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SSC Quantitative Aptitude Basic Calculations Question Bank Square and Cube Root (I)

  • question_answer
                \[3+\frac{1}{\sqrt{3}}+\frac{1}{3+\sqrt{3}}+\frac{1}{\sqrt{3}-3}\] is equal to

    A) 0

    B) 1

    C) 3

    D) \[3+\sqrt{3}\]

    Correct Answer: C

    Solution :

    [c]\[\frac{1}{\sqrt{3}}=\frac{1}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}}{3},\] \[\frac{1}{(3+\sqrt{3})}=\frac{1}{(3+\sqrt{3})}\times \frac{(3-\sqrt{3})}{(3-\sqrt{3})}\] \[=\frac{(3-\sqrt{3})}{(9-3)}=\frac{(3-\sqrt{3})}{6}\] \[\frac{1}{(\sqrt{3}-3}=\frac{1}{(\sqrt{3}-3)}\times \frac{(\sqrt{3}+3)}{(\sqrt{3}+3)}\] \[=\frac{(\sqrt{3}+3)}{3-9}=\frac{-\,\,(\sqrt{3}+3)}{6}\] \[\therefore \] Given expression \[3+\frac{\sqrt{3}}{3}+\frac{(3-\sqrt{3})}{6}-\frac{(\sqrt{3}+3)}{6}\] \[=3+\frac{\sqrt{3}}{3}+\frac{1}{2}-\frac{\sqrt{3}}{6}-\frac{\sqrt{3}}{6}-\frac{1}{2}\] \[=3+\frac{\sqrt{3}}{3}-\frac{2\sqrt{3}}{6}\] \[=3+\frac{\sqrt{3}}{3}-\frac{\sqrt{3}}{3}=3\]


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