SAT Mathematics Level 1 Practice Test - Math Olympus level 1.pdf · SAT Mathematics Level 1...

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


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


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

–


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±


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


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


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


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


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


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


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.


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.


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.


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.


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.


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.


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.


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.

SAT Mathematics Level 1 Practice Test - Math Olympus level 1.pdf · SAT Mathematics Level 1...

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