Both interest formulas and the question types a Grade 10 exam sets — finding the final amount, the interest earned, working back to the principal or rate, and inflation — each worked in full, with a worksheet. Depreciation is shown only as a preview of Grade 11.
Also known as: compound interest formula · simple interest · interest calculations
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Financial maths is among the most useful — and most reliably examined — maths you'll learn. Almost all of it starts with two formulas: simple interest and compound interest. This guide covers both, plus every question type built on them.
1The two formulas
A = final amount, P = principal (starting amount), i = interest rate as a decimal, n = number of years. The only difference: in simple interest n multiplies; in compound it is a power.
2Type 1: Simple interest (find the amount)
- 1A = P(1 + i·n) = 2000(1 + 0,08 × 3).i = 8% = 0,08, n = 3.
- 2= 2000(1,24) = R2 480.Interest earned = R480 (the same R160 each year).
3Type 2: Compound interest (find the amount)
- 1A = P(1 + i)n = 2000(1,08)3.Each year's interest earns interest the next year.
- 2= 2000 × 1,259712 ≈ R2 519,42.About R39 more than simple over the same 3 years.
Compound always beats simple after year 1, and the gap grows the longer the money is invested.
4Type 3: Find the interest earned
- 1A = 5 000(1,06)4 = 5 000 × 1,262477 ≈ R6 312,38.Find the final amount first.
- 2Interest = A − P = 6 312,38 − 5 000 = R1 312,38.Interest is the growth on top of the principal.
5Type 4: Work back to the principal
- 1A = P(1 + i)n → 10 000 = P(1,07)5.You know A, want P.
- 2P = 10 000 ÷ (1,07)5 = 10 000 ÷ 1,402552.Divide, don't subtract.
- 3P ≈ R7 129,86.This is the 'present value'.
6Type 5: Work back to the rate
- 1Interest = 960 − 800 = R160.The growth over the whole time.
- 2Simple: I = P·i·n → 160 = 800 × i × 2 = 1600i.Substitute into the interest formula.
- 3i = 160 ÷ 1 600 = 0,10 = 10%.Convert the decimal back to a percentage.
7Type 6: Inflation
Inflation is prices rising, and it rises compound — this year's increase is worked out on this year's price, not on the old one. So it uses the ordinary compound-growth formula, exactly as interest does. Population growth works the same way (that type is in Hire Purchase, Exchange Rates & Population Growth).
- 1A = P(1 + i)n = 18(1 + 0,06)4 = 18(1,06)4.Inflation is compound growth, so the rate is added: 1 + i.
- 2= 18 × 1,262477 ≈ R22,72.Keep the full value in the calculator and round only at the end.
Always convert the percentage to a decimal (9% → 0,09) — the single most common financial-maths slip. And read carefully: an amount that grows uses 1 + i, whether it is interest, inflation or a population.
To work back to P, divide — don't subtract. If R10 000 is the future value, the present value is 10 000 ÷ (1 + i)n, never 10 000 minus the interest.
8Looking ahead: Grade 11 — depreciation (decay)
Everything above is Grade 10 work: the Grade 10 ATP asks for the simple and compound growth formulae, applied to interest, hire purchase, inflation and population growth. Something that loses value — a car, a machine, a laptop — is decay, and decay is a Grade 11 topic. It is shown here because the formula is the growth one with a minus, and because you will meet it next year. It is not examined in Grade 10: nothing in this lesson's practice, quiz or paper uses it.
Straight from the ATP. Grade 10: “Use the simple and compound growth formulae … to solve problems, including interest, hire purchase, inflation, population growth and other real-life problems.” Grade 11 then adds: “Use simple and compound decay formulae …”, covering straight-line depreciation and depreciation on a reducing balance.
- 1A = P(1 − i)n = 180 000(1 − 0,15)3 = 180 000(0,85)3.Decay takes value away, so the rate is subtracted: 1 − i.
- 2= 180 000 × 0,614125 ≈ R110 542,50.In Grade 11 you also meet the straight-line version, which takes the same rand amount off every year and gives a different answer.
Financial maths grows into annuities and loans in Grade 12, all built on these formulas. Practise every type below.
Practice exercises
Work each one out, then click to reveal the answer.
- 1R1 000 at 10% simple interest for 2 years — total? (simple)
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1000(1 + 0,10 × 2) = R1 200 - 2R3 000 at 5% simple interest for 4 years — total? (simple)
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3000(1 + 0,20) = R3 600 - 3R1 000 at 10% compound for 2 years — total? (compound)
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1000(1,10)2 = R1 210 - 4R5 000 at 6% compound for 3 years (to cents). (compound)
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5000(1,06)3 ≈ R5 955,08 - 5Interest on R2 000 at 8% compound over 2 years. (interest)
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2000(1,08)2 − 2000 = 2332,80 − 2000 = R332,80 - 6What must you invest at 8% compound to get R5 000 in 3 years? (present value)
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5000 ÷ (1,08)3 ≈ R3 969,16 - 7R500 grows to R650 in 3 years (simple). Find the rate. (rate)
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150 = 500·i·3 = 1500i → i = 10% - 8A loaf of bread costs R22 today. Inflation has run at 6% p.a. for the past 4 years. What did it cost 4 years ago? (inflation, backwards)
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22 ÷ (1,06)4 = 22 ÷ 1,262477 ≈ R17,43 - 9Which is more after 5 years on R1 000 at 10% — simple or compound? (compare)
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Simple: 1 000(1 + 0,1 × 5) = R1 500
Compound: 1 000(1,1)5 = R1 610,51
Compound, by R110,51 — it earns interest on interest - 10A R60 item rises with 8% inflation for 1 year. New price? (inflation)
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60 × 1,08 = R64,80
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