Math 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.

The table below shows the 2010 human population and projected

2025 human population for different regions. In which region is the greatest percent increase predicted? Region

2010

2025

Africa

1,033,043

1,400,184

Asia

4,166,741

4,772,523

Latin America and the Caribbean

588,649

669,533

North America

351,659

397,522

Oceania

35,838

42,507

(A) Africa

(B) Asia

(D) North America 2.

(2,6) M (2,6)

(C) Latin America and the Caribbean (E) Oceania

is the midpoint of

AB .

If A has coordinates (10,12), the

coordinates of B are (A) (6,10) (D) (18,16)

(B) (–6,0) (E) (22,18)

(C) (–8,–4)

2

3.

MATH 1 & 2 SUBJECT TEST SAT MATH

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

of the solid formed will be

(A) 9π 4.

If f If f ( x ) =

(B) 36π x

2

(A) 0 5.

2

 x 

6

6 x  8

(C) 72π

(D) 144π

(E) 288π

(D) 24.5

(E) Undefined

, f (2) (2) =

(B) 5.75

(C) 6.25

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 purchases is 3:1, what is the average income per ticket sold? (A) $5.50 6.

(B) $5.75

(C) $6.50

(D) $7.25

(E) $14.50

Due to poor poor economic 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

1 1 , is subtracted from the result, and 3 2 the square root of the end result is 4. What was the original number? 1 (A) 1 (B) 5 1 (C) 15 1 (D) 16 (E) 49 2 2 2 6

7.

A value z is multiplied by

3

SAT MATHEMATICS LEVEL 1 PRACTICE TEST

8.

If two fair dice are rolled, what is the probability that the sum of the

dice is at most 5? (A) 5 (B) 6 36 36 9.

(C)

10 36

If 5 x + 2 = 7 , then 1 x = 8 3 12 2 –1 –2 (A) ___ (B) ___ (C) 1 15 15

(D)

26 36

(D) 2

(E)

30 36

(E) 4

 x 2  2x  8   6  3x  10. =  x 2  4   20  5x  –3 (A) ___ 5 (D)

3( x  4)(2  x ) (C) 5( x  2)(4  x )

(B) _3_ 5

3( x  4)( x  2) 5( x  2)( x  4)

(E)

3( x  4) 5(4  x )

11. If i2 = –1, then (5 + 6i)2 = (A) –11

(B) –11 + 11i

(C) –11 + 30i

(D) –11 + 60i

(E) 61 12. The mean of 48, 27, 36, 24, x , and 2 x is 37. x = 3 1 2 (A) 13 (B) 16 (C) 29 (D) 33 4 3 3 13.

3

(E) 40

32 x 6 y8 =

(A)

4 x y

3

4 3

2

(D)

2 x y

2

2 3

4y

(B) 2

2

2 x y

(E)

4 3

2

2 x y

4 3 3

2y

(C)

2

2 x y

3 3

4y

2

2

14. A circle is inscribed in a square of side length 6. The area of the region inside the square but outside the circle is (A) 36π (D) 36 − 9π

(B) 36π − 9 (E) 9π − 36

(C) 36π − 36

4

SAT MATH 1 & 2 SUBJECT TEST

__

15. If the binary operation a # b = a – √b, then (2 # 4) − (4 # 2) = b

__

__

(B) √2 – 2

(A) –32

(D) √2 – 2

(C) 0

(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 trapezoid ACDE with AC || DE is shown below. E is the midpoint of AB, and BD = DC and BC = DE.

The ratio of the area of triangle BDC to trapezoid ACDE is (A) 1:2 19. If 3

2 4

(A) –21

(B) 1:3 3 1

2

(C) 1:4

x

y

3

4

(B) –15

(D) 1:5

(E) 1:6

10 11 =   , then x + y − z = z 5   

(C) 0

(D) 15

(E) 21

20. If f ( x ) = 5 x + 3 and g ( x ) = x 2 − 1, then f ( g (2)) = (A) 3

(B) 13

(C) 18

(D) 39

(E) 168

5


21. Chords AB and CD of circle O intersect at point E. If CE = 3, ED = 12, and AE is 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 4 x − 5 y = 17 that passes through the point (5,2)? (A) 4 x − 5 y = 10

(B) 5 x + 4 y = 33

(D) 5 x − 4 y = 17

(E) y =

-5

4

x +

(C) 4 x + 5 y = 30 15 2

23. A stone is thrown vertically into the air from the edge of a building with height 12 meters. The height of the stone is given by the formula h = –4.9t 2 + 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 � ABC to the nearest integer is (A) 613

(B) 1024

(C) 1226

(D) 2240

(E) 2252

a+b = 4 , and a and b are non-negative integers, which of the 2 following cannot be a value of ab? 25. If

(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)

(D) (–5,5)

(E) (–5,6)

(C) (–5,4)

27. The difference between the product of the roots and the sum of the roots of the quadratic equation 6 x 2 − 12 x + 19 = 0 is 7 31 7 31 7 (A) (B) (C) (D) (E) – 6 6 12 12 6

6


28. In right triangle ABC, D E ||

B C,

The sine of angle θ is equal to 1 3 2 (A) (B) (C) 2 4 2

CD = 1.5, and BE = 2.0.

(D)

3 2

(E)

3 5

29. QUEST is a pentagon. The measure of angle Q = 3 x − 20, the measure of angle U = 2 x + 50, the measure of angle E = x + 30, the measure of angle S = 5 x − 90, and the measure of angle T = x + 90. Which two angles have equal measures? (A) E and S

(B) Q and U

(D) T and E

(E) U and E

(C) U and T

30. The vertices of triangle PQR are P (–3,2), Q(1,–4), and R(7,0). The altitude drawn from Q intersects the line PR at the point (A) (1,2) (D) (–3,2)

(B) (2,1)

(C) ( 1,–2)

(E) (7,0) 2

n

4

31. If q is a positive integer > 1 such that q3n –4 (A) 1 (B) –1 (C) 1, ___ (D) –1, _4_ 3 3 −

−

=

1,

n = (E)

1 ± i 47 6

7


32. The measure of arc AB in circle O is 108°.

a + b + c = 3 (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 34. If f ( x ) =

(B) 400 x x

2

2

 x  

(A) {–5, –1}

6

6x  8

(C) 489

(D) 1112

(E) 1600

, solve f ( x ) = 3.

(B) {2, 7.5}

17  73 17  73  , (D)   6 6  

  (C) 1  3 7 , 1  3 7  2   2  (E) ∅

QX 1 = . XR 4 The ratio of the area of �QXY to the area of trapezoid XYSR is 35. In �QRS, X is on QR and Y is on QS , so that X Y ||

(A) 1:4

(B) 1:15

(C) 1:16

(D) 1:24

R S and

(E) 1:25

8


36. In quadrilateral KLMN , KL = LM , KN = MN , and diagonals KM and NL intersect at P . If KP = PM , then which of the following statements is true?

I.

NP = PL.

II.

KLMN is a rhombus.

III.

The area of KLMN is

(A) I only

(B) II only

(D) II and III only

(E) I and III only

1 (KM )(NL). 2

(C) III only

37. If 7 x + 9 y = 86 and 4 x − 3 y = –19, x + 4 y = (A) – 31

18 19

1 (B) 22 3

(C) 31

18 19

(D) 35

(E) 105

38. The solution set to 10 x 2 + 11 x – 6 ≤ 0 is (A) –0.4 ≤ x ≤ 1.5

(B) –1.5 ≤ x ≤ 0.4

(E) –1.5 ≤ x ≤ –0.4

(D) x ≤ –1.5 or x ≥ 0.4

39. In simplest form,

2 1

2 x – 7 (A) ______ x – 2 2 x + 7 (D) ______ x – 2

(C) x ≤ –0.4 or x ≥ 1.5

1 x

 3 is equivalent to

1

3  x

7 – 2 x (B) ______ x – 2

2 x – 5 (C) ______ x – 2

(E) 1

40. In right triangle QRS, QR is perpendicular to RS, QR = 12, and RS = 12 3 . The area of the circle that circumscribes triangle QRS is (A) 108π

(B) 144π

(C) 288π

(D) 576π

(E) 1728π

9


41. The solution set for the equation |2 x − 1| − |x + 2| = 5 is (A) {–2}

(B) {3}

(C) {8}

(D) {3, –7}

(E) {–2, 8}

42. Given the graph of f ( x ) below, let g ( x ) = f ( x – 2) + 1. For what set of values of will g ( x ) = 0?

(A) {–2, 2, 4} (D) {–1, 5}

(B) {0, 4, 6}

(C) {–4, 2}

(E) ∅

43. Square ABCD has sides with length 20. Each of the smaller figures is 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

10


2 x _________ 44. Which of the statements about the graphs of f ( x ) = – x – 6 and x – 3 g ( x ) = x + 2 are true?

I. f ( x ) + g ( x ) = 2 x + 4 II. They intersect at one point III. They are the same except for one point (A) I only

(B) II only

(D) I and III only

(C) III only

(E) II and III only

45. A 25-foot ladder leans against a building. As the bottom of the ladder at point A slides 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 �DEW to the area of ABCD? (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 parallel to the x -axis and m∠ AOB = 120 degrees. To the nearest tenth, the y-coordinate of point B is (A) 3.5

(B) 4.0

(C) 5.5

(D) 6.6

(E) 6.9

11


� ABC

48. The vertices of

have coordinates A(–7,3), B(–1,0), and

C(–2,8). The coordinates of the center of the circumscribed circle are  5 5 (B)  2 , 3   6 6

 1 2 (A)  3 , 3   3 3

 1  (C)  1 , 4  2 

1 1 (E)  4 ,5   2 2

(D)  4,1 1   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 by x %, so that the volume of the pyramid is unchanged. x = (A) 20

(B) 25

(A)

x

(D)

2

2

(D) 50

(E) 64

2 x  1 , then f ( f ( x )) = x  2

50. If f ( x )  4 x

(C) 36





4x  1

(B)

(E)

3 x + 1 4 x + 3

4x  4

3 x 4 x + 3

3 x  4

4 x  3

(C)

4 x  3 3 x  4

Level 1 Practice Test Solutions

1. (A) The percent increase is computed as

D population

original population

. 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 America and the Caribbean

588,649

669,533

80,884

13.7

North America 351,659

397,522

45,863

13

Oceania

42,507

6,669

18.6

35,838

12


2. (B) Let B have coordinates ( x , y). The formula for finding the midpoint of a line segment is to average the x -coordinates and average 10 + x 12 + y = 2 and = 6. 2 2 Multiply each equation by 2. 10 + x = 4 gives x = –6 and 12 + y = 12 the y-coordinates. This gives the equations

gives y = 0. Point B must have coordinates (–6,0). 3. (E) The solid formed will be a hemisphere with radius 6. The volume of a sphere is given by the formula V = _4_πr 3. The volume of the 3 hemisphere will be half that number. With r = 6, the volume of the hemisphere is _2_π(6)3 = 288π. 3 (2)2  2  6 0  4. (A) f (2) = = 0. (2)2  6(2)  8 24

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 ticket raises $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 both sides of the equation to get equation to get

1

1 z

3



2



16 .

Add

1

1 z

3



2

= 4. Square

1 to both sides of the 2

1 1 1 z = 16 . Multiply by 3 for the answer: z = 49 . 3 2 2

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

13


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 subtracting and then multiplying by

8 5

to get x =

-

2 from both sides of the equation 3 1 2 . Half this amount is . 15 15 -

 x 2  2x  8   6  3x  ( x 4)( x 2) 3(2 x ) = =    2 ( 2)( 2) 5(4 ) x  4 x x x x  20 5    3(2 x ) . 2 − x and x – 2 are negatives of one another, so they 5(4 x )

10. (B) ( x 4) ( x 2)

reduce to be –1. The same is true for x − 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, so x + 2 x = 222 − 135 = 87 and x = 29.

13. (D) 32 = 8 3

y

8

=

^y h 8

1 3

=

y

8 3

=

4 = 23



y

2

2 3

=

y

2 3



4, so

y . 2

3

6

32 x y

3

8

32

=

=

23 4 .

2

2 x y

2 3

4y

3

2

x

6

=

x and 2

.

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

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

circle from the area of the square to get 36 – 9π. __

__

__

15. (B) 2 # 4 = 2 – √ 4 = 16 − 2 = 14. 4 # 2 = 4 – √2 = 16 – √2 . (2 # 4

2

__

__

4) − (4 # 2) = 14 – (16 – √2 ) = –2 + √2 .

14


16. (A) 25% of the gas used by all 45 European countries comes from Russia. Let x represent the percent usage by the remaining 32 countries. 7(1) + 6(.5) + 32 x = 0.3906. The = .25 is x 45 rest of Europe imports an average of 3.9% of its gas from Russia. The solution to the equation

17. (A) The center of the circle is the midpoint of A and B,

3 

2

9

=

3 and 9  ( 1) = 4. The equation of the circle is ( x − 3)2 + ( y − 4)2 = r 2. 2 Substituting either A or B into the equation for x and y gives r 2 = 61. 18. (B) Because A C ||

D E,

the heights of trapezoid ACDE and triangle

BDC are the same. Trapezoid ACDE is isosceles, so AE and BCD are equal, as are the measures of angles A and C. Consequently, BD and AE are equal, as are the measures of angles A and DBC, so ABDE is a parallelogram and AB = DE. The area of trapezoid ACDE is given by the 1 h( DE + AC ) . ED = BC and AC = 2BC, so this equa2 1 tion becomes Area = h( DE + AC ) . The ratio of the area of triangle 2 1 1 DBC to the area of trapezoid ACDE is h(BC) : h(3 BC ) or 1 : 3. 2 2 equation Area =

19. (D) matrices

3

6

2 4 2 x 18

3

2

1 9

x

y

3

4

2y 5

6

2x 18

10 z

11 5

9

2y 5

. The equality of the

means that 6 − 2 x = 10, so x = –2;

9 − 2 y = 11, so y = –1; and z = –18. Therefore, x + y − z = 15. 20. (C) To find f ( g (2)), first determine g (2), then substitute this value into f ( x ): g (2) = 22 – 1 = 3, so f ( g (2)) = f (3) = 5(3) + 3 = 18.

15


21. (D) When chords intersect inside a circle, the products of the segments 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 and AB has a length of 13. 22. (B) The slope of the line 4 x − 5 y = 17 is dicular line is

-

5 4

4 5

. The slope of the perpen-

. The equation of the line is y =

5 for x and 2 for y gives 2 =

-5

4

-5

^5h + b so that b =

4

x + b . Substituting

33 4

. y =

-5

4

x +

33 4

becomes 4 y = –5 x + 33 or 5 x + 4 y = 33. It is faster to know that a line perpendicular to Ax + By = C has the equation Bx − Ay = D. You could then have started the problem with 5 x + 4 y = D and determined that D = 33. 23. (C) The time the stone reaches its maximum height can be computed 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 � ABC can be calculated using trigonometry. Drop the height from A to BC and call the foot of the altitude AD so that AD = 40 sin(50). The area of the triangle 40 1 1 is ( BC )( AD ) = (80)(40) sin(50). 2 2 D. Then 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.

16


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 is y − 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 is – b

a

=

12 6

. The difference between these numbers is

28. (B) sin ^ ih

c 19 . The sum of the roots is = a 6

=

sin( ) DC EB

=

AD 

AE

1.5 2.0

=

. 3 4

With

DE

|| B C ,

AD AE

7 6

=

. DC . EB

Therefore,

.

29. (C) The sum of the measures of the angles is 12 x + 60. The sum of the interior angles of a pentagon is equal to 3(180) = 540. (Five sides implies three non-overlapping triangles.) Solve for x = 40. m∠U = m∠T = 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 ∠C has a measure of 54. Draw OC . ∠ A and ∠ ACO are congruent, as are ∠B and ∠BCO. m∠ A + m∠B = m∠ ACO + m∠BCO = 54. Therefore,

a+b+c 3

=

54 + 54 = 36. 3

17


33. (D) Alex’s distance from the base of Mount Colin is computed 800 . If the angle of elevation is doubled, then Alex must by _______ tan(23) 800 feet from the base. So Alex must move _______ 800 − _______ 800 be _______ tan(23) tan(46) tan(46) = 1112 feet closer to the base of Mount Colin. x + x – 6 so that 3( x 2 − 6 x + 8) = x 2 + x − 6. 34. (B) f ( x ) = 3 becomes 3 =__________ x 2 – 6 x + 8 2

Multiply through the left side of the equation to get 3 x 2 − 18 x + 24 = x 2 + x − 6 and then 2 x 2 − 19 x + 30 = 0. This factors to (2 x − 15)( x − 2) = 0 so x = 2, 15/2. 35. (D)

�QXY ~ �QRS,

so the ratio of the areas of the two trian-

gles will be equal to the square of the ratio of corresponding sides. Since QX 1 , QX 1 , so area QXY 1 . The area of trapezoid XYSR is = = area QRS 25 4 XR QR 5 1 the area of �QRS − the area of �QXY . Therefore, area QXY . 

areatrap XYSR

36. (C)



24

�KLM is

an isosceles triangle with legs KL and LM . P is the midpoint of KM because KP = PM , so LP is perpendicular to KM . LP is part of diagonal NL , so the diagonals of the quadrilateral are perpendicular. Whenever the diagonals of a quadrilateral are perpendicular, the area is equal to one-half the product of the diagonals (look at the areas of �KLM and �KNM and add them). There is no information to allow you to deduce that KN = LN , and therefore you cannot conclude that the quadrilateral is a rhombus nor is there any information to conclude the diagonals bisect each other to insure that NP = PL. 37. (D) Subtract the two equations to get 4 x + 12 y = 105. Divide by 3 to get x + 4 y = 35.

18


38. (B) y = 10 x 2 + 11 x – 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 10 x 2 + 11 x – 6 ≤ 0, is satisfied by –1.5 ≤ x ≤ 0.4. 39. (A) Because 3 − x = –( x − 3),

2− 1−

1 x

2−

− 3 = 1

1+

3 − x

1 x − 3

1 x −

. Multiply the

3

numerator and denominator by the common denominator x − 3 to get 1   2−  x − 3  = 2( x − 3) − 1 2 x − 7 . x − 3    = x − 3 + 1 x + 2  1 + 1   x − 3     x − 3 

40. (B) Triangle QRS must 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 144π. 41. (E) The graph of f ( x ) = |2 x − 1| − |x + 2| intersects the graph of y = 5 at x = –2 and at x = 8.

42. (D) Because the graph is translated right 2 and up 1, you need to find those points for which f ( x ) = –1. These are at x = –3 and 3. The graph of g ( x ) will cross the x -axis 2 points to the right of where f ( x ) = –1, so x = –1 and 5.

19


43. (D) Joining the midpoints of the sides of a quadrilateral forms parallelograms 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 EFGH is 25.

 x  6 ( x  2)( x  3) = = x + 2. With the exception x  3 x  3 of the point (3,5), which is removed from the graph of f ( x ), the two 44. (C) f ( x ) =

x

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 . WXCD is a parallelogram with one-half the area of ABCD because WD = XC and WD || XC . Therefore, E is the midpoint of the diagonals, and the distance from E to WD is half the distance from B to WD . If h is the distance from B to WD , the area of �DEW is 1 1   WD   h  2 2 



11 1  (WD )h =  AD  h 4 42 



1 ( AD)h = 8

1 8

area ABCD .

47. (C) Extend radius AO through O to intersect the circle at D. The altitude from B to the diameter forms a right triangle with m∠DOB = 60°. The length of the altitude is the opposite leg of the right triangle, s o its length is 4sin(60) = 3.46. Add this to 2, and the y-coordinate of B, to the nearest tenth, is 5.5.

20


48. (B) The circumcenter is the intersection of the perpendicular bisectors 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 is y = – 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 is y = 2 x + 9.5.

 5 5 Solve this system of equations to get the point  2 , 3  .  6 6 49. (C) If h is the height of the original pyramid and H is 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(1.25 s)2, where s represents the length of a side of the base of the original pyramid. Solve for x to get x = .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%.

 2 x − 1   −1 x 2 +  = 50. (B) f ( f ( x)) =   2 x − 1   +2  x + 2  2

2 2 x 1 2 x 1

1( x 2( x

2) 2)

=

  2 x − 1    2  −1  2 x +   x + 2 =     2 x − 1   x + 2 +2      x + 2 

4 x  2  x  2 2 x  1  2 x  4

=

3 x  4 4 x  3

.

Math Level 1

Recommend Documents