2012 Psat/nmsqt Practice Test Explanations - Section 2 g3722
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Mathematics: Section 2 Mathematics Question 1 Choice (C) is correct. If 2x + 4 = 8, then 2x = (2x + 4) – 4 = 8 – 4 = 4, and so x = 2. Therefore, 6x + 4 = (6)(2) + 4 = 12 + 4 = 16. Choice (A) is not correct. The value of x is 2, but the value of 6x + 4 is (6)(2) + 4 = 16. Choice (B) is not correct. The value of 2x + 4 is 8, but the value of 6x + 4 is 16. Choice (D) is not correct. Since 2x + 4 = 8, it follows that 3(2x + 4) = 6x + 12 = 24. Therefore, the value of 6x + 4 is 24 – 8 = 16, not 20. Choice (E) is not correct. Since 2x + 4 = 8, it follows that 3(2x + 4) = 6x + 12 = 24, but the value of 6x + 4 is 24 – 8 = 16. Mathematics Question 2 Choice (E) is correct. Since the veterinarian treated 2 mice, 3 cats, 6 dogs and no other animals, the total number of animals the veterinarian treated was 2 + 3 + 6 = 11. Therefore, the ratio of the number of cats treated to the total number of animals treated by the veterinarian was 3 to 11. Choice (A) is not correct. Had the veterinarian treated a total of 12 animals, the ratio of the number of cats treated to the total number of animals treated by the veterinarian would have been 3 to 12, or 1 to 4. However, the total number of animals the veterinarian treated was 2 + 3 + 6 = 11, and so the ratio of the number of cats treated to the total number of animals treated was 3 to 11. Choice (B) is not correct. The veterinarian treated 3 cats and a total of 2 + 3 + 6 = 11 animals. Therefore, the ratio of the number of cats treated to the total number of animals treated by the veterinarian was 3 to 11, not 1 to 6. Choice (C) is not correct. The veterinarian treated 3 cats and a total of 2 + 3 + 6 = 11 animals. Therefore, the ratio of the number of cats treated to the total number of animals treated by the veterinarian was 3 to 11, not 1 to 13. Choice (D) is not correct. The ratio of the number of cats treated by the veterinarian to the total number of other animals treated was 3 to 8, but the ratio of the number of cats treated to the total number of animals, including cats, treated was 3 to 11. Mathematics Question 3 Choice (E) is correct. Since the measure of an exterior angle of a triangle is equal to the sum of the measures of the interior angles at the other two vertices, it follows that 110 = 40 + x. Solving for x gives x = 70. Choice (A) is not correct. If the value of x were 30, then 40 + x would be equal to 70. However, 40 + x is equal to 110, not 70, so the value of x cannot be 30. Choice (B) is not correct. If the value of x were 40, then 40 + x would be equal to 80. However, 40 + x is equal to 110, not 80, so the value of x cannot be 40.
2012 PSAT/NMSQT Answer Explanations © 2012 The College Board. All Rights Reserved
Choice (C) is not correct. If the value of x were 50, then 40 + x would be equal to 90. However, 40 + x is equal to 110, not 90, so the value of x cannot be 50. Choice (D) is not correct. If the value of x were 60, then 40 + x would be equal to 100. However, 40 + x is equal to 110, not 100, so the value of x cannot be 60. Mathematics Question 4 Choice (D) is correct. For the 12 books sold, the three different sale prices represented were $2.50, $4.50 and $5.00. Alex sold 4 books for $2.50 each and Dana sold 0 books for $2.50, for a total of 4 + 0 = 4 books sold at $2.50 each. Alex sold 2 books for $4.50 each and Dana sold 3 books for $4.50 each, for a total of 2 + 3 = 5 books sold at $4.50 each. Alex sold 1 book for $5.00 and Dana sold 2 books for $5.00 each, for a total of 1 + 2 = 3 books sold at $5.00 each. Therefore, the most frequently occurring sale price for the 12 books was $4.50. Choice (A) is not correct. There were 4 books sold for $2.50 each. However, there were 5 books sold for $4.50 each. Therefore, the most frequently occurring sale price for the 12 books was not $2.50. Choice (B) is not correct. There were no books sold for the price of $4.00 each. Therefore, the most frequently occurring sale price for the 12 books was not $4.00. Choice (C) is not correct. There were no books sold for the price of $4.25 each. Therefore, the most frequently occurring sale price for the 12 books was not $4.25. Choice (E) is not correct. There were 3 books sold for the price of $5.00 each. However, there were 5 books sold for $4.50 each. Therefore, the most frequently occurring sale price for the 12 books sold was not $5.00. Mathematics Question 5 Choice (C) is correct. Since 1 yard is equal to 3 feet, it follows that 16 yards is equal to (16)(3) = 48 feet. Therefore, 16 yards is 48 – 16 = 32 feet longer than 16 feet. Choice (A) is not correct. Since 1 yard is equal to 3 feet, it follows that 16 yards is equal to (16)(3) = 48 feet. However, 16 yards is 48 – 16 = 32, not 48, feet longer than 16 feet. Choice (B) is not correct. Since 1 yard is equal to 3 feet, it follows that 16 yards is equal to (16)(3) = 48 feet. Thus 16 yards is 48 – 8 = 40 feet longer than 8 feet. However, 16 yards is 48 – 16 = 32 feet longer than 16 feet. Choice (D) is not correct. Since 1 yard is equal to 3 feet, it follows that 16 yards is equal to (16)(3) = 48 feet. Thus 16 yards is 48 – 24 = 24 feet longer than 24 feet. However, 16 yards is 48 – 16 = 32 feet longer than 16 feet. Choice (E) is not correct. Since 1 yard is equal to 3 feet, it follows that 16 yards is equal to (16)(3) = 48 feet. Thus 16 yards is 48 – 32 = 16 feet longer than 32 feet. However, 16 yards is 48 – 16 = 32 feet longer than 16 feet.
2012 PSAT/NMSQT Answer Explanations © 2012 The College Board. All Rights Reserved
Mathematics Question 6 2
2
2
2
Choice (E) is correct. For both rows of the table, r is equal to s: 2 = 4 and 3 =9. Hence r = ks, 2 where k = 1, for both rows of the table. Therefore, r could be directly proportional to s. Choice (A) is not correct. It could be true that could be proportional to s.
s is proportional to r, but it is r2, not
r , that
Choice (B) is not correct. If r were directly proportional to s, then it would be true that r = ks for some constant k. But 2 =
1 1 (4) and 3 = (9), so there is no constant k for which r = ks. 2 3
Therefore, r cannot be directly proportional to s. Choice (C) is not correct. If r + 1 were directly proportional to s, then it would be true that r + 1 = ks, for some constant k. But 2 + 1 =
3 4 (4) and 3 + 1 = (9), so there is no constant k for which r 4 9
+ 1 = ks. Therefore, r + 1 cannot be directly proportional to s. Choice (D) is not correct. If 2r were directly proportional to s, then it would be true that 2r = ks, for some constant k. But 2(2) = 1(4) and 2(3) =
2 (9), so there is no constant k for which 2r = ks. 3
Therefore, 2r cannot be directly proportional to s. Mathematics Question 7 Choice (D) is correct. Since 20 percent of the surveyed customers answered “black,” it follows that the probability is
20 100
1 that a customer chosen at random from those surveyed will be 5
one who answered “black.” Choice (A) is not correct. The probability cannot be
1 that a customer chosen at random from 20
those surveyed will be one who answered “black,” because the graph indicates that 20 percent, which is 1 in 5 (not 5 percent, which is 1 in 20), of the customers surveyed answered “black.” Choice (B) is not correct. The probability cannot be
1 that a customer chosen at random from 18
those surveyed will be one who answered “black,” because the graph indicates that 20 percent, which is 1 in 5 (not 1 in 18), of the customers surveyed answered “black.” Choice (C) is not correct. The probability cannot be
1 that a customer chosen at random from 10
those surveyed will be one who answered “black,” because the graph indicates that 20 percent, which is 1 in 5 (not 10 percent, which is 1 in 10), of the customers surveyed answered “black.” Choice (E) is not correct. The probability cannot be
1 that a customer chosen at random from 4
those surveyed will be one who answered “black,” because the graph indicates that 20 percent, which is 1 in 5 (not 25 percent, which is 1 in 4), of the customers surveyed answered “black.”
2012 PSAT/NMSQT Answer Explanations © 2012 The College Board. All Rights Reserved
Mathematics Question 8 Choice (A) is correct. Since point P lies on line segment RU , the sum of the measures of
SPU is 180°. Similarly, since point P lies on segment QT , the sum of the measures of QPS and SPT is 180°. The measure of SPU is 70°, so the measure of RPS is 180° − 70° = 110°. Since RPS has measure 110° and RPQ has measure 40°, the measure of SPT is 180° − (110° + 40°) = 30°. RPS and
Choice (B) is not correct. The measure of RPQ is 40°, so the measure of TPU is also 40°, since vertical angles are of equal measure. The measure of SPU is 70°, so the measure of RPS is 180° − 70° = 110°. If the measure of SPT were 35°, then the measure of RPU would be 110° + 35° + 40° = 185°. However, RPU is a straight angle, which has measure 180°, not 185°. Therefore, the measure of SPT cannot be 35°. Choice (C) is not correct. The measure of measure of SPT is 30°.
RPQ is 40°, as is the measure of
TPU, but the
Choice (D) is not correct. The measure of RPQ is 40°, so the measure of TPU is also 40°, since vertical angles are of equal measure. The measure of SPU is 70°, so the measure of RPS is 180° − 70° = 110°. If the measure of SPT were 45°, then the measure of RPU would be 110° + 45° + 40° = 195°. However, RPU is a straight angle, which has measure 180°, not 195°. Therefore, the measure of SPT cannot be 45°. Choice (E) is not correct. The measure of RPQ is 40°, so the measure of TPU is also 40°, since vertical angles are of equal measure. The measure of SPU is 70°, so the measure of RPS is 180° − 70° = 110°. If the measure of SPT were 50°, then the measure of RPU would be 110° + 50° + 40° = 200°. However, RPU is a straight angle, which has measure 180°, not 200°. Therefore, the measure of SPT cannot be 50°. Mathematics Question 9 Choice (D) is correct. The function f is defined by f(x) = 2x, so f(5t) = 2(5t) = 10t. Choice (A) is not correct. The function f is defined by f(x) = 2x, so f(t) = 2t, but f(5t) = 2(5t) = 10t. Choice (B) is not correct. The function f is defined by f(x) = 2x, so f(2.5t) = 2(2.5t) = 5t, but f(5t) = 2(5t) = 10t. Choice (C) is not correct. The function f is defined by f(x) = 2x, so f(5t) = 2(5t) = 10t, not (2 + 5)t = 7t. 2
Choice (E) is not correct. The function f is defined by f(x) = 2x, so f(5t) = 2(5t) = 10t, not 5 t = 25t. Mathematics Question 10 Choice (B) is correct. Adding the respective sides of the equations ax + by =14 and ax – by = 4 gives ax + by + ax – by =14 + 4, which simplifies to 2ax = 18. It follows that ax = 9. Choice (A) is not correct. Adding the respective sides of the given equations shows that the value of ax must be 9, not 5. 2012 PSAT/NMSQT Answer Explanations © 2012 The College Board. All Rights Reserved
Choice (C) is not correct. Adding the respective sides of the given equations shows that the value of ax, not by, must be 9. Choice (D) is not correct. Subtracting the respective sides of the given equations shows that the value of 2by, not by, must be 10. 2
2
Choice (E) is not correct. The value of (2ax)(2by), not ax – by , must be 180. Mathematics Question 11 Choice (A) is correct. Since the perimeter of the equilateral triangle is 18, it follows that the each side of the triangle is
18 3
6 units long. From the figure, the length of a radius of the circle is
equal to the length of a side of the equilateral triangle, 6 units. The length of a diameter of the circle is double the length of a radius, and therefore, it is (2)(6) = 12. Choice (B) is not correct. If the length of a diameter of the circle were 10, the length of a radius would be 5, from which it would follow that the equilateral triangle has perimeter (3)(5) = 15. But the perimeter of the triangle is 18, not 15, and so the length of a diameter of the circle cannot be 10. Choice (C) is not correct. If the length of a diameter of the circle were 9, the length of a radius would be 4.5, from which it would follow that the equilateral triangle has perimeter (3)(4.5) = 13.5. But the perimeter of the triangle is 18, not 13.5, and so the length of a diameter of the circle cannot be 9. Choice (D) is not correct. If the length of a diameter of the circle were 8, the length of a radius would be 4, from which it would follow that the equilateral triangle has perimeter (3)(4) = 12. But the perimeter of the triangle is 18, not 12, and so the length of a diameter of the circle cannot be 8. Choice (E) is not correct. The length of a radius of the circle is 6; however, the question asks for the length of a diameter of the circle, which is 12. Mathematics Question 12 Choice (E) is correct. The car travels at a rate of (2r + 4) miles per hour. The number of miles the car can travel in 2 hours can be found by multiplying the rate by the number of hours, which is 2. Therefore, in of r, the car can travel (2)(2r +4) = 4r + 8 miles in 2 hours. Choice (A) is not correct. The number of miles the car can travel in 2 hours can be found by multiplying the rate by the number of hours, which is 2. Therefore, in 2 hours, the car can travel (2)(2r +4) = 4r + 8 miles, not
2r
4 2
= r + 2 miles.
Choice (B) is not correct. The number of miles the car can travel in 2 hours can be found by multiplying the rate by the number of hours, which is 2. Therefore, in 2 hours, the car can travel (2)(2r +4) = 4r + 8 miles, not
2r + 4 = r + 4 miles. 2
Choice (C) is not correct. The number of miles the car can travel in 2 hours can be found by multiplying the rate by the number of hours, which is 2. Therefore, in 2 hours, the car can travel (2)(2r +4) = 4r + 8 miles, not (2)(2r) + 4 = 4r + 4 miles. 2012 PSAT/NMSQT Answer Explanations © 2012 The College Board. All Rights Reserved
Choice (D) is not correct. The number of miles the car can travel in 2 hours can be found by multiplying the rate by the number of hours, which is 2. Therefore, in 2 hours, the car can travel (2)(2r +4) = 4r + 8 miles, not (2)(2r) + (2 + 4) = 4r + 6 miles. Mathematics Question 13 Choice (B) is correct. The shaded region can be divided into two regions: a rectangle with vertices (0, 0), (0, 4), (3, 4) and (3, 0) and a right triangle with vertices (0, 4), (3, 10) and (3, 4). The rectangle has length 4 – 0 = 4 and width 3 – 0 = 3, so its area is (4)(3) = 12. The right triangle has one leg of length 3 – 0 = 3 and one leg of length 10 – 4 = 6. Thus the area of the triangle is
(3)(6) 2
9. Therefore, the area of the shaded region is 12 + 9 = 21.
Choice (A) is not correct. The shaded region is contained within a rectangle with vertices (0, 0), (0, 10), (3, 10) and (3, 0). This rectangle has length 10 and width 3, so its area is 30. However, the area of the shaded region is 30 – 9 = 21, which is the area of the rectangle minus the area of the unshaded triangular region inside of the rectangle. Choice (C) is not correct. The area of the shaded region is the sum of the area of a rectangle of area 12 and a right triangle of area 9. Thus the area of the shaded region is 12 + 9 = 21, not 9 + 9 = 18. Choice (D) is not correct. The shaded region can be divided into two regions: a rectangle with vertices (0, 0), (0, 4), (3, 4) and (3, 0) and a right triangle with vertices (0, 4), (3, 10) and (3, 4). The rectangle has area 12. However, the total area of the shaded region is the area of the rectangle plus the area of the triangle. The area of the triangle is 9, so the area of the shaded region is 12 + 9 = 21. Choice (E) is not correct. The shaded region can be divided into two regions: a rectangle with vertices (0, 0), (0, 4), (3, 4) and (3, 0) and a right triangle with vertices (0, 4), (3, 10) and (3, 4). The triangle has area 9. However, the total area of the shaded region is the area of the triangle plus the area of the rectangle. The area of the rectangle is 12, so the area of the shaded region is 9 + 12 = 21. Mathematics Question 14 Choice (E) is correct. Since renting video games from The Game Garage costs $3 per game, plus an additional $60 for an annual hip, the cost to rent n video games in a year from The Game Garage, in dollars, is 3n + 60. Since renting video games from The Video Vendor costs $6 per game with no annual hip fee, the cost to rent n video games in a year from The Video Vendor, in dollars, is 6n. Hence it costs less to rent n games in a year from The Game Garage than from The Video Vendor if and only if 3n + 60 < 6n. This inequality is equivalent to 60 < 3n, which is equivalent to n > 20. Therefore, n > 20 gives all values of n for which it costs less to rent n games in a year from The Game Garage than from The Video Vendor. Choice (A) is not correct. If n < 10, then it costs more, not less, to rent n games in a year from The Game Garage than from The Video Vendor. For example, if n = 5, it costs $75 to rent n = 5 games in a year from The Game Garage and $30 from The Video Vendor. Therefore, n < 10 cannot give all values of n for which it costs less to rent n games in a year from The Game Garage than from The Video Vendor. Choice (B) is not correct. If n = 10, then it costs more, not less, to rent n games in a year from The Game Garage than from The Video Vendor: it costs $90 to rent n = 10 games in a year from 2012 PSAT/NMSQT Answer Explanations © 2012 The College Board. All Rights Reserved
The Game Garage and $60 from The Video Vendor. Therefore, n = 10 cannot give all values of n for which it costs less to rent n games in a year from The Game Garage than from The Video Vendor. Choice (C) is not correct. If 10 < n < 20, then it costs more, not less, to rent n games in a year from The Game Garage than from The Video Vendor. For example, if n = 15, it costs $105 to rent n = 15 games in a year from The Game Garage and $90 from The Video Vendor. Therefore, 10 < n < 20 cannot give all values of n for which it costs less to rent n games in a year from The Game Garage than from The Video Vendor. Choice (D) is not correct. If n = 20, then it costs the same to rent n games in a year from The Game Garage or from The Video Vendor: it costs $120 to rent n = 20 games in a year from The Game Garage or from The Video Vendor. Therefore, n = 20 cannot give all values of n for which it costs less to rent n games in a year from The Game Garage than from The Video Vendor. Mathematics Question 15 Choice (C) is correct. The quadrilateral ABCD is composed of one rectangle and of two triangles of the same size and shape. Since DE = 3 and triangles BCF and CDE have the same size and shape, it follows that CF = DE = 3, and so CE = EF + CF = AB + CF = 2 + 3 = 5; furthermore, BF = 5. Thus, the area of ABCD is (2)(5) + 2
(5)(3) = 10 + 15 = 25. 2
Choice (A) is not correct. The sum of the areas of triangles BCF and CDE is 15; however, the question asks for the area of the entire quadrilateral ABCD, which is 25. Choice (B) is not correct. The area of ABCD is the sum of the area of the rectangle and the area of the two triangles, which is 10 + 7.5 +7.5 = 25, not 10 + 5 + 5 = 20. Choice (D) is not correct. The area of ABCD is the sum of the area of the rectangle and the area of the two triangles, which is 10 + 7.5 +7.5 = 25, not 10 + 10 + 10 = 30. Choice (E) is not correct. The area of ABCD is the sum of the area of the rectangle and the area of the two triangles, which is 10 + 7.5 +7.5 = 25, not 10 + 15 + 15 = 40. Mathematics Question 16 Choice (B) is correct. The sum of the five different positive integers is 100. The smallest of the integers is 10, so the sum of the other four is 90. One of these four integers will be largest when the other three are as small as possible. The other three must be different from each other and greater than 10, so the smallest possible values for the other three are 11, 12 and 13. Therefore, the largest possible value of one of the four integers other than 10 is 90 − (11 + 12 + 13) = 90 − 36 = 54. Choice (A) is not correct. The integers 10, 11, 12, 13 and 54 are five different positive integers whose sum is 100. The smallest of these five integers is 10, and the largest is 54. Therefore, the largest possible value of one of the four integers other than 10 cannot be 46, because 54 > 46. Choice (C) is not correct. The sum of the five different positive integers is 100, and the smallest of the integers is 10. If 64 were one of the integers, then the sum of the integers other than 10 and 64 would be 100 − (10 + 64) = 26. However, if the sum of three positive integers is 26, at least one of them must be less than 10, and then 10 would not be the smallest of the five integers. Therefore, 64 cannot be the maximum value of one of the four integers other than 10.
2012 PSAT/NMSQT Answer Explanations © 2012 The College Board. All Rights Reserved
Choice (D) is not correct. The sum of the five different positive integers is 100, and the smallest of the integers is 10. If 84 were one of the integers, then the sum of the integers other than 10 and 84 would be 100 − (10 + 84) = 6. However, if the sum of three positive integers is 6, all of them must be less than 10, and then 10 would not be the smallest of the five integers. Therefore, 84 cannot be the maximum value of one of the four integers other than 10. Choice (E) is not correct. The sum of the five different positive integers is 100, and the smallest of the integers is 10. If 90 were one of the integers, then the sum of the integers other than 10 and 90 would be 100 − (10 + 90) = 0. However, the sum of three positive integers cannot be 0. Therefore, 90 cannot be the maximum value of one of the four integers other than 10. Mathematics Question 17
2 1 x and z = x. Thus x + y + z = x + 3 2 2 1 2 1 x + x. It follows that 6(x + y + z) = 6x + 6 x + 6 x = 6x + 4x + 3x = 13x, and so x + 3 2 3 2 13 13 y+z= times the value of x. x. Therefore, the value of x + y + z is 6 6 Choice (A) is correct. Since 2x = 3y = 4z, it follows that y =
Choice (B) is not correct. Since 6(x + 2y + z) = 6x + 8x + 3x = 17x, the value of x + 2y + z is times the value of x, but the value of x + y + z is
13 times the value of x. 6
Choice (C) is not correct. The value of x + y + z is y + z is
17 6
x 13 times the value of , but the value of x + 3 2
13 times the value of x. 6
x 6x 8x 3x , the value of x + 2y + z = 17 3 3 x 17 13 is times the value of , but the value of x + y + z is times the value of x. 3 2 6 Choice (D) is not correct. Since 2(x + 2y + z) =
Choice (E) is not correct. It can be determined from the information given that the value of x + y + z is
13 times the value of x. 6
2012 PSAT/NMSQT Answer Explanations © 2012 The College Board. All Rights Reserved
Mathematics Question 18 Choice (D) is correct. When the point in the xy-plane with coordinates (a, b) is reflected about the line y = x, the image is the point with coordinates (b, a); that is, reflecting a point about the line y = x has effect of switching its coordinates. Hence when A (−2, −5) and B (3, 2) are reflected about y = x, their images are the points with coordinates (−5, −2) and (2, 3), respectively. The slope of the line through the points with coordinates (−5, −2) and (2, 3) is given by the quotient which is equal to
3 ( 2) , 2 ( 5)
5 . 7
Choice (A) is not correct. If A (−2, −5) and B (3, 2) are reflected about the x-axis, the points to which A and B are reflected have coordinates (−2, 5) and (3, −2), respectively, and the slope of the line through these two points is
7 . However, when A (−2, −5) and B (3, 2) are reflected 5
about y = x, their images are the points with coordinates (−5, −2) and (2, 3), respectively, and the slope of the line through these points is
5 . 7
Choice (B) is not correct. If A (−2, −5) and B (3, 2) are rotated 90 degrees clockwise, the points to which A and B are rotated have coordinates (−5, 2) and (2, −3), respectively, and the slope of the
5 . However, when A (−2, −5) and B (3, 2) are reflected about y 7
line through these two points is
= x, their images are the points with coordinates (−5, −2) and (2, 3), respectively, and the slope of the line through these points is
5 . 7
Choice (C) is not correct. When A (−2, −5) and B (3, 2) are reflected about the line y = x, their images are the points with coordinates (−5, −2) and (2, 3), respectively. The slope of the line through (−5, −2) and (2, 3) is given by the quotient
3 ( 2) 5 1 , which is equal to , not . 2 ( 5) 7 7
7 , but when A (−2, 5 5 −5) and B (3, 2) are reflected about y = x, the slope of the line through their images is . 7 Choice (E) is not correct. The line through A (−2, −5) and B (3, 2) has slope
Mathematics Question 19 Choice (A) is correct. Dividing both sides of 5 y
5
( 5 )( 5 ), it follows that
y 5 5x
1 y 5 5 x
1 (1) 5
5 5
1 . 5
2012 PSAT/NMSQT Answer Explanations © 2012 The College Board. All Rights Reserved
5. Thus
x 5 by x 5 yields
5y x 5
y 5 x
1, and so
5y x 5
1. Since
1 y 5 were equal to , then cross-multiplying would give 5y = 5x. 5x 5 1 y 5 x 5, not 5x. Therefore, is not equal to . 5x 5
Choice (B) is not correct. If But 5 y
y 5 were equal to 1, then it would follow that y 5 5x y 5 x 5. But x 5 5 y , not y. Therefore, is not equal to 1. 5x
Choice (C) is not correct. If equivalent to y
Choice (D) is not correct. If
y 5 were equal to 5x
5, then multiplying both expressions by
would give y = 5x, which is equivalent to 5y = 25x. But 5 y not equal to
5 x, which is
x 5, not 25x. Therefore,
5x 5
y 5 is 5x
5.
5x y 5 were equal to 5 5, then multiplying both expressions by 5x 5 y 5 would give y = 25x, which is equivalent to 5y = 125x. But 5 y x 5, not 125x. Therefore, 5x is not equal to 5 5. Choice (E) is not correct. If
Mathematics Question 20 Choice (B) is correct. Since the upper line graph shows the sum of the number of soft drinks and the number of containers of popcorn sold, and the lower line graph shows only the number of soft drinks sold, the number of popcorn containers sold each month can be found by subtracting the values of the lower line graph from the corresponding values of the upper line graph. Thus, in month 2, there were 2,750 − 1,750 = 1,000 containers of popcorn sold. Similarly, in month 3, there were 3,250 − 2,250 = 1,000; in month 4, there were 3,500 − 1,500 = 2,000; in month 5, there were 3,750 − 1,500 = 2,250; in month 6, there were 3,250 − 2,000 = 1,250; and in month 7, there were 3,750 − 2,250 = 1,500. Thus, from month 2 to month 3, there was no increase or decrease in the number of containers sold. From month 3 to month 4, there was an increase of 1,000. From month 4 to month 5, there was an increase of 250. From month 5 to month 6, there was a decrease of 1,000. From month 6 to month 7, there was an increase of 250. Therefore, month 4 showed the greatest increase in the number of containers of popcorn sold from the previous month. Choice (A) is not correct. From month 2 to month 3, there was no increase in the number of containers of popcorn sold. However, from month 3 to month 4, there was an increase of 1,000. Therefore, month 3 did not show the greatest increase in the number of containers of popcorn sold from the previous month. Choice (C) is not correct. From month 4 to month 5, there was an increase of 250 in the number of containers of popcorn sold. However, from month 3 to month 4, there was an increase of 1,000. Therefore, month 5 did not show the greatest increase in the number of containers of popcorn sold from the previous month. 2012 PSAT/NMSQT Answer Explanations © 2012 The College Board. All Rights Reserved
Choice (D) is not correct. From month 5 to month 6, there was a decrease of 1,000 in the number of containers of popcorn sold. However, from month 3 to month 4, there was an increase of 1,000. Therefore, month 6 did not show the greatest increase in the number of containers of popcorn sold from the previous month. Choice (E) is not correct. From month 6 to month 7, there was an increase of 250 in the number of containers of popcorn sold. However, from month 3 to month 4, there was an increase of 1,000. Therefore, month 7 did not show the greatest increase in the number of containers of popcorn sold from the previous month.
2012 PSAT/NMSQT Answer Explanations © 2012 The College Board. All Rights Reserved
2012 PSAT/NMSQT Answer Explanations © 2012 The College Board. All Rights Reserved
2012 PSAT/NMSQT Answer Explanations © 2012 The College Board. All Rights Reserved
Choice (C) is not correct. If the value of x were 50, then 40 + x would be equal to 90. However, 40 + x is equal to 110, not 90, so the value of x cannot be 50. Choice (D) is not correct. If the value of x were 60, then 40 + x would be equal to 100. However, 40 + x is equal to 110, not 100, so the value of x cannot be 60. Mathematics Question 4 Choice (D) is correct. For the 12 books sold, the three different sale prices represented were $2.50, $4.50 and $5.00. Alex sold 4 books for $2.50 each and Dana sold 0 books for $2.50, for a total of 4 + 0 = 4 books sold at $2.50 each. Alex sold 2 books for $4.50 each and Dana sold 3 books for $4.50 each, for a total of 2 + 3 = 5 books sold at $4.50 each. Alex sold 1 book for $5.00 and Dana sold 2 books for $5.00 each, for a total of 1 + 2 = 3 books sold at $5.00 each. Therefore, the most frequently occurring sale price for the 12 books was $4.50. Choice (A) is not correct. There were 4 books sold for $2.50 each. However, there were 5 books sold for $4.50 each. Therefore, the most frequently occurring sale price for the 12 books was not $2.50. Choice (B) is not correct. There were no books sold for the price of $4.00 each. Therefore, the most frequently occurring sale price for the 12 books was not $4.00. Choice (C) is not correct. There were no books sold for the price of $4.25 each. Therefore, the most frequently occurring sale price for the 12 books was not $4.25. Choice (E) is not correct. There were 3 books sold for the price of $5.00 each. However, there were 5 books sold for $4.50 each. Therefore, the most frequently occurring sale price for the 12 books sold was not $5.00. Mathematics Question 5 Choice (C) is correct. Since 1 yard is equal to 3 feet, it follows that 16 yards is equal to (16)(3) = 48 feet. Therefore, 16 yards is 48 – 16 = 32 feet longer than 16 feet. Choice (A) is not correct. Since 1 yard is equal to 3 feet, it follows that 16 yards is equal to (16)(3) = 48 feet. However, 16 yards is 48 – 16 = 32, not 48, feet longer than 16 feet. Choice (B) is not correct. Since 1 yard is equal to 3 feet, it follows that 16 yards is equal to (16)(3) = 48 feet. Thus 16 yards is 48 – 8 = 40 feet longer than 8 feet. However, 16 yards is 48 – 16 = 32 feet longer than 16 feet. Choice (D) is not correct. Since 1 yard is equal to 3 feet, it follows that 16 yards is equal to (16)(3) = 48 feet. Thus 16 yards is 48 – 24 = 24 feet longer than 24 feet. However, 16 yards is 48 – 16 = 32 feet longer than 16 feet. Choice (E) is not correct. Since 1 yard is equal to 3 feet, it follows that 16 yards is equal to (16)(3) = 48 feet. Thus 16 yards is 48 – 32 = 16 feet longer than 32 feet. However, 16 yards is 48 – 16 = 32 feet longer than 16 feet.
2012 PSAT/NMSQT Answer Explanations © 2012 The College Board. All Rights Reserved
Mathematics Question 6 2
2
2
2
Choice (E) is correct. For both rows of the table, r is equal to s: 2 = 4 and 3 =9. Hence r = ks, 2 where k = 1, for both rows of the table. Therefore, r could be directly proportional to s. Choice (A) is not correct. It could be true that could be proportional to s.
s is proportional to r, but it is r2, not
r , that
Choice (B) is not correct. If r were directly proportional to s, then it would be true that r = ks for some constant k. But 2 =
1 1 (4) and 3 = (9), so there is no constant k for which r = ks. 2 3
Therefore, r cannot be directly proportional to s. Choice (C) is not correct. If r + 1 were directly proportional to s, then it would be true that r + 1 = ks, for some constant k. But 2 + 1 =
3 4 (4) and 3 + 1 = (9), so there is no constant k for which r 4 9
+ 1 = ks. Therefore, r + 1 cannot be directly proportional to s. Choice (D) is not correct. If 2r were directly proportional to s, then it would be true that 2r = ks, for some constant k. But 2(2) = 1(4) and 2(3) =
2 (9), so there is no constant k for which 2r = ks. 3
Therefore, 2r cannot be directly proportional to s. Mathematics Question 7 Choice (D) is correct. Since 20 percent of the surveyed customers answered “black,” it follows that the probability is
20 100
1 that a customer chosen at random from those surveyed will be 5
one who answered “black.” Choice (A) is not correct. The probability cannot be
1 that a customer chosen at random from 20
those surveyed will be one who answered “black,” because the graph indicates that 20 percent, which is 1 in 5 (not 5 percent, which is 1 in 20), of the customers surveyed answered “black.” Choice (B) is not correct. The probability cannot be
1 that a customer chosen at random from 18
those surveyed will be one who answered “black,” because the graph indicates that 20 percent, which is 1 in 5 (not 1 in 18), of the customers surveyed answered “black.” Choice (C) is not correct. The probability cannot be
1 that a customer chosen at random from 10
those surveyed will be one who answered “black,” because the graph indicates that 20 percent, which is 1 in 5 (not 10 percent, which is 1 in 10), of the customers surveyed answered “black.” Choice (E) is not correct. The probability cannot be
1 that a customer chosen at random from 4
those surveyed will be one who answered “black,” because the graph indicates that 20 percent, which is 1 in 5 (not 25 percent, which is 1 in 4), of the customers surveyed answered “black.”
2012 PSAT/NMSQT Answer Explanations © 2012 The College Board. All Rights Reserved
Mathematics Question 8 Choice (A) is correct. Since point P lies on line segment RU , the sum of the measures of
SPU is 180°. Similarly, since point P lies on segment QT , the sum of the measures of QPS and SPT is 180°. The measure of SPU is 70°, so the measure of RPS is 180° − 70° = 110°. Since RPS has measure 110° and RPQ has measure 40°, the measure of SPT is 180° − (110° + 40°) = 30°. RPS and
Choice (B) is not correct. The measure of RPQ is 40°, so the measure of TPU is also 40°, since vertical angles are of equal measure. The measure of SPU is 70°, so the measure of RPS is 180° − 70° = 110°. If the measure of SPT were 35°, then the measure of RPU would be 110° + 35° + 40° = 185°. However, RPU is a straight angle, which has measure 180°, not 185°. Therefore, the measure of SPT cannot be 35°. Choice (C) is not correct. The measure of measure of SPT is 30°.
RPQ is 40°, as is the measure of
TPU, but the
Choice (D) is not correct. The measure of RPQ is 40°, so the measure of TPU is also 40°, since vertical angles are of equal measure. The measure of SPU is 70°, so the measure of RPS is 180° − 70° = 110°. If the measure of SPT were 45°, then the measure of RPU would be 110° + 45° + 40° = 195°. However, RPU is a straight angle, which has measure 180°, not 195°. Therefore, the measure of SPT cannot be 45°. Choice (E) is not correct. The measure of RPQ is 40°, so the measure of TPU is also 40°, since vertical angles are of equal measure. The measure of SPU is 70°, so the measure of RPS is 180° − 70° = 110°. If the measure of SPT were 50°, then the measure of RPU would be 110° + 50° + 40° = 200°. However, RPU is a straight angle, which has measure 180°, not 200°. Therefore, the measure of SPT cannot be 50°. Mathematics Question 9 Choice (D) is correct. The function f is defined by f(x) = 2x, so f(5t) = 2(5t) = 10t. Choice (A) is not correct. The function f is defined by f(x) = 2x, so f(t) = 2t, but f(5t) = 2(5t) = 10t. Choice (B) is not correct. The function f is defined by f(x) = 2x, so f(2.5t) = 2(2.5t) = 5t, but f(5t) = 2(5t) = 10t. Choice (C) is not correct. The function f is defined by f(x) = 2x, so f(5t) = 2(5t) = 10t, not (2 + 5)t = 7t. 2
Choice (E) is not correct. The function f is defined by f(x) = 2x, so f(5t) = 2(5t) = 10t, not 5 t = 25t. Mathematics Question 10 Choice (B) is correct. Adding the respective sides of the equations ax + by =14 and ax – by = 4 gives ax + by + ax – by =14 + 4, which simplifies to 2ax = 18. It follows that ax = 9. Choice (A) is not correct. Adding the respective sides of the given equations shows that the value of ax must be 9, not 5. 2012 PSAT/NMSQT Answer Explanations © 2012 The College Board. All Rights Reserved
Choice (C) is not correct. Adding the respective sides of the given equations shows that the value of ax, not by, must be 9. Choice (D) is not correct. Subtracting the respective sides of the given equations shows that the value of 2by, not by, must be 10. 2
2
Choice (E) is not correct. The value of (2ax)(2by), not ax – by , must be 180. Mathematics Question 11 Choice (A) is correct. Since the perimeter of the equilateral triangle is 18, it follows that the each side of the triangle is
18 3
6 units long. From the figure, the length of a radius of the circle is
equal to the length of a side of the equilateral triangle, 6 units. The length of a diameter of the circle is double the length of a radius, and therefore, it is (2)(6) = 12. Choice (B) is not correct. If the length of a diameter of the circle were 10, the length of a radius would be 5, from which it would follow that the equilateral triangle has perimeter (3)(5) = 15. But the perimeter of the triangle is 18, not 15, and so the length of a diameter of the circle cannot be 10. Choice (C) is not correct. If the length of a diameter of the circle were 9, the length of a radius would be 4.5, from which it would follow that the equilateral triangle has perimeter (3)(4.5) = 13.5. But the perimeter of the triangle is 18, not 13.5, and so the length of a diameter of the circle cannot be 9. Choice (D) is not correct. If the length of a diameter of the circle were 8, the length of a radius would be 4, from which it would follow that the equilateral triangle has perimeter (3)(4) = 12. But the perimeter of the triangle is 18, not 12, and so the length of a diameter of the circle cannot be 8. Choice (E) is not correct. The length of a radius of the circle is 6; however, the question asks for the length of a diameter of the circle, which is 12. Mathematics Question 12 Choice (E) is correct. The car travels at a rate of (2r + 4) miles per hour. The number of miles the car can travel in 2 hours can be found by multiplying the rate by the number of hours, which is 2. Therefore, in of r, the car can travel (2)(2r +4) = 4r + 8 miles in 2 hours. Choice (A) is not correct. The number of miles the car can travel in 2 hours can be found by multiplying the rate by the number of hours, which is 2. Therefore, in 2 hours, the car can travel (2)(2r +4) = 4r + 8 miles, not
2r
4 2
= r + 2 miles.
Choice (B) is not correct. The number of miles the car can travel in 2 hours can be found by multiplying the rate by the number of hours, which is 2. Therefore, in 2 hours, the car can travel (2)(2r +4) = 4r + 8 miles, not
2r + 4 = r + 4 miles. 2
Choice (C) is not correct. The number of miles the car can travel in 2 hours can be found by multiplying the rate by the number of hours, which is 2. Therefore, in 2 hours, the car can travel (2)(2r +4) = 4r + 8 miles, not (2)(2r) + 4 = 4r + 4 miles. 2012 PSAT/NMSQT Answer Explanations © 2012 The College Board. All Rights Reserved
Choice (D) is not correct. The number of miles the car can travel in 2 hours can be found by multiplying the rate by the number of hours, which is 2. Therefore, in 2 hours, the car can travel (2)(2r +4) = 4r + 8 miles, not (2)(2r) + (2 + 4) = 4r + 6 miles. Mathematics Question 13 Choice (B) is correct. The shaded region can be divided into two regions: a rectangle with vertices (0, 0), (0, 4), (3, 4) and (3, 0) and a right triangle with vertices (0, 4), (3, 10) and (3, 4). The rectangle has length 4 – 0 = 4 and width 3 – 0 = 3, so its area is (4)(3) = 12. The right triangle has one leg of length 3 – 0 = 3 and one leg of length 10 – 4 = 6. Thus the area of the triangle is
(3)(6) 2
9. Therefore, the area of the shaded region is 12 + 9 = 21.
Choice (A) is not correct. The shaded region is contained within a rectangle with vertices (0, 0), (0, 10), (3, 10) and (3, 0). This rectangle has length 10 and width 3, so its area is 30. However, the area of the shaded region is 30 – 9 = 21, which is the area of the rectangle minus the area of the unshaded triangular region inside of the rectangle. Choice (C) is not correct. The area of the shaded region is the sum of the area of a rectangle of area 12 and a right triangle of area 9. Thus the area of the shaded region is 12 + 9 = 21, not 9 + 9 = 18. Choice (D) is not correct. The shaded region can be divided into two regions: a rectangle with vertices (0, 0), (0, 4), (3, 4) and (3, 0) and a right triangle with vertices (0, 4), (3, 10) and (3, 4). The rectangle has area 12. However, the total area of the shaded region is the area of the rectangle plus the area of the triangle. The area of the triangle is 9, so the area of the shaded region is 12 + 9 = 21. Choice (E) is not correct. The shaded region can be divided into two regions: a rectangle with vertices (0, 0), (0, 4), (3, 4) and (3, 0) and a right triangle with vertices (0, 4), (3, 10) and (3, 4). The triangle has area 9. However, the total area of the shaded region is the area of the triangle plus the area of the rectangle. The area of the rectangle is 12, so the area of the shaded region is 9 + 12 = 21. Mathematics Question 14 Choice (E) is correct. Since renting video games from The Game Garage costs $3 per game, plus an additional $60 for an annual hip, the cost to rent n video games in a year from The Game Garage, in dollars, is 3n + 60. Since renting video games from The Video Vendor costs $6 per game with no annual hip fee, the cost to rent n video games in a year from The Video Vendor, in dollars, is 6n. Hence it costs less to rent n games in a year from The Game Garage than from The Video Vendor if and only if 3n + 60 < 6n. This inequality is equivalent to 60 < 3n, which is equivalent to n > 20. Therefore, n > 20 gives all values of n for which it costs less to rent n games in a year from The Game Garage than from The Video Vendor. Choice (A) is not correct. If n < 10, then it costs more, not less, to rent n games in a year from The Game Garage than from The Video Vendor. For example, if n = 5, it costs $75 to rent n = 5 games in a year from The Game Garage and $30 from The Video Vendor. Therefore, n < 10 cannot give all values of n for which it costs less to rent n games in a year from The Game Garage than from The Video Vendor. Choice (B) is not correct. If n = 10, then it costs more, not less, to rent n games in a year from The Game Garage than from The Video Vendor: it costs $90 to rent n = 10 games in a year from 2012 PSAT/NMSQT Answer Explanations © 2012 The College Board. All Rights Reserved
The Game Garage and $60 from The Video Vendor. Therefore, n = 10 cannot give all values of n for which it costs less to rent n games in a year from The Game Garage than from The Video Vendor. Choice (C) is not correct. If 10 < n < 20, then it costs more, not less, to rent n games in a year from The Game Garage than from The Video Vendor. For example, if n = 15, it costs $105 to rent n = 15 games in a year from The Game Garage and $90 from The Video Vendor. Therefore, 10 < n < 20 cannot give all values of n for which it costs less to rent n games in a year from The Game Garage than from The Video Vendor. Choice (D) is not correct. If n = 20, then it costs the same to rent n games in a year from The Game Garage or from The Video Vendor: it costs $120 to rent n = 20 games in a year from The Game Garage or from The Video Vendor. Therefore, n = 20 cannot give all values of n for which it costs less to rent n games in a year from The Game Garage than from The Video Vendor. Mathematics Question 15 Choice (C) is correct. The quadrilateral ABCD is composed of one rectangle and of two triangles of the same size and shape. Since DE = 3 and triangles BCF and CDE have the same size and shape, it follows that CF = DE = 3, and so CE = EF + CF = AB + CF = 2 + 3 = 5; furthermore, BF = 5. Thus, the area of ABCD is (2)(5) + 2
(5)(3) = 10 + 15 = 25. 2
Choice (A) is not correct. The sum of the areas of triangles BCF and CDE is 15; however, the question asks for the area of the entire quadrilateral ABCD, which is 25. Choice (B) is not correct. The area of ABCD is the sum of the area of the rectangle and the area of the two triangles, which is 10 + 7.5 +7.5 = 25, not 10 + 5 + 5 = 20. Choice (D) is not correct. The area of ABCD is the sum of the area of the rectangle and the area of the two triangles, which is 10 + 7.5 +7.5 = 25, not 10 + 10 + 10 = 30. Choice (E) is not correct. The area of ABCD is the sum of the area of the rectangle and the area of the two triangles, which is 10 + 7.5 +7.5 = 25, not 10 + 15 + 15 = 40. Mathematics Question 16 Choice (B) is correct. The sum of the five different positive integers is 100. The smallest of the integers is 10, so the sum of the other four is 90. One of these four integers will be largest when the other three are as small as possible. The other three must be different from each other and greater than 10, so the smallest possible values for the other three are 11, 12 and 13. Therefore, the largest possible value of one of the four integers other than 10 is 90 − (11 + 12 + 13) = 90 − 36 = 54. Choice (A) is not correct. The integers 10, 11, 12, 13 and 54 are five different positive integers whose sum is 100. The smallest of these five integers is 10, and the largest is 54. Therefore, the largest possible value of one of the four integers other than 10 cannot be 46, because 54 > 46. Choice (C) is not correct. The sum of the five different positive integers is 100, and the smallest of the integers is 10. If 64 were one of the integers, then the sum of the integers other than 10 and 64 would be 100 − (10 + 64) = 26. However, if the sum of three positive integers is 26, at least one of them must be less than 10, and then 10 would not be the smallest of the five integers. Therefore, 64 cannot be the maximum value of one of the four integers other than 10.
2012 PSAT/NMSQT Answer Explanations © 2012 The College Board. All Rights Reserved
Choice (D) is not correct. The sum of the five different positive integers is 100, and the smallest of the integers is 10. If 84 were one of the integers, then the sum of the integers other than 10 and 84 would be 100 − (10 + 84) = 6. However, if the sum of three positive integers is 6, all of them must be less than 10, and then 10 would not be the smallest of the five integers. Therefore, 84 cannot be the maximum value of one of the four integers other than 10. Choice (E) is not correct. The sum of the five different positive integers is 100, and the smallest of the integers is 10. If 90 were one of the integers, then the sum of the integers other than 10 and 90 would be 100 − (10 + 90) = 0. However, the sum of three positive integers cannot be 0. Therefore, 90 cannot be the maximum value of one of the four integers other than 10. Mathematics Question 17
2 1 x and z = x. Thus x + y + z = x + 3 2 2 1 2 1 x + x. It follows that 6(x + y + z) = 6x + 6 x + 6 x = 6x + 4x + 3x = 13x, and so x + 3 2 3 2 13 13 y+z= times the value of x. x. Therefore, the value of x + y + z is 6 6 Choice (A) is correct. Since 2x = 3y = 4z, it follows that y =
Choice (B) is not correct. Since 6(x + 2y + z) = 6x + 8x + 3x = 17x, the value of x + 2y + z is times the value of x, but the value of x + y + z is
13 times the value of x. 6
Choice (C) is not correct. The value of x + y + z is y + z is
17 6
x 13 times the value of , but the value of x + 3 2
13 times the value of x. 6
x 6x 8x 3x , the value of x + 2y + z = 17 3 3 x 17 13 is times the value of , but the value of x + y + z is times the value of x. 3 2 6 Choice (D) is not correct. Since 2(x + 2y + z) =
Choice (E) is not correct. It can be determined from the information given that the value of x + y + z is
13 times the value of x. 6
2012 PSAT/NMSQT Answer Explanations © 2012 The College Board. All Rights Reserved
Mathematics Question 18 Choice (D) is correct. When the point in the xy-plane with coordinates (a, b) is reflected about the line y = x, the image is the point with coordinates (b, a); that is, reflecting a point about the line y = x has effect of switching its coordinates. Hence when A (−2, −5) and B (3, 2) are reflected about y = x, their images are the points with coordinates (−5, −2) and (2, 3), respectively. The slope of the line through the points with coordinates (−5, −2) and (2, 3) is given by the quotient which is equal to
3 ( 2) , 2 ( 5)
5 . 7
Choice (A) is not correct. If A (−2, −5) and B (3, 2) are reflected about the x-axis, the points to which A and B are reflected have coordinates (−2, 5) and (3, −2), respectively, and the slope of the line through these two points is
7 . However, when A (−2, −5) and B (3, 2) are reflected 5
about y = x, their images are the points with coordinates (−5, −2) and (2, 3), respectively, and the slope of the line through these points is
5 . 7
Choice (B) is not correct. If A (−2, −5) and B (3, 2) are rotated 90 degrees clockwise, the points to which A and B are rotated have coordinates (−5, 2) and (2, −3), respectively, and the slope of the
5 . However, when A (−2, −5) and B (3, 2) are reflected about y 7
line through these two points is
= x, their images are the points with coordinates (−5, −2) and (2, 3), respectively, and the slope of the line through these points is
5 . 7
Choice (C) is not correct. When A (−2, −5) and B (3, 2) are reflected about the line y = x, their images are the points with coordinates (−5, −2) and (2, 3), respectively. The slope of the line through (−5, −2) and (2, 3) is given by the quotient
3 ( 2) 5 1 , which is equal to , not . 2 ( 5) 7 7
7 , but when A (−2, 5 5 −5) and B (3, 2) are reflected about y = x, the slope of the line through their images is . 7 Choice (E) is not correct. The line through A (−2, −5) and B (3, 2) has slope
Mathematics Question 19 Choice (A) is correct. Dividing both sides of 5 y
5
( 5 )( 5 ), it follows that
y 5 5x
1 y 5 5 x
1 (1) 5
5 5
1 . 5
2012 PSAT/NMSQT Answer Explanations © 2012 The College Board. All Rights Reserved
5. Thus
x 5 by x 5 yields
5y x 5
y 5 x
1, and so
5y x 5
1. Since
1 y 5 were equal to , then cross-multiplying would give 5y = 5x. 5x 5 1 y 5 x 5, not 5x. Therefore, is not equal to . 5x 5
Choice (B) is not correct. If But 5 y
y 5 were equal to 1, then it would follow that y 5 5x y 5 x 5. But x 5 5 y , not y. Therefore, is not equal to 1. 5x
Choice (C) is not correct. If equivalent to y
Choice (D) is not correct. If
y 5 were equal to 5x
5, then multiplying both expressions by
would give y = 5x, which is equivalent to 5y = 25x. But 5 y not equal to
5 x, which is
x 5, not 25x. Therefore,
5x 5
y 5 is 5x
5.
5x y 5 were equal to 5 5, then multiplying both expressions by 5x 5 y 5 would give y = 25x, which is equivalent to 5y = 125x. But 5 y x 5, not 125x. Therefore, 5x is not equal to 5 5. Choice (E) is not correct. If
Mathematics Question 20 Choice (B) is correct. Since the upper line graph shows the sum of the number of soft drinks and the number of containers of popcorn sold, and the lower line graph shows only the number of soft drinks sold, the number of popcorn containers sold each month can be found by subtracting the values of the lower line graph from the corresponding values of the upper line graph. Thus, in month 2, there were 2,750 − 1,750 = 1,000 containers of popcorn sold. Similarly, in month 3, there were 3,250 − 2,250 = 1,000; in month 4, there were 3,500 − 1,500 = 2,000; in month 5, there were 3,750 − 1,500 = 2,250; in month 6, there were 3,250 − 2,000 = 1,250; and in month 7, there were 3,750 − 2,250 = 1,500. Thus, from month 2 to month 3, there was no increase or decrease in the number of containers sold. From month 3 to month 4, there was an increase of 1,000. From month 4 to month 5, there was an increase of 250. From month 5 to month 6, there was a decrease of 1,000. From month 6 to month 7, there was an increase of 250. Therefore, month 4 showed the greatest increase in the number of containers of popcorn sold from the previous month. Choice (A) is not correct. From month 2 to month 3, there was no increase in the number of containers of popcorn sold. However, from month 3 to month 4, there was an increase of 1,000. Therefore, month 3 did not show the greatest increase in the number of containers of popcorn sold from the previous month. Choice (C) is not correct. From month 4 to month 5, there was an increase of 250 in the number of containers of popcorn sold. However, from month 3 to month 4, there was an increase of 1,000. Therefore, month 5 did not show the greatest increase in the number of containers of popcorn sold from the previous month. 2012 PSAT/NMSQT Answer Explanations © 2012 The College Board. All Rights Reserved
Choice (D) is not correct. From month 5 to month 6, there was a decrease of 1,000 in the number of containers of popcorn sold. However, from month 3 to month 4, there was an increase of 1,000. Therefore, month 6 did not show the greatest increase in the number of containers of popcorn sold from the previous month. Choice (E) is not correct. From month 6 to month 7, there was an increase of 250 in the number of containers of popcorn sold. However, from month 3 to month 4, there was an increase of 1,000. Therefore, month 7 did not show the greatest increase in the number of containers of popcorn sold from the previous month.
2012 PSAT/NMSQT Answer Explanations © 2012 The College Board. All Rights Reserved
2012 PSAT/NMSQT Answer Explanations © 2012 The College Board. All Rights Reserved
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