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Aptitude Questions: Quadratic Equations Set 14

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Hello Aspirants. Welcome to Online Quantitative Aptitude Section in AffairsCloud.com. Here we are creating sample questions in Quadratic Equations which is common for all the competitive exams. We have included Some questions that are repeatedly asked in bank exams !!!

Follow the link To solve Quadratic Equations with the help of Number Line

You have to solve equation I and II ,Give answer
A) If X > Y
B) If X < Y
C) If X ≥ Y
D) If X ≤ Y
E) If X = Y or cannot be established

  1. I.5/x – 4/y = -2
    II.2/x + 3/y = 13
    Answer – 1)If X>Y
    Explanation :
    5u-4v = -2………*3 =>15u-12v = -6
    2u+3v = 13……..*4 => 8u+12v = 52
    Solve this eq
    23u = 46
    U=2, v=3
    X=1/2,y=1/3
    1/2 > 1/3….x > y

  2. I.4X2 + 4X – 3= 0
    II.14Y2 – 17Y + 5 = 0
    Answer – 4)If X≤ Y
    Explanation :
    (2X-1)(2X+3) =0
    X=1/2, -3/2
    (2Y-1)(7Y-5) =0
    Y = 1/2, 5/7
    5/7, ½, ½, -3/2 => YYXX
    Y≥X

  3. I.2X2 – 17X + 36= 0
    II.Y2 = 16
    Answer – 3)If X≥ Y
    Explanation :
    (2X-9)(X-4) =0
    X=9/2, 4
    Y = ±4
    9/2, 4,4,-4 => XXYY
    X≥Y

  4. I.6X2 + 19X + 15= 0
    II.3Y2 + 11Y + 10 = 0
    Answer – 3)If X≥ Y
    Explanation :
    (3X+5)(2X+3) = 0
    X=-5/3, -3/2
    (Y+2)(3Y+5) =0
    Y = -2, -5/3
    -3/2, -5/3, -5/3, -2 => XXYY
    X≥ Y

  5. I.4X2 + 19X + 21= 0
    II.2Y2 – 23Y + 63 = 0
    Answer – 1)If X>Y
    Explanation :
    (4x+7)(x+3) = 0
    X =-7/4, -3
    (2y+9)(y+7) = 0
    Y = -9/2, -7
    -7/4,-3, -9/2,-7 => xxyy
    x>y

  6. I.3X2 – 19X + 28= 0
    II.4Y2 – 29Y + 45 = 0
    Answer – 5)If X=Y or cannot be established
    Explanation :
    (x-4)(3x-7) =0
    X=4,7/3
    (y-5)(4y-9) =0
    Y=5,9/4
    5,4,7/3,9/4 => yxyx

  7. I.X2 = 5625
    II.Y = 3√421875
    Answer – 4)If X≤ Y
    Explanation :
    X=±75
    Y =75
    x≤y

  8. I.8X2 + 22X + 15= 0
    II.4Y2 + 16Y + 15 = 0
    Answer – 3)If X≥ Y
    Explanation :
    (4x+5)(2x+3) =0
    X=-5/4, -3/2
    (2y+5)(2xy+3) =0
    y=-5/2, -3/2
    -5/4,-3/2,-3/2,-5/3 => xxyy
    X≥Y

  9. I.8X2 + 10X + 3= 0
    II.5Y2 – 8Y + 3 = 0
    Answer – 2)If X<Y
    Explanation :
    (2X+1)(4X+3) =0
    X= -1/2, -3/4
    (5Y-3)(Y-1) =0
    Y=1, 3/5
    3/5, 1, -1/2, -3/4
    YYXX => Y>X

  10. I.3X + 2Y – 9= 0
    II.4Y – 15Y – 65 = 0
    Answer – 1)If X>Y
    Explanation :
    45X+30Y = 135
    8X-30Y=130
    53X=265
    X=5, Y= -3
    X>Y