Quantitative Aptitude: Quadratic Equations Set 18

Quadratic Equations Practice Sets SBI PO, IBPS PO, NIACL, NICL, LIC, Dena Bank PO PGDBF, BOI, Bank of Baroda and other competitive 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. 3x2 – 13x – 10 = 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 A
    Solution:

    3x2 – 13x – 10 = 0
    3x2 – 15x + 2x – 10 = 0
    Gives x = -2/3, 5
    2y2 + 11y + 12 = 0
    2y2 + 8y + 3y + 12 = 0
    Gives y = -4, -3/2
    Put all values on number line and analyze the relationship
    -4…. -3/2….-2/3….. 5
  2. I. 3x2 – 8x – 16 = 0,
    II. 3y2 – 26y + 56 = 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 – 8x – 16 = 0
    3x2 – 12x + 4x – 16 = 0
    Gives x = -4/3, 4
    3y2 – 26y + 56 = 0
    3y2 – 26y + 56 = 0
    Gives y = 4, 14/3
  3. I. 3x2 + 10x – 8 = 0,
    II. 3y2 + 10y + 8 = 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:

    3x2 + 10x – 8 = 0
    3x2 + 12x – 2x – 8 = 0
    Gives x = -2, 2/3
    3y2 + 10y + 8 = 0
    3y2 + 6y + 4y + 8 = 0
    Gives y = -2, -4/3
  4. I. 3x2 – 25x + 52 = 0,
    II. 2y2 – 17y + 36 = 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:

    3x2 – 25x + 52 = 0
    3x2 – 12x – 13x + 52 = 0
    Gives x = 4, 13/3
    2y2 – 17y + 36 = 0
    2y2 – 8y – 9y + 36 = 0
    Gives y= 4, 9/2
  5. I. 3x2 + 7x – 6 = 0,
    II. 4y2 – 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 B
    Solution:

    3x2 + 7x – 6 = 0
    3x2 + 9x – 2x – 6 = 0
    Gives x = -3, 2/3
    4y2 – 11y + 6 = 0
    4y2 – 8y – 3y + 6 = 0
    Gives y= 3/4, 2
  6. I. 2x2 – 3x – 14 = 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 C
    Solution:

    2x2 – 3x – 14 = 0
    2x2 + 4x – 7x – 14 = 0
    Gives x = -2, 7/2
    3y2 + 16y + 20 = 0
    3y2 + 6y + 10y + 20 = 0
    Gives y = -10/3, -2
  7. I. 7x2 + 19x – 6 = 0,
    II. 3y2 – 8y – 16 = 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:

    7x2 + 19x – 6 = 0
    7x2 + 21x – 2x – 6 = 0
    Gives x = -3, 2/7
    3y2 – 8y – 16 = 0
    3y2 – 12y + 4y – 16 = 0
    So y = -4/3, 4
  8. I. 8x2 + (4 + 2√2)x + √2 = 0
    II. 3y2 – (6 + 2√3)y + 4√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:

    8x2 + (4 + 2√2)x + √2 = 0
    (8x2 + 4x) + (2√2x + √2) = 0
    4x (2x + 1) + √2 (2x + 1) = 0
    So x = -1/2 (-0.5), -√2/4 (-0.35)
    3y2 – (6 + 2√3)y + 4√3 = 0
    (3y2 – 6y) – (2√3y – 4√3) = 0
    3y (y – 2) – 2√3 (y – 2) = 0
    So, y = 2, 2√3/3
  9. I. 3x2 – (3 – 2√2)x – 2√2 = 0
    II. 5y2 + (2 + 5√2)y + 2√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:

    3x2 – (3 – 2√2)x – 3√2 = 0
    (3x2 – 3x) + (2√2x – 2√2) = 0
    3x (x – 1) + 2√2 (x – 1) = 0
    So x = 1, -2√2/3 (-0.9)
    5y2 + (2 + 5√2)y + 2√2 = 0
    (5y2 + 2y) + (5√2y + 2√2) = 0
    y (5y + 2) + √2 (5y + 2) = 0
    So, y = -2/5 (-0.4), -√2 (-1.4)
  10. I. 6x2 – (3 + 4√3)x + 2√3 = 0,
    II. y2 – (3 + √3)y + 3√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 B
    Solution:

    6x2 – (3 + 4√3)x + 2√3 = 0
    (6x2 – 3x) – (4√3x – 2√3) = 0
    3x (2x- 1) – 2√3 (x – 2) = 0,
    So x = 1/2, 2√3/3 (1.15)
    y2 – (3 + √3)y + 3√3 = 0
    (y2 – 3y) – (√3y – 3√3) = 0
    y (y – 3) – √3 (y – 3) = 0
    So x = 3, √3 (1.73)

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2 Thoughts to “Quantitative Aptitude: Quadratic Equations Set 18”

  1. Vivek Murugesan

    gd ques selection
    thanks

  2. Ayushi

    9th que is good … thanku

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