Question: Directions: In the following questions two equations numbered I and II are given. You have to solve both the equations and give answer. [Allahabad Bank (PO) 2011]
I. $ \frac{5}{7}-\frac{5}{21}=\frac{\sqrt{x}}{42} $
II. $ \frac{\sqrt{y}}{4}+\frac{\sqrt{y}}{16}=\frac{250}{\sqrt{y}} $
Options:
A) If $ x>y $
B) If $ x\ge y $
C) If $ x<y $
D) If $ x\le y $
E) If $ x=y $ or the relationship cannot be established
Show Answer
Answer:
Correct Answer: C
Solution:
- I. $ \frac{5}{7}-\frac{5}{21}=\frac{\sqrt{x}}{42} $
$ \Rightarrow $ $ \frac{15-5}{21}=\frac{\sqrt{x}}{42} $
$ \Rightarrow $ $ \frac{10}{21}=\frac{\sqrt{x}}{42} $
$ \Rightarrow $ $ \sqrt{x}=20 $
$ \Rightarrow $ $ x=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 $ $ \frac{5\sqrt{y}}{16}=\frac{250}{\sqrt{y}} $
$ \Rightarrow $ $ 5y=250\times 16 $
$ \Rightarrow $ $ y=50\times 16=800 $
$ \therefore $ $ x<y $
BETA
