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.
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Answer key: Algebra worksheet — GCSE Foundation
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- (b) 3p — p + p + p means three lots of p added together, and repeated addition of the same term is written as a coefficient: 3p. Writing p³ mistakes the repeated addition for repeated multiplication, as if the expression had been p × p × p. Writing 3 + p adds the number of terms (3) onto p as a separate constant, instead of writing 3 as a coefficient of p. Writing just p forgets to count the terms at all, as though repeating the same letter makes no difference. The simplified expression is 3p.
- (c) 5(x + y − 1) — Method: take out the highest common factor of all three terms and divide every term by it, the number term included. Working: the highest common factor of 5x, 5y and −5 is 5; dividing gives 5x ÷ 5 = x, 5y ÷ 5 = y and −5 ÷ 5 = −1, so the bracket holds x + y − 1. Answer: 5(x + y − 1), which multiplies back out to 5x + 5y − 5. The distractors: 5(x + y + 1) comes from dividing −5 by 5 and losing the minus sign; 5(x + y − 5) comes from dividing only the terms containing a letter by 5 and carrying the −5 into the bracket unchanged; 5(xy − 1) comes from collecting the unlike terms 5x and 5y as 5xy before factorising.
- (a) s/f — Sharing s sweets equally between f friends means dividing the total by the number of friends, written as a fraction: s/f. Writing f/s divides the wrong way round, sharing the number of friends between the sweets instead of the sweets between the friends. Writing s − f mistakes sharing for taking away, subtracting the number of friends from the number of sweets. Writing sf multiplies the two quantities together, which would make the total larger rather than splitting it into smaller equal parts. The number of sweets each friend receives is s/f.
- (b) x = 4 — Method: every point on a vertical line has the same x-coordinate, so its equation sets x equal to a number, and that number is the x-coordinate of any point known to be on the line. Working: the line is vertical and passes through (4, 7), whose x-coordinate is 4, so every point on the line has x-coordinate 4 and the equation is x = 4. Answer: x = 4. The distractors: y = 7 is the horizontal line through the same point, obtained by fixing the y-coordinate instead of the x-coordinate; x = 7 uses the right form but takes the y-coordinate of the point rather than its x-coordinate; y = 4x treats the 4 as a gradient and gives a sloping line through the origin, which does not pass through (4, 7) at all.
- (c) x = 3, y = 1 — Method: the y terms are +y and −y, so adding the two equations removes y and leaves an equation in x alone. Working: adding 2x + y = 7 and x − y = 2 gives 3x = 9, so x = 3; substituting x = 3 into x − y = 2 gives 3 − y = 2, so y = 1. Answer: x = 3, y = 1, which also satisfies 2 × 3 + 1 = 7. The distractors: x = 1, y = 3 comes from finding the two values correctly and then writing them against the wrong letters; x = 3, y = 2 comes from substituting x = 3 into 2x + y = 7 as 2 + 3 + y = 7, adding the coefficient instead of multiplying by it; x = 3, y = −1 comes from substituting into x − y = 2 as though it read x + y = 2.
- (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.
- (d) £40 — Method: add the two prices to make one expression in x, turn 'at least' into ≥, solve the inequality, then read the lowest possible price off the boundary of the solution set. Working: the two items cost x + (2x + 30) = 3x + 30 pounds, so 3x + 30 ≥ 150; subtracting 30 from both sides gives 3x ≥ 120; dividing both sides by 3 gives x ≥ 40, and the smallest value the shirt price is allowed to take is the boundary, £40. Answer: £40. The distractors: £50 comes from leaving the £30 out of the total and solving 3x ≥ 150; £60 comes from using the trousers expression on its own, 2x + 30 ≥ 150; £120 comes from stopping at 3x ≥ 120 and reading the 120 as the price of the shirt without dividing by 3.
- (a) 6b − 5 — Method: sort the expression into terms in b and number terms, then collect each kind, keeping the sign in front of every term. Working: the b terms give 2b + 2b + 2b = 6b; the number terms are −1 and −4, and −1 − 4 = −5. Answer: 6b − 5. The distractors: 6b − 3 comes from working out the difference 4 − 1 instead of adding two negative numbers; 6b + 5 comes from adding 1 and 4 while ignoring both minus signs; 8b − 5 comes from multiplying the coefficients 2, 2 and 2 instead of adding them.
- (c) y = −x + 30 — Method: in a linear model the value at the start is the constant term and the steady rate of change is the gradient, which is negative when the quantity is falling. Working: at x = 0 the candle is 30 cm tall, so the constant term is 30; it loses 1 cm every hour, so the gradient is −1 and the height after x hours is y = −x + 30. Answer: y = −x + 30. The distractors: y = x + 30 comes from taking the rate as +1 and making the candle grow rather than shrink; y = 30x − 1 comes from building the model correctly as 30 − 1x and then copying it into the form y = mx + c with the two numbers left where they stood, so the starting height 30 ends up multiplying x and the hourly 1 is left behind as the constant being taken away; y = −30x + 1 comes from reading the two numbers the other way round, taking 30 cm per hour as the rate and 1 cm as the starting height, which burns 30 cm an hour from a candle only 1 cm tall.
- (d) 6 — Method: rearrange the formula to make d the subject, then substitute C = 133. Working: C = 25 + 18d, so subtracting 25 from both sides gives C − 25 = 18d, then dividing by 18 gives d = (C − 25) / 18. Substituting C = 133: d = (133 − 25) / 18 = 108 / 18 = 6. The value 7.4 comes from dividing 133 by 18 without subtracting the fixed £25 first (133 / 18 ≈ 7.4). The value 4.6 comes from pairing the numbers the wrong way round, working out (133 − 18) / 25 = 4.6. The value 90 comes from subtracting both 25 and 18 from 133 instead of dividing by 18.
- (b) A straight line. — y = 3x + 2 is a linear function, since the highest power of x is 1, so its graph is a straight line with gradient 3 and y-intercept 2. Saying it is a U-shaped curve confuses a linear graph with a quadratic graph, which has an x² term. Saying it decreases then increases describes a curve with a turning point, which a straight line does not have. Saying it gets closer to an axis but never reaches it describes a reciprocal graph, y = k/x, not a linear one.
- (c) £9.70 — F = 2.50 + 1.20 × 6 = 2.50 + 7.20 = £9.70. £7.20 comes from forgetting to add the £2.50 fixed charge. £10.20 comes from adding the 2.50 to the number of miles before multiplying by 1.20, 1.20 × (6 + 2.50). £22.20 comes from adding the fixed charge and the rate together first and then multiplying by the miles, (2.50 + 1.20) × 6.
- (a) 6x + 12 — A regular hexagon has 6 equal sides, so the perimeter is 6(x + 2) = 6x + 12. A candidate who multiplies only the x-term by 6 and forgets to multiply the 2 gets 6x + 2. A candidate who multiplies only the number term by 6 and forgets to multiply the x gets x + 12. A candidate who adds 6 and 2 to make a single coefficient of x instead of expanding the brackets gets 8x.
- (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.
- (c) 32 — Using the rule 3n + 4: the 3rd term is 3 × 3 + 4 = 13, and the 5th term is 3 × 5 + 4 = 19, so their sum is 13 + 19 = 32. Forgetting to add the 4 for the 3rd term, 3 × 3 = 9, and adding the correct 5th term, gives 9 + 19 = 28. Rounding the 19 up to 20 to make the addition easier and then forgetting to take the extra 1 back off, 13 + 20 = 33, gives 33. Using the rule 4n + 3 instead of 3n + 4 gives 4 × 3 + 3 = 15 and 4 × 5 + 3 = 23, summing to 38.
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