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\[A = P(1+i)^n\]
Welcome to Grade 12 Financial Mathematics — one of the highest-weighted topics in your final exam. In this lesson you will master every formula and learn exactly when to use each one. Work through every step, then tackle the quiz at the end.
\[\text{6 formulas — all provided in the exam}\]
The exam gives you all six formulas. Your job is to identify which formula to use and substitute correctly. The six formulas cover: simple interest, compound growth, straight-line depreciation, reducing-balance depreciation, future value annuity, and present value annuity.
\[A,\; P,\; i,\; n,\; x\]
A = accumulated amount (what you end up with). P = principal (starting amount or loan). i = interest rate per period (as a decimal). n = number of periods. x = equal payment per period (annuities only). Always convert percentages to decimals before substituting.
\[A = P(1 + i \cdot n)\]
Simple interest: the interest is always calculated on the original principal. Growth is linear — the same rand amount is added every year. Keywords: simple interest, flat rate, rate per annum at simple interest.
\[A = 20\,000(1 + 0{,}12 \times 3) = R28\,000\]
Thabo invests R20 000 at 12% p.a. simple interest for 3 years. Substitute: \(A = 20\,000(1 + 0.12 \times 3) = 20\,000 \times 1.36 = \text{R28 000}\). He earns R2 000 interest per year — the same amount every year.
\[i = \frac{A/P - 1}{n}\]
You may need to find i or n by rearranging. Example: if \(A = 28\,000\), \(P = 20\,000\), \(n = 3\), then \(i = \frac{28000/20000 - 1}{3} = \frac{0.4}{3} \approx 13.3\%\). Always check your answer makes sense.
\[A = P(1 + i)^n\]
Compound interest: interest is earned on the previous interest as well — "interest on interest". Growth is exponential. Critical: adjust \(i\) and \(n\) to match the compounding period before substituting.
\[\text{Monthly: } i = \frac{r}{12},\; n = \text{years} \times 12\]
Annually: \(i = r\), \(n = \text{years}\).
Quarterly: \(i = r/4\), \(n = \text{years} \times 4\).
Monthly: \(i = r/12\), \(n = \text{years} \times 12\).
Semi-annually: \(i = r/2\), \(n = \text{years} \times 2\). More frequent compounding = higher return.
\[A = 50\,000(1{,}08)^5 = R73\,466{,}40\]
David invests R50 000 at 8% p.a. for 5 years compounded annually. \(i = 0.08\), \(n = 5\). \(A = 50\,000 \times 1.08^5 = \text{R73 466.40}\).
\[A = 50\,000\left(1 + \frac{0{,}08}{4}\right)^{20} = R74\,297{,}37\]
Same investment, but compounded quarterly. \(n = 5 \times 4 = 20\) quarters, \(i = 0.08/4 = 0.02\). The answer R74 297.37 is higher than the annual answer because interest compounds more often.
\[A = 50\,000\left(1 + \frac{0{,}08}{12}\right)^{60} = R74\,492{,}29\]
Compounded monthly: \(n = 5 \times 12 = 60\), \(i = 0.08/12\). Enter \(0.08 \div 12\) directly into your calculator — do not round the intermediate value. Tip: Annual < Quarterly < Monthly.
\[A = P(1 - i \cdot n)\]
Straight-line depreciation: the asset loses the same rand amount every year. Growth is linear (downward). Keywords: straight-line depreciation, linear reduction, flat depreciation. Note the minus sign — this is a decreasing amount.
\[A = 200\,000(1 - 0{,}12 \times 5) = R80\,000\]
A vehicle worth R200 000 depreciates at 12% p.a. straight-line for 5 years. Annual loss = \(200\,000 \times 0.12 = \text{R24 000}\) per year. Over 5 years: \(200\,000 - 5 \times 24\,000 = \text{R80 000}\).
\[2\,200 = 4\,000(1 - i \times 3) \implies i = 15\%\]
Sandy buys a printer for R4 000 and sells it 3 years later for R2 200. Find \(i\). Divide both sides by 4 000: \(0.55 = 1 - 3i\), so \(3i = 0.45\), giving \(i = 0.15 = 15\%\) p.a. Finding i or n by rearranging is a common exam question.
\[A = P(1 - i)^n\]
Reducing-balance depreciation: the asset loses a percentage of its current value each year. The depreciation slows as the asset becomes worth less. Keywords: reducing balance, diminishing balance, compound decay, compound decrease.
\[A = 200\,000(0{,}88)^5 = R105\,546{,}38\]
Same R200 000 vehicle at 12% p.a. reducing balance for 5 years. \(1 - 0.12 = 0.88\). \(A = 200\,000 \times 0.88^5 = \text{R105 546.38}\). Compare: straight-line gave R80 000. Reducing balance always gives a higher book value at the same rate.
\[A = P(1-i)^n \implies i = 1 - \left(\frac{A}{P}\right)^{\!1/n}\]
To find the depreciation rate: rearrange to \(\frac{A}{P} = (1-i)^n\), take both sides to the power \(1/n\), then subtract from 1. Example: \(P = 120\,000\), \(A = 65\,000\), \(n = 4\): \(i = 1 - (65000/120000)^{0.25} \approx 14.6\%\) p.a.
\[A = P(1 + i)^n\]
Inflation uses the same formula as compound growth. Prices increase by a percentage of the current price — so inflation compounds. A higher inflation rate means your money buys less over time. Always use the annual inflation rate.
\[A = 14(1{,}075)^5 = R20{,}10\]
Bread costs R14 now. Annual inflation rate = 7.5%. Cost in 5 years: \(A = 14 \times 1.075^5 = \text{R20.10}\). This means your money must grow faster than inflation to maintain purchasing power.
\[15\,000\!\left(1+\frac{0{,}08}{12}\right)^{\!36}\!\cdot\left(1+\frac{0{,}09}{4}\right)^{\!8} = R22\,765{,}78\]
Change in rate: draw a timeline. Sibusiso deposits R15 000 for 3 years at 8% p.a. monthly, then 2 more years at 9% p.a. quarterly. Chain the two compound calculations: grow at rate 1 for period 1, then grow at rate 2 for period 2.
\[12\,000\left(1{+}\tfrac{0{,}03}{12}\right)^{60} + 5\,000\left(1{+}\tfrac{0{,}03}{12}\right)^{48} - 8\,000\left(1{+}\tfrac{0{,}03}{12}\right)^{24}\]
Cheryl: opens with R12 000, deposits R5 000 after 1 year, withdraws R8 000 after 3 years. Rate: 3% p.a. monthly. Move each amount independently to T₅. Add deposits, subtract withdrawals. Answer: R11 081.99.
\[F = \frac{x[(1+i)^n - 1]}{i}\]
Use this formula when making equal deposits at equal intervals and asking "how much will I have saved?" F = future accumulated value. x = equal deposit per period. Payments are made at the end of each period.
\[F = \frac{500\left[\left(1+\frac{0{,}08}{12}\right)^{\!48}-1\right]}{\frac{0{,}08}{12}} = R28\,862{,}79\]
Mokgeseng saves R500/month at 8% p.a. compounded monthly for 4 years. \(x=500\), \(i=0.08/12\), \(n=48\). Total deposited = \(48 \times 500 = \text{R24 000}\). Interest earned = \(28\,862.79 - 24\,000 = \text{R4 862.79}\).
\[A = 28\,862{,}79\left(1+\frac{0{,}08}{12}\right)^{\!24} = R33\,852{,}82\]
Mokgeseng stops paying and leaves the money for 2 more years. Switch formulas: no more regular payments, so use compound growth \(A = P(1+i)^n\). The savings become the new principal \(P\). Two-phase problems are common in Grade 12 exams.
\[n = \frac{\log(1{,}51)}{\log(1{,}051)} = 8{,}28 \implies 9 \text{ payments}\]
Linda wants R30 000, saves R3 000 every 6 months at 10.2% p.a. semi-annually. Rearrange the formula, isolate \((1.051)^n\), then take \(\log\) of both sides. \(n = \log(1.51)/\log(1.051) = 8.28\). Always round UP for savings — rounding down means you fall short.
\[P = \frac{x[1-(1+i)^{-n}]}{i}\]
Use this formula when borrowing a lump sum and repaying in equal instalments. P = loan amount. x = equal instalment. The first payment is one period after the loan is taken. Keywords: loan, bond, hire purchase, monthly payment, instalment.
\[P = \frac{8\,000\left[1-\left(1+\frac{0{,}15}{12}\right)^{-60}\right]}{\frac{0{,}15}{12}} = R336\,276{,}73\]
Mpho can afford R8 000/month, 15% p.a. monthly, over 5 years. \(x=8000\), \(i=0.15/12\), \(n=60\). He can afford a car worth R336 276.73. Notice: total paid = \(60 \times 8000 = \text{R480 000}\) for a R336 000 car — the extra is interest.
\[x = \frac{150\,000 \times \frac{0{,}215}{4}}{1-\left(1+\frac{0{,}215}{4}\right)^{-16}} = R14\,212{,}35\]
Jenny repays R150 000 in 16 equal quarterly payments at 21.5% p.a. quarterly. Rearrange for \(x\). Total interest = \(14\,212.35 \times 16 - 150\,000 = \text{R77 397.60}\). Finding x by rearranging the PV formula is a standard exam question.
\[OB = P(1+i)^m - \frac{x[(1+i)^m-1]}{i}\]
The outstanding balance after \(m\) payments = how much the original loan has grown, minus the future value of all payments made. m = payments already made. Alternative: \(OB = \frac{x[1-(1+i)^{-r}]}{i}\) where \(r\) = payments remaining.
\[OB = 250\,000\left(1+\frac{0{,}15}{12}\right)^{36} - \frac{5\,000\left[\left(1+\frac{0{,}15}{12}\right)^{36}-1\right]}{\frac{0{,}15}{12}}\]
Sibusiso: R250 000 loan at 15% p.a. monthly, pays R5 000/month. Outstanding balance after 3 years (36 payments). First term: loan grown for 36 months. Second term: future value of 36 payments. Subtract: OB = R165 408.43.
\[\text{Last payment} = OB_{\text{final}} \times (1+i)\]
When \(n\) is not a whole number, the last payment is smaller than the regular payment. Calculate the outstanding balance after the last full payment, then multiply by \((1+i)\) to add one more period of interest. Example: George's last payment = \(3547.46 \times (1+0.14/12) = \text{R3 588.84}\).
\[1 + i_{\text{eff}} = \left(1 + \frac{i_{\text{nom}}}{n}\right)^n\]
Nominal rate: the stated annual rate (e.g. 12% p.a. compounded monthly).
Effective rate: the actual annual return after accounting for compounding within the year. The effective rate is always higher than the nominal rate (when \(n > 1\)).
\[1 + i_{\text{eff}} = \left(1 + \frac{0{,}18}{12}\right)^{12} = 1{,}1956\]
A bank offers 18% p.a. compounded monthly. Effective annual rate: \(i_{\text{eff}} = 1.015^{12} - 1 = 0.1956 = 19.56\%\) p.a. Although the nominal rate is 18%, you effectively earn 19.56% per year. Use effective rates to compare accounts with different compounding periods.
\[i_{\text{nom}} = n\left[(1 + i_{\text{eff}})^{1/n} - 1\right]\]
You may need to find the nominal rate given the effective rate. Rearrange the formula: \(i_{\text{nom}} = n \times [(1+i_{\text{eff}})^{1/n} - 1]\). Example: effective rate is 20% p.a. compounded quarterly → \(i_{\text{nom}} = 4 \times [(1.20)^{0.25}-1] \approx 18.56\%\) p.a.
\[\begin{aligned}A &= P(1+in) & A &= P(1+i)^n\\A &= P(1-in) & A &= P(1-i)^n\\F &= \tfrac{x[(1+i)^n-1]}{i} & P &= \tfrac{x[1-(1+i)^{-n}]}{i}\end{aligned}\]
Row 1: Simple interest (linear growth) | Compound growth (exponential).
Row 2: Straight-line depreciation | Reducing-balance depreciation.
Row 3: Future value annuity (savings) | Present value annuity (loans).
Plus: Outstanding balance formula and Nominal ↔ Effective conversion.