Question:A solid cube has a volume of 500 cubic centimeters. A second cube is constructed with side length 10% less...

1. TRANSLATE the problem information

• Given information:
  • Original cube volume: 500 cubic centimeters
  • Second cube has side length 10% less than original
• What this tells us: If original side length is \(\mathrm{s}\), second cube side length is \(\mathrm{0.90s}\)

2. INFER the most efficient approach

• Key insight: We don't need to find the actual side length of either cube
• We can work directly with the volume relationship
• Since volume depends on the cube of the side length, we need \(\mathrm{(0.90)^3}\)

3. SIMPLIFY the volume calculation

• Original cube: \(\mathrm{V_1 = s^3 = 500}\)
• Second cube: \(\mathrm{V_2 = (0.90s)^3 = (0.90)^3 \times s^3}\)
• Calculate \(\mathrm{(0.90)^3 = 0.90 \times 0.90 \times 0.90 = 0.729}\) (use calculator)
• Therefore: \(\mathrm{V_2 = 0.729 \times 500 = 364.5}\) (use calculator)

Answer: 364.5 cubic centimeters

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER reasoning: Students think that if the side length decreases by 10%, then the volume also decreases by 10%. They calculate \(\mathrm{500 - 0.10(500) = 450}\).

This linear thinking misses the crucial insight that volume scales with the cube of the side length, not linearly with it. This may lead them to select an answer of 450 if available among choices.

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly set up \(\mathrm{(0.9)^3 \times 500}\) but make calculation errors. Common mistakes include:

• Calculating \(\mathrm{(0.9)^3}\) as \(\mathrm{0.9 \times 3 = 2.7}\)
• Mishandling decimal multiplication in \(\mathrm{0.729 \times 500}\)

These calculation errors lead to various incorrect numerical answers and cause confusion when selecting from multiple choice options.

The Bottom Line:

This problem tests whether students understand that geometric scaling affects volume cubically, not linearly. The key insight is recognizing that a 10% reduction in side length results in a 27.1% reduction in volume (since \(\mathrm{0.729 = 1 - 0.271}\)).