Depreciation of an Industrial MachineYearValue of machine at end of year1$100{,}0002$80{,}0003$64{,}000An industrial machine depreciates at a constant...

1. TRANSLATE the problem information

• Given information:
  • Machine depreciates 20% per year
  • Starting value: $125,000
  • Table confirms: Year 1 = $100,000, Year 2 = $80,000, Year 3 = $64,000
  • Target value: $51,200

• What this tells us: If it loses 20% each year, it keeps 80% (or 0.8) of its value

2. INFER the approach

• This is exponential decay - the value decreases by a constant percentage each period
• We need to find the number of years (n) when the value reaches $51,200
• Strategy: Set up the exponential decay formula V = initial value × (retention rate)ⁿ

3. TRANSLATE into mathematical form

• Exponential decay formula: \(\mathrm{V = 125,000(0.8)^n}\)
• Our equation: \(\mathrm{51,200 = 125,000(0.8)^n}\)

4. SIMPLIFY to solve for n

• Divide both sides by 125,000:
  \(\mathrm{(0.8)^n = 51,200/125,000 = 0.4096}\) (use calculator)

• Find what power of 0.8 equals 0.4096:
  Check: \(\mathrm{0.8^4 = 0.4096}\) (use calculator)

• Therefore: \(\mathrm{n = 4}\)

5. INFER verification strategy

• Let's verify by calculating year 4 directly:
  \(\mathrm{Year\,4 = Year\,3 \times 0.8}\)
  \(\mathrm{= \$64,000 \times 0.8}\)
  \(\mathrm{= \$51,200}\) ✓

Answer: C (4)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students misinterpret "20% depreciation" and think the machine loses a fixed $25,000 each year (20% of original $125,000), treating this as linear decay instead of exponential decay.

They calculate:
\(\mathrm{Year\,4 = \$125,000 - 4(\$25,000)}\)
\(\mathrm{= \$25,000}\)
which doesn't match any answer choice. This leads to confusion and guessing.

Second Most Common Error:

Inadequate SIMPLIFY execution: Students correctly set up \(\mathrm{51,200 = 125,000(0.8)^n}\) but make calculation errors when computing 51,200/125,000 or when checking powers of 0.8.

They might incorrectly calculate \(\mathrm{(0.8)^3}\) or \(\mathrm{(0.8)^5}\) and select the wrong answer choice based on their miscalculation.

The Bottom Line:

The key insight is recognizing that percentage depreciation means exponential decay, not linear decrease. Students must understand that each year's depreciation is calculated from that year's current value, not the original purchase price.