7th Class Mathematics Integers Question Bank Integer

  • question_answer
    Fill in the blanks with >, < or =.
    (i) \[22\times (-36)+(-42)\div (-3)9\times 56\times (-8)\]
    (ii) \[26-22+21\div (-7)43-11-13\]
    (iii) \[(-16)\div [(-13)+(-3)]12\div [(-3)-0]\]

    A)
    (i) (ii) (iii)
    > < =
                   

    B)
    (i) (ii) (iii)
    <    <    >
                   

    C)
    (i) (ii) (iii)
    >    <    >
                   

    D)
    (i) (ii) (iii)
    =    >    <

    Correct Answer: C

    Solution :

              (i) \[22\times (-36)+(-42)\div (-3)\] \[=22\times (-36)-42\times \frac{1}{-3}\] \[=22\times (-36)+14=-792+14=-778\] Also,  \[9\times 50\times (-8)=-4032\] So,  \[-778\,\,\,-4032\] \[\therefore \] \[22\times (-36)+(-42)\div (-3)\,\,9\times 56\times (-8)\] (ii) \[26-22+21\div (-7)\] \[=26-22+21\times \frac{1}{-7}=26-22-3=26-25=1\]Also,  \[43-11-13=43-24=19\] So,       \[1\,\,19\] \[\therefore \]   \[26-22+21\div (-7)\,\,43-11-13\] (iii) \[-16\,\,\div \,(-13-3)+(-3)\] \[=-16\div (-13-3)=(-16)\div (-16)=1\] Also,\[12\div [(-3)-0]=12\div (-3)=12\times \frac{1}{-3}=-4\] \[\therefore \] \[-16\,\,\div \,[(-13)+(-3)]\,\,12\div [(-3)-0]\]


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