Mathematics (0580) | Ratio and proportion

Hello IGCSE Maths Students!

Welcome to the comprehensive study notes for Ratio and Proportion. This topic is fundamental to mathematics and appears everywhere, from cooking to financial planning. If you can master these skills, you'll be well-prepared not just for your exams, but also for many real-life situations!

We will cover how to simplify ratios, share money fairly, and understand how quantities relate to each other (proportion). Don't worry if this seems tricky at first—we'll break it down into simple, manageable steps.

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Section 1: Understanding Ratios (Core & Extended)

1.1 What is a Ratio?

A ratio compares two or more quantities of the same type. It shows how much of one quantity there is compared to another.

Example: If we mix 3 litres of yellow paint and 2 litres of blue paint, the ratio of yellow to blue is 3 : 2.

Key Rule: Ratios must always be given using the same units!

• If you compare 50 cm to 2 m, you must first convert 2 m to 200 cm. The comparison is 50 cm : 200 cm.

1.2 Simplifying Ratios

Just like fractions, ratios should be given in their simplest form (C1.10).

To simplify a ratio, divide all parts of the ratio by the Highest Common Factor (HCF).

Step-by-Step: Simplifying Ratios

1. Identify all the numbers in the ratio.
2. Find the largest number that divides exactly into all parts. (The HCF).
3. Divide each part by the HCF.

Example 1: Simplify the ratio 20 : 30 : 40. (This is a syllabus example!)

The HCF of 20, 30, and 40 is 10.
Divide all parts by 10: \(20 \div 10 : 30 \div 10 : 40 \div 10\)
Simplified ratio: 2 : 3 : 4.

Example 2: Simplify the ratio 4 kg : 500 g.

First, convert to the same unit (grams): 4 kg = 4000 g.
Ratio: 4000 : 500
The HCF is 500.
Divide by 500: \(4000 \div 500 : 500 \div 500\)
Simplified ratio: 8 : 1.

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Section 2: Dividing a Quantity in a Given Ratio (Core & Extended)

2.1 The "Total Parts" Method (C1.10)

One of the most common exam questions is to divide a total amount (like money or ingredients) according to a specific ratio. This requires calculating the value of one 'part'.

Step-by-Step: Dividing a Quantity

We want to divide $150 between Alex and Ben in the ratio 2 : 3.

1. Find the Total Number of Parts:
   Add the numbers in the ratio: \(2 + 3 = 5\) parts.
2. Find the Value of One Part:
   Divide the total quantity by the total parts: \[\text{Value of one part} = \frac{\text{Total Quantity}}{\text{Total Parts}} = \frac{\$150}{5} = \$30\]
3. Calculate Each Share:
   Multiply the value of one part by the number of parts each person gets:
   • Alex (2 parts): \(2 \times \$30 = \$60\)
   • Ben (3 parts): \(3 \times \$30 = \$90\)
4. Check Your Answer:
   Add the shares back up: \(\$60 + \$90 = \$150\). (It works!)

Common Mistake Alert!
Do not confuse the parts of the ratio with the actual quantities. The numbers in the ratio (2 and 3) are merely instructions for sharing.

Key Takeaway for Sharing

To divide a quantity, the most reliable method is to find the Total Parts, then find the value of One Part.

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Section 3: Proportional Reasoning and Context (Core & Extended)

3.1 Using Ratios in Context (C1.10)

Proportional reasoning involves scaling recipes, using maps, and determining the best value.

A. Map Scales

Map scales are often given as ratios, e.g., 1 : 50 000. This means:

1 unit on the map = 50 000 identical units in real life.

Example: A map has a scale of 1 : 20 000. If a path is 5 cm long on the map, how long is it in real life (in metres)?

Map distance: 5 cm
Real distance: \(5 \times 20\ 000 = 100\ 000\) cm
Convert to metres (divide by 100): \(100\ 000 \div 100 = 1000\) m.
The path is 1000 metres long.

B. Adapting Recipes

When you change the number of servings in a recipe, you must keep all ingredients in the same proportion.

Did you know? Proportional scaling is vital in chemistry, engineering, and architecture!

Example: A cake recipe for 8 people needs 200g of flour. How much flour is needed for 12 people?

1. Find the required scaling factor: \[\text{Factor} = \frac{\text{New People}}{\text{Old People}} = \frac{12}{8} = 1.5\]
2. Apply the factor to the ingredient: \[\text{Flour needed} = 200 \text{ g} \times 1.5 = 300 \text{ g}\]

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Section 4: Direct Proportion (Extended Content E2.8)

This section involves using algebra to describe and solve problems involving proportionality.

4.1 What is Direct Proportion?

Two quantities are in direct proportion if they increase or decrease at the same rate. When one quantity doubles, the other quantity also doubles.

Analogy: The more hours you work, the more money you earn.

We use the symbol \(\propto\) (pronounced "is proportional to").

If \(y\) is directly proportional to \(x\), we write:
\[y \propto x\]

To turn this proportion into a usable equation, we introduce the Constant of Proportionality, k:

\[y = kx\]

The goal in these problems is always to find the value of k first!

4.2 Direct Proportion with Powers and Roots (Extended E2.8)

The direct proportionality doesn't have to be linear (\(y = kx\)). The syllabus requires you to handle other variations, such as:

• Directly proportional to the square: \(y = kx^2\)
• Directly proportional to the cube: \(y = kx^3\)
• Directly proportional to the square root: \(y = k\sqrt{x}\)

Step-by-Step: Solving Extended Proportion Problems

Example: \(y\) is directly proportional to the square of \(x\). When \(x=3\), \(y=45\). Find \(y\) when \(x=5\).

1. Write the relationship as an equation:
   \(y \propto x^2\), so \(y = kx^2\).
2. Find the Constant \(k\):
   Substitute the known pair (\(x=3, y=45\)): \[45 = k(3^2)\] \[45 = 9k\] \[k = \frac{45}{9} = 5\]
3. Write the full formula:
   Now we know \(k=5\), the equation is: \(y = 5x^2\).
4. Solve for the unknown:
   Find \(y\) when \(x=5\): \[y = 5(5^2) = 5(25) = 125\]

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Section 5: Inverse Proportion (Extended Content E2.8)

5.1 What is Inverse Proportion?

Two quantities are in inverse proportion if, as one quantity increases, the other quantity decreases, such that their product remains constant.

Analogy: If more people share a task (increase), the time taken to finish it (decrease) goes down.

If \(y\) is inversely proportional to \(x\), we write:
\[y \propto \frac{1}{x}\]

To turn this into an equation, we again introduce the constant \(k\):

\[y = \frac{k}{x} \quad \text{or equivalently} \quad xy = k\]

5.2 Solving Inverse Proportion Problems

Example: The time \(T\) taken to build a wall is inversely proportional to the number of workers \(W\). If 4 workers take 12 hours, how long will 6 workers take?

1. Write the relationship as an equation:
   \(T \propto \frac{1}{W}\), so \(T = \frac{k}{W}\) (or \(TW = k\)).
2. Find the Constant \(k\):
   Substitute the known pair (\(W=4, T=12\)): \[12 = \frac{k}{4}\] \[k = 12 \times 4 = 48\]
3. Write the full formula:
   The equation is: \(T = \frac{48}{W}\). (The constant \(k=48\) represents the total "worker-hours" needed.)
4. Solve for the unknown:
   Find \(T\) when \(W=6\): \[T = \frac{48}{6} = 8 \text{ hours}\]

Extension: Inverse Square Proportion
You might also see inverse proportion to the square. If \(y\) is inversely proportional to the square of \(x\), the relationship is: \[y = \frac{k}{x^2}\]