Quantitative Aptitude: Quadratic Equations Set 13

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 – 5y + 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
    Option E
    Solution:

    20x2 – 31x + 12 = 0
    20x2 – 16x – 15x + 12 = 0
    So x = 3/4, 4/5
    3y2 – 5y + 2 = 0
    3y2 – 3y – 2y + 2 = 0
    So y = 1, 2/3
  2. I. 3x2 – 19x + 30 = 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
    Option C
    Solution:

    3x2 – 19x + 30 = 0
    3x2 – 9x – 10x + 30 = 0
    So x = 3, 10/3
    3y2 – 10y + 3 = 0
    3y2 – 9y – y + 3 = 0
    So y = 1/3, 3
  3. I. 3x2 – 25x + 52 = 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
    Option A
    Solution:

    3x2 – 25x + 52 = 0
    3x2 – 12x – 13x + 52 = 0
    So x = 4, 13/3
    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. 4x2 – 5x – 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
    Option E
    Solution:

    4x2 – 5x – 6 = 0
    4x2 – 8x + 3x – 6 = 0
    So x = -3/4, 2
    5y2 – 7y – 6 = 0
    5y2 – 10y + 3y – 6 = 0
    So y = -3/5, 2
  5. I. 3x2 – 10x + 8 = 0,
    II. 3y2 – 14y + 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
    Option D
    Solution:

    3x2 – 10x + 8 = 0
    3x2 – 6x – 4x + 8 = 0
    So x = 2, 4/3
    3y2 – 14y + 16 = 0
    3y2 – 6y – 8y + 16 = 0
    So y = 2, 8/3
  6. I. 2x2 + 17x + 30 = 0,
    II. 4y2 – 7y – 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
    Option B
    Solution:

    2x2 + 17x + 30 = 0
    2x2 + 12x + 5x + 30 = 0
    So x = -6, -5/2
    4y2 – 7y – 15 = 0
    4y2 – 12y + 5y – 15 = 0
    So y = -5/4, 3
  7. I. 3x2 + 16x + 20 = 0,
    II. 5y2 + 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
    Option D
    Solution:

    3x2 + 16x + 20 = 0
    3x2 + 6x + 10x + 20 = 0
    So x = -10/3, -2
    5y2 + 8y – 4 = 0
    5y2 + 10y – 2y – 4 = 0
    So y = -2, 2/5
  8. I. 2x2 + 17x + 21 = 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
    Option E
    Solution:

    2x2 + 17x + 21 = 0
    2x2 + 14x + 3x + 21 = 0
    So x = -7, -3/2
    2y2 + 13y + 15 = 0
    2y2 + 10y + 3y + 15 = 0
    So y = -5, -3/2
  9. I. 5x2 – 7x – 6 = 0,
    II. 3y2 – 19y + 28 = 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
    Option B
    Solution:

    5x2 – 7x – 6 = 0
    5x2 – 10x + 3x – 6 = 0
    So x = -3/5, 2
    3y2 – 19y + 28 = 0
    3y2 – 12y – 7y + 28 = 0
    So y = 7/3, 4
  10. I. 8x2 + 6x – 5 = 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
    Option E
    Solution:

    8x2 + 6x – 5 = 0
    8x2 – 4x + 10x – 5 = 0
    So x = -5/4, 1/2
    2y2 + 7y – 4 = 0
    2y2 + 8y – y – 4 = 0
    So y = -4, 1/2

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