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.It takes 8 painters 6 days to paint a fence. Working at the same rate, work out how many days 4 painters would take to paint the same fence.
- 2.In a mix of mortar the ratio of the mass of sand to the mass of cement is 5 : 2. Write the mass of cement as a fraction of the total mass of the mix.
- 3.A scale model of a bridge is built at a scale of 1 : 120. The real bridge is 84 m long. Work out the length of the model, in centimetres.
- 4.A baker uses 450 g of flour and 300 g of butter in a batch of pastry. Write the mass of butter as a fraction of the total mass of flour and butter, in its simplest form.
- 5.The volume of water in a paddling pool, in litres, is plotted against the time since the tap was turned on, in minutes. What are the units of the gradient of a tangent to this graph?
- 6.The concentration of a pollutant in a lake, in arbitrary units, follows the recurrence C_{n+1} = 0.75C_n + 40, with C_0 = 500: each week, 25% of the pollutant breaks down naturally, and then a further 40 units enter the lake from run-off. An ecologist classifies the lake as safe once C_n first drops below 300. Find the first whole number of weeks after which the lake is safe.
- 7.Two quantities x and y are in the ratio x : y = 2 : 5, and y = kx for a constant k. Work out the value of k.
- 8.In a bag, the ratio of green counters to yellow counters is 3 : 8. There are 24 yellow counters. A student says, 'There must be 9 green counters.' Is the student correct? Choose the correct verdict and reason.
- 9.A machine fills bottles at a constant rate. It fills 18 bottles in 3 minutes. Working at the same rate, work out how many bottles the machine fills in 8 minutes.
- 10.A cleaning solution is made by mixing disinfectant and water in the ratio 1 : 9. Priti wants to make 6 litres of the solution in total. Work out how much disinfectant she needs.
- 11.y is directly proportional to x. When x = 8, y = 20. Work out the value of y when x = 12.
- 12.A laptop bag has a mass of 2.4 kg. A school bag has a mass of 3.6 kg. Write the mass of the laptop bag as a fraction of the mass of the school bag, giving your answer in its simplest form.
- 13.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.
- 14.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.
- 15.A lorry is carrying a load of mass m tonnes. Given that 1 tonne = 1000 kg, write down an expression for the mass of the load in kilograms.
Answer key
- (b) 12 days — This is inverse proportion: fewer painters take longer. Multiply the original numbers to find the total painter-days needed: 8 × 6 = 48 painter-days. Divide by the new number of painters: 48 ÷ 4 = 12 days. Working out 6 × 4 ÷ 8 = 3 days treats it as direct proportion, as if fewer painters needed less time. Stopping at 48 gives the total painter-days, not the number of days. Working out 6 + (8 − 4) = 10 days adds the change in the number of painters straight onto the number of days, treating painters and days as the same kind of quantity. 4 painters take 12 days.
- (c) 2/7 — Method: each number in a ratio counts parts, so the whole is all the parts added together, and the fraction wanted is the cement parts over that total. Working: the mix has 5 + 2 = 7 parts altogether, and 2 of those parts are cement, so the cement is 2/7 of the mix. Answer: 2/7. The distractors: 2/5 comes from writing the cement parts over the sand parts, comparing one part with the other part instead of with the whole; 5/7 comes from writing the sand parts over the total, which answers about the wrong material; 7/2 comes from writing the total over the cement parts, turning the fraction upside down.
- (c) 70 cm — Method: convert the real length to centimetres, then divide by the scale factor. Working: 84 m = 8400 cm. 8400 ÷ 120 = 70 cm. Wrong options: 0.7 cm comes from dividing 84 by 120 without converting metres to centimetres; 1,008,000 cm comes from multiplying instead of dividing (8400 × 120); 7 cm comes from converting 84 m to 840 cm (using ×10 instead of ×100) before dividing.
- (b) 2/5 — Method: a fraction taken 'of the total' has the whole batch as its denominator, so add the two masses first and then write the butter over that total. Working: the total mass is 450 g + 300 g = 750 g, so the fraction is 300/750; the highest common factor of 300 and 750 is 150, and 300 ÷ 150 = 2 while 750 ÷ 150 = 5. Answer: 2/5 of the batch. The distractors: 2/3 comes from comparing the butter with the flour, 300/450, a part-to-part fraction when the question asks for a part compared with the whole; 3/5 comes from writing the flour over the total, 450/750, which answers about the wrong ingredient; 3/2 comes from writing the flour over the butter, 450/300, which both uses the wrong denominator and reverses the order.
- (b) litres per minute — The gradient of a tangent is the change in the quantity on the vertical axis divided by the change in the quantity on the horizontal axis, so its units come from both axes: litres on the vertical axis and minutes on the horizontal axis give litres per minute. Giving the units as minutes for each litre inverts the fraction, giving the units of the RECIPROCAL of the gradient, not the gradient itself. Writing just litres uses only the vertical axis's units and ignores that a gradient is a rate, not an amount. Writing just minutes uses only the horizontal axis's units. A gradient always combines both axes' units as one divided by the other.
- (a) 4 weeks — Apply the recurrence week by week. C_1 = 0.75 × 500 + 40 = 375 + 40 = 415. C_2 = 0.75 × 415 + 40 = 311.25 + 40 = 351.25. C_3 = 0.75 × 351.25 + 40 = 263.4375 + 40 = 303.4375. C_4 = 0.75 × 303.4375 + 40 = 227.578125 + 40 = 267.578125. C_3 = 303.4375 is still above 300, but C_4 = 267.58 has dropped below it, so the lake first becomes safe after 4 weeks. Taking 25% of the ORIGINAL 500 every week instead of 25% of the current amount, a flat 125 each time, gives 500 − 125 + 40 = 415, then 415 − 125 + 40 = 330, then 330 − 125 + 40 = 245, which crosses 300 a week too early and gives the wrong answer of 3 weeks. Continuing one extra step to C_5 = 0.75 × 267.578125 + 40 = 200.68 + 40 = 240.68 and calling it 5 weeks overshoots, since the concentration had already dropped below 300 at C_4. Forgetting the 40 units of run-off each week and only applying the decay gives C_1 = 0.75 × 500 = 375, then C_2 = 0.75 × 375 = 281.25 — this is already below 300 after only 2 weeks, because without the run-off the concentration falls much faster.
- (c) 2.5 — x : y = 2 : 5 means that for every matching pair of values, y ÷ x = 5 ÷ 2 = 2.5. So y = 2.5x, and comparing with y = kx gives k = 2.5. Dividing the other way round, 2 ÷ 5 = 0.4, gives x in terms of y — that is the constant for x = 0.4y, not for y = kx. Taking the y-part of the ratio on its own, 5, reads one number off the ratio instead of dividing the y-part by the x-part; 5 would only be right if the x-part were 1. Subtracting the two parts, 5 − 2 = 3, treats the ratio as a difference, but a ratio compares two quantities by multiplication, not by subtraction. The constant is k = 2.5.
- (c) Yes, correct, because 24 ÷ 8 × 3 = 9. — Method: divide the count you know by its ratio part to find the value of one part, then multiply by the part you want. Working: one part = 24 ÷ 8 = 3 counters, so green = 3 × 3 = 9 counters, and the student is correct. Dividing 24 by 3 instead of 8 gives 24 ÷ 3 × 8 = 64, the ratio parts the wrong way round. Stopping after 24 ÷ 8 = 3 gives 3, forgetting to multiply by the green ratio part. Adding 24 + 3 = 27 confuses adding a ratio part with scaling by it.
- (d) 48 — Find the rate first: 18 ÷ 3 = 6 bottles per minute. Then apply it to the new time: 6 × 8 = 48 bottles. Working out 18 + (8 − 3) = 23 adds the extra 5 minutes onto the number of bottles instead of scaling proportionally. Working out 18 × 8 = 144 multiplies the given number of bottles by the new number of minutes without finding the rate first. Writing 18 keeps the count the same, not realising it must change with the time. In 8 minutes the machine fills 48 bottles.
- (a) 0.6 litres — The ratio 1 : 9 means the solution has 1 + 9 = 10 equal parts in total. Each part is 6 ÷ 10 = 0.6 litres, and disinfectant is 1 part, so Priti needs 0.6 litres of disinfectant. Giving 0.667 litres divides by 9, the number of parts of water, instead of the total number of parts, 10 (6 ÷ 9 ≈ 0.667). Giving 6 litres is the total amount of solution, not just the disinfectant's share of it. Giving 5.4 litres works out the water's share (6 × 9 ÷ 10 = 5.4), not the disinfectant's.
- (c) 30 — Method: use y = kx and find k from the given pair of values, then substitute x = 12. Working: k = 20 ÷ 8 = 2.5, so y = 2.5 × 12 = 30. Answer: 30. 24 comes from treating the relationship as additive, adding the increase in x (12 − 8 = 4) straight onto y (20 + 4 = 24), instead of multiplying by k. 14.5 comes from finding k correctly (2.5) but then adding it to x instead of multiplying (12 + 2.5 = 14.5). 4.8 comes from finding k upside down, 8 ÷ 20 = 0.4, and multiplying by x: 12 × 0.4 = 4.8.
- (a) 2/3 — Put the laptop bag's mass over the school bag's mass: 2.4/3.6. Multiply both numbers by 10 to clear the decimals: 24/36. Divide both by their highest common factor, 12: 24÷12 = 2, 36÷12 = 3, giving 2/3. (3/2 comes from writing the masses the wrong way round. 1/3 comes from finding the difference in the masses, 3.6 − 2.4 = 1.2 kg, and writing it as a fraction of the school bag's mass, 1.2/3.6. 2/5 comes from comparing the laptop bag's mass to the total mass of both bags, 2.4/6.)
- (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.
- (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.
- (a) 1000m — Method: kilograms are a smaller unit than tonnes, so change tonnes into kilograms by multiplying by 1000. Working: m tonnes = m × 1000 kg = 1000m kg. So the expression is 1000m. Distractor m/1000 comes from dividing by 1000 instead of multiplying, which would make the number of kilograms smaller than the number of tonnes, the wrong way round. Distractor 1000 + m comes from adding the conversion factor instead of multiplying by it. Distractor m − 1000 comes from subtracting the conversion factor instead of multiplying by it.
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