1. Review of Ratios and Rates

Review of Ratios and Rates

Introduction

Ratios and rates are important mathematical tools that help us compare quantities, describe relationships, and solve everyday problems.

We use ratios and rates when:

• Comparing ingredients in recipes.

• Calculating the speed of a vehicle.

• Comparing prices while shopping.

• Mixing paint or other materials.

• Reading maps and scale drawings.

• Comparing the number of students in different groups.

• Calculating wages and production rates.

Although ratios and rates are closely related, they are not exactly the same.

A ratio compares two quantities using division, while a rate compares quantities measured in different units.

Understanding ratios and rates helps us make comparisons, recognise patterns, and solve practical problems.

What Is a Ratio?

A ratio is a comparison of two quantities using division.

For example, suppose a classroom contains:

• 12 boys

• 18 girls

The ratio of boys to girls is:

12 : 18

This means that for every 12 boys, there are 18 girls.

Ratios describe relative quantities rather than simply stating the total number.

Important Observation

The order of quantities in a ratio matters.

The ratio of boys to girls is different from the ratio of girls to boys.

Boys to girls:

12 : 18

Girls to boys:

18 : 12

These ratios compare the same groups but in opposite orders.

Writing Ratios in Different Forms

Ratios can be written in three common forms.

Suppose there are 3 red marbles and 5 blue marbles.

The ratio of red marbles to blue marbles can be written as:

All three represent the same comparison.

Important Observation

A ratio written as a fraction does not always represent a fraction of the whole.

For example, the ratio 3/5 compares red marbles to blue marbles.

The fraction of all marbles that are red is:

3/(3 + 5) = 3/8

These are different quantities.

Understanding Part-to-Part and Part-to-Whole Ratios

Ratios can compare parts of a group or compare one part with the entire group.

Example

A basket contains:

• 4 apples

• 6 oranges

Total fruit:

4 + 6 = 10

Part-to-Part Ratio

Apples to oranges:

4 : 6

Simplified:

2 : 3

Part-to-Whole Ratio

Apples to total fruit:

4 : 10

Simplified:

2 : 5

Comparison

Important Observation

Always identify which quantities the question asks you to compare.

Simplifying Ratios

Ratios can be simplified in the same way as fractions.

To simplify a ratio, divide both quantities by their greatest common factor.

Worked Example

Simplify:

18 : 24

The greatest common factor of 18 and 24 is 6.

Divide both terms by 6:

18 ÷ 6 = 3

24 ÷ 6 = 4

Therefore:

18 : 24 = 3 : 4

Another Example

Simplify:

45 : 60

Divide both terms by 15:

45 ÷ 15 = 3

60 ÷ 15 = 4

45 : 60 = 3 : 4

Important Observation

Dividing both terms of a ratio by the same nonzero number preserves the comparison.

Equivalent Ratios

Equivalent ratios represent the same relationship.

For example:

2 : 3 = 4 : 6 = 6 : 9 = 10 : 15

Each ratio compares quantities in the same proportion.

Creating Equivalent Ratios

Multiply or divide both terms by the same nonzero number.

Worked Example

Find an equivalent ratio to 3 : 5 where the first quantity is 18.

Determine the multiplier:

18 ÷ 3 = 6

Multiply the second quantity by 6:

5 × 6 = 30

Therefore:

3 : 5 = 18 : 30

Important Observation

Equivalent ratios are especially useful when scaling recipes, maps, models, and mixtures.

Comparing Quantities Using Ratios

Ratios allow us to compare groups of different sizes.

Example

Class A has 8 students who cycle to school and 12 who do not.

Class B has 15 students who cycle to school and 20 who do not.

Class A

Cyclists to non-cyclists:

8 : 12

Simplify:

2 : 3

Class B

Cyclists to non-cyclists:

15 : 20

Simplify:

3 : 4

To compare the proportion who cycle, use part-to-whole fractions.

Class A:

8/20 = 0.40 = 40%

Class B:

15/35 ≈ 0.429 = 42.9%

Interpretation

Class B has a slightly greater proportion of students who cycle to school.

Important Observation

When comparing groups of different sizes, percentages or equivalent part-to-whole ratios can make comparisons easier.

Ratios Involving Different Units

Before simplifying a ratio of quantities representing the same type of measurement, convert them to the same units.

Worked Example

Simplify:

2 metres : 50 centimetres

Convert 2 metres to centimetres:

2 m = 200 cm

The ratio becomes:

200 : 50

Divide both terms by 50:

4 : 1

Another Example

Simplify:

1.5 kg : 500 g

Convert kilograms to grams:

1.5 kg = 1500 g

1500 : 500

Divide both terms by 500:

3 : 1

Important Observation

Ignoring unit conversions can produce incorrect ratios.

What Is a Rate?

A rate compares two quantities measured in different units.

Examples include:

• Kilometres per hour

• Dollars per kilogram

• Litres per minute

• Words per minute

• Students per teacher

• Dollars per hour

Example

A car travels 180 kilometres in 3 hours.

The rate is:

180 km / 3 h

Simplify:

60 km/h

This means the car travels an average of 60 kilometres per hour.

Important Observation

Rates describe how one quantity changes or is measured relative to another.

What Is a Unit Rate?

A unit rate is a rate expressed per one unit of the second quantity.

For example:

240 kilometres in 4 hours

Unit rate:

240 ÷ 4 = 60

60 km/h

The rate is expressed per one hour.

Common Unit Rates

Important Observation

Unit rates make it easier to compare situations involving different quantities.

Calculating Unit Rates

Worked Example: Speed

A cyclist travels 72 kilometres in 4 hours.

Calculate the average speed.

Speed = distance ÷ time

Speed = 72 ÷ 4

Speed = 18 km/h

Worked Example: Wages

A worker earns $96 for 8 hours of work.

Hourly rate:

96 ÷ 8 = 12

Hourly wage = $12/h

Worked Example: Shopping

A 3 kg bag of rice costs $12.

Unit price:

12 ÷ 3 = 4

Unit price = $4/kg

Important Observation

To calculate a unit rate, divide the first quantity by the number of units in the second quantity.

Comparing Unit Prices

Unit rates are useful when comparing products with different package sizes.

Example

A supermarket sells two bags of rice.

• Bag A: 2 kg for $7.20

• Bag B: 5 kg for $16.50

Which is better value?

Bag A

Unit price:

7.20 ÷ 2 = 3.60

$3.60/kg

Bag B

Unit price:

16.50 ÷ 5 = 3.30

$3.30/kg

Comparison

Bag B costs $0.30 less per kilogram.

Therefore, Bag B offers the lower unit price.

Important Observation

The cheapest total price is not always the best value per unit.

However, package size, quality, and how much will actually be used may also affect a purchasing decision.

Ratios in Recipes

Recipes often use ratios to describe ingredient proportions.

Example

A recipe uses flour and milk in the ratio:

3 : 2

This means:

For every 3 parts flour, use 2 parts milk.

Scaling the Recipe

Suppose the original recipe uses:

• 300 g flour

• 200 mL milk

To double the recipe:

• Flour = 600 g

• Milk = 400 mL

The numerical ratio remains:

3 : 2

Important Observation

When scaling a recipe, multiply all ingredient quantities by the same factor.

The 3 : 2 comparison here is between the numerical measurements in the specified units; grams and millilitres are different kinds of units.

Sharing Quantities in a Given Ratio

Ratios can be used to divide a total quantity into proportional parts.

Worked Example

Three students share $120 in the ratio:

2 : 3 : 5

How much does each student receive?

Step 1: Find Total Parts

2 + 3 + 5 = 10 parts

Step 2: Find the Value of One Part

120 ÷ 10 = 12

One part represents $12.

Step 3: Calculate Each Share

First student:

2 × 12 = $24

Second student:

3 × 12 = $36

Third student:

5 × 12 = $60

Final Answer

The students receive:

$24, $36, and $60

Checking

24 + 36 + 60 = 120

The shares add to the original total.

Solving Missing-Value Ratio Problems

Equivalent ratios can be used to find unknown quantities.

Worked Example

A drink mixture uses juice and water in the ratio:

2 : 5

If 8 litres of juice are used, how much water is required?

Step 1: Write Equivalent Ratios

2 : 5 = 8 : x

Step 2: Find the Multiplier

8 ÷ 2 = 4

Step 3: Multiply the Water Quantity

5 × 4 = 20

Final Answer

20 litres of water are required.

Checking

8 : 20 simplifies to 2 : 5.

Using Proportions

A proportion is an equation showing that two ratios are equal.

For example:

3/4 = 6/8

A proportion can be used to solve missing-value problems.

Worked Example

Solve:

4/7 = x/21

Cross-multiply:

4 × 21 = 7x

84 = 7x

Divide by 7:

x = 12

Important Observation

Cross-multiplication is a useful method when solving equations involving equivalent ratios.

Rates Involving Time

Many everyday rates involve time.

Example: Water Flow

A tap fills a container with 45 litres of water in 3 minutes.

Calculate the flow rate.

Flow rate = volume ÷ time

45 ÷ 3 = 15

Flow rate = 15 L/min

Predicting Volume

At the same constant rate, how much water flows in 8 minutes?

Volume = rate × time

Volume = 15 × 8

Volume = 120 L

Important Observation

When the rate remains constant, we can use multiplication or division to calculate related quantities.

Distance, Speed, and Time

Speed is one of the most familiar applications of rates.

Speed = Distance ÷ Time

This can be rearranged:

Distance = Speed × Time

Time = Distance ÷ Speed

Worked Example

A train travels at an average speed of 80 km/h for 2.5 hours.

Distance:

d = 80 × 2.5

d = 200 km

Another Example

A car travels 150 km at an average speed of 60 km/h.

Time:

t = 150 ÷ 60

t = 2.5 hours

Important Observation

Always check that the distance and time units are consistent.

Understanding Ratios and Rates Graphically

Equivalent ratios and constant rates can be represented using graphs.

For a constant rate, the graph is a straight line passing through the origin.

Example

A cyclist travels at a constant speed of 15 km/h.

Distance is given by:

d = 15t

where:

• d = distance in kilometres

• t = time in hours

  

  Interpretation

  The slope of the graph represents the rate.

  A steeper line indicates a greater rate when the same axes and scales are used.

   

  Ratios in Maps and Scale Drawings

  Ratios are useful for representing large distances using smaller diagrams.

  

   

   

   

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  Example

  A map uses the scale:

  1 : 100,000

  This means 1 unit on the map represents 100,000 of the same units in reality.

  For example:

  1 cm on the map = 100,000 cm in reality

  Convert:

  100,000 cm = 1 km

  Therefore:

  1 cm on the map represents 1 km in reality.

  Worked Example

  Two towns are 7 cm apart on the map.

  Actual distance:

  7 × 1 = 7

  Actual distance = 7 km

  Important Observation

  Scale ratios require careful attention to units.

   

  Comparing Rates with Different Units

  Sometimes rates must be converted before they can be compared.

  Example

  Cyclist A travels at 18 km/h.

  Cyclist B travels at 6 m/s.

  Which cyclist is faster?

  Convert Cyclist B's speed to km/h.

  1 m/s = 3.6 km/h

  6 × 3.6 = 21.6

  Cyclist B travels at:

  21.6 km/h

  Therefore, Cyclist B is faster.

  Important Observation

  Rates must be expressed using equivalent units before making comparisons.

   

  Ratios, Rates, and Proportional Relationships

  Ratios and rates are closely connected to proportional relationships.

  A proportional relationship can be written as:

  y = kx

  where k is the constant of proportionality.

  Example

  A worker earns $15 per hour.

  Total earnings:

  E = 15h

  where:

  • E = earnings in dollars

  • h = hours worked

  The constant of proportionality is 15.

  Interpretation

  For every additional hour worked, earnings increase by $15.

  Important Observation

  A constant unit rate is the constant of proportionality in a proportional relationship.

   

  A General Method for Solving Ratio and Rate Problems

  1. Identify the quantities. Determine what is being compared.

  2. Check the order. Write the quantities in the order requested.

  3. Check the units. Convert units when necessary.

  4. Choose a method. Simplify a ratio, calculate a unit rate, or write a proportion.

  5. Calculate. Show the mathematical steps clearly.

  6. Interpret. Explain what the result means in context.

  7. Verify. Check that the answer is reasonable and satisfies the original relationship.

   

  Common Mistakes

  Reversing the order of a ratio

  The ratio of apples to oranges is not the same as oranges to apples.

  Confusing part-to-part and part-to-whole ratios

  A ratio of 2 : 3 does not mean that 2/3 of the total is the first quantity.

  Forgetting to simplify

  Ratios should usually be expressed in simplest whole-number form when appropriate.

  Ignoring units

  Ratios involving the same type of measurement must use consistent units before simplification.

  Calculating unit rates incorrectly

  Divide by the quantity that must become one unit.

  Comparing total prices instead of unit prices

  A larger package may cost more overall but less per kilogram.

  Multiplying only one term of a ratio

  Equivalent ratios require multiplying or dividing all terms by the same nonzero factor.

  Confusing rates and ratios

  A rate compares quantities with different units, while a ratio is a more general comparison.

  Failing to check totals

  When sharing a quantity in a ratio, the individual shares must add to the original total.

  Forgetting to interpret answers

  A numerical result should be explained using the context and appropriate units.

   

  Key Terms

  • Ratio: Comparison of two quantities using division.

  • Rate: Ratio comparing quantities measured in different units.

  • Unit Rate: Rate expressed per one unit of the second quantity.

  • Equivalent Ratios: Ratios representing the same comparison.

  • Simplest Form: Ratio reduced by dividing terms by their greatest common factor.

  • Part-to-Part Ratio: Comparison of two parts of a group.

  • Part-to-Whole Ratio: Comparison of one part with the entire group.

  • Proportion: Equation showing that two ratios are equal.

  • Unit Price: Cost per one unit of a product.

  • Speed: Distance travelled per unit of time.

  • Flow Rate: Volume of fluid passing per unit of time.

  • Scale: Ratio comparing dimensions on a representation with actual dimensions.

  • Constant of Proportionality: Fixed multiplier connecting two proportional quantities.

  • Proportional Relationship: Relationship in which one quantity is a constant multiple of another.

  • Cross-Multiplication: Method for solving proportions.

  • Conversion Factor: Multiplier used to convert between measurement units.

  • Greatest Common Factor: Largest positive whole number dividing two or more whole numbers.

  • Unit Conversion: Changing a measurement from one unit to another without changing its value.

   

  Key Takeaways

  • Ratios compare quantities using division.

  • Ratios can be written using colons, fractions, or words.

  • The order of terms in a ratio matters.

  • Part-to-part ratios differ from part-to-whole ratios.

  • Ratios can be simplified by dividing all terms by a common factor.

  • Equivalent ratios represent the same relationship.

  • Rates compare quantities measured in different units.

  • Unit rates describe quantities per one unit.

  • Unit prices allow fair comparisons between products.

  • Ratios are useful for scaling recipes and dividing quantities.

  • Proportions can be used to find unknown values.

  • Speed is a rate involving distance and time.

  • Scale drawings use ratios to represent actual dimensions.

  • Units must be consistent when comparing measurements.

  • Constant unit rates produce proportional relationships.

  • Answers should always be interpreted in context.

   

  Check Your Understanding

  Writing and Simplifying Ratios

  1. Write the ratio of 6 apples to 9 oranges in colon form.

  2. Simplify 12 : 18.

  3. Write 4 : 7 as a fraction.

  4. Write 5 : 8 in words.

  5. Simplify 25 : 40.

  Equivalent Ratios

  6. Find an equivalent ratio to 2 : 5 where the first term is 10.

  7. Complete 3 : 4 = 12 : x.

  8. Are 6 : 9 and 10 : 15 equivalent?

  9. Simplify 1 m : 25 cm.

  10. Simplify 2 kg : 500 g.

  Unit Rates

  11. A car travels 180 km in 3 hours. Find its average speed.

  12. A worker earns $84 in 7 hours. Find the hourly rate.

  13. A 4 kg bag of rice costs $14. Find the unit price.

  14. A tap fills 60 litres in 5 minutes. Find the flow rate.

  15. A cyclist travels 45 km in 2.5 hours. Find the average speed.

  Problem Solving

  16. Share $90 in the ratio 2 : 3.

  17. A recipe uses flour and sugar in a 4 : 1 ratio by mass. If 800 g of flour is used, how much sugar is required?

  18. A map scale is 1 : 50,000. What actual distance does 4 cm represent?

  19. Which is cheaper per kilogram: 2 kg for $9 or 5 kg for $20?

  20. Explain why unit rates are useful when comparing products.

  Answer Key

  Question	Answer
  1	6 : 9
  2	2 : 3
  3	4/7
  4	5 to 8
  5	5 : 8
  6	10 : 25
  7	x = 16
  8	Yes
  9	4 : 1
  10	4 : 1
  11	60 km/h
  12	$12/h
  13	$3.50/kg
  14	12 L/min
  15	18 km/h
  16	$36 and $54
  17	200 g
  18	2 km
  19	5 kg for $20
  20	They allow fair comparisons using the same units

   

  Extended Challenge

  1. Simplify the ratio 36 : 54 : 72.

  2. Share $240 in the ratio 3 : 5 : 7.

  3. A recipe uses flour, sugar, and butter in the ratio 5 : 2 : 3 by mass. If 750 g of flour is used, calculate the quantities of sugar and butter.

  4. A car travels 275 km in 3 hours 40 minutes. Calculate its average speed.

  5. A supermarket sells 750 g of cereal for $4.50 and 1.2 kg for $6.60. Determine the cheaper unit price.

  6. A map has scale 1 : 250,000. Calculate the actual distance represented by 8 cm.

  7. A machine produces 420 items in 35 minutes. Calculate its production rate per minute.

  8. A tank fills at 18 L/min. How long will it take to fill a 270 L tank?

  9. A cyclist travels at 8 m/s. Convert the speed to km/h.

  10. A school has a teacher-to-student ratio of 1 : 18. How many teachers are required for 540 students if the ratio is maintained exactly?

   

  Final Investigation: Planning a School Event

  A school is organising a sports event for 240 students.

  Students will be divided into three teams in the ratio:

  3 : 4 : 5

  The event organisers must also purchase refreshments and arrange transportation.

  Part A: Team Allocation

  1. Calculate the total number of ratio parts.

  2. Determine the number of students in each team.

  3. Check that the total is 240.

  4. Express the smallest team's size as a fraction of the total.

  5. Calculate the percentage of students in the largest team.

  Part B: Refreshments

  The school needs 360 bottles of water.

  Supplier A sells 24 bottles for $12.

  Supplier B sells 30 bottles for $13.50.

  6. Calculate the cost per bottle from Supplier A.

  7. Calculate the cost per bottle from Supplier B.

  8. Determine which supplier offers the lower unit price.

  9. Calculate the total cost of buying 360 bottles from Supplier A.

  10. Calculate the total cost from Supplier B.

  Part C: Transportation

  A bus travels 150 km in 2.5 hours.

  11. Calculate its average speed.

  12. Determine how far it would travel in 4 hours at the same speed.

  13. Calculate the time required to travel 210 km at the same speed.

  14. Explain why actual travel times may differ from predictions.

  15. Convert the average speed into metres per second.

  Part D: Scaling and Proportions

  A sports drink is mixed using concentrate and water in the ratio:

  1 : 4

  16. Calculate the amount of concentrate required for 20 litres of water.

  17. Calculate the total drink volume.

  18. Determine how much concentrate is required to make 30 litres of drink.

  19. Explain why the ratio must remain constant when scaling the mixture.

  20. Write a conclusion explaining how ratios and rates help organisers make efficient decisions.

  Investigation Check

  Team allocation

  • Total parts = 12

  • Team sizes = 60, 80, and 100 students

  • Largest team = approximately 41.7% of students

  Refreshments

  • Supplier A = $0.50 per bottle

  • Supplier B = $0.45 per bottle

  • Supplier A total = $180

  • Supplier B total = $162

  • Supplier B saves $18

  Transportation

  • Average speed = 60 km/h

  • Distance in 4 hours = 240 km

  • Time for 210 km = 3.5 hours

  • Speed ≈ 16.67 m/s

  Sports drink

  • 20 L water requires 5 L concentrate

  • Total mixture = 25 L

  • A 30 L mixture requires 6 L concentrate and 24 L water

2. Solving Proportions

Solving Proportions

Introduction

Proportions are used to compare quantities that maintain the same relationship. They help us solve problems involving recipes, maps, measurements, prices, scale drawings, and many other everyday situations.

For example, if 3 notebooks cost $12, how much would 8 notebooks cost at the same price per notebook?

We can solve this problem using a proportion.

A proportion is an equation stating that two ratios are equal.

Understanding proportions allows us to calculate missing values, recognise proportional relationships, and check whether quantities are being compared correctly.

What Is a Proportion?

A proportion is an equation showing that two ratios are equivalent.

For example:

2/3 = 4/6

Both fractions represent the same relationship.

The first ratio compares 2 parts to 3 parts.

The second ratio compares 4 parts to 6 parts.

Because both quantities were multiplied by 2, the ratios are equivalent.

Writing Proportions

A proportion can be written using fractions:

a/b = c/d

or using ratio notation:

a : b = c : d

For these expressions, b and d must not equal zero.

Example

3 : 5 = 9 : 15

This can also be written as:

3/5 = 9/15

Important Observation

The order of the quantities must remain consistent in both ratios.

Identifying Proportional Relationships

Two quantities have a proportional relationship when one quantity is always a constant multiple of the other.

For example, suppose pencils cost $2 each.

To determine whether the relationship is proportional, divide the total cost by the number of pencils.

2/1 = 2

4/2 = 2

6/3 = 2

8/4 = 2

10/5 = 2

The ratio is always 2.

Therefore, the quantities are proportional.

General Equation

A proportional relationship can be written as:

y = kx

where:

• x = independent variable

• y = dependent variable

• k = constant of proportionality

For the pencil example:

C = 2n

where C is the total cost and n is the number of pencils.

Important Observation

In a proportional relationship, the ratio y/x remains constant for all nonzero values of x in the domain.

Recognising Proportional Relationships in Tables

Tables can help us determine whether two quantities are proportional.

Example A

Calculate y/x:

6/2 = 3

12/4 = 3

18/6 = 3

24/8 = 3

The ratio is constant.

Therefore:

y = 3x

The relationship is proportional.

Example B

Calculate y/x:

7/2 = 3.5

13/4 = 3.25

19/6 ≈ 3.17

25/8 = 3.125

The ratios are not equal.

Therefore, the relationship is not proportional.

Important Observation

A relationship can increase at a constant rate without being proportional.

For example, y = 3x + 1 is linear but not proportional because it does not pass through the origin.

Recognising Proportions from Graphs

A proportional relationship produces a straight-line graph passing through the origin (0, 0).

Important Characteristics

A proportional graph:

• Is a straight line.

• Passes through the origin.

• Has a constant slope.

• Can be represented by y = kx.

A straight line that does not pass through the origin is not a proportional relationship.

Equivalent Ratios

Equivalent ratios represent the same comparison.

For example:

2 : 5 = 4 : 10 = 6 : 15 = 8 : 20

To create equivalent ratios, multiply or divide both terms by the same nonzero number.

Worked Example

Find an equivalent ratio to 3 : 7 where the first term is 15.

Step 1: Find the Multiplier

15 ÷ 3 = 5

Step 2: Multiply the Second Term

7 × 5 = 35

Final Answer

3 : 7 = 15 : 35

Checking

15/35 simplifies to 3/7.

Therefore, the ratios are equivalent.

Solving Missing-Value Problems Using Equivalent Ratios

A missing value can often be found by identifying the scale factor.

Worked Example

Solve:

4/9 = 20/x

Step 1: Compare the Numerators

4 × 5 = 20

The scale factor is 5.

Step 2: Apply the Same Factor

9 × 5 = 45

Final Answer

x = 45

Verification

4/9 = 20/45

Simplify:

20/45 = 4/9

The solution is correct.

Important Observation

Using equivalent ratios is often the quickest method when the scale factor is easy to recognise.

What Is Cross Multiplication?

Cross multiplication is a method used to solve equations containing two equal fractions.

For the proportion:

a/b = c/d

where b ≠ 0 and d ≠ 0:

a × d = b × c

This is called cross multiplication because the numerator of each fraction is multiplied by the denominator of the other fraction.

Example

3/4 = 6/8

Cross multiply:

3 × 8 = 4 × 6

24 = 24

The products are equal.

Therefore, the proportion is true.

Why Does Cross Multiplication Work?

Starting with:

a/b = c/d

Multiply both sides by bd:

ad = bc

Cross multiplication is therefore a consequence of multiplying both sides of an equation by the same nonzero quantity.

Visualising Cross Multiplication

Solving Proportions Using Cross Multiplication

Worked Example

Solve:

5/8 = x/24

Step 1: Cross Multiply

5 × 24 = 8 × x

120 = 8x

Step 2: Divide Both Sides by 8

x = 120 ÷ 8

x = 15

Step 3: Verify

5/8 = 15/24

Simplify 15/24:

15/24 = 5/8

The ratios are equivalent.

Cross Multiplication with an Unknown Denominator

Worked Example

Solve:

7/x = 14/18

Step 1: Cross Multiply

7 × 18 = 14x

126 = 14x

Step 2: Solve

x = 126 ÷ 14

x = 9

Step 3: Check

7/9 = 14/18

Simplify 14/18:

14/18 = 7/9

The solution is correct.

Important Observation

The unknown may appear in either the numerator or denominator.

The cross-multiplication procedure remains the same, provided the denominators are nonzero.

Cross Multiplication with Expressions

Sometimes the unknown is part of an algebraic expression.

Worked Example

Solve:

(x + 2)/5 = 4/3

Step 1: Cross Multiply

3(x + 2) = 5 × 4

3(x + 2) = 20

Step 2: Expand

3x + 6 = 20

Step 3: Solve

3x = 14

x = 14/3

Step 4: Verify

x + 2 = 14/3 + 6/3 = 20/3

Left side:

(20/3)/5 = 4/3

Right side:

4/3

Both sides are equal.

Solving Proportions with Decimals

Worked Example

Solve:

2.5/4 = x/12

Step 1: Cross Multiply

2.5 × 12 = 4x

30 = 4x

Step 2: Solve

x = 30 ÷ 4

x = 7.5

Verification

2.5/4 = 0.625

7.5/12 = 0.625

The solution is correct.

Important Observation

Proportions can involve whole numbers, fractions, decimals, and algebraic expressions.

Proportions in Recipes

Recipes frequently involve proportional relationships.

Worked Example

A recipe requires 3 cups of flour to make 12 muffins.

How much flour is needed to make 20 muffins, assuming the recipe scales proportionally?

Step 1: Define the Unknown

Let x = cups of flour required.

Step 2: Write the Proportion

3/12 = x/20

Step 3: Cross Multiply

3 × 20 = 12x

60 = 12x

Step 4: Solve

x = 5

Interpretation

The recipe requires 5 cups of flour to make 20 muffins.

Checking

12 muffins require 3 cups.

20 muffins require more flour.

Five cups is reasonable.

Proportions in Maps and Scale Drawings

Scale drawings use proportional relationships to represent actual dimensions.

Worked Example

A map uses a scale of:

1 cm : 5 km

Two towns are 7 cm apart on the map.

Find the actual distance.

Step 1: Set Up the Proportion

1/5 = 7/x

Step 2: Cross Multiply

x = 5 × 7

x = 35

Interpretation

The towns are 35 kilometres apart.

Important Observation

Always ensure that corresponding measurements occupy the same positions in the ratios.

Proportions in Shopping

Unit prices allow us to compare products and calculate costs.

Worked Example

A 3 kg bag of rice costs $9.

How much would 8 kg cost at the same price per kilogram?

Step 1: Write the Proportion

3/9 = 8/x

Step 2: Cross Multiply

3x = 9 × 8

3x = 72

Step 3: Solve

x = 24

Interpretation

Eight kilograms would cost $24 at the same unit price.

Checking with Unit Rate

9 ÷ 3 = $3/kg

8 × 3 = $24

Both methods agree.

Proportions in Speed and Distance

Proportions can be used when speed remains constant.

Worked Example

A cyclist travels 24 kilometres in 2 hours.

How far will the cyclist travel in 5 hours at the same average speed?

Step 1: Set Up the Proportion

24/2 = x/5

Step 2: Cross Multiply

24 × 5 = 2x

120 = 2x

Step 3: Solve

x = 60 km

Interpretation

The cyclist will travel 60 kilometres in 5 hours if the speed remains constant.

Important Observation

A proportion can only be used directly when the relevant relationship remains proportional.

If speed changes, the same proportion may no longer apply.

Proportional and Non-Proportional Situations

Not every relationship is proportional.

Example A: Proportional

A worker earns $15 per hour with no fixed bonus.

Earnings:

E = 15h

The ratio E/h is constant.

Therefore, the relationship is proportional.

Example B: Non-Proportional

A taxi charges a $5 starting fee plus $2 per kilometre.

Cost:

C = 2d + 5

The ratio C/d is not constant.

Therefore, the relationship is not proportional.

Comparison

Important Observation

A constant rate of change does not always mean that two quantities are proportional.

A proportional relationship must have the form y = kx.

Verifying Solutions to Proportions

After solving a proportion, substitute the answer into the original equation.

Example

Solve:

6/11 = 18/x

Cross multiply:

6x = 11 × 18

6x = 198

x = 33

Verify

6/11 = 18/33

Simplify 18/33:

18/33 = 6/11

The solution is correct.

Alternative Check

Cross products:

6 × 33 = 198

11 × 18 = 198

Both cross products are equal.

Important Observation

Checking solutions helps detect arithmetic mistakes, incorrect cross multiplication, and reversed ratios.

Choosing the Best Method

There are several ways to solve proportions.

Example

Solve:

4/7 = 12/x

Using equivalent ratios:

4 × 3 = 12

7 × 3 = 21

x = 21

Using cross multiplication:

4x = 7 × 12

4x = 84

x = 21

Both methods produce the same answer.

A General Method for Solving Proportions

Common Mistakes

Reversing one ratio

Corresponding quantities must appear in the same order in both ratios.

Cross multiplying incorrectly

Multiply each numerator by the opposite denominator.

Forgetting to solve the final equation

Cross multiplication produces an equation that must still be solved.

Ignoring units

Convert measurements when necessary before constructing proportions.

Assuming every linear relationship is proportional

A linear relationship with a nonzero starting value is not proportional.

Dividing only one term of a ratio

Equivalent ratios require the same multiplier or divisor for both terms.

Using a proportion when the relationship changes

Proportions require a constant multiplicative relationship.

Forgetting to verify

Substituting the answer helps confirm that the ratios are equivalent.

Confusing direct and inverse relationships

If one quantity doubles while another halves, the relationship may be inversely proportional rather than directly proportional.

Key Terms

• Proportion: Equation stating that two ratios are equal.

• Ratio: Comparison of quantities using division.

• Equivalent Ratios: Ratios representing the same comparison.

• Proportional Relationship: Relationship where one quantity is a constant multiple of another.

• Constant of Proportionality: Fixed ratio between corresponding quantities.

• Cross Multiplication: Method of solving proportions by multiplying diagonally across two fractions.

• Numerator: Top number of a fraction.

• Denominator: Bottom number of a fraction.

• Scale Factor: Multiplier used to produce an equivalent ratio or measurement.

• Unit Rate: Rate expressed per one unit of another quantity.

• Direct Proportion: Relationship where y = kx.

• Inverse Proportion: Relationship where the product of two quantities remains constant.

• Variable: Symbol representing an unknown or changing quantity.

• Equation: Mathematical statement showing equality.

• Verification: Process of checking whether a solution satisfies the original equation.

• Substitution: Replacing a variable with a known value.

• Scale Drawing: Drawing representing actual dimensions proportionally.

• Graph of a Proportion: Straight line passing through the origin.

• Cross Product: Product of a numerator and the opposite denominator in a proportion.

Key Takeaways

• A proportion states that two ratios are equal.

• Equivalent ratios can be created by multiplying or dividing both terms by the same nonzero number.

• Proportional relationships have a constant ratio between corresponding quantities.

• A proportional relationship can be written as y = kx.

• The graph of a proportional relationship is a straight line passing through the origin.

• Missing values can be calculated using equivalent ratios.

• Cross multiplication is useful when the scale factor is not obvious.

• The cross products of equal ratios are equal.

• Proportions are useful in recipes, shopping, maps, and constant-speed problems.

• Not every linear relationship is proportional.

• Units and corresponding quantities must be arranged consistently.

• Solutions should always be checked by substitution or equivalent ratios.

• Mathematical answers should be interpreted using the real-world context.

Check Your Understanding

Identifying Proportions

1. What is a proportion?

2. Are 2/3 and 6/9 equivalent ratios?

3. Is y = 4x a proportional relationship?

4. Is y = 4x + 2 a proportional relationship?

5. What must be true about y/x in a proportional relationship?

Equivalent Ratios

6. Complete 3 : 5 = 12 : x.

7. Complete 4 : 7 = x : 21.

8. Are 8 : 12 and 10 : 15 equivalent?

9. Find an equivalent ratio to 5 : 9 where the first term is 20.

10. Simplify 18 : 30.

Cross Multiplication

11. Solve 3/4 = x/20.

12. Solve 5/8 = 15/x.

13. Solve 7/9 = x/27.

14. Solve 2/5 = (x + 1)/15.

15. Solve 4/x = 12/21.

Practical Problems

16. A recipe uses 2 cups of flour for 8 muffins. How much flour is needed for 20 muffins?

17. A map uses a scale of 1 cm : 4 km. What actual distance does 6 cm represent?

18. Five notebooks cost $15. How much would 9 notebooks cost at the same unit price?

19. A cyclist travels 36 km in 3 hours. How far will the cyclist travel in 5 hours at the same speed?

20. Explain one method for verifying a proportion.

Answer Key

Extended Challenge

1. Solve 7/12 = x/36.

2. Solve 9/x = 15/25.

3. Solve (x + 3)/8 = 5/4.

4. Solve 2.5/6 = x/18.

5. Determine whether the table (2, 7), (4, 14), (6, 21), (8, 28) represents a proportional relationship.

6. A recipe requires 450 g of flour for 18 cookies. How much flour is required for 30 cookies?

7. A map uses a scale of 1 : 250,000. Find the actual distance represented by 6 cm.

8. A worker earns $135 for 9 hours. Calculate the earnings for 14 hours at the same hourly rate.

9. A machine produces 240 items in 8 minutes. How many items will it produce in 15 minutes at the same rate?

10. Explain why a taxi fare consisting of a fixed fee plus a distance charge is not directly proportional to distance.

Final Investigation: Planning a School Fundraiser

A school is organising a fundraiser and needs to purchase supplies, prepare food, and arrange transportation.

The organisers must use proportions to calculate quantities and costs.

Part A: Food Preparation

A recipe makes 12 servings using:

• 300 g flour

• 200 mL milk

• 4 eggs

The school needs 48 servings.

1. Determine the scale factor.

2. Calculate the amount of flour required.

3. Calculate the amount of milk required.

4. Calculate the number of eggs required.

5. Verify that all ingredient quantities have been scaled proportionally.

Part B: Purchasing Supplies

A supplier sells 5 notebooks for $12.50.

6. Calculate the unit price.

7. Write a proportion to determine the cost of 18 notebooks.

8. Solve the proportion using cross multiplication.

9. Verify the answer using the unit price.

10. Explain why the cost is proportional to the number of notebooks.

Part C: Transportation

A bus travels 120 kilometres in 2 hours at a constant average speed.

11. Calculate the average speed.

12. Write a proportion to determine the distance travelled in 3.5 hours.

13. Solve the proportion.

14. Calculate the time required to travel 300 kilometres at the same speed.

15. Explain why the proportion may not accurately predict travel time if traffic conditions change.

Part D: Scale Drawings

A school hall is represented on a plan using the scale:

1 cm : 2 m

The hall measures 9 cm by 6 cm on the plan.

16. Calculate the actual length.

17. Calculate the actual width.

18. Determine the actual floor area.

19. Explain why the scale factor must be applied to both dimensions.

20. Write a conclusion explaining how proportions help solve practical problems.

Investigation Check

Food preparation

• Scale factor = 4

• Flour = 1200 g

• Milk = 800 mL

• Eggs = 16

Supplies

• Unit price = $2.50

• Cost of 18 notebooks = $45

Transportation

• Average speed = 60 km/h

• Distance in 3.5 hours = 210 km

• Time for 300 km = 5 hours

Scale drawing

• Actual length = 18 m

• Actual width = 12 m

• Actual floor area = 216 m²

3. Direct Variation

Direct Variation

Introduction

Many quantities in everyday life change together in predictable ways. When one quantity increases, another may increase by the same factor. When one decreases, the other may decrease proportionally.

For example:

• Buying twice as many identical notebooks costs twice as much.

• Travelling for twice as long at a constant speed covers twice the distance.

• Working three times as many hours at a fixed hourly wage earns three times as much money.

• Doubling the volume of water doubles its mass when its density remains constant.

These situations are examples of direct variation, also called direct proportionality.

Direct variation helps us understand proportional relationships, create equations, interpret graphs, and make predictions.

What Is Direct Variation?

Direct variation is a relationship between two quantities in which one quantity is a constant multiple of the other.

If y varies directly as x, then:

y = kx

where:

• x is the independent variable.

• y is the dependent variable.

• k is the constant of variation.

The constant k describes how much y changes for each unit of x.

Example

Suppose a notebook costs $4.

The equation is:

y = 4x

Here, k = 4.

For every additional notebook purchased, the cost increases by $4.

Important Observation

In direct variation, multiplying x by a factor also multiplies y by the same factor.

If x doubles, y doubles.

If x triples, y triples.

If x is halved, y is halved.

This holds even when the constant of variation is negative, although the sign of y then depends on the sign of x.

The Constant of Variation

The constant of variation is the fixed ratio between y and x.

Starting with:

y = kx

Divide both sides by x, where x ≠ 0:

k = y/x

Worked Example

Suppose y varies directly as x.

When x = 6, y = 24.

Find k.

Step 1: Write the Formula

k = y/x

Step 2: Substitute

k = 24/6

k = 4

Step 3: Write the Equation

y = 4x

Interpretation

For every increase of 1 in x, y increases by 4.

The ratio y/x remains equal to 4 for every nonzero x.

Recognising Direct Variation in Tables

One way to identify direct variation is to calculate the ratio y/x for each pair of values.

Example: Direct Variation

The ratio is always 5.

Therefore, the relationship is directly proportional.

The equation is:

y = 5x

Example: Not Direct Variation

The ratios are not constant.

Therefore, the relationship is not direct variation.

These values follow:

y = 3x + 2

Although the relationship is linear, it is not directly proportional.

Important Observation

A constant difference in y does not automatically indicate direct variation. The ratio y/x must remain constant.

Recognising Direct Variation in Graphs

The graph of a direct variation relationship is a straight line passing through the origin.

The origin is:

(0, 0)

For a direct variation equation:

y = kx

the constant k is also the slope of the line.

Characteristics of Direct Variation Graphs

• The graph is a straight line.

• It passes through the origin.

• Its slope is constant.

• Its equation has the form y = kx.

• If k is positive, the line rises from left to right.

• If k is negative, the line falls from left to right.

Important Observation

A straight line that does not pass through the origin does not represent direct variation.

Exploring the Constant of Variation

Change k to investigate how the direct variation graph changes.

Questions to Consider

• What happens when k increases?

• What happens when k becomes negative?

• What happens when k = 0?

• Why does the line always pass through the origin?

Important Observation

When k = 0, y = 0 for all x. This is a degenerate direct-variation equation, sometimes excluded when a definition requires a nonzero constant.

Writing Direct Variation Equations

A direct variation equation can be constructed when one pair of corresponding values is known.

Worked Example

Suppose y varies directly as x.

When x = 8, y = 56.

Find the equation.

Step 1: Start with the General Equation

y = kx

Step 2: Substitute Known Values

56 = 8k

Step 3: Solve for k

k = 56/8

k = 7

Step 4: Write the Equation

y = 7x

Step 5: Verify

When x = 8:

y = 7(8) = 56

The equation agrees with the given information.

Using Direct Variation to Find Missing Values

Worked Example

Suppose y varies directly as x.

When x = 4, y = 18.

Find y when x = 10.

Step 1: Find k

k = y/x

k = 18/4

k = 4.5

Step 2: Write the Equation

y = 4.5x

Step 3: Substitute x = 10

y = 4.5(10)

y = 45

Final Answer

When x = 10, y = 45.

Alternative Method: Proportions

18/4 = y/10

Cross multiply:

4y = 180

y = 45

Both methods produce the same answer.

Direct Variation and Equivalent Ratios

Direct variation is closely related to equivalent ratios.

Suppose:

y = 3x

Some corresponding values are:

The ratios are:

3/1 = 6/2 = 12/4 = 24/8 = 3

Because the ratio remains constant, the relationship is proportional.

Important Observation

Equivalent ratios provide a useful way to solve direct variation problems without always writing the equation explicitly.

Direct Variation in Shopping

Shopping provides a familiar example of direct variation when each item has the same price.

Worked Example

A notebook costs $3.50.

Let:

n = number of notebooks

C = total cost in dollars

Step 1: Write the Equation

C = 3.5n

The constant of variation is 3.5.

Step 2: Calculate the Cost of 8 Notebooks

C = 3.5(8)

C = $28

Step 3: Determine How Many Notebooks Cost $42

42 = 3.5n

n = 42/3.5

n = 12

Interpretation

The total cost varies directly with the number of notebooks, assuming there are no fixed fees or discounts.

Direct Variation in Distance and Time

When speed is constant, distance varies directly with time.

d = vt

where:

• d = distance

• v = constant speed

• t = time

The constant of variation is the speed v.

Worked Example

A cyclist travels at a constant speed of 18 km/h.

Step 1: Write the Equation

d = 18t

Step 2: Find the Distance After 3.5 Hours

d = 18(3.5)

d = 63 km

Step 3: Find the Time Needed to Travel 90 km

90 = 18t

t = 90/18

t = 5 hours

Interpretation

The cyclist travels 18 kilometres for every hour of travel.

Important Observation

The relationship is directly proportional only when the speed remains constant and distance is measured from zero at the starting time.

Direct Variation in Wages

When a worker earns a fixed hourly wage without additional bonuses or fees, earnings vary directly with hours worked.

Worked Example

A worker earns $16 per hour.

Let:

E = earnings in dollars

h = hours worked

Equation

E = 16h

Calculate Earnings for 7.5 Hours

E = 16(7.5)

E = $120

Determine Hours Required to Earn $200

200 = 16h

h = 200/16

h = 12.5 hours

Important Observation

A fixed starting payment or bonus would change the equation and make total earnings non-proportional to hours worked.

Direct Variation in Science

Direct variation appears in many scientific relationships when relevant conditions remain constant.

Mass and Volume

For a substance with constant density:

m = ρV

where:

• m = mass

• ρ = density

• V = volume

The mass varies directly with volume.

Worked Example

A liquid has a density of 0.8 g/mL.

Find the mass of 250 mL of the liquid.

m = ρV

m = 0.8(250)

m = 200 g

Interpretation

Every additional millilitre adds 0.8 grams of mass.

The constant of variation is 0.8 g/mL.

Direct Variation in Geometry

Some geometric relationships involve direct variation.

Circumference and Diameter

For a circle:

C = πd

where:

• C = circumference

• d = diameter

• π ≈ 3.14159

Circumference varies directly with diameter.

The constant of variation is π.

Worked Example

A circular wheel has diameter 70 cm.

C = πd

C = π(70)

C ≈ 219.9 cm

Important Observation

When the diameter doubles, the circumference doubles.

However, the area of a circle does not vary directly with its radius because A = πr².

Direct Variation with Negative Constants

The constant of variation can be negative.

Example

y = −3x

The ratio y/x equals −3 for all nonzero x.

The graph is a straight line through the origin with negative slope.

Important Observation

Direct variation does not always mean that both quantities increase together.

When k is negative, an increase in x produces a decrease in y.

Direct Variation Compared with Inverse Variation

Direct variation and inverse variation describe different relationships.

Direct Variation

y = kx

The ratio y/x is constant.

Inverse Variation

y = k/x

The product xy is constant.

For positive x and k:

• Direct variation: doubling x doubles y.

• Inverse variation: doubling x halves y.

  

  Comparison

  Property	Direct variation	Inverse variation
  Equation	y = kx	y = k/x
  Constant relationship	y/x = k	xy = k
  Typical graph	Straight line through origin	Hyperbola
  If x doubles	y doubles	y halves
  Example	Cost at fixed unit price	Time for fixed distance at varying speed

  Important Observation

  Direct and inverse variation should not be confused. The type of relationship determines the appropriate equation and method.

   

  Distinguishing Direct Variation from Other Linear Relationships

  Not all linear relationships represent direct variation.

  Example A

  y = 5x

  This is direct variation.

  The graph passes through the origin.

  Example B

  y = 5x + 10

  This is not direct variation.

  The graph crosses the y-axis at 10.

  Real-World Interpretation

  A gym charges $5 per visit with no membership fee:

  C = 5v

  This is direct variation.

  Another gym charges a $10 membership fee plus $5 per visit:

  C = 5v + 10

  This is not direct variation.

  Important Observation

  A proportional relationship has no nonzero fixed starting amount.

   

  Solving Multi-Step Direct Variation Problems

  Worked Example

  A machine produces 180 items in 6 minutes at a constant rate.

  How many items will it produce in 15 minutes?

  Step 1: Define Variables

  Let:

  P = number of items produced

  t = time in minutes

  Step 2: Write the General Equation

  P = kt

  Step 3: Find k

  180 = 6k

  k = 180/6

  k = 30 items/min

  Step 4: Write the Equation

  P = 30t

  Step 5: Calculate Production After 15 Minutes

  P = 30(15)

  P = 450 items

  Step 6: Check Reasonableness

  The machine produces 30 items per minute.

  In 15 minutes, producing 450 items is consistent with this rate.

   

  A General Method for Solving Direct Variation Problems

  1. Identify the variables. Determine which quantities are being compared.

  2. Check proportionality. Confirm that the ratio between corresponding quantities remains constant.

  3. Write the equation. Begin with y = kx.

  4. Find the constant. Use k = y/x with a known pair of nonzero x and corresponding y.

  5. Construct the model. Substitute k into the equation.

  6. Solve the problem. Calculate the unknown quantity.

  7. Verify and interpret. Check the calculation, units, and reasonableness of the answer.

   

  Common Mistakes

  Assuming every increasing relationship is direct variation

  A relationship may increase without having a constant ratio.

  Confusing a constant difference with a constant ratio

  Direct variation requires y/x to remain constant, not merely equal increases in y.

  Forgetting that the graph must pass through the origin

  A line with a nonzero y-intercept is not direct variation.

  Calculating k incorrectly

  Use k = y/x, not x/y, when the equation is y = kx.

  Mixing up independent and dependent variables

  Identify which quantity depends on the other.

  Ignoring units

  The constant of variation may have units, such as dollars per item or kilometres per hour.

  Assuming k must be positive

  Negative constants are possible.

  Confusing direct and inverse variation

  Direct variation has a constant ratio; inverse variation has a constant product.

  Using a direct variation model when conditions change

  Constant price, speed, density, or wage assumptions must remain valid.

  Failing to verify

  Substitute the known values into the equation to confirm the model.

   

  Key Terms

  • Direct Variation: Relationship in which one quantity is a constant multiple of another.

  • Direct Proportionality: Another name for direct variation.

  • Proportional Relationship: Relationship represented by y = kx.

  • Constant of Variation: Fixed multiplier k connecting two directly proportional quantities.

  • Constant of Proportionality: Another name for the constant of variation.

  • Independent Variable: Input variable.

  • Dependent Variable: Output variable.

  • Ratio: Comparison of two quantities using division.

  • Unit Rate: Rate expressed per one unit of another quantity.

  • Equivalent Ratios: Ratios representing the same relationship.

  • Linear Relationship: Relationship represented by a straight-line graph.

  • Origin: Point (0, 0) on a coordinate plane.

  • Slope: Rate of change of a line.

  • Positive Slope: Line rises from left to right.

  • Negative Slope: Line falls from left to right.

  • Inverse Variation: Relationship represented by y = k/x.

  • Proportion: Equation showing that two ratios are equal.

  • Mathematical Model: Equation or representation describing a real-world relationship.

  • Verification: Checking whether a solution satisfies the original conditions.

   

  Key Takeaways

  • Direct variation describes a relationship in which one quantity is a constant multiple of another.

  • The general equation is y = kx.

  • The constant of variation is k = y/x for x ≠ 0.

  • A direct variation graph is a straight line passing through the origin.

  • The constant of variation equals the slope of the graph.

  • Direct variation can be identified using tables, graphs, and equations.

  • Equivalent ratios are a useful way to recognise proportional relationships.

  • A constant difference does not necessarily indicate direct variation.

  • A linear relationship with a nonzero y-intercept is not direct variation.

  • Direct variation can be used to calculate unknown values.

  • Common applications include shopping, wages, distance, mass, and circumference.

  • Negative constants of variation are possible.

  • Direct variation and inverse variation describe different mathematical relationships.

  • Solutions should be checked using substitution, units, and real-world reasoning.

   

  Check Your Understanding

  Understanding Direct Variation

  1. What is direct variation?

  2. Write the general equation for direct variation.

  3. What does k represent?

  4. What point must every direct variation graph pass through?

  5. Does y = 5x + 2 represent direct variation?

  Finding the Constant of Variation

  6. If y = 24 when x = 6, find k.

  7. If y = 35 when x = 7, write the direct variation equation.

  8. If y = −18 when x = 3, find k.

  9. If y = 12 when x = 8, find k.

  10. If y = 4.5x, identify the constant of variation.

  Solving Direct Variation Problems

  11. If y = 3x, find y when x = 12.

  12. If y = 7x, find x when y = 56.

  13. If y varies directly as x and y = 15 when x = 5, find y when x = 9.

  14. If y varies directly as x and y = 40 when x = 8, find x when y = 65.

  15. If y = −2x, find y when x = −6.

  Real-World Applications

  16. A notebook costs $4. Find the cost of 9 notebooks.

  17. A cyclist travels at 20 km/h. Find the distance travelled in 3.5 hours.

  18. A worker earns $18 per hour. Find the earnings for 6 hours.

  19. A liquid has density 1.2 g/mL. Find the mass of 150 mL.

  20. Explain why a taxi fare with a fixed starting fee is not directly proportional to distance.

  Answer Key

  Question	Answer
  1	One quantity is a constant multiple of another
  2	y = kx
  3	Constant of variation
  4	(0, 0)
  5	No
  6	4
  7	y = 5x
  8	−6
  9	1.5
  10	4.5
  11	36
  12	8
  13	27
  14	13
  15	12
  16	$36
  17	70 km
  18	$108
  19	180 g
  20	The fixed fee creates a nonzero y-intercept

   

  Extended Challenge

  1. Determine whether the table (2, 9), (4, 18), (6, 27), (8, 36) represents direct variation.

  2. Find the constant of variation for the relationship in Question 1.

  3. Write the equation of a direct variation relationship passing through (−4, 20).

  4. If y varies directly as x and y = 18 when x = 2.5, find y when x = 7.

  5. If y varies directly as x and y = −24 when x = 8, find x when y = 15.

  6. A worker earns $157.50 for 9 hours. Determine the hourly rate and write the direct variation equation.

  7. A vehicle travels 175 km in 2.5 hours at constant speed. Find the distance travelled in 6 hours.

  8. A liquid has mass 450 g and volume 375 mL. Find the constant of variation and predict the mass of 600 mL.

  9. Compare y = 6x and y = 6x + 12. Explain which relationship is directly proportional and why.

  10. A circle has circumference 94.2 cm. Using π ≈ 3.14, determine its diameter and explain why circumference varies directly with diameter.

   

  Final Investigation: Planning a School Supply Order

  A school is purchasing supplies for students.

  The cost of each item remains constant, and there are no additional fees or discounts.

  The school needs to investigate proportional relationships and predict total costs.

  Part A: Notebooks

  A supplier sells 8 notebooks for $28.

  1. Identify the independent variable.

  2. Identify the dependent variable.

  3. Calculate the cost per notebook.

  4. Determine the constant of variation.

  5. Write the direct variation equation.

  6. Calculate the cost of 20 notebooks.

  7. Determine how many notebooks can be purchased for $105.

  8. Construct a table for 0, 5, 10, 15, and 20 notebooks.

  9. Graph the relationship.

  10. Explain why the graph passes through the origin.

  Part B: Pencils

  A supplier sells pencils for $0.75 each.

  11. Write the direct variation equation.

  12. Calculate the cost of 48 pencils.

  13. Determine how many pencils can be purchased for $30.

  14. Explain the meaning of the constant of variation.

  15. Compare the slopes of the notebook and pencil cost graphs.

  Part C: Evaluating Relationships

  Consider these equations:

  A: y = 4x

  B: y = 4x + 5

  C: y = −2x

  16. Identify which equations represent direct variation.

  17. Determine the constant of variation for each direct variation equation.

  18. Explain why Equation B is not directly proportional.

  19. Describe the direction of the graph of Equation C.

  20. Write a conclusion explaining how direct variation can be used to plan purchases and predict costs.

  Investigation Check

  Notebooks

  • Unit price = $3.50

  • Constant of variation = 3.5

  • Equation: C = 3.5n

  • Cost of 20 notebooks = $70

  • Number of notebooks for $105 = 30

  Pencils

  • Equation: C = 0.75p

  • Cost of 48 pencils = $36

  • Number of pencils for $30 = 40

  Comparing Relationships

  • Equation A represents direct variation with k = 4.

  • Equation B is not direct variation because it has a nonzero y-intercept.

  • Equation C represents direct variation with k = −2.

  • Equation C has a negative slope.

4. Introduction to Linear Relationships

Introduction to Linear Relationships

Introduction

Mathematics helps us recognise patterns, describe relationships, and predict what will happen when quantities change.

Many everyday situations involve two quantities that are connected. For example:

• The number of hours worked and the money earned.

• The number of notebooks purchased and their total cost.

• The distance travelled and the time spent moving.

• The amount of water in a tank as it fills.

• The total cost of a taxi ride as the distance increases.

When one quantity changes at a constant rate compared with another, the relationship is called a linear relationship.

Linear relationships can be represented using tables, equations, graphs, and written descriptions.

Learning to recognise linear patterns allows us to create mathematical rules and use them to make predictions.

What Is a Linear Relationship?

A linear relationship is a relationship between two variables that has a constant rate of change.

A linear relationship can be represented by an equation of the form:

y = mx + b

where:

• x = independent variable or input

• y = dependent variable or output

• m = constant rate of change, or slope

• b = initial value, or y-intercept

Example

A taxi charges a $4 starting fee plus $2 for every kilometre travelled.

The total cost can be represented by:

C = 2d + 4

where:

• C = total cost in dollars

• d = distance travelled in kilometres

• 2 = cost per kilometre

• 4 = starting fee

This is a linear relationship because the total cost increases by $2 for every additional kilometre.

Important Observation

A linear relationship does not have to begin at zero.

The starting value may be positive, negative, or zero.

Understanding Input and Output

A mathematical relationship often connects an input to an output.

The input is the value we choose or measure.

The output is the value calculated from the input.

Example

Consider the rule:

y = 3x + 2

If x = 1:

y = 3(1) + 2 = 5

If x = 2:

y = 3(2) + 2 = 8

If x = 3:

y = 3(3) + 2 = 11

Input-Output Table

Observations

• Every increase of 1 in x produces an increase of 3 in y.

• The output starts at 2 when x = 0.

• The relationship has a constant rate of change.

Therefore, the relationship is linear.

Recognising Patterns in Input-Output Tables

An input-output table organises corresponding values of two variables.

Example

Examine the changes in the output values.

6 → 10 → 14 → 18 → 22

Each output increases by 4.

Since the inputs increase by 1 each time, the rate of change is:

m = 4

Finding the Rule

Start with:

y = 4x + b

Use the point (1, 6):

6 = 4(1) + b

6 = 4 + b

b = 2

Therefore:

y = 4x + 2

Checking

When x = 5:

y = 4(5) + 2

y = 22

The equation matches the table.

Constant Rate of Change

The rate of change describes how much the output changes compared with the input.

For a linear relationship:

Rate of change = Change in output ÷ Change in input

Using symbols:

m = Δy/Δx

where:

• Δy = change in y

• Δx = change in x

Worked Example

Consider:

Step 1: Calculate the Change in x

4 − 2 = 2

Step 2: Calculate the Change in y

13 − 7 = 6

Step 3: Calculate the Rate

m = 6/2

m = 3

Check Another Interval

(19 − 13)/(6 − 4) = 6/2 = 3

The rate of change remains constant.

Therefore, the relationship is linear.

Important Observation

When input values increase by more than 1, the output differences must be divided by the corresponding input differences.

Recognising Linear and Nonlinear Patterns

Not all numerical patterns are linear.

Linear Pattern

The output increases by 3 for every increase of 1 in x.

The rate of change is constant.

Nonlinear Pattern

The output differences are:

3, 5, 7

The differences are not constant.

Therefore, the relationship is nonlinear.

The equation is:

y = x²

Important Observation

A linear pattern has a constant rate of change, not simply an increasing or decreasing sequence of outputs.

Finding Algebraic Rules from Tables

An algebraic rule is an equation describing how the output depends on the input.

Worked Example

Determine the rule for:

Step 1: Identify the Rate of Change

The output increases by 3 whenever x increases by 1.

Therefore:

m = 3

Step 2: Identify the Initial Value

When x = 0:

y = 5

Therefore:

b = 5

Step 3: Write the Equation

y = 3x + 5

Step 4: Verify

When x = 4:

y = 3(4) + 5

y = 17

The rule matches the table.

Finding a Rule When the Table Does Not Include Zero

Sometimes a table does not contain x = 0.

In this situation, use a known input-output pair to find the initial value.

Worked Example

Step 1: Calculate the Rate of Change

m = (19 − 11)/(4 − 2)

m = 8/2

m = 4

Step 2: Write the General Equation

y = 4x + b

Step 3: Substitute a Known Point

Use (2, 11):

11 = 4(2) + b

11 = 8 + b

b = 3

Final Equation

y = 4x + 3

Verification

When x = 8:

y = 4(8) + 3

y = 35

The equation is correct.

Positive, Negative, and Zero Rates of Change

Linear relationships can increase, decrease, or remain constant.

Positive Rate of Change

y = 2x + 1

As x increases, y increases.

The graph rises from left to right.

Negative Rate of Change

y = −3x + 10

As x increases, y decreases.

The graph falls from left to right.

Zero Rate of Change

y = 5

The output remains 5 regardless of the input.

The graph is horizontal.

Comparison

Exploring Linear Relationships Graphically

A linear equation produces a straight-line graph.

The slope determines the direction and steepness of the line.

The y-intercept determines where the line crosses the y-axis.

Questions to Consider

• What happens when the slope increases?

• What happens when the slope becomes negative?

• What happens when the slope is zero?

• What happens when the initial value changes?

• Why does the graph remain a straight line?

Linear Relationships and Direct Variation

Direct variation is a special type of linear relationship.

Direct Variation

y = kx

The initial value is zero.

The graph passes through the origin.

General Linear Relationship

y = mx + b

The initial value may be nonzero.

Example

A worker earns $15 per hour without any starting bonus.

E = 15h

This is direct variation.

Another worker earns $15 per hour plus a $20 starting bonus.

E = 15h + 20

This is linear but not directly proportional.

Important Observation

All direct variation relationships are linear, but not all linear relationships are direct variation.

Predicting Values Using Linear Rules

Once a linear equation is known, it can be used to predict outputs for different inputs.

Worked Example

A machine produces 12 items per minute and has already produced 30 items.

The total number of items is:

P = 12t + 30

where t is the number of additional minutes.

Predict Production After 8 Minutes

P = 12(8) + 30

P = 96 + 30

P = 126 items

Determine the Time Needed to Reach 210 Items

210 = 12t + 30

180 = 12t

t = 15 minutes

Important Observation

Linear equations can be used to calculate either an output or an input.

Linear Relationships in Shopping

Many shopping situations can be represented using linear equations.

Worked Example

A store charges $6 for delivery plus $4 per item.

Let:

n = number of items

C = total cost in dollars

Equation

C = 4n + 6

Create a Table

Interpretation

• Rate of change = $4 per item.

• Initial value = $6 delivery charge.

Prediction

For 10 items:

C = 4(10) + 6

C = $46

Linear Relationships in Transportation

Transportation costs often contain both a fixed fee and a variable fee.

Worked Example

A taxi charges:

• $5 starting fee

• $2.50 per kilometre

Equation

C = 2.5d + 5

where:

• C = cost in dollars

• d = distance in kilometres

Calculate the Cost of a 12 km Journey

C = 2.5(12) + 5

C = 30 + 5

C = $35

Find the Distance for a $55 Fare

55 = 2.5d + 5

50 = 2.5d

d = 20 km

Interpretation

The slope represents the cost per kilometre.

The y-intercept represents the starting fee.

Linear Relationships in Science

Linear relationships appear in many scientific situations.

Temperature Conversion

The relationship between Celsius and Fahrenheit temperatures is:

F = 1.8C + 32

where:

• F = temperature in degrees Fahrenheit

• C = temperature in degrees Celsius

Worked Example

Convert 25°C to Fahrenheit.

F = 1.8(25) + 32

F = 45 + 32

F = 77°F

Interpretation

The slope is 1.8.

The initial value is 32, meaning 0°C corresponds to 32°F.

Important Observation

Temperature conversion is linear but not directly proportional because the equation contains a nonzero initial value.

Linear Relationships in Water Usage

Suppose a water tank initially contains 150 litres.

Water flows into the tank at a constant rate of 20 litres per minute.

The amount of water after t minutes is:

V = 20t + 150

Worked Example

Find the volume after 6 minutes.

V = 20(6) + 150

V = 120 + 150

V = 270 L

Finding the Time

How long will it take to reach 350 litres?

350 = 20t + 150

200 = 20t

t = 10 minutes

Important Observation

This model is valid while the water flows at a constant rate and the tank has sufficient capacity.

Using Two Points to Determine a Linear Rule

Sometimes we know only two points on a line.

We can use these points to calculate the slope.

Formula

m = (y₂ − y₁)/(x₂ − x₁)

Worked Example

A linear relationship passes through:

(2, 7) and (6, 19)

Step 1: Calculate the Slope

m = (19 − 7)/(6 − 2)

m = 12/4

m = 3

Step 2: Write the General Equation

y = 3x + b

Step 3: Substitute a Known Point

Use (2, 7):

7 = 3(2) + b

7 = 6 + b

b = 1

Final Equation

y = 3x + 1

Verification

When x = 6:

y = 3(6) + 1

y = 19

The equation matches the second point.

Comparing Linear Relationships

Two linear relationships can be compared by examining their slopes and initial values.

Example

Company A charges:

C = 5x + 10

Company B charges:

C = 7x + 2

where x is the number of items purchased.

Comparing Initial Values

Company A starts at $10.

Company B starts at $2.

Comparing Rates

Company A increases by $5 per item.

Company B increases by $7 per item.

Finding Equal Costs

5x + 10 = 7x + 2

8 = 2x

x = 4

At x = 4:

Company A:

5(4) + 10 = $30

Company B:

7(4) + 2 = $30

Interpretation

Both companies charge $30 for 4 items.

For fewer than 4 items, Company B is cheaper.

For more than 4 items, Company A is cheaper.

Checking Linear Rules

After finding an algebraic rule, it should be verified.

Example

Suppose a table suggests:

y = 4x − 3

Check the point (5, 17).

Substitute:

y = 4(5) − 3

y = 20 − 3

y = 17

The equation matches the point.

Additional Checks

• Does the equation match every point in the table?

• Is the rate of change constant?

• Does the initial value make sense?

• Are the units appropriate?

• Are predictions reasonable within the situation?

Important Observation

Checking more than one point reduces the chance of accepting an incorrect rule.

A General Method for Finding Linear Rules

Common Mistakes

Assuming every pattern is linear

Check that the rate of change remains constant.

Looking only at output differences

If the input intervals differ, calculate Δy/Δx.

Confusing slope with initial value

The slope describes the rate of change; the initial value describes the output when x = 0.

Assuming all linear relationships are proportional

A proportional relationship must pass through the origin.

Making sign errors

A decreasing relationship has a negative slope.

Forgetting to calculate b

When the table does not include x = 0, use a known point to determine the initial value.

Using the wrong variables

Define inputs and outputs clearly.

Ignoring units

Rates may be measured in dollars per item, litres per minute, or kilometres per hour.

Making unrealistic predictions

Linear models may only apply over a particular range.

Failing to verify

Substitute known values into the equation to check the rule.

Key Terms

• Linear Relationship: Relationship with a constant rate of change.

• Input: Independent variable used in a mathematical rule.

• Output: Dependent variable produced by a mathematical rule.

• Input-Output Table: Table displaying corresponding values of two variables.

• Independent Variable: Variable that is chosen or controlled.

• Dependent Variable: Variable whose value depends on another.

• Algebraic Rule: Equation describing a relationship between variables.

• Rate of Change: Change in output divided by change in input.

• Slope: Constant rate of change of a straight line.

• Initial Value: Output when the input equals zero.

• y-Intercept: Point where a graph crosses the y-axis.

• Slope-Intercept Form: y = mx + b.

• Positive Slope: Rate of change greater than zero.

• Negative Slope: Rate of change less than zero.

• Zero Slope: Rate of change equal to zero.

• Direct Variation: Linear relationship represented by y = kx.

• Constant of Proportionality: Fixed multiplier in a proportional relationship.

• Linear Equation: Equation whose graph is a straight line.

• Nonlinear Relationship: Relationship without a constant rate of change.

• Ordered Pair: Pair of coordinates written as (x, y).

• Prediction: Calculated estimate using a mathematical rule.

• Substitution: Replacing a variable with a known value.

• Verification: Checking whether an equation matches the given data.

Key Takeaways

• Linear relationships have a constant rate of change.

• Input-output tables help identify numerical patterns.

• A linear equation can be written as y = mx + b.

• The slope m represents the rate of change.

• The initial value b represents the output when x = 0.

• A positive slope produces an increasing relationship.

• A negative slope produces a decreasing relationship.

• A zero slope produces a constant output.

• Linear graphs are straight lines.

• Direct variation is a special type of linear relationship.

• A linear relationship does not have to pass through the origin.

• The rate of change can be calculated using Δy/Δx.

• Two points with different x-values determine a unique linear rule.

• Linear equations can be used to predict unknown values.

• Mathematical rules should be verified using known data.

• Real-world predictions must be interpreted using appropriate units and assumptions.

Check Your Understanding

Recognising Patterns

1. What is a linear relationship?

2. What does the slope represent?

3. What does the y-intercept represent?

4. Is y = 5x + 3 linear?

5. Is y = x² + 2 linear?

Finding Rules

6. Find the rule for x = 0, 1, 2, 3 and y = 4, 7, 10, 13.

7. Find the rule for x = 0, 1, 2, 3 and y = 10, 8, 6, 4.

8. Find the slope between (2, 5) and (4, 11).

9. Find the equation of a line with slope 4 and y-intercept −2.

10. Find the equation passing through (0, 6) and (3, 15).

Predicting Values

11. If y = 3x + 5, find y when x = 8.

12. If y = −2x + 12, find y when x = 4.

13. If y = 5x − 3, find x when y = 22.

14. If C = 4n + 6, find the cost when n = 10.

15. If V = 15t + 100, find the volume when t = 8.

Understanding Linear Relationships

16. Explain why y = 4x is directly proportional.

17. Explain why y = 4x + 5 is not directly proportional.

18. What does a negative slope indicate?

19. How can you determine whether a table represents a linear relationship?

20. Why is it important to verify a linear equation?

Answer Key

Extended Challenge

1. Find the linear rule for the points (2, 9), (4, 15), (6, 21), and (8, 27).

2. Determine whether the points (1, 4), (2, 7), (3, 10), and (4, 13) form a linear relationship.

3. Find the equation of the line through (3, 11) and (7, 27).

4. A taxi charges $6 initially and $2.50 per kilometre. Write an equation and find the cost of a 14 km trip.

5. A tank begins with 250 litres and loses 12 litres per minute. Write an equation for its volume and determine when it becomes empty, assuming the rate stays constant.

6. Company A charges $8 per item plus $15 delivery. Company B charges $10 per item plus $5 delivery. Determine when the costs are equal.

7. A worker earns $18 per hour plus a $50 bonus. Write an equation and determine earnings after 12 hours.

8. Determine the rate of change for the points (2, 20) and (8, 8).

9. Find the equation of a line with slope −3 passing through (4, 7).

10. Explain how you would distinguish a linear relationship from a quadratic relationship using an input-output table.

Final Investigation: Planning a School Event

A school is organising an event and needs to compare costs from two suppliers.

Supplier A

Supplier A charges:

• $5 per student

• $40 fixed booking fee

Supplier B

Supplier B charges:

• $7 per student

• $10 fixed booking fee

Part A: Constructing Linear Models

1. Define the independent variable.

2. Define the dependent variable.

3. Write an equation for Supplier A.

4. Write an equation for Supplier B.

5. Identify the slope and initial value of each equation.

Part B: Creating Tables

6. Calculate the cost for 0 students.

7. Calculate the cost for 10 students.

8. Calculate the cost for 20 students.

9. Calculate the cost for 30 students.

10. Organise the results in an input-output table.

Part C: Comparing Relationships

11. Graph both equations on the same coordinate plane.

12. Determine which supplier has the greater rate of change.

13. Determine which supplier has the lower starting fee.

14. Calculate the number of students at which both suppliers charge the same amount.

15. Determine the common cost at that number of students.

Part D: Predictions and Interpretation

16. Calculate the cost from each supplier for 50 students.

17. Determine which supplier is cheaper for 50 students.

18. Explain which supplier is better for smaller groups.

19. Explain which supplier is better for larger groups.

20. Write a conclusion explaining how linear equations can help organisers make informed decisions.

Investigation Check

Supplier A

C = 5n + 40

Supplier B

C = 7n + 10

Equal Costs

5n + 40 = 7n + 10

30 = 2n

n = 15 students

At 15 students:

C = 5(15) + 40

C = $115

For 50 Students

Supplier A:

5(50) + 40 = $290

Supplier B:

7(50) + 10 = $360

Supplier A is cheaper by $70.

Conclusion

Supplier B is cheaper for fewer than 15 students, both cost the same for 15 students, and Supplier A is cheaper for more than 15 students.

5. Modeling Relationships

Modeling Relationships

Introduction

Mathematics is not only about calculating answers. It is also about finding patterns, describing relationships, and using those relationships to understand the world.

A mathematical model is a mathematical representation of a real-world situation. Models can take the form of equations, tables, graphs, diagrams, or formulas.

For example, we can use mathematical models to:

• Calculate the cost of buying several items.

• Predict how far a vehicle will travel.

• Estimate the amount of water in a tank.

• Compare different mobile phone plans.

• Calculate earnings from working a certain number of hours.

• Predict the growth of a population.

• Determine whether a business is making a profit.

An algebraic model allows us to describe how one quantity depends on another and make predictions without calculating every possible value individually.

What Is a Mathematical Model?

A mathematical model uses mathematics to represent a situation or relationship.

A model may be an equation, graph, table, or other mathematical representation.

Example

A notebook costs $3.

If n represents the number of notebooks purchased and C represents the total cost, then:

C = 3n

This equation is a mathematical model.

It describes how the total cost changes with the number of notebooks purchased.

Using the Model

If 5 notebooks are purchased:

C = 3(5)

C = $15

If 12 notebooks are purchased:

C = 3(12)

C = $36

Important Observation

A useful mathematical model allows us to calculate unknown values and make predictions.

However, a model is only reliable when its assumptions match the real situation.

Understanding Variables

A variable is a symbol representing a quantity that can change or is unknown.

In mathematical models, variables often describe quantities that depend on one another.

Independent Variable

The independent variable is the input quantity.

It is usually represented by x.

Dependent Variable

The dependent variable is the output quantity.

It is usually represented by y.

Example

A worker earns $18 per hour.

Let:

• h = number of hours worked

• E = total earnings in dollars

The model is:

E = 18h

The number of hours worked is the independent variable.

The total earnings are the dependent variable.

Important Observation

The dependent variable changes according to the value of the independent variable.

Creating Algebraic Rules from Words

Real-world situations can often be translated into algebraic equations.

Common Mathematical Language

Worked Example

A swimming pool charges a $6 entrance fee plus $2 for each hour spent swimming.

Let:

h = number of hours

C = total cost

Step 1: Identify the Fixed Amount

Entrance fee = $6

Step 2: Identify the Variable Amount

Hourly charge = $2

Step 3: Write the Equation

C = 2h + 6

Step 4: Interpret

The total cost begins at $6 and increases by $2 for every additional hour.

Recognising Relationships from Tables

Tables can reveal patterns that help us construct algebraic rules.

Example

Step 1: Find the Pattern

The output increases by 3 whenever the input increases by 1.

Therefore, the rate of change is 3.

Step 2: Find the Initial Value

When x = 0:

y = 4

The initial value is 4.

Step 3: Write the Rule

y = 3x + 4

Step 4: Verify

When x = 4:

y = 3(4) + 4

y = 16

The rule matches the table.

Important Observation

A constant rate of change suggests a linear relationship.

Linear Mathematical Models

A linear model has the general form:

y = mx + b

where:

• m = constant rate of change

• b = initial value

Example

A taxi charges $5 to begin a journey and $2 per kilometre.

The model is:

C = 2d + 5

where:

• C = total cost in dollars

• d = distance in kilometres

Interpretation

• The slope is 2, representing $2 per kilometre.

• The initial value is 5, representing the starting fee.

Predicting Cost

For a 12 km journey:

C = 2(12) + 5

C = 24 + 5

C = $29

Important Observation

Linear models are appropriate when the rate of change remains constant.

Visualising a Linear Model

Consider the model:

C = 3x + 5

This could represent a $5 fixed fee plus $3 for each item.

Direct Variation Models

Some relationships are directly proportional.

A direct variation model has the form:

y = kx

where k is the constant of proportionality.

Example

A machine produces 25 items per minute.

Let:

P = number of items produced

t = time in minutes

The model is:

P = 25t

Predicting Production

After 8 minutes:

P = 25(8)

P = 200 items

Important Observation

A direct variation graph passes through the origin because the output is zero when the input is zero.

Comparing Direct Variation and General Linear Models

Example

Buying pencils at $2 each:

C = 2n

This is direct variation.

Buying pencils at $2 each with a $5 delivery fee:

C = 2n + 5

This is linear but not directly proportional.

Creating Models from Two Data Points

Sometimes we are given two observations rather than an equation.

If a linear relationship is appropriate, two points with different x-values determine the equation.

Worked Example

A water tank contains 120 litres after 2 minutes and 200 litres after 6 minutes.

Assume water enters at a constant rate.

Step 1: Identify the Points

(2, 120)

(6, 200)

Step 2: Calculate the Rate of Change

m = (200 − 120)/(6 − 2)

m = 80/4

m = 20 L/min

Step 3: Write the General Equation

V = 20t + b

Step 4: Find the Initial Volume

Use (2, 120):

120 = 20(2) + b

120 = 40 + b

b = 80

Final Model

V = 20t + 80

Interpretation

The tank initially contained 80 litres and gains 20 litres per minute.

Using Models to Make Predictions

Once a mathematical model has been constructed, it can be used to predict unknown values.

Worked Example

A company charges $15 per month plus $0.10 per text message.

The model is:

C = 15 + 0.10n

where n is the number of text messages.

Predict the Cost of 120 Messages

C = 15 + 0.10(120)

C = 15 + 12

C = $27

Determine the Number of Messages for a $35 Bill

35 = 15 + 0.10n

20 = 0.10n

n = 200 messages

Important Observation

A mathematical model can be used to find either the input or the output, depending on the information given.

Interpreting the Meaning of Slope

The slope describes how quickly one quantity changes relative to another.

Example

A tank loses 12 litres of water per minute.

The volume is modelled by:

V = 300 − 12t

where:

• V = volume in litres

• t = time in minutes

The slope is −12.

Interpretation

The negative slope indicates that the volume decreases by 12 litres every minute.

Prediction

After 10 minutes:

V = 300 − 12(10)

V = 180

180 litres remain.

Important Observation

A negative slope represents a decreasing relationship.

Interpreting the Initial Value

The initial value represents the output when the input is zero.

Example

A savings account begins with $250.

A person deposits $40 every week.

The model is:

S = 40w + 250

where:

• S = savings in dollars

• w = number of weeks

Interpretation

• Slope = 40 dollars per week.

• Initial value = $250.

Prediction

After 12 weeks:

S = 40(12) + 250

S = 480 + 250

S = $730

Important Observation

The initial value often represents a starting amount, fixed fee, or initial measurement.

Modeling Relationships in Shopping

Shopping situations often involve constant unit prices and fixed charges.

Worked Example

A store sells notebooks for $4 each and charges $6 for delivery.

Step 1: Define Variables

n = number of notebooks

C = total cost

Step 2: Write the Model

C = 4n + 6

Step 3: Predict the Cost of 15 Notebooks

C = 4(15) + 6

C = 60 + 6

C = $66

Step 4: Evaluate the Model

The model assumes:

• Each notebook costs $4.

• Delivery always costs $6.

• No discounts apply.

• No additional taxes or charges apply.

If any assumption changes, the model may need to be revised.

Modeling Relationships in Transportation

Transportation provides many examples of mathematical modeling.

Worked Example

A taxi company charges:

• $4 starting fee

• $2.50 per kilometre

Equation

C = 2.5d + 4

Prediction

For a 16 km journey:

C = 2.5(16) + 4

C = 40 + 4

C = $44

Model Limitations

The model may not account for:

• Waiting time.

• Traffic surcharges.

• Toll roads.

• Different rates at night.

• Special discounts.

Important Observation

A mathematical model simplifies reality by focusing on selected variables.

Modeling Relationships in Science

Mathematical models are widely used in science.

Example: Constant-Speed Motion

If an object travels at constant speed:

d = vt

where:

• d = distance

• v = speed

• t = time

Worked Example

A cyclist travels at 18 km/h.

The model is:

d = 18t

After 2.5 hours:

d = 18(2.5)

d = 45 km

Evaluating the Model

The model assumes constant speed.

In reality, the cyclist may:

• Stop for breaks.

• Travel uphill or downhill.

• Encounter traffic.

• Change speed due to fatigue.

The model is useful for estimating distance but may not describe every moment of the journey accurately.

Modeling Temperature Changes

Some temperature changes can be approximated by linear models over short periods.

Example

A liquid begins at 80°C and cools by approximately 3°C per minute during a short observation period.

The model is:

T = 80 − 3t

where:

• T = temperature in °C

• t = time in minutes

Prediction

After 5 minutes:

T = 80 − 3(5)

T = 65°C

Evaluating the Model

The model may work over the measured interval.

However, real cooling often slows as the liquid approaches room temperature.

Therefore, the model should not be assumed accurate indefinitely.

Important Observation

A model can be useful over a limited range even if it does not describe the entire process perfectly.

Comparing Mathematical Models

Sometimes two different equations describe competing choices.

Worked Example

Two internet providers offer different monthly plans.

Provider A:

C = 5d + 20

Provider B:

C = 3d + 36

where d represents additional data usage in GB and C represents the monthly cost in dollars.

Step 1: Compare Initial Costs

Provider A starts at $20.

Provider B starts at $36.

Step 2: Compare Rates

Provider A charges $5 per GB.

Provider B charges $3 per GB.

Step 3: Find Equal Costs

5d + 20 = 3d + 36

2d = 16

d = 8 GB

Step 4: Calculate the Common Cost

C = 5(8) + 20

C = $60

Interpretation

• Below 8 GB, Provider A is cheaper.

• At 8 GB, both cost $60.

• Above 8 GB, Provider B is cheaper.

Important Observation

Comparing models allows us to identify the best option under different conditions.

Recognising When a Linear Model Is Appropriate

A linear model is useful when the rate of change is constant.

Example: Linear Data

The volume increases by 15 litres each minute.

The model is:

V = 15t + 20

Example: Nonlinear Data

The output doubles each time.

The differences are not constant.

Therefore, a linear model is not appropriate.

This pattern can be represented by:

y = 2 × 2ˣ

Important Observation

A model should be selected based on the behaviour of the data, not simply because an equation is easy to use.

Evaluating Whether a Model Fits the Data

A mathematical model should be compared with real observations.

Example

Suppose a model predicts:

y = 3x + 2

Most predictions match the observations exactly.

At x = 3, the model predicts 11 while the actual value is 12.

The difference is:

12 − 11 = 1

This difference is called a residual.

Residual Formula

Residual = Actual value − Predicted value

Interpretation

Small residuals may indicate a good fit, especially when compared with the scale of the measurements.

However, a single small residual does not prove that a model is reliable.

Visualising Model Fit

Interpolation and Extrapolation

Mathematical models can be used to predict values within or beyond the observed data.

Interpolation

Interpolation means estimating a value within the range of observed data.

For example, if measurements were collected between 0 and 10 minutes, predicting a value at 6 minutes is interpolation.

Extrapolation

Extrapolation means predicting a value outside the observed range.

For example, using the same data to predict a value at 50 minutes is extrapolation.

Important Observation

Extrapolation is often less reliable because the relationship may change outside the observed range.

Evaluating the Reasonableness of Predictions

A mathematical model should produce answers that make sense in context.

Example

A water tank begins with 100 litres and loses 5 litres per minute.

The model is:

V = 100 − 5t

After 10 minutes:

V = 100 − 5(10)

V = 50 L

This is reasonable.

After 30 minutes:

V = 100 − 5(30)

V = −50 L

A negative volume is not physically possible.

Interpretation

The model is only valid until the tank becomes empty.

Set V = 0:

0 = 100 − 5t

5t = 100

t = 20 minutes

The physically meaningful domain is:

0 ≤ t ≤ 20

Important Observation

A mathematically correct calculation can produce an unrealistic answer if the model is used outside its valid domain.

Assumptions and Limitations of Mathematical Models

Mathematical models simplify real-world situations.

A model may assume:

• Prices remain constant.

• Speed does not change.

• There are no unexpected fees.

• Measurements are accurate.

• Conditions remain stable.

• Relationships continue in the same pattern.

Questions to Ask

• Does the model match the available data?

• Is the rate of change constant?

• Are important variables missing?

• Is the prediction within the observed range?

• Are the units correct?

• Does the answer make practical sense?

• Could changing conditions affect the result?

Important Observation

A useful model does not have to be perfect, but its assumptions and limitations should be understood.

A General Method for Creating Mathematical Models

Common Mistakes

Using the wrong type of model

Not every relationship is linear.

Confusing independent and dependent variables

Clearly identify which quantity depends on the other.

Ignoring the initial value

Some models include a fixed starting amount.

Calculating the rate of change incorrectly

Use change in output divided by change in input.

Assuming all linear models are directly proportional

A linear model may have a nonzero initial value.

Failing to check known values

A model should agree with the data used to construct it.

Using unrealistic predictions

A model may not remain valid indefinitely.

Ignoring units

A rate may represent dollars per item, litres per minute, or kilometres per hour.

Confusing interpolation and extrapolation

Predictions beyond observed data require additional caution.

Assuming models are exact

Real-world measurements and relationships may contain uncertainty.

Key Terms

• Mathematical Model: Mathematical representation of a real-world situation.

• Algebraic Rule: Equation describing a relationship between variables.

• Variable: Symbol representing an unknown or changing quantity.

• Independent Variable: Input quantity.

• Dependent Variable: Output quantity.

• Linear Model: Model with a constant rate of change.

• Direct Variation: Relationship represented by y = kx.

• Rate of Change: Change in output divided by change in input.

• Slope: Rate of change of a linear relationship.

• Initial Value: Output when the input equals zero.

• y-Intercept: Point where a graph crosses the y-axis.

• Prediction: Estimate calculated using a model.

• Interpolation: Estimating within the observed data range.

• Extrapolation: Predicting beyond the observed data range.

• Residual: Difference between an actual value and its predicted value.

• Model Fit: How closely a model represents observed data.

• Assumption: Condition accepted as true when constructing a model.

• Limitation: Restriction affecting the accuracy or usefulness of a model.

• Domain: Set of permitted input values.

• Range: Set of possible output values.

• Verification: Checking whether a model agrees with known information.

• Proportional Relationship: Relationship where one quantity is a constant multiple of another.

• Nonlinear Relationship: Relationship without a constant rate of change.

Key Takeaways

• Mathematical models describe relationships between quantities.

• Algebraic rules can represent real-world situations.

• Variables represent quantities that can change.

• Linear models have the form y = mx + b.

• The slope describes the rate of change.

• The initial value represents the output when the input is zero.

• Direct variation is a special type of linear relationship.

• Tables and graphs can help identify mathematical patterns.

• Models can be used to predict unknown values.

• Equations can be rearranged to calculate inputs or outputs.

• Real-world models often depend on simplifying assumptions.

• A model should be checked against observed data.

• Residuals measure differences between observed and predicted values.

• Interpolation estimates within the observed range.

• Extrapolation predicts beyond the observed range.

• A model may be valid only within a particular domain.

• Predictions should be evaluated for reasonableness.

Check Your Understanding

Understanding Mathematical Models

1. What is a mathematical model?

2. What is the difference between an independent and dependent variable?

3. What does the slope represent in a linear model?

4. What does the initial value represent?

5. Why are mathematical models useful?

Creating Algebraic Rules

6. A notebook costs $5. Write an equation for the cost of n notebooks.

7. A taxi charges $4 initially plus $3 per kilometre. Write the equation.

8. A worker earns $20 per hour plus a $50 bonus. Write an equation for earnings.

9. A tank begins with 200 litres and loses 10 litres per minute. Write an equation for volume.

10. Find the rule for the points (0, 3), (1, 7), (2, 11), and (3, 15).

Making Predictions

11. If C = 4n + 6, find C when n = 8.

12. If V = 100 − 5t, find V when t = 12.

13. If E = 18h + 30, find E when h = 10.

14. If d = 60t, find d when t = 2.5.

15. If y = 3x + 7, find x when y = 28.

Evaluating Models

16. Why might a linear model fail to predict long-term population growth?

17. What is interpolation?

18. What is extrapolation?

19. What is a residual?

20. Why should mathematical predictions be checked for reasonableness?

Answer Key

Extended Challenge

1. A delivery company charges $8 plus $2.50 per kilometre. Construct a model and calculate the cost of a 24 km delivery.

2. A tank contains 500 litres and loses 15 litres per minute. Write a model and determine when it becomes empty.

3. A worker earns $22 per hour plus a $75 bonus. Write a model and determine earnings after 16 hours.

4. Find the linear equation passing through (3, 14) and (7, 30).

5. A machine produces 180 items in 6 minutes at a constant rate. Construct a model for production.

6. Two companies charge C₁ = 4x + 25 and C₂ = 6x + 5. Determine when their costs are equal.

7. A model predicts y = 5x + 2. If the actual value at x = 4 is 25, calculate the residual.

8. Explain why a linear model may be unsuitable for a population that doubles every year.

9. A liquid cools from 90°C to 70°C in 5 minutes. Assuming a constant cooling rate over a short interval, construct a model.

10. Describe three factors that might limit the accuracy of a mathematical model.

Final Investigation: Comparing School Transport Options

A school is planning transportation for an educational trip.

Two bus companies offer different pricing structures.

Company A

• Fixed booking fee: $80

• Cost per student: $6

Company B

• Fixed booking fee: $140

• Cost per student: $4

Part A: Constructing Models

1. Identify the independent variable.

2. Identify the dependent variable.

3. Write an equation for Company A.

4. Write an equation for Company B.

5. Identify the slope and initial value of each equation.

Part B: Making Predictions

6. Calculate the cost for 10 students using each company.

7. Calculate the cost for 20 students.

8. Calculate the cost for 30 students.

9. Calculate the cost for 40 students.

10. Organise the results in a table.

Part C: Comparing Models

11. Graph both equations.

12. Determine which company has the lower starting fee.

13. Determine which company has the lower cost per student.

14. Calculate the number of students at which both companies charge the same amount.

15. Determine the cost at that number of students.

Part D: Evaluating the Models

16. Determine which company is cheaper for 50 students.

17. Calculate the difference in cost for 50 students.

18. Explain which company is more suitable for smaller groups.

19. Identify two assumptions made by these pricing models.

20. Write a conclusion explaining how mathematical models help schools make informed decisions.

Investigation Check

Company A

C = 6n + 80

Company B

C = 4n + 140

Equal Costs

6n + 80 = 4n + 140

2n = 60

n = 30 students

At 30 students:

C = 6(30) + 80

C = $260

For 50 Students

Company A:

C = 6(50) + 80 = $380

Company B:

C = 4(50) + 140 = $340

Company B is cheaper by $40.

Conclusion

Company A is cheaper for fewer than 30 students. Both companies charge the same amount for 30 students. Company B is cheaper for more than 30 students.