[Math Lair] Solutions for 2014 SAT Practice Test, Section 2

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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 2014–15 SAT practice test; you can find the test on the College Board's web site or in the Getting Ready for the SAT booklet. Note that this test is the same as the 2012–13 practice test. The following solutions illustrate faster, less formal methods that may work better than formal methods 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: List the first 12 terms of the sequence out:
    4, 11, 18, 25, 32, 39, 46, 53, 60, 67, 74, 81
    Select (B) 81.
    Solution 2: Look for a pattern: To get the second term of the sequence, you would start with 4 and add 7. To get the third term of the sequence, you would start with 4 and add 7 twice. To get the fourth term of the sequence, you would start with 4 and add 7 three times. Carrying on this pattern, to find the twelfth term you would start with 4 and add 7 11 times. So, the twelfth term would be 4 + 7 × 11 = 81. Select (B) 81.
    Solution 3: You might know that the formula for the nth term of an arithmetic sequence is tn = a + (n − 1)d, where a is the first term and d is the difference between the terms. In the given sequence, the first term is 4 and the difference between terms is 7. So:
    t12 = 4 + 11 × 7 = 81
    Select (B) 81.
  2. Solution 1: Solution 2: Guess and check: Try each of the answer choices as a value for x until you find one that makes the equation true. You will eventually find that the only one that works is (E) 9. Select that answer.
    Solution 3:
  3. Solution 1: If you're reasonably good at math, your instincts may tell you that the answer's obvious: It's 15. Since this is question 3, it's probably reasonable to trust your instincts. Select (B) 15 and move on without spending more time.
    Solution 2: Try a special case: Say that each of t and y is 15 (so their average is 15) and that each of w and x is 15 (so their average is 15). The average of t, y, w, and x is ¼(15 + 15 + 15 + 15) = 15. Select (B) 15.
  4. Solution 1: Look at the answer choices: Look at the answers, read them through carefully, and eliminate any that can't be correct: The only remaining choice is (A). Select (A) If Fred cannot swim, then he is not Kay's brother.
    Solution 2:
  5. Solution 1: Solution 2: Solution 3: We are told that AB = AO. Now, AO = BO, since both line segments are radii of the circle. So, all three sides of the triangle are equal; it must be equilateral. So, each angle of the triangle, including ∠ABO must measure 180° ÷ 3 = 60°. Select (D) 60°.
  6. Solution 2:
  7. Look at the diagram to find where the curve is higher than the straight line. The only places it is higher are between −3 and 0 (but not including the endpoints). Select (B) −3 < x < 0 only.
  8. This question isn't too difficult if you tackle it in an organized fashion:
    4 magazines × 12 issues per year =48
    2 magazines × 4 issues per year =8
    1 magazine × 52 issues per year =52
    Total =108
    Enter 108.
  9. This question isn't too hard, but all of the fractions can be confusing. You may find it easier to write the first equation as
    j ÷ k = 32
    Since k = 3/2:
    j ÷ (3/2) = 32
    Multiply both sides by 3/2:
    j = 48
    Multiply both sides by ½:
    ½j = 24
    Enter 24.
  10. You don't have to do this, but you might want to substitute k for x + y and then solve for k:
    k + 3z = 600
    k + z = 400
    To get rid of z, subtract 3 × the second equation from the first equation:
    3k + 3z = 1200
    k + 3z = 600
    2k = 600
    And so k = 300. Alternately, you can skip the substitution and solve for x + y directly using the exact same method as above.
  11. Solution 1: Solution 2: You've forgotten that the slope of the perpendicular is the negative reciprocal? Or do you hate coordinate geometry but love geometry? No sweat. Here's what to do:
  12. First, find the largest prime number less than 50. 49 is not a prime number, because 49 = 7 × 7. 47 is a prime number though. Next, find the smallest prime number larger than 50. 51 isn't prime; 51 = 3 × 17. 53 is prime, though. 53×47 = 2491.