Simple & Compound Interest • Topic 4 of 4

Growth & Depreciation

Population growth and machine depreciation are compound interest in disguise. Growth: P(1 + R/100)^T; depreciation: P(1 - R/100)^T. Mixed-rate problems (different rate each year) just chain the factors. Half-yearly compounding halves the rate and doubles the periods; quarterly divides rate by 4 and multiplies periods by 4.

✅ Solved examples

1. A town of 8000 grows 10% a year. Population after 2 years?
8000 x 1.1^2 = 9680.
2. A machine worth 50000 depreciates 20% a year. Value after 2 years?
50000 x 0.8^2 = 50000 x 0.64 = 32000.
3. CI on 8000 at 10% per annum, compounded half-yearly, for 1 year?
Rate 5% per half, 2 periods: 8000 x 1.05^2 = 8820; CI = 820.
4. Population grows 10% then falls 10% over two years from 5000. End value?
5000 x 1.1 x 0.9 = 4950.

✏️ Practice — try these, take hints as needed

1. Town 6000 grows 5% per year. After 2 years?
x 1.1025.
6000 x 1.1025.
—
6615
2. Machine 80000 depreciates 25% per year. After 2 years?
x 0.75^2.
80000 x 0.5625.
—
45000
3. CI on 10000 at 10% half-yearly for 1 year?
5% x 2 periods.
10000 x 1.1025.
CI = 1025.
1025
4. Value of 20000 after 20% rise then 20% fall?
1.2 x 0.8.
0.96.
—
19200
5. Population 4000 at 25% growth for 1 year?
x 1.25.
—
—
5000

📝 Topic test — 8 questions

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