Arithmetic Aptitude :: QE Quiz 3
Home » Arithmetic Aptitude » QE Quiz 3 » General Questions

Description: Free Online Test questions and answers on Quadratic Equation with explanation for various competitive exams,entrance test. Solved examples with detailed answer test 3

Exercise

 "In the middle of difficulty lies opportunity." - Albert Einstein
1 . Directions (Q. 1-10): 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. $2x^ 2$ + x - 1 = 0
II. $6y^ 2$ - 13y + 5 = 0
 $x > y$ $x < y$ $x \geq y$ x = y or no relation can be established between ‘x’ and ‘y’.
2 . I. $21x ^ 2$ - 122x + 165 = 0
II. $3y^2$ - 2y - 33 = 0
 $x > y$ $x \geq y$ $x < y$ x = y or no relation can be established between ‘x’ and ‘y’.
3 . I. $5x 2$ - 29x + 36 = 0
II. $10y 2$ - 3y - 27 = 0
 $x > y$ $x \leq y$ $x < y$ $x \geq y$
4 . I. 7x + 4y = 3
II. 5x + 3y = 3
 $x > y$ $x < y$ $x \leq y$ $x \geq y$
5 . I. $7x^ 2$ - 54x + 99 = 0
II. $4y^ 2$ - 16y + 15 = 0
 $x > y$ $x < y$ $x \geq y$ $x \leq y$
6 . I. $5x^ 2$ - 87x + 378 = 0
II. $3y^ 2$ - 49y + 200 = 0
 $x > y$ $x < y$ $x \geq y$ $x \leq y$
7 . I. $10x^ 2$ - x - 24 = 0
II. $y^ 2$ - 2y = 0
 x = y or no relation can be established between ‘x’ and ‘y’. $x < y$ $x \geq y$ $x \leq y$
8 . I. $x^ 2$ - 5x + 6 = 0
II. $2y^ 2$ - 15y + 27 = 0
 $x > y$ $x < y$ $x \geq y$ $x \leq y$
 $x > y$ $x < y$ $x \geq y$ $x \leq y$
10 . I. $14x ^ 2$ - 37x + 24 = 0
II. $28y^ 2$ - 53y + 24 = 0
 $x > y$ $x < y$ $x \geq y$ $x \leq y$