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A rectangular garden has a length that is 6 meters more than its width. If the perimeter of the garden...

GMAT Geometry & Trigonometry : (Geo_Trig) Questions

Source: Prism
Geometry & Trigonometry
Area and volume formulas
EASY
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A rectangular garden has a length that is 6 meters more than its width. If the perimeter of the garden is 68 meters, what is the area of the garden in square meters?

  1. 240
  2. 280
  3. 300
  4. 320
A
240
B
280
C
300
D
320
Solution

1. TRANSLATE the problem information

  • Given information:
    • Length is 6 meters more than width
    • Perimeter is 68 meters
    • Need to find area
  • What this tells us in math notation:
    • If width = \(\mathrm{w}\), then length = \(\mathrm{w + 6}\)
    • \(\mathrm{P = 68}\) meters

2. TRANSLATE the perimeter condition into an equation

  • Using the rectangle perimeter formula: \(\mathrm{P = 2(length + width)}\)
  • Substitute our expressions: \(\mathrm{68 = 2((w + 6) + w)}\)

3. SIMPLIFY to solve for width

  • \(\mathrm{68 = 2(2w + 6)}\)
  • \(\mathrm{68 = 4w + 12}\)
  • \(\mathrm{56 = 4w}\)
  • \(\mathrm{w = 14}\) meters

4. INFER what we need for the area calculation

  • We found width = 14 meters
  • But area needs both length and width
  • So we need to find length first: \(\mathrm{length = 14 + 6 = 20}\) meters

5. Calculate the final area

  • \(\mathrm{Area = length \times width = 20 \times 14 = 280}\) square meters

Answer: B. 280




Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE reasoning: Students misinterpret "length that is 6 meters more than its width" and set up the wrong relationship, such as \(\mathrm{length = 6w}\) instead of \(\mathrm{length = w + 6}\).

This leads to an equation like \(\mathrm{68 = 2(6w + w) = 14w}\), giving \(\mathrm{w \approx 4.86}\). Then they might calculate area as \(\mathrm{6(4.86) \times 4.86 \approx 142}\), which doesn't match any answer choice. This leads to confusion and guessing.

Second Most Common Error:

Poor INFER reasoning: Students correctly find \(\mathrm{w = 14}\) but forget they need to calculate the length before finding area. They might try to use \(\mathrm{area = 14 \times 6 = 84}\) (using width times the "6 more" instead of actual length).

Since 84 isn't an answer choice, this causes them to get stuck and guess.

The Bottom Line:

This problem tests whether students can correctly translate a verbal relationship into algebra and then systematically work through a multi-step solution, remembering to find all necessary dimensions before the final calculation.

Answer Choices Explained
A
240
B
280
C
300
D
320
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