Quantitative Aptitude: Quadratic Equations Set 15

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. 2y2 – 13y + 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 + 10x – 8 = 0
    3x2 + 12x – 2x – 8 = 0
    Gives x = -2, 2/3
    2y2 – 13y + 6 = 0
    2y2 – 12y – y + 6 = 0
    Gives y = 1/2, 6
  2. I. 16x2 + 8x – 15 = 0,
    II. 4y2 + 29y + 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 C
    Solution:

    16x2 + 8x – 15 = 0
    16x2 + 20x – 12x – 15 = 0
    Gives x = -5/4, 3/4
    4y2 + 29y + 30 = 0
    4y2 + 24y + 5y + 30 = 0
    Gives y = -6, -5/4
  3. I. 3x2 – 25x + 52 = 0,
    II. 15y2 – 38y – 40 = 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 – 25x + 52 = 0
    3x2 – 12x – 13x + 52 = 0
    Gives x = 4, 13/3
    15y2 – 38y – 40 = 0
    15y2 + 12y – 50y – 40 = 0
    Gives y = -4/5, 10/3
  4. I. 12x2 – 5x – 3 = 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 D
    Solution:

    12x2 – 5x – 3 = 0
    12x2 + 4x – 9x – 3 = 0
    Gives x = -1/3, 3/4
    4y2 – 11y + 6 = 0
    4y2 – 8y – 3y + 6 = 0
    Gives y= 3/4, 2
  5. I. 3x2 + 7x – 6 = 0,
    II. 6y2 – y – 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:

    Explanation:
    3x2 + 7x – 6 = 0
    3x2 + 9x – 2x – 6 = 0
    Gives x = -3, 2/3
    6y2 – y – 2 = 0
    6y2 + 3y – 4y – 2 = 0
    Gives y = -1/2, 2/3
    Put on number line
    -3… -1/2 … 2/3
  6. I. 4x2 + 15x + 9 = 0,
    II. 4y2 – 13y – 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:

    4x2 + 15x + 9 = 0
    4x2 + 12x + 3x + 9 = 0
    Gives x = -3, -3/4
    4y2 – 13y – 12 = 0
    4y2 – 16y + 3y – 12 = 0
    Gives y = -3/4, 4
  7. I. 2x2 – (6 + √3)x + 3√3 = 0,
    II. 3y2 – (9 + √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 E
    Solution:

    2x2 – 6x – √3x + 3√3 = 0
    2x (x- 3) – √3 (x – 3) = 0,
    So x = 3, √3/2 (0.7)
    3y2 – 9y – √3y + 3√3 = 0
    3y (y – 3) – √3 (y – 3) = 0
    So x = 3, √3/3 (0.6)
  8. I. 2x2 – (2 + 2√5)x + 2√5 = 0
    II. 4y2 – (6 + 2√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 – 2x – 2√5x + 2√5 = 0
    2x (x – 1) – 2√5 (x – 1) = 0
    So x = 1, √5 (2.2)
    4y2 – 6y – 2√2y + 3√2 = 0
    2y (2y – 3) – √2 (2y – 3) = 0
    So, y = 3/2, 1/√2 (0.7)
  9. I. 2x2 – 15√3x + 84 = 0,
    II. 3y2 – 2y – 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 A
    Solution:

    2x2 – 15√3x + 84 = 0
    Now multiply 2 and 84 = 168
    we have √3 in equation, so divide, 168/3 = 56
    Now make factors so as by multiply you get 56, and by addition or subtraction you get –15
    we have factors (-8) and (-7)
    So 2x2 – 15√3x + 84 = 0
    gives
    2x2 – 8√3x – 7√3x + 84 = 0
    2x (x – 4√3) – 7√3 (x – 4√3x) = 0
    So x = 3.5√3, 4√3
    3y2 – 2y – 8 = 0
    3y2 – 6y + 4y – 8 = 0
    So y = -4/3, 1
    Plot on number line
    -4/3……1….. 3.5√3….. 4√3
  10. I. 16x2 + 20x + 6 = 0
    II. 10y2 + 38y + 24 = 0
    A) x > y
    B) x < y
    C) x ≥ y
    D) x ≤ y
    E) x = y or relationship cannot be determined
    View Answer
    Option A
    Solution:

    Divide both equations by 2
    8x2 + 10x + 3 = 0
    8x2 + 4x + 6x + 3 = 0
    Gives x = -1/2, -3/4
    5y2 + 19y + 12 = 0
    5y2 + 15y + 4y + 12 = 0
    Gives y = -4, -4/5

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53 Thoughts to “Quantitative Aptitude: Quadratic Equations Set 15”

  1. (¯·.¸¸.-> °º åɱβȋ º° <-.¸¸.·¯)

    ty mam 🙂

  2. (¯·.¸¸.-> °º åɱβȋ º° <-.¸¸.·¯)

    @Shubhra_AspirantsZone:disqus mam…… in 10th ques ans shldn’t be A???

      1. (¯·.¸¸.-> °º åɱβȋ º° <-.¸¸.·¯)

        ty mam 🙂

  3. MITTHU

    10?? th how

  4. Shilpa Thakur

    10th?????????????????????//

      1. wont rest until i get success

        I. 2×2 – (2 + 2√5)x + 2√5 = 0
        II. 4y2 – (6 + 2√2)y + 3√2 = 0
        MAAM AISE QUESTIONS KO SIMPLIFY KARNA PADEGA USKE BAAD HI SOLVE KARNA HOGA
        YA KOI SHORT TRICK HAI SOLVE KARNE KA?

        1. U know the formula of discriminant method?

          1. wont rest until i get success

            NO MAAM BATA DJIYE
            SHAYAD PATA HO PAR NAAM SUNKE YAAD NHI AARHA

      2. wont rest until i get success

        MAAM Q-9 TYPE K AUR QSNS HAI?

          1. wont rest until i get success

            OKAY MAAM
            THANK YOU VERY MUCH:)

  5. purvi

    nice ques mam
    thanku

  6. Deepak sahu

    Thank you mam……….

  7. COLD BLOOD ;)

    tyy mam:)

  8. sachin shukla@ my turn 2017

    ty..

  9. Ayushi

    ty mam…. 7 n 8 are easy to solve but one has to noe about roots till 10

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