SAT Mathematics Level ­Practice Test There are 50 questions on this test You have hour (60 minutes) to complete it The measures of the angles of �QRS are m∠Q = 2x + 4, m∠R = 4x − 12, and m∠S = 3x + QR = y + 9, RS = 2y − 7, and QS = 3y − 13 The perimeter of �QRS is (A) 11   (B) 20   (C) 44   (D) 55   (E) 68 Given g(x) = 3x + , g c - m = 5x - –1 ​   1 ​   1 ​  1  ​   7  ​   (A) ​  _ (B) ​    (D) ​  _ (E) ​    (C) ​  _     11 16 81x y10 = (A) x y x    (B) 9x y 3 xy   (C) 3x y 3 9xy (D) 3x y 3 3xy     (E) x y x SAT Math & Subject teSt 10 y K G –6 x H –6 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 When the figure above is spun around its vertical axis, the total surface area of the solid formed will be (A) 144π   (B) 108π   (C) 72π   (D) 36π   (E) 9π + 12 If f(x) = 4x2 − and g(x) = 8x + 7, g  f(2) = (A) 15   (B) 23   (C) 127   (D) 345   (E) 2115 If p and q are positive integers with pq = 36, then (A)    (B)    (C) 1   (D) 2   (E) 9 p cannot be q SAT MATHEMATICS LEVEL ­PRACTICE TEST + x − = 1− 3x − 12 (A) 2x − 2x 2x −    (B)    (C) x − 12 3x − 3x − (D) 2x − 2x −    (E) 3x − 12 3x − 14 a +4ab 3a -10ab = ^3 625 h If ` j and if a and b not equal 0, a = 125 b (A) 76 76    (B) 2   (C)    (D) 4   (E) 21 21 _ 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 SAT Math & Subject teSt 11 The equation of the perpendicular bisector of the segment joining A(–9,2) to B(3,–4) is –1 ​ (x − 3)   (B) y + = ​  _ –1 ​ (x + 3)   (A) y − = ​  _ 2 (C) y + = 2(x + 3)   (D) y + = 2(x + 1)   (E) y − = 2(x − 3) 12 Tangent TB and secant TCA are drawn to circle O Diameter AB is drawn If TC = and CA = 10, then CB = (A)    (B)    (C) 15    (D) 10   (E) 33 13 Let p @ q = pq (5 @ 3) − (3 @ 5) = q-p (A) –184   (B) –59   (C) 0   (D) 59   (E) 184 SAT MATHEMATICS LEVEL ­PRACTICE TEST 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 standard deviation was 12 The teacher curves the score by raising each score by a total of points Which of the following statements is true? I The new mean is 75 II The new median is 71 III The new standard deviation is (A) I only   (B) III only   (C) I and II only (D) I, II, and III   (E) None of the statements are true 15 A set of triangles is formed by joining the midpoints of the larger triangles If the area of �ABC is 128, then the area of �DEF, the smallest triangle formed, is (A) 1    (B)    (C)    (D) 1   (E) SAT Math & Subject teSt 16 The graph of y = f(x) is shown above Which is the graph of g(x) = 2f(x − 2) + 1? (A) (B) (D) (C) (E) SAT MATHEMATICS LEVEL ­PRACTICE TEST 17 The number of bacteria, measured in thousands, in a culture is 380e 2.31t , where t is the number of 175 + e 3.21t days since the culture was formed According to this model, the culture can support a maximum population of modeled by the equation b (t ) = (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) = 3x − , g(g(x)) = 2x + (A) 9x − 7x − 12 x − 10    (B)    (C) 6x + 24x + 79 21x + 80 (D) 7x +    (E) 9x − 6x + 24x + 79 4x + 36x + 81 20 The area of �QED = 750 QE = 48 and QD = 52 To the nearest degree, what is the measure of the largest possible angle of �QED? (A) 76   (B) 77   (C) 78   (D) 143   (E) 145 21 Given log3(a) = c and log3(b) = 2c, a = (A) 3c   (B) c + 3   (C) b2   (D) b    (E) b SAT Math & Subject teSt 22 In isosceles trapezoid WTYH, WH || XZ || TY , m∠TWH = 120, and m∠HWE = 30 XZ passes through point E, the intersection of the diagonals If WH = 30, determine the ratio of XZ:TY (A) 1:2   (B) 2:3   (C) 3:4   (D) 4:5   (E) 5:6 23 The lengths of the sides of a triangle are 25, 29, and 34 To the nearest tenth of a degree, the measure of the largest angle is (A) 77.6°   (B) 77.7°   (C) 87.6°   (D) 87.7°   (E) 102.3° 24 One of the roots of a quadratic equation that has integral coefficients is −4 + i Which of the following describes the quadratic equation? (A) 800x2 + 1280x + 737 = 0   (B) 800x2 − 1280x + 737 = (C) 800x2 + 1280x + 287 = 0   (D) 800x2 − 1280x + 287 = (E) 800x2 + 1280x − 287 = 25 The parametric equations x = cos(2t) + and y = sin(t) + correspond to a subset of the graph (A) (x − 1)2 + ( y − 2)2 (C) x − = =    (B) (x − 1)2 − ( y − 2)2 2 ( y − 2)2    (D) x − = − ( y − 2) (E) x − = − ( y − 2) =1 SAT MATHEMATICS LEVEL ­PRACTICE TEST 26 A county commissioner will randomly select people to form a non-partisan committee to look into the issue of county services If there are Democrats and Republicans to choose from, what is the probability that the Democrats will have the more members than the Republicans on this committee? (A) 686 1316    (B) 2002 2002 (D) 2688 21336    (E) 24024 24024    (C) 1876 2002 27 Given the vectors u = [–5, 4] and v = [3, –1], |2u − 3v| = (A) [–19, 11]   (B) 502    (C) [19, 11] (D) 41 − 10    (E) 65 28 �ABC has vertices A(–11,4), B(–3,8), and C(3,–10) The coordinates of the center of the circle circumscribed about �ABC are (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? (A) 20   (B) 40   (C) 46.875   (D) 56.25   (E) 71.875 § 19S · 20cis ă â 18 = 30 Đ 2S à 5cis ă â (A) + 2i   (B) − − 2i   (C) + 2i (D) − + 2i 3   (E)  2i 10 SAT Math & Subject teSt 31 Given logb(a) = x and logb(c) = y, log a ( bc )= 5 + 4y 20 y (A) + y    (B)    (C) 3x 6x x (D) 2x+ 4y   (E) 2x + 20 y 32 The asymptotes of a hyperbola have equations y − =  (x + 3) If a focus of the hyperbola has coordinates (7,1), the equation of the hyperbola is (A) ( x + 3)2 ( y − 1)2 − = 16    (B) ( y − 1)2 ( x + 3)2 − =1 16 (C) ( x + 3)2 ( y − 1)2 − = 64 36    (D) ( y − 1)2 ( x + 3)2 − =1 36 64 (E) ( x + 3)2 ( y − 1)2 − =1 33 An inverted cone (vertex is down) with height 12 inches and base of radius inches is being filled with water What is the height of the water when the cone is half filled? (A) 6   (B)    (C)    (D)    (E) 34 Solve sin(t) = cos(2t) for –4π ≤ t ≤ –2π ­ S 5S S ẵ S 5S S ẵ (A) ưđ , , ¾    (B) ® , , ¾ ¯ ¿ ¿ ¯ ­ 23S 19S 5S ½    (D) ­ 23S , 19S , 5S ẵ (C) đ , , đ ắ ắ ¿ ¿ ¯ ¯ ­ 11S 7S 5S ẵ , , (E) đ ắ ¿ ¯ SAT MATHEMATICS LEVEL ­PRACTICE TEST 13 44 Diagonals AC and BD of quadrilateral ABCD are perpendicular AD = DC = 8, AC = BC = 6, m∠ADC = 60° The area of ABCD is (A) +    (B) 16    (C) 32 (D) + 16    (E) 48 45 cos c csc -1 c x + mm = (A) (D) x + 8x − 16 x + 8x − 34 x + 8x − 16    (B)    (C) x+4 x+4 (x + 4) x + 8x − 34 (x + 4)    (E) −16 − 8x − x 2 (x + 4) 46 Which of the following statements is true about the expression (a + b)n − (a − b)n? n terms if n is even n +1 II It has terms if n is odd I It has III The exponent on the last term is always n (A) I only   (B) II only   (C) I and II only (D) I and III only   (E) II and III only –3 ​  Find cos c V m = 47 In �VWX, sin(X) = and cos(W) = ​  _ 17 (A) - 77    (B) -    (C) 85 85    (D) 85    (E) 77 85 85 14 SAT Math & Subject teSt 48 The intersection of the hyperbola ellipse ( x + 1)2 ( y − 1)2 + = is 32 18 ( x + 1) ( y – 1) = and the – (A) (3,4), (3,–4), (–5,4), (–5,–4)   (B) (3,2), (3,–2), (–5,2), (–5,–2) (C) (3,4), (3,–2), (–5,4), (–5,–2)   (D) (–3,4), (–3,–2), (5,4), (5,–2) (E) (3,–4), (3,2), (–5,–4), (–5,2) 49 The equation 8x6 + 72x5 + bx4 + cx3 – 687x2 – 2160x – 1700 = 0, as shown in the figure, has two complex roots The product of these complex roots is –687 425 17 ​   (A) –4   (B) ​  _ (C) 9   (D) ​  _  ​    ​        (E) ​    2 3π ​ , 3π ​   < A < 2π, and cos(B) = ​  –8  ​, ​  _ –24 ​  , π < B < ​  _ 50 If sin(A) = ​  _ 17 25 cos(2A + B) = (A) 2184 −5544 −2184 −3696    (B)    (C)    (D) 7225 7225 7225 7225 (E) 5544 7225 SAT MATHEMATICS LEVEL ­PRACTICE TEST 15 Level Practice Test Solutions (D) The sum of the measures of the angles of a triangle is 180, so 2x + + 4x − 12 + 3x + = 180 Combine and solve: 9x = 180, or x = 20 m∠Q = 44, m∠R = 68, and m∠S = 68 �QRS is isosceles, and QR = QS Solve y + = 3y − 13 to get y = 11 The sum of the sides is 6y − 11, so the perimeter is 6(11) − 11 = 55 J N 3` - j + K 3` - j + O -9 + -1 4 (B) g ` - j = =K O = - 15 - = - 19 = 19 5` - j - K 5` - j - O 4 L P (D) 81x y10 = ^3 27x y h^3 3xy h = 3x y 3 3xy (B) The slope of HK is 10, so the slope of the altitude is - In stan10 dard form, this makes the equation 10x + y = C Substitute the coordinates of G to get 10x + y = –26 (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) With r = 6, the total surface area is 108π sq cm (C) f(2) = 4(2)2 − = 4(4) − = 15 The composition of functions g(f(2)) = g(15) = 8(15) + = 127 (D) The factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36 18 = 1, and = The ratio of the factors cannot be = , 12 16 SAT Math & Subject teSt (E) Multiply by the common denominator to change the problem into a simple fraction rather than a complex fraction J N K + x - O 3^x - 4h 2^x - 4h + 2x - K Oc ^ x - h m = 3x - 12 - = 3x - 14 K - 3x - 12 O L P a +4ab = (A) Rewrite the equation with a common base ` j 125 ^3 625 h 3a -10ab becomes 53 a2  ab / 3a 10 ab Set the exponents equal to get –3(a2+4ab) = (3a2–10ab) Gather like terms ab = 7a2 so 3 that a = b 21 _ _ 10 (D) ​NL​ is   parallel to the altitude from K to ​HJ​  As a consequence, �NLH ~ �KZH, and LN = 6KZ and HN = 6HZ With LN = and HN = 6(4) = 2.4, the area of �NLH is 5(6)(2.4) = 7.2 11 (C) The slope of AB is −4 − −6 −1 = = The slope of the perpen3 − ( −9) 12 dicular line is The midpoint of AB is (–3,–1), so the equation of the perpendicular bisector is y + = 2(x + 3) 17 SAT MATHEMATICS LEVEL ­PRACTICE TEST 12 (C) TB2 = (TC)(TA), so TB2 = (6)(16) and TB = ∠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 = 60 = 15 13 (A) @ = 53 35 = - 125 and @ = = 243 2 3-5 5-3 - 125 – 243 = –184 2 14 (C) Adding a constant to a set of data shifts the center of the data but does not alter the spread of the data 15 (C) The segment joining the midpoints of two sides of a triangle is half as long as the third side DE is the consequence of the 4th set of midpoints, so DE = c m = The ratio of the areas is the square of BC 16 ( ) area �FDE = the ratio of corresponding sides so area �ABC 16 of �FDE = = The area 256 area �ABC 128 = = 256 256 16 (B) The function g(x) = 2f(x − 2) + moves the graph of f(x) right 2, stretches the y-coordinates from the x-axis by a factor of 2, and moves the graph up unit Use the points (0,3), (3,0), and (5,0) from f(x) to 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.) 18 SAT Math & Subject teSt 18 (C) The radius of the sphere is 25 The distance from the center of the sphere to the intersecting plane lies along the perpendicular Use the Pythagorean Theorem to get r2 + 142 = 252 or r2 = 429 The area of the circle formed by the plane and sphere is πr2, or 429π 3x 2x 19 (B) g(g(x)) = 3x 2x 9 3x 2x = 3x 2x 9 2x = 2x 3(3 x 1) (2 x + 9) 9x − − 2x − 7x − 12 = = 2(3 x 1) + 9(2 x + 9) 6x − + 18x + 81 24x + 79 20 (D) The area of a triangle is given by the formula A= Substituting the given information, 750= absin(θ) (48)(52)sin(θ) Solving this equation for Q yields that Q = 37° or 143° If m∠Q = 37°, then ∠E would be the largest angle of the triangle, and that measure, found using the Law of Cosines, would be 78°, less than 143° 21 (D) log3(a) = c and log3(b) = 2c are equivalent to a = 3c and b = 32c = (3c)2 = a2 Therefore, a = b SAT MATHEMATICS LEVEL ­PRACTICE TEST 19 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 = and EH = 10 �EHZ is also 30-60-90, so HZ = 10 and EZ = 20 �EYZ is an isosceles triangle, making ZY = 20 In �THY, HZ EZ XZ = = Therefore, = EZ || TY , so HY TY 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, 2 so that cos ^i h = 25 + 29 - 34 or cos ^i h = 31 and θ = 77.7° 145 ^25 h^29 h 24 (A) If - + i is a root of the equation, then its conjugate - - i is also The sum of the roots is - = - b and the product a of the roots is c - m - c i m = 16 + 18 = 737 Getting a com25 64 800 mon denominator for the sum of the roots gives - 1280 = - b With 800 a a = 800, b = 1280, and c = 737, the equation is 800x2 + 1280x + 737 = y-2 Use the identity cos(2t) = y-2 m or x - =- ^ y - h2 − 2sin2(t) to get x - = - c 25 (D) cos(2t) = x − and sin(t) = 20 SAT Math & Subject teSt 26 (B) There is no indication in the problem that titles are associated with any of the positions on the committee, so this problem represents a combination (order is not important) There are 14 people to choose from, so the number of combinations is C5 = 2002 A 14 committee with more Democrats than Republicans could have Democrats and Republicans, Democrats and Republican, or Democrats and Republicans The number of ways this can happen is (8C5 )(6C0 ) + (8C4 )(6C1 ) + (8C2 )(6C2 ) = 1316 Therefore, the probability that the Democrats will have more members on the committee than the Republicans is 1316 2002 27 (B) 2u − 3v = [2(–5) − 3(3), 2(4) − 3(–1)] = [–19, 11] |2u − 3v |, the distance from the origin to the end of the vector, = ( 19)2 +112 = 502 28 (B) The circumcenter of a triangle is the intersection of the perpendicular bisectors of the sides of the triangle The slope of AB is , and its midpoint is (–7,6) The equation of the perpendicular bisector to AB is y = –2x − The slope of AC is –1, and its midpoint is (–4,–3) The equation of the perpendicular bisector of AC is y = x + The intersection of y = –2x − and y = x + is (–3,–2) 29 (D) The volume of the original pyramid is s h The volume of (8s ) 2H , where H is the new height Set these 1 expressions equal to get s h = (8s ) 2H Solve for H: 1.5625h The 3 the new pyramid is height needs to be increased by 56.25% to keep the volume the same 21 SAT MATHEMATICS LEVEL ­PRACTICE TEST 20cis c 19r m 18 30 (A) = 4cis ` 15r j = 4cis ` 5r j = cos ` 5r j + 4i sin ` 5r j r 18 6 5cis c m = c - m + 4i ` j =- + 2i 2 31 (B) Use the change-of-base formula to rewrite log log b ( bc ) as ( b c ) Apply the properties of logarithms to simplify the numerator a2 ( ) log b a 1 log b b + log b c log b b c and denominator of the fraction = log b (a) log b (a ) 4 + y log b (b ) + log b (c ) + 4y = = 3 = , recalling that logb(b) = 2x 6x log b (a ) ( ) ( ) ( ) 32 (C) The center of the hyperbola is (–3,1) Because the focus is at (7,1), the hyperbola is horizontal and takes the form ( x + 3) ( y – 1) = – a2 b2 The focus is 10 units from the center of the hyperbola The equation of the asymptotes for a horizontal hyperbola (found by replacing the with a zero) become y − =  b b (x + 3) The ratio must equal , and a2 + a a b2 = 102, so a = and b = 6, giving the equation ( x + 3)2 ( y − 1)2 − = 64 36 33 (B) The volume of the original cone is r ^8 h2 12 = 128r and the volume at the time in question is half this amount The ratio of the radius of the circle at the water’s surface to the height of the water is 2:3 At any given time, the volume of the water, in terms of the height of the water, is r ^ r h2 h = r c h m and h = rh Setting this equal to 27 3 22 SAT Math & Subject teSt one-half the original volume, the equation is rh = 64r Multiply by 27 27 to get h3 = 864 = – 96 = 27 – 32 = 27 – – Therefore, h = 4r 34 (C) cos(2t) = − sin2(t) Substituting this into the equation gives sin(t) = − sin2(t) or sin2(t) + sin(t) − 1= Factor and solve: (2 sin(t) − 1)(sin(t) + 1) = so sin(t) = , –1 In the interval ≤ t ≤ 2π, the answer to this problem would be ' r , 5r , 3r Subtract 4π from 6 each of these answer to satisfy the domain restriction of the problem, ' - 23r , - 19r , - 5r 6 35 (B) Sketch the graph of y = |2x − 3||x − 1| − on your graphing calculator and see that y(2) = 0, so this point needs to be eliminated from the interval from < x < 5 Algebraically: Determine those values for which |2x − 3||x − 1| = and then analyze the inequality |2x − 3||x − 1| = requires that 2x – 3|x – 1| = or 2x – 3|x – 1| = –2 If 2x – 3|x – 1| = then 2x – = 3|x – 1| If x > 1, this equation becomes 2x – = 3x – or x = If x < 1, the equation becomes 2x – = –3x + or x = If 2x – 3|x – 1| = –2, then 2x + = 3|x – 1| If x > 1, this equation becomes 2x + = 3x – or x = If x < 1, the equation becomes 2x + = –3x + or x = Given |2x − 3||x − 1| = when x = , 1, and 5, you can check that x = fails to solve the inequality, x = satisfies the inequality, and that x = fails to solve the inequality Therefore, all values between 5, with the exception of x = 1, satisfy the inequality and SAT MATHEMATICS LEVEL ­PRACTICE TEST 36 (C) / 12c 23 m k 23 is an infinite geometric series with |r| < and first k =0 12 = 36 In the same way, 1- = 12 The difference in these values is 24 term equal to 12 The sum of the series is / 18` -21 j k =0 k = 18 - `- j 37 (B) Use matrices to solve the equation [A]–1[B] = [C] where [A] = -4 33 3H so = 5, = - , and 8 and [B] = [C] = > > H > H y x -6 -3 11 1 = x − y + z = - - + = 31 Therefore, = 30 x-y+z 31 ` j 30 z 38 (E) f(x) = - cos c 3r x - 3r m = - cos c 3r ` x - jm + The 10 5 amplitude of the function is |–2| = 2, the graph is shifted to the right , 2r and the period of the function is 3r = 10 f c m = f c 31 m = + 2 39 (C) The polar coordinate form for 32i is 32cis ` r j DeMoivre’s Theorem states that if z = rcis(θ), then zn = rncis(nθ) The radius is for all the choices, so the leading coefficient is not the issue Of the five choices, only 2cis ` 11r j fails to be co-terminal with r + 2nr 10 10 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 24 SAT Math & Subject teSt 41 (D) The denominator of the function factors to (2x + 3)(2x − 3), so the function reduces The graph of f(x) has a hole at the point c , m f(2.5) = 2.25, so statement I is correct, and inspection of the graph of f(x) on your graphing calculator will verify that statement III is also true 42 (D) g–1(13) = a means that g(a) = 13 This leads to the equation 9log8(x − 3) − = 13 Add and divide by to get log8(x − 3) = Rewrite the logarithmic equation as an exponential equation: x − = 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 of the way along the median, so T is 15 units from RS �RST, �QST, and �QRT are all equal in area The area of �RST is ^5 15 h^30 h = 75 15 The area of �QRT = h(60) = 30h = 75 15 2 75 15 = 15 so h = 30 44 (D) The altitude to the base of an isosceles triangle also bisects the vertex angle, so m∠ADE = 30 With the hypotenuse of the triangle having a length of 8, AE = and DE = �AEC is a right angle with leg and hypotenuse Use the Pythagorean Theorem to determine that BE = The area of a quadrilateral with perpendicular diagonals 25 SAT MATHEMATICS LEVEL ­PRACTICE TEST is equal to half the product of the diagonals, so the area of ABCD is ^8 h^2 + h = + 16 45 (D) T Đ x4à csc 1 ă â ¹ implies that csc(T ) x4 and that Use the Pythagorean Theorem to determine that x4 2 the remaining leg of the right triangle has length ( x + 4) − = sin(T ) x + x − so that cos(T ) 2 x  8x  cos(2θ) = cos2(θ) − sin2(θ) x4 § x  x  · § · x  8x  25 ă  = ă = ă â x4ạ ( x  4) ( x  4)2 x4 â x  8x  34 ( x  4)2 46 (C) Both (a + b)n and (a − b)n contain n + terms when they are expanded All the odd terms in (a − b)n will be positive, and the even terms will be negative If n is even, there will be n + odd terms and n even terms in the 2 n expansion of (a − b) The odd terms will subtract out, leaving n terms The last terms in the expansions of both binomials, bn, will subtract out of the answer If n is odd, there will be an equal number of even and odd terms The odd terms will subtract out leaving n terms in the difference 47 (D) With sin(X) = and cos(W) = - , cos(X) = 15 and 17 17 sin(W) = V + W + X = 180, so V = 180 − (W + X) and cos(V) = cos(180 − (W + X)) = –cos(W + X) –cos(W + X) = sin(W)sin(X) − ... opposite the longest side) 3 42 = 25 2 + 29 2 − 2( 25) (29 ) cos(θ) 2( 25) (29 ) cos(θ) = 25 2 + 29 2 − 3 42, 2 so that cos ^i h = 25 + 29 - 34 or cos ^i h = 31 and θ = 77.7° 145 ^25 h ^29 h 24 (A) If - + i is a... ​  _ 17 25 cos(2A + B) = (A) 21 84 −5544 ? ?21 84 −3696    (B)    (C)    (D) 722 5 722 5 722 5 722 5 (E) 5544 722 5 SAT MATHEMATICS LEVEL ? ?PRACTICE TEST 15 Level Practice Test Solutions (D) The sum of... r2 + 1 42 = 25 2 or r2 = 429 The area of the circle formed by the plane and sphere is πr2, or 429 π 3x 2x 19 (B) g(g(x)) = 3x 2x 9 3x 2x = 3x 2x 9 2x = 2x 3(3 x 1) (2 x + 9) 9x − − 2x − 7x − 12