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📝 Cost limit inequality word problems (11 MCQs)

📖 From Digital SAT Algebra • 3. Mathematical Models in Algebra • 11 questions available

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What is Cost limit inequality word problems?

Definition:
Cost limit inequality word problems involve constraints where the total cost of goods or services must not exceed a specified budget or limit, expressed as $\text{Total Cost} \le \text{Maximum Cost}$ , and these are widely used in purchasing, project budgeting, and cost control to ensure financial constraints are respected.

Working:
To solve, identify the fixed and variable costs, set up the inequality with the total cost less than or equal to the limit, solve for the unknown (e.g., number of items or hours), and round appropriately (e.g., if the variable is discrete, round down for a maximum limit), and check the reasonableness of the solution in the real-world context.

Example:
A budget of $\<span class="katex-error" title="ParseError: KaTeX parse error: Can't use function '$ ' in math mode at position 5: 100 \̲)̲ for supplies, …" style="color:#cc0000">100 \) for supplies, where pencils cost $\</span>1.50$ each and a fixed cost of $\$10$ for shipping; the inequality for $x$ pencils is $1.5x + 10 \le 100$ , so $1.5x \le 90$ , $x \le 60$ , meaning you can buy at most 60 pencils within the budget.

Reason:
Cost limit inequalities are fundamental for financial planning, procurement, and cost management, helping individuals and organizations stay within budget and make informed purchasing decisions.

Easy

Medium

Hard

📝 All Cost limit inequality word problems MCQs

Q1. A phone plan charges a fixed monthly fee of $25 plus$ 0.08 per text message. If a customer can spend at most $41 per month, which inequality correctly models the maximum number $t$ of messages allowed?

25+0.08t<41

25+0.08t\le41

✅

25t+0.08\le41

0.08+25t\le41

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The phrase 'at most' means the total cost cannot exceed the budget, so the correct relationship is $25+0.08t\le41$ . The fixed charge is added once, while the message charge depends on $t$ .

Q2. A customer has a phone budget of $50. The plan costs$ 32 plus $0.06 per minute. Which interpretation is most accurate if $32+0.06m\le50$ ?

A.The customer must spend exactly $50

B.The customer can use no more than 300 minutes ✅

C.The customer must use at least 300 minutes

D.The customer can use exactly 18 minutes

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Subtracting the fixed cost gives $0.06m\le18$ , so $m\le300$ . Thus the inequality describes a maximum of 300 minutes rather than requiring the customer to use exactly 300 minutes.

Q3. A water utility charges a fixed service fee of $18 and$ 2.50 per cubic meter used. A household wants its bill to be no more than $68. Which statement best describes the feasible values of $w$ ?

w\le20

cubic meters ✅

w<20

cubic meters

w\ge20

cubic meters

w\le50

cubic meters

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The model is $18+2.50w\le68$ . Subtracting 18 gives $2.50w\le50$ , and dividing by 2.50 gives $w\le20$ . Therefore, usage cannot exceed 20 cubic meters.

Q4. A car rental company charges $45 per day plus a one-time insurance fee of$ 30. A renter has a maximum budget of $210. Which reasoning correctly determines the greatest whole number of rental days?

A.Solve

45d+30<210

, giving 3 days

B.Solve

45d+30\le210

, giving 4 days ✅

C.Solve

45d+30\ge210

, giving 4 days

D.Solve

45d+30\le210

, giving 5 days

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The budget condition is $45d+30\le210$ . Subtracting 30 gives $45d\le180$ , so $d\le4$ . Because rental days are whole numbers, the greatest allowable number is 4.

Q5. A student models a phone cost as $C=20+0.10m$ and concludes that a $35 budget permits 350 messages because $35\div0.10=350$ . What is the main modeling error?

A.The variable should represent dollars instead of messages

B.The fixed $20 charge was ignored before dividing by the per-message cost ✅

C.The inequality should always use

>

D.The per-message cost should be multiplied by 20

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The student incorrectly treats the entire $35 budget as usage money. The fixed$ 20 charge must first be removed, leaving $15 for messages. Thus $20+0.10m\le35$ gives $m\le150$ .

Q6. A water bill is modeled by $B=12+3.20w$ . A household claims that increasing the budget from $44 to$ 60 increases allowable water usage by 5 cubic meters. Is the claim correct?

A.Yes, because

60-44=16

and

16\div3.20=5

✅

B.No, because the increase is 4 cubic meters

C.No, because the fixed fee changes with usage

D.Yes, because the fixed fee is $12

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The difference in budgets is $60-44=16$ . Since each additional cubic meter costs $3.20, the additional allowable usage is $16\div3.20=5$ cubic meters. The fixed fee cancels when comparing the two budgets.

Q7. A customer has two phone plans. Plan A costs $18 plus$ 0.12 per minute, while Plan B costs $30 plus$ 0.06 per minute. With a maximum budget of $54, which plan allows more minutes?

A.Plan A allows 300 minutes, while Plan B allows 400 minutes

B.Plan A allows 400 minutes, while Plan B allows 300 minutes ✅

C.Both allow 300 minutes

D.Both allow 400 minutes

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: For Plan A, $18+0.12m\le54$ gives $m\le300$ . For Plan B, $30+0.06m\le54$ gives $m\le400$ . Although Plan B costs more initially, its lower variable rate permits greater usage.

Q8. A student solves the rental inequality $40+25d\le165$ by subtracting 40 and obtaining $25d\le125$ , but then writes $d\ge5$ . What caused the incorrect conclusion?

A.The student should have added 40

B.The inequality reverses whenever a positive number is divided

C.The student reversed the inequality unnecessarily after dividing by a positive number ✅

D.The fixed fee should be multiplied by

d

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: After $25d\le125$ , dividing by the positive number 25 preserves the inequality direction, giving $d\le5$ . Reversing the sign is required only when multiplying or dividing by a negative number.

Q9. A graph of a rental-cost inequality shows a solid point at $d=6$ on the horizontal axis and shading extending to the left. What does this graph imply about the rental period?

A.Exactly 6 days are required

B.At most 6 days are affordable ✅

C.More than 6 days are affordable

D.Fewer than 6 days are never affordable

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: A solid endpoint means the boundary value is included, while shading to the left represents values less than the endpoint. Therefore the graph represents $d\le6$ , meaning the renter can afford at most 6 days.

Q10. A renter compares two offers under a $300 limit. Offer A costs $60+40d$ , while Offer B costs $100+30d$ . Which conclusion follows from comparing their inequalities rather than merely comparing daily prices?

A.Offer A always permits more days

B.Offer B always permits more days

C.Both permit 6 days, but Offer B becomes relatively better for longer rentals ✅

D.Offer A and B have identical total costs for every rental period

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: For A, $60+40d\le300$ gives $d\le6$ . For B, $100+30d\le300$ gives $d\le6$ . Both reach six days under this budget, but B has the lower daily rate, so its advantage grows as the rental period increases.

Q11. A rental company charges $25 for the first day and$ 18 for each additional day. A customer has $115 available. Which maximum number of days can be rented?

A.5 days

B.6 days ✅

C.7 days

D.8 days

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The cost for $d$ days is $25+18(d-1)$ for $d\ge1$ . Requiring this to be at most $115 gives $18(d-1)\le90$ , so $d-1\le5$ , hence $d\le6$ . Therefore, 6 days is the maximum affordable rental period.

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