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

  • question_answer
                \[\frac{1}{(1+\sqrt{2}}+\frac{1}{(\sqrt{2}+\sqrt{3})}+\frac{1}{(\sqrt{3}+\sqrt{4})}+...+\frac{1}{(\sqrt{99}+\sqrt{100})}\]is equal to

    A) 1

    B) 5

    C) 9

    D) 10

    Correct Answer: C

    Solution :

    [c]\[\frac{1}{1+\sqrt{2}}=\frac{1}{(\sqrt{2}+1)}\times \frac{(\sqrt{2}-1)}{(\sqrt{2}-1)}\] \[=\frac{(\sqrt{2}-1)}{(2-1)}=(\sqrt{2}-1)\] Similarly, \[\frac{1}{\sqrt{2}+\sqrt{3}}=\frac{1}{\sqrt{3}+\sqrt{2}}\times \frac{(\sqrt{3}-\sqrt{2)}}{(\sqrt{3}-\sqrt{2})}\] \[=(\sqrt{3}-\sqrt{2}),\,\frac{1}{(\sqrt{3}+\sqrt{4})}\] \[=(\sqrt{4}-\sqrt{3}),\] \[...\frac{1}{(\sqrt{100}+\sqrt{99})}=(\sqrt{100}-\sqrt{99})\] \[\therefore \] Given expression \[=(\sqrt{2}-1)+(\sqrt{3}-\sqrt{2})+(\sqrt{4}-\sqrt{3})\] \[+...+(\sqrt{99}-\sqrt{98})+(\sqrt{100}-\sqrt{99})\] \[=(10-1)=9\]


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