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Difficulty: HardAbsolute Value Equations and Inequalities

A manufacturer of precision components produces cylindrical rods. The target diameter of the rods is 1.20 centimeters1.20\text{ centimeters}. A rod is classified as Grade A if its actual diameter, dd centimeters, satisfies the inequality ∣3−2.5d∣≤0.15|3 - 2.5d| \leq 0.15. To be used in a specific high-stress assembly, the rod's diameter must also satisfy the tolerance inequality ∣d−1.22∣<0.04|d - 1.22| < 0.04. Which of the following inequality expressions represents the complete set of all possible diameters, in centimeters, of rods that qualify as Grade A and are suitable for the assembly?

  1. A
    1.14≤d≤1.261.14 \le d \le 1.26
  2. B
    1.14≤d<1.261.14 \le d < 1.26
  3. 1.18<d<1.261.18 < d < 1.26Answer
  4. D
    1.18<d≤1.261.18 < d \le 1.26
  5. E
    1.14<d<1.181.14 < d < 1.18

Answer

The set of diameters satisfying both conditions is 1.18<d<1.261.18 < d < 1.26.
To satisfy both conditions, a rod's diameter must meet the Grade A requirement of 1.14≤d≤1.261.14 \le d \le 1.26 and the assembly requirement of 1.18<d<1.261.18 < d < 1.26. The intersection of these two intervals is the more restrictive range, which is 1.18<d<1.261.18 < d < 1.26. This ensures both inequalities are simultaneously true.

Step-by-Step Solution

1
Solve the first inequality representing Grade A rods: ∣3−2.5d∣≤0.15|3 - 2.5d| \le 0.15.
1.14≤d≤1.261.14 \le d \le 1.26
Rewrite the absolute value inequality as a compound inequality: −0.15≤3−2.5d≤0.15-0.15 \le 3 - 2.5d \le 0.15. Subtract 33 from all parts to get −3.15≤−2.5d≤−2.85-3.15 \le -2.5d \le -2.85. Divide all parts by −2.5-2.5, reversing the direction of the inequality signs: 1.26≥d≥1.141.26 \ge d \ge 1.14, which simplifies to 1.14≤d≤1.261.14 \le d \le 1.26.
2
Solve the second inequality representing suitability for the assembly: ∣d−1.22∣<0.04|d - 1.22| < 0.04.
1.18<d<1.261.18 < d < 1.26
Rewrite the absolute value inequality as a compound inequality: −0.04<d−1.22<0.04-0.04 < d - 1.22 < 0.04. Add 1.221.22 to all parts to isolate dd: 1.18<d<1.261.18 < d < 1.26.
3
Find the intersection of the two solution sets: [1.14,1.26]∩(1.18,1.26)[1.14, 1.26] \cap (1.18, 1.26).
1.18<d<1.261.18 < d < 1.26
For a rod to qualify for Grade A and be suitable for the assembly, its diameter must satisfy both conditions. The overlapping range is bounded below by the stricter lower bound of 1.181.18 (exclusive) and above by the stricter upper bound of 1.261.26 (exclusive).

Key Concept

Absolute Value Equations and Inequalities
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Topics

Absolute Value Equations and Inequalities

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