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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.

Explanation

  • A compound unit combines quantities, usually through division. Use speed=distancetime\text{speed}=\dfrac{\text{distance}}{\text{time}}, density=massvolume\text{density}=\dfrac{\text{mass}}{\text{volume}} and pressure=forcearea\text{pressure}=\dfrac{\text{force}}{\text{area}}.
  • Rates of pay and unit prices are totals divided by the relevant time or number of items.
  • Make the units compatible before substituting, and rearrange the formula if the unknown is in the numerator or denominator.
  • The units provide a useful check on the operation and show which quantity should be divided by which.
  • Examiners expect both the numerical value and the correct compound unit, such as kg/m3\text{kg}/\text{m}^3.

Worked example

A solid has mass 18.9kg18.9\,\text{kg} and volume 0.0075m30.0075\,\text{m}^3. Find its density.

  1. 1.Select density=massvolume\text{density}=\dfrac{\text{mass}}{\text{volume}}.
  2. 2.Substitute: density=18.90.0075\text{density}=\dfrac{18.9}{0.0075}.
  3. 3.Evaluate and attach the units: 2520kg/m32520\,\text{kg}/\text{m}^3.

Answer: 2520kg/m32520\,\text{kg}/\text{m}^3.

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.

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 156km156\,\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
  • 65km/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=65km/h156/2.4=65\,\text{km}/\text{h}.
Q3
Tier 3 · Hard

3

A force of 3.6kN3.6\,\text{kN} acts uniformly on a rectangular pad measuring 24cm24\,\text{cm} by 15cm15\,\text{cm}. Calculate the pressure in pascals, where 1Pa=1N/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
  • 100000Pa100\,000\,\text{Pa}
4Convert the force to 3600N3600\,\text{N} and the dimensions to 0.24m0.24\,\text{m} and 0.15m0.15\,\text{m}. The area is 0.24×0.15=0.036m20.24\times0.15=0.036\,\text{m}^2, so the pressure is 3600/0.036=100000Pa3600/0.036=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 5.46kg5.46\,\text{kg} and density 780kg/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
  • 0.007m30.007\,\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=0.007m35.46\div780=0.007\,\text{m}^3.
Q6
Tier 3 · Hard

6

A printer uses 0.450.45 litres of ink to print 1800018\,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 750g750\,\text{g} pack of cereal costs £3.30. A 1.2kg1.2\,\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 1.2kg1.2\,\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 1.2kg1.2\,\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 72km72\,\text{km} at 48km/h48\,\text{km}/\text{h}, stops for 1818 minutes, then travels 96km96\,\text{km} at 64km/h64\,\text{km}/\text{h}. Work out the average speed for the whole journey, including the stop. Round your answer to the nearest 0.1km/h0.1\,\text{km}/\text{h}.

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
QuestionAnswerMarkMark scheme
9
  • 50.9km/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 168km168\,\text{km}. The average speed is 168÷3.3=50.909168\div3.3=50.909\ldots, which rounds to 50.9km/h50.9\,\text{km}/\text{h}. This is 0.04090.0409\ldots from the upper rounding boundary 50.9550.95, a margin of more than 0.03km/h0.03\,\text{km}/\text{h}.
Q10
Tier 3 · Hard

10

A cuboid has dimensions 0.5m0.5\,\text{m} by 0.25m0.25\,\text{m} by 0.08m0.08\,\text{m} and density 2400kg/m32400\,\text{kg}/\text{m}^3. It rests on its 0.5m0.5\,\text{m} by 0.25m0.25\,\text{m} face. Work out the pressure it exerts on the floor. Use 10N/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
  • 1920Pa1920\,\text{Pa}
5The volume is 0.5×0.25×0.08=0.01m30.5\times0.25\times0.08=0.01\,\text{m}^3, so the mass is 2400×0.01=24kg2400\times0.01=24\,\text{kg}. Its weight is 24×10=240N24\times10=240\,\text{N}. The contact area is 0.5×0.25=0.125m20.5\times0.25=0.125\,\text{m}^2, so the pressure is 240÷0.125=1920Pa240\div0.125=1920\,\text{Pa}.

Verified exam appearances

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Other points in R Ratio, proportion and rates of change

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