Middle School Math Quiz: One Step Inequalities
7 questions · exam conditions
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One Step InequalitiesQuestion 1 of 7

A delivery truck can carry a maximum load of 24002400 pounds. It already has packages weighing 16501650 pounds loaded. If each additional package weighs 2525 pounds, what is the solution to the inequality representing how many more packages pp can be loaded?

p≤30p \leq 30 and the truck can carry exactly 30 more packages
p≤30p \leq 30 and the truck can carry at most 30 more packages
p<30p < 30 and the truck can carry at most 29 more packages
p≥30p \geq 30 and the truck can carry at least 30 more packages
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Middle School Math Quiz

Middle School Math Quiz: One Step Inequalities

Practice One Step Inequalities in Middle School Math with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on One Step Inequalities, giving you a quick way to practice the rules, question types, and explanations that matter most for Middle School Math.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A delivery truck can carry a maximum load of 24002400 pounds. It already has packages weighing 16501650 pounds loaded. If each additional package weighs 2525 pounds, what is the solution to the inequality representing how many more packages pp can be loaded?

  1. p≤30p \leq 30 and the truck can carry exactly 30 more packages
  2. p≤30p \leq 30 and the truck can carry at most 30 more packages (correct answer)
  3. p<30p < 30 and the truck can carry at most 29 more packages
  4. p≥30p \geq 30 and the truck can carry at least 30 more packages
Explanation: The inequality is 1650+25p≤24001650 + 25p \leq 2400. Subtracting 1650: 25p≤75025p \leq 750. Dividing by 25: p≤30p ≤ 30. This means at most 30 packages. Choice A incorrectly states 'exactly' instead of 'at most'. Choice C uses strict inequality when equality is allowed. Choice D reverses the inequality direction completely.

Question 2

A bakery needs to package at least 180180 cupcakes for a large order. They have boxes that hold 1515 cupcakes each and have already filled 88 boxes. The inequality 15(8+b)≥18015(8 + b) \geq 180 represents this situation, where bb is additional boxes needed. What is the minimum value of bb?

  1. b≥8b \geq 8, so the minimum is 8 boxes
  2. b≥12b \geq 12, so the minimum is 12 boxes
  3. b≥0b \geq 0, so the minimum is 0 boxes
  4. b≥4b \geq 4, so the minimum is 4 boxes (correct answer)
Explanation: When you encounter inequality word problems, start by understanding what the inequality represents and then solve it step by step to find what the variable must satisfy. Let's solve the inequality 15(8+b)≥18015(8 + b) \geq 180. First, divide both sides by 15: 8+b≥128 + b \geq 12. Then subtract 8 from both sides: b≥4b \geq 4. This means the bakery needs at least 4 additional boxes beyond the 8 they already filled. Let's verify: With 4 additional boxes, they'd have 8+4=128 + 4 = 12 total boxes, containing 15×12=18015 \times 12 = 180 cupcakes, which exactly meets their requirement. Since they need "at least" 180 cupcakes, 4 additional boxes is the minimum. Choice A incorrectly states b≥8b \geq 8, which would mean they need 8 additional boxes on top of the 8 already filled—that's 16 total boxes for 240 cupcakes, way more than necessary. Choice B claims b≥12b \geq 12, suggesting 12 additional boxes for 20 total boxes and 300 cupcakes—again, excessive. Choice C states b≥0b \geq 0, which would mean no additional boxes are needed, but 8 boxes only hold 120 cupcakes, falling short of the 180 requirement. When solving inequality word problems, always substitute your answer back into the original context to verify it makes sense. Here, 4 additional boxes gives exactly 180 cupcakes, confirming this is indeed the minimum needed to satisfy "at least 180."

Question 3

A swimming pool is being drained at a constant rate. The amount of water remaining after tt hours is given by 500−25t500 - 25t gallons. For safety reasons, the pool must maintain more than 150150 gallons at all times. What is the maximum number of complete hours the pool can be drained?

  1. 13 hours (correct answer)
  2. 14 hours
  3. 15 hours
  4. 16 hours
Explanation: We need 500−25t>150500 - 25t > 150. Subtracting 500: −25t>−350-25t > -350. Dividing by −25-25 and flipping the inequality: t<14t < 14. Since we want complete hours and tt must be less than 14, the maximum is 13 hours. Choice B would result in exactly 150 gallons (not more than). Choices C and D violate the constraint entirely.

Question 4

A temperature monitoring system records that the laboratory temperature must stay below −5°C-5°C to preserve certain samples. If the current temperature is 3x+73x + 7 degrees Celsius, which inequality represents the constraint on xx?

  1. x<−4x < -4 (correct answer)
  2. x>−4x > -4
  3. x<4x < 4
  4. x>4x > 4
Explanation: The temperature 3x+73x + 7 must be less than −5-5, so we have 3x+7<−53x + 7 < -5. Subtracting 7 from both sides gives 3x<−123x < -12. Dividing by 3 gives x<−4x < -4. Choice B reverses the inequality sign incorrectly. Choice C results from adding 7 instead of subtracting. Choice D combines both errors.

Question 5

A cell phone plan charges $35 per month plus $0.15 for each text message sent. If Jamie's monthly bill cannot exceed $50, what is the maximum number of text messages she can send?

  1. 99 text messages
  2. 101 text messages
  3. 100 text messages (correct answer)
  4. 333 text messages
Explanation: When you encounter a word problem involving a fixed cost plus a variable cost with a spending limit, you're looking at an inequality problem. You need to set up an equation that represents the total cost and find the maximum value that keeps you within the budget. Jamie's total monthly cost equals the fixed monthly charge plus the cost per text message times the number of messages: 35+0.15x≤5035 + 0.15x \leq 50, where xx is the number of text messages. To solve this inequality, subtract 35 from both sides: 0.15x≤150.15x \leq 15. Then divide both sides by 0.15: x≤100x \leq 100. This means Jamie can send at most 100 text messages while staying within her $50 budget. Let's verify: $35+0.15(100)=35+15=5035 + 0.15(100) = 35 + 15 = 50 $, which exactly meets her limit. Choice A (99 text messages) is incorrect because it's less than the maximum possible. While 99 messages would cost 35 + 0.15(99) = 49.85 , Jamie could actually afford one more message. Choice B (101 text messages) exceeds the budget: 35 + 0.15(101) = 50.15 , which is 15 cents over her limit. Choice D (333 text messages) represents a common error where students might divide 50 by 0.15, forgetting to subtract the fixed monthly charge first. Remember: when solving real-world inequality problems, always check whether your answer makes practical sense by substituting back into the original constraint. The maximum value should satisfy the inequality without exceeding the limit.

Question 6

Maria has $45 to spend on art supplies. After buying a sketchbook for $12, she wants to buy markers that cost $5.50 each. What is the maximum number of markers she can buy?

  1. 5 markers
  2. 6 markers (correct answer)
  3. 7 markers
  4. 8 markers
Explanation: After buying the sketchbook, Maria has 45−12=3345 - 12 = 33 dollars left. If she buys mm markers, she needs 5.50m≤335.50m \leq 33. Dividing by 5.50 gives m≤6m \leq 6. Since mm must be a whole number, the maximum is 6 markers. Choice A ignores that she can afford exactly 6. Choice C would cost 7×5.50=38.50>337 × 5.50 = 38.50 > 33. Choice D would cost 8×5.50=44>338 × 5.50 = 44 > 33.

Question 7

Which of the following represents the solution set for x−3>4\frac{x}{-3} > 4 when graphed on a number line?

  1. Open circle at −12-12, shaded to the right toward positive numbers
  2. Closed circle at −12-12, shaded to the left toward negative numbers
  3. Open circle at −12-12, shaded to the left toward negative numbers (correct answer)
  4. Open circle at 1212, shaded to the left toward negative numbers
Explanation: When solving inequalities involving fractions with negative denominators, you need to carefully track how the inequality sign changes during your algebraic steps. To solve x−3>4\frac{x}{-3} > 4, multiply both sides by −3-3. Since you're multiplying by a negative number, you must flip the inequality sign. This gives you: x<4×(−3)x < 4 \times (-3), which simplifies to x<−12x < -12. The solution x<−12x < -12 means all numbers less than −12-12. On a number line, this requires an open circle at −12-12 (since xx cannot equal −12-12) with shading extending left toward more negative numbers. This matches choice C. Let's examine why the other options are incorrect: Choice A places the circle correctly at −12-12 but shades right toward positive numbers, which would represent x>−12x > -12 instead of x<−12x < -12. Choice B uses a closed circle, which would only be correct if the original inequality included "or equal to" (≥ or ≤). Since we have a strict inequality (>), the circle must be open. Choice D places the circle at positive 1212 instead of negative 1212, likely resulting from forgetting to multiply 44 by the negative sign. Remember this key rule: whenever you multiply or divide both sides of an inequality by a negative number, always flip the inequality sign. This is the most common mistake students make with inequality problems, so double-check your work by substituting a test value from your solution region back into the original inequality.