Finance: Growth & Decay — Grade 11 Maths (CAPS)

Compound growth and decay, straight-line versus reducing-balance depreciation, different compounding periods, the difference between a nominal and an effective interest rate, and the Grade 10 hire-purchase revision the ATP still expects.

1The four formulae

Growth
Simple: A = P(1 + in) · Compound: A = P(1 + i)ⁿ

Decay
Straight-line: A = P(1 − in) · Reducing balance: A = P(1 − i)ⁿ

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Simple and straight-line are the ones without a power. Compound and reducing balance both have the ⁿ.

Worked ExampleR15 000 is invested at 8% p.a. compounded annually for 5 years

1
A = 15 000(1 + 0,08)⁵.
i = 8% = 0,08, n = 5.
2
= 15 000(1,08)⁵.
3
= 15 000 × 1,469328.
4
= R22 039,92.

2Revision from Grade 10: hire purchase

The Grade 11 ATP asks you to revise the growth formulae “including interest, hire purchase, inflation, population growth”, so it can appear in a Grade 11 paper without warning. Hire purchase is a deposit plus simple interest charged on what is still owed — not on the full cash price. The full treatment is in the Grade 10 lesson Hire Purchase, Exchange Rates & Population Growth; this is the shape of the question.

Worked ExampleA laptop's cash price is R14 500. On hire purchase: 15% deposit, the balance over 3 years at 12,5% p.a. simple interest. Find the monthly instalment.

1
Deposit = 0,15 × 14 500 = R2 175, so the loan is 14 500 − 2 175 = R12 325.
Interest is charged on the BALANCE, never on the cash price.
2
A = P(1 + in) = 12 325(1 + 0,125 × 3) = 12 325 × 1,375 = R16 946,88
Hire purchase is simple interest — no exponent.
3
Monthly instalment = 16 946,88 ÷ 36 = R470,75
Three years is 36 months.
4
Total paid = 2 175 + 16 946,88 = R19 121,88, which is R4 621,88 more than the cash price.
That difference is exactly the interest: 12 325 × 0,375 ✓

3Depreciation: two very different answers

Worked ExampleA car worth R240 000 depreciates at 15% p.a. for 4 years

1
Straight-line: A = 240 000(1 − 0,15 × 4) = 240 000(0,4) = R96 000.
The same rand amount is lost every year.
2
Reducing balance: A = 240 000(1 − 0,15)⁴ = 240 000(0,85)⁴.
Each year 15% of what is LEFT is lost.
3
= R125 281,50.
Higher — reducing balance never reaches zero.

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Read the question. 'Depreciates on a reducing balance' and 'depreciates on a straight line' give completely different answers from the same numbers.

4Compounding more often than yearly

A = P(1 + i/m)^mn

m = 12 monthly · 4 quarterly · 2 half-yearly

Worked ExampleR15 000 at 8% p.a. compounded monthly for 5 years

1
i = 0,08, m = 12, n = 5.
2
A = 15 000(1 + 0,08/12)⁶⁰.
mn = 12 × 5 = 60 periods.
3
= R22 347,69.
More than the R22 039,92 from annual compounding — interest earns interest sooner.

5Nominal versus effective rates

1 + i_eff = (1 + i_nom/m)^m

The nominal rate is the advertised yearly rate. The effective rate is what you actually earn once compounding is taken into account.

Worked ExampleConvert 8% p.a. compounded monthly to an effective annual rate

1
1 + i_eff = (1 + 0,08/12)¹².
2
= 1,083000.
3
i_eff = 8,30% p.a.
Always slightly higher than the nominal rate.

6Inflation, population and exchange rates

These all use the compound formula. Inflation makes prices grow; a weakening rand means you pay more rand for the same dollar.

Worked ExampleA loaf costs R18. Inflation is 6% p.a. What will it cost in 8 years?

1
A = 18(1,06)⁸.
Inflation is compound growth.
2
= R28,69.

7Fluctuating exchange rates

The ATP asks you to understand the implication of a changing exchange rate — on the petrol price, on imports and exports, and on overseas travel. The arithmetic is easy; the marks are in the explanation.

A weaker rand (R18 → R20 to the dollar) means you need MORE rands for each dollar. Imports and petrol get more expensive, and overseas travel costs more.
A stronger rand means fewer rands per dollar. Imports get cheaper — but exporters earn less for the same goods.
Petrol is priced in dollars per barrel, so a weakening rand pushes the pump price up even when the oil price has not moved at all.

Worked ExampleA tourist changes R24 000 into pounds when the rate is R23,50 to the pound, spends £600, and changes the rest back when the rate has moved to R22,80. How much does she get back?

1
24 000 ÷ 23,50 = £1 021,28
Rands to pounds: divide by the rate.
2
£1 021,28 − £600 = £421,28 left
3
421,28 × 22,80 = R9 605,18
Pounds back to rands: multiply by the new rate.
4
She receives about R9 605.
The rand strengthened, so each pound is worth less coming back — she loses on the swing.

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Read what ONE unit of the quoted currency costs first. When the rate is given as rands per one unit — R23,50 per £1, R18,75 per $1 — you divide the rand amount by the rate to get the foreign currency, and multiply to come back. A rate quoted the other way round (units per R1) reverses both operations, so always check the wording before you press a key.

Practice exercises

Work each one out, then click to reveal the answer.

📄 Worksheet ✅ Memo 📝 Exam paper 🗝️ Marking guide

1
R15 000 is invested at 8% p.a. compounded annually. Calculate the value after 5 years.
Show answer ▾Hide answer ▴
15 000(1,08)⁵ = R22 039,92
2
R15 000 is invested at 8% p.a. compounded monthly. Calculate the value after 5 years.
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15 000(1 + 0,08/12)⁶⁰ = R22 347,69
3
A fridge has a cash price of R9 200. On hire purchase a 20% deposit is paid and the balance is repaid over 2 years at 14% p.a. simple interest. Calculate the TOTAL amount paid. (hire purchase revision)
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Deposit = 0,20 × 9 200 = R1 840; loan = R7 360; A = 7 360(1 + 0,14 × 2) = 7 360 × 1,28 = R9 420,80; total = 1 840 + 9 420,80 = R11 260,80
4
How much MORE than its cash price does that fridge cost on hire purchase? (hire purchase revision)
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11 260,80 − 9 200 = R2 060,80 — which is exactly the interest, 7 360 × 0,28
5
A car worth R240 000 depreciates at 15% p.a. on a straight line. Calculate its value after 4 years.
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240 000(1 − 0,6) = R96 000
6
A car worth R240 000 depreciates at 15% p.a. on a reducing balance. Calculate its value after 4 years.
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240 000(0,85)⁴ = R125 281,50
7
Convert 8% p.a. compounded monthly to an effective annual rate.
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(1 + 0,08/12)¹² − 1 = 8,30%
8
A loaf of bread costs R18. If inflation is 6% p.a., what will it cost in 8 years?
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18(1,06)⁸ = R28,69
9
R50 000 grows to R80 000 in 6 years at compound interest, compounded annually. Calculate the annual interest rate.
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80 000 = 50 000(1 + i)⁶ → (1 + i)⁶ = 1,6
1 + i = 1,6^1/6 = 1,0815
i = 0,0815 → 8,15% p.a.
10
State the difference between a nominal and an effective interest rate.
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The nominal rate is the advertised annual rate before compounding; the effective rate is the true annual rate once compounding within the year is included.
11
The rate moves from R18,20 to R19,60 to the dollar. Has the rand strengthened or weakened, and what happens to the price of imported goods?
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Weakened ✓ (more rands are needed per dollar); imported goods become more expensive ✓
12
A camera costs $340. Calculate its cost in rands at R18,75 to the dollar, and then at R20,10 to the dollar.
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340 × 18,75 = R6 375 ✓; 340 × 20,10 = R6 834 ✓; an increase of R459 ✓
13
Explain why the petrol price can rise in South Africa even when the international oil price has not changed.
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Oil is priced in dollars ✓; if the rand weakens, more rands are needed to buy the same dollar amount of oil ✓, so the pump price rises without the oil price moving ✓