Finance: Growth & Decay (Grade 11)

Compound growth and decay, straight-line versus reducing-balance depreciation, different compounding periods, and the difference between a nominal and an effective interest rate.

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
= R22039,92.

2Depreciation: 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
= R125281,50.
Higher — reducing balance never reaches zero.

⚠️

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

3Compounding more often than yearly

A = P(1 + im)^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 + 0,0812)⁶⁰.
mn = 12 × 5 = 60 periods.
3
= R22347,69.
More than the R22039,92 from annual compounding — interest earns interest sooner.

4Nominal versus effective rates

1 + i_eff = (1 + i_nomm)^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 + 0,0812)¹².
2
= 1,083000.
3
i_eff = 8,30% p.a.
Always slightly higher than the nominal rate.

5Inflation, 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.

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.
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15 000(1,08)⁵ = R22039,92
2
R15 000 is invested at 8% p.a. compounded monthly. Calculate the value after 5 years.
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15 000(1 + 0,0812)⁶⁰ = R22347,69
3
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
4
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 = R125281,50
5
Convert 8% p.a. compounded monthly to an effective annual rate.
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(1 + 0,0812)¹² − 1 = 8,30%
6
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
7
R50 000 grows to R80 000 at 9% p.a. compounded annually. Calculate n.
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1,09ⁿ = 1,6 → n = 5,45 years
8
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.