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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- (d) 4 — Subtracting the second equation from the first: the y-terms cancel, and the x-terms combine as 3x − x = 2x; the right-hand sides give 14 − 6 = 8. This gives 2x = 8, so x = 4. A candidate who subtracts in the wrong order would get 2x = 6 − 14 = −8, so x = −4. A candidate who uses only the first equation, dividing 14 by 3 as if y were 0, would get x ≈ 4.67. A candidate who reports the value of y instead of x would get y = 6 − 4 = 2.
- (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).
- (c) 8a + 2 — The perimeter of a rectangle is twice the length plus twice the width: P = 2(3a + 2) + 2(a − 1) = (6a + 4) + (2a − 2) = 8a + 2. Answering 5a adds the length and width once each but doubles only one of them, missing that a rectangle has two of each side. Answering 8a + 6 distributes the 2 into (a − 1) correctly as far as 2a, but then adds 2 instead of subtracting it, as though the bracket had been (a + 1). Answering 12a + 8 uses the length for all four sides instead of using the length twice and the width twice, as if the field were a square with side (3a + 2). The perimeter of the field is (8a + 2) metres.
- (c) (x + 7)/4 — Method: dividing an expression by a number is written as a fraction, with the whole expression on top. Working: (x + 7) ÷ 4 = (x + 7)/4. Answer: (x + 7)/4. 4/(x + 7) comes from writing the numbers the wrong way round, putting 4 on top. 4(x + 7) comes from reading ÷ as ×, multiplying instead of dividing. (x + 7) − 4 comes from reading ÷ as −, subtracting instead of dividing.
- (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.
- (d) n/2 + 5 — Half of n is n ÷ 2, which is written as the fraction n/2. 'More than' means add, and the addition happens after the halving, so the expression is n/2 + 5. Writing (n + 5)/2 halves the 5 as well, because everything inside a bracket is divided; writing 2n + 5 doubles n instead of halving it; writing 5n/2 multiplies half of n by 5 instead of adding 5 to it.
- (b) 192 — This is a geometric sequence with first term 3 and common ratio 4: round 2 has 3 × 4 = 12, round 3 has 12 × 4 = 48, round 4 has 48 × 4 = 192. A candidate who applies the ×4 multiplier four times instead of three gets 3 × 4⁴ = 768. A candidate who wrongly treats the growth as arithmetic, taking the round 2 figure of 12 as a fixed amount added each round, gets 3, 15, 27, 39. A candidate who forgets the starting 3 people and just works out 4⁴ gets 256.
- (b) x ≤ 5 — Method: divide out the bracket first, then undo the number term; the inequality sign turns round only if both sides are multiplied or divided by a negative number. Working: dividing both sides of 2(x − 1) ≤ 8 by 2 gives x − 1 ≤ 4, and 2 is positive so the ≤ is unchanged; adding 1 to both sides gives x ≤ 5. Answer: x ≤ 5. The distractors: x ≤ 3 comes from subtracting 1 from 4 instead of adding 1 to both sides; x ≤ 4 comes from stopping at 8 ÷ 2 = 4 and never undoing the −1 inside the bracket; x < 5 comes from reading ≤ as a strict inequality, which wrongly leaves the boundary value out of the solution set.
- (a) 20 — Method: the paper has run out when S = 0, so substitute S = 0 into the equation and solve for t. Working: 0 = 480 − 24t, so 24t = 480, and t = 480 / 24 = 20. Answer: it takes 20 minutes. The value 480 comes from giving the starting number of sheets, the intercept of the equation, instead of solving for t. The value 456 comes from working out 480 − 24 and stopping after one step instead of solving the equation fully. The value 0.05 comes from inverting the division, working out 24 / 480 instead of 480 / 24.
- (a) 7n + 4 — Method: find the common difference, then find the constant that fits the first term. Working: 18 − 11 = 7, 25 − 18 = 7, 32 − 25 = 7, so the terms increase by 7 each time and the nth term has the form 7n + c. Substituting n = 1: 7(1) + c = 11, so c = 4. Answer: the correct nth term is 7n + 4. The value 7n is Ravi's value, which comes from using only the common difference and leaving out the constant. The value 7n + 11 comes from using the first term as the constant directly, without subtracting the common difference first. The value 11n + 7 comes from swapping the roles of the first term and the common difference — using the first term, 11, as the coefficient of n and the difference, 7, as the constant.
- (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.
- (c) x = 3, y = 1 — Method: the y terms are +y and −y, so adding the two equations removes y and leaves an equation in x alone. Working: adding 2x + y = 7 and x − y = 2 gives 3x = 9, so x = 3; substituting x = 3 into x − y = 2 gives 3 − y = 2, so y = 1. Answer: x = 3, y = 1, which also satisfies 2 × 3 + 1 = 7. The distractors: x = 1, y = 3 comes from finding the two values correctly and then writing them against the wrong letters; x = 3, y = 2 comes from substituting x = 3 into 2x + y = 7 as 2 + 3 + y = 7, adding the coefficient instead of multiplying by it; x = 3, y = −1 comes from substituting into x − y = 2 as though it read x + y = 2.
- (c) 7 — To reverse the rule, subtract the constant then divide by the coefficient: 30−2=28, then 28÷4=7, so n=7. A candidate who adds the constant instead of subtracting it, a sign error when rearranging, would compute (30+2)÷4=32÷4=8. A candidate who subtracts the constant correctly but then forgets to divide by the coefficient would stop at 30−2=28. A candidate who treats 4n+2 as a single term 6n, adding the coefficient and constant together, would compute 30÷6=5.
- (d) 6 — Method: rearrange the formula to make d the subject, then substitute C = 133. Working: C = 25 + 18d, so subtracting 25 from both sides gives C − 25 = 18d, then dividing by 18 gives d = (C − 25) / 18. Substituting C = 133: d = (133 − 25) / 18 = 108 / 18 = 6. The value 7.4 comes from dividing 133 by 18 without subtracting the fixed £25 first (133 / 18 ≈ 7.4). The value 4.6 comes from pairing the numbers the wrong way round, working out (133 − 18) / 25 = 4.6. The value 90 comes from subtracting both 25 and 18 from 133 instead of dividing by 18.
- (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.
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