[Math Lair] Solutions for Practice Test 10, The Official SAT Study Guide, Section 2

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Here are solutions for section 2 of practice test #10 in The Official SAT Study Guide, second edition, found on pages 948–953. 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. If 3 pencils cost $4.50, each pencil costs $1.50. If the price is increased by $0.50, they will cost $2 each, so 5 pencils will cost $10. Select (E) $10.00.
  2. Let x be the circumference of the circle with centre A. So, x/2 is the circumference of the circle with centre B. So, the radius of the circle with centre A is x/2π, and the radius of the circle with centre B is x/4π. The radius of the circle with centre A is twice the radius of the circle with centre B. So, if AC = 6, then AB = 4 and BC = 2. Select (B) 2.
  3. Guess and check: Try each point and see which satisfies |x| − |y| = 3. For points where this is the case, the x value must be further away from x = 0 than the y value is, so try a point that looks like it meets that criterion. B is such a point. B is located at (-4, -1). |x| − |y| = |-4| − |-1| = 4 − 1 = 3. select (B) B.
  4. When a number is divided by 3, the remainder must be either 0, 1, or 2. The only answer where all of the numbers are 0, 1, or 2 is (D) 0, 1, 2, 0. Select that answer.
  5. If y is inversely proportional to x and y = 15 when x = 5, then when x is 5 times greater than 5 (i.e. 25), y will be 1/5 of 15, or 3. Select (C) 3.
  6. Substitute 2x + z = 2y into the second equation:
    2x + 2y + z = 20
    2y + 2y = 20
    4y = 20
    y = 5
    Select (A) 5.
  7. Solve the equation:
    2(x − 3) = 7
    x − 3 = 3.5
    x = 6.5
    Enter 6.5.
  8. Substitute x = 4 into the given equation:
    y − 4 = 3(4 − 2)
    y = 4 + 3(2)
    y = 4 + 6
    y = 10
    Enter 10.
  9. The first term of the sequence is 20, the second is 8, the third is ½(20 + 8) = 14, the fourth is ½(8 + 14) = 11, and the fifth is ½(14 + 11) = 12.5. Enter 12.5.
  10. The median of the enrollment over the 5 years must be the third largest value, which we are told is 1351. Because the table already contains two values larger than 1351, the unknown value x must be no greater than 1351; otherwise, x would be the median, not 1351. Therefore, the greatest possible value for x is 1351. However, the question also says that no two years had the same enrolment, so x can't be 1351, so the largest possible value for it is 1350. Enter 1350.