Unit 2: Equations and Inequalities

Answer: D — This question tests the student's ability to translate a real-world constraint into an absolute value inequality and then solve it correctly. The phrase 'differs from 50 by no more than 2' means the distance between x and 50 should not exceed 2, which translates to |x - 50| ≤ 2. Solving this compound inequality: -2 ≤ x - 50 ≤ 2, adding 50 to all parts yields 48 ≤ x ≤ 52. Option A correctly identifies both the absolute value inequality and its solution set. Option B uses the wrong inequality symbol (≥ instead of ≤), which would represent measurements that differ by AT LEAST 2 mm—the opposite of what's required. Option C incorrectly places the 2 and 50 in the absolute value expression, resulting in a meaningless constraint in this context. Option D only represents the upper bound and fails to account for measurements below 50 mm that would also be acceptable (like 49 mm or 48.5 mm). The question requires students to both translate language into mathematical notation and solve a compound inequality, demonstrating conceptual understanding rather than routine calculation.

Answer: C — To solve 2x + 7 = x + 12, we need to collect all variable terms on one side and constants on the other. Subtracting x from both sides gives us: 2x - x + 7 = 12, which simplifies to x + 7 = 12. Then subtracting 7 from both sides yields x = 5. We can verify: 2(5) + 7 = 10 + 7 = 17, and 5 + 12 = 17 ✓. Option B (19) results from incorrectly adding 7 and 12 without properly isolating x. Option C (-5) comes from a sign error when moving terms. Option D (9.5) results from dividing instead of subtracting when combining like terms.

Answer: A — The correct answer is D because Tom starts with $120 and adds $5 each week. The inequality 5x + 120 >= 180 represents the situation where the total amount of money Tom has after x weeks is greater than or equal to the cost of the bike. Options A, B, and C are incorrect because they do not accurately represent Tom's situation. Option A is close, but it does not account for the fact that Tom can buy the bike when he has exactly $180, so it should be >=, not >.

Answer: B — This question requires students to translate a real-world constraint into an absolute value inequality and understand what the absolute value expression represents. The temperature must be between 15°C and 25°C (inclusive), which means 15 ≤ T ≤ 25. This is a compound inequality centered at 20 with a distance of 5 in either direction. The absolute value expression |T - 20| represents the distance from T to 20. For the compound inequality 15 ≤ T ≤ 25 to hold, this distance must be at most 5, giving us |T - 20| ≤ 5. Students can verify: if T = 15, then |15 - 20| = 5 ✓; if T = 25, then |25 - 20| = 5 ✓; if T = 20, then |20 - 20| = 0 ✓. Option B (|T - 20| ≥ 5) represents temperatures outside the acceptable range. Option C (|T - 20| < 5) excludes the boundary temperatures of 15°C and 25°C, which the problem states should be included ('inclusive'). Option D (|T - 20| ≤ 20) is too broad and would include temperatures down to 0°C and up to 40°C, far exceeding the acceptable range.

Answer: D — This question requires students to translate a real-world constraint into absolute value inequality notation and then determine the solution interval's width. The nominal value is 1000 ohms, and the specification allows for 5% tolerance. Five percent of 1000 is 0.05 × 1000 = 50 ohms. Therefore, the actual resistance x must satisfy 950 ≤ x ≤ 1050, which is equivalently expressed as |x - 1000| ≤ 50. The width of the interval [950, 1050] is 1050 - 950 = 100 ohms. Option A correctly identifies both the inequality and the width. Option B correctly identifies the inequality but incorrectly calculates the width as 50 instead of 100—this is a common error where students subtract the tolerance from the width rather than recognizing that tolerance extends in both directions. Option C misconstrues the center of the absolute value expression and dramatically overestimates the interval. Option D misplaces the center of the interval at 950 instead of 1000, which would only account for the lower bound. This question tests whether students can convert percentage constraints into mathematical form and understand that absolute value inequalities define symmetric intervals around a central value.