Equations And Inequations Ques 155
I. $x^{3}=(\sqrt[3]{216})^{3}$
II. $6 y^{2}=150$
Directions : In each of the following questions equation I and equation II have been given. You have to solve both of these equations and give answer If
(1) $x<y$
(2) $x>y$
(3) $x \leq y$
(4) $x \geq y$
(5) $x=y$ or no relation between two can be established.
(Corporation Bank Specialist Officer (Marketing) Exam. 22.02.2014)
Show Answer
Correct Answer: (2)
Solution: (2) I. $x^{3}=(216)^{\frac{1}{3} \times 3}=216$
$\Rightarrow x=\sqrt[3]{216}=6$
II. $6 y^{2}=150$
$\Rightarrow y^{2}=\frac{150}{6}=25$
$\Rightarrow y=\sqrt{25}= \pm 5$
Clearly, $x>y$
BETA
