IB Maths AI HLFinancial ApplicationsPaper 1 & 2FV & PV~7 min read
Compound Interest & Depreciation
Compound interest pays interest on the interest already earned, so an investment grows faster and faster. Depreciation is the mirror image — a value falling by a fixed percentage each year. One formula, with a sign change, handles both.
📘 What you need to know
Compound interest adds interest to the original amount and to interest already earned.
It needs a timeframe, a nominal annual rate r%, and a number of compounding periods.
Compound interest: FV = PV(1 + r100k)kn, with k = periods per year.
Compounding more often (larger k) earns more interest over the same time.
Depreciation is value falling at a fixed rate: FV = PV(1 − r100)n.
To find the time, set FV to the target and solve with logarithms or your GDC.
Compound interest
Compound interest pays interest on the initial amount and on all the interest added so far. That “interest on interest” is why it outgrows simple interest, where the same fixed amount is added each year.
From the same present value, a fixed factor above 1 gives accelerating growth; a factor below 1 gives a steadily falling value.
Compound interest
FV = PV × (1 + r100k)knFV = future value, PV = present value, r% = nominal annual rate, k = periods/year, n = years — in the formula booklet
Compounding periods
Interest need not be added just once a year. The number of compounding periods per year is k: annually k = 1, half-yearly k = 2, quarterly k = 4, monthly k = 12.
Each period earns rk% of the balance, and there are kn periods in n years — both appear in the formula.
More often means more interest: compounding monthly over the same timeframe earns more than compounding annually, because each new period also earns interest on the interest added just before it.
Depreciation
Depreciation is when a value falls over time at a constant rate — the price of a car or a machine, for example. The formula is the compound interest formula with k = 1 and the plus sign changed to a minus.
Compound depreciation
FV = PV × (1 − r100)nr% = rate of depreciation — not in the booklet, but derived from the interest formula
🧠Recipe — compound interest and depreciation
Identify the values — PV, rate r%, years n, and periods per year k.
For a growing value, use FV = PV(1 + r100k)kn.
For a falling value, use FV = PV(1 − r100)n.
Substitute carefully — divide r by 100k, and raise to the power kn (or n for depreciation).
To find the time, set FV to the target value and solve with logarithms; turn a decimal year into months by × 12.
Worked examples
WE 1
Compounded annually
$5000 is invested at a nominal annual interest rate of 4%, compounded annually. Find the value after 6 years, to the nearest dollar.
PV = 5000, r = 4, k = 1, n = 6FV = 5000(1 + 4100×1)1×6= 5000(1.04)6 = 5000 × 1.26532≈ $6327with k = 1 the formula is simply PV × (1 + r/100)n.
WE 2
Compounded monthly
USD $3000 is invested at a nominal annual interest rate of 3.6%, compounded monthly. Find the value after 4 years, to the nearest dollar.
PV = 3000, r = 3.6, k = 12, n = 4FV = 3000(1 + 3.6100×12)12×4= 3000(1.003)48 = 3000 × 1.15463≈ $3464monthly means k = 12, so the power is 12 × 4 = 48 periods.
WE 3
Compounded quarterly
€8000 is invested at a nominal annual interest rate of 5%, compounded quarterly. Find the value after 3 years, to the nearest euro.
PV = 8000, r = 5, k = 4, n = 3FV = 8000(1 + 5100×4)4×3= 8000(1.0125)12 = 8000 × 1.16076≈ €9286quarterly means k = 4 — four periods a year, twelve in three years.
WE 4
Depreciation: future value
A machine is bought for £24 000. It depreciates at 12% per year. Find its value after 7 years, to the nearest pound.
use the depreciation formula, PV = 24000, r = 12, n = 7FV = 24000(1 − 12100)7= 24000(0.88)7 = 24000 × 0.40868≈ £9808a 12% fall means multiplying by 0.88 — keeping 88% — each year.
WE 5
Finding the number of years
$2000 is invested at a nominal annual interest rate of 6%, compounded annually. After how many whole years does the investment first exceed $3000?
set FV > 30002000(1.06)n > 3000 ⇒ 1.06n > 1.5take logs to free nn > log1.061.5 = 6.959check: yr 6 gives $2837, yr 7 gives $30077 yearsround the inequality up — only at a whole year is the interest added.
WE 6
Full question: car depreciation
A car is bought for $30 000 and depreciates at 18% per year. (a) Find its value after 4 years, to the nearest dollar. (b) Find the number of years and months it takes for the value to fall to approximately $10 000.
(a) FV = 30000(1 − 0.18)4= 30000(0.82)4 = 30000 × 0.45212≈ $13564(b) set FV = 10000 and solve10000 = 30000(0.82)n ⇒ (0.82)n = 1⁄3n = log0.82(1⁄3) = 5.536 years(a) $13564 · (b) 5 years 6 months0.536 × 12 ≈ 6, so the decimal part of the year becomes 6 months.
💡 Top tips
Read the compounding period carefully: monthly → k = 12, quarterly → k = 4.
The exponent is kn — periods per year times the number of years.
For depreciation, multiply by 1 − r/100 — a 15% fall keeps 85%.
To find a time, rearrange to ration = number, then take logs.
Convert a decimal year to months by multiplying the decimal part by 12.
âš Common mistakes
Forgetting the k — using r/100 and power n when interest compounds more than once a year.
Dividing r by k but not raising to kn — both changes are needed together.
Using the depreciation rate as the multiplier — 18% gives a factor of 0.82, not 0.18.
Rounding the time the wrong way — for “exceeds”, round the number of years up.
Reading a decimal year as months directly — 5.5 years is 5 years 6 months, not 5 years 5 months.
Next up: Amortisation — paying off a loan with fixed regular repayments while interest accrues on the balance. The work shifts from a formula to your GDC’s finance (TVM) solver, where signs and period settings matter.
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