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JEE Main & Advanced AIEEE Solved Paper-2010

  • question_answer
      A person is to count 4500 currency notes. Let \[{{a}_{n}}\]denote the number of notes he counts in the\[{{n}^{th}}\]minute. If\[{{a}_{1}}={{a}_{2}}=....={{a}_{10}}=150\]and\[{{a}_{10}},{{a}_{11}},...\]are in an AP with common difference ? 2, then the time taken by him to count all notes is -       AIEEE  Solved  Paper-2010

    A) 24 minutes                         

    B) 34 minutes         

    C)        125 minutes

    D)        135 minutes

    Correct Answer: B

    Solution :

    \[{{a}_{1}}={{a}_{2}}={{a}_{3}}.....\text{ }{{a}_{9}}=150\] \[{{a}_{1}}+{{a}_{2}}+{{a}_{3}}+....+{{a}_{9}}=1350\] \[{{a}_{10}}+{{a}_{11}}+......+{{a}_{n}}=45001350=3150\] \[\frac{n}{2}[2\times 150+(n-1)(-2)]=3150\] \[150n-{{n}^{2}}+n=3150\] \[{{n}^{2}}-151n+3150=0\] \[n=25\text{ }min\] hence total time\[=25+9=34\text{ }min\]


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