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Sunday, 12 March 2017

SBI PO Quadratic Equation Questions and Answers Quiz 2

     
     
     
    1 . Directions (Q. Nos. 1-5) : In the following questions two equations numbered 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 the relationship cannot be established

    $Q.$
    I. $\sqrt{1225x} + \sqrt{4900} = 0$
    II. $(81)^{1\over 4} y + (343)^{1\over 3} = 0$

    A.   $ x > y $
    B.   $ x < y $
    C.   $ x \geq y $
    D.   $ x \leq y$
    2 . I. $18\over x^2$ + $6\over x^2$ - $12\over x^2$ = $8\over x^2$
    II. $y^ 3 $ + 9.68 + 5.64 = 16.95

    A.   $ x > y $
    B.   $ x \geq y $
    C.   $ x \leq y$
    D.   x = y or no relation can be established between ‘x’ and ‘y’.
    3 . I. $(2)^5 + (11)^3\over 6$ = $x^3$
    II. $4y^3 = - (589 \div 4 ) + 5 y^3$

    A.   $ x > y $
    B.   $ x \geq y $
    C.   $ x < y $
    D.   $ x \leq y$
    4 . I. $12x^ 2$ + llx + 12 = 10x 2 +22x
    II. $13y^ 2$ - 18y + 3 = 9y 2 - 10y

    A.   $ x > y $
    B.   $ x \geq y $
    C.   $ x < y $
    D.   $ x \leq y$
    5 . I. $(x^{7\over 5} \div 9)$ = $169 \div y{3\over 5}$
    II. $y^{1\over 4} \times y^{1\over 4} \times 7$ = $273 \div y^{1\over 2}$

    A.   $ x > y $
    B.   $ x \geq y $
    C.   $ x < y $
    D.   $ x \leq y$
    6 . Directions (Q. 6 - 10): Two equations (I) and (II) are given in each question. On the basis of these equations you have to decide the relation between x and y and give answer

    (1) if x > y
    (2) if x < y
    (3) if x $\geq$ y
    (4) if x $\leq$ y
    (5) if x = y, or no relation can be established between x and y.

    $Q.$
    I. x = $\sqrt[ 4] {2401}$
    II.$ 2y^ 2$ - 9y - 56 = 0

    A.   $ x > y $
    B.   $ x < y $
    C.   $ x \geq y $
    D.   x = y or no relation can be established between ‘x’ and ‘y’.
    7 . I. $5x^ 2$ + 3x - 14 = 0
    II.$ 2y^ 2$ - 9y + 10 = 0

    A.   $ x > y $
    B.   $ x < y $
    C.   $ x \geq y $
    D.   $ x \leq y$
    8 . I. $8x ^ 2$ + 31x + 21 = 0
    II. $5y ^2$ + 11y - 36 = 0

    A.   $ x > y $
    B.   $ x < y $
    C.   $ x \leq y$
    D.   x = y or no relation can be established between ‘x’ and ‘y’.
    9 . I. 3x - y = 12
    II. y = $\sqrt{ 1089}$

    A.   $ x > y $
    B.   $ x < y $
    C.   $ x \geq y $
    D.   $ x \leq y$
    10 . I. $15x^ 2$ + 68x + 77 = 0
    II. $3y^ 2$ + 29y + 68 = 0

    A.   $ x > y $
    B.   $ x < y $
    C.   $ x \geq y $
    D.   $ x \leq y$
      Answers & Solutions
       
      1 .    
      Answer : Option A
      Explanation :
      I. $\sqrt{1225x} + \sqrt{4900} = 0$
      35x + 70 = 0
      x = - $70\over 35$ = -2

      II. 3y + 7 = 0
      y = - $7\over 3$

      $x > y$
      2 .    
      Answer : Option D
      Explanation :
      I. $18 + 6 x - 12\over x^2$ = $8\over x^2$
      or , x = $1\over 3$ = 0.333

      II. $y^ 2$ = 16.95 - 9.68 - 5.64 = 1.63
      y = ±1.277
      3 .    
      Answer : Option A
      Explanation :
      I. $x^3 = $ $32 + 1331\over 6$ = $1363\over6$
      II. $5y^3 - 4y^3 =$ $589\over 4$
      or $y^3 =$ $589\over 4$

      $x > y$
      4 .    
      Answer : Option B
      Explanation :
      I. $2x ^2 $ - llx + 12 = 0
      or, x = 4, $3\over 2$

      II. $ 4y^ 2$ - 8y + 3 = 0
      y = $3\over 2$ , $1\over 2$

      $x \geq y$
      5 .    
      Answer : Option D
      Explanation :
      $(x^{7\over 5} \div 9)$ = $169 \div x{3\over 5}$
      $(x^{7\over 5} \times x{3\over 5}$ $ = 169 x 9
      $(x^{7+3\over 5} = 1521
      $x^2$ = 1521
      x = ± 39

      II. $y ^{1\over 4} $ + $y^{1\over 4} $ + $y^{1\over 2} $ = $273\over 7$
      or, y $1\over 4$ + $1\over 4$ + $1\over 2$ = 39
      y = 39

      x $\leq$ y
      6 .    
      Answer : Option D
      Explanation :
      I. x = $\sqrt[ 4] {2401}$
      $\Rightarrow$ x = 7

      II.$ 2y ^ 2$ - 16y + 7y - 56 = 0
      2y(y - 8) + 7(y - 8) = 0
      (2y + 7) (y - 8) = 0
      y = 8 , - $7\over 2$

      Hence, no relation exists between x and y.
      7 .    
      Answer : Option B
      Explanation :
      I. $ 5x^ 2$ + 10x - 7x - 14 = 0
      or, 5x(x + 2) - 7(x + 2) = 0
      or, (x + 2) (5x - 7) = 0
      x = - 2, $7\over 5$

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

      $x < y$
      8 .    
      Answer : Option D
      Explanation :
      I. $8x^ 2$ + 24x + 7x + 21 = 0
      or, 8x(x + 3) + 7(x + 3) = 0
      or, (x + 3) (8x + 7) = 0
      x = - 3, - $7\over 8$

      II. $5y^ 2$ + 20y - 9y - 36 = 0
      or, 5y(y + 4) - 9(y + 4) = 0
      or, (y + 4) (5y - 9) = 0
      y = -4, $9\over 5$

      Hence, no relation exists between x and y.
      9 .    
      Answer : Option B
      Explanation :
      I. y = $\sqrt{ 1089}$ $\Rightarrow$ y = 33

      II. x = $12 + y\over 3$ = $12 + 33 \over 3 $ = $45\over 3$ = 15

      $x < y$
      10 .    
      Answer : Option A
      Explanation :
      I. $15x^ 2$ + 68x + 77 = 0
      or, $15x^ 2$ + 35x + 33x + 77 = 0
      or, 5x(3x + 7) + 11(3x + 7) = 0
      or, (5x + 11) (3x + 7) = 0
      x = -$7\over 3$ , -$11\over 5$

      II. $3y^ 2$ + 29y + 68 = 0
      or, $ 3y^ 2$ + 12y + 17y + 68 = 0
      or, 3y(y + 4) + 17(y + 4) = 0
      or, (y + 4) (3y + 17) = 0
      y = -4, -$17\over 3$

      $x > y$

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