Arithmetic Aptitude :: QE Quiz 20
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Exercise

 "When ambition ends, happiness begins." - (Proverb)
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 or no relation can be established
(4) if x > y
(5) if x ≥ y

$Q.$
I. 7x + 3y = 77
II. 2x + 5y = $(2601)^{1\over 2}$ = 51
 x < y x ≤ y x = y or no relation can be established x > y
2 . I. $3x^2 - 6x - \sqrt{17}x + 2\sqrt{17}$ = 0
II. $10y^2 - 18y - 5\sqrt{17}y + 9\sqrt{17}$ = 0
 x < y x ≤ y x = y or no relation can be established x > y
3 . I . $(289)^{1\over 2} x$ - $\sqrt{324}$ = 203
II. $(484)^{1\over 2} y$ - $\sqrt{225}$ = 183
 x < y x ≤ y x = y or no relation can be established x > y
4 . I . $511 x^2$ = 3066
II. $12y^ 3$ - $9y^ 3$ = 1536
 x < y x ≤ y x > y x ≥ y
5 . Directions (Q. 5 - 7): In the following questions two equations numbered 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 relationship can be established

$Q.$
I. 3x + 4y = (4681)$^{1\over 2}$
II. 3x + 2y = $(961)^{1\over 2}$
 x < y x ≥ y x ≤ y x > y
6 . I. $3x^2 - 6x - \sqrt{17}x + 2\sqrt{17}$ = 0
II. $10y^2 - 15y + \sqrt{17}y - 3\sqrt{17}$ = 0
7 . I. $x^ 2$ - 16x + 63 = 0
II. $y^ 2$ - 2y - 35 = 0