Problem: The density of a certain solid substance is 813 kilograms per cubic meter. A sample of this substance is...

1. TRANSLATE the problem information

• Given information:
  • \(\mathrm{Density = 813\, kg/m^3}\)
  • Cube shape with \(\mathrm{edge\, length = 0.60\, meters}\)
  • Need mass to nearest whole number

2. INFER the solution strategy

• To find mass, we need: \(\mathrm{mass = density \times volume}\)
• First step: Calculate the cube's volume
• Second step: Multiply volume by density
• Third step: Round to nearest whole number

3. Calculate the volume of the cube

• \(\mathrm{Volume\, of\, cube = edge^3}\)
• \(\mathrm{V = (0.60)^3 = 0.216\, cubic\, meters}\) (use calculator)

4. SIMPLIFY using the density relationship

• \(\mathrm{mass = density \times volume}\)
• \(\mathrm{mass = 813\, kg/m^3 \times 0.216\, m^3 = 175.608\, kg}\) (use calculator)

5. Round to nearest whole number

• 175.608 rounds to 176 kg

Answer: A. 176

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students misinterpret "cube with edge length 0.60 meters" and use 0.60 as the volume directly instead of cubing it.

They calculate: \(\mathrm{mass = 813 \times 0.60 = 487.8 \approx 488\, kg}\)

This may lead them to select Choice B (488)

Second Most Common Error:

Poor SIMPLIFY execution: Students make arithmetic errors when cubing 0.60 or in the multiplication step, leading to significant calculation mistakes.

This causes confusion about which answer choice matches their work, leading them to guess among the remaining options.

The Bottom Line:

This problem tests whether students can systematically work through a multi-step density problem while correctly applying the cube volume formula - the key insight is recognizing that edge length must be cubed to get volume.