RRB Technician 2013 Ques 101
Question: If $ x+y=20 $ and $ xy=84; $ what is the value of $ {{(x)}^{2}}+(y^{2}) $ ?
Options:
A) 232
B) 400
C) 128
D) Cannot be determined
Show Answer
Answer:
Correct Answer: A
Solution:
- Given $ x+y=20 $
$ \Rightarrow $ $ {{(x+y)}^{2}}={{(20)}^{2}} $
$ \Rightarrow $ $ x^{2}+y^{2}+2xy=400 $
$ \Rightarrow $ $ x^{2}+y^{2}=400-2\times 84 $ $ (\because xy=84) $ $ =400-168=232 $
BETA
