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SBI PO Quadratic Equation Questions and Answers set - 8

IBPS EXAM Guru
     
     
     
    1 . Directions (Q. 1-5): 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 a relationship between x and y cannot be established.

    $Q.$
    I. $x^ 2$ + 3x = 28
    II. $y^ 2$ + 16y + 63 = 0

    A.   $ x > y$
    B.   $ x \geq y $
    C.   $ x < y$
    D.   $ x \leq y$
    2 . I. x = $\sqrt[3]{2197}$
    II. y$^2$ = 169

    A.   $ x > y$
    B.   $ x \geq y $
    C.   $ x < y$
    D.   $ x \leq y$
    3 . I. $8x^ 2 $ - 49x + 45 = 0
    II.$ 8y^ 2 $ - y - 9 = 0

    A.   $ x > y$
    B.   $ x \geq y $
    C.   $ x < y$
    D.   $ x \leq y$
    4 . I. 42x - 17y = -67
    II. 7x + 12y = -26

    A.   $ x > y$
    B.   $ x \geq y $
    C.   $ x < y$
    D.   $ x \leq y$
    5 . I. $x^ 2 $ - 8x + 15 = 0
    II. $2y^ 2 $ - 21y + 55 = 0

    A.   $ x > y$
    B.   $ x \geq y $
    C.   $ x < y$
    D.   $ x \leq y$
    6 . Directions (Q. 6-10): In each of these questions two equations (I) and (II) are given. You have to solve both the equations and give answer
    (1) if p > q
    (2) if p $\geq$ q
    (3) if p < q
    (4) if p $\leq$ q
    (5) if p = q or no relation can be established between p and q.

    $Q.$
    I. $2.3p - 20.01 = 0$
    II. $2.9q - p = 0$

    A.   $ p > q$
    B.   $ p \geq q $
    C.   $ p < q$
    D.   $ p \leq q$
    7 . I. p = $\sqrt{1764}$
    II. $q^2$ = 1764

    A.   $ p > q$
    B.   $ p \geq q $
    C.   $ p < q$
    D.   $ p \leq q$
    8 . I. $p^ 2 $ - 26p + 168 = 0
    II. $q^ 2 $ - 25q + 156 = 0

    A.   p = q or no relation can be established between ‘p’ and ‘q’.
    B.   $ p \leq q$
    C.   $ p < q$
    D.   $ p \geq q $
    9 . I. $p^ 2 $ - 13p + 42 = 0
    II. $q^ 2$ + q - 42 = 0

    A.   $ p > q$
    B.   $ p \geq q $
    C.   $ p < q$
    D.   $ p \leq q$
    10 . I. 6p - 5q = -47
    II. 5p + 3q = 11

    A.   $ p > q$
    B.   $ p \geq q $
    C.   $ p < q$
    D.   $ p \leq q$
      Answers & Solutions
       
      1 .    
      Answer : Option B
      Explanation :
      I. $x^ 2$ + 3x - 28 = 0
      or, $ x^ 2$ + 7x - 4x - 28 = 0
      or, x (x + 7) - 4 (x + 7) = 0
      or, (x - 4) (x + 7) = 0
      or, x = 4, -7

      II. $y^ 2$ + 9y + 7y + 63 = 0
      or, y(y + 9) + 7(y + 9) = 0
      or, (y + 7)(y + 9) = 0
      or, y = -7, -9

      $x \geq y$
      2 .    
      Answer : Option B
      Explanation :
      I. x = $\sqrt[3]{2197}$
      x = 13

      II. y$^2$ = 169
      y = ± 13

      x $\geq$ y
      3 .    
      Answer : Option B
      Explanation :
      I. $8x^ 2$ - 40x - 9x + 45 = 0
      or, 8x (x - 5) -9 (x - 5) = 0
      or, (8x - 9) (x - 5) = 0
      or, x = 5, 9/8

      II. $8y^ 2$ + 8y - 9y -9 = 0
      or, 8y (y + 1) -9 (y + 1) = 0
      or, (8y - 9) (y + 1) = 0
      y = $9\over 8$, -1

      x$\geq$y
      4 .    
      Answer : Option C
      Explanation :
      42x - 17y = -67
      42x + 72y = -156$\,\,\,\,\,$$\longrightarrow$ eqn (II) × 6
      -$\,\,\,\,\,\,\,$ -$\,\,\,\,\,\,\,\,\,\,\,\,\,$ +
      ----------------------
      -89y = 89

      y = -$89\over 89$ = -1 & x = -2

      x < y
      5 .    
      Answer : Option D
      Explanation :
      I. $x^ 2$ - 8x + 15 = 0
      or, $x^ 2 $ - 3x - 5x + 15 = 0
      or, x (x - 3) -5 (x - 3) = 0
      or, (x - 3) (x - 5) = 0
      or, x = 3, 5

      II. $2y^ 2 $ - 10y + 55 = 0
      or, 2y (y - 5) -11 (y - 5) = 0
      or, (y - 5)(2y - 11) = 0
      y = 5, $11\over 2$

      x $\leq$ y
      6 .    
      Answer : Option A
      Explanation :
      I. 2.3p - 20.01 = 0
      p = $20.01\over 2.3$ = 8.7

      II. $2.9q - p$ = 0
      or, p = 2.9q
      q = $8.7\over 2.9$ = 8.7
      ie,. p > q
      7 .    
      Answer : Option B
      Explanation :
      I. p = $\sqrt{1764}$
      p = 42

      II. $q^2$ = 1764
      q = ± 42

      P$\geq$ q
      8 .    
      Answer : Option A
      Explanation :
      I.$ p^ 2$ - 26p + 168 = 0
      $p^ 2$ - 12p - 14p + 168 = 0
      p(p - 12) - 14(p - 12) = 0
      (p - 12) (p - 14) = 0
      p = 12, 14

      II. $q^ 2$ - 25q + 156 = 0
      $q^ 2$ - 13q - 12q + 156 = 0
      q(q - 13) - 12(q - 13) = 0
      (q - 12) (q - 13) = 0
      q = 12, 13

      Hence, no relation can be established between p and q
      9 .    
      Answer : Option B
      Explanation :
      I. $p^ 2$ - l3q + 42 = 0
      $p^ 2$ - 6p - 7p + 42 = 0
      p(p - 6) - 7(p - 6) = 0
      (p - 6) (p - 7) = 0
      p = 6, 7
      II. $q^ 2$ + q - 42 = 0
      $q^ 2$ + 7q - 6p - 42 = 0
      q(q + 7) - 6(q + 7) = 0
      (q - 6)(q + 7) = 0
      q = 6, - 7
      ie p $\geq$ q
      10 .    
      Answer : Option C
      Explanation :
      eqn(I) × 3 $\,\,\,\,\,\,$ 18p - 15q = -141
      eqn (II) × 5 $\,\,\,\,$ 25p + 15q = 55
      $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$-----------------------
      $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ 43p = - 86
      p = -$86\over 43$ = -2

      5p + 3q = 11
      3q = 11 - 5p
      3q = 11 + 10
      3q = 21
      q = 7

      ie p < q

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