Quantitative Aptitude Ques 959
Question: If $ 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:
- Given, $ x^{2}=y+z $ On adding x both sides, we get $ x^{2}+x=x+y+z $
$ \Rightarrow $ $ x(x+1)=x+y+z $
$ \Rightarrow $ $ \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 addition 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 $
BETA
