Printable · GCSE Higher · ages 14-16
Ratio, proportion and rates of change worksheet — GCSE Higher
Fifteen questions across the ratio, proportion and rates of change statements at Higher 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 Higher
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- 1.5 decorators paint 3 rooms in 6 hours. All the decorators work at the same rate. Work out how long it takes 10 decorators to paint 5 rooms.
- 2.A straight-line graph passes through the origin and the point (4, 10). The line represents y = kx. Write down the ratio x : y in its simplest form.
- 3.A map has a scale of 1 : 20 000. A different map of the same area has a scale of 1 : 80 000. A lake's shoreline is drawn 5 cm long on the first map. Work out the length of the same shoreline on the second map, in centimetres.
- 4.A concert hall has 54 seats in total, of which 24 are reserved and the rest are free. Write the number of free seats as a fraction of the number of reserved seats. Give your answer in its simplest form.
- 5.Two quantities x and y are in direct proportion. When x = 8, the value of y is 20. Work out the value of y when x = 14.
- 6.The density of a type of solid plastic is 0.9 g/cm³. Work out the mass of 0.5 m³ of the plastic, in kilograms.
- 7.A post 2 metres tall casts a shadow 3 metres long. At the same time a nearby tree casts a shadow 12 metres long. Work out the height of the tree.
- 8.A cyclist's speed, in metres per second, is plotted against time, in seconds, since she sets off. Which statement correctly pairs the gradient of a tangent to this graph, and the area between the graph and the time-axis, with what each one represents?
- 9.A charity collects donations from adults and children in the ratio 5:2. Altogether, £238 is collected. Work out how much more the adults donate than the children.
- 10.A gardener mixes 300 ml of plant feed concentrate with 1.2 litres of water to make a spray. Write the ratio of concentrate to water in its simplest form.
- 11.A cyclist rides 30 km in 1 hour 30 minutes. Work out the average speed of the cyclist in km/h.
- 12.The time taken for a train journey is inversely proportional to the average speed of the train. At an average speed of 60 km/h the journey takes 2 hours. Work out the time taken at an average speed of 40 km/h.
- 13.Simplify the ratio 12 : 18 : 30 to its simplest form.
- 14.6 identical taps fill a paddling pool in 20 minutes. Each tap fills at the same steady rate. Work out how long 3 of these taps would take to fill the same pool.
- 15.The cost, in pounds, of hiring a minibus is modelled against the number of passengers booked. A tangent to the cost graph at 20 passengers passes through the points (16, 184) and (24, 216). Work out the instantaneous rate at which the cost increases with each extra passenger, at 20 passengers.
Answer key
- (c) 5 hours — Method: measure the job in decorator-hours, which is in the same ratio as the number of rooms, then share the decorator-hours between the decorators available. Working: 5 decorators × 6 hours = 30 decorator-hours for 3 rooms, so one room takes 30 ÷ 3 = 10 decorator-hours; 5 rooms take 5 × 10 = 50 decorator-hours; shared between 10 decorators that is 50 ÷ 10 = 5 hours. Answer: 5 hours. The distractors: 3 hours comes from halving the 6 hours because the number of decorators doubles, while forgetting that there are also more rooms to paint; 10 hours comes from scaling the 6 hours up for the rooms only, 6 × 5 ÷ 3, and leaving the workforce at 5 decorators; 6 hours comes from assuming that doubling the decorators and increasing the rooms cancel each other out, which they do not, because the rooms rise by a factor of 5/3 and the workforce by a factor of 2.
- (c) 2 : 5 — The point (4, 10) gives x = 4, y = 10, so x : y = 4 : 10. Dividing both parts by their highest common factor, 2, gives 2 : 5 in simplest form. Inverting the whole ratio gives 5 : 2, which is y : x instead of x : y. Dividing only the x-part by 2 and leaving the y-part as 10 gives 2 : 10, but scaling one part on its own changes the ratio: 2 : 10 is the same as 1 : 5, not 4 : 10. Dividing only the y-part by 2 and leaving the x-part as 4 gives 4 : 5, the same one-sided mistake made on the other part of the ratio.
- (c) 1.25 cm — Method: find the real length using the first map's scale, then use the second map's scale to find its drawn length. Working: real length = 5 × 20 000 = 100 000 cm. On the second map: 100 000 ÷ 80 000 = 1.25 cm. Wrong options: 20 cm comes from inverting the ratio of the two scales (5 × 80 000 ÷ 20 000); 5 cm comes from wrongly assuming the length looks the same on both maps; 12.5 cm comes from dropping a zero from the second scale factor and dividing by 8 000 instead of 80 000 (100 000 ÷ 8 000).
- (d) 5/4 — Work out the number of free seats: 54 − 24 = 30. Form the fraction 30/24; both numbers share a factor of 6, so 30 ÷ 6 = 5 and 24 ÷ 6 = 4, giving 5/4. 4/5 comes from writing the fraction the wrong way round, as reserved over free (24/30). 5/9 comes from comparing the free seats with the total number of seats (30/54), instead of with the reserved seats. 5/6 comes from miscalculating 54 − 24 as 20 instead of 30, then forming 20/24.
- (d) 35 — Method: in direct proportion the ratio y : x is the same for every pair, so find the constant and substitute the new value of x. Working: k = 20 ÷ 8 = 2.5, so y = 2.5x; when x = 14, y = 2.5 × 14 = 35. Answer: 35. The distractors: 26 comes from additive thinking — x rises by 6, so 6 is added to y — which would keep the difference constant rather than the ratio; 28 comes from rounding the constant 2.5 down to 2 and working out 2 × 14, which loses the half in the constant; 5.6 comes from using the constant upside down, 8 ÷ 20 = 0.4, and working out 0.4 × 14.
- (b) 450.00 kg — 1 m³ = 100 × 100 × 100 = 1,000,000 cm³, so 0.5 m³ = 500,000 cm³. Mass = density × volume = 0.9 × 500,000 = 450,000 g. Converting to kilograms by dividing by 1000 gives 450,000 ÷ 1000 = 450.00 kg. Skipping the m³-to-cm³ conversion and multiplying 0.9 × 0.5 = 0.45 treats the volume as if it were already 0.5 cm³, giving 0.45 kg. Finding the mass correctly in grams, 450,000 g, but not converting to kilograms leaves 450000.00 kg, out by a factor of 1000. Using the area conversion factor of 10,000, as if converting m² to cm², instead of the volume factor of 1,000,000 gives 0.5 × 10,000 = 5,000 'cm³', and a mass of 0.9 × 5,000 = 4,500 g, which is 4.50 kg.
- (c) 8 m — Method: in the same sunlight every object has its height and its shadow in the same ratio, so write 2:3 = h:12, find the multiplier that takes 3 to 12 and apply it to the height. Working: 12 ÷ 3 = 4, so the tree's shadow is 4 times the post's shadow; the height must be scaled by the same 4, giving 4 × 2 = 8 m. Answer: 8 m. The distractors: 18 m comes from setting up the proportion upside down, 12 ÷ 2 × 3, which scales by shadow over height instead of height over shadow; 24 m comes from multiplying the 12 m shadow by the post's height of 2 m and never dividing by the post's shadow of 3 m; 4 m is the scale factor 12 ÷ 3, given as a length instead of being used to scale the 2 m post.
- (d) Gradient = acceleration; area = distance travelled. — Method: on a speed–time graph, the gradient of the graph at an instant is the rate of change of speed with time, which is acceleration; the area between the graph and the time-axis over an interval is the total distance covered in that interval, because it accumulates speed × time. Working: gradient = acceleration and area = distance travelled is the correct pairing. Swapping the two quantities completely, gradient = distance travelled and area = acceleration, is the reverse of what each actually measures. Keeping gradient = acceleration correct but then also claiming area = acceleration too is wrong because the area is a different physical quantity, distance, not a second way of finding the same rate. Claiming the gradient itself gives the speed confuses the RATE OF CHANGE of the plotted quantity with the plotted quantity itself — the gradient is how fast the speed is changing, not the speed. On any rate graph, the gradient of the graph is always the RATE at that instant, and the area under the graph is always the TOTAL AMOUNT accumulated — keep straight which of the two questions each one answers.
- (a) £102 — Method: find the value of one part of the ratio, then work out each group's share before comparing them. Working: the ratio 5:2 has 5 + 2 = 7 parts, so one part is £238 ÷ 7 = £34. Adults donate 5 × £34 = £170 and children donate 2 × £34 = £68, so adults donate £170 − £68 = £102 more than children. So the difference is £102. Distractor £68 is only the children's donation, without finding the difference. Distractor £170 is only the adults' donation, without finding the difference. Distractor £136 comes from doubling the children's donation instead of subtracting it from the adults' donation.
- (c) 1 : 4 — Convert to the same unit: 1.2 litres = 1200 ml, since 1 litre = 1000 ml. This gives the ratio 300 : 1200. Divide both parts by their highest common factor, 300, to get 1 : 4. Giving 3 : 12 divides by 100 only, which is a common factor but not the highest one. Giving 1 : 1.2 has not converted 1.2 litres into millilitres, so the two parts are not in the same unit. Giving 4 : 1 swaps the order, comparing water to concentrate instead of concentrate to water.
- (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.
- (c) 3 hours — Method: inverse proportion means speed × time is constant for the journey, so find that constant and divide it by the new speed. Working: 60 × 2 = 120, which is the distance in kilometres; at 40 km/h the time is 120 ÷ 40 = 3 hours. Answer: 3 hours. The distractors: 1.5 hours is the ratio of the speeds, 60 ÷ 40, given as a time instead of being used to scale the original 2 hours; 1 hour 20 minutes comes from treating time as directly proportional to speed, 2 × 40 ÷ 60, which has the slower train arriving sooner; 2 hours comes from finding the constant 120 and then dividing it by the original 60 km/h again, so the time never changes.
- (d) 2:3:5 — The highest common factor of 12, 18 and 30 is 6. Divide every part by 6: 12 ÷ 6 = 2, 18 ÷ 6 = 3, 30 ÷ 6 = 5, giving 2 : 3 : 5. Dividing by 2 instead of 6 gives 6 : 9 : 15, which still shares a common factor of 3, so it is not fully simplified. Dividing by 3 instead of 6 gives 4 : 6 : 10, which still shares a common factor of 2, so it is not fully simplified either. Swapping the first two parts gives 3 : 2 : 5, the parts in the wrong order.
- (c) 40 minutes — Method: this is inverse proportion — fewer taps means longer, not shorter — so the number of taps × the time taken stays constant. Working: 6 × 20 = 120, and with 3 taps the time is 120 ÷ 3 = 40 minutes. So 3 taps take 40 minutes. Distractor 10 minutes comes from treating it as direct proportion instead of inverse, working out 20 × 3 ÷ 6. Distractor 30 minutes comes from halving the number of taps and adding half the original time, 20 + 10, instead of doubling the time. Distractor 17 minutes comes from subtracting the number of taps removed, 3, directly from the original time, 20.
- (d) £4 — The gradient of the tangent gives the instantaneous rate of change of cost with respect to the number of passengers, in pounds per passenger. The tangent passes through (16, 184) and (24, 216), so the change in cost is 216 − 184 = 32 and the change in passengers is 24 − 16 = 8. The gradient is 32 ÷ 8 = 4. Stopping after finding the change in cost, without dividing by the change in passengers, leaves 32, not a rate. Adding the two changes instead of dividing gives 32 + 8 = 40, which is not a rate either. Reading off only the change in passengers, 8, is not a rate at all — a rate needs the change in cost as well. The instantaneous rate is £4 per extra passenger.
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