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) 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.
- (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.
- (d) 47 — To reach the 8th term from the 1st term, the difference of 6 is added 7 times (8 − 1 = 7): 5 + 7 × 6 = 47. A candidate who multiplies by the term number itself, rather than one less, would compute 5 + 8 × 6 = 53. A candidate who uses one step too few (6 instead of 7) would reach 5 + 6 × 6 = 41. A candidate who forgets to include the first term at all would compute just 8 × 6 = 48.
- (d) x = 0 or x = 6 — Method: factorise by taking out the common factor x: x(x − 6) = 0, so x = 0 or x − 6 = 0, giving x = 0 or x = 6. Distractor origins: x = 6 comes from dividing both sides by x, which loses the solution x = 0; x = 0 stops after finding only one factor; x = 3 comes from halving the coefficient of x instead of factorising.
- (d) 12 — Method: write the nth term of the sequence, 18 + 4(n − 1), set it equal to 62, and solve for n. Working: 18 + 4(n − 1) = 62, so 4(n − 1) = 44, giving n − 1 = 11, so n = 12. Answer: row 12. 11 comes from using 18 + 4n = 62 instead of 18 + 4(n − 1) = 62, an off-by-one error, giving n = 11. 48 comes from correctly simplifying to 4n = 48 but stopping there, without dividing by 4 to find n. 15.5 comes from dividing 62 by 4 directly, ignoring the 18 seats already in the front row.
- (b) h = 2A/b — The height has been multiplied by the base and the result then divided by 2, so undo the division first. Multiplying both sides by 2 gives 2A = bh. Undoing the multiplication by the base comes next: dividing both sides by b gives 2A/b = h, so h = 2A/b. Dividing by 2 instead of multiplying gives A/(2b), a quarter of the correct height; multiplying by the base instead of dividing gives 2Ab; writing b/(2A) turns the final fraction upside down.
- (d) x > 4 — Method: collect the x terms on one side and the numbers on the other, then divide by the coefficient of x; dividing by a positive number leaves the sign as it is. Working: subtracting 2x from both sides of 4x + 1 > 2x + 9 gives 2x + 1 > 9; subtracting 1 from both sides gives 2x > 8; dividing both sides by 2 gives x > 4. Answer: x > 4. The distractors: x < 4 comes from turning the sign round while dividing by 2; x > 5 comes from adding the 1 to the 9 instead of subtracting it, giving 2x > 10; x < 5 comes from making both of those mistakes together.
- (a) (40, 0) — Method: a graph crosses the x-axis where y = 0, which in this context is the moment the tank holds no water, so substituting y = 0 and solving gives the time. Working: 0 = −5x + 200 gives 5x = 200, so x = 200 ÷ 5 = 40 and the crossing point is (40, 0); the tank is empty after 40 minutes. Answer: (40, 0). The distractors: (0, 200) is the y-axis crossing, the 200 litres in the tank at the start, found by substituting x = 0 instead of y = 0; (200, 0) comes from reading the constant 200 as the x-coordinate without dividing by the 5 litres lost each minute; (0, 40) has the right number in the wrong place, with the time written as a y-coordinate.
- (a) x = −1.8 and x = 2.8 — Method: the solutions of x² − x − 2 = 3 are the x-coordinates of the points where the curve y = x² − x − 2 meets the line y = 3, read off the grid to 1 decimal place. Working: the curve meets the line y = 3 at approximately x = −1.8 and at x = 2.8. Answer: x ≈ −1.8 and x ≈ 2.8. Distractor refutation: x = −1.0 and x = 2.0 comes from reading where the curve crosses the x-axis (y = 0), solving x² − x − 2 = 0, instead of where it meets the line y = 3. x = −3.0 and x = 4.0 comes from taking the two ends of the drawn curve as the intersection points, instead of finding where it actually crosses the line y = 3. x = 1.8 and x = 2.8 comes from a sign slip on the left-hand intersection, reading it as positive instead of negative.
- (a) An identity, true for every value of x — Expanding the bracket on the left gives 3x + 12, which matches the right-hand side exactly, so the statement is true for every value of x — this makes it an identity. A candidate who reasons that any statement with an equals sign must be an equation picks that option, missing that an equation is only true for particular value(s) of x, not all of them. A candidate who confuses an identity with a formula, because both relate two expressions, picks the formula option — but a formula connects two different quantities, such as area and side length, not two equivalent forms of the same expression. A candidate who assumes it can be solved for a single value of x, as with a normal equation, picks that option, not realising there is no single solution here.
- (b) £105 — The hourly charge is 25 × 3 = £75. Adding the call-out fee: £75 + £30 = £105. A candidate who forgets the call-out fee gives just the hourly charge, £75. A candidate who adds the call-out fee to the hourly rate before multiplying by the hours, (30 + 25) × 3, gets £165. A candidate who multiplies the call-out fee by the number of hours instead of the hourly rate, 30 × 3, gets £90.
- (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.
- (b) 1 — Method: replace the letter with its value, work out the multiplication first and the subtraction afterwards. Working: 3n = 3 × 1 = 3, so the expression becomes 3 − 2, which is 1. Answer: 1. The distractors: 3 comes from substituting into 3n but stopping before the 2 is taken away; −1 comes from subtracting the wrong way round and working out 2 − 3; 5 comes from adding the 2 instead of subtracting it, giving 3 + 2.
- (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.
- (c) y = 2 — Method: substituting Noah's value shows why it fails, and the equation is then solved by undoing the addition. Working: substituting y = 3 gives 3 + 10 = 13, which is not 12, so Noah's value is not a solution; subtracting 10 from both sides of y + 10 = 12 gives y = 2, and substituting back gives 2 + 10 = 12. Answer: y = 2. The distractors: y = 22 comes from adding 10 to both sides instead of subtracting it; y = −2 comes from carrying out the subtraction the wrong way round, 10 − 12 rather than 12 − 10; y = 12 comes from copying the right-hand side as the value of y and ignoring the 10 that is added to it.
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