Question

How many persons can be accommodated in a dining hall of dimensions $\mathbf{(}\mathbf{20}\mathbf{}\mathit{m}\mathbf{\times }\mathbf{16}\mathbf{}\mathit{m}\mathbf{\times }\mathbf{4}\mathbf{.}\mathbf{5}\mathbf{}\mathit{m}\mathbf{)}$, assuming that each person requires $\mathbf{5}\mathbf{}\mathit{c}\mathit{u}\mathit{b}\mathit{i}\mathit{c}\mathbf{}\mathit{m}\mathit{e}\mathit{t}\mathit{r}\mathit{e}\mathit{s}$ of air?

Answer 1

Step 1: Find the volume of the dinning hall

Given, the dimensions of dining hall

Length $l=20m$

Breadth $b=16m$

Height $h=4.5m$

Volume of air consumed for each person $=5m^3$.

Volume of the dinning hall $=l\times b\times h$

$=20\times 16\times 4.5$

$=1440m^3$

$\therefore$ Volume of the dinning hall $=1440m^3$

Step 2: Find the number of persons can be accommodated in a dining hall .

$\mathrm{Number}\mathrm{of}\mathrm{persons}=\frac{\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{dinning}\mathrm{hall}}{\mathrm{Volume}\mathrm{of}\mathrm{air}\mathrm{consumed}\mathrm{for}\mathrm{each}\mathrm{person}}$

$=\frac{1440}{5}$

$=288$

Therefore, $288$ persons can be accommodated in a dining hall.