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📝 Budget constraint inequality (14 MCQs)

📖 From Digital SAT Algebra • 3. Mathematical Models in Algebra • 14 questions available

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What is Budget constraint inequality?

Definition:
A budget constraint inequality is an algebraic expression that represents the limit on spending, where the total cost of items or services must be less than or equal to the available budget, expressed as $\text{Total Cost} \le \text{Budget}$ , and it is used in financial planning, purchasing decisions, and resource allocation to ensure spending does not exceed funds.

Working:
To set up a budget constraint, identify the costs of each item, multiply by the number of units, sum them, and set the sum less than or equal to the budget, then solve for the variable to find the maximum number of items affordable, or the maximum cost allowable, and this is often used with linear inequalities to model real-world financial limitations.

Example:
If a budget is $\<span class="katex-error" title="ParseError: KaTeX parse error: Can't use function '$ ' in math mode at position 5: 200 \̲)̲ for school sup…" style="color:#cc0000">200 \) for school supplies, and notebooks cost $\</span>3$ each and pens cost $\<span class="katex-error" title="ParseError: KaTeX parse error: Can't use function '$ ' in math mode at position 3: 2 \̲)̲ each, with $\dots" style="color:#cc0000">2$ each, with $x$ notebooks and $y$ pens, the inequality is $3x + 2y \le 200$ , meaning the total cost of notebooks and pens cannot exceed $\</span>200$ .

Reason:
Budget constraints are vital in personal finance, business, and project management, as they ensure financial sustainability and optimal use of resources, and they are a core concept in economics and decision-making.

Easy

Medium

Hard

📝 All Budget constraint inequality MCQs

Q1. A student has a monthly budget of 1200 units. Rent is fixed at 700, and the student wants to reserve 150 for emergencies. What inequality correctly models the maximum amount $x$ available for shopping and entertainment combined?

x\le350

✅

x\ge350

x\le1050

x\ge1050

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: After paying the fixed rent of 700 and reserving 150, the remaining amount is $1200-700-150=350$ . Since spending cannot exceed the remaining money, the correct budget constraint is $x\le350$ .

Q2. Which statement best describes a budget constraint when a person has fixed income and must decide how much to spend on rent, parties, and shopping?

A.Every possible combination of purchases must cost exactly the income

B.The total spending must not exceed the available income ✅

C.Each category must receive the same amount of money

D.Only essential expenses can be included in the model

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: A budget constraint limits total spending to the money available. Spending may be less than income, so requiring equality is unnecessarily restrictive, while equal category spending and excluding discretionary expenses are not required.

Q3. A person earns 2400 per month. Rent is 900, groceries are 450, and transportation is 250. If $p$ represents party spending and $s$ represents shopping, which constraint describes the feasible choices?

p+s\ge800

p+s=1600

p+s\le800

✅

p-s\le800

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The fixed expenses total $900+450+250=1600$ . Subtracting this from the income leaves $2400-1600=800$ for parties and shopping. Therefore their combined spending can be at most 800, giving $p+s\le800$ .

Q4. A shopper has 600 units remaining after rent and food. Shoes cost 80 per pair and shirts cost 40 each. If $x$ is pairs of shoes and $y$ is shirts, which choice is feasible?

x=5,y=6

x=4,y=7

✅

x=3,y=9

x=6,y=4

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: For each option, compare total cost with 600. Option B costs $4(80)+7(40)=320+280=600$ , so it exactly uses the available budget. The other combinations exceed 600, making them infeasible.

Q5. A student originally models a monthly budget as $700+20p+50s\le1200$ , where $p$ is the number of parties and $s$ is the number of shopping trips. Rent increases by 100. What is the new constraint?

800+20p+50s\le1200

✅

700+20p+50s\le1300

800+20p+50s\le1300

600+20p+50s\le1200

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The rent is a fixed expense, so increasing it by 100 changes 700 to 800. Income and variable costs remain unchanged. Therefore the revised constraint is $800+20p+50s\le1200$ .

Q6. A person has 1000 units after fixed bills. A party costs 125, while one shopping trip costs 75. If the person plans 4 parties, what is the greatest possible number of shopping trips?

A.5

B.6

C.7 ✅

D.8

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Four parties cost $4(125)=500$ , leaving $1000-500=500$ . Each shopping trip costs 75, so at most $500/75=6.66$ trips can be purchased. Because the number of trips must be whole, the maximum is 6, not 7.

Q7. A budget graph has $x$ representing shopping spending and $y$ representing party spending. The boundary line passes through $(0,600)$ and $(300,0)$ , and the feasible region lies below the line in the first quadrant. Which statement is correct?

A.A person can spend 400 on shopping and 300 on parties

B.A person can spend 200 on shopping and 150 on parties ✅

C.A person must spend exactly 600 in total

D.A person can spend 350 on shopping and 250 on parties

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The boundary decreases from $(0,600)$ to $(300,0)$ , so its equation is $y=600-2x$ . At $x=200$ , the maximum $y$ is 200. Thus $y=150$ is feasible, while the other listed combinations lie above the boundary.

Q8. Two roommates have 1800 units available after fixed obligations. They want to spend at least 300 on a social event but no more than 500 on shopping. Which reasoning correctly determines whether shopping $s=500$ and party spending $p=300$ is feasible?

A.It is infeasible because spending exactly 500 violates 'no more than'

B.It is feasible if

p+s\le1800

, because both category limits are satisfied ✅

C.It is infeasible because party spending must exceed shopping spending

D.It is feasible only if

p+s=1800

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The phrase 'no more than 500' allows shopping spending equal to 500, while 'at least 300' allows party spending equal to 300. Their total is 800, which is below the available 1800, so the choice is feasible.

Q9. A student claims that if $R+P+S\le2000$ , then increasing rent $R$ by 150 always makes the budget impossible. What is the flaw in this reasoning?

A.Rent is never part of a budget constraint

B.The original inequality may have unused money, so a 150 increase may still leave a feasible budget ✅

C.Increasing rent automatically decreases income

D.Party and shopping costs must increase whenever rent increases

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: A budget inequality allows spending below the maximum available amount. If the original plan used less than 2000, some unused money can absorb the 150 rent increase. Feasibility depends on the remaining slack, not merely on the increase.

Q10. A budget boundary is represented by $4x+2y=800$ . A student says that $(100,150)$ is feasible because $4(100)+2(150)=700<800$ , but then says $(150,100)$ is not feasible because $4(150)+2(100)=800$ . Which evaluation is correct?

A.Both statements are correct ✅

B.Only the first statement is correct

C.Only the second statement is correct

D.Neither statement is correct

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Both points satisfy the budget condition $4x+2y\le800$ . The first point costs 700, leaving 100 unused, while the second costs exactly 800. A boundary point is feasible when the constraint is less than or equal to the budget.

Q11. A family can allocate 1200 units between shopping $x$ and parties $y$ . Shopping receives a $20\%$ discount, while party costs remain unchanged. If the original prices are represented by $x$ and $y$ , which adjusted constraint correctly models the new spending?

0.8x+y\le1200

✅

1.2x+y\le1200

x+1.2y\le1200

0.2x+y\le1200

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: A $20\%$ discount means the shopper pays $80\%$ of the original shopping cost, represented by $0.8x$ . Party spending is unchanged, so the total adjusted spending is $0.8x+y$ , which cannot exceed 1200.

Q12. A student wants to choose party spending $p$ and shopping spending $s$ under $2p+3s\le600$ , with $p,s\ge0$ . Which strategy guarantees the greatest number of total activities $p+s$ , assuming the variables represent whole-number activity units?

A.Spend everything on parties

B.Spend everything on shopping

C.Compare the activities per unit of budget and prioritize parties ✅

D.Split the budget equally between parties and shopping

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: A party unit consumes 2 budget units, while a shopping unit consumes 3. Parties therefore provide more activity units per budget unit. Using the budget primarily on parties maximizes the total number of activities, subject to the whole-number restriction.

Q13. A person has 2400 units after income is received. Rent is 900, and they must spend at least 400 on essential shopping. They want to maximize party spending $p$ , but also save 300. What is the maximum party budget?

A.700

B.800 ✅

C.900

D.1200

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The maximum party amount is found by subtracting all fixed or required allocations: $2400-900-400-300=800$ . Because the problem asks for the maximum, the remaining amount can all be assigned to parties. Therefore $p\le800$ , giving 800.

Q14. Two budget plans are proposed. Plan A spends 900 on rent, 300 on parties, and 400 on shopping. Plan B spends 850 on rent, 450 on parties, and 350 on shopping. If income is 1700, which conclusion is valid?

A.Only Plan A is feasible

B.Only Plan B is feasible

C.Both plans are feasible ✅

D.Neither plan is feasible

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Plan A totals $900+300+400=1600$ , leaving 100. Plan B totals $850+450+350=1650$ , leaving 50. Since both totals are below the 1700 income, both plans satisfy the budget constraint, although they allocate money differently.

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