# SBI PO Quadratic Equation Questions and Answers set - 9

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 can be established between ‘x’ and ‘y’.

$Q.$
I. $2x^ 2$ + 13x - 7 = 0
II. $2y^ 2$ - 5y + 3 = 0

 A.   $x > y$ B.   $x \geq y$ C.   $x < y$ D.   $x \leq y$
2 . I. $2x^ 2$ -15x + 28 = 0
II. $4y^ 2$ - 16y + 15 = 0

 A.   $x > y$ B.   $x \geq y$ C.   $x < y$ D.   $x \leq y$
3 . I. $x^ 2$ + 8x + 16 = 0
II. $y^ 2$ = 16

 A.   $x > y$ B.   $x \geq y$ C.   $x < y$ D.   $x \leq y$
4 . I. $x^ 2$ - 2x - 24 = 0
II.$y^ 2$ + 8y = 0

 A.   $x > y$ B.   $x \geq y$ C.   $x < y$ D.   x = y or no relation can be established between ‘x’ and ‘y’.
5 . I. $x^ 2$ + 4x = 0
II.$y^ 2$ + 10y + 25 = 0

 A.   $x > y$ B.   $x \geq y$ C.   $x < y$ D.   $x \leq y$
6 . I. $2x^ 2$ + x – 1 = 0
II. $2y^ 2$ + 13y + 15 = 0

 A.   $x > y$ B.   $x \geq y$ C.   $x < y$ D.   $x \leq y$
7 . I. $x^ 2$ + 12x + 32 = 0
II. $2y^ 2$ + 15y + 27 = 0

 A.   $x > y$ B.   $x \geq y$ C.   $x < y$ D.   x = y or no relation can be established between ‘x’ and ‘y’.
8 . I. $6x^ 2$ – 17x + 12 = 0
II. $7y^ 2$ – 13y + 6 = 0

 A.   $x > y$ B.   $x \geq y$ C.   $x < y$ D.   $x \leq y$
9 . I. $x^ 2$ – 82x + 781 = 0
II. $y^ 2$ = 5041

 A.   $x > y$ B.   $x < y$ C.   x = y or no relation can be established between ‘x’ and ‘y’. D.   $x \geq y$
10 . I. $6x^ 2$ – 47x + 80 = 0
II. $2y^ 2$ – 9y + 10 = 0

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

1 .
 Answer : Option C Explanation : I.$2x^ 2$ + 13x - 7 = 0or $2x^ 2$ + 14x - x - 7 = 0or 2x (x + 7) - 1 (x + 7) = 0or (2x - 1) (x + 7) = 0x = $1\over 2$ , -7II. $2y^ 2$ - 5y + 3 = 0or $2y^ 2$ - 2y - 3y + 3 = 0or 2y(y - 1) - 3(y - 1) = 0or (2y - 3) (y - 1) = 0y = 1 , $3\over 2$Hence x < y
2 .
 Answer : Option A Explanation : I. $2x^ 2$ - 8x - 7x + 28 = 0or 2x (x - 4) - 7(x - 4) = 0or (2x - 7) (x - 4) = 0x = 4 , $7\over 2$II. $4y^ 2$ - 16y + 15 = 0or $4y^ 2$ - 6y - 10y + 15 = 0or 2y (2y - 3) - 5(2y - 3) = 0or (2y - 5) (2y - 3) = 0y = $5\over 2$, $3\over 2$Hence x > y
3 .
 Answer : Option D Explanation : I. $x 2$ + 8x + 16 = 0or $(x + 4)2$ = 0or x + 4 = 0 x = -4II. $y 2$ = 16 y = ±4Hence, x $\leq$ y
4 .
 Answer : Option D Explanation : I.$x 2$ - 2x - 24 = 0or $x 2$ + 4x - 6x - 24 = 0or x(x + 4) - 6(x + 4) = 0or (x - 6) (x + 4) = 0x = 6, - 4II. $y 2$ + 8y = 0or y(y + 8) = 0y = 0, - 8 ie No relation can be established between x and y.
5 .
 Answer : Option A Explanation : I. $x 2$ + 4x = 0or x(x + 4) = 0 x = 0, - 4II. $y 2$ + 10y + 25 = 0or $(y + 5)^ 2$ = 0or y + 5 = 0 y = - 5x > y
6 .
 Answer : Option A Explanation : I. $2x^ 2$ + 2x – x – 1 = 0or 2x(x + 1) – 1(x + 1) = 0or (2x – 1) (x + 1) = 0x = -1, $1\over 2$II. $2y^ 2$ + 3y + 10y + 15 = 0or y(2y + 3) + 5(2y + 3) = 0or (y + 5) (2y + 3) = 0y = -5 , -$3\over 2$x > y
7 .
 Answer : Option D Explanation : I. $x^ 2$ + 4x + 8x + 32 = 0or x(x + 4) + 8(x + 4) = 0or (x + 4) (x + 8) = 0x = – 4, – 8II. $2y^ 2$ + 6y + 9y + 27 = 0or 2y(y + 3) + 9(y + 3) = 0or (2y + 9) (y + 3) = 0y = -$9\over 2$, - 3No relation can be established between x and y.
8 .
 Answer : Option A Explanation : I. $6x^ 2$ – 9x – 8x + 12 = 0or 3x(2x – 3) – 4(2x – 3) = 0or (2x – 3) (3x – 4) = 0x = $3\over 2$, $4\over 3$II. $7y^ 2$ – 7y – 6y + 6 = 0or 7y(y – 1) – 6(y – 1) = 0or (7y – 6) (y – 1) = 0y = 1, $6\over 7$x > y
9 .
 Answer : Option C Explanation : I. $x^ 2$ – 11x – 71x + 781 =0or x(x – 11) – 71(x – 11) = 0or(x – 11)(x – 71) = 0x = 11, 71II. $y^ 2$ = 5041y = ± 71No relation between x and y.
10 .
 Answer : Option B Explanation : I. $6x^ 2$ – 15x – 32x + 80 = 0or 3x(2x – 5) – 16(2x – 5) = 0or (3x – 16) (2x – 5) = 0x = $16\over 3$, $5\over 2$II. $2y^ 2$ – 4y – 5y + 10 = 0or 2y(y – 2) – 5(y – 2) = 0or (y – 2) (2y – 5) = 0y = 2, $5\over 2$$x \geq y$