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Quants Questions : Quadratic Equations Set 7

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 !!!

You have to solve equation I and II ,Give answer
1)If X>Y
2)If X<Y
3)If X≥ Y
4)If X≤ Y 
5)If X=Y or cannot be established

  1. I. X2  + 3X + 2= 0
    II. Y2 + 5Y + 6 = 0
    3)If X≥ Y
    Explanation :
    (x+1)(x+2) = 0
    X = -1,-2
    (y+3)(y+2) = 0
    Y = -3,-2
    -1,-2,-2,-3
    x x  y y
    x≥y

  2. I. 6X2  – 7X + 2 = 0
    II. 12Y2 – 7Y + 1 = 0
    1)If X>Y
    Explanation :
    (2x-1)(3y-2) = 0
    X =1/2,2/3
    (4y-1)(3y-1) = 0
    Y =1/4,1/3
    0.7,0.5,0.33,0.25
    x x y y => x>y

  3. I. 8X2  – 22X – 21 = 0
    II. 4Y2 + 9Y = 0
    5)If X=Y or cannot be established
    Explanation :
    (2x-7)(4x+3) = 0
    X=7/3,-3/4
    Y(4y+9)=0
    Y =0,-9/4
    2.3,0,-0.75,-2.25,
    xyxy =no relation

  4. I. 2X2  – 15X + 25 = 0
    II. 5Y2 – 126Y + 25 = 0
    5)If X=Y or cannot be established
    Explanation :
    (2x-5)(x-5) = 0
    X =5,5/2
    (5y-1)(y-25)
    Y=25,1/5
    25,5,2.5,0.2
    yxxy =no relation

  5. I. 6X2  – X – 2 = 0
    II. 9Y2 – 22Y + 8 = 0
    5)If X=Y or cannot be established
    Explanation :
    (2x+1)(3x-2) = 0
    X = 2/3,-1/2
    (y-2)(9y-4) = 0
    Y=2,4/9
    2,0.7,0.4,-0.5
    xyxy = no relation

  6. I. 2X2  + 17X + 36 = 0
    II. 2Y2 + 9Y +9 = 0
    2)If X<Y
    Explanation :
    (x+4)(2x+9) = 0
    X= -4,-9/2
    (y+3)(2y+3) = 0
    Y = -3,-3/2
    -1.5,-3,-4,-4.5
    Yyxx => x<y

  7. I. 3X2  – 28X + 65 = 0
    II. 3Y2 – 16Y + 21 = 0
    1)If X>Y
    Explanation :
    (x-5)(3x-13) =0
    X=5,13/3
    (y-3)(3y-7) = 0
    Y=3,7/3
    5,4.3,3,2.3
    Xxyy => x>y

  8. I. X2  – 7X + 12 = 0
    II. Y2 – 10Y +24 = 0
    4)If X≤ Y
    Explanation :
    (X-4)(X-3) =0
    X =4,3
    (Y-4)(Y-6) = 0
    Y=6,4
    6,4,4,3
    YYXX => Y≥X

  9. I. 4X2  – 29X + 45 = 0
    II. 3Y2 – 19Y + 28 = 0
    5)If X=Y or cannot be established
    Explanation :
    (4X-9)(X-5) =0
    X=5,9/4
    (3Y-7)(Y-4) =0
    Y=4,7/4
    5,4,2.3,1.8
    yxxy => no relation

  10. I. X2  = 100
    II. Y2 – 20Y + 100 = 0
    4)If X≤ Y
    Explanation :
    X = +10,-10
    (y-10)(y-10) =0
    Y = 10,10
    10,10,10,-10
    Yyxx = y≥x

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