Equations And Inequations Ques 102
I. $\frac{5}{7}-\frac{5}{21}=\frac{\sqrt{x}}{42}$
II. $\frac{\sqrt{y}}{4}+\frac{\sqrt{y}}{16}=\frac{250}{\sqrt{y}}$
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) $x \geq y$
(3) $x<y$
(4) $x \leq y$
(5) $x=y$ or the relationship cannot be estab lished
(Allahabad Bank PO Exam. 17.04.2011)
Show Answer
Correct Answer: (3)
Solution: (3) I. $\frac{5}{7}-\frac{5}{21}=\frac{\sqrt{x}}{42}$
$\Rightarrow \frac{15-5}{21}=\frac{\sqrt{x}}{42}$
$\Rightarrow \sqrt{x}=\frac{10}{21} \times 42=20$
$\therefore x=20 \times 20=400$
II. $\frac{\sqrt{y}}{4}+\frac{\sqrt{y}}{16}=\frac{250}{\sqrt{y}}$
$\Rightarrow \frac{4 \sqrt{y}+\sqrt{y}}{16}=\frac{250}{\sqrt{y}}$
$\Rightarrow 5 \sqrt{y} \times \sqrt{y}=250 \times 16$
$\Rightarrow 5 y=250 \times 16$
$\Rightarrow y=\frac{250 \times 16}{5}=800$
Clearly, $x<y$
BETA
