A right circular cylinder has a volume of 432 cubic centimeters. The area of the base of the cylinder is...
GMAT Geometry & Trigonometry : (Geo_Trig) Questions
A right circular cylinder has a volume of \(432\) cubic centimeters. The area of the base of the cylinder is \(24\) square centimeters. What is the height, in centimeters, of the cylinder?
1. TRANSLATE the problem information
- Given information:
- \(\mathrm{Volume} = 432\) cubic centimeters
- \(\mathrm{Area\,of\,base} = 24\) square centimeters
- Find: height in centimeters
- What this tells us: We have two key pieces needed for the volume formula
2. INFER the solution approach
- Since we have volume and base area, we can use the cylinder volume formula directly
- Key insight: \(\mathrm{V} = \pi\mathrm{r}^2\mathrm{h}\), and \(\pi\mathrm{r}^2\) is exactly the area of the base
- So the formula becomes: \(\mathrm{V} = (\mathrm{Area\,of\,base}) \times \mathrm{h}\)
3. TRANSLATE the formula with our values
- Substitute known values: \(432 = 24 \times \mathrm{h}\)
4. SIMPLIFY to solve for height
- Divide both sides by 24: \(\mathrm{h} = 432 \div 24\)
- Calculate: \(\mathrm{h} = 18\)
Answer: A. 18
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak INFER skill: Students don't recognize that \(\pi\mathrm{r}^2\) in the volume formula equals the area of the base that's already given. Instead, they try to find the radius first by solving \(\pi\mathrm{r}^2 = 24\), then attempt to use \(\mathrm{V} = \pi\mathrm{r}^2\mathrm{h}\). This creates unnecessary complexity with π calculations and potential rounding errors.
This approach might lead them to get confused with decimal approximations and select Choice C (216) or abandon the systematic solution and guess.
Second Most Common Error:
Poor TRANSLATE reasoning: Students misread what they're solving for and confuse the area of the base (24) with the height, leading them to select Choice B (24) without performing any calculations.
The Bottom Line:
The key insight is recognizing that you don't need to find the radius separately - the area of the base is already provided and can substitute directly into the volume formula. This transforms a potentially complex problem into simple division.