[Math Lair] Solutions for Practice Test 5, The Official SAT Study Guide, Section 2

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Here are solutions for section 2 of practice test #5 in The Official SAT Study Guide, second edition, found on pages 638–643. 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. Solution 1: Solve the equation for x:
    3x + 9 = 5x + 1
    8 = 2x
    4 = x
    Select (D) 4.
    Solution 2: Guess and check: Try each of the ansswer choices to see which value of x makes the equation true. You'll find, that for x = 4, 3(4) + 9 = 5(4) + 1, so the answer is (D) 4.
  2. Look at the answer choices: Select (B) 2.
  3. Solution 1: To find the total number of combinations, multiply the number of colors by the number of sizes. There are 3 colors and 4 sizes. 3 × 4 = 12. Select (B) 12.
    Solution 2: Write each of the possibilities out, as shown below (where R = Red, W = White, B = blue, S = Small, M = Medium, L = Large, E = Extra-large):
    RS, RM, RL, RX, WS, WM, WL, WX, BS, BM, BL, BX
    There are 12 possibilities. Select (B) 12.
  4. Solution 1: Guess and check: Check each answer at −3 and 3 until you find one where f(−3) is the graeter. Start with the answers where the value of x appears to decrease as x increases: Select (D) f(x) = 4 − x³.
    Solution 2: Using your knowledge of graphs and transformations, quickly sketch the five functions. It should be apparent that (D) f(x) = 4 − x³ is the only one where f(−3) > f(3). Select that answer.
    Solution 3: Plot each function on your graphing calculator until you find one where f(−3) > f(3). You will find that (D) f(x) = 4 − x³ is the only such function. Select that answer.
  5. If 2r = 5s and 5s = 6t, then 2r = 6t. Solving that equation for r, we get r = 3t. Select (C) 3t.
  6. Solution 1: Solution 2: Guess and check: Try each of the first four answers. If none of those work, then the answer must be (E) For no value of x.
  7. Try a special case: Select a bunch of positive and negative numbers such that the probability of selecting a positive number is 3/5. If you select the following numbers, this is the case:
    1, 2, 3, −1, −2
    Now, there are 3 positive numbers (p) and 2 negative numbers (n). So, n/p = ⅔. Select (C) ⅔.
  8. Substitute c(x) = 640 and x = 20 into the given equation:
    640 = 600 · 20 − 20020 + k
    640 = 600 − 10 + k
    50 = k
    Select (B) 50.
  9. Guess and check: Try each pair of positive integers, starting with the smallest possible pairs: We found one ordered pair that satisfies the equation. Select (A) One.
  10. Solution 1: Solution 2:
  11. Try a special case: Assume that x = k and y = k. This is consistent with what we're given (that the average of the two is k). The average of k, k, and z is (2k + z)/3. Select (A) (2k + z)/3.
  12. Solution 1: Solution 2: Guess and check: If you have time remaining, find the remainder when 15 is divided by each number from 1 to 15. You'll find three numbers (4, 6, and 12) that leave remainder 3. Select (C) Three.