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College Algebra Quiz

College Algebra Quiz: Absolute Value Inequalities

Practice Absolute Value Inequalities in College Algebra with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

Question 1 / 20

0 of 20 answered

A conference budget centers at 15,00015{,}00015,000 dollars; it must satisfy ∣x−15000∣<1200|x-15000|<1200∣x−15000∣<1200. What is the solution set?

Select an answer to continue

What this quiz covers

This quiz focuses on Absolute Value Inequalities, giving you a quick way to practice the rules, question types, and explanations that matter most for College 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 conference budget centers at 15,00015{,}00015,000 dollars; it must satisfy ∣x−15000∣<1200|x-15000|<1200∣x−15000∣<1200. What is the solution set?

  1. x∈(13800,16200)x\in(13800,16200)x∈(13800,16200) (correct answer)
  2. x∈[13800,16200]x\in[13800,16200]x∈[13800,16200]
  3. x∈(−∞,13800]∪[16200,∞)x\in(-\infty,13800]\cup[16200,\infty)x∈(−∞,13800]∪[16200,∞)
  4. x∈(−∞,13800)∪(16200,∞)x\in(-\infty,13800)\cup(16200,\infty)x∈(−∞,13800)∪(16200,∞)

Explanation: This question tests understanding of absolute value inequalities and their solution sets. Absolute value inequalities involve expressions where the distance from zero is restricted by an inequality (e.g., |x| < a). In this specific problem, the inequality |x - 15000| < 1200 represents a range of acceptable budgets around 15000 dollars, bounded by a distance of 1200 dollars. The correct answer choice correctly identifies the solution set as the open interval (13800,16200), reflecting the real-world constraints described. A common distractor fails by misrepresenting the direction of the inequality, often due to misunderstanding the concept of distance in absolute terms. To help students: Emphasize the interpretation of absolute value as distance, practice graphing solutions on a number line, and reinforce the importance of checking solutions against the original inequality.

Question 2

A delivery zone is centered at x=40x=40x=40 on a straight highway map; service is offered when ∣x−40∣≤9|x-40|\le 9∣x−40∣≤9. What is the range?

  1. x∈[31,49]x\in[31,49]x∈[31,49] (correct answer)
  2. x∈(31,49)x\in(31,49)x∈(31,49)
  3. x∈(−∞,31)∪(49,∞)x\in(-\infty,31)\cup(49,\infty)x∈(−∞,31)∪(49,∞)
  4. x∈(−∞,31]∪[49,∞)x\in(-\infty,31]\cup[49,\infty)x∈(−∞,31]∪[49,∞)

Explanation: This question tests understanding of absolute value inequalities and their solution sets. Absolute value inequalities involve expressions where the distance from zero is restricted by an inequality (e.g., |x| < a). In this specific problem, the inequality |x - 40| ≤ 9 represents a range of acceptable positions around 40, bounded by a distance of 9 units. The correct answer choice correctly identifies the solution set as the closed interval [31,49], reflecting the real-world constraints described. A common distractor fails by misrepresenting the direction of the inequality, often due to misunderstanding the concept of distance in absolute terms. To help students: Emphasize the interpretation of absolute value as distance, practice graphing solutions on a number line, and reinforce the importance of checking solutions against the original inequality.

Question 3

A machined gear targets x=32x=32x=32 mm; tolerance is ∣x−32∣≤0.4|x-32|\le 0.4∣x−32∣≤0.4 to ensure fit. What is the solution set?

  1. x∈(31.6,32.4)x\in(31.6,32.4)x∈(31.6,32.4)
  2. x∈[31.6,32.4]x\in[31.6,32.4]x∈[31.6,32.4] (correct answer)
  3. x∈(−∞,31.6)∪(32.4,∞)x\in(-\infty,31.6)\cup(32.4,\infty)x∈(−∞,31.6)∪(32.4,∞)
  4. x∈(−∞,31.6]∪[32.4,∞)x\in(-\infty,31.6]\cup[32.4,\infty)x∈(−∞,31.6]∪[32.4,∞)

Explanation: This question tests understanding of absolute value inequalities and their solution sets. Absolute value inequalities involve expressions where the distance from zero is restricted by an inequality (e.g., |x| < a). In this specific problem, the inequality |x - 32| ≤ 0.4 represents a range of acceptable gear sizes around 32 mm, bounded by a distance of 0.4 mm. The correct answer choice correctly identifies the solution set as the closed interval [31.6,32.4], reflecting the real-world constraints described. A common distractor fails by misrepresenting the direction of the inequality, often due to misunderstanding the concept of distance in absolute terms. To help students: Emphasize the interpretation of absolute value as distance, practice graphing solutions on a number line, and reinforce the importance of checking solutions against the original inequality.

Question 4

A thermostat targets 22∘C22^\circ\text{C}22∘C; to prevent discomfort it must satisfy ∣T−22∣≤2|T-22|\le 2∣T−22∣≤2. What temperature range is acceptable?

  1. T∈(20,24)T\in(20,24)T∈(20,24)
  2. T∈[20,24]T\in[20,24]T∈[20,24] (correct answer)
  3. T∈(−∞,20]∪[24,∞)T\in(-\infty,20]\cup[24,\infty)T∈(−∞,20]∪[24,∞)
  4. T∈(−∞,20)∪(24,∞)T\in(-\infty,20)\cup(24,\infty)T∈(−∞,20)∪(24,∞)

Explanation: This question tests understanding of absolute value inequalities and their solution sets. Absolute value inequalities involve expressions where the distance from zero is restricted by an inequality (e.g., |x| < a). In this specific problem, the inequality |T - 22| ≤ 2 represents a range of acceptable temperatures around 22°C, bounded by a distance of 2°C. The correct answer choice correctly identifies the solution set as the closed interval [20,24], reflecting the real-world constraints described. A common distractor fails by misrepresenting the direction of the inequality, often due to misunderstanding the concept of distance in absolute terms. To help students: Emphasize the interpretation of absolute value as distance, practice graphing solutions on a number line, and reinforce the importance of checking solutions against the original inequality.

Question 5

A project estimates 800080008000 dollars; approval requires ∣x−8000∣≤600|x-8000|\le 600∣x−8000∣≤600. What is the permitted cost range?

  1. x∈[7400,8600]x\in[7400,8600]x∈[7400,8600] (correct answer)
  2. x∈(7400,8600)x\in(7400,8600)x∈(7400,8600)
  3. x∈(−∞,7400)∪(8600,∞)x\in(-\infty,7400)\cup(8600,\infty)x∈(−∞,7400)∪(8600,∞)
  4. x∈(−∞,7400]∪[8600,∞)x\in(-\infty,7400]\cup[8600,\infty)x∈(−∞,7400]∪[8600,∞)

Explanation: This question tests understanding of absolute value inequalities and their solution sets. Absolute value inequalities involve expressions where the distance from zero is restricted by an inequality (e.g., |x| < a). In this specific problem, the inequality |x - 8000| ≤ 600 represents a range of acceptable costs around 8000 dollars, bounded by a distance of 600 dollars. The correct answer choice correctly identifies the solution set as the closed interval [7400,8600], reflecting the real-world constraints described. A common distractor fails by misrepresenting the direction of the inequality, often due to misunderstanding the concept of distance in absolute terms. To help students: Emphasize the interpretation of absolute value as distance, practice graphing solutions on a number line, and reinforce the importance of checking solutions against the original inequality.

Question 6

A mobile clinic is centered at x=5x=5x=5 km and serves locations satisfying ∣x−5∣≤1.2|x-5|\le 1.2∣x−5∣≤1.2. Which interval is served?

  1. x∈[3.8,6.2]x\in[3.8,6.2]x∈[3.8,6.2] (correct answer)
  2. x∈(3.8,6.2)x\in(3.8,6.2)x∈(3.8,6.2)
  3. x∈(−∞,3.8)∪(6.2,∞)x\in(-\infty,3.8)\cup(6.2,\infty)x∈(−∞,3.8)∪(6.2,∞)
  4. x∈(−∞,3.8]∪[6.2,∞)x\in(-\infty,3.8]\cup[6.2,\infty)x∈(−∞,3.8]∪[6.2,∞)

Explanation: This question tests understanding of absolute value inequalities and their solution sets. Absolute value inequalities involve expressions where the distance from zero is restricted by an inequality (e.g., |x| < a). In this specific problem, the inequality |x - 5| ≤ 1.2 represents a range of acceptable locations around 5 km, bounded by a distance of 1.2 km. The correct answer choice correctly identifies the solution set as the closed interval [3.8,6.2], reflecting the real-world constraints described. A common distractor fails by misrepresenting the direction of the inequality, often due to misunderstanding the concept of distance in absolute terms. To help students: Emphasize the interpretation of absolute value as distance, practice graphing solutions on a number line, and reinforce the importance of checking solutions against the original inequality.

Question 7

A rescue boat must stay at least 444 nautical miles from reef at x=18x=18x=18; rule is ∣x−18∣≥4|x-18|\ge 4∣x−18∣≥4. What is the solution set?

  1. x∈[14,22]x\in[14,22]x∈[14,22]
  2. x∈(14,22)x\in(14,22)x∈(14,22)
  3. x∈(−∞,14)∪(22,∞)x\in(-\infty,14)\cup(22,\infty)x∈(−∞,14)∪(22,∞)
  4. x∈(−∞,14]∪[22,∞)x\in(-\infty,14]\cup[22,\infty)x∈(−∞,14]∪[22,∞) (correct answer)

Explanation: This question tests understanding of absolute value inequalities and their solution sets. Absolute value inequalities involve expressions where the distance from zero is restricted by an inequality (e.g., |x| < a). In this specific problem, the inequality |x - 18| ≥ 4 represents positions outside a range around 18, bounded by a distance of 4 nautical miles. The correct answer choice correctly identifies the solution set as the union of intervals (-∞,14] ∪ [22,∞), reflecting the real-world constraints described. A common distractor fails by misrepresenting the direction of the inequality, often due to misunderstanding the concept of distance in absolute terms. To help students: Emphasize the interpretation of absolute value as distance, practice graphing solutions on a number line, and reinforce the importance of checking solutions against the original inequality.

Question 8

The solution set of ∣3x+2∣−∣x−4∣>2x+6|3x + 2| - |x - 4| > 2x + 6∣3x+2∣−∣x−4∣>2x+6 includes which of the following intervals?

  1. (−∞,−4)(-\infty, -4)(−∞,−4) is entirely contained in the solution set (correct answer)
  2. (−1,3)(-1, 3)(−1,3) is entirely contained in the solution set
  3. (−∞,−4)(-\infty, -4)(−∞,−4) contains points both in and out of the solution set
  4. (5,∞)(5, \infty)(5,∞) is entirely contained in the solution set

Explanation: We solve ∣3x+2∣−∣x−4∣>2x+6|3x + 2| - |x - 4| > 2x + 6∣3x+2∣−∣x−4∣>2x+6 by considering cases based on the critical points x=−23x = -\frac{2}{3}x=−32​ (where 3x+2=03x + 2 = 03x+2=0) and x=4x = 4x=4 (where x−4=0x - 4 = 0x−4=0). Case 1: x<−23x < -\frac{2}{3}x<−32​. Then ∣3x+2∣=−(3x+2)=−3x−2|3x + 2| = -(3x + 2) = -3x - 2∣3x+2∣=−(3x+2)=−3x−2 and ∣x−4∣=−(x−4)=4−x|x - 4| = -(x - 4) = 4 - x∣x−4∣=−(x−4)=4−x. The inequality becomes (−3x−2)−(4−x)>2x+6(-3x - 2) - (4 - x) > 2x + 6(−3x−2)−(4−x)>2x+6, which simplifies to −3x−2−4+x>2x+6-3x - 2 - 4 + x > 2x + 6−3x−2−4+x>2x+6, giving −2x−6>2x+6-2x - 6 > 2x + 6−2x−6>2x+6. This leads to −4x>12-4x > 12−4x>12, so x<−3x < -3x<−3. Since we need x<−23≈−0.67x < -\frac{2}{3} \approx -0.67x<−32​≈−0.67, the solution in this case is x<−3x < -3x<−3. Case 2: −23≤x<4-\frac{2}{3} \leq x < 4−32​≤x<4. Then ∣3x+2∣=3x+2|3x + 2| = 3x + 2∣3x+2∣=3x+2 and ∣x−4∣=4−x|x - 4| = 4 - x∣x−4∣=4−x. The inequality becomes (3x+2)−(4−x)>2x+6(3x + 2) - (4 - x) > 2x + 6(3x+2)−(4−x)>2x+6, which simplifies to 3x+2−4+x>2x+63x + 2 - 4 + x > 2x + 63x+2−4+x>2x+6, giving 4x−2>2x+64x - 2 > 2x + 64x−2>2x+6. This leads to 2x>82x > 82x>8, so x>4x > 4x>4. But this contradicts our case assumption x<4x < 4x<4, so there are no solutions in this interval. Case 3: x≥4x \geq 4x≥4. Then ∣3x+2∣=3x+2|3x + 2| = 3x + 2∣3x+2∣=3x+2 and ∣x−4∣=x−4|x - 4| = x - 4∣x−4∣=x−4. The inequality becomes (3x+2)−(x−4)>2x+6(3x + 2) - (x - 4) > 2x + 6(3x+2)−(x−4)>2x+6, which simplifies to 3x+2−x+4>2x+63x + 2 - x + 4 > 2x + 63x+2−x+4>2x+6, giving 2x+6>2x+62x + 6 > 2x + 62x+6>2x+6. This simplifies to 0>00 > 00>0, which is false, so there are no solutions for x≥4x \geq 4x≥4. Therefore, the solution set is (−∞,−3)(-\infty, -3)(−∞,−3), which means (−∞,−4)(-\infty, -4)(−∞,−4) is entirely contained in the solution set.

Question 9

Consider the compound inequality ∣2x−3∣≥5|2x - 3| \geq 5∣2x−3∣≥5 AND ∣x+1∣<4|x + 1| < 4∣x+1∣<4. The solution set contains exactly how many integers?

  1. 222 integers
  2. 333 integers
  3. 444 integers (correct answer)
  4. 555 integers

Explanation: We solve each inequality separately, then find their intersection. For ∣2x−3∣≥5|2x - 3| \geq 5∣2x−3∣≥5: This means 2x−3≥52x - 3 \geq 52x−3≥5 or 2x−3≤−52x - 3 \leq -52x−3≤−5. From the first: 2x≥82x \geq 82x≥8, so x≥4x \geq 4x≥4. From the second: 2x≤−22x \leq -22x≤−2, so x≤−1x \leq -1x≤−1. Therefore, ∣2x−3∣≥5|2x - 3| \geq 5∣2x−3∣≥5 when x∈(−∞,−1]∪[4,∞)x \in (-\infty, -1] \cup [4, \infty)x∈(−∞,−1]∪[4,∞). For ∣x+1∣<4|x + 1| < 4∣x+1∣<4: This means −4<x+1<4-4 < x + 1 < 4−4<x+1<4, so −5<x<3-5 < x < 3−5<x<3. Therefore, ∣x+1∣<4|x + 1| < 4∣x+1∣<4 when x∈(−5,3)x \in (-5, 3)x∈(−5,3). The intersection is: ((−∞,−1]∪[4,∞))∩(−5,3)=(−5,−1]∪∅=(−5,−1]((-\infty, -1] \cup [4, \infty)) \cap (-5, 3) = (-5, -1] \cup \emptyset = (-5, -1]((−∞,−1]∪[4,∞))∩(−5,3)=(−5,−1]∪∅=(−5,−1]. The integers in the interval (−5,−1](-5, -1](−5,−1] are −4,−3,−2,−1-4, -3, -2, -1−4,−3,−2,−1. Let's verify: For x=−4x = -4x=−4: ∣2(−4)−3∣=∣−11∣=11≥5|2(-4) - 3| = |-11| = 11 \geq 5∣2(−4)−3∣=∣−11∣=11≥5 ✓ and ∣−4+1∣=3<4|-4 + 1| = 3 < 4∣−4+1∣=3<4 ✓. For x=−3x = -3x=−3: ∣2(−3)−3∣=∣−9∣=9≥5|2(-3) - 3| = |-9| = 9 \geq 5∣2(−3)−3∣=∣−9∣=9≥5 ✓ and ∣−3+1∣=2<4|-3 + 1| = 2 < 4∣−3+1∣=2<4 ✓. For x=−2x = -2x=−2: ∣2(−2)−3∣=∣−7∣=7≥5|2(-2) - 3| = |-7| = 7 \geq 5∣2(−2)−3∣=∣−7∣=7≥5 ✓ and ∣−2+1∣=1<4|-2 + 1| = 1 < 4∣−2+1∣=1<4 ✓. For x=−1x = -1x=−1: ∣2(−1)−3∣=∣−5∣=5≥5|2(-1) - 3| = |-5| = 5 \geq 5∣2(−1)−3∣=∣−5∣=5≥5 ✓ and ∣−1+1∣=0<4|-1 + 1| = 0 < 4∣−1+1∣=0<4 ✓. For x=0x = 0x=0: ∣2(0)−3∣=3≱5|2(0) - 3| = 3 \not\geq 5∣2(0)−3∣=3≥5 ✗. Therefore, there are exactly 4 integers in the solution set.

Question 10

A circuit board thickness targets 1.61.61.6 mm; specification is ∣x−1.6∣≤0.1|x-1.6|\le 0.1∣x−1.6∣≤0.1. What values satisfy this?

  1. x∈[1.5,1.7]x\in[1.5,1.7]x∈[1.5,1.7] (correct answer)
  2. x∈(1.5,1.7)x\in(1.5,1.7)x∈(1.5,1.7)
  3. x∈(−∞,1.5)∪(1.7,∞)x\in(-\infty,1.5)\cup(1.7,\infty)x∈(−∞,1.5)∪(1.7,∞)
  4. x∈(−∞,1.5]∪[1.7,∞)x\in(-\infty,1.5]\cup[1.7,\infty)x∈(−∞,1.5]∪[1.7,∞)

Explanation: This question tests understanding of absolute value inequalities and their solution sets. Absolute value inequalities involve expressions where the distance from zero is restricted by an inequality (e.g., |x| < a). In this specific problem, the inequality |x - 1.6| ≤ 0.1 represents a range of acceptable thicknesses around 1.6 mm, bounded by a distance of 0.1 mm. The correct answer choice correctly identifies the solution set as the closed interval [1.5,1.7], reflecting the real-world constraints described. A common distractor fails by misrepresenting the direction of the inequality, often due to misunderstanding the concept of distance in absolute terms. To help students: Emphasize the interpretation of absolute value as distance, practice graphing solutions on a number line, and reinforce the importance of checking solutions against the original inequality.

Question 11

A travel budget centers at 950950950 dollars; it stays on-plan if ∣x−950∣<75|x-950|<75∣x−950∣<75. Which interval represents acceptable xxx?

  1. x∈[875,1025]x\in[875,1025]x∈[875,1025]
  2. x∈(875,1025)x\in(875,1025)x∈(875,1025) (correct answer)
  3. x∈(−∞,875]∪[1025,∞)x\in(-\infty,875]\cup[1025,\infty)x∈(−∞,875]∪[1025,∞)
  4. x∈(−∞,875)∪(1025,∞)x\in(-\infty,875)\cup(1025,\infty)x∈(−∞,875)∪(1025,∞)

Explanation: This question tests understanding of absolute value inequalities and their solution sets. Absolute value inequalities involve expressions where the distance from zero is restricted by an inequality (e.g., |x| < a). In this specific problem, the inequality |x - 950| < 75 represents a range of acceptable budgets around 950 dollars, bounded by a distance of 75 dollars. The correct answer choice correctly identifies the solution set as the open interval (875,1025), reflecting the real-world constraints described. A common distractor fails by misrepresenting the direction of the inequality, often due to misunderstanding the concept of distance in absolute terms. To help students: Emphasize the interpretation of absolute value as distance, practice graphing solutions on a number line, and reinforce the importance of checking solutions against the original inequality.

Question 12

A factory machines rods with target length 505050 mm; quality requires ∣x−50∣≤0.2|x-50|\le 0.2∣x−50∣≤0.2. What is the acceptable range?

  1. x∈(49.8,50.2)x\in(49.8,50.2)x∈(49.8,50.2)
  2. x∈[49.8,50.2]x\in[49.8,50.2]x∈[49.8,50.2] (correct answer)
  3. x∈(−∞,49.8]∪[50.2,∞)x\in(-\infty,49.8]\cup[50.2,\infty)x∈(−∞,49.8]∪[50.2,∞)
  4. x∈(−∞,49.8)∪(50.2,∞)x\in(-\infty,49.8)\cup(50.2,\infty)x∈(−∞,49.8)∪(50.2,∞)

Explanation: This question tests understanding of absolute value inequalities and their solution sets. Absolute value inequalities involve expressions where the distance from zero is restricted by an inequality (e.g., |x| < a). In this specific problem, the inequality |x - 50| ≤ 0.2 represents a range of acceptable lengths around 50 mm, bounded by a distance of 0.2 mm. The correct answer choice correctly identifies the solution set as the closed interval [49.8,50.2], reflecting the real-world constraints described. A common distractor fails by misrepresenting the direction of the inequality, often due to misunderstanding the concept of distance in absolute terms. To help students: Emphasize the interpretation of absolute value as distance, practice graphing solutions on a number line, and reinforce the importance of checking solutions against the original inequality.

Question 13

A greenhouse targets 25∘C25^\circ\text{C}25∘C; plants thrive only if ∣T−25∣≤4|T-25|\le 4∣T−25∣≤4. What range of TTT works?

  1. T∈[21,29]T\in[21,29]T∈[21,29] (correct answer)
  2. T∈(21,29)T\in(21,29)T∈(21,29)
  3. T∈(−∞,21)∪(29,∞)T\in(-\infty,21)\cup(29,\infty)T∈(−∞,21)∪(29,∞)
  4. T∈(−∞,21]∪[29,∞)T\in(-\infty,21]\cup[29,\infty)T∈(−∞,21]∪[29,∞)

Explanation: This question tests understanding of absolute value inequalities and their solution sets. Absolute value inequalities involve expressions where the distance from zero is restricted by an inequality (e.g., |x| < a). In this specific problem, the inequality |T - 25| ≤ 4 represents a range of acceptable temperatures around 25°C, bounded by a distance of 4°C. The correct answer choice correctly identifies the solution set as the closed interval [21,29], reflecting the real-world constraints described. A common distractor fails by misrepresenting the direction of the inequality, often due to misunderstanding the concept of distance in absolute terms. To help students: Emphasize the interpretation of absolute value as distance, practice graphing solutions on a number line, and reinforce the importance of checking solutions against the original inequality.

Question 14

A courier operates near downtown at x=30x=30x=30 on a linear map; policy is ∣x−30∣<6|x-30|<6∣x−30∣<6. Which interval applies?

  1. x∈(24,36)x\in(24,36)x∈(24,36) (correct answer)
  2. x∈[24,36]x\in[24,36]x∈[24,36]
  3. x∈(−∞,24]∪[36,∞)x\in(-\infty,24]\cup[36,\infty)x∈(−∞,24]∪[36,∞)
  4. x∈(−∞,24)∪(36,∞)x\in(-\infty,24)\cup(36,\infty)x∈(−∞,24)∪(36,∞)

Explanation: This question tests understanding of absolute value inequalities and their solution sets. Absolute value inequalities involve expressions where the distance from zero is restricted by an inequality (e.g., |x| < a). In this specific problem, the inequality |x - 30| < 6 represents a range of acceptable positions around 30, bounded by a distance of 6 units. The correct answer choice correctly identifies the solution set as the open interval (24,36), reflecting the real-world constraints described. A common distractor fails by misrepresenting the direction of the inequality, often due to misunderstanding the concept of distance in absolute terms. To help students: Emphasize the interpretation of absolute value as distance, practice graphing solutions on a number line, and reinforce the importance of checking solutions against the original inequality.

Question 15

A club expects 450450450 dollars in dues; it remains solvent only if ∣x−450∣<40|x-450|<40∣x−450∣<40. Which interval fits xxx?

  1. x∈[410,490]x\in[410,490]x∈[410,490]
  2. x∈(410,490)x\in(410,490)x∈(410,490) (correct answer)
  3. x∈(−∞,410]∪[490,∞)x\in(-\infty,410]\cup[490,\infty)x∈(−∞,410]∪[490,∞)
  4. x∈(−∞,410)∪(490,∞)x\in(-\infty,410)\cup(490,\infty)x∈(−∞,410)∪(490,∞)

Explanation: This question tests understanding of absolute value inequalities and their solution sets. Absolute value inequalities involve expressions where the distance from zero is restricted by an inequality (e.g., |x| < a). In this specific problem, the inequality |x - 450| < 40 represents a range of acceptable dues around 450 dollars, bounded by a distance of 40 dollars. The correct answer choice correctly identifies the solution set as the open interval (410,490), reflecting the real-world constraints described. A common distractor fails by misrepresenting the direction of the inequality, often due to misunderstanding the concept of distance in absolute terms. To help students: Emphasize the interpretation of absolute value as distance, practice graphing solutions on a number line, and reinforce the importance of checking solutions against the original inequality.

Question 16

A monthly budget targets 120012001200 dollars; to stay flexible, ∣x−1200∣≤150|x-1200|\le 150∣x−1200∣≤150. What spending range is allowed?

  1. x∈[1050,1350]x\in[1050,1350]x∈[1050,1350] (correct answer)
  2. x∈(1050,1350)x\in(1050,1350)x∈(1050,1350)
  3. x∈(−∞,1050)∪(1350,∞)x\in(-\infty,1050)\cup(1350,\infty)x∈(−∞,1050)∪(1350,∞)
  4. x∈(−∞,1050]∪[1350,∞)x\in(-\infty,1050]\cup[1350,\infty)x∈(−∞,1050]∪[1350,∞)

Explanation: This question tests understanding of absolute value inequalities and their solution sets. Absolute value inequalities involve expressions where the distance from zero is restricted by an inequality (e.g., |x| < a). In this specific problem, the inequality |x - 1200| ≤ 150 represents a range of acceptable spending around 1200 dollars, bounded by a distance of 150 dollars. The correct answer choice correctly identifies the solution set as the closed interval [1050,1350], reflecting the real-world constraints described. A common distractor fails by misrepresenting the direction of the inequality, often due to misunderstanding the concept of distance in absolute terms. To help students: Emphasize the interpretation of absolute value as distance, practice graphing solutions on a number line, and reinforce the importance of checking solutions against the original inequality.

Question 17

A delivery hub is at mile marker 121212; drivers stay in-zone if ∣x−12∣≤5|x-12|\le 5∣x−12∣≤5. What mile markers qualify?

  1. x∈[7,17]x\in[7,17]x∈[7,17] (correct answer)
  2. x∈(7,17)x\in(7,17)x∈(7,17)
  3. x∈(−∞,7)∪(17,∞)x\in(-\infty,7)\cup(17,\infty)x∈(−∞,7)∪(17,∞)
  4. x∈(−∞,7]∪[17,∞)x\in(-\infty,7]\cup[17,\infty)x∈(−∞,7]∪[17,∞)

Explanation: This question tests understanding of absolute value inequalities and their solution sets. Absolute value inequalities involve expressions where the distance from zero is restricted by an inequality (e.g., |x| < a). In this specific problem, the inequality |x - 12| ≤ 5 represents a range of acceptable mile markers around 12, bounded by a distance of 5 miles. The correct answer choice correctly identifies the solution set as the closed interval [7,17], reflecting the real-world constraints described. A common distractor fails by misrepresenting the direction of the inequality, often due to misunderstanding the concept of distance in absolute terms. To help students: Emphasize the interpretation of absolute value as distance, practice graphing solutions on a number line, and reinforce the importance of checking solutions against the original inequality.

Question 18

If ∣x−a∣+∣x−b∣≤c|x - a| + |x - b| \leq c∣x−a∣+∣x−b∣≤c has no solution, where a<ba < ba<b and c>0c > 0c>0, then which condition must be true?

  1. c<b−ac < b - ac<b−a and a+b≠0a + b \neq 0a+b=0
  2. c<b−ac < b - ac<b−a regardless of the value of a+ba + ba+b (correct answer)
  3. c≤b−ac \leq b - ac≤b−a and a+b=0a + b = 0a+b=0
  4. c=0c = 0c=0 and a=ba = ba=b

Explanation: The function f(x)=∣x−a∣+∣x−b∣f(x) = |x - a| + |x - b|f(x)=∣x−a∣+∣x−b∣ represents the sum of distances from xxx to points aaa and bbb on the number line. The minimum value of this function occurs when xxx is between aaa and bbb (i.e., a≤x≤ba \leq x \leq ba≤x≤b), and this minimum value is b−ab - ab−a. Specifically, for any x∈[a,b]x \in [a, b]x∈[a,b], we have f(x)=∣x−a∣+∣x−b∣=(x−a)+(b−x)=b−af(x) = |x - a| + |x - b| = (x - a) + (b - x) = b - af(x)=∣x−a∣+∣x−b∣=(x−a)+(b−x)=b−a. For x<ax < ax<a, we have f(x)=(a−x)+(b−x)=a+b−2x>a+b−2a=b−af(x) = (a - x) + (b - x) = a + b - 2x > a + b - 2a = b - af(x)=(a−x)+(b−x)=a+b−2x>a+b−2a=b−a. For x>bx > bx>b, we have f(x)=(x−a)+(x−b)=2x−a−b>2b−a−b=b−af(x) = (x - a) + (x - b) = 2x - a - b > 2b - a - b = b - af(x)=(x−a)+(x−b)=2x−a−b>2b−a−b=b−a. Therefore, the minimum value of ∣x−a∣+∣x−b∣|x - a| + |x - b|∣x−a∣+∣x−b∣ is exactly b−ab - ab−a, achieved on the interval [a,b][a, b][a,b]. For the inequality ∣x−a∣+∣x−b∣≤c|x - a| + |x - b| \leq c∣x−a∣+∣x−b∣≤c to have no solution, we need ccc to be strictly less than the minimum possible value of ∣x−a∣+∣x−b∣|x - a| + |x - b|∣x−a∣+∣x−b∣. Since this minimum value is b−ab - ab−a, we need c<b−ac < b - ac<b−a. The value of a+ba + ba+b is irrelevant to this condition - what matters is only the distance b−ab - ab−a between the two points and how it compares to ccc. Therefore, the condition is c<b−ac < b - ac<b−a regardless of the value of a+ba + ba+b.

Question 19

A lens curvature measurement targets 121212 units; rework is needed when ∣x−12∣≥0.6|x-12|\ge 0.6∣x−12∣≥0.6. Which set needs rework?

  1. x∈(11.4,12.6)x\in(11.4,12.6)x∈(11.4,12.6)
  2. x∈[11.4,12.6]x\in[11.4,12.6]x∈[11.4,12.6]
  3. x∈(−∞,11.4)∪(12.6,∞)x\in(-\infty,11.4)\cup(12.6,\infty)x∈(−∞,11.4)∪(12.6,∞)
  4. x∈(−∞,11.4]∪[12.6,∞)x\in(-\infty,11.4]\cup[12.6,\infty)x∈(−∞,11.4]∪[12.6,∞) (correct answer)

Explanation: This question tests understanding of absolute value inequalities and their solution sets. Absolute value inequalities involve expressions where the distance from zero is restricted by an inequality (e.g., |x| < a). In this specific problem, the inequality |x - 12| ≥ 0.6 represents measurements outside a range around 12 units, bounded by a distance of 0.6 units, requiring rework. The correct answer choice correctly identifies the solution set as the union of intervals (-∞,11.4] ∪ [12.6,∞), reflecting the real-world constraints described. A common distractor fails by misrepresenting the direction of the inequality, often due to misunderstanding the concept of distance in absolute terms. To help students: Emphasize the interpretation of absolute value as distance, practice graphing solutions on a number line, and reinforce the importance of checking solutions against the original inequality.

Question 20

A drone must remain within 2.52.52.5 km of base at coordinate x=0x=0x=0; constraint is ∣x∣<2.5|x|<2.5∣x∣<2.5. What is the range?

  1. x∈[−2.5,2.5]x\in[-2.5,2.5]x∈[−2.5,2.5]
  2. x∈(−2.5,2.5)x\in(-2.5,2.5)x∈(−2.5,2.5) (correct answer)
  3. x∈(−∞,−2.5)∪(2.5,∞)x\in(-\infty,-2.5)\cup(2.5,\infty)x∈(−∞,−2.5)∪(2.5,∞)
  4. x∈(−∞,−2.5]∪[2.5,∞)x\in(-\infty,-2.5]\cup[2.5,\infty)x∈(−∞,−2.5]∪[2.5,∞)

Explanation: This question tests understanding of absolute value inequalities and their solution sets. Absolute value inequalities involve expressions where the distance from zero is restricted by an inequality (e.g., |x| < a). In this specific problem, the inequality |x - 0| < 2.5 represents a range of acceptable positions around 0 km, bounded by a distance of 2.5 km. The correct answer choice correctly identifies the solution set as the open interval (-2.5,2.5), reflecting the real-world constraints described. A common distractor fails by misrepresenting the direction of the inequality, often due to misunderstanding the concept of distance in absolute terms. To help students: Emphasize the interpretation of absolute value as distance, practice graphing solutions on a number line, and reinforce the importance of checking solutions against the original inequality.