[Math Lair] Solutions for Practice Test 7, The Official SAT Study Guide, Section 7

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Here are solutions for section 7 of practice test #7 in The Official SAT Study Guide, second edition, found on pages 785–790. 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. 60 is divisible by 3, so 60 − 3 = 57 is divisible by 3. Select (C) 57.
  2. Try drawing lines through the figures and see if you can find one through which two different lines can be drawn. You will find that the answer is (D) X.
  3. Looking at the graph, the only person showing a significant increase (i.e. the shaded bar being much higher than the unshaded bar) is Goldberg. Select (B) Goldberg.
  4. The average of 6, 6, 12, and 16 is ¼(6 + 6 + 12 + 16) = 10. If x were 10, the average of the 5 numbers would still be 10. Select (D) 10.
  5. If two expressions multiply to zero, either one expression must be zero, or the other must be zero. So, either 2x − 2 = 0 (Equation 1) or 2 − x = 0 (Equation 2). Solving equation 1, x = 1. Solving equation 2, x = 2. Select (D) 1 and 2 only.
  6. Take the cube root of both sides of the equation:
    x = y³
    Select (C) y³
  7. Look at the answer choices: Look at each answer, and eliminate any answers where any number does not satisfy exactly one criterion: Select (A) 14-20-13.
  8. If x > 3, then both sides of the equation are positive. We can then square both sides:
    x + 9 = (x − 3)²
    x + 9 = x² − 6x + 9
    x = x² − 6x
    Select (C) x = x² − 6x.
  9. To find the number of integers between 1 and 100 that are not squares, we can find the number of squares and subtract that from 100. Now, 1² = 1 and 10² = 100, so there are 10 squares in total between 1 and 100. Subtracting 10 from 100, 100 − 10 - 90. Select (E) 90.
  10. The larger circle's area is π(1)² = π. The smaller circle's area is π(½)² = ¼π. The ratio of the area of the larger to the area of the smaller is 1:¼ = 4:1. Select (D) 4:1.
  11. This question is a lot easier if you realize that the sum of the integers from −22 to 22 is 0. Now, summing the next few integers after 22, we get 23 + 24 + 25 = 72. So, the sum of the integers between −22 and 25 is 72. Select (B) 25.
  12. The second graph is the same as the first graph of y = f(x), shifted 2 units down and 3 units to the left. So, if the first graph is the graph of y = f(x), the second graph would be the graph of y = f(x − 2) −3. So, h = −2 and k = −3, and hk = 6. Select (E) 6.