Finance, Growth & Decay
Table of Contents
Finance: Growth & Decay#
In Grade 10 you learned the basic formulas. In Grade 11, the problems get more realistic — different compounding periods, depreciation vs growth, and multi-step timelines where rates or conditions change partway through.
The Two Master Formulas#
Everything in finance comes from these two formulas:
| Situation | Formula | Use when… |
|---|---|---|
| Growth (getting bigger) | $A = P(1 + i)^n$ | Savings, investments, inflation, appreciation |
| Decay (getting smaller) | $A = P(1 - i)^n$ | Depreciation, reducing balance, population decline |
Where:
- $A$ = final amount (what you end up with)
- $P$ = principal / starting amount
- $i$ = interest rate per period (as a decimal)
- $n$ = number of periods
⚠️ The #1 Rule: $i$ and $n$ must match. If you compound monthly, then $i$ must be the monthly rate and $n$ must be in months.
Different Compounding Periods#
This is the big new idea in Grade 11. Instead of compounding once per year, interest can compound more frequently:
| Compounding | Times per year | $i$ becomes | $n$ becomes |
|---|---|---|---|
| Annually | 1 | $\frac{r}{1}$ | $t \times 1$ |
| Semi-annually | 2 | $\frac{r}{2}$ | $t \times 2$ |
| Quarterly | 4 | $\frac{r}{4}$ | $t \times 4$ |
| Monthly | 12 | $\frac{r}{12}$ | $t \times 12$ |
| Daily | 365 | $\frac{r}{365}$ | $t \times 365$ |
Where $r$ is the annual (nominal) rate and $t$ is the time in years.
Worked Example 1: Monthly Compounding#
R15 000 is invested at 9% p.a. compounded monthly for 4 years. Find the final amount.
Step 1: Identify the variables:
- $P = 15\,000$
- $r = 0.09$ (annual rate)
- Compounding: monthly → divide by 12
Step 2: Adjust $i$ and $n$:
- $i = \frac{0.09}{12} = 0.0075$
- $n = 4 \times 12 = 48$
Step 3: Substitute:
$$A = 15\,000(1 + 0.0075)^{48} = 15\,000(1.0075)^{48}$$$$A = 15\,000 \times 1.4314 = R21\,470.79$$Effective vs Nominal Interest Rate#
The nominal rate is the advertised annual rate. The effective rate is what you ACTUALLY earn after compounding.
$$i_{\text{eff}} = \left(1 + \frac{i_{\text{nom}}}{m}\right)^m - 1$$Where $m$ = number of compounding periods per year.
Worked Example 2: Finding the Effective Rate#
A bank offers 8.4% p.a. compounded monthly. What is the effective annual rate?
$$i_{\text{eff}} = \left(1 + \frac{0.084}{12}\right)^{12} - 1 = (1.007)^{12} - 1 = 1.0873 - 1 = 0.0873$$Effective rate = 8.73% p.a.
This means the investment actually grows by 8.73% per year, not just 8.4%.
Depreciation: Two Types#
| Type | Formula | How it works |
|---|---|---|
| Straight-line | $A = P(1 - in)$ | Same amount lost every year. Value drops linearly. |
| Reducing balance | $A = P(1 - i)^n$ | Percentage of current value lost each year. Drops fast then slows. |
💡 Key insight: Straight-line depreciation uses $in$ (simple), reducing balance uses $(1-i)^n$ (compound). The exam will tell you which type — read carefully!
Worked Example 3: Depreciation#
A car worth R280 000 depreciates at 15% p.a. on the reducing balance. What is it worth after 5 years?
$$A = 280\,000(1 - 0.15)^5 = 280\,000(0.85)^5 = 280\,000 \times 0.4437 = R124\,236.28$$Straight-line comparison: $A = 280\,000(1 - 0.15 \times 5) = 280\,000(0.25) = R70\,000$
The reducing balance gives a higher value because each year’s depreciation is smaller (it’s 15% of a shrinking amount).
Multi-Step Timeline Problems#
In Grade 11, you get problems where conditions change partway through. The strategy:
- Draw a timeline marking every change point.
- Work in stages — calculate $A$ at each change point.
- The $A$ from one stage becomes the $P$ for the next stage.
Worked Example 4: Rate Change#
R50 000 is invested at 10% p.a. compounded annually for 3 years, then the rate changes to 12% p.a. compounded semi-annually for 2 more years.
Stage 1 (years 0–3):
$$A_1 = 50\,000(1.10)^3 = 50\,000 \times 1.331 = R66\,550$$Stage 2 (years 3–5): $A_1$ becomes the new $P$:
$$A_2 = 66\,550\left(1 + \frac{0.12}{2}\right)^{2 \times 2} = 66\,550(1.06)^4 = 66\,550 \times 1.2625 = R84\,019.27$$Solving for Unknown Variables#
You may need to find $P$, $i$, or $n$ instead of $A$.
Finding $n$ (how long?)#
Use logarithms (or trial and error in Grade 11):
$$A = P(1 + i)^n \Rightarrow (1 + i)^n = \frac{A}{P} \Rightarrow n = \frac{\log\left(\frac{A}{P}\right)}{\log(1 + i)}$$Worked Example 5: Finding the Time#
How long will it take for R20 000 to double at 8% p.a. compounded annually?
$$40\,000 = 20\,000(1.08)^n$$$$(1.08)^n = 2$$$$n = \frac{\log 2}{\log 1.08} = \frac{0.3010}{0.0334} = 9.01 \text{ years}$$So it takes approximately 9 years to double.
🚨 Common Mistakes#
- Not adjusting $i$ and $n$ for compounding: If it says “quarterly”, you MUST divide the rate by 4 and multiply the time by 4. This is the most common error.
- Confusing growth and decay: Growth uses $(1 + i)$, decay uses $(1 - i)$. Read the problem carefully — “depreciation” = decay.
- Straight-line vs reducing balance: Straight-line is $P(1 - in)$, reducing balance is $P(1 - i)^n$. Don’t mix the formulas.
- Rounding too early: Keep all decimals during calculation, only round the final answer to 2 decimal places.
- Timeline problems — using wrong $P$: Each stage’s starting value is the PREVIOUS stage’s final value, not the original $P$.
💡 Pro Tip: The “Double Check” — Growth vs Decay#
Before you calculate, ask yourself: “Should this number be BIGGER or SMALLER than what I started with?”
- Investment / savings → bigger (growth)
- Depreciation / car value → smaller (decay)
If your answer goes the wrong direction, you’ve used the wrong formula.
🔗 Related Grade 11 topics:
- Exponential Functions — compound growth IS an exponential function
- Surds & Exponential Equations — solving for $n$ uses logarithms/exponent skills
📌 Grade 10 foundation: Simple & Compound Interest — the basic formulas
📌 Grade 12 extension: Future Value & Annuities and Present Value & Loans — regular payments
⏮️ Functions | 🏠 Back to Grade 11 | ⏭️ Probability
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