# Financial mathematics: compound interest and depreciation

> IB Mathematics Analysis and Approaches SL · IB AA SL
> Source: https://www.owlsprep.com/study/ib-math-aa-sl-u1-financial-mathematics-compound-interest-annuities/

We cover the two financial models in IB AA SL 1.4 — compound interest and reducing-balance depreciation — including different compounding periods and how to rearrange each formula to find an unknown rate or number of years.

**Prerequisites:** [Exponential functions and percentage change](https://www.owlsprep.com/study/ib-math-aa-sl-u1-exponential-functions-intro/); [Geometric sequences and series](https://www.owlsprep.com/study/ib-math-aa-sl-u1-geometric-sequences-series/)

## Learning objectives

- Derive and apply the compound interest formula for different discrete compounding periods (annual, semi-annual, quarterly, monthly)
- Calculate reducing-balance (annual) depreciation for tangible assets
- Rearrange the compound interest and depreciation formulas to solve for the rate or the number of years
- Select and apply the correct growth or decay formula in IB AA SL structured exam questions

## Compound Interest Fundamentals

Compound interest differs from simple interest, which only earns returns on the initial principal sum. For compounding periods shorter than 1 year (monthly, quarterly, weekly), you must adjust the interest rate and total number of periods to match the compounding frequency.

**Discrete Compound Interest** — Where $k$ is the number of compounding periods per year, $r$ is the annual nominal percentage rate, and $n$ is the total number of years of investment.

*Notation:* $A = P\left(1 + \frac{r}{100k}\right)^{kn}$

**Derivation:** Derive the compound interest formula

*Starting from:* After 1 compounding period, the value is $P \times \left(1 + \frac{r}{100k}\right)$

1. After 2 periods, value = $P \times \left(1 + \frac{r}{100k}\right)^2$
2. After $k$ periods (1 full year), value = $P \times \left(1 + \frac{r}{100k}\right)^k$
3. After $kn$ total periods over $n$ years, the final value is the standard compound interest formula

*Conclusion:* This formula is directly provided in the IB AA SL formula booklet (topic 1.4)

**Worked example:** You invest \$2500 at a 4.2% annual nominal rate compounded monthly. Calculate the final value after 6 years.

1. Identify all given values: $P = 2500$, $r = 4.2$, $k = 12$, $n = 6$
2. $$A = 2500 \times \left(1 + \frac{4.2}{100 \times 12}\right)^{12 \times 6}$$
3. Simplify the terms inside the bracket and exponent, then compute the final value to get \$3215.08

**Check your understanding**

1. What is the total number of compounding periods for a 3 year investment compounded quarterly?

   *Why:* 4 periods per year multiplied by 3 years gives 12 total periods

*Calculator:* allowed

## Reducing Balance Depreciation

Depreciation describes the loss in value of a physical asset over time, such as a car, electronics, or machinery. Reducing balance depreciation is the exponential decay equivalent of compound interest growth, using a decay factor instead of a growth factor.

**Reducing Balance Depreciation** — Where $V_0$ is the initial purchase value, $r$ is the annual percentage depreciation rate, and $n$ is the number of full years of ownership.

*Notation:* $V_n = V_0\left(1 - \frac{r}{100}\right)^n$

> **Not in the booklet**
>
> This depreciation formula is not printed in the IB formula booklet, but it is just compound interest with a decay factor $(1 - r/100)$ in place of the growth factor $(1 + r/100)$, so it is easy to reconstruct.

**Worked example:** A car purchased for \$32000 depreciates at 15% per year. Find its value after 5 full years.

1. Identify values: $V_0 = 32000$, $r = 15$, $n =5$
2. $$V_5 = 32000 \times (1 - 0.15)^5 = 32000 \times 0.85^5$$
3. Calculate to get a final depreciated value of \$14199 (to the nearest dollar)

*Calculator:* allowed

## Solving for the Rate or the Number of Years

Exam questions often give you the start and end values and ask for the interest rate or the number of years. Because the unknown is inside the exponent (for time) or the base (for rate), you rearrange the same formula: take a root to unlock the rate, or take logarithms to unlock the time.

**Worked example:** An investment of \$6000 grows to \$9000 under annual compound interest over 5 years. Find the annual interest rate $r$.

1. Set up the equation with $k=1$: $9000 = 6000\left(1 + \frac{r}{100}\right)^5$
2. $$\left(1 + \frac{r}{100}\right)^5 = \frac{9000}{6000} = 1.5$$
3. Take the 5th root of both sides, then solve for $r$

   $$1 + \frac{r}{100} = 1.5^{1/5} = 1.0845 \implies r \approx 8.45\%$$

**Worked example:** A \$5000 investment earns 6% annual compound interest. Find the least whole number of years for its value to first exceed \$8000.

1. Set up the inequality: $5000(1.06)^n > 8000$, so $(1.06)^n > 1.6$
2. $$n > \frac{\ln 1.6}{\ln 1.06} \approx 8.07$$
3. Since $n$ must be a whole number of years, the least value is $n = 9$ years

**Check your understanding**

1. To solve for the number of years when the unknown is in the exponent, which tool do you use?

   *Why:* Taking logs of both sides brings the exponent down so you can solve for n.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Using the annual percentage rate directly for monthly compounding without dividing by 12
  - Why it fails: Forgets that the rate must match the length of the compounding period
  - Correct: Always divide the nominal annual rate by the number of compounding periods per year before substituting
- **Wrong:** Adding the depreciation rate to 1 instead of subtracting it
  - Why it fails: Confuses exponential growth (compound interest) with exponential decay (depreciation)
  - Correct: For reducing balance depreciation, use a factor of $(1 - r/100)$ not $(1 + r/100)$
- **Wrong:** Using number of years as the total number of periods for monthly compounding
  - Why it fails: Fails to convert time units to match the compounding frequency
  - Correct: Multiply the number of years by the number of periods per year to get total $kn$
- **Wrong:** Rounding a 'how many years' answer down when the target has not yet been reached
  - Why it fails: If $n > 8.07$, then 8 whole years is still below the target — only 9 whole years reaches it
  - Correct: For a 'first exceeds' question, round the solved value up to the next whole year

## Cheatsheet

| Scenario | Formula | IB Formula Booklet |
| --- | --- | --- |
| Compound Interest | $A = P\left(1 + \frac{r}{100k}\right)^{kn}$ | Given in booklet (topic 1.4) |
| Reducing Balance Depreciation | $V_n = V_0\left(1 - \frac{r}{100}\right)^n$ | Not in booklet — use $(1 - r/100)^n$ |
| Solve for rate $r$ | $1 + \frac{r}{100k} = \left(\frac{A}{P}\right)^{1/(kn)}$ | Rearranged form (take a root) |
| Solve for years $n$ | $n = \dfrac{\ln(A/P)}{k\,\ln\!\left(1 + \frac{r}{100k}\right)}$ | Rearranged form (take logs) |

## What's next

Now that you can apply the two IB AA SL financial models — compound interest and reducing-balance depreciation — and rearrange each to solve for an unknown rate or number of years, you can connect them back to the geometric sequences and series they come from. Compound growth and decay are geometric sequences in disguise, so the same reasoning powers population, radioactive decay and other exponential-model questions on Paper 1 and Paper 2. Practice deciding, from the wording alone, whether a scenario is growth or decay before you pick your factor $(1 + r/100)$ or $(1 - r/100)$.

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