sat mathematics level 2 practice test w/ answers from mymaxscore
Post on 16-Mar-2016
273 Views
Preview:
DESCRIPTION
TRANSCRIPT
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
SAT MATh 1 & 2 SubjecT TeST4
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
SAT MATheMATIcS LeVeL 2 PRAcTIce TeST 5
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
SAT MATh 1 & 2 SubjecT TeST6
16. Thegraphofy=f(x)isshownabove.Whichisthegraphofg(x)=
2f(x−2)+1?
(A)
(B)
(C)
(D)
(E)
SAT MATheMATIcS LeVeL 2 PRAcTIce TeST 7
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
SAT MATh 1 & 2 SubjecT TeST8
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
SAT MATheMATIcS LeVeL 2 PRAcTIce TeST 9
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
SAT MATh 1 & 2 SubjecT TeST10
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
SAT MATheMATIcS LeVeL 2 PRAcTIce TeST 11
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
SAT MATh 1 & 2 SubjecT TeST12
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
SAT MATheMATIcS LeVeL 2 PRAcTIce TeST 13
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
SAT MATh 1 & 2 SubjecT TeST14
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
SAT MATheMATIcS LeVeL 2 PRAcTIce TeST 15
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.
SAT MATh 1 & 2 SubjecT TeST16
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).
SAT MATheMATIcS LeVeL 2 PRAcTIce TeST 17
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.)
SAT MATh 1 & 2 SubjecT TeST18
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 .
SAT MATheMATIcS LeVeL 2 PRAcTIce TeST 19
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 .
SAT MATh 1 & 2 SubjecT TeST20
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.
SAT MATheMATIcS LeVeL 2 PRAcTIce TeST 21
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
SAT MATh 1 & 2 SubjecT TeST22
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.
SAT MATheMATIcS LeVeL 2 PRAcTIce TeST 23
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.
SAT MATh 1 & 2 SubjecT TeST24
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
SAT MATheMATIcS LeVeL 2 PRAcTIce TeST 25
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)−
SAT MATh 1 & 2 SubjecT TeST26
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
SAT MATheMATIcS LeVeL 2 PRAcTIce TeST 27
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.
top related