By Shiva
May 05, 2017

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 ≥ y
(3) if x < y
(4) if x ≤ y
(5) if x = y or no relation can be established between x and y.

$Q.$
I.$63x - 94\sqrt{x} + 35 = 0$
II. $32y - 52\sqrt{y} + 21 = 0$

 A.    $x > y$ B.    $x \geq y$ C.    $x < y$ D.    x = y or no relation can be established between ‘x’ and ‘y’.
2 . I.$x^2 - 7\sqrt{3}x - 35\sqrt{15} = 5\sqrt{5}x$
II.$y^2 - 5\sqrt{5}y + 30 = 0$

 A.    $x > y$ B.    $x \geq y$ C.    $x < y$ D.    $x \leq y$
3 . I. $14x^ 2$ + 11x - 15 = 0
II. $20y^ 2$ - 31y + 12 = 0

 A.    $x > y$ B.    $x \geq y$ C.    $x < y$ D.    $x \leq y$
4 . I. 5x + 4y = 41
II. 4x + 5y = 40

 A.    $x > y$ B.    $x \geq y$ C.    $x < y$ D.    $x \leq y$
5 . I. $\sqrt{x}$ - $(18)^{15\over 2}\over x ^{2}$ = 0
II. $\sqrt{y}$ - $(19)^{9\over 2}\over y$ = 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. Solve both the equations and give answer
(1) if x > y
(2) if x < y
(3) if x ≥ y
(4) if x ≤ y
(5) if x = y or no relation can be established between ‘x’ and ‘y’.

$Q.$
I . $63x - 194\sqrt{x} + 143 = 0$
II .$99y - 225\sqrt{y} + 150 = 0$

 A.    $x > y$ B.    $x \geq y$ C.    $x < y$ D.    x = y or no relation can be established between ‘x’ and ‘y’.
7 . I. $16x^ 2$ - 40x - 39 = 0
II. $12y^ 2$ - 113y + 255 = 0

 A.    $x > y$ B.    x < y C.    x ≥ y D.    $x \leq y$
8 . I. $x - 7\sqrt{3}x + 36 = 0$
II. $y - 5\sqrt{3}y - 7\sqrt{2}y + 70 = 0$

 A.    $x > y$ B.    x < y C.    x ≥ y D.    $x \leq y$
9 . I . $x^2 - 7\sqrt{x} + 84$ = 0
II . $y^2 - 5\sqrt{5} + 30$ = 0

 A.    $x > y$ B.    x < y C.    x ≥ y D.    $x \leq y$
10 . I. 10x + 6y = 13
II. 45x + 24y = 56

 A.    $x > y$ B.    x < y C.    x ≥ y D.    $x \leq y$
1 .
 Answer : Option D Explanation : I.$63x - 94\sqrt{x} + 35 = 0$ or, $63x - 49\sqrt{x} - 45\sqrt{x} + 35 = 0$ or, $(9\sqrt{x} - 7)(7\sqrt{x} - 5) = 0$ x = $49\over 81$, $25\over 49$ II. $32y - 52\sqrt{y} + 21 = 0$ or, $32x - 28\sqrt{x} - 24\sqrt{x} + 21 = 0$ or, $(4\sqrt{y} - 3)(8\sqrt{y} - 7) = 0$ y = $9\over 16$, $49\over 64$ Therefore relation can't be established between x and y.
2 .
 Answer : Option A Explanation : I.$x^2 - 7\sqrt{3}x - 35\sqrt{15} = 5\sqrt{5}x$ or,$x^2 - 7\sqrt{3}x - 5\sqrt{3}x - 35\sqrt{15} = 0$ or, $(x - 7\sqrt{5})(x - 5\sqrt{5}) = 0$ x = 7$\sqrt{3}$, $5\sqrt{5}$ II.$y^2 - 5\sqrt{5}y + 30 = 0$ or,$y^2 - 3\sqrt{5}y - 2\sqrt{5}y + 30 = 0$ or, $(y - 3\sqrt{5})(y - 2\sqrt{5}) = 0$ y = 3$\sqrt{5}$, $2\sqrt{5}$ x > y
3 .
 Answer : Option C Explanation : I. $14x^ 2$ + 11x - 15 = 0 or (7x - 5) (2x + 3) = 0 x = $5\over 7$, - $3\over 2$ II. $20y^ 2$ - 31y + 12 = 0 or (4y - 3), (5y - 4) = 0 y = $3\over 4$ , $4\over 5$ x < y
4 .
 Answer : Option A Explanation : I. 5x + 4y = 41 ... (i) II. 4x + 5y = 40 ... (ii) On solving both equations, we have x = 5 and y = 4 x > y
5 .
 Answer : Option C Explanation : I. $\sqrt{x}$ - $(18)^{15\over 2}\over x ^{2}$ = 0 or $x ^ {5\over 2}$ = $(18)^{15\over 2}$ x = $(18)^3$ II. $\sqrt{y}$ - $(19)^{9\over 2}\over y$ = 0 or $y ^ {3\over 2}$ = $(19)^{9\over 2}$ y = $(19)^3$ x < y
6 .
 Answer : Option D Explanation : I . $63x - 194\sqrt{x} + 143 = 0$ or $63x - 117\sqrt{x} - 77\sqrt{x} + 143 = 0$ or $(7\sqrt{x} - 13)(9\sqrt{x} - 11) = 0$ x = $169\over 49$, $121\over 81$ II .$99y - 225\sqrt{y} + 150 = 0$ or $99y - 90\sqrt{x} - 165\sqrt{x} + 143 = 0$ or $(11\sqrt{y} - 10)(9\sqrt{y} - 15) = 0$ x = $100\over 121$, $225\over 81$ Therefore relation cannot be established between x and y.
7 .
 Answer : Option B Explanation : I. $16x^ 2$ - 40x - 39 = 0 or $16x^ 2$ - 52x + 12x - 39 = 0 or (4x- 13) (4x + 3) = 0 x = $13\over 4$, - $3\over 4$ II. $12y^ 2$ - 113y + 255 = 0 or $12y^ 2$ - 45y - 68y + 255 = 0 or (4y - 15) (3y - 17) = 0 y = $15\over 4$, $17\over 3$ Therefore x < y
8 .
 Answer : Option B Explanation : I. $x - 7\sqrt{3}x + 36 = 0$ or$x - 7\sqrt{3}x \sqrt{x} + 36 = 0$ or$x - 3\sqrt{3}x \sqrt{x} - 4\sqrt{x} \sqrt{x} + 36 = 0$ or $(\sqrt{x} - 3\sqrt{3})( \sqrt{x} - 7\sqrt{3}) = 0$ x = 27 , 48 II. $y - 5\sqrt{3}y - 7\sqrt{2}y + 70 = 0$ or $y - 5\sqrt{2}y \sqrt{y} - 7\sqrt{2}y \sqrt{y} + 70 = 0$ or $(\sqrt{y} - 5\sqrt{2})( \sqrt{y} - 7\sqrt{2}) = 0$ y = 50 , 98 x < y
9 .
 Answer : Option A Explanation : I . $x^2 - 7\sqrt{x} + 84$ = 0 or $(x - 4\sqrt{7})(x - 3\sqrt{7})$ = 0 x = $4\sqrt{7}$, $3\sqrt{7}$ II . $y^2 - 5\sqrt{5} + 30$ = 0 or $(y - 2\sqrt{5})(x - 3\sqrt{5})$ = 0 y = $2\sqrt{5}$, $3\sqrt{5}$ x > y
10 .
 Answer : Option B Explanation : I. 10x + 6y = 13 II. 45x + 24y = 56 On solving both eqns, x = $4\over 5$, y = $5\over 6$ x < y

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