SAT Mathematics Level 2 Practice Test

By what percent must the height be increased so that the volume of the new pyramid is the same as the volume of the original pyramid?... Diagonals AC and BD of quadrilateral ABCD are per

Practice Test

There are 50 questions on this test You have 1 hour (60 minutes) to

complete it.

1 The measures of the angles of �QRS are m∠Q = 2x + 4, m∠R = 4x − 12, and m∠S = 3x + 8 QR = y + 9, RS = 2y − 7, and QS = 3y − 13 The perimeter of �QRS is

4 The vertices of �GHK above have coordinates G(–3,4), H(1,–3), and K(2,7) The equation of the altitude to HK is

(A) 10x + y = –26 (B) 10x + y = 7 (C) 10x + y = 27

(D) x + 10y = 37 (E) x + 10y = 72

5 face area of the solid formed will be

When the figure above is spun around its vertical axis, the total sur-(A) 144π (B) 108π (C) 72π (D) 36π (E) 9π + 12

6 If f(x) = 4x2 − 1 and g(x) = 8x + 7, g  f(2) =

(A) 15 (B) 23 (C) 127 (D) 345 (E) 2115

7 If p and q are positive integers with pq = 36, then q p cannot be(A) 1

4 (B)

4

9 (C) 1 (D) 2 (E) 9

10 In isosceles �KHJ, _HJ = 8, NL HJ ⊥ , and MP HJ ⊥ If K is

10 cm from base HJ and KL = 4KH, the area of �LNH is

(A) 4 (B) 4.8 (C) 6 (D) 7.2 (E) 16

12 Tangent TB and secant TCA are drawn to circle O Diameter AB is drawn If TC = 6 and CA = 10, then CB =

(A) 2 6 (B) 4 6 (C) 2 15 (D) 10 (E) 2 33

13 Let p @ q = q p p q

- (5 @ 3) − (3 @ 5) = (A) –184 (B) –59 (C) 0 (D) 59 (E) 184

14 Grades for the test on proofs did not go as well as the teacher had hoped The mean grade was 68, the median grade was 64, and the stan-dard deviation was 12 The teacher curves the score by raising each score

16 The graph of y = f(x) is shown above Which is the graph of g(x) = 2f(x − 2) + 1?

17 The number of bacteria, measured in thousands, in a culture is modeled by the equation ( ) 2.313.21

380175

t t

e

b t

e

=+ , where t is the number of

days since the culture was formed According to this model, the culture can support a maximum population of

(A) 2.17 (B) 205 (C) 380 (D) 760 (E) ∞

18 A sphere with diameter 50 cm intersects a plane 14 cm from the center of the sphere What is the number of square centimeters in the area of the circle formed?

(A) 49π (B) 196π (C) 429π (D) 576π (E) 2304π

19 Given g(x) =

92

13+

49

−+ (C)

10

21 80

x x

− +

22 In isosceles trapezoid WTYH, WH XZ TY|| || , m∠TWH = 120, and

m nals If WH = 30, determine the ratio of XZ:TY.

∠HWE = 30 XZ passes through point E, the intersection of the diago-(A) 1:2 (B) 2:3 (C) 3:4 (D) 4:5 (E) 5:6

23 est tenth of a degree, the measure of the largest angle is

=

− +

26 A county commissioner will randomly select 5 people to form a non-partisan committee to look into the issue of county services If there are 8 Democrats and 6 Republicans to choose from, what is the probability that the Democrats will have the more members than the Republicans on this committee?

27 Given the vectors u = [–5, 4] and v = [3, –1], |2u − 3v| =

(A) [–19, 11] (B) 502 (C) [19, 11]

(D) 2 41 3 10− (E) 2 65

28 nates of the center of the circle circumscribed about �ABC are

�ABC has vertices A(–11,4), B(–3,8), and C(3,–10) The coordi-(A) (–2,–3) (B) (–3,–2) (C) (3,2)

(D) (2,3) (E) (–1,–1)

29 Each side of the base of a square pyramid is reduced by 20% By what percent must the height be increased so that the volume of the new pyramid is the same as the volume of the original pyramid?

38 Which of the following statements is true about the function

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

value of n for which |a n| > 1,000,000?

(A) 11 (B) 12 (C) 13 (D) 14 (E) 15

41 Which of the following statements is true about the graph of the function ( ) (2 3)( 2 2)(2 1)

T to QR?

(A) 2 15 (B) 5

2 15 (C) 3 15 (D) 7

2 15 (E) 5 15

44 Diagonals AC and BD of quadrilateral ABCD are perpendicular

16 84

x x x

(E) (3,–4), (3,2), (–5,–4), (–5,2)

49 The equation 8x6 + 72x5 + bx4 + cx3 – 687x2 – 2160x – 1700 = 0, as

plex roots is

7225

− (D) 2184

7225

(E) 5544

7225

Level 2 Practice Test Solutions

KKKK

dard form, this makes the equation 10x + y = C Substitute the coordi-5 (B) The figure formed when the figure is rotated about the vertical axis is a hemisphere The total surface area of the figure is the area of the hemisphere (2πr2) plus the area of the circle that serves as the base (πr2)

6=1, and

18

2 =9 The ratio of the factors cannot be 2.

9 (A) Rewrite the equation with a common base `125j =

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,

∠TCB is a right angle and �TCB a right triangle Using the Pythagorean Theorem, TC2 + CB2 = TB2 yields 36 + CB2 = 96 CB2 = 60, so CB =

2

area FDE area ABC= =

follow the motions

17 (C) The end behavior of a rational function is not impacted by any constants that are added or subtracted to a term Consequently, the 175

in the denominator of b(t) will have no impact on the value of b(t) when

the values of t get sufficiently large The function b(t) reduces to 380 when t

is sufficiently large (Graphing this function with your graphing calculator gives a very clear picture of the maximum value of the function.)

1 3 2

1 9 2

1 3 3

x x x

x

=

92

92992

132

192

133

x

x x

x x

x

=

) 9 2

+

x x

21 (D) log3(a) = c and log3(b) = 2c are equivalent to a = 3 c and b = 3 2c

= (3c)2 = a2 Therefore, a = b

22 (B) Drop the altitude from E to WH with the foot of the altitude being L LH = 15 Use the 30-60-90 relationship to determine that

LE = 5 3 and EH = 10 3 �EHZ is also 30-60-90, so HZ = 10 and

EZ = 20 �EYZ is an isosceles triangle, making ZY = 20 In �THY,

XZ

TY =

23 (B) With SSS information, use the Law of Cosines to determine the measure of the largest angle (which is located opposite the longest side)

342 = 252 + 292 − 2(25)(29) cos(θ) 2(25)(29) cos(θ) = 252 + 292 − 342,

so that cos^i =h 252 25 292+^ 29h^2-34h 2 or cos^ hi = 14531 and θ = 77.7°.

24 (A) If -54 + 3 28 i is a root of the equation, then its conjugate

Getting a com-a = 800, b = 1280, and c = 737, the equation is 800x2 + 1280x + 737 = 0.

25 (D) cos(2t) = x − 1 and sin(t) = y-32 Use the identity cos(2t) =

1 − 2sin2(t) to get x-1=1 2- cy-3 2m2 or x-2=-92^y-2h2

29 (D) The volume of the original pyramid is s2h

3

1

The volume of the new pyramid is (8s)2H

3

1 , where H is the new height Set these

expressions equal to get s2h

a

c b

b

b

) ( log 2

log 3

1 log

3

a

c b

2

log 3

4 log

3

5

a

c b

b

y a

The focus is 10 units from the center of the hyperbola The equation of the asymptotes for a horizontal hyperbola (found by replacing the 1 with

the answer to this problem would be 'r6 6,5r,32r1 Subtract 4π from each of these answer to satisfy the domain restriction of the problem,

(B) Sketch the graph of y = |2x − 3||x − 1| − 2 on your graphing cal-the interval from

5

1 < x < 5.

Algebraically: Determine those values for which |2x − 3||x − 1| = 2 and then analyze the inequality |2x − 3||x − 1| = 2 requires that 2x – 3|x – 1| = 2 or 2x – 3|x – 1| = –2.

If 2x – 3|x – 1| = 2 then 2x – 2 = 3|x – 1| If x comes 2x – 2 = 3x – 3 or x = 1 If x < 1, the equation becomes 2x – 2 = –3x + 3 or x = 1.

> 1, this equation be-If 2x – 3|x – 1| = –2, then 2x + 2 = 3|x comes 2x + 2 = 3x – 3 or x = 5 If x < 1, the equation becomes 2x + 2 = –3x + 3 or x = 1

> H [C] =

5

2 36-

39 (C) The polar coordinate form for 32i is 32cis`r2j DeMoivre’s

Theorem states that if z = rcis( θ), then z n = r n cis(n θ) The radius is 2 for

all the choices, so the leading coefficient is not the issue Of the five choices, only 2cis 10`11rj fails to be co-terminal with 10r + 2n5r

40 (D) The terms of the sequence generated by this recursion formula are: 4, –3, 17, 43, –95, –319, 767, 2491, –5939, –19351, 46175, 150403, –358859, –1168927, 2789063, 9084907

9log8(x − 3) − 5 = 13 Add 5 and divide by 9 to get log8(x − 3) = 2 Rewrite the logarithmic equation as an exponential equation: x − 3 = 82

= 64 so x = 67.

43 (C) The centroid is the intersection of the medians of the triangle

The median from Q to RS is also the altitude to RS Use the Pythagorean

Theorem to show that this altitude has length 602 −152 =15 15 The centroid lies 2

vertex angle, so m∠ADE = 30 With the hypotenuse of the triangle hav-leg 4 and hypotenuse 6 Use the Pythagorean Theorem to determine

that BE = 2 5 The area of a quadrilateral with perpendicular diagonals

is equal to half the product of the diagonals, so the area of ABCD is

T cos(2θ) = cos2(θ) − sin2(θ)

=

2 2

2

4

5 4

9 8

If n is even, there will be n2 + 1 odd terms and n2 even terms in the

expansion of (a − b) n The odd terms will subtract out, leaving n2 terms

) 1 ( x  2  y  2

Solve

so that (y − 1)2 = 9 or y − 1 = 3 and y = 4 or –2 Check that these are the same values of y you get when x = –5 The coordinates of the

four points of intersection for these two graphs are (3,4), (3,–2), (–5,4), and (–5,–2)

49 (B) The graph of the function shows that x = –2.5 is a ble root, and x = –2 and x = 2 are single roots This implies that (2x + 5)2(x + 2)(x – 2) is a factor of 8x6 + 72x5 + bx4 + cx3 − 687x2 −

dou-2160x – 1700 = 0 When expanded, (2x + 5)2(x + 2)(x – 2) is a

be of the form 2x2 + bx + 17 in order for the leading coefficient to be 8

and the constant to be –1700 The product of the complex roots from the quadratic is 17