[Math Lair] Solutions for Practice Test 9, The Official SAT Study Guide, Section 5

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Here are solutions for section 5 of practice test #9 in The Official SAT Study Guide, second edition, found on pages 903–908. The solutions below 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. The fifth term of the sequence is (30 + 1) × 2 = 62. The sixth term of the sequence is (62 + 1) × 2 = 126. Select (E) 126.
  2. Expand the first equation and substitute ax = 15:
    a(x + y) = 45
    ax + ay = 45
    15 + ay = 45
    ay = 30
    Select (E) 30.
  3. List out all of the possibilities:
    456, 465, 546, 564, 645, 654
    There are six possibilities. Select (C) Six.
  4. There are three faces, each with area r. There are two triangular faces, each with area t. Therefore, the total surface area is 3r + 2t. Select (B) 3r + 2t.
  5. Guess and check: Try each answer until you find one that works:
  6. Try a special case: Say that each of the 14 books weighs p pounds. Then, the average weight will also be p pounds. The total weight of the books will be 14p pounds. Select (E) 14p.
  7. As point B is halfway between points C and A, its y-coordinate (t) will be halfway between the y-coordinate of C (5) and the y-coordinate of A (−1). The number halfway between 5 and −1 is (5 + (−1))/2 = 2. Select (C) 2.
  8. If you multiply a bunch of terms together and the result is 0, then one (or more) of the terms must be 0. So, either k = 0 or 2x + 3 = 0 or x − 1 = 0. However, 2x + 3 cannot equal 0 because then x would equal −3/2, which violates the condition that x > 1. Similarly, x − 1 cannot equal 0 becuse x > 1. So, k must equal 0. Select (B) 0.
  9. From the reference information, the circumference of a circle is 2πr, where r is the radius of a circle. Setting the circumference equal to π:
    π = 2πr
    1 = 2r
    r = ½
    Select (B) ½.
  10. Try a special case: Say that x = 3 and y = 5. In this case:
    1. xy = 15, which is divisible by 15.
    2. 3x + 5y = 9 + 25 = 34, which is not divisible by 15.
    3. 5x + 3y = 15 + 15 = 30, which is divisible by 15.
    If desired, you can try additional cases or reason about the values of x and y to convince yourself that I and III will always be divisible by 15. Select (D) I and III only.
  11. If n is the smallest of the three integers, then:
    n + (n + 2) + (n + 4) = 111
    3n + 6 = 111
    Select (D) 3n + 6 = 111.