Arithmetic Aptitude :: QE Quiz 19
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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
(4) if x ≤ y
(5) if x = y or no relation can be established between x and y.

$Q.$
I. $x^ 2$ - 2x - 15 = 0
II. $y^ 2$ + 5y + 6 = 0
 $x > y$ x ≥ y $x < y$ $x ≤ y$
2 . I. $x^ 2$ - x - 12 = 0
II. $y^ 2$ -3y + 2 = 0
 $x > y$ $x \geq y$ $x < y$ x = y or no relation can be established between ‘x’ and ‘y’.
3 . I.$x - \sqrt{169} = 0$
II.$y^2 - \sqrt{169} = 0$
 $x > y$ x ≥ y $x < y$ $x < y$
4 . I. $x^2 - 32 = 112$
II. y - $\sqrt{256}$ = 0
 $x > y$ $x \geq y$ $x < y$ $x \leq y$
5 . I. $x^ 2$ - 25 = 0
II. $y^ 2$ - 9y + 20 = 0
 $x > y$ $x \geq y$ $x < y$ x = y or no relation can be established between ‘x’ and ‘y’.
6 . Directions (Q. 6 - 10): In the following questions, three equations numbered I, II and III are given. You have to solve all the equations either together or separately, or two together and one separately or by any other method and give answer
(1) if x = y > z
(2) if x < y = z
(3) if x < y > z
(4) if x = y = z or if none of the above relationship can be established.
(5) if x ≤ y < z

$Q.$
I. 3x + 5y = 69
II. 9x + 4y = 108
III. x + z = 12
 x = y > z x < y = z x < y > z x = y = z or if none of the above relationship can be established.
7 . I . $y = \sqrt{9^{3 \times {1\over3}} \times {9^{3 \times {1\over3}}}}$ = $\sqrt{9 \times 9}$ = 9
II. 2x + 5z = 54
III. 6x + 4z = 74
 x = y > z x < y = z x < y > z x ≤ y < z
8 . I. 2x + 3y + 4z = 66
II. 2x + y + 3z = 42
III. 3x + 2y + 4z = 63
 x = y > z x < y = z x < y > z x ≤ y < z
9 . I. $(x + z)^ 3$ = 1728 = $12^ 3$
II. 2x + 3y = 35
III. x - z = 2
 x = y > z x < y = z x < y > z x ≤ y < z
10 . I. 4x + 5y = 37
II. x + z = 8
III. 7x + 3y = 36
 x = y > z x < y = z x < y > z x ≤ y < z