A parcel service charges a fixed handling fee of $4 per shipment.In addition, the service charges $6 for each kilogram...

1. TRANSLATE the problem information

• Given information:
  • Fixed handling fee: \(\$4\) per shipment (regardless of weight)
  • Variable charge: \(\$6\) per kilogram of weight
  • Box weight: 9 kilograms
• What this tells us: We have a fixed cost plus a variable cost based on weight

2. INFER the mathematical relationship

• Total shipping cost = Fixed cost + Variable cost
• Variable cost depends on weight: (rate per kg) × (weight in kg)
• This gives us the cost function: \(\mathrm{Total\ Cost} = 4 + 6\mathrm{w}\), where w is weight

3. SIMPLIFY by substituting and calculating

• Substitute w = 9 kilograms:
  \(\mathrm{Total\ Cost} = 4 + 6(9)\)
• Calculate the variable cost:
  \(6 \times 9 = 54\)
• Add the fixed cost:
  \(54 + 4 = 58\)

Answer: 58

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students calculate only the variable cost portion and forget about the fixed handling fee.

They see "\(\$6\) for each kilogram" and 9 kilograms, so they calculate \(6 \times 9 = 54\), then submit 54 as their final answer. They miss that there's an additional \(\$4\) fixed fee that must be added regardless of weight. This incomplete understanding of cost structure leads them to provide 54 instead of the correct 58.

Second Most Common Error:

Poor SIMPLIFY execution: Students set up the problem correctly but make arithmetic errors in calculation.

They correctly identify that \(\mathrm{Total\ Cost} = 4 + 6(9)\), but then make errors like calculating \(6 \times 9 = 45\) (instead of 54) or adding incorrectly (\(54 + 4 = 57\)). This leads to various incorrect numerical answers.

The Bottom Line:

This problem tests whether students can identify and combine different types of costs (fixed vs. variable) in real-world contexts. The key insight is recognizing that shipping costs have two distinct components that must both be included in the total.