Rates, unit prices, density and pressure · Higher
| English | 中文 | Pinyin |
|---|---|---|
| density/ˈdensɪti/ | 密度 | mì dù |
Two packages have different sizes and prices. Comparing the sticker prices alone cannot tell which gives more material for each yuan.
- Two packages have different sizes and prices. Comparing the sticker prices alone cannot tell which gives more material for each yuan.
- This lesson studies density 密度: Mass per unit volume, with both units stated.
Choose the mathematical structure
- A compound rate divides one quantity by another: speed=d/t, pay=earnings/time, unit price=cost/amount, density=mass/volume and pressure=force/area. Rearrange these equations before substituting. Convert each dimension separately; converting cm² to m² uses a squared length factor. Comparisons must use the same units and conditions.
- State the allowed inputs and units before calculating. An equation should express the relationship, not just record a calculator entry.
Which description correctly defines density?
Mass per unit volume, with both units stated.
Work through a checked case
- Check the result against the starting quantities. Substitute into the original relation, or compare the graph and numerical answer where appropriate.
A 750 g pack costing 18 yuan has unit price 18/0.75=24 yuan/kg. A 1.2 kg pack at 30 yuan costs 25 yuan/kg, so the first is better value if quality and waste are equal. A block of mass 540 g and volume 200 cm³ has density 2.7 g/cm³; a 50 cm³ piece of that material has mass 135 g. Since 1 g=0.001 kg and 1 cm³=0.000001 m³, 2.7 g/cm³=2700 kg/m³. A 120 N force over 0.03 m² gives pressure 4000 N/m²=4000 Pa. At fixed force, halving the area doubles pressure. An hourly pay rate of 48 yuan/hour gives 120 yuan for 2.5 hours.
Rates, unit prices, density and pressure
A compound rate divides one quantity by another: speed=d/t, pay=earnings/time, unit price=cost/amount, density=mass/volume and pressure=force/area
Classify the worked-case statements, then explain the units or invariant that justifies each decision.
Find density for mass 540 g and volume 200 cm³.
Density=540/200=2.7 g/cm³.
Test a tempting shortcut
- Volume conversions cube the length factor and area conversions square it. A density of 2.7 g/cm³ is not 2.7 kg/m³. A cheaper package may cost more per kilogram; include only comparable products.
- When a shortcut fails, identify the assumption it breaks. Keep an exact value until the requested final rounding.
Converting density from g/cm³ to kg/m³ leaves its numerical value unchanged. This claim is false. Explain which definition or assumption it violates.
Find pressure in Pa for 120 N on 0.03 m².
Pressure=force/area=120/0.03=4000.
Converting density from g/cm³ to kg/m³ leaves its numerical value unchanged.
Volume conversions cube the length factor and area conversions square it. A density of 2.7 g/cm³ is not 2.7 kg/m³. A cheaper package may cost more per kilogram; include only comparable products.
Interpret a new situation
- AQA R1/R11 includes speed, pay, pricing, density and pressure in numerical and algebraic contexts. Show the rearrangement with named quantities and carry units through the answer; use an inverse check such as density×volume=mass.
- A complete solution gives the mathematical result and explains what it means. Check that it is possible in the stated context.
Find the unit price in yuan/kg for 750 g at 18 yuan.
750 g=0.75 kg, so 18/0.75=24.
Match each part of a complete solution to its purpose.
An assumption justifies the model; a check tests the result; interpretation connects it to the question.
Use this in your course
- 8300 · Higher · 3.3. Match the target tier and specification before assigning extensions.
- Give the method before the final answer, and use the paper's calculator and formula rules. Review a wrong answer by locating the first invalid step.
Mass per unit volume, with both units stated. Choose the relationship, show the method, check its assumptions and interpret the result.