Arithmetic Aptitude :: QE Quiz 2
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Exercise

 "Loneliness is the most terrible poverty." - Mother Teresa
1 . Directions (Q. Nos. 1-5) : In the following questions two equations numbered I and II are given. You have to solve both the equations and—
(1) if x > y
(2) if x $\geq$ y
(3) if x < y
(4) if x $\leq$ y
(5) if x = y or the relationship cannot be established

$Q.$
I. $\sqrt{1225x} + \sqrt{4900} = 0$
II. $(81)^{1\over 4} y + (343)^{1\over 3} = 0$
 $x > y$ $x < y$ $x \geq y$ $x \leq y$
2 . I. $18\over x^2$ + $6\over x^2$ - $12\over x^2$ = $8\over x^2$
II. $y^ 3$ + 9.68 + 5.64 = 16.95
 $x > y$ $x \geq y$ $x \leq y$ x = y or no relation can be established between ‘x’ and ‘y’.
3 . I. $(2)^5 + (11)^3\over 6$ = $x^3$
II. $4y^3 = - (589 \div 4 ) + 5 y^3$
 $x > y$ $x \geq y$ $x < y$ $x \leq y$
4 . I. $12x^ 2$ + llx + 12 = 10x 2 +22x
II. $13y^ 2$ - 18y + 3 = 9y 2 - 10y
 $x > y$ $x \geq y$ $x < y$ $x \leq y$
5 . I. $(x^{7\over 5} \div 9)$ = $169 \div y{3\over 5}$
II. $y^{1\over 4} \times y^{1\over 4} \times 7$ = $273 \div y^{1\over 2}$
 $x > y$ $x \geq y$ $x < y$ $x \leq y$
6 . Directions (Q. 6 - 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. x = $\sqrt[ 4] {2401}$
II.$2y^ 2$ - 9y - 56 = 0
 $x > y$ $x < y$ $x \geq y$ x = y or no relation can be established between ‘x’ and ‘y’.
7 . I. $5x^ 2$ + 3x - 14 = 0
II.$2y^ 2$ - 9y + 10 = 0
 $x > y$ $x < y$ $x \geq y$ $x \leq y$
8 . I. $8x ^ 2$ + 31x + 21 = 0
II. $5y ^2$ + 11y - 36 = 0
 $x > y$ $x < y$ $x \leq y$ x = y or no relation can be established between ‘x’ and ‘y’.
II. y = $\sqrt{ 1089}$
 $x > y$ $x < y$ $x \geq y$ $x \leq y$
10 . I. $15x^ 2$ + 68x + 77 = 0
II. $3y^ 2$ + 29y + 68 = 0
 $x > y$ $x < y$ $x \geq y$ $x \leq y$