Both interest formulas and every question type an exam sets — finding the final amount, the interest earned, working back to the principal or rate, depreciation, and inflation — each worked in full, with a worksheet.
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) = R2480.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 ≈ R2519,42.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 = 5000(1,06)4 = 5000 × 1,262477 ≈ R6312,38.Find the final amount first.
- 2Interest = A − P = 6312,38 − 5000 = R1312,38.Interest is the growth on top of the principal.
5Type 4: Work back to the principal
- 1A = P(1 + i)n → 10000 = P(1,07)5.You know A, want P.
- 2P = 10000 ÷ (1,07)5 = 10000 ÷ 1,402552.Divide, don't subtract.
- 3P ≈ R7129,86.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 ÷ 1600 = 0,10 = 10%.Convert the decimal back to a percentage.
7Type 6: Depreciation and inflation
Depreciation (something losing value) uses the compound formula with a minus: A = P(1 − i)n. Inflation (rising prices) uses the ordinary compound-growth formula.
- 1A = P(1 − i)n = 180000(1 − 0,15)3 = 180000(0,85)3.Depreciation reduces the value, so use 1 − i.
- 2= 180000 × 0,614125 ≈ R110 542,50.
Always convert the percentage to a decimal (9% → 0,09) — the single most common financial-maths slip. And read carefully: growth uses 1 + i, depreciation uses 1 − i.
To work back to P, divide — don't subtract. If R10 000 is the future value, the present value is 10000 ÷ (1+i)n, never 10000 minus the interest.
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.
- 1R1000 at 10% simple interest for 2 years — total? (simple)
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1000(1 + 0,10×2) = R1200 - 2R3000 at 5% simple interest for 4 years — total? (simple)
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3000(1 + 0,20) = R3600 - 3R1000 at 10% compound for 2 years — total? (compound)
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1000(1,10)2 = R1210 - 4R5000 at 6% compound for 3 years (to cents). (compound)
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5000(1,06)3 ≈ R5955,08 - 5Interest on R2000 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 R5000 in 3 years? (present value)
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5000 ÷ (1,08)3 ≈ R3969,16 - 7R500 grows to R650 in 3 years (simple). Find the rate. (rate)
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150 = 500·i·3 = 1500i → i = 10% - 8A R20 000 asset depreciates at 10%/yr. Value after 2 years? (depreciation)
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20000(0,90)2 = R16 200 - 9Which is more after 5 years on R1000 at 10% — simple or compound? (compare)
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Compound (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
Now practise it
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