What you need to know
Simple interest is paid on the original amount (the principal) only, so the same dollar amount is added every period: I = Prn, where P is the principal, r is the interest rate per period as a decimal and n is the number of periods. The final amount is A = P + I.
Compound interest is paid on the principal and on interest already earned, so the balance grows faster each period: A = P(1 + r)[sup n], and the interest earned is I = A − P. The key skill is matching r and n to the compounding period: for 6% per year compounded monthly, r = 0.06 ÷ 12 = 0.005 per month and n counts months.
| Nominal rate | Compounded | r per period | n for 2 years |
|---|
| 6% per year | annually | 0.06 | 2 |
| 6% per year | half-yearly | 0.03 | 4 |
| 6% per year | quarterly | 0.015 | 8 |
| 6% per year | monthly | 0.005 | 24 |
Simple interest: $5000 at 4% per year simple interest for 3 years
- I = Prn = 5000 × 0.04 × 3 = $600
- A = 5000 + 600 = $5600
Compound interest: $5000 at 4% per year compounded annually for 3 years
- A = P(1 + r)n = 5000 × 1.043 = 5000 × 1.124864 = $5624.32
- Interest earned: 5624.32 − 5000 = $624.32, which is $24.32 more than simple interest over the same 3 years. The gap grows quickly with more years.
Monthly compounding: $8000 at 6% per year compounded monthly for 2 years
- r = 0.06 ÷ 12 = 0.005 per month; n = 2 × 12 = 24 months
- A = 8000 × 1.00524 = 8000 × 1.12716… = $9017.28
- Interest: 9017.28 − 8000 = $1017.28
Find the principal: how much must be invested now at 5% per year compounded annually to have $20,000 in 5 years?
- 20,000 = P × 1.055
- P = 20,0001.055 = 20,0001.27628… = $15,670.52
- This is sometimes called the present value of $20,000.
Find the time: how long does $1000 take to double at 7% per year compounded annually?
- 1000 × 1.07n = 2000, so 1.07n = 2
- Take logs: n ln 1.07 = ln 2, so n = ln 2ln 1.07 = 0.6931…0.0677… = 10.24…
- Interest is only added at the end of each year, so it takes 11 years to be at least double. (You can also find this by trial and error with the table function on a calculator.)
Find the rate: $3000 grows to $3600 in 4 years with annual compounding. What is the rate?
- 3000(1 + r)4 = 3600, so (1 + r)4 = 1.2
- 1 + r = 1.21/4 = 1.0466…
- r = 0.0466…, so the rate is 4.66% per year (to 2 decimal places)
Common mistake: using the annual rate with a number of months. The rate and the number of periods must describe the same period. 6% per year monthly for 2 years is r = 0.005 and n = 24, never r = 0.06 and n = 24. Another common mistake: rounding too early. Keep 1.00524 in your calculator (or at least 6 decimal places) until the final step, then round money to the nearest cent. Rounding 1.12716 to 1.13 changes the answer by more than $20.
Check yourself
- Find the simple interest on $2000 at 5% per year for 4 years.
- Find the final amount when $2000 is invested at 5% per year compounded annually for 4 years.
- Find the final amount when $10,000 is invested at 3% per year compounded monthly for 1 year.
- How much must be invested now at 6% per year compounded annually to have $5000 in 3 years?
- How many whole years does it take money to at least double at 10% per year compounded annually?
- Find the interest earned on $1500 at 8% per year compounded quarterly for 2 years.
Answers: 1. 2000 × 0.05 × 4 = $400 2. 2000 × 1.054 = $2431.01 3. 10,000 × 1.002512 = $10,304.16 4. 5000 ÷ 1.063 = $4198.10 5. n = ln 2 ÷ ln 1.1 = 7.27…, so 8 years 6. 1500 × 1.028 = $1757.49, so interest = $257.49