Question

Sai is ordering new shelving units for his store. Each unit is $7$ feet in length and extends from floor to ceiling. Sai's store has $119$ feet of wall space that includes $21$ feet of windows along the walls. If the shelving units cannot be placed in front of the windows, which of the following inequalities includes all possible values of $r$, the number of shelving units that Sai could use?

• A. $r\le \frac{119-21}{7}$
• B. $r\ge \frac{119+21}{7}$
• C. $r\le 119-21+7r$
• D. $r\ge 119+21-7r$

Answer 1

The correct option is A $r\le \frac{119-21}{7}$

To figure out the total number of shelving units Sai could use, find the total available wall space and divide by the length of the units.

The total amount of wall space can be calculated as $119-21$. Because the length of each unit is $7$ feet, the maximum number of units Sai could put up can be calculated as$\frac{119-21}{7}$ .

Because this is the maximum number of units Sai could put up, $r$ has to be less than or equal to this number. Therefore, the correct answer is (A) is $r\le \frac{119-21}{7}$.