SAT SUBJECT TEST MATH LEVEL 2
PART 3
MODEL TESTS
Answer Sheet
MODEL TEST 1
Model Test 1
The following directions are for the print book only. Since this is an eBook, record all answers and selfevaluations separately. 
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 selfevaluation chart in order to pinpoint the topics that give you the most difficulty.
*Note: All Model Tests contain hyperlinks between questions and answers. Click on the question numbers to navigate between questions and answers.
50 questions: 1 hour Directions: Decide which answer choice is best. If the exact numerical value is not one of the answer choices, select the closest approximation. Fill in the oval on the answer sheet that corresponds to your choice. Notes: (1) You will need to use a scientific or graphing calculator to answer some of the questions. (2) You will have to decide whether to put your calculator in degree or radian mode for some problems. (3) All figures that accompany problems are plane figures unless otherwise stated. Figures are drawn as accurately as possible to provide useful information for solving the problem, except when it is stated in a particular problem that the figure is not drawn to scale. (4) Unless otherwise indicated, the domain of a function is the set of all real numbers for which the functional value is also a real number. 
Reference Information. The following formulas are provided for your information. Volume of a right circular cone with radius r and height Lateral area of a right circular cone if the base has circumference C and slant height is l : Volume of a sphere of radius Surface area of a sphere of radius Volume of a pyramid of base area B and height 
1. The slope of a line perpendicular to the line whose equation is
(A) −3
(B)
(C)
(D)
(E)
2. What is the range of the data set 8, 12, 12, 15, 18?
(A) 10
(B) 12
(C) 13
(D) 15
(E) 18
3. If , for what value(s) of x does the graph of y = f(x) have a vertical asymptote?
(A) –7
(B) 0
(C) –7,0,7
(D) –7,7
(E) 7
4. If and g(x) = x^{2} + 1, then f(g(2)) =
(A) 2.24
(B) 3.00
(C) 3.61
(D) 6.00
(E) 6.16
5.
(A) –0.25
(B) –0.16
(C) 0.16
(D) 6.35
(E) The value is not a real number.
6. The circumference of circle x^{2} + y^{2} – 10y – 36 = 0 is
(A) 38
(B) 49
(C) 54
(D) 125
(E) 192
7. Twentyfive percent of a group of unrelated students are only children. The students are asked one at a time whether they are only children. What is the probability that the 5th student asked is the first only child?
(A) 0.00098
(B) 0.08
(C) 0.24
(D) 0.25
(E) 0.50
8. If f(x) = 2 for all real numbers x, then f(x + 2) =
(A) 0
(B) 2
(C) 4
(D) x
(E) The value cannot be determined.
9. The volume of the region between two concentric spheres of radii 2 and 5 is
(A) 28
(B) 66
(C) 113
(D) 368
(E) 490
10. If a, b, and c are real numbers and if then a could equal
(A)
(B)
(C) 9
(D) 3
(E) 9b^{6}
11. In right triangle ABC, AB = 10, BC = 8, AC = 6. The sine of A is
(A)
(B)
(C)
(D)
(E)
12. If 16^{x} = 4 and 5^{x+y }= 625, then y =
(A) 1
(B) 2
(C)
(D) 5
(E)
13. If the parameter is eliminated from the equations x = t^{2} + 1 and y = 2t , then the relation between x and y is
(A) y = x – 1
(B) y = 1 – x
(C) y^{2} = x – 1
(D) y^{2} = (x – 1)^{2}
(E) y^{2} = 4x – 4
14. Let f(x) be a polynomial function: f(x) = x^{5} + · · · . If f(1) = 0 and f(2) = 0, then f(x) is divisible by
(A) x – 3
(B) x^{2} – 2
(C) x^{2} + 2
(D) x^{2} – 3x + 2
(E) x^{2} + 3x + 2
15. If x – y = 2, y – z = 4, and x – y – z = –3, then y =
(A) 1
(B) 5
(C) 9
(D) 11
(E) 13
16. If z > 0, a = z cos, and b = z sin, then =
(A) 1
(B) z
(C) 2z
(D) z cos sin
(E) z (cos + sin
17. If the vertices of a triangle are (u,0), (v,8), and (0,0), then the area of the triangle is
(A) 4u 
(B) 2v 
(C) uv 
(D) 2uv 
(E) uv 
18. If what must the value of k be in order for f(x) to be a continuous function?
(A) –2
(B) 0
(C) 2
(D) 5
(E) No value of k will make f(x) a continuous function.
19. What is the probability that a prime number is less than 7, given that it is less than 13?
(A)
(B)
(C)
(D)
(E)
20. The ellipse 4x^{2} + 8y^{2} = 64 and the circle x^{2} + y^{2} = 9 intersect at points where the y coordinate is
(A) ±
(B) ±
(C) ±
(D) ±
(E) ± 10.00
21. Each term of a sequence, after the first, is inversely proportional to the term preceding it. If the first two terms are 2 and 6, what is the twelfth term?
(A) 2
(B) 6
(C) 46
(D) 2 · 3^{11}
(E) The twelfth term cannot be determined.
22. A company offers you the use of its computer for a fee. Plan A costs $6 to join and then $9 per hour to use the computer. Plan B costs $25 to join and then $2.25 per hour to use the computer. After how many minutes of use would the cost of plan A be the same as the cost of plan B?
(A) 18,052
(B) 173
(C) 169
(D) 165
(E) 157
23. If the probability that the Giants will win the NFC championship is p and if the probability that the Raiders will win the AFC championship is q , what is the probability that only one of these teams will win its respective championship?
(A) pq
(B) p + q –2pq
(C) p – q
(D) 1 – pq
(E) 2pq – p – q
24. If a geometric sequence begins with the terms , 1, · · · , what is the sum of the first 10 terms?
(A) 9841
(B) 6561
(C) 3280
(D) 33
(E) 6
25. The value of is
(A) greater than 10^{100}
(B) between 10^{10} and 10^{100}
(C) between 10^{5} and 10^{10}
(D) between 10 and 10^{5}
(E) less than 10
26. If A is the angle formed by the line 2y = 3x + 7 and the x axis, then A equals
(A) –45°
(B) 0°
(C) 56°
(D) 72°
(E) 215°
27. A U.S. dollar equals 0.716 European euros, and a Japanese yen equals 0.00776 European euros. How many U.S. dollars equal a Japanese yen?
(A) 0.0056
(B) 0.011
(C) 0.71
(D) 94.2
(E) 179.98
28. If (x – 4)^{2 }+ 4(y – 3)^{2 }= 16 is graphed, the sum of the distances from any fixed point on the curve to the two foci is
(A) 4
(B) 8
(C) 12
(D) 16
(E) 32
29. In the equation x^{2} + kx + 54 = 0, one root is twice the other root. The value(s) of k is (are)
(A) –5.2
(B) 15.6
(C) 22.0
(D) ± 5.2
(E) ± 15.6
30. The remainder obtained when 3x^{4} + 7x^{3} + 8x^{2} – 2x – 3 is divided by x + 1 is
(A) –3
(B) 0
(C) 3
(D) 5
(E) 13
31. If f(x) = e^{x} and g(x) = f(x) + f ^{–1}(x ), what does g (2) equal?
(A) 5.1
(B) 7.4
(C) 7.5
(D) 8.1
(E) 8.3
32. If x_{0} = 3 and , then x_{3} =
(A) 2.65
(B) 2.58
(C) 2.56
(D) 2.55
(E) 2.54
33. For what values of k does the graph of pass through the origin?
(A) only 0
(B) only 1
(C) ±1
(D) ±
(E) no value
34. If
(A) 15°
(B) 30°
(C) 45°
(D) 60°
(E) 75°
35. If x^{2} + 3x + 2 < 0 and f(x) = x^{2} – 3x + 2, then
(A) 0 < f(x) < 6
(B)
(C) f(x) > 12
(D) f(x) > 0
(E) 6 < f(x) < 12
36. If f(x) = x  + [x ], the value of f(–2.5) + f(1.5) is
(A) −2
(B) 1
(C) 1.5
(D) 2
(E) 3
37. If (sec x)(tan x) < 0, which of the following must be true?
I. tan x < 0
II. csc x cot x < 0
III. x is in the third or fourth quadrant
(A) I only
(B) II only
(C) III only
(D) II and III
(E) I and II
38. At the end of a meeting all participants shook hands with each other. Twentyeight handshakes were exchanged. How many people were at the meeting?
(A) 7
(B) 8
(C) 14
(D) 28
(E) 56
39. Suppose the graph of f(x) = 2x^{2} is translated 3 units down and 2 units right. If the resulting graph represents the graph of g(x ), what is the value of g (–1.2)?
(A) –1.72
(B) –0.12
(C) 2.88
(D) 17.48
(E) 37.28
40. Four points on the graph of a polynomial P are shown in the table above. If P is a polynomial of degree 3, then P(x) could equal
(A) (x – 5)(x – 2)(x + 1)
(B) (x – 5)(x + 2)(x + 1)
(C) (x + 5)(x – 2)(x – 1)
(D) (x + 5)(x + 2)(x – 1)
(E) (x + 5)(x + 2)(x + 1)
41. If f(x) = ax + b, which of the following make(s) f(x) = f^{–1}(x )?
I. a = –1, b = any real number
II. a = 1, b = 0
III. a = any real number, b = 0
(A) only I
(B) only II
(C) only III
(D) only I and II
(E) only I and III
42. In the figure above, A = 110°, a = and b = 2. What is the value of C ?
(A) 50°
(B) 25°
(C) 20°
(D) 15°
(E) 10°
43. If vector and vector = (3,–2), find the value of
(A) 5.4
(B) 6
(C) 7
(D) 7.2
(E) 52
44. If and , then g(f(3)) =
(A) 0.2
(B) 1.7
(C) 2.1
(D) 3.5
(E) 8.7
45. In ABC above, a = 2x, b = 3x + 2, , and C = 60°. Find x.
(A) 0.50
(B) 0.64
(C) 0.77
(D) 1.64
(E) 1.78
46. If log_{a} 5 = x and log_{a} 7 = y , then log_{a}
(A) xy
(B) x – y
(C) (x + y )
(D) (y – x )
(E)
47. If f(x) = 3x^{2} + 4x + 5, what must the value of k equal so that the graph of f(x – k) will be symmetric to the yaxis?
(A) – 4
(B) –
(C) –
(D)
(E)
48. If f(x) = cos x and g(x) = 2x + 1, which of the following are even functions?
I. f(x) · g(x )
II. f(g(x ))
III. g(f(x ))
(A) only I
(B) only II
(C) only III
(D) only I and II
(E) only II and III
49. A cylinder whose base radius is 3 is inscribed in a sphere of radius 5. What is the difference between the volume of the sphere and the volume of the cylinder?
(A) 88
(B) 297
(C) 354
(D) 448
(E) 1345
50. Under which conditions is negative?
(A) 0 < y < x
(B) x < y < 0
(C) x < 0 < y
(D) y < x < 0
(E) none of the above
If there is still time remaining, you may review your answers.
Answer Key
MODEL TEST 1



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 SelfEvaluation Chart at the end of the test.
An asterisk appears next to those solutions in which a graphing calculator is necessary.
1. (C) Solve for y. Slope Slope of perpendicular
2. (A) Range = largest value – smallest value = 18 – 8 = 10. [4.1]
3. (A) Vertical asymptotes occur where the denominator is zero but the numerator is not. The denominator, x^{2} – 49, factors into (x + 7)(x – 7). Since both numerator and denominator are zero when x = 7, a vertical asymptote occurs only at x = –7. [1.2]
4. * (C) Enter the function f into Y_{1 }and the function g into Y_{2}. Evaluate Y_{1}(Y_{2}(2)) to get the correct answer choice C.
An alternative solution is to evaluate g(2) = 5 and f(5) = , and either use your calculator to evaluate or observe that 3 < < 4, indicating 3.61 as the only feasible answer choice. [1.1]
5. * (C) Enter the expression into your graphing calculator. [1.4]
6. * (B) Complete the square to get x^{2} + (y – 5)^{2 }= 61.
Radius
7. * (B) Whether or not students in the group have siblings are independent events. The probability that each of the first four is not an only child is (0.75)^{4}. The probability that the fifth student is an only child is 0.25, so the probability of seeing the first four children with siblings and the fifth an only child is (0.75)^{4}(0.25) ≈ 0.08 [4.2]
8. (B) Regardless of what is substituted for x, f(x) still equals 2
Alternative Solution: f(x + 2) causes the graph of f(x) to be shifted 2 units to the left. Since f(x) = 2 for all x, f(x + 2) will also equal 2 for all x. [1.1]
9. * (E) Enter the formula for the volume of a sphere (in the reference list of formulas) into Y_{1}. Return to the Home Screen, and enter Y_{1}(5) – Y_{1}(2) to get the correct answer choice E.
An alternative solution is to evaluate directly. [2.2]
10. * (D) Since , divide out b^{3}c^{8}, leaving a^{5} = 9a^{3}. Therefore a = ±3. [1.4]
11. (C) sin
12. (C) Since the power is the square root, x = because the square root of 16 is 4.
Since 5^{4} = 625, x + y = 4, so that
13. (E) . Eliminate the parameter and get or y^{2} = 4x – 4. [1.6]
14. (D) f(1) = 0 and f(2) = 0 imply that x – 1 and x – 2 are factors of f(x). Their product, x^{2} – 3x + 2, is also a factor. [1.2]
15. (C) Add the first two equations to get x – z = 6. Substitute this in the third equation to get 6 – y = –3, and solve for y. [algebra]
16. (B) a^{2} = z^{2} cos^{2} and b^{2} = z^{2} sin^{2}, so a^{2} + b^{2} = z^{2}(cos^{2} + sin^{2}) = z^{2} because cos^{2} + sin^{2} = 1. Since when z > 0, the correct answer choice is B. [1.3]
17. (A) Sketch a graph of the three vertices. The base is u  and the altitude is 8. Therefore, the area is 4 u . [2.1]
18. * (E) Plot the graph of in the standard window, and observe the asymptote at x = 2. This says that no value of k can make f(x) continuous at x = 2.
An alternative solution is to observe that x = 2 makes the denominator of f(x) equal to zero, thereby implying that x = 2 is a vertical asymptote. Thus, f(x) cannot be made continuous at the point with that x value. [1.6]
19. (D) There are 5 prime numbers less than 13: 2, 3, 5, 7, 11. Three of these are less than 7, so the correct probability is .
20. * (D) Substituting for x^{2} and solving for y gives 4(9 – y^{2}) + 8y^{2} = 64. 4y^{2} = 28, and so y^{2} = 7 and
21. (B) t_{n }· t_{n}_{ + 1} = K. 2 · 6 = K = 12. Therefore, 6 · t_{3 }= 12, and so t_{3 }= 2. Continuing this process gives all odd terms to be 2 and all even terms to be 6. [3.4]
22. * (C) Graph the cost of Company A y = 6 + 9x and the cost of Company B y = 25 + 2.25x in a window xε[0,10] and yε[0,50]. Use CALC/intersect to find the xcoordinate of the point of intersection at 2.8148 hours, the “breakeven” point. Multiply by 60 to convert this time to the correct answer choice.
An alternative solution is to solve the equation 6 + 9x = 25 + 2.25x and multiply the solution by 60 to get the answer of about 169 minutes. [1.2]
23. (B) The probability that both teams will win is pq. The probability that both will lose is (1 – p)(1 – q). The probability that only one will win is 1 – [pq + (1 – p)(1 – q )] = 1 – (pq + 1 – p – q + pq) = p + q – 2pq.
Alternative Solution: The probability that the Giants will win and the Raiders will lose is p (1 – q). The probability that the Raiders will win and the Giants will lose is q (1 – p ). Therefore, the probability that either one of these results will occur is p (1 – q) + q (1 – p) = p + q – 2pq. [4.2]
24. * (A) Calculate the common ratio as _{}. The first term is so the n^{th} term is . Use the sum and sequence features of your calculator to evaluate the sum of the first 10 terms in the generated sequence:
The range is 0 to 9 instead of 1 to 10 because the formula for t_{n }uses the exponent n – 1.
An alternative solution is to use the formula for the sum of a geometric series:
25. * (C) No calculator currently on the market can compute 453!, so doing this problem requires some knowledge of factorial arithmetic. The easiest solution to the problem is to observe that is the number of combinations of 453 taken 3 at a time (_{453}C_{3}). Enter 453MATH/PRB/nCr3 into your calculator to find that the correct answer choice is C.
An alternative solution is to simplify
26. * (C) Solve for y: . Slope . Tan A also equals . Therefore, ≈ 56°. [1.3]
27. * (B) 1 yen equals 0.0076 euros, and 1 euro equals dollars. Therefore, 1 yen equals 0.0076 × 1.40 = 0.011 dollars. [algebra]
28. (B) Divide the equation through by 16 to get . This is the equation of an ellipse with a^{2} = 16. The sum of the distances to the foci = 2a = 8. [2.1]
29. * (E) If the roots are r and 2r, their sum and their product . Therefore, and .
Alternative Solution: If the roots are r and 2r, (x – r )(x – 2r) = 0. Multiply to obtain x^{2} – 3r + 2r^{2} = 0, which represents x^{2} + kx + 54 = 0. Thus, –3r = k and 2r^{2} = 54.
Since , then and
30. (C) Substituting –1 for x gives 3.
Alternative Solution: Use synthetic division to get
31. * (D) The inverse of f(x) = e^{x} is f^{–1}(x) = ln x. g(2) = e^{2} + ln 2 8.1. [1.4]
32. * (C) This is a recursively defined sequence. Press 3 ENTER on your calculator. Then enter and press ENTER 3 times to get x_{3 } 2.56. [3.4]
33. (C) If the graph passes through the origin, x = 0 and y = 0, then . k^{2} = 1, and so k = ±1. [2.1]
34. * (D) Graph using Ztrig in degree mode. Find the point of intersection with CALC/intersect to arrive at the correct answer choice D. An alternative solution uses the identities tan and tan 30º = to deduce , so = 60º. [1.3]
35. * (E) The problem is asking for the range of f(x) values for values of x that satisfy the inequality. First graph the inequality in Y_{1}, starting with the standard window and zooming in until the x values for the portion of the graph that falls below the xaxis can be identified as the interval (–2,–1). Then enter the formula for f(x) in Y_{2}. Although it can be done graphically, the simplest way to find the range of values of f(x) that correspond to xε(–2,–1) is to use the TABLE function. Deselect Y_{1} and enter TBLSET and set TblStart to –2, Tbl = 0.1, and Indpnt and Depend to Auto. Then enter TABLE and observe that the Y_{2 }values range from 12 to 6 as x ranges from –2 to –1, yielding the correct answer choice D.
An alternative solution is to solve the inequality algebraically by solving the associated equation x^{2} + 3x + 2 = 0 and testing points. The left side of the equation factors as (x + 2)(x + 1), and the Zero Product Property implies that x = –2 or x = –1. Points inside the interval (–2,–1) satisfy the inequality, while those outside it do not. Since the graph of f(x) is a parabola and f(–2) = 12 and f(–1) = 6, f(x) takes the range of values between 12 and 6. [1.2]
36. * (D) Recall that the notation [x] means the greatest integer less than or equal to x. Enter abs(x) + int(x) into Y_{1}. Return to the Home Screen, and enter Y_{1}(–2.5) + Y_{1}(1.5) to get the correct answer choice D.
An alternative solution evaluates –2.5 + [–2.5] + 1.5 + [1.5] without the aid of a calculator. Of these 4 values, only [–2.5] is tricky since [–2.5] = –3, not –2. Thus, –2.5 + [–2.5] + 1.5 + [1.5] = 2.5 – 3 + 1.5 + 1 = 2. [1.6]
37. (C) Set up the following table.
Q1 
Q2 
Q3 
Q4 

sec x 
+ 
– 
– 
+ 
tan x 
+ 
– 
+ 
– 
cot x 
+ 
– 
+ 
– 
csc x 
+ 
+ 
– 
– 
The product secx tanx is negative only when its factors have different signs, so III is the only true statement. [1.3]
38. (B)
39. * (D) Since the function g is f translated 3 down and 2 right, g(x) = f(x – 2) – 3. Therefore, g(–1.2) = f(–3.2) – 3 = 2(–3.2)^{2 }– 3 = 17.48. [2.1]
40. (D) Since –5 and 1 are both zeros, (x + 5) and (x – 1) are factors of P(x ). Since P(x) changes sign between x = –3 and x = –1, there is a zero between these two values. Choice D is the only one that meets all three criteria. [1.2]
41. * (D) The graph of f must be symmetric about the line y = x. In other words, interchanging x and y must leave the graph unchanged. In I, x = –y + b, which is equivalent to y = –x + b, which is symmetric about y = x. In II, x = y. In III, x = ay, or
42. * (C) Law of sines:
43. * (D)
44. * (C)
45. * (B) Law of Cosines:
Use program QUADFORM to get x = ±0.64.
Since a side of a triangle must be positive, x can equal only 0.64. [1.3]
46. (D)
47. * (D) Graph y = 3x^{2} + 4x + 5 in the standard window, and observe that the graph must be moved slightly to the right to be symmetric to the yaxis. Therefore, k must be positive. Use CALC/minimum to find the vertex of the parabola and observe that its xcoordinate is –0.66666. . . . If the function entered into and graph Y_{2 }to verify this answer. [2.1]
48. * (C) Use ZTrig to plot the graphs of y = (cos x) · (2x + 1), y = cos(2x + 1), and y = 2(cos x) + 1 to see that only the third graph is symmetric about the yaxis and thus represents an even function.
An alternative solution is to use your knowledge of transformations. Although f is an even function, g is not; therefore, (I) f · g is not even. Also, f(g(x )) = cos(2x + 1), which is a cosine curve shifted less than to the left. Thus, f(g(x )) (II) is not even. However, g(f(x )) = 2 cos x + 1 is a cosine curve with period 2, amplitude 2, shifted 1 unit up. Thus, g(f(x )) (III) is even. [1.1]
49. * (B) Height of cylinder is 8.
Volume of sphere
Volume of cylinder
Difference
50. * (B) In answer choice B, x and y have the same sign, and x is less than y. Therefore, xy is positive, x – y is negative, and the quotient is negative. The numerators and denominators in answer choices A, C, and D both have the same sign, so the quotients are positive. [algebra]
SelfEvaluation Chart for Model Test 1


Rating 
Number Right 
Excellent 
41–50 
Very good 
33–40 
Above average 
25–32 
Average 
15–24 
Below average 
Below 15 
Calculating Your Score
Raw score R = number right – (number wrong), rounded = ______________
Approximate scaled score S = 800 – 10(44 – R) = ______________
If R 44, S = 800.