Question

In the cafeteria, there is one large table with 10 seats and many smaller tables with 4 seats. The cafeteria has a fire code restriction that the room can hold a maximum of 70 students. If a student is seated at every seat at each table, how many small tables can be in the cafeteria without breaking the fire code?
Use your table to help you write the solution to the inequality.
x=14 x=15 (boundary) x=16
10+4x≤70 True True False


x

Answer 1

Let \( x \) represent the number of smaller tables in the cafeteria. Each small table has 4 seats, and there is one large table with 10 seats. The total seating capacity can be expressed with the following inequality based on the fire code restriction:

\[
10 + 4x \leq 70
\]

To solve for \( x \), we start by isolating \( x \):

\[
4x \leq 70 - 10
\]
\[
4x \leq 60
\]
\[
x \leq \frac{60}{4}
\]
\[
x \leq 15
\]

This means that the maximum number of small tables that can be in the cafeteria is 15.

Now let's evaluate the boundary points you provided (x = 14, x = 15, x = 16) using the inequality \( 10 + 4x \leq 70 \):

- For \( x = 14 \):
\[
10 + 4(14) = 10 + 56 = 66 \quad \text{(True)}
\]

- For \( x = 15 \):
\[
10 + 4(15) = 10 + 60 = 70 \quad \text{(True)}
\]

- For \( x = 16 \):
\[
10 + 4(16) = 10 + 64 = 74 \quad \text{(False)}
\]

Based on this evaluation, the values of \( x \) satisfying the inequality are \( x \leq 15 \). Therefore, the maximum number of small tables that can be in the cafeteria without breaking the fire code is **15**.