Practise 20 exam-style questions on Standard units and compound measures for Edexcel GCSE Maths (Calculator). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.
20 questions
1–5 marks each
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20 questions, marked as you go
Sample questions with model answers
Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.
Question 1
2 marks
A jug holds 1.35 litres of juice. Each glass holds 180 ml.
How many glasses can be filled completely from the jug?
Model answer
1.35 litres = 1350 ml \(1350 \div 180 = 7.5\) Only whole glasses count, so 7 glasses can be filled completely.
Explanation
Change both amounts into the same unit (1 litre = 1000 ml), then divide. The answer 7.5 means 7 full glasses and half a glass, so the number of glasses filled completely is 7 - do not round up to 8.
Mark scheme
1 mark
M1 for \(1350 \div 180\) (= 7.5), or \(1.35 \div 0.18\)
oe, e.g. repeated addition of 180 ml; correct answer with no working scores full marks
1 mark
A1 for 7
cao; do not accept 7.5 or 8
Question 2
2 marks
Show that a density of 19.3 g/cm³ is the same as a density of 19 300 kg/m³.
Model answer
1 m³ = 100 × 100 × 100 cm³ = 1 000 000 cm³ Mass of 1 m³ = \(19.3 \times 1000000 = 19300000\) g \(19300000 \div 1000 = 19300\) kg So 19.3 g/cm³ = 19 300 kg/m³
Explanation
Find the mass of 1 m³: there are 1 000 000 cm³ in 1 m³, each with mass 19.3 g. Then change grams into kilograms by dividing by 1000. Overall, g/cm³ to kg/m³ is × 1000, a handy fact for checking.
Mark scheme
1 mark
M1 for a correct first step, e.g. 1 m³ = 1 000 000 cm³ used to give 19 300 000 g in 1 m³, or 19.3 g = 0.0193 kg in 1 cm³
oe
1 mark
A1 for a complete method leading to 19 300 kg/m³, e.g. \(19.3 \times 1000000 \div 1000 = 19300\)
exact conversion; all the working must be shown; answer only scores 0
Question 3
3 marks
Kai runs 5 km in 23 minutes 15 seconds. Lena runs at an average speed of 3.5 m/s.
Who runs faster, Kai or Lena? You must show how you get your answer.
Model answer
Kai's time: 23 × 60 + 15 = 1395 seconds Kai's speed: \(5000 \div 1395 = 3.584...\) m/s Lena's speed: 3.5 m/s 3.584... > 3.5, so Kai runs faster.
Explanation
To compare speeds they must be in the same units. Change 5 km into 5000 m and 23 minutes 15 seconds into 1395 seconds, then divide. Writing the time as 23.15 minutes is a common error - 15 seconds is 0.25 of a minute.
Mark scheme
1 mark
M1 for Kai's time in seconds (1395 s) or in hours (0.3875 h)
oe
1 mark
M1 for Kai's speed in units that can be compared, e.g. \(5000 \div 1395\) (= 3.58... m/s), or Lena's speed changed to km/h (12.6) with Kai's speed in km/h (12.9...)
oe
1 mark
C1 for Kai, with correct comparable speeds, e.g. 3.58 m/s and 3.5 m/s, or 12.9 km/h and 12.6 km/h
accept 3.58 to 3.59 m/s or 12.9 km/h; or comparing times: Lena takes 1428.5... s for 5 km
Question 4
4 marks
A lorry can carry a maximum load of 18 tonnes. A builder needs to move 75 m³ of sand. The density of the sand is 1.6 g/cm³.
1 tonne = 1000 kg
What is the least number of trips the lorry must make to move all the sand? You must show your working.
Model answer
1 m³ = 1 000 000 cm³, so 75 m³ = 75 000 000 cm³ Mass = \(1.6 \times 75000000 = 120000000\) g = 120 000 kg = 120 tonnes \(120 \div 18 = 6.66...\) 6 trips is not enough, so the lorry must make 7 trips.
Explanation
Mass = density × volume, but the units must match: either change 75 m³ into cm³ (× 1 000 000), or change 1.6 g/cm³ into 1600 kg/m³. Then divide the total mass by the maximum load. The answer 6.66... must be rounded up, because 6 trips would leave some sand behind.
Mark scheme
1 mark
P1 for a correct unit conversion, e.g. 1.6 g/cm³ = 1600 kg/m³, or 75 m³ = 75 000 000 cm³
oe
1 mark
P1 for the mass of sand: 120 000 kg (120 tonnes or 120 000 000 g)
oe
1 mark
P1 for their \(120 \div 18\) (= 6.66...), or \(120000 \div 18000\)
oe
1 mark
A1 for 7 trips
cao; do not accept 6 or 6.67
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