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SSC Quantitative Aptitude Algebra Question Bank Elementary Algebra (II)

  • question_answer
    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

    A) \[-\,1\]

    B) \[1\]

    C) \[2\]

    D) \[4\]

    Correct Answer: B

    Solution :

    [b] \[{{x}^{2}}=y+z\] \[\Rightarrow \]   \[{{x}^{2}}+x=x+y+z\] \[\Rightarrow \]   \[x\,(x+1)=x+y+z\]                   … (i) Similarly, \[y\,(y+1)=x+y+z\]                   … (ii) and       \[z\,(z+1)=x+y+z\]                   … (iii) \[\therefore \]      \[\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\] \[=\frac{x}{x+y+z}+\frac{y}{x+y+z}+\frac{z}{x+y+z}\] \[=\frac{x+y+z}{x+y+z}=1\]


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