Practice Test 1

Section 2

Directions: Each of Questions 1-9 consists of two quantities. Compare the quantities in Quantities A and B and choose

e9780738668499_i1773.jpg

Questions 7-9 use the following inequality: − 15 < x < − 11

e9780738668499_i1774.jpg

Directions: Questions 10-25 have several different formats. Unless otherwise indicated, select a single answer choice. For numeric entry questions, follow the instructions below.

 

Numeric Entry Questions

Enter your answer in the answer box(es) below the question.

For Questions 16, 17, and 18, use the following chart.

Specialty Store Larger Shoe Sizes

e9780738668499_i1812.jpg

For Questions 23 and 24, use the following diagram of a Norman window, which consists of a rectangle surmounted by a semicircle.

e9780738668499_i1847.jpg

ANSWERS AND EXPLANATIONS

  1. (C) e9780738668499_i1872.jpg, which means that 3M = 5 + x + 7. Then e9780738668499_i1873.jpg.
  2. (D) Either of m and e9780738668499_i1874.jpg can be positive and the other will be negative.
  3. (B) N2 = (yx)2 = y2 − 2xy + x2. Since x, y > 2, both x, y are positive. Thus, y2 − 2xy + x2 must be less than x2 + y2.
  4. . (D) The value of q is unknown. If e9780738668499_i1875.jpg, then e9780738668499_i1876.jpg. If q = 0.1, then (q)(1 − q) = (0.1)(0.9) = 0.09 < 0.1.
  5. (A) Since 2’ = 2, mn = 1. Thus, m = n + 1, which means that m > n.
  6. (C) By definition, 2y = y2 − 2y. Then y2 − 2y = −1, which can be written as y2 − 2y + 1 = 0. Factor the left side to get (y − 1 )(y − 1) = 0. Thus, y = 1.
  7. (B) For −15 < x < −11, x2 < x4. If the reciprocal of each quantity is used, the inequality sign is reversed. Thus, e9780738668499_i1877.jpg.
  8. (B) For a negative value of x, e9780738668499_i1878.jpg has a negative value and e9780738668499_i1879.jpg has a positive value.
  9. (A) For −15 < x < − 11, x5 < x3. By taking the reciprocal of each quantity, the inequality sign is reversed. Thus, e9780738668499_i1880.jpg.
  10. (D) Let 2x represent the number of men and 3x represent the number of women. Then 2x + 3x = 365, which becomes 5x = 365. This means that x = 73. Thus, the number of women is (3)(73) = 219.
  11. (D) 1 = πr2, so e9780738668499_i1881.jpg. Then the diameter must be (2) e9780738668499_i1882.jpg.
  12. (E) The surface area is (2)(2)(3x) + (2)(2x)(5x) + (2)(3x)(5x) = 12x2 + 20x2 + 30x2 = 62x2.
  13. (C) If the x-intercept is the negative reciprocal of the y-intercept, then one of x and y is negative and the other is positive. A line containing these two intercepts must be positive. None of the other answer choices gives sufficient information for determining a positive slope.
  14. (B, D) (x + 10)2 = x2 + 20x + 100. For choice (B), (x − 10)2 + 40x = x2 − 20x + 100 + 40x = x2 + 20x + 100. For choice (D), (x − 5)(x + 5) + (5)(4x + 25) = x2 − 25 + 20x + 125 = x2 + 20x + 100. Choice (A) is obviously incorrect. The simplified expressions for choices (C) and (E) are x2 + 10x + 25 and x2 + 20x + 105.
  15. (A) 2n + 2n = (2)(2n) = (21)(2n) = 2n+1.
  16. (E) There are 60 pairs of women’s shoes in size 12. Then (0.40)(60) = 24 pairs of size 12 are black. Of these, 100% − 25% = 75% do not have high heels. Thus, the required number is (0.75)(24) = 18.
  17. 13 There are 340 pairs of men’s shoes and 360 pairs of women’s shoes. The number of pairs of men’s shoes that are larger than size 13 is 90. Thus, the required percent is e9780738668499_i1883.jpg .
  18. 13,360 There are 30 pairs of size 8 shoes, for which the total selling price is (30($200) = $6,000. The average price of each pair of size 9 shoes is ($200)(1.10) = $220, and the average price of each size 10 shoes is ($220)(1.10) = $242. Since there are 80 pairs of size 10 shoes, the total selling price is (80)($242) = $19,360. Thus, the difference in total price is $19,360 − $6,000 = $13,360.
  19. (A) The number of ways to choose three men is e9780738668499_i1884.jpg . The number of ways to choose two women is e9780738668499_i1885.jpg . Therefore, the number of different committees is (35)(36) = 1,260.
  20. (E) A scatter plot is a graph of points that represent ordered pairs of variables.
  21. (B) We are given that e9780738668499_i1886.jpg, so e9780738668499_i1887.jpg. Thus, e9780738668499_i1888.jpg.
  22. (D) Whenever a fraction has a value of zero, the numerator must equal zero and the denominator must not equal zero. Then x − 6 = 0 and y − 4 ≠ 0, which means that x = 6 and y ≠ 4.
  23. (C) The base of the rectangle is 4y and there are two heights whose combined lengths are 2x. The semicircle at the top contains a diameter of length 4y, so its length is e9780738668499_i1889.jpg. Thus, the perimeter is 2x + 4y + 2πy.
  24. (C) The area of the rectangular portion is the product of the base and height, which is 4xy. For the semicircle, its radius is e9780738668499_i1890.jpg. Then the area of the semicircle is e9780738668499_i1891.jpg . Thus, the area of the figure is 4xy + 2πy2.
  25. (D) Redraw the figure and connect the center of the circle (point O) with points B and C. Also, draw e9780738668499_i1892.jpg perpendicular to e9780738668499_i1893.jpgat point E.
    e9780738668499_i1894.jpg

    We know that mABC = 60° and it can be proven that e9780738668499_i1895.jpg bisects ∠ABC. This means that ΔBOE is a 30°-60°-90° right triangle, with mOBE = 30°.

    It can be shown that E is the midpoint of e9780738668499_i1896.jpg, so BE = 3. e9780738668499_i1897.jpg is a radius and also the hypotenuse of ΔBOE We determine that mBOE = 60° and since BE = 3, e9780738668499_i1898.jpg Then the area of the circle is (π)(2√3)2 = 12π. The area of ΔABC is e9780738668499_i1899.jpg.The area of the combined shaded regions is the difference between the areas of the circle and the triangle, which is 12π − 9√3.