[Math Lair] Solutions for Practice Test 2, The Official SAT Study Guide, Section 2

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Here are solutions for section 2 of the second practice test in The Official SAT Study Guide, second edition, found on pages 452–457. The following solutions demonstrate faster, more informal methods that might work better for you on a fast-paced test such as the SAT. To learn more about these methods, see my e-book Succeeding in SAT Math or the SAT math tips page.

  1. Cathy sold 48 cars in May. She sold 20 + 18 cars in the months of January and February combined. So, she sold 48 − (20 + 18) = 10 more cars in May than in January and February combined. Select (A) 10.
  2. Since the figure is being rotated 90° counterclockwise, you can see what it looks like by rotating your paper 90° counterclockwise (in the direction of the arrow; the right edge of your paper will now be at the top of your desk). Now, rotate your paper back and find the shape that looks like the one you just saw (it may help to just rotate your paper back 45°, allowing you to switch your view between the figure and the answers). The answer is (D).
  3. Convert the sentence into an equation. The first part of the sentence results in:
    3 + 2x = 10
    The second part of the sentence asks us to find 4x. So, solve the equation for 4x:
    2x = 7
    4x = 14
    Therefore, the answer is (D) 14.
  4. Solution 1: Try a special case: Say that a = − 1. Then a = −1, 2a = −2, 4a = −4, and 8a = −8. a is the largest, so we can eliminate (B), (C), and (D). If you aren't convinced that a must always be the largest, you may want to try another special case or two. You will always find that a is the largest, so select (A) a.

    Solution 2: Using a graphing calculator, plot y = x, y = 2x, y = 4x, and y = 8x. Examine the area of the graph where y < 0. You will find that y = x is always the largest. Therefore, a must be the largest of the four and the answer is (A) a.
  5. Solution 1: The shape can be thought of as a large rectangle and a small piece that appears square. We are not given the height of the square, but since the height of the entire figure is 6 and the height of the figure below the square is 4, the height of the square is 2 (or, if you don't see that, you can use a finger or your pencil to measure the height and the width; you'll find they're the exact same). So:
    Area = area of large rectangle + area of square
      = 4(6) + 2(2)
      = 24 + 4
      = 28
    Select (B) 28.

    Solution 2: This shape can be thought of as a 4×4 square plus a long rectangle. The height of the rectangle is 6 − 4 = 2, and the width of the rectangle is 2 + 4 = 6. So:
    Area = area of square + area of long rectangle
      = 4(4) + 2(6)
      = 16 + 12
      = 28
    Select (B) 28.

    Solution 3: This shape can be thought of as a 6 × (2 + 4) square minus a 2×4 rectangle. Similar to the above solutions, the area is 36 − 8 = 28. Select (B) 28.
  6. Solution 1: Solution 2: Guess and check:
  7. We are given that the only mode is 5. Therefore, since there are four 5's, there cannot be four of any other number. There are already three 6's. So, n cannot be another 6, or else there would be two modes. Select (A) 6.
  8. Find where circles Y and Z overlap, and sum all of the numbers found there. The two circles overlap in two places, one with a "3" and one with a "7". The number of elements in the intersection of Y and Z is 3 + 7 = 10. Select (C) 10.
  9. Solution 1: Solution 2: Substitute m = t³ into w = m² + m:
    w = m² + m
    w = (t³)² + t³
    w = t6 + t³
    (don't forget laws of exponents!)
    Therefore, the answer is (E) t6 + t³.
  10. Solution 1: Draw a diagram: Plot y = |−2x + 6| on your graphing calculator. You should find that the graph corresponds with that of (B). Select that answer.

    Solution 2: Draw a diagram: The graph of y = −2x + 6 is given. The graph of y = |−2x + 6| would be the same, except that the negative portion of the graph would be reflected in the x-axis. Draw that in the given graph and compare with the answer choices. The only answer choice that looks similar is (B). Select that answer.

    Solution 3:
  11. This question is not difficult computationally, but it does require a fair bit of insight.

    Solution 1: Draw a diagram: Draw a box that just surrounds the cylinder. The width of the box is d, the height of the box is h, and the depth of the box is also d (since the depth would be equal to the circle's diameter). So, the volume of the box is d²h. Select (B) d²h.

    Solution 2:

  12. The sentences in the question can be converted into equations:
    x² = 4y² (sentence 1)
    x = 1 + 2y (sentence 2)
    We can rewrite sentence 1 as
    x² = (2y
    and sentence 2 as
    x − 1 = 2y

    From sentence 2, substitute x − 1 for 2y in sentence 1:

    x² = (x − 1)²

    Now, take the square root of both sides of the equation. Important: Don't forget about the negative square root, or else the equation will have no solution.

    x = ±(x − 1)
    x = x − 1 or x = 1 − x
    The first equation has no solution. Looking at the second equation:
    2x = 1
    x = ½
    Select (E) ½.

    This was a complex question, so if you have time it's a good idea to check your work. From sentence 2:

    ½ = 1 + 2y
    −½ = 2y
    y = −¼
    and substitute that value for y into the first sentence:
    x² = 4(¼)²
    x² = ¼
    x = ±½
    Of course, the negative solution doesn't satisfy the second sentence, so the answer x = ½ is correct.