[Math Lair] Solutions for Practice Test 4, The Official SAT Study Guide, Section 3

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Here are solutions for section 3 of the fourth practice test in The Official SAT Study Guide, second edition, found on pages 581–586. The following solutions 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. Guess and check: Since 3b + 1 < 10, probably the largest value of b would make the expression false. So, try (E) 3. 3(3) + 1 = 10, which is not < 10. Select (E) 3.
  2. Solution 1: Guess and check: Try each answer until you find one that works. Say that you started with (B) 2. 24 · 2 = 28 = 256, so the answer must be smaller. Try (A) 1. 24 · 1 = 16. So, that is the correct answer. Select (A) 1.

    Solution 2: You may know (or could determine) that 24 = 16. Since we are given that 24x = 16, then 4 = 4x, or x = 1. Select (A) 1.

  3. Solution 1: Try a special case: Say that r = 2. Then the question would become "How much greater than 0 is 7?" The answer, of course, is 7. Select (E) 7.

    Solution 2: Convert the sentence into an equation:
    How muchx (or another variable)
    greater than+
    r − 2r − 2
    is=
    r + 5r + 5
    The resulting equation is:

    x + r − 2 = r + 5
    x = 7
    Select (E) 7.
  4. Solution 1: Solution 2:
  5. Solution 1: Solution 2:
  6. Solution 1: Solution 2: If a and b are odd numbers, then (a + 1) is even and b is odd. Remembering the properties of even and odd numbers:
    1. even × odd = even
    2. even + odd = odd
    3. even − odd = odd
    So, II and III are odd. Select (E) II and III.
  7. Substitute x = 2 in f(x) = (3 − 2x²)/x:
    f(2) = (3 − 2(2)²) / 2
    f(2) = −5 / 2
    Select (D) −5 / 2.
  8. Solution 1: Solution 2: Draw a diagram: Usually, when a diagram is not drawn to scale when it could easily have been, this indicates that the question would be really simple if the diagram were drawn accurately. So, on the diagram you are given, re-draw line m such that x > 90. After drawing the line, if you look at the diagram, it is pretty obvious that (B) through (E) are false and (A) is true. Select (A) y < 90.
  9. Solution 1: Where a line crosses the x-axis, y = 0. Set y = 0 and solve for x:
    0 = 5x − 10
    10 = 5x
    x = 2
    So, the coordinates of the point are (2, 0). Select (D) 2.

    Solution 2: Draw a diagram: On your graphing calculator, graph y = 5x + 10. You can then determine that the line crosses the x-axis at (2, 0). Select (D) 2.

    Solution 3:

  10. Solution 1: Draw a diagram: Draw a graph with a line matching the description of l, and a few lines matching the description of k. Your graph might look something like the following:
    [diagram for question 16]
    Looking at the diagram, not all of the possible lines for k satisfy (A), (D), or (E), none of them satisfy (B), and all of them satisfy (C). Select (C) Line k has a negative slope.

    Solution 2: If k is perpendicular to l, then the slope of k must be the negative reciprocal of l. Since line l has a positive slope, then the slope of k must be negative. Select (C) Line k has a negative slope.

  11. Solution 1: Substitute a ^ b = (a + b)/(ab) in 1 ^ 2 = 2 ^ x:
    (1 + 2)/(1 − 2) = (2 + x)/(2 − x)
    Cross-multiplying and simplifying, being very careful with the negative signs, we get:
    3(2 − x) = −1(2 + x)
    6 − 3x = −2 − x
    −2x = −8
    x = 4
    Select (A) 4.

    Solution 2:

  12. Solution 1: Look for a pattern: List the total cost of 1, 2, 3, ... shirts, and see if you can find a pattern:
    ShirtsCost
    1x
    2x + xz
    3x + 2(xz)
    4x + 3(xz)
    You can list more entries if desired until you see that the general term must be:
    nx + (n − 1)(xz)
    Select (A) x + (n − 1)(xz).

    Solution 2:

  13. Solution 1: Solution 2: