Equations And Inequations Ques 63
I. $\frac{18}{x^{2}}+\frac{6}{x}-\frac{12}{x^{2}}=\frac{8}{x^{2}}$
II. $y^{2}+9.68+5.64=16.95$
Directions : In the following questions two equations numbered I and II are given. You have to solve both the equations and give answer. If
(1) $x>y$
(2) $\quad x \geq y$
(3) $x<y$
(4) $x \leq y$
(5) $x=y$ or the relationship cannot be established
(Bank Of Baroda PO Exam.13.03.2011)
Show Answer
Correct Answer: 3
Solution: (3) I. $\frac{18}{x^{2}}-\frac{12}{x^{2}}-\frac{8}{x^{2}}=\frac{-2}{x^{2}}$
$\Rightarrow \frac{18-12-8}{x^{2}}=\frac{-6}{x}$
$\Rightarrow \frac{-2}{x^{2}}=\frac{-6}{x} \Rightarrow \frac{2}{x}=6 \Rightarrow x=\frac{1}{3}$
$\Rightarrow x=\frac{2}{6}=\frac{1}{3} \Rightarrow x^{3}=\frac{1}{27}=0.037$
II. $y^{3}=16.95-9.68-5.64$ $=1.63$
Clearly, $x<y$
BETA
