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JCECE Engineering JCECE Engineering Solved Paper-2015

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

    A) \[12.5\,\,\min \]                              

    B) \[135\,\,\min \]

    C) \[34\,\,\min \]                  

    D) \[24\,\,\min \]

    Correct Answer: C

    Solution :

    The number of notes counted in first 10 minutes\[=150\times 10=1500\] Suppose, the person counts the remaining \[3000\] currency notes in n minutes. Then, \[3000=\] Sum of n terms of an \[AP\] with first term \[148\]and common difference\[-2\] \[\Rightarrow \]               \[3000=\frac{n}{2}\{2\times 148+(n-1)\times (-2)\}\] \[\Rightarrow \]               \[3000=n(149-n)\] \[\Rightarrow \]               \[{{n}^{2}}-149n+3000=0\] \[\Rightarrow \]               \[(n-125)(n-24)=0\] \[\Rightarrow \]               \[n=125,\,\,24\] Clearly, \[n=125\] is not possible. Total, time taken\[=(10+24)=34\min \].


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