Quantitative Aptitude: Quadratic Equations Set 22

Quadratic Equations Practice Sets for IBPS PO, NICL, NIACL, 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 + 10x – 8 = 0,
    II. 3y2 – 20y + 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 D
    Solution: 

    3x2 + 10x – 8 = 0
    3x2 + 12x  – 2x + 8 = 0
    Gives x = -4, 2/3
    3y2 – 20y + 12 = 0
    3y2 – 18y – 2y + 12 = 0
    Gives y = 2/3, 6
  2. I. 4x2 – 23x + 15 = 0,
    II. 4y2 + 9y – 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 C
    Solution: 

    4x2 – 23x + 15 = 0
    4x2 – 20x -3x + 15 = 0
    Gives x = 3/4, 5
    4y2 + 9y – 9 = 0
    4y2 + 12y – 3y – 9 = 0
    Gives y = -3, 3/4
  3. I. 5x2 – 13x – 6 = 0,
    II. 5y2 – 18y – 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: 

    x = -2/5, 3
    y = -2/5, 4
  4. I. 2x2 + 7x – 4 = 0,
    II. 3y2 – 19y + 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 B
    Solution: 

    2x2 + 7x – 4 = 0
    Gives x = -4, 1/2
    3y2 – 19y + 20 = 0
    Gives y= 4/3, 5
  5. I. 2x2 – 3x – 9 = 0,
    II. 3y2 + 13y + 14 = 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: 

    2x2 – 3x – 9 = 0
    Gives x = -3/2, 3
    3y2 + 13y + 14 = 0
    Gives y= -7/3, -2 
  6. I. 3x2 – x – 10 = 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 E
    Solution: 

    3x2 – x – 10 = 0
    Gives x = -5/3, 2
    3y2 – 11y + 6 = 0
    Gives y = 2/3, 3
  7. I. 3x2 – (3 – 2√2)x – 2√2 = 0
    II. 3y2 – (1 + 3√3)y + √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 E
    Solution:

    3x2 – (3 – 2√2)x – 2√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)
    3y2 – (1 + 3√3)y + √3 = 0
    (3y2 – y) – (3√3y – √3) = 0
    y (3y – 1) – √3 (3y – 1) = 0
    So, y = 1/3, √3 (1.7)
  8. I. x2 + (4 + √2)x + 4√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 D
    Solution:

    x2 + (4 + √2)x + 4√2 = 0
    (x2 + 4x) + (√2x + 4√2) = 0
    x (x + 4) + √2 (x + 4) = 0
    So x = -4, -√2 (-1.4)
    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)
  9. I. 6x2 – (3 + 4√3)x + 2√3 = 0,
    II. 3y2 – (6 + 2√3)y + 4√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 D
    Solution:

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

    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)
    y2 – (3 + √3)y + 3√3 = 0
    (y2 – 3y) – (√3y – 3√3) = 0
    y (y – 3) – √3 (y – 3) = 0
    So y = 3, √3 (1.73)

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