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 shop's number of online orders is proportional to its number of in-store orders. Last month there were 45 in-store orders and 180 online orders. Write the ratio of online orders to in-store orders, in its simplest form.
- 2.x × y is used to test whether two quantities are in inverse proportion. For the pairs x = 4, y = 15 and x = 6, y = 10, which statement is correct?
- 3.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.
- 4.The strength of a radio signal, in units, is inversely proportional to the square of the distance from the transmitter, in km. At a distance of 2 km the signal strength is 20 units. Construct the equation connecting signal strength S and distance d, then work out the distance at which the signal strength is 5 units.
- 5.A train journey takes 2 hours 15 minutes. A bus journey to the same place takes 3 hours 45 minutes. Write the time for the train journey as a fraction of the time for the bus journey, giving your answer in its simplest form.
- 6.In a class, 50% of the students study French, 30% study Spanish and the rest study German. Write down the ratio of French : Spanish : German students, in its simplest form.
- 7.A scale drawing of a garden uses a scale of 1 : 20. Write down a formula for the real length, L metres, in terms of the length on the drawing, d centimetres, given that 1 metre = 100 centimetres.
- 8.A printer prints at a constant rate. The time taken to print a batch of forms is inversely proportional to the printer's speed, in pages per minute. Printing at 20 pages per minute takes 15 minutes. Work out how long the same batch takes to print at 25 pages per minute. Give your answer in minutes.
- 9.A market stall sells apples at a fixed price per kilogram. Priya buys 3 kg for £5.40. Write the ratio of the mass in kilograms to the cost in pounds, in its simplest form.
- 10.A curve has equation y = x² + 2x. Work out the average rate of change of y with respect to x over the interval from x = 1 to x = 4.y = x² + 2x
- 11.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.
- 12.A fruit squash is made by mixing squash and water. In a jug, 2/9 of the total volume is squash. A caterer makes up 4.5 litres of the squash mixture. Work out how many litres of squash are in the mixture.
- 13.A candidate wants to estimate the instantaneous rate of change of a reservoir's water level, in metres, at t = 5 days after heavy rain began. The reservoir's water level is plotted against time, in days, since the rain began. The candidate uses the chord joining the points at t = 0 days and t = 20 days to estimate the rate of change at t = 5 days. Give a reason why this is likely to be a poor estimate.
- 14.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?
- 15.Work out the value of x when 4/x = 2/5.
Answer key
- (d) 4:1 — Write the ratio online : in-store using the numbers in the question: 180 : 45. Divide both parts by their highest common factor, 45, to give 4 : 1. (1:4 comes from writing the ratio the wrong way round, in-store : online. 3:1 comes from subtracting the in-store orders from the online orders, 180 − 45 = 135, and comparing that to the in-store orders, 135:45, instead of dividing. 4:5 comes from comparing the online orders to the total number of orders, 180:225.)
- (b) They are in inverse proportion, because x × y = 60 for both pairs. — Testing inverse proportion means checking that x × y is the same for every pair: 4 × 15 = 60 and 6 × 10 = 60, so the quantities are in inverse proportion. Saying they are not in inverse proportion because x + y differs uses addition, which is not the correct test. Saying they are not in inverse proportion because y ÷ x differs uses the test for direct proportion, and finding that it differs tells us nothing about inverse proportion. Saying x × y = 40 for both pairs is an arithmetic slip: 4 × 15 = 60, not 40.
- (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.
- (c) 4 km — Since signal strength is inversely proportional to the square of the distance, S = k/d². Using d = 2, S = 20: 2² = 4, so 20 = k ÷ 4, giving k = 20 × 4 = 80. The equation is S = 80/d². When S = 5: d² = 80 ÷ 5 = 16, so d = 4 (taking the positive root, since distance cannot be negative). Stopping at d² = 16 without taking the square root leaves 16, the square of the distance, not the distance itself. Treating the relationship as inversely proportional to distance itself, rather than to its square, gives k = 20 × 2 = 40 and then d = 40 ÷ 5 = 8, a different relationship. Multiplying by S instead of dividing by it when isolating d² gives d² = 80 × 5 = 400 and d = 20, the wrong operation. The distance at which the signal strength is 5 units is 4 km.
- (d) 3/5 — Convert both times to minutes: 2 hours 15 minutes = 135 minutes; 3 hours 45 minutes = 225 minutes. Put the train time over the bus time: 135/225. Divide both numbers by their highest common factor, 45: 135÷45 = 3, 225÷45 = 5, giving 3/5. (5/3 comes from writing the times the wrong way round. 2/5 comes from finding the difference, 225 − 135 = 90 minutes, and writing it as a fraction of the bus time, 90/225. 3/8 comes from comparing the train time to the total time for both journeys, 135/360.)
- (d) 5:3:2 — German = 100% − 50% − 30% = 20%. The ratio 50 : 30 : 20 simplifies by dividing every part by 10 to give 5 : 3 : 2.
- (c) L = d/5 — The scale 1 : 20 means each cm on the drawing represents 20 cm in real life, so the real length in cm is 20d. Converting to metres by dividing by 100: L = 20d/100 = d/5.
- (a) 12 — Speed × time is constant: k = 20 × 15 = 300. At 25 pages per minute, the time is 300 ÷ 25 = 12 minutes. Getting 18.75 comes from treating speed and time as directly proportional and working out 15 × 25 ÷ 20 instead of dividing k by the new speed. Getting 20 comes from adding the increase in speed (25 − 20 = 5) onto the time (15 + 5 = 20). Getting 10 comes from subtracting that same increase in speed from the time (15 − 5 = 10).
- (c) 5 : 9 — Write the ratio mass : cost = 3 : 5.4. Multiply both parts by 10 to clear the decimal: 30 : 54. Both numbers share a factor of 6, so 30 ÷ 6 = 5 and 54 ÷ 6 = 9, giving 5 : 9. 3 : 5 comes from ignoring the decimal point and treating £5.40 as £5. 9 : 5 comes from writing the ratio the wrong way round, cost to mass instead of mass to cost. 1 : 18 comes from multiplying only the cost by 10 instead of both parts, giving 3 : 54, and then cancelling that correctly to 1 : 18 — the cancelling is fine, but the ratio being cancelled is not the right one.
- (a) 7 — The average rate of change of y with respect to x over an interval is the change in y divided by the change in x between its two endpoints — the gradient of the chord joining them, not a rate at a single point. At x = 1: 2 × 1 = 2, so y = 1 + 2 = 3. At x = 4: 2 × 4 = 8, so y = 16 + 8 = 24. The change in y is 24 − 3 = 21 and the change in x is 4 − 1 = 3, so the average rate of change is 21 ÷ 3 = 7. Reporting the change in y, 21, on its own is not a rate of change, because it has not been divided by the 3 units of x over which it happened. Reporting 3 is not a rate either — 3 is the width of the interval, and also the value of y at x = 1, and neither of those measures how fast y is changing. Subtracting in the wrong order gives −21 ÷ 3 = −7, the wrong sign. The average rate of change of y with respect to x over the interval is 7.
- (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.
- (d) 1.00 litres — Squash is 2/9 of the mixture, so the squash volume is 4.5 × 2/9 = 1.00 litres. Using the water's fraction, 7/9, instead of squash's gives 4.5 × 7/9 = 3.50 litres — the volume of water, not squash. Dividing 4.5 by 9 but forgetting to multiply by the numerator 2 gives 4.5 ÷ 9 = 0.50 litres, which is only 1/9 of the mixture. Halving the total volume instead of applying the fraction 2/9 gives 4.5 ÷ 2 = 2.25 litres, which assumes the mixture is half squash.
- (a) It's an average over 20 days, which may miss the day-5 rate. — Method: a chord's gradient is the AVERAGE rate of change across the whole interval it spans; it only closely approximates the INSTANTANEOUS rate of change at a point inside that interval when the rate of change is roughly constant across the interval, which usually means the interval needs to be short. Working: here the chord spans 20 days while the point of interest, t = 5, is only a quarter of the way along it, so if the reservoir's level rose or fell at different rates over that time, the chord's gradient will not be close to the true gradient of the curve at t = 5 — this is the correct reason. Claiming the chord's gradient needs the water level at every day in between is wrong: a chord's gradient needs only the two endpoint values, at t = 0 and t = 20. Claiming a chord can only estimate the rate at its own endpoints is wrong: a chord between two points can be used to estimate the instantaneous rate of change at any point inside the interval, including one that is not an endpoint — that is exactly the technique being used here, and it is the SIZE of the interval that makes the estimate poor, not the fact that t = 5 is an interior point. Claiming the units do not match is wrong: the chord's gradient and the instantaneous rate of change are both measured in metres per day, so the units are the same. A chord is only a good estimate of an instantaneous rate when the interval it spans is short enough that the rate does not change much within it — always check how long the interval is compared with how far it is to the point you actually want.
- (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) 10 — Method: two equal fractions can be rearranged by cross-multiplying, multiplying each numerator by the other denominator. Working: 4 × 5 = 2 × x, so 2x = 20 and x = 20 ÷ 2 = 10. Answer: 10. The distractors: 20 comes from cross-multiplying to 4 × 5 = 20 and stopping there, without dividing by the 2; 8 comes from multiplying the two numerators, 4 × 2; 2.5 comes from working only with the right-hand fraction, 5 ÷ 2, and ignoring the 4.
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