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Aptitude Questions: Quadratic Equations Set 11

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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 !!!

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.4/√x + 3/√x = √x
    II. Y2 – (7 5/2/√y) = 0
    Answer – 5)If X=Y or cannot be established
    Explanation :
    7/√x = /√x
    X =7
    (Y35/2 – 7 5/2 )/√y=0
    Y5/2 =7 5/2
    y = 7   x=y

  2. I. X2/3  = 49
    II.Y =(-4)4
    Answer – 1)If X>Y
    Explanation :
    X = 73 = 343
    Y = -4*-4*-4*-4 = 256
    x>y

  3. I. X2  – 18X + 72= 0
    II.Y2 – 20Y + 99 = 0
    Answer – 5)If X=Y or cannot be established
    Explanation :
    (x-12)(x-6) = 0
    X=6,12
    (y-11)(y-9) = 0
    Y=9,11
    12,11,9,6=> xyyx

  4. I.X = ±√441
    II.Y = (9261)1/3
    Answer – 4)If X≤ Y
    Explanation :
    X=±21
    Y = 21

  5. I. X2  – 698  =  326
    II.Y2 + 528 = 1552
    Answer – 5)If X=Y or cannot be established
    Explanation :
    X2  = 326+698 =1024
    Y2=1552 – 528 = 1024
    X=y = ±32

  6. I. 2X2 + 11X + 12= 0
    II.2Y2 + 19Y + 45 = 0
    Answer – 1)If X>Y
    Explanation 
    (2x+8)(2x+3) = 0
    X= -8/2, -3/2 = -4,-1.5
    (2y+10)(2y+9) = 0
    Y= -10/2, -9/2 = -5, -4.5
    -1.5,-4,-4.5,-5 => xxyy
    x>y

  7. I. 21X2  – 23X + 6= 0
    II.2Y2 – 11Y + 15 = 0
    Answer – 2)If X<Y
    Explanation :
    (7x-3)(3x-2) = 0
    X=3/7, 2/3 = 0.43, 0.67
    (2y-5)(y-3) = 0
    Y=3,5/2=3,2.5
    3,5/2,2/3,3/7 => yyxx
    X<y

  8. I. 3X2  + 10X + 8= 0
    II.3Y2 + 7Y + 4 = 0
    Answer – 4)If X≤ Y
    Explanation :
    (3x+4)(x+2) = 0
    X= -2,-4/3
    (y+1)(3y+4) = 0
    Y= -1,-4/3
    -1,-1.3,-1.3,-2 => yyxx
    x≤y

  9. I. 25X2  + 25X + 6= 0
    II.5Y2 + 20Y + 20 = 0
    Answer – 1)If X>Y
    Explanation :
    (5x+2)(5x+3) =0
    X= -2/5, -3/5 = -0.4, -0.6
    (5x+10)(x+2) = 0
    X= -10/5, -2 = -2, -2
    -0.4, -0.6, -2, -2 => Xxyy
    X>y

  10. I. 2X2 + 5X + 2= 0
    II.4Y2 – 1 = 0
    Answer – 4)If X≤ Y
    Explanation :
    (2x+1)(x+2) =0
    X= -1/2, -2
    Y2  = ±1/2 = ½, -1/2
    ½,-1/2,-1/2, -2
    yxxy

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