A freight elevator can carry at most 80 kilograms on a single trip. Each small crate weighs 2.5 kilograms, and...

1. TRANSLATE the problem information

• Given information:
  • Elevator capacity: at most 80 kg
  • Small crate weight: 2.5 kg each
  • Large crate weight: 5 kg each
  • Already loaded: 16 large crates
  • Find: maximum number of small crates (let's call this s)

• The constraint becomes: weight of small crates + weight of large crates \(\leq\) 80

2. TRANSLATE this into a mathematical inequality

• Current load from large crates: \(\mathrm{16 \times 5 = 80}\) kg
• Total weight equation: \(\mathrm{2.5s + 80 \leq 80}\)
• This captures our constraint perfectly

3. SIMPLIFY the inequality algebraically

• Start with: \(\mathrm{2.5s + 80 \leq 80}\)
• Subtract 80 from both sides: \(\mathrm{2.5s \leq 0}\)
• Divide by 2.5: \(\mathrm{s \leq 0}\)

4. APPLY CONSTRAINTS to select the final answer

• Since we cannot have a negative number of crates in reality
• The maximum value for s is 0
• Therefore, the elevator cannot carry any additional small crates

Answer: A (0)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students may misinterpret "at most 80 kg" as "exactly 80 kg" or fail to account for the weight already on the elevator. They might set up something like \(\mathrm{2.5s = 80}\) or ignore the existing 80 kg load entirely.

This faulty reasoning could lead them to think \(\mathrm{s = 32}\) (from \(\mathrm{80 \div 2.5}\)), causing them to select Choice D (32).

Second Most Common Error:

Inadequate SIMPLIFY execution: Students correctly set up \(\mathrm{2.5s + 80 \leq 80}\) but make an algebraic error. They might subtract incorrectly or forget to account for the inequality direction, potentially arriving at \(\mathrm{s = 16}\) or \(\mathrm{s = 8}\).

This may lead them to select Choice B (8) or Choice C (16).

The Bottom Line:

This problem tricks students because the elevator is already at maximum capacity with just the large crates. The key insight is recognizing that "at most" creates a ceiling that's already reached, leaving no room for additional items.