Ratio, proportion & rates of change

Percentages: Repeated Percentage Change, Compound Interest & Depreciation

Year 10 · Year 11

✓Calculate the result of repeated percentage change over multiple periods.
✓Solve problems involving compound interest, calculating future values or interest earned.
✓Solve problems involving depreciation, calculating future values of assets.
✓Understand and apply the multiplier method for repeated percentage increases and decreases.

Key concepts

Repeated Percentage Change

Repeated percentage change occurs when a percentage increase or decrease is applied to a value, and then subsequent changes are applied to the new value after each period. This is different from simple percentage change where the change is always based on the original amount.

Final Value = Original Value × (Multiplier)^n, where 'n' is the number of periods. The multiplier for an increase of r% is (1 + r/100). The multiplier for a decrease of r% is (1 - r/100).

Compound Interest

Compound interest is a form of repeated percentage increase. It is interest calculated on the initial principal and also on the accumulated interest from previous periods. This means your money grows faster than with simple interest because you earn interest on your interest.

A = P(1 + r/100)^n, where A is the final amount, P is the principal (original amount invested), r is the annual interest rate as a percentage, and n is the number of years.

Depreciation

Depreciation is a form of repeated percentage decrease. It refers to the reduction in the value of an asset over time, often due to wear and tear, age, or becoming obsolete. Each year, the value decreases by a percentage of its current value.

A = P(1 - r/100)^n, where A is the final value, P is the original value, r is the annual rate of depreciation as a percentage, and n is the number of years.

Key facts to remember

1Repeated percentage change means the percentage is applied to the new value each time, not the original.
2The multiplier for an r% increase is (1 + r/100); for an r% decrease, it's (1 - r/100).
3The general formula for repeated percentage change is Final Value = Original Value × (Multiplier)^n.
4Compound interest is a repeated percentage increase, calculated on the principal plus accumulated interest.
5Depreciation is a repeated percentage decrease, reducing an asset's value over time.
6For monetary answers, always round to two decimal places (the nearest penny).
7Always keep full calculator accuracy during intermediate steps to avoid premature rounding errors.

Worked examples

Example 1

Amelia invests £4500 into a savings account that pays 2.5% compound interest per annum. How much will her investment be worth after 3 years? Give your answer to two decimal places.

IIdentify the principal (P), interest rate (r), and number of years (n): P = £4500, r = 2.5%, n = 3.

IICalculate the multiplier for a 2.5% increase: 1 + (2.5/100) = 1 + 0.025 = 1.025.

IIIApply the compound interest formula: A = P(Multiplier)^n.

IVA = 4500 × (1.025)^3.

VA = 4500 × 1.076890625.

VIA = 4846.0078125.

VIIRound to two decimal places for money: £4846.01.

Answer

£4846.01

Remember to round to two decimal places for monetary values.

Example 2

A new car costs £22000. Its value depreciates by 12% in the first year and then by 8% for each of the next 4 years. Calculate the value of the car after 5 years. Give your answer to the nearest penny.

ICalculate the value after the first year (12% depreciation):

IIMultiplier for 12% decrease = 1 - (12/100) = 1 - 0.12 = 0.88.

IIIValue after 1 year = £22000 × 0.88 = £19360.

IVCalculate the value for the next 4 years (8% depreciation):

VNew principal (P) = £19360, rate (r) = 8%, number of years (n) = 4.

VIMultiplier for 8% decrease = 1 - (8/100) = 1 - 0.08 = 0.92.

VIIValue after 4 more years = £19360 × (0.92)^4.

VIIIValue after 4 more years = £19360 × 0.71639296.

9Value after 5 years = £13859.9079616.

10Round to two decimal places: £13859.91.

Answer

£13859.91

When different rates apply, calculate the changes sequentially.

Example 3

A rare stamp increases in value by 6% each year. If its current value is £150, after how many whole years will its value first exceed £200?

IIdentify original value (P) = £150, percentage increase (r) = 6%.

IICalculate the multiplier: 1 + (6/100) = 1.06.

IIIWe want to find 'n' such that 150 × (1.06)^n > 200.

IVDivide by 150: (1.06)^n > 200/150 = 4/3 ≈ 1.333...

VUse trial and improvement for 'n':

VIn = 1: 1.06^1 = 1.06 (Not > 1.333)

VIIn = 2: 1.06^2 = 1.1236 (Not > 1.333)

VIIIn = 3: 1.06^3 = 1.191016 (Not > 1.333)

9n = 4: 1.06^4 = 1.26247696 (Not > 1.333)

10n = 5: 1.06^5 = 1.3382255776 (This is > 1.333)

11So, after 5 whole years, the value will first exceed £200.

Answer

5 whole years

For finding 'n' at GCSE, trial and improvement is the expected method unless logarithms are explicitly taught.

Common mistakes

✗Using simple interest calculations instead of compound interest for repeated changes.
✗Calculating the percentage change on the original amount each time, rather than the new, adjusted amount.
✗Incorrectly forming the multiplier (e.g., using 0.05 for 5% increase instead of 1.05).
✗Rounding intermediate calculations, which can lead to an inaccurate final answer.
✗Forgetting to subtract the original principal from the final amount when asked for the 'interest earned' rather than the 'total amount'.

Exam tips

★Identify whether the problem involves an increase (compound interest, growth) or a decrease (depreciation, decay) and choose the correct multiplier.
★Always use the multiplier method: Final Value = Original Value × (Multiplier)^n. It is more efficient and less prone to error than calculating year by year.
★Show your working clearly, including the multiplier used and the power 'n'. This helps gain method marks even if a calculation error occurs.
★For questions involving money, ensure your final answer is correctly rounded to two decimal places, representing pounds and pence.
★If the number of periods ('n') is unknown, use trial and improvement to find the smallest integer 'n' that satisfies the condition.

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