A large wheel of cheese is in the shape of a cylinder with a height of 5 inches. An employee...

1. TRANSLATE the problem information

• Given information:
  • Cylindrical cheese wheel (height 5 inches - note this for later)
  • Wedge cut with central angle = \(80°\)
  • Wedge volume = 20 cubic inches
  • Need to find: Original total volume
• What we're looking for: The volume of the entire wheel before the wedge was removed

2. INFER the key relationship

• The crucial insight: This wedge represents a fraction of the entire wheel
• That fraction is determined by the angle: \(80°\) out of \(360°\) total
• Since the wedge and wheel have the same height and radius, the volume ratio equals the angle ratio
• Set up the proportion: (Wedge volume)/(Total volume) = (Wedge angle)/(Full circle)

3. TRANSLATE this relationship into mathematics

• Let V = original volume of the entire wheel
• Write the proportion: \(\frac{20}{\mathrm{V}} = \frac{80°}{360°}\)

4. SIMPLIFY the fraction on the right side

• Reduce \(\frac{80}{360}\):
  • \(\frac{80}{360} = \frac{8}{36} = \frac{2}{9}\)
• Now we have: \(\frac{20}{\mathrm{V}} = \frac{2}{9}\)

5. SIMPLIFY by solving the proportion

• Cross multiply: \(2 \times \mathrm{V} = 20 \times 9\)
• Calculate: \(2\mathrm{V} = 180\)
• Divide by 2: \(\mathrm{V} = 90\)

Answer: C. 90

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize this as a proportional relationship problem. Instead, they try to use the cylinder volume formula \(\mathrm{V = πr^2h}\), getting stuck because they don't have the radius. They may attempt to work backwards using the height (5 inches) and wedge volume (20 cubic inches), leading to complicated calculations that don't yield any of the answer choices. This leads to confusion and guessing.

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly set up the proportion \(\frac{20}{\mathrm{V}} = \frac{80}{360}\) but make arithmetic errors. A common mistake is incorrectly simplifying \(\frac{80}{360}\) (perhaps getting \(\frac{8}{36}\) but then reducing incorrectly to \(\frac{1}{4}\) instead of \(\frac{2}{9}\)), or making errors in cross multiplication. For example, if they use \(\frac{20}{\mathrm{V}} = \frac{1}{4}\), they get \(\mathrm{V} = 80\), leading them to select Choice B (80).

The Bottom Line:

The key challenge is recognizing that the height information is irrelevant—this is purely about proportional relationships between angles and volumes, not about calculating cylinder volumes from scratch.