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SBI PO Quadratic Equation Questions and Answers set - 10

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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 between ‘x’ and ‘y’ can be established.

    $Q.$
    I. $3x^ 2$ – 7x – 20 = 0
    II. $y^ 2$ – 8y + 16 = 0

    A.   $ x > y$
    B.   $ x \geq y $
    C.   $ x < y$
    D.   $ x \leq y$
    2 . I. $x^ 2$ – 72 = 0
    II. $y^ 2$ – 9y + 8 = 0

    A.   $ x > y$
    B.   $ x \geq y $
    C.   $ x < y$
    D.   x = y or no relation can be established between ‘x’ and ‘y’.
    3 . I. $9x^ 2$ – 114x + 361 = 0
    II. $y^ 2$ = 36

    A.   $ x > y$
    B.   $ x \geq y $
    C.   $ x < y$
    D.   $ x \leq y$
    4 . I. 13x + 17y = 107
    II. x – 11y = – 41

    A.   $ x > y$
    B.   $ x \geq y $
    C.   $ x < y$
    D.   $ x \leq y$
    5 . I. $9x^ 2$ + 18x + 9 = 0
    II. $y^ 2$ – 3y + 2 = 0

    A.   $ x > y$
    B.   $ x \geq y $
    C.   $ x < y$
    D.   $ x \leq y$
    6 . I. 4x + 7y = 42
    II. 3x - 11y = – l

    A.   $ x > y$
    B.   $ x \geq y $
    C.   $ x < y$
    D.   $ x \leq y$
    7 . I. $9x^ 2 $ – 29x + 22 = 0
    II.$ y^ 2 $ – 7y + 12 = 0

    A.   $ x > y$
    B.   $ x \geq y $
    C.   $ x < y$
    D.   $ x \leq y$
    8 . I. $3x^ 2 $ – 4x – 32 = 0
    II. $2y^ 2 $ – 17y + 36 = 0

    A.   $ x > y$
    B.   $ x \geq y $
    C.   $ x < y$
    D.   $ x \leq y$
    9 . I. $3x^ 2 $ – 19x – 14 = 0
    II. $2y^ 2$ + 5y + 3 = 0

    A.   $ x > y$
    B.   $ x \geq y $
    C.   $ x < y$
    D.   $ x \leq y$
    10 . I. $x^ 2 $ + 14x + 49 = 0
    II. $y^ 2$ + 9y = 0

    A.   $ x > y$
    B.   $ x \geq y $
    C.   $ x < y$
    D.   x = y or no relation can be established between ‘x’ and ‘y’.
      Answers & Solutions
       
      1 .    
      Answer : Option D
      Explanation :
      I. $3x^ 2$ - 12x + 5x - 20 = 0
      or 3x(x - 4) + 5(x - 4) = 0
      or (3x + 5) (x - 4) = 0
      x = - $5\over 3$, 4

      II. $y^ 2$ - 8y + 16 = 0
      or $(y - 4)^ 2 $ = 0
      (y - 4) = 0
      or y = 4

      $x \leq y$
      2 .    
      Answer : Option D
      Explanation :
      I. $x^ 2$ - 72 = 0
      or $x^ 2$ = 72
      x = + 8.485

      II.$y^ 2$ - y - 8y + 8 = 0
      or y(y - 1) - 8(y - 1) = 0
      or (y - 1) (y - 8) = 0
      y = 1, 8

      3 .    
      Answer : Option A
      Explanation :
      I. $9x^ 2$ - l14x + 361 =0
      or $(3x - 19)^ 2$ = 0
      3x - 19 = 0
      x = $19\over 3$ = 6.33

      II. $y^ 2$ = 36
      y = ±6

      x > y
      4 .    
      Answer : Option C
      Explanation :
      I. 13x + 17y = 107 $\longrightarrow$eqn (II) × 13
      13x ± 143y = ± 533
      --------------------------
      160y = 640
      y = $640\over 160$ = 4 x = 11y - 41
      x = 44 - 41 = 3
      x < y
      5 .    
      Answer : Option C
      Explanation :
      I. $9x^ 2$ + 18x + 9 = 0
      or $x^ 2$ + 2x + 1 = 0
      or $(x + l)^ 2$ = 0
      x + 1 = 0, or x = -1

      II.$y
      2$ - y - 2y + 2 = 0
      or y(y - 1) -2(y - 1) = 0
      or (y - 1) (y - 2) = 0
      y = 1 , 2
      x < y
      6 .    
      Answer : Option A
      Explanation :
      eqn (I) ×3 - eqn (II) × 4
      12x + 21y = 126
      12x - 44y = -4
      -$\,\,\,\,\,\,$ +$\,\,\,\,\,\,\,\,\,\,$ + .
      ----------------------
      65y = 130

      y = 2
      and x = 7
      X > Y
      7 .    
      Answer : Option C
      Explanation :
      I. $ 9x^ 2$ - 18x - 1 lx + 22 = 0
      or 9x(x - 2)- 11(x - 2) = 0
      or (x - 2)(9x - 11) = 0
      x = 2, $11\over 9$

      II. $y^ 2$ - 3y - 4y + 12 - 0
      or y(y - 3) - 4(y - 3) = 0
      or (y - 3) (y - 4) = 0
      y = 3, 4
      x < y
      8 .    
      Answer : Option D
      Explanation :
      I. $3x^ 2$ - 4x - 32 = 0
      or $3x^ 2$ - 12x + 8x - 32 = 0
      or 3x(x - 4) + 8(x - 4) = 0
      or (3x + 8) (x - 4) = 0

      x = 4, -$8\over 3$

      II. $2y^ 2$ - 8y - 9y + 36 = 0
      or 2y(y - 4) - 9(y - 4) = 0
      or (2y - 9) (y - 4) = 0
      or (2y - 9) (y - 4) = 0
      y = 4, $9\over 2$

      $x \leq y$
      9 .    
      Answer : Option A
      Explanation :
      I. $3x^ 2$ - 21x + 2x - 14 = 0
      or 3x(x - 7) + 2(x - 7) = 0
      or (3x + 2) (x - 7) = 0
      x = 7, -$2\over 3$

      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 = -$3\over 2$, -1

      x > y
      10 .    
      Answer : Option D
      Explanation :
      I. $x^ 2$ + 14x + 49 = 0
      or $(x + 7)^ 2$ = 0
      x + 7 = 0
      or, x = -7

      II. $y^ 2$ + 9y = 0
      or y(y + 9) = 0
      y = 0, -9

      ie,. no relation between x and y.



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