Recipes and Scaling — Proportion in Everyday Cooking

Introduction

Recipes and scaling questions are one of the most practical applications of ratio and proportion. They test your ability to adjust quantities fairly when the number of servings, people, or items changes. The idea is simple: if all ingredients in a recipe are in fixed ratio, then doubling (or halving) the number of servings means doubling (or halving) every ingredient.

For example, a recipe might say:

• 200 g flour
• 100 g sugar
• 2 eggs

Scaling recipes is essentially a ratio problem. The ratio of ingredients must remain constant to keep the recipe balanced. That’s why you cannot increase one ingredient without adjusting the others in the same proportion — doing so would change the taste or texture.

In GCSE exams, recipe and scaling questions often appear as word problems, sometimes mixed with unit conversions (grams ↔ kilograms, millilitres ↔ litres). You may be asked to:

• Scale a recipe up or down for a different number of people.
• Compare actual amounts used with the correct proportion and spot an error.
• Use ratios to decide how much of each ingredient is required.
• Work with units like grams, kilograms, millilitres, and litres correctly.

Key Vocabulary

Before solving recipe and scaling problems, it’s important to understand the common terms and concepts. These are frequently used in exam questions and in real-life situations.

• Ratio: A way of comparing quantities. Example: flour : sugar : butter = 2 : 1 : 1 means for every 2 parts flour, there is 1 part sugar and 1 part butter.
• Scaling: Increasing or decreasing all ingredients by the same factor. Example: If a recipe for 4 people uses 200 g flour, then for 6 people you multiply by \(\tfrac{6}{4} = 1.5\), giving 300 g flour.
• Proportion: The “fair share” idea. All ingredients must increase or decrease in proportion to one another so the balance remains the same.
• Factor / Multiplier: The number you multiply each ingredient by when scaling. Example: To scale from 2 people to 10 people, the factor is \(10 \div 2 = 5\).
• Unit conversion: Converting between measurement units while scaling. Common conversions: - 1000 g = 1 kg - 1000 mL = 1 L - 250 mL ≈ 1 cup (in some contexts, but GCSE sticks to metric).
• Servings / Portions: The number of people a recipe is designed to feed. Scaling always compares desired servings to original servings.
• Exact vs. approximate amounts: Some exam questions require rounding (e.g. half an egg might not be practical). Be prepared to interpret answers sensibly.
• Best buy link: Sometimes recipe scaling is combined with price or shopping questions, where you need to work out how much of each ingredient to buy at given pack sizes.

Core Ideas

Recipe and scaling problems revolve around the principle of proportional change. Once you know how many people the original recipe serves, you can scale up or down by multiplying all ingredients by the same factor. Here are the essential concepts:

• Scaling factor: \[ \text{Factor} = \frac{\text{New servings}}{\text{Original servings}} \] Multiply every ingredient by this factor to get the new amounts.
• Keeping ratios constant: If a recipe says flour : sugar : butter = 4 : 2 : 1, doubling flour means doubling sugar and butter too. The “shape” of the ratio never changes.
• Working with units: Always check if units need converting. For example: - If the recipe calls for 250 g butter but you only have packs of 0.5 kg, convert 0.5 kg = 500 g. - This avoids miscalculations when scaling.
• Division before multiplication: To scale down (e.g. from 12 servings to 3), find the factor first (\(3 ÷ 12 = 0.25\)), then multiply. This ensures accuracy.
• Reverse problems: Sometimes exams ask: “If you used 900 g flour, how many people does this serve?” Solve by dividing: \(900 ÷ \text{original flour}\), then multiplying by the original servings.
• Compound context: Recipe questions may be linked with cost. Example: “If 2 cakes need 600 g flour, how much would it cost to buy enough flour at £1.20 per kg?”
• Graphs and proportion tables: A linear graph through the origin or a simple table can represent scaling. For instance, doubling servings doubles quantities.

Step-by-Step Method

Here’s a reliable method for solving any recipe and scaling problem, whether it’s increasing, decreasing, or comparing ingredient amounts.

1. Identify the original servings. Check how many people the recipe is designed for. This is your starting point.
2. Note the desired servings. How many people (or items) are needed in the problem? This is the target.
3. Find the scaling factor. \[ \text{Factor} = \frac{\text{New servings}}{\text{Original servings}} \] Example: From 4 servings to 10 servings → factor = \(10 ÷ 4 = 2.5\).
4. Multiply every ingredient by the factor. Apply the same multiplier to all quantities so the ratios remain unchanged. Example: 200 g flour × 2.5 = 500 g flour.
5. Convert units if needed. - 1000 g = 1 kg - 1000 mL = 1 L - 60 minutes = 1 hour Convert to the units asked for in the exam.
6. Check practicality. If your answer involves fractions of eggs or packets, round sensibly and explain. Some exam questions expect you to say “3 eggs” instead of “2.5 eggs”.
7. Sence-check the result. - If servings increase, ingredient amounts should also increase. - If servings decrease, amounts should decrease. - The ratio between ingredients should be identical.

Worked Examples

Let’s apply the method to some recipe and scaling problems, starting simple and moving to exam-style contexts.

Example 1 (Foundation — Doubling Servings)

A recipe for 4 people needs:

• 200 g flour
• 100 g sugar
• 2 eggs

Scaling factor = \(8 ÷ 4 = 2\). Flour = 200 × 2 = 400 g Sugar = 100 × 2 = 200 g Eggs = 2 × 2 = 4 Answer: 400 g flour, 200 g sugar, 4 eggs.

Example 2 (Foundation — Reducing Servings)

A recipe for 12 cakes uses 300 g butter. How much butter is needed for 5 cakes?

Scaling factor = \(5 ÷ 12 ≈ 0.416\). Butter = 300 × 0.416 ≈ 125 g. Answer: 125 g butter.

Example 3 (Foundation — With Units)

A recipe for 6 drinks uses 1.2 L of juice. How much juice is needed for 15 drinks?

Scaling factor = \(15 ÷ 6 = 2.5\). Juice = 1.2 × 2.5 = 3.0 L. Answer: 3 L juice.

Example 4 (Higher — Finding Servings from Ingredients)

A recipe for 5 people uses 250 g pasta. You have 900 g pasta. How many people can you serve?

Scaling factor = \(900 ÷ 250 = 3.6\). Servings = 5 × 3.6 = 18 people. Answer: 18 people.

Example 5 (Higher — Mixed Ingredients)

A recipe for 10 biscuits requires:

• 300 g flour
• 120 g sugar

Scaling factor = \(25 ÷ 10 = 2.5\). Flour = 300 × 2.5 = 750 g Sugar = 120 × 2.5 = 300 g Answer: 750 g flour, 300 g sugar.

Example 6 (Higher — Combined with Cost)

A recipe for 8 servings uses 600 g potatoes. A supermarket sells potatoes in 2 kg bags for £3.60. What is the cost to buy enough potatoes for 20 servings?

Scaling factor = \(20 ÷ 8 = 2.5\). Potatoes = 600 × 2.5 = 1500 g = 1.5 kg. Cost per kg = £3.60 ÷ 2 = £1.80. Cost = 1.5 × £1.80 = £2.70. Answer: £2.70.

Common Mistakes & Fixes

Recipe and scaling problems can seem simple, but many students lose marks through small errors. Here are the most common mistakes and how to fix them.

• Forgetting to scale all ingredients. Mistake: Only adjusting one ingredient and leaving the others the same. Fix: Write a list of every ingredient and multiply each by the scaling factor.
• Using the wrong scaling factor. Mistake: Dividing the wrong way round, e.g. \(4 ÷ 10\) instead of \(10 ÷ 4\). Fix: Always do: \[ \text{Factor} = \frac{\text{New servings}}{\text{Original servings}}. \]
• Mixing units. Mistake: Treating 1 kg as 100 g, or 1 L as 100 mL. Fix: Convert carefully: 1000 g = 1 kg, 1000 mL = 1 L.
• Not simplifying ratios first. Mistake: Working with big numbers that could be simplified (e.g. 400:200 instead of 2:1). Fix: Reduce ratios to their simplest form before scaling — this reduces calculation errors.
• Rounding incorrectly. Mistake: Writing 2.5 eggs as “2 eggs” without comment. Fix: Round sensibly and explain (“use 3 eggs instead of 2.5”).
• Not checking logic. Mistake: Ending up with fewer ingredients for more servings. Fix: Ask: “If I make more, do I need more ingredients?” If not, something went wrong.
• Overlooking costs. Mistake: Forgetting to scale up the cost in recipe + price problems. Fix: Treat cost like any other ingredient — multiply by the scaling factor.

Practice Questions — Foundation

These problems focus on basic recipe scaling. Use the scaling factor method to solve each one. Remember to multiply every ingredient by the same factor.

1. A recipe for 4 people uses 300 g rice. How much rice is needed for 10 people?
2. A recipe for 12 biscuits uses 240 g flour. How much flour is needed for 18 biscuits?
3. A fruit drink for 5 servings uses 1.5 L of juice. How much juice is needed for 20 servings?
4. A recipe for 8 pancakes requires: - 400 g flour - 2 eggs - 300 mL milk Work out the amounts needed for 2 pancakes.
5. A recipe for 6 people uses 750 g potatoes. You only want to cook for 2 people. How many grams of potatoes do you need?
6. A recipe for 10 cakes uses 200 g butter. If you only have 50 g butter, how many cakes can you make?

Practice Questions — Higher

These questions involve more complex recipe scaling, including unit conversions, finding servings, and linking to costs.

1. A recipe for 5 people uses 250 g flour. (a) Find the amount needed for 14 people. (b) If you have 1.5 kg of flour, how many people can you serve?
2. A smoothie recipe for 3 servings uses 900 mL of milk. (a) How much milk is needed for 8 servings? (b) If you only have 2.25 L of milk, how many servings can you make?
3. A cake recipe for 12 cakes uses: - 480 g flour - 240 g sugar - 3 eggs Work out the quantities needed to make 30 cakes.
4. A recipe for 4 people needs 600 g chicken. Chicken costs £5.20 per kg. (a) How much chicken is needed for 10 people? (b) How much will it cost?
5. A recipe for 6 loaves of bread uses 2.4 kg flour. (a) Find the amount of flour needed for 15 loaves. (b) If you only have 4.8 kg flour, how many loaves can you make?
6. A soup recipe for 8 people uses 2 L of stock. (a) Write the formula linking stock \(S\) (litres) to people \(P\). (b) Use the formula to find how much stock is needed for 25 people.
7. A recipe for 10 biscuits requires 350 g flour. Packs of flour are sold in 1.4 kg bags. How many bags are needed to make 200 biscuits?

Challenge Questions

These tougher problems combine scaling with cost, unit conversions, and multi-step reasoning. They are similar to high-mark GCSE exam questions.

1. A recipe for 12 people uses 1.8 kg of beef. (a) How much beef is needed for 35 people? (b) Beef costs £9.60 per kg. What is the total cost?
2. A recipe for 8 servings uses: - 320 g pasta - 200 g vegetables - 400 mL sauce (a) Write the ratio pasta : vegetables : sauce. (b) Find the quantities needed for 30 servings.
3. A recipe for 10 people requires 2.5 L of stock. (a) Write a formula linking stock \(S\) (litres) to servings \(n\). (b) Use the formula to calculate stock for 24 servings. (c) If stock is sold in 1.5 L cartons costing £2.70 each, what is the total cost?
4. A cake recipe for 20 cakes uses 500 g sugar. (a) How much sugar is needed for 75 cakes? (b) If sugar is sold in 2 kg bags, how many bags are needed? (c) What is the cost if each bag costs £3.20?
5. A recipe for 4 pizzas uses 1.2 kg flour. (a) How much flour is needed for 15 pizzas? (b) If you only have 3.6 kg flour, how many pizzas can you make? (c) Write the direct proportion equation linking flour \(F\) (kg) and pizzas \(P\).
6. A recipe for 5 smoothies uses 1.5 L of milk and 0.5 kg fruit. (a) Find the quantities needed for 18 smoothies. (b) Milk costs £0.80 per litre and fruit costs £3.20 per kg. What is the total cost? (c) If you only have £10, what is the maximum number of smoothies you can make?
7. A recipe for 6 portions of soup uses 750 mL stock and 450 g vegetables. (a) How much of each ingredient is needed for 25 portions? (b) If stock is sold in 500 mL cartons costing £1.10 each and vegetables in 1.5 kg bags costing £4.50, what is the total cost? (c) How many people can be served with £15 budget?

Quick Revision Sheet

Here’s a one-page summary for recipe and scaling problems — ideal for quick revision before exams.

Key Formulas

• Scaling factor: \[ \text{Factor} = \frac{\text{New servings}}{\text{Original servings}} \]
• Scaled ingredient: \[ \text{New amount} = \text{Original amount} \times \text{Factor} \]
• Reverse calculation: \[ \text{Servings} = \frac{\text{Available amount}}{\text{Original amount}} \times \text{Original servings} \]

How to Spot Scaling Problems

• Recipe mentions a fixed number of servings/people/items.
• Question asks for more or fewer servings.
• Ratios between ingredients must stay the same.

5-Step Method

1. Note original servings.
2. Note desired servings.
3. Calculate scaling factor (new ÷ original).
4. Multiply each ingredient by the factor.
5. Convert units (g ↔ kg, mL ↔ L) if needed.

Unit Conversions

• 1000 g = 1 kg
• 1000 mL = 1 L
• 60 min = 1 h

Common Traps

• Scaling one ingredient but forgetting the others.
• Mixing up scaling factor (divide wrong way).
• Confusing grams and kilograms (1000 g = 1 kg).
• Not rounding sensibly (e.g. half an egg).

Quick Practice

1. A recipe for 6 people needs 900 g potatoes. How much for 4 people? Answer: \(900 \times \tfrac{4}{6} = 600\) g.
2. A recipe for 8 people uses 2.4 L stock. How much for 20 people? Answer: \(2.4 \times \tfrac{20}{8} = 6.0\) L.

Answers

Foundation

1. Factor = \(10 \div 4 = 2.5\). Rice = \(300 \times 2.5 = \mathbf{750 \text{ g}}\).
2. Factor = \(18 \div 12 = 1.5\). Flour = \(240 \times 1.5 = \mathbf{360 \text{ g}}\).
3. Factor = \(20 \div 5 = 4\). Juice = \(1.5 \times 4 = \mathbf{6 \text{ L}}\).
4. Factor = \(2 \div 8 = 0.25\). Flour = \(400 \times 0.25 = 100\) g Eggs = \(2 \times 0.25 = 0.5\) → practically 1 egg Milk = \(300 \times 0.25 = 75\) mL Answer: 100 g flour, 1 egg, 75 mL milk.
5. Factor = \(2 \div 6 = 0.333...\). Potatoes = \(750 \times 0.333... = \mathbf{250 \text{ g}}\).
6. Factor = \(50 \div 200 = 0.25\). Servings = \(10 \times 0.25 = \mathbf{2.5 \text{ cakes}}\) → practically 2 cakes.

Higher

1. (a) Factor = \(14 \div 5 = 2.8\). Flour = \(250 \times 2.8 = 700\) g. (b) Factor = \(1500 \div 250 = 6\). Servings = \(5 \times 6 = \mathbf{30 \text{ people}}\).
2. (a) Factor = \(8 \div 3 = 2.666...\). Milk = \(900 \times 2.666... = 2400\) mL = 2.4 L. (b) Factor = \(2250 \div 900 = 2.5\). Servings = \(3 \times 2.5 = \mathbf{7.5 \approx 7 \text{ servings}}\).
3. Factor = \(30 \div 12 = 2.5\). Flour = \(480 \times 2.5 = 1200\) g Sugar = \(240 \times 2.5 = 600\) g Eggs = \(3 \times 2.5 = 7.5\) → practically 8 eggs.
4. (a) Factor = \(10 \div 4 = 2.5\). Chicken = \(600 \times 2.5 = 1500\) g = 1.5 kg. (b) Cost = \(1.5 \times 5.20 = \mathbf{£7.80}\).
5. (a) Factor = \(15 \div 6 = 2.5\). Flour = \(2.4 \times 2.5 = 6.0\) kg. (b) Factor = \(4.8 \div 2.4 = 2\). Servings = \(6 \times 2 = \mathbf{12 \text{ loaves}}\).
6. (a) \(S = \tfrac{2}{8} \times P = 0.25P\). (b) For \(P=25\): \(S = 0.25 \times 25 = \mathbf{6.25 \text{ L}}\).
7. Factor = \(200 \div 10 = 20\). Flour = \(350 \times 20 = 7000\) g = 7.0 kg. Bags = \(7.0 \div 1.4 = \mathbf{5 \text{ bags}}\).

Challenge

1. (a) Factor = \(35 \div 12 = 2.916...\). Beef = \(1.8 \times 2.916... = 5.25\) kg. (b) Cost = \(5.25 \times 9.60 = \mathbf{£50.40}\).
2. (a) Ratio = 320 : 200 : 400 = 16 : 10 : 20 = 8 : 5 : 10. (b) Factor = \(30 \div 8 = 3.75\). Pasta = 320 × 3.75 = 1200 g Veg = 200 × 3.75 = 750 g Sauce = 400 × 3.75 = 1500 mL = 1.5 L
3. (a) Formula: \(S = \tfrac{2.5}{10}n = 0.25n\). (b) For 24 servings: \(S = 0.25 \times 24 = 6\) L. (c) Cartons = \(6 \div 1.5 = 4\). Cost = 4 × 2.70 = \(\mathbf{£10.80}\).
4. (a) Factor = \(75 \div 20 = 3.75\). Sugar = 500 × 3.75 = 1875 g = 1.875 kg. (b) Bags = 1.875 ÷ 2 = 0.9375 → 1 bag. (c) Cost = £3.20.
5. (a) Factor = \(15 \div 4 = 3.75\). Flour = 1.2 × 3.75 = 4.5 kg. (b) Factor = \(3.6 \div 1.2 = 3\). Pizzas = 4 × 3 = 12 pizzas. (c) Equation: \(F = 0.3P\).
6. (a) Factor = \(18 \div 5 = 3.6\). Milk = 1.5 × 3.6 = 5.4 L Fruit = 0.5 × 3.6 = 1.8 kg (b) Cost = (5.4 × 0.80) + (1.8 × 3.20) = £4.32 + £5.76 = \(\mathbf{£10.08}\). (c) Budget £10 → slightly less than 18 smoothies. Maximum = 17 smoothies (by cost limit).
7. (a) Factor = \(25 \div 6 = 4.166...\). Stock = 750 × 4.166... = 3125 mL = 3.125 L Vegetables = 450 × 4.166... = 1875 g = 1.875 kg (b) Stock: need 7 cartons (3.5 L). Cost = £7.70 Vegetables: 2 bags (3 kg). Cost = £9.00 Total cost = £16.70 (c) Budget £15 → about 22 servings possible.

Conclusion & Next Steps

Recipe and scaling problems are pure ratio in action: once you know how many servings the original recipe makes, everything scales by the same factor. The golden rule is \[ \text{New amount} \;=\; \text{Original amount} \times \frac{\text{New servings}}{\text{Original servings}}. \] If you keep this factor consistent across all ingredients (and convert units cleanly), the rest is routine arithmetic. Graphs of quantity against servings go through the origin, reinforcing that “double the servings → double every ingredient”.

• What you’ve mastered: spotting scaling problems, finding a clean factor, applying it to every ingredient, handling unit conversions, and linking to cost/pack-size questions.
• Exam habits: write the factor first; line up ingredients in a small table; convert grams↔kilograms and millilitres↔litres before comparing; round practically (e.g. whole eggs/packets) and state the rounding.
• Sanity checks: more servings must give more of each ingredient; ratios between ingredients remain identical; results should be in realistic units and quantities.

Where this connects next:

1. Direct Proportion: Interpreting “per 1” rates (price per kg, speed per hour) and using \(y=kx\) alongside recipe scaling.
2. Best-Buy & Cost Optimisation: After scaling, compare pack sizes and prices fairly (always to a common unit like £/kg or £/L).
3. Compound/Two-Step Scaling: E.g. scale to servings, then apply a waste percentage, or convert units after scaling.
4. Inverse & Mixed Proportion: Contrast “more means less” scenarios (workers–time) so you pick the right model quickly in mixed-topic questions.
5. Graphs of Proportionality: Reinforce that proportional relationships plot as straight lines through \((0,0)\); use gradients to read “per 1” values.