Australian Curriculum v9 / ACiQ Year 8 Mathematics - Unit 3 - Rates, ratios and practical modelling

Rates, ratios and practical modelling

Use ratios, rates and unit rates to model scale, speed, concentration and other practical relationships.

Updated 2026-07-26 - 12 min read

Rates, ratios and practical modelling is taught here as a connected set of decisions, not a list of facts. Work through the prerequisite recall, explicit models, carefully faded examples, misconception repairs and transfer task before using the target in Check, Practice, Review or Rapid Revision.

This note is designed to work with the guided lessons, curated practice, flashcards, Tutor context, Review and Rapid Revision for the same canonical target. The same three evidence checks are used throughout, so feedback can route a learner back to the precise idea that needs repair.

Simplify and scale ratios

Equivalent ratios multiply or divide every term by the same non-zero factor while preserving order and meaning. This relationship must be selected from the quantities and conditions in the problem, then checked against the context.

A dependable reasoning routine

  1. Name the unknowns, units and constraints before calculating.
  2. Choose the relationship represented by simplify and scale ratios and state why it applies.
  3. Keep exact values for as long as possible, show substitutions and preserve units through each step.
  4. Check the result by substitution, estimation, an alternative representation or the original context.

Repair: A ratio describes a relationship; 2:3 could represent 4:6, 10:15 or any common scaling.

The repair matters because the shortcut may appear to work in one familiar example while failing when the method, representation, scale, constraint or accuracy requirement changes. Use the routine above to make the reasoning visible enough for another learner to verify.

Example 1.1

Simplify 18:30.

Step 1 - identify the governing idea: Equivalent ratios multiply or divide every term by the same non-zero factor while preserving order and meaning.

Step 2 - apply it to this evidence: Dividing both terms by their greatest common factor 6 gives 3:5.

Result: 3:5

The relationship is visible in the working: Dividing both terms by their greatest common factor 6 gives 3:5. Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • 9:10 — It does not agree with the required relationship: Dividing both terms by their greatest common factor 6 gives 3:5.
  • 18:5 — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • 3:30 — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Example 1.2

Red:blue paint is mixed 2:5. How much blue is needed with 8 L red?

Step 1 - identify the governing idea: Equivalent ratios multiply or divide every term by the same non-zero factor while preserving order and meaning.

Step 2 - apply it to this evidence: The scale factor from 2 to 8 is 4, so 5 × 4 = 20.

Result: 20 L

The relationship is visible in the working: The scale factor from 2 to 8 is 4, so 5 × 4 = 20. Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • 11 L — It does not agree with the required relationship: The scale factor from 2 to 8 is 4, so 5 × 4 = 20.
  • 32 L — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • 3.2 L — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Example 1.3

A class ratio of laptops to students is 1:3. Which count is possible?

Step 1 - identify the governing idea: Equivalent ratios multiply or divide every term by the same non-zero factor while preserving order and meaning.

Step 2 - apply it to this evidence: Both terms are 10 times the base ratio.

Result: 10 laptops and 30 students

The relationship is visible in the working: Both terms are 10 times the base ratio. Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • 10 laptops and 3 students — It does not agree with the required relationship: Both terms are 10 times the base ratio.
  • 30 laptops and 10 students — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • 11 laptops and 30 students — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Calculate and compare unit rates

A unit rate expresses one quantity per one unit of another; divide in the order demanded by the units. This relationship must be selected from the quantities and conditions in the problem, then checked against the context.

A dependable reasoning routine

  1. Name the unknowns, units and constraints before calculating.
  2. Choose the relationship represented by calculate and compare unit rates and state why it applies.
  3. Keep exact values for as long as possible, show substitutions and preserve units through each step.
  4. Check the result by substitution, estimation, an alternative representation or the original context.

Repair: Rates compare quantities on a common per-one basis; totals alone do not account for different amounts or times.

The repair matters because the shortcut may appear to work in one familiar example while failing when the method, representation, scale, constraint or accuracy requirement changes. Use the routine above to make the reasoning visible enough for another learner to verify.

Example 2.1

A car travels 180 km in 3 h. Find average speed.

Step 1 - identify the governing idea: A unit rate expresses one quantity per one unit of another; divide in the order demanded by the units.

Step 2 - apply it to this evidence: Distance divided by time is 180 ÷ 3 = 60 km/h.

Result: 60 km/h

The relationship is visible in the working: Distance divided by time is 180 ÷ 3 = 60 km/h. Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • 540 km/h — It does not agree with the required relationship: Distance divided by time is 180 ÷ 3 = 60 km/h.
  • 177 km/h — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • 0.017 km/h — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Example 2.2

750 g of cereal costs $6. Which is the unit price per 100 g?

Step 1 - identify the governing idea: A unit rate expresses one quantity per one unit of another; divide in the order demanded by the units.

Step 2 - apply it to this evidence: There are 7.5 lots of 100 g, and $6 ÷ 7.5 = $0.80.

Result: $0.80

The relationship is visible in the working: There are 7.5 lots of 100 g, and $6 ÷ 7.5 = $0.80. Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • $8.00 — It does not agree with the required relationship: There are 7.5 lots of 100 g, and $6 ÷ 7.5 = $0.80.
  • $1.25 — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • $4.50 — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Example 2.3

Runner A covers 5 km in 24 min; Runner B covers 8 km in 40 min. Who is faster?

Step 1 - identify the governing idea: A unit rate expresses one quantity per one unit of another; divide in the order demanded by the units.

Step 2 - apply it to this evidence: A runs 0.2083 km/min while B runs 0.2 km/min.

Result: Runner A

The relationship is visible in the working: A runs 0.2083 km/min while B runs 0.2 km/min. Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • Runner B — It does not agree with the required relationship: A runs 0.2083 km/min while B runs 0.2 km/min.
  • They have equal speed — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • It cannot be decided — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Model scale, concentration and practical rates

Interpret the units and relationship first, then apply a consistent scale factor or rate to the required quantity. This relationship must be selected from the quantities and conditions in the problem, then checked against the context.

A dependable reasoning routine

  1. Name the unknowns, units and constraints before calculating.
  2. Choose the relationship represented by model scale, concentration and practical rates and state why it applies.
  3. Keep exact values for as long as possible, show substitutions and preserve units through each step.
  4. Check the result by substitution, estimation, an alternative representation or the original context.

Repair: A scale is multiplicative: one drawing unit corresponds to 50 real units in the same measurement dimension.

The repair matters because the shortcut may appear to work in one familiar example while failing when the method, representation, scale, constraint or accuracy requirement changes. Use the routine above to make the reasoning visible enough for another learner to verify.

Example 3.1

On a 1:200 map, a path is 6 cm. What is the real length?

Step 1 - identify the governing idea: Interpret the units and relationship first, then apply a consistent scale factor or rate to the required quantity.

Step 2 - apply it to this evidence: 6 × 200 = 1200 cm, which is 12 m.

Result: 12 m

The relationship is visible in the working: 6 × 200 = 1200 cm, which is 12 m. Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • 1200 m — It does not agree with the required relationship: 6 × 200 = 1200 cm, which is 12 m.
  • 3 m — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • 206 cm — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Example 3.2

A drink uses 30 mL concentrate per 750 mL water. How much concentrate for 2.5 L water?

Step 1 - identify the governing idea: Interpret the units and relationship first, then apply a consistent scale factor or rate to the required quantity.

Step 2 - apply it to this evidence: 2500/750 = 10/3, and 30 × 10/3 = 100 mL.

Result: 100 mL

The relationship is visible in the working: 2500/750 = 10/3, and 30 × 10/3 = 100 mL. Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • 75 mL — It does not agree with the required relationship: 2500/750 = 10/3, and 30 × 10/3 = 100 mL.
  • 625 mL — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • 250 mL — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Example 3.3

A machine produces 84 parts in 7 minutes at a constant rate. How many in 25 minutes?

Step 1 - identify the governing idea: Interpret the units and relationship first, then apply a consistent scale factor or rate to the required quantity.

Step 2 - apply it to this evidence: The unit rate is 12 parts/min, so 25 × 12 = 300.

Result: 300 parts

The relationship is visible in the working: The unit rate is 12 parts/min, so 25 × 12 = 300. Check the units and substitute or estimate where possible. A plausible-looking number is not enough unless it satisfies the original conditions.

Why the alternatives fail:

  • 2100 parts — It does not agree with the required relationship: The unit rate is 12 parts/min, so 25 × 12 = 300.
  • 175 parts — It changes or misses a condition in the question. Reapply the relationship and retain the stated units or accuracy.
  • 12 parts — A substitution, estimate, ordering or unit check rejects this result; it does not reproduce the conditions in the prompt.

Retrieval check

Try these without looking back at the examples.

  1. Red:blue paint is mixed 2:5. How much blue is needed with 8 L red?
  2. 750 g of cereal costs $6. Which is the unit price per 100 g?
  3. A drink uses 30 mL concentrate per 750 mL water. How much concentrate for 2.5 L water?

Answers

  1. 20 L — The scale factor from 2 to 8 is 4, so 5 × 4 = 20.
  2. $0.80 — There are 7.5 lots of 100 g, and $6 ÷ 7.5 = $0.80.
  3. 100 mL — 2500/750 = 10/3, and 30 × 10/3 = 100 mL.

Transfer task

Find an unfamiliar example from school, daily life, a credible news source or another subject. Explain which of the three evidence checks applies. Complete the task, then audit your own response: identify the evidence used, the relationship applied, one plausible misconception and the final reasonableness check. If a peer could not reproduce your reasoning, add the missing step.

Sources