Quantitative Aptitude Ques 1383

Question: If $ x+y+z=12, $ $ x^{2}+y^{2}+z^{2}=44, $ then find $ xy+yz+zx. $

Options:

A) 26

B) 60

C) 36

D) 40

Show Answer

Answer:

Correct Answer: B

Solution:

  • Given expression in the form of $ {{(a+b+c)}^{2}}=a^{2}+b^{2}+c^{2}+2(ab+bc+ca) $ $ 2(ab+bc+ca)={{(a+b+c)}^{2}}-(a^{2}+b^{2}+c^{2}) $ $ ab+bc+ca=\frac{{{(a+b+c)}^{2}}-(a^{2}+b^{2}+c^{2})}{2} $

$ \therefore $ $ xy+yz+zx=\frac{{{(x+y+z)}^{2}}-(x^{2}+y^{2}+z^{2})}{2} $ $ =\frac{{{(12)}^{2}}-44}{2}=\frac{144-44}{2}=50 $