Mathematics · AQA 8300 §R4
Growth and Decay
- Apply the compound growth and decay formula to real-world problems
- Model population and financial growth/decay over multiple periods
- Calculate depreciated values of assets after a given number of years
- Find original values given compound interest outcomes (reverse problems)
- Interpret and sketch exponential graphs including asymptotes
🔑 Core Concepts
The Multiplier — Why It Works
Before learning formulae, understand the multiplier principle: instead of adding or subtracting a percentage each time, we multiply by a fixed factor. This factor is applied repeatedly, which is the essence of compounding.
DEFINITION — Multiplier
For a percentage rate $r$%:Growth multiplier: $1 + \dfrac{r}{100}$ (always > 1)
Decay multiplier: $1 - \dfrac{r}{100}$ (always between 0 and 1)
EXAM TIP — Spotting the multiplier
A 12% increase → multiply by $1.12$. A 12% decrease → multiply by $0.88$. Applying the multiplier $n$ times gives the compound formula.
Exponential Growth
Exponential growth occurs when a quantity increases by the same percentage of its current value each period. Each period the increase is larger in absolute terms because the base is growing — this is why compound interest earns more than simple interest over time.
Exponential Growth Formula
$$A = P\!\left(1 + \frac{r}{100}\right)^{\!n}$$
$A$ = final amount
$P$ = principal / initial value
$r$ = percentage rate per period
$n$ = number of periods
QUICK EXAMPLE — Compound Interest
£5,000 invested at 3% per annum compound interest for 4 years:$A = 5000 \times (1.03)^4 = 5000 \times 1.12551 = £5,627.55$
COMMON MISTAKE — Simple vs Compound
Simple interest would give $5000 \times 0.03 \times 4 = £600$ extra. Compound gives £627.55. The difference grows dramatically over many years — do not use simple interest when compound is specified.
Exponential Decay & Depreciation
Decay occurs when a quantity decreases by the same percentage of its current value each period. The quantity approaches zero but never actually reaches it — this is the defining feature of exponential decay.
Exponential Decay Formula
$$A = P\!\left(1 - \frac{r}{100}\right)^{\!n}$$
$A$ = final (reduced) amount
$P$ = initial value
$r$ = percentage decrease per period
$n$ = number of periods
DEFINITION — Depreciation
Depreciation is the loss of value of an asset over time due to wear, age, or obsolescence. Car depreciation is a classic exam context. A car worth £18,000 depreciating at 15% per year after 5 years: $A = 18000 \times (0.85)^5 = 18000 \times 0.4437 = £7,987$
EXAM TIP — Decay multiplier check
The decay multiplier must be between 0 and 1. If $r = 20$%, multiplier $= 0.80$. If your multiplier is greater than 1, you have used the wrong formula — you are modelling growth, not decay.
Population Growth & Bacterial Doubling
The same compound formula applies to populations and bacteria. Bacterial growth is often given as a doubling time — the time it takes for the population to double. This is a special case where the multiplier is exactly 2 and $r = 100$%.
DEFINITION — Doubling Time
If a bacterial colony doubles every $d$ hours, then after $t$ hours the number of doublings is $n = \dfrac{t}{d}$ and the population is $A = P \times 2^{t/d}$. This is a special case of the growth formula with multiplier 2.
MEMORY TRICK — Doubling
Doubling every $d$ hours ↔ 100% growth rate, so multiplier = $1 + 1 = 2$. Start with 500 bacteria doubling every 3 hours: after 12 hours, $n = 4$ doublings, so $A = 500 \times 2^4 = 500 \times 16 = 8{,}000$.
Half-Life
Half-life is the time taken for a radioactive substance (or drug concentration, etc.) to reduce to half its current amount. It is the decay analogue of doubling time.
Half-Life Formula
$$A = P \times \left(\frac{1}{2}\right)^{\!n}$$
$n$ = number of half-lives elapsed
Multiplier = $0.5$ (50% decay each period)
QUICK EXAMPLE — Radioactive Decay
A 200 mg sample has a half-life of 6 hours. How much remains after 24 hours?Number of half-lives: $n = 24 \div 6 = 4$
$A = 200 \times (0.5)^4 = 200 \times 0.0625 = 12.5$ mg
Comparing Simple and Compound Growth
A key Grade 9 skill is comparing the two models over varying time periods and identifying at what point compound overtakes simple.
DEFINITION — Simple Interest
Simple interest: $A = P(1 + \dfrac{rn}{100})$ — interest is always calculated on the original principal only. The amount grows linearly (straight line graph). Compound interest grows exponentially (curved graph) and will always eventually exceed simple interest.
| Feature | Simple Interest | Compound Interest |
|---|---|---|
| Formula | $P(1 + \tfrac{rn}{100})$ | $P(1 + \tfrac{r}{100})^n$ |
| Graph shape | Straight line | Exponential curve |
| Interest base | Always original $P$ | Growing total each year |
| After 1 period | Same result | Same result |
| After many periods | Lower total | Higher total (always) |
| Asymptote | None | None (growth) / $y = 0$ (decay) |
Finding the Original Value (Reverse Exponential)
Sometimes the final amount and the multiplier are given and you must find the initial value $P$. This is a reverse calculation — divide the final amount by the multiplier raised to the power $n$.
Finding Original Value
$$P = \frac{A}{\left(1 \pm \dfrac{r}{100}\right)^{\!n}}$$
Rearrange the growth/decay formula for $P$
Use $+$ for growth contexts, $-$ for decay contexts
EXAM TIP — Reverse problems
"A car is worth £9,000 after depreciating at 18% per year for 3 years. Find its original value." This is a divide problem: $P = 9000 \div (0.82)^3 = 9000 \div 0.5514 = £16{,}323$ (to the nearest pound).
Finding the Number of Periods (Higher Tier)
At Grade 9, you may need to find $n$ — the number of periods for a quantity to reach a target value. Without logarithms, use trial and improvement (systematic substitution of integer values). With logarithms:
Finding n Using Logarithms (HT)
$$n = \frac{\log(A/P)}{\log\!\left(1 \pm \dfrac{r}{100}\right)}$$
From $A = P \cdot m^n$, take logs of both sides: $\log A = \log P + n\log m$
Rearrange: $n = \dfrac{\log(A/P)}{\log m}$
EXAM TIP — Trial and Improvement
If logs are not needed (the question says "find the minimum number of years"), use a table: calculate $A$ for $n = 1, 2, 3, \ldots$ until $A$ exceeds (or drops below) the target. Write each line of working to earn method marks.
Exponential Graphs — Features and Asymptotes
The graphs of $y = ka^x$ have distinctive features that examiners test. Understanding the shape, intercepts, and behaviour is essential for Grade 9.
DEFINITION — Exponential Graph Features
Growth ($a > 1$): passes through $(0, k)$, always increasing, asymptote $y = 0$ as $x \to -\infty$, rises steeply for large $x$.Decay ($0 < a < 1$): passes through $(0, k)$, always decreasing, asymptote $y = 0$ as $x \to +\infty$, never touches the $x$-axis.
IMPORTANT — The Asymptote
The horizontal asymptote of a basic exponential is $y = 0$ (the $x$-axis). The graph approaches but never reaches this line. In transformed graphs like $y = 2^x + 3$, the asymptote shifts to $y = 3$.
MEMORY TRICK — Graph Shape
Growth graph: like a hockey stick curving upward. Decay graph: like a skateboard ramp falling steeply then flattening. Both pass through $(0, k)$ where $k$ is the initial value.
🗺️ Visual Notes
Growth & Decay
📈 Growth Formula
- $A = P(1 + r/100)^n$
- Multiplier $> 1$
- Compound interest
- Population increase
📉 Decay Formula
- $A = P(1 - r/100)^n$
- Multiplier between 0 and 1
- Car depreciation
- Radioactive half-life
🔍 Reverse Problems
- Divide $A$ by multiplier$^n$
- Find original price
- Find rate: rearrange formula
- Find $n$: logs or trial/improve
📊 Graphs
- Asymptote at $y = 0$
- $y$-intercept = initial value
- Growth: $a > 1$, rising curve
- Decay: $0 < a < 1$, falling curve
🦠 Special Cases
- Doubling: multiplier = 2
- Half-life: multiplier = 0.5
- Bacterial growth models
- Drug concentration decay
💡 Grade 9 Skills
- Multi-step compound problems
- Comparing growth models
- Trial and improvement for $n$
- Sketch and annotate graphs
Growth vs Decay — At a Glance
| Context | Formula | Multiplier | Example Rate | After 10 periods on £1,000 |
|---|---|---|---|---|
| 5% compound growth | $1000 \times 1.05^{10}$ | 1.05 | 5% increase | £1,628.89 |
| 5% simple interest | $1000 \times (1 + 0.5)$ | linear | 5% on original | £1,500.00 |
| 10% depreciation | $1000 \times 0.90^{10}$ | 0.90 | 10% decrease | £348.68 |
| 50% half-life (decay) | $1000 \times 0.5^{10}$ | 0.5 | 50% decrease | £0.98 |
Which Formula? — Decision Process
Read the problem: is the quantity increasing or decreasing each period?
→
Increasing → Growth multiplier $= 1 + r/100$
→
Do you know $P$, $r$, $n$? Apply $A = P \cdot m^n$ directly.
→
Missing $P$? Divide: $P = A \div m^n$
→
Missing $n$? Try successive values (trial & improve) or use logs: $n = \log(A/P) \div \log m$
Exponential Graph — Key Features
| Feature | Growth $y = Pa^x$ ($a>1$) | Decay $y = Pa^x$ ($0<a<1$) |
|---|---|---|
| $y$-intercept | $(0, P)$ | $(0, P)$ |
| As $x \to +\infty$ | $y \to +\infty$ | $y \to 0^+$ |
| As $x \to -\infty$ | $y \to 0^+$ | $y \to +\infty$ |
| Asymptote | $y = 0$ (as $x \to -\infty$) | $y = 0$ (as $x \to +\infty$) |
| Shape | Increasing, concave up | Decreasing, concave up |
| Crosses $x$-axis? | Never | Never |
✏️ Worked Examples
Grade 4–5
A car is bought for £12,500. It depreciates at 14% per year. Calculate its value after 3 years. Give your answer to the nearest pound.
1
Identify the type
The value is decreasing each year by 14%. This is exponential decay (depreciation).
2
Write the decay multiplier
Decay multiplier $= 1 - \dfrac{14}{100} = 1 - 0.14 = 0.86$
3
Apply the formula
$$A = 12500 \times (0.86)^3$$
$$A = 12500 \times 0.636056 = 7950.70$$
4
State the answer with units
Value after 3 years $= £7{,}951$ (to the nearest pound).
Value after 3 years = £7,951
Grade 6–7
$A = 45000 \times (1.024)^n$ where $n$ is the number of years after 2020.
A town's population grows at 2.4% per year compound. In 2020 the population was 45,000. In which year will the population first exceed 55,000? Use trial and improvement to find the minimum number of complete years.
1
Set up the growth model
Growth multiplier $= 1 + \dfrac{2.4}{100} = 1.024$$A = 45000 \times (1.024)^n$ where $n$ is the number of years after 2020.
2
Systematic trial and improvement
| $n$ (years) | Calculation | Population | Exceeds 55,000? |
|---|---|---|---|
| 5 | $45000 \times 1.024^5$ | 50,667 | No |
| 8 | $45000 \times 1.024^8$ | 54,395 | No |
| 9 | $45000 \times 1.024^9$ | 55,700 | Yes ✓ |
3
Interpret the result
After $n = 8$ years (2028) the population is still below 55,000. After $n = 9$ years (2029) it first exceeds 55,000.
The population first exceeds 55,000 in 2029 (after 9 complete years).
Grade 9
Solve for $m$: $m = (0.375)^{1/12}$
$m = 0.375^{0.08\overline{3}} = 0.9124...$ (using calculator: $0.375^{1/12}$)
$240 \times (0.9124)^n \geq 50$
$(0.9124)^n \geq \dfrac{50}{240} = 0.2083\overline{3}$
Check: $n=17$: $240 \times (0.9124)^{17} = 240 \times 0.2153 = 51.7$ g ✓ (above 50)
$n=18$: $240 \times (0.9124)^{18} = 240 \times 0.1964 = 47.1$ g ✗ (below 50)
A radioactive isotope decays so that after 12 years only 37.5% of the original mass remains. A scientist has a 240 g sample. She needs at least 50 g for an experiment. Find the number of complete years she has before the sample falls below 50 g.
1
Find the annual decay multiplier
After 12 years, 37.5% remains, so $(m)^{12} = 0.375$.Solve for $m$: $m = (0.375)^{1/12}$
$m = 0.375^{0.08\overline{3}} = 0.9124...$ (using calculator: $0.375^{1/12}$)
2
Set up the inequality
We need $A \geq 50$, so solve:$240 \times (0.9124)^n \geq 50$
$(0.9124)^n \geq \dfrac{50}{240} = 0.2083\overline{3}$
3
Trial and improvement (or use logs)
Using logs: $n = \dfrac{\log(50/240)}{\log(0.9124)} = \dfrac{\log(0.2083)}{\log(0.9124)} = \dfrac{-0.6812}{-0.03975} = 17.13...$Check: $n=17$: $240 \times (0.9124)^{17} = 240 \times 0.2153 = 51.7$ g ✓ (above 50)
$n=18$: $240 \times (0.9124)^{18} = 240 \times 0.1964 = 47.1$ g ✗ (below 50)
4
Interpret with context
After 17 complete years, the sample is just above 50 g (51.7 g). After 18 years it has fallen below 50 g. The scientist has 17 complete years.
The scientist has 17 complete years before the sample falls below 50 g.
❓ Exam Questions
Q1
1 mark
Write down the decimal multiplier for a 7.5% decrease.
Answer: $1 - 0.075 = \mathbf{0.925}$
Mark scheme: B1 for 0.925
Mark scheme: B1 for 0.925
Q2
2 marks
£3,200 is invested at a compound interest rate of 3.5% per annum. Calculate the value of the investment after 6 years. Give your answer to the nearest penny.
Working:
Multiplier $= 1.035$
$A = 3200 \times (1.035)^6 = 3200 \times 1.22926 = £3{,}933.64$
Mark scheme:
M1 for $3200 \times (1.035)^6$ (or sight of multiplier 1.035 used correctly)
A1 for £3,933.64
Multiplier $= 1.035$
$A = 3200 \times (1.035)^6 = 3200 \times 1.22926 = £3{,}933.64$
Mark scheme:
M1 for $3200 \times (1.035)^6$ (or sight of multiplier 1.035 used correctly)
A1 for £3,933.64
Q3
3 marks
A motorbike is bought for £8,750. Its value decreases by 22% in the first year and by 15% each subsequent year. Calculate the value of the motorbike after 4 years. Give your answer to the nearest pound.
Working:
After year 1: $8750 \times 0.78 = £6{,}825$
After years 2–4 (3 more years at 15%): $6825 \times (0.85)^3 = 6825 \times 0.614125 = £4{,}191.40$
Value $= £4{,}191$ (nearest pound)
Mark scheme:
M1 for $8750 \times 0.78$ seen (or = 6825)
M1 for multiplying by $(0.85)^3$ (or applying 15% decrease 3 times)
A1 for £4,191 (accept £4,191 to £4,192)
After year 1: $8750 \times 0.78 = £6{,}825$
After years 2–4 (3 more years at 15%): $6825 \times (0.85)^3 = 6825 \times 0.614125 = £4{,}191.40$
Value $= £4{,}191$ (nearest pound)
Mark scheme:
M1 for $8750 \times 0.78$ seen (or = 6825)
M1 for multiplying by $(0.85)^3$ (or applying 15% decrease 3 times)
A1 for £4,191 (accept £4,191 to £4,192)
Q4
4 marks
A house was worth £185,000 after its value increased by an average of 4.2% per year compound for 5 years. Calculate the original value of the house to the nearest £1,000.
Working:
Multiplier $= 1.042$
$A = P \times (1.042)^5$
$(1.042)^5 = 1.22918...$
$P = \dfrac{185000}{1.22918} = 150{,}507...$
Original value $\approx £151{,}000$ (nearest £1,000)
Mark scheme:
M1 for identifying reverse exponential: $P = A \div m^n$
M1 for $(1.042)^5$ evaluated correctly (= 1.2292 or better)
A1 for correct unrounded value (150,507)
A1 for £151,000 (or £150,000 if correctly rounded to nearest £1,000)
Multiplier $= 1.042$
$A = P \times (1.042)^5$
$(1.042)^5 = 1.22918...$
$P = \dfrac{185000}{1.22918} = 150{,}507...$
Original value $\approx £151{,}000$ (nearest £1,000)
Mark scheme:
M1 for identifying reverse exponential: $P = A \div m^n$
M1 for $(1.042)^5$ evaluated correctly (= 1.2292 or better)
A1 for correct unrounded value (150,507)
A1 for £151,000 (or £150,000 if correctly rounded to nearest £1,000)
Q5
3 marks
A bacterial colony initially contains 500 bacteria. The colony doubles every 2.5 hours. How many bacteria will there be after 10 hours? Give your answer in standard form to 3 significant figures.
Working:
Number of doubling periods in 10 hours: $n = 10 \div 2.5 = 4$
$A = 500 \times 2^4 = 500 \times 16 = 8{,}000$
In standard form: $8.00 \times 10^3$
Mark scheme:
M1 for $n = 10 \div 2.5 = 4$ (or equivalent)
M1 for $500 \times 2^4$
A1 for $8000$ or $8.00 \times 10^3$
Number of doubling periods in 10 hours: $n = 10 \div 2.5 = 4$
$A = 500 \times 2^4 = 500 \times 16 = 8{,}000$
In standard form: $8.00 \times 10^3$
Mark scheme:
M1 for $n = 10 \div 2.5 = 4$ (or equivalent)
M1 for $500 \times 2^4$
A1 for $8000$ or $8.00 \times 10^3$
Q6
6 marks
Amelia invests £5,000 in Account A which pays 2% per annum compound interest. Ben invests £5,000 in Account B which pays 2.5% per annum simple interest.
(a) Show that after 10 years, Ben's account has more money than Amelia's.
(b) After how many complete years will Amelia's account first have more money than Ben's? Show all your working.
Part (a):
Amelia (compound): $5000 \times (1.02)^{10} = 5000 \times 1.21899 = £6{,}094.97$
Ben (simple): $5000 + 5000 \times 0.025 \times 10 = 5000 + 1250 = £6{,}250$
Ben: £6,250 > Amelia: £6,095 ✓ (shown)
Part (b) — Trial and Improvement:
After 23 years: Amelia ≈ Ben (Amelia slightly behind). After 24 years Amelia first exceeds Ben.
Mark scheme:
M1 for setting up correct compound expression for Amelia
M1 for correct simple interest expression for Ben
A1 for correct numerical comparison at $n = 10$ [part a]
M1 for systematic trial and improvement comparing both accounts
A1 for identifying the crossover between $n = 23$ and $n = 24$
A1 for 24 years
Amelia (compound): $5000 \times (1.02)^{10} = 5000 \times 1.21899 = £6{,}094.97$
Ben (simple): $5000 + 5000 \times 0.025 \times 10 = 5000 + 1250 = £6{,}250$
Ben: £6,250 > Amelia: £6,095 ✓ (shown)
Part (b) — Trial and Improvement:
| $n$ years | Amelia (compound) | Ben (simple) | Amelia ahead? |
|---|---|---|---|
| 20 | $5000 \times 1.02^{20} = £7{,}430$ | $5000 + 5000 \times 0.025 \times 20 = £7{,}500$ | No |
| 30 | $5000 \times 1.02^{30} = £9{,}056$ | $5000 + 3750 \times \frac{30}{10} = £8{,}750$ | Yes ✓ |
| 25 | $5000 \times 1.02^{25} = £8{,}203$ | $5000 + 3125 = £8{,}125$ | Yes ✓ |
| 23 | $5000 \times 1.02^{23} = £7{,}874$ | $5000 + 2875 = £7{,}875$ | No (equal approx) |
| 24 | $5000 \times 1.02^{24} = £8{,}031$ | $5000 + 3000 = £8{,}000$ | Yes ✓ |
Mark scheme:
M1 for setting up correct compound expression for Amelia
M1 for correct simple interest expression for Ben
A1 for correct numerical comparison at $n = 10$ [part a]
M1 for systematic trial and improvement comparing both accounts
A1 for identifying the crossover between $n = 23$ and $n = 24$
A1 for 24 years
⭐ Grade 9 Model Answers
Full Annotated Solution — Q6 (6 marks)
GRADE 9 ANNOTATION — What earns every mark
Part (a) — 3 marks:
The key is to clearly show both calculations and state the comparison explicitly. A Grade 9 student writes:
"Amelia's account after 10 years: $A = 5000 \times (1.02)^{10}$. Multiplier $= 1.02$ (using $1 + 2/100$). $1.02^{10} = 1.21899$. $A = £6{,}094.97$." — This earns M1+A1.
"Ben's account: Simple interest $= 5000 \times 0.025 \times 10 = £1{,}250$. Total $= £6{,}250$." — This earns M1.
"£6,250 > £6,094.97, so Ben's account has more money. ✓" — The explicit comparison earns the final A1.
Part (b) — 3 marks:
M1: Setting up the comparison to solve — even writing "$5000 \times 1.02^n$ vs $5000(1 + 0.025n)$" earns the method mark.
M1: At least 3 correct trials shown in a table or list. Must show values for both accounts to earn this mark — not just one account.
A1: Answer of 24 years with supporting evidence of the crossover (showing $n = 23$ gives Amelia behind and $n = 24$ gives Amelia ahead).
Grade 9 tip: When comparing two models, always show both the year where A is BEHIND and the year where A is AHEAD to prove the crossover. Writing "therefore after 24 complete years, Amelia's account first exceeds Ben's" earns the conclusion mark.
EXAM TIP — Why compound always overtakes simple
Mathematically: as $n \to \infty$, exponential growth $P(1+r)^n$ grows faster than linear growth $P(1+rn)$. The crossover point depends on the rates. If the compound rate equals the simple rate, compound always eventually overtakes — but if compound rate is lower (as here, 2% vs 2.5%), the crossover takes longer. This is a classic Grade 9 comparison that tests conceptual understanding.
📋 Revision Sheet
Key Definitions
| Term | Meaning |
|---|---|
| Compound growth | % applied to current (growing) amount |
| Compound decay | % applied to current (shrinking) amount |
| Multiplier | Factor applied each period (e.g. 1.05 or 0.87) |
| Depreciation | Reduction in asset value over time |
| Half-life | Time for a quantity to halve |
| Asymptote | Line the graph approaches but never reaches |
Essential Formulae
Growth: $A = P(1+r/100)^n$
Decay: $A = P(1-r/100)^n$
Original value: $P = A \div (m)^n$
Doubling: $A = P \times 2^n$ ($n$ = number of doublings)
Half-life: $A = P \times (0.5)^n$ ($n$ = number of half-lives)
Find $n$ (logs): $n = \dfrac{\log(A/P)}{\log m}$
Simple interest: $A = P(1 + rn/100)$
Memory Hooks
- GROWTH = BIGGER multiplier: $1 + r/100 > 1$
- DECAY = SMALLER multiplier: $0 < 1 - r/100 < 1$
- Reverse? DIVIDE by multiplier$^n$
- "Compound" = power: the $n$ goes as an exponent
- Doubling → ×2 each period: bacteria, populations
- Half-life → ×0.5 each period: radioactivity
- Asymptote at $y = 0$: decay never hits zero
- $y$-intercept = $P$: always the initial value
Exam Tips
- Always write the multiplier explicitly before substituting
- Use brackets around the multiplier in your calculator: $(1.035)^6$
- Round only at the final step — keep full precision throughout
- For "find $n$": show at least two values either side of the answer
- For reverse problems: divide (not subtract) by the multiplier power
- Check decay answers are less than $P$ and growth answers are more
- In comparison questions: always state which is larger and by how much
- State units (£, g, years) in your final answer
🔄 Flashcards
Click any card to reveal the answer. Use these to self-test before an exam.
✗ Common Mistakes
MISTAKE 1 — Using Simple Instead of Compound Interest
What students do wrong: They add the percentage each year using the original amount (e.g. "5% of £1,000 = £50 every year") instead of applying the multiplier.Why marks are lost: The method is fundamentally wrong — this earns 0 marks when compound is specified.
How to avoid it: When you see "compound" in a question, immediately write the multiplier $(1 \pm r/100)$ and raise it to the power $n$. Never multiply rate × time when compound is involved.
MISTAKE 2 — Multiplier Greater Than 1 for Decay
What students do wrong: Writing $A = P(1 + 0.15)^n$ when the quantity is decreasing by 15%.Why marks are lost: The answer will be larger than $P$, which is impossible for decay — it implies the quantity grew, losing all accuracy marks.
How to avoid it: Decrease → subtract from 1. A 15% decrease → multiplier $= 1 - 0.15 = 0.85$. Always check: is your multiplier less than 1 for decay problems?
MISTAKE 3 — Rounding Too Early
What students do wrong: Rounding intermediate steps, e.g. computing $(1.035)^6 = 1.23$ instead of $1.22926$, then multiplying.Why marks are lost: The final answer will not match the mark scheme's accepted range, losing the accuracy mark (A1) even though the method is correct.
How to avoid it: Keep at least 5 significant figures in all intermediate calculations. Only round your final answer to the precision stated in the question.
MISTAKE 4 — Reverse Problem: Subtracting Instead of Dividing
What students do wrong: To find the original value, students subtract the compound increase rather than dividing by the multiplier: e.g. "$£6,094 - £1,094 = £5,000$" (accidentally correct here, but wrong method for most cases).Why marks are lost: No method marks for incorrect approach; wrong answer in most cases.
How to avoid it: The correct reverse formula is $P = A \div (m)^n$. This is dividing by the multiplier raised to the power, not subtracting the interest.
MISTAKE 5 — Forgetting the "Complete Years" Requirement
What students do wrong: When asked "after how many complete years", students calculate the exact decimal answer (e.g. 17.13 years) and give 17.13 as their answer, or round to 17 without checking.Why marks are lost: The question asks for complete years — you must verify whether the condition is met at $n = 17$ and $n = 18$. Giving a decimal answer loses marks.
How to avoid it: After finding the decimal, always check the integer either side. If you need the quantity to exceed a value, round up; if it must stay above a threshold, check both sides.
MISTAKE 6 — Misidentifying the Asymptote
What students do wrong: Stating the asymptote is $x = 0$ (the $y$-axis) instead of $y = 0$ (the $x$-axis), or saying "the graph touches $y = 0$".Why marks are lost: Confusing $x$ and $y$ axes loses the mark; saying the graph "reaches" or "crosses" the asymptote loses the mark.
How to avoid it: The asymptote of a standard exponential decay $y = Pa^x$ is always the horizontal line $y = 0$. The graph approaches but never touches it. For shifted graphs (e.g. $y = 2^x + 3$), the asymptote shifts to $y = 3$.
✅ Final Checklist
Click each item to mark it as done. Your progress is saved automatically.
- I can write the growth multiplier for any percentage increase
- I can write the decay multiplier for any percentage decrease
- I can apply $A = P(1+r/100)^n$ to compound interest problems
- I can apply $A = P(1-r/100)^n$ to depreciation/decay problems
- I can find the original value using $P = A \div (m)^n$
- I can model bacterial doubling using $A = P \times 2^n$
- I can calculate remaining amount after multiple half-lives
- I can find the minimum number of periods using trial and improvement
- I can compare compound and simple interest over the same period
- I know that compound always eventually overtakes simple for the same rate
- I can identify the asymptote of an exponential graph as $y = 0$
- I can state the $y$-intercept of an exponential graph as the initial value $P$
- I can sketch and label growth vs decay exponential curves correctly
- I know not to round intermediate calculations — only the final answer
- I can solve a multi-step decay problem finding both the rate and the time