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Finance, Growth & Decay

Grade 12 CAPS: solve for time using logarithms, build annuities from a geometric series, find an outstanding balance, and critically analyse a loan or investment—including spotting a pyramid scheme. Learn it as one connected route—not a list of formulas.

1. Guided Course 2. Mastery Bank 3. Quick Test
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Grade 12 CAPS Mathematics

Finance, Growth & Decay

From a growth formula to a full annuity; from an annuity to a loan; from a loan to the decision of whether it is actually a good one. Work through the slides in order, then move to the practice bank.

Scope check:Official CAPS requirement

Your 5-hour learning route

A focused plan: revise the Grade 10/11 tools tightly, then build the new Grade 12 skills on top of them.

45
min
1. Reset growth & decay

Simple/compound formulas, hire purchase, depreciation, effective rates.

30
min
2. Solve for n

Bring in logarithms, then round the time period correctly.

90
min
3. Build annuities

Future value, present value, deferred deposits, outstanding balance.

45
min
4. Judge the deal

Compare loan options honestly and spot a pyramid scheme.

45
min
5. Recall under pressure

Complete the Mastery Bank, then the short self-test.

Study rule
Draw a timeline for every annuity or loan question before writing a single formula. It is the one habit that prevents almost every mistake in this topic.

What CAPS actually asks in Grade 12

Three genuinely new skills, built on Grade 10/11 growth and decay.

  • 1

    Use logarithms to solve for \(n\). Find the time period in \(A=P(1+i)^n\) or \(A=P(1-i)^n\).

  • 2

    Apply geometric series knowledge to annuity and bond repayment problems. Derive and use the future and present value annuity formulas.

  • 3

    Critically analyse investment and loan options and make an informed decision—including recognising a pyramid scheme.

Assumed, not re-taught
Simple and compound growth (Grade 10), simple and compound decay, straight-line and reducing-balance depreciation, and nominal vs effective rates (Grade 11) are examined cumulatively in Grade 12. The next few slides revise them tightly before building on top.

Revision: growth and decay, side by side

Four formulas, one pattern: a multiplier vs an exponent, a plus vs a minus.

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

Interest on the original amount only, every period. Used for basic savings and hire purchase.

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

Interest earns interest. Used for real investments and population/inflation growth.

Simple decay
\[A=P(1-in)\]

The same rand amount is lost every period. This is straight-line depreciation.

Compound decay
\[A=P(1-i)^n\]

Each period's loss is a percentage of what's left. This is reducing-balance depreciation.

Reading the words
"Every year, on the original amount" → simple (\(n\) is a multiplier). "Compounded" or "on the reducing balance" → compound (\(n\) is an exponent).

Worked example: simple vs compound growth

Same principal, same rate—watch the two formulas diverge.

Compare after 3 years Level 1–2

R8 000 is invested at 8% p.a. Compare the value after 3 years under simple interest and under compound interest.

Show both calculations
  1. 1Simple: \(A=8\,000(1+0{,}08\times3)=8\,000(1{,}24)=\boxed{R9\,920{,}00}\)
  2. 2Compound: \(A=8\,000(1{,}08)^3=\boxed{R10\,077{,}70}\)
  3. 3Both give the same growth in year 1. By year 3, compounding has earned an extra \(R157{,}70\) because year 2 and 3's interest is calculated on a growing balance, not the original R8 000.

Revision: hire purchase

Deposit, then simple interest on the balance only—never on the full cash price.

1
Deposit

Subtract it from the cash price first.

2
Interest

Simple interest, on the balance only.

3
Instalment

(Balance + interest) ÷ number of months.

Full hire purchase calculation Level 2

A fridge has a cash price of R12 000. A 10% deposit is required, with the balance repaid at 12% p.a. simple interest over 2 years in equal monthly instalments.

Show the complete working
  1. 1Deposit \(=0{,}10\times12\,000=R1\,200\); balance financed \(=R10\,800\).
  2. 2Interest \(=10\,800\times0{,}12\times2=R2\,592\).
  3. 3Total to repay \(=10\,800+2\,592=R13\,392\).
  4. 4Monthly instalment over 24 months: \(\boxed{R558{,}00}\).
Common mistake
Hire purchase interest is always simple interest, and it is always calculated on the balance after the deposit—never on the full cash price, and never compounded.

Revision: depreciation, two methods

Straight-line loses the same rand amount every year. Reducing balance loses a percentage of what's left.

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

A straight line—the same amount every year. Can reach R0.

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

A curve that flattens out—smaller losses each year. Never actually reaches R0.

Worked example: comparing the two methods

Same asset, same rate, same 5 years—very different book values.

Straight-line vs reducing balance Level 2

A machine costs R180 000 and both methods use a rate of 15% p.a. Compare the book value after 5 years.

Show both calculations
  1. 1Straight-line: \(A=180\,000(1-0{,}15\times5)=180\,000(0{,}25)=\boxed{R45\,000{,}00}\)
  2. 2Reducing balance: \(A=180\,000(1-0{,}15)^5=180\,000(0{,}85)^5=\boxed{R79\,866{,}96}\)
  3. 3Both methods depreciate the asset by the same R27 000 in year 1—they only diverge from year 2 onward, once reducing balance starts working off a smaller base.

Revision: nominal vs effective interest rates

The rate you're quoted is not always the rate you actually earn.

Converting nominal to effective
\[1+i_{\text{eff}}=\left(1+\dfrac{i_{\text{nom}}}{m}\right)^{m}\]

\(m\) is the number of times the rate compounds per year (12 for monthly, 4 for quarterly, 2 for semi-annually).

Why it matters
The more often interest compounds, the higher the true (effective) annual rate—even if the quoted (nominal) rate stays the same. Never compare two nominal rates directly if they compound at different frequencies.

Worked example: which rate actually wins?

A higher compounding frequency does not automatically mean a better return.

Compare two real offers Level 2–3

Option A: 8,9% p.a. compounded monthly. Option B: 9,1% p.a. compounded quarterly. Which is the better investment?

Show both effective rates
  1. 1Option A: \(\left(1+\dfrac{0{,}089}{12}\right)^{12}-1\approx\boxed{9{,}27\%}\)
  2. 2Option B: \(\left(1+\dfrac{0{,}091}{4}\right)^4-1\approx\boxed{9{,}42\%}\)
  3. 3Option B is the better investment, even though A compounds more often—its higher nominal rate wins once both are converted fairly.
Quick check: revision

A savings account pays interest that itself earns interest every year. Which formula applies?

Which depreciation method can reduce an asset's book value all the way to R0?

Two accounts both advertise "9% p.a." One compounds monthly, the other compounds annually. What is true?

The timeline: your most useful tool

CAPS itself names this technique—draw one before setting up any annuity or loan equation.

Ordinary annuity rule
Both annuity formulas assume the FIRST deposit happens one period from now (not at "Now" itself), and the LAST deposit happens exactly at the final period. No deposit is ever made at time zero.

Solving for n: bringing in logarithms

The new Grade 12 skill: find the TIME, not the final value.

Isolate the power, then take logs
\[A=P(1+i)^n\ \Rightarrow\ \dfrac{A}{P}=(1+i)^n\ \Rightarrow\ n=\dfrac{\log\left(\dfrac{A}{P}\right)}{\log(1+i)}\]

The same rearrangement works for decay, using \((1-i)\) instead of \((1+i)\).

The rounding trap
A logarithm almost never gives a whole number. Interest is only credited at the end of a full period, so you must check the year below AND the year above your raw answer, then choose the correct whole period—never leave a decimal year as your final answer.

Worked example: solve for n, both directions

Growth rounds up to reach a target. Decay rounds up to drop below one.

Growth

How long to reach R25 000?

R15 000 invested at 9% p.a. compounded annually.

\[n=\dfrac{\log(25\,000/15\,000)}{\log1{,}09}\approx5{,}93\]

At \(n=5\): R23 079{,}36 (short). At \(n=6\): R25 156{,}50 (reached). \(\boxed{6\text{ years}}\)

Decay

How long to drop to R60 000?

A car costing R220 000 depreciates on a reducing balance at 18% p.a.

\[n=\dfrac{\log(60\,000/220\,000)}{\log0{,}82}\approx6{,}55\]

At \(n=6\): R66 881{,}47 (still above). At \(n=7\): R54 842{,}80 (below). \(\boxed{7\text{ years}}\)

Quick check: solving for n

You solve for \(n\) with logarithms and get \(n=4{,}18\). The question asks "after how many complete years". What should you do?

Which rearrangement of \(A=P(1-i)^n\) correctly isolates \(n\)?

From geometric series to annuities

An annuity is just a series of deposits—so the geometric series formula builds it.

Every deposit grows for a different number of periods

The LAST deposit earns no interest yet; the deposit before that earns one period of interest; and so on back to the FIRST deposit, which earns \(n-1\) periods of interest.

\[F=x+x(1+i)+x(1+i)^2+\cdots+x(1+i)^{n-1}\]

This is a geometric series with first term \(x\), ratio \((1+i)\), and \(n\) terms.

Sum it, and the annuity formula appears
\[F=x\cdot\dfrac{(1+i)^n-1}{i}\]

This is exactly the geometric series sum formula \(S_n=\dfrac{a\left((1+i)^n-1\right)}{(1+i)-1}\), with \(a=x\).

Worked example: future value annuity

First one year of deposits, then a full 6-year savings goal.

Direct substitution, one year Level 1

R500 is deposited at the end of every month into an account earning 9% p.a. compounded monthly. Calculate the value of the account after 1 year.

Show the working
  1. 1\(i=\dfrac{0{,}09}{12}=0{,}0075\), \(n=12\) monthly deposits.
  2. 2\(F=x\cdot\dfrac{(1+i)^n-1}{i}=500\times\dfrac{(1{,}0075)^{12}-1}{0{,}0075}\)
  3. 3\(\boxed{R6\,253{,}79}\)
Saving toward a goal Level 2–3

R750 is deposited at the end of every month into an account earning 8,4% p.a. compounded monthly. Calculate the value of the account after 6 years.

Show the complete working
  1. 1\(i=\dfrac{0{,}084}{12}=0{,}007\), \(n=6\times12=72\) monthly deposits.
  2. 2\(F=x\cdot\dfrac{(1+i)^n-1}{i}=750\times\dfrac{(1{,}007)^{72}-1}{0{,}007}\)
  3. 3\(\boxed{R69\,902{,}73}\)
Rate and period must match
Deposits are monthly, so both the rate AND the number of periods must be converted to months—never mix an annual n with a monthly i.

Present value annuities: a loan is a promise of future payments

Same geometric series, but discounting backward instead of growing forward.

A loan equals the present value of every instalment
\[P=x(1+i)^{-1}+x(1+i)^{-2}+\cdots+x(1+i)^{-n}=x\cdot\dfrac{1-(1+i)^{-n}}{i}\]

The bank hands over \(P\) today because it's worth exactly the same, in today's money, as all \(n\) future instalments discounted back.

Same skill, opposite direction
Future value asks "what will regular deposits grow into?" Present value asks "what lump sum today is equivalent to these regular payments?" Both use the identical timeline and the identical \(x\), \(i\), \(n\).

Worked example: finding a loan instalment

First a short 1-year loan, then the exact method behind every car and home loan repayment.

Direct substitution, one year Level 1

A loan of R20 000 is taken out at 15% p.a. compounded monthly, to be repaid in equal monthly instalments over 1 year. Calculate the monthly instalment.

Show the working
  1. 1\(i=\dfrac{0{,}15}{12}=0{,}0125\), \(n=12\).
  2. 2\(x=\dfrac{Pi}{1-(1+i)^{-n}}=\dfrac{20\,000\times0{,}0125}{1-(1{,}0125)^{-12}}\)
  3. 3\(\boxed{R1\,805{,}17}\) per month.
Solve for the instalment Level 2–3

A loan of R180 000 is taken out at 13% p.a. compounded monthly, to be repaid in equal monthly instalments over 5 years. Calculate the monthly instalment.

Show the complete working
  1. 1\(i=\dfrac{0{,}13}{12}\approx0{,}010833\), \(n=5\times12=60\).
  2. 2From \(P=x\cdot\dfrac{1-(1+i)^{-n}}{i}\), make \(x\) the subject: \(x=\dfrac{Pi}{1-(1+i)^{-n}}\).
  3. 3\(x=\dfrac{180\,000\times0{,}010833}{1-(1{,}010833)^{-60}}\)
  4. 4\(\boxed{R4\,095{,}55}\) per month.

When the timeline doesn't line up neatly

Both annuity formulas need the first deposit exactly one period from now. When it isn't, adjust the timeline first.

The fix
Never assume the number of deposits equals the total number of years. Count the actual deposit arrows on your own timeline—that count is your \(n\) for the annuity formula.

Worked example: a deferred annuity

The single most common Level 4 trap in this topic.

Deposits start late Level 4

A parent wants R500 000 saved in exactly 10 years' time. They can only start depositing money 2 years from now, and will then deposit an equal amount at the end of every month until the 10-year mark, into an account earning 9,6% p.a. compounded monthly. Calculate the required monthly deposit.

Show the complete working
  1. 1Deposits run from month 25 to month 120: that is \(120-24=96\) deposits, not 120.
  2. 2\(i=\dfrac{0{,}096}{12}=0{,}008\).
  3. 3\(x=\dfrac{500\,000\times i}{(1{,}008)^{96}-1}\)
  4. 4\(\boxed{R3\,481{,}67}\) per month.
Quick check: annuities

Which formula would you use to find the monthly amount a company must save now to have a lump sum available in 5 years?

A loan's monthly instalments are being calculated. What does \(P\) represent in \(P=x\cdot\dfrac{1-(1+i)^{-n}}{i}\)?

Outstanding balance: the present value of what's left

This is the bridge to bond and loan repayment problems.

Never a simple fraction of the loan

A loan is NOT paid off in a straight line. Early instalments are mostly interest; later instalments are mostly capital. The only correct way to find what's still owed is:

\[\text{Outstanding balance}=x\cdot\dfrac{1-(1+i)^{-(\text{remaining periods})}}{i}\]

Using the SAME instalment \(x\) and rate \(i\) as the original loan.

Worked example: outstanding balance

Continuing the R180 000 loan from earlier.

How much is still owed? Level 3

Using the same loan (R180 000 at 13% p.a. compounded monthly, R4 095,55 per month over 5 years), calculate the outstanding balance immediately after the 36th instalment has been paid.

Show the complete working
  1. 1Total instalments \(n=60\); after 36 payments, \(60-36=24\) instalments remain.
  2. 2Balance \(=4\,095{,}55\times\dfrac{1-(1{,}010833)^{-24}}{0{,}010833}\)
  3. 3\(\boxed{R86\,146{,}32}\)
Common mistake
Do not assume paying 36 of 60 instalments (60% of the payments) means only 40% of the loan remains. That naive, straight-line assumption would predict R72 000 still owed — but the true remaining balance is R86 146,32, noticeably MORE than that guess, because early instalments are weighted toward interest, not capital.

Critically analysing investment and loan options

CAPS explicitly asks you to make an informed decision—not just calculate a number.

For an investment

Always compare EFFECTIVE annual rates, never nominal rates. Check fees, access restrictions, and whether the return is realistic.

For a loan

A lower monthly instalment usually means a LONGER term, which usually means MORE total interest paid. Always calculate the total amount actually repaid, not just the monthly figure.

Two traps, same lesson
"Lower rate" and "lower instalment" both sound safer, but neither guarantees the cheaper deal on its own. Only the total amount repaid tells the full story.

Worked example: comparing two loan offers

First the same idea with smaller numbers, then a case where the lower interest rate does not win.

Same loan, two structures (smaller amounts) Level 2–3

A learner needs to borrow R50 000. Option A: 14% p.a. compounded monthly, over 2 years. Option B: 11% p.a. compounded monthly, over 3 years. Which option costs less overall?

Show both totals
  1. 1Option A: instalment \(\approx R2\,400{,}64\) over 24 months, total interest \(\approx R7\,615{,}46\).
  2. 2Option B: instalment \(\approx R1\,636{,}94\) over 36 months, total interest \(\approx R8\,929{,}69\).
  3. 3\(\boxed{\text{Option A still costs less overall}}\), despite its higher rate and higher instalment.
Same loan, two structures Level 4

A learner needs to borrow R150 000. Option A: 14% p.a. compounded monthly, over 4 years. Option B: 11,5% p.a. compounded monthly, over 6 years. Which option costs less overall?

Show both totals
  1. 1Option A: instalment \(\approx R4\,098{,}97\) over 48 months, total repaid \(\approx R196\,750{,}63\), total interest \(\approx R46\,750{,}63\).
  2. 2Option B: instalment \(\approx R2\,893{,}67\) over 72 months, total repaid \(\approx R208\,344{,}49\), total interest \(\approx R58\,344{,}49\).
  3. 3\(\boxed{\text{Option A costs R11\,593{,}86 less overall}}\), even though Option B has the lower rate and the lower monthly instalment—its longer term costs more in accumulated interest.

Recognising a pyramid scheme

CAPS names this explicitly as part of critically analysing investment options.

The red flags

Guaranteed, very high, short-term returns. Profit comes mainly from recruiting new members, not from a real product or service. Pressure to recruit friends and family quickly.

Why it always collapses

Every level needs exponentially more new recruits than the last—a geometric sequence with ratio greater than 1. The pool of possible recruits is finite, so the scheme mathematically must run out of new members and collapse.

The law
Pyramid and multiplication schemes are illegal in South Africa. Most participants lose their money, and joining early does not make it safe or legal—it only changes who loses last.
Quick check: judging the deal

Loan X has a lower monthly instalment than Loan Y for the same amount borrowed. What can you conclude?

An "investment" pays returns mainly from money brought in by new recruits rather than real trade or products. What is this?

Mistake clinic: repair the exact error

These are small slips, but each one can break an otherwise correct finance question.

Rate

Nominal ≠ period rate

Always divide the nominal rate by the number of compounding periods per year before substituting.

Rounding

A raw \(n\) is never final

After solving for \(n\) with logarithms, check the two surrounding whole periods directly.

Timeline

Count deposits, don't guess

A deferred or irregular annuity needs its own timeline—the number of years is not automatically the number of deposits.

Balance

Balance is never linear

Outstanding balance is the present value of what's left—not a simple fraction of payments made.

Comparison

Rate alone doesn't decide

A lower rate or lower instalment can still cost more overall once the full term is considered.

Your Paper 1 checklist

Use this before submitting an exam answer.

Avoid this

Comparing nominal rates directly.

Leaving a decimal year as a final answer.

Assuming years = number of deposits.

Do this

Draw the timeline first, every time.

Convert to effective rates before comparing.

Calculate the TOTAL repaid before judging a loan.

Apply simple and compound growth and decay, including hire purchase and both depreciation methods.
Convert nominal to effective interest rates.
Solve for the time period \(n\) using logarithms, in both growth and decay.
Derive and apply both the future and present value annuity formulas.
Handle a deferred annuity and calculate an outstanding loan balance.
Critically compare loan and investment options, and recognise a pyramid scheme.
More explanation and exercises:Siyavula Grade 12 Future Value Annuities
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 7 real DBE/provincial exam questions, from a direct growth calculation through to a loan re-amortized after missed payments. 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.

1 / 31

Learn It in Short Videos

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

Future & Present Value Examples

Worked annuity examples covering both directions of the formula.

Online Maths by Miss Pythagoras · Future and Present Value Examples

Future Value Annuities

A second worked walkthrough of the future value annuity formula.

Online Maths by Miss Pythagoras · Financial Mathematics Grade 12

Solving for n With Logarithms

Finding the exact time period of an investment when n sits in the exponent.

Online Maths by Miss Pythagoras · L2 Using Logarithms to find the period of an investment

Outstanding Balance

Finding what's still owed on a loan as the present value of the remaining instalments.

Online Maths by Miss Pythagoras · L5 Outstanding Balance on a Loan

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 Mastery Bank

17 questions by skill, including 7 real DBE/provincial exam questions, with concise reveal answers and methods.

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 12 Finance

Use its annuity derivation and sinking-fund sequence for extra explanation and exercises.

Free • CAPS aligned
Open Siyavula
DBE
Official free books
DBE Grade 12 Textbooks

Official state-owned learner books and teacher support for Grade 12 Mathematics.

Official • free access
Open DBE Books
PDF
Printable worksheet + memo
Maths At Sharp: Worksheet 4

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

Free • download & print
Open Worksheet
NSC
Official past papers
DBE NSC Examination Archive

Use official papers after the course and original practice are secure.

Past papers • memos
Open DBE Archive
x

Related Topics

These skills strengthen the algebra and series work underneath every finance question.

Σ

Patterns & Sequences

Geometric series behind every annuity

Study this →
x

Algebra

Rearranging equations to solve for x or n

Study this →
f

Functions & Graphs

Exponential growth and logarithms

Study this →
Student Questions

Frequently Asked Questions

Short answers for the checks learners make while preparing for the Grade 12 CAPS exam.

What is genuinely new in Grade 12 Finance, Growth & Decay?

Solving for the time period \(n\) using logarithms, deriving and applying both annuity formulas via geometric series (including bond/loan repayment problems), and critically analysing investment and loan options, including recognising a pyramid scheme.

Do I still need simple interest, hire purchase and depreciation?

Yes. Those are Grade 10 and 11 skills that Grade 12 exams test cumulatively, and several Grade 12 questions (like a sinking fund) combine them directly with the new annuity work.

How do I know how many deposits are in an annuity?

Draw a timeline. Mark today, mark when the first deposit actually happens, and mark the final date, then count the real number of deposit arrows—never assume it equals the number of years.

Why isn't the outstanding balance just a fraction of the payments made?

Early instalments on a loan are mostly interest, and later instalments are mostly capital. The only correct method is to find the present value of the instalments still owed, at the loan's own interest rate.

Where should I practise next?

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