Understanding compound percentages
🎯 In this topic you will
- Use and understand compound percentages
🧠 Key Words
- compound percentage
Show Definitions
- compound percentage: A percentage change applied repeatedly over time, where each change is based on the new value after the previous change.
You already know how to work out a percentage increase and decrease using a multiplier. For example:
When a price of $300$ is increased by 20%, the new price is $300 \times 1.2 = 360$.
The multiplier is 1.2 because $100\% + 20\% = 120\% = \tfrac{120}{100} = 1.2$
When a price of $300$ is decreased by 20%, the new price is $300 \times 0.8 = 240$.
The multiplier is 0.8 because $100\% - 20\% = 80\% = \tfrac{80}{100} = 0.8$
A compound percentage change is when a percentage increase or decrease is followed by another percentage increase or decrease.
❓ EXERCISE
1. Copy and complete the workings for these compound percentage changes:
a. $200 increased by 10%, then increased by 15%.
b. $200 decreased by 10%, then decreased by 15%.
c. $200 increased by 20%, then decreased by 5%.
👀 Show answer
1a.
$200 × 1.1 = 220
220 × 1.15 = $253
1b.
$200 × 0.9 = 180
180 × 0.85 = $153
1c.
$200 × 1.2 = 240
240 × 0.95 = $228
🔎 Reasoning Tip
Remember:
- For an increase, add the percentage onto 100%.
- For a decrease, subtract the percentage from 100%.
🧠 Think like a Mathematician
Task: Investigate what happens to the value of an item when it increases by a percentage and then decreases by the same percentage.
Discussion Statements:
Questions:
👀 show answer
- a) Sofia is correct. After an increase and then a decrease by the same percentage, the final value is lower than the starting value.
- b) Calculation: Start = $800. Increase by 10%: $800 \times 1.1 = 880$. Decrease by 10%: $880 \times 0.9 = 792$. Final value = $792. Marcus’s mistake: he assumed the effects of +10% and -10% cancel each other out, but the base values are different.
- c) If the coin first decreases by 10%, $800 \times 0.9 = 720$. Then increase by 10%: $720 \times 1.1 = 792$. The final value is still less than $800.
- d) In both cases (increase then decrease, or decrease then increase), the final value = $792, which is less than $800.
❓ EXERCISE
3a. Work out these compound percentage changes:
- i) 60 increased by 20%, then decreased by 20%
- ii) 60 decreased by 20%, then increased by 20%
3b. Which sign, <, > or = is missing from this sentence?
60 increased by 20%, then decreased by 20% □ 60 decreased by 20%, then increased by 20%
3c. Without doing any calculations, decide which sign (<, > or =) is missing from each sentence:
- i) 72 decreased by 15%, then increased by 15% □ 72 increased by 15%, then decreased by 15%
- ii) 140 increased by 8%, then decreased by 8% □ 140 decreased by 8%, then increased by 8%
👀 Show answer
3a i)
60 × 1.2 = 72
72 × 0.8 = 57.6
3a ii)
60 × 0.8 = 48
48 × 1.2 = 57.6
3b.
Both results are 57.6, so the missing sign is =
3c i)
If you decrease then increase by the same percentage, the final value is less than the original. Missing sign: <
3c ii)
Same reasoning: increasing then decreasing (or vice versa) leaves the value smaller. Missing sign: <
🧠 Think like a Mathematician
Task: Compare three different methods for working out compound percentage increases and reflect on which approach is most effective.
Methods:
Questions:
👀 show answer
- a) Anil increases the number step by step, while Raj combines the two percentage increases in one calculation. Both give the same result.
- b) Raj directly multiplies by both factors, while Mari first simplifies the combined multiplier $(1.2 \times 1.1 = 1.32)$ before multiplying once. Both methods are equivalent.
- c) Raj’s or Mari’s methods are easiest with a calculator, since they involve fewer separate steps.
- d) Anil’s step-by-step method is easier without a calculator because it uses simpler numbers at each stage.
- e) Preference may vary: - If you want fewer steps, Mari’s method is simplest. - If you want to check progress after each stage, Anil’s method is clearer. - Raj’s method is a balance between the two.
❓ EXERCISE
5a. Work out the final value after these compound percentage increases (Raj’s method):
- i) 120 increased by 25%, then increased by 30%
- ii) 40 increased by 15%, then increased by 40%
5b. Work out the final value after these compound percentage increases (Mari’s method):
- i) 400 increased by 50%, then increased by 5%
- ii) 90 increased by 12%, then increased by 8%
🔎 Reasoning Tip
Remember that in part bi, the multiplier for a 5% increase is 1.05.
👀 Show answer
6. Mari uses the train company ‘GoRail’ to travel to work.
- In 2018, GoRail increased the cost of a ticket by 7%.
- In 2019, GoRail increased the cost of a ticket by 5%.
6a. Use Mari’s method to work out the multiplier for the compound percentage increase.
6b. In 2017, Mari buys a ticket for $60. How much will this ticket cost her in 2019?
👀 Show answer
6a.
Multiplier = 1.07 × 1.05 = 1.1235
6b.
60 × 1.1235 = $67.41
7a. Work out the final value after these compound percentage decreases (Raj’s method):
i) 100 ↓10%, then ↓20% ii) 80 ↓25%, then ↓12%
7b. Work out the final value after these compound percentage decreases (Mari’s method):
i) 600 ↓50%, then ↓5% ii) 76 ↓30%, then ↓9%
👀 Show answer
8. Mari buys a new motorbike. In year 1 the value decreases by 18%. In year 2 it decreases by 12%.
8a. Find the multiplier for the compound percentage decrease.
8b. If the bike cost $6400 in 2017, what will it be worth after two years?
👀 Show answer
8a. Multiplier = (1 − 0.18)(1 − 0.12) = 0.82 × 0.88 = 0.7216
8b. Value after two years = 6400 × 0.7216 = $4618.24
9. Match each percentage change (left) with its multiplier (right).
Changes
- 20% increase then 10% decrease
- 18% increase then 30% decrease
- 12% decrease then 20% increase
- 40% decrease then 35% increase
- 15% increase then 32% decrease
- 5% decrease then 8% increase
Multipliers
- 1.056
- 0.782
- 1.08
- 0.826
- 1.026
👀 Show answer
Tip: Convert each change to a multiplier and multiply them.
- A: 1.2 × 0.9 = 1.08 → matches iii
- B: 1.18 × 0.7 = 0.826 → matches iv
- C: 0.88 × 1.2 = 1.056 → matches i
- D: 0.6 × 1.35 = 0.81 → leftover card (not listed)
- E: 1.15 × 0.68 = 0.782 → matches ii
- F: 0.95 × 1.08 = 1.026 → matches v
Answer map: A→iii, B→iv, C→i, D→0.81 (extra), E→ii, F→v.
10.
Context: $5000 invested at 4% interest per year (compound, added at year end).
10a. Zara says: for the end of year 2, instead of writing 5000 × 1.04 × 1.04, Pieter could write 5000 × (1.04)². Is she correct? Explain.
10b. For the end of year 3, write a single calculation instead of 5000 × 1.04 × 1.04 × 1.04.
10c. For the end of year 4, write a single calculation instead of multiplying four 1.04s.
10d. If the investor writes 5000 × (1.04)^8, for how many years has the money been invested? Explain.
10e. Write a calculation to find the amount at the end of: i) 12 years ii) 20 years iii) n years.
10f. After how many years will the balance be more than $9000? Show working.
👀 Show answer
10a.Yes. By index laws, multiplying by 1.04 twice is the same as raising 1.04 to the power 2: 1.04 × 1.04 = (1.04)². So 5000 × 1.04 × 1.04 = 5000 × (1.04)².
10b.5000 × (1.04)³.
10c.5000 × (1.04)^4.
10d.5000 × (1.04)^8 corresponds to 8 years, because the exponent counts how many yearly 4% growth factors are applied.
10e.
i) 5000 × (1.04)^12
ii) 5000 × (1.04)^20
iii) 5000 × (1.04)^n
10f. Solve 5000(1.04)^t > 9000 → (1.04)^t > 1.8. Using logs: t > ln(1.8)/ln(1.04) ≈ 14.99. Therefore after 15 years it first exceeds $9000: 5000 × (1.04)^15 ≈ $9004.72.
11.
Context: The population of a town is 10,000. It decreases at a steady rate of 10% per year.
11a. Write a calculation to work out the population of the town after:
i) 1 year ii) 2 years iii) 3 years
11b. What does the calculation 10000 × (0.9)⁵ represent?
11c. What does the calculation 10000 × (0.9)¹⁰ represent?
11d. After how many years does the population of the town first fall below 6000 people? Show how you worked it out.
11e. Write a calculation to work out the population of the town after n years.
👀 Show answer
11a.
i) After 1 year: 10000 × 0.9 = 9000
ii) After 2 years: 10000 × (0.9)² = 8100
iii) After 3 years: 10000 × (0.9)³ = 7290
11b. This represents the town’s population after 5 years.
11c. This represents the town’s population after 10 years.
11d. Solve 10000 × (0.9)ᵗ < 6000.
(0.9)ᵗ < 0.6 → t > ln(0.6)/ln(0.9) ≈ 5.13.
So after 6 years, the population first falls below 6000.
11e. General formula: 10000 × (0.9)ⁿ
⚠️ Be careful! Compound Percentages
- Chain changes by multiplying multipliers, not adding percentages.
Example: +20% then −15% ⇒ 1.20 × 0.85 = 1.02 (overall +2%), not +5%. - Order matters when the rates differ. +20% then −10% ≠ +10% then −20% because 1.20×0.90 ≠ 1.10×0.80.
- +P% then −P% is a net decrease. Net factor =(1+P/100)(1−P/100)=1−(P/100)² < 1.
- Build one-step multipliers. Overall multiplier = product of step multipliers.
Example: ↓18% then ↓12% ⇒ 0.82 × 0.88 = 0.7216. - Use powers for repeated equal changes. After n periods at rate r: initial × (1±r)n.
- Use the correct base for % change between two values. Multiplier = new ÷ original; % change = (multiplier−1)×100%.
- Distinguish “increase by A%” vs “is A% of”.
“Increase by 132%” ⇒ factor 2.32; “is 132% of” ⇒ factor 1.32. - Delay rounding. Keep extra decimals in the multiplier product; round only the final answer.
- Sense-check. Multiplier > 1 ⇒ bigger; < 1 ⇒ smaller; = 1 ⇒ unchanged. Compare against easy benchmarks (10%, 25%, 50%).