Barron's SAT Subject Test Math Level 2, 10th Edition (2012)

Part 3. MODEL TESTS

Answer Sheet
MODELTEST 4

Model Test 4

Tear out the preceding answer sheet. Decide which is the best choice by rounding your answer when appropriate. Blacken the corresponding space on the answer sheet. When finished, check your answers with those at the end of the test. For questions that you got wrong, note the sections containing the material that you must review. Also, if you do not fully understand how you arrived at some of the correct answers, you should review the appropriate sections. Finally, fill out the self-evaluation chart in order to pinpoint the topics that give you the most difficulty.

1. If point (a,b) lies on the graph of function f, which of the following points must lie on the graph of the inverse f ?

 (A) (a,b)

 (B) (–a,b)

 (C) (a,–b)

 (D) (b,a)

 (E) (–b,–a)

2. Harry had grades of 70, 80, 85, and 80 on his quizzes. If all quizzes have the same weight, what grade must he get on his next quiz so that his average will be 80?

 (A) 85

 (B) 90

 (C) 95

 (D) 100

 (E) more than 100

3. Which of the following is an asymptote of · tan πx ?

 (A) x = –2

 (B) x = –1

 (C)

 (D) x = 1

 (E) x = 2

4. If log_b x = p and log_b y = q, then log_b xy =

 (A) pq

 (B) p + q

 (C)

 (D) p – q

 (E) p^q

5. The sum of the roots of 3x³ + 4x² – 4x = 0 is

 (A)

 (B)

 (C) 0

 (D)

 (E) 4

6. If then

 (A) 0

 (B)

 (C)

 (D)

 (E) 1

7. If f(x) = log(x + 1), what is f ^–1(3)?

 (A) 0.60

 (B) 4

 (C) 999

 (D) 1001

 (E) 10,000

8. If f(x) 0 for all x, then f(2 – x) is

 (A) –2

 (B) 0

 (C) 2

 (D) 0

 (E) 2

9. How many four-digit numbers can be formed from the numbers 0, 2, 4, 8 if no digit is repeated?

 (A) 18

 (B) 24

 (C) 27

 (D) 36

 (E) 64

10. If x –1 is a factor of x² + ax – 4, then a has the value

 (A) 4

 (B) 3

 (C) 2

 (D) 1

 (E) none of the above

11. If 10 coins are to be flipped and the first 5 all come up heads, what is the probability that exactly 3 more heads will be flipped?

 (A) 0.0439

 (B) 0.1172

 (C) 0.1250

 (D) 0.3125

 (E) 0.6000

12. If and n is a positive integer, which of the following statements is FALSE?

 (A) i ⁴ⁿ = 1

 (B) i ^{4n + 1} = –i

 (C) i ^{4n + 2} = –1

 (D) i ^{n + 4} = iⁿ

 (E) i ^{4n + 3} = –i

13. If log_r 3 = 7.1, then log_r

 (A) 2.66

 (B) 3.55

 (C)

 (D)

 (E)

14. If f(x) = 4x² and g(x) = f(sin x) + f(cos x), then g(23°) is

 (A) 1

 (B) 4

 (C) 4.29

 (D) 5.37

 (E) 8

15. What is the sum of the roots of the equation

 (A) – 0.315

 (B) – 0.318

 (C) 1.414

 (D) 3.15

 (E) 4.56

16. Which of the following equations has (have) graphs consisting of two perpendicular lines?

  I. xy = 0

 II. |y | = |x |

  III. |xy | = 1

 (A) only I

 (B) only II

 (C) only III

 (D) only I and II

 (E) I, II, and III

17. A line, m, is parallel to a plane, X, and is 6 inches from X. The set of points that are 6 inches from m and 1 inch from X form

 (A) a line parallel to m

 (B) two lines parallel to m

 (C) four lines parallel to m

 (D) one point

 (E) the empty set

18. In the figure above, if VO = VY, what is the slope of segment VO?

 (A)

 (B)

 (C)

 (D)

 (E) Cannot be determined from the given information.

19. A cylindrical bar of metal has a base radius of 2 and a height of 9. It is melted down and reformed into a cube. A side of the cube is

 (A) 2.32

 (B) 3.84

 (C) 4.84

 (D) 97.21

 (E) 113.10

20. The graph of y = (x + 2)(2x – 3) can be expressed as a set of parametric equations. If x = 2t – 2 and y = f(t), then f(t) =

 (A) 2t(4t – 5)

 (B) (2t – 2)(4t – 7)

 (C) 2t(4t – 7)

 (D) (2t – 2)(4t – 5)

 (E) 2t(4t + 1)

21. If points and lie on the graph of y = x³ + ax² + bx + c, and y₁ – y₂ = 3, then b =

 (A) 1.473

 (B) 1.061

 (C) –0.354

 (D) –0.939

 (E) –2.167

22. Rent-a-Rek has 27 cars available for rental. Twenty of these are compact, and 7 are midsize. If two cars are selected at random, what is the probability that both are compact?

 (A) 0.0576

 (B) 0.0598

 (C) 0.481

 (D) 0.521

 (E) 0.541

23. If a and b are real numbers, with a > b and |a| < |b|, then

 (A) a > 0

 (B) a < 0

 (C) b > 0

 (D) b < 0

 (E) none of the above

24. If [x] is defined to represent the greatest integer less than or equal to x, and the maximum value of f(x) is

 (A) –1

 (B) 0

 (C)

 (D) 1

 (E) 2

25.

 (A) 0

 (B) 1

 (C) 2

 (D) 3

 (E)

26. A right circular cone whose base radius is 4 is inscribed in a sphere of radius 5. What is the ratio of the volume of the cone to the volume of the sphere?

 (A) 0.222 : 1

 (B) 0.256 : 1

 (C) 0.288 : 1

 (D) 0.333 : 1

 (E) 0.864 : 1

27. If x₀ = 1 and then x₃ =

 (A) 1.260

 (B) 1.361

 (C) 1.396

 (D) 1.408

 (E) 1.412

28. The y-intercept of is

 (A) 0.22

 (B) 0.67

 (C) 1.41

 (D) 1.49

 (E) 4.58

29. If the center of the circle x² + y² + ax + by + 2 = 0 is point (4,–8), then a + b =

 (A) –8

 (B) –4

 (C) 4

 (D) 8

 (E) 24

30. If p(x) = 3x² + 9x + 7 and p(a) = 2, then a =

 (A) only 0.736

 (B) only –2.264

 (C) 0.736 or 2.264

 (D) 0.736 or –2.264

 (E) –0.736 or –2.264

31. If i is a root of x⁴ + 2x³ – 3x² + 2x – 4 = 0, the product of the real roots is

 (A) –4

 (B) –2

 (C) 0

 (D) 2

 (E) 4

32. If and 270° B 360°, sin(A + B) =

 (A) –0.832

 (B) –0.554

 (C) –0.333

 (D) 0.733

 (E) 0.954

33. If a family has three children, what is the probability that at least one is a boy?

 (A) 0.875

 (B) 0.67

 (C) 0.5

 (D) 0.375

 (E) 0.25

34. If sec 1.4 = x, find the value of csc(2 tan⁻¹ x).

 (A) 0.33

 (B) 0.87

 (C) 1.00

 (D) 1.06

 (E) 3.03

35. The graph of |y – 1| = |x + 1| forms an X. The two branches of the X intersect at a point whose coordinates are

 (A) (1,1)

 (B) (–1,1)

 (C) (1,–1)

 (D) (–1,–1)

 (E) (0,0)

36. For what value of x between 0° and 360° does cos 2x = 2 cos x ?

 (A) 68.5° or 291.5°

 (B) only 68.5°

 (C) 103.9° or 256.1°

 (D) 90° or 270°

 (E) 111.5° or 248.5°

37. For what value(s) of x will the graph of the function have a maximum?

 (A)

 (B)

 (C)

 (D)

 (E)

38. For each positive integer n, let S_n = the sum of all positive integers less than or equal to n. Then S₅₁ equals

 (A) 50

 (B) 51

 (C) 1250

 (D) 1275

 (E) 1326

39. If the graphs of 3x² + 4y² – 6x + 8y – 5 = 0 and (x – 2)² = 4(y + 2) are drawn on the same coordinate system, at how many points do they intersect?

 (A) 0

 (B) 1

 (C) 2

 (D) 3

 (E) 4

40. If log_x 2 = log₃x is satisfied by two values of x, what is their sum?

 (A) 0

 (B) 1.73

 (C) 2.35

 (D) 2.81

 (E) 3.14

41. Which of the following lines are asymptotes for the graph of

  I. x = –1

 II. x = 5

 III. y = 3

 (A) I only

 (B) II only

 (C) I and II

 (D) I and III

 (E) I, II, and III

42. If and 0° 180°, then =

 (A) 0°

 (B) 0° or 180°

 (C) 80.5°

 (D) 0° or 80.5°

 (E) 99.5°

43. If f(x,y) = 2x² – y² and g(x) = 2^x, which one of the following is equal to 2^2x?

 (A) f(x,g(x))

 (B) f(g(x),x)

 (C) f(g(x),g(x))

 (D) f(g(x),0)

 (E) g(f(x,x))

44. Two positive numbers, a and b, are in the sequence 4, a, b, 12. The first three numbers form a geometric sequence, and the last three numbers form an arithmetic sequence. The difference b – a equals

 (A) 1

 (B)

 (C) 2

 (D)

 (E) 3

45. A sector of a circle has an arc length of 2.4 feet and an area of 14.3 square feet. How many degrees are in the central angle?

 (A) 63.4°

 (B) 20.2°

 (C) 14.3°

 (D) 12.9°

 (E) 11.5°

46. The y- coordinate of one focus of the ellipse 36x² + 25y² + 144x – 50y – 731 = 0 is

 (A) –2

 (B) 1

 (C) 3.32

 (D) 4.32

 (E) 7.81

47. In the figure above, ABCD is a square. M is the point one-third of the way from B to C. N is the point one-half of the way from D to C. Then =

 (A) 50.8°

 (B) 45.0°

 (C) 36.9°

 (D) 36.1°

 (E) 30.0°

48. If f is a linear function such that f(7) = 5, f(12) = –6, and f(x) = 23.7, what is the value of x?

 (A) –3.2

 (B) –1.5

 (C) 1

 (D) 2.4

 (E) 3.1

49. Under which of the following conditions is negative?

 (A) x < y < 0

 (B) y < x < 0

 (C) 0 < y < x

 (D) x < 0 < y

 (E) all of the above.

50. The binary operation * is defined over the set of real numbers to be . What is the value of 2 * (5 * 3)?

 (A) 1.84

 (B) 2.14

 (C) 2.79

 (D) 3.65

 (E) 4.01

If there is still time remaining, you may review your answers.

Answer Key
MODEL TEST 4

ANSWERS EXPLAINED

The following explanations are keyed to the review portions of this book. The number in brackets after each explanation indicates the appropriate section in the Review of Major Topics (Part 2). If a problem can be solved using algebraic techniques alone, [algebra] appears after the explanation, and no reference is given for that problem in the Self-Evaluation Chart at the end of the test.

An asterisk appears next to those solutions in which a graphing calculator is necessary.

1. (D) Since inverse functions are symmetric about the line y = x, if point (a,b) lies on f, point (b,a) must lie on f ^–1. [1.1]

2. * (A) Average . Therefore, x = 85. [4.1]

3. * (C) Plot the graph y = (x² + 3x + 2)/(x + 2)tan(x) in a [–2.5,2.5] by [–5,5] window and use TRACE to approximate the location of asymptotes. The only answer choice that could be an asymptote occurs at .

An alternative solution is to factor the numerator of f(x) and observe that the factor x + 2 divides out: tan x=(x + 1) tan x. The only asymptotes occur because of tan x, when x is a multiple of . Setting yields . [1.2, 1.3]

4. (B) The bases are the same, so the log of a product equals the sum of the logs. [1.4]

5. * (A) Plot the graph of y = 3x³ + 4x² – 4x in the standard window and use CALC/zero to find the three zeros of this function. Sum these three values to get the correct answer choice.

An alternative solution is first to observe that x factors out, so that x = 0 is one zero.

The other factor is a quadratic, so the sum of its zeros is . [1.2]

6. (A) and Therefore, . [1.1]

7. (C) By the definition of f, log(x + 1) = 3. Therefore, x + 1 = 10³ = 1000 and x = 999. [1.4]

8. (B) The f(2 – x) just shifts and reflects the graph horizontally; it does not have any vertical effect on the graph. Therefore, regardless of what is substituted for x, f(x) 0. [2.1]

9. (A) Only 3 of the numbers can be used in the thousands place, 3 are left for the hundreds place, 2 for the tens place, and only one for the units place. 3 · 3 · 2 · 1 = 18. [3.1]

10. (B) Substituting 1 for x gives 1 + a – 4 = 0, and so a = 3. [1.2]

11. * (D) The first 5 flips have no effect on the next 5 flips, so the problem becomes “What is the probability of getting exactly 3 heads in 5 flips of a coin?” outcomes contain 3 heads out of a total of 2⁵ = 32 possible outcomes. P(3H) = 0.3125. [4.2]

12. (B) i ⁴ⁿ = 1; i ^{4n + 1} = i; i ^{4n + 2} = –1; i ^{4n + 3} = –i; i ^{4n + 4} = (i ⁴ⁿ)(i ⁴) = (1)(1). [3.2]

13. (B) Since .[1.4]

14. (B) The 23° is superfluous because g(x) = f(sin x) + f(cos x) = 4 sin²x + 4 cos²x = 4(sin²x + cos²x) = 4(1) = 4. [1.3]

15. * (D) This is a tricky problem. If you just plot the graph of y = (x – )(x² – + ), you will see only one real zero, at approximately 1.414( ). This is because the zeros of the quadratic factor are imaginary. Since, however, they are imaginary conjugates, their sum is real—namely twice the real part. Therefore, graphing the function, using CALC/zero to find the zeros, and summing them will give you the wrong answer.

To get the correct answer, you must use the fact that the sum of the zeros of the quadratic factor is . Since the zero of the linear factor is 1.414, the sum of the zeros is about 1.732 + 1.414 3.15. [1.2]

16. (D) Graph I consists of the lines x = 0 and y = 0, which are the coordinate axes and are therefore perpendicular. Graph II consists of y = |x| and y = –|x|, which are at ±45° to the coordinate axes and are therefore perpendicular. Graph III consists of the hyperbolas xy = 1 and xy = –1. Therefore, the correct answer choice is D.

There are two reasons why a graphing calculator solution is not recommended here. One is that equations, not functions, are given, and solving these equations so that they can be graphed involves two branches each. The other reason is that even with graphs, you would have to make judgments about perpendicularity. At a minimum, this would require you to graph the equations in a square window. [1.6]

17. (B) Points 6 in. from m form a cylinder, with m as axis, which is tangent to plane X. Points 1 in. from X are two planes parallel to X, one above and one below X. The cylinder intersects only one of the planes in two lines parallel to m. [2.2]

18. (E) Since the y-coordinate of the point V could be at any height, the slope of VO could be any value. [2.1]

19. * (C) Volume of cylinder = πr²h = 36π = volume of cube = s³. Therefore, 4.84. [2.2]

20. (C) Substitute 2t – 2 for x. [1.6]

21. * (D) y₁ = 2^3/2 + 2a + b + c and y₂ = –(2)^3/2 + 2a – 2b + c. So, y₁ – y₂ = (2^3/2 + 2^3/2) + 2b = 3. Therefore, 5.65685 + 2.828b 3 and –0.939. [1.2]

22. * (E) The probability that the first car selected is compact is . There are 26 cars left, of which 19 are compact. The probability that the second car is also compact is . [4.2]

23. (D) Here, a could be either positive or negative. However, b must be negative. [algebra]

24. * (C) Plot the graph of y = abs(x – int(x) – 1/2) in an xε[–5,5] and yε[–2,2] window and observe that the maximum value is .

An alternative solution is to sketch a portion of the graph by hand and observe the maximum value. [1.6]

25. * (D) Plot the graph y = (x³ – 8)(x² – 4) in the standard window. Using CALC/value, observe that y is not defined when x = 2. Therefore, enter 1.999 for x and observe that y is approximately equal to 3.

An alternative solution is to factor the numerator and denominator, divide out x – 2, and substitute the limiting value 2 into the resulting expression:

[1.5]

26. * (B) A sketch will help you see that the height of the cone is 5 + 3 = 8.

The volume of the cone is , and the volume of the sphere is . The desired ratio is V_c :V_s = 0.256 : 1. [2.2]

27. * (C) Enter 1 into your calculator. Then enter three times, to accomplish three iterations that result in x₃ to get the correct answer choice E.

An alternative solution is to use the formula to evaluate x₁, x₂, and x₃, in turn. [3.4]

28. * (D) With your calculator in radian mode, plot the graph of y=abs( (1/sin(x + π/5))) in an xε[–1,1] and yε[0,5] window. Use VALUE/X = 0 to determine that the y-intercept is approximately 1.49. [1.3]

29. (D) The equation of the circle is (x – 4)² + (y + 8)² = r². Multiplying out indicates that a = –8 and b = 16, and so a + b = 8. [2.1]

30. (E) Since p(a) = 2, 3a² + 9a + 5 = 0. Solve by using the Quadratic Formula to get the correct answer choice.

31. * (A) Since i is a root of the equation, so is –i, so there are 2 real roots. Plot the graph of y = x⁴ + 2x³ – 3x² + 2x – 4 in the standard window. Use CALC/zero to find one of the roots, and store the answer (X) in A. Then use CALC/zero again to find the other root, and multiply A and X to get the correct answer choice. [3.2]

32. * (E) First find sin^–1(3/5) 36.87°, the reference angle for angle A. Since A is in the second quadrant, A 180° – 36.87° = 143.13°. Store 180-Ans in A. Similarly, find cos^–1(1/3) 70.53°, the reference angle for angle B. Since B is in the fourth quadrant, store 360-Ans in B. Then evaluate sin(A + B) to get the correct answer choice. [1.3]

33. * (A) The probability that at least one is a boy is (1 – the probability that all 3 are girls) or 1 – (0.5)³ = 0.875. [4.2]

34. * (E) Since sec1.4 = x, . Therefore, csc(2tan^–1 x) = 1/sin(2 tan^–1(1/cos 1.4)) 3.03. [1.3]

35. * (B) The equation |y – 1| = |x + 1| defines two functions: . Plot these graphs in the standard window and observe that they intersect at (–1,1).

An alternative solution is to recall that the important point of an absolute value occurs when the expression within the absolute value sign equals zero. The important point of this absolute value problem occurs when y – 1 = 0 and x + 1 = 0, i.e., at (–1,1). [1.6]

36. * (E) With your calculator in degree mode, plot the graphs of y = cos2x and y = 2 cos x in an xε[0,360] and yε[–2,2] window. Use CALC/intersect to find the correct answer choice E. [1.3]

37. (E) Since sin has a maximum at Thus, and Therefore, . [1.3]

38. * (E) Enter LIST/MATH/sum(LIST/OPS/seq(X,X,1,51)) to compute the desired sum.

An alternative solution is to observe that the sequence is arithmetic with t₁ = 1 and d = 1. Using the formula for the sum of the first n terms of an arithmetic sequence, [3.4]

39. * (C) Complete the square in the first equation to get 3(x – 1)² + 4(y + 1)² = 12. Solving this equation for y yields . Solving for y in the second equation, . Plot the graphs of these three equations in the standard window to see that the graphs intersect in two places.

An alternative solution can be found by completing the square in the first equation and dividing by 12 to get the standard form equation of an ellipse, .

The second equation is the standard form of a parabola. Sketch these two equations and observe the number of points of intersection. [2.1]

40. * (D) Let y = log_x 2 = log₃ x. Converting to exponential form gives x^y = 2 and 3^y = x. Substitute to get 3^y2 = 2, which can be converted into . Thus, y ±0.7943. Therefore, 3^0.7943 = x 2.393 or 3^–0.7943 = x 0.4178. Therefore, the sum of two x’s is 2.81. [1.4]

41. (D) Factor the numerator and denominator: . Since the x – 5 divide out, the only vertical asymptote is at x = –1. Since the degree of the numerator and denominator are equal, y approaches 3 as x approaches , so y= 3 is a horizontal asymptote. [1.5]

42. * (C) With your calculator in degree mode, plot the graphs of y = 1/2 and y = (3 sin(2x))/(1 – cos(2x)) in the window [0,180] by [–2,2]. Use CALC/intersect to find the one point of intersection in the specified interval, at 80.5°.

An alternative solution is to cross-multiply the original equation and use the double angle formulas for sine and cosine, to get

6sin2 = 1 – cos2
12 sincos = 1 – (1 – 2sin²) = 2sin²
6sincos – sin² = 0
sin(6cos – sin) = 0

Therefore, sin = 0 or tan = 6. It follows that = 0°,180°, or 80.5°. The first two solutions make the denominator of the original equation equal to zero, so 80.5° is the only solution. [1.3]

43. (C) Backsolve until you get f(g(x),g(x)) = 2(2^x)² – (2^x)² = (2^x)² = 2^2x. [1.1]

44. (E) From the geometric sequence From the arithmetic sequence, b – a = 12 – b, or 2b – a = 12. Substituting gives

Solving gives a = 6 or –4. Eliminate –4 since a is given to be positive. Substituting the 6 gives 2b – 6 = 12, giving b = 9. Therefore, b – a = 3. [3.4]

45. * (E) Since s = r, then 2.4 = r, which implies that and so which implies that Therefore, which implies that 2.4² = 28.6. Therefore, . [1.3]

46. * (D) Complete the square and put the equation of the ellipse in standard form:


The center of the ellipse is at (–2,1), with a² = 36 and b² = 25, and the major axis is parallel to the y-axis. Each focus is units above and below the center.

Therefore, the y-coordinates of the foci are and –2.32. [2.1]

47. * (B) Because you are bisecting one side and trisecting another side, it is convenient to let the length of the sides be a number divisible by both 2 and 3. Let AB = AD = 6. Thus BM = 2, MC = 4, and CN = ND = 3. Let NAD = x, so that, using right triangle NAD, tan , which implies that x = tan^–1 0.5 26.6°. Let MAB = y, so that, using right triangle MAB, tan , which implies that . Therefore, θ 90° – 26.6° – 18.4° 45°. [1.3]

48. * (B) The slope of the line is . An equation of the line is therefore . Substitute 23.7 for y and solve for x to get x = –1.5. [1.2]

49. (A) You must check each answer choice, one at a time. In A, x < 0, y < 0, x – y < 0, so the expression is negative. In B, x < 0, y < 0, x – y > 0, so the expression is positive. At this point you know that the correct answer choice must be A. [algebra]

50. * (B) Put your calculator in radian mode: 5 * 3 = .

2 * (5 * 3) = 2 * 2.823 2.823 cos . [1.1]

Self-Evaluation Chart for Model Test 4

Calculating Your Score

 Raw score R = number right – (number wrong), rounded = ______________

 Approximate scaled score S = 800 – 10(44 – R) = ______________

 If R 44, S = 800.