Quantitative Aptitude: Quadratic Equations Set 17

Quadratic Equations 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 – 25x + 52 = 0,
    II. 3y2 – 16y + 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 C
    Solution:

    3x2 – 25x + 52 = 0
    3x2 – 12x – 13x + 52 = 0
    Gives x = 4, 13/3
    3y2 – 16y + 16 = 0
    3y2 – 14y – 4y + 16 = 0
    Gives y = 4, 4/3
  2. I. 3x2 – 8x – 16 = 0,
    II. 2y2 – 5y – 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:

    3x2 – 8x – 16 = 0
    3x2 – 12x + 4x – 16 = 0
    Gives x = -4/3, 4
    2y2 – 5y – 12 = 0
    2y2 – 8y + 3y – 12 = 0
    Gives y = -3/2, 4
  3. I. 3x2 + 22x + 24 = 0,
    II. 2y2 + 7y – 4 = 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 + 22 x + 24 = 0
    3x2 + 18x + 4x + 24 = 0
    So x = -4/3, -6
    2y2 + 7y – 4 = 0
    2y2 + 8y – y – 4 = 0
    Gives y = -4, 1/2
  4. I. 2x2 – 9x + 4 = 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 D
    Solution:

    2x2 – 9x + 4 = 0
    2x2 – 8x – x + 4 = 0
    So x = 4 , 1/2
    2y2 – 17y + 36 = 0
    2y2 – 8y – 9y + 36 = 0
    Gives y= 4, 9/2
  5. I. 3x2 + 7x – 6 = 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:

    Explanation:
    3x2 + 7x – 6 = 0
    3x2 + 9x – 2x – 6 = 0
    Gives x = -3, 2/3
    3y2 – 19y + 20 = 0
    3y2 – 15y – 4y + 20 = 0
    Gives y = 4/3, 5
  6. I. 3x2 – 4x – 4 = 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:

    3x2 – 4x – 4 = 0
    3x2 – 6x + 2x – 4 = 0
    Gives x = -2/3, 2
    3y2 + 16y + 20 = 0
    3y2 + 6y + 10y + 20 = 0
    Gives y = -10/3, -2
  7. I. 7x2 + 19x – 6 = 0,
    II. 2y2 + 13y + 21 = 0
    A) x > y
    B) x < y
    C) x ≥ y
    D) x ≤ y
    E) x = y or relationship cannot be determined
    View Answer
    Option C
    Solution:

    7x2 + 19x – 6 = 0
    7x2 + 21x – 2x – 6 = 0
    Gives x = -3, 2/7
    2y2 + 13y + 21 = 0
    2y2 + 6y + 7y + 21 = 0
    So y = -7/2, -3
  8. I. x2 + (4 + √2)x + 4√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 B
    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)
    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)
  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
    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, -√2
  10. I. 4x2 – 12x + 5 = 0,
    II. 2y2 + 3y – 20 = 0
    A) x > y
    B) x < y
    C) x ≥ y
    D) x ≤ y
    E) x = y or relation cannot be established
    View Answer
    Option E
    Solution:

    4x2 – 12x + 5 = 0
    4x2 – 2x – 10x + 5 = 0
    x = 1/2, 5/2
    2y2 + 3y – 20 = 0
    2y2 + 8y – 5y – 20 = 0
    So y = -3, 5/2

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7 Thoughts to “Quantitative Aptitude: Quadratic Equations Set 17”

  1. ?@shh...ishhh..⛷?

    thnx..

  2. Pawan

    please check question 6 eq-1

  3. AVI™ (Sinchen loveR)

    8/8
    Q8 ans may be E ?

    1. Both values of x are negative. And that if y positive. So why E ??

  4. sachin shukla@ my turn 2017

    ty…

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