A new car is purchased for $25,000. The value of the car depreciates by 20% at the end of each...

1. TRANSLATE the problem information

• Given information:
  • Initial car value: $25,000
  • Car depreciates by 20% each year
  • Need value after 3 years

• TRANSLATE the key phrase: "depreciates by 20%" means the car retains \(\mathrm{100\% - 20\% = 80\%}\) of its value each year

2. INFER the mathematical pattern

• This creates a repeated multiplication situation
• Each year: \(\mathrm{New\ value = Previous\ value \times 0.8}\)
• We can either calculate year-by-year or use the pattern \(\mathrm{Value = Initial \times (0.8)^n}\)

3. SIMPLIFY using year-by-year calculation

• End of Year 1: \(\mathrm{\$25,000 \times 0.8 = \$20,000}\)
• End of Year 2: \(\mathrm{\$20,000 \times 0.8 = \$16,000}\)
• End of Year 3: \(\mathrm{\$16,000 \times 0.8 = \$12,800}\)

OR using the exponential pattern:

• Value after 3 years = \(\mathrm{\$25,000 \times (0.8)^3}\)
• Calculate \(\mathrm{(0.8)^3 = 0.512}\) (use calculator)
• \(\mathrm{\$25,000 \times 0.512 = \$12,800}\) (use calculator)

Answer: B) $12,800

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students misinterpret "depreciates by 20%" as meaning the car is worth 20% of its original value each year, rather than losing 20% each year.

This leads them to calculate: \(\mathrm{\$25,000 \times 0.2 = \$5,000}\) per year, thinking the car loses $5,000 annually for a final value of \(\mathrm{\$25,000 - 3(\$5,000) = \$10,000}\).

This may lead them to select Choice A ($10,000).

Second Most Common Error:

Poor INFER reasoning: Students correctly understand the 80% retention but apply it incorrectly by calculating \(\mathrm{\$25,000 \times 0.8 \times 3 = \$60,000}\), treating it as simple multiplication rather than repeated multiplication.

This leads to confusion since $60,000 exceeds the original value, causing them to get stuck and guess.

The Bottom Line:

The key challenge is correctly translating percentage depreciation into mathematical operations and recognizing that each year's calculation builds on the previous year's result, not the original value.