A right circular cone has a height of 22 centimeters (cm) and a base with a diameter of 6 cm....
GMAT Geometry & Trigonometry : (Geo_Trig) Questions
A right circular cone has a height of \(22\) centimeters (cm) and a base with a diameter of \(6\) cm. The volume of this cone is \(n\pi\) cm³. What is the value of n?
1. TRANSLATE the problem information
- Given information:
- Right circular cone with height = 22 cm
- Base diameter = 6 cm
- Volume is expressed as \(\mathrm{nπ\ cm^3}\)
- Need to find: The value of n
2. INFER what formula to use
- For cone volume problems, we need the formula \(\mathrm{V = \frac{1}{3}πr^2h}\)
- Since we have diameter (6 cm), we need radius: \(\mathrm{r = diameter \div 2 = 3\ cm}\)
- Our strategy: Calculate the volume, then identify the coefficient of π
3. SIMPLIFY by substituting into the volume formula
- \(\mathrm{V = \frac{1}{3}πr^2h}\)
- \(\mathrm{V = \frac{1}{3}π(3)^2(22)}\)
- \(\mathrm{V = \frac{1}{3}π(9)(22)}\)
- \(\mathrm{V = \frac{1}{3}π(198)}\)
- \(\mathrm{V = 66π\ cm^3}\)
4. INFER the final answer
- The volume equals \(\mathrm{66π\ cm^3}\)
- Since the problem states volume = \(\mathrm{nπ\ cm^3}\)
- Therefore: \(\mathrm{n = 66}\)
Answer: 66
Why Students Usually Falter on This Problem
Most Common Error Path:
Conceptual confusion about radius vs diameter: Students often use the diameter (6 cm) directly in the volume formula instead of converting to radius (3 cm).
Using \(\mathrm{V = \frac{1}{3}π(6)^2(22)}\)
\(\mathrm{= \frac{1}{3}π(36)(22)}\)
\(\mathrm{= \frac{1}{3}π(792)}\)
\(\mathrm{= 264π}\)
This leads them to conclude \(\mathrm{n = 264}\), which would be an incorrect answer.
Second Most Common Error:
Missing conceptual knowledge about cone volume formula: Students might forget the \(\mathrm{\frac{1}{3}}\) factor and use the cylinder volume formula \(\mathrm{V = πr^2h}\) instead.
Using \(\mathrm{V = π(3)^2(22)}\)
\(\mathrm{= π(9)(22)}\)
\(\mathrm{= 198π}\)
This leads them to conclude \(\mathrm{n = 198}\), another incorrect result.
The Bottom Line:
This problem tests whether students can correctly recall and apply the cone volume formula while carefully distinguishing between diameter and radius. Success requires both accurate formula knowledge and careful attention to the given measurements.