Quant Questions Bank PO

Quantitative Aptitude: Quadratic Equations Set 24

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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. 5x2  – 18x + 9 =0
    II. 3y2 + 5y – 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 A
    Solution:
    5x2   – 18x + 9  = 0
    => 5x2  – 15x – 3x  + 9 =0
    => (5x – 3 )(x-3 )= 0
    => x=  3/ 5 or  x= 3
     3y2 + 5y – 2 = 0
    => 3y2  + 6y – y -2 = 0
    => (3y-1)(y + 2) = 0
    => y = 1/3  or  -2
  2. I. √x  – √6 / √x = 0
    II. y3 – 63/2  = 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: 
    √x  –   √6 / √x = 0
    x – √6 = 0
    x = √6
    y3   – 6 (3/2 )  = 0
    =>y3  = (√6)3
    => y = √6
  3. I. (625)1/4x + √1225  = 155
    II. √196y + 13  = 279
    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: 
    5x + 35 = 155
    => 5x = 155 – 35
    => x = 120/ 5 = 24√196 y  + 13 = 279
    => 14y = 279 -13
    => y = 266/14 =19
  4. I. 3x2 – 17x + 24 =0
    II. 4y2  – 15y + 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: 
    3x2  -17x  + 24 =0
    => 3x2 – 9x  – 8x +24 = 0
    => (3x- 8)(x-3) = 0
    =>  x=  8/3 or 3
    4y2  – 15y + 14 = 0
    => 4y2  – 8y -7y +14 = 0
    => (4y -7 )(y-2)=0
    => y = 7 /4 or  2
  5. I. x2 – 2x – √5 x + 2√5 = 0
    II. y2 – √3 y – √2 y + √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 A
    Solution: 
    x2  – 2x- √5 x + 2√5 = 0
    => x(x-2 ) – √5 (x-2 )= 0
    => (x-2)(x-√5)=0
    => x= 2 0r √5
    y2  – √3 y  – √2 y + √6 =0
    => y(y-√3) – √2(y – √3) = 0
    => (y – √2)(y- √3) =0
    => y = √2 or √3
  6. I. 5x2 – 23x + 12 = 0,
    II. 5y2 – 28y + 15 = 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 = 3/5, 4
    y = 3/5, 5
  7. I. 6x2 + 5x – 6 = 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 D
    Solution:
    x = 2/3, 3
    y = -3/2, 2/3
  8. I. 3x2 – 5x – 12 = 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 B
    Solution:
    x = -4/3, 3
    y = 4, 9/2
  9. I. 8x2 – (4 + 4√3)x + 2√3 = 0,
    II. 3y2 – (4 + 3√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 B
    Solution:
    8x2 – (4 + 4√3)x + 2√3 = 0
    (8x2 – 4x) – (4√3x – 2√3) = 0
    4x (2x- 1) – 2√3 (2x – 1) = 0,
    So x = 1/2 (0.5), 2√3/4 (0.87)
    3y2 – (4 + 3√3)y + 4√3 = 0
    (3y2 – 4y) – (3√3y – 4√3) = 0
    y (3y – 4) – √3 (3y – 4) = 0
    So, y = √3 (1.732), 4/3
    put on number line
    0.5………..0.87……..4/3………1.732
  10. I. x2 + (4 + 2√2)x + 8√2 = 0
    II. 3y2 – (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√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)y + √3 = 0
    (3y2 – 3y) – (√3y – √3) = 0
    3y (y – 1) – √3 (y – 1) = 0
    So y = 1, √3/3 (0.57)

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29 thoughts on “Quantitative Aptitude: Quadratic Equations Set 24”

      1. thinker

        Mam question 6me ,x=3/5,4 and y=3/5,5 to eska answer no relation kese Hua .Y bda ya equal hona chahiye.ye question mujhse hmesha wrong ho hate h.why no relation?

      2. Pawan

        value of x is same but value of y coming 2/3,-1 please check my eq can i make it as 3y^2+3y+2y-2=0

  4. Jellyfish

    8/10 Thx az
    Queno 5 option should be A
    Que 6 2nd statement 5y^2-28y+15=0 he chahiye

    1. Shubhra

      What doubt ?
      See 9th solution.
      U have to take without root values together
      And with root s values together
      And then take common like in 2nd step in 9th.

      See again. Ask if doubt

      1. वैम्पायर हंटर

        Wo 6th ke 2nd equtn me sign minus minus hh
        To fr value dono plus me likh rkhe ho ye bol rha hun…

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