Printable · GCSE Foundation · ages 14-16
Compound units: speed, density, pressure worksheet — GCSE Foundation
Fifteen questions on "compound units: speed, density, pressure" — DfE statement R11. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
Compound units: speed, density, pressure worksheet — GCSE Foundation
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- 1.A material has a density of 2 g/cm³. Work out the mass of 5 cm³ of this material.
- 2.A metal sample has a mass of 342.6 g and a volume of 18 cm³. Work out the density of the sample, in g/cm³, to 1 decimal place.
- 3.A car travels 150 km in 3 hours at a constant speed. Work out the average speed of the car, in km/h.
- 4.Jamal is paid £54 for working 6 hours. Work out his rate of pay, in £ per hour.
- 5.Work out the time taken to travel 120 km at an average speed of 60 km/h.
- 6.Two cars leave the same point at the same time and travel in opposite directions. One travels at 80 km/h and the other at 60 km/h. Work out how long it takes until the cars are 350 km apart.
- 7.A cyclist rides 30 km in 1 hour 30 minutes. Work out the average speed of the cyclist in km/h.
- 8.Maya is paid £105.30 for working 9 hours. Work out her rate of pay, in £ per hour.
- 9.A metal block has a mass of 270 g and a volume of 30 cm³. Work out the density of the block, in g/cm³.
- 10.Ffion is paid £11.20 per hour. She works 6 hours on Monday and 4.5 hours on Tuesday. Work out her total pay for the two days.
- 11.Noah runs 2 km in 10 minutes. Work out how long he takes to run 5 km at the same speed.
- 12.A 5 kg bag of potatoes costs £4.50. Work out the cost per kilogram.
- 13.A cyclist travels 45 km in 3 hours at a constant speed. Work out the cyclist's average speed, in km/h.
- 14.A train travels at a constant speed of 90 km/h. Work out the speed in m/s.
- 15.A force of 126 N acts on an area of 3.5 m². Work out the pressure on the area, in N/m².
Answer key
- (c) 10 g — Method: mass = density × volume. Working: 2 × 5 = 10, in grams because the density is in grams per cubic centimetre. Answer: 10 g. The distractors: 2.5 g comes from dividing the volume by the density, 5 ÷ 2, instead of multiplying; 0.4 g comes from dividing the density by the volume, 2 ÷ 5; 7 g comes from adding the density and the volume.
- (b) 19.0 g/cm³ — Density = mass ÷ volume. 342.6 ÷ 18 = 19.0333…, which rounds to 19.0 g/cm³ (1 d.p.). 6166.8 g/cm³ comes from multiplying the mass by the volume instead of dividing (342.6 × 18). 324.6 g/cm³ comes from subtracting the volume from the mass (342.6 − 18) instead of dividing. 0.1 g/cm³ comes from dividing the volume by the mass instead of the mass by the volume (18 ÷ 342.6 = 0.0525…, rounded to 1 d.p.).
- (d) 50 — Speed = distance ÷ time, so 150 ÷ 3 = 50 km/h. (450 comes from multiplying 150 by 3 instead of dividing. 153 comes from adding 150 and 3. 147 comes from subtracting 3 from 150 — none of these use the correct compound-unit calculation, distance ÷ time.)
- (a) £9.00 — Rate of pay = total pay ÷ number of hours. £54 ÷ 6 = £9.00 per hour. £324.00 comes from multiplying the pay by the hours instead of dividing (£54 × 6). £0.11 comes from dividing the hours by the pay instead of the pay by the hours (6 ÷ 54). £48.00 comes from subtracting the hours from the pay (£54 − 6) instead of dividing.
- (c) 2 hours — Method: time = distance ÷ speed, and the units of the speed fix the unit of the time. Working: 120 ÷ 60 = 2, and because the speed is in kilometres per hour the time is in hours. Answer: 2 hours. The distractors: 0.5 hours comes from dividing the speed by the distance, 60 ÷ 120, the division the wrong way round; 2 minutes comes from the correct division but the wrong unit, reading kilometres per hour as kilometres per minute; 180 minutes comes from adding 120 and 60 instead of dividing.
- (b) 2.5 hours — Method: when two objects move in opposite directions the gap between them grows at the sum of their speeds, so divide the required gap by that combined rate. Working: 80 + 60 = 140 km of gap each hour, and 350 ÷ 140 = 2.5. Answer: 2.5 hours. The distractors: 17.5 hours comes from subtracting the speeds, 80 − 60 = 20, which is the rate for two cars travelling in the same direction; 5 hours comes from using the mean of the two speeds, 70 km/h, instead of their sum; 4.375 hours comes from dividing 350 by 80 and ignoring the second car altogether.
- (d) 20 km/h — Method: average speed = total distance ÷ total time, with the time written in hours. Working: 1 hour 30 minutes = 1.5 hours, and 30 ÷ 1.5 = 20. Answer: 20 km/h. The distractors: 45 km/h comes from multiplying 30 by 1.5 instead of dividing; 15 km/h comes from dividing by 2, as if the ride had taken 2 hours; 30 km/h comes from dividing by the whole hour only and ignoring the extra 30 minutes.
- (a) £11.70 — Rate of pay = total pay ÷ hours worked, so £105.30 ÷ 9 = £11.70 per hour. Working out £105.30 − 9 = £96.30 subtracts the number of hours from the total pay instead of dividing. Working out £105.30 × 9 = £947.70 multiplies total pay by hours worked instead of dividing. Misplacing the decimal point in the correct answer gives £117.00 instead of £11.70. Maya's rate of pay is £11.70 per hour.
- (c) 9 g/cm³ — Density = mass ÷ volume, so 270 ÷ 30 = 9 g/cm³. Working out 270 − 30 = 240 subtracts the volume from the mass instead of dividing. Writing 270 gives the mass on its own, without dividing by the volume at all. Working out 270 × 30 = 8100 multiplies mass and volume together instead of dividing. The density of the block is 9 g/cm³.
- (d) £117.60 — Add the hours worked over the two days: 6 + 4.5 = 10.5 hours. Multiply by the rate of pay: 10.5 × £11.20 = £117.60. (£67.20 is Monday's pay only. £50.40 is Tuesday's pay only. £106.40 comes from mistakenly adding the hours as 6 + 3.5 = 9.5 — misreading Tuesday's 4.5 hours as 3.5 — and then multiplying by £11.20.)
- (d) 25 minutes — Method: find the time for one kilometre, then multiply by the number of kilometres — the unitary method with a rate. Working: 10 ÷ 2 = 5 minutes per km, and 5 × 5 = 25. Answer: 25 minutes. The distractors: 20 minutes comes from multiplying the 10 minutes by 2, the distance in the given rate, instead of by the scale factor 2.5; 50 minutes comes from multiplying 10 by 5, treating the 10 minutes as the time for a single kilometre; 15 minutes comes from adding the 5 km on to the 10 minutes, adding quantities that are not the same kind.
- (c) £0.90 — Cost per kilogram = total cost ÷ mass. £4.50 ÷ 5 = £0.90. £1.11 comes from dividing the mass by the cost instead of the cost by the mass (5 ÷ 4.50). £22.50 comes from multiplying the cost by the mass instead of dividing (£4.50 × 5). £0.50 comes from subtracting the two numbers (5 − 4.50) instead of dividing.
- (a) 15 km/h — Speed = distance ÷ time, so 45 ÷ 3 = 15 km/h. Working out 45 × 3 = 135 multiplies distance and time together instead of dividing. Working out 45 + 3 = 48 and 45 − 3 = 42 both combine the two values by addition or subtraction, which does not give a speed at all. The cyclist's average speed is 15 km/h.
- (b) 25 m/s — Convert km/h to m/s by multiplying by 1000 (km to m) and dividing by 3600 (hours to seconds): 90 × 1000 ÷ 3600 = 25 m/s. Working out 90 ÷ 60 = 1.5 converts using 60, as if going from hours to minutes rather than to seconds. Working out 90 × 3.6 = 324 multiplies by 3.6 instead of dividing by it, going the wrong way between the units. Working out 90 × 1000 = 90000 converts kilometres to metres but forgets to convert hours to seconds at all. The train's speed is 25 m/s.
- (a) 36 — Pressure = force ÷ area, so 126 ÷ 3.5 = 36 N/m². (0.03 comes from dividing the area by the force instead of the force by the area, the wrong way round. 129.5 comes from adding 126 and 3.5 instead of dividing. 441 comes from multiplying 126 by 3.5 instead of dividing.)
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