Quantitative Aptitude: Quadratic Equations Questions Set 66

  1. 102-x-2=0
    6y2+33y+15=0
    x<y
    x<=y
    x>y
    x>=y
    the relation cannot be determined
    Option E
    10x2-x-2=0
    10x2 – 5x+ 4x-2=0
    5x(2x -1) + 2(2x -1)=0
    (2x-1) ( 5x+2)=0
    x= 1/2 , -2/5
    6y2+33y+15=0
    6y2 +30y+ 3y +15=0
    6y( y+5) + 3(y+5)=0
    (y+5) ( 6y+3)=0
    y= -5, -3/6
    Hence the relation cannot be determined

     

  2. 35x2-13x-4=0
    9y2 +9y-2=0
    x<y
    x<=y
    x>y
    x>=y
    the relation cannot be determined
    Option C
    35x2-13x-4=0
    35x2 – 20x+ 7x-4=0
    5x(7x -4) + 1(7x -4)=0
    (7x-4) ( 5x-1)=0
    x= 4/7 , 1/5
    9y2 +9y-2=0
    9y2 +6y+ 3y -2=0
    3y( 3y+2) + 1(3y+2)=0
    (3y+2) ( 3y+1)=0
    y= -2/3,- 1/3
    Hence x>y

     

  3. 35x2-46x+15=0
    10y2+131y+13=0
    x<y
    x<=y
    x>y
    x>=y
    the relation cannot be determined
    Option C
    35x2-46x+15=0
    35x2 – 25x- 21x+15=0
    5x(7x -5) – 3(7x -5)=0
    (7x-5) ( 5x-3)=0
    x= 5/7 , 3/5
    10y2+131y+13=0
    10y2 +y+ 130y +13=0
    y( 10y+1) + 13(10y+1)=0
    (10y+1) ( y+13)=0
    y= -1/10,- 13
    Hence x>y

     

  4. 10x2+59x+45=0
    20y2-13y+2=0
    x<y
    x<=y
    x>y
    x>=y
    the relation cannot be determined
    Option A
    10x2+59x+45=0
    10x2 +50x+9x+45=0
    10x(x +5) + 9(x +5)=0
    (x+5) ( 10x+9)=0
    x= -5 , -9/10
    20y2-13y+2=0
    20y2 -5y- 8y +2=0
    5y( 4y-1) – 2(4y-1)=0
    (4y-1) ( 5y-2)=0
    y= 1/4, 2/5
    Hence y>x

     

  5. 2x2-31x= -120
    8y2-74y=-143
    x<y
    x<=y
    x>y
    x>=y
    the relation cannot be determined
    Option C
    2x2-31x= -120
    2x2 -16x-15x+120=0
    2x(x -8) -15(x -8)=0
    (x-8) ( 2x-15)=0
    x= 8 , 15/2
    8y2-74y=-143
    8y2 -22y-52y+143 =
    2y( 4y-11) -13(4y-11)=0
    (4y-11) ( 2y-13)=0
    y= 11/4, 13/2
    Hence x>y

     

  6. 8x2-41x-5=0
    49y2-49y-18=0
    x<y
    x<=y
    x>y
    x>=y
    the relation cannot be determined
    Option C
    8x2-41x-5=0
    8x2 -40x-x-5=0
    8x(x -5) – 1(x -5)=0
    (x-5) ( 8x-1)=0
    x= 5 , 1/8
    49y2-49y-18=0
    49y2 -63y+ 14y -18=0
    7y( 7y-9) + 2(7y-9)=0
    (7y-9) ( 7y+2)=0
    y= 9/7, -2/7
    Hence x>y

     

  7. 40x2-59x+21=0
    14y2+53y+14=0
    x<y
    x<=y
    x>y
    x>=y
    the relation cannot be determined
    Option C
    40x2-59x+21=0
    40x2 -24x-35x+21=0
    8x(5x -3) -7(5x -3)=0
    (5x-3) ( 8x-7)=0
    x= 3/5 , 7/8
    14y2+53y+14=0
    14y2 +49y+4y +14=0
    7y( 2y+7) + 2(2y+7)=0
    (2y+7) ( 7y+2)=0
    y= -7/2, -2/7
    Hence x>y

     

  8. 40x2-13x+1=0
    10y2+19y+6=0
    x<y
    x<=y
    x>y
    x>=y
    the relation cannot be determined
    Option C
    40x2-13x+1=0
    40x2 -8x-5x+1=0
    8x(5x -1) -1(5x -1)=0
    (5x-1) ( 8x-1)=0
    x= 1/5 , 1/8
    10y2+19y+6=0
    10y2 +15y+4y -6=0
    5y( 2y+3) +2(2y+3)=0
    (2y=3) ( 5y+2)=0
    y= -3/2, -2/5
    x>y

     

  9. 10x2-9x+2=0
    6y2+ 27y-15=0
    x<y
    x<=y
    x>y
    x>=y
    the relation cannot be determined
    Option D
    10x2-9x+2=0
    10x2 – 5x- 4x+2=0
    5x(2x -1) – 2(2x -1)=0
    (2x-1) ( 5x-2)=0
    x= 1/2 , 2/5
    6y2+ 27y-15=0
    6y2 +30y-3y -15=0
    6y( y+5) – 3(y+5)=0
    (y+5) ( 6y-3)=0
    y= -5, +3/6
    Hence x≥y

     

  10. 35x2+51x+18=0
    16y2 =1
    x<y
    x<=y
    x>y
    x>=y
    the relation cannot be determined
    Option A
    35x2+51x+18=0
    35x2 + 30x+ 21x+18=0
    5x(7x +6) + 3(7x +6)=0
    (7x+6) ( 5x+3)=0
    x= -6/7 , -3/5
    16y2 =1
    16y2-1=0
    (4y-1) ( 4y+1)=0
    y= 1/4,- 1/4
    Hence y>x

     

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