Printable · GCSE Foundation · ages 14-16
Ratio, proportion and rates of change worksheet — GCSE Foundation
Fifteen questions across the ratio, proportion and rates of change statements at Foundation tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
Ratio, proportion and rates of change worksheet — GCSE Foundation
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- 1.An Ordnance Survey map has a scale of 1 : 50 000. A cycle route measures 9.4 cm on the map. Work out the real length of the route, in kilometres.
- 2.A kettle uses 2.1 kWh of energy. A toaster uses 0.9 kWh of energy. Write the energy used by the kettle as a fraction of the energy used by the toaster, giving your answer in its simplest form.
- 3.A worker is paid £13.20 per hour. Work out this rate of pay in pence per minute.
- 4.A statue exerts a downward force of 68.4 N on its base, which has an area of 2.85 m². Work out the pressure that the statue exerts on its base, in N/m².
- 5.A metal has a density of 7.8 g/cm³. Work out the density of the metal in kg/m³.
- 6.Two broadband providers show their monthly cost on straight-line graphs, with the cost in pounds on the vertical axis and the data used in gigabytes on the horizontal axis. Provider X's line passes through (0, 15) and (100, 35). Provider Y's line passes through (0, 5) and (100, 45). Work out which provider is cheaper for a customer who uses 40 gigabytes in a month.
- 7.A cyclist travels 45 km in 3 hours at a constant speed. Work out the cyclist's average speed, in km/h.
- 8.A model railway is built at a scale where the ratio of model length to real length is 1 : 76. Given that a real carriage is 20.6 metres long, work out the length of the model carriage in centimetres, to 1 decimal place.
- 9.A plan of a garden is drawn to a scale of 1 : 200. A path measures 3 cm on the plan. Work out the real length of the path, in centimetres.
- 10.A cyclist rides at an average speed of 24 km/h for 1 hour 15 minutes. Work out the distance travelled, in km.
- 11.Line P passes through the origin and the point (3, 21). Line Q passes through the origin and the point (5, 45). Both lines represent the distance, in metres, run by an athlete against the time, in seconds. Work out the gradient of the steeper line.
- 12.A machine is bought for £8500. Its value depreciates by 6% each year. Work out the value of the machine after 3 years, to the nearest pound.
- 13.A runner's speed is 8 m/s. Given that there are 3600 seconds in an hour and 1000 metres in a kilometre, work out the runner's speed in km/h.
- 14.At a youth club the ratio of juniors to seniors is 3:5. There are 40 members altogether. Work out how many seniors there are.
- 15.A hosepipe fills a paddling pool at a constant rate. The volume of water in the pool increases by 15 litres every 3 minutes. Write down the gradient of the graph of volume against time.
Answer key
- (d) 4.7 km — Multiply the map length by the scale factor: 9.4 × 50 000 = 470 000 cm. Convert to kilometres: 470 000 cm = 4700 m = 4.7 km. Converting only to metres and calling the answer 4700 kilometres mistakes metres for kilometres. Misreading the scale as 1 : 5000 instead of 1 : 50 000, 9.4 × 5000 = 47 000 cm = 0.47 km, is ten times too small. Misplacing the decimal point in 9.4 and effectively using 94, 94 × 50 000 = 4 700 000 cm = 47 km, is ten times too big.
- (d) 7/3 — Put the kettle's energy over the toaster's energy: 2.1/0.9. Multiply both numbers by 10 to clear the decimals: 21/9. Divide both by their highest common factor, 3: 21÷3 = 7, 9÷3 = 3, giving 7/3. (3/7 comes from writing the energy values the wrong way round. 4/3 comes from finding the difference, 2.1 − 0.9 = 1.2 kWh, and writing it as a fraction of the toaster's energy, 1.2/0.9. 7/10 comes from comparing the kettle's energy to the total energy used by both appliances, 2.1/3.0.)
- (a) 22p — First convert the hourly rate to pence: £13.20 = 1320p. Then convert from per hour to per minute by dividing by 60, since there are 60 minutes in an hour: 1320 ÷ 60 = 22p per minute. Working out 1320 × 60 = 79200p multiplies by 60 instead of dividing, going the wrong way between per hour and per minute. Working out 13.20 × 10 = 132p converts pounds to pence using the wrong power of ten, and dividing that by 60 carries the error through to give 2.2p. Working out 13.20 × 100 = 1320p converts the currency correctly but stops there, leaving the rate as pence per hour rather than completing the second conversion to pence per minute. The rate of pay is 22p per minute.
- (a) 24 N/m² — Pressure = force ÷ area. 68.4 ÷ 2.85 = 24 N/m². 194.94 N/m² comes from multiplying the force by the area instead of dividing (68.4 × 2.85). 65.55 N/m² comes from subtracting the area from the force (68.4 − 2.85) instead of dividing. 0.04 N/m² comes from dividing the area by the force instead of the force by the area (2.85 ÷ 68.4).
- (c) 7800 kg/m³ — Method: build the conversion factor from the two unit changes separately — one for the mass, one for the volume. Working: 1 kg = 1000 g, so the mass figure is divided by 1000; 1 m = 100 cm, so 1 m³ = 100 × 100 × 100 = 1000000 cm³ and the volume figure is multiplied by 1000000. The density figure is therefore multiplied by 1000000 ÷ 1000 = 1000, giving 7.8 × 1000 = 7800. So the density of the metal is 7800 kg/m³. Distractor 780 kg/m³ comes from multiplying by 100 instead of 1000. Distractor 78000 kg/m³ comes from multiplying by 10000, an extra zero. Distractor 7.8 kg/m³ comes from not converting the units at all.
- (b) Provider Y — £21.00 against Provider X's £23.00 — Provider X's gradient is (35 − 15) ÷ 100 = 0.2, so cost = 15 + 0.2 × 40 = 15 + 8 = £23.00. Provider Y's gradient is (45 − 5) ÷ 100 = 0.4, so cost = 5 + 0.4 × 40 = 5 + 16 = £21.00. £21.00 is less than £23.00, so Provider Y is cheaper: 'Provider Y — £21.00 against Provider X's £23.00'. Getting both costs right but naming Provider X as cheaper compares the two numbers the wrong way round — £23.00 is more than £21.00, not less. Comparing only the fixed fees, £15.00 and £5.00, ignores the cost of the 40 gigabytes actually used. Reading off the costs at 100 gigabytes, £35.00 and £45.00, directly from the graph answers a different usage from the 40 gigabytes the question asks about.
- (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.
- (c) 27.1 cm — The model length is 20.6 ÷ 76 = 0.271052... metres. Converting to centimetres by multiplying by 100 gives 27.1052..., which rounds to 27.1 cm. Forgetting to convert metres to centimetres leaves the answer as 0.271052... metres, which rounds to 0.3 cm if the unit is simply relabelled. Multiplying by 1000 instead of 100 when converting metres to centimetres gives 271.052..., which rounds to 271.1 cm. Multiplying by 76 instead of dividing, 20.6 × 76 = 1565.6, uses the scale factor the wrong way round — that would be the real length if the model were 20.6 units long, not the other way round.
- (a) 600 cm — Method: a scale of 1 : n means each 1 cm on the plan stands for n cm in real life, so multiply the plan length by the scale. Working: 3 × 200 = 600. Answer: 600 cm. The distractors: 60 cm comes from multiplying by 20 instead of 200; 203 cm comes from adding the scale to the plan length instead of multiplying; 6000 cm comes from reading the scale as 1 : 2000.
- (c) 30 km — First convert 1 hour 15 minutes to hours: 15 minutes is a quarter of an hour, so the time is 1.25 hours. Then multiply by the speed: 24 × 1.25 = 30 km. Reading 15 minutes as 0.15 hours (writing the minutes after the decimal point instead of as a fraction of 60) gives 24 × 1.15 = 27.6 km. Working out 24 × 75 = 1800 multiplies by the number of minutes instead of converting to hours first. Working out 24 ÷ 1.25 = 19.2 divides by the time instead of multiplying. The cyclist travels 30 km.
- (b) 9 m/s — The gradient of line P is 21 ÷ 3 = 7, so P has a rate of 7 m/s. The gradient of line Q is 45 ÷ 5 = 9, so Q has a rate of 9 m/s. Because 9 is greater than 7, line Q is the steeper line, with gradient 9 m/s. Taking line P's gradient instead of Q's gives 7 m/s, the less steep line. Subtracting the two lines' coordinates directly, (45 − 21) ÷ (5 − 3) = 24 ÷ 2 = 12 m/s, mixes points from different lines rather than using one line's own two points. Adding the two gradients, 7 + 9 = 16 m/s, treats 'steeper' as a total rather than a comparison.
- (d) £7060 — A 6% decrease each year means the value becomes 100% − 6% = 94% of the previous year's value, and 94% = 0.94, so the multiplier is 0.94. Apply it once for each of the 3 years: £8500 × 0.94 = £7990 after 1 year, £7990 × 0.94 = £7510.60 after 2 years, £7510.60 × 0.94 = £7059.96 after 3 years, which rounds to £7060 to the nearest pound. (£6970 comes from using simple depreciation instead of compound, taking 6% of the original £8500 three times: £8500 − 3 × £510 = £6970. £7990 is the value after only 1 year, forgetting the remaining 2 years. £7511 is the value after only 2 years, £8500 × 0.94² = £7510.60, forgetting the third year.)
- (a) 28.8 km/h — Method: first change metres per second into metres per hour, then change metres into kilometres. Working: 8 × 3600 = 28800 metres per hour, then 28800 ÷ 1000 = 28.8 km/h. So the runner's speed is 28.8 km/h. Distractor 28800 km/h comes from stopping after the first step and forgetting to change metres into kilometres. Distractor 2.22 km/h comes from dividing by 3600 instead of multiplying, then multiplying by 1000. Distractor 2.88 km/h comes from using 360 instead of 3600 seconds in an hour, missing a zero.
- (d) 25 — Method: add the parts of the ratio, divide the total membership by the number of parts to find the value of one part, then multiply by the parts belonging to the group asked for. Working: 3 + 5 = 8 parts, 40 ÷ 8 = 5 members in one part, and the seniors are 5 parts, so 5 × 5 = 25. Answer: 25 seniors. The distractors: 15 is the number of juniors, which is the 3-part group; 5 is the size of one part only; 24 comes from dividing the 40 by 5, the seniors' number in the ratio, to get 8 and then multiplying that by 3.
- (d) 5 — The gradient equals the amount gained divided by the time taken: 15 ÷ 3 = 5 litres per minute.
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