Quantitative Aptitude: Quadratic Equations Set 1

Quadratic Equation Bank PO

Directions: In the following questions, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly-

  1. I. 20x2 – 31x + 12 = 0,
    II. 6y2 – 7y + 2 = 0

    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established
    View Answer
    A) If x > y
    Solution:

    20x2 – 31x + 12 = 0
    20x2 – 16x – 15x + 12 = 0
    So x = 3/4, 4/5
    6y2 – 7y + 2 = 0
    6y2 – 3y – 4y + 2 = 0
    So y = 1/2, 2/3
    Put on number line
    1/2… 2/3… 3/4… 4/5
  2. I. 3x2 + 22 x + 24 = 0,
    II. 3y2 – 10y + 3 = 0

    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established
    View Answer
    B) If x < y
    Solution:

    3x2 + 22 x + 24 = 0
    3x2 + 18x + 4x + 24 = 0
    So x = -4/3, -6
    3y2 – 10y + 3 = 0
    3y2 – 9y – y + 3 = 0
    So y = 1/3, 3
    Put on number line
    -6… -4/3… 1/3… 3
  3. I. 6x2 – x – 2 = 0,
    II. 5y2 – 18y + 9 = 0

    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established
    View Answer
    E) If x = y or relation cannot be established
    Solution:

    6x2 – x – 2 = 0
    6x2 + 3x – 4x – 2 = 0
    So x = -1/2, 2/3
    5y2 – 18y + 9 = 0
    5y2 – 15y – 3y + 9 = 0
    So y = 3/5, 3
    Put on number line
    -1/2 …. 3/5 ….2/3 …. 3
  4. I. x2 – x – 6 = 0,
    II. 5y2 – 7y – 6 = 0

    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established
    View Answer
    E) If x = y or relation cannot be established
    Solution:

    x2 – x – 6 = 0
    x2 – 2x + 3x – 6 = 0
    So x = -3, 2
    5y2 – 7y – 6 = 0
    5y2 – 10y + 3y – 6 = 0
    So y = -3/5, 2
    Put on number line
    -3 …. -3/5….. 2
  5. I. 3x2 – 10x + 8 = 0,
    II. 3y2 + 8y – 16 = 0

    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established
    View Answer
    C) If x ≥ y
    Solution:

    3x2 – 10x + 8 = 0
    3x2 – 6x – 4x + 8 = 0
    So x = 2, 4/3
    3y2 + 8y – 16 = 0
    3y2 + 12y – 4y – 16 = 0
    So y = -4, 4/3
    Put on number line
    -4 …. 4/3…. 2
  6. I. 2x2 + 17x + 30 = 0,
    II. 2y2 + 13y + 18 = 0

    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established
    View Answer
    E) If x = y or cannot be established
    Solution:

    2x2 + 17x + 30 = 0
    2x2 + 12x + 5x + 30 = 0
    So x = -6, -5/2
    2y2 + 13y + 18 = 0
    2y2 + 4y + 9y + 18 = 0
    So y = -9/2, -2
    Put on number line
    -6 … -9/2 …. -5/2 …. -2
  7. I. 3x2 + 16x + 20 = 0,
    II. 3y2 + 8y + 4 = 0

    A) x > y
    B) x < y
    C) x ≥ y
    D) x ≤ y
    E) x = y or relationship cannot be determined
    View Answer
    D) If x ≤ y
    Solution:

    3x2 + 16x + 20 = 0
    3x2 + 6x + 10x + 20 = 0
    So x = -10/3, -2
    3y2 + 8y + 4 = 0
    3y2 + 6y + 2y + 4 = 0
    So y = -2, -2/3
    put on number line
    -10/3…. -2…. -2/3
  8. I. x2 + x – 20 = 0,
    II. 2y2 + 13y + 15 = 0

    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established
    View Answer
    E) If x = y or relation cannot be established
    Solution:

    x2 + x – 20 = 0
    (x+5)(x-4) = 0
    So x = -5, 4
    2y2 + 13y + 15 = 0
    2y2 + 10y + 3y + 15 = 0
    So y = -5, -3/2
    Put on number line
    -5…. -3/2…. 4
  9. I. 5x2 – 7x – 6 = 0,
    II. 5y2 + 23y + 12 = 0

    A) If x > y
    B) If x < y
    C) If x ≥ y
    D) If x ≤ y
    E) If x = y or relation cannot be established
    View Answer
    C) If x ≥ y
    Solution:

    5x2 – 7x – 6 = 0
    5x2 – 10x + 3x – 6 = 0
    So x = -3/5, 2
    5y2 + 23y + 12 = 0
    5y2 + 20y + 3y + 12 = 0
    So y = -4, -3/5
    Put on number line
    -4….. -3/5…. 2
  10. I. 2x2 – 9x + 4 = 0,
    II. 2y2 + 7y – 4 = 0

    A) x > y
    B) x < y
    C) x ≥ y
    D) x ≤ y
    E) x = y or relationship cannot be determined
    View Answer
    C) If x ≥ y
    Solution:

    2x2 – 9x + 4 = 0
    2x2 – 8x – x + 4 = 0
    So x = 4 , 1/2
    2y2 + 7y – 4 = 0
    2y2 + 8y – y – 4 = 0
    So y = -4, 1/2
    Put on number line
    -4……. 1/2…… 4

 

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