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📝 Linear inequality formulas and word problems (12 MCQs)

📖 From Digital SAT Algebra • 3. Mathematical Models in Algebra • 12 questions available

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What is Linear inequality formulas and word problems?

Definition:
Linear inequality formulas and word problems involve mathematical expressions where two linear expressions are compared using inequality symbols, and they represent real-world constraints such as minimum or maximum limits, and solving them requires translating the verbal description into a linear inequality, solving for the variable, and interpreting the solution within the given context.

Working:
To solve, identify the unknown, set up the inequality using keywords (e.g., less than" for $<$ "greater than or equal to" for $\ge$ ) simplify both sides if necessary isolate the variable and present the solution as an interval or set; for example if $3x + 5 \le 20$ solve to get $x \le 5$ and if the problem involves a real-world constraint like budget round appropriately and check.

Example:
A company produces $x$ units of a product with a production cost of $\<span class="katex-error" title="ParseError: KaTeX parse error: Can't use function '$ ' in math mode at position 3: 5 \̲)̲ per unit plus …" style="color:#cc0000">5 \) per unit plus a fixed cost of $\</span>100$ and they have a budget of $\$500$ ; the inequality is $5x + 100 \le 500$ so $5x \le 400$ $x \le 80$ meaning they can produce at most 80 units.

Reason:
Linear inequalities are vital in business budgeting and planning as they help set realistic limits and make decisions based on constraints and mastering them enhances analytical skills applicable in many disciplines."

Easy

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Hard

📝 All Linear inequality formulas and word problems MCQs

Q1. A school club has a budget of $240 and must spend$ 18 per participant plus a fixed $42 fee. Which inequality correctly represents the greatest number $x$ of participants the club can support?

18x-42\le240

18x+42\le240

✅

42x+18\le240

18(x+42)\le240

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The variable cost is $18x$ , while the fixed fee is $42 regardless of participation. Since total spending cannot exceed$ 240, the correct model is $18x+42\le240$ . The other choices incorrectly subtract or multiply the fixed fee.

Q2. A delivery service charges a $35 base fee and$ 4.50 per kilometer. If a customer can spend at most $80, which statement correctly describes the feasible distances $d$ ?

d\ge10

d\le10

✅

d<10

d>10

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The cost model is $35+4.50d\le80$ . Subtracting 35 gives $4.50d\le45$ , so $d\le10$ . Because spending exactly $80 is allowed, the endpoint is included, making the non-strict inequality essential.

Q3. A student wants to buy notebooks costing $6 each and pens costing$ 2 each. She has $40 and needs at least 3 notebooks. If $n$ represents notebooks and $p$ represents pens, which model captures both restrictions?

6n+2p\le40,\ n\ge3,\ n,p\ge0

✅

6n+2p\ge40,\ n\le3,\ p\ge0

6n+2p\le40,\ n\le3,\ p>0

6n+2p=40,\ n\ge3,\ p\ge0

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The budget creates $6n+2p\le40$ , because spending cannot exceed $40. The requirement of at least three notebooks gives $n\ge3$ , while quantities cannot be negative. Equality would incorrectly require spending every dollar.

Q4. A factory produces tables and chairs. Each table requires 4 hours and each chair requires 2 hours. With at most 40 labor hours, a manager also requires at least 6 tables. If $t$ and $c$ are the numbers produced, which condition must be satisfied?

4t+2c\ge40,\ t\ge6

4t+2c\le40,\ t\ge6

✅

4t+2c\le40,\ t\le6

4t+2c=40,\ t\ge6

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: At most 40 hours means the labor requirement cannot exceed 40, so $4t+2c\le40$ . The manager requires at least six tables, giving $t\ge6$ . Combining these conditions identifies feasible production plans.

Q5. A farmer has 100 meters of fencing for a rectangular vegetable area. The length is 8 meters more than the width. If the farmer wants the perimeter to be no more than 100 meters, which inequality determines the possible widths $w$ ?

2w+8\le100

2(w+w+8)\le100

✅

w(w+8)\le100

2w+(w+8)\le100

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The length is $w+8$ , so the perimeter is $2w+2(w+8)$ . Because the available fencing is at most 100 meters, the correct model is $2(w+w+8)\le100$ . This requires recognizing both pairs of sides.

Q6. A theater sells adult tickets for $12 and student tickets for$ 8. It must collect at least $720 from 70 tickets sold. If $s$ students are among the 70 tickets, which inequality should be solved to determine possible values of $s$ ?

12(70-s)+8s\ge720

✅

12(70+s)+8s\ge720

12s+8(70-s)\le720

12(70-s)+8s\le720

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: If $s$ students are sold, then $70-s$ adult tickets are sold. Revenue is therefore $12(70-s)+8s$ . Since the theater needs at least $720, revenue must be greater than or equal to 720.

Q7. A student claims that $5x-7\le18$ means $x\le2.2$ . Which explanation best identifies the error?

A.The student should add 7 before dividing by 5. ✅

B.The student should subtract 7 from both sides before dividing by 5.

C.The student divided by a positive number incorrectly.

D.The inequality symbol must reverse when dividing by 5.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Adding 7 to both sides gives $5x\le25$ , and dividing by positive 5 gives $x\le5$ . The student likely subtracted 7 instead of adding it. The inequality does not reverse because 5 is positive.

Q8. A worker models a weekly earning requirement as $15h+60\ge300$ , where $h$ is hours worked. He concludes that $h\ge24$ . What mistake did he make?

A.He multiplied 300 by 15.

B.He subtracted 60 incorrectly.

C.He divided before subtracting the fixed amount. ✅

D.He should divide by 60 instead of 15.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Subtracting the fixed $60 first gives $15h\ge240$ . Dividing by 15 gives $h\ge16$ . The incorrect result of 24 comes from treating the total requirement as if the fixed earning were not part of the model.

Q9. A graph shows a shaded region below the solid line $y=2x+3$ , with the restriction $x\ge0$ . A business uses $x$ for advertising units and $y$ for sales units. Which interpretation best matches the graph?

A.Sales must exceed

2x+3

B.Sales can be at most

2x+3

, with nonnegative advertising. ✅

C.Advertising must be at most zero.

D.Sales must equal exactly

2x+3

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: A shaded region below a solid boundary represents values satisfying $y\le2x+3$ , while the solid line means boundary points are included. The restriction $x\ge0$ further requires nonnegative advertising.

Q10. A café has 60 kilograms of ingredients. Sandwiches use 0.5 kg each and salads use 0.75 kg each. The café must prepare at least 40 meals. Which pair of constraints correctly models $s$ sandwiches and $a$ salads?

0.5s+0.75a\le60,\ s+a\ge40,\ s,a\ge0

✅

0.5s+0.75a\ge60,\ s+a\le40

0.5s+0.75a=60,\ s+a=40

0.5s+0.75a\le40,\ s+a\ge60

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The ingredient supply creates $0.5s+0.75a\le60$ , while the minimum meal requirement gives $s+a\ge40$ . Nonnegative quantities are also necessary. This combines a resource limit with a production minimum.

Q11. A company must ship at least 240 boxes using small trucks carrying 30 boxes each and large trucks carrying 50 boxes each. The company wants to minimize the number of trucks while using at least 2 large trucks. Which strategy guarantees the minimum number of trucks?

A.Use 2 large trucks and 4 small trucks.

B.Use 3 large trucks and 2 small trucks.

C.Use 4 large trucks and 1 small truck.

D.Use 5 large trucks only. ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: Five large trucks carry $5(50)=250$ boxes, satisfying the requirement with only five trucks. Four large and one small also uses five trucks, but the question asks for a strategy that guarantees minimum count; fewer than five trucks cannot carry 240 boxes because four large trucks carry only 200 boxes.

Q12. A rectangular garden has width $w$ meters and length $w+6$ meters. Its area must be at least 40 square meters, while its width cannot exceed 10 meters. Which interval contains all feasible widths?

0<w<4

w\ge4

and

w\le10

✅

w\le4

and

w\ge10

4<w<10

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The area condition is $w(w+6)\ge40$ . Since $w^2+6w-40=(w+10)(w-4)$ , positive widths satisfying the condition have $w\ge4$ . Combining this with the maximum width of 10 gives $4\le w\le10$ , including both endpoints.

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