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 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.
- 2.Four sequences are shown below. Sequence P: 4, 8, 16, 32, ... Sequence Q: 4, 8, 12, 16, ... Sequence R: 4, 7, 12, 19, ... Sequence S: 4, 9, 16, 25, ... Work out which of the sequences P, Q, R and S is geometric.
- 3.f(x) = x³ − 3x − 20, and the equation f(x) = 0 has exactly one solution. Work out the pair of consecutive integers between which that solution lies.y = x
- 4.A plumber charges a call-out fee plus an hourly rate. The total cost, y in pounds, of a job lasting x hours is given by y = 45x + 60. Work out the total cost of a job that lasts 3 hours.y = 45x + 60
- 5.The equation x² − 7 = 0 has a positive root. Let f(x) = x² − 7. Given that x₁ = 2.6 and x₂ = 2.65, work out which of these is correct.y = x² − 7
- 6.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.
- 7.A company's weekly profit, P thousand pounds, when it makes x thousand items is modelled by P = x² − 8x + 12, for x ≥ 0. Work out the values of x for which the company makes a loss.
- 8.The equation x³ = 6x + 20 can be solved using the iterative formula xₙ₊₁ = ∛(6xₙ + 20). Taking x₀ = 3, x₁ = 3.3620 correct to 4 decimal places. Using the full unrounded value of x₁, work out x₂ correct to 3 decimal places.
- 9.The equation x² − 3x − 7 = 0 can be solved using the iterative formula xₙ₊₁ = √(3xₙ + 7). The starting value is x₀ = 4, so x₁ is the value after the formula has been used once. Work out x₃ correct to 3 decimal places.
- 10.A graph shows the total monthly cost of a mobile phone tariff against the number of minutes of calls. The line starts at £12 and stays level until 200 minutes, then rises by 5p for each further minute of calls. Use the graph to work out the total cost of a month in which 250 minutes of calls are made.
- 11.A gym charges a joining fee plus a monthly fee. Anna paid £100 in total after 3 months of membership. Ben paid £160 in total after 6 months of membership (same joining fee and monthly fee as Anna). Work out the monthly fee.
- 12.A gas company charges a standing charge plus a rate per unit used. The total cost, y in pounds, for using x units is given by y = 0.15x + 20. Work out the cost of using 200 units.y = 0.15x + 20
- 13.The table shows the velocity, in m/s, of a train at various times, in seconds: 0 at t = 0, 20 at t = 4, 35 at t = 10, 27 at t = 14. Assuming the velocity changes at a constant rate between each pair of readings, estimate the distance the train travels between t = 0 and t = 12, using the trapezium rule.
- 14.The tangent to a curve at the point (5, 2) is the line y = mx + c. This tangent crosses the y-axis at (0, −8). Work out the gradient, m, of the tangent.
- 15.Two numbers have a sum of 45. The larger number is 9 more than the smaller number. Form an equation using n for the smaller number, and solve it to find the larger number.
Answer key
- (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.
- (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) 3 and 4 — Method: the graph of f(x) is continuous, so where it crosses the x-axis the value of f(x) changes sign; substitute consecutive integers until one value is negative and the next is positive. Working: f(2) = 8 − 6 − 20 = −18, f(3) = 27 − 9 − 20 = −2 and f(4) = 64 − 12 − 20 = 32. The sign changes from negative to positive between x = 3 and x = 4, so the solution lies there. Answer: 3 and 4. The distractors: 2 and 3 comes from ignoring the −3x term and solving x³ = 20, whose root is 2.71, one interval to the left; 6 and 7 comes from reading x³ as x² and solving x² − 3x − 20 = 0, whose positive root is 6.22; 4 and 5 is the interval immediately after the change of sign, named by a candidate who finds f(4) positive and quotes the interval beginning there instead of the one across which the sign actually turned.
- (d) £195 — Substituting x = 3 into y = 45x + 60 gives y = 45 × 3 + 60 = 135 + 60 = 195. A candidate who forgets to add the call-out fee would get only 45 × 3 = £135. A candidate who adds the hours to the fee and the rate instead of multiplying would get 45 + 60 + 3 = £108. A candidate who multiplies both the hourly rate and the call-out fee by the number of hours would get 45 × 3 + 60 × 3 = £315.
- (d) x₂ closer: f(x₂) = 0.0225, nearer to 0 — f(2.6) = 2.6² − 7 = 6.76 − 7 = −0.24, and f(2.65) = 2.65² − 7 = 7.0225 − 7 = 0.0225. The closer a value of x is to the root, the closer f(x) is to zero — regardless of sign. Since |0.0225| = 0.0225 is much smaller than |−0.24| = 0.24, x₂ = 2.65 is closer to the root. 'x₁ closer: −0.24 is the smaller value' comes from comparing the SIGNED values of f(x) rather than their distances from zero — −0.24 is indeed less than 0.0225 as a number, but that does not mean x₁ is closer to the root. 'x₁ closer: f(x₁) negative ⇒ nearer root' invents a rule that a negative f(x) means x is closer to the root; the sign of f(x) only tells you which side of the root x is on, not how close it is. 'x₂ is the exact root, since f(x₂) ≈ 0' misreads f(x₂) = 0.0225 as zero; the true root is √7 ≈ 2.6458, so f(2.65) is close to zero but not equal to it, and x₂ is an approximation, not the exact root.
- (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.
- (b) 2 < x < 6 — Factorise: x² − 8x + 12 = (x − 2)(x − 6), giving roots at x = 2 and x = 6. Since the coefficient of x² is positive, the graph is a U-shape that is negative, below the axis, between its roots. So P < 0 for 2 < x < 6. Distractor routes: x < 2 or x > 6 takes the region outside the roots, which is where P is positive, a profit, the opposite of a loss. 2 ≤ x ≤ 6 includes the endpoints, where P = 0 exactly, break-even rather than a loss, since the inequality is strict. −6 < x < −2 comes from factorising as (x + 2)(x + 6), reversing the sign of both roots.
- (d) 3.425 — x₁ = ∛(6 × 3 + 20) = ∛38 = 3.3620 (unrounded, 3.36198...). x₂ = ∛(6 × 3.3620 + 20) = ∛40.172 = 3.425 (3 d.p.). Choosing 3.362 stops at x₁ instead of continuing to x₂. Choosing 2.722 leaves out the '+ 20' inside the root, working out ∛(6 × 3.3620) = ∛20.172 = 2.722. Choosing 0.556 subtracts 20 instead of adding it, working out ∛(6 × 3.3620 − 20) = ∛0.172 = 0.556.
- (d) 4.521 — x₁ = √(3 × 4 + 7) = √19 = 4.358898944. x₂ = √(3 × 4.358898944 + 7) = √20.076696833 = 4.480702716. x₃ = √(3 × 4.480702716 + 7) = √20.442108148 = 4.521294964, which rounds to 4.521. Mislabelling the starting value x₀ as x₁, so that the working stops one iteration too early, reports the true x₂ = 4.480702716, which rounds to 4.481. Working out one iteration too many reports the true x₄ = √(3 × 4.521294964 + 7) = 4.534741987, which rounds to 4.535. Applying the formula in the wrong order, calculating √(3xₙ) + 7 at every step instead of √(3xₙ + 7), gives, from x₀ = 4: √12 + 7 = 10.464101615, then √(3 × 10.464101615) + 7 = 12.602883619, then √(3 × 12.602883619) + 7 = 13.148873950, which rounds to 13.149.
- (a) £14.50 — Method: the first 200 minutes are already covered by the £12, so only 250 − 200 = 50 minutes are charged extra. Extra charge = 50 × £0.05 = £2.50. Total cost = £12 + £2.50 = £14.50. Distractor origins: £24.50 charges 5p for all 250 minutes instead of only the 50 minutes over 200 (250 × £0.05 = £12.50, plus £12 = £24.50); £12.50 makes that same slip of charging all 250 minutes but then forgets to add the £12 monthly fee; £13.50 works out 250 − 200 wrongly as 30 extra minutes instead of 50 (30 × £0.05 = £1.50, plus £12 = £13.50).
- (d) £20 — Let f be the joining fee and m the monthly fee: f + 3m = 100 and f + 6m = 160. Subtracting the first equation from the second eliminates f: 3m = 60, so m = 20. A candidate who finds the joining fee instead of the monthly fee would get f = 100 − 3(20) = £40. A candidate who divides Ben's total by his number of months, ignoring that part of the cost is a fixed joining fee, would get 160 ÷ 6 ≈ £26.67. A candidate who divides the difference in cost by the total number of months instead of the difference in months would get (160 − 100) ÷ 9 ≈ £6.67.
- (a) £50 — Substituting x = 200 into y = 0.15x + 20 gives y = 0.15 × 200 + 20 = 30 + 20 = £50. A candidate who forgets to add the standing charge would get only 0.15 × 200 = £30. A candidate who misplaces the decimal point in the rate, using 1.5 instead of 0.15, would get 1.5 × 200 + 20 = £320. A candidate who swaps the roles of the rate and the number of units would work out 0.15 × 20 + 200 = £203.
- (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) 2 — The gradient between two points on a line is the change in y divided by the change in x. Using (5, 2) and (0, −8): (2 − (−8)) ÷ (5 − 0) = 10 ÷ 5 = 2. Inverting the fraction, dividing the change in x by the change in y instead, gives 5 ÷ 10 = 0.5. Reversing the order of the x-values in the denominator, giving (2 − (−8)) ÷ (0 − 5) = 10 ÷ (−5), gives −2. Misreading the y-intercept as 8 instead of −8, giving (2 − 8) ÷ 5 = −6 ÷ 5, gives −1.2.
- (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.
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