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SBI PO Quadratic Equation Questions and Answers set - 9

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    1 . Directions(Q. 1-10): In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer
    (1) if x > y
    (2) if x $\geq$ y
    (3) if x < y
    (4) if x $\leq$ y
    (5) if x = y or no relation can be established between ‘x’ and ‘y’.

    $Q.$
    I. $2x^ 2$ + 13x - 7 = 0
    II. $2y^ 2 $ - 5y + 3 = 0

    A.   $ x > y$
    B.   $ x \geq y $
    C.   $ x < y$
    D.   $ x \leq y$
    2 . I. $2x^ 2$ -15x + 28 = 0
    II. $4y^ 2 $ - 16y + 15 = 0

    A.   $ x > y$
    B.   $ x \geq y $
    C.   $ x < y$
    D.   $ x \leq y$
    3 . I. $x^ 2$ + 8x + 16 = 0
    II. $y^ 2$ = 16

    A.   $ x > y$
    B.   $ x \geq y $
    C.   $ x < y$
    D.   $ x \leq y$
    4 . I. $x^ 2 $ - 2x - 24 = 0
    II.$ y^ 2$ + 8y = 0

    A.   $ x > y$
    B.   $ x \geq y $
    C.   $ x < y$
    D.   x = y or no relation can be established between ‘x’ and ‘y’.
    5 . I. $x^ 2$ + 4x = 0
    II.$ y^ 2$ + 10y + 25 = 0

    A.   $ x > y$
    B.   $ x \geq y $
    C.   $ x < y$
    D.   $ x \leq y$
    6 . I. $2x^ 2$ + x – 1 = 0
    II. $2y^ 2$ + 13y + 15 = 0

    A.   $ x > y$
    B.   $ x \geq y $
    C.   $ x < y$
    D.   $ x \leq y$
    7 . I. $x^ 2$ + 12x + 32 = 0
    II. $2y^ 2$ + 15y + 27 = 0

    A.   $ x > y$
    B.   $ x \geq y $
    C.   $ x < y$
    D.   x = y or no relation can be established between ‘x’ and ‘y’.
    8 . I. $6x^ 2$ – 17x + 12 = 0
    II. $7y^ 2$ – 13y + 6 = 0

    A.   $ x > y$
    B.   $ x \geq y $
    C.   $ x < y$
    D.   $ x \leq y$
    9 . I. $x^ 2$ – 82x + 781 = 0
    II. $y^ 2$ = 5041

    A.   $ x > y$
    B.   $ x < y$
    C.   x = y or no relation can be established between ‘x’ and ‘y’.
    D.   $ x \geq y $
    10 . I. $6x^ 2$ – 47x + 80 = 0
    II. $2y^ 2$ – 9y + 10 = 0

    A.   $ x > y$
    B.   $ x \geq y $
    C.   $ x < y$
    D.   $ x \leq y$
      Answers & Solutions
       
      1 .    
      Answer : Option C
      Explanation :
      I.$ 2x^ 2$ + 13x - 7 = 0
      or $2x^ 2$ + 14x - x - 7 = 0
      or 2x (x + 7) - 1 (x + 7) = 0
      or (2x - 1) (x + 7) = 0
      x = $1\over 2$ , -7

      II. $2y^ 2$ - 5y + 3 = 0
      or $ 2y^ 2$ - 2y - 3y + 3 = 0
      or 2y(y - 1) - 3(y - 1) = 0
      or (2y - 3) (y - 1) = 0
      y = 1 , $3\over 2$

      Hence x < y
      2 .    
      Answer : Option A
      Explanation :
      I. $2x^ 2$ - 8x - 7x + 28 = 0
      or 2x (x - 4) - 7(x - 4) = 0
      or (2x - 7) (x - 4) = 0
      x = 4 , $7\over 2$

      II. $4y^ 2$ - 16y + 15 = 0
      or $ 4y^ 2$ - 6y - 10y + 15 = 0
      or 2y (2y - 3) - 5(2y - 3) = 0
      or (2y - 5) (2y - 3) = 0
      y = $5\over 2$, $3\over 2$

      Hence x > y
      3 .    
      Answer : Option D
      Explanation :
      I. $x 2$ + 8x + 16 = 0
      or $(x + 4)
      2$ = 0
      or x + 4 = 0
      x = -4

      II. $y 2$ = 16
      y = ±4

      Hence, x $\leq$ y
      4 .    
      Answer : Option D
      Explanation :
      I.$ x 2$ - 2x - 24 = 0
      or $ x 2$ + 4x - 6x - 24 = 0
      or x(x + 4) - 6(x + 4) = 0
      or (x - 6) (x + 4) = 0
      x = 6, - 4

      II. $y 2$ + 8y = 0
      or y(y + 8) = 0
      y = 0, - 8

      ie No relation can be established between x and y.
      5 .    
      Answer : Option A
      Explanation :
      I. $x 2$ + 4x = 0
      or x(x + 4) = 0
      x = 0, - 4

      II. $y 2$ + 10y + 25 = 0
      or $ (y + 5)^ 2$ = 0
      or y + 5 = 0
      y = - 5

      x > y
      6 .    
      Answer : Option A
      Explanation :
      I. $2x^ 2$ + 2x – x – 1 = 0
      or 2x(x + 1) – 1(x + 1) = 0
      or (2x – 1) (x + 1) = 0
      x = -1, $1\over 2$

      II. $2y^ 2$ + 3y + 10y + 15 = 0
      or y(2y + 3) + 5(2y + 3) = 0
      or (y + 5) (2y + 3) = 0
      y = -5 , -$3\over 2$

      x > y
      7 .    
      Answer : Option D
      Explanation :
      I. $x^ 2$ + 4x + 8x + 32 = 0
      or x(x + 4) + 8(x + 4) = 0
      or (x + 4) (x + 8) = 0
      x = – 4, – 8

      II. $2y^ 2$ + 6y + 9y + 27 = 0
      or 2y(y + 3) + 9(y + 3) = 0
      or (2y + 9) (y + 3) = 0
      y = -$9\over 2$, - 3

      No relation can be established between x and y.
      8 .    
      Answer : Option A
      Explanation :
      I. $6x^ 2$ – 9x – 8x + 12 = 0
      or 3x(2x – 3) – 4(2x – 3) = 0
      or (2x – 3) (3x – 4) = 0
      x = $3\over 2$, $4\over 3$

      II. $7y^ 2$ – 7y – 6y + 6 = 0
      or 7y(y – 1) – 6(y – 1) = 0
      or (7y – 6) (y – 1) = 0
      y = 1, $6\over 7$

      x > y
      9 .    
      Answer : Option C
      Explanation :
      I. $x^ 2$ – 11x – 71x + 781 =0
      or x(x – 11) – 71(x – 11) = 0
      or(x – 11)(x – 71) = 0
      x = 11, 71

      II. $y^ 2 $ = 5041
      y = ± 71

      No relation between x and y.
      10 .    
      Answer : Option B
      Explanation :
      I. $6x^ 2$ – 15x – 32x + 80 = 0
      or 3x(2x – 5) – 16(2x – 5) = 0
      or (3x – 16) (2x – 5) = 0
      x = $16\over 3$, $5\over 2$

      II. $2y^ 2$ – 4y – 5y + 10 = 0
      or 2y(y – 2) – 5(y – 2) = 0
      or (y – 2) (2y – 5) = 0
      y = 2, $5\over 2$

      $x \geq y$



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