Financial mathematics: compound interest and depreciation

IB Mathematics Analysis and Approaches SL· Number and Algebra, SL 1.4 Financial applications of geometric sequences and series· 12 min read

1. Compound Interest Fundamentals★★☆☆☆⏱ 10 min

✓ Calculator OK

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.

📘 Definition

Discrete Compound Interest

$A = P (1 + \frac{r}{100 k})^{k n}$

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.

🔬 Derivation

Goal:

Derive the compound interest formula

Starting from:

After 1 compounding period, the value is $P \times (1 + \frac{r}{100 k})$

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

Result:

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

✓ Quick check

What is the total number of compounding periods for a 3 year investment compounded quarterly?
Reveal answer
12 —
4 periods per year multiplied by 3 years gives 12 total periods

2. Reducing Balance Depreciation★★☆☆☆⏱ 8 min

✓ Calculator OK

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.

📘 Definition

Reducing Balance Depreciation

$V_{n} = V_{0} (1 - \frac{r}{100})^{n}$

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.

📐 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)

3. Solving for the Rate or the Number of Years★★★☆☆⏱ 12 min

✓ Calculator OK

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 $k = 1$ : $9000 = 6000 (1 + \frac{r}{100})^{5}$
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

✓ Quick check

To solve for the number of years when the unknown is in the exponent, which tool do you use?
Reveal answer
Logarithms —
Taking logs of both sides brings the exponent down so you can solve for n.

4. Common Pitfalls

Wrong move:

Using the annual percentage rate directly for monthly compounding without dividing by 12

Why:

Forgets that the rate must match the length of the compounding period

Correct move:

Always divide the nominal annual rate by the number of compounding periods per year before substituting

Wrong move:

Adding the depreciation rate to 1 instead of subtracting it

Why:

Confuses exponential growth (compound interest) with exponential decay (depreciation)

Correct move:

For reducing balance depreciation, use a factor of $(1 - r /100)$ not $(1 + r /100)$

Wrong move:

Using number of years as the total number of periods for monthly compounding

Why:

Fails to convert time units to match the compounding frequency

Correct move:

Multiply the number of years by the number of periods per year to get total $k n$

Wrong move:

Rounding a 'how many years' answer down when the target has not yet been reached

Why:

If $n > 8.07$ , then 8 whole years is still below the target — only 9 whole years reaches it

Correct move:

For a 'first exceeds' question, round the solved value up to the next whole year

5. Quick Reference Cheatsheet

Scenario	Formula	IB Formula Booklet
Compound Interest	$A = P (1 + \frac{r}{100 k})^{k n}$	Given in booklet (topic 1.4)
Reducing Balance Depreciation	$V_{n} = V_{0} (1 - \frac{r}{100})^{n}$	Not in booklet — use $(1 - r /100)^{n}$
Solve for rate $r$	$1 + \frac{r}{100 k} = (\frac{A}{P})^{1/ (k n)}$	Rearranged form (take a root)
Solve for years $n$	$n = \frac{ln ( A / P )}{k ln ( 1 + \frac{r}{100 k} )}$	Rearranged form (take logs)

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2025 · Paper 2
3-mark monthly compound interest calculation
2023 · Paper 2
4-mark multi-part depreciation question

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 $(1 - r /100)$ .