[Math Lair] Solutions for Practice Test 4, The Official SAT Study Guide, Section 6

Math Lair Home > Test Preparation > Solutions for Practice Test 4, The Official SAT Study Guide, Section 6
SAT Practice Test Solutions:
2014–15 SAT Practice Test
2013–14 SAT Practice Test
The Official SAT Study Guide, second edition
SAT Math Tips

Here are solutions for section 6 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. Solution 1: Solution 2: Guess and check. Try each answer, starting with the middle answer, to see if it works. Does (C) 9 work? (3 + 9)/2 = 6, which is not 7½. So, try a larger answer, say (D). (3 + 12)/2 = 15/2 = 7½, so this is the correct answer. Select (D) 12.
  2. Solution 1: From the diagram, angle 2 is an obtuse angle and angle 4 is another obtuse angle. So, look for two obtuse angles formed by m. Angles 6 and 8 are obtuse. Therefore, the answer is (D) 6 and 8.

    Solution 2: Using Z, F, and C patterns (or however you remember congruent angles when a line intersects a pair of parallel lines), you can determine that angle 2 is congruent to angles 6 and 8, and angle 4 is congruent to angles 6 and 8. Based on this information, the only answer that makes sense is (D) 6 and 8. Select that answer.

  3. There are a few ways of solving this problem; I'll only illustrate the most direct route.
  4. Substitute k = 4 in A(k) = 4k − 30:
    A(15) = 4(15) − 30
    A(15) = 30
    Select (E) $30.
  5. Solution 1: Substitute xr = v in v = kr:
    xr = kr
    x = k
    So, x is equal to k. Select (D) x.

    Solution 2:

  6. Solution 1: This question is pretty easy if you notice that the three triangles on the left-hand side form a quadrilateral, and you remember that the interior angles of a quadrilateral sum to 360°. The four angles of the quadrilateral are: The sum of these must be equal to 360°:
    a° + b° + b° + a° + c° + b° = 360°
    2a + 3b + c = 360
    c = 360 − 2a −3b
    Select (E) 360 − 2a −3b.

    Solution 2: If the above solution didn't occur to you, it's still possible to solve the problem; it just takes a bit longer.

  7. Don't try to solve for t. Instead, just multiply both sides of the equation by 4:
    4t³ = 1404
    Enter 1404.
  8. It's possible to solve this problem in a more intuitive manner, but I've chosen to include a solution with all of the steps.
  9. Solution 1: Try a special case: Say that Tameka cut the pieces so that the angle is 24°. There would be 360/24 = 15 pieces. So, 15 is a possible answer. Enter 15.

    Solution 2: Consider a different problem: Say that we were asked to find the number of pieces if Tameka cut the pizza into slices with a 30° angle. There would then be 360/30 = 12 pieces. However, the actual question specifies that the angle is less than 30°, so there must be more than 12 pieces. Enter the next integer higher than 12, which is 13.

  10. Solution 1: Guess and check: Say that a = 1. The sum of the terms would be 1 + 3 + 9 + 27 + 81 = 121. This isn't right. More importantly, though, it is 121/605 = 1/5 of the correct answer.
    Since it appears that the value should be 5 times larger, try a = 5. 5 + 15 + 45 + 135 + 405 = 605. This answer is correct. Enter 5.
  11. Solution 1: There are three different places in which the shaded card can go. Having placed the shaded card, there are four different places where the next card can go, three different places for the card after that, two different places for the card after that, and one place for the final card. This makes 3 × 4 × 3 × 2 × 1 = 72 possible arrangements. Enter 72.

    Solution 2: To find the number of arrangements, find the total number of arrangements if there were no conditions, and subtract the number of arrangements where the shaded card is at either end. If there were no conditions, there would be 5! = 120 arrangements. However, there are 4! = 24 ways to arrange the cards so that the shaded card is on the left end, and 24 ways to arrange the cards so that the shaded card is on the right end. So, the total number of valid possibilities is 120 − 24 − 24 = 72. Enter 72.