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.The price of a cycling helmet rises from £80 to £116. Work out the percentage increase.
- 2.40% of a number is 12 more than 25% of the same number. Work out the number.
- 3.A water tank is a cuboid measuring 30 cm × 20 cm × 10 cm. The tank is full of water. Work out how many litres of water it holds.
- 4.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.
- 5.The depth of water in a tank, in cm, is recorded every 10 seconds: at t = 10, depth = 32; at t = 20, depth = 45; at t = 30, depth = 56. Use the most appropriate chord from these readings to estimate the instantaneous rate of change of depth at t = 20.
- 6.3 builders put up a fence in 12 days. All the builders work at the same rate, so the number of days is inversely proportional to the number of builders. Work out how long 9 builders take to put up the same fence.
- 7.A tangent to a distance–time graph, with distance in kilometres and time in minutes, has gradient 0.78 kilometres per minute at a particular point. Work out the instantaneous rate of change of distance with time, in kilometres per hour.
- 8.Last week Priya worked 5 shifts of 7 hours. This week she worked 4 shifts of 8 hours. Write the number of hours she worked last week as a fraction of the number of hours she worked this week.
- 9.300 g of a salt solution of concentration 10% is poured into 200 g of a salt solution of concentration 25%. Work out the concentration of salt in the mixture.
- 10.A water butt is being filled from a hosepipe at a constant rate while a small leak drains water out at a constant rate, giving a constant net rate of change. The volume of water in the butt, V litres, is shown on a straight-line graph against time, t minutes. The line passes through the points (5, 20) and (15, 60). The butt is empty at t = 0. Work out how many minutes it takes to reach a volume of 100 litres.
- 11.In a fruit drink, cranberry juice and apple juice are mixed in the ratio 3:7. Write the amount of apple juice as a fraction of the amount of cranberry juice, in its simplest form.
- 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.The total cost, in pounds, of manufacturing a batch of components is plotted against the number of components produced. A tangent to this graph at a particular point has gradient 3.60. What does the value 3.60 represent?
- 14.A gym charges members a joining fee plus a fixed amount per month. The total amount paid, C pounds, is shown on a straight-line graph against the number of months, m. The line passes through the points (3, 130) and (7, 210). Work out the total amount a member will have paid after 12 months.
- 15.y is inversely proportional to x². When x = 2, y = 8. Construct the equation connecting x and y, then work out the value of y when x = 4.
Answer key
- (a) 45% — Method: percentage increase = increase ÷ original amount × 100. Working: the increase is £116 − £80 = £36, and 36 ÷ 80 = 0.45, so 0.45 × 100 = 45%. Answer: 45%. The distractors: 36% comes from quoting the £36 increase as though pounds and per cent were the same thing; 31% comes from dividing the £36 increase by the new price £116 instead of by the original £80, which gives 31% to the nearest per cent; 145% is the new price written as a percentage of the original price, which is the whole of the new price rather than the increase.
- (b) 80 — Method: the difference between 40% and 25% of the number is 15% of the number, and that difference is 12. Working: 15% of the number is 12, so 1% of the number is 12 ÷ 15 = 0.8, and the number is 0.8 × 100 = 80. Check: 40% of 80 is 32, 25% of 80 is 20, and 32 − 20 = 12. Answer: 80. The distractors: 30 comes from solving 40% of the number = 12; 48 comes from solving 25% of the number = 12; 15 is the percentage difference written as the answer.
- (b) 6 litres — Method: find the volume of the cuboid in cm³, then change cm³ into litres using 1 litre = 1000 cm³. Working: 30 × 20 × 10 = 6000 cm³, and 6000 ÷ 1000 = 6. Answer: 6 litres. The distractors: 60 litres comes from using 1 litre = 100 cm³; 600 litres comes from using 1 litre = 10 cm³; 0.6 litres comes from using 1 litre = 10 000 cm³.
- (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.
- (c) 1.2 cm/s — To estimate an instantaneous rate of change at a point from a table of readings, use the chord that spans the point symmetrically — equal steps either side — because the over-estimate on one side and the under-estimate on the other largely cancel. Here that is the chord from t = 10 to t = 30. The change in depth is 56 − 32 = 24 and the change in time is 30 − 10 = 20, so the estimate is 24 ÷ 20 = 1.2 cm/s. The one-sided chord from t = 20 to t = 30 gives (56 − 45) ÷ (30 − 20) = 11 ÷ 10 = 1.1 cm/s, which estimates the rate somewhere between t = 20 and t = 30 rather than at t = 20 itself. Dividing the 20-second change in depth by the 10-second gap between consecutive readings gives 24 ÷ 10 = 2.4, mixing the change from one interval with the time from another. Reporting the change in depth, 24, on its own is not a rate at all, because it has not been divided by a time. The best estimate of the instantaneous rate of change of depth at t = 20 is 1.2 cm/s.
- (a) 4 days — Method: the fence is a fixed amount of work, so builders × days is constant; find that product and divide it by the new number of builders. Working: 3 × 12 = 36 builder-days of work, so with 9 builders the time is 36 ÷ 9 = 4 days. Answer: 4 days. The distractors: 6 days comes from halving the 12 days because there are more builders, rather than dividing by the factor of 3 by which the workforce has grown; 36 days is the constant product of builders and days, given as a number of days instead of being shared between the builders; 9 days comes from taking 3 days off the 12, treating three extra builders as three fewer days, which is additive rather than proportional.
- (a) 46.8 km/h — Method: 1 hour = 60 minutes, so a rate given in kilometres per minute is converted to kilometres per hour by multiplying by 60. Working: 0.78 × 60 = 46.8, so the instantaneous rate of change is 46.8 kilometres per hour. Keeping the given value unchanged and only relabelling the unit gives 0.78 km/h, which ignores that the time unit has changed. Dividing by 60 instead of multiplying — as you would when converting to a larger length unit — gives 0.78 ÷ 60 = 0.013 km/h, the wrong direction for a rate measured against a larger time unit. Adding 60 to the given rate instead of multiplying by it gives 0.78 + 60 = 60.78 km/h. Converting a rate always means multiplying or dividing by the conversion factor between the units, never adding it, and the direction depends on whether the new time unit is bigger or smaller than the old one.
- (a) 35/32 — Work out each weekly total first. Last week: 5 × 7 = 35 hours. This week: 4 × 8 = 32 hours. Last week's total is being written as a fraction of this week's total, so last week goes on the top and this week goes on the bottom, giving 35/32. The two totals share no common factor, so the fraction cannot be cancelled. It is greater than 1, which says that Priya worked more hours last week than this week.
- (c) 16% — Method: scale each concentration to its own mass to find the salt it contains, add the two masses of salt, then write the ratio of salt to mixture per 100 g. Working: 10:100 = x:300 gives 30 g of salt, and 25:100 = y:200 gives 50 g of salt; the mixture holds 30 + 50 = 80 g of salt in 300 + 200 = 500 g of solution; 80:500 = 16:100. Answer: 16%. The distractors: 17.5% is the mean of 10% and 25%, which ignores that there is more of the weaker solution than of the stronger one; 19% comes from swapping the two concentrations over, working out (300 × 25% + 200 × 10%) ÷ 500; 26.7% comes from dividing the 80 g of salt by the 300 g of the first solution rather than by the 500 g of mixture.
- (d) 25 — Gradient = (60 − 20) ÷ (15 − 5) = 40 ÷ 10 = 4 litres per minute. Since the butt is empty at t = 0, V = 4t. Setting V = 100 gives t = 100 ÷ 4 = 25 minutes.
- (d) 7/3 — The ratio cranberry : apple is 3:7, so apple juice is 7 parts and cranberry juice is 3 parts. Write apple over cranberry: 7/3. (3/7 comes from writing the ratio the wrong way round, cranberry over apple. 7/10 comes from comparing the apple juice to the total amount of the mixture, 7 parts out of 10. 3/10 comes from comparing the cranberry juice to the total amount of the mixture, 3 parts out of 10.)
- (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) £3.60 per component — The gradient of a cost-against-components graph has units of pounds per component, since cost is measured in pounds and the horizontal axis counts components. So 3.60 means it costs an extra £3.60 to produce one more component at that point. Calling it '£3.60 total cost' confuses the gradient, a rate, with the y-value on the graph, which is the total cost itself. Giving it as 3.60 components per pound swaps which axis is on top, giving the units of the reciprocal gradient, not the gradient itself. Turning 3.60 into a percentage invents a unit that has no basis in the graph's axes — a gradient here is a number of pounds, not a percentage. Always build the gradient's units from the two axes' own units, in the order y-axis over x-axis.
- (c) £310 — Gradient = (210 − 130) ÷ (7 − 3) = 80 ÷ 4 = 20, so the monthly rate is £20. Using C = 20m + c with the point (3, 130): 130 = 60 + c, so c = 70. After 12 months: C = 20 × 12 + 70 = 240 + 70 = £310.
- (a) 2 — Since y is inversely proportional to x², y = k/x². Using x = 2, y = 8: 2² = 4, so 8 = k ÷ 4, giving k = 8 × 4 = 32. The equation is y = 32/x². When x = 4: 4² = 16, so y = 32 ÷ 16 = 2. Treating the relationship as inversely proportional to x itself, rather than to x², gives k = 8 × 2 = 16 and then y = 16 ÷ 4 = 4, a different relationship. Using x instead of x² in the new calculation gives y = 32 ÷ 4 = 8, skipping the square. Multiplying by x² instead of dividing by it gives y = 32 × 16 = 512, the wrong operation for an inverse relationship. When x = 4, y = 2.
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