Ratio, Proportion and Rates of ChangeAQA GCSE Mathematics (8300)

Ratio, Proportion and Rate of Change

Work through the notes, try the practice questions, then take the quiz. The report tells you exactly what to revise next. (2026)

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Master the topic

Worked examples4 Key formulae4 Definitions6 Common mistakes4

Step-by-step worked examples

01.Dividing in a ratio and reverse ratio
Foundation
Question
Two brothers share an inheritance in the ratio $5 : 3$ . The elder brother receives £4 500 more than the younger. Find the total inheritance.
Solution
1. 1
  The difference between the two shares corresponds to $5 - 3 = 2$ parts.
  $2 \text{ parts} = £4\,500$
2. 2
  Find the value of one part.
  $1 \text{ part} = £2\,250$
3. 3
  Total parts $= 5 + 3 = 8$ .
  $\text{Total} = 8 \times £2\,250 = £18\,000$
Answer
Total inheritance $= £18\,000$ .
02.Direct proportion with a power
Higher
Question
$F$ is directly proportional to $v^2$ . When $v = 10$ , $F = 250$ . Find $F$ when $v = 14$ .
Solution
1. 1
  Write the proportionality equation.
  $F = kv^2$
2. 2
  Substitute the known values to find $k$ .
  $250 = k(10)^2 \implies k = 250 \div 100 = 2.5$
3. 3
  Write the complete equation.
  $F = 2.5v^2$
4. 4
  Substitute $v = 14$ .
  $F = 2.5 \times 196 = 490$
Answer
$F = 490$
03.Reverse percentage and VAT
Higher
Question
A television costs £552 including 15% VAT. Find the price before VAT.
Solution
1. 1
  Identify the multiplier: adding 15% VAT multiplies the price by $1.15$ .
  $\text{price with VAT} = \text{price without VAT} \times 1.15$
2. 2
  Divide by the multiplier to reverse the operation.
  $\text{price without VAT} = 552 \div 1.15$
3. 3
  Calculate.
  $552 \div 1.15 = £480$
Answer
Price before VAT $= £480$ .
04.Compound interest and finding time
Challenge
Question
£6 000 is invested at 3.5% compound interest per year. After how many complete years does the investment first exceed £8 000?
Solution
1. 1
  Write the compound interest formula with the known values.
  $A = 6000 \times 1.035^n$
2. 2
  Set up the inequality.
  $6000 \times 1.035^n > 8000 \implies 1.035^n > \tfrac{8000}{6000} = 1.\overline{3}$
3. 3
  Use trial and improvement.
  $1.035^8 \approx 1.3168; \quad 1.035^9 \approx 1.3629$
4. 4
  $1.035^8 \approx 1.317 < 1.\overline{3}$ but $1.035^9 \approx 1.363 > 1.\overline{3}$ , so the investment exceeds £8 000 after 9 complete years.
Answer
After 9 complete years.

Key formulae

Percentage Change Multiplier
$\text{new value} = \text{original} \times \left(1 \pm \frac{r}{100}\right)$
- $r$
  — percentage increase or decrease
- $+$
  — use for an increase
- $-$
  — use for a decrease
When to use
Applying a percentage increase or decrease in one step; the reverse (finding the original) uses division by the same multiplier.
Compound Interest / Exponential Growth & Decay
$A = P\left(1 \pm \frac{r}{100}\right)^n$
- $P$
  — initial principal (or starting amount)
- $r$
  — percentage rate per period
- $n$
  — number of periods
- $A$
  — final amount
When to use
Compound interest (growth, $+$ ) and depreciation/decay ( $-$ ); any situation where the same percentage change is applied repeatedly.
Speed, Distance and Time
$S = \frac{D}{T}, \quad D = ST, \quad T = \frac{D}{S}$
- $S$
  — speed (e.g. km/h, m/s)
- $D$
  — distance (e.g. km, m)
- $T$
  — time (must match speed units: h or s)
When to use
Any speed, distance or time calculation; also used to find the gradient of a distance–time graph.
Density and Pressure
$\text{Density} = \frac{M}{V}, \qquad \text{Pressure} = \frac{F}{A}$
- $M$
  — mass (g or kg)
- $V$
  — volume (cm³ or m³)
- $F$
  — force (N)
- $A$
  — area (cm² or m²)
When to use
Problems involving material properties (density) or forces acting over a surface (pressure); rearrange as needed using the formula triangle.

Practice with flashcards

Card 1 of 60 known · 0 to review

Key definitions and keywords

Ratio: A comparison of two or more quantities of the same type, written as $a : b$ . It shows how many times one quantity contains another.
Direct proportion: Two quantities are in direct proportion if their ratio is constant: $y/x = k$ . The graph is a straight line through the origin.
Inverse proportion: Two quantities are in inverse proportion if their product is constant: $xy = k$ . As one increases, the other decreases in the same ratio.
Compound measure: A measure formed by combining two other measures, e.g. speed (distance/time), density (mass/volume), pressure (force/area).
Multiplier: The number you multiply by to carry out a percentage change in one step. A 20% increase has multiplier 1.20; a 15% decrease has multiplier 0.85.
Exponential growth/decay: A process where a quantity increases (growth) or decreases (decay) by a fixed percentage each time period. Modelled by $A = A_0 \times m^t$ where $m > 1$ for growth and $0 < m < 1$ for decay.

Common mistakes and misconceptions

Mistake
Reverse percentage: subtracting the percentage from the final value instead of dividing by the multiplier
Why it happens
Students think 'undo a 20% increase' means 'take 20% off the final price', not realising that 20% of the final price is larger than 20% of the original.
How to avoid it
Always identify the multiplier first ( $+20\% \to 1.20$ ) and divide the final value by it. Check: the original should be smaller than the final value for an increase.
Mistake
Confusing direct and inverse proportion and using the wrong equation
Why it happens
Students see a proportionality statement and default to $y = kx$ without checking whether it is direct or inverse.
How to avoid it
Read the question carefully: 'directly proportional to' → $y = kx^n$ ; 'inversely proportional to' → $y = k/x^n$ . Check the graph shape: direct → passes through origin; inverse → hyperbola.
Mistake
Using simple interest formula for a compound interest question
Why it happens
Students apply $I = PRT/100$ for all interest problems without checking whether the interest is compounded.
How to avoid it
Check the question: 'compound interest' means use $A = P(1 + r/100)^n$ . If 'simple interest' is stated, use $I = PRT/100$ . For more than one period, compound always gives a higher final amount.
Mistake
Units mismatch in compound measure calculations
Why it happens
Students substitute values with inconsistent units (e.g. time in minutes into a km/h speed formula) without converting first.
How to avoid it
Write the units next to every value before substituting. If distance is in km and time in minutes, either convert time to hours ( $\div 60$ ) or convert to metres and seconds. The units of the answer should match the units of the formula.

Ratio, Proportion and Rates of ChangeAQA GCSE Mathematics (8300)

Ratio, Proportion and Rate of Change

Work through the notes, try the practice questions, then take the quiz. The report tells you exactly what to revise next. (2026)

PreviousNext

Master the topic

Worked examples4 Key formulae4 Definitions6 Common mistakes4

Step-by-step worked examples

01.Dividing in a ratio and reverse ratio
Foundation
Question
Two brothers share an inheritance in the ratio $5 : 3$ . The elder brother receives £4 500 more than the younger. Find the total inheritance.
Solution
1. 1
  The difference between the two shares corresponds to $5 - 3 = 2$ parts.
  $2 \text{ parts} = £4\,500$
2. 2
  Find the value of one part.
  $1 \text{ part} = £2\,250$
3. 3
  Total parts $= 5 + 3 = 8$ .
  $\text{Total} = 8 \times £2\,250 = £18\,000$
Answer
Total inheritance $= £18\,000$ .
02.Direct proportion with a power
Higher
Question
$F$ is directly proportional to $v^2$ . When $v = 10$ , $F = 250$ . Find $F$ when $v = 14$ .
Solution
1. 1
  Write the proportionality equation.
  $F = kv^2$
2. 2
  Substitute the known values to find $k$ .
  $250 = k(10)^2 \implies k = 250 \div 100 = 2.5$
3. 3
  Write the complete equation.
  $F = 2.5v^2$
4. 4
  Substitute $v = 14$ .
  $F = 2.5 \times 196 = 490$
Answer
$F = 490$
03.Reverse percentage and VAT
Higher
Question
A television costs £552 including 15% VAT. Find the price before VAT.
Solution
1. 1
  Identify the multiplier: adding 15% VAT multiplies the price by $1.15$ .
  $\text{price with VAT} = \text{price without VAT} \times 1.15$
2. 2
  Divide by the multiplier to reverse the operation.
  $\text{price without VAT} = 552 \div 1.15$
3. 3
  Calculate.
  $552 \div 1.15 = £480$
Answer
Price before VAT $= £480$ .
04.Compound interest and finding time
Challenge
Question
£6 000 is invested at 3.5% compound interest per year. After how many complete years does the investment first exceed £8 000?
Solution
1. 1
  Write the compound interest formula with the known values.
  $A = 6000 \times 1.035^n$
2. 2
  Set up the inequality.
  $6000 \times 1.035^n > 8000 \implies 1.035^n > \tfrac{8000}{6000} = 1.\overline{3}$
3. 3
  Use trial and improvement.
  $1.035^8 \approx 1.3168; \quad 1.035^9 \approx 1.3629$
4. 4
  $1.035^8 \approx 1.317 < 1.\overline{3}$ but $1.035^9 \approx 1.363 > 1.\overline{3}$ , so the investment exceeds £8 000 after 9 complete years.
Answer
After 9 complete years.

Key formulae

Percentage Change Multiplier
$\text{new value} = \text{original} \times \left(1 \pm \frac{r}{100}\right)$
- $r$
  — percentage increase or decrease
- $+$
  — use for an increase
- $-$
  — use for a decrease
When to use
Applying a percentage increase or decrease in one step; the reverse (finding the original) uses division by the same multiplier.
Compound Interest / Exponential Growth & Decay
$A = P\left(1 \pm \frac{r}{100}\right)^n$
- $P$
  — initial principal (or starting amount)
- $r$
  — percentage rate per period
- $n$
  — number of periods
- $A$
  — final amount
When to use
Compound interest (growth, $+$ ) and depreciation/decay ( $-$ ); any situation where the same percentage change is applied repeatedly.
Speed, Distance and Time
$S = \frac{D}{T}, \quad D = ST, \quad T = \frac{D}{S}$
- $S$
  — speed (e.g. km/h, m/s)
- $D$
  — distance (e.g. km, m)
- $T$
  — time (must match speed units: h or s)
When to use
Any speed, distance or time calculation; also used to find the gradient of a distance–time graph.
Density and Pressure
$\text{Density} = \frac{M}{V}, \qquad \text{Pressure} = \frac{F}{A}$
- $M$
  — mass (g or kg)
- $V$
  — volume (cm³ or m³)
- $F$
  — force (N)
- $A$
  — area (cm² or m²)
When to use
Problems involving material properties (density) or forces acting over a surface (pressure); rearrange as needed using the formula triangle.

Practice with flashcards

Card 1 of 60 known · 0 to review

Key definitions and keywords

Ratio: A comparison of two or more quantities of the same type, written as $a : b$ . It shows how many times one quantity contains another.
Direct proportion: Two quantities are in direct proportion if their ratio is constant: $y/x = k$ . The graph is a straight line through the origin.
Inverse proportion: Two quantities are in inverse proportion if their product is constant: $xy = k$ . As one increases, the other decreases in the same ratio.
Compound measure: A measure formed by combining two other measures, e.g. speed (distance/time), density (mass/volume), pressure (force/area).
Multiplier: The number you multiply by to carry out a percentage change in one step. A 20% increase has multiplier 1.20; a 15% decrease has multiplier 0.85.
Exponential growth/decay: A process where a quantity increases (growth) or decreases (decay) by a fixed percentage each time period. Modelled by $A = A_0 \times m^t$ where $m > 1$ for growth and $0 < m < 1$ for decay.

Common mistakes and misconceptions

Mistake
Reverse percentage: subtracting the percentage from the final value instead of dividing by the multiplier
Why it happens
Students think 'undo a 20% increase' means 'take 20% off the final price', not realising that 20% of the final price is larger than 20% of the original.
How to avoid it
Always identify the multiplier first ( $+20\% \to 1.20$ ) and divide the final value by it. Check: the original should be smaller than the final value for an increase.
Mistake
Confusing direct and inverse proportion and using the wrong equation
Why it happens
Students see a proportionality statement and default to $y = kx$ without checking whether it is direct or inverse.
How to avoid it
Read the question carefully: 'directly proportional to' → $y = kx^n$ ; 'inversely proportional to' → $y = k/x^n$ . Check the graph shape: direct → passes through origin; inverse → hyperbola.
Mistake
Using simple interest formula for a compound interest question
Why it happens
Students apply $I = PRT/100$ for all interest problems without checking whether the interest is compounded.
How to avoid it
Check the question: 'compound interest' means use $A = P(1 + r/100)^n$ . If 'simple interest' is stated, use $I = PRT/100$ . For more than one period, compound always gives a higher final amount.
Mistake
Units mismatch in compound measure calculations
Why it happens
Students substitute values with inconsistent units (e.g. time in minutes into a km/h speed formula) without converting first.
How to avoid it
Write the units next to every value before substituting. If distance is in km and time in minutes, either convert time to hours ( $\div 60$ ) or convert to metres and seconds. The units of the answer should match the units of the formula.

Ratio, Proportion and Rate of Change

Master the topic

Step-by-step worked examples

01.Dividing in a ratio and reverse ratio

02.Direct proportion with a power

03.Reverse percentage and VAT

04.Compound interest and finding time

Key formulae

Practice with flashcards

Key definitions and keywords

Common mistakes and misconceptions

Ratio, Proportion and Rate of Change

Master the topic

Step-by-step worked examples

01.Dividing in a ratio and reverse ratio

02.Direct proportion with a power

03.Reverse percentage and VAT

04.Compound interest and finding time

Key formulae

Practice with flashcards

Key definitions and keywords

Common mistakes and misconceptions