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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- (c) y = x² — Method: test a candidate rule against every pair given, not just one — a rule that fits one pair and fails another is not the rule. Working: the outputs 1, 4, 9 rise by 3 and then by 5, so they are not going up in equal steps and the input is not simply multiplied by a fixed number; comparing each output with its own input gives 1 × 1 = 1, 2 × 2 = 4 and 3 × 3 = 9, and all three pairs fit. Answer: y = x². The distractors: y = 3x comes from fitting only the last pair, where 3 × 3 = 9, and reading that 3 as a multiplier; y = 3x − 2 comes from assuming a multiply-then-add rule and using the first step in the outputs, 4 − 1 = 3, as the multiplier — it fits the first two pairs and fails the third; y = 2x comes from fitting only the pair 2 and 4 and reading every output as double its input.
- (a) 33 — Substitute n=4 into 2n²+1: 2×4²+1=2×16+1=33. A candidate who computes n² as 2×n instead of n×n would compute 2×(2×4)+1=2×8+1=17. A candidate who correctly finds 2×16 but forgets to add the constant 1 would stop at 32. A candidate who squares the whole term 2n, rather than squaring n before multiplying by 2, would compute (2×4)²+1=64+1=65.
- (c) a formula — A statement that shows how to calculate one quantity from others, using an equals sign, is a formula. P = 2l + 2w tells us how to work out the perimeter, P, from the length and width, so it is a formula. Distractor origins: 'an equation' confuses a formula linking several letters with an equation solved for one unknown value; 'an identity' confuses a formula with a statement that is true for every value of a single variable; 'an expression' forgets that an expression has no equals sign at all.
- (c) (3, 9) — Reflecting in the horizontal line y = 2 keeps the x-coordinate the same and maps y to 2 × 2 − y = 4 − (−5) = 9, so S = (3, 9). A candidate who uses k − y instead of 2k − y gets 2 − (−5) = 7, giving (3, 7). A candidate who reflects in the x-axis instead of the line y = 2, simply changing the sign of y, gets (3, 5). A candidate who also changes the sign of the x-coordinate, as if reflecting in both axes, gets (−3, 9).
- (c) 72 m — The first section is a triangle with base 4 s and height 8 m/s, area = 1/2 × 4 × 8 = 16 m. The middle section is a rectangle with base 6 s and height 8 m/s, area = 6 × 8 = 48 m. The final section is a triangle with base 2 s and height 8 m/s, area = 1/2 × 2 × 8 = 8 m. Total distance = 16 + 48 + 8 = 72 m. Treating both sloping sections as full rectangles instead of triangles — forgetting the factor of a half — gives 4 × 8 + 6 × 8 + 2 × 8 = 96 m, which is wrong because a sloping speed-time section is a triangle, not a rectangle. Leaving out the final, decreasing section entirely gives 16 + 48 = 64 m, which is wrong because the cyclist is still moving and covering distance while slowing down. Leaving out the middle, constant-speed section gives 16 + 8 = 24 m, which is wrong because the cyclist covers the most distance of all during the 6 seconds of constant speed.
- (d) x = 5y + 4 — To make x the subject of y = (x − 4)/5, first multiply both sides by 5 to clear the fraction: 5y = x − 4, then add 4 to both sides: x = 5y + 4. Writing x = 5y − 4 multiplies correctly but keeps the minus sign on the 4 instead of changing it to a plus when moving it across. Writing x = y/5 + 4 divides by 5 instead of multiplying, the wrong inverse of the fraction. Writing x = 5(y + 4) adds 4 before multiplying by 5, reversing the correct order of the two steps. The correct rearrangement is x = 5y + 4.
- (a) 4 — a₅ = a₄ + a₃, so a₄ = a₅ − a₃ = 29 − 11 = 18. a₄ = a₃ + a₂, so a₂ = a₄ − a₃ = 18 − 11 = 7. a₃ = a₂ + a₁, so a₁ = a₃ − a₂ = 11 − 7 = 4. Checking forwards: 4, 7, 11, 18, 29. Answering 7 stops one step early, reporting a₂ = 7 instead of continuing back one more step to a₁ — wrong, because the question asks for a₁, not a₂. Answering 18 reports a₄ = 18, an intermediate value found along the way, instead of a₁ — wrong, because a₄ is a term used to reach the answer, not the term the question asks for. Answering 3 takes one backward step too many, working out a further term a₀ = a₂ − a₁ = 7 − 4 = 3 — wrong, because the sequence starts at a₁, so a₁ = 4 is as far back as the question goes.
- (d) 8 — Since the mean of the four numbers is 12.5, their total is 4 × 12.5 = 50. The three known numbers add up to 8 + 15 + 19 = 42, so n = 50 − 42 = 8. A candidate who multiplies the mean by 3 instead of 4 gets a total of 37.5, giving n = 37.5 − 42 = −4.5. A candidate who forgets to subtract the three known numbers and gives the total itself as n states n = 50. A candidate who subtracts the mean from the total of the three known numbers instead of the other way round gets n = 42 − 12.5 = 29.5.
- (a) x = 3.5 — Method: clear the fractions by multiplying both sides by 15, expand both brackets, then collect the x terms on one side and the numbers on the other. Working: multiplying both sides by 15 gives 3(3x + 2) = 5(x + 4), which expands to 9x + 6 = 5x + 20; subtracting 5x and 6 from both sides gives 4x = 14, and dividing both sides by 4 gives x = 3.5. Answer: x = 3.5. The distractors: x = 1 comes from collecting the x terms by adding the 5x instead of subtracting it, giving 9x + 5x = 20 − 6 and so 14x = 14; x = 4.5 comes from expanding 3(3x + 2) as 9x + 2, multiplying only the x term by the 3, which leads to 4x = 18; x = 10 comes from reaching 4x = 14 correctly and then subtracting 4 instead of dividing by 4.
- (d) 23 — Method: let the smallest integer be n, so the three consecutive integers are n, n + 1 and n + 2. Form the equation n + (n + 1) + (n + 2) = 72. Working: simplify the left side: 3n + 3 = 72, so 3n = 69, giving n = 23. Answer: 23. 24 comes from dividing 72 by 3 directly, 72 ÷ 3 = 24, which finds the middle integer rather than realising the three numbers differ. 25 comes from correctly finding n = 23 but reading off the largest integer, n + 2, instead of the smallest as asked. 21 comes from dividing first and subtracting after, (72 ÷ 3) − 3, instead of subtracting 3 before dividing by 3.
- (a) (40, 0) — Method: a graph crosses the x-axis where y = 0, which in this context is the moment the tank holds no water, so substituting y = 0 and solving gives the time. Working: 0 = −5x + 200 gives 5x = 200, so x = 200 ÷ 5 = 40 and the crossing point is (40, 0); the tank is empty after 40 minutes. Answer: (40, 0). The distractors: (0, 200) is the y-axis crossing, the 200 litres in the tank at the start, found by substituting x = 0 instead of y = 0; (200, 0) comes from reading the constant 200 as the x-coordinate without dividing by the 5 litres lost each minute; (0, 40) has the right number in the wrong place, with the time written as a y-coordinate.
- (b) £14 — Method: read the fixed charge (the cost at 0 miles) and the rate (the cost per extra mile) from the graph, then use them to work out the cost for a distance beyond the part that is plotted. Working: the graph shows a fixed charge of £2 at 0 miles, and the cost rises by £2 for every extra mile, so for 6 miles the cost is £2 + (£2 × 6) = £2 + £12 = £14. Answer: £14. Distractor refutation: £12 comes from multiplying the rate by the distance and leaving out the £2 fixed charge. £8 comes from misreading the rate as £1 per mile instead of £2 per mile. £24 comes from adding the fixed charge to the rate first and then multiplying the total by the distance, instead of multiplying the rate by the distance and then adding the fixed charge.
- (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) −3 — Method: rewrite the equation in the form y = mx + c, then read off the gradient. Working: y = 4 − 3x can be written as y = −3x + 4, so comparing with y = mx + c gives m = −3. Answer: the gradient is −3. The value 3 comes from ignoring the negative sign on the x term. The value 4 comes from reading off the y-intercept instead of the gradient. The value −4 comes from a sign error, applying the negative sign to the intercept instead of the gradient.
- (a) 19 — Each time x increases by 1, y increases by 3 (4, 7, 10, 13 — a constant difference of 3). So at x = 4, y = 13 + 3 = 16, and at x = 5, y = 16 + 3 = 19. A candidate who stops one step early, giving the value for x = 4 instead of x = 5, answers 16. A candidate who overcounts and adds three steps of 3 instead of two from x = 3 gets 13 + 9 = 22. A candidate who mistakes the y-intercept (4) for the common difference and adds 4 twice from x = 3 gets 13 + 8 = 21.
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