IB Maths AI SLTopic 1 — Financial ApplicationsPaper 1 & 2In formula booklet~7 min read
Compound Interest & Depreciation
Compound interest grows money by multiplying each period; depreciation shrinks it the same way. One formula does both — just flip + to − for depreciation.
📘 What you need to know
Compound interest = interest paid on both the original amount AND previously earned interest. Grows faster than simple interest.
Formula booklet: FV = PV × (1 + r/(100k))kn
FV = future value, PV = present value, r = nominal annual rate (%), k = compounding periods per year, n = years.
Why two formulae? Compound depreciation usually happens annually, so k = 1 and the (1 − …) shows things going DOWN. If your GDC finance solver does depreciation, just enter the rate as a NEGATIVE number.
🧭 Recipe — any compound interest / depreciation problem
Substitute and evaluate on the GDC. Keep at least 4 d.p. in working.
Round to the question’s requirement: money 2 d.p., otherwise 3 s.f. by default.
Worked examples
WE 1
Compound interest — annual compounding
$5000 is invested at 4% per annum, compounded annually. Find the value after 10 years.
List the valuesPV = 5000, r = 4, k = 1, n = 10Apply formulaFV = 5000 × (1 + 4/100)^10 = 5000 × 1.04^10 = 5000 × 1.4802…FV = $7401.22
WE 2
Find PV — work backwards
Sarah wants $20 000 in 6 years. Her bank pays 4% nominal annual interest, compounded quarterly. How much should she invest now?
List the valuesFV = 20 000, r = 4, k = 4, n = 6Rearrange formula for PV20 000 = PV × (1 + 4/400)^(4×6)20 000 = PV × 1.01^24PV = 20 000 / 1.26973…PV = $15 751.32smaller PV needed because compound growth does the work.
WE 3
Monthly compounding
¥500 000 is invested at 1.8% nominal annual interest, compounded monthly. Find the value after 3 years.
List valuesPV = 500 000, r = 1.8, k = 12, n = 3Apply formulaFV = 500 000 × (1 + 1.8/(100×12))^(12×3) = 500 000 × (1.0015)^36 = 500 000 × 1.05544…FV = ¥527 720.95monthly compounding: divide r by 12 AND multiply n by 12 in the exponent.
WE 4
Compound depreciation
A laptop costs $1200 new and depreciates 18% per year. Find its value after 4 years.
List valuesPV = 1200, r = 18, n = 4Apply depreciation formulaFV = 1200 × (1 − 18/100)^4 = 1200 × 0.82^4 = 1200 × 0.4521…FV = $542.55
WE 5
Find n for depreciation
A machine worth €25 000 depreciates at 12% per year. After how many full years does its value first drop below €10 000?
Set up inequality25 000 × 0.88^n < 10 0000.88^n < 0.4Take logs (sign flips because log 0.88 < 0)n > log(0.4) / log(0.88)n > 7.168Check whole-year valuesat n=7: 25 000 × 0.88^7 = €10 217 (still above)at n=8: 25 000 × 0.88^8 = €8991 (below ✓)n = 8 yearsalways check the integer either side — exam answer must be a whole year here.
WE 6
Compare compounding frequencies
$10 000 is invested at 5% per annum for 5 years. Compare the future value when interest is compounded (a) annually and (b) monthly.
(a) Annual (k = 1)FV = 10 000 × (1 + 5/100)^5 = 10 000 × 1.05^5 = $12 762.82(b) Monthly (k = 12)FV = 10 000 × (1 + 5/1200)^60 = 10 000 × 1.00417^60 = $12 833.59monthly wins by $70.77same rate, same time, same start — but more frequent compounding always gives MORE money.
💡 Top tips
k matters: monthly (k=12) gives slightly more than annual (k=1) at the same nominal rate.
Use GDC’s TVM solver as a check — most have one. For depreciation, enter rate as negative.
Round at the END: keep 4+ d.p. in working so the final 2 d.p. (or 3 s.f.) answer is correct.
Money in (PV) is positive; on a finance solver, money OUT (a loan) would be entered as negative.
⚠ Common mistakes
Forgetting to divide r by k: at 6% compounded monthly, each month gets 0.5%, not 6%.
Forgetting to multiply n by k in the exponent: 5 years monthly = exponent 60, not 5.
Using compound interest formula for depreciation with + sign: depreciation uses (1 − r/100), not (1 + r/100).
Confusing “loses 18%” with r = 0.18: r is the rate (=18), not the decimal. Then the formula uses 1 − 18/100 = 0.82.
Up next: Amortisation — when you take a loan and pay it back in fixed monthly chunks. Uses the GDC’s finance solver (TVM) rather than a single formula.
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