How many persons can be accommodated in a big - hall of dimensions 40 m . 25m and 15 m. Assuming that each person requires ` 5 m ^(3) ` of air?

Answer

To solve the problem of how many persons can be accommodated in a big hall with dimensions 40 m, 25 m, and 15 m, given that each person requires 5 m³ of air, we will follow these steps:

### Step-by-Step Solution:

1. **Calculate the Volume of the Hall**:
The volume \( V \) of a rectangular hall can be calculated using the formula:
\[ V = \text{length} \times \text{width} \times \text{height} \]
Here, the dimensions of the hall are:
- Length = 40 m
- Width = 25 m
- Height = 15 m

So, the volume of the hall is:
\[ V = 40 \, \text{m} \times 25 \, \text{m} \times 15 \, \text{m} \]

2. **Perform the Multiplication**:
First, multiply the length and width:
\[ 40 \times 25 = 1000 \, \text{m}^2 \]
Now, multiply this result by the height:
\[ 1000 \, \text{m}^2 \times 15 \, \text{m} = 15000 \, \text{m}^3 \]
Therefore, the volume of the hall is \( 15000 \, \text{m}^3 \).

3. **Determine the Volume Required for One Person**:
Each person requires \( 5 \, \text{m}^3 \) of air.

4. **Calculate the Number of Persons that can be Accommodated**:
Let \( N \) be the number of persons that can be accommodated in the hall. The total volume of air required for \( N \) persons is:
\[ N \times 5 \, \text{m}^3 \]
Setting this equal to the volume of the hall, we have:
\[ 15000 \, \text{m}^3 = N \times 5 \, \text{m}^3 \]

5. **Solve for \( N \)**:
To find \( N \), divide both sides of the equation by \( 5 \, \text{m}^3 \):
\[ N = \frac{15000 \, \text{m}^3}{5 \, \text{m}^3} = 3000 \]

### Final Answer:
The number of persons that can be accommodated in the hall is **3000**.

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