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  3. CEE Kerala Engineering

CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2008

  • question_answer
    The HM of two numbers is 4. Their AM is A and GM is G. If\[2A+{{G}^{2}}=27,\]then A is equal to

    A)  9                                            

    B)  \[\frac{9}{2}\]

    C)  18                         

    D)         27

    E)  15

    Correct Answer: B

    Solution :

    Let the two numbers be a and b. Since, \[\frac{2ab}{2A}=4,A=\frac{a+b}{2}\]and\[G=\sqrt{ab}\] \[\therefore \]  \[\frac{2ab}{2A}=4\Rightarrow A=\frac{ab}{4}\] and        \[{{G}^{2}}=4A\] Given,   \[2A+{{G}^{2}}=27\] \[\therefore \]  \[2A+4A=27\] \[\Rightarrow \]               \[A=\frac{9}{2}\]


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