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SSC Sample Paper Mock Test-5 SSC CGL Tear-II Paper-1

  • question_answer
    If\[a+\frac{1}{b}=1,\]\[b+\frac{1}{c}=1,\]then\[c+\frac{1}{a}\] is equal to

    A)  0                                

    B)  \[\frac{1}{2}\]             

    C)  1        

    D)  2

    Correct Answer: C

    Solution :

    \[a+\frac{1}{b}=1\]\[\Rightarrow \]\[a=\left( 1-\frac{1}{b} \right)\]\[\Rightarrow \]\[a=\left( \frac{b-1}{b} \right)\]\[\Rightarrow \]\[\frac{1}{a}=\frac{b}{b-1}\] \[b+\frac{1}{c}=2\]\[\Rightarrow \]\[\frac{1}{c}=1-b=c=\frac{1}{1-b}\] \[c+\frac{1}{b}=\frac{1}{(1-b)}+\frac{b}{(b-1)}=\frac{1}{(1-b)}-\frac{b}{(1-b)}=\left( \frac{1-b}{1-b} \right)=1\]                        


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