The Fundamental Idea#
Interest is the cost of using money. If you borrow money, you pay interest. If you save money, you earn interest. Grade 10 introduces the two basic types.
1. Simple Interest#
Interest is calculated only on the original amount (principal). The amount of interest is the same every year.
$$ A = P(1 + ni) $$- $A$: Final amount (what you end up with)
- $P$: Principal (the starting amount)
- $n$: Number of years
- $i$: Interest rate as a decimal ($8\% = 0.08$)
Why $(1 + ni)$?#
After $n$ years, you have the original amount PLUS the interest:
- Interest = $P \times n \times i$
- Total = $P + Pni = P(1 + ni)$
Worked Example 1#
R5 000 is invested at 8% simple interest for 3 years. Find the total amount.
$A = 5000(1 + 3 \times 0.08) = 5000(1.24) = \text{R}6\,200$
Interest earned: $6200 - 5000 = \text{R}1\,200$
Worked Example 2: Finding the rate#
R2 000 grows to R2 600 in 4 years with simple interest. What is the rate?
$2600 = 2000(1 + 4i)$
$1.3 = 1 + 4i$
$4i = 0.3$
$i = 0.075 = 7.5\%$
2. Compound Interest#
Interest is calculated on the new total each year (interest on interest). The amount grows faster and faster.
$$ A = P(1 + i)^n $$Why $(1 + i)^n$?#
Each year, the amount is multiplied by $(1 + i)$:
- After 1 year: $P(1 + i)$
- After 2 years: $P(1 + i)(1 + i) = P(1 + i)^2$
- After $n$ years: $P(1 + i)^n$
Worked Example 1#
R5 000 is invested at 8% compound interest for 3 years. Find the total amount.
$A = 5000(1.08)^3 = 5000(1.259712) = \text{R}6\,298.56$
Compare: Simple interest gave R6 200. Compound interest gives R6 298.56 — that’s R98.56 MORE because you earned “interest on interest.”
Worked Example 2: Finding the time#
How many years for R10 000 to double at 12% compound interest?
$20000 = 10000(1.12)^n$
$2 = (1.12)^n$
Using trial and improvement or logarithms:
$(1.12)^6 = 1.974$ (not quite)
$(1.12)^7 = 2.211$ (passed it)
It takes approximately 7 years to double.
The Rule of 72: A quick estimate — divide 72 by the interest rate. $72 \div 12 = 6$ years. This gives a rough answer fast.
3. Simple vs Compound: The Comparison#
| Feature | Simple Interest | Compound Interest |
|---|---|---|
| Formula | $A = P(1 + ni)$ | $A = P(1 + i)^n$ |
| Growth type | Linear (constant) | Exponential (accelerating) |
| Interest earned | Same each year | Increases each year |
| Graph shape | Straight line | Curve (gets steeper) |
| Better for saving? | No | Yes (earns more) |
| Better for borrowing? | Yes (costs less) | No (costs more) |
4. Hire Purchase#
Hire purchase is a way to buy expensive items (like a TV or car) by paying monthly. It ALWAYS uses simple interest on the cash price.
Worked Example#
A laptop costs R12 000 cash. You buy it on hire purchase at 15% p.a. simple interest over 2 years.
Total: $A = 12000(1 + 2 \times 0.15) = 12000(1.3) = \text{R}15\,600$
Monthly payment: $\frac{15600}{24} = \text{R}650$ per month
You pay R3 600 MORE than the cash price — that’s the cost of “buying time.”
Usually a deposit is required first. If the deposit is 10%: Deposit = $12000 \times 0.10 = \text{R}1\,200$ Loan = $12000 - 1200 = \text{R}10\,800$ Then apply the interest to R10 800, not R12 000.
5. Inflation (The “Shrinking Rand”)#
Inflation uses the compound interest formula to show how prices INCREASE over time.
$$ A = P(1 + i)^n $$Worked Example#
A loaf of bread costs R18 today. If inflation is 6% per year, what will it cost in 5 years?
$A = 18(1.06)^5 = 18(1.3382) = \text{R}24.09$
6. Population Growth and Decay#
The same formula works for any quantity that grows or shrinks:
- Growth: $A = P(1 + i)^n$ (population, prices)
- Decay: $A = P(1 - i)^n$ (depreciation, radioactive decay)
Worked Example: Depreciation#
A car worth R250 000 depreciates at 15% per year. What is it worth after 4 years?
$A = 250000(1 - 0.15)^4 = 250000(0.85)^4 = 250000(0.5220) = \text{R}130\,502$
7. Exchange Rates#
Converting between currencies:
- Buying foreign currency: You pay MORE rands (use the higher rate)
- Selling foreign currency: You get FEWER rands (use the lower rate)
Worked Example#
Exchange rate: R18.50/$ (buy) and R18.00/$ (sell). You want to buy $500.
Cost = $500 \times 18.50 = \text{R}9\,250$
🚨 Common Mistakes#
- Using the wrong formula: Simple interest has $n$ OUTSIDE the bracket: $P(1 + ni)$. Compound interest has $n$ as a POWER: $P(1 + i)^n$. Mixing these up is devastating.
- Forgetting to convert the percentage: $8\% = 0.08$, not $8$. If you use $i = 8$, your answer will be astronomically wrong.
- Hire purchase deposit: Calculate interest on the LOAN amount (after deposit), not the full cash price.
- Depreciation uses minus: $A = P(1 - i)^n$, NOT $(1 + i)^n$. The value is DECREASING.
- Rounding too early: Keep all decimal places during calculation. Only round the final answer.
💡 Pro Tip: The “Growth or Decay?” Check#
Before you start, ask: “Is this thing getting bigger or smaller?”
- Getting bigger → use $(1 + i)$
- Getting smaller → use $(1 - i)$
🔗 Related Grade 10 topics:
- Exponent Laws — compound interest uses the power law $(1+i)^n$
- Sketching Graphs — compound growth IS an exponential function
📌 Where this leads in Grade 12:
- Future Value & Sinking Funds — annuities and regular payments
- Present Value & Loans — how home loans and car payments work