A rectangular storage box has interior dimensions 50text{ cm} by 50text{ cm} by 80text{ cm}. A solid sphere of the...
GMAT Geometry & Trigonometry : (Geo_Trig) Questions
A rectangular storage box has interior dimensions \(50\text{ cm}\) by \(50\text{ cm}\) by \(80\text{ cm}\). A solid sphere of the largest possible size is placed inside so that it touches the interior of the box where possible. To the nearest cubic centimeter, what is the volume of the space in the box not occupied by the sphere?
1. TRANSLATE the problem information
- Given information:
- Rectangular box: \(\mathrm{50\,cm \times 50\,cm \times 80\,cm}\) interior dimensions
- Need largest possible sphere that fits inside
- Find volume of remaining space
2. INFER the sphere size constraint
- The largest sphere that can fit must have its diameter limited by the smallest box dimension
- Box dimensions: 50, 50, 80 → smallest is 50 cm
- Therefore: sphere diameter = 50 cm, so radius = 25 cm
3. SIMPLIFY the volume calculations
- Box volume: \(\mathrm{V_{box} = 50 \times 50 \times 80 = 200,000\,cm^3}\)
- Sphere volume:
\(\mathrm{V_{sphere} = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi(25)^3 = \frac{4}{3}\pi(15,625)}\)
\(\mathrm{= \frac{62,500\pi}{3}}\)
- Using \(\mathrm{\pi \approx 3.1416}\) (use calculator): \(\mathrm{\frac{62,500 \times 3.1416}{3} = 65,450\,cm^3}\)
4. SIMPLIFY the final subtraction
- Space not occupied = \(\mathrm{200,000 - 65,450 = 134,550\,cm^3}\)
Answer: (C) 134,550
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak INFER skill: Students may think the sphere diameter should equal the largest dimension (80 cm) instead of the smallest dimension (50 cm).
Using radius = 40 cm instead of 25 cm would give:
\(\mathrm{V_{sphere} = \frac{4}{3}\pi(40)^3 = \frac{4}{3}\pi(64,000) \approx 268,000\,cm^3}\)
Since this exceeds the box volume, they realize something's wrong and may guess randomly, or they might subtract anyway getting a negative result and become confused.
Second Most Common Error:
Poor SIMPLIFY execution: Students correctly identify \(\mathrm{r = 25\,cm}\) but make calculation errors with \(\mathrm{\frac{4}{3}\pi(25)^3}\).
Common mistakes include:
- Forgetting to cube the radius: using 25 instead of 15,625
- Calculation errors with the fraction \(\mathrm{\frac{4}{3}}\) or \(\mathrm{\pi}\) approximation
- Order of operations errors
These calculation mistakes typically lead to sphere volumes that don't match any answer choice exactly, causing students to select Choice (A) 86,900 or Choice (B) 133,333 as "close enough" estimates.
The Bottom Line:
This problem requires spatial reasoning to determine sphere constraints, then careful multi-step calculations. The key insight about diameter limitation is what separates systematic solvers from guessers.