Year 9 - Direct proportion, rates and scale - Free practice

Year 9 Direct Proportion, Rates and Scale Worksheets & Practice

Model real situations with ratios, rates and scale. Worked solutions every time, free.

Quick answer

How do you solve problems involving direct proportion, rates and scale?

Direct proportion means two quantities increase together at a constant ratio, written y = kx, where k is the constant of proportionality. Find k by dividing a known y by its matching x, then use it to solve for unknowns. For example, if 4 litres of paint covers 24 m², then k = 6 m²/L, so 7 litres covers 42 m².

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Six questions across Foundation, Standard and Extension, with full worked solutions, written to the Australian Curriculum.

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Worked examples

Year 9 direct proportion, rates and scale: worked examples

Three examples, from a standard question to a worded problem. Have a go at each one before you open the solution.

  1. Example 1 · Finding the constant of proportionality

    A tap fills a tank at a constant rate. After 3 hours the tank holds 480 litres. Write an equation in the form y=kxy = kx relating the volume of water yy (in litres) to the time xx (in hours), and find the volume after 7 hours.

    Show worked solution

    Key idea: In a direct proportion relationship y=kxy = kx, the constant kk is found by dividing a known yy-value by the matching xx-value.

    1. Find kk: k=yx=4803=160k = \frac{y}{x} = \frac{480}{3} = 160
    2. Write the equation: y=160xy = 160x
    3. Substitute x=7x = 7: y=160×7=1120y = 160 \times 7 = 1120

    Answer: The equation is y=160xy = 160x and the volume after 7 hours is 1120 litres.

  2. Example 2 · Converting and comparing rates

    Runner A completes 10 km in 48 minutes. Runner B runs at 13.5 km/h. Which runner is faster, and by how much (in km/h)?

    Show worked solution

    Key idea: Convert both speeds to the same units before comparing.

    1. Convert Runner A's time to hours: 48 min=4860 h=0.8 h48 \text{ min} = \frac{48}{60} \text{ h} = 0.8 \text{ h}
    2. Find Runner A's speed: Speed=100.8=12.5 km/h\text{Speed} = \frac{10}{0.8} = 12.5 \text{ km/h}
    3. Compare: Runner B runs at 13.5 km/h and Runner A at 12.5 km/h, so Runner B is faster.
    4. Find the difference: 13.5−12.5=1 km/h13.5 - 12.5 = 1 \text{ km/h}

    Answer: Runner B is faster by 1 km/h.

  3. Example 3 · Scale diagrams in a real-life context

    A landscape architect draws a plan of a rectangular garden using a scale of 1 : 250. On the plan, the garden measures 8 cm long and 5 cm wide. Paving costs $34 per square metre. Find the total cost to pave the entire garden.

    Show worked solution

    Key idea: Use the scale factor to convert plan measurements to real measurements, then calculate area and cost.

    1. Convert plan dimensions to real dimensions (1 cm on the plan = 250 cm = 2.5 m in real life): Length=8×2.5=20 m,Width=5×2.5=12.5 m\text{Length} = 8 \times 2.5 = 20 \text{ m}, \text{Width} = 5 \times 2.5 = 12.5 \text{ m}
    2. Find the real area: Area=20×12.5=250 m2\text{Area} = 20 \times 12.5 = 250 \text{ m}^2
    3. Calculate the cost: Cost=250×34=8500\text{Cost} = 250 \times 34 = 8500

    Answer: The total cost to pave the garden is $8500.

Common mistakes

Where students usually lose marks

Dividing in the wrong order to find k

Given y=480y = 480 when x=3x = 3, a student writes k=3÷480=0.00625k = 3 \div 480 = 0.00625.

Always divide the yy-value by the xx-value: k=y÷xk = y \div x. Check that kk makes sense in context; here, kk should be the number of litres per hour, which must be greater than 1.

Forgetting to convert units before calculating a rate

A runner completes 10 km in 48 minutes, so speed =10÷48≈0.208= 10 \div 48 \approx 0.208 km/h.

Convert time to hours before dividing: 48 minutes =0.8= 0.8 hours, giving 10÷0.8=12.510 \div 0.8 = 12.5 km/h. Always check that your units match the rate units required.

Squaring the scale factor incorrectly for area

Scale is 1 : 250, plan area is 8×5=408 \times 5 = 40 cm², so real area =40×250=10 000= 40 \times 250 = 10\,000 cm².

When scaling an area, the scale factor must be applied to each dimension separately before multiplying, or the scale factor must be squared: real area =40×2502= 40 \times 250^2 cm². It is safer to convert each length first, then find the area.

Assuming a relationship is proportional when it is not

A taxi charges a $3.50 flag fall plus $2.10 per km, so a student writes the direct proportion equation y=2.10xy = 2.10x.

A direct proportion relationship must pass through the origin (when x=0x = 0, y=0y = 0). A fixed starting fee means the relationship is linear but not directly proportional. The correct equation here is y=2.10x+3.50y = 2.10x + 3.50.

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Your turn

A car travels at a constant speed and covers 270 km in 3 hours. A map of the route uses a scale of 1 : 4 000 000. How long, in centimetres correct to 1 decimal place, is the route on the map?

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About this worksheet

Why we built it

Direct proportion, rates and scale sit at the heart of Year 9 mathematical modelling, connecting classroom maths to everyday situations like speed, unit pricing, map reading and mixing solutions. These worksheets build your student's skill from recognising proportional relationships all the way to setting up and solving multi-step problems, with fully worked solutions so they can follow the reasoning at every stage.

What's covered

Sub-skills your student will practise

  • ✓Identifying direct proportion relationships and writing the equation y = kx
  • ✓Finding the constant of proportionality k from tables, graphs and word problems
  • ✓Solving practical rate problems including speed, density and unit cost
  • ✓Converting and comparing rates expressed in different units
  • ✓Interpreting and applying map and scale diagram ratios
  • ✓Solving multi-step problems involving ratio, rate and proportion together

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Frequently asked

Questions parents ask about direct proportion, rates and scale

What is direct proportion and how is it taught in Year 9?

Two quantities are in direct proportion when they increase or decrease at the same rate, written as y = kx where k is the constant of proportionality. In Year 9, students move beyond simple ratio to set up and solve these equations in practical contexts like speed, fuel consumption and ingredient scaling.

How are rates different from ratios at Year 9 level?

A ratio compares two quantities of the same kind, such as 3 : 1. A rate compares quantities of different kinds, such as 60 km per hour or $4.50 per kilogram. Year 9 students work with both, including converting between units and choosing the most useful form for a given problem.

What kinds of scale problems appear in Year 9 maths?

Students read and apply scale ratios on maps and technical drawings, calculate real distances or dimensions from scaled measurements, and work backwards to find what scale was used. Problems often involve unit conversion as an extra step.

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