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You are serving bratwurst and hamburgers at your annual picnic. You want at least three bratwursts or hamburgers for each of your 50 guests. Bratwursts cost [tex]$1.25 each and hamburgers $[/tex]0.95 each. Your budget is [tex]$200.

Let $[/tex]x[tex]$ be the number of bratwursts and $[/tex]y[tex]$ be the number of hamburgers. Which system of inequalities represents this situation?

A. $[/tex]x + y \geq 150 \quad 1.25x + 0.95y \geq 200[tex]$

B. $[/tex]x + y \leq 50 \quad 1.25x + 0.95y \geq 200[tex]$

C.
$[/tex]
\begin{array}{ll}
x + y \geq 150 & 1.25x + 0.95y \leq 200 \\
x + y \leq 50 & 1.25x + 0.95y \leq 200
\end{array}
$


Sagot :

To solve this problem, let's break down the constraints and requirements one by one.

1. Food Quantity Requirement:
- We need at least three bratwursts or hamburgers per guest.
- There are 50 guests.
- Therefore, we need at least [tex]\(3 \times 50 = 150\)[/tex] items in total.
- This translates to the inequality [tex]\( x + y \geq 150 \)[/tex], where [tex]\(x\)[/tex] is the number of bratwursts and [tex]\(y\)[/tex] is the number of hamburgers.

2. Budget Constraint:
- Bratwursts cost \[tex]$1.25 each and hamburgers cost \$[/tex]0.95 each.
- We have a total budget of \[tex]$200. - Therefore, the total cost of purchasing \(x\) bratwursts and \(y\) hamburgers should not exceed \$[/tex]200.
- This translates to the inequality [tex]\( 1.25x + 0.95y \leq 200 \)[/tex].

Putting these constraints together, we get the system of inequalities:

[tex]\[ \begin{cases} x + y \geq 150 \\ 1.25x + 0.95y \leq 200 \end{cases} \][/tex]

This matches one of the provided options. Therefore, the correct system of inequalities representing this situation is:

[tex]\[ \begin{array}{ll} x+y \geq 150 & 1.25x + 0.95y \leq 200 \end{array} \][/tex]