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) −5 — Method: apply the machine's operation to the input, and carry the sign of the input through the multiplication. Working: the machine gives 5 × (−1); a positive number multiplied by a negative number gives a negative result, and 5 × 1 = 5, so the output is 5 below zero. Answer: −5, which is one lot of −1 taken five times. The distractors: 5 comes from working out 5 × 1 and losing the minus sign of the input; 4 comes from adding 5 to −1 instead of multiplying; −6 comes from subtracting 5 from −1 instead of multiplying.
- (d) 8(7x − 3) — Method: find the highest common factor of the two terms, write it in front of a bracket and divide each term by it. Working: 56 = 8 × 7 and 24 = 8 × 3, so the highest common factor is 8; dividing gives 56x ÷ 8 = 7x and 24 ÷ 8 = 3, and the subtraction sign stays between them. Answer: 8(7x − 3), which multiplies back out to 56x − 24. The distractors: 8(7x + 3) comes from dropping the minus sign of −24 while dividing; 8(56x − 3) comes from dividing only the number term by 8 and leaving 56x untouched inside the bracket; 8(7x − 16) comes from subtracting 8 from 24 instead of dividing 24 by 8.
- (d) Yes, since expanding 6(x − 1) gives 6x − 6. — Expand the bracket: 6(x − 1) = 6 × x − 6 × 1 = 6x − 6, which matches the second expression exactly, so the student is correct for every value of x. Saying 6(x − 1) means 6x − 1 comes from multiplying only the x and forgetting to multiply the 1 by 6. Saying brackets always change an expression's value is not true — expanding here gives back an equivalent expression, not a different one. The identity holds for every value of x, not just whole numbers, since both sides are expanded algebraically, not tested by substitution.
- (a) 1 — Method: the value of y when x = 0 is where the line meets the y-axis, which is the constant c in y = mx + c, so the gradient is worked out from the two given points first and the constant follows by substituting one of them. Working: m = (16 − 7) ÷ (5 − 2) = 9 ÷ 3 = 3, so the line is y = 3x + c; substituting x = 2 and y = 7 gives 7 = 3 × 2 + c, so c = 7 − 6 = 1, and the value of y when x = 0 is that constant. Answer: 1. The distractors: 3 comes from stopping at the gradient and offering it as the intercept; 4 comes from stepping back from x = 2 to x = 0 by one unit of x instead of two, 7 − 3 = 4; −1 comes from working the constant out as mx − y, 3 × 2 − 7 = −1, instead of y − mx.
- (c) 4 — Method: subtract 3 from both sides, then divide by 4. Working: 4x = 19 − 3 = 16, so x = 16 ÷ 4 = 4. Answer: 4. 4.75 comes from dividing 19 by 4 directly, without subtracting 3 first. 3.75 comes from subtracting 4 instead of 3 before dividing. 5.5 comes from a sign error, adding 3 to 19 instead of subtracting it.
- (d) (−1, 4) — In parallelogram ABCD the side DC is parallel and equal to the side AB, so D = C − AB. The vector from A to B is (2 − (−3), 1 − 1) = (5, 0), so D = (4 − 5, 4 − 0) = (−1, 4). A candidate who adds this vector to C instead of subtracting it gets (4 + 5, 4 + 0) = (9, 4). A candidate who subtracts A's coordinates from C's rather than the vector AB, and drops the minus sign on −3 while doing so, works out (4 − 3, 4 − 1) and gets (1, 3). A candidate who makes only the y-part of that slip, working out 4 − 1 instead of 4 − 0, gets (−1, 3).
- (c) P — Sequence P has a common ratio of 2 (4 × 2 = 8, 8 × 2 = 16, 16 × 2 = 32), so it is geometric. Sequence Q has a common difference of 4, which is arithmetic, not geometric — a candidate who confuses a constant difference with a constant ratio picks Q. Sequence R has increasing differences of 3, 5, 7, a quadratic sequence, not geometric — a candidate who assumes any fast-growing sequence must be geometric picks R. Sequence S is the square numbers from 2² onwards (2², 3², 4², 5²), which is also quadratic, not geometric — a candidate who thinks squaring always means geometric growth picks S.
- (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.
- (a) 23 — The number of logs increases by 4 for each layer down, starting from 3 in the top layer, so the nth layer has 3+(n−1)×4 logs. For the 6th layer: 3+5×4=3+20=23. A candidate who adds the difference of 4 six times instead of five, treating the top layer as needing an addition too, would compute 3+6×4=27. A candidate who uses 4n instead of 4n−1, omitting the adjustment for the first layer, would compute 4×6=24. A candidate who subtracts the common difference, 4, instead of 1 when adjusting the multiplier would compute 4×6−4=20.
- (a) x = 3 — Method: factorise x² + 2x − 15 as (x + 5)(x − 3), since 5 × (−3) = −15 and 5 + (−3) = 2. Setting each bracket equal to zero gives x + 5 = 0 or x − 3 = 0, so x = −5 or x = 3. Only x = 3 is offered here. Distractor origins: x = −3 reverses the sign of the factor pair, treating the bracket (x − 3) as giving x = −3 instead of x = 3; x = 5 takes the number from the other factor, (x + 5), but with the wrong sign, giving x = 5 instead of x = −5; x = 15 takes the constant term of the original expression as if it were a root, without factorising at all.
- (a) £14.50 — Method: the first 200 minutes are already covered by the £12, so only 250 − 200 = 50 minutes are charged extra. Extra charge = 50 × £0.05 = £2.50. Total cost = £12 + £2.50 = £14.50. Distractor origins: £24.50 charges 5p for all 250 minutes instead of only the 50 minutes over 200 (250 × £0.05 = £12.50, plus £12 = £24.50); £12.50 makes that same slip of charging all 250 minutes but then forgets to add the £12 monthly fee; £13.50 works out 250 − 200 wrongly as 30 extra minutes instead of 50 (30 × £0.05 = £1.50, plus £12 = £13.50).
- (d) 3n + 2 — The common difference is 3 (8−5=3), so the expression starts 3n. To match the first term when n=1, 3×1+c=5, so c=2: the nth term is 3n+2. A candidate who uses the first term itself as the constant, instead of first term minus the common difference, would write 3n+5 (giving 8, 11, 14, 17 — one term too high throughout). A candidate who omits the constant term altogether would write just 3n (giving 3, 6, 9, 12, not matching the sequence at all). A candidate who adds the common difference to n instead of multiplying would write n+3 (giving 4, 5, 6, 7, far too small).
- (a) 2(3x + 1) = 6x + 2 — Expanding 2(3x + 1) = 6x + 2 gives an expression that matches the right-hand side exactly for every value of x — it is an identity. 4x − 3 = 3x + 5 is an ordinary equation with one solution, x = 8. 7 − x = x − 7 is also an ordinary equation with one solution, x = 7. 5x + 1 = 5(x + 1) never holds for any value of x at all, since expanding the right-hand side gives 5x + 5, and 5x + 1 = 5x + 5 would require 1 = 5, which is impossible.
- (d) 7 — Method: set up the equation 2.50 + 1.80m = 15.10, then subtract the booking fee and divide by the cost per mile. Working: 1.80m = 15.10 − 2.50 = 12.60; m = 12.60 ÷ 1.80 = 7. Answer: 7 miles. 8.39 comes from dividing the whole £15.10 by £1.80 without first subtracting the booking fee: 15.10 ÷ 1.80 ≈ 8.39. 5.32 comes from swapping the two amounts round, subtracting £1.80 and dividing by £2.50: (15.10 − 1.80) ÷ 2.50 ≈ 5.32. 9.78 comes from adding the booking fee instead of subtracting it: (15.10 + 2.50) ÷ 1.80 ≈ 9.78.
- (b) 27 — 27 = 3³ (3 × 3 × 3), so it is a cube number. The other three, 9 = 3², 16 = 4² and 25 = 5², are all square numbers, not cube numbers — each is reached by multiplying a whole number by itself only twice, not three times.
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