A total of 2 squares each have side length r. A total of 6 equilateral triangles each have side length...
GMAT Algebra : (Alg) Questions
A total of 2 squares each have side length \(\mathrm{r}\). A total of 6 equilateral triangles each have side length \(\mathrm{t}\). None of these squares and triangles shares a side. The sum of the perimeters of all these squares and triangles is \(210\). Which equation represents this situation?
\(\mathrm{6r + 24t = 210}\)
\(\mathrm{2r + 6t = 210}\)
\(\mathrm{8r + 18t = 210}\)
\(\mathrm{6r + 2t = 210}\)
1. TRANSLATE the problem information
- Given information:
- 2 squares, each with side length \(\mathrm{r}\)
- 6 equilateral triangles, each with side length \(\mathrm{t}\)
- Total perimeter of all shapes = 210
- What this tells us: We need to find total perimeter from squares + total perimeter from triangles
2. INFER the approach
- Strategy: Calculate the perimeter contribution from each shape type separately, then add them
- We'll need to use perimeter formulas for squares and triangles
3. SIMPLIFY to find square perimeter contribution
- Perimeter of one square = \(4\mathrm{r}\)
- Total from 2 squares = \(2 \times 4\mathrm{r} = 8\mathrm{r}\)
4. SIMPLIFY to find triangle perimeter contribution
- Perimeter of one equilateral triangle = \(3\mathrm{t}\)
- Total from 6 triangles = \(6 \times 3\mathrm{t} = 18\mathrm{t}\)
5. INFER the final equation
- Total perimeter = \(8\mathrm{r} + 18\mathrm{t}\)
- Since total perimeter = 210: \(8\mathrm{r} + 18\mathrm{t} = 210\)
Answer: C
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak TRANSLATE skills: Students misinterpret what quantities need to be calculated, focusing on individual side lengths rather than total perimeters.
They might think the equation should just use the side lengths directly (\(\mathrm{r}\) and \(\mathrm{t}\)) without calculating perimeters, leading them to create equations like \(2\mathrm{r} + 6\mathrm{t} = 210\) based on the number of shapes times their side lengths.
This may lead them to select Choice B (\(2\mathrm{r} + 6\mathrm{t} = 210\))
Second Most Common Error:
Poor SIMPLIFY execution: Students correctly identify the need for perimeters but make calculation errors when multiplying.
Common mistakes include calculating \(2 \times 4\mathrm{r} = 6\mathrm{r}\) instead of \(8\mathrm{r}\), or \(6 \times 3\mathrm{t} = 24\mathrm{t}\) instead of \(18\mathrm{t}\). These arithmetic errors lead to incorrect coefficients in the final equation.
This may lead them to select Choice A (\(6\mathrm{r} + 24\mathrm{t} = 210\)) or other incorrect options
The Bottom Line:
This problem requires careful translation of word quantities into mathematical expressions, plus accurate application of basic perimeter formulas. Students must distinguish between individual measurements and total contributions from multiple shapes.
\(\mathrm{6r + 24t = 210}\)
\(\mathrm{2r + 6t = 210}\)
\(\mathrm{8r + 18t = 210}\)
\(\mathrm{6r + 2t = 210}\)