Quantitative Aptitude: Quadratic Equations Questions Set 63

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  1. I. x2 -4x -21 = 0
    II. y2 = 64
    x
    x<=y
    x>y
    x>=y
    Hence the relation cannot be determined
    Option E
    x2 -4x -21 = 0
    x2 -7x + 3x -21 = 0
    x( x-7) + 3( x-7)=0
    (x-7) (x+3)=0
    x=7 or x= -3
    y 2 = 64
    y= ±8
    Hence the relation cannot be determined

     

  2. I. x2 +7x +10 = 0
    II. y2 -5y + 6=0
    x>y
    x>=y
    x< y
    x<=y
    it cannot be determined
    Option C
    x2 + 7x +10 = 0
    x2 +2x + 5x + 10 = 0
    x( x+ 2) + 5( x+2)=0
    (x+2) (x+5)=0
    x=-2 or x= -5
    y2 -5y + 6=0
    y2 -2y -3y+ 6=0
    y(y-2) -3( y-2)=0
    (y-2) ( y-3) =0
    y= 2 or 3
    so x< y

     

  3. I. x2 -8x +16 = 0
    II. y2 +4y + 4=0
    x> y
    x>=y
    x
    x<=y
    it cannot be determined
    Option A
    x2 -8x +16 = 0
    (x-4)2= 0
    =>x=4
    y2 +4y + 4=0
    ( y+ 2) 2 = 0
    y= -2
    Hence x> y

     

  4. I. x2 -12x +36 = 0
    II. y2 +22y + 121=0
    x> y
    x>=y
    x
    x<=y
    it cannot be determined
    Option A
    x2 -12x +36 = 0
    (x-6)2= 0
    =>x=6
    y2 +22y+121=0
    ( y+ 11) 2 = 0
    y= -11
    Hence x> y

     

  5. I. x2 +8x +15 = 0
    II. y2 -10y + 24=0
    x>y
    x>=y
    x< y
    x<=y
    it cannot be determined
    Option C
    x2 + 8x +15 = 0
    x2 +3x + 5x + 15 = 0
    x( x+ 3) + 5( x+3)=0
    (x+3) (x+5)=0
    x=-3 or x= -5
    y2 -10y + 24=0
    y2 -4y -6y+ 24=0
    y(y-4) -6( y-4)=0
    (y-4) ( y-6) =0
    y= 4 or 6
    so x< y

     

  6. I. x2 -2x +1 = 0
    II. y2 +8y + 16=0
    x
    x<=y
    x> y
    x>=y
    it cannot be determined
    Option C
    x2 -2x +1 = 0
    (x-1)2= 0
    =>x=1
    y2 +8y + 16=0
    ( y+ 4) 2 = 0
    y= -4
    Hence x> y

     

  7. I. x2 -3x -10 = 0
    II. y2 = 4
    x
    x<=y
    x>y
    x ≥ y
    it cannot be determined
    Option D
    x2 -3x -10 = 0
    x2 -5x + 2x -10 = 0
    x( x-5) + 2( x-5)=0
    (x-5) (x+2)=0
    x=5 or x= -2
    y 2 = 4
    y= ± 2
    Hence x ≥ y

     

  8. I. x2 +24x + 144=0
    II. y2 -18y +81 = 0
    x
    x<=y
    x>y
    x>=y
    it cannot be determined
    Option A
    x2 +24x + 144=0
    ( x+ 12) 2 = 0
    x= -12
    y2 -18y +81 = 0
    (y-9)2= 0
    =>y=9
    Hence y> x

     

  9. I.2x2 +13x +15 = 0
    II. y2 -6y -7=0
    x
    x<=y
    x>y
    x>=y
    it cannot be determined
    Option A
    2x2 +13x +15 = 0
    2x2 +3x +10x +15 = 0
    x( 2x + 3) + 5( 2x + 3)=0
    (2x+3) ( x+5)=0
    x= -3/2 or -5
    y2 -6y -7=0
    y2 -7y + y-7=0
    y( y-7) + 1( y-7)=0
    ( y-7) ( y+1)=0
    y= 7 or -1
    Hence y> x

     

  10. I. x2 -2x -8=0
    II. 4y2 +14y +6 = 0
    x
    x<=y
    x>y
    x>=y
    it cannot be determined
    Option C
    x2 -2x -8=0
    x2 -4x + 2x-8=0
    x( x-4) + 2( x-4)=0
    ( x-4) ( x+2)=0
    x= 4 or -2
    4y2 +14y +6 = 0
    4y2 +2y +12y +6 = 0
    y( 4y + 2) + 3( 4y + 2)=0
    (4y+2) ( y+3)=0
    y= -2/4 or -3
    Hence x>y

     

  11. I. 3x2 +16x +5 = 0
    II. 5y2+ 37y +60=0
    x
    x<=y
    x>y
    x≥y
    it cannot be determined
    Option D
    3x2 +16x +5 = 0
    3x2 +x +15x +5 = 0
    x( 3x + 1) + 5( 3x + 1)=0
    (3x+1) ( x+5)=0
    x= -1/3 or -5
    5y2+ 37y +60=0
    5y2 + 25y +12y +60
    5y(y+5)+12(y+5)=0
    ( y+5)( 5y +12) =0
    y= -5 or -12/5
    Hence x≥y

     

  12. I. 2x2 -13x +20 = 0
    II. 3y2-10y +7=0
    x
    x<=y
    x>y
    x<=y
    it cannot be determined
    Option C
    2x2 -13x +20 = 0
    2x2 -8x -5x +20 = 0
    2x(x-4) – 5(x-4) = 0
    (x-4) ( 2x-5)=0
    x= 4 or 5/2
    3y2 -10y +7 = 0
    3y2 -3y -7y +7 = 0
    3y(y-1) – 7(y-1) = 0
    (y-1) ( 3y-7)=0
    y= 1 or 7/3
    Hence x>y

     

  13. I. x2 -12x +20 = 0
    II. 6y2 -29y +20 = 0
    x
    x<=y
    x>y
    x>=y
    it cannot be determined
    Option C
    x2 -12x +20 = 0
    x2 -10x -2x +20 = 0
    x(x-10) – 2(x-10) = 0
    (x-10) ( x-2)=0
    x= 10 or 2
    6y2 -29y +20 = 0
    6y2 -24y -5y +20 = 0
    6y(y-4) – 5(y-4) = 0
    (y-4) ( 6y-5)=0
    y= 4 or 5/6
    Hence x>y

     

  14. I. x2 -6x +8 = 0
    II. y2 -6y +5 = 0
    x
    x<=y
    x>y
    x>=y
    Hence the relation cannot be determined
    Option E
    x2 -6x +8 = 0
    x2 -4x -2x +8 = 0
    x(x-4) – 2(x-4) = 0
    (x-4) ( x-2)=0
    x= 4 or 2
    y2 -6y +5 = 0
    y2 -y -5y +5 = 0
    y(y-1) – 5(y-1) = 0
    (y-1) ( y-5)=0
    y= 1 or 5
    Hence the relation cannot be determined

     

  15. I. x2 -12x +32 = 0
    II. y2 -20y +100 = 0
    x
    x<=y
    x>y
    x>=y
    it cannot be determined
    Option A
    x2 -12x +32 = 0
    x2 -8x -4x +32 = 0
    x(x-8) – 4(x-8) = 0
    (x-8) ( x-4)=0
    x= 8 or 4
    y2 -20y +100 = 0
    (y-10) ^2=0
    y= 10
    Hence y> x

     

  16. I. x2 -x -30 = 0
    II. y 2 = 169
    x
    x<=y
    x>y
    x>=y
    Hence the relation cannot be determined
    Option E
    x2 -x -30 = 0
    x2 -6x + 5x -30 = 0
    x( x-6) + 5( x-6)=0
    (x-6) (x+5)=0
    x=6 or x= -5
    y 2 = 169
    y= ±13
    Hence the relation cannot be determined

     

  17. I.x2 + 3x -10 = 0
    II. y2 -13y/2 + 3=0
    x< y
    x<=y
    x>y
    x>=y
    it cannot be determined
    Option A
    x2 + 3x -10 = 0
    x2 -2x + 5x – 10 = 0
    x( x- 2) + 5( x-2)=0
    (x-2) (x+5)=0
    x=2 or x= -5
    y2 -13y/2 + 3=0
    y2 –y/2 -6y+ 3=0
    y(y-1/2) -6( y-1/2)=0
    (y-1/2) ( y-6) =0
    y= 1/2 or 6
    so x< y

     

  18. I. x2 -10x +25 = 0
    II. y2 +14y + 49=0
    x
    x<=y
    x> y
    x>=y
    it cannot be determined
    Option C
    x2 -10x +25 = 0
    (x-5)2= 0
    =>x=5
    y2 +14y + 49=0
    ( y+ 7) 2 = 0
    y= -7
    Hence x> y

     

  19. I. x2 +30x +225 = 0
    II. y2 -30y+225=0
    x
    x<=y
    x>y
    x>=y
    it cannot be determined
    Option A
    x2 +30x +225 = 0
    (x+15)2= 0
    =>x=-15
    y2 -30y+225=0
    ( y- 15) 2 = 0
    y= 15
    Hence y>x

     

  20. I. x2 + 5x -36 = 0
    II. y2 +3y + 2=0
    x
    x<=y
    x>y
    x>=y
    cannot be determined
    Option E
    x2 + 5x -36 = 0
    x2 -4x + 9x -36 = 0
    x( x-4) + 9( x-4)=0
    (x-4) (x+9)=0
    x=4 or x= -9
    y2 +3y + 2=0
    y2 +2y +y+ 2=0
    y(y+2) + 1( y+2)=0
    (y+2) ( y+1) =0
    y= -2 or -1
    so cannot be determined

     

  21. I. x2 +32x +256 = 0
    II. y2 -28y + 196=0
    x
    x<=y
    x>y
    x>=y
    it cannot be determined
    Option A
    x2 +32x +256 = 0
    (x+16)2= 0
    =>x= -16
    y2 -28y + 196=0
    ( y-14 ) 2 = 0
    y= 14
    Hence y>x

     

  22. I. x2 – 64=0
    II. y1/2 = 5
    x
    x<=y
    x>y
    x>=y
    it cannot be determined
    Option A
    x2 – 64=0
    (x-8) (x+8)=0
    x=8 or x= -8
    y 1/2 = 5
    y= 25
    Hence y>x

     

  23. I. x2 -18x +77 = 0
    II. y2 +8y +15=0
    x
    x<=y
    x>y
    x>=y
    it cannot be determined
    Option C
    x2 -18x +77 = 0
    x2 -7x -11x +77 = 0
    x( x-7) -11( x-7)=0
    (x-7) ( x-11)=0
    x= 7 or 11
    y2 +8y +15=0
    y2 +5y +3 y+15=0
    y( y+5) + 3( y+5)=0
    ( y+5) ( y+3)=0
    y= -5or -3
    Hence x> y

     

  24. I. x2 -12x +11 = 0
    II. y2 -5y -150=0
    x
    x<=y
    x>y
    x>=y
    Hence it cannot be determined
    Option E
    x2 -12x +11 = 0
    x2 -x -11x +11 = 0
    x( x-1) -11( x-1)=0
    (x-1) ( x-11)=0
    x= 1 or 11
    y2 -5y -150=0
    y2 +10y -15 y-150=0
    y( y+10) -15( y+10)=0
    ( y+10) ( y-15)=0
    y= -10 or +15
    Hence it cannot be determined

     

  25. I. x2 +3x +2 = 0
    II. y2 + y =0
    x
    x<=y
    x>y
    x>=y
    it cannot be determined
    Option B
    x2 +3x +2 = 0
    x2 +x +2x +2 = 0
    x( x+1) +2( x+1)=0
    (x+1) ( x+2)=0
    x= -1 or -2
    y2 + y =0
    y ( y+1)=0
    y= 0 or -1
    Hence x ≤ y

     

  26. I. x2 -24x +119 = 0
    II. y2+ 6y + 8 =0
    x
    x<=y
    x>y
    x>=y
    it cannot be determined
    Option C
    x2 -24x +119 = 0
    x2 -17x-7x +119 = 0
    x( x-7) -17( x-7)
    (x – 7)(x – 17) = 0
    x = 7 or 17
    y2+ 6y + 8 =0
    y2 +4y +2y+8 = 0
    y( y+4) +2( y+4)
    (y+2)(y +4) = 0
    y= -2 or -4
    So, x >y

     

  27. I. 6x2 + x – 2 = 0
    II. 20y2 – 13y + 2 = 0
    x
    x<=y
    x>y
    x>=y
    the relation cannot be determined.
    Option E
    6x2 + x – 2 = 0
    (2x – 1)(3x + 2) = 0
    x = 1/2 or -2/3
    20y2 – 13y + 2 = 0
    20y2 – 8y -5y+ 2 = 0
    4y(5y -2) -1( 5y-2)=0
    (4y– 1)(5y – 2) = 0
    y = 1/4 or 2/5
    So the relation cannot be determined.

     

  28. I. 3x2 + 13x+ 14 = 0
    II. 2y4 – 3y2 + 2 = 0
    x
    x<=y
    x>y
    x>=y
    it cannot be determined
    Option B
    3x2 + 13x+ 14 = 0
    3x2 + 7x + 6x + 14 = 0
    (x + 2)(3x + 7) = 0
    x = -2, -7/3
    2y4 – 3y2 + 2 = 0
    y4 -3y2/2 +1=0
    y4 + y2/2 -2y2 +1=0
    (y2 – 2)(y2 – 1/2) = 0
    y= ±√ 2, , √1/2 or -√1/2
    So, y≥ x

     

  29. I. x2 – 2x – 48= 0
    II. 6y2 + 35y + 25 = 0
    x
    x<=y
    x>y
    x>=y
    it cannot be determined
    Option E
    x2 – 2x – 48= 0
    x2 +6x –8x-48 = 0
    (x – 8)(x + 6) = 0
    x = -6 or + 8
    6y2 + 35y + 25 = 0
    6y2 +15y +10y + 25 = 0
    (3y + 5)(2y + 5) = 0
    y = -5/3 or -5/2
    Hence it cannot be determined

     

  30. I. x2 -24x + 143 = 0
    II. y2– 6y + 8 = 0
    x
    x<=y
    x>y
    x>=y
    it cannot be determined
    Option C
    x2 -24x + 143 = 0
    x2 -13x-11x + 143 = 0
    (x – 11)(x – 13) = 0
    x = 11 or 1 3
    y2– 6y + 8 = 0
    y2– 2y -4y+ 8 = 0
    (y– 4)(y– 2) = 0
    y = 4 or 2
    So y < x

     

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