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.
Algebra worksheet — GCSE Foundation
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- 1.Amelia's mother is three times as old as Amelia. Four years ago her mother was five times as old as Amelia was then. Work out Amelia's age now.
- 2.Grace has t books. Noah has 1 book fewer than Grace. Write down an expression for the number of books Noah has.
- 3.The first four terms of a sequence are 9, 14, 19, 24. Work out an expression, in terms of n, for the nth term.
- 4.To solve 5x − 4 = 2x + 8, Chloe's working is shown. Line 1: 5x − 4 = 2x + 8. Line 2: 3x − 4 = 8. Line 3: 3x = 4. Line 4: x = 4/3. Chloe has made a mistake in her working. Which line contains the mistake?
- 5.Priya buys b balloons for a party. Each balloon costs 15p. Write down an expression for the total cost, in pence.
- 6.Solve the inequality 3x + 6 ≤ 0.
- 7.The graph of y = x² − 2x − 8 takes these values: when x = −2, y = 0; when x = −1, y = −5; when x = 0, y = −8; when x = 3, y = −5; when x = 4, y = 0. Use these values to write down the two solutions of x² − 2x − 8 = 0.y = x² − 2x − 8
- 8.A number machine multiplies its input by 2 and then subtracts 5. Work out the output when the input is 6.
- 9.Simplify k + k + k + k + k.
- 10.The first four terms of a sequence are 2, 6, 10, 14. Priya says the nth term is 4n. Work out the correct expression for the nth term.
- 11.a = 2 and b = 5. Work out the value of a²b.
- 12.Work out the value of a² − 2b when a = −3 and b = 4.
- 13.Factorise x² − 64.
- 14.A sequence has the position-to-term rule n² + 2, where n is the position number. Work out the 6th term.
- 15.A rectangular photo has length (x + 3) cm and width x cm. Its area is 40 cm². Work out the value of x.
Answer key
- (d) 8 — Method: write both present ages in terms of one letter, take four years off each of them, and form an equation from the comparison four years ago. Working: let Amelia be y, so her mother is 3y; four years ago they were y − 4 and 3y − 4, giving 3y − 4 = 5(y − 4); expanding gives 3y − 4 = 5y − 20; adding 20 and subtracting 3y from both sides gives 16 = 2y, and dividing by 2 gives y = 8. Checking: Amelia is 8 and her mother 24; four years ago they were 4 and 20, and 20 = 5 × 4. Answer: 8. The distractors: 4 comes from solving correctly but giving Amelia's age four years ago instead of her age now; 16 comes from stopping at 2y = 16 and giving 16 as her age; 2 comes from collecting the y terms by adding the 5y instead of subtracting it, giving 3y + 5y = 20 − 4 and so 8y = 16.
- (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) 5n + 4 — Method: find the common difference, then find the constant that fits the first term. Working: 14 − 9 = 5, 19 − 14 = 5, 24 − 19 = 5, so the terms increase by 5 each time and the nth term has the form 5n + c. Substituting n = 1: 5(1) + c = 9, so c = 4. Answer: the nth term is 5n + 4. The value 5n comes from leaving out the constant. The value 5n + 9 comes from using the first term as the constant directly, without subtracting the common difference first. The value 9n + 5 comes from swapping the roles of the first term and the common difference — using the first term, 9, as the coefficient of n and the difference, 5, as the constant.
- (d) Line 3 — Method: check each line of Chloe's working against the correct algebraic step. Working: Line 1 to Line 2 is correct, subtracting 2x from both sides gives 3x − 4 = 8. But Line 2 to Line 3 should add 4 to both sides, giving 3x = 12, not 3x = 4 — the constant −4 has been dropped rather than removed correctly. Line 4 follows correctly from Chloe's own, incorrect, Line 3. Answer: Line 3. Line 1 is simply the original equation, copied out correctly. Line 2 correctly subtracts 2x from both sides of Line 1. Line 4 divides Chloe's own Line 3 by 3 correctly — the arithmetic there is fine, the mistake happened one line earlier.
- (d) 15b — Method: each balloon costs 15p, so b balloons cost 15 × b pence. Working: 15 × b = 15b. Answer: 15b. b + 15 comes from treating the 15p as a one-off addition rather than a rate applied to every balloon. b − 15 comes from misreading the situation as a discount rather than a cost per balloon. b/15 comes from dividing instead of multiplying, swapping the two operations.
- (b) x ≤ −2 — Method: take the number term off both sides, then divide by the coefficient of x; the sign turns round only when you divide BY a negative number, and here you divide by 3. Working: subtracting 6 from both sides of 3x + 6 ≤ 0 gives 3x ≤ −6; dividing both sides by 3, which is positive, gives x ≤ −2. Answer: x ≤ −2. The distractors: x ≥ −2 comes from turning the sign round because the right-hand side has become negative, which is not the rule; it is the sign of the divisor that matters; x ≤ 2 comes from moving the 6 across without changing its sign, giving 3x ≤ 6; x ≤ −18 comes from multiplying both sides by 3 instead of dividing by it.
- (c) x = −2 or x = 4 — The solutions of x² − 2x − 8 = 0 are the x-values where y = 0. From the table, y = 0 when x = −2 and when x = 4, so those are the two solutions. Distractor origins: x = −1 or x = 3 picks the pair of x-values that give equal (but non-zero) y-values instead of y = 0; x = 0 or x = −8 mixes up an x-value with its corresponding y-value; x = −2 only reads off one of the two roots and misses the other.
- (a) 7 — Multiply the input by 2: 6 × 2 = 12. Then subtract 5: 12 − 5 = 7. A candidate who does the operations in the wrong order, subtracting 5 first and then multiplying by 2, gets (6 − 5) × 2 = 2. A candidate who only carries out the multiplication and forgets to subtract gets 12. A candidate who adds 5 instead of subtracting gets 6 × 2 + 5 = 17.
- (a) 5k — Method: adding a letter to itself repeatedly means counting how many of that letter you have, written as a coefficient in front of the letter. Working: k + k + k + k + k is five lots of k, written 5k. Answer: 5k. k⁵ comes from treating repeated addition as repeated multiplication (raising to a power) instead of counting copies. 5 + k comes from adding the count of terms (5) to a single k instead of multiplying. k/5 comes from dividing instead of counting how many k's there are.
- (c) 4n − 2 — Method: find the common difference, then find the constant that fits the first term. Working: 6 − 2 = 4, 10 − 6 = 4, 14 − 10 = 4, so the terms increase by 4 each time and the nth term has the form 4n + c. Substituting n = 1: 4(1) + c = 2, so c = −2. Answer: the correct nth term is 4n − 2. The value 4n is Priya's value, which comes from using only the common difference and leaving out the constant. The value 4n + 2 comes from a sign error when finding the constant. The value 2n + 2 comes from using the first term, 2, as the coefficient of n instead of the common difference, and then attaching +2 rather than working the constant out.
- (b) 20 — a²b means a² multiplied by b: a² = 2² = 4, and 4 × 5 = 20. Reading the expression as (ab)² instead of a²b gives (2 × 5)² = 100. Squaring b instead of a, 2 × 5² = 50, squares the wrong letter. Adding a² and b instead of multiplying them, 2² + 5 = 9, uses the wrong operation.
- (a) 1 — a² − 2b = (−3)² − 2(4) = 9 − 8 = 1. A candidate who squares −3 but keeps the negative sign gets −9 − 8 = −17. A candidate who forgets to square a and substitutes it as −3 gets −3 − 8 = −11. A candidate who adds instead of subtracting 2b gets 9 + 8 = 17.
- (d) (x − 8)(x + 8) — x² − 64 = x² − 8², a difference of two squares, which factorises as (x − 8)(x + 8). A candidate who treats it as a perfect square with a repeated negative factor gets (x − 8)(x − 8), which expands to x² − 16x + 64 — wrong on both the middle and constant terms. A candidate who uses a repeated positive factor gets (x + 8)(x + 8), which expands to x² + 16x + 64. A candidate who picks a different factor pair of 64, such as 4 and 16, without checking that the middle term cancels, gets (x − 4)(x + 16), which expands to x² + 12x − 64 — the wrong middle term.
- (a) 38 — Method: substitute the position number into the rule and follow the order of operations, so the squaring is carried out before the 2 is added. Working: n = 6 gives 6² + 2; 6² means 6 × 6 = 36, and then 2 is added to 36. Answer: 38. The distractors: 14 comes from multiplying the position by 2 instead of squaring it, 6 × 2 + 2; 36 comes from squaring correctly and then forgetting to add the 2; 64 comes from adding the 2 first and squaring afterwards, (6 + 2)².
- (b) x = 5 — Area = length × width, so x(x + 3) = 40, which rearranges to x² + 3x − 40 = 0. This factorises as (x + 8)(x − 5) = 0: the two numbers in the brackets must multiply to −40 and add to +3, and the pair 8 and −5 does both. This gives x = −8 or x = 5. Since x is a length, it cannot be negative, so x = 5. A candidate who gives both solutions without rejecting the negative one, which cannot be a length, answers x = 5 or x = −8. A candidate who picks the wrong factor pair of 40, such as 10 and −4 instead of 8 and −5, gets (x + 10)(x − 4) = 0 and answers x = 4. A candidate who rejects the wrong root, keeping the negative solution instead of the positive one, answers x = −8.
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