Everything covered in this lesson, in one place - useful for revision or printing.
\( A = P(1+i)^{n} \)
Finance, Growth and Decay is one of the most predictable topics in the Grade 11 exam — every question is built from a handful of formulas. This lesson works through all of them:
\( A = P(1 + i\,n) \)
Simple interest is worked out on the ORIGINAL amount (the principal) only, so you earn the same rand amount every year.
Notice \(n\) sits OUTSIDE the bracket — the growth is linear.
\( A = P(1 + i)^{n} \)
Compound interest is worked out on the amount at the START of each year — so you earn interest on your interest.
Here \(n\) is in the POWER, which is what makes it grow faster and faster.
\( \text{Simple } R11\,600 \;<\; \text{Compound } R12\,308{,}99 \)
Invest R8 000 at 9% for 5 years. Simple: \(8\,000(1 + 0{,}09 \times 5) = R11\,600\). Compound: \(8\,000(1{,}09)^{5} = R12\,308{,}99\).
\( \text{Compound grows faster over time} \)
Over the same time at the same rate, compound interest always ends higher than simple. The longer the term, the bigger the gap.
\( A = P(1 - i\,n) \)
Straight-line (simple) depreciation loses the SAME rand amount each year. It can reach zero — a book value of nothing.
Same shape as simple interest, but with a minus.
\( A = P(1 - i)^{n} \)
Reducing-balance depreciation takes a PERCENTAGE of a shrinking value each year, so it falls fast at first, then slows — and never quite reaches zero.
\( 180\,000(0{,}88)^{4} = R107\,945{,}16 \)
A car worth R180 000 depreciates at 12% per annum on a reducing-balance basis. After 4 years: \(180\,000(1 - 0{,}12)^{4} = R107\,945{,}16\).
\( \text{Growth } (1+i) \qquad \text{Decay } (1-i) \)
The only difference between a growth and a decay formula is the sign in the bracket. Read the words: 'decreases', 'depreciates', 'loses value' all mean minus.
\( i^{(m)} \text{ nominal} \qquad i_{\text{eff}} \text{ effective} \)
The NOMINAL rate is the advertised one (say 11,5% compounded quarterly). The EFFECTIVE rate is what you actually earn over a year once the compounding is taken into account.
\( 1 + i_{\text{eff}} = \left(1 + \dfrac{i^{(m)}}{m}\right)^{m} \)
\(m\) is the number of compounding periods a year: monthly \(m=12\), quarterly \(m=4\).
\( (1{,}02875)^{4} - 1 = 12{,}01\% \)
Nominal 11,5% compounded quarterly: \(\left(1 + \dfrac{0{,}115}{4}\right)^{4} - 1 = 12{,}01\%\).
\( i_{\text{eff}} > i^{(m)} \)
Because interest compounds during the year, the effective rate is always a little higher than the nominal rate. That is exactly why banks quote the nominal one.
\( i \to \dfrac{i}{m} \qquad n \to n \times m \)
If interest compounds monthly, do BOTH: divide the annual rate by 12 AND multiply the number of years by 12. Miss one and the answer is wrong.
\( 15\,000(1{,}00575)^{96} = R26\,009{,}69 \)
R15 000 at 6,9% per annum compounded monthly for 8 years: \(i = \dfrac{0{,}069}{12},\ n = 8 \times 12 = 96\).
\( \text{Round only on the last line} \)
Do not round the rate or intermediate values — carry the full number in your calculator and round only the final rand amount.
\( \text{Move every amount to the same date} \)
When money goes in or comes out at different times, you cannot just add it. Grow each amount to a chosen focal date first, THEN combine.
\( \text{deposits } + \qquad \text{withdrawals } - \)
A deposit adds to the final value; a withdrawal subtracts. Each one is grown from the date it happened to the focal date.
\( P(1{,}01)^{60} - 2\,000(1{,}01)^{42} = 23\,564 \)
James invests P for 5 years at 12% monthly and withdraws R2 000 after 18 months. The withdrawal grows for \(60 - 18 = 42\) months.
\( P = \dfrac{23\,564 + 2\,000(1{,}01)^{42}}{(1{,}01)^{60}} = R14\,642{,}83 \)
Make P the subject: bring the withdrawal across, then divide by the growth factor for the full term.
\( A = P(1 + i)^{n} \)
Inflation and population growth are just compound GROWTH. A R1 200 bag at 6,1% inflation costs \(1\,200(1{,}061)^{5} = R1\,613{,}46\) in 5 years.
\( P = \dfrac{A}{(1 + i)^{n}} \)
To find what to invest NOW for a future goal, divide instead of multiply. For R50 000 in 6 years at 9%: \(P = \dfrac{50\,000}{(1{,}09)^{6}} = R29\,813{,}37\).
\( \text{Words} \to \text{formula} \)
Every question tells you which formula it wants: 'simple' vs 'compounded', 'increases' vs 'depreciates', 'monthly' vs 'annually'. Translate the words first, then substitute.
\[ \begin{gathered} A = P(1+i\,n) \qquad A = P(1+i)^{n} \ A = P(1-i\,n) \qquad A = P(1-i)^{n} \ 1 + i_{\text{eff}} = \left(1 + \tfrac{i^{(m)}}{m}\right)^{m} \end{gathered} \]
The whole topic on one board.