Financial Mathematics

Full lesson notes

Everything covered in this lesson, in one place - useful for revision or printing.

\[ A=P(1+i)^n \qquad F=x\left[\frac{(1+i)^n-1}{i}\right] \qquad P=x\left[\frac{1-(1+i)^{-n}}{i}\right] \]

Finance, Growth and Decay is the most practical topic in Grade 12 Mathematics. It answers questions people face for the rest of their lives.

Suggested duration: 6 lessons of 60 minutes, plus an assessment lesson.
Calculator: scientific calculator; keep full accuracy until the final line.
Currency: round money to the nearest cent unless told otherwise.

This lesson uses timelines throughout, because the value of money depends on when it is deposited, borrowed, paid or withdrawn.

1. Foundations

\[ \text{money} + \text{time} + \text{interest rate} \;\longrightarrow\; \text{a decision} \]

Financial mathematics answers practical questions such as:

• How long will an investment take to reach a target?
• How much should a person save every month?
• What loan can a person afford?
• How much is still owed after several repayments?
• How can a business save to replace equipment?
• Which investment or loan option is financially better?
• Why are pyramid schemes mathematically unsustainable?

This lesson develops the concepts from familiar growth and decay models to Grade 12 annuities and multi-step financial decisions.

\[ \textbf{13 outcomes} \]

By the end of the lesson sequence, you should be able to:

1. distinguish simple growth, compound growth, straight-line depreciation and reducing-balance depreciation;
2. convert a quoted nominal annual rate into the correct rate per period;
3. calculate an effective annual interest rate;
4. solve for time using logarithms and interpret the result in context;
5. distinguish a future-value annuity from a present-value annuity;
6. use the future-value formula for regular savings and sinking funds;
7. use the present-value formula for loans and regular withdrawals;
8. calculate repayments, interest earned, interest paid and total cost;
9. calculate the outstanding balance of a loan by two methods;
10. calculate a smaller final payment when equal payments do not settle a loan exactly;
11. adapt a model for immediate, delayed, missed or stopped payments;
12. compare financial options critically, considering cost, affordability, risk and return; and
13. explain, using exponential growth, why a pyramid scheme must collapse.

\[ \text{5 minutes — not for marks} \]

Answer these five questions before you begin. They are not for marks; they show which route to take.

1. Write \(12.6\%\) as a decimal.
2. An account pays \(12\%\) p.a. compounded monthly. State the monthly rate and the number of periods in 3 years.
3. Which process is linear: simple interest or compound interest?
4. Solve \(2=(1.08)^n\) for \(n\), correct to two decimal places.
5. Explain the difference between a deposit and a withdrawal.

\[ n=\frac{\log 2}{\log(1.08)}\approx 9.01 \]

1. \(0.126\)
2. \(i=\dfrac{0.12}{12}=0.01\) per month and \(n=3(12)=36\)
3. Simple interest
4. \(n=\dfrac{\log 2}{\log(1.08)}\approx 9.01\)
5. A deposit adds money to an account; a withdrawal removes money.

Symbol or term Meaning \(P\) Principal or present value: the value at the starting date \(A\) Accumulated or final value of a single amount \(F\) Future value of a series of regular payments \(x\) Equal regular deposit, withdrawal or repayment \(i\) Interest rate per compounding/payment period , as a decimal \(n\) Number of compounding periods or payments \(m\) Number of compounding periods per year nominal rate Quoted annual rate before allowing for compounding within the year effective rate Actual percentage growth over a full year annuity Equal payments made at equal time intervals at a fixed interest rate amortise Repay a loan gradually through regular payments outstanding balance Amount needed to settle the loan at a stated date sinking fund Regular savings accumulated for a known future replacement cost inflation Increase in price or value over time depreciation Decrease in the value of an asset over time

Learn this vocabulary now — every question uses it.

\[ \text{now} \;\rightarrow\; \text{present value} \qquad \text{later} \;\rightarrow\; \text{future value} \]

Say this sentence out loud before choosing any formula:

This frame is especially useful if English is not your first language — it forces you to identify the three things that decide the whole solution.

\[ \textbf{R} \rightarrow \textbf{A} \rightarrow \textbf{T} \rightarrow \textbf{E} \rightarrow \textbf{S} \]

Use RATES for every single question. It is non-negotiable.

1. R – Read: underline amounts, rates, dates and payment words.
2. A – Align: make the interest period, payment period and value of \(n\) use the same unit.
3. T – Timeline: place every cash flow at the correct point in time.
4. E – Equation: select and substitute into the correct formula.
5. S – Sense-check: interpret, round only at the end and ask whether the answer is realistic.

Compounding \(m\) Rate per period Periods in \(t\) years Annually 1 \(i=\frac{j}{1}\) \(n=t\) Half-yearly 2 \(i=\frac{j}{2}\) \(n=2t\) Quarterly 4 \(i=\frac{j}{4}\) \(n=4t\) Monthly 12 \(i=\frac{j}{12}\) \(n=12t\) Daily 365* \(i=\frac{j}{365}\) \(n=365t\)

Here \(j\) is the quoted nominal annual rate as a decimal. *Unless stated otherwise.

\[ \text{RATES: Read, Align, Timeline, Equation, Sense-check} \]

Key idea: before any calculation, align your units and draw the timeline. Most lost marks in this topic come from mismatched \(i\) and \(n\), not from difficult algebra.

2. Growth & Decay

Situation Formula Pattern Simple interest \(A=P(1+in)\) Linear growth Compound interest or inflation \(A=P(1+i)^n\) Exponential growth Straight-line depreciation \(A=P(1-in)\) Linear decay Reducing-balance depreciation \(A=P(1-i)^n\) Exponential decay

These four models cover every single-amount question. Notice the pattern: \(n\) as a multiplier is linear; \(n\) as an exponent is exponential.

\[ \text{same rand amount} \;\Rightarrow\; \text{linear} \qquad \text{percentage of a changing balance} \;\Rightarrow\; \text{exponential} \]

• Simple interest / straight line: the same rand amount is added or removed each period.
• Compound interest / reducing balance: the percentage acts on the changing balance each period.
• Words such as compounded, reducing balance, inflation and appreciates usually indicate an exponential model.

\[ A=10\,000(1+0.08\times 3)=\text{R}12\,400.00 \]

R10 000 is invested for 3 years at \(8\%\) p.a.

Compound interest is higher because interest is earned on earlier interest.

\[ \text{colour-code } P,\; i,\; n \]

\[ A=P(1\pm i)^n \quad\text{vs}\quad A=P(1\pm in) \]

Key idea: decide linear or exponential first, and growth or decay second. That single decision picks the formula.

3. Rates & Logarithms

\[ 1+i_{\text{eff}}=\left(1+\frac{j}{m}\right)^m \qquad\Longrightarrow\qquad i_{\text{eff}}=\left(1+\frac{j}{m}\right)^m-1 \]

A nominal rate \(j\) is the quoted annual rate; the effective rate is the actual growth over a full year once compounding inside the year is taken into account.

\[ i_{\text{eff}}=\left(1+\frac{0.12}{12}\right)^{12}-1=0.126825\ldots \]

A bank quotes \(12\%\) p.a. compounded monthly.

The effective rate is higher than the nominal rate because interest is added during the year and then also earns interest.

\[ (1+i_1)^{m_1}=(1+i_2)^{m_2} \]

When the payment and compounding intervals differ, you must convert properly.

For example, to convert an effective annual rate \(i_{\text{eff}}\) to an equivalent monthly rate \(i_m\):

\[ n=\frac{\log(A/P)}{\log(1+i)} \]

From \(A=P(1+i)^n\), isolate the exponential term and take logarithms of both sides.

\[ n=\frac{\log 1.5}{\log 1.095}=4.4677\ldots \]

How long will R50 000 take to grow to R75 000 at \(9.5\%\) p.a. compounded annually?

This is approximately 4 years and \(0.4677(12)=5.61\) months, so about 4 years and 6 months.

\[ 4.47 \text{ years} \;\ne\; 4\text{ years and }47\text{ months} \]

• If the question asks when a value will first be at least a target and transactions occur only at full periods, round up.
• If it asks for years and months, multiply only the decimal part of the years by 12.
• If \(n\) represents months, do not multiply it by 12 again.
• For depreciation below a threshold after full years, test the next integer year.

\[ i_{\text{eff}}=\left(1+\frac{j}{m}\right)^m-1 \qquad n=\frac{\log(A/P)}{\log(1+i)} \]

Key idea: the effective rate lets you compare accounts fairly, and logarithms are the only way to bring \(n\) down out of the exponent.

4. Annuity Basics

\[ \text{equal payments} + \text{equal intervals} + \text{constant rate} + \text{matching periods} \]

The standard annuity formulas require all four of these:

1. equal payments;
2. equal time intervals between payments;
3. a constant interest rate; and
4. a compounding period that matches the payment interval, or an equivalent rate conversion.

Question clue Model Formula Small regular deposits first; large accumulated amount later Future value annuity \(F=x\left[\frac{(1+i)^n-1}{i}\right]\) Large loan or amount now; small payments later Present value annuity \(P=x\left[\frac{1-(1+i)^{-n}}{i}\right]\)

This single table decides most annuity questions.

Time 0 Regular intervals End no large amount deposits \(x,x,x,\ldots\) large \(F\): future value large \(P\) received repayments \(x,x,x,\ldots\) loan becomes \(0\): present value

\[ \text{Where is the large amount — at the start or at the end?} \]

Key idea: find the large amount on the timeline. At the start it is a present value; at the end it is a future value.

5. Future Value & Sinking Funds

\[ F=x\left[\frac{(1+i)^n-1}{i}\right] \]

For \(n\) equal end-of-period deposits, this gives the accumulated value immediately after the last deposit.

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

At the end, the final deposit has earned no interest, the previous deposit has earned one period of interest, and so on.

This is a geometric series with ratio \((1+i)\), which simplifies to the boxed formula. You are not memorising a random rule — it is the geometric series sum you already know.

\[ F=1\,500\left[\frac{(1.007)^{60}-1}{0.007}\right]=\text{R}111\,372.06 \]

Nomvula deposits R1 500 at the end of every month for 5 years. The account earns \(8.4\%\) p.a. compounded monthly. Find the final balance and the interest earned.

\[ F_{\text{due}}=x\left[\frac{(1+i)^n-1}{i}\right](1+i) \]

The standard formula assumes the first payment is after one period. If every payment is shifted one period earlier and the value is required one period after the last deposit, multiply by \((1+i)\).

\[ \text{stage 1: } F \qquad \text{stage 2: } A=P(1+i)^k \]

Use two stages:

1. find the annuity value when the last deposit is made;
2. grow that lump sum with \(A=P(1+i)^k\) for the remaining \(k\) periods.

\[ \text{Target fund}=\text{future cost of new asset}-\text{future sale value of old asset} \]

A sinking fund is a future-value annuity used to accumulate money to replace an asset.

Once you have the target, solve \(F=x\left[\frac{(1+i)^n-1}{i}\right]\) for \(x\).

\[ F=669\,112.79-97\,615.17=\text{R}571\,497.62 \]

A business owns equipment worth R220 000, depreciating at \(15\%\) p.a. on the reducing balance. New equipment costs R500 000 now and will inflate at \(6\%\) p.a. Replacement is in 5 years. The sinking fund earns \(8.4\%\) p.a. compounded monthly, with deposits beginning one month from now.

\[ x=\frac{571\,497.62(0.007)}{(1.007)^{60}-1}=\text{R}7\,697.14 \]

Note the rearrangement: when you know \(F\) and want \(x\), multiply by \(i\) and divide by \((1+i)^n-1\).

\[ \text{new cost} \;-\; \text{old sale value} \;=\; \text{amount to save} \]

\[ F=x\left[\frac{(1+i)^n-1}{i}\right] \]

Key idea: a sinking fund is just a future-value annuity with the target worked out first. Always compute the target before touching the annuity formula.

6. Present Value & Loans

\[ P=x\left[\frac{1-(1+i)^{-n}}{i}\right] \]

For a loan repaid by \(n\) equal end-of-period payments. The same formula applies when a lump sum is invested now to fund equal future withdrawals.

\[ x=\frac{350\,000(0.0095)}{1-(1.0095)^{-60}}=\text{R}7\,679.85 \]

A R350 000 loan is repaid monthly over 5 years at \(11.4\%\) p.a. compounded monthly. The first payment is one month after the loan is granted.

\[ \text{Total interest}\approx 460\,791.19-350\,000=\text{R}110\,791.19 \]

Using the unrounded repayment:

\[ \text{longer term} \;\Rightarrow\; \text{lower instalment, higher total interest} \]

A longer term usually lowers the monthly repayment but increases the total interest paid. “Best” therefore depends on both affordability and total cost — a distinction the examiners test directly.

\[ P=x\left[\frac{1-(1+i)^{-n}}{i}\right] \]

Key idea: a loan is a present-value annuity. The money arrives now; the equal payments settle it later.

7. Balances & Final Payments

\[ B=P(1+i)^k-x\left[\frac{(1+i)^k-1}{i}\right] \]

Suppose a loan \(P\) is planned for \(n\) payments of \(x\), and \(k\) payments have just been made. Move the original loan and all payments already made forward to time \(k\).

\[ B=x\left[\frac{1-(1+i)^{-(n-k)}}{i}\right] \]

Find the present value, at time \(k\), of the remaining \((n-k)\) payments.

\[ B\approx \text{R}233\,228.38 \]

Use the R350 000 loan from Worked Example 5 and find the balance immediately after payment 24. Use full calculator accuracy for \(x\).

The two answers should agree, apart from minor rounding differences. Using both is the best way to check your work in an examination.

\[ B \;\ne\; 350\,000-24x \]

Interest is added to the unpaid balance every month. Early repayments contain a relatively large interest component, so the principal falls more slowly at first.

\[ 100\,000=5\,000\left[\frac{1-(1.01)^{-n}}{0.01}\right] \]

To solve a present-value annuity for \(n\), isolate \((1+i)^{-n}\) and use logarithms. A non-integer \(n\) means the equal payments do not settle the loan exactly.

\[ \text{Final payment}=2\,113.66(1.01)=\text{R}2\,134.79 \]

A R100 000 loan earns \(12\%\) p.a. compounded monthly. The borrower pays R5 000 each month.

This means 22 full payments of R5 000 are made, followed by a smaller payment one month later.

\[ x\le Pi \;\Rightarrow\; \text{the debt never decreases} \]

If the regular repayment is less than or equal to the interest added during the first period, the debt will not decrease.

\[ B=P(1+i)^k-x\left[\frac{(1+i)^k-1}{i}\right] \;=\; x\left[\frac{1-(1+i)^{-(n-k)}}{i}\right] \]

Key idea: two methods, one answer. Use the second as a check, and always test feasibility before assuming a loan can be repaid.

8. Irregular Timing

\[ P_d=P(1+i)^d \]

If \(d\) ordinary payments are missed before repayments start:

1. grow the loan through the no-payment period: \(P_d=P(1+i)^d\);
2. calculate how many payments remain by the final date;
3. apply the present-value formula to \(P_d\).

\[ \text{value} = \text{normal annuity} \;-\; \text{value of the missed payments} \]

Choose a common valuation date. Then:

• calculate the value of the normal annuity; and
• subtract the value, at that same date, of each missed deposit; or
• add the accumulated effect of each missed repayment to the debt.

\[ \text{stage 1} \;\longrightarrow\; \text{stage 2 (new rate)} \]

End the first calculation on the change date. Use that result as the principal for the next stage with the new rate. Never average the two rates.

\[ (1+i_c)^{\text{compounding periods per year}}=(1+i_p)^{\text{payment periods per year}} \]

Convert to an equivalent rate matching the payment interval before using any annuity formula.

\[ \text{move it to the same valuation date, then add or subtract} \]

Treat it as a separate single amount and move it to the same valuation date as the annuity before adding or subtracting.

\[ \text{draw the timeline} \;\rightarrow\; \text{split into stages} \]

Key idea: whenever conditions change, the safest method is always a timeline divided into stages. Do not try to force one formula onto a changing situation.

9. Financial Decisions

\[ \text{same date} \;+\; \text{same total deposits} \]

Compare options at the same date and using the same total deposits. Consider:

• accumulated value;
• effective annual return;
• fees and penalties;
• tax implications if information is provided;
• access to the money;
• fixed versus variable return;
• risk and guarantees; and
• the effect of inflation on purchasing power.

\[ \text{equal payments} \;\ne\; \text{equal annual contributions} \]

First make the total yearly deposits equal, or calculate each option using its actual cash flow.

\[ \text{lowest instalment} \;\ne\; \text{cheapest loan} \]

For each option calculate or discuss:

1. deposit required;
2. principal borrowed;
3. monthly repayment;
4. repayment period;
5. total repayments;
6. total interest and fees;
7. balloon or residual payment, if any;
8. affordability; and
9. risk if interest rates change.

\[ \text{people at level } k = r^k \]

A pyramid scheme depends mainly on fees from new recruits rather than value created through genuine goods, services or investment returns.

If every participant must recruit \(r\) new people, the number of people at level \(k\) is \(r^k\).

Level New recruits required 1 \(6\) 2 \(6^2=36\) 3 \(6^3=216\) 6 \(6^6=46\,656\) 10 \(6^{10}=60\,466\,176\)

If each person recruits 6 people, the required number grows exponentially and soon exceeds the available population. Most people must therefore lose money.

\[ \text{recruitment} \;>\; \text{real value created} \;\Rightarrow\; \text{walk away} \]

• earnings depend mainly on recruiting others;
• guaranteed unusually high returns;
• pressure to join quickly;
• no clear, sustainable source of profit;
• unclear or fake product sales; and
• money from new members pays earlier members.

\[ \text{cost} + \text{affordability} + \text{risk} + \text{return} \]

Key idea: a financial recommendation needs a calculation and a justification. Never judge an option on the instalment alone.

10. Practice & Assessment

Lesson Focus Suggested learner evidence 1 Diagnostic; growth, decay, rates and logarithmic time Completed formula-sort and time calculation 2 Annuity conditions, timelines and future value Annotated timeline and savings calculation 3 Sinking funds and irregular saving patterns Three-stage sinking-fund solution 4 Present value, repayments and total interest Loan calculation and explanation of affordability 5 Outstanding balance, final payment and delayed payments Both balance methods used as a check 6 Comparing options and pyramid schemes Written recommendation supported by calculations 7 Mixed assessment and correction Test plus error-analysis reflection

The recommended order for teaching or self-study.

\[ \text{scaffold the process, not the mathematics} \]

Learners needing substantial support

• Provide the formula decision table and RATES checklist.
• Use pre-drawn timelines with only the amounts missing.
• Begin with annual compounding before monthly or quarterly compounding.
• Let learners label units beside every value of \(i\) and \(n\).
• Provide partially completed substitutions.
• Pair numerical calculation with a verbal sentence explaining the answer.
• Allow a vocabulary bank and bilingual discussion before written English responses.

Learners working at grade level

• Require independent formula selection.
• Mix future- and present-value questions.
• Include total interest, outstanding balance and comparison questions.
• Require both a calculation and a contextual conclusion.

\[ \text{one topic, many representations} \]

Learners ready for extension

• Use changing rates, missed payments and different payment/compounding intervals.
• Ask for algebraic rearrangement before substitution.
• Require both outstanding-balance methods and an explanation of any difference.
• Analyse whether a proposed repayment can ever settle a loan.
• Compare options in which the cheapest total cost is not the most affordable monthly option.

Multiple ways to learn

• Visual: timelines, colour-coded cash flows and growth tables.
• Auditory: learners explain to a partner why a question is future or present value.
• Hands-on: groups act as months on a timeline and hold deposit/payment cards.
• Reading/writing: learners annotate financial advertisements and write a recommendation.
• Collaborative: mixed-readiness groups assign the roles of reader, timeline builder, calculator and checker.

\[ \text{Questions 1–4} \]

1. R24 000 is invested for 4 years at \(7\%\) p.a. simple interest. Find \(A\).
2. A vehicle costing R320 000 depreciates at \(16\%\) p.a. on the reducing balance for 3 years. Find its value.
3. An account pays \(10.8\%\) p.a. compounded monthly. State \(i\) and \(n\) for 5 years.
4. Decide whether each is future value or present value:
   • monthly savings for university fees;
   • a home loan repaid monthly;
   • a lump sum invested now to provide monthly withdrawals.

\[ \text{Questions 5–7} \]

5. R800 is deposited monthly for 4 years at \(9.6\%\) p.a. compounded monthly. Calculate the final value and interest earned.
6. A R500 000 home loan is repaid monthly over 20 years at \(10.8\%\) p.a. compounded monthly. Calculate: the monthly repayment; total interest over 20 years; and the balance immediately after payment 60.
7. A machine worth R300 000 depreciates at \(18\%\) p.a. on reducing balance. After how many complete years will it first be worth less than R100 000?

\[ \text{Questions 8–10} \]

8. A learner deposits R1 000 monthly for 6 years, then stops contributing while the accumulated amount grows for a further 4 years. The rate is \(9\%\) p.a. compounded monthly. Find the value after 10 years.
9. A loan payment begins after a six-month grace period. Explain precisely how a timeline determines the growth period and the number of repayments.
10. A loan of R600 000 charges \(12\%\) p.a. compounded monthly. A borrower proposes R5 000 per month. Decide, with a calculation, whether the loan can ever be repaid under these conditions.

\[ \text{Check your work} \]

1. \(\text{R}30\,720.00\)
2. \(320\,000(0.84)^3=\text{R}189\,665.28\)
3. \(i=0.108/12=0.009\); \(n=60\)
4. Future value; present value; present value
5. \(\text{R}46\,590.40\); interest \(=46\,590.40-38\,400=\text{R}8\,190.40\)

\[ \text{Check your work} \]

6. Repayment \(\approx\text{R}5\,093.06\); total interest \(\approx\text{R}722\,334.50\); balance after payment 60 \(\approx\text{R}453\,091.11\)
7. \(n\approx5.54\), so after 6 complete years
8. First find the future value after 72 deposits, then compound that amount for 48 more months: \(F_{72}\approx\text{R}95\,007.03\), and the value after 10 years is approximately \(\text{R}135\,993.57\).
9. An ordinary first payment is at month 1. If the first actual payment is at month 6, five ordinary payments were missed; grow the loan for 5 months, and count the payments from month 6 to the final date.
10. First-month interest is \(600\,000(0.12/12)=\text{R}6\,000\), which exceeds the R5 000 payment. The balance grows, so the loan cannot be repaid with that payment.

Misconception Correction Using \(12\) instead of \(0.12\) Divide a percentage by 100 before substituting. Dividing \(i\) by 12 but leaving \(n\) in years If the rate is monthly, \(n\) must be months. Choosing future value because the question mentions “future” Identify the cash-flow pattern: regular savings lead to \(F\); a loan received now leads to \(P\). Counting time periods instead of payments Mark every actual payment on the timeline and count the marks. Rounding the monthly rate or repayment early Store values in calculator memory and round only the final answer.

These account for the majority of lost marks in this topic.

Misconception Correction Outstanding balance \(=P-kx\) The unpaid balance earns compound interest. Use a balance formula. A decimal value of \(n\) means that many equal payments Use the integer number of full payments, then calculate the smaller final payment. Lower monthly repayment means cheaper loan Compare total repayment and total interest as well as affordability. Guaranteed high returns mean a good investment Investigate risk, source of return, fees, regulation and recruitment dependence.

Read this table again the night before the examination.

\[ \textbf{Mixed assessment — 40 marks} \]

Question 1: growth, rates and time [8]

• 1.1 R85 000 is invested at \(8.2\%\) p.a. compounded quarterly for 6 years. Calculate the accumulated value. (3)
• 1.2 Calculate the effective annual interest rate. (2)
• 1.3 How long would the investment take to double if the same nominal rate continued? Give the answer in years and months. (3)

Question 2: future-value annuity [8]
Karabo deposits R1 200 at the end of every month for 7 years into an account earning \(9.3\%\) p.a. compounded monthly.

• 2.1 Draw or describe a correctly labelled timeline. (2)
• 2.2 Calculate the final value. (3)
• 2.3 Calculate the interest earned. (2)
• 2.4 State one effect of making the first deposit immediately instead. (1)

\[ \textbf{Mixed assessment — continued} \]

Question 3: loan and balance [12]
A R420 000 loan is amortised through monthly repayments over 8 years at \(10.5\%\) p.a. compounded monthly. The first payment is after one month.

• 3.1 Calculate the monthly repayment. (4)
• 3.2 Calculate the total interest if all repayments remain equal. (2)
• 3.3 Calculate the outstanding balance immediately after payment 36. (4)
• 3.4 Explain why the principal has not decreased by \(36x\). (2)

Question 4: sinking fund [8]
A school will replace equipment in 4 years. New equipment costs R180 000 today and prices inflate at \(5.5\%\) p.a. The old equipment will have a sale value of R28 000 in 4 years. The school deposits monthly into a sinking fund earning \(7.8\%\) p.a. compounded monthly, starting one month from now.

• 4.1 Find the expected replacement price. (2)
• 4.2 Find the required fund after the old equipment is sold. (2)
• 4.3 Find the monthly sinking-fund deposit. (4)

Question 5: critical analysis [4]
A scheme requires each member to recruit five new members and promises payment once those recruits join.

• 5.1 How many new members are needed at level 8? (2)
• 5.2 Give two mathematical or financial reasons not to join. (2)

\[ A=85\,000\left(1+\frac{0.082}{4}\right)^{24}=\text{R}138\,334.70 \]

The decimal part gives \(0.5394(12)=6.47\ldots\) months, so approximately 8 years and 6 months.

\[ F=1\,200\left[\frac{(1+0.093/12)^{84}-1}{0.093/12}\right]=\text{R}141\,313.73 \]

\[ x=\frac{420\,000(0.105/12)}{1-(1+0.105/12)^{-96}}=\text{R}6\,484.81 \]

\[ x=\frac{F(0.078/12)}{(1+0.078/12)^{48}-1}=\text{R}3\,474.53 \]

\[ \textbf{40 marks} \]

Key idea: method marks are awarded for correct formula selection, the periodic rate and the number of periods. Always show these three things, even if your arithmetic slips.

11. Summary & Checklist

\[ \text{five minutes} \]

Complete in the final five minutes:

1. Write one sentence that distinguishes future value from present value.
2. A rate is monthly. What must be true about \(n\)?
3. Name the two outstanding-balance methods.
4. Explain one reason a long loan term can be expensive.
5. Rate your confidence from 1 to 4 and state the next skill you need to practise.

\[ \begin{aligned} \text{Simple growth: }&A=P(1+in)\\ \text{Compound growth/inflation: }&A=P(1+i)^n\\ \text{Straight-line depreciation: }&A=P(1-in)\\ \text{Reducing-balance depreciation: }&A=P(1-i)^n \end{aligned} \]

Effective annual rate

\[ F=x\left[\frac{(1+i)^n-1}{i}\right] \qquad P=x\left[\frac{1-(1+i)^{-n}}{i}\right] \]

Outstanding loan balance after \(k\) payments

Last payment one period after the final full paymentInterest

\[ \text{CAPS-aligned} \]

This resource synthesises the explanations, sequencing, formula conventions and problem types in three Grade 12 textbooks. Examples and assessment questions in this lesson are newly written.

1. Mind Action Series Mathematics Grade 12, New Edition, Chapter 9: Financial Mathematics, pp. 275–295. Prior-grade formulae, annuity conditions, future and present value, outstanding balances, final payments, sinking funds, deferred and missed payments, and differing intervals.
2. Siyavula Mathematics Grade 12, Chapter 3: Finance, pp. 146–170. Logarithmic time calculations, future- and present-value annuities, sinking funds, investment and loan comparisons, grace periods, and end-of-chapter applications.
3. Platinum Mathematics Grade 12 Learner's Book, Topic 4: Finance, Growth and Decay, pp. 60–85. Effective rates, derivation of annuity formulae from geometric series, two outstanding-balance methods, delayed repayments, repayment feasibility, comparing options and pyramid schemes.

\[ \text{before you submit} \]

Before submitting a financial-mathematics solution, ask:

• Did I convert the percentage to a decimal?
• Do the rate, payments and value of \(n\) use the same time unit?
• Did I draw or mentally check the timeline?
• Did I choose future value or present value from the cash-flow pattern?
• Did I count payments, not merely time intervals?
• Did I keep full calculator accuracy until the end?
• Did I round money to cents?
• Did I state what the answer means in context?
• Does the answer make financial sense?

\[ A=P(1+i)^n \quad F=x\left[\frac{(1+i)^n-1}{i}\right] \quad P=x\left[\frac{1-(1+i)^{-n}}{i}\right] \quad B=x\left[\frac{1-(1+i)^{-(n-k)}}{i}\right] \]

You have now covered the complete Finance, Growth and Decay syllabus:

• the four single-amount growth and decay models;
• nominal, effective and equivalent interest rates;
• solving for time with logarithms and interpreting the answer;
• the four annuity conditions and how to choose future or present value;
• future-value annuities, annuity due and sinking funds;
• present-value annuities, repayments and total cost;
• outstanding balances by two methods and the smaller final payment;
• delayed, missed and changed payment conditions;
• comparing investments and loans critically; and
• why pyramid schemes must collapse.