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.
Calculator
Answer key: Algebra worksheet — GCSE Higher
MathsUKwww.geekhero.co.uk
- (a) 27 — Method: let the smaller number be n, so the larger number is n + 9. Form the equation n + (n + 9) = 45. Working: simplify: 2n + 9 = 45, so 2n = 36, giving n = 18. The larger number is 18 + 9 = 27. Answer: 27. 18 is the smaller number, found correctly, but given as the answer to the question asking for the larger number. 22.5 comes from ignoring the "9 more" part and simply halving the total, 45 ÷ 2. 36 comes from a sign error when isolating the constant, treating the equation as 2n = 45 + 9 instead of 2n = 45 − 9.
- (a) 2.154 — Continuing xₙ₊₁ = ∛(3xₙ + 4) from x₁ = 2.154 gives x₂ = 2.187, x₃ = 2.195, settling towards the root near 2.196 — this is the formula that converges. Continuing xₙ₊₁ = (xₙ³ − 4) ÷ 3 from x₁ = 1.333 gives x₂ = −0.543 and then x₃ = −1.387, moving further from the root each time. Continuing xₙ₊₁ = 4 ÷ (xₙ² − 3) from x₁ = 4 gives x₂ = 0.308 and then x₃ = −1.377, swinging wildly rather than settling. Continuing xₙ₊₁ = xₙ³ − 2xₙ − 4 from x₁ = 0 gives x₂ = −4 and then x₃ = −60, running away from the root entirely.
- (c) P — Sequence P has a common ratio of 2 (4 × 2 = 8, 8 × 2 = 16, 16 × 2 = 32), so it is geometric. Sequence Q has a common difference of 4, which is arithmetic, not geometric — a candidate who confuses a constant difference with a constant ratio picks Q. Sequence R has increasing differences of 3, 5, 7, a quadratic sequence, not geometric — a candidate who assumes any fast-growing sequence must be geometric picks R. Sequence S is the square numbers from 2² onwards (2², 3², 4², 5²), which is also quadratic, not geometric — a candidate who thinks squaring always means geometric growth picks S.
- (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.
- (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) 48 litres — Split the area into three sections. The rectangle (t = 0 to 2) has area 2 × 12 = 24. The first trapezium (t = 2 to 5, parallel sides 12 and 3, width 3) has area 1/2 × (12 + 3) × 3 = 22.5. The second trapezium (t = 5 to 6, parallel sides 3 and 0, width 1) has area 1/2 × (3 + 0) × 1 = 1.5. Total volume: 24 + 22.5 + 1.5 = 48 litres. Forgetting the rectangle and adding only the two trapeziums gives 22.5 + 1.5 = 24 litres. Using a single trapezium across the whole 6 minutes, with parallel sides 12 and 0, ignoring that the first 2 minutes are constant, gives 1/2 × (12 + 0) × 6 = 36 litres. Correctly finding the rectangle and the first trapezium but forgetting to halve the second trapezium, using (3 + 0) × 1 = 3 instead of 1.5, gives 24 + 22.5 + 3 = 49.5 litres.
- (a) 4 — Rectangle 1: P = 2(9 + 4) = 2 × 13 = 26 cm. Rectangle 2: P = 2(6 + 5) = 2 × 11 = 22 cm. Difference = 26 − 22 = 4 cm. 2 comes from finding the difference between (l + w) for each rectangle, 13 − 11 = 2, but forgetting to double it. 26 comes from giving rectangle 1's whole perimeter instead of the difference between the two. 6 comes from doubling only the change in length, 2 × (9 − 6), and ignoring that the width also changed.
- (b) x ≥ 5 — Expand the bracket: 2(3x − 1) = 6x − 2, so the inequality is 6x − 2 ≥ 4x + 8. Subtract 4x from both sides and add 2 to both sides: 2x ≥ 10. Divide both sides by 2: x ≥ 5. A candidate who only multiplies the 3x by 2 and forgets to multiply the −1 gets 6x − 1 ≥ 4x + 8, leading to x ≥ 4.5. A candidate who adds 4x instead of subtracting it gets 10x ≥ 10, leading to x ≥ 1. A candidate who multiplies by 2 instead of dividing gets x ≥ 20.
- (c) 50 — Method: the edge of the track is a circle centred on the origin, so it has equation x² + y² = r²; use the floodlight whose coordinates are given to find r², then substitute x = 0 to find where that circle meets the positive y-axis. Working: 30² + 40² = 900 + 1600 = 2500, so the edge of the track is x² + y² = 2500. Substituting x = 0 gives k² = 2500, so k = √2500 = 50, taking the positive root because k is positive. Answer: k = 50. The distractors: 2500 is r² used as though it were the radius, which would place the second floodlight fifty times too far from the centre; 70 adds the two coordinates, 30 + 40, instead of adding their squares; 40 takes the y-coordinate of the first floodlight to be the radius, which would only be true if that floodlight stood on the y-axis itself.
- (a) 271 m — The reading at t = 12 isn't in the table, so it must be interpolated between t = 10 (35 m/s) and t = 14 (27 m/s): the velocity falls by 8 m/s over the 4-second gap, so over 2 seconds it falls by 4 m/s, giving 35 − 4 = 31 m/s. The trapezium rule then uses three strips of unequal width, 4, 6 and 2 seconds, with areas of 40, 165 and 66: 40 + 165 + 66 = 271 m. Writing 267 m comes from skipping the interpolation and using the table's t = 14 reading, 27, directly as the height at t = 12. Writing 282 m comes from treating all three strips as though they were 4 seconds wide, instead of using the true gaps of 4, 6 and 2 seconds between the readings. Writing 275 m comes from using the t = 10 reading, 35, at both ends of the final strip, instead of interpolating a new value at t = 12.
- (d) £16.70 — The mileage charge is 2.20 × 6 = £13.20. Adding the fixed fee: £13.20 + £3.50 = £16.70. A candidate who forgets the fixed fee gives just the mileage charge, £13.20. A candidate who adds the fixed fee to the per-mile rate before multiplying by the number of miles, (3.50 + 2.20) × 6, gets £34.20. A candidate who rounds £2.20 down to £2 gets 2 × 6 + 3.50 = £15.50.
- (d) 110 — Method: form the equation 15 + 0.08m = 23.80, where m is the number of minutes, then solve for m. Working: subtract the fixed fee: 0.08m = 23.80 − 15 = 8.80. Divide by the cost per minute: m = 8.80 ÷ 0.08 = 110. Answer: 110 minutes. 1.1 comes from using 8 instead of 0.08 as the cost per minute, forgetting to convert pence to pounds. 297.5 comes from dividing the whole bill by the cost per minute without subtracting the fixed fee first, 23.80 ÷ 0.08. 485 comes from adding the fixed fee to the bill instead of subtracting it, before dividing by the cost per minute, (23.80 + 15) ÷ 0.08.
- (b) 4 — Method: form the equation 30 + 25h = 130, where h is the number of hours, then solve for h. Working: subtract the call-out fee from the total bill: 25h = 130 − 30 = 100. Divide by the hourly rate: h = 100 ÷ 25 = 4. Answer: 4 hours. 5.2 comes from dividing the whole bill by the hourly rate without subtracting the fixed fee first, 130 ÷ 25. 3.5 comes from swapping the fee and the rate, subtracting the rate from the bill and dividing by the fee, (130 − 25) ÷ 30. 6.4 comes from adding the call-out fee to the bill instead of subtracting it, before dividing by the rate, (130 + 30) ÷ 25.
- (d) 8 — Method: write both present ages in terms of one letter, take four years off each of them, and form an equation from the comparison four years ago. Working: let Amelia be y, so her mother is 3y; four years ago they were y − 4 and 3y − 4, giving 3y − 4 = 5(y − 4); expanding gives 3y − 4 = 5y − 20; adding 20 and subtracting 3y from both sides gives 16 = 2y, and dividing by 2 gives y = 8. Checking: Amelia is 8 and her mother 24; four years ago they were 4 and 20, and 20 = 5 × 4. Answer: 8. The distractors: 4 comes from solving correctly but giving Amelia's age four years ago instead of her age now; 16 comes from stopping at 2y = 16 and giving 16 as her age; 2 comes from collecting the y terms by adding the 5y instead of subtracting it, giving 3y + 5y = 20 − 4 and so 8y = 16.
- (c) 4 years — The yearly multiplier is 1 − 0.12 = 0.88. After 1 year: 2400 × 0.88 = 2112. After 2 years: 2112 × 0.88 = 1858.56. After 3 years: 1858.56 × 0.88 = 1635.5328. After 4 years: 1635.5328 × 0.88 = 1439.27 (2 d.p.), which is the first value below £1,500 — so it takes 4 complete years. Stopping after 3 years, where the value is still £1,635.53, above £1,500, gives 3 years — wrong, because the threshold has not been crossed yet at that point. Continuing for one year longer than needed gives 5 years — wrong, because the value has already fallen below £1,500 by year 4, so year 5 overcounts. Multiplying by the percentage itself instead of by the multiplier — using 0.12 in place of 0.88 — gives 2400 × 0.12 = £288 after a single year, apparently below £1,500 straight away, so 1 year — wrong, because 12% is the part lost each year, and the part left is 88%, so the multiplier is 0.88.
Build your own mix at the worksheet builder.