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Level 4 and Level 5 SAT Subject Questions


Math SAT 800: How To Master the Toughest Problems


  • Question: If 2x < 100 and x is an integer, how many of the 2x + 2 integers will be divisible by 3 and by 2?

    (a) 1

    (b) 2

    (c) 3

    (d) 4

    (d) 5

    • Answer: 26 = 64 and 27 = 128. If 2x < 100, then the highest x is x = 6
      Possible values for x: 0 , 1 , 2 , 3 , 4, , 5 , 6. 2x + 2 can take the values 3 , 4 , 6 , 10 , 18 , 34 , 66
      Out of these values, only 6 , 18 and 66 are divisible by 3 and by 2. The correct answer is (c).

  • Question: The inequality |2x - 1| > 5 must be true in which one of the following cases?
    I. x < -5
    II. x > 7
    III. x > 0

    (a) II only

    (b) I, II and II

    (c) I and II only

    (d) I and III only

    (e) I only

    • Answer: |2x - 1| > 5,
      -5 > 2x - 1 or 2x - 1 > 5
      -4 > 2x or 2x > 6
      -4 > 2x results in x < -2
      2x > 6 results in x > 3
      I answer is true, II answer is also true, but III answer is false, so the correct answer is (c)

  • Question: What is the closest approximation of the solution of the equation 2x - 1 = 3x + 1?

    (a) -4.42

    (b) -5.81

    (c) -3.22

    (d) 4.93

    (e) 3.33

    • log(2x - 1) = 3log(x + 1)
      (x - 1)log2 = (x + 1)log3
      x(log2 - log3) = log3 + log2
      x = (log3 + log2)/(log2 - log3)
      x is aprox. = -4.42


  • Question: What is the range of (x - y) if 3 < x < 4 and -2 < y< -1?

    (a) 4< x-y <5

    (b) 1< x-y <3

    (c) 1< x-y <5

    (d) 4< x-y <6

    (e) 3< x-y <6

    • Answer: We can determine the range of -y:
      1<-y<2
      We determine the range of x-y by adding the ranges of x and -y:
      Therefore, 4< x-y <6

  • Question: For some positive real number ‘a’, the first 3 terms of a geometric progression are a - 1, a + 3 and 3a + 1. What is the numerical value of the fourth term?

    (a) 25

    (b) 36

    (c) 32

    (d) 100

    (e) 9

    • Answer:
      a + 3 = k(a - 1)
      3a + 1 = k(a + 3)

      (a + 3)(a + 3) = (3a + 1)(a - 1)
      a2 + 6a + 9 = 3a2 - 2a - 1
      2a2 - 8a -10 = 0
      a2 - 4a -5 = 0
      the solutions of this equation are 5 and -1. The only positive solution is 5, so the progression is 4, 8, 16. The fourth term will be 16·2 = 32


  • Question: What is the volume of the geometric solid produced by the equilateral triangle in the figure below when it is rotated 360o about the altitude m?

    (a) ¶m3/9

    (b) ¶m2/9

    (c) ¶m3/4

    (d) ¶m3

    (e) m3/9

    • Answer:
      The solid produced by the triangle rotation will be a cone with a radius equal to half the side of the triangle.

      We apply Pythagoras Theorem in one of the 2 right triangles created by the altitude m:

      r2 + m2 = (2r)2
      r = m/√3
      The volume of the cone:
      Vcone = ¶m(m/√3)2/3
      Vcone = ¶m3/9


  • Question: In the (x,y) plane, which of the following statements are true?

    I. Line y + x = 5 is perpendicular to line y - x = 5.
    II. Lines y + x = 5 and y - x = 5 intersect each other on the y axis.
    III. Lines y + x = 5 and y - x = 5 intersect each other on the x axis.

    (a) I is the only true statement

    (b) II is the only true statement

    (c) I and II are both true

    (d) I and III are both true

    (e) II and III are both true

    • Answer:
      y + x = 5 can be written as y = -x + 5. The slope of this equation is m1 = -1.
      y - x = 5 can be written as y = x + 5. The slope of this equation is m2 = 1.
      m2 = -1/m1 so the 2 lines are perpendicular.
      We also need to find where the 2 lines intersect. If we add the 2 equations, 2·y = 10, y = 5
      From the first equation, x = 5 - y = 5 - 5 = 0. In conclusion the lines intersect at (0, 5) and this point is on the y axis.
      In conclusion I and II statements are correct.

  • Question: Find the domain of the function f(x) = √( -x) / [(x - 2)(x + 2)]:

    (a) (-∞ , -2) U ( -2 , 0)

    (b) (-∞ , -2) U ( -2 , 0]

    (c) (-∞ , 2) U ( 2 , 0]

    (d) (-∞ , 2) U ( 2 , 0)

    (e) (-∞ , -2) U ( -2 , 2)

    • Answer:
      From the numerator of the fraction, -x should be positive or equal to zero, so x<=0.
      x can’t take the values x = 2 and x = -2, so the domain of the function is (-∞ , -2) U ( -2 , 0]
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