Quantitative Aptitude: Quadratic Equations Set 12

Type of Quadratic Equations Questions Asked in All Exams – Bank PO, Insurance and Other related Exams.

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. 3y2 – 16y + 16 = 0
    A) x > y
    B) x < y
    C) x ≥ y
    D) x ≤ y
    E) x = y or relation cannot be established
    View Answer
    Option B
    Solution:

    20x2 – 31x + 12 = 0
    20x2 – 16x – 15x + 12 = 0
    So x = 3/4, 4/5
    3y2 – 16y + 16 = 0
    3y2 – 14y – 4y + 16 = 0
    Gives y = 4, 4/3
  2. I. 3x2 + 22 x + 24 = 0
    II. 2y2 – 5y – 12 = 0
    A) x > y
    B) x < y
    C) x ≥ y
    D) x ≤ y
    E) x = y or relation cannot be established
    View Answer
    Option E
    Solution:

    3x2 + 22 x + 24 = 0
    3x2 + 18x + 4x + 24 = 0
    So x = -4/3, -6
    2y2 – 5y – 12 = 0
    2y2 – 8y + 3y – 12 = 0
    Gives y = -3/2, 4
  3. I. 2x2 – 9x + 4 = 0
    II. 4y2 – 13y – 12 = 0
    A) x > y
    B) x < y
    C) x ≥ y
    D) x ≤ y
    E) x = y or relation cannot be established
    View Answer
    Option E
    Solution:

    Solution:
    2x2 – 9x + 4 = 0
    2x2 – 8x – x + 4 = 0
    So x = 4 , 1/2
    4y2 – 13y – 12 = 0
    4y2 – 16y + 3y – 12 = 0
    Gives y = -3/4, 4
  4. I. 5x2 + 23x + 12 = 0
    II. 5y2 – 7y – 6 = 0
    A) x > y
    B) x < y
    C) x ≥ y
    D) x ≤ y
    E) x = y or relation cannot be established
    View Answer
    Option D
    Solution:

    5x2 + 23x + 12 = 0
    5x2 + 20x + 3x + 12 = 0
    So x = -4, -3/5
    5y2 – 7y – 6 = 0
    5y2 – 10y + 3y – 6 = 0
    So y = -3/5, 2
    Put all values on number line and analyze the relationship
    -4….. -3/5…. 2
  5. I. 7x2 + 19x – 6 = 0,
    II. 2y2 – 7y + 3 = 0
    A) x > y
    B) x < y
    C) x ≥ y
    D) x ≤ y
    E) x = y or relation cannot be established
    View Answer
    Option B
    Solution:

    7x2 + 19x – 6 = 0
    7x2 + 21x – 2x – 6 = 0
    Gives x = -3, 2/7
    2y2 – 7y + 3 = 0
    2y2 – 6y – y + 3 = 0
    So y = 1/2, 3
  6. I. 4x2 – 12x + 5 = 0,
    II. 2y2 – 19y + 35 = 0
    A) x > y
    B) x < y
    C) x ≥ y
    D) x ≤ y
    E) x = y or relation cannot be established
    View Answer
    Option D
    4x2 – 12x + 5 = 0
    4x2 – 2x – 10x + 5 = 0
    x = 1/2, 5/2
    2y2 – 19y + 35 = 0
    2y2 – 14y – 5y + 35 = 0
    So y = 5/2, 7
  7. I. 2x2 + 5x ¬– 12 = 0,
    II. 4y2 + 13y – 12 = 0
    A) x > y
    B) x < y
    C) x ≥ y
    D) x ≤ y
    E) x = y or relation cannot be established
    View Answer
    Option E
    2x2 + 5x ¬– 12 = 0
    2x2 + 8x ¬– 3x – 12 = 0
    So x = -4 , 3/2
    4y2 + 13y – 12 = 0
    4y2 + 16y – 3y – 12 = 0
    y = -4, 3/4
  8. I. 3x2 + 22x + 24 = 0,
    II. 4y2 – 9y – 9 = 0
    A) x > y
    B) x < y
    C) x ≥ y
    D) x ≤ y
    E) x = y or relation cannot be established
    View Answer
    Option B
    3x2 + 22 x + 24 = 0
    3x2 + 18x + 4x + 24 = 0
    Gives x = -4/3, -6
    4y2 – 9y – 9 = 0
    4y2 – 12y + 3y – 9 = 0
    y = -3/4, 3
  9. I. 20x2 – 31x + 12 = 0,
    II. 4y2 + 9y – 9 = 0
    A) x > y
    B) x < y
    C) x ≥ y
    D) x ≤ y
    E) x = y or relation cannot be established
    View Answer
    Option C
    20x2 – 31x + 12 = 0
    20x2 – 16x – 15x + 12 = 0
    Gives x = 3/4, 4/5
    4y2 + 9y – 9 = 0
    4y2 + 12y – 3y – 9 = 0
    y = 3/4, -3
  10. I. 6x2 – 7x – 3 = 0,
    II. 4y2 + 5y – 6 = 0
    A) x > y
    B) x < y
    C) x ≥ y
    D) x ≤ y
    E) x = y or relation cannot be established
    View Answer
    Option E
    Solution:

    6x2 – 7x – 3 = 0
    6x2 + 2x – 9x – 3 = 0
    Gives x = -1/3, 3/2
    4y2 + 5y – 6 = 0
    4y2 + 8y – 3y – 6 = 0
    Gives y = -2, 3/4

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