Printable · GCSE Foundation · ages 14-16
Algebra worksheet — GCSE Foundation
Fifteen questions across the algebra statements at Foundation tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
Answer key: Algebra worksheet — GCSE Foundation
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- (c) 5n + 7 — Method: find the common difference between the rows, then find the constant by adjusting the first row's total. Working: each row has 5 more seats than the last, so the coefficient of n is 5. The constant is the first row's total minus the common difference: 12 − 5 = 7. Answer: the nth term is 5n + 7. 5n + 12 comes from using row 1's total, 12, as the constant without subtracting the common difference. 5n + 2 comes from a slip in working out the constant, subtracting the common difference twice (12 − 5 − 5 = 2) instead of once. 7n + 5 comes from swapping the common difference and the constant.
- (d) The fixed call-out fee, charged before the hourly rate. — Method: compare the formula with C = (fixed charge) + (rate) × h, where the fixed charge is the part that does not depend on h. Working: in C = 40 + 25h, the term 25h depends on the number of hours, h, but 40 does not change however many hours the job takes. Answer: 40 is the fixed call-out fee. 'The hourly rate' confuses the fixed term with the coefficient of h, which is actually 25. 'The total cost for a job lasting 1 hour' comes from substituting h = 1 into the formula (40 + 25 = 65) rather than reading off the constant term. 'The number of hours worked before charging starts' wrongly treats the constant, which is in pounds, as if it were measured in hours.
- (a) 4n + 1 — Method: find the increase in cost per hour, then find the constant by adjusting the 1-hour cost. Working: the cost goes up by £4 for each extra hour (9 − 5 = 4), so the coefficient of n is 4. The constant is the 1-hour cost minus the common difference: 5 − 4 = 1. Answer: the nth term is 4n + 1. 4n + 5 comes from using the 1-hour cost, 5, as the constant without subtracting the common difference. 4n − 3 comes from a slip in working out the constant, subtracting the common difference twice (5 − 4 − 4 = −3) instead of once. n + 4 comes from swapping the hourly increase and the constant.
- (a) 52 — Test successive terms: n=6 gives 6²+3=39, which is not greater than 50. n=7 gives 7²+3=52, which is greater than 50, so the first term greater than 50 is 52. A candidate who stops at n=6, before checking whether 39 actually exceeds 50, would give 39. A candidate who solves n²>50 instead of n²+3>50, ignoring the +3 in the search, would find n=8 is the first value with n²>50 (since 7²=49) and compute 8²+3=67. A candidate who computes n² by doubling n instead of squaring it would compute 2×7+3=17.
- (d) 8(7x − 3) — Method: find the highest common factor of the two terms, write it in front of a bracket and divide each term by it. Working: 56 = 8 × 7 and 24 = 8 × 3, so the highest common factor is 8; dividing gives 56x ÷ 8 = 7x and 24 ÷ 8 = 3, and the subtraction sign stays between them. Answer: 8(7x − 3), which multiplies back out to 56x − 24. The distractors: 8(7x + 3) comes from dropping the minus sign of −24 while dividing; 8(56x − 3) comes from dividing only the number term by 8 and leaving 56x untouched inside the bracket; 8(7x − 16) comes from subtracting 8 from 24 instead of dividing 24 by 8.
- (b) 6x − 3 = 4x + 9 — Method: multiply every term inside the bracket by the number outside it; the right-hand side stays as it is given. Working: 3 × 2x = 6x and 3 × (−1) = −3, so 3(2x − 1) = 6x − 3. Answer: 6x − 3 = 4x + 9. 6x − 1 = 4x + 9 comes from multiplying only the 2x by 3 and leaving the −1 unchanged. 5x − 3 = 4x + 9 comes from adding the 3 to the 2 instead of multiplying, treating 3 × 2x as (3 + 2)x = 5x. 6x − 4 = 4x + 9 comes from working out 3 × (−1) as −1 − 3 = −4 instead of 3 × (−1) = −3.
- (a) 23 — The number of logs increases by 4 for each layer down, starting from 3 in the top layer, so the nth layer has 3+(n−1)×4 logs. For the 6th layer: 3+5×4=3+20=23. A candidate who adds the difference of 4 six times instead of five, treating the top layer as needing an addition too, would compute 3+6×4=27. A candidate who uses 4n instead of 4n−1, omitting the adjustment for the first layer, would compute 4×6=24. A candidate who subtracts the common difference, 4, instead of 1 when adjusting the multiplier would compute 4×6−4=20.
- (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.
- (c) S = P + C — Method: S has C subtracted from it, so undo that subtraction by adding C to both sides. Working: P + C = S − C + C, so P + C = S, which is written S = P + C. Answer: S = P + C. S = P − C repeats the formula's own operation instead of its inverse. S = C − P subtracts the wrong way round. S = PC multiplies the two given quantities instead of adding them.
- (a) (6, 0) — Method: a pair of coordinates records movement from the origin, the across movement written first and the up or down movement second. Working: C is 6 units to the right of the origin, so the across number is 6; it is 0 units up, so the up number is 0; written in order that gives (6, 0), a point on the x-axis. Answer: (6, 0). The distractors: (0, 6) comes from writing the two movements the wrong way round; (0, 0) comes from reading '0 units up' as meaning the point never left the origin at all, which ignores the movement across; (7, 0) comes from counting the origin itself as the first unit while counting 6 units to the right.
- (d) t = 5 — The maximum height occurs halfway between the two times when the ball is at ground level: the midpoint of t = 1 and t = 9 is (1 + 9) ÷ 2 = 5, so the ball reaches its maximum height at t = 5 seconds. Getting t = 4 comes from halving the DIFFERENCE between the times, 9 − 1 = 8, then 8 ÷ 2 = 4, instead of finding their midpoint. Getting t = 8 uses that difference, 9 − 1 = 8, as if the gap between the two times were itself the time of the maximum. Getting t = 10 adds the two times, 1 + 9 = 10, but forgets to divide by 2.
- (a) 7 — Method: substitute the given value of x into the rule and carry out the operation written in it. Working: y = x + 4 with x = 3 gives y = 3 + 4, an addition of two whole numbers. Answer: y = 7. The distractors: 12 comes from multiplying by 4 instead of adding 4; 4 comes from writing down the number in the rule and never using the input at all; 1 comes from subtracting, 4 − 3, instead of adding.
- (c) 8a + 2 — The perimeter of a rectangle is twice the length plus twice the width: P = 2(3a + 2) + 2(a − 1) = (6a + 4) + (2a − 2) = 8a + 2. Answering 5a adds the length and width once each but doubles only one of them, missing that a rectangle has two of each side. Answering 8a + 6 distributes the 2 into (a − 1) correctly as far as 2a, but then adds 2 instead of subtracting it, as though the bracket had been (a + 1). Answering 12a + 8 uses the length for all four sides instead of using the length twice and the width twice, as if the field were a square with side (3a + 2). The perimeter of the field is (8a + 2) metres.
- (d) x = 3 — Method: undo the operations on the left in reverse order — deal with the subtraction of 1 first, then with the multiplication by 3. Working: adding 1 to both sides gives 3x = 9, and dividing both sides by 3 gives x = 3. Answer: x = 3. The distractors: x = 9 comes from stopping at 3x = 9 and writing 9 as the value of x; x = 6 comes from subtracting 3 from 9 instead of dividing by 3; x = 27 comes from multiplying 9 by 3 instead of dividing by 3.
- (a) False — 4(2x − 3) = 8x − 12, not 8x − 3. — Expand the bracket by multiplying both terms by 4: 4 × 2x = 8x and 4 × (−3) = −12, so 4(2x − 3) = 8x − 12, which is not 8x − 3 — the student is wrong. Saying 4(2x − 3) = 8x − 3 comes from multiplying only the 2x by 4 and copying the −3 across unchanged. Saying 4(2x − 3) = 2x − 12 comes from multiplying only the −3 by 4 and leaving 2x unmultiplied. Claiming it is true because both expressions are linear ignores that equivalence depends on the actual coefficients, not the type of expression.
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