A rectangular garden has a perimeter of 48 meters. The length of the garden is 5 times its width. What...

1. TRANSLATE the problem information

• Given information:
  • Perimeter = 48 meters
  • Length = 5 × width
  • Need to find both length and width

• What this tells us: We have two conditions that must both be satisfied simultaneously

2. TRANSLATE into mathematical equations

• Let \(\mathrm{W}\) = width and \(\mathrm{L}\) = length
• Perimeter condition: \(\mathrm{2L + 2W = 48}\)
• Length-width relationship: \(\mathrm{L = 5W}\)

3. INFER the solution strategy

• We have two equations with two unknowns - this is a system we can solve
• Since we already have \(\mathrm{L}\) expressed in terms of \(\mathrm{W}\), substitution is the most direct approach
• First simplify the perimeter equation to make substitution easier

4. SIMPLIFY the perimeter equation

• \(\mathrm{2L + 2W = 48}\)
• Divide everything by 2: \(\mathrm{L + W = 24}\)

5. SIMPLIFY using substitution

• Substitute \(\mathrm{L = 5W}\) into \(\mathrm{L + W = 24}\):
• \(\mathrm{5W + W = 24}\)
• \(\mathrm{6W = 24}\)
• \(\mathrm{W = 4}\) meters

6. SIMPLIFY to find the length

• \(\mathrm{L = 5W = 5(4) = 20}\) meters

7. APPLY CONSTRAINTS by checking the answer

• Perimeter check: \(\mathrm{2(20) + 2(4) = 40 + 8 = 48}\) ✓
• Relationship check: \(\mathrm{20 = 5 × 4}\) ✓

Answer: C (20, 4)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY execution: Students correctly set up \(\mathrm{L + W = 24}\) and \(\mathrm{L = 5W}\), but make arithmetic errors when solving \(\mathrm{6W = 24}\), incorrectly calculating \(\mathrm{W = 6}\) or \(\mathrm{W = 3}\).

If \(\mathrm{W = 6}\), then \(\mathrm{L = 30}\), giving \(\mathrm{(30, 6)}\). If \(\mathrm{W = 3}\), then \(\mathrm{L = 15}\), giving \(\mathrm{(15, 3)}\). Neither matches the answer choices, leading to confusion and guessing.

Second Most Common Error:

Inadequate TRANSLATE reasoning: Students mix up which dimension should be 5 times the other, setting up \(\mathrm{W = 5L}\) instead of \(\mathrm{L = 5W}\).

This leads to \(\mathrm{W + 5W = 24}\), so \(\mathrm{6W = 24}\) and \(\mathrm{W = 4}\), but then \(\mathrm{L = W/5 = 4/5 = 0.8}\). The result \(\mathrm{(0.8, 4)}\) doesn't match any choice, leading them to abandon systematic solution and guess.

The Bottom Line:

This problem tests whether students can translate word relationships accurately into algebra and then execute multi-step algebraic solutions without arithmetic mistakes. The key insight is that "length is 5 times width" means \(\mathrm{L = 5W}\), not \(\mathrm{W = 5L}\).