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.
Calculator
Ratio, proportion and rates of change worksheet — GCSE Higher
MathsUKwww.geekhero.co.uk
- 1.A trust fund contains £8000. Each year, 5% interest is added, and then £600 is paid out to a charity. Work out the amount remaining in the fund after 2 years.
- 2.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.
- 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.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.
- 5.y is directly proportional to √x. When x = 4, y = 8. Construct the equation connecting x and y, then work out the value of y when x = 9.
- 6.A concrete block exerts a force of 45 N on the ground through a base with an area of 0.05 m². Work out the pressure on the ground, in N/m².
- 7.A conversion graph between kilometres and miles passes through the origin and the point (50, 31), with the number of kilometres on the horizontal axis and the number of miles on the vertical axis. Work out the number of miles equivalent to 1 kilometre.
- 8.A cartographer is choosing a scale for a map to show a small park in as much detail as possible. Which of these scales would show the MOST detail — 1 : 500, 1 : 5000, or 1 : 50 000? Give the correct scale and reason.
- 9.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.
- 10.A coach journey is 372 miles in total. After a stop, the coach has travelled 217 miles. Write the distance still to travel as a fraction of the total journey. Give your answer in its simplest form.
- 11.A company's turnover this year is £180,000. Last year's turnover was £120,000. Write down this year's turnover as a percentage of last year's turnover.
- 12.A charity shop and a school share collection-box money in the ratio 5 : 8. The charity shop receives £47.50. Work out how much the school receives.
- 13.In a fruit bowl, the ratio of apples to oranges is 5 : 3. Work out what percentage of the fruit in the bowl is apples.
- 14.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.
- 15.A train travels at a constant speed of 90 km/h. Work out the speed in m/s.
Answer key
- (c) £7590 — Apply interest, then subtract the payment, once for each year. Year 1: 8000 × 1.05 = 8400, then 8400 − 600 = 7800. Year 2: 7800 × 1.05 = 8190, then 8190 − 600 = 7590, so £7590 remains after 2 years. Forgetting the payments altogether and only compounding the interest gives 8000 × 1.05 = 8400, then 8400 × 1.05 = 8820 — this ignores that £600 leaves the fund every year. Subtracting the £600 BEFORE adding interest each year, instead of after, gives (8000 − 600) × 1.05 = 7770, then (7770 − 600) × 1.05 = 7528.50, which changes the order the two operations happen in and so changes the amount that earns interest each year. Subtracting the two payments as one lump sum of £1200 at the very end, from the no-withdrawal total 8820 − 1200 = 7620, ignores that the first £600 withdrawal also stops earning interest during the second year. Always apply interest, then the withdrawal, in that order, once for every single year.
- (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.
- (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) 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.
- (b) 12 — Since y is directly proportional to √x, y = k√x. Using x = 4, y = 8: √4 = 2, so 8 = k × 2, giving k = 8 ÷ 2 = 4. The equation is y = 4√x. When x = 9: √9 = 3, so y = 4 × 3 = 12. Treating the relationship as if y were proportional to x itself, rather than to √x, gives k = 8 ÷ 4 = 2 and then y = 2 × 9 = 18, which is a different relationship. Multiplying k by the new x-value instead of by its square root gives y = 4 × 9 = 36, skipping the square root altogether. Reporting √9 on its own, without multiplying by k, gives only 3, not the value of y. When x = 9, y = 12.
- (b) 900 N/m² — Pressure = force ÷ area, so 45 ÷ 0.05 = 900 N/m². Working out 45 ÷ 5 = 9 misplaces the decimal point in the area, treating 0.05 m² as 5 m². Working out 45 × 0.05 = 2.25 multiplies force and area together instead of dividing. Working out 45 + 0.05 = 45.05 simply adds the two given numbers, which does not give a pressure at all. The pressure on the ground is 900 N/m².
- (a) 0.62 miles — The gradient of the line is the change in miles divided by the change in kilometres: 31 ÷ 50 = 0.62, so 1 kilometre converts to 0.62 miles. Dividing the wrong way round, 50 ÷ 31 = 1.612..., rounds to 1.61 miles — that finds how many kilometres are in 1 mile, not the reverse. Doubling the gradient, 1.24 miles, comes from using 62 ÷ 50 instead of 31 ÷ 50. Reading off the y-coordinate of the given point without dividing by the x-coordinate gives 31.00 miles, which is the number of miles for 50 kilometres, not for 1 kilometre.
- (b) 1 : 500 — smallest real distance per cm (5 m), most detail — Method: compare what one centimetre represents in real life for each scale — the scale with the smallest real distance per cm shows the most detail. Working: for 1 : 500, 1 cm represents 500 cm (5 m); for 1 : 5000, 1 cm represents 50 m; for 1 : 50 000, 1 cm represents 500 m. Since 5 m is the smallest, 1 : 500 shows the most detail. Wrong options: '1 : 50 000 — covers the largest real area' wrongly assumes covering more area means more detail, when it is the opposite; '1 : 5000 — the middle value' wrongly assumes the middle scale is automatically the most balanced; '1 : 500 — covers the largest real distance' picks the correct scale but states an incorrect fact, since 1 : 500 actually covers the smallest real distance per cm.
- (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.
- (a) 5/12 — Work out the distance still to travel: 372 − 217 = 155 miles. Form the fraction 155/372; both numbers share a factor of 31, so 155 ÷ 31 = 5 and 372 ÷ 31 = 12, giving 5/12. 7/12 comes from writing the distance already travelled as the fraction of the journey (217/372 = 7/12), instead of the distance still to travel. 145/372 comes from miscalculating 372 − 217 as 145 instead of 155. 5/7 comes from comparing the remaining distance with the distance already travelled (155/217 = 5/7), instead of with the total journey.
- (c) 150% — Percentage = (180,000 ÷ 120,000) × 100 = 150%.
- (b) £76.00 — One part of the ratio is £47.50 ÷ 5 = £9.50. The school receives 8 parts, so its share is 9.50 × 8 = £76.00. Dividing £47.50 by 8 instead of 5, treating the charity's amount as if it were 8 parts, gives 47.50 ÷ 8 = 5.9375, then × 5 = £29.69. Adding the charity's amount to the school's amount instead of stopping at the school's own share gives the total collected, 9.50 × 13 = £123.50. Adding one part to the charity's amount instead of multiplying one part by 8 gives 47.50 + 9.50 = £57.00.
- (b) 62.5% — Total parts = 5 + 3 = 8. Apples make up 5 parts, so the percentage is 5/8 × 100 = 62.5%. A student who finds the oranges' share instead gets 3/8 × 100 = 37.5%. A student who assumes an even split gets 50%. A student who inverts the fraction gets 8/5 × 100 = 160%.
- (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) 25 m/s — Convert km/h to m/s by multiplying by 1000 (km to m) and dividing by 3600 (hours to seconds): 90 × 1000 ÷ 3600 = 25 m/s. Working out 90 ÷ 60 = 1.5 converts using 60, as if going from hours to minutes rather than to seconds. Working out 90 × 3.6 = 324 multiplies by 3.6 instead of dividing by it, going the wrong way between the units. Working out 90 × 1000 = 90000 converts kilometres to metres but forgets to convert hours to seconds at all. The train's speed is 25 m/s.
Build your own mix at the worksheet builder.