How many persons can be accommodated in a dining hall of dimensions $(20 m \times 16 m \times 4.5 m)$ , assuming that each person requires $5$ cubic meters of air.

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Solution

Soln:
Volume of the dinning hall = $(20 \times 16 \times 4.5) m^{3}$
air consumed of each person = $5 m^{3}$
So air consumed of total no. of persons = $5 x m^{3}$
so air filled in the dinning hall = volume of dinning hall
$5 x = 20 \times 16 \times 4.5$
$x = \frac{20 \times 16 \times 4.5}{5} = 288$
Total no. of persons in dinning hall = 288

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How many persons can be accommodated in a hall of length $16 m$ . breadth $12.5 m$ and height $4.5 m$ , assuming that $3.6 m^{3}$ of air is required for each person?

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How many persons can be accommodated in a dining hall of dimensions (20 m×16 m×4.5 m), assuming that each person requires 5 cubic meters of air.

Soln:Volume of the dinning hall = (20×16×4.5)m3air consumed of each person = 5m3So air consumed of total no. of persons = 5xm3so air filled in the dinning hall = volume of dinning hall5x=20×16×4.5x=20×16×4.55=288Total no. of persons in dinning hall = 288

How many persons can be accommodated in a dining hall of dimensions $(20 m \times 16 m \times 4.5 m)$ , assuming that each person requires $5$ cubic meters of air.