Maya bought silk ribbon and cotton ribbon for a craft project. The silk ribbon costs $1.20 per yard, and the...

1. TRANSLATE the problem information

• Given information:
  • Silk ribbon: \(\$1.20\) per yard
  • Cotton ribbon: \(\$0.80\) per yard
  • Total spent: \(\$10.80\)
  • Cotton ribbon quantity: 3 more yards than twice the silk ribbon yards
• Set up variables: Let \(\mathrm{s}\) = yards of silk ribbon, \(\mathrm{c}\) = yards of cotton ribbon

2. TRANSLATE relationships into equations

• Quantity relationship: \(\mathrm{c = 2s + 3}\)
• Cost relationship: \(\mathrm{1.20s + 0.80c = 10.80}\)

3. INFER the solution strategy

• Since we already have c expressed in terms of s, substitution is the most direct approach
• Substitute the quantity equation into the cost equation

4. SIMPLIFY through substitution and algebra

• Substitute: \(\mathrm{1.20s + 0.80(2s + 3) = 10.80}\)
• Distribute: \(\mathrm{1.20s + 1.60s + 2.40 = 10.80}\)
• Combine like terms: \(\mathrm{2.80s + 2.40 = 10.80}\)
• Solve: \(\mathrm{2.80s = 8.40}\), so \(\mathrm{s = 3}\) yards

5. Find the cotton ribbon quantity

• \(\mathrm{c = 2(3) + 3 = 9}\) yards

Answer: C (9)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students misinterpret "3 more than twice the yards of silk ribbon" and write \(\mathrm{c = 2s - 3}\) instead of \(\mathrm{c = 2s + 3}\).

The phrase structure can be confusing - students might focus on the subtraction implied by "more than" rather than recognizing it means addition. With this incorrect constraint, they would get a different value for s and subsequently a wrong value for c.

This may lead them to select Choice B (6) or another incorrect answer.

Second Most Common Error:

Inadequate SIMPLIFY execution: Students make arithmetic errors when distributing \(\mathrm{0.80(2s + 3)}\) or when combining like terms \(\mathrm{(2.80s)}\).

Small calculation mistakes compound through the multi-step algebraic process. For instance, incorrectly calculating \(\mathrm{0.80 \times 3 = 2.40}\) as 3.20 would throw off the entire solution.

This leads to confusion and potentially guessing among the remaining choices.

The Bottom Line:

This problem requires careful parsing of complex word relationships and methodical algebraic execution. The combination of interpreting "3 more than twice" correctly AND executing several algebraic steps without error creates multiple failure points for students.