Creating Equations: Modeling with Equations and Inequalities as Constraints (CCSS.A-CED.3)
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Common Core High School Algebra › Creating Equations: Modeling with Equations and Inequalities as Constraints (CCSS.A-CED.3)
A school club is ordering tacos and burritos for a fundraiser. Tacos cost 2 dollars and take 2 minutes to prepare; burritos cost 3 dollars and take 4 minutes to prepare. They can spend at most 180 dollars, have at most 240 minutes of prep time, and want at least 70 total items. Let $x$ be the number of tacos and $y$ be the number of burritos.
Which system of inequalities models the situation?
$2x + 3y \le 180,; 2x + 4y \le 240,; x + y \le 70$
$2x + 3y \le 180,; x + y \ge 70$
$2x + 3y \le 180,; 2x + 4y \le 240,; x + y \ge 70$
$2x + 3y \le 180,; 2x + 3y \le 240,; x + y \ge 70$
Explanation
- Budget at most 180 dollars: $2x + 3y \le 180$.
- Prep time at most 240 minutes: $2x + 4y \le 240$ (2 minutes per taco, 4 per burrito).
- At least 70 items: $x + y \ge 70$. Choice A reverses the last inequality ($\le$ instead of $\ge$). Choice B omits the prep-time constraint entirely. Choice D miscounts burrito prep time (uses $3y$ instead of $4y$).
A theater sells regular tickets for 12 dollars and student tickets for 8 dollars. The auditorium seats at most 150 people, they want to sell at least 140 tickets, and they need at least 1500 dollars in revenue. Let $x$ be the number of regular tickets and $y$ be the number of student tickets.
Which solution is viable in context?
$(90,,60)$
$(100,,30)$
$(120,,40)$
$(90,,50)$
Explanation
Constraints: seats $x+y \le 150$, minimum tickets $x+y \ge 140$, revenue $12x + 8y \ge 1500$.
- For $(90,60)$: $90+60=150$ satisfies both $\le 150$ and $\ge 140$; revenue $12\cdot90+8\cdot60=1080+480=1560 \ge 1500$ ✓ (viable).
- $(100,30)$: $100+30=130$ fails $x+y \ge 140$.
- $(120,40)$: $120+40=160$ exceeds $150$ seats.
- $(90,50)$: $90+50=140$ meets ticket count, but revenue $1080+400=1480<1500$ fails the revenue constraint.
A bakery makes muffins and scones. Each muffin uses 0.2 pounds of flour and 10 minutes of oven time; each scone uses 0.3 pounds of flour and 12 minutes of oven time. The bakery has at most 25 pounds of flour, at most 1200 minutes of oven time, and must bake at least 40 muffins. Let $x$ be the number of muffins and $y$ be the number of scones.
Which system of inequalities models the situation?
$0.2x + 0.3y \ge 25,; 10x + 12y \le 1200,; x \ge 40$
$0.2x + 0.3y \le 25,; 10x + 12y \le 1200$
$0.2x + 0.4y \le 25,; 10x + 12y \le 1200,; x \ge 40$
$0.2x + 0.3y \le 25,; 10x + 12y \le 1200,; x \ge 40$
Explanation
- Flour at most 25 lb: $0.2x + 0.3y \le 25$.
- Oven time at most 1200 min: $10x + 12y \le 1200$.
- At least 40 muffins: $x \ge 40$. Choice A reverses the flour inequality sign. Choice B omits the minimum-muffins requirement. Choice C miscounts scone flour (uses $0.4y$ instead of $0.3y$).
A workshop builds chairs and stools. Each chair uses 3 board-feet of wood and 1.5 labor hours; each stool uses 2 board-feet and 1 labor hour. They have at most 600 board-feet of wood, at most 350 labor hours, and need to produce at least 120 total pieces. Let $x$ be chairs and $y$ be stools.
Which solution is viable in context?
$(220,,10)$
$(100,,50)$
$(140,,260)$
$(-5,,130)$
Explanation
Constraints: wood $3x+2y \le 600$, labor $1.5x + y \le 350$, total pieces $x+y \ge 120$.
- $(100,50)$: wood $300+100=400 \le 600$, labor $150+50=200 \le 350$, total $150 \ge 120$ ✓ (viable).
- $(220,10)$: wood $660+20=680>600$ (fails wood).
- $(140,260)$: wood $420+520=940>600$ and labor $210+260=470>350$ (fails both).
- $(-5,130)$: negative chairs are unrealistic in context.
A delivery company can rent small and large vans for a day. Each small van holds 80 boxes and costs 120 dollars; each large van holds 140 boxes and costs 200 dollars. They must carry at least 900 boxes, can spend at most 1600 dollars, and can rent at most 12 vans in total. Let $x$ be small vans and $y$ be large vans.
Which system of inequalities models the situation?
$80x + 140y \le 900,; 120x + 200y \le 1600,; x + y \le 12$
$80x + 140y \ge 900,; 120x + 200y \le 1600,; x + y \le 12$
$80x + 140y \ge 900,; 120x + 200y \le 1600$
$80x + 140y \ge 900,; 100x + 200y \le 1600,; x + y \le 12$
Explanation
- Capacity at least 900 boxes: $80x + 140y \ge 900$.
- Cost at most 1600 dollars: $120x + 200y \le 1600$.
- At most 12 vans: $x + y \le 12$. Choice A reverses the capacity inequality. Choice C omits the van-count constraint. Choice D misstates the small-van cost (uses $100x$ instead of $120x$).
A smoothie stand will blend berry smoothies ($x$) and green smoothies ($y$). Each berry uses 2 cups of fruit and 1 cup of yogurt, and each green uses 1 cup of fruit and 2 cups of yogurt. They have at most 40 cups of fruit and at most 30 cups of yogurt.
Which system of inequalities models the situation?
$2x + y \ge 40;\ x + 2y \ge 30$
$2x + y \le 40;\ x + 2y \le 30$
$2x + 3y \le 40;\ x + 2y \le 30$
$2x + y \le 40$
Explanation
Fruit limit: each berry uses 2 and each green 1, so $2x + y \le 40$. Yogurt limit: each berry 1 and each green 2, so $x + 2y \le 30$. Choice B states both correctly. Choice A reverses the direction ($\ge$) even though the resources are limited. Choice C miscounts the fruit for greens as $3y$ instead of $y$. Choice D omits the yogurt constraint entirely.
A community movie night sells adult tickets ($x$) and student tickets ($y$). The theater seats at most 100 people, and they must collect at least 900 dollars to cover costs; adult tickets are 12 dollars and student tickets are 8 dollars.
Which solution is viable in context?
$(x,y)=(60,20)$
$(x,y)=(70,40)$
$(x,y)=(60,40)$
$(x,y)=(0,100)$
Explanation
Constraints: seats $x+y \le 100$ and revenue $12x+8y \ge 900$. Test A: $60+20=80$ seats ok, but $12(60)+8(20)=720+160=880<900$ fails revenue. Test B: $70+40=110>100$ exceeds seating. Test C: $60+40=100$ seats ok and $12(60)+8(40)=720+320=1040\ge 900$ revenue ok, so viable. Test D: $0+100=100$ seats ok, but $8(100)=800<900$ fails revenue.
A workshop builds benches ($x$) and stools ($y$). Each bench uses 4 boards and 3 labor hours, and each stool uses 2 boards and 1 labor hour. The shop has at most 80 boards and at most 40 labor hours available.
Which system of inequalities models the situation?
$4x + 2y \le 80;\ 3x + y \le 40$
$4x + 2y \ge 80;\ 3x + y \ge 40$
$4x + 2y \le 80$
$4x + 3y \le 80;\ 3x + y \le 40$
Explanation
Boards: $4x+2y \le 80$ because boards are limited. Labor: $3x+y \le 40$ because hours are limited. Choice A has both correct. Choice B reverses the direction to $\ge$ as if they must use at least that many resources, which is not stated. Choice C omits the labor constraint. Choice D miscounts stool boards as $3y$ instead of $2y$.
A student is choosing protein bars ($x$) and fruit cups ($y$) for the week. Each bar costs 3 dollars and provides 8 grams of protein, and each fruit cup costs 2 dollars and provides 1 gram of protein. They can spend at most 18 dollars and need at least 24 grams of protein.
Which solution is viable in context?
$(x,y)=(2,8)$
$(x,y)=(1,10)$
$(x,y)=(2,5)$
$(x,y)=(3,4)$
Explanation
Constraints: cost $3x+2y \le 18$ and protein $8x+y \ge 24$. Test A: cost $3(2)+2(8)=6+16=22>18$ fails budget (protein $=24$ is fine). Test B: cost $3+20=23>18$ and protein $8+10=18<24$, fails both. Test C: cost $6+10=16$ ok, protein $16+5=21<24$ fails protein. Test D: cost $9+8=17\le 18$ and protein $24+4=28\ge 24$, so viable.
A club will rent small vans ($x$) and minibuses ($y$) for a trip. Each van seats 7 and costs 80 dollars, and each minibus seats 18 and costs 200 dollars. They must seat at least 90 people and have at most 900 dollars to spend.
Which system of inequalities models the situation?
$7x + 18y \le 90;\ 80x + 200y \ge 900$
$7x + 18y \ge 90;\ 80x + 200y \le 900$
$7x + 18y \ge 90$
$8x + 20y \le 900;\ 7x + 18y \ge 90$
Explanation
Seating at least 90 gives $7x+18y \ge 90$. Budget at most 900 gives $80x+200y \le 900$. Choice B has both correctly. Choice A reverses both directions. Choice C omits the budget constraint. Choice D uses incorrect vehicle costs ($8$ and $20$ instead of $80$ and $200$), a common place-value slip.