Printable · GCSE Higher · ages 14-16
Algebra worksheet — GCSE Higher
Fifteen questions across the algebra 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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Algebra worksheet — GCSE Higher
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- 1.The mass of a chemical sample decays by 15% every hour. It starts at 500 g. Work out the mass remaining after 3 hours, correct to 1 decimal place.
- 2.A bacteria colony doubles every 4 hours and its size is recorded only every 4 hours. The recorded population 20 hours after the start is 9,600. After how many hours did the recorded population first exceed 1,000?
- 3.An exponential model has equation y = A × bˣ, where b > 1. Its graph passes through the points (1, 135) and (2, 405). Work out the value of A.
- 4.A scout leader is 44 years old and one of the scouts is 16 years old. Work out how many years it will be until the leader is exactly twice as old as the scout.
- 5.A runner's journey is shown on a distance-time graph: she runs 8 km in the first hour, rests for 30 minutes with no distance gained, then runs a further 4 km in the next 30 minutes. Work out her average speed for the whole journey, in kilometres per hour.
- 6.The equation x³ − 5x − 3 = 0 can be rearranged to give an iterative formula of the form xₙ₊₁ = ∛(…). Work out which one of these is a correct rearrangement.
- 7.A rectangular vegetable plot has an area of 30 m² and its length is 4 m greater than its width, x metres. This gives x² + 4x − 30 = 0, which can be solved using the iterative formula xₙ₊₁ = √(30 − 4xₙ). The starting value is x₀ = 3, so x₁ is the value after the formula has been used once. Work out x₃, and use it to find an estimate for the length of the plot, giving your answer correct to 1 decimal place.
- 8.f(x) = x³ − 3x² − 4. Work out the pair of consecutive integers between which the solution of f(x) = 0 lies.y = x
- 9.A colony of bacteria doubles in number every hour. At 9am there are 5 bacteria in the colony. Work out how many bacteria there will be at 12 noon.
- 10.A population of bacteria doubles every 3 hours. It starts at 800. Work out the population after 9 hours.
- 11.A car's journey is represented by a velocity-time graph. The total distance travelled, found from the area under the graph, is 350 metres, and the total time for the journey is 25 seconds. Work out the car's average speed for the whole journey, in m/s.
- 12.f(x) = x³ − 5x − 6. Given that f(2.6) = −1.424 and f(2.7) = 0.183, work out what this shows about the equation x³ − 5x − 6 = 0.y = x
- 13.On Monday, a runner covers 15 km in 2.5 hours. On Tuesday, she covers 12 km in 1.5 hours. Using the formula speed = distance ÷ time, work out on which day she ran faster, and by how much.
- 14.The equation x² + 2x − 5 = 0 can be solved using the iterative formula xₙ₊₁ = 5/(xₙ + 2). The starting value is x₀ = 1, so x₁ is the value after the formula has been used once. Work out x₂ correct to 2 decimal places.
- 15.A speed-time graph shows a constant speed of 15 m/s for 20 seconds. Work out the distance travelled, using the area under the graph.
Answer key
- (c) 307.1 g — The multiplier for one hour is 1 − 0.15 = 0.85. After 3 hours the mass is 500 × 0.85³. Since 0.85 × 0.85 = 0.7225 and 0.85 × 0.7225 = 0.614125, the mass is 500 × 0.614125 = 307.0625 g, which rounds to 307.1 g. Stopping after only 2 hours instead of 3 gives 500 × 0.7225 = 361.25 g, which rounds to 361.3 g — wrong, because the question asks for 3 hours, not 2. Treating the decay as simple (15% × 3 = 45% lost in total, applied once) gives 500 × 0.55 = 275.0 g, which is wrong because the decay compounds hour by hour rather than adding up. Working out the AMOUNT LOST instead of the mass remaining gives 500 − 307.0625 = 192.9375 g, which rounds to 192.9 g — wrong, because the question asks what remains, not what has decayed away. Whenever a percentage decreases repeatedly, multiply by the same factor each period rather than adding the percentages together.
- (d) 8 — Working backwards by halving (undoing the doubling), one 4-hour step at a time: 9,600 (20 h) → 4,800 (16 h) → 2,400 (12 h) → 1,200 (8 h) → 600 (4 h) → 300 (0 h). Reading these in time order — 300, 600, 1,200, 2,400, 4,800, 9,600 at 0, 4, 8, 12, 16, 20 hours — the population is still at or below 1,000 at 4 hours (600) and first goes above 1,000 at 8 hours (1,200). So the recorded population first exceeds 1,000 at 8 hours. Answering 20 just reads off the time stated in the question, without working out when the threshold was actually first crossed — wrong, because 9,600 is only the value AT 20 hours, not necessarily the first time the population passed 1,000. Answering 0 comes from recovering the starting population by dividing 9,600 by 5 (the number of 4-hour gaps up to 20 hours) instead of by 2⁵ = 32 (the correct number of halvings), giving a wrongly-inflated starting value of 9,600 ÷ 5 = 1,920 — already above 1,000 at 0 hours — wrong, because doubling means the value must be halved five times, dividing by 2 five times (2⁵ = 32), not divided once by the number of gaps. Answering 12 comes from halving back only twice, from 9,600 to 4,800 (16 h) to 2,400 (12 h), and stopping there because 2,400 is already above 1,000, without checking that 1,200 at 8 hours is also above 1,000 and occurs earlier — wrong, because the FIRST recorded time above 1,000 is the earliest such time, not the first one reached while working backwards from 20 hours.
- (a) 45 — The ratio between the two given points, one power of x apart, gives b: 405 ÷ 135 = 3, so b = 3. Substituting back, at x = 1, y = A × b, so 135 = A × 3, giving A = 45. Giving the common ratio b itself instead of A confuses which unknown was asked for and produces 3 — wrong, because the question asks for A, not b. Giving 135 instead treats the first given point as the y-intercept and reads A off it directly — wrong, because that point is at x = 1, not x = 0, so 135 is A × b, not A. Assuming A equals b⁰ = 1 by itself, rather than substituting a known point to solve for A, gives 1 — wrong, because b⁰ is always 1 regardless of A; A must be found using an actual (x, y) pair from the graph.
- (b) 12 years — Method: call the number of years t, add t to both ages, and form an equation from the comparison at that future time. Working: in t years the leader will be 44 + t and the scout will be 16 + t, so 44 + t = 2(16 + t); expanding gives 44 + t = 32 + 2t, and subtracting t and 32 from both sides gives t = 12. Checking: in 12 years the leader will be 56 and the scout 28, and 56 = 2 × 28. Answer: 12 years. The distractors: 6 years comes from halving the leader's present age instead, so that the scout has to reach 22, which takes 6 years; 14 years comes from halving the 28-year gap between the two ages; 28 years comes from giving the age gap itself as the number of years.
- (c) 6 km/h — Total distance = 8 + 4 = 12 km. Total time = 1 hour + 0.5 hours resting + 0.5 hours = 2 hours. Average speed = total distance ÷ total time = 12 ÷ 2 = 6 km/h. A speed of 8 km/h comes from leaving the resting time out of the total time: 12 ÷ 1.5 = 8. A speed of 4 km/h comes from dividing by too much time, such as double-counting the rest period: 12 ÷ 3 = 4. A speed of 12 km/h simply gives the total distance and forgets to divide by the total time at all.
- (a) xₙ₊₁ = ∛(5xₙ + 3) — Starting from x³ − 5x − 3 = 0, add 5x and 3 to both sides to get x³ = 5x + 3, then take the cube root of both sides: x = ∛(5x + 3), giving the iterative formula xₙ₊₁ = ∛(5xₙ + 3). A sign error when moving the constant term across, treating x³ − 5x − 3 = 0 as x³ = 5x − 3, gives xₙ₊₁ = ∛(5xₙ − 3). Swapping the coefficient of x with the constant term gives xₙ₊₁ = ∛(3xₙ + 5), which does not come from x³ = 5x + 3 at all. Treating cubing as meaning multiply by 3 rather than raise to the power 3, and so undoing it by dividing by 3 instead of taking a cube root, gives xₙ₊₁ = (5xₙ + 3) ÷ 3.
- (a) 7.9 m — Method: the iteration converges on the width of the plot, so run the formula three times from the starting value and then add 4 m, because the length is 4 m greater than the width. Working: x₁ = √(30 − 4 × 3) = √18 = 4.24264…; x₂ = √(30 − 4 × 4.24264…) = √13.02943… = 3.60963…; x₃ = √(30 − 4 × 3.60963…) = √15.56147… = 3.94480…. The estimate for the length is 3.94480… + 4 = 7.94480…, which is 7.9 m correct to 1 decimal place. Answer: 7.9 m. The iteration is still oscillating at x₃, so this is the estimate that three uses of the formula give, not a settled value. The distractors: 3.9 m is x₃ itself, the width, given by a candidate who runs the iteration correctly and then stops before the step the question actually asks for; 7.6 m uses x₂ in place of x₃, one use of the formula short, and then adds the 4 m correctly; 15.8 m multiplies the width by 4 instead of adding 4 m to it, reading greater than as a multiplier.
- (d) 3 and 4 — f(3) = 3³ − 3 × 3² − 4 = 27 − 27 − 4 = −4, and f(4) = 4³ − 3 × 4² − 4 = 64 − 48 − 4 = 12. Since f(3) is negative and f(4) is positive, there is a change of sign, so the solution lies between 3 and 4. A sign slip when expanding −3x², treating it as +3x², gives f(3) = 27 + 27 − 4 = 50, a positive value; taken with the correctly negative f(2) = 8 − 12 − 4 = −8, this reads as a change of sign between 2 and 3. The same sign slip applied at x = 1 gives f(1) = 1 + 3 − 4 = 0, read as the point the sign changes, against f(0) = 0 + 0 − 4 = −4, giving 0 and 1. Shifting the correctly found interval up by one integer, an indexing slip, reports the change as lying between 4 and 5 instead of 3 and 4.
- (a) 40 — From 9am to 12 noon is 3 hours, so the population doubles three times: 5 × 2³ = 40. A candidate who counts the elapsed time as 4 hours (an off-by-one counting error) would reach 5 × 2⁴ = 80. A candidate who counts it as only 2 hours would reach 5 × 2² = 20. A candidate who misreads 'doubles' as 'increases by 2' each hour would compute 5 + 3 × 2 = 11.
- (c) 6,400 — 9 hours contains 9 ÷ 3 = 3 whole periods of doubling, so the population is 800 × 2³. Since 2³ = 8, the population after 9 hours is 800 × 8 = 6,400. Adding 100% growth three times instead of compounding it — treating the growth as simple, not repeated doubling — gives 800 × 4 = 3,200, which is wrong because each period doubles the CURRENT population, not the original one. Using 9 as the power instead of dividing by the 3-hour period first gives 800 × 2⁹ = 409,600, which is wrong because the exponent counts periods, not hours. Giving the growth factor 2³ = 8 on its own, without multiplying by the starting population 800, leaves the answer as 8, which is wrong because the question asks for the population, not the multiplier. Always check that your final number of periods matches the total time divided by the period length.
- (b) 14 m/s — Average speed for a whole journey is total distance divided by total time: 350 ÷ 25 = 14 m/s. Multiplying the distance and time instead of dividing gives 350 × 25 = 8750 m/s. Adding the distance and time instead of dividing gives 350 + 25 = 375 m/s. Inverting the division, working out time divided by distance, gives 25 ÷ 350, which rounds to 0.07 m/s.
- (a) It has a solution between x = 2.6 and x = 2.7 — f(2.6) is negative and f(2.7) is positive, so the graph of f crosses the x-axis between x = 2.6 and x = 2.7, meaning the equation has a solution there. Choosing 'x = 2.6 is a solution' reads an end of the interval as the root itself, but f(2.6) = −1.424, which is not zero — the change of sign locates a root between the two values, it does not land on either of them. Choosing 'between x = −2.6 and x = −2.7' confuses the negative f-VALUE at 2.6 with a negative x-value. Choosing 'no root in this interval' misapplies the rule, which needs a CHANGE of sign — and a change of sign is exactly what these two values show.
- (a) Tuesday, by 2 km/h — Monday's speed is 15 ÷ 2.5 = 6 km/h and Tuesday's speed is 12 ÷ 1.5 = 8 km/h, so Tuesday was faster, by 8 − 6 = 2 km/h. A candidate who works out the correct speeds but mislabels which day is faster gets Monday, by 2 km/h. A candidate who divides 15 ÷ 2.5 incorrectly as 5 instead of 6 gets a difference of 8 − 5 = 3 km/h, still crediting Tuesday. A candidate who forgets to find Monday's speed and gives Tuesday's speed itself as the difference states Tuesday, by 8 km/h.
- (a) 1.36 — Method: put the starting value into the right-hand side to get x₁, feed that value back in to get x₂, and round only once the second value has been found. Working: x₁ = 5 ÷ (1 + 2) = 5 ÷ 3 = 1.66666…; x₂ = 5 ÷ (1.66666… + 2) = 5 ÷ 3.66666… = 1.36363…. The digit in the third decimal place is 3, so x₂ = 1.36 correct to 2 decimal places. Answer: 1.36. The distractors: 1.67 is x₁, the value after a single use of the formula, given by a candidate who counts the starting value itself as x₁; 1.49 is x₃ = 1.48648…, one use of the formula too many; 1.37 comes from writing x₁ down as 1.66, truncating the display instead of keeping it in full, and then working out 5 ÷ 3.66 = 1.36612…, which rounds up to 1.37.
- (d) 300 m — For a constant speed, the speed-time graph is a horizontal line, and the area underneath is a rectangle: distance = speed × time = 15 × 20 = 300 m. Adding the two numbers instead of multiplying gives 35 m; dividing instead of multiplying gives 1.33 m; halving the product, as you would for a triangle, gives 7.5 m — but this section of the graph is a rectangle, not a triangle, so there is no halving to do.
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