Quantitative Aptitude Ques 2089

Question: Directions: In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer. [IBPS (RRB) Grade A 2012]

I. $ (6x^{2}+17)-(3x^{2}+20)=0 $ II. $ (5y^{2}-12)-(9y^{2}-16)=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. $ (6x^{2}+17)-(3x^{2}+20)=0 $

$ \Rightarrow $ $ 6x^{2}-3x^{2}+17-20=0 $

$ \Rightarrow $ $ 3x^{2}-3=0 $

$ \Rightarrow $ $ x^{2}=1 $
$ \Rightarrow $ $ x=\pm 1 $ II. $ (5y^{2}-12)-(9y^{2}-16)=0 $

$ \Rightarrow $ $ 5y^{2}-9y^{2}-12+16=0 $

$ \Rightarrow $ $ -4y^{2}+4=0 $

$ \Rightarrow $ $ y^{2}=1 $
$ \Rightarrow $ $ y=\pm 1 $ Hence, $ x=y $

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