The area A, in square centimeters, of a rectangular painting can be represented by the expression \(\mathrm{w(w + 29)}\), where...
GMAT Advanced Math : (Adv_Math) Questions
The area \(\mathrm{A}\), in square centimeters, of a rectangular painting can be represented by the expression \(\mathrm{w(w + 29)}\), where \(\mathrm{w}\) is the width, in centimeters, of the painting. Which expression represents the length, in centimeters, of the painting?
\(\mathrm{w}\)
\(\mathrm{29}\)
\(\mathrm{(w + 29)}\)
\(\mathrm{w(w + 29)}\)
1. TRANSLATE the problem information
- Given information:
- Area = \(\mathrm{w(w + 29)}\) square centimeters
- \(\mathrm{w}\) = width in centimeters
- Need to find: expression for length in centimeters
2. INFER the approach using area relationships
- For any rectangle: \(\mathrm{Area = length \times width}\)
- Since we know the area expression and the width, we can solve for length
- Strategy: Rearrange \(\mathrm{Area = length \times width}\) to get \(\mathrm{length = Area \div width}\)
3. SIMPLIFY by substituting and dividing
- Substitute known values: \(\mathrm{w(w + 29) = length \times w}\)
- Divide both sides by \(\mathrm{w}\): \(\mathrm{length = w(w + 29) \div w}\)
- Perform the division: \(\mathrm{length = (w + 29)}\)
Answer: C. \(\mathrm{(w + 29)}\)
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak INFER skill: Students don't recognize they need to use the rectangle area formula to work backwards from area to length.
Instead, they might look at the answer choices and guess based on what "looks right" or seems to match parts of the given expression. This leads to confusion and guessing among all choices.
Second Most Common Error:
Poor SIMPLIFY execution: Students recognize the correct approach but make algebraic errors when dividing \(\mathrm{w(w + 29)}\) by \(\mathrm{w}\).
They might incorrectly think \(\mathrm{w(w + 29) \div w = w + 29/w}\) instead of \(\mathrm{(w + 29)}\), or they might get confused about which terms cancel. This may lead them to select Choice A \(\mathrm{(w)}\) if they think the width terms somehow cancel completely.
The Bottom Line:
This problem tests whether students can work backwards from a given area expression using the fundamental rectangle relationship. The key insight is recognizing that when you know \(\mathrm{Area = length \times width}\), you can always solve for any unknown variable by rearranging the formula.
\(\mathrm{w}\)
\(\mathrm{29}\)
\(\mathrm{(w + 29)}\)
\(\mathrm{w(w + 29)}\)