Quantitative Aptitude: Quadratic Equations Set 27

Quadratic Equations Practice Sets for IBPS PO, IBPS RRB, LIC, UIICL, OICL, 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. 4x2 – x – 14 = 0,
    II. 2y2 – 13y + 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: 

    4x2 – x – 14 = 0
    4x2 – 8x + 7x – 14 = 0
    Gives x = -7/4, 2
    2y2 – 13y + 20 = 0
    2y2 – 8y – 5y + 20 = 0
    Gives y = 5/2, 4
  2. I. 2x2 – 11x + 14 = 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 – 11x + 14 = 0
    2x2 – 4x – 7x + 14 = 0
    Gives x = 2, 7/2
    3y2 + 13y + 14 = 0
    3y2 + 6y + 7y+ 14 = 0
    Gives y = -7/3, -2
  3. I. 3x2 + 14x + 15 = 0,
    II. 3y2 – 13y – 30 = 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 + 14x + 15 = 0
    3x2 + 9x + 5x + 15 = 0
    Gives x = -5/3, 6
    3y2 – 13y – 30 = 0
    3y2 – 18y + 5y – 30 = 0
    Gives y = -3, -5/3
  4. I. 3x2 + 28x + 60 = 0,
    II. 2y2 – 3y – 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: 

    3x2 + 28x + 60 = 0
    3x2 + 18x + 10x + 60 = 0
    Gives x = -6, -10/3
    2y2 – 3y – 20 = 0
    2y2 – 8y + 5y – 20 = 0
    Gives y= -5/2, 4
  5. I. 3x2 – 8x – 35 = 0,
    II. 3y2 + 37y + 104 = 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 – 8x – 35 = 0
    3x2 – 15x + 7x – 35 = 0
    Gives x = 5, -7/3
    3y2 + 37y + 104 = 0
    3y2 + 24y + 13y + 104 = 0
    Gives y= -8, -13/3
  6. I. 3x2 – 5x – 78 = 0,
    II. 3y2 + 28y + 65 = 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 – 78 = 0
    3x2 – 18x + 13x – 78 = 0
    Gives x = -13/3, 6
    3y2 + 28y + 65 = 0
    3y2 + 15y + 13y + 65 = 0
    Gives y = -5, -13/3
  7. I. 3x2 – 7x – 40 = 0,
    II. 3y2 + 26y + 48 = 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 – 7x – 40 = 0
    3x2 – 15x + 8x – 40 = 0
    Gives x = -8/3, 5
    3y2 + 26y + 48 = 0
    3y2 + 18y + 8y + 48 = 0
    Gives y = -6, -8/3
  8. I. x2 + (4 + 2√2)x + 8√2 = 0
    II. 3y2 + (3 + 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 B
    Solution:

    x2 + (4 + 2√2)x + 8√2 = 0
    (x2 + 4x) + (2√2x + 8√2) = 0
    x (x + 4) + 2√2 (x + 4) = 0
    So x = -4, -2√2 (-2.8)
    3y2 + (3 + 3√2)y + 3√2 = 0
    (3y2 + 3y) + (3√2y + 3√2) = 0
    3y (y + 1) + 3√2 (y + 1) = 0
    So, y = -1, -√2 (-1.41)
  9. I. 6x2 – (3 + 4√3)x + 2√3 = 0,
    II. 4y2 – (2 + 4√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:

    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 (0.5), 2√3/3 (1.16)
    4y2 – (2 + 4√3)y + 2√3 = 0
    (4y2 – 2y) – (4√3y – 2√3) = 0
    2y (2y – 1) – 2√3 (2y – 1) = 0
    So, y = 1/2 (0.5), √3 (1.73)
  10. I. x2 + (4 + 2√2)x + 8√2 = 0
    II. y2 – (2 + 3√3)y + 6√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:

    x2 + (4 + 2√2)x + 8√2 = 0
    (x2 + 4x) + (2√2x + 8√2) = 0
    x (x + 4) + 2√2 (x + 4) = 0
    So x = -4, -2√2 (-2.82)
    y2 – (2 + 3√3)y + 6√3 = 0
    (y2 – 2y) – (3√3y – 6√3) = 0
    y (y – 2) – 3√3 (y – 2) = 0
    So y = 2, 3√3 (5.2)

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10 Thoughts to “Quantitative Aptitude: Quadratic Equations Set 27”

  1. Sandy

    @MOD
    qsn 8
    2nd equation mai y ki value -1 & -root2 ayegi….
    So overall X<Y ans hoga…
    Do check

  2. ?Jab Harry Met Sejal?

    sir
    1st Ques me
    2*14 how we get factors 8 /2 or 5/2
    may be instead of 14 there is 20
    or instead of 13 there is 11
    2y^2 -13Y +14 = 0

  3. ครђ

    Thanks ma’am nice set :))

  4. sonali

    ty AZ…nice ques..

  5. cute smart innocent bacha

    tq mam

  6. swati

    thanks a lot team AZ

  7. ~lonely~hanker~nitrous oxide~

    tq mam

  8. kach Kach Baccha

    tq az

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

    done

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