Direct proportion and scale - Year 9 Mathematics

What you will learn

identify direct proportion and find the constant of proportionality $k$ ,
recognise proportional relationships from tables and graphs (straight line through the origin),
use scale factors to interpret and create scale drawings and maps,
solve problems involving rates such as speed, density, flow rate, and exchange rates,
build and apply proportional models in practical contexts.

Worked example 0 Real-world example: currency exchange

An exchange rate is 1 AUD $= 0.65$ USD. You want to buy an item priced at 120 USD. How many Australian dollars do you need?

The relationship is directly proportional: USD $= 0.65 \times$ AUD, so $k = 0.65$ .
Rearrange: AUD $= \dfrac{\text{USD}}{0.65}$ .
AUD $= \dfrac{120}{0.65} \approx 184.62$ .

Key idea: the exchange rate is the constant of proportionality. As long as the rate stays fixed, you can convert any amount by multiplying or dividing by $k$ .

1. Direct proportion

Two quantities $x$ and $y$ are directly proportional if:

Direct proportion

y = kx

where $k$ is the constant of proportionality. Equivalently, $\dfrac{y}{x} = k$ for all pairs of values.

How to find $k$ : given any pair $(x, y)$ where $x \neq 0$ , calculate $k = \dfrac{y}{x}$ .

Worked example 1 Finding the constant k

The cost of petrol is directly proportional to the number of litres purchased. 25 litres costs $47.50. Find $k$ and write the equation.

$k = \dfrac{47.50}{25} = 1.90$ .
The equation is $C = 1.90L$ , where $C$ is the cost in dollars and $L$ is litres.
Check: when $L = 25$ , $C = 1.90 \times 25 = 47.50$ . Correct.

Worked example 2 Testing for direct proportion from a table

Does the following table show direct proportion?

$x$	2	5	8	10
$y$	6	15	24	30

Calculate $\dfrac{y}{x}$ for each pair: $\dfrac{6}{2} = 3$ , $\dfrac{15}{5} = 3$ , $\dfrac{24}{8} = 3$ , $\dfrac{30}{10} = 3$ .
The ratio is constant ( $k = 3$ ), so $y$ is directly proportional to $x$ .
The equation is $y = 3x$ .

2. Recognising proportional relationships from graphs

A directly proportional relationship produces a straight line through the origin on a Cartesian plane. The gradient of the line equals $k$ .

Proportional (through origin) vs non-proportional (y-intercept not zero).

Worked example 3 Reading k from a graph

A graph of distance vs time passes through the origin and through the point $(4, 20)$ . Find the speed.

Since the graph is a straight line through the origin, distance is directly proportional to time.
$k = \dfrac{20}{4} = 5$ .
The speed is 5 units of distance per unit of time (e.g. 5 km/h).

3. Scale drawings and maps

A scale drawing represents an object where all lengths are multiplied by the same scale factor.

Scale factor

\text{Scale factor} = \frac{\text{drawing length}}{\text{actual length}}

A scale of 1 : $n$ means 1 unit on the drawing represents $n$ units in reality.

Worked example 4 Using a map scale

A map has a scale of 1 : 50 000. Two towns are 8.4 cm apart on the map. Find the actual distance.

Actual distance $= 8.4 \times 50\,000 = 420\,000$ cm.
Convert: $420\,000$ cm $= 4\,200$ m $= 4.2$ km.

Worked example 5 Creating a scale drawing

A room is 6 m by 4.5 m. Draw it at a scale of 1 : 100.

Drawing length $= \dfrac{6 \times 100}{100} = 6$ cm (since 6 m $= 600$ cm, and $600 \div 100 = 6$ cm).
Drawing width $= \dfrac{4.5 \times 100}{100} = 4.5$ cm.
Draw a rectangle 6 cm $\times$ 4.5 cm and label the scale.

4. Rates in practical contexts

A rate compares two quantities with different units. When one quantity is directly proportional to another, their rate is constant.

Common rates

Speed

\text{speed} = \frac{\text{distance}}{\text{time}}

Density

\text{density} = \frac{\text{mass}}{\text{volume}}

Flow rate

\text{flow rate} = \frac{\text{volume}}{\text{time}}

Worked example 6 Density as a rate

A block of aluminium has a mass of 540 g and a volume of 200 cm $^3$ . Find the density and use it to predict the mass of a 350 cm $^3$ block of the same material.

Density $= \dfrac{540}{200} = 2.7$ g/cm $^3$ .
Mass is directly proportional to volume for a uniform material: $m = 2.7V$ .
For $V = 350$ : $m = 2.7 \times 350 = 945$ g.

Worked example 7 Comparing rates

Tap A fills a tank at 12 litres per minute. Tap B fills at 8 litres per minute. How long does it take to fill a 300-litre tank if both taps run together?

Combined flow rate $= 12 + 8 = 20$ litres per minute.
Time $= \dfrac{300}{20} = 15$ minutes.

Worked example 8 Modelling with proportion

The distance a spring stretches is directly proportional to the force applied (Hooke’s law). A force of 10 N stretches the spring by 4 cm. What force is needed for a 7 cm stretch?

$k = \dfrac{d}{F} = \dfrac{4}{10} = 0.4$ cm/N.
For $d = 7$ : $F = \dfrac{7}{0.4} = 17.5$ N.

Practice

Fluency

Tier 1: basic skills

$y$ is directly proportional to $x$ . When $x = 4$ , $y = 20$ . Find $k$ and write the equation.
Using the equation from Q1, find $y$ when $x = 7$ .
Does this table show direct proportion? $x$ : 3, 6, 9, 12; $y$ : 9, 18, 27, 36.
Does this table show direct proportion? $x$ : 2, 4, 6, 8; $y$ : 5, 9, 13, 17.
A map has a scale of 1 : 25 000. Two points are 6.4 cm apart on the map. Find the actual distance in metres.
An actual length of 3 km needs to be drawn on a map at 1 : 50 000 scale. What is the drawing length in cm?
A car travels at a constant speed of 80 km/h. How far does it travel in 2.5 hours?
A liquid flows at 5 litres per minute. How long does it take to fill a 120-litre container?
Convert 1 AUD $= 0.70$ USD. How many USD do you get for 250 AUD?
A block of iron has density 7.87 g/cm $^3$ . Find the mass of a 15 cm $^3$ piece.

Reasoning

Tier 2: mixed practice

The graph of $y$ against $x$ is a straight line through the origin with gradient 2.5. Write the equation and state whether $y$ is directly proportional to $x$ .
A recipe for 6 serves needs 450 g of flour. How much flour is needed for 10 serves? What assumption are you making?
On a 1 : 200 floor plan, a room measures 3.5 cm by 2.8 cm. Find the actual area of the room in square metres.
A cyclist covers 36 km in 1.5 hours. A runner covers 15 km in 1.25 hours. Express each as a rate in km/h and determine who is faster.
Gold has a density of 19.3 g/cm $^3$ . A gold bar has dimensions 25 cm $\times$ 5 cm $\times$ 2 cm. Find its mass in kilograms.
The extension of a spring is directly proportional to the load. A 6 kg load produces an extension of 9 cm. Find the extension for a 10 kg load and the load needed for a 15 cm extension.
Water flows into a pool at 40 litres per minute and drains out at 15 litres per minute simultaneously. How long until the pool, which holds 5000 litres, is full?
On a map at 1 : 10 000 scale, a park has an area of 12 cm $^2$ . What is the actual area in square metres?

Reasoning

Tier 3: explain and apply

Explain how you can tell from a table of values whether two quantities are directly proportional, and give an example of a table that almost looks proportional but is not.
A delivery company charges a $5 base fee plus $2 per kilogram. Is the total cost directly proportional to the weight? Justify your answer and sketch the graph.
Two maps of the same region have scales 1 : 20 000 and 1 : 50 000. A road is 8 cm long on the first map. How long is it on the second map?
The power output of a solar panel is directly proportional to its area. A 2 m $^2$ panel produces 600 W. A rooftop can fit 14 m $^2$ of panels. What is the maximum power output? Discuss one real-world factor that might cause the actual output to differ.
A student claims that since $y = 3x + 1$ has a constant rate of change (gradient 3), it must be a proportional relationship. Identify and explain the student’s error.

Challenge

Reasoning

Harder reasoning

A model car is built at a scale of 1 : 18. The model weighs 1.2 kg. If the model and the real car are made of materials with the same density, estimate the mass of the real car. (Hint: mass scales with volume, which scales as the cube of the linear scale factor.)
Two quantities are related by $y = kx^2$ . Is $y$ directly proportional to $x$ ? Is $y$ directly proportional to $x^2$ ? A ball is dropped and falls $d = 4.9t^2$ metres in $t$ seconds. Find the distance fallen in 3 seconds and the time to fall 100 m.
Water leaks from a tank at a rate proportional to the depth of water. When the depth is 2 m, the leak rate is 0.5 litres per minute. Write the rate as an equation and find the leak rate when the depth is 3.5 m.
A photographer enlarges a 10 cm $\times$ 15 cm photo to fit inside a frame that is 35 cm wide. What scale factor is used? What is the height of the enlarged photo? If the print costs $0.08 per cm $^2$ , find the printing cost of the enlargement.

Answers

Answer key

Attempt the practice first. When you're ready to check, expand the answers below.

Show the full answer key

Tier 1

$k = \dfrac{20}{4} = 5$ . Equation: $y = 5x$ .
$y = 5 \times 7 = 35$ .
Yes. $\dfrac{y}{x} = 3$ for every pair, so $y = 3x$ .
No. $\dfrac{5}{2} = 2.5$ , $\dfrac{9}{4} = 2.25$ , $\dfrac{13}{6} \approx 2.17$ , $\dfrac{17}{8} = 2.125$ . The ratio is not constant (this is $y = 2x + 1$ ).
Actual distance $= 6.4 \times 25\,000 = 160\,000$ cm $= 1\,600$ m.
$3$ km $= 300\,000$ cm. Drawing length $= \dfrac{300\,000}{50\,000} = 6$ cm.
Distance $= 80 \times 2.5 = 200$ km.
Time $= \dfrac{120}{5} = 24$ minutes.
$250 \times 0.70 = 175$ USD.
Mass $= 7.87 \times 15 = 118.05$ g.

Tier 2

$y = 2.5x$ . Yes, $y$ is directly proportional to $x$ because the line passes through the origin and $k = 2.5$ .
Flour per serve $= \dfrac{450}{6} = 75$ g. For 10 serves: $75 \times 10 = 750$ g. Assumption: the amount of flour is directly proportional to the number of serves (the recipe scales linearly).
Drawing dimensions: $3.5$ cm $\times$ $2.8$ cm. Actual: $3.5 \times 200 = 700$ cm $= 7$ m and $2.8 \times 200 = 560$ cm $= 5.6$ m. Area $= 7 \times 5.6 = 39.2$ m $^2$ .
Cyclist: $\dfrac{36}{1.5} = 24$ km/h. Runner: $\dfrac{15}{1.25} = 12$ km/h. The cyclist is faster.
Volume $= 25 \times 5 \times 2 = 250$ cm $^3$ . Mass $= 19.3 \times 250 = 4\,825$ g $= 4.825$ kg.
$k = \dfrac{9}{6} = 1.5$ cm/kg. Extension for 10 kg: $1.5 \times 10 = 15$ cm. Load for 15 cm: $\dfrac{15}{1.5} = 10$ kg.
Net flow rate $= 40 - 15 = 25$ litres per minute. Time $= \dfrac{5000}{25} = 200$ minutes $= 3$ hours $20$ minutes.
Scale factor for length $= 10\,000$ . Area scale factor $= 10\,000^2 = 10^8$ . Actual area $= 12 \times 10^8$ cm $^2 = 12 \times 10^4$ m $^2 = 120\,000$ m $^2$ (or $0.12$ km $^2$ ).

Tier 3

Calculate $\dfrac{y}{x}$ for every pair. If the ratio is the same each time, the relationship is directly proportional. Example of a near-miss: $x$ : 1, 2, 3, 4; $y$ : 3, 6, 9, 13. The first three ratios are 3, but the last is 3.25, so it is not proportional.
No. The total cost $C = 5 + 2w$ has a non-zero $y$ -intercept ($5 base fee), so $C$ is not directly proportional to $w$ . The graph is a straight line crossing the $C$ -axis at 5, not through the origin. Doubling the weight does not double the cost.
Road actual length $= 8 \times 20\,000 = 160\,000$ cm. On the second map: $\dfrac{160\,000}{50\,000} = 3.2$ cm.
$k = \dfrac{600}{2} = 300$ W/m $^2$ . Maximum output $= 300 \times 14 = 4\,200$ W. Real-world factors: panels may not all face the sun at the optimal angle; shading, temperature, and panel efficiency losses reduce actual output.
The student confuses a constant rate of change (gradient) with direct proportionality. Direct proportion requires $y = kx$ (the line passes through the origin). Since $y = 3x + 1$ has $y = 1$ when $x = 0$ , it does not pass through the origin and is not a proportional relationship.

Challenge

Linear scale factor $= 18$ . Volume scale factor $= 18^3 = 5\,832$ . Estimated real car mass $= 1.2 \times 5\,832 = 6\,998.4$ kg $\approx 7\,000$ kg. (In practice, real cars are not solid like models, so the actual mass would be much less — around 1,200—1,800 kg. The calculation shows what would happen if density were identical throughout.)
$y$ is not directly proportional to $x$ (doubling $x$ quadruples $y$ ). However, $y$ is directly proportional to $x^2$ with constant $k$ . Distance in 3 s: $d = 4.9 \times 9 = 44.1$ m. Time to fall 100 m: $100 = 4.9t^2$ , so $t^2 = \dfrac{100}{4.9} \approx 20.41$ and $t \approx 4.52$ s.
Let rate $= k \times \text{depth}$ . When depth $= 2$ : $0.5 = k \times 2$ , so $k = 0.25$ litres per minute per metre. Equation: rate $= 0.25d$ . When $d = 3.5$ : rate $= 0.25 \times 3.5 = 0.875$ litres per minute.
Scale factor $= \dfrac{35}{15} = \dfrac{7}{3} \approx 2.333$ . Height $= 10 \times \dfrac{7}{3} = \dfrac{70}{3} \approx 23.3$ cm. Enlarged area $= 35 \times \dfrac{70}{3} = \dfrac{2450}{3} \approx 816.7$ cm $^2$ . Cost $= 816.7 \times 0.08 \approx 65.33$ dollars.

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