The dimensions of a room are 15 m, 10 m and 8 m. The volume of a bag is 2.25 m³. The maximum number of bags that can be accommodated in the room is

### Concept & Maximizing Space (Volume Division) To find the maximum number of identical items that can fit into a larger space, divide the total volume of the space by the volume of a single item. $$\text{Number of items} = \frac{\text{Total Volume}}{\text{Volume of one item}}$$ ### Step-by-Step Solution * **Given:** Dimensions of room = $15\text{ m} \times 10\text{ m} \times 8\text{ m}$. Volume of one bag = $2.25\text{ m}^3$. * **Step 1: Calculate total volume of the room** $\text{Volume} = 15 \times 10 \times 8 = 1200\text{ m}^3$. * **Step 2: Divide by the volume of a single bag** $\text{Number of bags} = \frac{1200}{2.25}$ * **Step 3: Simplify the fraction** Multiply numerator and denominator by 100 to remove decimals: $\frac{120000}{225}$ Divide by 25: $\frac{4800}{9}$ Divide by 3: $\frac{1600}{3} = 533.33...$ * Since we can only accommodate whole bags, we take the integer part (floor value). Maximum bags = $533$. ### Exam Strategy & Shortcut Instead of multiplying $15 \times 10 \times 8$ fully, keep it as a product and write $2.25$ as a fraction ($\frac{9}{4}$). $\text{Number of bags} = \frac{15 \times 10 \times 8}{\frac{9}{4}} = \frac{15 \times 10 \times 8 \times 4}{9}$. Cancel out common factor $3$: $\frac{5 \times 10 \times 8 \times 4}{3} = \frac{1600}{3} \approx 533$. ### Common Pitfall Rounding up to 534 is a common mistake. Even if the decimal was $.9$, you cannot fit a whole bag into a partial space. Always round down (take the floor) for physical packing problems. ### Final Answer Therefore, the correct answer is **533**.