Quantitative Aptitude: Quadratic Equations Set 25

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 – 5x – 12 = 0,
    II. 3y2 + 22y + 24 = 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 – 5x – 12 = 0
    3x2 – 9x + 4x – 12 = 0
    Gives x = -4/3, 3
    3y2 + 22y + 24 = 0
    3y2 + 18y + 4y + 24 = 0
    Gives y = -6, -4/3
  2. I. 3x2 – 2x – 8 = 0,
    II. 3y2 – 8y – 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 E
    Solution: 

    3x2 – 2x – 8 = 0
    3x2 – 6x + 4x – 8 = 0
    Gives x = -4/3, 2 
    3y2 – 8y – 16 = 0
    3y2 – 12y + 4y – 16 = 0
    Gives y = -4/3, 1
  3. I. 3x2 – 7x – 6 = 0,
    II. 3y2 + 20y + 25 = 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 – 7x – 6 = 0
    3x2 – 9x + 2x – 6 = 0
    Gives x = -2/3, 3
    3y2 + 20y + 25 = 0
    3y2 + 15y + 5y + 25 = 0
    Gives y = -5, -5/3
  4. I. 4x2 + 13x – 12 = 0,
    II. 3y2 – 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 + 13x – 12 = 0
    4x2 + 16x – 3x – 12 = 0
    Gives x = -4, 3/4
    3y2 – 7y – 6 = 0
    3y2 – 9y + 2y – 6 = 0
    Gives y= -2/3, 3
  5. I. 3x2 + 20x + 32 = 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 + 20x + 32 = 0
    3x2 + 12x + 8x + 32 = 0
    Gives x = -4, -8/3
    3y2 + 14y + 16 = 0
    3y2 + 6y + 8y + 16 = 0
    Gives y= -8/3, -2
  6. I. 2x2 – x – 15 = 0,
    II. 3y2 – 25y + 52 = 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 – x – 15 = 0
    2x2 – 6x + 5x – 15 = 0
    Gives x = -5/2, 3
    3y2 – 25y + 52 = 0
    3y2 – 12y – 13y + 52 = 0
    Gives y = 4, 13/3
  7. I. 4x2 – 9x – 28 = 0,
    II. 4y2 + 19y + 21 = 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 – 9x – 28 = 0
    4x2 – 16x + 7x – 28 = 0
    Gives x = -7/4, 4
    4y2 + 19y + 21 = 0
    4y2 + 12y +7y + 21 = 0
    Gives y = -3, -7/4
  8. I. 2x2 + (4 + 2√2)x + 4√2 = 0
    II. y2 + (3 + √2)y + 3√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:

    2x2 + (4 + 2√2)x + 4√2 = 0
    (2x2 + 4x) + (2√2x + 4√2) = 0
    2x (x + 2) + 2√2 (x + 2) = 0
    So x = -2, -√2 (-1.4)
    y2 + (3 + √2)y + 3√2 = 0
    (y2 + 3y) + (√2y + 3√2) = 0
    y (y + 3) + √2 (y + 3) = 0
    So, y = -3 (-0.4), -√2 (-1.4)
  9. I. 3x2 – (1 + 6√3)x + 2√3 = 0,
    II. 4y2 – (2 + 2√3)y + √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 – (1 + 6√3)x + 2√3 = 0
    (3x2 – x) – (6√3x – 2√3) = 0
    x (3x- 1) – 2√3 (3x – 1) = 0,
    So x = 1/3, 2√3 (3.5)
    4y2 – (2 + 2√3)y + √3 = 0
    (4y2 – 2y) – (2√3y – √3) = 0
    2y (2y – 1) – √3 (2y – 1) = 0
    So, y = 1/2, √3/2 (0.86)
  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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14 Thoughts to “Quantitative Aptitude: Quadratic Equations Set 25”

  1. Sachin Shukla

    3×2 – 7x – 6 = 0,
    II. 3y2 + 20y + 25 = 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:
    3×2 – 7x – 6 = 0
    3×2 – 9x + 2x – 6 = 0
    Gives x = -2/3, 3
    3y2 + 20y + 25 = 0
    3y2 + 15y + 5y + 25 = 0
    Gives y = -5, -5/3

    1. thinker

      X bda hona chahiye na question wrong h

  2. Sachin Shukla

    10/10 ty:))

  3. M@nish...

    in Q 3 ans should be x>y

  4. Jeetesh Chandra

    9/10..

  5. AVI™ (Sinchen loveR)

    in Q2 y≥x hona chaheye
    X = +2, -4/3
    Y=+4,-4/3

    in Q3 X≥Y hona chaheye
    x=+3, -2/3
    y=-5,-5/3
    9/10 ty,,ty,,

  6. Randy Orton™????

    100% thnx az

  7. Bahot ho gaya

    Thanks 🙂 love this website

  8. ~lonely~hanker~nitrous oxide~

    tq mam

  9. Chotu D(mind) !! {NONSENSE}

    done

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