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SSC Sample Paper Mock Test-15 SSC CGL Tear-II Paper-1

  • question_answer
    The dimensions of a field are \[12\,\,\text{m}\times 10\,\,\text{m}\]. A pit 5 m long, 4 m wide and 2 m deep is dug in one corner of the field and the Earth removed has been evenly spread over the remaining area of the field. The level of the field is raised by

    A) 30 cm

    B)  35 cm  

    C)  38 cm

    D)  40 cm

    Correct Answer: D

    Solution :

    Area of the field = length \[\times \]breadth \[=12\times 10=120\,\,{{\text{m}}^{2}}\] Area of the pit's surface \[=5\times 4=20\,\,{{\text{m}}^{2}}\] Area on which the Earth is to be spread \[=120-20=100\,\,{{\text{m}}^{2}}\] Volume of Earth dug out \[=5\times 4\times 2=40\,\,{{\text{m}}^{3}}\] \[\therefore \] Level of field raised\[=\frac{40}{100}=\frac{2}{5}\,\,\text{m}\] \[=\frac{2}{5}\times 100=40\,\,\text{cm}\]


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