Pre-Algebra Quiz: Compound Inequalities
8 questions · exam conditions
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Compound InequalitiesQuestion 1 of 8

A temperature control system maintains safe operation when the temperature TT (in degrees Celsius) satisfies −15<T≤40-15 < T \leq 40. However, an alarm sounds when T<−10T < -10 or T>35T > 35. For which values of TT does the system operate safely but without an alarm?

−15<T≤−10-15 < T \leq -10
−10≤T≤35-10 \leq T \leq 35
−10<T≤35-10 < T \leq 35
−15<T<−10-15 < T < -10 or 35<T≤4035 < T \leq 40
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Pre-Algebra Quiz

Pre-Algebra Quiz: Compound Inequalities

Practice Compound Inequalities in Pre-Algebra 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 Compound Inequalities, giving you a quick way to practice the rules, question types, and explanations that matter most for Pre-Algebra.

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 temperature control system maintains safe operation when the temperature TT (in degrees Celsius) satisfies −15<T≤40-15 < T \leq 40. However, an alarm sounds when T<−10T < -10 or T>35T > 35. For which values of TT does the system operate safely but without an alarm?

  1. −15<T≤−10-15 < T \leq -10
  2. −10≤T≤35-10 \leq T \leq 35 (correct answer)
  3. −10<T≤35-10 < T \leq 35
  4. −15<T<−10-15 < T < -10 or 35<T≤4035 < T \leq 40
Explanation: For safe operation without alarm, we need the intersection of safe operation (−15<T≤40-15 < T \leq 40) and no alarm (−10≤T≤35-10 \leq T \leq 35). The alarm sounds when T<−10T < -10 or T>35T > 35, so no alarm means −10≤T≤35-10 \leq T \leq 35. The intersection gives −10≤T≤35-10 \leq T \leq 35. Choice A gives only the unsafe region below -10. Choice C incorrectly excludes T = -10. Choice D gives the regions where the system is safe but the alarm IS sounding.

Question 2

A manufacturing process requires the width ww of a component to satisfy ∣w−12∣≤0.5|w - 12| \leq 0.5. Additionally, quality control rejects components where w<11.7w < 11.7 or w>12.4w > 12.4. What range of widths passes both requirements?

  1. 11.5≤w≤12.511.5 \leq w \leq 12.5
  2. 11.7≤w≤12.411.7 \leq w \leq 12.4 (correct answer)
  3. 11.7≤w≤12.511.7 \leq w \leq 12.5
  4. 11.5≤w≤12.411.5 \leq w \leq 12.4
Explanation: First, solve ∣w−12∣≤0.5|w - 12| \leq 0.5: this gives −0.5≤w−12≤0.5-0.5 \leq w - 12 \leq 0.5, so 11.5≤w≤12.511.5 \leq w \leq 12.5. Quality control accepts components where it's NOT true that w<11.7w < 11.7 or w>12.4w > 12.4, meaning 11.7≤w≤12.411.7 \leq w \leq 12.4. The intersection of both requirements is 11.7≤w≤12.411.7 \leq w \leq 12.4. Choice A gives only the absolute value constraint. Choice C incorrectly takes the left bound from quality control and right bound from absolute value. Choice D incorrectly takes the left bound from absolute value and right bound from quality control.

Question 3

A company's profit PP (in thousands of dollars) satisfies P≤50P \leq 50 or P≥80P \geq 80. If the profit must also satisfy P>30P > 30, which compound inequality represents all possible profit values?

  1. 30<P≤5030 < P \leq 50 or P≥80P \geq 80 (correct answer)
  2. 30<P≤5030 < P \leq 50 and P≥80P \geq 80
  3. P≤30P \leq 30 or 50<P<8050 < P < 80
  4. 30<P<5030 < P < 50 or P>80P > 80
Explanation: We need the intersection of (P≤50P \leq 50 or P≥80P \geq 80) with P>30P > 30. For the first part: P≤50P \leq 50 intersected with P>30P > 30 gives 30<P≤5030 < P \leq 50. For the second part: P≥80P \geq 80 intersected with P>30P > 30 gives P≥80P \geq 80 (since 80 > 30). The final answer uses 'or' to combine these regions. Choice B incorrectly uses 'and' between disjoint intervals. Choice C represents the complement of what we want. Choice D incorrectly excludes P = 50 and P = 80.

Question 4

A parking garage charges $3 per hour and has a maximum daily charge of $18. If someone parks for $hh hoursandpayshours and pays PP $ dollars, which compound inequality correctly describes the relationship when the maximum daily charge applies?

  1. h≥6h \geq 6 or P=18P = 18
  2. h>6h > 6 and P=18P = 18
  3. h≤6h \leq 6 and P=3hP = 3h
  4. h≥6h \geq 6 and P=18P = 18 (correct answer)
Explanation: When you encounter parking fee problems with maximum charges, you need to identify the breakpoint where the hourly rate stops applying and the maximum kicks in. Here, at $3 per hour with an $18 maximum, that breakpoint occurs at $18÷3=618 ÷ 3 = 6 $ hours. The key insight is understanding when the maximum daily charge applies. For any parking duration of 6 hours or more, you would normally pay 3 × 6 = 18 dollars or more, but the maximum caps it at exactly $18. So when someone parks for 6 or more hours, they always pay the maximum of $18. Choice D correctly captures this: $h ≥ 6$$ and $$P = 18$$. The "greater than or equal to" symbol includes exactly 6 hours (where you'd pay 18 anyway) and any longer duration where the maximum applies. Choice A uses "or" instead of "and," which creates a logical error. The relationship requires both conditions to be true simultaneously, not either one or the other. Choice B excludes exactly 6 hours by using h>6h > 6, but at exactly 6 hours, you'd still pay $18, so the maximum effectively applies. Choice C describes the opposite scenario—when someone parks for 6 hours or less and pays the regular hourly rate without hitting the maximum. Remember that compound inequalities with "and" mean both conditions must be satisfied at the same time. Look for the breakpoint where pricing structures change, and always check whether the boundary value should be included or excluded.

Question 5

If 2x−7>12x - 7 > 1 and 5x+3≤235x + 3 \leq 23, what is the solution set for xx?

  1. x>4x > 4 and x≤4x \leq 4
  2. 4<x≤44 < x \leq 4
  3. No solution exists (correct answer)
  4. x>4x > 4 or x≤4x \leq 4
Explanation: Solving the first inequality: 2x−7>12x - 7 > 1 gives 2x>82x > 8, so x>4x > 4. Solving the second inequality: 5x+3≤235x + 3 \leq 23 gives 5x≤205x \leq 20, so x≤4x \leq 4. For a compound inequality with 'and', we need both conditions satisfied simultaneously. However, no number can be both greater than 4 AND less than or equal to 4. Choice A incorrectly lists both conditions separately. Choice B shows the impossible intersection notation. Choice D incorrectly uses 'or' instead of 'and'.

Question 6

Solve the compound inequality: −3≤2x+1<7-3 \leq 2x + 1 < 7. What values of xx satisfy this inequality?

  1. −2≤x<3-2 \leq x < 3 (correct answer)
  2. −1≤x<4-1 \leq x < 4
  3. −4≤x<6-4 \leq x < 6
  4. −2<x≤3-2 < x \leq 3
Explanation: This is a compound inequality of the form a≤expression<ba \leq expression < b. We solve by isolating x: −3≤2x+1<7-3 \leq 2x + 1 < 7. Subtract 1 from all parts: −4≤2x<6-4 \leq 2x < 6. Divide all parts by 2: −2≤x<3-2 \leq x < 3. Choice B incorrectly subtracts 1 instead of adding it in the first step. Choice C forgets to divide by 2 in the final step. Choice D has the wrong inequality symbols (should be ≤ on the left and < on the right, not the reverse).

Question 7

The solution set for the inequality 2x−3>52x - 3 > 5 OR x+4≤1x + 4 \leq 1 can be written as (−∞,a]∪(b,∞)(-\infty, a] \cup (b, \infty) for some values aa and bb. What are the values of aa and bb?

  1. a=1,b=5a = 1, b = 5
  2. a=−3,b=8a = -3, b = 8
  3. a=4,b=−3a = 4, b = -3
  4. a=−3,b=4a = -3, b = 4 (correct answer)
Explanation: When you encounter compound inequalities with "OR," you're looking for values that satisfy either condition. The solution will be the union of two separate intervals. Start by solving each inequality individually. For 2x−3>52x - 3 > 5, add 3 to both sides: 2x>82x > 8, then divide by 2: x>4x > 4. This gives you the interval (4,∞)(4, \infty). For x+4≤1x + 4 \leq 1, subtract 4 from both sides: x≤−3x \leq -3. This gives you the interval (−∞,−3](-\infty, -3]. Since this is an "OR" statement, you take the union of these intervals: (−∞,−3]∪(4,∞)(-\infty, -3] \cup (4, \infty). Comparing this to the form (−∞,a]∪(b,∞)(-\infty, a] \cup (b, \infty), you can see that a=−3a = -3 and b=4b = 4. Answer choice A gives a=1,b=5a = 1, b = 5, which would represent (−∞,1]∪(5,∞)(-\infty, 1] \cup (5, \infty) - this comes from solving the inequalities incorrectly. Answer choice B has a=−3,b=8a = -3, b = 8, suggesting someone found the correct first interval but made an error when solving 2x>82x > 8, perhaps forgetting to divide by 2. Answer choice C reverses the values with a=4,b=−3a = 4, b = -3, which would create the impossible interval (−∞,4]∪(−3,∞)(-\infty, 4] \cup (-3, \infty) that overlaps incorrectly. Remember: with "OR" compound inequalities, solve each part separately, then unite the intervals. Always double-check that your intervals make sense when written in the requested format.

Question 8

A student incorrectly solved −2<3x+1≤7-2 < 3x + 1 \leq 7 and got −1<x≤2.67-1 < x \leq 2.67. What error did the student likely make?

  1. Failed to flip inequality signs when dividing by 3
  2. Forgot to subtract 1 from all three parts before dividing
  3. Added 1 instead of subtracting 1 from all three parts (correct answer)
  4. Divided by 3 before isolating the 3x3x term completely
Explanation: The correct solution: −2<3x+1≤7-2 < 3x + 1 \leq 7. Subtract 1: −3<3x≤6-3 < 3x \leq 6. Divide by 3: −1<x≤2-1 < x \leq 2. The student got −1<x≤2.67-1 < x \leq 2.67, suggesting they worked with −2<3x+1≤7-2 < 3x + 1 \leq 7, added 1 to get −1<3x+2≤8-1 < 3x + 2 \leq 8, then somehow got to their final answer. The student likely added 1 instead of subtracting 1. Choice A is wrong because dividing by positive 3 doesn't flip signs. Choice B would give a different error pattern. Choice D would lead to algebraic mistakes but not this specific error.