Compound Inequalities - Math 1

1. 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 $m$ would allow her to meet her goal?
   Answer: Sarah needs $m ≥ 30$ active minutes since her step count alone doesn't meet the high activity requirement. 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 $m ≥ 30$ 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 $m ≥ 10,000$ 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 $m ≥ 5,000$ 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.
2. A delivery truck's fuel efficiency $E$ (miles per gallon) varies with speed $s$ (mph) according to $E = -0.02s^2 + 1.6s + 20$. Company policy requires $E \geq 35$ mpg, but traffic laws limit speed to $s \leq 50$ mph. Which speeds satisfy both constraints?
   Answer: $30 \leq s \leq 50$, since both constraints together create this feasible range Solve $-0.02s^2 + 1.6s + 20 \geq 35$: $-0.02s^2 + 1.6s - 15 \geq 0$. Multiplying by -50: $s^2 - 80s + 750 \leq 0$. Using the quadratic formula: $s = 30$ or $s = 50$, so $30 \leq s \leq 50$. Combined with $s \leq 50$, the answer is $30 \leq s \leq 50$. Choice A uses wrong quadratic roots. Choice C misapplies the constraint. Choice D uses incorrect calculations.
3. A marketing survey finds that customer satisfaction $S$ (on a scale of 1-10) relates to price $p$ (in dollars) by the condition: customers are satisfied when $S \geq 7$ AND either $p \leq 15$ OR $p \geq 25$. If $S = 11 - 0.2p$, which prices result in satisfied customers?
   Answer: $p \leq 15$, since higher prices cannot achieve the satisfaction threshold 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 $S \geq 7$. Since $S = 11 - 0.2p$, we need: $11 - 0.2p \geq 7$ $4 \geq 0.2p$ $p \leq 20$ So satisfaction requires $p \leq 20$. But there's an additional constraint: customers must be satisfied AND either $p \leq 15$ OR $p \geq 25$. Now we need the intersection of $p \leq 20$ with the condition "$p \leq 15$ OR $p \geq 25$." For $p \leq 15$: These prices satisfy both $p \leq 20$ and the price constraint, so they work. For $p \geq 25$: These prices satisfy the price constraint but violate $p \leq 20$ (since you can't have $p \geq 25$ and $p \leq 20$ simultaneously), so they don't work. Therefore, only $p \leq 15$ results in satisfied customers. Choice A ($5 \leq p \leq 15$) is too restrictive—prices below $5 also work. Choice B ($$p \leq 20$$) ignores the additional price constraint entirely. Choice C ($$p \leq 15$$ or $$p \geq 25$$) fails to account for the satisfaction threshold—while $p \geq 25$ satisfies the price constraint, it makes $S < 7$. 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.
4. A storage tank's pressure $P$ (in psi) must satisfy safety protocols. Normal operation requires $12 \leq P \leq 28$, but during system startup, an additional constraint $|P - 20| \leq 6$ must also be met. What pressures are acceptable during startup?
   Answer: $14 \leq P \leq 26$, representing the intersection of normal and startup constraints Normal operation: $12 \leq P \leq 28$. Startup constraint: $|P - 20| \leq 6$ means $14 \leq P \leq 26$. During startup, both constraints apply (AND), so we need the intersection: $\max(12,14) \leq P \leq \min(28,26)$, giving $14 \leq P \leq 26$. Choices B, C, and D each ignore one part of the intersection calculation.
5. 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 $m$ each day, which compound inequality represents the valid range for daily miles driven?
   Answer: $50 < 70 + 0.30m ≤ 80$, which gives $-66.67 < m ≤ 33.33$, so the valid range is $0 < m ≤ 33.33$ For a 2-day rental driving m miles each day: Cost = 2($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), 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.
6. A quality control system rejects products when the measurement $m$ satisfies $|2m - 8| \geq 6$ AND $m \neq 1$. Which values of $m$ will cause product rejection?
   Answer: $m < 1$ or $m \geq 7$, since the exclusion condition modifies the lower bound $|2m - 8| \geq 6$ means $2m - 8 \leq -6$ OR $2m - 8 \geq 6$, so $m \leq 1$ OR $m \geq 7$. With the additional condition $m \neq 1$, rejection occurs when $m < 1$ OR $m \geq 7$. Choice A doesn't account for $m \neq 1$. Choice C has incorrect inequality directions. Choice D describes acceptance, not rejection.
7. An optimization algorithm converges when the error $e$ satisfies $|3e - 6| < 12$ OR $e^2 \leq 9$. For numerical stability, the algorithm also requires $e \neq 0$. Which error values allow convergence?
   Answer: $-3 \leq e \leq 6$ and $e \neq 0$, combining both conditions with the stability constraint First condition: $|3e - 6| < 12$ means $-12 < 3e - 6 < 12$, so $-6 < 3e < 18$, giving $-2 < e < 6$. Second condition: $e^2 \leq 9$ means $-3 \leq e \leq 3$. The convergence condition is ($-2 < e < 6$) OR ($-3 \leq e \leq 3$), which equals $-3 \leq e < 6$. With $e \neq 0$, the answer is $-3 \leq e \leq 6$ and $e \neq 0$. Choice A has wrong bounds. Choice B uses AND instead of OR. Choice C ignores the first condition's contribution.
8. A satellite's orbit is stable when its altitude $h$ (in km) satisfies $2h - 300 \leq h + 100$ AND $3h + 50 > 2h + 200$. Additionally, for communication purposes, either $h \leq 500$ OR $h \geq 800$ must hold. Which altitudes satisfy all requirements?
   Answer: $150 < h \leq 400$, since this represents the stable range within communication limits 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 $2h - 300 \leq h + 100$, subtract $h$ from both sides: $h - 300 \leq 100$, so $h \leq 400$. For $3h + 50 > 2h + 200$, subtract $2h$ from both sides: $h + 50 > 200$, so $h > 150$. The orbital stability range is $150 < h \leq 400$. Next, the communication constraint requires either $h \leq 500$ OR $h \geq 800$. 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 $150 < h \leq 400$ falls entirely within the low communication range $h \leq 500$. For the high communication range $h \geq 800$, there's no overlap with the stability range since 800 > 400. Therefore, only $150 < h \leq 400$ satisfies all requirements. Answer A incorrectly includes $h \geq 800$, which fails the orbital stability constraints. Answer B claims only $h \geq 800$ works, ignoring that high altitudes violate stability requirements. Answer D gives $200 \leq h \leq 600$, which violates the communication constraint for $500 < h < 600$ 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.
9. The temperature $T$ (in °F) inside a greenhouse must satisfy $65 ≤ T ≤ 80$ during the day and $T < 65$ OR $T > 80$ 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?
   Answer: The alarm would be triggered at exactly 2 of the reading times. 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.
10. 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 $m$ the customer can travel?
    Answer: $25 < 8 + 2.5m \leq 40$, which simplifies to $6.8 < m \leq 12.8$ 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.
11. A temperature monitoring system triggers an alarm when the recorded temperature $T$ satisfies $|T - 72| > 8$ OR $T < 55$. Which of the following temperatures would NOT trigger the alarm?
    Answer: $T = 68$ degrees, since it falls within normal operating range The alarm triggers when $T < 64$ OR $T > 80$ OR $T < 55$, which simplifies to $T < 64$ OR $T > 80$. Only $T = 68$ falls in the safe range $[64, 80]$. Choice B: $52 < 64$ triggers alarm. Choice C: $82 > 80$ triggers alarm. Choice D: $45 < 64$ triggers alarm.
12. A scientist studying bacterial growth finds that the population $P$ (in thousands) after $h$ hours follows the model $P = 2 + 3h$. 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?
    Answer: $2 ≤ h ≤ 6$ hours 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 $8 ≤ P ≤ 20$. Substituting the model $P = 2 + 3h$: $8 ≤ 2 + 3h ≤ 20$. Solving the left inequality: $8 ≤ 2 + 3h$ gives $6 ≤ 3h$, so $h ≥ 2$. Solving the right inequality: $2 + 3h ≤ 20$ gives $3h ≤ 18$, so $h ≤ 6$. Additionally, evacuation occurs when $P > 25$, meaning $2 + 3h > 25$, which gives $h > 7.67$. Since we want to avoid evacuation, we need $h ≤ 7.67$. Combining all constraints: $2 ≤ h ≤ 6$ AND $h ≤ 7.67$. The most restrictive condition gives us $2 ≤ h ≤ 6$. Choice A ($8 ≤ h ≤ 20$) uses the population values instead of solving for time. Choice B ($2 ≤ h < 7.67$) correctly finds the lower bound but uses the evacuation threshold as the upper bound, ignoring that the experiment stops being successful at $h = 6$. Choice D ($6 ≤ h < 25$) 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.
13. A water reservoir's capacity is being monitored. The water level $h$ (in feet) must satisfy $15 \leq h \leq 25$ for normal operations, but during maintenance periods, the additional constraint $h - 3 \leq 2(h - 8)$ must also be met. What values of $h$ satisfy both the normal operating range AND the maintenance constraint?
    Answer: $15 \leq h \leq 25$, since maintenance doesn't further restrict the range Solving $h - 3 \leq 2(h - 8)$: $h - 3 \leq 2h - 16$, so $13 \leq h$. The compound inequality is $h \geq 13$ AND $15 \leq h \leq 25$, which gives $15 \leq h \leq 25$. Choice B is impossible. Choice C ignores the normal operating minimum. Choice D ignores the normal operating maximum.
14. A manufacturing company's daily profit $P$ (in thousands of dollars) is modeled by $P = -2x^2 + 16x - 24$, where $x$ is the number of units produced (in hundreds). For the company to remain profitable while meeting production constraints, they need $P \geq 6$ AND $x \leq 5$. Which compound inequality represents the valid production levels?
    Answer: $3 \leq x \leq 5$ First solve $-2x^2 + 16x - 24 \geq 6$: $-2x^2 + 16x - 30 \geq 0$, or $x^2 - 8x + 15 \leq 0$. Factoring: $(x-3)(x-5) \leq 0$, so $3 \leq x \leq 5$. Combined with the constraint $x \leq 5$, the answer is $3 \leq x \leq 5$. Choice A uses the wrong quadratic roots. Choice B incorrectly solves the profit inequality. Choice D ignores the production constraint.
15. A water tank's depth $d$ (in feet) varies according to the equation $d = 12 - 0.5t$, where $t$ is time in hours after 6 AM. The tank's pump automatically turns on when $d ≤ 4$ feet, and an emergency alarm sounds when $d ≤ 2$ feet. During which time interval will the pump be running but the emergency alarm will NOT be sounding?
    Answer: $16 ≤ t < 20$ hours after 6 AM 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).
16. A physics experiment requires the velocity $v$ to satisfy either $3v + 7 > 2v + 15$ OR $5 - v \geq 2v - 4$. Additionally, for safety reasons, $v$ must be positive. Which compound inequality represents all acceptable velocities?
    Answer: $0 < v \leq 3$ or $v > 8$, covering both experimental conditions with safety constraint First inequality: $3v + 7 > 2v + 15$ gives $v > 8$. Second inequality: $5 - v \geq 2v - 4$ gives $v \leq 3$. The experimental condition is $v > 8$ OR $v \leq 3$. With $v > 0$, this becomes $0 < v \leq 3$ OR $v > 8$. Choice B misses the $v > 8$ case. Choice C ignores the safety constraint. Choice D takes the wrong complement.
17. A manufacturing company produces widgets with weights that must satisfy the compound inequality $12.5 \leq w < 18.2$ 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?
    Answer: The batch fails quality control since only 40% of widgets meet the weight requirements. The compound inequality $12.5 \leq w < 18.2$ 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.

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