Ratio and Proportion Revision Notes for Grade 12 NSC Matric Mathematics

Ratio and Proportion

Understanding ratios

A ratio is a way of comparing two quantities that have the same units. When you see a ratio, you're looking at how one measurement relates to another measurement of the same type. Ratios are incredibly useful in mathematics because they help us understand relationships and make comparisons.

The key thing to remember about ratios is that they show relative size, not actual size. Think of a ratio as telling you "for every this much of one thing, there's that much of another thing."

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Critical Properties of Ratios:

When working with ratios, keep these important properties in mind:

Ratios have no units - once you create a ratio, the units cancel out
Ratios should be written in simplest form - always reduce to the lowest terms
Ratios compare quantities of the same kind - you can't compare apples to metres!

Writing ratios

You can express the same ratio in several different ways. For example, if a rectangle has a length of 20 cm and width of 60 cm, you can write the ratio of length to width as:

length to width = 20 to 60
length : width = 20 : 60
length/width = 20/60 = 1/3

All of these expressions tell us the same thing: for every 1 unit of length, there are 3 units of width.

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Worked Example: Simplifying Ratios

Question: A rectangle has length 50 cm and width 150 cm. Express the ratio of length to width in simplest form.

Solution: Step 1: Write the ratio as a fraction

$\frac{\text{length}}{\text{width}} = \frac{50}{150}$

Step 2: Find the highest common factor (HCF) of 50 and 150

HCF of 50 and 150 = 50

Step 3: Divide both numbers by the HCF

$50 ÷ 50 = 1$

$150 ÷ 50 = 3$

Step 4: Write the final answer

length : width = 1 : 3

This means for every 1 unit of length, there are 3 units of width.

Understanding proportions

Proportion describes the equality of ratios. When two or more ratios are equal to each other, we say they are in the same proportion. This concept is fundamental to solving many geometry problems.

The diagram above shows parallel lines cut by transversals, creating proportional segments. This is a common setup in geometry problems.

When ratios are equal

If $\frac{w}{x} = \frac{y}{z}$ , then we say that w and x are in the same proportion as y and z.

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Four Equivalent Forms of Proportion:

This relationship can be expressed in four equivalent ways:

$wz = xy$ (cross multiplication)
$\frac{x}{w} = \frac{z}{y}$ (reciprocal proportion)
$\frac{w}{y} = \frac{x}{z}$ (alternative proportion)
$\frac{y}{w} = \frac{z}{x}$ (another reciprocal form)

Different forms of proportional relationships

Understanding the various ways to express proportions is crucial for solving problems efficiently.

There are three equivalent forms:

Basic proportion: $\frac{AB}{BC} = \frac{DE}{EF}$
Reciprocal proportion: $\frac{BC}{AB} = \frac{EF}{DE}$
Cross multiplication: $AB × EF = BC × DE$

Cross multiplication method

Cross multiplication is one of the most powerful tools for solving proportion problems. When you have a proportion $\frac{a}{b} = \frac{c}{d}$ , you can cross multiply to get $ad = bc$ .

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Worked Example: Solving Proportions

Question: Solve for p: $\frac{8}{40} = \frac{p}{25}$

Solution: Step 1: Cross multiply

$8 × 25 = 40 × p$

$200 = 40p$

Step 2: Solve for p

$p = 200 ÷ 40 = 5$

Step 3: Check your answer

$\frac{8}{40} = \frac{1}{5}$ and $\frac{5}{25} = \frac{1}{5}$ ✓

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Worked Example: Geometric Proportions

Question: In the diagram below, if AB = 12, BC = 18, DE = 54, and EF = 36, show that $\frac{AB}{BC} = \frac{FE}{ED}$ .

Solution: Step 1: Calculate $\frac{AB}{BC}$

$\frac{AB}{BC} = \frac{12}{18} = \frac{2}{3}$

Step 2: Calculate $\frac{FE}{ED}$ \

$\frac{FE}{ED} = \frac{36}{54} = \frac{2}{3}$

Step 3: Compare the ratios

Since $\frac{2}{3} = \frac{2}{3}$ , we have shown that $\frac{AB}{BC} = \frac{FE}{ED}$

This demonstrates that the segments are proportional.

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Worked Example: Real-World Proportions

Question: A mixture contains 2 parts of substance A for every 5 parts of substance B. If the total weight is 50 kg, how much of substance B is in the mixture?

Solution: Step 1: Understand the ratio

Substance A : Substance B = 2 : 5

Total parts = 2 + 5 = 7 parts

Step 2: Find the weight of one part

Weight of one part = $50 \text{ kg} ÷ 7 = \frac{50}{7} \text{ kg}$

Step 3: Calculate substance B

Substance B = $5 \text{ parts} × \frac{50}{7} \text{ kg} = \frac{250}{7} \text{ kg} ≈ 35.7 \text{ kg}$

Practical applications

Proportions appear everywhere in real life and geometry. They help us:

Scale drawings and maps
Calculate missing measurements in similar figures
Solve mixture problems
Work with rates and ratios
Understand growth patterns

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Exam Tips:

Always check if your ratios are in simplest form
When cross multiplying, be careful with your arithmetic
Look for parallel lines in geometry diagrams - they often create proportional segments
Remember that ratios have no units, but the original quantities do
Practice identifying which quantities should be compared in word problems

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Key Points to Remember:

Ratios compare quantities of the same type and have no units once formed
Proportions are equal ratios - when two ratios are equivalent, they're proportional
Cross multiplication is your main tool for solving proportion equations: if $\frac{a}{b} = \frac{c}{d}$ , then $ad = bc$
Always simplify ratios to their lowest terms for the clearest representation
Check your work by substituting back into the original proportion to verify your solution

Ratio and Proportion (Grade 12 NSC Matric Mathematics): Revision Notes