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 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.
- 2.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
- 3.The number of euros, e, received is directly proportional to the number of pounds, p, exchanged. Exchanging £40 gives 46 euros. Work out how many euros are received for £65, giving your answer to the nearest euro.
- 4.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?
- 5.Two quantities, P and Q, are in inverse proportion. A graph is drawn with P on the vertical axis and Q on the horizontal axis. Write down which description fits the shape of this graph.
- 6.A machine is bought for £8500. Its value depreciates by 6% each year. Work out the value of the machine after 3 years, to the nearest pound.
- 7.There are 2500 electric cars registered in a town. The number is predicted to increase by 6% each year. Work out the predicted number of electric cars after 3 years, to the nearest whole number.
- 8.A savings account starts with £500. Each year, 4% interest is added, and then £30 is withdrawn from the account. Which recurrence correctly models the balance, £S_n, after n years, with S_0 = 500?
- 9.6 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.
- 10.The height of a ball, in metres, above the ground is plotted against time, in seconds, after it is thrown. The tangent to the graph at t = 1.5 seconds has gradient 2. Which of these four statements about the ball at t = 1.5 seconds is correct?
- 11.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.
- 12.y is inversely proportional to x², and x is positive. When x = 2, y = 45. Construct the equation connecting x and y, then work out the value of x when y = 5.
- 13.A scale drawing of a park has a scale of 1 : 2500. On the drawing, the distance between the entrance and the lake is 4.4 cm. A jogger runs from the entrance to the lake and then back to the entrance. Work out the total distance the jogger runs, in kilometres.
- 14.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.
- 15.A cyclist rides 17.5 km on Saturday and 12.25 km on Sunday. Write the distance ridden on Saturday as a fraction of the distance ridden on Sunday, giving your answer in its simplest form.
Answer key
- (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) 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.
- (b) 75 — The exchange rate is constant: k = 46 ÷ 40 = 1.15 euros per pound. For £65, the number of euros is 1.15 × 65 = 74.75, which rounds to 75 euros. Getting 74 comes from rounding 74.75 down instead of to the nearest whole number. Getting 57 comes from using the reciprocal rate (40 ÷ 46) instead of 46 ÷ 40. Getting 71 comes from adding the difference between 65 and 40 (25) onto 46 instead of using the proportional rate.
- (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.
- (d) A falling curve that never touches either axis — Method: inverse proportion means the product of the two quantities is constant, so P = k ÷ Q; as Q grows P shrinks, and P can never reach zero because k divided by a number is never zero. Working: taking k = 12 as an example, the pairs (1, 12), (2, 6), (3, 4), (6, 2) and (12, 1) drop steeply at first and then flatten out, so the graph is a curve that approaches both axes without meeting either of them. Answer: a falling curve that never touches either axis. The distractors: 'a straight line through the origin' is the graph of direct proportion, P = kQ, which is the opposite relationship; 'a straight line with a negative gradient' is the commonest error, reading 'P falls as Q rises' as a straight line, but on such a line P would drop by the same amount for every increase in Q and would cross the horizontal axis into negative values; 'a straight line crossing the vertical axis above zero' is a relationship of the form P = mQ + c, in which P and Q are not proportional at all.
- (d) £7060 — A 6% decrease each year means the value becomes 100% − 6% = 94% of the previous year's value, and 94% = 0.94, so the multiplier is 0.94. Apply it once for each of the 3 years: £8500 × 0.94 = £7990 after 1 year, £7990 × 0.94 = £7510.60 after 2 years, £7510.60 × 0.94 = £7059.96 after 3 years, which rounds to £7060 to the nearest pound. (£6970 comes from using simple depreciation instead of compound, taking 6% of the original £8500 three times: £8500 − 3 × £510 = £6970. £7990 is the value after only 1 year, forgetting the remaining 2 years. £7511 is the value after only 2 years, £8500 × 0.94² = £7510.60, forgetting the third year.)
- (a) 2978 — A rise of 6% is a multiplier of 1.06, applied once for each year. After year 1: 2500 × 1.06 = 2650. After year 2: 2650 × 1.06 = 2809. After year 3: 2809 × 1.06 = 2977.54, which is 2978 to the nearest whole number. Multiplying by 1.18 in one go would be wrong, because the second and third years grow from larger numbers than the first.
- (a) S_{n+1} = 1.04S_n − 30 — Adding 4% interest multiplies the balance by 1 + 0.04 = 1.04. Withdrawing £30 afterwards subtracts a fixed 30, giving S_{n+1} = 1.04S_n − 30. Writing +30 instead of −30 mistakes a withdrawal for a deposit — the £30 leaves the account, so it must be subtracted. Writing 0.96 instead of 1.04 treats the 4% as a decrease rather than an increase, as if the interest were shrinking the balance instead of growing it. Writing 1.4 instead of 1.04 turns 4% into 40%, a common slip when converting a percentage to a multiplier — 4% as a decimal is 0.04, so the multiplier is 1.04, not 1.4. Always convert the percentage to a decimal first, then add 1 for growth or subtract from 1 for decay, before applying any fixed amount that is added or removed.
- (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.
- (c) Height rising at 2 m/s at t = 1.5 s — A tangent's gradient on a height-time graph is the instantaneous rate of change of height, in metres per second, so gradient 2 means the ball's height is increasing at 2 m/s at t = 1.5 s. Saying the height 'is 2 m' confuses the gradient, a rate, with the y-value on the graph, which is the ball's height itself. Saying the ball 'travelled 2 m from t = 1 to t = 2' treats the instantaneous gradient at one instant as if it were the total distance risen over a whole one-second interval, which is a different quantity found from two height readings, not from one tangent. Saying the speed 'is 2 m/s²' uses the wrong units — m/s² measures acceleration, the rate of change of speed, not speed itself. Always check that the units quoted match what a height-time graph's gradient can actually give you: metres per second.
- (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) 6 — Since y is inversely proportional to x², y = k/x². Using x = 2, y = 45: 2² = 4, so 45 = k ÷ 4, giving k = 45 × 4 = 180. The equation is y = 180/x². When y = 5: x² = 180 ÷ 5 = 36, so x = 6, taking the positive root because the question states that x is positive. Treating the relationship as inversely proportional to x itself gives k = 45 × 2 = 90 and then x = 90 ÷ 5 = 18, a different relationship. Multiplying by y instead of dividing by it when isolating x² gives x² = 180 × 5 = 900 and x = 30, the wrong operation. Stopping at x² = 36 without taking the square root leaves 36, not the value of x. When y = 5, x = 6.
- (b) 0.22 km — The real one-way distance is 4.4 × 2500 = 11000 cm. Converting units: 11000 ÷ 100 = 110 m, and 110 ÷ 1000 = 0.11 km. Since the jogger runs there and back, the total distance is 0.11 × 2 = 0.22 km. 0.11 km comes from working out only the one-way distance and forgetting the return journey. 220 km comes from correctly doubling the one-way distance in metres, 110 × 2 = 220, but leaving it mislabelled as kilometres instead of converting metres to kilometres. 110 km comes from working out only the one-way distance in metres, 110, and mislabelling it as kilometres.
- (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.
- (a) 10/7 — Put Saturday's distance over Sunday's distance: 17.5/12.25. Multiply both numbers by 100 to clear the decimals: 1750/1225. Divide both by their highest common factor, 175: 1750÷175 = 10, 1225÷175 = 7, giving 10/7. (7/10 comes from writing the distances the wrong way round. 3/7 comes from finding the difference, 17.5 − 12.25 = 5.25 km, and writing it as a fraction of Sunday's distance, 5.25/12.25. 10/17 comes from comparing Saturday's distance to the total distance ridden, 17.5/29.75.)
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