Quantitative Aptitude Ques 781

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. $ \sqrt{500}x+\sqrt{402}=0 $ II. $ \sqrt{360}y+{{(200)}^{1/2}}=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: C

Solution:

  • I. $ \sqrt{500}x+\sqrt{402}=0 $

$ \Rightarrow $ $ 10\sqrt{5}x+\sqrt{402}=0 $

$ \Rightarrow $ $ 10\sqrt{5x}+20.04=0 $

$ \Rightarrow $ $ x=\frac{-20.04}{22.36}=-0.896 $ II. $ \sqrt{360}y+\sqrt{200}=0 $

$ \Rightarrow $ $ 6\sqrt{10}y+10\sqrt{2}=0 $

$ \Rightarrow $ $ y=\frac{-10\sqrt{2}}{6\times \sqrt{5}\times \sqrt{2}}=\frac{-10}{13.41}=-0.745 $

$ \therefore $ $ x<y $