A field is $80 m$ long and $50 m$ broad. In one corner of field, a pit which is $10 m$ long, $7.5 m$ broad and $8 m$ deep has been dug out. The earth taken out of it is evenly spread over the remaining part of field. Find the rise in level of field.

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Solution

Length of field $= 80 m$
Breadth of field $= 50 m$
Area of rectangular field $= 80 m \times 50 m = 4000 m^{2}$
Length of pit $(l) = 10 m$
Breadth of pit $(b) = 7.5 m$
Height of pit $(h) = 8 m$
Volume of earth taken out $=$ Volume of pit
$= l \times b \times h$
$= 10 m \times 7.5 m \times 8 m$
$= 600 m^{3}$
Area of the base of pit $= 10 m \times 7.5 m = 75 m^{2}$
Area of remaining portion of rectangular field
$=$ Area of rectangular field $-$ Area of the base of pit
$= 4000 m^{2} - 75 m^{2}$
$= 3925 m^{2}$
Now rise in the level of remaining part of the field $= \frac{600 m^{3}}{3925 m^{2}}$
$= 0.1528 m$
$= 0.1528 \times 100 c m$
$= 15.28 c m$
So, rise in the remaining part of the field $= 15.28 c m$

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