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📝 At most inequality meaning and examples (12 MCQs)

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

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What is At most inequality meaning and examples?

Definition:
At most" in inequality problems means the quantity is less than or equal to a certain value represented by $\le$ and it is synonymous with "no more than indicating a maximum limit, meaning the value cannot exceed the stated amount, and this is often used for constraints like spending limits, time limits, or inventory limits.

Working:
To translate at most use $\le$ and place the variable accordingly; for example, at most 30" is $x \le 30$ and when solving the solution includes all values up to and including the boundary and this is essential for respecting upper bounds and the graph shows a closed circle at the boundary with an arrow pointing left.

Example:
A school can accommodate at most 500 students; if currently 420 students are enrolled and $x$ new students apply then the inequality is $420 + x \le 500$ so $x \le 80$ meaning the school can accept at most 80 new students.

Reason:
"At most" is a common phrase in real-world constraints such as capacity budget and time management and understanding it helps in planning resource allocation and adherence to limits in various professional and personal contexts."

Easy

Medium

Hard

📝 All At most inequality meaning and examples MCQs

Q1. A school club has a budget of $480. Each student attending a workshop costs$ 32 for materials. Which inequality correctly represents the greatest possible number $x$ of students who can attend?

32x<480

32x\le480

✅

32x\ge480

32+x\le480

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The phrase 'at most' means the quantity cannot exceed the stated limit, so the cost must be less than or equal to $480. Since each student costs$ 32, the model is $32x\le480$ , allowing exactly the maximum budget to be used.

Q2. A delivery service charges a fixed fee of $18 plus$ 7 per package. A customer can spend at most $74. Which inequality and maximum number of packages correctly model the situation?

18+7x<74,\ x=7

18+7x\le74,\ x=8

✅

18+7x\ge74,\ x=8

18x+7\le74,\ x=3

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The fixed charge is $' in math mode at position 29: …riable cost is \̲(̲7x$ , while the…" style="color:#cc0000">18 and the variable cost is $7x$ , while the total may be at most74. Thus $18+7x\le74$ . Solving gives $7x\le56$ , so $x\le8$ , making 8 packages the greatest feasible whole number.

Q3. A worker earns $15 per hour and wants to work at most 26 hours this week. If $h$ represents hours worked, which statement best describes the solution set?

h\ge26

h<26

h\le26

✅

h=26

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: 'At most 26 hours' includes 26 hours but prohibits anything greater. Therefore the correct inequality is $h\le26$ . The distinction matters because using $h<26$ would incorrectly exclude the allowed endpoint.

Q4. A student solves $5x+12\le47$ and concludes $x\le5$ . What is the best evaluation of the student's reasoning?

A.Correct, because 12 is subtracted from 47 to get 35 and

35/5=5

B.Incorrect, because

x\ge5

after dividing by 5.

C.Incorrect, because the correct result is

x\le7

. ✅

D.Correct, because 'at most' always produces

x=5

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Subtracting 12 from both sides gives $5x\le35$ . Dividing by the positive number 5 preserves the inequality direction and gives $x\le7$ . The student's arithmetic incorrectly treats $35/5$ as 5, causing the wrong boundary.

Q5. A graph represents the inequality $y\le-2x+6$ . Which description should a student use to identify the solution region?

A.The region above a dashed line

B.The region below a solid line ✅

C.The region above a solid line

D.The region below a dashed line

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Because the inequality contains $\le$ , points on the boundary are included, so the boundary must be solid. The symbol also indicates values of $y$ less than or equal to the line, meaning the shaded region lies below the line.

Q6. A fitness center allows a maximum of 5 visitors per instructor. If 7 instructors are present, a manager claims that 36 visitors are allowed because $7\times5+1=36$ . Which assessment is correct?

A.The manager is correct because one extra visitor is always allowed.

B.The manager is incorrect because the maximum is 35 visitors. ✅

C.The manager is incorrect because the maximum is 34 visitors.

D.The manager is correct because 'at most' means slightly more than the calculated value.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Each instructor can supervise at most 5 visitors, so 7 instructors allow $7(5)=35$ visitors. Adding one more visitor violates the stated maximum. The phrase 'at most' requires the total to remain less than or equal to the calculated capacity.

Q7. A café sells sandwiches for $6 each and charges a one-time delivery fee of$ 9. A customer has $65 and wants to buy the greatest possible number of sandwiches. Which modelling step is most important before choosing the answer?

A.Ignore the delivery fee because it is fixed.

B.Set

6s+9\le65

and require

s

to be a whole number. ✅

C.Set

6s+9<65

because the entire budget must not be reached.

D.Set

9s+6\le65

because the fee is paid for every sandwich.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The fixed fee must be included once, while the sandwich cost depends on $s$ . Since the customer can spend at most $65, $6s+9\le65$ . Finally, $s$ must be a nonnegative whole number because partial sandwiches are not meaningful.

Q8. A student says that solving $40-3x\le16$ gives $x\le8$ . Another student says the answer is $x\ge8$ . Which reasoning is correct?

A.The first student is correct because subtracting 40 never changes an inequality.

B.The second student is correct because dividing by

-3

reverses the inequality. ✅

C.Both are correct because

x=8

is the only boundary value.

D.Neither is correct because the inequality has no solution.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Subtracting 40 gives $-3x\le-24$ . Dividing by the negative number $-3$ reverses the inequality, producing $x\ge8$ . The second student's answer correctly accounts for this crucial sign reversal.

Q9. A graph of a feasible region is bounded by $x=2$ , $y=5$ , and $x+y=9$ , with the boundary lines included. Which point is feasible if the conditions require $x\ge2$ , $y\le5$ , and $x+y\le9$ ?

(1,5)

(2,5)

✅

(3,7)

(5,5)

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The point $(2,5)$ satisfies all three conditions: $2\ge2$ , $5\le5$ , and $2+5=7\le9$ . The other choices violate at least one restriction, so interpreting the intersection of the graph's boundaries is essential.

Q10. A mobile data plan allows at most 12 GB per month. A customer has already used 7.5 GB and expects daily use of 0.3 GB for the next $d$ days. What is the greatest whole number of days that can be predicted without exceeding the limit?

A.12 days

B.14 days ✅

C.15 days

D.16 days

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The model is $7.5+0.3d\le12$ . Subtracting 7.5 gives $0.3d\le4.5$ , so $d\le15$ . Thus 15 days is actually feasible, making option C correct; using 14 would unnecessarily underestimate the maximum.

Q11. A manufacturer must keep the total weight of a package at most 50 kg. The package contains a fixed 8 kg container and identical units weighing 3.5 kg each. A worker calculates $3.5n+8<50$ and reports $n=11$ . Which correction is most accurate?

A.The strict inequality should be replaced by

\le

, and the maximum is 12 units. ✅

B.The inequality is correct, and 11 units is the maximum.

C.The fixed weight should be multiplied by

n

, giving 10 units.

D.The inequality should be

3.5n+8\ge50

, giving 12 units.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: 'At most 50 kg' means the package may weigh exactly 50 kg, so the correct model is $3.5n+8\le50$ . This gives $3.5n\le42$ , hence $n\le12$ . Therefore 12 whole units are allowed, whereas 11 is not the maximum.

Q12. A competition awards 4 points for each successful task and subtracts 3 points for each failed task. A contestant may attempt at most 10 tasks and must score at least 20 points. Which pair of conditions correctly models $s$ successes and $f$ failures?

s+f\le10,\ 4s-3f\ge20

✅

s+f\ge10,\ 4s+3f\ge20

s+f=10,\ 4s-3f\le20

s-f\le10,\ 4s+3f\ge20

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: The phrase 'at most 10 tasks' gives $s+f\le10$ , while 'at least 20 points' gives $4s-3f\ge20$ . Together these constraints capture both the attempt limit and the minimum score, demonstrating how different inequality phrases combine in one model.

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