An asset (a car, sound gear, a laptop) loses the same percentage of its value every year. Because the value is shrinking, the dollar amount lost gets smaller each year too. It is exactly compound interest, but running downhill.
V = value after n years · P = starting price · r = depreciation rate (as a decimal) · n = number of years.
The multiplier is (1 − r). Lose 20% a year and you keep 80%, so you multiply by 0.8 each year.
Sound gear worth $8,000 loses 20% of its value each year. What is it worth after 2 years?
| End of year | Calculation | Value |
|---|---|---|
| Start | opening | $8,000.00 |
| 1 | 8000 × 0.8 | $6,400.00 |
| 2 | 6400 × 0.8 | $5,120.00 |
| 3 | 5120 × 0.8 | $4,096.00 |
The drop shrinks each year: $1,600, then $1,280, then $1,024. That is what "reducing balance" means.
The same question, written the QCAA way. Each line earns its own tick.
If a question adds a claim, finish with a reasonableness line, e.g. "5120 > 5000, so it still beats the $5000 trade-in." ✓ reasonableness
• This is not straight-line depreciation (that loses the same dollars each year). Here you lose a percentage of a shrinking value.
• Multiply by (1 − r), the amount you keep, not by r.
• Round only at the very end.
Keep 80% each year, so multiply by 0.8, once per year.
A percentage of a smaller number is a smaller drop, so the curve falls fast at first, then flattens. (Straight-line would be a ruler-straight slope.)
Don't read yet, just have a go in your head:
Keep 90% → 10000 × 0.9 = $9,000 → × 0.9 = $8,100 → × 0.9 = $7,290
10000 × 0.9 = $9,000 → × 0.9 = ? → × 0.9 = ?
A $5,000 laptop loses 25% a year. What is it worth after 2 years? Work it out, then check.
In one sentence, out loud: why does the dollar value lost get smaller each year? If you can say it, you've got it.