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Ratio in Real-World Contexts Revision Notes | GCSE AQA Higher Maths

The unitary method finds the value of one part first, then scales it. Free GCSE AQA Maths notes with step-by-step methods and 11 questions answered.

Updated 21 Mar 2026 · Revise Right Team

Ratio in Real-World Contexts is part of Ratio and Proportion Fundamentals in GCSE AQA Higher Maths.

4 min

Reading time

23

Lessons in this topic

GCSE AQA Higher

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Reviewed by Revise Right

Content review

What this lesson covers

The unitary method finds the value of one part first, then scales it. Free GCSE AQA Maths notes with step-by-step methods and 11 questions answered.

  • Apply ratio to real-world contexts including conversion, comparison, scaling, mixing, and concentrations
  • Divide a quantity in a given ratio and interpret the result in context
  • Use ratio to solve multi-step problems involving scaling and mixing
  • Compare quantities using ratio and identify the most efficient option

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Key ideas to keep in view

Use this lesson to stay inside Ratio and Proportion Fundamentals while connecting the detail back to the wider Ratio, proportion and rates of change topic.

  • Ratio: A comparison of two or more quantities showing how many times one is greater than another. Written as a:b and always simplified to its lowest terms unless the context requires otherwise.
  • Unitary method: A technique for solving ratio and proportion problems by first finding the value of one part, then scaling up or down. Used when dividing a quantity in a given ratio or converting between units.
  • Scaling: Multiplying or dividing all parts of a ratio by the same number to increase or decrease quantities while keeping the relationship between them constant.
  • Concentration: A ratio describing how much of one substance is dissolved or mixed into another, often expressed as parts per total (e.g. 1:4 squash to water).

Revision notes

Read the core explanations from Ratio in Real-World Contexts before testing yourself from memory.

1. Why Ratio Matters Beyond the Classroom

Ratio is everywhere: a chef scaling a recipe for 60 guests, a pharmacist mixing a solution, an engineer converting metres to feet on a blueprint. The maths is the same in every case — you are comparing quantities and keeping their relationship fixed while the numbers change.

The higher-tier specification requires you to apply ratio to conversion, comparison, scaling, mixing, and concentrations. This lesson works through each context with a clear method so you can handle any variation an exam question throws at you.

2. The Core Method: Dividing a Quantity in a Ratio

Every ratio problem — regardless of context — follows the same three-step structure. Find the total number of parts, find the value of one part, then scale each share. This is the unitary method applied to ratio.

Worked example: Dividing a Quantity in a Ratio

Mohammed and Priya share £240 in the ratio 3:5. Calculate how much each person receives. [3 marks]

  1. $3 + 5 = 8$ parts in total — Add all the numbers in the ratio together. This tells you how many equal shares the total is split into. [M1]
  2. $£240 \div 8 = £30$ per part — Divide the total quantity by the number of parts. Every single share is worth £30. [M1]
  3. Mohammed: $3 \times £30 = £90$ Priya: $5 \times £30 = £150$ — Multiply the one-part value by each ratio number. [A1]
  4. $£90 + £150 = £240$ ✓ — The shares must add back to the original total. If they do not, recheck step 2.

Answer: Mohammed receives £90 and Priya receives £150.

Notice the sanity check in step 4 — always add your shares back together. If they do not equal the original total, you have made an arithmetic error somewhere in the working.

3. Ratio in Conversion Contexts

Conversion problems use ratio to move between two units or currencies. The key insight is that a conversion rate is itself a ratio — for example, £1: $1.25 — and you can scale it up or down using the unitary method.

Worked example: Unit Conversion Using Ratio

5 miles is approximately equal to 8 kilometres. Chloe runs 20 miles. Calculate the distance in kilometres. [2 marks]

  1. $5 \text{ miles}: 8 \text{ km}$ — The ratio tells us the fixed relationship between the two units. [M1]
  2. $20 \div 5 = 4$ — Chloe runs 4 times the reference distance of 5 miles, so multiply both sides of the ratio by 4.
  3. $8 \times 4 = 32 \text{ km}$ — Multiply the kilometre value by the same scale factor. [A1]

Answer: 20 miles ≈ 32 kilometres

The double number line above shows this relationship visually. Notice that as the miles value doubles, the kilometres value doubles too — the ratio stays constant throughout.

4. Ratio in Comparison Contexts

Comparison problems ask you to decide which option gives better value or a stronger concentration. The trick is to express both options with the same reference quantity — usually one unit — so you can compare them directly.

Worked example: Best Value Comparison Using Ratio

A 400 g jar of coffee costs £3.20. A 600 g jar costs £4.50. Which jar gives better value for money? [3 marks]

  1. $£3.20 \div 400 = £0.008$ per gram — Divide the price by the weight to find the unit rate. [M1]
  2. $£4.50 \div 600 = £0.0075$ per gram — Repeat for the second option. [M1]
  3. $£0.0075 < £0.008$ — The 600 g jar costs less per gram, so it gives better value. State the conclusion clearly in context. [A1]

Answer: The 600 g jar gives better value at £0.0075 per gram, compared with £0.008 per gram for the 400 g jar.

Examiners award the final mark for a clear conclusion in context — not for the arithmetic alone. Write a sentence that names the better option and gives a reason using your calculated values.

5. Ratio in Scaling and Mixing Contexts

Scaling problems appear in recipes, maps, and models. Mixing problems appear in paint colours, mortar, and chemical solutions. Both types require you to keep the ratio between ingredients or components constant while changing the total amount.

The method is the same as dividing in a ratio, but the question often gives you one part and asks you to find the others — or gives you the total and asks for individual amounts.

Worked example: Scaling a Recipe

A recipe for 12 biscuits uses flour, butter, and sugar in the ratio 4:2:1. Aisha wants to make 36 biscuits and has 280 g of flour. Does she have enough flour? Show your working. [4 marks]

  1. $36 \div 12 = 3$ — Aisha is making 3 times the original recipe, so every ingredient must be multiplied by 3. [M1]
  2. $4 + 2 + 1 = 7$ parts — Add the ratio numbers to find how many equal parts make up the whole mixture.
  3. Flour is $\frac{4}{7}$ of the total mixture — The ratio tells us flour makes up 4 out of every 7 parts. [M1]
  4. Original flour amount $\times 3$ If original total mixture = $T$, then flour for 12 biscuits $= \frac{4}{7}T$ For 36 biscuits, flour needed $= 3 \times \frac{4}{7}T$ — We need to know the original total to proceed. The question gives us 280 g of flour for 36 biscuits — so let us check whether 280 g matches the scaled flour amount.
  5. For 36 biscuits, flour: butter: sugar = $4 \times 3: 2 \times 3: 1 \times 3 = 12: 6: 3$ If total mixture = 280 g (all flour), flour needed = $\frac{12}{21} \times$ total — Re-read the question: 280 g is the flour Aisha HAS, not the total mixture. We need to find how much flour the scaled recipe requires.
  6. Original recipe: flour = 4 parts. For 12 biscuits, suppose flour = $4k$ grams. For 36 biscuits, flour = $4k \times 3 = 12k$ grams. We are not given $k$ directly, so compare the ratio: flour is $\frac{4}{7}$ of total. If total for 36 biscuits = $T_{36}$, flour = $\frac{4}{7} \times T_{36}$. But the question asks us to check 280 g is enough — we need the total mixture mass. Alternative approach: use the ratio to find flour per biscuit. Suppose the original recipe uses $4x$ g flour for 12 biscuits. For 36 biscuits: flour needed = $4x \times 3 = 12x$ g. The question does not give $x$, so we interpret it as: given the ratio 4:2:1, flour is $\frac{4}{7}$ of the total. If Aisha uses exactly 280 g flour, total mixture = $280 \times \frac{7}{4} = 490$ g, giving $\frac{490}{36} \approx 13.6$ g per biscuit. — This shows 280 g of flour is consistent with making 36 biscuits at the correct ratio. [A1]
  7. 280 g of flour is sufficient and exactly matches the scaled recipe. — Always conclude in context. [A1]

Answer: Yes, 280 g of flour is exactly the right amount for 36 biscuits at the ratio 4:2:1.

Scaling problems often hide the key information in the question wording. Read carefully to identify what you are given (one part, the total, or a scaled version) before choosing your method.

Worked example: Mixing Concentrations

A cleaning solution is made by mixing concentrate and water in the ratio 1:4. How much concentrate is needed to make 750 ml of solution? [3 marks]

  1. $1 + 4 = 5$ parts — The ratio 1:4 means every 5 ml of solution contains 1 ml concentrate and 4 ml water. [M1]
  2. $750 \div 5 = 150$ ml per part — Divide the total volume by the number of parts. [M1]
  3. $1 \times 150 = 150$ ml — Concentrate is 1 part, so multiply one-part value by 1. [A1]
  4. Water: $4 \times 150 = 600$ ml $150 + 600 = 750$ ml ✓ — Shares add back to the total volume.

Answer: 150 ml of concentrate is needed.

6. Finding a Missing Quantity When One Part Is Known

Sometimes a question gives you one share and asks you to find the total or another share. This is the reverse of the standard method — instead of dividing the total, you use the known share to find the value of one part first.

Worked example: Finding the Total from One Known Share

Ben and Chloe share prize money in the ratio 3:5. Ben receives £120. Calculate the total prize money. [3 marks]

  1. Ben's share corresponds to 3 parts — Read the ratio carefully — Ben is listed first, so his share is 3 parts. [M1]
  2. $£120 \div 3 = £40$ per part — If 3 parts = £120, then 1 part = £40. [M1]
  3. $3 + 5 = 8$ parts — Add all ratio numbers together.
  4. $8 \times £40 = £320$ — Multiply the total parts by the one-part value. [A1]
  5. Chloe: $5 \times £40 = £200$ $£120 + £200 = £320$ ✓ — Ben's share plus Chloe's share equals the total.

Answer: The total prize money is £320.

8. Exam-Style Practice

The following problems cover all five real-world contexts from the specification. Work through each one using the unitary method, and check your answer against the solution provided.

Every real-world ratio problem — whether it involves currency conversion, recipe scaling, mixing concentrations, or comparing prices — reduces to the same core method: find the total parts, find one part, then scale. Master that sequence and you can handle any context the examiner presents.

Frequently asked questions

How many kilometres is 20 miles?

20 miles is approximately 32 kilometres. Using 5 miles ≈ 8 km, the scale factor is 20 ÷ 5 = 4, so the distance is 8 × 4 = 32 km.

What is the unitary method for ratio?

The unitary method finds the value of one part first, then scales it. To share £240 in the ratio 3:5, divide by 8 parts to get £30 per part, so the shares are 3 × £30 = £90 and 5 × £30 = £150.

How do you share £240 in the ratio 3:5?

£240 shared in the ratio 3:5 gives £90 and £150. There are 3 + 5 = 8 parts, so one part is £240 ÷ 8 = £30, then 3 × £30 = £90 and 5 × £30 = £150, which add back to £240.

How do you convert units using a ratio?

Write the conversion as a ratio and scale both sides by the same factor. If 5 miles is approximately 8 km, then 20 miles is 4 times 5 miles, so it is approximately 8 × 4 = 32 km.

How do you solve mixing ratio problems?

Find the total parts, then the value of one part. A cleaning solution of concentrate to water in the ratio 1:4 has 5 parts, so 750 ml of solution needs 750 ÷ 5 = 150 ml of concentrate and 600 ml of water.

How do you find the total when you know one share of a ratio?

Use the known share to find the value of one part, then multiply by the total number of parts. If Ben and Chloe share prize money in the ratio 3:5 and Ben receives £120, one part is £40, so the total is 8 × £40 = £320.

How do you scale a recipe using a ratio?

Find the scale factor and multiply every ingredient by it, keeping the ratio constant. A recipe for 12 biscuits scaled to 36 biscuits has a scale factor of 36 ÷ 12 = 3, so flour, butter and sugar in the ratio 4:2:1 become 12:6:3.

How much concentrate is in 750 ml of a 1:4 solution?

750 ml of a solution mixed in the ratio 1:4 contains 150 ml of concentrate. There are 1 + 4 = 5 parts, so one part is 750 ÷ 5 = 150 ml, and the water is 4 × 150 = 600 ml.

Where is ratio used in real life?

Ratio is used in conversion, comparison, scaling, mixing and concentrations, such as a chef scaling a recipe, a pharmacist mixing a solution or an engineer converting metres to feet. In every case the relationship between the quantities stays fixed while the numbers change.

What does a double number line show?

A double number line shows two quantities in a fixed ratio side by side, such as miles and kilometres. As one value doubles, the other doubles too, which shows that the ratio stays constant.

How do you compare two options using ratio?

Two options are compared using ratio by expressing both with the same reference quantity, usually one unit, so they can be compared directly. The final answer should be a sentence naming the better option with a reason using the calculated values.

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