Financial Mathematics • Topic 4 of 6

Depreciation

Depreciation is the loss of value over time by a fixed percentage each period — the mirror image of growth. The model is A = P(1 − r)^t, where r is the rate of decline as a decimal. A machine worth $5,000 that loses 20% a year is worth 5000 × 0.80 = $4,000 after one year, and 5000(0.8)² = $3,200 after two. A base between 0 and 1 (like 0.8) signals decay. Subtract the rate from 1 before multiplying, and apply the factor once per period. The SAT uses depreciation for vehicles, equipment and electronics, and asks you to tell decay from growth by the base.

Value versus years: a single rose curve falling steeply at first then flattening as it approaches the horizontal axisDepreciationYearsValuePValue falls by a fixed percent each year

✅ Solved examples

1. A machine worth $5,000 loses 20% per year. Value after 1 year?
5000 × (1 − 0.20) = 5000 × 0.80 = $4,000.
2. A $2,000 laptop loses 25% in a year. Value after 1 year?
2000 × 0.75 = $1,500.
3. A $5,000 asset loses 20% per year. Value after 2 years?
5000(0.8)² = 5000 × 0.64 = $3,200.
4. Does y = 800(0.85)^t show growth or decay?
Base 0.85 < 1, so decay.

✏️ Practice — try these, take hints as needed

1. A $3,000 car loses 10% in a year. Value after 1 year?
Multiply by (1 − 0.10).
3000 × 0.90.
—
$2,700.
2. A $4,000 machine loses 25% in a year. Value after 1 year?
4000 × 0.75.
—
—
$3,000.
3. A $1,000 phone loses 20% per year. Value after 2 years?
1000 × 0.8².
1000 × 0.64.
—
$640.
4. Does y = 500(0.9)^t represent growth or decay?
Base < 1?
0.9 < 1.
—
Decay.
5. A $2,000 asset loses 50% per year. Value after 1 year?
2000 × 0.50.
—
—
$1,000.

📝 Topic test — 8 questions

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