Question

Steve and Michelle Neely are celebrating their 30th anniversary by having a reception at a local reception hall They have budgeted 52,500 for their reception. If the reception hall charges a $80 cleanup lee plus $33 per person, find the greatest number of people that they may invite and still stay within their budget. Steve and Michelle can invite at most people to the reception (Round down to the nearest whole person)

Answer 1

73 people

Explanation

Step 1

the reception hall charges a $80 cleanup lee plus $33 per person

let x represents the number of invited people,

Let C represents the total cost,so

$C=80+33x$

Now, if the budget is $2500, the cost must be equal or smaller than C, hence

$\begin{gathered} \text{Cost}\leq2500 \\ \text{Cost}=80+33x \\ \text{therefore} \\ \text{2500 }\leq80+33x\rightarrow inequality \end{gathered}$

Step 2

now, solve the inequality

$\begin{gathered} \text{Cost }\leq80+33x \\ \text{2500}\leq80+33x \\ \text{subtract 80 in both sides} \\ \text{2500-80}\leq80+33x-80 \\ 2420\leq33x \\ \text{divide both sides by 33} \\ \frac{2420}{33}\leq\frac{33x}{33} \\ 73.33\leq x \\ \text{rounded to the nearest whole person} \\ 73\leq x \end{gathered}$

it means the greatest number of people Steve and MIchelle can invite is 73 people.

I hope this helps you