Question

Question 24We are planning a reception for our daughter. The hall that we are renting charges a $80 cleanup fee. Inaddition to the cleanup fee, they charge $29 per guest for food and drinks.If we have $2,100.00 available, what is the maximum number of guest we can invite without going over ourbudget?The maximum number of guests is

Answer 1

Answer:

69 guests

Explanation:

• Cleanup Fee = $80

,

• Cost of food and drinks per guest = $29

Let the number of guests = g

• Cost of food and drinks for g guest = $29g

The total expense will then be:

$80+29g$

Since a maximum of $2,100 is available, it means the total expense is either exactly $2,100 or less than $2,100.

Thus, we have the inequality:

$80+29g\le2100$

Next, solve for g:

$\begin{gathered} 29g\le2100-80 \\ 29g\le2020 \\ \frac{29g}{29}\le\frac{2020}{29} \\ g\le69\frac{19}{29} \end{gathered}$

Therefore, the maximum number of guests is 69.