A solid cylindrical metal rod has a height of 30 centimeters and a radius of 8 centimeters. From this rod,...
GMAT Geometry & Trigonometry : (Geo_Trig) Questions
A solid cylindrical metal rod has a height of 30 centimeters and a radius of 8 centimeters. From this rod, two identical cones are carved out, creating hollow spaces. The base of each cone is one of the circular faces of the cylinder, and the vertices of the two cones meet at the center point of the cylinder. To the nearest cubic centimeter, what is the volume of the metal remaining in the rod after the two cones have been removed?
1. TRANSLATE the problem setup
- Given information:
- Cylinder: height = \(30\) cm, radius = \(8\) cm
- Two identical cones carved out
- Each cone's base sits on one end of the cylinder
- The cone vertices meet at the center point of the cylinder
- What this tells us: We need to find cylinder volume minus the volume of two cones
2. INFER the strategy and cone dimensions
- Strategy: Calculate total cylinder volume, then subtract the volume of both carved-out cones
- Key insight: If the cone vertices meet at the center point of the cylinder, then each cone extends from one end to the middle
- Each cone height = \(30 \div 2 = 15\) cm
- Each cone radius = \(8\) cm (same as cylinder radius)
3. Calculate cylinder volume
- Using \(\mathrm{V} = \pi\mathrm{r}^2\mathrm{h}\):
\(\mathrm{V_{cylinder}} = \pi(8)^2(30)\)
\(= \pi(64)(30)\)
\(= 1920\pi\) cm³
4. Calculate volume of carved-out cones
- Volume of one cone using \(\mathrm{V} = \frac{1}{3}\pi\mathrm{r}^2\mathrm{h}\):
\(\mathrm{V_{one\_cone}} = \frac{1}{3}\pi(8)^2(15)\)
\(= \frac{1}{3}\pi(64)(15)\)
\(= 320\pi\) cm³
- Total volume of both cones:
\(\mathrm{V_{two\_cones}} = 2 \times 320\pi\)
\(= 640\pi\) cm³
5. SIMPLIFY to find remaining volume
- Remaining volume = Original volume - Carved-out volume
\(\mathrm{V_{remaining}} = 1920\pi - 640\pi\)
\(= 1280\pi\) cm³
- Convert to decimal (use calculator):
\(1280 \times 3.14159 \approx 4021.24\) cm³
- Round to nearest cubic centimeter: \(4021\) cm³
Answer: B (4,021)
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak TRANSLATE skill: Students misinterpret "vertices meet at center point" and assume each cone has the full cylinder height (\(30\) cm) instead of half the height (\(15\) cm).
Using height = \(30\) cm for each cone:
\(\mathrm{V_{one\_cone}} = \frac{1}{3}\pi(8)^2(30)\)
\(= 640\pi\) cm³
Total cone volume = \(2 \times 640\pi = 1280\pi\) cm³
Remaining volume = \(1920\pi - 1280\pi = 640\pi \approx 2011\) cm³
This leads them to select Choice A (2,011)
Second Most Common Error:
Poor SIMPLIFY execution: Students correctly find that the remaining volume is \(1280\pi\) cm³ but make arithmetic errors when converting to a decimal, either using an incorrect value of π or making calculation mistakes.
This causes confusion and may lead to guessing among the remaining answer choices.
The Bottom Line:
The key challenge is correctly interpreting the geometric constraint that vertices meeting at the center means each cone only extends halfway through the cylinder, not the full height.