Arithmetic Aptitude :: QE Quiz 14
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Exercise

 "In the middle of difficulty lies opportunity." - Albert Einstein
1 . Directions (Q. 1-5) : In the following questions, two equations numbered I and H 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 $\geq$ y
(5) if x = y or relationship cannot be established

$Q.$
I. 4x + 3y = $(1600)^ {1\over 2}$
II. 6x – 5y = $(484) ^{1\over2}$
 $x > y$ $x \geq y$ $x < y$ $x \leq y$
2 . I. $2x^2 - (4\div \sqrt{13})x + 2\sqrt{13}$ = 0
I. $10y^2 - (18 + 5\sqrt{13})y + 9\sqrt{13}$ = 0
 $x > y$ x ≥ y $x < y$ $x \leq y$
3 . I. $(6x^ 2x + l7) – (3x^ 2 + 20) = 0$
II. $(5y^ 2 – 12) – (9y^ 2 – 16) = 0$
 $x > y$ $x < y$ $x \geq y$ x = y or no relation can be established between ‘x’ and ‘y’.
4 . I. $(169)^{1\over 2}x + sqrt{289} = 134$
II. $(361)^{1\over 2}y^2 - 270 = 1269$
 $x > y$ x ≥ y $x < y$ $x \leq y$
5 . I. 821$x^2 - 75 x^2 = 256$
II. $\sqrt{196}y^3 - 12y^3 = 16$
 $x > y$ $x \geq y$ $x < y$ x ≤ y
6 . I.5x - 7y = -24
II.13x + 3y = 86
 $x > y$ $x \geq y$ $x < y$ $x \leq y$
7 . I. $x^ 2$ - 13x + 40 = 0
II. $y^ 2$ + 3y - 40 = 0
 $x > y$ x ≥ y $x < y$ $x \leq y$
8 . I. $8x^ 2$ -26x + 15 = 0
II. $2y^ 2$ -17y + 30 = 0
 $x > y$ $x \geq y$ $x < y$ x ≤ y
9 . I. $x^ 2$ = 484
II. $y^ 2$ - 45y + 506 = 0
 $x > y$ $x \geq y$ $x < y$ x ≤ y
II. y = $\sqrt[3]{2197}$
 $x > y$ $x < y$ $x \leq y$ x = y or no relation can be established between ‘x’ and ‘y’.