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.
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Ratio, proportion and rates of change worksheet — GCSE Higher
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- 1.A recipe uses flour, sugar and butter in the ratio 8 : 3 : 5. Write the ratio of flour to the rest of the mixture (sugar and butter combined) in its simplest form.
- 2.For the pairs x = 6, y = 15 and x = 10, y = 25, which statement is correct?
- 3.A cyclist rides 19.3 km in 47 minutes. Work out her average speed, in km/h, to 1 decimal place.
- 4.A crate exerts a force of 84 N on the floor through a base with an area of 2.4 m². Work out the pressure on the floor, in N/m².
- 5.The exchange rate is £1 = €1.15. Convert £200 to euros.
- 6.6 identical printers can print a batch of exam papers in 40 minutes, all working at the same rate. Working at the same rate, work out how many minutes 4 of these printers would take to print the same batch.
- 7.A recipe for 8 muffins needs 200 g of sugar. Sam wants to make 20 muffins for a bake sale, and he already has 350 g of sugar. Work out how many more grams of sugar he needs to buy.
- 8.Leah puts £4000 into a savings account paying 3% compound interest each year. At the end of 2 years she takes out all of the money and spends £1500 of it on a laptop. Work out how much of the money she has left.
- 9.A garden centre sells two mathematically similar sacks of grass seed. The amount of lawn a sack can treat is proportional to the volume of seed inside it. The smaller sack is 20 cm tall and treats a lawn of area 30 m². The larger sack is 40 cm tall. A gardener needs to treat a lawn with an area of 500 m² using only the larger sacks. Work out the minimum number of larger sacks needed.
- 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.A map has a scale of 1 : 25 000. Two towns are 5 km apart in real life. Work out the distance between the towns on the map, in centimetres.
- 12.The exchange rate is £1 = 1.28 US dollars. Convert £350 into US dollars.
- 13.A population of fish in a lake is modelled by the recurrence P_{n+1} = 1.1P_n − 30, where P_n is the population after n years, 30 fish are removed by fishing each year, and P_0 = 400. After how many complete years does the population first exceed 460?
- 14.The cost of manufacturing a spherical container is proportional to the cube of its radius. A container of radius 3 cm costs £54 to manufacture. Construct the equation connecting cost C and radius r, then work out the cost of a container of radius 5 cm.
- 15.A delivery driver travels 45 km in the first 30 minutes of a journey, and then a further 75 km in the next 1 hour. Work out her average speed for the whole journey, in km/h.
Answer key
- (d) 1 : 1 — Sugar and butter together make 3 + 5 = 8 parts of the mixture. Comparing flour to this, 8 : 8, simplifies to 1 : 1. Giving 1 : 2 compares flour with the whole mixture (8 + 3 + 5 = 16 parts, giving 8 : 16 = 1 : 2) instead of with the rest of the mixture. Giving 3 : 5 is the ratio of sugar to butter, not of flour to the rest of the mixture. Giving 8 : 3 compares flour only with sugar, leaving butter out altogether.
- (c) They are in direct proportion, because y ÷ x = 2.5 for both pairs. — Testing direct proportion means checking that y ÷ x is the same for every pair: 15 ÷ 6 = 2.5 and 25 ÷ 10 = 2.5, so the quantities are in direct proportion. Saying they are not in proportion because x + y differs uses addition, which is not the correct test for proportion. Saying they are not in proportion because y − x differs also uses the wrong test — subtraction, not division. Saying they are in proportion because x × y is 90 and 250 uses multiplication, which is the test for inverse proportion, and the two products are not even equal to each other, so this option also contradicts itself.
- (d) 24.6 km/h — Average speed = distance ÷ time, with time in hours. 47 minutes = 47 ÷ 60 hours. 19.3 ÷ (47 ÷ 60) = 19.3 ÷ 47 × 60 = 24.638…, which rounds to 24.6 km/h (1 d.p.). 0.4 km/h comes from dividing the distance by 47 and treating the result as km/h directly, without converting the minutes to hours at all. 41.1 km/h comes from converting minutes to hours by dividing by 100 instead of 60 (19.3 ÷ 47 × 100). 0.3 km/h comes from dividing the distance by 60 instead of converting the 47 minutes to hours first.
- (b) 35 N/m² — Pressure = force ÷ area. 84 ÷ 2.4 = 35 N/m². 201.6 N/m² comes from multiplying the force by the area instead of dividing (84 × 2.4). 81.6 N/m² comes from subtracting the area from the force (84 − 2.4) instead of dividing. 0.03 N/m² comes from dividing the area by the force instead of the force by the area (2.4 ÷ 84).
- (b) €230.00 — Multiply the amount in pounds by the exchange rate: 200 × 1.15 = 230, so £200 = €230.00. Working out 200 + 1.15 = 201.15 treats the exchange rate as an amount to add rather than a multiplier. Working out 200 × 0.15 = 30 finds only the extra amount earned for every pound and forgets to add it back to the original £200. Working out 200 × 11.5 = 2300.00 misplaces the decimal point in the exchange rate, multiplying by 11.5 instead of 1.15. £200 converts to €230.00.
- (a) 60 minutes — Method: for inverse proportion, printers × time is constant. Working: 6 × 40 = 240 (the constant). With 4 printers: 240 ÷ 4 = 60 minutes. Wrong options: 26.7 minutes comes from treating the relationship as direct proportion, scaling the time down as printers decrease (40 × 4 ÷ 6); 24 minutes comes from multiplying the two printer counts together instead of using the constant; 40 minutes comes from not adjusting the time at all for the change in printers.
- (a) 150 g — Method: scale the recipe to find the total sugar needed, then subtract the sugar Sam already has. Working: 200 ÷ 8 × 20 = 500, so 500 g is needed in total; 500 − 350 = 150, so 150 g still to buy. Stopping after finding the total, 500, without subtracting what he has gives 500 g. Scaling the wrong way round, 200 × 8 ÷ 20 = 80, wrongly suggests he already has enough, giving 0 g. Adding the amount he has instead of subtracting it, 500 + 350 = 850, gives 850 g.
- (c) £2743.60 — Each year the balance is multiplied by 1.03. After the first year: 4000 × 1.03 = 4120. After the second year: 4120 × 1.03 = 4243.60, so that is what Leah takes out. She then spends £1500 of it, which leaves 4243.60 − 1500 = 2743.60. She has £2743.60 left.
- (c) 3 — Method: find the height scale factor, cube it to find the volume (and coverage) scale factor, use it to find one large sack's coverage, then divide the total lawn area by this and round up to a whole number of sacks. Working: height scale factor = 40 ÷ 20 = 2, so coverage scale factor = 2³ = 8, and each large sack covers 30 × 8 = 240 m². 500 ÷ 240 = 2.08…, which rounds UP to 3 whole sacks. Answer: 3. 2 comes from correctly finding that each large sack covers 240 m², but then rounding 500 ÷ 240 down instead of up, which would leave part of the lawn untreated. 5 comes from squaring the height scale factor (2² = 4) instead of cubing it, giving a coverage of only 30 × 4 = 120 m² per sack. 17 comes from forgetting to scale the coverage at all and dividing 500 by the smaller sack's coverage of 30 m².
- (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.
- (c) 20 cm — Convert 5 km to centimetres: 5 km = 5000 m = 500 000 cm. Divide by the scale factor: 500 000 ÷ 25 000 = 20, giving 20 cm. Converting only as far as metres, 5000 ÷ 25 000 = 0.2, gives 0.2 cm — the conversion to centimetres was never finished. Dropping a zero in the division gives 2 cm, ten times too small. Misreading the scale as 1 : 2500 instead of 1 : 25 000 gives 500 000 ÷ 2500 = 200 cm, ten times too big.
- (c) 448.00 US dollars — Method: multiply the amount in pounds by the exchange rate. Working: £350 × 1.28 = 448.00 US dollars. Wrong options: 273.44 US dollars comes from dividing by the rate instead of multiplying (350 ÷ 1.28); 351.28 US dollars comes from adding the rate to the amount instead of multiplying; 4,480.00 US dollars comes from a decimal-point slip, using 12.8 instead of 1.28.
- (b) 5 years — The recurrence P_{n+1} = 1.1P_n − 30 must be applied once per year, checking after each application whether the population has passed 460. Starting from P_0 = 400: 400 × 1.1 − 30 = 410, so P_1 = 410. Then 410 × 1.1 − 30 = 421, so P_2 = 421. Then 421 × 1.1 − 30 = 433.1, so P_3 = 433.1. Then 433.1 × 1.1 − 30 = 446.41, so P_4 = 446.41, which is still below 460. Then 446.41 × 1.1 − 30 = 461.051, so P_5 = 461.051, the first value above 460. The population first exceeds 460 after 5 complete years. Stopping at P_4 = 446.41 and reporting 4 years reports the last year the population was still below 460, not the first year it was above. Counting the starting value P_0 = 400 as a year of growth makes P_5 the sixth number in the list and gives 6 years, but P_0 is the population before any year has passed, so P_5 is reached after 5 years, not 6. Reading "exceed 460" as "exceed the starting population of 400" instead gives P_1 = 410, already above 400, and 1 year — but the threshold named in the question is 460, not the starting value, so always check every value against the number actually stated in the question.
- (b) £250 — Since cost is proportional to the cube of the radius, C = kr³. Using r = 3, C = 54: 3³ = 27, so 54 = k × 27, giving k = 54 ÷ 27 = 2. The equation is C = 2r³. When r = 5: 5³ = 125, so C = 2 × 125 = 250. Treating the relationship as proportional to r² instead of r³ gives k = 54 ÷ 9 = 6 and then C = 6 × 25 = 150, which models area scaling, not volume scaling. Treating it as proportional to r itself gives k = 54 ÷ 3 = 18 and then C = 18 × 5 = 90. Finding k correctly from the cube but then multiplying it by the radius instead of by the cube of the radius gives 2 × 5 = 10, which applies the right constant to the wrong power of r. The cost of a container of radius 5 cm is £250.
- (a) 80 km/h — Average speed = total distance ÷ total time. Total distance = 45 + 75 = 120 km. Total time = 30 minutes + 1 hour = 1.5 hours. 120 ÷ 1.5 = 80 km/h. 60 km/h comes from treating the 30 minutes as a whole hour, giving a total time of 2 hours instead of 1.5 (120 ÷ 2). 82.5 km/h comes from averaging the two separate speeds (45 ÷ 0.5 = 90 km/h and 75 ÷ 1 = 75 km/h, then (90 + 75) ÷ 2) instead of using total distance over total time. 75 km/h comes from using only the second part of the journey (75 km in 1 hour) and ignoring the first part.
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