math level 1

8/12/2019 Math Level 1

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SAT MATH 1 & 2 SUBJECT TEST2

3. When the figure below is spun around its vertical axis, the volume

of the solid formed will be

(A) 9 (B) 36 (C) 72 (D) 144 (E) 288

4. Iff(x) =86

6

2

2

x

xx

,f(2) =

(A) 0 (B) 5.75 (C) 6.25 (D) 24.5 (E) Undefined

5. A high school musical production sells student tickets for $5 each

and adult tickets for $8 each. If the ratio of adult to student tickets pur-

chases is 3:1, what is the average income per ticket sold?

(A) $5.50 (B) $5.75 (C) $6.50 (D) $7.25 (E) $14.50

6. Due to poor economic conditions, a company had to lay off 20% of

its workforce. When the economy improved, it was able to restore the

number of employees to its original number. By what percent was the

depleted workforce increased in order to return to the original number

of employees?

(A) 20 (B) 25 (C) 80 (D) 120 (E) 125

7. A value zis multiplied by1

3,

1

2is subtracted from the result, and

the square root of the end result is 4. What was the original number?

(A) 1

2 (B) 15

6 (C) 115

2 (D) 16 (E)

149

2


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15. If the binary operation a# b= ab __b, then (2 # 4) (4 # 2) =

(A) 32 (B) __2 2 (C) 0 (D) __2 2 (E) 32

16. Of the 45 countries in Europe, 7 get 100% of their natural gas from

Russia, and 6 get 50% of their natural gas from Russia. If 25% of all the

natural gas imported into Europe comes from Russia, what is the average

percent of imported natural gas from Russia for the remaining countries

in Europe?

(A) 3.9% (B) 20% (C) 25% (D) 75% (E) 78.1%

17. A(3,9) and B(9,1) are the endpoints of the diameter of a circle.

The equation for this circle is

(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. Isosceles trapezoidACDEwith ||AC DE is shown below. Eis the

midpoint ofAB, and BD= DCandBC = DE.

The ratio of the area of triangle BDCto trapezoidACDEis

(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+ 3 andg(x) =x2 1, thenf(g(2)) =

(A) 3 (B) 13 (C) 18 (D) 39 (E) 168


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SAT MATHEMATICS LEVEL 1 PRACTICE TEST 5

21. Chords AB and CD of circle O intersect at point E. If CE= 3,

ED= 12, andAEis 5 units longer than EB,AB=

(A) 4 (B) 9 (C) 11 (D) 13 (E) 18

22. Which is the equation of the line perpendicular to 4x 5y= 17 that

passes through the point (5,2)?

(A) 4x 5y= 10 (B) 5x+ 4y= 33 (C) 4x+ 5y= 30

(D) 5x 4y= 17 (E) y x4

5

2

15=-

+

23. A stone is thrown vertically into the air from the edge of a build-

ing with height 12 meters. The height of the stone is given by the for-

mula h= 4.9t2+ 34.3t+ 12. What is the maximum height, in meters,

of the stone?

(A) 3.5 (B) 12 (C) 72.025 (D) 114.9 (E) 468.2

24. In ABC,AB= 40, the measure of angle B= 50, and BC= 80. The

area of ABCto the nearest integer is

(A) 613 (B) 1024 (C) 1226 (D) 2240 (E) 2252

25. If 42

a b+= , and aand bare non-negative integers, which of the

following cannot be a value of ab?

(A) 0 (B) 7 (C) 14 (D) 15 (E) 16

26. The perpendicular bisector of the segment with endpoints (3,5) and

(1,3) passes through

(A) (5,2) (B) (5,3) (C) (5,4)

(D) (5,5) (E) (5,6)

27. The difference between the product of the roots and the sum of the

roots of the quadratic equation 6x2 12x+ 19 = 0 is

(A)

7

6 (B)

31

6 (C)

7

12 (D)

31

12 (E)7

6


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28. In right triangleABC, ||D E B C, CD= 1.5, and BE= 2.0.

The sine of angle is equal to

(A)1

2 (B)

3

4 (C)

2

2 (D)

3

2 (E)

3

5

29. QUESTis a pentagon. The measure of angle Q= 3x 20, the mea-

sure of angle U= 2x+ 50, the measure of angle E=x+ 30, the measure

of angle S= 5x 90, and the measure of angle T= x+ 90. Which two

angles have equal measures?(A) Eand S (B) Qand U (C) Uand T

(D) Tand E (E) Uand E

30. The vertices of triangle PQRare P(3,2), Q(1,4), and R(7,0). The

altitude drawn from Qintersects the line PRat the point

(A) (1,2) (B) (2,1) (C) ( 1,2)

(D) (3,2) (E) (7,0)

31. If qis a positive integer > 1 such that2 4

31

nn

q

= , n=

(A) 1 (B) 1 (C) 1, 4___3 (D) 1, 4__

3 (E)

1 47

6

i


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32. The measure of arcABin circle Ois 108.

3

a 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

Mount Colin was 23. To the nearest foot, how much closer to the base

of Mount Colin must Alex move so that his angle of elevation is doubled?

(A) 200 (B) 400 (C) 489 (D) 1112 (E) 1600

34. Iff(x) =2

2

6

6 8

x x

x 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. InQRS,Xis on QR and Yis on QS , so that ||X Y R Sand1

4

QX

XR = .

The ratio of the area of QXYto the area of trapezoidXYSRis

(A) 1:4 (B) 1:15 (C) 1:16 (D) 1:24 (E) 1:25


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36. In quadrilateral KLMN, KL= LM, KN=MN, and diagonals KM

and NL intersect at P. If KP= PM, then which of the following state-

ments is true?

I. NP= PL.

II. KLMNis a rhombus.

III. The area of KLMNis1

2(KM)(NL).

(A) I only (B) II only (C) III only

(D) II and III only (E) I and III only

37. If 7x+ 9y= 86 and 4x 3y= 19,x+ 4y=

(A)18

3119

(B)1

223

(C)18

3119

(D) 35 (E) 105

38. The solution set to 10x2+ 11x 6 0 is

(A) 0.4 x 1.5 (B) 1.5 x 0.4 (C)x 0.4 orx 1.5

(D)x 1.5 orx 0.4 (E) 1.5 x 0.4

39. In simplest form,

12

3

11

3

x

x

is equivalent to

(A) 2x 7______x 2

(B) 7 2x______x 2

(C) 2x 5______x 2

(D) 2x+ 7______x 2

(E) 1

40. In right triangle QRS, QRis perpendicular to RS, QR= 12, and RS

= 12 3 . The area of the circle that circumscribes triangle QRSis

(A) 108 (B) 144 (C) 288 (D) 576 (E) 1728


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41. The solution set for the equation |2x 1| |x+ 2| = 5 is

(A) {2} (B) {3} (C) {8} (D) {3, 7} (E) {2, 8}

42. Given the graph off(x) below, letg(x) =f(x 2) + 1. For what set of

values of willg(x) = 0?

(A) {2, 2, 4} (B) {0, 4, 6} (C) {4, 2}

(D) {1, 5} (E)

43. SquareABCDhas sides with length 20. Each of the smaller figuresis formed by connecting midpoints of the next larger figure.

What is the area of EFGH?

(A)25

64

(B)25

16

(C)25

4

(D) 25 (E) 100


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44. Which of the statements about the graphs off(x) = x2x 6_________x 3

and

g(x) =x+ 2 are true?

I. f(x) +g(x) = 2x+ 4

II. They intersect at one point

III. They are the same except for one point

(A) I only (B) II only (C) III only

(D) I and III only (E) II and III only

45. A 25-foot ladder leans against a building. As the bottom of the lad-

der at pointAslides away from the building, the top of the ladder, B,

slides from a height of 24 feet above the ground to a height of 16 feet.

How many feet did the bottom of the ladder slide?

(A) 7 (B) 8 (C) 9 (D) 12.2 (E) 19.2

46. In parallelogram ABCD, W is the midpoint of AD , and X is the

midpoint of BC . CW and DX are drawn and intersect at point E.

What is the ratio of the area of DEWto the area ofABCD?

(A) 1:2 (B) 1:4 (C) 1:6 (D) 1:8 (E) 1:16

47. Point O(3,2) is the center of a circle with radius 4. OA is paral-

lel to thex-axis and mAOB= 120 degrees. To the nearest tenth, the

y-coordinate of point Bis

(A) 3.5 (B) 4.0 (C) 5.5 (D) 6.6 (E) 6.9


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48. The vertices of ABC have coordinates A(7,3), B(1,0), and

C(2,8). The coordinates of the center of the circumscribed circle are

(A)1 2

3 , 33 3

(B)5 5

2 , 36 6

(C)1

1 , 42

(D) 14,12

(E)1 1

4 , 52 2

49. Each side of the base of a square pyramid is increased in length by

25%, and the height of the pyramid is decreased byx%, so that the vol-

ume of the pyramid is unchanged.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 1

4 4

x x

x x

(B)3 4

4 3

x

x

(C)4 3

3 4

x

x

(D) 34 3

xx +

(E) 3 14 3xx

+

+

Level 1 Practice Test Solutions

1. (A) The percent increase is computed aspopulation

original populationD

. The

greatest percent change will occur when the numerator is large and the

denominator is small.

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

Latin Americaand theCaribbean

588,649 669,533 80,884 13.7

North America 351,659 397,522 45,863 13

Oceania 35,838 42,507 6,669 18.6


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2. (B) Let Bhave coordinates (x,y). The formula for finding the mid-

point of a line segment is to average the x-coordinates and average

the y-coordinates. This gives the equations10

22

x+= and

126

2

y+= .

Multiply each equation by 2. 10 + x= 4 gives x= 6 and 12 + y= 12

givesy= 0. Point Bmust have coordinates (6,0).

3. (E) The solid formed will be a hemisphere with radius 6. The vol-

ume of a sphere is given by the formula V = 4__3

r3. The volume of the

hemisphere will be half that number. With r= 6, the volume of the

hemisphere is 2__3(6)3 = 288.

4. (A)f(2) =

2

2

(2) 2 6 0

(2) 6(2) 8 24

= 0.

5. (D) Because the tickets are sold in the ratio of 3:1 you can work with

just 4 tickets. The three adult tickets raise $24 while the student ticketraises $5. The 4 tickets bring in $29 or an average of $7.25 each.

6. (B) It may help to think of the business as having 100 employees.

After the layoffs, the workforce is 80 employees. To bring the workforce

back to 100, 20 people must be hired. 20 out of the current level of 80

is a 25% increase.

7. (E) The equation described by the sentence is 1 1

3 2z = 4. Square

both sides of the equation to get 1 1

163 2z . Add

1

2to both sides of the

equation to get1 1

163 2

z = . Multiply by 3 for the answer:1

492

z = .

8. (C) At most 5 means 5 or less. There is one way to get a 2 (1 on

each die). There are two ways to get a 3 (1 and 2 or 2 and 1), three ways


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to get a 4 (1,3 or 2,2, or 3,1), and four ways to get a 5 (1,4 or 2,3 or 3,2

or 4,1). This is a total of 10 outcomes out of the possible 36 outcomes

when two dice are rolled.

9. (A) Solve the equation by subtracting3

2 from both sides of the equation

and then multiplying by5

8 to getx=15

2- . Half this amount is

15

1- .

10. (B)2

2

2 8 6 3

4 20 5

x x x

x x

=

( 4)( 2) 3(2 )

( 2)( 2) 5(4 )

x x x

x x x

=

( 4) (2 )( 2) (4 )

x

x

x

x

35

. 2 xandx 2 are negatives of one another, so they

reduce to be 1. The same is true forx 4 and 4 x. The two factors of

1 multiply to 1 so the answer is 3/5.

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) The mean of the 6 numbers is 37, so the sum of the six numbers

is 222. The four given numbers sum to 135, 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 8 31

3

82

3

22 23

= = = =^ h . x y32 6 83 = 2x y y42 232 .

14. (D) The radius of the inscribed circle is 3. The area of the circle is

32, or 9. The area of the square is 62= 36. Subtract the area of the

circle from the area of the square to get 36 9.

15. (B) 2 #4 = 24 __4 = 16 2 = 14. 4 #2 = 42 __2 = 16 __2 . (2 #

4) (4 #2) = 14 (16

__

2 ) = 2 + __

2 .


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21. (D) When chords intersect inside a circle, the products of the seg-

ments formed by the chords are equal. That is, (AE)(EB) = (CE)(ED).

Solve for EB: (EB+ 5)(EB) = (3)(12), so (EB)2+ 5EB 36 = 0 or (EB 4)

(EB+ 9) = 0 and EB= 4.AE= EB+ 5 = 9 andABhas a length of 13.

22. (B) The slope of the line 4x 5y= 17 is5

4 . The slope of the perpen-

dicular line is4

5- . The equation of the line is y x b

4

5=-

+ . Substituting

5 forxand 2 forygives b24

55=

-+^ h so that b=

4

33 . y x4

5

4

33=-

+

becomes 4y = 5x + 33 or 5x + 4y = 33. It is faster to know that a

line perpendicular toAx+ By= Chas the equation BxAy= D. You

could then have started the problem with 5x+ 4y= Dand determined

that D= 33.

23. (C) The time the stone reaches its maximum height can be com-

puted using the axis of symmetry for the equation. Maximum height is

reached at t= 34.3/(2)(4.9) = 3.5 seconds. Substitute this number

into the equation for the height of the stone 4.9(3.5)2+ 34.3(3.5) +

12 = 72.025 meters.

24. (C) The altitude to side BC in ABCcan be calculated using trigo-

nometry. Drop the height fromAto BC and call the foot of the altitude

D. Then sin(50)40

AD= so thatAD= 40 sin(50). The area of the triangle

is1

( )( )2

BC AD =1

2(80)(40) sin(50).

25. (C) a+ b= 8 so the values for (a,b) could be (0,8), (1,7), (2,6),

(3,5), (4,4), (5,3), (6,2), (7,1), and (8,0). The only product not available

from the choices listed is 14.


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26. (C) The segment with endpoints (3,5) and (1,3) has its midpoint

(1,1) and its slope is 2. The slope of the perpendicular line is 1/2 and

the equation of the perpendicular bisector isy 1 = 1/2 (x 1). Of the

points available as choices, only (5,4) lies on this line.

27. (A) The product of the roots is19

6

c

a = . The sum of the roots is

12

6

b

a

= . The difference between these numbers is

6

7 .

28. (B) sin( )

AD

AE . With ||D E B C,

AD DC

AE EB= . Therefore,

.

.sin

EB

DC

2 0

1 5

4

3i = = =^ h .

29. (C) The sum of the measures of the angles is 12x+ 60. The sum

of the interior angles of a pentagon is equal to 3(180) = 540. (Five

sides implies three non-overlapping triangles.) Solve forx= 40. mU=

mT= 130.

30. (B) The slope of PR is 1/5, so the slope of the altitude to PR is 5.

The only point among the choices that satisfies that the slope to point Q

will be 2/3 is (2,1).

31. (D) The exponent must be equal to 0. 3n2 n 4 = 0 becomes

(3n 4) (n+ 1) = 0 so n= 4/3, or 1.

32. (C) Inscribed Chas a measure of 54. Draw OC . Aand ACO

are congruent, as areBandBCO. mA+ mB= mACO+ mBCO

= 54. Therefore,3

cba ++=

54 54

3

+

= 36.


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38. (B)y= 10x2+ 11x 6 is a parabola that opens up and crosses the

x-axis at x= 1.5 and x= 0.4. The parabola will be below the x-axis

between these values, so y 0, or 10x2+ 11x 6 0, is satisfied by

1.5 x 0.4.

39. (A) Because 3 x= (x 3),

x

x

3

11

3

12

=

3

11

3

12

+

x

x . Multiply the

numerator and denominator by the common denominatorx 3 to get

3

3

3

11

3

12

+

x

x

x

x

=2

72

13

1)3(2

+

=

+

x

x

x

x.

40. (B) Triangle QRSmust be a 30-60-90 triangle because the longer

leg is 3 times longer than the shorter leg. The hypotenuse of the right

triangle has length 24. The circumscribed circle about a right triangle

must have its center at the midpoint of the hypotenuse, so the radius of

the circle is 12, and the area of the circle is 144.

41. (E) The graph off(x) = |2x 1| |x+ 2| intersects the graph ofy=

5 atx= 2 and atx= 8.

42. (D) Because the graph is translated right 2 and up 1, you need to

find those points for whichf(x) = 1. These are atx= 3 and 3. The graph

ofg(x) will cross thex-axis 2 points to the right of wheref(x) = 1, so

x= 1 and 5.


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43. (D) Joining the midpoints of the sides of a quadrilateral forms paral-

lelograms whose lengths are 1/2 the length of the diagonal of the larger

square. The lengths of the sides of the squares, in reduced order, are

AB= 20, 10 2 , 10, 5 2 , 5 = EF. The area of EFGHis 25.

44. (C)f(x) =6

3

x x

x

=( 2)( 3)

3

x x

x

=x + 2. With the exception

of the point (3,5), which is removed from the graph of f(x), the two

graphs are exactly the same.

45. (D) When the 25-foot ladder originally was at a height of 24 feet,

the foot of the ladder was 7 feet from the base of the building. This can

be determined using the Pythagorean Theorem: 242+ 72= 252. When

the top of the ladder falls to a height of 16 feet, the distance from the

foot of the ladder to the base of the building must satisfy the equation

162+ b2= 252, so that b2= 252 162= 369, so b= 19.2 ft. The ladder

slipped 12.2 feet further from the building.

46. (D) Draw WX. WXCDis a parallelogram with one-half the area of

ABCD because WD=XCand ||WD XC . Therefore, Eis the midpoint

of the diagonals, and the distance from E to WD is half the distance

from Bto WD . If his the distance from Bto WD , the area ofDEW is

1 1 1

( )2 2 4

WD h WD h

=1 1 1

( )4 2 8

AD h AD h

=1

8areaABCD.

47. (C) Extend radius AO through Oto intersect the circle at D. The

altitude from Bto the diameter forms a right triangle with mDOB=

60. The length of the altitude is the opposite leg of the right triangle, so

its length is 4sin(60) = 3.46. Add this to 2, and they-coordinate of B, to

the nearest tenth, is 5.5.


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48. (B) The circumcenter is the intersection of the perpendicular bisec-

tors of the sides of the triangle. The midpoint of AC is (4.5,5.5), and

the slope of AC is 1. The equation of the perpendicular bisector to AC

isy= x+ 1. The midpoint of AB is (4,1.5), and the slope of AB is

0.5. The equation of the perpendicular bisector to AB isy= 2x+ 9.5.

Solve this system of equations to get the point5 5

2 , 36 6

.

49. (C) If his the height of the original pyramid and His the height of

the new pyramid, then the volumes of the two pyramids are related by

the equation (1/3)hs2= (1/3)(x)h(1.25s)2, wheresrepresents the length

of a side of the base of the original pyramid. Solve forxto getx= .64.

The height of the new pyramid must be 64% of the original pyramid, so

the height of the original pyramid was reduced by 36%.

50. (B)2

2

12

12

122

))((+

+

+

=

x

x

x

x

xff ) = 2

2

22

12

12

122

+

+

+

+

+

x

x

x

x

x

x

=

)2(212

)2(1122

xx

xx

=4212

224

xx

xx

=34

43

x

x

.

math level 1

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