Edexcel GCSE Maths (1MA1), Higher tier · Ratio, proportion and rates of change
Practise Speed, density and pressure. 4 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy3 marks
Ella runs 5 km in 24 minutes. Jon runs at an average speed of 12 km/h.
Who runs at the greater average speed? You must show your working.[3]
Show the answer and mark scheme
Answer: Ella: her speed is 12.5 km/h, which is more than 12 km/h.
M1 for 24 minutes = 0.4 hours, or \(5 \div 24\) km per minute
A1 for 12.5 km/h (or Jon runs 4.8 km in 24 minutes, or Jon takes 25 minutes for 5 km)
C1 for Ella, from a correct comparison
Worked solution: 24 minutes \(= \frac{24}{60} = 0.4\) hours. Ella's speed \(= 5 \div 0.4 = 12.5\) km/h. 12.5 km/h > 12 km/h, so Ella is faster.
Question 2Medium3 marks
A car travels 60 km at an average speed of 40 km/h. It then travels a further 60 km at an average speed of 60 km/h. Ali says, 'The average speed for the whole journey is 50 km/h.'
Show that Ali is wrong.[3]
Show the answer and mark scheme
Answer: Times: 1.5 h and 1 h. Average speed \(= \frac{120}{2.5} = 48\) km/h, not 50 km/h.
M1 for the time of one section: \(60 \div 40 = 1.5\) h or \(60 \div 60 = 1\) h
M1 for total distance ÷ total time: \(120 \div 2.5\)
C1 for 48 km/h with the conclusion that Ali is wrong
Worked solution: First part: \(60 \div 40 = 1.5\) h. Second part: \(60 \div 60 = 1\) h. Average speed \(= \frac{120}{2.5} = 48\) km/h. The car spends longer at the slower speed, so the average is less than 50 km/h.
Question 3Hard4 marks
220 g of liquid X is mixed with 80 g of liquid Y. The density of liquid X is 1.1 g/cm³ and the density of liquid Y is 0.8 g/cm³. Assume that the volume of the mixture is the total volume of the two liquids. Work out the density of the mixture.[4]
Show the answer and mark scheme
Answer: 1 g/cm³
M1 for the volume of X, \(220 \div 1.1 = 200\)
M1 for the volume of Y, \(80 \div 0.8 = 100\)
M1 for \((220 + 80) \div (200 + 100)\)
A1 for 1 cao
Worked solution: Volume of X \(= 220 \div 1.1 = 200\) cm³. Volume of Y \(= 80 \div 0.8 = 100\) cm³. Density \(= \frac{220 + 80}{200 + 100} = \frac{300}{300} = 1\) g/cm³.