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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- (a) 16 — k = 8m ÷ 5 = (8 × 10) ÷ 5 = 80 ÷ 5 = 16. A candidate who multiplies 8 by 10 but forgets to divide by 5 gets 80. A candidate who divides 10 by 8 and multiplies by 5, using the fraction upside down, gets 10 × 5 ÷ 8 = 6.25. A candidate who adds instead of substituting into the formula gets 8 + 10 = 18.
- (d) (4, 3) — The sides are parallel to the axes: the missing vertex must share the y-coordinate 3 with (−2, 3) and the x-coordinate 4 with (4, −3), giving (4, 3). (−4, 3) comes from a sign error on the x-coordinate. (4, −9) comes from continuing the pattern of the given points by subtracting 6 from the y-coordinate again instead of matching it to (−2, 3). (3, 4) comes from swapping the x- and y-coordinates.
- (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).
- (c) Second and fourth — If x + y = 0 then y = −x, so x and y always have opposite signs, one positive and one negative. A point with a negative x and a positive y lies in the second quadrant, and a point with a positive x and a negative y lies in the fourth quadrant, so the point lies in the second or the fourth. A candidate who reads x + y = 0 as meaning x and y have the same sign picks First and third, which is where x × y is positive, not where x + y = 0. A candidate who decides that y must be the positive coordinate picks the two quadrants above the x-axis, First and second. A candidate who decides that y must be the negative coordinate picks the two quadrants below the x-axis, Third and fourth.
- (a) 33 — Substitute n=4 into 2n²+1: 2×4²+1=2×16+1=33. A candidate who computes n² as 2×n instead of n×n would compute 2×(2×4)+1=2×8+1=17. A candidate who correctly finds 2×16 but forgets to add the constant 1 would stop at 32. A candidate who squares the whole term 2n, rather than squaring n before multiplying by 2, would compute (2×4)²+1=64+1=65.
- (c) 5 — Method: call the number of years ago t, take t off both ages, and form an equation from the comparison at that time. Working: t years ago the father was 40 − t and Ethan was 10 − t, so 40 − t = 7(10 − t); expanding gives 40 − t = 70 − 7t, adding 7t to both sides gives 40 + 6t = 70, subtracting 40 gives 6t = 30, and dividing by 6 gives t = 5. Checking: five years ago the father was 35 and Ethan was 5, and 35 = 7 × 5. Answer: 5. The distractors: 35 comes from finding the right moment but giving the father's age at that time instead of the number of years; 30 comes from using Ethan's present age on the right-hand side, 40 − t = 7 × 10, which gives t = −30 and is then written as 30; 24 comes from reaching 6t = 30 correctly and subtracting 6 instead of dividing by 6.
- (b) It crosses the x-axis, since the minimum is below it. — A minimum turning point at (3, −4) means the lowest value the curve reaches is y = −4, which is below the x-axis (y = 0); since the curve opens upward from there, it must rise up through y = 0 on both sides, crossing the x-axis twice. Saying it does not cross confuses 'the minimum is negative' with 'the whole curve stays negative' — a minimum below the axis guarantees the curve rises above it elsewhere. Saying it touches the x-axis once at (3, −4) mistakes the turning point itself for a root — the turning point is not on the x-axis at all here, since its y-coordinate is −4, not 0. Saying it is impossible to tell ignores that the two facts given — that the turning point is a minimum, and that its y-coordinate is negative — are together enough to decide the number of crossings without knowing the equation.
- (b) 4x − 4 = 20 — Method: subtract 2x from both sides of the equation, and simplify each side separately. Working: left side: 6x − 4 − 2x = 4x − 4. Right side: 2x + 20 − 2x = 20. Answer: 4x − 4 = 20. 4x = 20 drops the −4 from the left side, as though subtracting 2x also removes the constant term. 8x − 4 = 20 comes from moving the 2x across to the left without changing its sign: it is taken off the right side correctly, leaving 20, but added to the left side instead of subtracted, giving 6x + 2x = 8x. 4x − 4 = 2x + 20 comes from subtracting 2x from the left-hand side only and leaving the right-hand side unchanged; whatever is done to one side must be done to the other.
- (a) 5 — A factor of the whole expression must divide every term exactly. 15x + 20 = 5(3x + 4), so 5 is a factor. Distractor origins: 3 divides 15x exactly but does not divide 20 exactly; 4 divides 20 exactly but does not divide 15x exactly; 10 also divides 20 exactly but does not divide 15x exactly, so it is a factor of only one term, not of the whole expression.
- (b) 1215 — Method: generate the terms one at a time with the term-to-term rule and compare each with 1000 as you go, stopping at the first one that passes it. Working: the terms are 5, then 5 × 3 = 15, then 45, then 135, then 405, and 405 × 3 = 1215; 405 is still below 1000 while 1215 is above it. Answer: 1215. The distractors: 405 comes from stopping at the last term that is still below 1000 instead of giving the first one above it; 3645 comes from carrying on one term too far, past the first term that passes 1000; 2187 comes from using the multiplier 3 as the first term as well, generating 3, 9, 27, 81, 243, 729, 2187 instead of the sequence described.
- (c) 6 km/h — Total distance = 8 + 4 = 12 km. Total time = 1 hour + 0.5 hours resting + 0.5 hours = 2 hours. Average speed = total distance ÷ total time = 12 ÷ 2 = 6 km/h. A speed of 8 km/h comes from leaving the resting time out of the total time: 12 ÷ 1.5 = 8. A speed of 4 km/h comes from dividing by too much time, such as double-counting the rest period: 12 ÷ 3 = 4. A speed of 12 km/h simply gives the total distance and forgets to divide by the total time at all.
- (b) 12 — Method: substitute the total cost into the formula, then subtract the fixed cost and divide by the cost per person. Working: 92 = 6p + 20, so 6p = 92 − 20 = 72, p = 72 ÷ 6 = 12. Answer: 12 people. 15.33 comes from dividing the whole £92 by £6 without first subtracting the £20 fixed cost: 92 ÷ 6 ≈ 15.33. 4.3 comes from swapping the two amounts round, subtracting £6 and dividing by £20: (92 − 6) ÷ 20 = 4.3. 18.67 comes from adding the fixed cost instead of subtracting it: (92 + 20) ÷ 6 ≈ 18.67.
- (c) x = (y + 24)/4 — Method: expand the bracket first, then undo the operations done to x in reverse order. Working: y = 4(x − 6) = 4x − 24, so y + 24 = 4x, so x = (y + 24)/4. Answer: x = (y + 24)/4. x = (y + 6)/4 comes from expanding the bracket incorrectly, treating 4(x − 6) as 4x − 6 instead of 4x − 24. x = 4y + 96 comes from multiplying by 4 instead of dividing by 4 to undo the multiplication, giving 4(y + 24) = 4y + 96. x = y/4 − 6 comes from dividing by 4 first without expanding the bracket, then subtracting 6 as if the bracket's operation still applied afterwards.
- (d) 36 — (2n)² means the whole of 2n is squared, so with n = 3: (2n)² = (2 × 3)² = 6² = 36. Answering 18 instead works out 2n² — squaring only the n and then multiplying by 2 — which is a different expression because the brackets around 2n are missing. Answering 12 squares only the coefficient, treating (2n)² as 2² × n = 4 × 3 = 12, and forgets to square the n as well. Answering 9 ignores the coefficient of 2 altogether and works out n² on its own. The value of (2n)² when n = 3 is 36.
- (a) y = −2x + 7 — Using y = −2x + c, and substituting the point (3, 1): 1 = −2(3) + c, so 1 = −6 + c, and c = 7, giving y = −2x + 7. A candidate who uses the y-coordinate of the point directly as the y-intercept, instead of solving for c, would write y = −2x + 1. A candidate who drops the negative sign on the gradient would write y = 2x + 7. A candidate who makes a sign error when isolating c, writing c = 1 − 6 = −5 instead of c = 1 + 6 = 7, would write y = −2x − 5.
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