SAT Mathematics Level 2 Practice Test - EXAMPLANET MATH LEVEL2 PRACTICE...SAT Mathematics Level 2 Practice Test There are 50 questions on this test. You have 1 hour (60 minutes) to

SAT Mathematics Level 2 Practice Test

There are 50 questions on this test. You have 1 hour (60 minutes) to

complete it.

1. Themeasuresoftheanglesof�QRSarem∠Q=2x+4,m∠R=

4x−12,andm∠S=3x+8.QR=y+9,RS=2y−7,andQS=3y−13.

Theperimeterof�QRSis

(A)11 (B)20 (C)44 (D)55 (E)68

2. Giveng(x)=xx

5 13 2-+ , g

43-c m=

(A)–1___2 (B)1__

9 (C)1___

11 (D)7___

16 (E)1__

2

3.7 103 81x y =

(A) 3539 xyx (B) 2 3 39x y xy (C) 2 3 33 9x y xy

(D) 2 3 33 3x y xy (E) 3533 xyx

SAT MATh 1 & 2 SubjecT TeST2

–6

–6 1

1

10y

K

H

G

x

6

4. Theverticesof�GHKabovehavecoordinatesG(–3,4),H(1,–3),

andK(2,7).Theequationofthealtitudeto____

HKis

(A)10x+y=–26 (B)10x+y=7 (C)10x+y=27

(D)x+10y=37 (E)x+10y=72

5. Whenthefigureaboveisspunarounditsverticalaxis,thetotalsur-

faceareaofthesolidformedwillbe

(A)144π (B)108π (C)72π (D)36π (E)9π+12

6. Iff(x)=4x2−1andg(x)=8x+7,g f(2)=

(A)15 (B)23 (C)127 (D)345 (E)2115

7. Ifpandqarepositiveintegerswithpq=36,then qp

cannotbe

(A)14

(B)49

(C)1 (D)2 (E)9

SAT MATheMATIcS LeVeL 2 PRAcTIce TeST 3

8.

2 13 4

21

3 12

x

x

+−

−−

=

(A)23

2−xx

(B)2382

−−

xx

(C)2 83 12

xx−

−

(D)2 73 12

xx−

− (E)

2 53 14

xx−

−

9. If 1251 625

a a abab43 3 102

2

=-

+

` ^j h andifaandbdonotequal0,ba =

(A)421

(B)2 (C)7621

(D)4 (E)763

10. In isosceles �KHJ, ___

HJ = 8, NL HJ⊥ , and MP HJ⊥ . If K is

10cmfrombaseHJandKL=.4KH,theareaof�LNHis

(A)4 (B)4.8 (C)6 (D)7.2 (E)16


11. Theequationoftheperpendicularbisectorofthesegmentjoining

A(–9,2)toB(3,–4)is

(A)y−1=–1___2(x−3) (B)y+1=–1___

2(x+3)

(C)y+1=2(x+3) (D)y+3=2(x+1) (E)y−1=2(x−3)

12. TangentTBandsecantTCA aredrawntocircleO.DiameterABis

drawn.IfTC=6andCA=10,thenCB=

(A)2 6 (B)4 6 (C)2 15 (D)10 (E)2 33

13. Letp@q= q ppq

-.(5@3)−(3@5)=

(A)–184 (B)–59 (C)0 (D)59 (E)184


14. Gradesforthetestonproofsdidnotgoaswellastheteacherhad

hoped.Themeangradewas68,themediangradewas64,andthestan-

darddeviationwas12.Theteachercurvesthescorebyraisingeachscore

byatotalof7points.Whichofthefollowingstatementsistrue?

I.Thenewmeanis75.

II.Thenewmedianis71.

III.Thenewstandarddeviationis7.

(A)Ionly (B)IIIonly (C)IandIIonly

(D)I,II,andIII (E)Noneofthestatementsaretrue

15. Asetoftrianglesisformedbyjoiningthemidpointsofthelarger

triangles.Iftheareaof�ABCis128,thentheareaof�DEF,thesmall-

esttriangleformed,is

(A)18

(B)14

(C)12

(D)1 (E)4


16. Thegraphofy=f(x)isshownabove.Whichisthegraphofg(x)=

2f(x−2)+1?

(A)

(B)

(C)

(D)

(E)


17. Thenumberofbacteria,measuredinthousands,inacultureis

modeledbytheequation ( )2.31

3.21

380175

t

t

eb t

e=

+,where t isthenumberof

dayssincetheculturewasformed.Accordingtothismodel,theculture

cansupportamaximumpopulationof

(A)2.17 (B)205 (C)380 (D)760 (E)∞

18. Aspherewithdiameter50cmintersectsaplane14cmfromthe

centerofthesphere.Whatisthenumberofsquarecentimetersinthe

areaofthecircleformed?

(A)49π (B)196π (C)429π (D)576π (E)2304π

19. Giveng(x)=9213

+−

xx

,g(g(x))=

(A)7649

+−xx

(B) 7 1224 79

xx−

+ (C) 10

21 80xx−

+

(D) 7 624 79

xx+

+ (E)

2

2

9 6 14 36 81

x xx x

− +

+ +

20. Theareaof�QED=750.QE=48andQD=52.Tothenearest

degree,whatisthemeasureofthelargestpossibleangleof�QED?

(A)76 (B)77 (C)78 (D)143 (E)145

21. Givenlog3(a)=candlog3(b)=2c,a=

(A)3c (B)c +3 (C)b2 (D) b (E)2b


22. InisoscelestrapezoidWTYH, || ||WH XZ TY ,m∠TWH=120,and

m∠HWE=30.XZpassesthroughpointE,theintersectionofthediago-

nals.IfWH=30,determinetheratioofXZ:TY.

(A)1:2 (B)2:3 (C)3:4 (D)4:5 (E)5:6

23. Thelengthsofthesidesofatriangleare25,29,and34.Tothenear-

esttenthofadegree,themeasureofthelargestangleis

(A)77.6° (B)77.7° (C)87.6° (D)87.7° (E)102.3°

24. Oneoftherootsofaquadraticequationthathasintegralcoeffi-

cientsis i823

54+

−.Whichofthefollowingdescribesthequadratic

equation?

(A)800x2+1280x+737=0 (B)800x2−1280x+737=0

(C)800x2+1280x+287=0 (D)800x2−1280x+287=0

(E)800x2+1280x−287=0

25. Theparametricequationsx=cos(2t)+1andy=3sin(t)+2cor-

respondtoasubsetofthegraph

(A)( ) ( ) 1

92

11 22

=−

+− yx

(B)( ) ( ) 1

92

11 22

=−

−− yx

(C) ( )22922 −=− yx (D) ( )22

922 −−=− yx

(E) ( )2322 −−=− yx


26. A county commissioner will randomly select 5 people to form a

non-partisan committee to look into the issue of county services. If

thereare8Democratsand6Republicanstochoosefrom,whatisthe

probabilitythattheDemocratswillhavethemoremembersthanthe

Republicansonthiscommittee?

(A)6862002

(B)13162002

(C)18762002

(D)268824024

(E)2133624024

27. Giventhevectorsu=[–5,4]andv=[3,–1],|2u−3v|=

(A)[–19,11] (B) 502 (C)[19,11]

(D)2 41 3 10− (E)2 65

28. �ABChasverticesA(–11,4),B(–3,8),andC(3,–10).Thecoordi-

natesofthecenterofthecirclecircumscribedabout�ABCare

(A)(–2,–3) (B)(–3,–2) (C)(3,2)

(D)(2,3) (E)(–1,–1)

29. Eachsideofthebaseofasquarepyramidisreducedby20%.By

whatpercentmusttheheightbeincreasedsothatthevolumeofthenew

pyramidisthesameasthevolumeoftheoriginalpyramid?

(A)20 (B)40 (C)46.875 (D)56.25 (E)71.875

30.

1920cis

182

5cis9

=

(A) i232 +− (B) i232 −− (C) i232 +

(D) 322 i+− (E) 322 i


31. Givenlogb(a)=xandlogb(c)=y, ( )3 452log cba

=

(A)2

4

35

x

y+ (B)xy

645+

(C)203

yx

(D)2x+4y (E)20

23

yx +

32. Theasymptotesofahyperbolahaveequationsy−1= 43 (x+3).

If a focusof thehyperbolahascoordinates (7,1), theequationof the

hyperbolais

(A)2 2( 3) ( 1)

116 9

x y+ −− = (B)

2 2( 1) ( 3)1

9 16y x− +

− =

(C)2 2( 3) ( 1)

164 36

x y+ −− = (D)

2 2( 1) ( 3)1

36 64y x− +

− =

(E)2 2( 3) ( 1)

14 3

x y+ −− =

33. Aninvertedcone(vertexisdown)withheight12inchesandbase

ofradius8inchesisbeingfilledwithwater.Whatistheheightofthe

waterwhentheconeishalffilled?

(A)6 (B)6 43 (C)8 63 (D)9 43 (E)9 63

34. Solvesin(t)=cos(2t)for–4π≤ t ≤–2π.

(A) 5, ,

6 6 2 (B)

5, ,

3 3 2

(C)23 19 5

, ,6 6 2

(D)23 19 5

, ,3 3 2

(E)11 7 5

, ,3 3 2


35. |2x−3||x−1|<2when

(A)15

<x <5 (B)15

<x <2and2<x <5 (C)x <15

orx >5

(D)x <5 (E)x >15

36. 12 32 1 28 1

k k

kk

0 0

=- -3 3

= =

c `m j/ /

(A)–5 (B)5 (C)24 (D)27 (E)30

37. If x y z3 4 2 3- + =

x y z2 8 1 8- =- -

x y z4 6 3 1- - =

thenzyx +−

1 =

(A)225

(B)3031

(C)3130

(D)192

(E)252

38. Which of the following statements is true about the function

3 3( ) 3 2cos

5 10f x x ?

I.Thegraphhasanamplitudeof2.

II.Thegraphisshiftedtotheright2.

III.Thefunctionsatisfiestheequation f f25

631=c cm m.

(A)Ionly (B)IIonly (C)IIIonly

(D)IandIIonly (E)IandIIIonly


39. Allofthefollowingsolvetheequationz5=32iEXCEPT

(A)2cis(π __2) (B)2cis(9π ___

10) (C)2cis(11π ____

10)

(D)2cis(13π ____10

) (E)2cis(17π ____10

)

40. Givena1=4,a2=–2,andan=2an–2−3an–1,whatisthesmallest

valueofnforwhich|an|>1,000,000?

(A)11 (B)12 (C)13 (D)14 (E)15

41. Whichofthefollowingstatements is trueaboutthegraphofthe

function2

(2 3)( 2)(2 1)( )

4 9x x x

f xx

− + −=

−?

I.f(x)=94

hastwosolutions.

II.f(x)=76

hastwosolutions.

III.Therangeofthefunctionisthesetofrealnumbers.

(A)IIIonly (B)IandIIonly (C)IIandIIIonly

(D)IandIIIonly (E)I,II,andIII

42. Giveng(x)=9log8(x−3)−5,g –1(13)=

(A)313

(B)6 (C)61 (D)67 (E)259

43. Isosceles�QRShasdimensionsQR=QS=60andRS=30.The

centroidof�QRS is locatedatpointT.What is thedistance from

TtoQR?

(A)2 15 (B)5215 (C)3 15

(D)7215 (E)5 15


44. Diagonals AC and BD of quadrilateral ABCD are perpendicular.

AD=DC=8,AC=BC=6,m∠ADC=60°.TheareaofABCDis

(A)4 5 +8 3 (B)16 3 (C)32 3

(D)8 5 +16 3 (E)48

45. cos csc x25

41 +- cc mm=

(A)2 8 16

4x x

x+ −

+ (B)

( )

2

2

8 16

4

x x

x

+ −

+ (C)

2 8 344

x xx+ −

+

(D)( )

2

2

8 34

4

x x

x

+ −

+ (E)

( )

2

2

16 8

4

x x

x

− − −

+

46. Which of the following statements is true about the expression

(a+b)n−(a−b)n?

I.Ithas2n termsifniseven.

II.Ithas21+n termsifnisodd.

III.Theexponentonthelasttermisalwaysn.

(A)Ionly (B)IIonly (C)IandIIonly

(D)IandIIIonly (E)IIandIIIonly

47. In�VWX,sin(X)= 178 andcos(W)=–3___

5.Findcos V

2c m=

(A) 8577- (B)

852- (C)

852 (D)

859 (E) 85

77


48. The intersection of the hyperbola 19)1(

8)1( 22

=+ yx –– and the

ellipse2 2( 1) ( 1)

132 18

x y+ −+ = is

(A)(3,4),(3,–4),(–5,4),(–5,–4)

(B)(3,2),(3,–2),(–5,2),(–5,–2)

(C)(3,4),(3,–2),(–5,4),(–5,–2)

(D)(–3,4),(–3,–2),(5,4),(5,–2)

(E)(3,–4),(3,2),(–5,–4),(–5,2)

49. Theequation8x6+72x5+bx4+cx3–687x2–2160x–1700=0,as

showninthefigure,hastwocomplexroots.Theproductofthesecom-

plexrootsis

(A)–4 (B)17___2 (C)9 (D)–687_____

2 (E)425____

2

50. If sin(A) = –8___17

, 3π ___2 < A < 2π, and cos(B) 0 = –24____

25, π < B < 3π ___

2,

cos(2A+B)=

(A)5544

7225−

(B)2184

7225−

(C)3696

7225−

(D)21847225

(E)55447225


Level 2 Practice Test Solutions1. (D)Thesumof themeasuresof theanglesofa triangle is180, so

2x+4+4x−12+3x+8=180.Combineand solve:9x=180,or

x=20.m∠Q=44,m∠R=68,andm∠S=68.�QRSisisosceles,and

QR=QS.Solvey+9=3y−13togety=11.Thesumofthesidesis

6y−11,sotheperimeteris6(11)−11=55.

2.(B)g 43

5 43 1

3 43 2

5 43 1

3 43 2

191

44

15 49 8

191- =

- -

- +=

- -

- +=- -- + =

-- =

J

L

KKKK

``

`

`

`N

P

OOOO

jj

j

j

j.

3.(D) 7 103 81x y = x y xy27 36 93 3^ ^h h= 2 3 33 3x y xy.

4.(B)TheslopeofHKis10,sotheslopeofthealtitudeis 101- .Instan-

dardform,thismakestheequation10x+y=C.Substitutethecoordi-

natesofGtoget10x+y=–26.

5.(B)Thefigureformedwhenthefigureisrotatedaboutthevertical

axisisahemisphere.Thetotalsurfaceareaofthefigureistheareaofthe

hemisphere(2πr2)plustheareaofthecirclethatservesasthebase(πr2).

Withr=6,thetotalsurfaceareais108πsqcm.

6.(C)f(2)=4(2)2−1=4(4)−1=15.Thecompositionoffunctions

g(f(2))=g(15)=8(15)+7=127.

7.(D)Thefactorsof36are1,2,3,4,6,9,12,18,and36.312

=14

,

66=1,and

182= 9 .Theratioofthefactorscannotbe2.


8. (E) Multiply by the common denominator to change the

problem into a simple fraction rather than a complex fraction.

2 5

x

xxx x

xx

x1 3 122

32

41

3 12 24

3 43 4 2 3

3 14--

+-

- -

-= -

-

-=

+

-

J

L

KKKK ^

^c ^N

P

OOOO h

h m h.

9. (A) Rewrite the equation with a common base. 1251 a ab42

=+

` j

625 a ab3 3 102-^ h becomes2 24 3 103 4/35 5

a ab a ab.Settheexponentsequal

to get –3(a2+4ab) = 34 (3a2–10ab). Gather like terms 3

4 ab = 7a2 so

that 421

ab= .

10. (D)___

NLisparalleltothealtitudefromKto___

HJ.Asaconsequence,

�NLH~�KZH,andLN=.6KZandHN=.6HZ.WithLN=6andHN

=.6(4)=2.4,theareaof�NLHis.5(6)(2.4)=7.2.

11. (C)TheslopeofABis4 2 6 1

3 ( 9) 12 2− − − −

= =− −

.Theslopeoftheperpen-

dicularlineis2.ThemidpointofABis(–3,–1),sotheequationofthe

perpendicularbisectorisy+1=2(x+3).


12. (C) TB2 = (TC)(TA), so TB2 = (6)(16) and TB = 2 6 . ∠ACB

is inscribed in a semicircle, so ∠ACB is a right angle. Consequently,

∠TCBisarightangleand�TCBarighttriangle.UsingthePythagorean

Theorem,TC2+CB2=TB2yields36+CB2=96.CB2=60,soCB=

60 2 15= .

13. (A) 5 @ 3 = 3 553

- = 2

125- and 3 @ 5 = 5 335

- = 2

243 .

2125- – 2

243 =–184.

14. (C)Addingaconstanttoasetofdatashiftsthecenterofthedata

butdoesnotalterthespreadofthedata.

15. (C)The segment joining themidpointsof twosidesofa triangle

ishalfaslongasthethirdside.DEistheconsequenceofthe4thsetof

midpoints,soBCDE

21

1614

= =c m .Theratiooftheareasisthesquareof

theratioofcorrespondingsidesso2

1 1 16 256

area FDEarea ABC

= =�

� ( ) .Thearea

of�FDE=128 1

256 256 2area ABC

= =�

.

16. (B)Thefunctiong(x)=2f(x−2)+1movesthegraphoff(x)right

2,stretchesthey-coordinatesfromthex-axisbyafactorof2,andmoves

thegraphup1unit.Usethepoints(0,3),(3,0),and(5,0)fromf(x)to

followthemotions.

17. (C)Theendbehaviorofarationalfunctionisnotimpactedbyany

constantsthatareaddedorsubtractedtoaterm.Consequently,the175

inthedenominatorofb(t)willhavenoimpactonthevalueofb(t)when

the values of t get sufficiently large.Thefunctionb(t)reducesto380whent

issufficientlylarge.(Graphingthisfunctionwithyourgraphingcalculator

givesaveryclearpictureofthemaximumvalueofthefunction.)


18. (C)Theradiusofthesphereis25.Thedistancefromthecenterof

thespheretotheintersectingplaneliesalongtheperpendicular.Usethe

PythagoreanTheoremtogetr2+142=252orr2=429.Theareaofthe

circleformedbytheplaneandsphereisπr2,or429π.

19. (B)g(g(x))=9

92132

192133

xxxx

=9292

992132

192133

xx

xxxx

=

)92(9)13(2)92()13(3

+++xxxx

=9 3 2 9

6 2 18 81x x

x x− − −

− + +=

7 1224 79

xx−

+.

20. (D)TheareaofatriangleisgivenbytheformulaA=12

absin(θ).

Substitutingthegiveninformation,750=12

(48)(52)sin(θ).Solvingthis

equation forQ yields thatQ=37°or143°. Ifm∠Q=37°, then∠E

wouldbethelargestangleofthetriangle,andthatmeasure,foundusing

theLawofCosines,wouldbe78°,lessthan143°.

21. (D)log3(a)=candlog3(b)=2careequivalenttoa=3candb=32c

=(3c)2=a2.Therefore,a= b .


22. (B)DropthealtitudefromEtoWHwiththefootofthealtitude

being L. LH = 15. Use the 30-60-90 relationship to determine that

LE=5 3 andEH= 310 .�EHZ isalso30-60-90,soHZ=10and

EZ = 20. �EYZ is an isosceles triangle, making ZY = 20. In �THY,

||EZ TY ,so13

HZ EZHY TY

= = .Therefore,23

XZTY

= .

23. (B)WithSSSinformation,usetheLawofCosinestodeterminethe

measureofthelargestangle(whichislocatedoppositethelongestside).

342=252+292−2(25)(29)cos(θ).2(25)(29)cos(θ)=252+292−342,

sothatcos 2 25 2925 29 342 2 2

i = + -^ ^ ^h h h orcos 14531i =^ h andθ=77.7°.

24. (A) If i54

83 2- + is a root of the equation, then its conjugate

i54

83 2- - isalso.Thesumoftherootsis a

b58- = - andtheproduct

oftheroots is 54

83 2 i

2 2- -c cm m = 25

166418+ = 800

737 .Gettingacom-

mondenominatorforthesumoftherootsgives ab

8001280- = - .With

a=800,b=1280,andc=737,theequationis800x2+1280x+737=0.

25. (D)cos(2t)=x−1andsin(t)=y

32-

.Usetheidentitycos(2t)=

1−2sin2(t)togetxy

1 1 2 32 2

- = --c m orx y2 9

2 2 2- =- -^ h .


26. (B) There is no indication in the problem that titles are associ-

atedwithanyofthepositionsonthecommittee,sothisproblemrep-

resents a combination (order is not important). There are 14 people

to choose from, so the number of combinations is 14C5 = 2002. A

committee with more Democrats than Republicans could have 5

Democrats and 0 Republicans, 4 Democrats and 1 Republican, or 3

Democrats and 2 Republicans. The number of ways this can happen

is(8C5)(6C0)+(8C4)(6C1)+(8C2)(6C2)=1316.Therefore,theprob-

abilitythattheDemocratswillhavemoremembersonthecommittee

thantheRepublicansis13162002

.

27. (B)2u−3v=[2(–5)−3(3),2(4)−3(–1)]=[–19,11].|2u−3v|,the

distancefromtheorigintotheendofthevector,= ( 19)2 +112 = 502 .

28. (B)Thecircumcenterofatriangleistheintersectionoftheperpen-

dicularbisectorsofthesidesofthetriangle.TheslopeofABis 12

,and

itsmidpointis(–7,6).TheequationoftheperpendicularbisectortoAB

isy=–2x−8.TheslopeofACis–1,anditsmidpointis(–4,–3).The

equationoftheperpendicularbisectorofACisy=x+1.Theintersec-

tionofy=–2x−8andy=x+1is(–3,–2).

29. (D)Thevolumeof theoriginalpyramid is hs 231

.Thevolumeof

the new pyramid is Hs 2)8(31

, where H is the new height. Set these

expressionsequal toget hs 231

= Hs 2)8(31

.Solve forH:1.5625h.The

heightneedstobeincreasedby56.25%tokeepthevolumethesame.


30. (A)cis

cis4cis 4cis 4 4cos sini

5 92

20 1819

1815

65

65

65

r

rr r r r= = = +

c

c` ` ` `

m

mj j j j

i i4 23 4 2

1 2 3 2= - + =- +c `m j .

31. (B) Use the change-of-base formula to rewrite loga2

b5c43( ) as

( )( )2

3 45

loglog

acb

b

b .Applythepropertiesoflogarithmstosimplifythenumerator

anddenominatorofthefraction.( ))(log2

log31 45

a

cb

b

b=

( ) ( ))(log2

log31log

31 45

a

cb

b

bb +

=( ) ( )

)(log2

log34log

35

a

cb

b

bb +=

xy

x

y

645

234

35

+=

+,recallingthatlogb(b)=1.

32. (C)Thecenterofthehyperbolais(–3,1).Becausethefocusisat(7,1),

thehyperbola ishorizontal and takes the form 1)1()3(2

2

2

2

=+

by

ax –– .

Thefocusis10unitsfromthecenterofthehyperbola.Theequationof

theasymptotesforahorizontalhyperbola(foundbyreplacingthe1with

azero)becomey−1=ab (x+3).Theratio

ab mustequal

43 ,anda2+

b2=102,soa=8andb=6,givingtheequation2 2( 3) ( 1)

164 36

x y+ −− = .

33. (B)Thevolumeoftheoriginalconeis 31 8 12 1282r r=^ h andthe

volume at the time in question is half this amount. The ratio of the

radiusofthecircleatthewater’ssurfacetotheheightofthewateris2:3.

Atanygiventime,thevolumeofthewater,intermsoftheheightofthe

water, is r hh31

31

322

2

r r=^ ch m and h h274 3r= .Setting thisequal to


one-halftheoriginalvolume,theequationis h274 643r r= .Multiplyby

427r

togeth3=864=9�96=27�32=27�8�4.Therefore,h=6 43 .

34. (C) cos(2t) = 1 − 2 sin2(t). Substituting this into the equation

givessin(t)=1−2sin2(t)or2sin2(t)+sin(t)−1=0.Factorandsolve:

(2sin(t)−1)(sin(t)+1)=0sosin(t)=12

,–1.Intheinterval0≤ t ≤2π,

theanswertothisproblemwouldbe , ,6 65 3

2r r r' 1.Subtract4πfrom

eachoftheseanswertosatisfythedomainrestrictionoftheproblem,

, ,623

619 5

2r r r- - -' 1.

35. (B)Sketchthegraphofy=|2x−3||x−1|−2onyourgraphingcal-

culatorandseethaty(2)=0,sothispointneedstobeeliminatedfrom

theintervalfrom51 <x <5.

Algebraically:Determinethosevaluesforwhich|2x−3||x−1|=2

andthenanalyzetheinequality. |2x−3||x−1|=2requiresthat2x–

3|x –1|=2or2x–3|x –1|=–2.

If2x–3|x–1|=2then2x–2=3|x–1|.Ifx >1,thisequationbe-

comes2x–2=3x–3orx=1.Ifx<1,theequationbecomes2x–2=

–3x+3orx=1.

If2x–3|x –1|=–2,then2x+2=3|x –1|.Ifx>1,thisequationbe-

comes2x+2=3x–3orx=5.Ifx<1,theequationbecomes2x+2=

–3x+3orx= 15

.

Given|2x−3||x−1|=2whenx= 15

,1,and5,youcancheckthat

x=0failstosolvetheinequality,x=2satisfiestheinequality,andthat

x=6failstosolvetheinequality.Therefore,allvaluesbetween 15

and

5,withtheexceptionofx=1,satisfytheinequality.


36. (C) 12 32

k

k

0

3

=

c m/ isaninfinitegeometricserieswith|r|<1andfirst

termequalto12.Thesumoftheseriesis1 3

212

-=36.Inthesameway,

18 21

k

k

0

-3

=

` j/ =1 2

118

- -` j=12.Thedifferenceinthesevaluesis24.

37. (B)Usematricestosolvetheequation[A]–1[B]=[C]where[A]=3

4

486

213

2-

-

-

-

> H and [B] = 833

11-> H. [C] =

52 36

-> H so x1 5= , y

132= - , and

z1 6= .x−y+z= 5

123

61

3031- - + =` j .Therefore, x y z

13130

- += .

38. (E) f(x) = cos x3 2 53 3

10r r- -c m = cos x2 5

321 3r- - +`c jm . The

amplitudeofthefunctionis|–2|=2,thegraphisshiftedtotheright12

,

andtheperiodofthefunctionis5

32

310r

r

= . f f25

631

25

27= = +c cm m .

39. (C) The polar coordinate form for 32i is 2cis3 2r` j. DeMoivre’s

Theoremstatesthatifz=rcis(θ),thenzn=rncis(nθ).Theradiusis2for

all thechoices, so the leadingcoefficient isnot the issue.Of the five

choices,only2cis 1011r` jfailstobeco-terminalwith n

102

5r r+ .

40. (D)Thetermsofthesequencegeneratedbythisrecursionformula

are:4,–3,17,43,–95,–319,767,2491,–5939,–19351,46175,150403,

–358859,–1168927,2789063,9084907.


41. (D)Thedenominatorofthefunctionfactorsto(2x+3)(2x−3),so

thefunctionreduces.Thegraphoff(x)hasaholeatthepoint ,23

67c m.

f(2.5)=2.25,sostatementIiscorrect,andinspectionofthegraphof

f(x)onyourgraphingcalculatorwillverifythatstatementIIIisalsotrue.

42. (D)g–1(13)=ameans thatg(a)=13.This leads to theequation

9log8(x−3)−5=13.Add5anddivideby9 toget log8(x−3)=2.

Rewritethelogarithmicequationasanexponentialequation:x−3=82

=64sox=67.

43. (C)Thecentroidistheintersectionofthemediansofthetriangle.

ThemedianfromQtoRSisalsothealtitudetoRS.UsethePythagorean

Theoremtoshowthatthisaltitudehaslength 2 260 15 15 15− = .The

centroidlies23

ofthewayalongthemedian,soTis5 15unitsfromRS.

�RST,�QST, and�QRT are all equal in area.Theareaof�RST is

21 5 15 30 75 15=^ ^h h .Theareaof�QRT=

1(60) 30 75 15

2h h= =

soh=75 5

15 1530 2

= .

44. (D)Thealtitudetothebaseofanisoscelestrianglealsobisectsthe

vertexangle,som∠ADE=30.Withthehypotenuseofthetrianglehav-

inga lengthof8,AE=4andDE= 4 3 .�AEC isarightanglewith

leg 4 and hypotenuse 6. Use the Pythagorean Theorem to determine

thatBE=2 5 .Theareaofaquadrilateralwithperpendiculardiagonals


is equal tohalf theproductof thediagonals, so theareaofABCD is

21 8 2 5 4 3 8 5 16 3+ = +^ ^h h .

45. (D)5

4csc 1 x

implies that5

4)csc(

x and that

45

)sin(x

. Use the Pythagorean Theorem to determine that

the remaining leg of the right triangle has length 22 5)4( −+x =

982 −+ xx sothat4

98)cos(

2

x

xx.cos(2θ)=cos2(θ)−sin2(θ)

=2

22

45

498

xx

xx=

2 2

2 2 2

8 9 25 8 34( 4) ( 4) ( 4)

x x x xx x x

.

46. (C)Both(a+b)nand(a−b)ncontainn+1termswhentheyare

expanded.Alltheoddtermsin(a−b)nwillbepositive,andtheeven

termswillbenegative.

Ifniseven,therewillbe n2 +1oddtermsand n

2 eventermsinthe

expansionof(a−b)n.Theoddtermswillsubtractout,leaving n2 terms.

Thelasttermsintheexpansionsofbothbinomials,bn,willsubtractout

oftheanswer.

Ifnisodd,therewillbeanequalnumberofevenandoddterms.

Theoddtermswillsubtractoutleaving n2 termsinthedifference.

47. (D) With sin(X) = 178 and cos(W) = 5

3- , cos(X) = 1715 and

sin(W)= 54 .V+W+X=180,soV=180−(W+X)andcos(V)=

cos(180− (W+X))=–cos(W+X). –cos(W+X)= sin(W)sin(X)−


cos(W)cos(X)= 54

5178 3

1715

8577- - =c ` c `m j m j .cos V

2` j= 2

1 8577+

=

162170 85

81859= = .

Let a = sin–1

178` j and b = cos–1

53-c m. Compute cos(.5(180 −

(a+b)))=.976187andcomparethisdecimaltothechoicesavailable.

48. (C)Multiplybothsidesof theequation2 2( 1) ( 1)

132 18

x yby

2toget2 2( 1) ( 1)

216 9

x y+ −+ = .Addthisto 1

9)1(

8)1( 22 yx

toget

23( 1)3

16x +

= sothat(x+1)2=16,x+1=4andx=3or–5.Substitute

x=3intotheequation2 2( 1) ( 1)

216 9

x y++ =

-toget 1

9)1( 2y

.Solve

sothat(y−1)2=9ory−1=3andy=4or–2.Checkthatthese

arethesamevaluesofyyougetwhenx=–5.Thecoordinatesofthe

fourpointsofintersectionforthesetwographsare(3,4),(3,–2),(–5,4),

and(–5,–2).

49. (B) The graph of the function shows that x = –2.5 is a dou-

ble root, and x = –2 and x = 2 are single roots. This implies that

(2x+5)2(x +2)(x –2)isafactorof8x6+72x5+bx4+cx3−687x2−

2160x – 1700 = 0. When expanded, (2x + 5)2(x + 2)(x – 2) is a


fourth-degreepolynomialwithaleadingcoefficientof4andendswitha

constantof–100.Thismeansthattheremainingquadraticfactormust

beoftheform2x2+bx+17inorderfortheleadingcoefficienttobe8

andtheconstanttobe–1700.Theproductofthecomplexrootsfrom

thequadraticis 172

ca= .

50. (A) sin(A) =8

17,

32

< A < 2π, gives cos(A) =1517

while

cos(B) =24

25−

, π < B <32

, gives sin(B) = 257- . cos(2A) = cos2(A) −

sin2(A) =1715

178

2891612 2

- - =c cm m and sin(2A) = 2 sin(A) cos(A) =

1715

178

2892 240- = -c cm m .

cos(2A + B) = cos(2A)cos(B) − sin(2A)sin(B) =289161

2524- -c cm m

289240

257

72255544- - = -c cm m .

An alternative solution: Use your calculator to store angle A

and angle B. ∠A is a fourth quadrant angle and is equal to 2π −

sin–1(8/17) [note that sin–1(8/17) is the reference (acute) angle].

∠B=π+cos–1(24/25).Storeeachoftheseexpressionsintoyourcal-

culator’smemoryandhavethecalculatorcomputecos(2A+B).There

is agoodchance thatyourcalculatorwillnotbeable toconvert the

decimalresponsetoafraction,butyoucanchangethefractionsfrom

thechoicestodecimals.

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SAT Mathematics Level 2 Practice Test - EXAMPLANET MATH LEVEL2 PRACTICE...SAT Mathematics Level 2 Practice Test There are 50 questions on this test. You have 1 hour (60 minutes) to