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.
Answer key: Algebra worksheet — GCSE Higher
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- (b) 4x − 4 = 20 — Method: subtract 2x from both sides of the equation, and simplify each side separately. Working: left side: 6x − 4 − 2x = 4x − 4. Right side: 2x + 20 − 2x = 20. Answer: 4x − 4 = 20. 4x = 20 drops the −4 from the left side, as though subtracting 2x also removes the constant term. 8x − 4 = 20 comes from moving the 2x across to the left without changing its sign: it is taken off the right side correctly, leaving 20, but added to the left side instead of subtracted, giving 6x + 2x = 8x. 4x − 4 = 2x + 20 comes from subtracting 2x from the left-hand side only and leaving the right-hand side unchanged; whatever is done to one side must be done to the other.
- (b) 6x + 6 — Perimeter = 2[(2x + 5) + (x − 2)] = 2(3x + 3) = 6x + 6. A candidate who adds the length and width but forgets to double for the perimeter gets 3x + 3. A candidate who makes a sign error and adds 2 instead of subtracting it before doubling gets 2[(2x + 5) + (x + 2)] = 6x + 14. A candidate who multiplies the length and width instead of adding them, confusing the perimeter formula with the area formula, and then doubles that product, gets 2(2x + 5)(x − 2) = 4x² + 2x − 20.
- (b) 5 — Method: expand the bracket first, then solve the resulting linear equation. Working: 3(2x − 1) = 6x − 3, so 6x − 3 = 27. Add 3 to both sides: 6x = 30. Divide by 6: x = 5. Answer: 5. 4 2/3 comes from multiplying only the 2x inside the bracket by 3 and forgetting to multiply the −1 as well, giving 6x − 1 = 27. 4 comes from a sign error when expanding, writing 6x + 3 instead of 6x − 3. 4.5 comes from dividing 27 by 3 to get 2x − 1 = 9, then forgetting to add 1 back before dividing by 2, using 2x = 9 instead of 2x = 10.
- (c) 15 and 9 — Method: write the two facts as two equations in the same pair of letters and add them, because the letter with opposite signs cancels. Working: with x the larger number and y the smaller, x + y = 24 and x − y = 6; adding gives 2x = 30, so x = 15, and substituting into x + y = 24 gives y = 9. Answer: 15 and 9, which add to 24 and differ by 6. The distractors: 18 and 6 come from halving 24 to 12 and then adding and subtracting the whole difference of 6 instead of half of it, which leaves a difference of 12; 15 and 21 come from finding the larger number correctly and then adding 6 to it instead of subtracting; 15 and 6 come from finding the larger number and then taking the 6 in the question to be the smaller number itself.
- (a) x = 6 — Method: a minus sign in front of a bracket changes the sign of every term inside it, so expand the bracket first and then simplify. Working: expanding gives 4x − 2x + 6 = 18, which simplifies to 2x + 6 = 18; subtracting 6 from both sides gives 2x = 12, and dividing both sides by 2 gives x = 6. Answer: x = 6. The distractors: x = 12 comes from leaving the −6 unchanged when the bracket is removed, giving 4x − 2x − 6 = 18 and so 2x = 24; x = 24 comes from reaching 2x = 12 correctly and then multiplying by 2 instead of dividing by 2; x = 3 comes from dividing by 2 too early, at 2x + 6 = 18, and dividing only the 2x and the 18 while leaving the 6 untouched, which gives x + 6 = 9.
- (a) Litres — The area under a graph is found by multiplying a y-value by an x-value, so its units are the y-axis units multiplied by the x-axis units: litres per minute × minutes = litres, since the 'per minute' cancels with the 'minutes'. Answering 'litres per minute' keeps the y-axis units unchanged, as if multiplying by time did nothing to the units at all. Answering 'litres per minute squared' treats the x-axis as also being measured 'per minute', squaring a unit that should instead cancel. Answering 'minutes' keeps only the x-axis units and drops the rate altogether.
- (a) y = −2x − 1 — Method: the gradient is the change in y divided by the change in x with both differences taken in the same order, and the constant then comes from substituting either point into y = mx + c. Working: m = (−9 − 3) ÷ (4 − (−2)) = (−12) ÷ 6 = −2, so the line is y = −2x + c; substituting (−2, 3) gives 3 = −2 × (−2) + c = 4 + c, so c = 3 − 4 = −1. Answer: y = −2x − 1. The distractors: y = −2x + 1 comes from rearranging 3 = 4 + c the wrong way round and taking the constant as 4 − 3; y = 2x + 7 comes from losing the minus sign when −12 is divided by 6 and then substituting correctly, 3 = 2 × (−2) + c; y = −(1/2)x + 2 comes from writing the gradient upside down as the change in x over the change in y, 6 ÷ (−12).
- (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) x = 3 — The table is symmetrical about the turning point: y = 0 at both x = 1 and x = 5, and the lowest value, y = −4, occurs exactly halfway between them, at x = 3. Choosing x = 5 picks one of the roots rather than the midpoint between them. Choosing x = 1 picks the other root for the same reason. Choosing x = 6 picks the x-value where y returns to its starting value of 5, which is not the turning point.
- (d) L/6 − 4 — Method: find the length of one equal piece first (divide by 6), then apply the later change (subtract 4) to that piece. Working: one piece is L/6 metres; removing 4 metres from it gives L/6 − 4. Answer: L/6 − 4. L/6 + 4 comes from adding the 4 metres instead of removing it. (L − 4)/6 comes from removing the 4 metres from the whole rope before cutting it into pieces, the wrong order. 4 − L/6 comes from subtracting the piece length from 4 instead of the other way round.
- (d) 5n + 4 — Method: find the common difference, then find the constant that fits the first term. Working: 14 − 9 = 5, 19 − 14 = 5, 24 − 19 = 5, so the terms increase by 5 each time and the nth term has the form 5n + c. Substituting n = 1: 5(1) + c = 9, so c = 4. Answer: the nth term is 5n + 4. The value 5n comes from leaving out the constant. The value 5n + 9 comes from using the first term as the constant directly, without subtracting the common difference first. The value 9n + 5 comes from swapping the roles of the first term and the common difference — using the first term, 9, as the coefficient of n and the difference, 5, as the constant.
- (d) 10 — The right angle is at (1, 1). The vertical side has length 5 − 1 = 4 and the horizontal side has length 6 − 1 = 5, so the area is (4 × 5) ÷ 2 = 20 ÷ 2 = 10. A candidate who forgets to halve the product of the two sides gets 4 × 5 = 20. A candidate who forgets to subtract the shared vertex's coordinate and uses the raw coordinates 6 and 5 as the side lengths gets (6 × 5) ÷ 2 = 30 ÷ 2 = 15. A candidate who uses only one side length as the area gets 5.
- (c) 21.6 — m² = (−2.4)² = 5.76, since a negative number squared is positive, so 5m² = 28.8. Then 3m = 3 × (−2.4) = −7.2. Total: 28.8 + (−7.2) = 21.6. −36 comes from taking m² as negative, 5 × (−5.76) − 7.2 = −36, forgetting that squaring removes the negative sign. 36 comes from ignoring the negative sign on m altogether and using m = 2.4 throughout. 29.4 comes from treating 3m as 3 + m instead of 3 × m, giving 28.8 + (3 + (−2.4)).
- (c) 8n + 7 — Method: find the weekly increase, then find the constant by adjusting the week 1 total. Working: the total goes up by £8 each week (23 − 15 = 8), so the coefficient of n is 8. The constant is the week 1 total minus the common difference: 15 − 8 = 7. Answer: the nth term is 8n + 7. 8n + 15 comes from using the week 1 total, 15, as the constant without subtracting the common difference. 8n − 1 comes from a slip in working out the constant, subtracting the common difference twice (15 − 8 − 8 = −1) instead of once. 7n + 8 comes from swapping the weekly increase and the constant.
- (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.
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