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NotesMath AA SLTopic 1.4
Unit 1 · Number & Algebra · Topic 1.4

IB Math AA SL — Financial applications

Topic 1.4 of IB Mathematics: Analysis and Approaches covers Financial applications, which is part of Unit 1: Number & Algebra. Students explore key concepts including Compound interest, Depreciation, GDC finance solver. A strong understanding of financial applications is essential for IB Math AA SL exams and builds the foundation for connected topics across the syllabus.

Exam technique guidePractice questions

Key concepts in Financial applications

Key Idea: Money that grows or shrinks by the same percentage each period — savings, loans, depreciating cars. On Paper 2 the GDC's TVM solver does the heavy lifting; on Paper 1 you work it by hand.

📈 Compound interest

FV=PV(1+r100k)knFV = PV\left(1 + \frac{r}{100k}\right)^{kn}FV=PV(1+100kr​)kn
FVFVFV
future (final) value
PVPVPV
present value — the amount invested
rrr
annual interest rate as a percent
kkk
times compounded per year
nnn
number of years
Compound interest pays interest on the interest, so the balance grows geometrically. Per-period rate = r ÷ k, and the number of periods = k × n. Half-yearly k = 2, quarterly k = 4, monthly k = 12 — and annual is just k = 1, giving FV = PV(1 + r/100)ⁿ.

📉 Depreciation = decay

FV=PV(1−r100)nFV = PV\left(1 - \frac{r}{100}\right)^{n}FV=PV(1−100r​)n
PVPVPV
starting value
rrr
annual rate lost as a percent
nnn
number of years

✏️ IB-style worked examples

IB-style question — compound interest, compounded quarterly

$6000 is invested at 5% per year, compounded quarterly. Find its value after 4 years, to the nearest dollar.

Step by step:

  1. Quarterly → k = 4. Per-period rate = 5 ÷ 4 %; periods = 4 × 4 = 16.

    FV=6000(1+5100×4)4×4FV = 6000\left(1 + \frac{5}{100 \times 4}\right)^{4 \times 4}FV=6000(1+100×45​)4×4
  2. Simplify inside the bracket.

    =6000(1.0125)16= 6000(1.0125)^{16}=6000(1.0125)16
  3. Work it out on the GDC.

    =6000×1.21989=7319.36= 6000 \times 1.21989 = 7319.36=6000×1.21989=7319.36
Final answer:

≈ $7319.

IB-style question — depreciation from a model

A van is worth V = 28 000(0.85)ᵗ dollars, t years after purchase. (a) Write down the annual rate of depreciation. (b) Find its value after 5 years, to the nearest dollar.

Step by step:

  1. (a) The multiplier is 0.85, so the rate lost is 1 − 0.85.

    1−0.85=0.15=15%1 - 0.85 = 0.15 = 15\%1−0.85=0.15=15%
  2. (b) Substitute t = 5.

    V=28000×0.855V = 28000 \times 0.85^{5}V=28000×0.855
  3. Work it out on the GDC.

    =28000×0.443705=12423.7= 28000 \times 0.443705 = 12423.7=28000×0.443705=12423.7
Final answer:

(a) 15% per year. (b) ≈ $12 424.

IB-style question — TVM solver, find how long (Paper 2)

$9000 is invested at 3.6% per year, compounded monthly. Find the number of whole years until it first exceeds $11 000.

Step by step:

  1. Enter I% = 3.6, PV = −9000, PMT = 0, FV = 11000, P/Y = C/Y = 12. Solve for N.

    N=66.9 monthsN = 66.9\ \text{months}N=66.9 months
  2. Round up to a whole month, then convert to years.

    67 months=6 years67 \text{ months} = 6 \text{ years}67 months=6 years
  3. N is in months (P/Y = 12), so divide by 12 — and round up, a part-period doesn't count.

    67÷12=5.58→667 \div 12 = 5.58 \to 667÷12=5.58→6
Final answer:

6 years.

IB-style question — TVM solver, find the rate (Paper 2)

$4000 grows to $5000 in 5 years, compounded quarterly. Find the annual interest rate.

Step by step:

  1. List the TVM inputs — that list is your working. 5 years quarterly → N = 5 × 4 = 20.

    N=20, PV=−4000, PMT=0, FV=5000, P/Y=C/Y=4N = 20,\ PV = -4000,\ PMT = 0,\ FV = 5000,\ P/Y = C/Y = 4N=20, PV=−4000, PMT=0, FV=5000, P/Y=C/Y=4
  2. Leave I% blank, put the cursor on it and solve.

    I%≈4.49% (2 dp)I\% \approx 4.49\%\ \text{(2 dp)}I%≈4.49% (2 dp)
Final answer:

Annual rate ≈ 4.49%.

Important: In the TVM solver, money you pay out (invest) is negative — PV = −4000, not 4000 — or the solver errors. Match P/Y = C/Y to the question (quarterly → 4, monthly → 12), and remember N counts periods: N = years × frequency. When the answer is N, round up for "how long until".

Tap each card to reveal the answer.

Exam Tips

  • Compound interest: FV = PV(1 + r/(100k))ᵏⁿ — divide the rate by k, raise to k × n.
  • Depreciation multiplies by (1 − rate) each year; rate lost = 1 − multiplier.
  • TVM signs: money invested → PV negative; PMT = 0 when there are no regular deposits.
  • Set P/Y = C/Y = compounding frequency, and N = years × frequency.
  • Paper 1 → work by hand; Paper 2 → leave the unknown blank, solve with ALPHA + ENTER (round N up).

What you'll learn in Topic 1.4

  • 1.4.1 Compound interest
  • 1.4.2 Depreciation
  • 1.4.3 GDC finance solver
Suggested study order: Read the notes for each sub-topic below → test yourself with flashcards → attempt practice questions → review exam technique.

Study resources — 1.4 Financial applications

1.4.1

Compound interest

Notes
1.4.2

Depreciation

Notes
1.4.3

GDC finance solver

Notes

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Topic 1.4 Financial applications forms a core part of Unit 1: Number & Algebra in IB Math AA SL. Mastering these concepts will strengthen your understanding of connected topics across the syllabus and prepare you for exam questions that require analysis, evaluation, and real-world application.

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