a couple are celebrating their 30th anniversary by having a reception at a local reception hall. they have…
a couple are celebrating their 30th anniversary by having a reception at a local reception hall. they have budgeted $3,000 for their reception. if the reception hall charges a $50 cleanup fee plus $34 per person, find the greatest number of people that they may invite and still stay within their budget. the couple can invite at most people to the reception. (round down to the nearest whole person.)
Answer
Explanation:
Step1: Set up an inequality
Let $x$ be the number of people. The total cost is the cleanup fee plus the cost per - person. The cleanup fee is $50$ and the cost per person is $34x$. The total budget is $3000$. So the inequality is $34x + 50\leq3000$.
Step2: Solve the inequality for $x$
First, subtract $50$ from both sides of the inequality: $34x\leq3000 - 50$, which simplifies to $34x\leq2950$. Then divide both sides by $34$: $x\leq\frac{2950}{34}\approx86.76$.
Step3: Round down to the nearest whole person
Since we need to round down to the nearest whole person, $x = 86$.
Answer:
86