8th Class Mathematics Rational Numbers Question Bank Rational Numbers

  • question_answer
    If \[\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\]\[\frac{1}{90}+\frac{1}{110}+\frac{1}{132}+\frac{1}{156}+=\frac{a}{b}\] and HCF \[\mathbf{(a,}\,\,\mathbf{b)=1}\] then (a + b) = ?

    A)  156 

    B)  162            

    C)  165                             

    D)  126

    Correct Answer: A

    Solution :

    (a): \[1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}=\frac{a}{b}\] \[\frac{a}{b}=\left( 1-\frac{1}{13} \right)=\frac{12}{13}\] \[\therefore a\times b=12\times 13=156\] 


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