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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2010

  • question_answer
    The sum of all two digit natural numbers which leave a remainder 5 when they are divided by 7 is equal to

    A)  715                       

    B)         702

    C)  615                       

    D)         602

    E)  589

    Correct Answer: B

    Solution :

    The required natural numbers are \[12,19,26,33,40,47,54,61,68,75,82,89,96\] \[\therefore \] \[Sum=12+19+26+33+40+47+54\] \[+61+68+75+82+89+96\] \[=702\] Alternate The required natural numbers are \[12,\text{ }19,\text{ }26,\text{ }...,96\] Let there are n terms in this series. Then last term \[=96=12+(n-1)7\] \[\Rightarrow \]               \[n-1=\frac{84}{7}=12\] \[\Rightarrow \]               \[n=13\] Now, sum of these terms \[{{S}_{13}}=\frac{13}{2}[24+12\times 7]=\frac{13}{2}\times 108\]                 \[=13\times 54=702\]


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