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SAT Mathematics Level 1 Practice Test
There are 50 questions on this test. You have 1 hour (60 minutes) to
complete it.
1. Thetablebelowshowsthe2010humanpopulationandprojected
2025 human population for different regions. In which region is the
greatestpercentincreasepredicted?
Region 2010 2025
Africa 1,033,043 1,400,184
Asia 4,166,741 4,772,523
LatinAmericaandtheCaribbean
588,649 669,533
NorthAmerica 351,659 397,522
Oceania 35,838 42,507
(A)Africa (B)Asia (C)LatinAmericaandtheCaribbean
(D)NorthAmerica (E)Oceania
2. M(2,6) is themidpointof AB . IfAhascoordinates (10,12), the
coordinatesofBare
(A)(6,10) (B)(–6,0) (C)(–8,–4)
(D)(18,16) (E)(22,18)
SAT MATh 1 & 2 SubjecT TeST2
3. Whenthefigurebelowisspunarounditsverticalaxis,thevolume
ofthesolidformedwillbe
(A)9π (B)36π (C)72π (D)144π (E)288π
4. Iff(x)=86
62
2
xx
xx,f(2)=
(A)0 (B)5.75 (C)6.25 (D)24.5 (E)Undefined
5. Ahighschoolmusicalproductionsellsstudentticketsfor$5each
andadultticketsfor$8each.Iftheratioofadulttostudentticketspur-
chasesis3:1,whatistheaverageincomeperticketsold?
(A)$5.50 (B)$5.75 (C)$6.50 (D)$7.25 (E)$14.50
6. Duetopooreconomicconditions,acompanyhadtolayoff20%of
itsworkforce.Whentheeconomyimproved,itwasabletorestorethe
numberofemployeestoitsoriginalnumber.Bywhatpercentwasthe
depletedworkforceincreasedinordertoreturntotheoriginalnumber
ofemployees?
(A)20 (B)25 (C)80 (D)120 (E)125
7. Avaluezismultipliedby 13
, 12
issubtractedfromtheresult,and
thesquarerootoftheendresultis4.Whatwastheoriginalnumber?
(A) 12
(B) 15
6 (C) 1
152
(D)16 (E) 149
2
SAT MATheMATIcS LeVeL 1 PRAcTIce TeST 3
8. Iftwofairdicearerolled,whatistheprobabilitythatthesumofthe
diceisatmost5?
(A) 536
(B) 636
(C) 1036
(D) 2636
(E) 3036
9. If 5 2 7
8 3 12x + = ,then 1
2x =
(A)–1___15
(B)–2___15
(C)1 (D)2 (E)4
10.2
2
2 8 6 34 20 5
x x xx x
=
(A)–3___5 (B)3__
5 (C)
3( 4)(2 )5( 2)(4 )
x xx x
(D)3( 4)( 2)5( 2)( 4)
x xx x
(E)3( 4)5(4 )
xx
11. Ifi2=–1,then(5+6i)2=
(A)–11 (B)–11+11i (C)–11+30i (D)–11+60i
(E)61
12. Themeanof48,27,36,24,x,and2xis37.x=
(A)1
133
(B)2
163
(C)29 (D)3
334
(E)40
13. 6 83 32x y =
(A) x y4 23 4 3 (B) x y2 42 4 3 (C) x y y2 42 3 23
(D) x y y2 42 2 23 (E) x y y2 22 3 23
14. Acircle is inscribed ina squareof side length6.Theareaof the
regioninsidethesquarebutoutsidethecircleis
(A)36π (B)36π−9 (C)36π−36
(D)36−9π (E)9π−36
SAT MATh 1 & 2 SubjecT TeST4
15. Ifthebinaryoperationa#b=ab–√__
b ,then(2#4)−(4#2)=
(A)–32 (B) √__
2–2 (C)0 (D) √__
2–2 (E)32
16. Ofthe45countriesinEurope,7get100%oftheirnaturalgasfrom
Russia,and6get50%oftheirnaturalgasfromRussia.If25%ofallthe
naturalgasimportedintoEuropecomesfromRussia,whatistheaverage
percentofimportednaturalgasfromRussiafortheremainingcountries
inEurope?
(A)3.9% (B)20% (C)25% (D)75% (E)78.1%
17. A(–3,9)andB(9,–1)aretheendpointsofthediameterofacircle.
Theequationforthiscircleis
(A)(x–3)2+(y–4)2=61 (B)(x–7)2+(y+4)2=269
(C)(x+7)2+(y−4)2=61 (D)(x+3)2+(y+4)2=169
(E)(x+3)2+(y–4)2=25
18. IsoscelestrapezoidACDEwith ||AC DE isshownbelow.Eisthe
midpointofAB,andBD=DCand BC = DE.
TheratiooftheareaoftriangleBDCtotrapezoidACDEis
(A)1:2 (B)1:3 (C)1:4 (D)1:5 (E)1:6
19. If2 3
3 24 1 3 4
x y=
10 11
5z,thenx+y−z=
(A)–21 (B)–15 (C)0 (D)15 (E)21
20. Iff(x)=5x+3andg(x)=x2−1,thenf(g(2))=
(A)3 (B)13 (C)18 (D)39 (E)168
SAT MATheMATIcS LeVeL 1 PRAcTIce TeST 5
21. Chords AB and CD ofcircleO intersectatpointE. IfCE=3,
ED=12,andAEis5unitslongerthanEB,AB=
(A)4 (B)9 (C)11 (D)13 (E)18
22. Whichistheequationofthelineperpendicularto4x−5y=17that
passesthroughthepoint(5,2)?
(A)4x−5y=10 (B)5x+4y=33 (C)4x+5y=30
(D)5x−4y=17 (E) y x45
215= - +
23. Astoneisthrownverticallyintotheairfromtheedgeofabuild-
ingwithheight12meters.Theheightofthestoneisgivenbythefor-
mulah=–4.9t2+34.3t+12.Whatisthemaximumheight,inmeters,
ofthestone?
(A)3.5 (B)12 (C)72.025 (D)114.9 (E)468.2
24. In�ABC,AB=40,themeasureofangleB=50°,andBC=80.The
areaof�ABCtothenearestintegeris
(A)613 (B)1024 (C)1226 (D)2240 (E)2252
25. If 42
a b+= , anda andb arenon-negative integers,whichof the
followingcannotbeavalueofab?
(A)0 (B)7 (C)14 (D)15 (E)16
26. Theperpendicularbisectorofthesegmentwithendpoints(3,5)and
(–1,–3)passesthrough
(A)(–5,2) (B)(–5,3) (C)(–5,4)
(D)(–5,5) (E)(–5,6)
27. Thedifferencebetweentheproductoftherootsandthesumofthe
rootsofthequadraticequation6x2−12x+19=0is
(A)76
(B)316
(C)7
12 (D)
3112
(E)76
–
SAT MATh 1 & 2 SubjecT TeST6
28. InrighttriangleABC, || DE BC,CD=1.5,andBE=2.0.
Thesineofangleθisequalto
(A)12
(B)34
(C)2
2 (D)
32
(E)35
29. QUESTisapentagon.ThemeasureofangleQ=3x−20,themea-
sureofangleU=2x+50,themeasureofangleE=x+30,themeasure
ofangleS=5x−90,andthemeasureofangleT=x+90.Whichtwo
angleshaveequalmeasures?
(A)EandS (B)QandU (C)UandT
(D)TandE (E)UandE
30. TheverticesoftrianglePQRareP(–3,2),Q(1,–4),andR(7,0).The
altitudedrawnfromQintersectsthelinePRatthepoint
(A)(1,2) (B)(2,1) (C)(1,–2)
(D)(–3,2) (E)(7,0)
31. Ifqisapositiveinteger>1suchthat2 43 1
nnq− −
= ,n=
(A)1 (B)–1 (C)1,–4___3
(D)–1,4__3 (E)
1 476i±
SAT MATheMATIcS LeVeL 1 PRAcTIce TeST 7
32. ThemeasureofarcABincircleOis108°.
3a b c+ +
=
(A)18 (B)27 (C)36 (D)45 (E)54
33. Alex observed that the angle of elevation to the top of 800-foot
MountColinwas23°.Tothenearestfoot,howmuchclosertothebase
ofMountColinmustAlexmovesothathisangleofelevationisdoubled?
(A)200 (B)400 (C)489 (D)1112 (E)1600
34. Iff(x)=2
2
66 8
x xx x
,solvef(x)=3.
(A){–5,–1} (B){2,7.5} (C) 1 3 7 1 3 7,
2 2
(D)17 73 17 73
, 6 6
(E)∅
35. In�QRS,XisonQR andYisonQS ,sothat || X Y RSand14
QXXR
= .
Theratiooftheareaof�QXYtotheareaoftrapezoidXYSRis
(A)1:4 (B)1:15 (C)1:16 (D)1:24 (E)1:25
SAT MATh 1 & 2 SubjecT TeST8
36. InquadrilateralKLMN,KL=LM,KN=MN,anddiagonals KM
and NL intersectatP.IfKP=PM,thenwhichofthefollowingstate-
mentsistrue?
I. NP=PL.
II. KLMNisarhombus.
III. TheareaofKLMNis12
(KM)(NL).
(A)Ionly (B)IIonly (C)IIIonly
(D)IIandIIIonly (E)IandIIIonly
37. If7x+9y=86and4x−3y=–19,x+4y=
(A)18
3119
– (B)1
223
(C)18
3119
(D)35 (E)105
38. Thesolutionsetto10x2+11x–6≤0is
(A)–0.4≤ x ≤1.5 (B)–1.5≤ x ≤0.4 (C)x ≤–0.4orx ≥1.5
(D)x ≤–1.5orx ≥0.4 (E)–1.5≤ x ≤–0.4
39. Insimplestform,
12
31
13
x
x
isequivalentto
(A)2x–7______x–2
(B)7–2x ______x–2
(C)2x–5______x–2
(D)2x+7______x–2
(E)1
40. InrighttriangleQRS,QRisperpendiculartoRS,QR=12,andRS
=12 3 .TheareaofthecirclethatcircumscribestriangleQRSis
(A)108π (B)144π (C)288π (D)576π (E)1728π
SAT MATheMATIcS LeVeL 1 PRAcTIce TeST 9
41. Thesolutionsetfortheequation|2x−1|−|x+2|=5is
(A){–2} (B){3} (C){8} (D){3,–7} (E){–2,8}
42. Giventhegraphoff(x)below,letg(x)=f(x–2)+1.Forwhatsetof
valuesofwillg(x)=0?
(A){–2,2,4} (B){0,4,6} (C){–4,2}
(D){–1,5} (E) ∅
43. SquareABCDhassideswithlength20.Eachofthesmallerfigures
isformedbyconnectingmidpointsofthenextlargerfigure.
WhatistheareaofEFGH?
(A)2564
(B)2516
(C)254
(D)25 (E)100
SAT MATh 1 & 2 SubjecT TeST10
44. Whichofthestatementsaboutthegraphsoff(x)=x2–x–6_________
x–3 and
g(x)=x+2aretrue?
I. f(x)+g(x)=2x+4
II.Theyintersectatonepoint
III.Theyarethesameexceptforonepoint
(A)Ionly (B)IIonly (C)IIIonly
(D)IandIIIonly (E)IIandIIIonly
45. A25-footladderleansagainstabuilding.Asthebottomofthelad-
deratpointAslidesawayfromthebuilding,thetopoftheladder,B,
slidesfromaheightof24feetabovethegroundtoaheightof16feet.
Howmanyfeetdidthebottomoftheladderslide?
(A)7 (B)8 (C)9 (D)12.2 (E)19.2
46. InparallelogramABCD,W is themidpointof AD ,andX is the
midpoint of BC . CW and DX are drawn and intersect at pointE.
Whatistheratiooftheareaof�DEWtotheareaofABCD?
(A)1:2 (B)1:4 (C)1:6 (D)1:8 (E)1:16
47. PointO(3,2)isthecenterofacirclewithradius4.OA isparal-
leltothex-axisandm∠AOB=120degrees.Tothenearesttenth,the
y-coordinateofpointBis
(A)3.5 (B)4.0 (C)5.5 (D)6.6 (E)6.9
SAT MATheMATIcS LeVeL 1 PRAcTIce TeST 11
48. The vertices of �ABC have coordinates A(–7,3), B(–1,0), and
C(–2,8).Thecoordinatesofthecenterofthecircumscribedcircleare
(A)1 2
3 , 33 3
(B)5 5
2 , 36 6
(C)1
1 , 42
(D) 14,1
2 (E) 1 1
4 , 52 2
49. Eachsideofthebaseofasquarepyramidisincreasedinlengthby
25%,andtheheightofthepyramidisdecreasedbyx%,sothatthevol-
umeofthepyramidisunchanged.x=
(A)20 (B)25 (C)36 (D)50 (E)64
50. If2 1
( )2
xf x
x,thenf(f(x))=
(A)2
2
4 4 14 4
x xx x
(B)3 44 3xx
(C) 4 33 4
xx
(D) 34 3
xx +
(E) 3 14 3
xx+
+
Level 1 Practice Test Solutions
1. (A) The percent increase is computed aspopulation
original populationD
. The
greatestpercentchangewilloccurwhenthenumeratorislargeandthe
denominatorissmall.
2010 2025 � Pct�
Africa 1,033,043 1,400,184 367,141 35.5
Asia 4,166,741 4,772,523 605,782 14.5
LatinAmericaandtheCaribbean
588,649 669,533 80,884 13.7
NorthAmerica 351,659 397,522 45,863 13
Oceania 35,838 42,507 6,669 18.6
SAT MATh 1 & 2 SubjecT TeST12
2.(B)LetBhavecoordinates(x,y).Theformulafor findingthemid-
point of a line segment is to average the x-coordinates and average
the y-coordinates.This gives the equations 102
2x+= and 12
62
y+= .
Multiplyeachequationby2.10+x=4givesx=–6and12+y=12
givesy=0.PointBmusthavecoordinates(–6,0).
3.(E)Thesolid formedwillbeahemispherewith radius6.Thevol-
umeofasphereisgivenbytheformulaV=4__3
πr3.Thevolumeofthe
hemispherewill behalf thatnumber.With r =6, the volumeof the
hemisphereis2__3π(6)3=288π.
4.(A)f(2)=2
2
(2) 2 6 0(2) 6(2) 8 24=0.
5.(D)Becausetheticketsaresoldintheratioof3:1youcanworkwith
just4tickets.Thethreeadultticketsraise$24whilethestudentticket
raises$5.The4ticketsbringin$29oranaverageof$7.25each.
6.(B) Itmayhelp to thinkof thebusiness ashaving100 employees.
Afterthelayoffs,theworkforceis80employees.Tobringtheworkforce
backto100,20peoplemustbehired.20outofthecurrentlevelof80
isa25%increase.
7.(E)Theequationdescribedbythesentenceis 1 13 2
z =4.Square
bothsidesoftheequationtoget1 1
163 2
z .Add12
tobothsidesofthe
equationtoget1 1
163 2
z = .Multiplyby3fortheanswer:1
492
z = .
8.(C)“Atmost5”means5orless.Thereisonewaytogeta2(1on
eachdie).Therearetwowaystogeta3(1and2or2and1),threeways
SAT MATheMATIcS LeVeL 1 PRAcTIce TeST 13
togeta4(1,3or2,2,or3,1),andfourwaystogeta5(1,4or2,3or3,2
or4,1).Thisisatotalof10outcomesoutofthepossible36outcomes
whentwodicearerolled.
9.(A)Solvetheequationbysubtracting 32 frombothsidesoftheequation
andthenmultiplyingby 58 togetx= 15
2- .Halfthisamountis 151- .
10. (B)2
2
2 8 6 34 20 5
x x xx x
=( 4)( 2) 3(2 )( 2)( 2) 5(4 )x x xx x x
=
( 4) (2 )
( 2) (4 )xx
xx
35
.2−xandx–2arenegativesofoneanother,sothey
reducetobe–1.Thesameistrueforx−4and4−x.Thetwofactorsof
–1multiplyto1sotheansweris3/5.
11. (D)(a+b)2=a2+2ab+b2,so(5+6i)2=52+2(5)(6i)+(6i)2=25
+60i+36i2=25+60i−36,or–11+60i.
12. (C)Themeanofthe6numbersis37,sothesumofthesixnumbers
is222.Thefourgivennumberssumto135,sox+2x=222−135=87
andx=29.
13. (D) 32 = 8 � 4 = 23 � 4, so 32 2 43 3= . x x63 2= and
y y y y y y83 831
38 2
32 2 23= = = =^ h . x y32 6 83 =2x y y42 232 .
14. (D)Theradiusoftheinscribedcircleis3.Theareaofthecircleis
π×32,or9π.Theareaofthesquareis62=36.Subtracttheareaofthe
circlefromtheareaofthesquaretoget36–9π.
15. (B)2#4=24–√__
4=16−2=14.4#2=42–√__
2=16–√__
2.(2#
4)−(4#2)=14–(16–√__
2)=–2+√__
2.
SAT MATh 1 & 2 SubjecT TeST14
16. (A)25%ofthegasusedbyall45Europeancountriescomesfrom
Russia.Letxrepresentthepercentusagebytheremaining32countries.
Thesolutiontotheequation7(1) 6(.5) 32
.2545
x+ += isx=0.3906.The
restofEuropeimportsanaverageof3.9%ofitsgasfromRussia.
17. (A)ThecenterofthecircleisthemidpointofAandB, 3 92
=
3and9 ( 1)2
=4.Theequationofthecircleis(x−3)2+(y−4)2=r2.
SubstitutingeitherAorBintotheequationforxandygivesr2=61.
18. (B)Because || AC DE,theheightsoftrapezoidACDEandtriangle
BDCarethesame.TrapezoidACDE is isosceles,soAEandBCDare
equal, as are themeasuresof anglesA andC.Consequently,BD and
AEareequal,asarethemeasuresofanglesAandDBC,soABDEisa
parallelogramandAB=DE.TheareaoftrapezoidACDEisgivenbythe
equationArea=1
( )2
h DE AC+ .ED =BCandAC=2BC,sothisequa-
tionbecomesArea=1
( )2
h DE AC+ .Theratioof theareaof triangle
DBCtotheareaoftrapezoidACDEis12
h(BC):1
(3 )2
h BC or1:3.
19. (D)2 3 6 2 9 2
3 2 4 1 3 4 18 5
x y x y. The equality of the
matrices6 2 9 2 10 11
18 5 5
x yz
meansthat6−2x=10,sox=–2;
9−2y=11,soy=–1;andz=–18.Therefore,x+y−z=15.
20. (C)Tofindf(g(2)),firstdetermineg(2),thensubstitutethisvalue
intof(x):g(2)=22–1=3,sof(g(2))=f(3)=5(3)+3=18.
SAT MATheMATIcS LeVeL 1 PRAcTIce TeST 15
21. (D)Whenchordsintersectinsideacircle,theproductsoftheseg-
mentsformedbythechordsareequal.Thatis,(AE)(EB)=(CE)(ED).
SolveforEB:(EB+5)(EB)=(3)(12),so(EB)2+5EB–36=0or(EB–4)
(EB+9)=0andEB=4.AE=EB+5=9andABhasalengthof13.
22. (B)Theslopeoftheline4x−5y=17is 54 .Theslopeoftheperpen-
dicularlineis 45- .Theequationofthelineisy x b
45= - + .Substituting
5forxand2forygives b245 5= - +^ h sothatb=
433 . y x
45
433= - +
becomes 4y = –5x + 33 or 5x + 4y = 33. It is faster to know that a
lineperpendiculartoAx+By=ChastheequationBx−Ay=D.You
couldthenhavestartedtheproblemwith5x+4y=Danddetermined
thatD=33.
23. (C)Thetimethestonereachesitsmaximumheightcanbecom-
putedusingtheaxisofsymmetryfortheequation.Maximumheightis
reachedat t=–34.3/(2)(–4.9)=3.5seconds.Substitute thisnumber
intotheequationfortheheightofthestone–4.9(3.5)2+34.3(3.5)+
12=72.025meters.
24. (C)ThealtitudetosideBC in�ABCcanbecalculatedusingtrigo-
nometry.DroptheheightfromAtoBC andcallthefootofthealtitude
D.Then sin(50)40AD
= sothatAD=40sin(50).Theareaofthetriangle
is1
( )( )2
BC AD =12
(80)(40)sin(50).
25. (C)a+b=8sothevaluesfor(a,b)couldbe(0,8),(1,7),(2,6),
(3,5),(4,4),(5,3),(6,2),(7,1),and(8,0).Theonlyproductnotavailable
fromthechoiceslistedis14.
SAT MATh 1 & 2 SubjecT TeST16
26. (C)Thesegmentwithendpoints(3,5)and(–1,–3)hasitsmidpoint
(1,1)anditsslopeis2.Theslopeoftheperpendicularlineis–1/2and
theequationoftheperpendicularbisectorisy−1=–1/2(x−1).Ofthe
pointsavailableaschoices,only(–5,4)liesonthisline.
27. (A)Theproductof the roots is 196
ca= .The sumof the roots is
126
ba–
= .Thedifferencebetweenthesenumbersis 67 .
28. (B) sin( )ADAE
. With || DE BC, AD DCAE EB
= . Therefore,
.
.sin EBDC
2 01 5
43i = = =^ h .
29. (C)Thesumofthemeasuresoftheanglesis12x+60.Thesum
of the interior angles of a pentagon is equal to 3(180) = 540. (Five
sidesimpliesthreenon-overlappingtriangles.)Solveforx=40.m∠U=
m∠T=130.
30. (B)Theslopeof PR is–1/5,sotheslopeofthealtitudeto PR is5.
TheonlypointamongthechoicesthatsatisfiesthattheslopetopointQ
willbe2/3is(2,1).
31. (D)Theexponentmustbeequal to0.3n2−n−4=0becomes
(3n−4)(n+1)=0son=4/3,or–1.
32. (C)Inscribed∠Chasameasureof54.DrawOC .∠Aand∠ACO
arecongruent,asare∠Band∠BCO.m∠A+m∠B=m∠ACO+m∠BCO
=54.Therefore,3cba ++
=54 54
3+
=36.
SAT MATheMATIcS LeVeL 1 PRAcTIce TeST 17
33. (D) Alex’s distance from the base of Mount Colin is computed
by 800_______tan(23)
. If the angle of elevation is doubled, then Alex must
be 800_______tan(46)
feetfromthebase.SoAlexmustmove 800_______tan(23)
− 800_______tan(46)
=1112feetclosertothebaseofMountColin.
34. (B)f(x)=3becomes3=x2+x–6__________
x2–6x+8sothat3(x2−6x+8)=x2+x−6.
Multiplythroughtheleftsideoftheequationtoget3x2−18x+24=x2
+x−6andthen2x2−19x+30=0.Thisfactorsto(2x−15)(x−2)=0
sox=2,15/2.
35. (D) �QXY ~ �QRS, so the ratio of the areas of the two trian-
gleswillbeequaltothesquareoftheratioofcorrespondingsides.Since
14
QXXR
= , 15
QXQR
= ,so 125
area QXYarea QRS
.TheareaoftrapezoidXYSRis
theareaof�QRS−theareaof�QXY.Therefore, 124
area QXYarea trap XYSR
.
36. (C)�KLM isan isoscelestrianglewith legsKLandLM.P is the
midpointofKM becauseKP=PM,soLP isperpendiculartoKM .LP
ispartofdiagonal NL ,sothediagonalsofthequadrilateralareperpen-
dicular.Wheneverthediagonalsofaquadrilateralareperpendicular,the
areaisequaltoone-halftheproductofthediagonals(lookattheareas
of�KLMand�KNMandaddthem).Thereisnoinformationtoallow
youtodeducethatKN=LN,andthereforeyoucannotconcludethat
thequadrilateralisarhombusnoristhereanyinformationtoconclude
thediagonalsbisecteachothertoinsurethatNP=PL.
37. (D)Subtractthetwoequationstoget4x+12y=105.Divideby3
togetx+4y=35.
SAT MATh 1 & 2 SubjecT TeST18
38. (B)y=10x2+11x–6isaparabolathatopensupandcrossesthe
x-axisatx=–1.5andx=0.4.Theparabolawillbebelowthex-axis
between thesevalues, soy ≤0,or10x2+11x –6≤0, is satisfiedby
–1.5≤ x ≤0.4.
39. (A)Because3−x=–(x−3),
x
x
−−
−−
311
312
=
311
312
−+
−−
x
x .Multiplythe
numeratoranddenominatorbythecommondenominatorx−3toget
33
311
312
−−
−+
−−
xx
x
x
=272
131)3(2
+−
=+−−−
xx
xx .
40. (B)TriangleQRSmustbea30-60-90trianglebecausethelonger
legis 3 timeslongerthantheshorterleg.Thehypotenuseoftheright
trianglehas length24.Thecircumscribedcircleabouta right triangle
musthaveitscenteratthemidpointofthehypotenuse,sotheradiusof
thecircleis12,andtheareaofthecircleis144π.
41. (E)Thegraphoff(x)=|2x−1|−|x+2|intersectsthegraphofy=
5atx=–2andatx=8.
42. (D)Becausethegraphistranslatedright2andup1,youneedto
findthosepointsforwhichf(x)=–1.Theseareatx=–3and3.Thegraph
ofg(x)willcrossthex-axis2pointstotherightofwheref(x)=–1,so
x=–1and5.
SAT MATheMATIcS LeVeL 1 PRAcTIce TeST 19
43. (D)Joiningthemidpointsofthesidesofaquadrilateralformsparal-
lelogramswhoselengthsare1/2thelengthofthediagonalofthelarger
square.The lengthsof the sidesof the squares, in reducedorder, are
AB=20,10 2 ,10,5 2 ,5=EF.TheareaofEFGHis25.
44. (C)f(x)=2 6
3x x
x =
( 2)( 3)3
x xx
=x +2.Withtheexception
of thepoint (3,5),which is removed from thegraphof f(x), the two
graphsareexactlythesame.
45. (D)Whenthe25-footladderoriginallywasataheightof24feet,
thefootoftheladderwas7feetfromthebaseofthebuilding.Thiscan
bedeterminedusingthePythagoreanTheorem:242+72=252.When
thetopoftheladderfallstoaheightof16feet,thedistancefromthe
footoftheladdertothebaseofthebuildingmustsatisfytheequation
162+b2=252,sothatb2=252−162=369,sob=19.2ft.Theladder
slipped12.2feetfurtherfromthebuilding.
46. (D)DrawWX .WXCDisaparallelogramwithone-halftheareaof
ABCD becauseWD=XCand ||WD XC .Therefore,Eisthemidpoint
of thediagonals, and thedistance fromE to WD ishalf thedistance
fromBtoWD .IfhisthedistancefromBtoWD ,theareaof �DEW is
1 1 1( )
2 2 4WD h WD h =
1 1 1( )
4 2 8AD h AD h =
18
area ABCD.
47. (C)Extendradius AO throughOtointersectthecircleatD.The
altitudefromBtothediameterformsarighttrianglewithm∠DOB=
60°.Thelengthofthealtitudeistheoppositelegoftherighttriangle,so
itslengthis4sin(60)=3.46.Addthisto2,andthey-coordinateofB,to
thenearesttenth,is5.5.
SAT MATh 1 & 2 SubjecT TeST20
48. (B)Thecircumcenteristheintersectionoftheperpendicularbisec-
torsofthesidesofthetriangle.Themidpointof AC is(–4.5,5.5),and
theslopeof AC is1.Theequationoftheperpendicularbisectorto AC
isy=–x+1.Themidpointof AB is(–4,1.5),andtheslopeof AB is
–0.5.Theequationoftheperpendicularbisectorto AB isy=2x+9.5.
Solvethissystemofequationstogetthepoint5 5
2 , 36 6
.
49. (C)IfhistheheightoftheoriginalpyramidandHistheheightof
thenewpyramid,thenthevolumesofthetwopyramidsarerelatedby
theequation(1/3)hs2=(1/3)(x)h(1.25s)2,where srepresentsthelength
ofasideofthebaseoftheoriginalpyramid.Solveforxtogetx=.64.
Theheightofthenewpyramidmustbe64%oftheoriginalpyramid,so
theheightoftheoriginalpyramidwasreducedby36%.
50. (B)
2212
12122
))((+
+−
−
+−
=
xxxx
xff ) =22
2212
12122
++
+
+−
−
+−
xx
xxxx
=
)2(212)2(1122
xxxx
=4212224
xx
xx=
3443
x
x.