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R8: Ratio & Linear Functions

Foundation Higher AQAEdexcelOCREduqasCCEA

Relate ratios to fractions and linear functions; represent ratios graphically

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📋 Key Concepts

Definition: A ratio relationship can be represented as a linear function y = kx, where k is the constant of proportionality (gradient).

Key Relationships

y = kx (direct proportion)
k = y/x (gradient of the line)

📝 Graphs of Direct Proportion

Features: The graph is a straight line that passes through the origin (0,0). The gradient is the constant of proportionality.
Example 1

y is directly proportional to x. When x = 4, y = 12.

a) Find the equation connecting y and x.

b) Sketch the graph.

Solution:

a) y = kx

12 = k × 4, so k = 3

Equation: y = 3x

b) The graph is a straight line through (0,0) with gradient 3.

Points: (1,3), (2,6), (4,12)

📝 Finding the Gradient

Method: The gradient (k) can be found from any point on the line using k = y/x.
Example 2

A graph shows the relationship between cost (£) and number of items. The line passes through (5, 15). Find the cost per item.

Solution:

Gradient = y/x = 15/5 = 3

Cost per item = £3

Equation: Cost = 3 × Number of items

📝 Ratio and Linear Functions

Connection: A ratio a:b corresponds to a gradient of b/a when plotting y against x.
Example 3

The ratio of blue paint to red paint is 2:5. Express this as a linear relationship.

Solution:

If we plot Blue (y) against Red (x):

Blue/Red = 2/5

Blue = (2/5) × Red

y = 0.4x (gradient = 0.4)

For every 5 units of red, you need 2 units of blue.

📝 Interpreting Graphs

Example 4

A graph shows distance (km) against time (hours). The line passes through (2, 80). What is the speed?

Solution:

Speed = gradient = distance/time = 80/2 = 40

Speed = 40 km/h

Equation: Distance = 40 × Time

Example 5

A conversion graph shows pounds (£) against dollars ($). £50 = $65. Write the equation.

Solution:

Gradient = $65/£50 = 1.3

Equation: $ = 1.3 × £

📝 From Graph to Ratio

Method: Read coordinates from the graph and express as a ratio in simplest form.
Example 6

A graph of y against x passes through (6, 8). Express y:x as a ratio.

Solution:

Ratio y:x = 8:6

Simplify: 8:6 = 4:3 (divide by 2)

Answer: y:x = 4:3

❓ Practice Questions

Q1: y is proportional to x. When x = 3, y = 18. Find the equation.

Q2: A graph passes through (4, 10). What is the gradient?

Q3: The ratio of flour to butter is 3:2. Write this as a linear function (flour in terms of butter).

Q4: A distance-time graph passes through (3, 120). Find the speed in km/h.

Q5: The ratio x:y is 5:7. What is the gradient of the line y plotted against x?

✅ Answers

  1. y = 6x
  2. 2.5 (or 5/2)
  3. Flour = (3/2) × Butter or F = 1.5B
  4. 40 km/h
  5. 7/5 = 1.4

🎯 Exam Tips

🧠 Problem-Solving Strategies

Problem-Solving

A ratio expressed as a linear function takes the form y = kx where k is the constant of proportionality. The ratio y:x corresponds to the gradient of the line. Use a table to find the multiplier between corresponding values. Check your answer by substituting back into the relationship.
Multi-Step Problem

The ratio of y to x is 3:5. Express y as a function of x. When x = 35, find y. When y = 27, find x.

Solution: y/x = 3/5, so y = 3x/5. When x = 35: y = 3(35)/5 = 21. When y = 27: 27 = 3x/5, so x = 27 × 5/3 = 45.

⚠️ Common Errors

Watch Out!

1. Wrong: If y:x = 2:3, writing y = 2x instead of y = ⅔x Correct: y/x = 2/3, so y = (2/3)x. The ratio 2:3 means y is ⅔ of x, not 2 times x.

2. Wrong: Confusing the ratio y:x with x:y when writing the function Correct: y:x = 3:5 means y = 3k and x = 5k for some k, giving y = (3/5)x. Read the ratio order carefully.

3. Wrong: Assuming the linear function must pass through the origin Correct: Only direct proportion relationships pass through the origin. If the relationship has an offset (e.g. y = 2x + 3), it is linear but NOT directly proportional.

✍️ 6-Mark Exam Question

Extended Answer

6 marks: The cost (£C) of hiring a van is directly proportional to the number of days (d). Hiring for 5 days costs £225. Write C as a function of d. A customer has a budget of £315. How many full days can they hire the van for? The company adds a £40 insurance fee. Write the new cost function and find how many full days the customer can now afford.

Step 1: C = kd. 225 = k × 5, so k = 45. Therefore C = 45d.

Step 2: For £315: 315 = 45d, so d = 7 days.

Step 3: New cost: C = 45d + 40.

Step 4: 315 = 45d + 40, so 45d = 275, d = 6.11... So 6 full days.

Mark scheme: M1 for setting up proportion, A1 for k = 45, A1 for C = 45d, M1 for solving, A1 for 7 days, M1 for new function, A1 for 6 full days

📊 AO3: Reason & Interpret

Reasoning and Interpretation

Two taxi companies charge differently. Company A: C = 1.80d + 3 (d = distance in miles). Company B: C = 2.40d.

(a) For what distance do both companies charge the same amount?

(b) Which company is cheaper for a 5-mile journey?

(c) A customer says "Company B is always better value because there's no booking fee." Is this correct? Explain.

Answers: (a) 1.80d + 3 = 2.40d → 3 = 0.60d → d = 5 miles (b) At 5 miles both charge £12 — they're equal (c) No — for journeys under 5 miles, Company A is cheaper (e.g. 2 miles: A = £6.60, B = £4.80... wait: A = 1.80×2+3 = £6.60, B = 2.40×2 = £4.80, so B is actually cheaper for short trips). For journeys over 5 miles, A is cheaper. The customer is wrong.

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