A square garden has its side length increased by a factor that makes the new area 1.69 times the original...

1. TRANSLATE the problem information

• Given information:
  • Square garden has side length increased by some factor
  • New area is 1.69 times the original area
  • Original area is p% of the new area
• What this tells us: We need to find what percent p represents

2. TRANSLATE the mathematical setup

• Let original side length = \(\mathrm{s}\)
• Original area = \(\mathrm{s^2}\)
• New area = \(\mathrm{1.69 \times original\,area = 1.69s^2}\)
• The key phrase: "original area is p% of new area" means:
  \(\mathrm{original\,area = \frac{p}{100} \times new\,area}\)

3. SIMPLIFY by setting up the equation

• Substitute our expressions:
  \(\mathrm{s^2 = \frac{p}{100} \times 1.69s^2}\)
• Divide both sides by \(\mathrm{s^2}\) to eliminate the variable:
  \(\mathrm{1 = \frac{p}{100} \times 1.69}\)
• This gives us:
  \(\mathrm{\frac{p}{100} = \frac{1}{1.69}}\)

4. SIMPLIFY to solve for p

• Multiply both sides by 100:
  \(\mathrm{p = \frac{100}{1.69}}\)
• Calculate this division (use calculator):
  \(\mathrm{p = 59.171...}\)
• Round to nearest whole number:
  \(\mathrm{p = 59}\)

Answer: D) 59

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students misinterpret "original area is p% of new area" and instead set up new area = (p/100) × original area, essentially flipping the relationship.

Using this backwards setup: \(\mathrm{1.69s^2 = \frac{p}{100} \times s^2}\), they get \(\mathrm{p = 169}\), which doesn't match any answer choice. This leads to confusion and guessing.

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly set up the equation but make arithmetic errors when calculating \(\mathrm{\frac{100}{1.69}}\), potentially getting values like \(\mathrm{\frac{100}{1.7} \approx 58.8}\) or \(\mathrm{\frac{100}{1.8} \approx 55.6}\).

This may lead them to select Choice C (56) or get confused between close answer choices.

The Bottom Line:

This problem tests whether students can accurately translate percent language into mathematical equations - the word order in "A is p% of B" is crucial and directly translates to \(\mathrm{A = \frac{p}{100} \times B}\).