## Sunday, 19 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 $\geq$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’.

1 .
 Answer : Option D Explanation : I. $3x^ 2$ - 12x + 5x - 20 = 0or 3x(x - 4) + 5(x - 4) = 0or (3x + 5) (x - 4) = 0x = - $5\over 3$, 4II. $y^ 2$ - 8y + 16 = 0or $(y - 4)^ 2$ = 0(y - 4) = 0or y = 4$x \leq y$
2 .
 Answer : Option D Explanation : I. $x^ 2$ - 72 = 0or $x^ 2$ = 72x = + 8.485II.$y^ 2$ - y - 8y + 8 = 0or y(y - 1) - 8(y - 1) = 0or (y - 1) (y - 8) = 0y = 1, 8
3 .
 Answer : Option A Explanation : I. $9x^ 2$ - l14x + 361 =0or $(3x - 19)^ 2$ = 03x - 19 = 0x = $19\over 3$ = 6.33II. $y^ 2$ = 36y = ±6x > y
4 .
 Answer : Option C Explanation : I. 13x + 17y = 107 $\longrightarrow$eqn (II) × 1313x ± 143y = ± 533--------------------------160y = 640y = $640\over 160$ = 4 x = 11y - 41 x = 44 - 41 = 3x < y
5 .
 Answer : Option C Explanation : I. $9x^ 2$ + 18x + 9 = 0or $x^ 2$ + 2x + 1 = 0or $(x + l)^ 2$ = 0x + 1 = 0, or x = -1II.$y2$ - y - 2y + 2 = 0or y(y - 1) -2(y - 1) = 0or (y - 1) (y - 2) = 0y = 1 , 2 x < y
6 .
 Answer : Option A Explanation : eqn (I) ×3 - eqn (II) × 412x + 21y = 12612x - 44y = -4-$\,\,\,\,\,\,$ +$\,\,\,\,\,\,\,\,\,\,$ + .----------------------65y = 130y = 2and x = 7X > Y
7 .
 Answer : Option C Explanation : I. $9x^ 2$ - 18x - 1 lx + 22 = 0or 9x(x - 2)- 11(x - 2) = 0or (x - 2)(9x - 11) = 0x = 2, $11\over 9$II. $y^ 2$ - 3y - 4y + 12 - 0or y(y - 3) - 4(y - 3) = 0or (y - 3) (y - 4) = 0y = 3, 4x < y
8 .
 Answer : Option D Explanation : I. $3x^ 2$ - 4x - 32 = 0or $3x^ 2$ - 12x + 8x - 32 = 0or 3x(x - 4) + 8(x - 4) = 0or (3x + 8) (x - 4) = 0x = 4, -$8\over 3$II. $2y^ 2$ - 8y - 9y + 36 = 0or 2y(y - 4) - 9(y - 4) = 0or (2y - 9) (y - 4) = 0or (2y - 9) (y - 4) = 0y = 4, $9\over 2$$x \leq y$
9 .
 Answer : Option A Explanation : I. $3x^ 2$ - 21x + 2x - 14 = 0or 3x(x - 7) + 2(x - 7) = 0or (3x + 2) (x - 7) = 0x = 7, -$2\over 3$II. $2y^ 2$ + 5y + 3 = 0or $2y^ 2$ + 2y + 3y + 3 = 0or 2y(y + 1) + 3(y + 1) = 0or (2y + 3) (y + 1) = 0y = -$3\over 2$, -1x > y
10 .
 Answer : Option D Explanation : I. $x^ 2$ + 14x + 49 = 0or $(x + 7)^ 2$ = 0x + 7 = 0or, x = -7II. $y^ 2$ + 9y = 0or y(y + 9) = 0y = 0, -9ie,. no relation between x and y.