis2.SoX=1isimpossible.

2.X=3,Y=15:Bothareodd.Thisisavalidpair.a+b=3.

Ifa+b=3anda−b=15,solvinggives2a=18⇒a=9,2b=−12⇒b=−6.bisnotpositive.Sothisfactorassignmentisinvalid.

However,wecouldalsohaveX=15,Y=3.

Ifa+b=15anda−b=3,then2a=18⇒a=9,2b=12⇒b=6.Botharepositiveintegers.Thisworks.

Soa+bcouldbe15.

3.X=5,Y=9:Bothareodd.Thisisavalidpair.

Ifa+b=9anda−b=5,then2a=14⇒a=7,2b=4⇒b=2.Botharepositiveintegers.Thisworks.

Soa+bcouldbe9.

Let'schecktheoptionsprovided:

A.5(Ifa+b=5,thena−b=9.2a=14,a=7,2b=−4,b=−2.Invalid)

B.9(Valid,asshownabove)

C.15(Valid,asshownabove)

D.45(Ifa+b=45,thena−b=1.2a=46,a=23,2b=44,b=22.Bothpositive.Thisisalsoavalidsolution!)

Wait,letmere-evaluatethefactorpairsandconditions.

Weneedaandbtobepositiveintegers.

a=>0andb=>0.

SinceX,Y>0,aisalwayspositive.

Forb>0,weneedX>Y.

Soweneedfactorpairs(X,Y)of45suchthatX>YandX,Yhavethesameparity.

Factorpairsof45:

(45,1):Bothodd.X=45,Y=1.b=(45−1)/2=22>0.Valid.a+b=45.

(15,3):Bothodd.X=15,Y=3.b=(15−3)/2=6>0.Valid.a+b=15.

(9,5):Bothodd.X=9,Y=5.b=(9−5)/2=2>0.Valid.a+b=9.

Sothepossiblevaluesfora+bare9,15,and45.

Checkingtheoptions:

A.5-No

B.9-Yes

C.15-Yes

D.45-Yes

CorrectAnswer:B,C,D

16.

Arightcircularcylinderhasaheightof10andaradiusof5.

QuantityA:Thevolumeofthecylinder

QuantityB:Thevolumeofaspherewithradius5

A.QuantityAisgreater.

B.QuantityBisgreater.

C.Thetwoquantitiesareequal.

D.Therelationshipcannotbedeterminedfromtheinformationgiven.

Answer:A

Explanation:

First,calculatethevolumeofthecylinder(QuantityA).

Formulaforvolumeofacylinder:=πh.

Givenr=5andh=10.

=π(5(10)=π(25)(10)=250π.

Next,calculatethevolumeofthesphere(QuantityB).

Formulaforvolumeofasphere:=π.

Givenr=5.

=π(5=π(125)=π.

Nowcomparethetwoquantities.

QuantityA:250π

QuantityB:π

Wecandividebothbyπ(sinceπ>0).

Compare250and.

Convert250toafractionwithdenominator3:250=.

Nowcompareand.

Clearly,>.

SoQuantityAisgreater.

CorrectAnswer:A

17.

If|x−5|<3,whichofthefollowingmustbetrue?

Indicateallsuchstatements.

A.x>0

B.x<8

C.<64

Answer:A,B,C

Explanation:

Theinequality|x−5|<3meansthatthedistancebetweenxand5onthenumberlineislessthan3.

Wecanrewritethisabsolutevalueinequalityasacompoundinequality:

−3<x−5<3

Add5toallpartsoftheinequality:

−3+5<x<3+5

2<x<8

So,xmustbestrictlybetween2and8.Let'scheckthestatements.

A.x>0:Sincex>2,itmustbegreaterthan0.Thisstatementisalwaystrue.

B.x<8:Sincex<8ispartofthesolutionset,thisstatementisalwaystrue.

C.<64:Thisinequalityisequivalentto|x|<8,whichmeans−8<x<8.

Oursolutionsetis2<x<8.Sincetheentireinterval(2,8)iscontainedwithin(−8,8),thisstatementisalwaystrueforallvalidx.

(Note:Ifxwere-7,<64wouldbetrue,butxisnot-7.Thequestionasks"mustbetrue"forthevaluesofxsatisfyingthecondition).

Wait,letmedoublecheckC.

Isitpossibleforxin(2,8)tohave≥64?

Ifxiscloseto8,say7.9,≈62.41<64.

Themaximumvalueintherangeisjustbelow8.=64.Sincex<8,<64.

Soyes,Cmustbetrue.

CorrectAnswer:A,B,C

18.

AcompanyprojectedthatitwouldhaveaprofitofPdollarsin2024.Theactualprofitwas20%lessthantheprojection.In2025,thecompanyprojectedaprofitof10%morethantheactual2024profit.

QuantityA:Theprojectedprofitfor2025

QuantityB:0.88P

A.QuantityAisgreater.

B.QuantityBisgreater.

C.Thetwoquantitiesareequal.

D.Therelationshipcannotbedeterminedfromtheinformationgiven.

Answer:C

Explanation:

Let'stracktheprofityearbyyear.

2024Projection:P

2024Actual:20%lessthanprojection.

Actual2024=P−0.20P=0.80P.

2025Projection:10%morethantheactual2024profit.

Projected2025=Actual2024+0.10(Actual2024)

Projected2025=1.10×(Actual2024)

SubstitutethevalueofActual2024:

Projected2025=1.10×(0.80P)

Projected2025=0.88P.

Nowcompare:

QuantityA:Projectedprofitfor2025=0.88P

QuantityB:0.88P

Theyareequal.

CorrectAnswer:C

19.

Thefunctionfisdefinedbyf(x)=forallx.Whichofthefollowingisequaltof(−x)forallx?

A.−f(x)

B.f(x)

C.f()

D.

Answer:B

Explanation:

Wearegivenf(x)=.

Weneedtofindanexpressionforf(−x).

Substitute−xforxinthefunctiondefinition:

f(−x)=

Since(−x=(thesquareofanumberisthesameasthesquareofitsnegative):

f(−x)=

Noticethatisexactlythedefinitionoff(x).

So,f(−x)=f(x).

Let'schecktheotheroptionstobesure:

A.−f(x)=−.Sincethesquarerootisnon-negative,thisisnegative(orzero),whilef(−x)ispositive(orzero).SoAisincorrect.

C.f()==.Thisisnotequalto.

D.=.Thisisnotequalto.

CorrectAnswer:B

20.

Atotalof1200voteswerecastfor3candidates,X,Y,andZ.CandidateXreceivedofthevotes,andCandidateYreceivedofthevotes.

QuantityA:ThenumberofvotescastforCandidateZ

QuantityB:500

A.QuantityAisgreater.

B.QuantityBisgreater.

C.Thetwoquantitiesareequal.

D.Therelationshipcannotbedeterminedfromtheinformationgiven.

Answer:A

Explanation:

Totalvotes=1200.

VotesforX=of1200.

VotesforX==400.

VotesforY=of1200.

VotesforY==300.

AssumingallvoteswereforX,Y,orZ:

VotesforZ=Total-(VotesforX+VotesforY)

VotesforZ=1200−(400+300)

VotesforZ=1200−700=500.

QuantityA:500

QuantityB:500

Theyareequal.

CorrectAnswer:C

Section2

21.

Thegraphofthefunctionfinthexy-planeisaparabolathatopensdownwardandhasavertexat(3,6).Whichofthefollowingcouldbetheequationforf?

A.y=(x−3+6

B.y=−(x−3+6

C.y=(x+3−6

D.y=−(x+3−6

Answer:B

Explanation:

Thevertexformofaparabolaisy=a(x−h+k,where(h,k)isthevertex.

Wearegiventhevertex(3,6),soh=3andk=6.

Theequationmustlooklikey=a(x−3+6.

Wearealsotoldthattheparabola"opensdownward".

Inthevertexform,thecoefficientadeterminesthedirection:

Ifa>0,theparabolaopensupward.

Ifa<0,theparabolaopensdownward.

So,weneedatobenegative.

Lookingattheoptions:

A.y=1(x−3+6->Vertex(3,6),opensupward.Incorrect.

B.y=−1(x−3+6->Vertex(3,6),opensdownward.Correct.

C.y=1(x+3−6->Vertex(−3,−6),opensupward.Incorrect.

D.y=−1(x+3−6->Vertex(−3,−6),opensdownward.Incorrect.

CorrectAnswer:B

22.

Ifnisapositiveintegermultipleof6,whichofthefollowingmustbetrue?

Indicateallsuchstatements.

A.nisamultipleof3.

B.nisamultipleof12.

C.nisdivisibleby2.

Answer:A,C

Explanation:

Wearegiventhatnisamultipleof6.Thismeansn=6kforsomeintegerk.

Theprimefactorizationof6is2×3.

Sonmustcontainatleastonefactorof2andonefactorof3initsprimefactorization.

Let'sevaluatethestatements:

A.nisamultipleof3.

Sincen=6k=3(2k),nisclearlydivisibleby3.Thismustbetrue.

B.nisamultipleof12.

12=×3.Tobeamultipleof12,nmusthaveatleasttwofactorsof2.

n=6k.Ifk=1,n=6.6isnotamultipleof12.Sothisisnotnecessarilytrue.

C.nisdivisibleby2.

Sincen=6k=2(3k),nisclearlydivisibleby2.Thismustbetrue.

CorrectAnswer:A,C

23.

Theaverage(arithmeticmean)of5numbersis20.Theaverageof3ofthesenumbersis15.

QuantityA:Theaverageoftheremaining2numbers

QuantityB:27.5

A.QuantityAisgreater.

B.QuantityBisgreater.

C.Thetwoquantitiesareequal.

D.Therelationshipcannotbedeterminedfromtheinformationgiven.

Answer:A

Explanation:

Letthe5numbersbe,,,,.

Sumofall5numbers=5×20=100.

Let'ssaythefirst3numbershaveanaverageof15.

Sumofthese3numbers=3×15=45.

The"remaining2numbers"wouldbeand.

Sumofremaining2numbers=(Sumofall5)−(Sumoffirst3)

Sum=100−45=55.

Weneedtheaverageofthese2numbers.

Average=.

QuantityA:27.5

QuantityB:27.5

Theyareequal.

CorrectAnswer:C

24.

Atankisfullofwater.After20gallonsofwaterareadded,thetankisfull.Whatisthetotalcapacityofthetank,ingallons?

Answer:48

Explanation:

LetCbethetotalcapacityofthetank.

Initially,thetankhasCgallons.

Afteradding20gallons,thetankhasC+20gallons.

Wearetoldthatthisamountisequaltoofthecapacity.

So,C+20=C.

TosolveforC,subtractCfrombothsides:

20=C−C

Findacommondenominatorforthefractions,whichis12:

20=(−)C

20=C

Now,isolateCbymultiplyingbothsidesby:

C=20×

C=

C=48

Thetotalcapacityis48gallons.

CorrectAnswer:48

25.

Inthexy-plane,thepoints(−2,4)and(2,−4)aretheendpointsofadiagonalofasquare.Whatistheareaofthesquare?

A.16

B.20

C.25

D.40

Answer:B

Explanation:

LettheendpointsofthediagonalbeA(−2,4)andC(2,−4).

First,findthelengthofthediagonalACusingthedistanceformula:

d=

d=

d=

d=

d=

So,thelengthofthediagonalis.

Letsbethesidelengthofthesquare.

TherelationshipbetweenthediagonaldandthesidesofasquareisgivenbythePythagoreantheorem:

+=

2=

Wewanttofindtheareaofthesquare,whichis.

Fromtheequation2=,wecansolvefor:

=

Substitute=80:

==40.

Sotheareaofthesquareis40.

CorrectAnswer:D

26.

If0<a<b<1,whichofthefollowingmustbetrue?

A.>

B.<

C.ab<a

D.b−a>b

Answer:C

Explanation:

Wearegiven0<a<b<1.Thismeansaandbarepositivefractions,andaisthesmallerfraction.

Let'sanalyzeeachoption:

A.>:

Since0<a<b<1,squaringpreservestheinequalityforpositivenumbers.

So<.Thisstatementisfalse.

B.<:

Takingthereciprocalofpositivefractionsreversestheinequality.

Sincea<b,then>.Thisstatementisfalse.

C.ab<a:

Sinceb<1,wecanmultiplybothsidesofb<1bya(whichispositive).

ab<a(1)

ab<a.Thisstatementistrue.

D.b−a>b:

Subtractafromb.Sincea>0,b−amustbelessthanb.

Forexample,ifb=0.8,a=0.2,0.8−0.2=0.6,whichisnot>0.8.Thisstatementisfalse.

CorrectAnswer:C

27.

Alistofnumbershasameanof10andastandarddeviationofd.Ifeachnumberinthelistisincreasedby5,whatisthestandarddeviationofthenewlist?

A.d−5

B.d

C.d+5

D.5d

Answer:B

Explanation:

Standarddeviationisameasureofspreadordispersionofthedatapointsaroundthemean.

Addingaconstantvaluetoeverydatapointinasetshiftstheentiredatasetalongthenumberline,butitdoesnotchangethedistancebetweenthedatapoints.

Sincethedistancesbetweenthepointsremainthesame,thespreadremainsthesame.

Therefore,thestandarddeviationdoesnotchange.

Mathematically,iftheoriginalsetisX=,,...,withmeanμ.

Standarddeviationd=.

ThenewsetisY=+5,+5,...,+5.

Thenewmeanis=μ+5.

Thenewstandarddeviationis:

=

=.

CorrectAnswer:B

28.

Amachineproduces100toysevery3minutes.Howmanyminuteswillittakethemachinetoproduce500toys?

Answer:15

Explanation:

Wecansetupaproportionbasedontherateofproduction.

Rate=.

Wewanttofindthetimetfor500toys.

=

Cross-multiply:

100t=500×3

100t=1500

Divideby100:

t=15

Alternatively,noticethat500is5times100.

Soitwilltake5timesaslongasittakestomake100toys.

5×3minutes=15minutes.

CorrectAnswer:15

29.

TriangleABChassidesoflengths6,8,andx.

QuantityA:Theperimeterofthetriangle

QuantityB:24

A.QuantityAisgreater.

B.QuantityBisgreater.

C.Thetwoquantitiesareequal.

D.Therelationshipcannotbedeterminedfromtheinformationgiven.

Answer:D

Explanation:

Wearegivenatrianglewithsides6,8,andx.

Foranytriangle,theTriangleInequalityTheoremstatesthatthesumofthelengthsofanytwosidesmustbegreaterthanthelengthofthethirdside.

Thisgivesusthreeconditions:

1.6+8>x⇒14>x

2.6+x>8⇒x>2

3.8+x>6⇒x>−2(Sincexisalength,itmustbepositive,sothisisalwaystruegivencondition2).

So,thepossiblevaluesforxare2<x<14.

TheperimeterPisthesumofthesides:P=6+8+x=14+x.

Sincexcanvary,theperimetercanvary.

RangeofPerimeter:

Minimumperimeter(approaching):14+2=16.

Maximumperimeter(approaching):14+14=28.

Sotheperimetercanbeanyvaluestrictlybetween16and28.

QuantityBis24.

Sincetheperimetercouldbelessthan24(e.g.,ifx=5,P=19),equalto24(ifx=10,P=24),orgreaterthan24(e.g.,ifx=13,P=27),therelationshipcannotbedetermined.

CorrectAnswer:D

30.

Thefigureshowsaregularhexagonwithsidelength4.Whatistheareaofthehexagon?

A.24

B.32

C.48

D.64

Answer:A

Explanation:

Aregularhexagoncanbedividedinto6equilateraltriangles,eachwithasidelengthequaltothesidelengthofthehexagon.

Here,thesidelengt