Compound interest is repeated percentage change. The amount changes, then the next percentage is applied to the new amount. This makes the process multiplicative rather than additive and connects financial mathematics directly to percentages, indices and exponential growth.
This Secondary 3 Mathematics Learning Guide develops compound interest, repeated growth, depreciation, reverse growth, comparison of financial models and calculator discipline. It complements the earlier guide on ratio, percentage, rate and speed by concentrating on repeated multiplicative change across several periods.
The current 2027 SEC G3 Mathematics syllabus listing from SEAB identifies Mathematics as K310, and the official formula list includes the compound-interest relationship A=P(1+r/100)ⁿ. This article uses that formula as a model of repeated percentage growth while keeping the interpretation explicit.
Simple Growth Versus Compound Growth
If $1000 increases by 5% once, the increase is $50 and the new amount is $1050. If it increases by another 5%, the second increase is 5% of $1050, not 5% of the original $1000.
After two periods, amount = 1000×1.05×1.05 = 1000(1.05)² = $1102.50. Adding two lots of $50 would give $1100 and would describe a different model.
The Compound-Interest Formula
A=P(1+r/100)ⁿ, where P is the initial principal, r is the percentage rate per compounding period, n is the number of such periods and A is the final amount under the stated model.
The multiplier 1+r/100 represents 100% of the existing amount plus r% growth. The exponent n records how many times the same multiplier is applied.
Worked Example 1: Annual Compound Interest
$2500 is invested at 4% per year compounded annually for 3 years. Find the final amount.
A=2500(1.04)³ = 2500×1.124864 = $2812.16.
The interest earned is 2812.16−2500=$312.16. The final amount and the interest are different quantities, so read the question carefully.
Why Repeated Percentage Change Uses Powers
One 3% increase multiplies by 1.03. Two increases multiply by 1.03². Ten increases multiply by 1.03¹⁰. The exponent is therefore a compact record of repeated identical multipliers.
This is also why repeated growth produces an exponential model: the variable number of periods appears in the exponent.
Worked Example 2: Population-Style Growth Model
A simplified model begins at 12,000 and grows by 2.5% per year for 6 years. Find the modelled value after 6 years.
Value=12000(1.025)⁶≈13,916 to the nearest whole number.
This is a mathematical model, not a claim that a real population must grow at exactly 2.5% every year. The multiplier is an assumption supplied by the model.
Depreciation Uses a Multiplier Below 1
A decrease of r% leaves 100−r percent of the current value. The multiplier is 1−r/100.
Thus 15% depreciation uses multiplier 0.85. Repeated annual depreciation for n years gives V=P(0.85)ⁿ under the constant-rate model.
Worked Example 3: Depreciation
A machine is valued at $18,000 and depreciates by 12% per year. Find its modelled value after 4 years.
V=18000(0.88)⁴≈$10,794.52.
The value lost is about $7205.48. Notice that the annual dollar decrease is not constant because 12% is applied to a shrinking base.
Repeated Increase and Decrease Do Not Cancel
An increase of 20% followed by a decrease of 20% has multiplier 1.2×0.8=0.96. The final amount is 96% of the original, an overall 4% decrease.
The reason is that the two percentages use different bases. Equal percentage numbers do not imply equal absolute changes.
Worked Example 4: Two Different Changes
A price rises by 8% and later falls by 5%. Find the overall percentage change.
Net multiplier=1.08×0.95=1.026. Therefore the final price is 102.6% of the original, giving an overall 2.6% increase.
Reverse Compound Growth
If the final amount, rate and number of periods are known, divide by the growth factor to recover the original:
P=A/(1+r/100)ⁿ.
This is the repeated-change version of reverse percentage.
Worked Example 5: Recover the Principal
An investment is worth $5324 after 2 years at 10% annual compound growth. Find the initial amount.
P=5324/(1.10)² = 5324/1.21 = $4400.
Check: 4400×1.21=5324.
Find the Number of Periods by Testing Powers
At this level, some questions may be designed so the number of periods can be identified by repeated multiplication or calculator testing rather than logarithms.
If $1000 grows at 10% per period, values are 1100 after 1 period, 1210 after 2, 1331 after 3 and 1464.10 after 4. A target can be matched against these powers.
Worked Example 6: Compare Two Growth Plans
Plan A grows $5000 by 4% per year for 5 years. Plan B adds a fixed $220 per year for 5 years. Which produces more?
Plan A: 5000(1.04)⁵≈$6083.26.
Plan B: 5000+5(220)=$6100.
Plan B is higher by about $16.74 after five years under these stated models. Compound growth does not automatically beat every linear-growth plan over every time horizon.
Annual Rate and Compounding Period Must Match
The rate r in the formula must correspond to one compounding period, and n must count those same periods. If a question explicitly gives a monthly rate, n counts months. If it gives an annual rate compounded annually, n counts years.
Do not convert an annual rate to a monthly rate unless the problem states how the conversion should be modelled. Financial products can use conventions beyond the simple school model.
Worked Example 7: Monthly Model With a Stated Monthly Rate
$3000 grows at a stated rate of 0.5% per month for 12 months. Find the final amount.
A=3000(1.005)¹²≈$3185.03.
This example uses a monthly rate because that is explicitly given. It should not be confused with simply dividing some unrelated annual rate by 12 unless the model says to do so.
Growth Factor and Percentage Change Are Inverse Descriptions
A multiplier of 1.07 means a 7% increase. A multiplier of 0.82 means an 18% decrease. A multiplier of 1.245 means a 24.5% increase.
This conversion is useful when several multipliers have already been combined.
Worked Example 8: Overall Change From Multipliers
A value changes by multipliers 1.12, 0.95 and 1.03 in three successive periods. Find the overall percentage change.
Net multiplier=1.12×0.95×1.03=1.09592. Therefore the overall change is a 9.592% increase.
Rounding Financial Results
For currency, the final answer is often rounded to the nearest cent unless the question specifies otherwise. Keep full calculator precision through intermediate stages and round only at the end.
Rounding the amount after every year can create a slightly different final result from applying the compound formula directly. Follow the stated convention in the problem.
Financial Mathematics Is Still Modelling
A school compound-interest model usually assumes a fixed percentage rate, regular compounding and no extra deposits, withdrawals, fees or taxes unless stated. Real financial products may behave differently.
The mathematical task is to answer the model that was given, while keeping clear which assumptions make that model work.
Common Errors
Adding the same interest amount each year: repair by applying the percentage to the current amount.
Using 4 instead of 1.04 for a 4% growth multiplier: repair by translating 104% into decimal form.
Confusing final amount with interest earned: repair by subtracting the principal only when interest itself is requested.
Rounding every intermediate stage: repair by keeping calculator precision until the final answer.
Independent Practice
1. Find the final amount for $1800 at 5% compounded annually for 4 years.
2. Find the interest earned in Question 1.
3. A value of 9000 grows by 3% per year for 5 years. Find the final value.
4. A machine worth $25,000 depreciates 10% per year for 3 years. Find its modelled value.
5. A price rises 15% then falls 10%. Find the overall percentage change.
6. A final amount of $5832 results after 2 years at 8% annual compound growth. Find the principal.
7. Compare $4000 growing 6% annually for 4 years with $4000 plus $270 each year for 4 years.
8. A value changes by multipliers 1.05, 1.04 and 0.98. Find the overall percentage change.
9. $2500 grows at a stated monthly rate of 0.4% for 18 months. Find the final amount.
10. Explain why a 30% increase followed by a 30% decrease is an overall decrease.
Explained Answers
1. 1800(1.05)⁴≈$2187.91.
2. $2187.91−$1800=$387.91.
3. 9000(1.03)⁵≈$10,433.47.
4. 25000(0.90)³=$18,225.
5. 1.15×0.90=1.035, so overall increase=3.5%.
6. P=5832/(1.08)²=5832/1.1664=$5000.
7. Compound plan: 4000(1.06)⁴≈$5049.91. Linear plan: 4000+1080=$5080, so the linear plan is higher by about $30.09 after four years.
8. Net multiplier=1.05×1.04×0.98=1.07016, so increase=7.016%.
9. 2500(1.004)¹⁸≈$2686.25.
10. Multipliers give 1.3×0.7=0.91, leaving 91% of the original, a 9% decrease.
Continue the Secondary 3 Learning Route
Continue with Number Structure, Prime Factorisation, HCF and LCM, Algebraic Patterns, nth-Term Rules and Identities, and Simultaneous Linear Equations and Modelling.
Financial growth becomes mathematically clear when every percentage is converted into the multiplier applied to the current value. Return to the Secondary Mathematics Hub.
