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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Answer key: Algebra worksheet — GCSE Higher
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- (b) £47, and T=15+8h is a formula (relates T and h). — The charge is £15 fixed plus £8 for each hour: T=15+8h, so for h=4, T=15+8×4=15+32=47. Because T=15+8h relates two different quantities, the total charge T and the number of hours h, it is a formula, not an equation — an equation is solved for one particular value of an unknown, but this relationship holds for every value of h a job might last. A candidate who forgets to include the fixed £15 fee would compute only 8×4=32. A candidate who adds the three numbers in the formula together instead of multiplying the hourly rate by the number of hours would compute 15+8+4=27. A candidate who correctly finds the charge but mistakes the formula for an equation, treating it as something to be solved for one specific value of h rather than a general relationship between T and h, would pick the correct charge with the wrong classification.
- (c) xₙ₊₁ = ∛(9 − 4xₙ) — Method: a formula xₙ₊₁ = f(xₙ) is a correct rearrangement when the equation x = f(x) turns back into the equation you started with, so rearrange x³ + 4x − 9 = 0 by making the cube the subject. Working: x³ + 4x − 9 = 0 gives x³ = 9 − 4x, because the 4x and the 9 each change sign as they cross the equals sign; taking the cube root of both sides gives x = ∛(9 − 4x), which is the formula xₙ₊₁ = ∛(9 − 4xₙ). Answer: xₙ₊₁ = ∛(9 − 4xₙ). The distractors: ∛(9 + 4xₙ) comes from writing x³ = 9 + 4x, moving the 4x across the equals sign without changing its sign; (9 + xₙ³)/4 comes from making the linear term the subject but keeping the sign of the cube, writing 4x = 9 + x³ when the equation gives 4x = 9 − x³; ∛(9 − 4xₙ³) cubes the x in the linear term as well, changing a term the original equation never cubed.
- (b) f(1) = −3 and f(2) = 3 — f(1) = 1³ − 1 − 3 = 1 − 1 − 3 = −3. f(2) = 2³ − 2 − 3 = 8 − 2 − 3 = 3. Since f(1) is negative and f(2) is positive, there is a change of sign, so a solution lies between x = 1 and x = 2. Dropping the −x term entirely gives f(1) = 1 − 3 = −2 and f(2) = 8 − 3 = 5. Using x² instead of x³ throughout gives f(1) = 1 − 1 − 3 = −3, which happens to coincide with the correct value at x = 1, but f(2) = 4 − 2 − 3 = −1, which does not show a change of sign at all. Working out −f(x) instead of f(x), a sign-flip error, gives f(1) = 3 and f(2) = −3, the correct sizes but with both signs reversed.
- (b) 4.39 — x₁ = ∛(150 − 3 × 4²) = ∛(150 − 48) = ∛102 = 4.672 (unrounded). x₂ = ∛(150 − 3 × 4.672²) = ∛(150 − 65.49) = ∛84.51 = 4.39 (2 d.p.). Choosing 4.67 stops after only one iteration, giving x₁ instead of x₂. Choosing 84.51 finds the value inside the cube root for x₂ but never takes the cube root. Choosing 6.32 comes from adding 3xₙ² instead of subtracting it inside the root, which does not match the given formula.
- (a) 0 — the line does not intersect the circle — Substitute y = 2x + 7 into x² + y² = 4: x² + (2x + 7)² = 4, which expands to x² + 4x² + 28x + 49 = 4, giving 5x² + 28x + 45 = 0. The discriminant is b² − 4ac = 28² − 4 × 5 × 45. Since 28² = 784 and 4 × 5 × 45 = 900, the discriminant is 784 − 900 = −116. Since the discriminant is negative, the quadratic has no real solutions, so the line does not meet the circle at all: 0 intersection points. Distractor routes: "1 — the line is a tangent" confuses a negative discriminant with a zero one; a discriminant of exactly zero gives one point, a tangent, but −116 is not zero. "2 — the line crosses the circle at two points" assumes a positive discriminant without actually working it out. "It cannot be found without solving the quadratic" is the whole point the discriminant exists to avoid — its sign alone, without finding x, tells you the number of real solutions.
- (c) 496 — d = 0.4v² + 6. First v² = 35 × 35 = 1225. Then 0.4 × 1225 = 490. Add 6: 490 + 6 = 496 m. 490 comes from forgetting to add the 6 at the end. 202 comes from squaring 0.4v together instead of squaring only v, (0.4 × 35)² + 6 = 14² + 6 = 202. 20 comes from using v instead of v², 0.4 × 35 + 6.
- (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) 12 — Method: call the younger brother's age x, write the elder brother's age in terms of x, and form an equation from the total. Working: the elder brother is x + 6, so x + (x + 6) = 30; simplifying gives 2x + 6 = 30, subtracting 6 from both sides gives 2x = 24, and dividing by 2 gives x = 12. Checking: 12 and 18 add up to 30 and differ by 6. Answer: 12. The distractors: 18 comes from solving correctly and then giving the elder brother's age, which is not the age asked for; 15 comes from halving 30 and ignoring the 6-year difference altogether; 24 comes from taking 6 off the total, 30 − 6 = 24, and giving that as an age.
- (c) x = 0 or x = 3 — Method: take out the common factor 5x: 5x(x − 3) = 0, so 5x = 0 or x − 3 = 0, giving x = 0 or x = 3. Distractor origins: x = 0 or x = 15 comes from forgetting to divide the second term by the common factor 5x correctly, leaving 15 instead of 3; x = 3 loses the solution x = 0 by dividing both sides by x; x = 5 or x = 3 mistakes the coefficient 5 itself for one of the solutions.
- (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.
- (b) x = 5 — Area = length × width, so x(x + 3) = 40, which rearranges to x² + 3x − 40 = 0. This factorises as (x + 8)(x − 5) = 0: the two numbers in the brackets must multiply to −40 and add to +3, and the pair 8 and −5 does both. This gives x = −8 or x = 5. Since x is a length, it cannot be negative, so x = 5. A candidate who gives both solutions without rejecting the negative one, which cannot be a length, answers x = 5 or x = −8. A candidate who picks the wrong factor pair of 40, such as 10 and −4 instead of 8 and −5, gets (x + 10)(x − 4) = 0 and answers x = 4. A candidate who rejects the wrong root, keeping the negative solution instead of the positive one, answers x = −8.
- (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) 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.
- (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.
- (d) 6.0 — Subtract 1.8 from both sides: 6.3x ≤ 38.2. Divide both sides by 6.3: x ≤ 6.0634… . The question asks for the greatest value to 1 decimal place that still satisfies the inequality. Testing 6.1: 6.3 × 6.1 + 1.8 = 40.23, which is more than 40, so 6.1 fails. Testing 6.0: 6.3 × 6.0 + 1.8 = 39.6, which is no more than 40, so 6.0 works and is the greatest such value. A candidate who simply rounds 6.0634… to 1 decimal place answers 6.1, without checking that it satisfies the inequality. A candidate who adds 1.8 instead of subtracting works out 41.8 ÷ 6.3 and answers 6.6. A candidate who forgets the 1.8 altogether divides 40 by 6.3 and answers 6.3.
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