What is compound interest?
Interest is extra money paid for using someone's money. The money you start with is the principal (P). The rate (r) is the percentage paid each year (per annum, p.a.).
Simple interest is paid only on the principal. It is the same every year: SI = P × r × n ÷ 100.
Compound interest is paid on the principal and on the interest already earned. At the end of each period the interest joins the principal, and the next interest is worked out on this bigger amount. So the money grows faster.
Example: 1000 at 10% p.a. Simple: 1100, 1200, 1300. Compound: 1100, 1210, 1331.
The compound interest formula
Each year the amount is multiplied by the same number, the multiplier (1 + r/100). For 10% it is 1.1. After n years:
A = P(1 + r/100)ⁿ
A is the final amount (also called future value). The compound interest earned is CI = A − P.
Step by step (year by year)
For small n you can also work one year at a time: interest = amount × r/100, then add it on. The formula just does all the years at once.
Finding P, r or n
Rearrange the same formula. To find P: P = A ÷ (1 + r/100)ⁿ (this is the present value). To find r: take the n-th root of A/P. To find n: try whole years, or use logarithms: n = log(A/P) ÷ log(1 + r/100).
Compounding more often: half-yearly, quarterly, monthly
Banks may add interest more than once a year. If it is added k times a year:
- rate per period = r ÷ k
- number of periods = k × n
A = P(1 + r/(100k))^(kn)
Half-yearly: k = 2. Quarterly: k = 4. Monthly: k = 12. The more often interest is added, the bigger the final amount, but the gain gets smaller and smaller. 1000 at 10% for 1 year gives 1100 (yearly), 1102.50 (half-yearly), 1103.81 (quarterly), 1104.71 (monthly).
Effective annual rate
To compare offers, find the real yearly growth: effective rate = (1 + r/(100k))^k − 1. 12% p.a. compounded monthly is the same as 12.68% compounded once a year.
Continuous compounding
Imagine interest added every second, then every instant. The amount does not become infinite. It reaches a limit:
A = Pe^(rt), with r as a decimal (5% → 0.05) and t in years.
The number e ≈ 2.71828 comes from this exact idea: 1 at 100% compounded continuously for 1 year grows to e. Continuous compounding gives the largest possible amount for a given rate.
Compound interest is exponential growth (and depreciation is decay)
The amounts 1000, 1100, 1210, 1331, … form a geometric sequence with common ratio 1.1. As a function of time, A(n) = P × 1.1ⁿ is an exponential function. Its graph curves upward, while simple interest gives a straight line.
The same rule works for things that grow or shrink by a fixed percent each year: population, bacteria, or a value that loses worth.
Depreciation: a car or phone that loses r% of its value each year is worth A = P(1 − r/100)ⁿ after n years. The multiplier is less than 1.
Rule of 72
Money doubles in about 72 ÷ r years. At 8% it doubles in about 9 years. It is a quick estimate, not exact.
Try it: your own savings plan
Pick an amount you might save, like 2000. Guess how much it becomes at 7% for 10 years. Write your guess. Now set the sliders in the 3D to P = 2000, r = 7, n = 10 and check. Then change to monthly. Who wins: a higher rate, or more frequent compounding? (Hint: the rate matters much more.)
Key formulas and definitions
- Simple interest: SI = P × r × n ÷ 100, amount = P + SI
- Compound amount (yearly): A = P(1 + r/100)ⁿ
- Compound interest: CI = A − P
- k times a year: A = P(1 + r/(100k))^(kn)
- Continuous: A = Pe^(rt) (r as a decimal)
- Depreciation: A = P(1 − r/100)ⁿ
- Effective annual rate = (1 + r/(100k))^k − 1
- For 2 years: CI − SI = P(r/100)²
Worked examples
1. Find the amount and the compound interest on 5000 at 8% p.a. for 2 years, compounded yearly.
Multiplier = 1.08. Year 1: 5000 × 1.08 = 5400. Year 2: 5400 × 1.08 = 5832. Or A = 5000 × 1.08² = 5000 × 1.1664 = 5832. CI = 5832 − 5000 = 832.
2. Find SI and CI on 2000 at 5% p.a. for 3 years. How much more is CI?
SI = 2000 × 5 × 3 ÷ 100 = 300. A = 2000 × 1.05³ = 2000 × 1.157625 = 2315.25, so CI = 315.25. CI is 15.25 more, because interest earned interest.
3. 10 000 is invested at 6% p.a. compounded half-yearly for 3 years. Find the amount.
Rate per half-year = 6 ÷ 2 = 3%. Periods = 2 × 3 = 6. A = 10 000 × 1.03⁶ = 10 000 × 1.194052 = 11 940.52.
4. A car bought for 800 000 loses 15% of its value every year. What is it worth after 3 years?
Multiplier = 1 − 0.15 = 0.85. Value = 800 000 × 0.85³ = 800 000 × 0.614125 = 491 300.
5. 1000 grows to 1210 in 2 years with yearly compounding. Find the rate.
(1 + r/100)² = 1210 ÷ 1000 = 1.21. Square root: 1 + r/100 = 1.1, so r = 10%.
6. How many whole years does it take money to double at 6% p.a. compounded yearly?
We need 1.06ⁿ ≥ 2. 1.06¹¹ ≈ 1.898 (not yet), 1.06¹² ≈ 2.012 (yes). So 12 years. Check with the Rule of 72: 72 ÷ 6 = 12. Using logs: n = log 2 ÷ log 1.06 ≈ 11.9.
7. Find the amount when 1000 is invested at 5% p.a. compounded continuously for 10 years.
A = Pe^(rt) = 1000 × e^(0.05 × 10) = 1000 × e^0.5 = 1000 × 1.648721 = 1648.72.
8. A card charges 12% p.a. compounded monthly. What is the effective annual rate?
Monthly rate = 1%. Effective rate = 1.01¹² − 1 = 1.126825 − 1 = 0.1268 = 12.68%.
Common mistakes
- Using the simple interest rule for every year, so the interest is the same each year. In compound interest each year's interest is on the new, bigger amount.
- Forgetting to divide the rate AND multiply the time when compounding half-yearly or monthly. 8% half-yearly for 3 years means 4% for 6 periods.
- Giving A when the question asks for the compound interest. CI = A − P.
- For depreciation, writing (1 + r/100) instead of (1 − r/100). Losing value means the multiplier is less than 1.