**Question:**

A field is 150 m long and 100 m wide. A plot (outside the field) 50 m long and 30 m wide is dug to a depth of 8 m and the earth taken out from the plot is spread evenly in the field. By how much is the level of field raised?

**Solution:**

The dimensions of the plot dug outside the field are $50 \mathrm{~m} \times 30 \mathrm{~m} \times 8 \mathrm{~m}$.

Hence, volume of the earth dug - out from the plot $=50 \times 30 \times 8=12000 \mathrm{~m}^{3}$

Suppose that the level of the earth rises by $\mathrm{hm}$.

When we spread this dug - out earth on the field of length $150 \mathrm{~m}$, breadth $100 \mathrm{~m}$ and height $\mathrm{h} \mathrm{m}$, we have:

Volume of earth dug $-$ out $=150 \times 100 \times \mathrm{h}$

$\Rightarrow 12000=15000 \times \mathrm{h}$

$\Rightarrow \mathrm{h}=\frac{12000}{15000}=0.8 \mathrm{~m}$

$\Rightarrow \mathrm{h}=80 \mathrm{~cm} \quad(\because 1 \mathrm{~m}=100 \mathrm{~cm})$

$\therefore$ The level of the field will rise by $80 \mathrm{~cm}$.