Mathematics · AQA 8300 §N4
Percentages
- Calculate a percentage of a quantity using both fraction and decimal methods
- Find percentage increase and decrease using a multiplier efficiently
- Calculate reverse percentages to find the original value before a change
- Apply the compound interest formula to growth and depreciation problems
- Solve multi-step percentage problems in real-world contexts (profit, VAT, salary)
🔑 Core Concepts
1. Percentage of a Quantity
A percentage is a number expressed as a fraction of 100. The word comes from the Latin per centum — "out of one hundred". To find a percentage of a quantity, convert the percentage to a decimal multiplier and multiply.
DEFINITION — Percentage of a Quantity
To find $p\%$ of a value $Q$: convert $p\%$ to a decimal $\left(\dfrac{p}{100}\right)$, then multiply: $\dfrac{p}{100} \times Q$.
Finding a Percentage of a Quantity
$$\text{Result} = \frac{p}{100} \times Q$$
$p$ = percentage value
$Q$ = original quantity
EXAM TIP
Always convert percentages to decimals before multiplying. $35\% = 0.35$, $7.5\% = 0.075$. On a calculator, this is always the fastest method.
COMMON MISTAKE
Do not find $15\%$ of £80 by doing $15 \div 80$. You must do $0.15 \times 80 = £12$.
2. Percentage Increase and Decrease — Multipliers
Rather than finding the percentage amount and adding/subtracting separately, a multiplier applies the change in a single step. This is the preferred method at GCSE and is essential for compound calculations.
DEFINITION — Multiplier
The multiplier for a percentage change is the single number you multiply the original by to get the new value after the change. It combines the "original" and the "change" into one factor.
Multiplier for Percentage Increase
$$\text{Multiplier} = 1 + \frac{r}{100}$$
$r$ = percentage increase
e.g. $20\%$ increase → $\times 1.20$
Multiplier for Percentage Decrease
$$\text{Multiplier} = 1 - \frac{r}{100}$$
$r$ = percentage decrease
e.g. $15\%$ decrease → $\times 0.85$
EXAM TIP
Write out your multiplier explicitly before calculating. "Increase by 6%" → multiplier is $1.06$. "Decrease by 6%" → multiplier is $0.94$. These add to 2, which is a useful check.
COMMON MISTAKE
A $20\%$ increase followed by a $20\%$ decrease does not return to the original. $\times 1.20 \times 0.80 = \times 0.96$, which is a $4\%$ overall decrease.
3. Percentage Change Formula
When you need to express a change as a percentage, you always divide by the original value — not the new value. This is the most commonly tested formula in percentage problems.
Percentage Change
$$\% \text{ change} = \frac{\text{change}}{\text{original}} \times 100$$
change = new value − original value
Positive result = increase; Negative = decrease
MEMORY TRICK
"Change Over Original" — C.O.O. Like a COO (chief operating officer) operating on changes. Always divide by the original, never the new value.
COMMON MISTAKE
A price rises from £40 to £50. Students write $\frac{10}{50} \times 100 = 20\%$. Correct answer: $\frac{10}{40} \times 100 = 25\%$. Always divide by the original (£40).
4. Reverse Percentages
A reverse percentage problem gives you the result after a percentage change and asks you to find the original value before the change. The key insight is that the result equals the original multiplied by the multiplier — so we divide by the multiplier to reverse it.
DEFINITION — Reverse Percentage
If a value has been changed by a percentage, the original value can be recovered by dividing the final value by the multiplier for that change.
Reverse Percentage — Finding the Original
$$\text{Original} = \frac{\text{Final value}}{\text{Multiplier}}$$
After $r\%$ increase: divide by $(1 + r/100)$
After $r\%$ decrease: divide by $(1 - r/100)$
EXAM TIP
The key phrase in reverse percentage questions is "after a X% increase/decrease". Identify the multiplier used, then divide. Never subtract the percentage from the final value directly — that gives the wrong original.
COMMON MISTAKE
A jacket costs £68 after a 15% reduction. Wrong method: $68 + (15\% \text{ of } 68) = £78.20$. Correct method: $68 \div 0.85 = £80$. The 15% was of the original price, not the sale price.
5. Simple vs Compound Interest
Simple interest is calculated only on the principal (the original amount). Compound interest is calculated on the principal plus any previously accumulated interest — interest earns interest. Over time, compound interest grows much faster.
DEFINITION — Simple Interest
Interest calculated only on the original principal each period. The interest amount is the same every year.
Simple Interest
$$I = \frac{P \times r \times t}{100}$$
$P$ = principal (initial amount)
$r$ = annual interest rate (%)
$t$ = time in years
$I$ = total interest earned
DEFINITION — Compound Interest
Interest calculated on the growing balance each period. Each year's interest is added to the principal, so the next year's interest is larger. The total amount grows exponentially.
Compound Interest Formula
$$A = P\left(1 + \frac{r}{100}\right)^n$$
$A$ = final amount (principal + interest)
$P$ = principal (initial investment)
$r$ = annual interest rate (%)
$n$ = number of years (periods)
EXAM TIP
The compound interest formula gives the total amount $A$, not just the interest. To find the interest earned, calculate $A - P$. Always state clearly which you are finding.
COMMON MISTAKE
For compound interest, do not multiply the annual interest by the number of years. That gives simple interest. You must use the formula $A = P(1 + r/100)^n$ or repeatedly multiply by the multiplier.
6. Depreciation
Depreciation is compound decrease — assets like cars lose value over time. The same compound formula applies, but the multiplier is less than 1 (a percentage decrease).
Depreciation Formula
$$V = P\left(1 - \frac{r}{100}\right)^n$$
$V$ = value after depreciation
$P$ = initial value (purchase price)
$r$ = annual depreciation rate (%)
$n$ = number of years
MEMORY TRICK
Growth uses $(1 + r/100)^n$ — the bracket is greater than 1, so values GROW. Depreciation uses $(1 - r/100)^n$ — the bracket is less than 1, so values SHRINK. Plus means grow, minus means shrink.
7. Percentages in Context — Profit, Loss, VAT, Salary
Percentage calculations appear throughout business and finance contexts. Understanding the vocabulary is essential for interpreting exam questions correctly.
DEFINITION — Profit and Loss
Profit = Selling price − Cost price (when selling price > cost price)Loss = Cost price − Selling price (when cost price > selling price)
Profit/Loss % = $\dfrac{\text{Profit or Loss}}{\text{Cost price}} \times 100$
DEFINITION — VAT (Value Added Tax)
VAT is added to the pre-tax price. In the UK, the standard rate is 20%. VAT-inclusive price = pre-tax price $\times 1.20$. To find the pre-tax price from a VAT-inclusive price: divide by $1.20$ (reverse percentage).
EXAM TIP
For profit/loss percentage: always divide by the cost price (what you paid), not the selling price. The percentage tells you how efficient the trade was relative to the original investment.
🗺️ Visual Notes
Percentages
% of a Quantity
- Convert to decimal first
- Multiply by the quantity
- $p\% = p \div 100$
- e.g. $35\%$ of £200 = £70
% Increase/Decrease
- Use a multiplier
- Increase: $1 + r/100$
- Decrease: $1 - r/100$
- One-step calculation
% Change Formula
- Change ÷ Original × 100
- Always divide by original
- Positive = increase
- Negative = decrease
Reverse Percentages
- Given final, find original
- Divide by multiplier
- Undo the change
- Key word: "after"
Compound Interest
- $A = P(1 + r/100)^n$
- Interest on interest
- Exponential growth
- Depreciation: $(1 - r/100)^n$
Real-World Contexts
- Profit and loss
- VAT (add 20%)
- Salary increases
- Hire purchase deals
Simple Interest vs Compound Interest
The table below shows £1000 invested at 5% per year for 4 years under each method, illustrating how they diverge over time.
| Feature | Simple Interest | Compound Interest |
|---|---|---|
| Formula | $I = Prt/100$ | $A = P(1 + r/100)^n$ |
| Interest basis | Original principal only | Growing balance each year |
| After Year 1 | £1050 | £1050.00 |
| After Year 2 | £1100 | £1102.50 |
| After Year 3 | £1150 | £1157.63 |
| After Year 4 | £1200 | £1215.51 |
| Growth type | Linear | Exponential |
| Better for investor? | No | Yes (always) |
Reverse Percentage — Process Chain
Identify the final value given
→
Identify the % change applied
→
Write the multiplier (e.g. 0.85 for −15%)
→
Divide final ÷ multiplier
→
Check: apply % change to your answer to verify
Common Multiplier Reference
| Change | Multiplier | Reverse (original from final) |
|---|---|---|
| 5% increase | ×1.05 | ÷1.05 |
| 10% increase | ×1.10 | ÷1.10 |
| 15% increase | ×1.15 | ÷1.15 |
| 20% increase (VAT) | ×1.20 | ÷1.20 |
| 25% increase | ×1.25 | ÷1.25 |
| 5% decrease | ×0.95 | ÷0.95 |
| 10% decrease | ×0.90 | ÷0.90 |
| 15% decrease | ×0.85 | ÷0.85 |
| 20% decrease | ×0.80 | ÷0.80 |
| 25% decrease | ×0.75 | ÷0.75 |
✏️ Worked Examples
Grade 4–5 · Foundation
A coat costs £85. It is reduced by 30% in a sale. What is the sale price?
1
Identify the multiplier
A 30% decrease uses the multiplier $1 - \dfrac{30}{100} = 1 - 0.30 = 0.70$.
2
Apply the multiplier
$$\text{Sale price} = 85 \times 0.70 = £59.50$$
3
Check the answer makes sense
30% of £85 = $0.30 \times 85 = £25.50$. Then $£85 - £25.50 = £59.50$. ✓
Sale price = £59.50
Grade 6–7 · Higher
A laptop's price after a 12% increase is £560. What was the original price?
1
Recognise this as a reverse percentage
The question gives the price after a change and asks for the price before. We need to reverse the 12% increase.
2
Write the multiplier that was applied
A 12% increase uses multiplier $1 + \dfrac{12}{100} = 1.12$.So: $\text{Original} \times 1.12 = £560$
3
Divide by the multiplier to reverse
$$\text{Original} = \frac{560}{1.12} = £500$$
4
Verify the answer
$£500 \times 1.12 = £560$ ✓
Original price = £500
Grade 9 · Multi-Step Compound
Aisha invests £4000 in Account A at 3.5% per annum compound interest for 3 years. Ben invests £4200 in Account B at 2.8% per annum simple interest for 4 years. Who has more money at the end of their investment period, and by how much? Give your answer to the nearest penny.
1
Calculate Aisha's compound interest amount
Using $A = P\left(1 + \dfrac{r}{100}\right)^n$ with $P = 4000$, $r = 3.5$, $n = 3$:
$$A = 4000 \times (1.035)^3$$
$$(1.035)^3 = 1.035 \times 1.035 \times 1.035 = 1.108717875$$
$$A = 4000 \times 1.108717875 = £4434.87\ (\text{to nearest penny})$$
2
Calculate Ben's simple interest amount
Using $I = \dfrac{P \times r \times t}{100}$ with $P = 4200$, $r = 2.8$, $t = 4$:
$$I = \frac{4200 \times 2.8 \times 4}{100} = \frac{47040}{100} = £470.40$$
$$\text{Total} = 4200 + 470.40 = £4670.40$$
3
Compare the two final amounts
Aisha: £4434.87Ben: £4670.40
Difference: $4670.40 - 4434.87 = £235.53$
4
State the conclusion clearly
Ben has more money. Although compound interest gives a better percentage return, Ben invested more money for more years, so his total is greater.
Ben has more money, by £235.53.
❓ Exam Questions
Q1
1 mark
Write down the multiplier for a 7% increase.
Answer: 1.07
Mark scheme: B1 for 1.07.
Mark scheme: B1 for 1.07.
Q2
2 marks
A mobile phone costs £320 before VAT. VAT is charged at 20%. Calculate the total price including VAT.
Answer: £384
Working:
$320 \times 1.20 = £384$
Mark scheme:
M1 for $320 \times 1.20$ or equivalent (finding 20% and adding).
A1 for £384.
Working:
$320 \times 1.20 = £384$
Mark scheme:
M1 for $320 \times 1.20$ or equivalent (finding 20% and adding).
A1 for £384.
Q3
3 marks
A car was bought for £18 000. After one year it had depreciated in value by 22%. What is the value of the car after one year?
The car continues to depreciate at 22% per year. What will be its value after 3 years in total? Give your answer to the nearest pound.
Answer: After 1 year: £14 040; After 3 years: £8512
Working:
Multiplier = $1 - 0.22 = 0.78$
After 1 year: $18000 \times 0.78 = £14040$
After 3 years: $18000 \times (0.78)^3 = 18000 \times 0.474552 = £8541.94 \approx £8542$
Note: $(0.78)^3 = 0.78 \times 0.78 \times 0.78 = 0.474552$
Mark scheme:
M1 for using multiplier 0.78.
A1 for £14 040 (after 1 year).
A1 for £8542 (after 3 years, follow through from their multiplier).
Working:
Multiplier = $1 - 0.22 = 0.78$
After 1 year: $18000 \times 0.78 = £14040$
After 3 years: $18000 \times (0.78)^3 = 18000 \times 0.474552 = £8541.94 \approx £8542$
Note: $(0.78)^3 = 0.78 \times 0.78 \times 0.78 = 0.474552$
Mark scheme:
M1 for using multiplier 0.78.
A1 for £14 040 (after 1 year).
A1 for £8542 (after 3 years, follow through from their multiplier).
Q4
3 marks
A shop sells a pair of trainers for £91 after a 30% reduction. What was the original price of the trainers?
Answer: £130
Working:
A 30% reduction means the multiplier applied was 0.70.
Original $\times$ 0.70 = £91
Original = $91 \div 0.70 = £130$
Mark scheme:
M1 for recognising the multiplier is 0.70 (or equivalent: "£91 represents 70%").
M1 for dividing by 0.70.
A1 for £130.
Working:
A 30% reduction means the multiplier applied was 0.70.
Original $\times$ 0.70 = £91
Original = $91 \div 0.70 = £130$
Mark scheme:
M1 for recognising the multiplier is 0.70 (or equivalent: "£91 represents 70%").
M1 for dividing by 0.70.
A1 for £130.
Q5
4 marks
James buys an antique vase for £240 and sells it for £300.
(a) Calculate his percentage profit.
(b) James then buys a painting for £150 and makes a 35% profit. What does he sell the painting for?
Answers: (a) 25% (b) £202.50
Working:
(a) Profit = £300 − £240 = £60
Percentage profit = $\dfrac{60}{240} \times 100 = 25\%$
(b) Selling price = $£150 \times 1.35 = £202.50$
Mark scheme:
(a) M1 for $\frac{60}{240} \times 100$; A1 for 25%.
(b) M1 for $150 \times 1.35$; A1 for £202.50.
Working:
(a) Profit = £300 − £240 = £60
Percentage profit = $\dfrac{60}{240} \times 100 = 25\%$
(b) Selling price = $£150 \times 1.35 = £202.50$
Mark scheme:
(a) M1 for $\frac{60}{240} \times 100$; A1 for 25%.
(b) M1 for $150 \times 1.35$; A1 for £202.50.
Q6
6 marks
Sarah invests £5000 at 4% per annum compound interest.
(a) How much does Sarah have after 5 years? Give your answer to the nearest penny.
(b) How much interest has she earned in total?
(c) In which year does her total amount first exceed £6000? Show all your working.
Answers: (a) £6083.26 (b) £1083.26 (c) Year 5
Working:
(a) $A = 5000 \times (1.04)^5 = 5000 \times 1.2166529... = £6083.26$
(b) Interest = $6083.26 - 5000 = £1083.26$
(c) Year by year:
Year 1: $5000 \times 1.04 = £5200$
Year 2: $5200 \times 1.04 = £5408$
Year 3: $5408 \times 1.04 = £5624.32$
Year 4: $5624.32 \times 1.04 = £5849.29$
Year 5: $5849.29 \times 1.04 = £6083.26$ ✓
She first exceeds £6000 after 5 years.
Mark scheme:
(a) M1 for $(1.04)^5$; A1 for £6083.26.
(b) B1 for subtracting correctly from 5000.
(c) M1 for systematic year-by-year method; M1 for reaching Year 5; A1 for correct year with working shown.
Working:
(a) $A = 5000 \times (1.04)^5 = 5000 \times 1.2166529... = £6083.26$
(b) Interest = $6083.26 - 5000 = £1083.26$
(c) Year by year:
Year 1: $5000 \times 1.04 = £5200$
Year 2: $5200 \times 1.04 = £5408$
Year 3: $5408 \times 1.04 = £5624.32$
Year 4: $5624.32 \times 1.04 = £5849.29$
Year 5: $5849.29 \times 1.04 = £6083.26$ ✓
She first exceeds £6000 after 5 years.
Mark scheme:
(a) M1 for $(1.04)^5$; A1 for £6083.26.
(b) B1 for subtracting correctly from 5000.
(c) M1 for systematic year-by-year method; M1 for reaching Year 5; A1 for correct year with working shown.
⭐ Grade 9 Model Answers
Below is a full, annotated model answer for Question 6 — demonstrating the level of clarity and precision needed for full marks on a 6-mark compound interest problem.
IMPORTANT — What examiners reward at Grade 9
At Grade 9, marks are awarded for: (1) Correct method shown; (2) Accurate arithmetic; (3) Clear communication; (4) Appropriate rounding with justification; (5) Logical structure. Never skip steps — show all intermediate values.
Grade 9 · Full Model Answer — Q6
Sarah invests £5000 at 4% per annum compound interest. (a) Amount after 5 years. (b) Interest earned. (c) First year exceeding £6000.
a
Part (a) — Identify the formula and substitute [M1]
The compound interest formula is $A = P\left(1 + \dfrac{r}{100}\right)^n$.Here $P = 5000$, $r = 4$, $n = 5$.
$$A = 5000 \times \left(1 + \frac{4}{100}\right)^5 = 5000 \times (1.04)^5$$
a
Part (a) — Evaluate the power carefully [A1]
$(1.04)^5 = 1.04 \times 1.04 \times 1.04 \times 1.04 \times 1.04$Step by step: $1.04^2 = 1.0816$; $1.04^3 = 1.124864$; $1.04^4 = 1.16985856$; $1.04^5 = 1.2166529...$
$$A = 5000 \times 1.2166529... = 6083.2647...$$ Rounding to the nearest penny: $A = £6083.26$ (2 d.p. since money)
b
Part (b) — Interest = Total − Principal [B1]
Total interest = $A - P = £6083.26 - £5000 = £1083.26$Note: the formula gives total amount, not just interest. Always subtract the principal to find interest alone.
c
Part (c) — Systematic year-by-year table [M1+M1]
Rather than solving $(1.04)^n \gt 1.2$ algebraically (which requires logarithms — not required at GCSE), build a table:| End of Year | Amount | Exceeds £6000? |
|---|---|---|
| 1 | $5000 \times 1.04 = £5200.00$ | No |
| 2 | $5200 \times 1.04 = £5408.00$ | No |
| 3 | $5408 \times 1.04 = £5624.32$ | No |
| 4 | $5624.32 \times 1.04 = £5849.29$ | No |
| 5 | $5849.29 \times 1.04 = £6083.26$ | Yes ✓ |
c
Part (c) — State the conclusion [A1]
Sarah's investment first exceeds £6000 at the end of Year 5.
(a) £6083.26 | (b) £1083.26 | (c) Year 5
EXAM TIP — Why this answer earns full marks
- The formula is written out with all values substituted before calculating (shows method clearly for M marks even if arithmetic slips)
- Intermediate steps of $(1.04)^5$ are shown — guards against losing all marks if the final answer is wrong
- Part (b) explicitly states that $A - P$ gives interest — shows understanding of what the formula produces
- Part (c) uses a table with a running total — systematic, easy for examiners to follow, and all M marks are visible
- Rounding is done at the final step, not throughout (prevents accumulated rounding errors)
📋 Revision Sheet
Key Definitions
| Term | Meaning |
|---|---|
| Percentage | A fraction with denominator 100 |
| Multiplier | Single factor combining original + change |
| Principal | Original amount invested/borrowed |
| Compound interest | Interest calculated on growing balance |
| Depreciation | Compound decrease in value over time |
| Reverse percentage | Finding original value before a % change |
| Profit % | Profit ÷ cost price × 100 |
| VAT | Value Added Tax (UK standard: 20%) |
Essential Formulae
$\%\text{ of quantity}: \dfrac{p}{100} \times Q$
$\%\text{ change}: \dfrac{\text{change}}{\text{original}} \times 100$
Increase multiplier: $1 + \dfrac{r}{100}$
Decrease multiplier: $1 - \dfrac{r}{100}$
Reverse: $\text{original} = \dfrac{\text{final}}{\text{multiplier}}$
Simple interest: $I = \dfrac{Prt}{100}$
Compound: $A = P\left(1 + \dfrac{r}{100}\right)^n$
Depreciation: $V = P\left(1 - \dfrac{r}{100}\right)^n$
Memory Hooks
- COO — Change Over Original (for % change formula)
- ÷ not − — For reverse %, always DIVIDE by multiplier, never subtract the %
- +1 grows, −1 shrinks — $(1+r/100)$ for growth, $(1-r/100)$ for decay
- Compound = interest on interest — the defining property
- A minus P — total Amount minus Principal gives Interest earned
- Profit over COST — % profit uses cost price as denominator, not selling price
- Two steps not one — 20% up then 20% down ≠ 0% change (it's −4%)
Exam Tips
- Show the multiplier you use before calculating — earns M marks even with arithmetic errors
- For compound interest, write the formula first with values substituted
- Reverse % questions always contain the word "after" — use divide, not subtract
- For "how many years" questions, build a year-by-year table (no logs needed at GCSE)
- Always check: does your answer make sense? (bigger after increase, smaller after decrease)
- Round only at the very end to avoid accumulated rounding errors
- State units (£) in every answer involving money
- For multi-step problems, label each step clearly so the examiner can follow your method
🔄 Flashcards
Click any card to flip it and reveal the answer. Test yourself on all 15 cards.
✗ Common Mistakes
MISTAKE 1 — Dividing by the new value in percentage change
What students do wrong: A price rises from £50 to £60. Students write $\frac{10}{60} \times 100 = 16.7\%$.Why marks are lost: This gives the wrong answer and the M mark is not awarded for incorrect method.
How to avoid it: Always divide by the original value (the one you started with). Here: $\frac{10}{50} \times 100 = 20\%$.
MISTAKE 2 — Subtracting the percentage in reverse percentage
What students do wrong: A price after a 20% increase is £120. Students write $120 - (20\% \text{ of } 120) = £96$.Why marks are lost: This finds 20% of the new value, not the original — the answer is wrong and no method marks awarded.
How to avoid it: Always divide by the multiplier: $120 \div 1.20 = £100$. The 20% was of the original (£100), not the new value (£120).
MISTAKE 3 — Using simple interest instead of compound
What students do wrong: For compound interest, students multiply the annual interest by the number of years. E.g. for £1000 at 5% for 3 years: $1000 + (50 \times 3) = £1150$.Why marks are lost: This gives £1150, not the correct £1157.63. The word "compound" in the question signals that the formula $A = P(1+r/100)^n$ must be used.
How to avoid it: Whenever you see "compound", use the formula. The key difference is that each year's interest earns further interest.
MISTAKE 4 — Treating consecutive percentage changes as additive
What students do wrong: A price increases by 10% then decreases by 10%. Students conclude the net change is 0%.Why marks are lost: The net change is actually $1.10 \times 0.90 = 0.99$, a 1% decrease overall. The percentages are NOT additive because they apply to different values.
How to avoid it: Always multiply the multipliers: $(1+a/100)\times(1-b/100)$. Only add/subtract the percentages if they apply to the same original value.
MISTAKE 5 — Confusing "total amount" and "interest earned"
What students do wrong: After applying $A = P(1+r/100)^n$, students report $A$ as the interest earned.Why marks are lost: $A$ is the total amount (principal + interest). The interest earned is $A - P$. This distinction is frequently tested in part (b) of compound interest questions.
How to avoid it: Re-read the question. "How much is in the account?" → give $A$. "How much interest was earned?" → give $A - P$.
MISTAKE 6 — Using selling price as denominator in profit percentage
What students do wrong: A guitar bought for £80 is sold for £100. Students write $\frac{20}{100} \times 100 = 20\%$ profit.Why marks are lost: Percentage profit is always expressed as a fraction of the cost (what you paid). Correct answer: $\frac{20}{80} \times 100 = 25\%$ profit.
How to avoid it: "Profit percent" = profit divided by what you paid (cost price), not what you received.
✅ Final Checklist
Click each item to mark it as done. Your progress is saved automatically.
- I can find a percentage of a quantity by converting to a decimal and multiplying
- I know the multiplier for a percentage increase is $1 + r/100$
- I know the multiplier for a percentage decrease is $1 - r/100$
- I can apply a multiplier in a single step to find a new value after a % change
- I can calculate percentage change using (change ÷ original) × 100
- I always divide by the original value in the % change formula, not the new value
- I can identify a reverse percentage question (given "after" a change, find "before")
- I can solve reverse percentage problems by dividing the final value by the multiplier
- I know the compound interest formula: $A = P(1 + r/100)^n$
- I know the depreciation formula: $V = P(1 - r/100)^n$
- I understand the difference between simple interest (linear) and compound interest (exponential)
- I can find the interest earned by calculating $A - P$
- I can solve profit/loss percentage problems and use the cost price as the denominator
- I can apply VAT by multiplying by 1.20 and remove VAT by dividing by 1.20
- I can solve multi-step percentage problems combining increase, decrease, and reverse methods