[Math Lair] Solutions for Practice Test 2, The Official SAT Study Guide, Section 8

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Here are solutions for section 8 of the second practice test in The Official SAT Study Guide, second edition, found on pages 481–486. The following solutions illustrate faster, less formal methods that may work better than formal methods 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. 15 minutes after the show begins, 15⁄90 of it is complete. 15⁄90 = 1⁄6. Select (B) 1⁄6.
  2. Solution 1: We can assume that the diagram is drawn to scale. Just by glancing at the diagram, you can tell that the answer must be either (B) HK or (D) JK. Using your finger or your pencil, carefully measure these two. You will find that the longest side is (D) JK. Select that answer.

    Solution 2:

    Solution 3: Looking at the diagram, it's obvious that the answer must be either (B) HK or (D) JK. Both lines form part of triangle JHK. The largest angle in this triangle is angle H. Since the largest angle is opposite the largest side, JK must be larger than HK. Select (D) JK.

    Solution 4:

  3. Solution 1: Look for a pattern: Each number in the f(n) column increases by 6 for every box you go to the right. So, p must equal 19 + 6 = 25. Select (C) 25.

    Solution 2:

  4. Solution 1: Convert the sentence into an equation as follows:
    5 years5
    less than[subtract the previous term from the following term]
    twice as long as2 ×
    Maly hasn
    The result is 2n − 5. Select (C) 2n − 5.

    Solution 2:

  5. Try a special case: Let x = 1. Then, the next odd number greater than x is 3. Next, evaluate each answer choice to see which are equal to 3:
    (A)1 − 1 = 0
    (B)1 + 1 = 2
    (C)1 + 2 = 3
    (D)1 + 3 = 4
    (E)2(1) − 1 = 1
    The only answer yielding 3 when x = 1 is (C) x + 2. Select that answer.

    Note: If you chose a different number as a special case, such as 3, then both (C) and (E) might match. In that case, try a different special case,

  6. Solution 2:
  7. There are a few ways to look at this question, which doesn't require any calculation, but it requires some insight. A circle has rotational symmetry of infinite order. So, you could rotate the given figure any number of degrees and still have the same circle, but with two differently-aligned rectangles in it. So, there must be an infinite number of possibilities. Select (E) More than four.
  8. This question, while not too difficult, can trip you up if you don't realize there are two possibilities for the length of BC.
    Solution 1: Solution 2:
  9. Solution 1: Solution 2: