Question

A field is in the form of a rectangle of length 18 m and width 15 m. A pit, 7.5 m long, 6 m broad and 0.8 m deep, is dug in a corner of the field and the earth taken out is spread over the remaining area of the field. Find out the extent to which the level of the field has been raised.

Answer 1

$\text{Length of a field (L)}=18m$

$\text{and width (B)}=15m$

$\text{Length of pit (l)}=7.5m$

$\text{Breadth (b)}=6m$

$\text{and depth (h)}=0.8m$

$\therefore \text{Volume of earth dugout}=lbh$

$=7.5\times 6\times 0.8m^3$

$=45\times 0.8=\frac{45\times 4}{5}=36m^3$

$\text{Total area of the field}=L\times B$

$=18\times 15=270m^2$

$\text{and area of pit}=lb=7.5\times 6=45m^2$

$\therefore \text{Remaining area of the field excluding pit}$

$=270-45=225m^2$

$\text{Let by spreading the earth on the remaining part of the field, the increase in height}=h^{\prime }$

$\Rightarrow 225\times h^{\prime }=36$

$\Rightarrow h^{\prime }=\frac{36}{225}=\frac{4}{25}=0.16m=16cm$

$\therefore \text{Level of field raised}=16cm$