Question 545·Easy·Algebra

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A bakery sells each cookie for $2 and wants to make at least $50 in cookie sales today. If the bakery has already made $34 from cookie sales, what is the minimum number of additional cookies it must sell to meet its goal?

An inequality fits a sales goal that can be met or exceeded: at least includes the goal itself. Add the sales already made to the price times the number of new items, then use a Desmos regression on the equality to find the cutoff. Don’t stop at the dollars still needed when the question asks for a number of items.

Hints

Hint 1
Let $n$ be the number of additional cookies. Each new cookie adds $2$ in sales, so new sales are $2n$ . How would you combine that with the amount the bakery has already made?
Hint 2
The boundary is where sales exactly reach the goal. Find that value by changing the inequality to an equality. Since at least includes equality, would the bakery need one more cookie if the boundary is a whole number?

Step-by-step

Model the goal and find the cutoff
Step 1
Turn the sales goal into an inequality
Let $n$ count additional cookies. They bring in $2n$ in new sales, which you add to the $34$ already made. The total must be at least $50$ , so $34+2n\ge 50$ . The symbol $\ge$ means “greater than or equal to.” Don’t multiply the $34$ by $n$ ; those sales have already happened.
Step 2
Find the boundary and choose the minimum
To find where sales exactly reach $50$ , type $34+2n_1\sim50$ in Desmos. The subscript in $n_1$ makes it a value Desmos can find, and $\sim$ asks Desmos to fit the equality. Under PARAMETERS, it shows $n_1=8$ . An exact whole-number boundary counts when the goal says “at least.” So the bakery must sell 8 additional cookies. Choice D.

Lessons that teach this

SAT AlgebraIntermediateCoreSolve one-variable inequalities

Question 545·Easy·Algebra

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