Equations And Inequations Ques 38
I. $\left(x^{1 / 4} \div 16\right)^{2}=144 \div x^{3 / 2}$
II. $y^{1 / 3} \times y^{2 / 3} \times 3104=16 \times y^{2}$
Directions : In the following questions two equations numbered I and II 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 $x<y$
(5) if $x=y$ or the relationship cannot be established
(UCO Bank PO Exam. 30.01.2011)
Show Answer
Correct Answer: (3)
Solution: (3) I. $\left(\frac{x^{\frac{1}{4}}}{16}\right)^{2}=\frac{144}{x^{\frac{3}{2}}}$
$\Rightarrow \frac{x^{\frac{1}{2}}}{256}=\frac{144}{x^{\frac{3}{2}}}$
$\Rightarrow x^{\frac{1}{2}} \times x^{\frac{3}{2}}=256 \times 144$
$\Rightarrow x^{2}=256 \times 144$
$\therefore x=\sqrt{256 \times 144}$
$= \pm(16 \times 12)= \pm 192$
II. $y^{\frac{1}{3}} \times y^{\frac{2}{3}} \times 3104=16 \times y^{2}$
$\Rightarrow y \times 3104=16 \times y^{2}$
$\Rightarrow 3104=16 y$
$\Rightarrow y=\frac{3104}{16}=194$
Clearly, $x<y$
BETA
