[Math Lair] Solutions for 2013 SAT Practice Test, Section 9

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Here are solutions for the section 9 of the 2013-14 SAT practice test; you can find the test at the College Board's web site or in the Getting Ready for the SAT booklet. 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. Solution 1: Guess and check: Try each answer and see which one solves the equation. The one that does is (A) −27.
    Solution 2: Solve the equation:
    (1/3)y + 9 = 0
    (1/3)y = −9
    y = −27
    Therefore, the answer is (A) −27.
    It's worthwhile noting here that you shouldn't rush through easy questions such as this one, because they're worth as much as hard questions. It's easy to make a careless error on this one and miss easy marks.
  2. Solution 1: Solution 2: PR is the radius of the larger circle and the diameter of the smaller circle. Since the radius of a circle is twice its diameter, the radius of the smaller circle is ½ that of the larger, and ½(4) = 2. Therefore, the answer is (B) 2.
  3. Try a special case: We aren't told what the distance between the curve and either R or S is, so presumably it doesn't matter. So, since RS is 12, we'll assume that the radius of each circle is 6. Now, the perimeter of two semicircles of radius 6 is the same as the perimeter of one circle of radius 6, which is 2π(6) = 12π. Therefore, the answer is (C) 12π. (6) Try a special case: Assume that, say, k = 3. Then h = 4. So, (7 − k)⁄h becomes (7 − 3)⁄4 = 4⁄4 = 1. Look at the choices: The only answer choice that matches our calculated value is (A) 1. Select that answer.
  4. Try a special case: Say that h = 4 and k = 3. Then the expression becomes (7 − 3)/4 = 1. The only answer that corresponds is (A) 1. Select that answer.
  5. This is the type of question that you could waste a lot of time on, but with the correct insight, the question is quite easy. The easiest way to solve the question is to note that the equation is x² = x + 6. In words, this equation states that x² is six greater than x. Therefore, x² must be greater than x. So, the answer is (E) x² > x.

    However, if you miss the above insight, you could solve for x, find that x = −2 or x = 3, eliminate choices (A), (B), and (C), and then substitute −2 and 3 in (D) and (E) to see which is true.

  6. Substitute the values of f(10) and f(5) in the given equation. This produces an equation in a. We can then solve for a:
    f(10) + f(5) = 55
    5(10) − 2a + 5(5) − 2a = 55
    −4a = −20 a = 5
    Therefore, the answer is (C) 5.
  7. Solution 1: Estimate the answer: Since the graph is drawn to scale, you can look at the bars to see for which pair the ratio would be the largest. For 1994, the ratio appears to be much larger than for any other year. So, the answer is (E) 1994.
    Solution 2: Read the numbers off of the graph and compare them to see which is greatest. We really only need to where the enrollment at School B is greater than that at school A: The largest ratio is 2:1. So, the answer is (E) 1994.