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SSC Quantitative Aptitude Algebra Question Bank Elementary Algebra (II)

  • question_answer
    If \[a+\frac{1}{b}=b+\frac{1}{c}=c+\frac{1}{a},\]where \[a\ne b\ne c\ne 0,\]then the value of \[{{a}^{2}}{{b}^{2}}{{c}^{2}}\]is [SSC CGL Tier II, 2015]

    A) \[abc\]

    B) \[1\]

    C) \[-\,1\]

    D) \[0\]

    Correct Answer: B

    Solution :

    [b] \[a+\frac{1}{b}=b+\frac{1}{c}=c+\frac{1}{a}\]                         \[a-b=\frac{1}{c}-\frac{1}{b}\]             \[\Rightarrow \]   \[a-b=\frac{b-c}{cb}\]             \[\therefore \]      \[cb=\frac{b-c}{a-b}\] Similarly, \[ac=\frac{c-1}{b-c}\]and\[ab=\frac{a-b}{c-a}\] Then, \[{{a}^{2}}{{b}^{2}}{{c}^{2}}=\frac{(b-c)}{(a-b)}\times \frac{(c-a)}{(b-c)}\times \frac{(a-b)}{(c-a)}=1\]


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