A rectangular storage container has no top cover. The interior dimensions of the container are 2 feet by 3 feet...
GMAT Geometry & Trigonometry : (Geo_Trig) Questions
A rectangular storage container has no top cover. The interior dimensions of the container are 2 feet by 3 feet by 4 feet. What is the total surface area of the interior bottom and four walls, in square feet?
- \(24\)
- \(40\)
- \(46\)
- \(52\)
- \(80\)
24
40
46
52
80
1. TRANSLATE the problem information
- Given information:
- Rectangular container: 2 feet by 3 feet by 4 feet
- No top cover
- Need interior surface area of bottom and walls
- What this tells us: We calculate 5 surfaces total (not 6, since no top)
2. VISUALIZE the container structure
- Picture an open rectangular box
- Identify the surfaces: 1 bottom + 4 walls
- INFER the dimensions of each surface:
- Bottom: uses the base dimensions \(\mathrm{2 × 3}\)
- Walls: each uses one base dimension × height (4)
3. Calculate the bottom area
- Bottom surface: \(\mathrm{2 × 3 = 6}\) square feet
4. INFER and calculate wall dimensions
- The four walls come in two pairs:
- Two walls: 2 feet wide × 4 feet tall
- Two walls: 3 feet wide × 4 feet tall
5. Calculate wall areas
- Two walls of \(\mathrm{2 × 4}\): \(\mathrm{2 × (2 × 4) = 2 × 8 = 16}\) square feet
- Two walls of \(\mathrm{3 × 4}\): \(\mathrm{2 × (3 × 4) = 2 × 12 = 24}\) square feet
6. Find total surface area
- Total = Bottom + All walls = \(\mathrm{6 + 16 + 24 = 46}\) square feet
Answer: C. 46
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak VISUALIZE skill: Students include the top surface in their calculation, treating this like a closed rectangular box instead of recognizing "no top cover."
They calculate all 6 surfaces: bottom (6) + four walls (40) + top (6) = \(\mathrm{52}\) square feet.
This leads them to select Choice D (52).
Second Most Common Error:
Poor INFER reasoning: Students correctly exclude the top but miscalculate wall areas by using wrong dimension pairings or forgetting that walls come in pairs.
For example, they might calculate walls as: \(\mathrm{2×4 + 3×4 + 2×4 + 3×4 = 8+12+8+12 = 40}\), then add bottom: \(\mathrm{40+6 = 46}\). While this gives the right answer, the reasoning shows they're adding individual walls rather than recognizing the paired structure, which could lead to errors on similar problems.
The Bottom Line:
This problem tests spatial visualization more than calculation skills. Students who can't mentally picture the open container structure will include surfaces that shouldn't be counted or miss the paired nature of opposite walls.
24
40
46
52
80