A cylindrical vase has a constant cross-sectional area and an interior height of 12text{ cm}. Stones are placed in the...

1. TRANSLATE the problem information

• Given information:
  • Cylindrical vase with \(\mathrm{height = 12\text{ cm}}\)
  • Stones occupy \(\mathrm{60\%}\) of total interior volume
  • Water volume used = \(\mathrm{270\text{ cm}^3}\)
  • Water fills the remaining space to the top

2. INFER the volume relationship

• If stones occupy \(\mathrm{60\%}\) of the total volume, then water must occupy the remaining \(\mathrm{40\%}\)
• Let \(\mathrm{V}\) = total interior volume of the vase
• Water volume = \(\mathrm{40\%\text{ of }V = 0.4V}\)
• Since \(\mathrm{270\text{ cm}^3}\) of water was used: \(\mathrm{0.4V = 270}\)

3. SIMPLIFY to find total volume

• Solve for \(\mathrm{V}\): \(\mathrm{V = 270 \div 0.4 = 675\text{ cm}^3}\)

4. INFER how to find base area

• For a cylinder: \(\mathrm{Volume = base\text{ }area \times height}\)
• We have \(\mathrm{V = 675\text{ cm}^3}\) and \(\mathrm{height = 12\text{ cm}}\)
• Therefore: \(\mathrm{base\text{ }area = V \div height}\)

5. SIMPLIFY to get final answer

• \(\mathrm{Base\text{ }area = 675 \div 12 = 56.25\text{ cm}^2}\)

Answer: D) 56.25

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students misinterpret the percentage relationship, thinking that \(\mathrm{270\text{ cm}^3}\) represents \(\mathrm{60\%}\) of the volume (the stone volume) rather than the \(\mathrm{40\%}\) water volume.

This leads them to set up: \(\mathrm{0.6V = 270}\), giving \(\mathrm{V = 450\text{ cm}^3}\), and \(\mathrm{base\text{ }area = 450 \div 12 = 37.5\text{ cm}^2}\). Since \(\mathrm{37.5}\) isn't an answer choice, this leads to confusion and guessing.

Second Most Common Error:

Poor INFER reasoning: Students correctly identify that water occupies \(\mathrm{40\%}\) but then try to find the stone volume first instead of working directly with the water volume equation.

They might calculate stone volume as \(\mathrm{270 \div 0.4 \times 0.6 = 1012.5}\), creating unnecessary complexity and potential arithmetic errors. This causes them to get stuck and randomly select an answer.

The Bottom Line:

The key insight is recognizing that the \(\mathrm{270\text{ cm}^3}\) represents the complement of the stone volume (the \(\mathrm{40\%}\) that remains), not the stone volume itself. Students who miss this percentage relationship interpretation cannot set up the correct equation to begin the solution.