Learn in your language
Browse this course
Digital SAT · Math · Problem-Solving and Data Analysis

Rates, proportions and derived units

Use proportional relationships and cancel units through single-step and multidimensional conversions.

What you will tackle

  • Reason with ratios, unit rates, proportional relationships and scale drawings
  • Convert simple, compound, squared and cubed units
  • Combine product and quotient units across multistep contexts

Units are part of the calculation

A rate of 18 km in 30 minutes is 36 km/h because 30 minutes is 0.5 h. For an average speed over several legs, divide total distance by total time rather than averaging unequal-duration speeds. A proportional relationship has the form y = kx and passes through the origin.

Apply a conversion in every dimension

Since 1 m = 100 cm, 1 m² = 10,000 cm² and 1 m³ = 1,000,000 cm³. For a speed or density, convert both numerator and denominator as needed. A map scale 1:25,000 compares matching length units; 4 cm represents 100,000 cm, or 1 km.

Build your notes, then practise

Work on paper before opening an explanation. Guided tasks provide a starting point; independent tasks remove the scaffolding; application tasks ask you to justify a method or interpretation in context.

Guided practice

Guided 1

Convert 72 km/h to m/s, 0.4 m² to cm², and a 6 cm map length at 1:25,000 into kilometres.

Show a starting hint

Write conversion fractions so unwanted units cancel.

Compare with the explained answer
  1. 72 × 1000/3600 = 20 m/s.
  2. 0.4 × 100² = 4000 cm².
  3. 6 × 25,000 = 150,000 cm = 1.5 km.

Answer / response criteria: 20 m/s; 4000 cm²; 1.5 km.

Independent practice

Independent 1

A material has density 2.4 g/cm³. Convert it to kg/m³. Then find the mass of 0.003 m³.

Compare with the explained answer
  1. One gram is 0.001 kg and one cubic centimetre is 0.000001 m³.
  2. 2.4 × 0.001/0.000001 = 2400 kg/m³.
  3. Mass = density × volume = 2400 × 0.003 = 7.2 kg.

Answer / response criteria: 2400 kg/m³; 7.2 kg.

Application practice

Application 1

A cyclist travels 30 km at 15 km/h and returns 30 km at 30 km/h. Find the average speed and explain why 22.5 km/h is incorrect.

Compare with the explained answer
  1. The outward journey takes 2 h and the return 1 h.
  2. Total distance is 60 km in 3 h, so average speed is 20 km/h.
  3. A simple average of speeds weights each speed equally, but the slower leg lasts twice as long.

Answer / response criteria: 20 km/h; total distance divided by total time is required.

Recall without your notes

What can you explain now?

A cyclist travels 30 km at 15 km/h and returns 30 km at 30 km/h. Find the average speed and explain why 22.5 km/h is incorrect.

Compare with the explanation

20 km/h; total distance divided by total time is required.

Explain the reasoning before comparing. This review records your self-assessment rather than an official score or mastery decision.

After trying it yourself, choose your next review. This is your self-assessment.

Your review choice appears on Today. Sign in to sync it across devices.

Source and review status

Core coverage follows the detailed SAT skills in Appendix B of the September 2026 framework. The conditional School Day Essay is separate. The framework is maintained rather than issued as an annual subject syllabus.

Original teaching and explained self-check practice are available. Independent subject-specialist review and difficulty calibration remain pending; this is not an exam-board endorsement or a mastery certificate.

Official source specification / guide