[Math Lair] Solutions for 2014 SAT Practice Test, Section 2

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SAT Practice Test Solutions:
2014–15 SAT Practice Test
2013–14 SAT Practice Test
The Official SAT Study Guide, second edition
SAT Math Tips

Here are solutions for section 2 of the 2014–15 SAT practice test; you can find the test on the College Board's web site or in the Getting Ready for the SAT booklet. Note that this test is the same as the 2012–13 practice test. 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. Keep in mind that vertical angles are equal. Since m and the combination of p and x are vertical, they must be equal. So:
    m = p + x
    40 = 25 + x
    x = 15
    Select (A) 15.
  2. Solution 1: Solution 2:
  3. Try a special case: Say that n = 1. Then, k = 3 (since 81 = 2³). So, n/k = ⅓. Select (B) ⅓.
  4. Solution 1: Solution 2: If f(x) = 3x + 4, then:
      2f(x) + 4
    = 2(3x + 4) + 4
    = 6x + 12
    Select (E) 6x + 12.
  5. It's possible to solve this problem in a less formal way than the following three solutions illustrate, but I've chosen to present these solutions in order to illustrate the thought processes that might go into solving this problem:
    Solution 1: Solution 2: Solution 3:
  6. Using the diagram, count the number of blocks (intersections) you need to travel to reach W from F. The answer is (D) 3½. Select that answer.
  7. Solution 1: Solution 2:
  8. Solution 1: This is actually a very easy question if you remember the laws of exponents and don't get intimidated by all of the letters. Factor the expression by (2x)y:
    (2x)3y − (2x)y = (2x)y((2x)2y − 1)
    Select (C) (2x)y((2x)2y − 1).
    Solution 2: This is a bit more time-consuming, but if you have some time left it's still a good way of approaching the problem:
  9. If the product of two integers ends with 9, then either both integers end in 3 or one integer ends in 9 and the other ends in 1. However, since j, k, and n are consecutive, both j and n can't both end in 3. So, j ends in 9 and n ends in 1, and so k ends in 0. Select (A) 0.