Quantitative Aptitude Ques 331

Question: Directions: In the following questions two equations numbered I and II are given. You have to solve both the equations and give answer.

I. $ \frac{3}{\sqrt{x}}+\frac{4}{\sqrt{x}}=\sqrt{x} $
II. $ y^{3}-\frac{{{(7)}^{7/2}}}{\sqrt{y}}=0 $

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: E

Solution:

  • I. $ \frac{3}{\sqrt{x}}+\frac{4}{\sqrt{x}}=\sqrt{x} $

$ \Rightarrow $ $ \frac{3+4}{\sqrt{x}}=\sqrt{x} $
$ \Rightarrow $ $ x=7 $ II. $ y^{3}-\frac{{{(7)}^{7/2}}}{\sqrt{y}}=0 $

$ \Rightarrow $ $ \frac{{{(y)}^{3}}{{(y)}^{1/2}}-{{(7)}^{7/2}}}{\sqrt{y}}=0 $
$ \Rightarrow $ $ {{(y)}^{7/2}}-{{(7)}^{7/2}}=0 $

$ \Rightarrow $ $ {{(y)}^{7/2}}={{(7)}^{7/2}} $

$ \Rightarrow $ $ y=7 $ Hence, $ x=y $