• Question #1: The sum of the two solutions of the quadratic equation f(x) = 0 is equal to 1 and the product of the solutions is equal to -20. What are the solutions of the equation f(x) = 16 - x ? (a) x₁ = 3 and x₂ = -3 (b) x₁ = 6 and x₂ = -6 (c) x₁ = 5 and x₂ = -4 (d) x₁ = -5 and x₂ = 4 (e) x₁ = 6 and x₂ = 0
  • Answer: If a, b are the solutions of the equation f(x) = 0, ab = -20 a + b = 1

    We solve this system of equations, and a = 5, b = -4. Then f(x) = (x-5)(x+4) f(x) = x² - x - 20.

    f(x) = 16 - x x² - x - 20 = 16 - x

    x² = 36, x₁ = 6 and x₂ = -6

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• Question #2: In the (x, y) coordinate plane, three lines have the equations: l₁: y = ax + 1 l₂: y = bx + 2 l₃: y = cx + 3

  Which of the following may be values of a, b and c, if line l₃ is perpendicular to both lines l₁ and l₂?

  (a) a = -2, b = -2, c = .5 (b) a = -2, b = -2, c = 2 (c) a = -2, b = -2, c = -2 (d) a = -2, b = 2, c = .5 (e) a = 2, b = -2, c = 2

  • Answer: If line l₃ is perpendicular to both lines l₁ and l₂, then l₁ must be parallel to l₂. The slopes of l₁ and l₂ should be equal, slope_l1 = slope_l2, and the slope_l3 = -1/slope_l1. a = -2, b = -2, c = .5 is the only answer that satisfies these conditions.

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• Question #3: The management team of a company has 250 men and 125 women. If 200 of the managers have a master degree, and 100 of the managers with the master degree are women, how many of the managers are men without a master degree?

  (a) 125 (b) 150 (c) 175 (d) 200 (e) 225

  • Answer: If 200 of the managers have a master degree, and 100 of the managers with a master degree are women, then 100 of the managers with a master degree are men. The company has 250 men managers and 100 of the managers with a master degree are men, so the number of managers that are men without a master degree is 250 - 100 = 150.

• Question #4: In the figure below, the area of square ABCD is equal to the sum of the areas of triangles ABE and DCE. If AB = 6, then CE =

  (a) 5 (b) 6 (c) 2 (d) 3 (e) 4

  • Answer: Area_ABCD = AB² Area_ABE = AB·BE/2 Area_DCE = DC·CE/2 = AB·CE/2

    The problem states that the area of square ABCD is equal to the sum of the areas of triangles ABE and DCE. AB² = AB·BE/2 + AB·CE/2 AB = BE/2 + CE/2 BE = CE + AB AB = CE/2 + AB/2 + CE/2 AB/2 = CE/2 + CE/2 CE = AB/2 CE = 3

• Question #5:

  If α and β are the angles of the right triangle shown in the figure above, then sin²α + sin²β is equal to:

  (a) cos(β) (b) sin(β) (c) 1 (d) cos²(β) (e) -1

  • Answer: sin²α + sin²β = sin²α + sin²(90^o - α) = sin²α + cos²α = 1.

• Question #6: The average of numbers (a + 9) and (a - 1) is equal to b, where a and b are integers. The product of the same two integers is equal to (b - 1)². What is the value of a?

  (a) a = 9 (b) a = 1 (c) a = 0 (d) a = 5 (e) a = 11

  • Answer: 1/2(a + 9 + a - 1) = b (a + 9)(a - 1) = (b - 1)² We need to solve this system of equations. 1/2(2a + 8 ) = b a + 4 = b (a + 9)(a - 1) = (a + 3)² a² + 9a - a - 9 = a² + 6a + 9 2a = 18 a = 9

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