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PSAT/NMSQT ® Practice Test #1 Answer Explanations Confidential: For Michigan Use Only

CONFIDENTIAL: For Michigan Use Only PSAT/NMSQT Practice Test #1 Math Test – No Calculator Answer Explanations Choice C is the correct answer. Using the distributive property to expand the given expression gives ( ) ( ) 2 s ss st s t tt - =- . Choices A, B, and D are incorrect . In each of these choices, at least one of the products in the expansion is not correct. For example ( ) 2 ss s tt = , not ,s t and ( ) s ts t = , not st or . s t Question 12 p(x ) = 3( x 2 + 10x + 5)  5( x  k) In the polynomial p(x) defined above, k is a constant. If p (x ) is divisible by x, what is the value of k? A) 3 B) 2 C) 0 D) 3 Item Difficulty: Medium Content: Passport to Advanced Math Correct Answer: A Choice A is the correct answer. If polynomial p(x ) is divisible by x, then x must be a factor of the polynomial, or equivalently, the constant term o f the polynomial must be zero. Multiplying out on the right side of the equation gives p (x ) = 3x 2 + 30 x + 15 − 5 x + 5k, which can be rewritten as p (x ) = 3x 2 + 25 x + (5 k + 15) . Hence, 5k + 15 = 0, and so k = −3. Choices B, C, and D are the not correct answers because if the value of k were as indicated in those choices, then x would not be a factor of the polynomial p(x) , and so p(x) would not be divisible by x. Page 62

CONFIDENTIAL: For Michigan Use Only PSAT/NMSQT Practice Test #1 Math Test – No Calculator Answer Explanations Question 13 In the xy-plane, if the parabola with equation y = ax 2 + bx + c , where a, b, and c are constants, passes through the point (1, 1), which of the following must be true? A) a  b = 1 B) b + c = 1 C) a + b + c = 1 D) a  b + c = 1 Item Difficulty: Hard Content: Passport to Advanced Math Correct Answer: D Choice D is the correct answer. If the graph of a parabola passes through the point ( − 1 , 1), then the ordered pair ( −1 , 1) must satisfy the equation of the parabola. Thus, ( ) ( ) 2 1 1 1, a bc = - +-+ which is equivalent to 1. abc - += Choices A, B, and C are incorrect and could result from misinterpreting what it means for the point (?1, 1) to be on the parabola or from common calculation errors while expressing this fact algebraically. These are the directions students will see in the test for the Student -Produced Response questions. Question 1 4 Item Difficulty: Easy Content: Heart of Algebra Correct Answer: 300 The correct answer is 300. To solve the given equation for h, first add 6 to both sides of the equation to get 30 10 h = . Then multiply both sides of this equation by 10 to yield h = 300 . Page 63

CONFIDENTIAL: For Michigan Use Only PSAT/NMSQT Practice Test #1 Math Test – No Calculator Answer Explanations Question 1 5 What is the value of a if (2a + 3)  (4a  8) = 7 ? Item Difficulty: Medium Content: Heart of Algebra Correct Answer: 2 The correct answer is 2. The equation given can be rewritten as 2a + 3 ? 4a + 8 = 7, which is equivalent to ? 2a + 11 = 7, and so a = 2. Question 1 6 Item Difficulty: Medium Content: Passport to Advanced Math Correct Answer: 9 The correct answer is 9. Mul tiplying out the given expression gives 2 2 4(9 ) . 4 x x Since x  0 , dividing both the numerator and the denominator of the fraction by 2 4x simplifies the expression to 9. Question 1 7 If x  2 is a factor of x 2  bx + b, where b is a constant, what is the value of b ? Item Difficulty: Hard Content: Passport to Advanced Math Correct Answer: 4 The correct answer is 4. If x ? 2 is a factor of x2 ? bx + b, where b is a constant, then x2 ? bx + b can be written as the product (x ? 2)(x ? a) for some real number a. Expanding ( x ? 2)( x ? a) gives x2 ? 2x ? ax + 2a, which can be rewritten as x2 ? (2 + a )x + 2a. Hence, x2 ? (2 + a) x + 2a = x 2 ? bx + b is true for all values of x. Consequently, the coefficients of like terms on each side of the equation must be the same: 2 + a = b and 2a = b . Solving this system gives 4. b = Page 64

CONFIDENTIAL: For Michigan Use Only PSAT/NMSQT Practice Test #1 Math Test – Calculator Answer Explanations Question 21 What is the minimum value of the function graphed on the xy -plane above, for 4  x  6 ? A)   B) 4 C) 2 D) 1 Item Difficulty: Hard Content: Passport to Advanced Math Correct Answer: C Choice C is the correct answer. The minimum value of a graphed fun ction is the minimum y- value of all the points on the graph. For the graph shown, the minimum is at the left endpoint of the graph, the y- value of which is − 2. Choice A is incorrect. If the graph would continue indefinitely downward, then the minimum value of the function would be negative infinity. However, the domain of the function is restricted ( − 4  x  6) and the minimum value of the graph occurs at point ( 4 , 2 ). Choice B is incorrect ; 4 is the x- value of the point on the graph where the minimum value of the function occurs. Choice D is incorrect because there are points of the graph below the x- axis; therefore, the minimum value of the function cannot be positive. Page 84

CONFIDENTIAL: For Michigan Use Only PSAT/NMSQT Practice Test #1 Math Test – Calculator Answer Explanations ▼ Questions 22 -24 refer to the following information. In 1929, the astronomer Edwin Hubble published the data shown. The graph plots the velocity of galaxies relative to Earth against the distances of galaxies from Earth. Hubble’s data can be modeled by t he equation v = 500 d, where v is the velocity, in kilometers per second, at which the galaxy is moving away from Earth and d is the distance, in megaparsecs, of the galaxy from Earth. Assume that the relationship is valid for larger distances than are show n in the graph. (A megaparsec (Mpc) is 3.1 ? 10 19 kilometers.) Question 22 According to Hubble?s data, how fast, in meters per second, is G alaxy Q moving away from Earth? A) 2 ? 10 6 m/s B) 5 ? 10 5 m/s C) 5 ? 10 2 m/s D) 2.5 ? 10 2 m/s Item Difficulty: Hard Content: Probability and Data Analysis Correct Answer: B Choice B is the correct answer. The coordinates of the data point that represent Galaxy Q on the scatterplot are ( 2.0, 500 ), which means that Galaxy Q is at a distance of about 2.0 Mpc from E arth and moves away from Earth at a velocity of approximately 500 km/s. The question asks for the velocity in meters per second; therefore, kilometers (km) need to be converted into meters ( m ). Since 1 km is Page 85

CONFIDENTIAL: For Michigan Use Only PSAT/NMSQT Practice Test #1 Math Test – Calculator Answer Explanations equal to 1,000 m , it follows that Galaxy Q is mo ving away from Earth at a velocity of 500 ? 1,000 m/s, or 5 ? 10 5 m/s. Choices A, C, and D are incorrect and may result from an incorrect interpretation of the coordinates of the point that represents Galaxy Q on the scatterplot or an incorrect conversion of the units. Question 23 There are four galaxies shown in the graph at approximately 0.9 Mpc from Earth. Which of the following is clos est to the range of velocities of these four galaxies, in kilometers per second? A) 100 B) 200 C) 450 D) 700 Item Difficulty: Hard Content: Probability and Data Analysis Correct Answer: D Choice D is the correct answer. The velocities, in km/s, of the f our galaxies shown in the graph at approximately 0.9 Mpc from Earth are about ?50, +200, +500, and +650 . Thus, the range of the four velocities is approximately 650 ? ( − 50) = 700 km/s. Choices A, B, and C are incorrect . The range of velocities is the dif ference between the largest and smallest velocity. Each of the answer choices A, B, and C are too small compared to the real value of the range. Question 24 Based on the model, what is the velocity, in kilometers per second, of a galaxy that is 15 Mpc from Earth? A) 7,500 km/s B) 5,000 km/s C) 1,100 km/s D) 750 km/s Item Difficulty: Medium Content: Heart of Algebra Correct Answer: A Choice A is the correct answer. The model indicates that the relationship between the velocities of the galaxies, in km/s, and their distance from Earth, in Mpc, is v = 500d . Therefore, the velocity of a galaxy that is 15 Mpc from Earth is v = 500(15) km/s, or 7,500 km/s. Page 86

CONFIDENTIAL: For Michigan Use Only PSAT/NMSQT Practice Test #1 Math Test – Calculator Answer Explanations Based on the model, the other choices are incorrect: Choice B is the speed of a galaxy that is 10 Mpc from Earth. Choice C is the speed of a galaxy that is 2.2 Mpc from Earth. Choice D is the speed of a galaxy that is 1.5 Mpc from Earth. ▲ Question 25 Janice puts a fence around her rectangular garden. The garden has a length that is 9 feet less than 3 times its width. What is the perimeter of Janice?s fence if the area of her garden is 5,670 square feet? A) 342 feet B) 318 feet C) 300 feet D) 270 feet Item Difficulty: Hard Content: Passport to Advanced Math Correct Answer: A Choice A is the correct answer. Let w represent the width of Janice’s garden and 3w ? 9 represent the length of Janice’s garden. Since the area of Janice’s garden is 5,670 square feet, it follows that w(3w ? 9 ) = 5,670 , which after dividing by 3 on both sides simplifies to w(w ? 3) = 1,890 . From this point on, different ways could be used to solve this equation. One could rewrite this quadratic equation in the standard form and use the quadratic formula to solve it. Another approach would be to look among integer factors of 1,890 and try to find two that differ from each other by 3 and whose product is 1,890 . The prime factorization of 1,890 (23 357) can help with this. Two factors that satisfy the conditions above are 42 and 45 (note that 42 = 237 and 45 = 3 25) . The numbers ?45 and ?42 also satisfy the above conditions (w = ?42) , but since w represents the width of Janice’s garden, the negative values of w can be rejected. Thus w = 45 feet, and so the length of the garden must be 3(45) ? 9 = 126 feet. Therefore , the perimeter of Janice’s garden is 2(45 + 126) = 2(171) = 342 feet. Choice B is incorrect. This answer choice could result from incorrectly identifying the width of the garden as 42 feet instead of 45 feet. Choices C and D are incorrect ; both answers would result in an area of the garden that is significantly smaller than 5,670 square feet. For example, if the perimeter of the garden were 270 feet, as in choice D, then w + l = 135 feet, where w represents the width and l represents the length of the garden. So l = 135 ? w . It is also given that l = 3w ? 9 , which Page 87

CONFIDENTIAL: For Michigan Use Only PSAT/NMSQT Practice Test #1 Math Test – Calculator Answer Explanations implies that 135 ? w = 3 w ? 9 . Solving this for w gives w = 36, and so l = 99 . The area of the garden would then be 36 ? 99 square feet, which is clearly less than 5,600 square feet. Question 26 A) sin A B) sin B C) tan A D) tan B Item Difficulty: Hard Content: Additional Topic in Math Correct Answer: D Choice D is the correct answer. Since the ratio b a involves only the legs of the right triangle, it follows that , of the given choices, the ratio can be equal to the tangent of one of the angles. In a right triangle, the tangent of an acute angle is defined as the ratio of the opposite side to the adjacent side of the angle. Side b is opposite to angle B and side a is adjacent to angle B. Therefore, tan . b B a = Choices A and B cannot be correct ; the sine of an acute angle in a right triangle is defined as the ratio of the opposite side to the hypotenuse, and the ratio shown involves only the legs of the triangle. Choice C is incorrect . In the triangle ABC shown, tan , a A b = not b a. Page 88

CONFIDENTIAL: For Michigan Use Only PSAT/NMSQT Practice Test #1 Math Test – Calculator Answer Explanations Question 27 A system of inequalities and a graph are shown above. Which section or sections of the graph could represent all of the solutions to the system? A) Section R B) Sections Q and S C) Sections Q and P D) Sections Q, R, and S Item Difficulty: Hard Content: Heart of Algebra Correct Answer: A Choice A is the correct answer. The solution set of the inequality y  ?x is the union of sections R and S of the graph. The solution set of the inequality 2y > 3 x + 2 is the union of sections R and Q of the graph. The solutions of the system consist of the coordinates of all the points that satisfy both inequalities, and therefore, section R represents all the solutions to the system since it is common to the solut ions of both inequalities. Choices B, C, and D are incorrect because they contain ordered pairs that do not satisfy both of the inequalities. Page 89

CONFIDENTIAL: For Michigan Use Only PSAT/NMSQT Practice Test #1 Math Test – Calculator Answer Explanations Question 28 The xy-plane above shows one of the two points of intersection of the graphs of a linear func tion and a quadratic function. The shown point of intersection has coordinates ( v,w ). If the vertex of the graph of the quadratic function is at (4, 19), what is the value of v ? Item Difficulty: Medium Content: Passport to Advanced Math The correct answer is 6. Since the vertex of the graph of the quadratic function is at (4, 19) , the equation of the parabola is of the form y = a(x ? 4)2 + 19 . It is also given that the parabola passes through point (0, 3) . This means that a( 3 ? 4) 2 + 19 , and so a = ?1 . So the graph of the parabola is y = ?( x ? 4)2 + 19 . Since the line passes through the points (0, ?9) and (2, ?1) , one can calculate the slope of the line ( ) 1 ( 9) 4 20 - -- = - that passes through these points and write the equation of the line in t he slope-int ercept form as y = 4x ? 9 . The coordinates of the intersection points of the line and the parabola satisfy both the equation of the parabola and the equation of the line. Therefore, these coordinates are the solutions to the system of equations below: y = 4x ? 9 y = ?( x ? 4) 2 + 19 Substituting 4x ? 9 for y into the second equation gives 4x ? 9 = ? (x ? 4) 2 + 19 , which is equivalent to x 2 ? 4x ? 12 = 0 . After factoring, this equation can be rewritten as Page 90

CONFIDENTIAL: For Michigan Use Only PSAT/NMSQT Practice Test #1 Math Test – Calculator Answer Explanations ( x ? 6)( x + 2) = 0 , and so x = 6 or x = ?2 . Since point (v, w ) is on the right side of the y -axis, it follows that v cannot be ?2 . Therefore, v = 6 . Question 29 In a college archaeology class, 78 students are going to a dig site to find and study artifacts. The dig site has been divided into 24 sections, and each section will be studied by a group of either 2 or 4 students. How many of the sections will be studied by a group of 2 students? Item Difficulty: Hard Content: Heart of Algebra The correct answer is 9. Let x be the number of sections that will be studied by 2 students and y be the number of sections that will be studied by 4 students. Since there are 24 sections that will be studied by 78 students, it follows that x + y = 24 and 2x + 4 y = 78 . Solving this system gives x = 9 and y = 15 . Therefore, 9 of the sections will be studied by a group of 2 students. Alternatively, if all 24 sections were studied by a group of 4 students, then the total number of students required would be 24 ? 4 = 96 . Since the actual number of students is 78 , the difference 96 ? 78 = 18 represents the number of “missing” students, and each pair of these “missing” students represents one of the sections that will be studied by 2 students. Hence, the number of sections that will be studied by 2 students is equal to the number of pairs that 18 students can form, which is 18 9. 2 = ▼ Questions 30 and 31 refer to the following information. An arrow is launched upward with an initial speed of 100 meters per second (m/s). The equations above describe the constant -acceleration motion of the arrow, where v 0 is the initial speed of the arrow, v is the speed of the arrow as it is moving up in the air, h is the height of the arrow above the ground, t is the time elapsed since the arrow was projected upward, and g is the acceleration due to gravity (9.8 m/s 2). Page 91

CONFIDENTIAL: For Michigan Use Only PSAT/NMSQT Practice Test #1 Math Test – Calculator Answer Explanations Question 30 What is the maximum height from the ground the arrow will rise to the nearest m eter? Item Difficulty: Hard Content: Passport to Advanced Math The correct answer is 510 . As the arrow moves upward, its speed decreases continuously and it becomes 0 when the arrow reaches its maximum height. Using the position- speed equation and the fact that 0 v= when h is maximum gives 0 = 100 2 − 2gh. Solving for h gives 2 100 2(9.8) h = meters, which to the nearest meter is 510 . Al ternatively, the maximum height can be found using the position -time equation. Substituting 100 for v0 and 9.8 for g into this equation gives 2 1 100 (9.8) . 2 ht t =- Completing the square gives the equivalent equation ( ) ( ) 22 100 100 4.9 4.9 . 9.8 9.8 ht =- - + Therefore, the maximum height from the ground the arrow will rise is ( ) 2 100 4.9 9.8 meters, which to the nearest meter is 510 . Question 31 How long will it take for the arrow to reach its maximum height to the nearest tenth of a second? Item Difficulty: Hard Content: Passport to Advanced Math The correct answer is 10.2 seconds (or 51/5 seconds) . As the arrow moves upward, its speed decreases continuously, and it becomes 0 when the arrow reaches its maximum height. Using the speed-time equation and the fact that 0 v= when h is maximum, we get 0 = 100 ? 9.8t. Solving this equation for t gives 100 10.2041 9.8 t== seconds, which to the nearest tenth of a second is 10.2 . ▲ Page 92