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8th Class Mathematics Exponents and Power Question Bank Exponents and Powers

  • question_answer
    The value of \[\frac{1}{2}\left[ \frac{-7}{4}+\frac{5}{3}+\frac{-5}{6}+\frac{1}{3}+\frac{-1}{2} \right]\]\[+\frac{1}{3}\left[ \frac{-12}{5}+\frac{-7}{20}+\frac{3}{14}+\frac{1}{7}+\frac{-1}{10} \right]\] is

    A)  \[\frac{1153}{840}\]     

    B)  \[\frac{128}{105}\]         

    C)  \[\frac{128}{105}\]                  

    D)  \[\frac{-125}{256}\]

    Correct Answer: A

    Solution :

    (a): Let \[A=\frac{1}{2}\left[ \frac{-7}{4}+\frac{5}{3}-\frac{5}{6}+\frac{1}{3}-\frac{1}{2} \right]\] \[=\frac{1}{2}\left[ \frac{-21+20-10+4-6}{12} \right]\] \[=\frac{1}{2}\left[ \frac{-13}{12} \right]=\frac{-13}{24}\] Let B\[=\frac{1}{3}\left[ \frac{-12}{5}+\frac{7}{20}-\frac{3}{14}+\frac{1}{7}-\frac{1}{10} \right]\] \[=\frac{1}{3}\left[ \frac{-336-49+30+20-14}{140} \right]\] \[=\frac{1}{3}\left[ \frac{-349}{140} \right]=\frac{-349}{420}\] Now \[A+B=\frac{-13}{24}-\frac{349}{420}\] \[=\frac{-1820-2094}{3360}=\frac{-3904}{3360}\]             \[=\frac{-976}{840}=\frac{-122}{105}\].           


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