Quantitative Aptitude: Quadratic Equations Set 16

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. 4x2 + 5x – 6 = 0,
    II. 2y2 + 11y + 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
    Option E
    Solution:

    4x2 + 5x – 6 = 0
    4x2 + 8x – 3x – 6 = 0
    Gives x = -2, 3/4
    2y2 + 11y + 12 = 0
    2y2 + 8y + 3y + 12 = 0
    Gives y = -4, -3/2
  2. I. 12x2 – 49x + 30 = 0,
    II. 6y2 – 35y + 50 = 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:

    12x2 – 49x + 30 = 0
    12x2 – 9x – 40x + 30 = 0
    Gives x = 3/4, 10/3
    6y2 – 35y + 50 = 0
    6y2 – 15y – 20y + 50 = 0
    Gives y = 5/2, 10/3
  3. I. 4x2 + 13x + 10 = 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 D
    Solution:

    4x2 + 13x + 10 = 0
    4x2 + 8x + 5x + 10 = 0
    Gives x = -2, -5/4
    4y2 – 7y – 15 = 0
    4y2 – 12y + 5y – 15 = 0
    Gives y = -5/4, 3
  4. I. 12x2 – 5x – 3 = 0,
    II. 6y2 + y – 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:

    12x2 – 5x – 3 = 0
    12x2 + 4x – 9x – 3 = 0
    Gives x = -1/3, 3/4
    6y2 + y – 2 = 0
    6y2 – 3y + 4y – 2 = 0
    Gives y= -2/3, 1/2
  5. I. 3x2 + 7x – 6 = 0,
    II. 3y2 – 11y + 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 D
    Solution:

    Explanation:
    3x2 + 7x – 6 = 0
    3x2 + 9x – 2x – 6 = 0
    Gives x = -3, 2/3
    3y2 – 11y + 6 = 0
    3y2 – 9y – 2y + 6 = 0
    Gives y = 2/3, 3
  6. I. 5x2 – 36x – 32 = 0,
    II. 3y2 + 16y + 20 = 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:

    5x2 – 36x – 32 = 0
    5x2 + 4x – 40x – 32 = 0
    Gives x = -4/5, 8
    3y2 + 16y + 20 = 0
    3y2 + 6y + 10y + 20 = 0
    Gives y = -10/3, -2
  7. I. 3x2 – (6 + 2√3)x + 4√3 = 0,
    II. 3y2 – (2 + 3√3)y + 2√3 = 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:

    3x2 – (6 + 2√3)x + 4√3 = 0
    (3x2 – 6x) – (2√3x – 4√3) = 0
    3x (x- 2) – 2√3 (x – 2) = 0,
    So x = 2, 2√3/3 (1.15)
    3y2 – (2 + 3√3)y + 2√3 = 0
    (3y2 – 2y) – (3√3y – 2√3) = 0
    y (3y – 2) – √3 (3y – 2) = 0
    So x = 2/3, √3 (1.73)
  8. I. 2x2 + (4 + √2)x + 2√2 = 0
    II. y2 – (1 + 3√3)y + 3√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 B
    Solution:

    2x2 + (4 + √2)x + 2√2 = 0
    (2x2 + 4x) + (√2x + 2√2) = 0
    2x (x + 2) + √2 (x + 2) = 0
    So x = -2, -√2/2 (-0.7)
    y2 – (1 + 3√3)y + 3√3 = 0
    (y2 – y) – (3√3y – 3√3) = 0
    y (y – 1) – 3√3 (y – 1) = 0
    So, y = 1, 3√3 (5.2)
  9. I. x2 + (3 + 2√2)x + 6√2 = 0
    II. 5y2 – (1 + 5√2)y + √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 B
    Solution:

    x2 + (3 + 2√2)x + 6√2 = 0
    (x2 + 3x) + (2√2x + 6√2) = 0
    x (x + 3) + 2√2 (x + 3) = 0
    So x = -3, -2√2
    5y2 – (1 + 5√2)y + √2 = 0
    (5y2 – y) – (5√2y – √2) = 0
    y (5y – 1) – 3√2 (5y – 1) = 0
    So, y = 1/5, 3√2
  10. I. 2x2 + (4 + 2√6)x + 4√6 = 0
    II. 5y2 + (10 + √6)y + 2√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 D
    Solution:

    2x2 + (4 + 2√6)x + 4√6 = 0
    (2x2 + 4x) + (2√6x + 4√6) = 0
    2x (x + 2) + 2√6 (x + 2) = 0
    So x = -2, -√6
    5y2 + (10 + √6)y + 2√6 = 0
    (5y2 + 10y) + (√6y + 2√6) = 0
    5y (y + 2) + √6 (y + 2) = 0
    So, y = -2, -√6/5

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38 Thoughts to “Quantitative Aptitude: Quadratic Equations Set 16”

  1. Shilpa Thakur

    q 1 ????????// kya answer hi hoga?????????// -2 is < -1.5 so e nhi hoga kya

    1. Yes its E
      Alternate roots

      1. Shilpa Thakur

        okk////////// thank you

  2. Divaker

    some questions are confusing but solvable thx

  3. Sunil Sharma

    question 10 . answer should be D part. x<=y

  4. sachin shukla@ my turn 2017

    ty)))

  5. Deepak sahu

    Mam 10nth question me E Option hoga plz chech kariye

    1. D hi hai

      Root 6 is -2.4
      -2 will come in middle

  6. MiMi

    thnk u mam :)))

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