IB Maths AA HL Topic 1 โ€” Number & Algebra Paper 1 & 2 ~10 min read

Compound Interest & Depreciation

This is the financial maths note. Compound interest is when interest is added to both the original amount and the interest already accumulated โ€” so growth speeds up over time. Depreciation is the same idea in reverse: a value that loses a fixed percentage each period, like a car or a laptop. Both follow geometric-style formulas, with one twist (the compounding-periods variable k) that catches students out.

๐Ÿ“˜ What you need to know

Simple vs compound interest

The two interest types behave very differently over long time periods. Understanding the difference is the conceptual heart of this topic.

๐Ÿ“
Simple interest
Same fixed amount of interest each year โ€” based only on the original amount.
FV = PV(1 + rn100)
Linear growth โ†’ arithmetic in flavour
๐Ÿ“ˆ
Compound interest
Interest grows on interest โ€” the balance at year-end becomes the new “principal” for next year.
FV = PV(1 + r100)n
Exponential growth โ†’ geometric in flavour
$1000 invested at 8%/year โ€” simple vs compound (10 years)
1000 1300 1600 1900 2200 0 1 2 3 4 5 6 7 8 9 10 Year $1800 $2159 Simple Compound growing gap
After 10 years on $1000 at 8%, simple interest gives you $1800 (a flat +$80 each year). Compound interest gives you $2159 โ€” the gap of nearly $360 is “interest on interest”. Over 30+ years the compound curve absolutely dominates. This is why early saving matters.

Compounding periods โ€” the k variable

Banks don’t always wait a full year before adding interest. Many add it monthly, quarterly, or even daily. The nominal annual rate stays the same, but it gets divided into smaller chunks and applied more frequently.

Frequency
k
Periods/year
Rate per period
Annually
1
once per year
r%
Half-yearly
2
every 6 months
r/2 %
Quarterly
4
every 3 months
r/4 %
Monthly
12
every month
r/12 %
Example:   6% per annum compounded monthly means 0.5% interest is added each month โ€” and there are 12 of these per year, so 60 of them in 5 years.

๐Ÿค” Why does monthly compounding give more than annual?

Because each month’s interest is added to the balance, and then next month’s interest is calculated on the new (slightly larger) balance. With annual compounding, you wait the full 12 months before getting any growth-on-growth boost. The more compounding periods, the more “interest on interest” gets squeezed in.

The compound interest formula

Compound interest FV = PV ยท (1 + r100k)kn โœ“ in formula booklet
What each piece means:
  FV = future value (what you end up with)  |  PV = present value (your starting deposit)
  r% = nominal annual interest rate  |  k = compounding periods per year
  n = number of years (not periods!)
Watch the units carefully. n is in years, not periods. The formula handles the conversion automatically โ€” kn in the exponent gives you the total number of compounding events.

๐Ÿงญ Recipe โ€” compound interest question

  1. Identify each variable: starting amount (PV), nominal rate r, compounding frequency k, number of years n.
  2. Substitute into the formula. Don’t simplify r/100k in your head โ€” leave it for the calculator.
  3. Use your GDC. The formula booklet’s compound interest formula is also accessible through most graphing calculators’ Finance package.
  4. Round at the end, to whatever precision the question asks for.

Compound depreciation

Depreciation is the same idea as compound interest, but the value decreases by a fixed percentage each year instead of increasing. Cars, laptops, and machinery typically follow this pattern in the early years of their lifespan.

Compound depreciation (annual) FV = PV ยท (1 โˆ’ r100)n โœ— NOT in formula booklet โ€” but easy to derive

๐Ÿค” How to derive it from the booklet

Take the compound interest formula, set k = 1 (depreciation almost always compounds annually), and replace the + with a โˆ’ because the value is shrinking instead of growing. That gives you FV = PV(1 โˆ’ r/100)n. If you ever forget the depreciation formula in the exam, it’s a one-line reconstruction from the compound interest formula in your booklet.

GDC tip: if you’re using your calculator’s Finance package for a depreciation question, just enter the interest rate as a negative number (e.g. โˆ’15 instead of 15). The same formula handles both growth and decay this way.

Worked examples

WE 1

Compound interest โ€” compounded annually

Maria deposits $5000 into an account that pays 4% per year compounded annually. Find the value of the account after 8 years, to the nearest dollar.

Step 1: Identify variables PV = 5000,   r = 4,   k = 1,   n = 8 Step 2: Substitute into formula FV = 5000 ร— (1 + 4100 ร— 1)1 ร— 8 = 5000 ร— (1.04)8 Step 3: Compute on GDC = 5000 ร— 1.36857… = 6842.85… โ‰ˆ $6843
WE 2

Compound interest โ€” compounded monthly

Daniel invests $3000 in an account paying a nominal annual interest rate of 6% compounded monthly. How much will be in the account after 4 years, to the nearest cent?

Step 1: Identify variables PV = 3000,   r = 6,   k = 12,   n = 4 Step 2: Substitute FV = 3000 ร— (1 + 6100 ร— 12)12 ร— 4 = 3000 ร— (1.005)48 Step 3: Compute on GDC = 3000 ร— 1.27049… = 3811.47… โ‰ˆ $3811.47 k = 12 means 0.5% added every month, 48 times total over 4 years
WE 3

Compare compounding frequencies

Two banks both offer a nominal annual rate of 4%. Bank A compounds monthly; Bank B compounds annually. If $10,000 is invested in each for 5 years, how much more does Bank A return?

Step 1: Bank A โ€” monthly compounding FVA = 10000 ร— (1 + 41200)60 = 10000 ร— 1.22099… = $12,209.97 Step 2: Bank B โ€” annual compounding FVB = 10000 ร— (1.04)5 = 10000 ร— 1.21665… = $12,166.53 Step 3: Find the difference FVA โˆ’ FVB = 12209.97 โˆ’ 12166.53 Bank A returns $43.44 more small but real โ€” and grows larger for higher rates and longer periods
WE 4

Depreciation โ€” value after n years

A new laptop costs $1500. It depreciates by 12% each year. Find its value after 6 years, to the nearest dollar.

Step 1: Identify variables PV = 1500,   r = 12,   n = 6 Step 2: Apply depreciation formula FV = 1500 ร— (1 โˆ’ 12100)6 = 1500 ร— (0.88)6 Step 3: Compute = 1500 ร— 0.46437… = 696.55… โ‰ˆ $697
WE 5

Depreciation โ€” find the time using logs

A piece of machinery is bought for $25,000 and depreciates at 18% per year. Find the time, in years and months, for the value to fall to $12,500 (half its original value).

Step 1: Set up the equation 12500 = 25000 ร— (0.82)n 0.5 = (0.82)n Step 2: Take logs of both sides ln(0.5) = n ยท ln(0.82) n = ln(0.5)ln(0.82) = โˆ’0.6931โˆ’0.1985 = 3.491… Step 3: Convert decimal years โ†’ years and months 3 years + 0.491 ร— 12 months = 3 years + 5.89… months โ‰ˆ 3 years 6 months always state both signs of the logs are negative โ€” they cancel and give a positive n

๐Ÿ’ก Top tips

โš  Common mistakes

Compound interest is a beautiful demonstration of geometric growth at work โ€” every formula in this note is just u1 ยท rn in disguise. The “growth factor” (1 + r/100k) is the common ratio; kn is the number of multiplications. If you ever forget a formula, fall back on geometric reasoning.

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