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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Algebra worksheet — GCSE Foundation

Maths · GCSE Foundation · Algebra · 15 questions

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1.Which of these numbers is a cube number?
(a)9
(b)27
(c)16
(d)25
2.The nth term of a sequence is 3n + 2. Work out the 4th term of the sequence.
(a)12
(b)14
(c)9
(d)11
3.Tom buys 2 notebooks and 6 pencils on Monday, priced at p pence each for a notebook and q pence each for a pencil. On Tuesday he buys 3 more notebooks and 1 more pencil. Write a simplified expression, in pence, for the total amount Tom has spent on notebooks and pencils.
(a)5p + 7q
(b)5p + 6q
(c)2p + 7q
(d)12pq
4.Write down the expression that means the same as m × m × m × n.
(a)3mn
(b)mn³
(c)m³ + n
(d)m³n
5.A geometric sequence begins 80, 40, 20, 10, ... Work out the next term in the sequence.
(a)2.5
(b)0
(c)20
(d)5
6.A rectangular field has length (3a + 2) metres and width (a − 1) metres. Write a simplified expression for the perimeter of the field.
(a)12a + 8
(b)8a + 6
(c)8a + 2
(d)5a
7.Solve 6x − 5 = 3x + 13.
(a)18
(b)2
(c)2.67
(d)6
8.Noah says that y = 3 is the solution of the equation y + 10 = 12. Noah is wrong. Work out the correct value of y.
(a)y = 22
(b)y = −2
(c)y = 2
(d)y = 12
9.Solve 7x − 5 = 3x + 11
(a)−4
(b)1.5
(c)0.6
(d)4
10.The point F has x-coordinate −1 and lies 6 units above the x-axis. Write down the coordinates of F.
(a)(1, 6)
(b)(−1, 6)
(c)(−1, −6)
(d)(6, −1)
11.A cycle route is 84 km long. Freya sets off along it at a steady 14 km/h. Write down the function for the distance y, in kilometres, that is still to be cycled after x hours.
(a)y = 84 + 14x
(b)y = 84 − 14x
(c)y = 14 − 84x
(d)y = 14x − 84
12.A cinema sells adult tickets for £10 and child tickets for £6. A family group buys 9 tickets in total, spending £74. Work out how many adult tickets were bought.
(a)7.4
(b)1.25
(c)5
(d)4
13.A tree is 2 metres tall. Each year its height increases by 10% of its height at the start of that year. Work out the height of the tree after 3 years, giving your answer to 1 decimal place.
(a)2.6
(b)2.9
(c)2.4
(d)2.7
14.The graph of y = 24/x is drawn for x > 0. Work out the value of y when x = 6.
(a)y = 18
(b)y = 30
(c)y = 144
(d)y = 4
15.A number, w, satisfies the statement "three times w, minus 5, is at least 16". Which of these is the correct inequality for w?
(a)3(w − 5) ≥ 16
(b)3w − 5 ≥ 16
(c)3w − 5 > 16
(d)3w − 5 ≤ 16

Answer key

(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.
(b) 14 — Substitute n=4 into 3n+2: 3×4+2=14. A candidate who adds 3 and n instead of multiplying would compute 3+4+2=9. A candidate who substitutes the wrong term number, n=3, would reach 3×3+2=11. A candidate who forgets to add the constant term would compute just 3×4=12.
(a) 5p + 7q — Tom's total spend combines like terms: (2p + 6q) + (3p + q) = 5p + 7q, adding the notebook terms (2p + 3p = 5p) and the pencil terms (6q + q = 7q) separately. Answering 12pq adds every coefficient together (2 + 6 + 3 + 1 = 12) and multiplies the letters, combining unlike terms as though notebooks and pencils were the same item. Answering 5p + 6q correctly combines the notebook terms but forgets to add Tuesday's extra pencil to the 6q. Answering 2p + 7q correctly combines the pencil terms but forgets to add Tuesday's 3 extra notebooks to the 2p. The total amount Tom has spent is 5p + 7q pence.
(d) m³n — m × m × m × n means three m's multiplied together, then multiplied by n, which is written as m³n. Putting the power of 3 on n instead of m, mn³, puts the power on the wrong letter. Writing 3mn treats the repeated multiplication as if it were repeated addition, as if it meant 3 × m × n. Writing m³ + n adds the n instead of multiplying it in.
(d) 5 — Each term is found by multiplying the previous term by the common ratio, 0.5: 80, 40, 20, 10, and the next term is 10 × 0.5 = 5. A candidate who instead subtracts the same amount each time (repeating the last difference of 10) would reach 10 − 10 = 0. A candidate who divides by 4 instead of by 2 would reach 10 ÷ 4 = 2.5. A candidate who multiplies by 2 instead of dividing (reversing the direction of the sequence) would reach 10 × 2 = 20.
(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.
(d) 6 — Method: collect the x-terms on one side and the constants on the other, then divide by the remaining coefficient of x. Working: 6x − 3x = 13 + 5, so 3x = 18, x = 18 ÷ 3 = 6. Answer: x = 6. 2.67 comes from a sign error when moving the 5, subtracting instead of adding: 3x = 13 − 5 = 8, x = 8 ÷ 3 ≈ 2.67. 2 comes from a sign error when moving the x-term, adding instead of subtracting: 9x = 18, x = 2. 18 comes from correctly finding 3x = 18 but forgetting to divide by 3.
(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.
(d) 4 — Method: collect the x terms on one side and the number terms on the other. Working: subtract 3x from both sides: 4x − 5 = 11. Add 5 to both sides: 4x = 16. Divide by 4: x = 4. Answer: 4. 0.6 comes from adding the x terms instead of subtracting when collecting them, 7x + 3x = 10x, and also subtracting the constants the wrong way round, 11 − 5 = 6, giving 10x = 6. 1.5 comes from correctly collecting the x terms as 4x but subtracting the constants the wrong way round, 11 − 5 instead of 11 + 5. −4 comes from moving the x terms to the wrong side, giving 3x − 7x instead of 7x − 3x, along with a matching sign error on the constants.
(b) (−1, 6) — '6 units above the x-axis' gives a y-coordinate of 6, and the x-coordinate is given as −1, so F = (−1, 6). (6, −1) comes from swapping the x- and y-coordinates. (−1, −6) comes from treating 'above' as a negative direction instead of positive. (1, 6) comes from dropping the negative sign on the given x-coordinate.
(b) y = 84 − 14x — Method: the distance still to go is the whole route minus the distance already covered, and at a steady speed the distance covered after x hours is the speed multiplied by x. Working: after x hours Freya has cycled 14x km, so the distance remaining is the 84 km route take away 14x, giving y = 84 − 14x, a straight line with intercept 84 and gradient −14. Answer: y = 84 − 14x. The distractors: y = 84 + 14x comes from adding the distance cycled to the length of the route rather than taking it away, so the ride would grow longer the further she goes; y = 14x − 84 comes from carrying out the subtraction the wrong way round, which is negative for the whole of the ride; y = 14 − 84x comes from swapping the two numbers over, treating 84 km as the hourly speed and 14 km as the length of the route.
(c) 5 — Method: let a be the number of adult tickets, so the number of child tickets is 9 − a. Form the equation 10a + 6(9 − a) = 74. Working: expand the bracket: 10a + 54 − 6a = 74, so 4a = 20, giving a = 5. Answer: 5 adult tickets. 4 comes from correctly finding the number of child tickets but reporting it instead of the number of adult tickets asked for. 1.25 comes from a sign error expanding the bracket, writing 10a + 54 + 6a = 74 instead of subtracting, which gives 16a = 20. 7.4 comes from dividing the total takings by the adult ticket price only, ignoring the children entirely, 74 ÷ 10.
(d) 2.7 — The height after 3 years of 10% compound growth is 2 × 1.1³ = 2.662 m, which rounds to 2.7 m. A candidate who adds 10% of the original height (0.2 m) in each of the 3 years, instead of compounding on the new height each time, would reach 2 + 3×0.2 = 2.6 m. A candidate who compounds for only 2 years would reach 2 × 1.1² = 2.42 m, rounding to 2.4 m. A candidate who compounds for 4 years instead of 3 would reach 2 × 1.1⁴ = 2.928 m, rounding to 2.9 m.
(d) y = 4 — Method: substitute x = 6 into y = 24/x: y = 24 ÷ 6 = 4. Distractor origins: y = 18 subtracts instead of dividing (24 − 6 = 18); y = 144 multiplies instead of dividing (24 × 6 = 144); y = 30 adds instead of dividing (24 + 6 = 30).
(b) 3w − 5 ≥ 16 — "Three times w, minus 5" translates to 3w − 5, and "is at least 16" means it must be 16 or more, giving 3w − 5 ≥ 16. A candidate who reads "at least" as a strict inequality writes 3w − 5 > 16. A candidate who misreads the wording and applies the subtraction before the multiplication writes 3(w − 5) ≥ 16. A candidate who reverses the direction of the inequality writes 3w − 5 ≤ 16.

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