Math 1 Quiz: Compound Inequalities
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Compound InequalitiesQuestion 1 of 17

A fitness tracker monitors daily step counts and categorizes activity levels as follows:

  • Low activity: fewer than 5,000 steps
  • Moderate activity: 5,000 to 9,999 steps
  • High activity: 10,000 steps or more

The tracker also measures active minutes, where a day is considered 'active' if it has at least 30 minutes of moderate-to-vigorous activity.

Sarah's goal is to have either high step activity OR be considered an active day (or both). If Sarah has walked 8,750 steps on a particular day, what range of active minutes mm would allow her to meet her goal?

Sarah needs m<30m < 30 active minutes since she already has moderate step activity.
Sarah needs m≥10000m ≥ 10000 active minutes to compensate for her moderate step activity level.
Sarah needs m≥30m ≥ 30 active minutes since her step count alone doesn't meet the high activity requirement.
Sarah needs m≥5000m ≥ 5000 active minutes to maintain her current moderate activity classification.
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Math 1 Quiz

Math 1 Quiz: Compound Inequalities

Practice Compound Inequalities in Math 1 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 Math 1.

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 fitness tracker monitors daily step counts and categorizes activity levels as follows:

  • Low activity: fewer than 5,000 steps
  • Moderate activity: 5,000 to 9,999 steps
  • High activity: 10,000 steps or more

The tracker also measures active minutes, where a day is considered 'active' if it has at least 30 minutes of moderate-to-vigorous activity.

Sarah's goal is to have either high step activity OR be considered an active day (or both). If Sarah has walked 8,750 steps on a particular day, what range of active minutes mm would allow her to meet her goal?

  1. Sarah needs m<30m < 30 active minutes since she already has moderate step activity.
  2. Sarah needs m≥10000m ≥ 10000 active minutes to compensate for her moderate step activity level.
  3. Sarah needs m≥30m ≥ 30 active minutes since her step count alone doesn't meet the high activity requirement. (correct answer)
  4. Sarah needs m≥5000m ≥ 5000 active minutes to maintain her current moderate activity classification.
Explanation: When you encounter logic problems involving "OR" conditions, break down what each part of the condition means and check if either can be satisfied. Sarah's goal requires meeting at least one of two conditions: (1) high step activity (≥10,000 steps) OR (2) an active day (≥30 active minutes). With 8,750 steps, Sarah has moderate activity but hasn't reached the 10,000-step threshold for high activity. This means condition (1) is not met, so she must satisfy condition (2) by getting at least 30 active minutes. Choice C correctly identifies that Sarah needs m≥30m ≥ 30 active minutes since her step count alone doesn't meet the high activity requirement. Choice A incorrectly suggests Sarah needs fewer than 30 minutes because she has moderate steps. This misunderstands that moderate activity doesn't automatically satisfy her goal - she specifically needs either high steps OR active minutes. Choice B claims she needs m≥10,000m ≥ 10,000 minutes to "compensate" for moderate steps. This confuses the step count threshold (10,000 steps) with active minutes and misunderstands that the conditions are alternatives, not compensatory requirements. Choice D suggests m≥5,000m ≥ 5,000 minutes to "maintain" moderate classification. This misinterprets the goal entirely - Sarah isn't trying to maintain moderate activity but to achieve either high step activity or active day status. Remember: In "OR" logic problems, you only need to satisfy one condition. If one part fails, focus on what's needed for the other part to succeed.

Question 2

A delivery truck's fuel efficiency EE (miles per gallon) varies with speed ss (mph) according to E=−0.02s2+1.6s+20E = -0.02s^2 + 1.6s + 20. Company policy requires E≥35E \geq 35 mpg, but traffic laws limit speed to s≤50s \leq 50 mph. Which speeds satisfy both constraints?

  1. 20≤s≤5020 \leq s \leq 50, since the efficiency constraint provides the lower bound
  2. 30≤s≤5030 \leq s \leq 50, since both constraints together create this feasible range (correct answer)
  3. 20≤s≤3020 \leq s \leq 30, since higher speeds violate the legal constraint
  4. 25≤s≤4525 \leq s \leq 45, since this balances efficiency and legal requirements optimally
Explanation: Solve −0.02s2+1.6s+20≥35-0.02s^2 + 1.6s + 20 \geq 35: −0.02s2+1.6s−15≥0-0.02s^2 + 1.6s - 15 \geq 0. Multiplying by -50: s2−80s+750≤0s^2 - 80s + 750 \leq 0. Using the quadratic formula: s=30s = 30 or s=50s = 50, so 30≤s≤5030 \leq s \leq 50. Combined with s≤50s \leq 50, the answer is 30≤s≤5030 \leq s \leq 50. Choice A uses wrong quadratic roots. Choice C misapplies the constraint. Choice D uses incorrect calculations.

Question 3

A satellite's orbit is stable when its altitude hh (in km) satisfies 2h−300≤h+1002h - 300 \leq h + 100 AND 3h+50>2h+2003h + 50 > 2h + 200. Additionally, for communication purposes, either h≤500h \leq 500 OR h≥800h \geq 800 must hold. Which altitudes satisfy all requirements?

  1. 150<h≤400150 < h \leq 400 or h≥800h \geq 800, covering both viable communication altitude ranges
  2. h≥800h \geq 800, since higher altitudes satisfy all orbital and communication constraints
  3. 150<h≤400150 < h \leq 400, since this represents the stable range within communication limits (correct answer)
  4. 200≤h≤600200 \leq h \leq 600, representing the intersection of stability and communication needs
Explanation: When you encounter compound inequalities with additional constraints, you need to solve each condition separately and then find where all conditions overlap. Let's solve the orbital stability requirements first. For 2h−300≤h+1002h - 300 \leq h + 100, subtract hh from both sides: h−300≤100h - 300 \leq 100, so h≤400h \leq 400. For 3h+50>2h+2003h + 50 > 2h + 200, subtract 2h2h from both sides: h+50>200h + 50 > 200, so h>150h > 150. The orbital stability range is 150<h≤400150 < h \leq 400. Next, the communication constraint requires either h≤500h \leq 500 OR h≥800h \geq 800. This means acceptable altitudes fall in two separate ranges: low altitudes (up to 500 km) or high altitudes (800 km and above). Now find where orbital stability intersects with communication needs. The stability range 150<h≤400150 < h \leq 400 falls entirely within the low communication range h≤500h \leq 500. For the high communication range h≥800h \geq 800, there's no overlap with the stability range since 800 > 400. Therefore, only 150<h≤400150 < h \leq 400 satisfies all requirements. Answer A incorrectly includes h≥800h \geq 800, which fails the orbital stability constraints. Answer B claims only h≥800h \geq 800 works, ignoring that high altitudes violate stability requirements. Answer D gives 200≤h≤600200 \leq h \leq 600, which violates the communication constraint for 500<h<600500 < h < 600 and incorrectly restricts the lower bound. When solving compound inequalities, always check that your final answer satisfies every original constraint—don't assume broader ranges are automatically better.

Question 4

A water reservoir's capacity is being monitored. The water level hh (in feet) must satisfy 15≤h≤2515 \leq h \leq 25 for normal operations, but during maintenance periods, the additional constraint h−3≤2(h−8)h - 3 \leq 2(h - 8) must also be met. What values of hh satisfy both the normal operating range AND the maintenance constraint?

  1. 15≤h≤2515 \leq h \leq 25, since maintenance doesn't further restrict the range (correct answer)
  2. 15≤h≤1315 \leq h \leq 13, since the intersection creates this narrow band
  3. 13≤h≤2513 \leq h \leq 25, since the maintenance constraint provides the lower bound
  4. 13≤h≤1513 \leq h \leq 15, since both constraints must be simultaneously satisfied
Explanation: Solving h−3≤2(h−8)h - 3 \leq 2(h - 8): h−3≤2h−16h - 3 \leq 2h - 16, so 13≤h13 \leq h. The compound inequality is h≥13h \geq 13 AND 15≤h≤2515 \leq h \leq 25, which gives 15≤h≤2515 \leq h \leq 25. Choice B is impossible. Choice C ignores the normal operating minimum. Choice D ignores the normal operating maximum.

Question 5

A physics experiment requires the velocity vv to satisfy either 3v+7>2v+153v + 7 > 2v + 15 OR 5−v≥2v−45 - v \geq 2v - 4. Additionally, for safety reasons, vv must be positive. Which compound inequality represents all acceptable velocities?

  1. 0<v≤30 < v \leq 3 or v>8v > 8, covering both experimental conditions with safety constraint (correct answer)
  2. 0<v≤30 < v \leq 3, since the intersection of all conditions yields this range
  3. v≤3v \leq 3 or v>8v > 8, since safety constraint doesn't affect the experimental bounds
  4. 3<v≤83 < v \leq 8, since this represents the complement of the experimental conditions
Explanation: First inequality: 3v+7>2v+153v + 7 > 2v + 15 gives v>8v > 8. Second inequality: 5−v≥2v−45 - v \geq 2v - 4 gives v≤3v \leq 3. The experimental condition is v>8v > 8 OR v≤3v \leq 3. With v>0v > 0, this becomes 0<v≤30 < v \leq 3 OR v>8v > 8. Choice B misses the v>8v > 8 case. Choice C ignores the safety constraint. Choice D takes the wrong complement.

Question 6

A storage tank's pressure PP (in psi) must satisfy safety protocols. Normal operation requires 12≤P≤2812 \leq P \leq 28, but during system startup, an additional constraint ∣P−20∣≤6|P - 20| \leq 6 must also be met. What pressures are acceptable during startup?

  1. 14≤P≤2614 \leq P \leq 26, representing the intersection of normal and startup constraints (correct answer)
  2. 12≤P≤2812 \leq P \leq 28, since normal operation constraints remain unchanged during startup
  3. 14≤P≤2814 \leq P \leq 28, since startup constraint only affects the lower bound
  4. 12≤P≤2612 \leq P \leq 26, since startup constraint only affects the upper bound
Explanation: Normal operation: 12≤P≤2812 \leq P \leq 28. Startup constraint: ∣P−20∣≤6|P - 20| \leq 6 means 14≤P≤2614 \leq P \leq 26. During startup, both constraints apply (AND), so we need the intersection: max⁡(12,14)≤P≤min⁡(28,26)\max(12,14) \leq P \leq \min(28,26), giving 14≤P≤2614 \leq P \leq 26. Choices B, C, and D each ignore one part of the intersection calculation.

Question 7

An optimization algorithm converges when the error ee satisfies ∣3e−6∣<12|3e - 6| < 12 OR e2≤9e^2 \leq 9. For numerical stability, the algorithm also requires e≠0e \neq 0. Which error values allow convergence?

  1. −3≤e<0-3 \leq e < 0 or 0<e≤60 < e \leq 6, representing the union with zero excluded
  2. −2<e<6-2 < e < 6 and e≠0e \neq 0, representing the intersection of both convergence conditions
  3. −3≤e≤3-3 \leq e \leq 3 and e≠0e \neq 0, since the quadratic condition is more restrictive
  4. −3≤e≤6-3 \leq e \leq 6 and e≠0e \neq 0, combining both conditions with the stability constraint (correct answer)
Explanation: First condition: ∣3e−6∣<12|3e - 6| < 12 means −12<3e−6<12-12 < 3e - 6 < 12, so −6<3e<18-6 < 3e < 18, giving −2<e<6-2 < e < 6. Second condition: e2≤9e^2 \leq 9 means −3≤e≤3-3 \leq e \leq 3. The convergence condition is (−2<e<6-2 < e < 6) OR (−3≤e≤3-3 \leq e \leq 3), which equals −3≤e<6-3 \leq e < 6. With e≠0e \neq 0, the answer is −3≤e≤6-3 \leq e \leq 6 and e≠0e \neq 0. Choice A has wrong bounds. Choice B uses AND instead of OR. Choice C ignores the first condition's contribution.

Question 8

A marketing survey finds that customer satisfaction SS (on a scale of 1-10) relates to price pp (in dollars) by the condition: customers are satisfied when S≥7S \geq 7 AND either p≤15p \leq 15 OR p≥25p \geq 25. If S=11−0.2pS = 11 - 0.2p, which prices result in satisfied customers?

  1. 5≤p≤155 \leq p \leq 15, since this balances satisfaction requirements with price constraints
  2. p≤20p \leq 20, since this represents the maximum price for adequate satisfaction
  3. p≤15p \leq 15 or p≥25p \geq 25, since both price ranges can theoretically satisfy customers
  4. p≤15p \leq 15, since higher prices cannot achieve the satisfaction threshold (correct answer)
Explanation: When you encounter problems involving compound conditions (multiple requirements connected by AND/OR), you need to work systematically through each constraint to find where they all overlap. First, let's find where S≥7S \geq 7. Since S=11−0.2pS = 11 - 0.2p, we need: 11−0.2p≥711 - 0.2p \geq 7 4≥0.2p4 \geq 0.2p p≤20p \leq 20 So satisfaction requires p≤20p \leq 20. But there's an additional constraint: customers must be satisfied AND either p≤15p \leq 15 OR p≥25p \geq 25. Now we need the intersection of p≤20p \leq 20 with the condition "p≤15p \leq 15 OR p≥25p \geq 25." For p≤15p \leq 15: These prices satisfy both p≤20p \leq 20 and the price constraint, so they work. For p≥25p \geq 25: These prices satisfy the price constraint but violate p≤20p \leq 20 (since you can't have p≥25p \geq 25 and p≤20p \leq 20 simultaneously), so they don't work. Therefore, only p≤15p \leq 15 results in satisfied customers. Choice A (5≤p≤155 \leq p \leq 15) is too restrictive—prices below 5alsowork.ChoiceB(5 also work. Choice B ( p≤20p \leq 20 )ignorestheadditionalpriceconstraintentirely.ChoiceC() ignores the additional price constraint entirely. Choice C ( p≤15p \leq 15 oror p≥25p \geq 25 )failstoaccountforthesatisfactionthreshold—while) fails to account for the satisfaction threshold—while p≥25p \geq 25 satisfiesthepriceconstraint,itmakessatisfies the price constraint, it makes S<7S < 7 $. Study tip: In compound condition problems, solve each constraint separately, then find their intersection. Don't assume all theoretically possible ranges will work—always check against every requirement.

Question 9

A manufacturing company's daily profit PP (in thousands of dollars) is modeled by P=−2x2+16x−24P = -2x^2 + 16x - 24, where xx is the number of units produced (in hundreds). For the company to remain profitable while meeting production constraints, they need P≥6P \geq 6 AND x≤5x \leq 5. Which compound inequality represents the valid production levels?

  1. 1≤x≤51 \leq x \leq 5
  2. 2≤x≤52 \leq x \leq 5
  3. 3≤x≤53 \leq x \leq 5 (correct answer)
  4. 1≤x≤31 \leq x \leq 3
Explanation: First solve −2x2+16x−24≥6-2x^2 + 16x - 24 \geq 6: −2x2+16x−30≥0-2x^2 + 16x - 30 \geq 0, or x2−8x+15≤0x^2 - 8x + 15 \leq 0. Factoring: (x−3)(x−5)≤0(x-3)(x-5) \leq 0, so 3≤x≤53 \leq x \leq 5. Combined with the constraint x≤5x \leq 5, the answer is 3≤x≤53 \leq x \leq 5. Choice A uses the wrong quadratic roots. Choice B incorrectly solves the profit inequality. Choice D ignores the production constraint.

Question 10

The temperature TT (in °F) inside a greenhouse must satisfy 65≤T≤8065 ≤ T ≤ 80 during the day and T<65T < 65 OR T>80T > 80 triggers an alarm system. On a particular day, temperature readings were taken every 2 hours starting at 6 AM: 63°F, 68°F, 75°F, 82°F, 79°F, 71°F, 66°F. At how many of these reading times would the alarm system be triggered?

  1. The alarm would be triggered at exactly 2 of the reading times. (correct answer)
  2. The alarm would be triggered at exactly 3 of the reading times.
  3. The alarm would be triggered at exactly 1 of the reading times.
  4. The alarm would be triggered at exactly 4 of the reading times.
Explanation: The alarm triggers when T < 65 OR T > 80. Checking each reading: 63°F (triggers - less than 65), 68°F (no trigger), 75°F (no trigger), 82°F (triggers - greater than 80), 79°F (no trigger), 71°F (no trigger), 66°F (no trigger). The alarm triggers at exactly 2 times. Choice B might result from incorrectly including a borderline case. Choice C undercounts the triggers. Choice D overcounts significantly.

Question 11

A delivery service charges a base fee of $8 plus $2.50 per mile. A customer wants to spend more than $25 but no more than $40 on delivery. Which compound inequality correctly represents the range of miles $mm $ the customer can travel?

  1. 25<8+2.5m≤4025 < 8 + 2.5m \leq 40, which simplifies to 6.8<m≤12.86.8 < m \leq 12.8 (correct answer)
  2. 25<8+2.5m≤4025 < 8 + 2.5m \leq 40, which simplifies to 6.8≤m<12.86.8 ≤ m < 12.8
  3. 25≤8+2.5m<4025 ≤ 8 + 2.5m < 40, which simplifies to 6.8≤m<12.86.8 ≤ m < 12.8
  4. 25<8+2.5m<4025 < 8 + 2.5m < 40, which simplifies to 6.8<m<12.86.8 < m < 12.8
Explanation: The cost function is C = 8 + 2.5m. The customer wants to spend 'more than $25' (C > 25) and 'no more than $40' (C ≤ 40). This gives us 25 < 8 + 2.5m ≤ 40. Subtracting 8 from all parts: 17 < 2.5m ≤ 32. Dividing by 2.5: 6.8 < m ≤ 12.8. Choice B has the wrong inequality symbol for the left side. Choice C uses ≤ instead of < for the left side. Choice D uses < instead of ≤ for the right side.

Question 12

A scientist studying bacterial growth finds that the population PP (in thousands) after hh hours follows the model P=2+3hP = 2 + 3h. The experiment is considered successful if the population is between 8 and 20 thousand bacteria, inclusive, but the lab must be evacuated if the population exceeds 25 thousand. What is the time interval during which the experiment is successful AND evacuation is not required?

  1. 8≤h≤208 ≤ h ≤ 20 hours
  2. 2≤h<7.672 ≤ h < 7.67 hours
  3. 2≤h≤62 ≤ h ≤ 6 hours (correct answer)
  4. 6≤h<256 ≤ h < 25 hours
Explanation: This problem tests your ability to work with compound inequalities—situations where you need to satisfy multiple conditions simultaneously. When you see "between X and Y" combined with another constraint, you'll need to set up and solve a system of inequalities. The experiment is successful when the population is between 8 and 20 thousand (inclusive), so you need 8≤P≤208 ≤ P ≤ 20. Substituting the model P=2+3hP = 2 + 3h: 8≤2+3h≤208 ≤ 2 + 3h ≤ 20. Solving the left inequality: 8≤2+3h8 ≤ 2 + 3h gives 6≤3h6 ≤ 3h, so h≥2h ≥ 2. Solving the right inequality: 2+3h≤202 + 3h ≤ 20 gives 3h≤183h ≤ 18, so h≤6h ≤ 6. Additionally, evacuation occurs when P>25P > 25, meaning 2+3h>252 + 3h > 25, which gives h>7.67h > 7.67. Since we want to avoid evacuation, we need h≤7.67h ≤ 7.67. Combining all constraints: 2≤h≤62 ≤ h ≤ 6 AND h≤7.67h ≤ 7.67. The most restrictive condition gives us 2≤h≤62 ≤ h ≤ 6. Choice A (8≤h≤208 ≤ h ≤ 20) uses the population values instead of solving for time. Choice B (2≤h<7.672 ≤ h < 7.67) correctly finds the lower bound but uses the evacuation threshold as the upper bound, ignoring that the experiment stops being successful at h=6h = 6. Choice D (6≤h<256 ≤ h < 25) incorrectly interprets the success conditions and uses population values mixed with time values. When solving compound inequalities, always identify each constraint separately, solve for the variable, then find the intersection of all valid ranges.

Question 13

A water tank's depth dd (in feet) varies according to the equation d=12−0.5td = 12 - 0.5t, where tt is time in hours after 6 AM. The tank's pump automatically turns on when d≤4d ≤ 4 feet, and an emergency alarm sounds when d≤2d ≤ 2 feet. During which time interval will the pump be running but the emergency alarm will NOT be sounding?

  1. 16≤t≤2016 ≤ t ≤ 20 hours after 6 AM
  2. 16≤t<2016 ≤ t < 20 hours after 6 AM (correct answer)
  3. 16<t≤2016 < t ≤ 20 hours after 6 AM
  4. 4≤t<24 ≤ t < 2 hours after 6 AM
Explanation: We need pump running (d ≤ 4) AND alarm not sounding (d > 2), so 2 < d ≤ 4. Substituting d = 12 - 0.5t: 2 < 12 - 0.5t ≤ 4. For the left inequality: 2 < 12 - 0.5t gives 0.5t < 10, so t < 20. For the right inequality: 12 - 0.5t ≤ 4 gives 8 ≤ 0.5t, so t ≥ 16. Combined: 16 ≤ t < 20. Choice A includes t = 20, when the alarm would sound. Choice C excludes t = 16, when the pump first turns on. Choice D has impossible bounds (4 ≥ 2 is always false for time intervals).

Question 14

A manufacturing company produces widgets with weights that must satisfy the compound inequality 12.5≤w<18.212.5 \leq w < 18.2 grams to meet quality standards. If a batch contains widgets weighing 11.8, 12.5, 15.3, 18.2, and 19.1 grams, and the company needs at least 60% of widgets in each batch to pass quality control to avoid penalties, what is the status of this batch?

  1. The batch passes quality control since exactly 60% of widgets meet the weight requirements.
  2. The batch fails quality control since only 40% of widgets meet the weight requirements. (correct answer)
  3. The batch passes quality control since 80% of widgets meet the weight requirements.
  4. The batch fails quality control since only 20% of widgets meet the weight requirements.
Explanation: The compound inequality 12.5≤w<18.212.5 \leq w < 18.2 means weights must be greater than or equal to 12.5 AND less than 18.2. Checking each widget: 11.8 (fails - too light), 12.5 (passes), 15.3 (passes), 18.2 (fails - not less than 18.2), 19.1 (fails - too heavy). Only 2 out of 5 widgets pass, which is 40%. Since 40% < 60%, the batch fails quality control. Choice A incorrectly counts 18.2 as passing. Choice C incorrectly excludes 11.8 only. Choice D miscounts entirely.

Question 15

A rental car company charges $35 per day plus $0.15 per mile driven. A customer's budget allows for spending more than $50 but at most $80 for a 2-day rental. If the customer drives the same number of miles $mm $ each day, which compound inequality represents the valid range for daily miles driven?

  1. 50<70+0.30m≤8050 < 70 + 0.30m ≤ 80, which gives −66.67<m≤33.33-66.67 < m ≤ 33.33
  2. 50<70+0.30m≤8050 < 70 + 0.30m ≤ 80, which gives −66.67<m≤33.33-66.67 < m ≤ 33.33, but since m≥0m ≥ 0, the answer is 0≤m≤33.330 ≤ m ≤ 33.33
  3. 50<35+0.15m≤8050 < 35 + 0.15m ≤ 80, which gives 100<m≤300100 < m ≤ 300
  4. 50<70+0.30m≤8050 < 70 + 0.30m ≤ 80, which gives −66.67<m≤33.33-66.67 < m ≤ 33.33, so the valid range is 0<m≤33.330 < m ≤ 33.33 (correct answer)
Explanation: For a 2-day rental driving m miles each day: Cost = 2(35)+2(m)(35) + 2(m)(0.15) = $70 + $0.30m. Budget constraint: $50 < Cost ≤ $80, so 50 < 70 + 0.30m ≤ 80. Solving: -20 < 0.30m ≤ 10, so -66.67 < m ≤ 33.33. Since miles driven must be positive and the budget requires spending MORE than 50(not≥50 (not ≥50), we need m > 0, giving 0 < m ≤ 33.33. Choice A shows the calculation but doesn't address the positivity constraint. Choice B incorrectly uses ≥ instead of > for the lower bound. Choice C uses the wrong cost formula for 2 days.

Question 16

A quality control system rejects products when the measurement mm satisfies ∣2m−8∣≥6|2m - 8| \geq 6 AND m≠1m \neq 1. Which values of mm will cause product rejection?

  1. m≤1m \leq 1 or m≥7m \geq 7, since this covers all values outside the tolerance band
  2. m<1m < 1 or m≥7m \geq 7, since the exclusion condition modifies the lower bound (correct answer)
  3. m≤7m \leq 7 or m≥1m \geq 1, since the absolute value creates this broader range
  4. 1<m≤71 < m \leq 7, since this represents the acceptable range between extremes
Explanation: ∣2m−8∣≥6|2m - 8| \geq 6 means 2m−8≤−62m - 8 \leq -6 OR 2m−8≥62m - 8 \geq 6, so m≤1m \leq 1 OR m≥7m \geq 7. With the additional condition m≠1m \neq 1, rejection occurs when m<1m < 1 OR m≥7m \geq 7. Choice A doesn't account for m≠1m \neq 1. Choice C has incorrect inequality directions. Choice D describes acceptance, not rejection.

Question 17

A temperature monitoring system triggers an alarm when the recorded temperature TT satisfies ∣T−72∣>8|T - 72| > 8 OR T<55T < 55. Which of the following temperatures would NOT trigger the alarm?

  1. T=68T = 68 degrees, since it falls within normal operating range (correct answer)
  2. T=52T = 52 degrees, since it satisfies the secondary condition only
  3. T=82T = 82 degrees, since it exceeds the upper temperature threshold
  4. T=45T = 45 degrees, since it falls below both critical thresholds
Explanation: The alarm triggers when T<64T < 64 OR T>80T > 80 OR T<55T < 55, which simplifies to T<64T < 64 OR T>80T > 80. Only T=68T = 68 falls in the safe range [64,80][64, 80]. Choice B: 52<6452 < 64 triggers alarm. Choice C: 82>8082 > 80 triggers alarm. Choice D: 45<6445 < 64 triggers alarm.