A wire that is 4{,650} millimeters long is to be cut into as many 120-millimeter pieces as possible. Each cut...

1. TRANSLATE the problem information

• Given information:
  • Total wire length: 4,650 mm
  • Desired piece length: 120 mm each
  • Waste per cut: 15 mm
  • Find: Maximum number of pieces possible

2. INFER the cutting relationship

• Key insight: To get k pieces from one wire, you need exactly (k-1) cuts
• Think about it: 1 cut → 2 pieces, 2 cuts → 3 pieces, etc.
• Total wire consumed = (pieces × length) + (cuts × waste per cut)
• Total wire consumed = \(120\mathrm{k} + 15(\mathrm{k}-1)\)

3. TRANSLATE the constraint

• The wire consumed cannot exceed what we have:
  \(120\mathrm{k} + 15(\mathrm{k}-1) \leq 4,650\)

4. SIMPLIFY the inequality

• Expand: \(120\mathrm{k} + 15\mathrm{k} - 15 \leq 4,650\)
• Combine: \(135\mathrm{k} - 15 \leq 4,650\)
• Add 15: \(135\mathrm{k} \leq 4,665\)
• Divide by 135: \(\mathrm{k} \leq 34.555...\) (use calculator)

5. APPLY CONSTRAINTS to find the answer

• Since we can't cut a fractional piece, k must be a whole number
• Maximum whole number ≤ 34.555 is \(\mathrm{k} = 34\)

6. Verify the answer

• For \(\mathrm{k} = 34\): \(120(34) + 15(33) = 4,575 \leq 4,650\) ✓
• For \(\mathrm{k} = 35\): \(120(35) + 15(34) = 4,710 \gt 4,650\) ✗

Answer: A. 34

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students often assume you need k cuts to get k pieces, leading to the equation \(120\mathrm{k} + 15\mathrm{k} \leq 4,650\). This gives \(135\mathrm{k} \leq 4,650\), so \(\mathrm{k} \leq 34.44\), leading them to conclude \(\mathrm{k} = 34\). While this happens to give the right answer, it's based on incorrect reasoning about the cutting process.

Second Most Common Error:

Poor TRANSLATE reasoning: Students might ignore the waste entirely and simply divide \(4,650 \div 120 = 38.75\), concluding they can get 38 pieces. This may lead them to select Choice C (38).

The Bottom Line:

This problem requires careful attention to the relationship between cuts and pieces. The cutting waste creates a constraint that's more complex than simple division, and students must translate the physical cutting process accurately into mathematics.