[Math Lair] Solutions for Practice Test 3, The Official SAT Study Guide, Section 2

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Here are solutions for section 2 of the third practice test in The Official SAT Study Guide, second edition, found on pages 514–519. 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. If 20y − 5y = 15, then:
    15y = 15
    y = 1
    Given that y = 1 and y = x − 5, then 6 = x. Select (A) 6.
  2. Draw a diagram: Draw two lines, from top to bottom and from left to right, dividing the circle graph into quarters. Now, looking at the diagram, the Vanilla piece occupies an entire quarter and a bit more, so it represents more than 25%. However, pecan, peach, mint, and strawberry each occupy less than one of the quarters, so there are four that represent less than 25%. Select (D) Four.
  3. Solution 1: Solution 2: You may know that the external angles of a polygon sum to 360°. The three angles denoted by x, y, and z are the external angles of a triangle. So, they must sum to 360°. Select (E) 360.

    Solution 3:

  4. We are given that 6x + 4 = 7. Subtract 8 from both sides of the equation gives us:
    6x − 4 = − 1
    Select (B) −1.

    Note: In a question such as the above, where you are given an equation and asked to find the value of an expression, it's usually a lot slower to solve for x than it is to try to turn one side of the equation into the expression you're looking for.

  5. Solution 1: The distance between A and B is the same as the distances between B and C, C and D, and so on. You can tell this either by looking at the diagram (you can assume it's drawn to scale) and noticing that the distances appear the exact same, or by noting that the figure has rotational symmetry, so all of the distances have to be the same. Either way, ABC contains 2 equal segments, while AEC contains 3 equal segments. The ratio of their lengths is 2 to 3. Select (B) 2 to 3.

    Solution 2: I wouldn't recommend a formal solution here due to the amount of time required, but if you wanted to do that, here is an outline of how you could do so: Draw a radius from the centre of the circle to each of A, B, C, D, and E, prove that each of the five triangles thus created are congruent (since two of the sides of each triangle are radii and the other side is that of the equilateral pentagon, all three sides of each triangle are equal). Since each triangle is congruent, the angle formed by the two radii must be equal in all five triangles, so each of the arcs AB, BC, ... must be equal. ABC contains two equal segments and AEC 3, so the ratio must be 2 to 3. Select (B) 2 to 3.

  6. Solution 1: You can turn the sentence into an equation as follows:
    how much greaterWhat we're looking for (you can call it x if you like)
    is=
    the sum of[add the two following things together]
    ss
    and+
    tt
    than− (
    the sum of[add the two following things together]
    ss
    and+
    ww
    )
    Putting it all together, we get:
    x = s + t − (s + w)
    x = s + tsw
    x = tw
    Select (C) tw

    Solution 2: Try a special case: If you find all the letters to be confusing, substitute numbers for them.

  7. When t = 4:
    P(4) = 3,000 · 21
    =6,000
    When t = 16:
    P(16) = 3,000 · 24
    = 48,000
    To find the population increase, subtract the number of organisms at t = 16 (48,000) from the number at t = 4 (6,000). The result is 48,000 − 6,000 = 42,000. Select (D) 42,000.
  8. The elements common to sets A and B will be the numbers that are both in circle A and circle B. There is a 5 and a 2 that are in both circles, so the answer is 5 + 2 = 7. Select (D) 7.
  9. Solution 1: Solution 2: Solution 3:
  10. Break the problem up and solve each part. First:
    g(x) = f(3x + 1)
    g(2) = f(3 · 2 + 1)
    g(2) = f(7)
    Next, find the value of f(7). From the table, this is −5. Select (A) −5.
  11. This problem may appear intimidating because you've probably never been taught how to find the area of little swirly things in class. However, with the right technique, it's not too difficult.
  12. [diagram for question 18]Solution 1: Draw a diagram: Draw six points, draw all lines between them, and count the number of lines very carefully. If you are sufficiently careful, you will find there are 15 lines. Select (A) 15.

    Solution 2: A line extends from any point to any other point. Since there are 6 points, there are 6 × 5 = 30 lines. However, this counts each line twice (e.g. it counts the line from point 1 to point 2, and the line from point 2 to point 1). So, we have to divide this result by 2, giving 15. Select (A) 15.

  13. Solution 1: Since the area of the pool is 4,000 square meters, xy = 4000, or x = 4000⁄y. The total length of the rope is y + 4x = y + 4(4000⁄y = y + 16000⁄y. Select (B) y + 16000⁄y.

    Solution 2: