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) 20 — F = 4s, so when s = 5, F = 4 × 5 = 20. 9 comes from adding 4 and 5 instead of multiplying, the mistake described in the question. 25 comes from squaring s (5²) instead of multiplying by 4. 45 comes from writing the digits 4 and 5 next to each other instead of carrying out the multiplication.
- (b) x = 2 and x = −5 — Set each factor equal to zero: x − 2 = 0 gives x = 2, and x + 5 = 0 gives x = −5, so the graph crosses the x-axis at x = 2 and x = −5. Writing x = −2 and x = 5 flips the sign of both roots. Writing x = 2 and x = 5 keeps the first root correct but forgets to flip the sign for the second factor, using +5 instead of solving x + 5 = 0. Writing x = −2 and x = −5 flips the sign of the first root only, from solving x − 2 = 0 as x = −2.
- (a) 3.5 km/h — Average speed is total distance ÷ total time. Total distance = 5 + 9 = 14 km. Total time, including the rest, = 1 + 1 + 2 = 4 hours. So average speed = 14 ÷ 4 = 3.5 km/h. Leaving out the 1 hour rest and dividing by only the 3 hours of walking gives 14 ÷ 3 = 4.67 km/h. Averaging the two separate speeds, 5 km/h and 4.5 km/h, instead of using total distance over total time, gives (5 + 4.5) ÷ 2 = 4.75 km/h. Dividing only the second leg's distance by the total time, 9 ÷ 4 = 2.25 km/h, ignores the first leg's distance.
- (a) 12 km/h — Method: convert 15 minutes to hours: 15 ÷ 60 = 0.25 h. Speed = distance ÷ time = 3 ÷ 0.25 = 12 km/h. Distractor origins: 9 km/h is the speed for the second section (3 km in 20 minutes) instead of the first; 8 km/h is the average speed for the whole journey (6 km in 45 minutes) instead of just the first section; 0.2 km/h divides 3 km by 15 without converting the minutes into hours.
- (b) 3 — Method: rearrange the equation into the form y = mx + c first, then read off the gradient. Working: dividing 4y = 12x + 20 by 4 gives y = 3x + 5, so the gradient is 3. The value 12 comes from reading the coefficient of x before dividing the equation by 4. The value 20 comes from using the constant term of the unsimplified equation instead of the gradient. The value 5 comes from finding the y-intercept, 20 / 4 = 5, and giving that instead of the gradient.
- (a) 8 — Both points share the x-coordinate, so the distance between them is the difference between the y-coordinates: 3 − (−5) = 8. A candidate who mistakenly uses the equal x-coordinates instead of the y-coordinates gets 4 − 4 = 0. A candidate who adds the y-coordinates instead of subtracting them gets 3 + (−5) = −2. A candidate who reads off only point B's y-coordinate as the distance gets 3.
- (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.
- (a) 14 — Substituting x = 3: y = 5(3) − 1 = 15 − 1 = 14. A candidate who stops after the multiplication and forgets to subtract 1 gets 15. A candidate who adds 1 instead of subtracting gets 16. A candidate who incorrectly treats the expression as 5 × (3 − 1) gets 10.
- (a) x + 14 — Expand each bracket separately: 3(x + 4) = 3x + 12, and −2(x − 1) = −2x + 2 (multiply −2 by both x and −1). Combine: 3x + 12 − 2x + 2 = x + 14. Writing x + 10 comes from taking −2(x − 1) as −2x − 2, not flipping the sign of the −1 inside the bracket. Writing 5x + 10 comes from treating the second bracket as +2(x − 1) instead of subtracting it, so the x-terms are added rather than subtracted. Writing x + 13 comes from only multiplying the 2 by the x, and carrying the −1 across unmultiplied.
- (d) (4, 8) — The other endpoint is found from 2 × midpoint − known endpoint: x = 2 × (−1) − (−6) = −2 + 6 = 4, y = 2 × 5 − 2 = 10 − 2 = 8, giving (4, 8). (−3.5, 3.5) comes from averaging the given endpoint and the midpoint as if they were the two endpoints of a segment, instead of working backwards from the midpoint. (5, 3) comes from working out (−1 − (−6), 5 − 2) instead of doubling the midpoint before subtracting. (4, 5) comes from correctly finding the x-coordinate but copying the midpoint's y-coordinate of 5 instead of doubling it.
- (a) The equation has one solution; the inequality has many. — Method: solve each statement. From 2x + 3 = 11, 2x = 8, so x = 4 — a single value. From 2x + 3 > 11, 2x > 8, so x > 4 — every number greater than 4 makes the inequality true, so there are many solutions. So the equation has one solution and the inequality has many. Distractor origins: swapping the two round gives the range to the equation and the single value to the inequality; saying both have exactly one solution treats the > sign as if it were an = sign; saying both have many solutions treats the equation as if it were an inequality.
- (d) Formula A gives £39 and Formula B gives £39, and since 3(2n + 5) expands to 6n + 15 for every value of n, the stallholder is correct. — Formula A: 3(2 × 4 + 5) = 3 × 13 = £39. Formula B: 6 × 4 + 15 = 24 + 15 = £39. Expanding Formula A algebraically gives 3(2n + 5) = 6n + 15, which is identical to Formula B for every value of n, not just n = 4, so the stallholder is correct — this is an identity, not a coincidence. The option giving £29 for Formula A comes from multiplying only the 2n by 3 and forgetting to also multiply the 5, then adding the unmultiplied 5: 3 × 2 × 4 = 24, + 5 = 29. The two options that reach the correct numbers but reject the stallholder's claim both use faulty reasoning — matching values at one value of n, or counting terms, does not decide whether two expressions are identical for every n; expanding the bracket does.
- (a) 11 — 2n − 7 = 15, so adding 7 to both sides gives 2n = 22, and dividing by 2 gives n = 11. A candidate who makes a sign error and forms the equation 2n + 7 = 15 gets 2n = 8 and n = 4. A candidate who correctly adds 7 to get 2n = 22 but forgets to divide by 2 gets n = 22. A candidate who ignores the −7 altogether and solves 2n = 15 gets n = 7.5.
- (d) −2 — Method: substitute the value into both terms, working out the index and the multiplication before the addition. Working: m² = (−2) × (−2) = 4 and 3m = 3 × (−2) = −6, so the expression becomes 4 + (−6), which is −2. Answer: −2. The distractors: −10 comes from squaring −2 as −4, giving −4 + (−6); 10 comes from working out 3m as +6 and losing the minus sign, giving 4 + 6; −14 comes from working from left to right instead of multiplying first, giving (4 + 3) × (−2).
- (a) An identity, because it is true for every value of x — Expanding the brackets multiplies both terms inside by 4, giving 4x + 12, which is exactly the right-hand side. The two sides are therefore equal whatever x is, and a statement true for every value of the letter is an identity. An equation is true only for particular values, and trying to solve this one leads to 0 = 0, which places no restriction on x at all. Being able to expand brackets is not what makes a statement an identity, since any equation with brackets can be expanded. Nor is it a formula: a formula links two different quantities, and only one letter appears here.
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