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SSC Quantitative Aptitude Algebra Question Bank Elementary Algebra (II)

  • question_answer
    If \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}=xy+yz+zx,\]then the value of \[\frac{3{{x}^{4}}+7{{y}^{4}}+5{{z}^{4}}}{5{{x}^{2}}{{y}^{2}}+7{{y}^{2}}{{z}^{2}}+3{{z}^{2}}{{x}^{2}}}\]is [SSC CGL Tier II, 2015]

    A) 0

    B) -1

    C) 2

    D) 1

    Correct Answer: D

    Solution :

    [d] \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}=xy+yz+zx\]             \[\Rightarrow \]   \[2{{x}^{2}}+2{{y}^{2}}+2{{z}^{2}}=2xy+2yz+2zx\]             \[\Rightarrow \]   \[{{x}^{2}}-2xy+{{y}^{2}}+{{y}^{2}}-2yz\]                                     \[+{{z}^{2}}+{{z}^{2}}-2zx+{{x}^{2}}=0\] \[\Rightarrow \]   \[{{(x-y)}^{2}}+{{(y-z)}^{2}}+{{(z-x)}^{2}}=0\] \[\therefore \]      \[x=y=z\] Then,\[\frac{3{{x}^{4}}+7{{y}^{4}}+5{{z}^{4}}}{5{{x}^{2}}{{y}^{2}}+7{{y}^{2}}+3{{z}^{2}}{{x}^{2}}}\] \[=\frac{{{x}^{4}}(3+7+5)}{5{{x}^{4}}+7{{x}^{4}}+3{{x}^{4}}}\][\[\because \]x = y= z] \[=\frac{15}{5+7+3}=\frac{15}{15}=1\]


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