# Simple Interest, Compound Interest and Depreciation

> How money grows and how assets lose value. Simple and compound interest, compounding periods and depreciation.

From HSC Maths Advanced on WhizzNotes, free study notes for NSW students: https://whizznotes.com/mathematics-advanced/interest-and-depreciation/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: Two ways to grow
Jamal gets 1000 dollars for his 16th birthday. Bank A offers 6% simple interest. Bank B offers 6% compound interest. They sound the same.

After one year, both give him 1060 dollars. But after 30 years, Bank A gives 2800 dollars and Bank B gives over 5700 dollars. The difference is that Bank B pays interest **on the interest**. That one idea, compounding, is behind savings, loans, superannuation and credit card debt.

**In short:** Simple interest grows by the same amount each period, $I = Prn$, while compound interest grows by the same percentage, $A = P(1 + r)^n$. Depreciation is the same idea in reverse: an item loses value by a fixed percentage each year, $S = V_0(1 - r)^n$.

## Starting point: Years 9 and 10

### Simple interest

Simple interest is paid only on the original amount, the **principal** $P$.

$$I = Prn$$

$r$ is the rate per period as a decimal and $n$ is the number of periods.

### Worked example: Simple interest
5000 dollars at 4% per year simple interest for 3 years: $I = 5000 \times 0.04 \times 3 = 600$ dollars. Total $= 5600$ dollars.

### Compound interest

With compound interest, each period's interest is added to the balance, and the next period's interest is calculated on the new balance.

$$A = P(1 + r)^n$$

$A$ is the final amount. The interest earned is $A - P$.

### Worked example: Jamal's Bank B
$A = 1000(1.06)^{30} \approx 5743.49$ dollars. Interest earned: about 4743 dollars.

### Calculator tip: Type the formula in one go
For 5000 dollars at 4.2% a year, compounded monthly for 3 years, type `5000 × ( 1 + 0.042 ÷ 12 ) ^ 36`. Keep the whole bracket in one line. Rounding $1.0035$ part way changes the cents. For depreciation, use a minus inside the bracket.

### Compounding periods

If interest is compounded more than once a year, divide the yearly rate and multiply the years.

- Monthly: $r = \dfrac{\text{annual rate}}{12}$, $n = 12 \times \text{years}$.
- Quarterly: divide by 4, multiply by 4.

8% per year compounded monthly for 5 years: $r = \dfrac{0.08}{12}$ and $n = 60$.

### Depreciation

Cars, phones and machines lose value. **Declining balance depreciation** takes a fixed percentage off each year:

$$S = V_0(1 - r)^n$$

$V_0$ is the starting value and $S$ is the **salvage value** after $n$ years.

### Worked example: A laptop
A 2400 dollar laptop depreciates at 25% per year. After 3 years: $S = 2400(0.75)^3 = 1012.50$ dollars.

**Straight-line depreciation** takes the same **amount** off each year, a percentage of the purchase price:
$$S = V_0 - Dn$$
where $D$ is the amount lost each year. A 30 000 dollar car losing 10% of its purchase price each year loses 3000 dollars a year, so after 4 years it is worth $30\,000 - 4 \times 3000 = 18\,000$ dollars. Compare declining balance: $30\,000(0.9)^4 \approx 19\,683$ dollars.

### Key idea: Simple versus compound
- Simple interest grows by the same amount each year: a straight line.
- Compound interest grows by the same percentage each year: an exponential curve.

### Watch out: Rate per period
Always match $r$ and $n$ to the same period. A yearly rate with months will give a hugely wrong answer.

**Mind map**

```text
Interest
  Simple
    I = Prn
    Linear growth
  Compound
    A = P(1 + r)ⁿ
    Interest on interest
    Exponential growth
  Periods
    Divide rate
    Multiply time
  Depreciation
    S = V₀(1 - r)ⁿ
    Value falls by a percent
```

### Year 11: Finding the time or the rate
**How long to double?** At 5% per year compounded yearly, solve $2P = P(1.05)^n$. Then $1.05^n = 2$, so

$$n = \frac{\ln 2}{\ln 1.05} \approx 14.2$$

so it takes 15 full years.

**What rate is needed?** To turn 4000 dollars into 5000 dollars in 4 years: $(1 + r)^4 = 1.25$, so $1 + r = 1.25^{\frac{1}{4}} \approx 1.0574$, a rate of about 5.74% per year.

**Which compounding is better?** Work out both. For 10 000 dollars over 3 years: 6% p.a. compounded monthly gives $10\,000\left(1 + \frac{0.06}{12}\right)^{36} \approx 11\,966.81$, while 6.1% compounded yearly gives $10\,000(1.061)^3 \approx 11\,943.90$. More frequent compounding can beat a slightly higher rate.

**A rate of depreciation.** If a 40 000 dollar machine is worth 25 000 dollars after 4 years (declining balance), $(1 - r)^4 = 0.625$, so $1 - r = 0.625^{\frac{1}{4}} \approx 0.889$, a rate of about 11.1% per year.

### Year 12: Compound interest as a geometric sequence
The balance at the end of each year, $P(1 + r)$, $P(1 + r)^2$, $P(1 + r)^3$, ..., is a geometric sequence with ratio $1 + r$. This is the key to annuities in the next chapter, where regular deposits each grow by compound interest and their total is a geometric series.

**Present value.** To have 10 000 dollars in 5 years at 6% per year, you need to invest

$$P = \frac{A}{(1 + r)^n} = \frac{10\,000}{1.06^5} \approx 7472.58 \text{ dollars today}$$

Money now is worth more than the same money later, because it can earn interest.

**Recurrence rules.** A rule like $P_n = 1.2P_{n-1} - 40$ with $P_0 = 300$ describes growth of 20% each step, minus 40. Step through it: $P_1 = 320$, $P_2 = 344$, $P_3 = 372.8$.

Writing it out shows a geometric series:
$$P_n = 1.2^n P_0 - 40(1 + 1.2 + \dots + 1.2^{n-1}) = 300(1.2)^n - 40 \times \frac{1.2^n - 1}{0.2}$$
which simplifies to $P_n = 200 + 100(1.2)^n$. Check: $n = 1$ gives 320.

## Quick summary

- Simple interest: $I = Prn$. Grows by the same amount each period.
- Compound interest: $A = P(1 + r)^n$. Grows by the same percentage each period.
- For monthly compounding, divide the yearly rate by 12 and multiply the years by 12.
- Depreciation: $S = V_0(1 - r)^n$.
- Use logs to find how long an investment takes to reach a target.

## Quiz

Try these questions from the chapter quiz.

1. Find the simple interest on 2000 dollars at 5% per year for 4 years.
   - A. 100 dollars
   - B. 400 dollars
   - C. 431 dollars
   - D. 2400 dollars
   - Answer: B. 400 dollars. $2000 \times 0.05 \times 4 = 400$.

2. 3000 dollars is invested at 3% compound interest per year for 2 years. The final amount is:
   - A. 3180.00 dollars
   - B. 3182.70 dollars
   - C. 3090.00 dollars
   - D. 3300.00 dollars
   - Answer: B. 3182.70 dollars. $3000(1.03)^2 = 3182.70$.

3. Simple interest grows:
   - A. by the same amount each year
   - B. by the same percentage each year
   - C. faster and faster
   - D. slower and slower
   - Answer: A. by the same amount each year. It is only paid on the principal.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/mathematics-advanced/interest-and-depreciation/#quiz