Secondary Mathematics: Financial Arithmetic — Growth, Depreciation and Interest

Secondary Mathematics · Worked Repair Guide 16

Financial arithmetic is percentage mathematics with a timeline. The same ideas behind ratios, growth factors and reverse percentages reappear in discounts, mark-ups, simple interest, compound growth, depreciation and instalment calculations. The main challenge is identifying the correct base at each stage.

This guide develops one central habit: write the multiplier and the time step before pressing the calculator. A 6% increase means multiply by 1.06. A 6% decrease means multiply by 0.94. Repeated change means repeated multiplication, not repeated addition of the original percentage amount.

All money examples are invented teaching models. This material is for Mathematics learning only and is not financial, investment, credit or purchasing advice.

1. Percentages need a base

Ten per cent of $200 is $20 because 10/100×200=20. The number 200 is the base quantity.

If a price rises from $200 to $220, the increase is $20 and the percentage increase relative to the original price is 20/200×100%=10%.

Entry check: an increase from $80 to $100 is $20, which is 25% of the original $80. It is not 20% merely because the increase is 20 dollars.

The denominator determines the interpretation. Always ask “percentage of what?”

2. Growth multipliers compress percentage increase

An increase of r% multiplies the original amount by 1+r/100.

A 7% increase uses multiplier 1.07. Thus $500 becomes 500×1.07=$535.

The multiplier contains both the original 100% and the added 7%.

Writing the multiplier first reduces errors because it records the direction of change before any arithmetic.

3. Depreciation multipliers represent percentage decrease

A decrease of r% multiplies by 1−r/100.

A 12% decrease uses multiplier 0.88. An invented value of $2500 becomes 2500×0.88=$2200 after one decrease.

Do not subtract 12 from the dollar value. The percentage acts on the current base quantity.

A 100% decrease gives multiplier zero. A decrease greater than 100% may be mathematically expressible in some abstract models but usually requires careful contextual interpretation.

4. Repeated percentage change is multiplicative

If a quantity grows by 5% per period for three periods, the multiplier is 1.05³.

An initial $1000 becomes 1000×1.05³=$1157.625, or $1157.63 if rounded to cents.

The total increase is 15.7625%, not exactly 15%, because each later 5% is applied to a new, larger base.

This is the central difference between repeated percentage growth and repeated addition of a fixed amount.

5. Repeated depreciation also compounds

A 20% annual depreciation means multiply by 0.8 each year.

An invented asset valued at $5000 becomes 5000×0.8²=$3200 after two years under this model.

Subtracting 20% of the original $5000 twice would give $3000, which describes a different model.

The phrase “20% of its value each year” normally signals a changing base and therefore repeated multiplication.

6. An increase followed by the same percentage decrease does not return to the start

Start with 100. Increase by 20%: 100×1.2=120. Then decrease the new amount by 20%: 120×0.8=96.

The combined multiplier is 1.2×0.8=0.96.

The changes do not cancel because the second percentage uses a different base.

This is a useful example of why percentage language must always carry its reference quantity.

7. Reverse percentage divides by the multiplier

Suppose a price after a 15% increase is $460. Let the original price be P. Then 1.15P=460, so P=460/1.15=$400.

Do not find 15% of 460 and subtract it. The 15% increase was based on the original amount, not the final amount.

Worked example: after a 20% discount, a price is $144. Original price=144/0.8=$180.

Reverse percentage is an equation-solving problem disguised as a shopping or finance context.

8. Discount and mark-up are opposite directions, not inverse percentages

A 25% mark-up uses multiplier 1.25. A 25% discount uses multiplier 0.75.

Starting from $80, a 25% mark-up gives $100. Discounting $100 by 25% gives $75, not $80.

To return from $100 to $80 requires a 20% discount because 80/100=0.8.

Equal percentage labels do not generally reverse one another when their bases differ.

9. Simple interest adds a fixed percentage of the original principal

In a simple-interest model, interest each period is calculated on the original principal P.

Simple interest I=Prt

where r is the rate per period written as a decimal and t is the number of periods under the model.

Worked example: P=$2000, r=0.04 per year, t=3 years. Interest=2000×0.04×3=$240. Final amount=$2240.

The model assumes the same $80 interest each year because the base remains the original $2000.

10. Compound interest changes the base each period

In a compound-growth model with annual rate r, the amount after n periods is A=P(1+r)ⁿ where the rate and period units match.

For P=$2000 at 4% annually for three years, A=2000(1.04)³≈$2249.73.

This is slightly larger than the simple-interest result because later growth is applied to previous growth as well as the original principal.

Real financial products may include fees, taxes, different compounding conventions and contractual rules. School models simplify those details unless explicitly included.

11. The period of the rate must match the exponent

If a model gives 2% growth per month for six months, use 1.02⁶.

Do not use an annual exponent unless the rate is annual or has been converted according to the model.

Worked example: $1500 growing at 1% per month for 8 months gives 1500×1.01⁸≈$1624.28.

Write “1% per month × 8 months → 8 multiplications by 1.01” before calculating.

12. Comparing simple and compound models reveals the changing base

Take $1000 at 10% for two years.

Simple model: interest=$100 each year, final amount=$1200.

Compound model: $1000×1.1²=$1210.

The $10 difference is the second-year growth on the first year’s $100 growth.

This small example makes the structural difference visible without large numbers.

13. Instalment arithmetic separates total cost from timing

In a simplified school model, a product might require a deposit plus equal monthly payments.

If an invented purchase has $300 deposit plus 18 payments of $75, total paid=300+18×75=$1650.

If the cash price is $1500, the simplified difference is $150. As a percentage of the cash price, that is 150/1500×100%=10%.

This arithmetic does not by itself describe the true cost of credit or compare real financing products. It is a school Mathematics calculation using stated values only.

14. Profit, loss and margin use different bases

If an item costs $80 and sells for $100, profit is $20.

Profit as a percentage of cost is 20/80×100%=25%.

Profit as a percentage of selling price is 20/100×100%=20%.

The numerator is the same; the denominator changes. Terms such as mark-up and margin can have specific conventions, so use the definition supplied by the problem.

15. Inflation-style models are repeated percentage growth models

If a simplified price index grows by 3% each year, multiply by 1.03 each year.

An invented basket costing $200 today would be modelled at 200×1.03⁵≈$231.85 after five years if the 3% rate remained constant.

This is a mathematical model, not a forecast. Actual prices may change at different rates and the composition of a real basket may change.

The distinction between model and reality is part of responsible quantitative interpretation.

16. Capstone: combine discount, tax-style percentage and reverse calculation

An invented listed price receives a 20% discount, then a 9% charge is applied to the discounted price. The final amount is $872. Find the listed price.

Let listed price be P. Discount multiplier=0.8. Charge multiplier=1.09. Final multiplier=0.8×1.09=0.872.

Therefore 0.872P=872, so P=$1000.

Notice that the net change is a 12.8% decrease, not simply “20%−9%=11% decrease”. The two percentages act sequentially on different bases.

This problem is solved cleanly by composing multipliers before using the final value.

17. Independent practice

1. Find 15% of $240.
2. A value rises from 80 to 92. Find the percentage increase.
3. Increase $600 by 8%.
4. Decrease $450 by 12%.
5. Find the result of increasing 1000 by 5% for two periods.
6. Find the result of depreciating 2000 by 10% for three periods.
7. Start with 100, increase by 30%, then decrease by 30%. Find the final value.
8. After a 25% increase, a value is 500. Find the original.
9. After a 20% discount, a price is $96. Find the original price.
10. Find simple interest on $3000 at 5% per year for 4 years.
11. Find the final amount under the same simple-interest model.
12. Find the amount after 4 years if $3000 compounds annually at 5%.
13. Find $1200 after 6 months at 2% growth per month.
14. An invented purchase requires $250 deposit plus 12 payments of $90. Find the total paid.
15. If the cash price in Question 14 is $1250, find the excess paid as a percentage of cash price.
16. An item costs $60 and sells for $75. Find profit as a percentage of cost.
17. Find profit as a percentage of selling price.
18. A value grows by 4% annually for 5 years. Write the multiplier without calculating the final value.
19. A value falls by 7% annually for 4 years. Write the multiplier.
20. Explain why adding 4% five times is not generally identical to multiplying by 1.04⁵.

18. Worked answers

1. $36.

2. 15%. Increase 12 divided by original 80.

3. $648.

4. $396.

5. 1102.5. Use 1000×1.05².

6. 1458. Use 2000×0.9³.

7. 91. Combined multiplier 1.3×0.7=0.91.

8. 400. Divide by 1.25.

9. $120. Divide by 0.8.

10. $600. 3000×0.05×4.

11. $3600.

12. Approximately $3646.52. 3000×1.05⁴.

13. Approximately $1351.40. 1200×1.02⁶.

14. $1330.

15. 6.4%. Excess $80 divided by $1250.

16. 25%. Profit $15 divided by cost $60.

17. 20%. Profit $15 divided by selling price $75.

18. 1.04⁵.

19. 0.93⁴.

20. Repeated multiplication changes the base after each period, so later percentage amounts are calculated from new values.

19. Diagnose the first percentage-base error

Common failures include using the final value as the base for a percentage increase, subtracting the same percentage of the original amount during depreciation, confusing simple and compound models, mismatching monthly rates with annual exponents, or assuming equal percentage increases and decreases cancel.

A useful correction note writes the multiplier explicitly: “8% increase → 1.08”, “12% decrease → 0.88”, “reverse percentage → divide by multiplier”, or “new period → new base”.

Then vary the timeline. A repaired skill should survive one-stage, multi-stage and reverse calculations without needing a memorised story template.

20. Continue through the BTT Mathematics library

Return to the BTT Mathematics Hub. Use Ratio, Percentage and the Correct Base for denominator discipline, Indices, Roots and Standard Form for repeated powers, and Algebraic Fractions, Formulae and Substitution for symbolic financial models.

The BTT Mathematical Lab is the diagnostic route when repeated percentage or base-selection errors persist.

21. Sources and scope

All monetary values, rates, purchase examples and growth scenarios in this guide are invented for Mathematics teaching. The formulas illustrate simplified school models and do not model the full contractual, regulatory, tax or fee structure of real financial products.

For the current Singapore Secondary curriculum doorway, see MOE: Curriculum for secondary schools. Match financial-arithmetic depth to the learner’s actual subject level and school programme.