Quantitative Aptitude Ques 2108

Question: $ x^{2}=y+z, $ $ y^{2}=z+x $ and $ z^{2}=x+y, $ then the value of $ \frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1} $ is

Options:

A) $ -1 $

B) $ 1 $

C) $ 2 $

D) $ 4 $

Show Answer

Answer:

Correct Answer: B

Solution:

  • $ x^{2}=y+z $ On adding x both sides, we get $ x^{2}+x=x+y+z $
    $ \Rightarrow $ $ x(x+1)=x+y+z $ $ \frac{1}{x+1}=\frac{x}{x+y+z} $ … (i) Similarly, $ \frac{1}{y+1}=\frac{y}{x+y+z} $ … (ii) and $ \frac{1}{z+1}=\frac{z}{x+y+z} $ … (iii) On adding Eqs. (i), (ii) and (iii), we get
    $ \frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=\frac{x}{x+y+z}+\frac{y}{x+y+z} $ $ +,\frac{z}{x+y+z} $ $ =\frac{x+y+z}{x+y+z}=1 $
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