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📝 How Solve Applications with Linear Inequalities (14 MCQs)

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

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What is How Solve Applications with Linear Inequalities?

Definition:
Applications with linear inequalities involve using mathematical statements with inequality symbols ( $<, \le, >, \ge$ ) to represent constraints or limits in real-world situations, such as budgets, capacities, minimum requirements, and work hours, and solving these applications requires translating word problems into inequalities, solving them algebraically, and interpreting the solution set in the context of the problem.

Working:
To solve, first define the variable, identify the inequality symbol based on the phrasing (e.g., at least" means $\ge$ "no more than" means $\le$ ) write the inequality solve it using inverse operations (remembering that multiplying or dividing by a negative flips the inequality) and graph or describe the solution set then check the reasonableness of the answer in the original context especially for discrete quantities like people or items.

Example:
A student needs at least 60 points to pass a test where each correct answer gives 2 points and there are no penalties; if $x$ is the number of correct answers then $2x \ge 60$ so $x \ge 30$ meaning the student must get at least 30 questions correct to pass.

Reason:
Linear inequalities are used extensively in fields like economics engineering and daily life to model constraints make decisions and optimize resources making them a critical tool for problem-solving and critical thinking."

Easy

Medium

Hard

📝 All How Solve Applications with Linear Inequalities MCQs

Q1. A school club has Rs. 4,800 for an event. Fixed expenses are Rs. 1,200, and each participant costs Rs. 150. Which inequality correctly represents the maximum number $x$ of participants the club can accommodate?

1200+150x<4800

1200+150x\le4800

✅

150x-1200\le4800

1200-150x\le4800

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The total cost consists of the fixed Rs. 1,200 plus Rs. 150 for each participant. Since the budget cannot be exceeded, the total must be less than or equal to Rs. 4,800. Therefore, $1200+150x\le4800$ correctly models the situation.

Q2. A delivery company charges Rs. 250 as a base fee and Rs. 40 per kilometer. A customer can spend at most Rs. 1,050. What is the greatest whole number of kilometers the customer can travel?

A.19 km

B.20 km ✅

C.21 km

D.26 km

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The cost inequality is $250+40x\le1050$ . Subtracting 250 gives $40x\le800$ , so $x\le20$ . Because distance is measured in whole kilometers in this model, the greatest possible distance is 20 km.

Q3. A student wants to buy notebooks costing Rs. 85 each and has Rs. 700. The student must keep at least Rs. 105 for transportation. Which conclusion follows from the situation?

A.The student can buy at most 6 notebooks. ✅

B.The student can buy at most 7 notebooks.

C.The student must buy exactly 6 notebooks.

D.The student can buy exactly 7 notebooks.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The amount available for notebooks is $700-105=595$ rupees. Thus $85x\le595$ , giving $x\le7$ . However, $85(7)=595$ , so 7 notebooks actually fits exactly. Therefore option B is mathematically correct, while option A incorrectly limits the purchase.

Q4. A fitness center charges Rs. 900 monthly plus Rs. 120 for each personal-training session. A member wants the total monthly cost to remain below Rs. 1,500. Which statement is correct?

A.At most 4 sessions are possible.

B.At most 5 sessions are possible. ✅

C.Exactly 5 sessions are required.

D.At least 5 sessions are possible.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The inequality is $900+120x<1500$ . This gives $120x<600$ , so $x<5$ . Since the number of sessions must be a whole number, the greatest possible number is 4. Thus option A is correct; choosing 5 would make the cost exactly Rs. 1,500, which is not below the limit.

Q5. A factory can operate a machine for no more than 72 hours per week. Machine A uses 3 hours for each batch, while Machine B uses 2 hours for each batch. If 10 batches of A are planned, which inequality determines the maximum number $y$ of B batches?

30+2y<72

30+2y\le72

✅

3(10+y)\le72

30-2y\le72

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Ten batches of Machine A require $3(10)=30$ hours. The remaining batches of B require $2y$ hours. Since the total operating time cannot exceed 72 hours, the correct model is $30+2y\le72$ . This also shows why subtracting the B time is inappropriate.

Q6. A charity has Rs. 6,000 to purchase food packages. Each package costs Rs. 180, but a sponsor contributes Rs. 600 toward the purchase. If $x$ packages are bought, which inequality gives the feasible number of packages?

180x+600\le6000

180x-600\le6000

180x\le6600

✅

6000x+600\le180

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The sponsor's Rs. 600 contribution reduces the amount the charity must pay from its own budget. Therefore $180x\le6000+600=6600$ . This gives the correct model and avoids the common error of adding the contribution to the actual purchase cost.

Q7. A student has 12 hours available for studying and recreation. Studying requires at least 2 hours, while recreation must be no more than 5 hours. If $s$ represents study hours and $r$ represents recreation hours, which system best represents these restrictions?

s\le2,\ r\ge5,\ s+r\le12

s\ge2,\ r\le5,\ s+r\le12

✅

s\ge2,\ r\ge5,\ s+r\ge12

s\le2,\ r\le5,\ s+r\ge12

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The phrase 'at least 2 hours' translates to $s\ge2$ , while 'no more than 5 hours' translates to $r\le5$ . Together, the available time requires $s+r\le12$ . Each inequality represents a different real-world restriction, so all three are needed.

Q8. A theater's seating plan requires at least 120 adults and at least 80 students for a special event. The total number of seats available is 250. If $a$ and $s$ represent adults and students, which additional inequality is necessary?

a+s\ge250

a+s\le250

✅

a-s\le250

a+s=200

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The minimum requirements are $a\ge120$ and $s\ge80$ , but the theater cannot seat more than 250 people. Therefore the total must satisfy $a+s\le250$ . The equality $a+s=200$ would incorrectly force the event to have exactly the minimum number.

Q9. A student solves $4x+7\le31$ as follows: $4x\le24$ , so $x\le6$ . Another student says the answer should be $x\ge6$ . Which evaluation is correct?

A.The second student is correct because subtraction reverses the sign.

B.The first student is correct because dividing by positive 4 preserves the inequality. ✅

C.Both are correct because 6 is the boundary value.

D.Neither is correct because the inequality should become

x<6

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Subtracting 7 gives $4x\le24$ . Dividing by positive 4 does not reverse the inequality, so $x\le6$ . Inequality reversal occurs only when multiplying or dividing by a negative number. The boundary value 6 is included because the original sign is $\le$ .

Q10. A worker needs at least Rs. 3,000 in earnings. The worker already earns Rs. 1,200 and receives Rs. 150 per completed task. A solution claims $x\le12$ . What is the error?

A.The worker should subtract the task payment from Rs. 3,000.

B.The phrase 'at least' requires a greater-than-or-equal inequality, giving

x\ge12

. ✅

C.The fixed earnings should be multiplied by

x

D.The worker needs exactly 12 tasks, not at least 12.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The earnings condition is $1200+150x\ge3000$ . Subtracting 1,200 gives $150x\ge1800$ , and dividing by 150 gives $x\ge12$ . The error is reversing the direction of the inequality even though the situation requires reaching a minimum amount.

Q11. A graph of a feasible region shows a solid boundary line and shading below it. The boundary line represents $y=3x+2$ . Which inequality describes the shaded region?

y>3x+2

y<3x+2

y\ge3x+2

y\le3x+2

✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: Shading below a boundary line represents values less than the line. Because the boundary is solid, points on the line are included. Combining these two observations gives $y\le3x+2$ . A dashed line would instead indicate that equality is excluded.

Q12. A catering service charges Rs. 2,000 plus Rs. 350 per guest. Another service charges Rs. 800 plus Rs. 500 per guest. For how many guests is the first service no more expensive than the second?

A.At most 7 guests

B.At least 8 guests ✅

C.Exactly 8 guests

D.Fewer than 7 guests

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Let $x$ be the number of guests. The comparison is $2000+350x\le800+500x$ . Subtracting 800 and $350x$ gives $1200\le150x$ , so $x\ge8$ . Thus the first service becomes no more expensive starting at 8 guests.

Q13. A farmer has 60 meters of fencing to enclose a rectangular garden against a straight wall, so fencing is needed for only three sides. If the width is $x$ meters and the length is $y$ meters, which condition correctly models the fencing limit and captures the fact that the garden may use less than the available fencing?

2x+y=60

2x+y\ge60

2x+y\le60

✅

x+2y\ge60

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Two widths require $2x$ meters and one length requires $y$ meters, so the fencing used is $2x+y$ . Because the farmer has 60 meters available and does not have to use every meter, the total cannot exceed 60. Hence $2x+y\le60$ .

Q14. A company allows an employee to spend at most 8 hours on two tasks. Task A takes 30 minutes per unit and Task B takes 20 minutes per unit. If at least 10 units of Task A must be completed, which inequality describes the possible number $b$ of Task B units?

300+20b\le480

✅

300+20b\ge480

30+20b\le8

10(30)+20b\ge480

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Eight hours equals 480 minutes. At least 10 units of Task A require 300 minutes, leaving at most 180 minutes for Task B. Therefore $300+20b\le480$ , which gives $b\le9$ . The model correctly combines the minimum requirement with the maximum available time.

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