**Question:**

*Tick (✓) the correct answer:*

The dimensions of a room are (10 m × 8 m × 3.3 m). How many men can be accommodated in this room if each man requires 3 m3 of space?

(a) 99

(b) 88

(c) 77

(d) 75

**Solution:**

(b) 88

Volume of the room $=10 \times 8 \times 3.3=264 \mathrm{~m}^{3}$

One person requires $3 \mathrm{~m}^{3}$.

$\therefore$ Total no. of people that can be accommodated $=\frac{264}{3}=88$