C1.11 — Ratio and Proportion #
IGCSE Mathematics | Core Topic
A ratio compares two or more quantities. Ratios appear in everyday life — in cooking recipes, map reading, mixing colours, and comparing prices. In this topic you will learn how to simplify ratios, share quantities using a ratio, and apply proportional reasoning to real-world situations.
VIDEO NEEDED: Introduction video for this topic
YouTube Search: “IGCSE O Level maths ratio and proportion introduction”
1. What is a Ratio? #
A ratio shows how two or more quantities compare to each other. It is written with a colon ( : ).
Example: a bag with 3 red balls and 5 blue balls has a red-to-blue ratio of 3 : 5.
- Order matters. The ratio 3:5 (red:blue) is not the same as 5:3 (blue:red). Always match the ratio to the order written in the question.
- Ratios have no units. Both quantities must be in the same unit before you write the ratio.
2. Simplifying Ratios #
A ratio is in its simplest form when all the numbers share no common factor other than 1. To simplify, divide every number by the highest common factor (HCF).
The HCF (Highest Common Factor) is the largest number that divides exactly into all parts of the ratio with no remainder.
How to find it: Ask yourself — what is the biggest number that goes into every part exactly? A useful shortcut is to test common factors in order: try 2, then 5, then 10, and so on. Pick the largest one that works for all parts.
- Find the HCF of 20, 30, and 40.
Test 10: $20 \div 10 = 2$ ✓ $30 \div 10 = 3$ ✓ $40 \div 10 = 4$ ✓
Test 20 (next candidate): $30 \div 20 = 1.5$ ✗ not a whole number, so 20 does not work.
HCF = 10 - Divide every part by 10: $20 \div 10 \ : \ 30 \div 10 \ : \ 40 \div 10$
- Simplest form: 2 : 3 : 4
- Convert to the same unit: $2\text{ m} = 200\text{ cm}$
- Write the ratio: $50 : 200$
- HCF of 50 and 200 = 50 → divide each: $50 \div 50 \ : \ 200 \div 50$
- Simplest form: 1 : 4
3. Dividing a Quantity in a Given Ratio #
Think of the ratio as dividing a quantity into equal groups. In ratio 3:5, there are 3 groups of one quantity and 5 groups of another — 8 equal groups in total. The table method uses this idea without drawing the groups out.
Case A — The total is given #
- Add the ratio parts to get the total: $3 + 5 = 8$ → place 40 in the Total column
- Find the multiplier: $40 \div 8 = 5$
- Multiply all ratio parts by 5:
| Red | Green | Total | |
|---|---|---|---|
| Ratio | 3 | 5 | 8 |
| × 5 | 15 | 25 | 40 |
Answer: 15 red pens and 25 green pens.
- Add the ratio parts: $8 + 1 + 3 = 12$ → place 300 in the Total column
- Find the multiplier: $300 \div 12 = 25$
- Multiply all ratio parts by 25:
| Petrol | Diesel | Electric | Total | |
|---|---|---|---|---|
| Ratio | 8 | 1 | 3 | 12 |
| × 25 | 200 | 25 | 75 | 300 |
Answer: 75 electric cars.
Case B — One part is given (not the total) #
The 45 is only the blue quantity — place it in the Blue column, not the Total column.
- Find the multiplier using the known part: $45 \div 3 = 15$
- Multiply all ratio parts by 15:
| Blue | Orange | Total | |
|---|---|---|---|
| Ratio | 3 | 2 | 5 |
| × 15 | 45 | 30 | 75 |
Answer: 30 orange counters.
Case C — The difference between two parts is given #
The 120 min is the difference between their times. Use a Difference column. To find the difference in ratio parts, subtract (do not add).
- Difference in ratio parts: $7 – 2 = 5$ → place 120 in the Difference column
- Find the multiplier: $120 \div 5 = 24$
- Multiply all ratio parts by 24:
| Millie | Lydia | Difference | |
|---|---|---|---|
| Ratio | 7 | 2 | 5 |
| × 24 | 168 | 48 | 120 |
Answer: Millie studied 168 minutes, Lydia studied 48 minutes.
- Total given → add ratio parts → place number in Total column
- One part given → place number in that part’s column
- Difference given → subtract ratio parts → place number in Difference column
In all cases: multiplier = given number ÷ its ratio number. Then multiply all parts by the multiplier.
4. Proportional Reasoning in Context #
Adapting Recipes #
To scale a recipe up or down, find the scale factor and multiply every ingredient by it.
- Scale factor: $10 \div 4 = 2.5$
- Flour needed: $200 \times 2.5 = 500\text{ g}$
Mixing with Limited Amounts #
When you have a fixed amount of each ingredient and want the maximum you can make while keeping the correct ratio:
- For each ingredient: divide the available amount by its ratio number
- Choose the lowest result as your multiplier — a higher number would mean running out of that ingredient
- Multiply all ratio numbers by this multiplier, then add the results for the total
- Water: $450 \div 9 = 50$
- Juice: $80 \div 2 = 40$
- Lowest value is 40 → multiplier = 40 (using 50 would require 100 ml juice, but only 80 ml is available)
- Water used: $9 \times 40 = 360\text{ ml}$ Juice used: $2 \times 40 = 80\text{ ml}$
- Total drink: $360 + 80 = \mathbf{440}\text{ ml}$
All 80 ml of juice is used. 90 ml of water is left over ($450 – 360 = 90\text{ ml}$).
Map Scales #
A map scale is a ratio comparing map distance to real distance, written as 1 : n. One unit on the map equals $n$ units in real life.
- $1\text{ cm on map} = 50{,}000\text{ cm in real life}$
- $3\text{ cm} \times 50{,}000 = 150{,}000\text{ cm}$
- Convert: $150{,}000\text{ cm} = 1{,}500\text{ m} = 1.5\text{ km}$
Best Value #
To find the best value, calculate the price per unit (e.g. per gram, per ml). The lower the price per unit, the better the value.
| Pack | Size | Price | Price per 100 g |
|---|---|---|---|
| Pack A | 200 g | 80p | $80 \div 2 = 40\text{p}$ |
| Pack B | 500 g | 175p | $175 \div 5 = 35\text{p}$ |
Pack B is better value — it costs less per 100 g.
Syllabus Reference — C1.11 Ratio and Proportion #
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.
e.g. 20:30:40 in its simplest form is 2:3:4. e.g. adapt recipes; use map scales; determine best value.
