The mass of an object is the product of its density and its volume. An object is molded from a...

1. TRANSLATE the problem information

• Given information:
  • Two cubes fused together
  • Larger cube side length = \(2 \times \mathrm{smaller\ cube\ side\ length}\)
  • Density = 5.0 g/cm³
  • Total mass = 450 grams
  • Need to find: mass of larger cube

• What this tells us: We need to work with cube volumes and use the mass = density × volume relationship

2. TRANSLATE relationships into variables

• Let \(\mathrm{s}\) = side length of smaller cube (cm)
• Then larger cube side length = \(2\mathrm{s}\) (cm)

3. INFER the volume relationships

• Smaller cube volume = \(\mathrm{s}^3\)
• Larger cube volume = \((2\mathrm{s})^3 = 8\mathrm{s}^3\)
• Total volume = \(\mathrm{s}^3 + 8\mathrm{s}^3 = 9\mathrm{s}^3\)

4. INFER the strategic approach

• Use total mass to find the unknown side length s
• Then calculate mass of just the larger cube

5. SIMPLIFY to find the side length

• Mass = Density × Volume for the entire object:

\(450 = 5.0 \times (9\mathrm{s}^3)\)

• Simplify:

\(450 = 45\mathrm{s}^3\)

• Divide both sides by 45:

\(\mathrm{s}^3 = 10\)

6. SIMPLIFY to find mass of larger cube

• Mass of larger cube = Density × Volume of larger cube
• Mass of larger cube = \(5.0 \times 8\mathrm{s}^3\)
• Substitute \(\mathrm{s}^3 = 10\):

Mass = \(5.0 \times 8 \times 10 = 400\)

Answer: 400

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Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY skills: Students incorrectly calculate \((2\mathrm{s})^3\), thinking it equals \(2\mathrm{s}^3\) instead of \(8\mathrm{s}^3\)

When students make this error, they get:

• Total volume = \(\mathrm{s}^3 + 2\mathrm{s}^3 = 3\mathrm{s}^3\) (wrong)
• From \(450 = 5.0 \times (3\mathrm{s}^3)\): \(\mathrm{s}^3 = 30\)
• Mass of larger cube = \(5.0 \times 2 \times 30 = 300\) grams

This leads to an incorrect answer of 300 grams instead of 400.

Second Most Common Error:

Poor INFER reasoning: Students calculate the total volume correctly but then find the mass of the smaller cube instead of the larger cube

After correctly finding \(\mathrm{s}^3 = 10\), they calculate:

• Mass of smaller cube = \(5.0 \times \mathrm{s}^3 = 5.0 \times 10 = 50\) grams

This gives them 50 grams, completely missing what the question actually asked for.

The Bottom Line:

This problem tests whether students can correctly handle exponent relationships (especially cubing expressions) and maintain focus on what quantity they're solving for throughout a multi-step calculation process.