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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- (b) 4, 12, 36, 108 — Method: multiply the previous term by 3 each time, starting from the first term. Working: 4 × 3 = 12, 12 × 3 = 36, 36 × 3 = 108. Answer: 4, 12, 36, 108. 4, 7, 10, 13 comes from adding 3 each time instead of multiplying by 3, confusing this with an arithmetic sequence. 4, 12, 15, 18 comes from multiplying correctly to get the second term, then switching to adding 3 for the rest. 12, 36, 108, 324 comes from listing the terms after the first term, missing off the starting value of 4.
- (d) $y = x^2 + 3$ — Method: a point lies on a curve only if substituting its x-coordinate into the equation gives back its y-coordinate, so read one or two points off curve B and test each equation. Working: curve B crosses the y-axis at (0, 3), and its lowest point is also (0, 3); substituting x = 0 into y = x² + 3 gives 0² + 3 = 3, which matches. Checking a second point: at x = 2 curve B is at y = 7, and 2² + 3 = 4 + 3 = 7, which matches as well. Answer: curve B has equation y = x² + 3. Distractor refutation: y = x² − 3 comes from reading the 3 as a move down instead of a move up; substituting x = 0 gives −3, so that curve would cross the y-axis three squares below the origin, while curve B crosses it three squares above. y = (x − 3)² comes from putting the 3 inside the brackets; substituting x = 0 gives (−3)² = 9, and that curve's lowest point is at (3, 0), three squares to the right along the x-axis, whereas curve B has its lowest point on the y-axis. y = x² + 3x comes from attaching the 3 to the x term instead of writing it on its own; substituting x = 0 gives 0² + 3 × 0 = 0, so that curve passes through the origin, and curve B does not pass through the origin.
- (b) y = 84 − 14x — Method: the distance still to go is the whole route minus the distance already covered, and at a steady speed the distance covered after x hours is the speed multiplied by x. Working: after x hours Freya has cycled 14x km, so the distance remaining is the 84 km route take away 14x, giving y = 84 − 14x, a straight line with intercept 84 and gradient −14. Answer: y = 84 − 14x. The distractors: y = 84 + 14x comes from adding the distance cycled to the length of the route rather than taking it away, so the ride would grow longer the further she goes; y = 14x − 84 comes from carrying out the subtraction the wrong way round, which is negative for the whole of the ride; y = 14 − 84x comes from swapping the two numbers over, treating 84 km as the hourly speed and 14 km as the length of the route.
- (a) 11 — Substitute n=6 into 2n−1: 2×6−1=11. A candidate who adds 2 and 6 and then subtracts 1, instead of multiplying 2 by 6 first, would compute 2+6−1=7. A candidate who substitutes the wrong term number, n=5, would reach 2×5−1=9. A candidate who forgets to subtract 1 would compute just 2×6=12.
- (d) 3, 6, 12, 24 — Method: a term-to-term rule is applied to the term just written, so start with the first term and use the rule three more times. Working: the first term is 3; 3 × 2 = 6 is the second; 6 × 2 = 12 is the third; 12 × 2 = 24 is the fourth. Answer: 3, 6, 12, 24. The distractors: 3, 5, 7, 9 comes from adding 2 each time instead of multiplying by 2; 3, 9, 27, 81 comes from multiplying by the first term, 3, instead of by the multiplier 2 the rule gives; 6, 12, 24, 48 comes from doubling before writing anything down, so the list starts one term too late and the given first term is missing.
- (b) −5 — Method: a point whose x-coordinate is 0 lies on the y-axis, so its y-coordinate is the constant c; once m and c are both known the equation can be written down and x substituted into it. Working: the line passes through (0, 1), so c = 1 and the equation is y = −3x + 1; substituting x = 2 gives y = −3 × 2 + 1 = −6 + 1 = −5. Answer: −5. The distractors: 7 comes from ignoring the minus sign on the gradient and working out 3 × 2 + 1; 5 comes from working out 3 × 2 = 6 and then using the minus sign to take the constant off the product, 6 − 1 = 5; −7 comes from subtracting the constant instead of adding it, −6 − 1 = −7.
- (d) 20 km/h — Method: total distance = 9 + 21 = 30 km. Total time = 15 + 45 + 30 = 90 minutes = 1.5 hours. Average speed = total distance ÷ total time = 30 ÷ 1.5 = 20 km/h. Distractor origins: 40 km/h ignores the 45-minute stop, dividing the total distance by the travel time only (30 km ÷ 45 minutes = 40 km/h); 39 km/h averages the two section speeds, 36 km/h and 42 km/h, instead of using total distance over total time; 36 km/h uses only the first section's speed (9 km in 15 minutes), ignoring the second section and the stop.
- (d) $y = x^2 - 2x - 3$ — Method: read the two x-intercepts (roots) from the graph, write the quadratic as the product of the corresponding factors, then expand. Working: the curve crosses the x-axis at x = −1 and x = 3, so the equation factorises as (x + 1)(x − 3), which expands to x² − 2x − 3. Answer: y = x² − 2x − 3. Distractor refutation: y = x² − x − 6 comes from misreading the left-hand crossing point as x = −2 instead of x = −1, giving factors (x + 2)(x − 3). y = x² − x − 2 comes from misreading the right-hand crossing point as x = 2 instead of x = 3, giving factors (x + 1)(x − 2). y = x² + 2x − 3 comes from writing the factors as (x − 1)(x + 3), swapping which root gets the plus sign and which gets the minus sign, giving the wrong sign on the x term.
- (d) 4n + 2 — Method: find the common difference, then find the constant by adjusting the first term. Working: 10 − 6 = 4, 14 − 10 = 4, so the common difference is 4 and the coefficient of n is 4. The constant is the first term minus the common difference: 6 − 4 = 2. Answer: the nth term is 4n + 2. 4n + 6 comes from using the first term, 6, as the constant without subtracting the common difference. 4n − 2 comes from working out the constant the wrong way round, as the common difference minus the first term (4 − 6 = −2) instead of the first term minus the common difference. 6n + 4 comes from swapping the common difference and the first term.
- (a) 17 — Substitute a = 3 into 5a + 2: 5 × 3 + 2 = 15 + 2 = 17. 10 comes from treating 5a as 5 + a instead of 5 × a, giving 5 + 3 + 2. 25 comes from adding the 2 to a before multiplying by 5, 5 × (3 + 2). 15 comes from working out 5 × 3 correctly but forgetting to add the 2.
- (b) t − 1 — Method: "fewer than" means a subtraction, and the amount taken away is taken away from the quantity being compared with. Working: Grace has t books and Noah has one book fewer, so Noah has t with 1 taken off, written t − 1. Answer: t − 1. The distractors: t + 1 comes from reading "fewer" as "more" and adding instead of subtracting; 1 − t comes from writing the subtraction the wrong way round, taking Grace's books from the 1 rather than the 1 from Grace's books; t comes from giving the number of books Grace has, which is not the person the question asks about.
- (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.
- (c) −8 — The nth term is the first term plus (n − 1) lots of the common difference: 40 + 8 × (−6) = 40 − 48 = −8. A candidate who uses 9 lots of the common difference instead of 8 gets 40 + 9 × (−6) = −14. A candidate who treats the common difference as +6 instead of −6 gets 40 + 8 × 6 = 88. A candidate who uses only 7 lots of the common difference gets 40 + 7 × (−6) = −2.
- (b) y decreases towards zero but never reaches it — Method: as x gets larger, dividing 20 by a bigger number gives a smaller result, so y decreases; because 20/x can never be exactly zero for any positive x, the curve gets closer to zero without ever reaching it, so y decreases towards zero but never reaches it. Distractor origins: 'y increases towards a limit but never reaches it' has the relationship backwards, treating y as increasing when it is actually decreasing; 'y decreases at a steady rate and reaches zero' wrongly assumes the graph behaves like a straight line that eventually hits zero; 'y stays the same however large x becomes' wrongly assumes there is no change in y at all.
- (a) 50 — Method: in ab², only the b is squared, so square b first, then multiply by a. Working: b² = 5² = 25, then a × b² = 2 × 25 = 50. Answer: 50. 100 comes from squaring the product ab instead of just b: (2 × 5)² = 100. 20 comes from squaring a instead of b: a² × b = 4 × 5 = 20. 10 comes from ignoring the square altogether and working out a × b = 2 × 5 = 10.
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