Question:A new lightweight alloy is developed by a materials scientist. For a given volume, the mass of a sample of...

1. TRANSLATE the problem information

• Given information:
  • Alloy mass = 85% of aluminum mass (for same volume)
  • Aluminum mass = k% of alloy mass
  • Need to find k, rounded to nearest whole number
• What this tells us: We have two percentage relationships connecting the same two quantities

2. TRANSLATE into mathematical equations

• Let \(\mathrm{M_A}\) = mass of aluminum sample and \(\mathrm{M_L}\) = mass of alloy sample
• First relationship: \(\mathrm{M_L = 0.85M_A}\)
• Second relationship: \(\mathrm{M_A = \frac{k}{100}M_L}\)

3. INFER the solution strategy

• We have two equations with three variables (\(\mathrm{M_A}\), \(\mathrm{M_L}\), and \(\mathrm{k}\))
• Since we only care about k, we can eliminate \(\mathrm{M_A}\) and \(\mathrm{M_L}\) using substitution
• Substitute the first equation into the second to solve for k

4. SIMPLIFY by substitution

• Start with: \(\mathrm{M_A = \frac{k}{100}M_L}\)
• Substitute \(\mathrm{M_L = 0.85M_A}\):
  \(\mathrm{M_A = \frac{k}{100}(0.85M_A)}\)
• Divide both sides by \(\mathrm{M_A}\):
  \(\mathrm{1 = \frac{k}{100}(0.85)}\)
  \(\mathrm{1 = \frac{0.85k}{100}}\)

5. SIMPLIFY to solve for k

• Multiply both sides by 100: \(\mathrm{100 = 0.85k}\)
• Divide by 0.85: \(\mathrm{k = \frac{100}{0.85} = \frac{2000}{17} ≈ 117.647}\) (use calculator)
• Round to nearest whole number: \(\mathrm{k = 118}\)

Answer: 118

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students misinterpret which quantity is the percentage of which. They might set up \(\mathrm{M_A = 0.85M_L}\) instead of \(\mathrm{M_L = 0.85M_A}\), confusing "alloy is 85% of aluminum" with "aluminum is 85% of alloy."

When they then substitute incorrectly, they get \(\mathrm{k = \frac{100}{1.15} ≈ 87}\), leading them to select an answer around 85-90 or causes confusion and guessing.

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly set up the equations but make calculation errors when computing \(\mathrm{\frac{100}{0.85}}\). They might incorrectly calculate this as \(\mathrm{100 × 0.85 = 85}\) or struggle with the fraction arithmetic.

This leads to selecting an incorrect answer choice or abandoning the systematic solution and guessing.

The Bottom Line:

This problem challenges students to carefully track the direction of percentage relationships and work with reciprocal percentages, where being 85% of something means that something is about 118% of you.