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R11

Use compound units such as speed, rates of pay, unit pricing, density and pressure

Compound measures

Worked answers, methods and verified real exam appearances for R11 on Edexcel GCSE Maths 1MA1.

Definition

Specification R11: use compound units such as speed, rates of pay, unit pricing, density and pressure. A compound unit is a rate formed from two quantities, so its name states the division: metres per second, grams per cubic centimetre, newtons per square metre.

Explanation

  • Write the formula before substituting: speed=distancetime\text{speed}=\dfrac{\text{distance}}{\text{time}}, density=massvolume\text{density}=\dfrac{\text{mass}}{\text{volume}}, pressure=forcearea\text{pressure}=\dfrac{\text{force}}{\text{area}}. Rates of pay and unit prices are a total divided by the hours or the number of items.
  • Make the units match the unit you must answer in before you calculate: 180180 km per hour becomes metres per second by multiplying by 10001000 and dividing by 36003600. Rearrange when the unknown is not the subject; mass is density multiplied by volume, and time is distance divided by speed.
  • Multi-step problems chain these: mass and density give a volume, subtracting a known volume gives the volume of the remaining material, and dividing its mass by that volume gives its density.
  • The unit in the question tells you which quantity is divided by which.
  • State the unit with the value; some schemes award it a separate mark.

Worked example

1

A solid metal block has mass 18.918.9 kg and volume 0.00750.0075 m3^3. Work out the density of the metal. Give the units with your answer.

(3)

(Total for Question 1 is 3 marks)

Working

  1. 1.Density is mass divided by volume: density=18.90.0075\text{density}=\dfrac{18.9}{0.0075}.
  2. 2.Multiply numerator and denominator by 10 00010\,000 to clear the decimal: 189 00075\dfrac{189\,000}{75}.
  3. 3.Divide: 189 000÷75=2520189\,000\div75=2520.
  4. 4.The mass is in kilograms and the volume in cubic metres, so the unit is kg/m3^3.

Answer: 25202520 kg/m3^3

Mark scheme

Mark scheme for question 1
QuestionAnswerMarkAdditional guidance
1for a process to find the density, eg 18.9÷0.007518.9\div0.0075P1oe
for 25202520A1Ignore incorrect or absent units for this mark
(indep) for kg/m3^3B1Ignore incorrect or absent numerical answer for this mark

Common mistakes

  • Don't divide volume by mass when calculating density.
  • Don't substitute minutes into a formula when the requested speed is per hour.
  • Don't give a numerical answer without the required compound unit.

Mark-scheme wording to use

  • “for conversion from km per hour to metres per second, … for a complete process eg 180 × 1000 ÷ 3600 … cao” — Pearson 1MA1, November 2022 Paper 3H, Q7(b)
  • “for process to find volume of tin eg 600 ÷ 0.6 (= 1000) … for process to find volume of salt eg “1000” – 700 (= 300) … for a process to find density of the salt eg 600 ÷ “300”” — Pearson 1MA1, June 2024 Paper 3H, Q5
  • “P1 start to process eg. 3 × 80 (=240) P1 ‘240’ ÷ 5” — Pearson 1MA1, Sample Paper 1H, Q4(a)

What the examiner wants

Each stage of the process written down, with the right quantity on top. The June 2024 Paper 3H report says most students divided mass by density to find a volume, but many did not see that the remaining volume was that answer minus 700700, and a surprising number wrote 600/300600/300 and gave 0.50.5 or 0.20.2 instead of 22.

Exam tip

Write the compound-unit formula first, because a correct formula can earn a method mark despite arithmetic error.

Worked practice

Q1
Tier 1 · Easy

1

A shift lasting 77 hours pays £52.50. Work out the hourly rate of pay.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
QuestionAnswerMarkMark scheme
1
  • £7.50 per hour
2Divide total pay by time: £52.50/7=£7.5052.50/7=£7.50 per hour.
Q2
Tier 2 · Standard

2

A coach travels 156 km156\,\text{km} in 22 hours 2424 minutes. Calculate its average speed in km/h\text{km}/\text{h}.

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
QuestionAnswerMarkMark scheme
2
  • 65 km/h65\,\text{km}/\text{h}
32424 minutes is 24/60=0.424/60=0.4 hours, so the time is 2.42.4 hours. Average speed is 156/2.4=65 km/h156/2.4=65\,\text{km}/\text{h}.
Q3
Tier 3 · Hard

3

A force of 3.6 kN3.6\,\text{kN} acts uniformly on a rectangular pad measuring 24 cm24\,\text{cm} by 15 cm15\,\text{cm}. Calculate the pressure in pascals, where 1 Pa=1 N/m21\,\text{Pa}=1\,\text{N}/\text{m}^2.

(4)

(Total for Question 3 is 4 marks)

Mark scheme

Mark scheme for question 3
QuestionAnswerMarkMark scheme
3
  • 100 000 Pa100\,000\,\text{Pa}
4Convert the force to 3600 N3600\,\text{N} and the dimensions to 0.24 m0.24\,\text{m} and 0.15 m0.15\,\text{m}. The area is 0.24×0.15=0.036 m20.24\times0.15=0.036\,\text{m}^2, so the pressure is 3600/0.036=100 000 Pa3600/0.036=100\,000\,\text{Pa}.
Q4
Tier 1 · Easy

4

A machine packs 5454 cartons in 66 minutes. Work out the rate in cartons per minute.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
QuestionAnswerMarkMark scheme
4
  • 99 cartons per minute
2Divide the number of cartons by the time: 54÷6=954\div6=9 cartons per minute.
Q5
Tier 2 · Standard

5

A material has mass 5.46 kg5.46\,\text{kg} and density 780 kg/m3780\,\text{kg}/\text{m}^3. Work out its volume.

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
QuestionAnswerMarkMark scheme
5
  • 0.007 m30.007\,\text{m}^3
3From density=massvolume\text{density}=\dfrac{\text{mass}}{\text{volume}}, volume =massdensity=\dfrac{\text{mass}}{\text{density}}. Therefore the volume is 5.46÷780=0.007 m35.46\div780=0.007\,\text{m}^3.
Q6
Tier 3 · Hard

6

A printer uses 0.450.45 litres of ink to print 18 00018\,000 pages, and the ink costs £28 per litre. Work out the ink cost per 10001000 pages.

(4)

(Total for Question 6 is 4 marks)

Mark scheme

Mark scheme for question 6
QuestionAnswerMarkMark scheme
6
  • £0.70 per 10001000 pages
4The ink used per 10001000 pages is 0.45÷18=0.0250.45\div18=0.025 litres. The cost is 0.025×£28=£0.700.025\times£28=£0.70 per 10001000 pages.
Q7
Tier 2 · Standard

7

A 750 g750\,\text{g} pack of cereal costs £3.30. A 1.2 kg1.2\,\text{kg} pack costs £5.16. Work out which pack is better value and by how much per kilogram.

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
QuestionAnswerMarkMark scheme
7
  • The 1.2 kg1.2\,\text{kg} pack, by £0.10 per kilogram
3The smaller pack costs £3.30÷0.75=£4.403.30\div0.75=£4.40 per kilogram. The larger pack costs £5.16÷1.2=£4.305.16\div1.2=£4.30 per kilogram. The 1.2 kg1.2\,\text{kg} pack is cheaper by £4.40−£4.30=£0.104.40-£4.30=£0.10 per kilogram.
Q8
Tier 3 · Hard

8

A technician works 4040 hours in one week. The first 3636 hours are paid at £12.40 per hour and the remaining hours are paid at one and a half times this rate. Work out the technician's average rate of pay for the week.

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
QuestionAnswerMarkMark scheme
8
  • £13.02 per hour
4The overtime rate is 1.5×£12.40=£18.601.5\times£12.40=£18.60 per hour. The total pay is 36×£12.40+4×£18.60=£446.40+£74.40=£520.8036\times£12.40+4\times£18.60=£446.40+£74.40=£520.80. The average rate is £520.80÷40=£13.02520.80\div40=£13.02 per hour.
Q9
Tier 3 · Hard

9

A driver travels 72 km72\,\text{km} at 48 km/h48\,\text{km}/\text{h}, stops for 1818 minutes, then travels 96 km96\,\text{km} at 64 km/h64\,\text{km}/\text{h}. Work out the average speed for the whole journey, including the stop. Round your answer to the nearest 0.1 km/h0.1\,\text{km}/\text{h}.

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
QuestionAnswerMarkMark scheme
9
  • 50.9 km/h50.9\,\text{km}/\text{h}
5The travel times are 72÷48=1.572\div48=1.5 hours and 96÷64=1.596\div64=1.5 hours. The stop is 18÷60=0.318\div60=0.3 hours, so the total time is 3.33.3 hours and the distance is 168 km168\,\text{km}. The average speed is 168÷3.3=50.909…168\div3.3=50.909\ldots, which rounds to 50.9 km/h50.9\,\text{km}/\text{h}. This is 0.0409…0.0409\ldots from the upper rounding boundary 50.9550.95, a margin of more than 0.03 km/h0.03\,\text{km}/\text{h}.
Q10
Tier 3 · Hard

10

A cuboid has dimensions 0.5 m0.5\,\text{m} by 0.25 m0.25\,\text{m} by 0.08 m0.08\,\text{m} and density 2400 kg/m32400\,\text{kg}/\text{m}^3. It rests on its 0.5 m0.5\,\text{m} by 0.25 m0.25\,\text{m} face. Work out the pressure it exerts on the floor. Use 10 N/kg10\,\text{N}/\text{kg} for gravitational field strength.

(5)

(Total for Question 10 is 5 marks)

Mark scheme

Mark scheme for question 10
QuestionAnswerMarkMark scheme
10
  • 1920 Pa1920\,\text{Pa}
5The volume is 0.5×0.25×0.08=0.01 m30.5\times0.25\times0.08=0.01\,\text{m}^3, so the mass is 2400×0.01=24 kg2400\times0.01=24\,\text{kg}. Its weight is 24×10=240 N24\times10=240\,\text{N}. The contact area is 0.5×0.25=0.125 m20.5\times0.25=0.125\,\text{m}^2, so the pressure is 240÷0.125=1920 Pa240\div0.125=1920\,\text{Pa}.

Verified exam appearances

SeriesPaperQuestionMarksCalculatorTierLinks
2023-063HQ63AllowedHigherQPMS
2019-063HQ134AllowedHigherQPMS
2024-112FQ223AllowedFoundationQPMS
2024-111HQ83Non-calculatorHigherQPMS
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2023-113FQ273AllowedFoundationQPMS
2019-113HQ134AllowedHigherQPMS
2019-112FQ103AllowedFoundationQPMS
2019-111HQ94Non-calculatorHigherQPMS
2024-112HQ125AllowedHigherQPMS
2021-113HQ44AllowedHigherQPMS
2022-062FQ274AllowedFoundationQPMS
2022-063HQ64AllowedHigherQPMS
2019-063FQ224AllowedFoundationQPMS
2024-061HQ54Non-calculatorHigherQPMS
2022-113HQ255AllowedHigherQPMS
2023-112FQ284AllowedFoundationQPMS
2024-111FQ97Non-calculatorFoundationQPMS
2022-062HQ93AllowedHigherQPMS
2022-063HQ44AllowedHigherQPMS
2024-063HQ54AllowedHigherQPMS
2023-111FQ134Non-calculatorFoundationQPMS
2024-062FQ145AllowedFoundationQPMS
2024-061FQ163Non-calculatorFoundationQPMS
2024-111FQ224Non-calculatorFoundationQPMS
2022-112HQ235AllowedHigherQPMS
2021-111FQ144Non-calculatorFoundationQPMS
2024-112HQ33AllowedHigherQPMS
2023-063HQ82AllowedHigherQPMS
2023-063FQ253AllowedFoundationQPMS
2023-061HQ83Non-calculatorHigherQPMS
2022-113FQ155AllowedFoundationQPMS
2019-061FQ94Non-calculatorFoundationQPMS
2022-061FQ293Non-calculatorFoundationQPMS
2023-062FQ186AllowedFoundationQPMS
2022-061HQ73Non-calculatorHigherQPMS
2019-112FQ143AllowedFoundationQPMS
2024-112FQ283AllowedFoundationQPMS
2019-111FQ163Non-calculatorFoundationQPMS
2023-062FQ163AllowedFoundationQPMS
2022-111HQ72Non-calculatorHigherQPMS
2023-112FQ62AllowedFoundationQPMS
2019-111FQ275Non-calculatorFoundationQPMS
2023-063FQ144AllowedFoundationQPMS
2023-113HQ83AllowedHigherQPMS
2021-112FQ143AllowedFoundationQPMS
2022-113FQ274AllowedFoundationQPMS
2023-063FQ135AllowedFoundationQPMS
2019-111HQ85Non-calculatorHigherQPMS
2022-111FQ145Non-calculatorFoundationQPMS
2022-063FQ274AllowedFoundationQPMS
2023-111FQ252Non-calculatorFoundationQPMS
2019-113HQ64AllowedHigherQPMS
2021-113FQ224AllowedFoundationQPMS
2022-062FQ245AllowedFoundationQPMS
2022-111FQ252Non-calculatorFoundationQPMS
2023-113FQ173AllowedFoundationQPMS
2021-113HQ24AllowedHigherQPMS
2023-113HQ73AllowedHigherQPMS
2023-063FQ272AllowedFoundationQPMS
2019-063HQ113AllowedHigherQPMS
2022-113FQ194AllowedFoundationQPMS
2024-062FQ94AllowedFoundationQPMS
2024-063FQ284AllowedFoundationQPMS
2023-061FQ273Non-calculatorFoundationQPMS
2024-113FQ144AllowedFoundationQPMS
2021-112FQ235AllowedFoundationQPMS
2021-113FQ113AllowedFoundationQPMS
2021-113FQ244AllowedFoundationQPMS
2019-113FQ274AllowedFoundationQPMS
2022-062HQ45AllowedHigherQPMS
2024-061HQ35Non-calculatorHigherQPMS
2022-113HQ74AllowedHigherQPMS
2019-062FQ184AllowedFoundationQPMS
2024-063FQ264AllowedFoundationQPMS
2021-112HQ73AllowedHigherQPMS
2022-112FQ134AllowedFoundationQPMS
2024-063HQ74AllowedHigherQPMS
2022-062HQ74AllowedHigherQPMS
2022-061FQ164Non-calculatorFoundationQPMS
2024-111HQ54Non-calculatorHigherQPMS
2023-113FQ263AllowedFoundationQPMS
2021-112HQ35AllowedHigherQPMS
2023-112FQ122AllowedFoundationQPMS
2022-063FQ254AllowedFoundationQPMS

Other points in R Ratio, proportion and rates of change

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