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Finance, Growth & Decay

Grade 11 CAPS: simple and compound depreciation, nominal vs effective interest rates, the effect of compounding periods, and multi-stage investment problems. Includes real DBE and provincial exam questions.

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Grade 11 CAPS Mathematics

Finance, Growth & Decay

From straight-line depreciation to reducing balance; from nominal rates to effective rates; from a single compounding period to a full multi-stage timeline. Work through the slides in order, then move to the practice bank.

Scope check:Official CAPS requirement

Your 4-hour learning route

Decay first, then two ways rates and time can complicate a calculation.

50
min
1. Simple & compound decay

Straight-line vs reducing balance.

45
min
2. Solving for the rate

Working backward from two known values.

45
min
3. Nominal vs effective rates

Comparing rates fairly across compounding frequencies.

40
min
4. Compounding periods

How frequency changes growth AND decay.

40
min
5. Multi-stage problems

Deposits, withdrawals and rate changes on one timeline.

Study rule
Every slide from here on builds on Grade 10's simple/compound growth formulae. If those two formulas aren't automatic yet, revisit Grade 10 Finance & Growth first.

What CAPS actually asks in Grade 11

Decay, fair rate comparison, and problems with more than one stage.

  • 1

    Solve problems involving simple and compound depreciation for calculating the depreciation of value of an asset.

  • 2

    Calculate the value of \(n\) in the equation \(x=\left(1+\dfrac{i}{n}\right)^n\), and use it to calculate effective and nominal interest rates and convert between them, for varying compounding periods.

  • 3

    Make sensible decisions regarding the use and impact of the effect of different periods of compounding growth and decay.

Not yet
Solving for \(n\) using logarithms, sinking funds, and annuities are Grade 12 content — this page stops short of those.

Quick revision: Grade 10 growth

The two formulas every decay formula below is built to mirror.

Simple growth (revision)
\[A=P(1+in)\]

The original amount grows by the same RAND amount every year.

Compound growth (revision)
\[A=P(1+i)^n\]

Each year's growth is a percentage of the ALREADY-GROWN previous total.

The pattern
Every decay formula below simply flips the sign inside the bracket: growth ADDS a percentage, decay SUBTRACTS one.

Simple decay: straight-line depreciation

The mirror image of simple growth.

Straight-line depreciation
\[A=P(1-in)\]

The asset loses the SAME rand amount every year, calculated from the ORIGINAL cost \(P\). This is why it can reach exactly R0 if depreciated for long enough.

Compound decay: reducing-balance depreciation

The mirror image of compound growth.

Reducing-balance depreciation
\[A=P(1-i)^n\]

Each year's loss is a PERCENTAGE of the ALREADY-REDUCED previous book value. This is why it can never reach exactly R0 — a percentage of a positive number is never the whole amount.

Worked example: the two methods side by side

Same asset, same rate — very different remaining values.

R20 000 over 4 years Level 1–2

An asset costing R20 000 depreciates for 4 years at 12% p.a. Calculate how much MORE the reducing-balance value is than the straight-line value.

Show both calculations
  1. 1Straight-line: \(20\,000(1-0{,}12\times4)=\boxed{R10\,400{,}00}\)
  2. 2Reducing balance: \(20\,000(1-0{,}12)^4\approx\boxed{R11\,993{,}91}\)
  3. 3Difference: \(11\,993{,}91-10\,400{,}00=\boxed{R1\,593{,}91}\) — reducing balance always retains MORE value beyond the first year.

Why reducing balance retains more value

A straight line falling at a constant rate vs a curve that flattens out.

R50 000 depreciated at 8% p.a. for 10 years: R21 719 remaining under reducing balance vs only R10 000 under straight-line. Once a straight-line asset gets close to R0, reducing balance is still shedding a shrinking rand amount each year.

Worked example: solving for the straight-line rate

Rearranging the formula to make \(i\) the subject.

A computer's depreciation rate Level 2

A computer cost R18 000 and has a book value of R9 000 after 3 years, depreciated on a straight-line basis. Calculate the annual rate of depreciation.

Show the complete working
  1. 1\(9\,000=18\,000(1-3i)\)
  2. 2\(1-3i=0{,}5\Rightarrow3i=0{,}5\)
  3. 3\(\boxed{i\approx16{,}67\%}\) — don't forget the final division by \(n=3\).

Worked example: solving for the reducing-balance rate

A genuine DBE exam question — this time an nth root is unavoidable.

A car's depreciation rate Level 2–3

A car bought for R175 000 sells 4 years later for R79 120. Calculate the annual reducing-balance depreciation rate.

Show the complete exam working
  1. 1\(79\,120=175\,000(1-i)^4\)
  2. 2\((1-i)=\left(\dfrac{79\,120}{175\,000}\right)^{\frac14}\)
  3. 3\(\boxed{i\approx18{,}00\%}\) — solving for a rate INSIDE an exponent always needs a root, never a straight division by \(n\).
Real exam question:North West DBE, Paper 1, November 2025 (Q8.2.3) — independently re-derived and confirmed to match the official memo.

Worked example: a six-year school bus

A genuine provincial topic-test question — straightforward substitution once the method is clear.

A depreciating school bus Level 2

A school bought a school bus for R850 000. The bus depreciates at 12,5% p.a. on the reducing-balance method. Calculate the book value 6 years after it was bought.

Show the working
\(A=850\,000(1-0{,}125)^6=850\,000(0{,}875)^6\approx\boxed{R381\,476{,}02}\)
Real exam question:KwaZulu-Natal Provincial Topic Test, Finance, Growth & Decay, September 2025 (Q1.2) — independently re-derived and confirmed to match the official memo.
Quick check: simple & compound decay

An asset loses the same PERCENTAGE of its remaining (already-reduced) value every year. Which method is this?

Solving for the rate \(i\) inside \(A=P(1-i)^n\) requires:

Worked example: depreciation and a trade-in decision

First a simple trade-in, then a real-life case where the trade-in value is never the original price.

A simple trade-in Level 2

A car costing R200 000 depreciates on a reducing balance at 20% p.a. After 3 years, it is traded in against a new car costing R280 000. Calculate the amount still to be paid.

Show the working
  1. 1Trade-in value: \(200\,000(1-0{,}20)^3=\boxed{R102\,400{,}00}\)
  2. 2Amount still to pay: \(280\,000-102\,400=\boxed{R177\,600{,}00}\)
Trading in a car Level 4

A car costing R320 000 depreciates on a reducing balance at 19% p.a. After 4 years, it is traded in against a new car costing R450 000. Calculate the amount still to be paid after the trade-in value is deducted.

Show the complete working
  1. 1Trade-in value: \(320\,000(1-0{,}19)^4\approx R137\,749{,}51\)
  2. 2Amount still to pay: \(450\,000-137\,749{,}51=\boxed{R312\,250{,}49}\)
Common trap
Never assume the trade-in equals the original purchase price — always calculate the actual depreciated value first.

Nominal vs effective interest rates

Why "9,5% p.a." doesn't always mean what it seems to.

Nominal rate

The quoted annual rate, e.g. "9,5% p.a. compounded quarterly." On its own it does NOT tell you the true annual growth.

Effective rate
\[1+i_{\text{eff}}=\left(1+\dfrac{i_{\text{nom}}}{m}\right)^m\]

The TRUE annual growth once compounding is accounted for, where \(m\) is the number of compounding periods per year.

Why it matters
Two investments with different nominal rates AND different compounding frequencies can only be fairly compared once both are converted to effective annual rates.

Worked example: converting nominal to effective

A genuine provincial topic-test question.

Quarterly compounding Level 2

Calculate the effective annual interest rate if an investment earns interest at a rate of 9,5% p.a. compounded quarterly.

Show the working
  1. 1\(1+i_{\text{eff}}=\left(1+\dfrac{0{,}095}{4}\right)^{4}\)
  2. 2\(\boxed{i_{\text{eff}}\approx9{,}84\%}\) — higher than the quoted 9,5%, because compounding within the year adds extra growth.
Real exam question:KwaZulu-Natal Provincial Topic Test, Finance, Growth & Decay, September 2025 (Q1.1) — independently re-derived and confirmed to match the official memo.

Worked example: comparing two rates fairly

A genuine DBE exam question — the higher nominal rate does NOT automatically win.

Daily vs monthly compounding Level 3

Option A pays 8,2% p.a. compounded daily (365 days). Option B pays 8,3% p.a. compounded monthly. Which option is the better investment?

Show the complete exam working
  1. 1Option A: \(\left(1+\dfrac{0{,}082}{365}\right)^{365}-1\approx8{,}54\%\)
  2. 2Option B: \(\left(1+\dfrac{0{,}083}{12}\right)^{12}-1\approx8{,}62\%\)
  3. 3\(\boxed{\text{Option B is better}}\) — daily compounding (Option A) does not automatically win just because it compounds more often.
Real exam question:North West DBE, Paper 1, November 2025 (Q8.1.1 & Q8.1.2) — independently re-derived and confirmed to match the official memo.
Quick check: nominal & effective rates

"Compounded quarterly" means which value of \(m\) in the effective-rate formula?

Two investment options have different nominal rates AND different compounding frequencies. To fairly compare them, you should:

The effect of compounding periods

More frequent compounding always helps growth — but it HURTS an asset's value under decay.

Counter-intuitive but true
For GROWTH, more frequent compounding of the same nominal rate always increases the effective rate — you earn more. For DECAY, more frequent compounding of the same nominal rate means the asset loses value MORE slowly overall, so it's worth MORE at the end, not less.

Worked example: monthly vs annual decay

First the predictable case (growth), then the counter-intuitive one (decay).

Growth, two ways — the predictable case Level 2

R100 000 grows at a NOMINAL rate of 12% p.a. for 2 years. Calculate its value if the growth is compounded (a) monthly and (b) annually, and state which method leaves it worth MORE.

Show the working
  1. 1Monthly: \(i=\dfrac{0{,}12}{12}\), \(n=24\): \(A=100\,000(1+i)^{24}\approx\boxed{R126\,973{,}46}\)
  2. 2Annually: \(A=100\,000(1{,}12)^2=\boxed{R125\,440{,}00}\)
  3. 3\(\boxed{\text{Monthly compounding leaves it worth more}}\) — matching the intuition that more frequent compounding always helps growth.
Equipment depreciating two ways Level 3–4

Equipment costing R200 000 depreciates at a NOMINAL rate of 16% p.a. on a reducing balance for 3 years. Calculate its book value if the depreciation is calculated (a) monthly and (b) annually, and state which method leaves the equipment worth MORE.

Show the complete working
  1. 1Monthly: \(i=\dfrac{0{,}16}{12}\), \(n=36\): \(A=200\,000(1-i)^{36}\approx\boxed{R123\,357{,}75}\)
  2. 2Annually: \(A=200\,000(1-0{,}16)^3\approx R118\,540{,}80\)
  3. 3\(\boxed{\text{Monthly compounding leaves it worth more}}\) — the opposite of what happens with growth.

Worked example: compound decay in reverse

A genuine DBE exam question — the same formula, a non-financial context.

A declining bee population Level 3

The number of bees in an apiary is declining each year at a rate of 2,5% of the total the previous year. There are currently 20 416 bees. Calculate how many bees there were 8 years ago.

Show the complete exam working
  1. 1\(A=P(1-i)^n\Rightarrow P=\dfrac{A}{(1-i)^n}\)
  2. 2\(P=\dfrac{20\,416}{(0{,}975)^8}\approx\boxed{25\,000\text{ bees}}\)
Real exam question:KwaZulu-Natal DBE, Paper 1, November 2024 (Q8.2) — independently re-derived and confirmed to match the official memo.
Key idea
Compound decay applies to ANY quantity that shrinks by a constant percentage each period — not just money or asset values.
Quick check: compounding periods

The SAME nominal decay rate is compounded monthly instead of annually. The asset's final book value will be:

A population declines by 2,5% p.a. To find the population 8 years ago from today's value, you should:

Multi-stage problems: draw the timeline first

Deposits, withdrawals and rate changes, all on one investment.

Before calculating anything

Mark every year on a timeline where something CHANGES: a deposit, a withdrawal, or a new interest rate. Calculate the balance stage by stage, left to right — never try to jump straight to the final answer with one formula.

Common trap
Misreading WHEN a withdrawal happens (e.g. year 2 vs year 4) changes every calculation after it. Read the timing of every event twice before starting.

Worked example: a full multi-stage investment

Start with just a rate change, then a genuine DBE exam question with four stages on one timeline.

A simpler 2-stage investment: just a rate change Level 2–3

R100 000 is invested at 8% p.a. compounded quarterly for 3 years. The rate then changes to 10% p.a. compounded monthly for a further 2 years. Calculate the balance after 5 years in total.

Show the working
  1. 1Grow at 8% quarterly for 3 years: \(100\,000(1{,}02)^{12}\approx R126\,824{,}18\)
  2. 2Switch to 10% monthly for 2 more years: \(126\,824{,}18\left(1+\dfrac{0{,}10}{12}\right)^{24}\approx\boxed{R154\,775{,}08}\)
A 6-year investment with a withdrawal and a rate change Level 4

R300 000 is invested at 9% p.a. compounded quarterly. After exactly 2 years, R175 000 is withdrawn. The remaining balance continues at 9% p.a. compounded quarterly until year 4, when the rate changes to 8,5% p.a. compounded monthly for a further 2 years. Calculate the balance after 6 years in total.

Yr 0-2
Grow at 9% quarterly\(300\,000(1{,}0225)^8\approx R358\,449{,}34\)
Yr 2
Withdraw R175 000\(358\,449{,}34-175\,000=R183\,449{,}34\)
Yr 2-4
Continue at 9% quarterly\(183\,449{,}34(1{,}0225)^8\approx R219\,190{,}99\)
Yr 4-6
Switch to 8,5% monthly\(219\,190{,}99\left(1+\dfrac{0{,}085}{12}\right)^{24}\approx\boxed{R259\,652{,}50}\)
Real exam question:North West DBE, Paper 1, November 2025 (Q8.2.1 & Q8.2.2, adapted) — independently re-derived and confirmed to match the official memo.
Quick check: multi-stage problems

Before calculating a multi-stage investment problem, the first thing you should do is:

A withdrawal happens partway through an investment. What must you do to the balance at that point?

Mistake clinic: repair the exact error

Small slips that break an otherwise correct answer.

Formula choice

Straight-line ≠ reducing balance

"Percentage of what's LEFT" is reducing balance. "Same rand amount, based on the ORIGINAL cost" is straight-line.

Solving for i

Root, not division

A rate trapped inside \((1-i)^n\) needs an nth root — dividing by \(n\) only works for the straight-line formula.

Comparing rates

Convert before comparing

Never compare nominal rates directly if their compounding frequencies differ — convert both to effective rates first.

Compounding & decay

More frequent ≠ faster decay

For decay, more frequent compounding of the same nominal rate leaves MORE value at the end, not less.

Multi-stage

Timing is everything

Misreading WHEN a withdrawal or rate change happens changes every calculation after it.

Your Paper 1 checklist

Use this before submitting an exam answer.

Avoid this

Confusing straight-line and reducing-balance decay.

Comparing nominal rates without converting to effective first.

Losing track of when a withdrawal or rate change happens.

Do this

Ask "original cost, or already-reduced value?" first.

Convert every rate to an effective annual rate before comparing.

Draw a timeline for any problem with more than one stage.

Apply both straight-line and reducing-balance depreciation formulas.
Solve for an unknown depreciation rate, including using an nth root.
Convert between nominal and effective interest rates.
Explain the effect of compounding periods on both growth and decay.
Solve a multi-stage problem involving deposits, withdrawals and rate changes.
More explanation and exercises:Siyavula Grade 11 Finance, Growth and Decay
Summary complete

You now have the route.

Use the Mastery Bank to build fluency. Then take the test without notes and use the result to choose the exact slide to revisit.

Next step 1

Mastery Bank

17 questions, including 6 real DBE/provincial exam questions, from a direct decay calculation through to a full multi-stage investment. Answers reveal only after your attempt.

Next step 2

Quick Test

A short exam-style self-check. Mark it, then revisit the exact slide that matches a missed skill.

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Learn It in Short Videos

Four free, independent videos — not made by Equation Station SA.

Depreciation: Grade 11

Straight-line and reducing-balance depreciation problems, worked through.

Kevinmathscience · Financial maths grade 11, Depreciation

Effective & Nominal Rates

Converting between nominal and effective interest rates, step by step.

Kevinmathscience · Financial Maths Grade 11, Effective to Nominal

Depreciation: A Second Explainer

Another full walkthrough of straight-line vs reducing-balance depreciation.

Online Maths by Miss Pythagoras · Financial Mathematics Grade11: L3 Depreciation

Past Paper Practice

Real Grade 11 exam-style finance questions, worked from scratch.

The Maths Shack · FINANCIAL MATHS (Grade 11) South Africa (GDE) past paper finance questions

Practise in the right order

The core teaching is above. These are the next steps, not a replacement for it.

01
Built-in practice
Finance, Growth & Decay Mastery Bank

17 questions by skill, including 6 real DBE/provincial exam questions, with concise reveal answers.

Start after the slides
Open Mastery Bank
02
Built-in check
Test Your Knowledge

Use the short exam-style self-check when you want a fast confidence check.

Then target one weak skill
Take the Quick Test
CAPS
Free textbook chapter
Siyavula: Grade 11 Finance

Use its own worked examples for extra explanation and exercises.

Free • CAPS aligned
Open Siyavula
DBE
Official free books
DBE Grade 11 Textbooks

Official state-owned learner books and teacher support for Grade 11 Mathematics.

Official • free access
Open DBE Books
PDF
Printable worksheet + memo
Maths At Sharp: Worksheet 9

A free downloadable CAPS worksheet on finance, growth and decay, with a fully worked memorandum.

Free • download & print
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Related Topics

This topic continues from Grade 10 into Grade 12.

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Finance & Growth (Grade 10)

Simple & compound growth, hire purchase

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Finance, Growth & Decay (Grade 12)

Annuities, loans, critical analysis

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Algebra (Grade 11)

Surds, quadratics, inequalities

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Student Questions

Frequently Asked Questions

Short answers for the checks learners make while preparing for the Grade 11 CAPS exam.

What does CAPS require for Grade 11 Finance, Growth and Decay?

Solve problems involving simple and compound depreciation; calculate nominal and effective interest rates and convert between them; understand the effect that different periods of compounding growth and decay have on a given investment or loan; and solve multi-stage financial problems.

What is the difference between straight-line and reducing-balance depreciation?

Straight-line depreciation subtracts the same rand amount from the original cost every year: A = P(1 - in). Reducing-balance depreciation subtracts a percentage of the previous year's already-reduced value every year: A = P(1 - i)^n, and can never reach exactly R0.

Is solving for n using logarithms part of Grade 11?

No. Calculating the value of n in A = P(1 + i)^n or A = P(1 - i)^n using logarithms is Grade 12 content.

Where should I practise next?

Finish the interactive slides, open the Mastery Bank, then take the short self-test without notes.