Question 2·200 Super-Hard SAT Math Questions·Algebra

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A community center has a budget of $5,000 to buy meal kits for a program. The center will buy two types of kits:

Standard kits cost $12.50 each.
Deluxe kits cost $22 each.

To meet program requirements, the center must buy at least 280 total kits, and the number of standard kits must be at least twice the number of deluxe kits.

Which choice is the maximum number of deluxe kits the center can buy while meeting all requirements?

Several purchase limits form a system of inequalities: every kit combination must satisfy all of them. To maximize the deluxe count, use the ratio rule to find the fewest standard kits it allows for a given deluxe count, then use a Desmos regression to find where that cost reaches the budget. Don't treat a minimum total as an exact total; extra kits may still fit.

Hints

Hint 1
Let $x$ count standard kits and $y$ count deluxe kits. “At least twice” puts a minimum on $x$ : what inequality says you need at least two standard kits for every deluxe kit?
Hint 2
For a fixed $y$ , buying more standard kits raises the cost. Start with the fewest standards the ratio allows, $x=2y$ , to find a lower bound on cost for that deluxe count. Where does this cost reach the budget?
Hint 3
A decimal cutoff from Desmos isn't a kit count. Round down to get the largest whole-number candidate, then check whether that purchase also meets the separate minimum for total kits.

Step-by-step

Find the cheapest way to buy each deluxe count
Step 1
Write the required ratio
Let $x$ be the number of standard kits and $y$ the number of deluxe kits. “At least twice” puts a lower bound, or minimum, on standard kits, so $x\ge2y$ . The standard count must be twice the deluxe count or greater, not the other way around.
Step 2
Write the budget limit
The standard kits cost $12.50x$ dollars and the deluxe kits cost $22y$ dollars. Their combined cost has a budget cap of $5,000, so $12.50x+22y\le5000$ .
Step 3
Write the total-kit requirement
Adding both kinds gives the total number of kits. At least $280$ means $x+y\ge280$ . Don't set the total equal to $280$ : more kits are allowed if the other requirements still hold.
Step 4
Find the lowest cost for a fixed deluxe count
For a fixed $y$ , use the fewest standard kits the ratio allows to find a lower bound on cost. That's $x=2y$ ; any extra standard kits raise the bill. So even this ratio-based purchase must fit the budget: $12.50(2y)+22y\le5000$ .
Step 5
Find where that cost reaches the budget
Type $12.5(2y_1)+22y_1\sim5000$ in Desmos. The $\sim$ tells Desmos to find the unknown $y_1$ that makes both sides equal. Under PARAMETERS, it shows $y_1\approx106.383$ . This is a ceiling: above it, even the cheapest allowed purchase exceeds the budget.
Step 6
Check the largest whole-number candidate
Kit counts are whole numbers, so the largest candidate below that ceiling is $106$ deluxe kits. Type $2(106)$ in Desmos to get $212$ standard kits. Then $12.5(212)+22(106)$ gives a cost of $4,982, and $212+106$ gives $318$ total kits. The ratio holds, the cost is within budget, and $318\ge280$ . The maximum is $106$ deluxe kits. Choice B.

Question 2·200 Super-Hard SAT Math Questions·Algebra

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