A rectangular tank measuring  $5\mathrm{m}\times 4.5\mathrm{m}\times 2.1\mathrm{m}$ is dug in the centre of the field measuring $13.5\mathrm{m}\times 2.5\mathrm{m}.$  The earth dug out is spread evenly over the remaining portion of the field. How much is the level of the field raised?

We have
Area of the field $\text{=}\left(13.5\times 2.5\right){\mathrm{m}}^2=33.75{\mathrm{m}}^2$
Volume of the earth dug out =  vol. of the tank $=\left(5\times 4.5\times 2.1\right){\mathrm{m}}^3=47.25{\mathrm{m}}^3$
Area of the tank $=\left(5\times 4.5\right){\mathrm{m}}^2=22.5{\mathrm{m}}^2$
Area  on  which the earth dug out has been spread
= Area of the field  - Area of the tank $=\left(33.75-22.5\right){\mathrm{m}}^2=11.25{\mathrm{m}}^2$
$\therefore \mid \text{ Level raised }=\frac{\text{ volume of the earth dug out }}{\text{ Areaof whichtheearthis spread }}=\frac{47.25}{11.25}=4.2\mathrm{m}$