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 kettle uses 2.1 kWh of energy. A toaster uses 0.9 kWh of energy. Write the energy used by the kettle as a fraction of the energy used by the toaster, giving your answer in its simplest form.
- 2.A car travels at a constant speed. It travels 150 miles in 3 hours. At the same speed, work out how far it travels in 5 hours.
- 3.The volume of fuel remaining in a storage tank, in thousands of litres, is plotted against time, in hours, since a leak was detected. A tangent to the graph at t = 2 hours has gradient −8.75 thousand litres per hour. A tangent at t = 8 hours has gradient −3.5 thousand litres per hour. Work out by what factor the instantaneous rate of change is greater in size at t = 2 hours than at t = 8 hours.
- 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.A straight line passes through the points (1, 20) and (5, 8). Work out the gradient of the line.
- 6.The density of a type of wood is 0.8 g/cm³. Work out the mass of a piece of this wood with a volume of 150 cm³.
- 7.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?
- 8.It takes 2 identical pumps 10 hours to empty a flooded basement. Working at the same rate, work out how many hours 5 of these pumps would take to empty the same basement.
- 9.Two cars travel at constant speeds. Car A travels 150 km in 3 hours. Car B travels 180 km in 4 hours. Which car is faster, and what is its speed?
- 10.Two broadband providers show their monthly cost on straight-line graphs, with the cost in pounds on the vertical axis and the data used in gigabytes on the horizontal axis. Provider X's line passes through (0, 15) and (100, 35). Provider Y's line passes through (0, 5) and (100, 45). Work out which provider is cheaper for a customer who uses 40 gigabytes in a month.
- 11.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.
- 12.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.
- 13.Two mathematically similar polygons have perimeters in the ratio 2 : 5. Write the ratio of their areas in its simplest form.
- 14.A sequence is defined by the iterative rule x_{n+1} = 0.8x_n + 50, with x_0 = 200. Work out x_2, the value of the sequence after two iterations.
- 15.Map A has a scale of 1 : 25000 and Map B has a scale of 1 : 50000, both showing the same area. A real distance of 10 km is measured on each map. On which map does this distance appear as the longer length, and how long is it on that map, in centimetres?
Answer key
- (d) 7/3 — Put the kettle's energy over the toaster's energy: 2.1/0.9. Multiply both numbers by 10 to clear the decimals: 21/9. Divide both by their highest common factor, 3: 21÷3 = 7, 9÷3 = 3, giving 7/3. (3/7 comes from writing the energy values the wrong way round. 4/3 comes from finding the difference, 2.1 − 0.9 = 1.2 kWh, and writing it as a fraction of the toaster's energy, 1.2/0.9. 7/10 comes from comparing the kettle's energy to the total energy used by both appliances, 2.1/3.0.)
- (c) 250 miles — Find the distance travelled in 1 hour: 150 ÷ 3 = 50 miles. Multiply by 5 hours: 50 × 5 = 250 miles. Giving 300 miles doubles the original distance (150 × 2 = 300) using a scale factor of 2 instead of the correct 5 ÷ 3. Giving 200 miles adds only one extra hour's distance, 50, instead of the two extra hours actually needed (150 + 50 = 200, rather than 150 + 100). Giving 90 miles divides by the scale factor instead of multiplying (150 × 3 ÷ 5 = 90).
- (a) 2.5 — Method: the factor by which the SIZE of one rate is greater than the size of another is found by dividing the larger magnitude by the smaller magnitude, ignoring their signs, so here the two magnitudes to work with are 8.75 and 3.5. Working: 8.75 ÷ 3.5 = 2.5, so the size of the instantaneous rate of change at t = 2 hours was 2.5 times the size of the instantaneous rate of change at t = 8 hours. Subtracting the two magnitudes, 8.75 − 3.5 = 5.25, gives how many thousand litres per hour greater one rate is than the other, not how many times greater — that is a difference, not a factor. Dividing the magnitudes the wrong way round, 3.5 ÷ 8.75 = 0.4, gives the factor by which the rate at t = 8 hours is smaller than at t = 2 hours, the reciprocal of what was asked for. Adding the magnitudes, 8.75 + 3.5 = 12.25, combines the two rates instead of comparing them, and does not answer a 'by what factor' question at all. A question that asks 'by what factor' is always answered by a division, in the order the question states it — check which rate is on top before you divide.
- (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.
- (b) −3 — Method: the gradient is the change in the vertical value divided by the change in the horizontal value, with both changes taken in the same direction along the line. Working: going from (1, 20) to (5, 8) the change in y is 8 − 20 = −12 and the change in x is 5 − 1 = 4, so the gradient is −12 ÷ 4 = −3. Answer: −3, and the negative sign is expected because the line falls from left to right. The distractors: 3 comes from subtracting the smaller y from the larger, 20 − 8 = 12, while still taking the x values from left to right, which loses the minus sign that says the line falls; −12 is the change in y left undivided by the change in x of 4; −1/3 comes from dividing the change in x by the change in y, 4 ÷ (−12), turning the gradient upside down.
- (d) 120 g — Mass = density × volume, so 0.8 × 150 = 120 g. Working out 150 ÷ 0.8 = 187.5 divides by the density instead of multiplying, the wrong way round for finding a mass. Working out 150 × 8 = 1200 misplaces the decimal point in the density, treating 0.8 g/cm³ as 8 g/cm³. Working out 150 − 0.8 = 149.2 simply subtracts the density from the volume, which does not give a mass. The piece of wood has a mass of 120 g.
- (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.
- (a) 4 hours — This is inverse proportion: more pumps take less time. Multiply the original numbers to find the total pump-hours needed: 2 × 10 = 20 pump-hours. Divide by the new number of pumps: 20 ÷ 5 = 4 hours. Working out 10 × 5 ÷ 2 = 25 hours treats it as direct proportion, as if more pumps needed more time. Stopping at 20 gives the total pump-hours, not the number of hours. Working out 10 − (5 − 2) = 7 hours subtracts the extra number of pumps straight from the number of hours, treating pumps and hours as the same kind of quantity. 5 pumps take 4 hours.
- (b) Car A, 50 km/h — Method: speed = distance ÷ time for each car, then compare. Working: Car A = 150 ÷ 3 = 50 km/h. Car B = 180 ÷ 4 = 45 km/h. Since 50 > 45, Car A is faster, travelling at 50 km/h. Wrong options: Car B, 45 km/h correctly finds Car B's speed but wrongly names the slower car as faster; Car A, 45 km/h picks the correct car but uses Car B's speed by mistake; Car B, 50 km/h picks the wrong car but uses Car A's correct speed value.
- (b) Provider Y — £21.00 against Provider X's £23.00 — Provider X's gradient is (35 − 15) ÷ 100 = 0.2, so cost = 15 + 0.2 × 40 = 15 + 8 = £23.00. Provider Y's gradient is (45 − 5) ÷ 100 = 0.4, so cost = 5 + 0.4 × 40 = 5 + 16 = £21.00. £21.00 is less than £23.00, so Provider Y is cheaper: 'Provider Y — £21.00 against Provider X's £23.00'. Getting both costs right but naming Provider X as cheaper compares the two numbers the wrong way round — £23.00 is more than £21.00, not less. Comparing only the fixed fees, £15.00 and £5.00, ignores the cost of the 40 gigabytes actually used. Reading off the costs at 100 gigabytes, £35.00 and £45.00, directly from the graph answers a different usage from the 40 gigabytes the question asks about.
- (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.
- (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.
- (b) 4 : 25 — For similar shapes, the ratio of areas is the ratio of lengths squared: 2² : 5² = 4 : 25. 2 : 5 comes from using the perimeter ratio itself as the area ratio, without squaring it at all. 8 : 125 comes from cubing each part instead of squaring (2³ : 5³) — cubing is the rule for volume, not area. 4 : 5 comes from squaring only the first part of the ratio (2² = 4), and leaving the second part unsquared.
- (d) 218 — The rule x_{n+1} = 0.8x_n + 50 must be applied once for each step, using the result of the previous step every time — not the same starting value repeated. Starting from x_0 = 200: 0.8 × 200 = 160, so x_1 = 160 + 50 = 210. Then 0.8 × 210 = 168, so x_2 = 168 + 50 = 218. Stopping after one iteration leaves x_1 = 210, not x_2. Applying only the multiplier twice without adding 50 at each step uses 0.8² = 0.64, and 0.64 × 200 = 128, which drops the 50 completely. Adding 50 twice at the end instead of once per step, 128 + 100 = 228, still does not reproduce the actual recurrence, because the 50 added at the first step is itself multiplied by 0.8 at the second step. After two iterations, x_2 = 218.
- (c) Map A, where the distance is 40 cm — 10 km = 1,000,000 cm. On Map A: 1000000 ÷ 25000 = 40 cm. On Map B: 1000000 ÷ 50000 = 20 cm. Since 40 cm is longer than 20 cm, the same real distance appears longer on Map A, the map with the smaller scale number. 'Map B, where the distance is 20 cm' has the correct working for Map B but names the wrong map as the one with the longer length. 'Map A, where the distance is 20 cm' correctly identifies Map A but pairs it with Map B's length. 'Map B, where the distance is 40 cm' correctly identifies Map A's length but attaches it to the wrong map.
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