Ratio, Proportion & Rates · Notes

1.Lesson overview

Syllabus focus

Cambridge IGCSE syllabus reference

1.11 Ratio and proportion
1.12 Rates

Edexcel IGCSE syllabus reference

1.7 Ratio and proportion

AQA IGCSE syllabus reference

1.3 Ratio and proportion

A ratio compares quantities in compatible units, whereas a rate links different units. Decide whether the problem calls for total parts, a unitary value or a constant multiplier, and carry the units through every line.

By the end of this lesson

Understand and use ratio and proportion to:
• give ratios in their simplest form
• divide a quantity in a given ratio
• use proportional reasoning and ratios in context.
1 Use common measures of rate.
2 Apply other measures of rate.
3 Solve problems involving average speed.

2.Learning map

Separate three ideas that are often confused: a ratio compares parts, a proportion preserves a multiplicative relationship, and a rate compares quantities measured in different units.

Ratio and rate reasoning

Part–whole and rate structure

Use fraction equivalence and unit conversion from earlier Number lessons. Big idea. A rate compares quantities with different units; convert units before substituting into a rate formula.

Dividing in a Ratio

Dividing (or sharing) a quantity in a given ratio is one of the most common exam questions in this topic.

Direct Proportion

Two quantities are in direct proportion if when one increases, the other increases at the same rate . The ratio between them stays constant.

Inverse Proportion

Two quantities are in inverse proportion if when one increases, the other decreases at the same rate. Their product stays constant.

3.Core idea

Use fraction equivalence and unit conversion from earlier Number lessons.

Big idea. A rate compares quantities with different units; convert units before substituting into a rate formula.

Key relationship

speed = \frac{distance}{time}, density = \frac{mass}{volume}

Worked example

Example. Divide 360 in the ratio $2 : 3 : 4$ .

The total number of parts is $2 + 3 + 4 = 9$ , so one part is $360 \div 9 = 40$ . The shares are 80, 120 and 160.

4.Dividing in a Ratio

Dividing (or sharing) a quantity in a given ratio is one of the most common exam questions in this topic.

A bar of eight equal parts representing £120, with three parts forming a share of £45 and five parts forming a share of £75. — Figure 1: The ratio 3 : 5 makes 8 equal parts of £15 each, so the shares are 3 × £15 = £45 and 5 × £15 = £75.

Step-by-Step Method

1
Add the parts of the ratio to find the total number of parts
2
Divide the quantity by the total to find the value of one part
3
Multiply each ratio number by the value of one part

Dividing in a Ratio

One part = \frac{Total quantity}{Sum of ratio parts}

Example 1: Divide $200 in the ratio 3 : 5

Total $p a r t s = 3 + 5 = 8$

One part $ $200 \div 8$ = $25

1
First share 3 $\times$ $25 = $75
2
Second share 5 $\times$ $25 = $125

Check: $75 + $125 $200 Correct:

Example 2: Divide 540 kg in the ratio 2 : 3 : 4

Total $p a r t s = 2 + 3 + 4 = 9$

One part $540 \div 9 = 60$ kg

1
First = $2 \times 60 = 120$ kg
2
Second = $3 \times 60 = 180$ kg
3
Third = $4 \times 60 = 240$ kg

Check: $120 + 180 + 240 = 540$ Correct:

Example 3: Ali and Ben share money in the ratio 5 : 3. Ali gets $40 more than Ben. How much does each get?

Difference in $p a r t s = 5 - 3 = 2$ parts

2 parts $40, so 1 part = $20

1
$A l i = 5$ $\times$ $20 = $100
2
$B e n = 3$ $\times$ $20 = $60

Check: $100 - $60 $40 Correct:

Exam Tip

Always check your answer by adding the shares back together — they must equal the original total.

Share a Total in a Given RatioOpen full screen

5.Direct Proportion

Two quantities are in direct proportion if when one increases, the other increases at the same rate. The ratio between them stays constant.

Direct Proportion

\frac{y}{x} = k (constant) If y is directly proportional to x : y = k x

The unitary method means finding the value of one unit first, then multiplying to find the required amount.

Example 1: 5 pens cost $20. How much do 8 pens cost?

1
Cost of 1 pen: $ $20 \div 5 =$ $4
2
Cost of 8 pens: $8 \times$ $4 = $32

Example 2: A car uses 12 litres of fuel to travel 180 km. How far can it travel on 20 litres?

1
Distance per litre: $180 \div 12 = 15$ km
2
Distance for 20 litres: $20 \times 15 = 300$ km

Example 3: A recipe for 4 people needs 300 g of flour. How much flour for 10 people?

1
Flour for 1 person: $300 \div 4 = 75$ g
2
Flour for 10 people: $10 \times 75 = 750$ g

Recognising Direct Proportion

If you double one quantity and the other also doubles, it is direct proportion. A graph of directly proportional quantities is a straight line through the origin.

Find k and Predict a New ValueOpen full screen

6.Inverse Proportion

Two quantities are in inverse proportion if when one increases, the other decreases at the same rate. Their product stays constant.

Inverse Proportion

x \times y = k (constant) If y is inversely proportional to x : y = \frac{k}{x}

Example 1: 6 workers can complete a job in 10 days. How long will it take 4 workers?

1
Total worker-days $6 \times 10 = 60$
2
Time for 4 workers $60 \div 4 = 15$ days
3
More workers → less time. Fewer workers → more time.

Example 2: A car travelling at 60

km/h

takes 3 hours. How long at 90

km/h

1
Distance = $60 \times 3 = 180$ km
2
Time at $90 km/h = 180 \div 90 = 2$ hours

Example 3: 8 taps fill a tank in 3 hours. How long for 12 taps?

1
Total tap-hours $8 \times 3 = 24$
2
Time for 12 taps $24 \div 12 = 2$ hours

Direct vs Inverse Proportion

Quantity	Direct Proportion	Inverse Proportion
Relationship	Both increase together	One increases, the other decreases
Constant	$\frac{y}{x} = k$	$x \times y = k$
Example	More items → higher cost	More workers → less time
Graph	Straight line through origin	Curve (hyperbola)

Common Mistake

Students often use direct proportion when inverse is needed. Ask yourself: "If I increase one quantity, does the other go up or down?" If it goes down, it's inverse proportion.

Planned interactivecore

Ratio, Proportion and Rates Lab

Share equal parts, vary a constant and build rate journeys.

Connect ratio bars, direct/inverse graphs and compound measures with units.

Inputs & controls

ratio parts
total
direct or inverse
constant
journey distances and times
units

Learner actions

change total
drag x
build journey
choose a model

Visible outputs

shares and unit part
table and graph
constant test
rate
dimensional check

Implementation note

Use total distance divided by total time for average speed and visibly reject averaging speeds without weighting.

Direct vs Inverse Proportion GraphsOpen full screen

7.Proportional Reasoning in Real Contexts

What is proportional reasoning?

Two quantities are in proportion when their ratio stays constant. You can scale one quantity up or down and the other changes by the same factor.

Example 1 — Recipes

A cake for 4 people uses 200 g of flour. How much flour is needed for 10 people?

1
Ratio of people: $4 : 10 = 2 : 5$ , so flour scales by $\times$ $\frac{5}{2}$ .
2
Flour needed = $200 \times 5 \div 2$ = 500 g.

Example 2 — Map scales

1
On a map with scale 1 : 25 000, two towns are 8 cm apart. Find the real distance.
2
Real distance = $8 \times 25000 = 200000 cm = 2 km$ .

Example 3 — Best value

Pack A: 500 g for $2.00. Pack B: 800 g for $3.00. Which is better value?

1
Pack A: $200 \div 500$ $0.40 g. Pack B: $300 \div 800$ = $0.375 g.
2
Pack B is better value (lower price per 100 g).

Exam tip

Always compare like with like: cost per unit, distance per second, etc. Decide which quantity to scale to 1 before comparing.

8.Simplifying Ratios

A ratio compares two or more quantities of the same kind. We write a ratio using a colon, e.g. 3:5. Ratios have no units.

Simplifying Ratios

To write a ratio in its simplest form, divide all parts by their highest common factor (HCF).

If the ratio involves decimals, multiply to make whole numbers first. If it involves different units, convert to the same unit first.

Example 1: Simplify 12 : 18

1
HCF of 12 and 18 is 6.
2
12 : 18 $\frac{12}{6}$ : $\frac{18}{6} = 2$ : 3

Example 2: Simplify 0.4 : 1.2

1
Multiply both by 10 to remove decimals: 4 : 12
2
HCF of 4 and 12 is 4: 4 : 12 1 : 3

Example 3: Simplify 40 cm : 1.2 m (different units)

Convert to the same unit: $1.2 m = 120 cm$

1
40 : 120
2
The HCF is 40, so $40 : 120 = 1 : 3$

Example 4: Simplify

\frac{1}{3}

\frac{1}{2}

1
Multiply both by the LCM of the denominators (6):
2
$\frac{1}{3} \times 6$ : $\frac{1}{2} \times 62$ : 3

Writing a Ratio in the Form 1 : n

Ratios are sometimes required in the form $1 : n$ (or $n : 1$ ), where $n$ need not be a whole number — this is different from simplifying to lowest integer terms, and is especially useful for comparing map scales.

Example: Write the ratio $4 : 9$ in the form $1 : n$ .

1
Divide both parts by the first number, 4: $\frac{4}{4} : \frac{9}{4}$ .
2
$1 : 2.25$ .

Just like equivalent fractions, you can multiply or divide all parts of a ratio by the same number to get an equivalent ratio.

The Ratio 2:3 is Equivalent to:

$4:6, 6:9, 8:12, 10:15, ...$

All of these simplify back to 2:3.

Common Mistake

Always check that the quantities are in the same units before writing a ratio. "500 g to 2 kg" is NOT 500 : 2. Convert first: 500 g : 2000 $g = 1$ : 4.

9.Challenge Questions

Three stepping-stone questions spanning ratio, rate and proportion — each goes one step further than the worked examples above.

Challenge 1 — working from one share back to the total

Question. A prize fund is split between three charities in the ratio $2 : 5 : 7$ . The smallest charity receives $150. Find the total prize fund.

1
The smallest charity's share corresponds to the smallest ratio number, $2$ parts, so $2$ parts represent $150.
2
One part is $75.
3
Total parts $= 2 + 5 + 7 = 14$ , so $14 \times 75 = 1050$ and the total fund is $1050.

Every dividing-ratio question uses the same idea — find one part — but here you're given a single share instead of the total, so work out one part from that share first.

Challenge 2 — a rate question with an inverse-proportion twist

Question. A tap fills a $60$ -litre tank at a steady flow rate of $4$ litres per minute. (a) How long does it take to fill the tank? (b) A second, identical tap is opened alongside the first. Assuming the flow rates add together, how long would the two taps take?

1
(a) Time $= \frac{volume}{flow rate} = 60 \div 4 = 15$ minutes.
2
(b) Combined flow rate $= 4 + 4 = 8$ litres per minute.
3
New time $= 60 \div 8 = 7.5$ minutes.

Flow rate (litres per minute) is a rate just like speed or density — it links two different units. Doubling the number of taps is exactly the same idea as the "more workers, less time" inverse-proportion examples above.

Challenge 3 — the same amount, two different ratios

Question. Paint Shade A is mixed as yellow : blue $= 5 : 3$ . To make $40$ litres of Shade A, (a) how much yellow paint is used? (b) If a second batch uses the same amount of yellow but is mixed as yellow : blue $= 5 : 4$ instead, how much blue paint and total paint does that batch contain?

1
(a) Total parts for Shade A $= 5 + 3 = 8$ , so one part $= 40 \div 8 = 5$ litres. Yellow $= 5 \times 5 = 25$ litres.
2
(b) In the new ratio $5 : 4$ , yellow is still $5$ parts, and it's still $25$ litres — but that means one part is now $25 \div 5 = 5$ litres in this ratio too (it happens to match here, since the yellow-part number didn't change).
3
Blue $= 4 \times 5 = 20$ litres. Total $= 9 \times 5 = 45$ litres.

The trap: you cannot reuse "one part" from the first ratio in the second — a different ratio means a different total number of parts, so recompute one part from the fixed quantity (the yellow) each time.

10.Exam-Style Worked Examples

Solving a ratio problem

1
Read what is given
Is it the total, the difference, or one share?
2
Count the parts
Add the ratio numbers for a total; subtract them for a difference.
3
Find one part
Divide the given amount by that number of parts.
4
Scale up each share
Multiply one part by each ratio number.
5
Check
The shares must reproduce the original total or difference.

Worked example 1 — sharing in a ratio (3 marks)

Question. $£4200$ is shared between Amir, Beth and Chen in the ratio $3 : 5 : 6$ . How much does each receive?

1
Total number of parts $= 3 + 5 + 6 = 14$ .
2
One part $= \frac{4200}{14} = 300$ .
3
Amir $= 3 \times 300 = £900$ ; Beth $= 5 \times 300 = £1500$ ; Chen $= 6 \times 300 = £1800$ . Check: they total $£4200$ .

Marking. 1 mark per point. Always check the parts add back to the total — it catches arithmetic slips instantly.

Worked example 2 — a given difference (4 marks)

Question. Two numbers are in the ratio $7 : 4$ . Their difference is $18$ . Find both numbers.

1
The difference in parts is $7 - 4 = 3$ parts.
2
So $3$ parts $= 18$ , giving one part $= 6$ .
3
The numbers are $7 \times 6 = 42$ and $4 \times 6 = 24$ .
4
Check: $42 - 24 = 18$ , and $42 : 24$ simplifies to $7 : 4$ .

Marking. 1 mark per point. Read carefully whether the question gives the total or the difference — the number of parts used differs.

Worked example 3 — best value and rates (3 marks)

Question. Rice is sold as $750 g$ for $£2.40$ or $2 kg$ for $£6.20$ . Which is better value? Show your working.

1
Work out the price per gram for each, using the same units throughout.
2
Small: $\frac{240}{750} = 0.32$ pence per gram. Large: $\frac{620}{2000} = 0.31$ pence per gram.
3
The $2 kg$ pack is better value, at $0.31$ p per gram compared with $0.32$ p.

Marking. 1 mark per point. Convert to consistent units first, and state which is better value with the figures to justify it.

11.Exam Tips & Common Misconceptions

Exam Tips

Always simplify ratios by dividing by the HCF: 12 : 18 2 : 3.
For "share in ratio", first add the parts (e.g. 2:3:5 → 10 parts), find one part, then multiply.
Direct proportion: y = kx. If x doubles, y doubles. Inverse: y = $\frac{k}{x}$ . If x doubles, y halves.
Best-buy: divide price by quantity to get unit price — the smaller, the better.
Map scales: 1 : 50 000 means 1 cm on the map 50 000 $c m = 500$ $m = 0.5$ km in reality.

Common Misconceptions

"3 : 5 means three-fifths" — it means 3 parts out of 8 total parts.
Multiplying ratios elementwise: scaling 2 : 3 by a factor of 4 gives 8 : 12, not 8 : 3 or 2 : 12.
Using direct proportion when inverse is intended — "more workers, less time" is inverse, not direct.
Forgetting units in scale problems — convert cm to $m$ or km before stating the final length.

12.Worked method — Share 35 in the ratio 2 : 3 using equal units

A ratio bar works because every unit has the same value; always check the shares return the total.

Transfer the method

Reasoning prompt. A journey time is given in hours and minutes. What must happen before calculating an average speed?

1
Convert the whole time into one unit, usually hours or seconds.
2
Match the distance unit to the speed unit required.
3
Use the rate relationship and give the result with its unit.

Self-check

The shares of a ratio must add back to the original total.

13.Rate denominators and weighted journeys

Read the unit before choosing the operation

Speed

distance per time

$speed = distance \div time$ ; convert minutes to hours before using km/h.

Density

mass per volume

$density = mass \div volume$ ; a unit such as g/cm³ reveals the denominator.

Unit price

cost per item or mass

Divide each comparable cost by the same quantity before deciding which offer is better value.

Worked Examples: Density and Population Density

Like speed, density and population density are rates that link two different units — but unlike speed, the formula is usually given in the question.

Density. A metal block has mass 540 g and volume 60 cm³. Using $density = \frac{mass}{volume}$ , find its density.

1
$density = 540 \div 60 = 9$
2
The density is $9 g/cm^{3}$ .

Population density. A city has a population of 850 000 and covers an area of 340 km². Using $population density = \frac{population}{area}$ , find its population density.

1
$population density = 850000 \div 340 = 2500$
2
The population density is $2500$ people per km².

Converting between km/h and m/s

Speed is often given in km/h but needed in m/s (or the reverse). Since $1 km = 1000 m$ and $1 hour = 3600 s$ :

1 km/h = \frac{1000 m}{3600 s} = \frac{1}{3.6} m/s

So: km/h → m/s: divide by 3.6. m/s → km/h: multiply by 3.6.

Worked examples: km/h ⇄ m/s

1
A car's speedometer reads $72$ km/h. In m/s: $72 \div 3.6 = 20$ m/s.
2
A sprinter runs at $8$ m/s. In km/h: $8 \times 3.6 = 28.8$ km/h.

Average speed uses total distance divided by total time

1
Leg 1
60 km in 1 h · speed 60 km/h
2
Leg 2
60 km in 2 h · speed 30 km/h
3
Totals
120 km in 3 h · combine first
4
Average speed
$120 \div 3 = 40$ km/h

The mean of 60 and 30 is 45, but that is wrong here because the times are unequal.

Worked example: average speed is not usually the mean of two speeds

A vehicle travels 60 km at 30 km/h, then 60 km at 60 km/h.

First stage time: $60 \div 30 = 2$ h. Second stage time: $60 \div 60 = 1$ h.

Total distance $= 120$ km and total time $= 3$ h, so average speed $= 120 \div 3 = 40 km/h$ .

The arithmetic mean, 45 km/h, is wrong because the vehicle spends different amounts of time at the two speeds.

Memory device: the DST triangle

Picture a triangle split into three cells: D alone on top, S and T side by side underneath. Cover up whichever letter you want to find, and the two that remain show you what to do:

Cover D: S and T are side by side, so $D = S \times T$ .
Cover S: D sits above T, so $S = D \div T$ .
Cover T: D sits above S, so $T = D \div S$ .

The same triangle works for any "per" rate — swap D and S for mass and density, or cost and unit price.

14.Proportion equations: find $k$ before predicting

Direct and inverse proportion are equations, not just verbal patterns. A single known pair of values lets you find the constant $k$ ; the equation can then predict a new value. The relationship is valid only while the relevant conditions stay fixed, such as the number of identical workers, the rate of work or the material being measured.

Direct proportion

Write $y = k x$ . The quotient $\frac{y}{x}$ remains constant, so scaling $x$ by a factor also scales $y$ by that factor.

Inverse proportion

Write $y = \frac{k}{x}$ . The product $x y$ remains constant, so multiplying $x$ by a factor divides $y$ by that factor.

Use units as a check

If $m = k L$ , then $k$ has units of mass per length. Units help expose a reversed equation or an inappropriate model.

Worked example — direct and inverse models

Direct proportion. The mass $m$ kg of a uniform cable is directly proportional to its length $L$ m. A cable of length $18 m$ has mass $1.35 kg$ . Find the mass of $42 m$ .

1
Write $m = k L$ . Using the known cable, $1.35 = 18 k$ , so $k = \frac{1.35}{18} = 0.075$ kg per metre.
2
For $42 m$ , $m = 0.075 \times 42 = 3.15 kg$ .
3
Inverse proportion. Six identical workers take $15$ hours to finish a task. With the same rate of work, $w t = k$ , so $6 \times 15 = 90$ worker-hours. For $10$ workers, $t = \frac{90}{10} = 9$ hours.

A trend alone does not prove proportion

More workers may reduce the time, but it is inversely proportional only when the total work and each worker's rate stay constant. Setup time, crowding or a changing task can break the model.

15.Summary

Key Points

Central principle. A rate compares quantities with different units; convert units before substituting into a rate formula
Dividing (or sharing) a quantity in a given ratio is one of the most common exam questions in this topic.
Add the parts of the ratio to find the total number of parts
Divide the quantity by the total to find the value of one part
Multiply each ratio number by the value of one part
Key relationship: $speed = \frac{distance}{time}, density = \frac{mass}{volume}$
Dividing in a Ratio: $One part = \frac{Total quantity}{Sum of ratio parts}$
Direct Proportion: $\frac{y}{x} = k (constant) If y is directly proportional to x : y = k x$
Inverse Proportion: $x \times y = k (constant) If y is inversely proportional to x : y = \frac{k}{x}$