Ratio, Proportion, and Rates of Change

Ratio, Proportion, and Rates of Change

:::info Board Coverage AQA Paper 1 & 2 | Edexcel Paper 1 & 2 | OCR Paper 1, 2, 3 | WJEC Unit 1 & 2

1. Ratio

1.1 Writing and Simplifying Ratios

A ratio compares quantities of the same kind. Ratios are written in their simplest form by Dividing all parts by their highest common factor.

Worked Example. Simplify the ratio $24 : 36 : 48$.

\mathrm{HCF(24, 36, 48) = 12

$24 : 36 : 48 = 2 : 3 : 4$

Worked Example. Express the ratio $0.4 : 1.2$ in its simplest form.

Multiply both by 10: $4 : 12 = 1 : 3$.

Worked Example (Higher Tier). Express the ratio $2\frac{1}{2} : 3\frac{3}{4}$ as a ratio of Whole numbers.

Convert to improper fractions: $\frac{5}{2} : \frac{15}{4}$.

Multiply both by 4: $10 : 15 = 2 : 3$.

1.2 Sharing in a Ratio

To share a quantity $Q$ in the ratio $a : b$:

\mathrm{Each share = \frac{Q}{a + b}

First share $= \frac{aQ}{a + b}$Second share $= \frac{bQ}{a + b}$.

Worked Example. Share 360 pounds in the ratio $2 : 3 : 4$.

\mathrm{Total parts = 2 + 3 + 4 = 9 \mathrm{One part = \frac{360}{9} = 40 \mathrm{Shares = 80, 120, 160

Worked Example (Higher Tier). Share 560 pounds in the ratio $\frac{1}{2} : \frac{1}{3} : \frac{1}{4}$.

Multiply each part by 12: $6 : 4 : 3$.

Total parts: 13. One part: $\frac{560}{13}$.

Shares: $\frac{3360}{13}, \frac{2240}{13}, \frac{1680}{13}$Approximately 258.46, 172.31, 129.23 Pounds.

1.3 Ratios and Fractions

The ratio $a : b$ means:

• The first quantity is $\frac{a}{a + b}$ of the total
• The second quantity is $\frac{b}{a + b}$ of the total

Worked Example. In a class, the ratio of boys to girls is $3 : 5$. What fraction of the class Are girls?

$\frac{5}{3 + 5} = \frac{5}{8}$

1.4 Combining Ratios

Worked Example. The ratio of $A$ to $B$ is $2 : 3$ and the ratio of $B$ to $C$ is $4 : 5$. Find The ratio $A : B : C$.

Make $B$ the same in both ratios. LCM of 3 and 4 is 12.

$A : B = 8 : 12$ and $B : C = 12 : 15$.

Therefore $A : B : C = 8 : 12 : 15$.

Worked Example (Higher Tier). The ratio of $A$ to $B$ is $3 : 4$ and the ratio of $B$ to $C$ is $6 : 7$. Find the ratio $A : B : C$.

$A : B = 3 : 4 = 9 : 12$ and $B : C = 6 : 7 = 12 : 14$.

$A : B : C = 9 : 12 : 14$.

1.5 Unit Ratios

Worked Example. 8 chocolate bars cost 3.60 pounds. Find the cost of 15 bars.

\mathrm{Cost per bar = \frac{3.60}{8} = 0.45 \mathrm{ pounds \mathrm{Cost of 15 bars = 15 \times 0.45 = 6.75 \mathrm{ pounds

Worked Example (Higher Tier). It takes 5 people 8 hours to paint a fence. How long would it take 3 people?

This is an inverse proportion problem. Total person-hours $= 5 \times 8 = 40$.

Time for 3 people $= \frac{40}{3} \approx 13.3$ hours.

2. Direct and Inverse Proportion

2.1 Direct Proportion

$y$ is directly proportional to $x$ when $y = kx$ for a constant $k$.

Key features:

• As $x$ increases, $y$ increases
• The ratio $\frac{y}{x}$ is constant
• The graph is a straight line through the origin with gradient $k$

Worked Example. The cost $C$ of petrol is directly proportional to the number of litres $L$. 12 Litres costs 16.80 pounds. Find the cost of 20 litres.

$16.80 = 12k \implies k = 1.40$ $C = 1.40L$ C = 1.40 \times 20 = 28.00 \mathrm{ pounds

2.2 Inverse Proportion

$y$ is inversely proportional to $x$ when $y = \frac{k}{x}$ for a constant $k$.

Key features:

• As $x$ increases, $y$ decreases
• The product $xy$ is constant
• The graph is a reciprocal curve

Worked Example. The time $t$ to complete a job is inversely proportional to the number of Workers $n$. 6 workers take 10 hours. How long do 15 workers take?

$10 = \frac{k}{6} \implies k = 60$ t = \frac{60}{15} = 4 \mathrm{ hours

2.3 Proportionality with Powers

$y$ is directly proportional to $x^2$: $y = kx^2$.

$y$ is inversely proportional to $\sqrt{x}$: $y = \frac{k}{\sqrt{x}}$.

Worked Example. $y$ is directly proportional to $x^3$. When $x = 2$, $y = 24$. Find $y$ when $x = 5$.

$24 = 8k \implies k = 3$ $y = 3 \times 125 = 375$

Worked Example (Higher Tier). The kinetic energy $E$ of an object is directly proportional to The square of its velocity $v$. When $v = 10$ m/s, $E = 500$ J. Find $E$ when $v = 15$ m/s.

$E = kv^2 \implies 500 = 100k \implies k = 5$

E = 5 \times 225 = 1125 \mathrm{ J

3. Real-Life Graphs

3.1 Distance-Time Graphs

Feature	Meaning
Gradient	Speed
Horizontal section	Stationary
Steeper section	Faster speed
Downhill section	Returning towards start

Worked Example. A person walks at 2 m/s for 10 seconds, then stops for 5 seconds, then runs at 5 M/s for 4 seconds. Draw and interpret the distance-time graph.

• After 10 s: distance = 20 m (gradient = 2 m/s)
• After 15 s: distance = 20 m (horizontal, stationary)
• After 19 s: distance = 20 + 20 = 40 m (gradient = 5 m/s)

Worked Example (Higher Tier). A car accelerates uniformly from rest. The distance-time graph is A curve. How would you find the instantaneous speed at $t = 4$ s?

Draw the tangent to the curve at $t = 4$ s and find its gradient. Alternatively, differentiate the Distance function if given.

3.2 Speed-Time (Velocity-Time) Graphs

Feature	Meaning
Gradient	Acceleration
Area under the graph	Distance travelled
Below the time axis	Moving in the opposite direction

Key equations of motion (constant acceleration):

$v = u + at$ $s = ut + \frac{1}{2}at^2$ $v^2 = u^2 + 2as$ $s = \frac{1}{2}(u + v)t$

Where $u$ = initial velocity, $v$ = final velocity, $a$ = acceleration, $t$ = time, $s$ = Displacement.

Worked Example. A car accelerates from rest at 3 m/s$^2$ for 8 seconds. Find the distance Travelled.

s = 0 \times 8 + \frac{1}{2} \times 3 \times 64 = 96 \mathrm{ m

Worked Example (Higher Tier). A car travels at 20 m/s for 5 s, then decelerates uniformly to Rest in 4 s. Find the total distance travelled.

Phase 1: $d_1 = 20 \times 5 = 100$ m.

Phase 2: $a = \frac{0 - 20}{4} = -5$ m/s$^2$. $d_2 = 20 \times 4 + \frac{1}{2} \times (-5) \times 16 = 80 - 40 = 40$ m.

Total: 140 m.

3.3 Rates of Change

The rate of change of a quantity is how fast it changes per unit of another quantity ( Time).

\mathrm{Rate of change = \frac{\mathrm{change in quantity}{\mathrm{change in time}

This is the gradient of the graph of that quantity against time.

Worked Example. The population of a town was 45000 in 2020 and 54000 in 2025. Find the average Rate of change.

\mathrm{Rate = \frac{54000 - 45000}{2025 - 2020} = \frac{9000}{5} = 1800 \mathrm{ people per year

3.4 Instantaneous Rate of Change

The instantaneous rate of change at a point is the gradient of the tangent to the curve at That point.

Worked Example. The volume of water in a tank is given by $V = 2t^2 + 3t$ litres, where $t$ is In minutes. Find the rate at which the volume is increasing when $t = 4$.

Draw the tangent at $t = 4$ (or differentiate):

$\frac{dV}{dt} = 4t + 3$ \mathrm{At t = 4: \frac{dV}{dt} = 16 + 3 = 19 \mathrm{ litres/minute

4. Growth and Decay

4.1 Exponential Growth

$y = a(1 + r)^t$

Where $a$ is the initial amount, $r$ is the growth rate (as a decimal), and $t$ is time.

4.2 Exponential Decay

$y = a(1 - r)^t$

Worked Example. A radioactive substance decays at a rate of 8% per hour. If there are initially 500 g, how much remains after 12 hours?

y = 500 \times 0.92^{12} = 500 \times 0.3677\ldots = 183.8 \mathrm{ g

4.3 Half-Life

The half-life is the time taken for a quantity to reduce to half its initial value.

For exponential decay with rate $r$:

$t_{1/2} = \frac{\ln 2}{-\ln(1 - r)}$

Worked Example. A substance has a half-life of 6 hours. If the initial mass is 80 g, find the Mass after 24 hours.

\mathrm{Number of half-lives = \frac{24}{6} = 4 \mathrm{Mass = 80 \times \left(\frac{1}{2}\right)^4 = 80 \times \frac{1}{16} = 5 \mathrm{ g

Worked Example (Higher Tier). The mass of a radioactive substance after $t$ years is given by $m = 200 \times 0.95^t$ grams. Find the half-life.

$100 = 200 \times 0.95^t$ $0.5 = 0.95^t$ t = \frac{\ln 0.5}{\ln 0.95} = \frac{-0.693}{-0.0513} \approx 13.5 \mathrm{ years

5. Compound Measures

5.1 Speed, Density, and Pressure

Measure	Formula	Units
Speed	\mathrm{speed = \frac{\mathrm{distance}{\mathrm{time}	m/s, km/h
Density	\rho = \frac{\mathrm{mass}{\mathrm{volume}	kg/m$^3$G/cm$^3$
Pressure	P = \frac{\mathrm{force}{\mathrm{area}	Pa (N/m$^2$)

Worked Example. A block of metal has mass 1.8 kg and volume 200 cm$^3$. Find its density in G/cm$^3$.

\rho = \frac{1800 \mathrm{ g}{200 \mathrm{ cm^3} = 9 \mathrm{ g/cm^3

Worked Example. A force of 600 N acts on an area of 0.75 m$^2$. Find the pressure.

P = \frac{600}{0.75} = 800 \mathrm{ Pa

Worked Example (Higher Tier). A pressure of 5000 Pa is exerted on an area of 0.04 m$^2$. Find The force.

F = P \times A = 5000 \times 0.04 = 200 \mathrm{ N

5.2 Unit Conversions

Conversion	Factor
1 km = 1000 m	$\times 1000$
1 m = 100 cm	$\times 100$
1 hour = 60 min = 3600 s	$\times 60$, $\times 3600$
1 litre = 1000 cm$^3$	$\times 1000$
1 kg = 1000 g	$\times 1000$

Worked Example. Convert 72 km/h to m/s.

72 \mathrm{ km/h = \frac{72000 \mathrm{ m}{3600 \mathrm{ s} = 20 \mathrm{ m/s

Shortcut: To convert km/h to m/s, divide by 3.6. To convert m/s to km/h, multiply by 3.6.

5.3 Currency Conversions

Worked Example. The exchange rate is 1 pound = 1.25 euros. Convert 240 pounds to euros.

240 \times 1.25 = 300 \mathrm{ euros

Worked Example (Higher Tier). The exchange rate is 1 dollar = 0.82 pounds. A laptop costs 899 Dollars. How much is this in pounds?

899 \times 0.82 = 737.18 \mathrm{ pounds

5.4 Best Buy Problems

Compare the unit cost (cost per gram, per litre, etc.) of different products.

Worked Example. Brand A: 400 g for 3.20 pounds. Brand B: 650 g for 4.55 pounds. Which is better Value?

\mathrm{Brand A: \frac{320}{400} = 0.80 \mathrm{ p/g \mathrm{Brand B: \frac{455}{650} = 0.70 \mathrm{ p/g

Brand B is better value (lower cost per gram).

6. Interpretation of Graphs in Context

6.1 Gradient and Area

When interpreting real-life graphs:

• The gradient represents a rate of change
• The area under the curve represents a total or accumulated quantity

6.2 Financial Graphs

• Straight line with positive gradient: Simple interest or constant rate of saving
• Curve getting steeper: Compound interest or accelerating growth
• Curve flattening out: Diminishing returns

6.3 Conversion Graphs

A conversion graph between two units is always a straight line through the origin (if the conversion Is proportional).

Worked Example. A conversion graph between miles and kilometres passes through $(0, 0)$ and $(5, 8)$. Find the conversion factor.

\mathrm{Gradient = \frac{8 - 0}{5 - 0} = 1.6

So 1 mile = 1.6 km.

7. Ratio Problems with Algebra

7.1 Expressing Ratios as Algebraic Fractions

Worked Example. Two numbers are in the ratio $3 : 5$. If 4 is added to each number, the ratio Becomes $2 : 3$. Find the numbers.

Let the numbers be $3x$ and $5x$.

$\frac{3x + 4}{5x + 4} = \frac{2}{3}$ $3(3x + 4) = 2(5x + 4)$ $9x + 12 = 10x + 8$ $x = 4$

The numbers are 12 and 20.

7.2 Changing Ratios

Worked Example. A bag contains red and blue counters in the ratio $3 : 5$. When 8 red counters Are added, the ratio becomes $5 : 6$. How many blue counters are there?

Let red $= 3x$ and blue $= 5x$.

$\frac{3x + 8}{5x} = \frac{5}{6}$ $6(3x + 8) = 25x$ $18x + 48 = 25x$ $7x = 48$

Since $x$ must be an integer, there is no solution with whole counters. Let us re-check: $7x = 48$ Gives $x = 48/7$Which is not an integer. This suggests the problem has no integer solution, which Would need to be re-examined in context.

7.3 More Complex Ratio Problems (Higher Tier)

Worked Example. The ratio of the angles in a triangle is $2 : 3 : 4$. Find the largest angle.

Total parts: $2 + 3 + 4 = 9$.

Each part $= \frac{180}{9} = 20^{\circ}$.

Largest angle $= 4 \times 20 = 80^{\circ}$.

Worked Example (Higher Tier). A metal alloy is made from copper, zinc, and tin in the ratio $7 : 3 : 2$. If the alloy weighs 480 g, how much copper does it contain?

Total parts: $7 + 3 + 2 = 12$.

Copper $= \frac{7}{12} \times 480 = 280$ g.

Worked Example (Higher Tier). Two quantities are in the ratio $a : b$. When each quantity is Increased by 15, the ratio becomes $5 : 4$. The original ratio is $3 : 2$. Find $a$ and $b$.

This is consistent with the earlier worked example: $a = 12$ and $b = 20$.

Let me verify: $12 : 20$ simplifies to $3 : 2$. After adding 15: $27 : 35$. But $27 : 35 \neq 5 : 4$.

So the numbers $12$ and $20$ with added $4$ gave $5 : 6$Not $5 : 4$. Let me solve from scratch.

$\frac{a + 15}{b + 15} = \frac{5}{4}$ and $\frac{a}{b} = \frac{3}{2}$.

From the second equation: $a = \frac{3b}{2}$.

Substituting: $\frac{3b/2 + 15}{b + 15} = \frac{5}{4}$.

$4(3b/2 + 15) = 5(b + 15)$.

$6b + 60 = 5b + 75$.

$b = 15$, $a = 22.5$.

Verify: $22.5 : 15 = 3 : 2$. After adding 15: $37.5 : 30 = 5 : 4$. $\checkmark$

8. Density Problems (Higher Tier)

Worked Example. A solid metal cone has base radius 4 cm and height 9 cm. Its density is 8.4 G/cm$^3$. Find its mass.

Volume = \frac{1}{3}\pi \times 16 \times 9 = 48\pi \approx 150.80 \mathrm{ cm^3.

Mass = 8.4 \times 150.80 \approx 1266.7 \mathrm{ g.

Worked Example. A cuboid measures 5 \mathrm{ cm \times 3 \mathrm{ cm \times 2 \mathrm{ cm. It has Mass 180 g. Find its density in kg/m$^3$.

Volume = 30 \mathrm{ cm^3 = 30 \times 10^{-6} \mathrm{ m^3.

Mass = 0.180 \mathrm{ kg.

Density = \frac{0.180}{30 \times 10^{-6}} = \frac{0.180}{0.00003} = 6000 \mathrm{ kg/m^3.

Common Pitfalls

• Mixing up the order of a ratio. The ratio $A : B = 3 : 5$ means $A/B = 3/5$Not $B/A$.
• Forgetting to add all parts when sharing in a ratio. The total number of parts is $a + b$Not just $a$ or $b$.
• Confusing direct and inverse proportion graphs. Direct proportion is a straight line through the origin; inverse proportion is a curve.
• Incorrect unit conversions, especially km/h to m/s (divide by 3.6, not multiply).
• Reading the wrong axis on distance-time or speed-time graphs. Always check what each axis represents.
• Forgetting that the area under a speed-time graph gives distance, even when the graph goes below the time axis (in which case it represents returning).
• Forgetting to use consistent units in density/pressure problems. Convert everything to SI units before calculating.
• Assuming that a ratio of $a : b : c$ with three unknowns has a unique solution without additional constraints.

9.1 Compound Measures Summary

Measure	Formula	Units
Speed	\mathrm{speed = \frac{d}{t}	m/s, km/h
Density	\mathrm{density = \frac{m}{V}	kg/m$^3$G/cm$^3$
Pressure	\mathrm{pressure = \frac{F}{A}	Pa (N/m$^2$)

9.2 Proportionality: A Quick Reference

Type	Equation	Graph shape
Direct	$y = kx$	Straight line through origin
Inverse	$y = k/x$	Reciprocal curve
Direct sq.	$y = kx^2$	Quadratic through origin
Direct sq.rt.	$y = k\sqrt{x}$	Square root curve

How to determine the type of proportionality from a table of values:

1. Check if $y/x$ is constant (direct proportion).
2. Check if $xy$ is constant (inverse proportion).
3. Check if $y/x^2$ is constant (direct square proportion).
4. Check if $y/\sqrt{x}$ is constant (direct square root proportion).

Worked Example. Given the table:

$x$	2	4	8
$y$	6	12	24

Check $y/x$: 6/2 = 3$$12/4 = 3$$24/8 = 3. Constant, so $y$ is directly proportional to $x$. The equation is $y = 3x$.

9.3 Graphical Interpretation of Rates

Distance-time graphs:

• Gradient = speed.
• A horizontal line means stationary.
• A steeper gradient means faster speed.
• A curve means the speed is changing (acceleration or deceleration).

Speed-time graphs:

• Gradient = acceleration.
• Area under the graph = distance travelled.
• A horizontal line means constant speed (zero acceleration).
• Area below the time axis represents motion in the opposite direction.

Worked Example. A car accelerates from rest at 2\mathrm{ m/s^2 for 10 seconds, then travels at Constant speed for 20 seconds, then decelerates uniformly to rest in 5 seconds.

Acceleration phase: $v = 2 \times 10 = 20$ m/s. Distance = $\frac{1}{2} \times 20 \times 10 = 100$ M.

Constant speed phase: Distance = $20 \times 20 = 400$ m.

Deceleration phase: a = -20/5 = -4\mathrm{ m/s^2. Distance = $\frac{1}{2} \times 20 \times 5 = 50$ M.

Total distance = $100 + 400 + 50 = 550$ m.

9.4 Reverse Percentage Problems

A reverse percentage problem asks: “A price after VAT is 120 pounds. The VAT rate is 20%. What was The original price?”

The original price is NOT $120 - 24 = 96$ pounds. Instead:

\mathrm{Original \times 1.20 = 120 \implies \mathrm{Original = \frac{120}{1.20} = 100 \mathrm{ pounds

Worked Example. A population grows by 15% to reach 4600. What was the original population?

\mathrm{Original \times 1.15 = 4600 \implies \mathrm{Original = \frac{4600}{1.15} = 4000

9.5 Best Buy Problems

When comparing products of different sizes, calculate the unit price for each.

Worked Example. Brand A: 400 g for 2.40 pounds. Brand B: 750 g for 3.90 pounds. Which is better Value?

Brand A: $2.40/400 = 0.0060$ pounds/g = 0.60 pounds per 100 g.

Brand B: $3.90/750 = 0.0052$ pounds/g = 0.52 pounds per 100 g.

Brand B is better value.

9.6 Exchange Rates

If 1 pound = 1.18 euros, then to convert pounds to euros, multiply by 1.18. To convert euros to Pounds, divide by 1.18.

Worked Example. A meal costs 45 euros. How much is this in pounds?

45 \div 1.18 \approx 38.14 \mathrm{ pounds

Worked Example. A tourist exchanges 500 pounds for euros at a rate of 1 pound = 1.15 euros, with A commission of 3 pounds. How many euros do they receive?

Euros before commission: $500 \times 1.15 = 575$ euros. Commission: 3 pounds = $3 \times 1.15 = 3.45$ euros. Net: $575 - 3.45 = 571.55$ euros.

Practice Questions

1. Divide 420 pounds in the ratio $3 : 5 : 8$.

2. The ratio of $A$ to $B$ is $5 : 2$ and $B$ to $C$ is $3 : 4$. Find $A : B : C$.

3. $y$ is inversely proportional to $x^2$. When x = 3$$y = 12. Find $y$ when $x = 6$.

4. A car travels 45 km in 30 minutes, then 60 km in 45 minutes. Calculate the average speed for the whole journey.

5. The density of a material is 7.8 g/cm$^3$. Find the mass of a block of dimensions 5 \mathrm{ cm \times 3 \mathrm{ cm \times 2 \mathrm{ cm.

6. A population grows by 4% each year. If the initial population is 50000, find the population after 7 years.

7. On a speed-time graph, a car accelerates uniformly from 10 m/s to 30 m/s over 8 seconds. Calculate the distance travelled during this acceleration.

8. Two numbers are in the ratio $4 : 7$. When 10 is subtracted from each, the ratio becomes $1 : 2$. Find the numbers.

9. A 500 g box of cereal costs 2.40 pounds and a 750 g box costs 3.30 pounds. Which is better value per 100 g?

10. The half-life of carbon-14 is 5730 years. A sample contains 12.5% of its original carbon-14. How old is the sample?

11. A cylinder has radius 5 cm, height 12 cm, and mass 1.2 kg. Find its density in g/cm$^3$.

12. Water flows from a tap at a rate of 0.5 litres per second. How long does it take to fill a tank of volume 450 litres?

13. $y$ is directly proportional to $x^n$. When x = 2$$y = 24. When x = 4$$y = 192. Find the value of $n$.

14. The ratio of boys to girls in a school is $3 : 5$. If there are 240 more girls than boys, how many students are there in total?

15. A car decelerates uniformly from 25 m/s to 5 m/s in 4 seconds. Sketch the speed-time graph and find the distance travelled.

Extended Practice (Higher Tier)

16. The ratio of the interior angles of a triangle is $2 : 3 : 4$. Find the size of each angle.

17. $y$ is directly proportional to $\sqrt{x}$. When x = 9$$y = 18. Find $y$ when $x = 25$.

18. A train travels at 120 km/h for 45 minutes, then at 80 km/h for 1 hour 15 minutes. Calculate the average speed for the entire journey.

19. Three friends share a pizza in the ratio of their ages. If their ages are 12, 15, and 18, what fraction of the pizza does the oldest friend receive?

20. The population of a town was 80000 in 2020 and 88000 in 2025. Assuming exponential growth, find the annual growth rate as a percentage.

21. Water flows into a tank at 12 litres per minute and out at 5 litres per minute. If the tank initially contains 200 litres, how long until it contains 500 litres?

22. A rectangular field has dimensions in the ratio $3 : 5$. If the perimeter is 320 m, find the area of the field.

23. On a journey, a cyclist accelerates from rest to 8 m/s over 10 seconds, maintains this speed for 2 minutes, then decelerates to rest over 5 seconds. Draw the speed-time graph and find the total distance.

24. The cost of petrol increases by 8% one year and then by 5% the next year. What is the overall percentage increase over the two years?

25. A map has a scale of $1 : 25000$. Two towns are 8 cm apart on the map. Find the actual distance between them in kilometres.

Extended Content

10.1 Simple Interest vs Compound Interest

Simple interest is calculated only on the original principal:

$$I = Prt$$

Where $P$ is the principal, $r$ is the annual rate, and $t$ is the time in years. The total amount Is $A = P + Prt = P(1 + rt)$.

Compound interest is calculated on the principal plus accumulated interest:

$$A = P(1 + r)^t$$

Worked Example. EUR 2000 is invested at 5% per annum for 3 years. Compare simple and compound Interest.

Simple interest: $I = 2000 \times 0.05 \times 3 = 300$. Total = EUR 2300.

Compound interest: $A = 2000(1.05)^3 = 2000 \times 1.157625 = 2315.25$. Total = EUR 2315.25.

The compound interest earns EUR 15.25 more than simple interest over 3 years.

10.2 Percentage Change Calculations

Percentage increase: \frac{\mathrm{new - \mathrm{original}{\mathrm{original} \times 100\%

Percentage decrease: \frac{\mathrm{original - \mathrm{new}{\mathrm{original} \times 100\%

Worked Example. A shirt originally costs 40 pounds. It is reduced to 32 pounds in a sale. What Is the percentage decrease?

$$\frac{40 - 32}{40} \times 100\% = \frac{8}{40} \times 100\% = 20\%$$

Successive percentage changes. If a price increases by 10% then decreases by 10%, it does NOT Return to the original. A 10% increase on 100 gives 110. A 10% decrease on 110 gives $110 \times 0.9 = 99$. The net effect is a 1% decrease.

Worked Example. A salary of 30000 pounds increases by 4% one year and 3% the next. What is the New salary?

30000 \times 1.04 \times 1.03 = 30000 \times 1.0712 = 32136 \mathrm{ pounds

The overall percentage increase is $7.12\%$ (not $4\% + 3\% = 7\%$).

10.3 Pressure

Pressure is the force per unit area:

$$P = \frac{F}{A}$$

Worked Example. A person weighing 600 N stands on the ground. If each foot has area 180\mathrm{ cm^2Find the pressure.

Total area = 360\mathrm{ cm^2 = 0.036\mathrm{ m^2.

P = \frac{600}{0.036} = 16667 \mathrm{ Pa \approx 16.7 \mathrm{ kPa

Worked Example. The pressure at the bottom of a swimming pool is 30 kPa. The pool has a Rectangular base of 10\mathrm{ m \times 5\mathrm{ m. Find the total force on the base.

F = P \times A = 30000 \times 50 = 1500000 \mathrm{ N = 1500 \mathrm{ kN

10.4 Speed, Distance, Time Problems with Multiple Stages

Worked Example. A cyclist travels from A to B at 20 km/h and returns from B to A at 12 km/h. The Total journey takes 4 hours. Find the distance from A to B.

Let $d$ be the distance. Then:

$$\frac{d}{20} + \frac{d}{12} = 4$$ $$D\left(\frac{1}{20} + \frac{1}{12}\right) = 4$$ $$D\left(\frac{3 + 5}{60}\right) = 4$$ \frac{8d}{60} = 4 \implies d = 30 \mathrm{ km

Note: The average speed is NOT $(20 + 12)/2 = 16$ km/h. The average speed is:

\frac{\mathrm{total distance}{\mathrm{total time} = \frac{60}{4} = 15 \mathrm{ km/h

The harmonic mean of 20 and 12 is $\frac{2}{1/20 + 1/12} = 15$Which equals the average speed for a Return journey.

Worked Examples

Example 1:

A typical exam question on Ratio, Proportion, and Rates of Change requires you to apply your knowledge to an unfamiliar context. Read the question carefully, identify the key concept being tested, and structure your answer using the appropriate terminology.

Example 2:

Multi-step problems in Ratio, Proportion, and Rates of Change often combine two or more concepts. Break the problem down: identify what you need to find, recall the relevant formula or principle, substitute values, and state your answer with correct units or formatting.

Summary

This topic covers the mathematical techniques and concepts related to ratio, proportion, and rates of change, including key theorems, methods, and problem-solving approaches.

Key concepts include:

• quadratic equations and the discriminant
• simultaneous equations
• polynomial division and the factor theorem
• partial fractions
• binomial expansion

Regular practice with a variety of question types is essential to build fluency and confidence in applying these mathematical techniques.

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