## Saturday, 18 March 2017

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 $\leq$y
(3) if x < y
(4) if x $\leq$
5) if x = y or the relationship between x and y cannot be established.

$Q.$
I. $x^ 2$ + 12x + 36 = 0
II. $y^ 2$ + 15y + 56 = 0

 A.   $x > y$ B.   $x \geq y$ C.   $x < y$ D.   $x \leq y$
2 . I. $x^ 2$ = 35
II. $y^ 2$ + 13y + 42 = 0

 A.   $x > y$ B.   $x \geq y$ C.   $x < y$ D.   $x \leq y$
3 . I. $2x^ 2$ - 3x - 35 = 0
II. $y^ 2$ - 7y + 6 = 0

 A.   $x > y$ B.   $x \geq y$ C.   $x < y$ D.   x = y or no relation can be established between ‘x’ and ‘y’.
4 . I. $6x^ 2$ - 29x + 35 = 0
II. $2y^ 2$ - 19y + 35 = 0

 A.   $x > y$ B.   $x \geq y$ C.   $x < y$ D.   $x \leq y$
5 . I.$12x^ 2$ - 47x + 40 = 0
II. $4y^ 2$ + 3y - 10 = 0

 A.   $x > y$ B.   $x \geq y$ C.   $x < y$ D.   $x \leq y$
6 . I. $x^ 2$ + 3x - 28 = 0
II.$y^ 2$ - 11y + 28 = 0

 A.   $x > y$ B.   $x \geq y$ C.   $x < y$ D.   $x \leq y$
7 . I. $6x^ 2$ - 17x + 12 = 0
II.$6y^ 2$ - 7y + 2 = 0

 A.   $x > y$ B.   $x \geq y$ C.   $x < y$ D.   $x \leq y$
8 . I. x = $\sqrt{256}\over \sqrt{576}$
II. $3y^ 2$ + y-2 = 0

 A.   $x > y$ B.   $x \geq y$ C.   $x < y$ D.   $x \leq y$
9 . I. $x^ 2$ = 64
II. y$^ 2$ = 9y

 A.   $x > y$ B.   $x \geq y$ C.   $x \leq y$ D.   x = y or no relation can be established between ‘x’ and ‘y’.
10 . I. $x^ 2$ + 6x - 7 = 0
II. 41y + 17 = 140

 A.   $x > y$ B.   $x \geq y$ C.   $x < y$ D.   $x \leq y$

1 .
 Answer : Option A Explanation : I. $x^ 2$ + 12x + 36 = 0or, $(x + 6)^ 2$ = 0or, x + 6 = 0or, x = - 6II. $y^ 2$ + 15y + 56 = 0or, $y^ 2$ + 7y + 8y + 56 = 0or, y(y + 7) + 8(y + 7) = 0or, (y + 7) (y + 8) = 0y = -7, -8x > y
2 .
 Answer : Option A Explanation : I. $x^ 2$ = 35 $x^ 2$ = $\sqrt{35}$II. $y 2$ + 13y + 42 = 0or, $y 2$ + 6y + 7y + 42 = 0or, y(y + 6) + 7(y + 6) = 0or, (y + 6) (y + 7) = 0y = -6, - 7x > y
3 .
 Answer : Option D Explanation : I. $2x^ 2$ - 3x - 35 = 0or, $2x^ 2$ - 10x + 7x - 35 = 0or, 2x(x - 5) + 7(x - 5) = 0or, (2x + 7) (x - 5) = 0x = -$7\over 2$ , 5II. $y^ 2$ - 7y + 6 = 0or, $y^ 2$ - y - 6y + 6 = 0or, y(y - 1) - 6(y - 1)or, (y - 1)(y - 6) = 0y = 1, 6No relation can be established between x and y.
4 .
 Answer : Option D Explanation : I. $6x^ 2$ - 29x + 35 = 0or, $6x^ 2$ - 15x - 14x + 35 = 0or, 3x(2x - 5) -7(2x - 5) = 0or, (3x - 7) (2x - 5) = 0x = $7\over 3$, $5\over 2$II. $2y^ 2$ - 19y + 35 = 0or, $2y^ 2$ - 14y - 5y + 35 = 0or, 2y(y - 7) -5 (y - 7) = 0or, (2y - 5)(y - 7) = 0y = $5\over 2$, 7X $\leq$ y
5 .
 Answer : Option B Explanation : I. $12x^ 2$ - 47x + 40 = 0or, $12x^ 2$ - 32x - 15x + 40 = 0or, 4x(3x - 8) -5(3x - 8) = 0or, (4x - 5) (3x - 8) = 0x = $5\over 4$, $8\over 3$II. $4y^ 2$ + 3y - 10 = 0or, $4y^ 2$ + 8y - 5y - 10 = 0or, 4y(y + 2) -5(y + 2) = 0or, (4y - 5) (y + 2) = 0y = $5\over 4$, - 2 x $\geq$ y
6 .
 Answer : Option D Explanation : I. $x^ 2$ + 7x - 4x - 28 = 0or, x(x + 7) - 4 (x + 7) = 0or, (x - 4)(x + 7) = 0x = 4, - 7II. $y^ 2$ - 11y + 28 = 0or, $y^ 2$ - 7y - 4y + 28 = 0or, y (y - 7) -4(y - 7) = 0or, (y - 4) (y - 7) = 0y = 4, 7x $\leq$ y
7 .
 Answer : Option A Explanation : I. $6x^ 2$ - 17x + 12 = 0or, $6x^ 2$ - 9x - 8x + 12 = 0or, 3x (2x - 3) - 4 (2x - 3) = 0or, (3x - 4) (2x - 3) = 0x = $4\over 3$, $3\over 2$II. $6y^ 2$ - 3y - 4y + 2 = 0or, 3y (2y - 1) - 2 (2y - 1) =0or, (3y - 2) (2y - 1) = 0y = $2\over 3$, $1\over 2$x > y
8 .
 Answer : Option B Explanation : I. x = $\sqrt{256}\over \sqrt{576}$x = $16\over 24$ = $2\over 3$II. $3y^ 2$ + y - 2 = 0or, $3y^ 2$ + 3y - 2y - 2 = 0or, 3y (y + 1) - 2(y + 1) = 0or, (3y - 2) (y + 1) = 0y = $2\over 3$, -1$x \geq y$
9 .
 Answer : Option D Explanation : I. $x^ 2$ = 64x = ± 8II.$y^ 2$ = 9yor, $y^ 2$ - 9y = 0or, y (y - 9) = 0y = 0, 9no relationship can be established between x and y
10 .
 Answer : Option C Explanation : I. $x^ 2$ + 6x - 7 = 0or, $x^ 2$ + 7x - x - 7 = 0or, x(x + 7) -1 (x + 7)= 0or, (x - 1) (x + 7) = 0x = 1, -7II. 41y + 17 = 140or, 41y = 140 - 17 = 123y = $123\over 41$ = 3x < y