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
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- 1.A geometric sequence has first term 5, and each term after that is 3 times the term before it. Work out the first term of the sequence that is greater than 1000.
- 2.The graph of y = x² − 2x − 8 crosses the x-axis at x = −2 and x = 4. Work out the x-coordinate of the turning point of the curve, using the symmetry of the graph.y = x² − 2x − 8
- 3.The nth term of a sequence is n² + 3. Work out the first term of the sequence that is greater than 50.
- 4.Jack is x years old. His brother Kwame is 3 years older than Jack. Write down an expression for Kwame's age.
- 5.A sequence has the position-to-term rule: the nth term is 3n. Write down the first four terms of the sequence.
- 6.The diagram shows the graph of a quadratic function. Write down the values of x where the graph crosses the x-axis.
- 7.A formula is F = 4s. A pupil says that when s = 5 the value of F is 9, because they added 4 and 5 instead of multiplying. Work out the correct value of F when s = 5.
- 8.A plumber charges a call-out fee of £30 plus £25 for each hour worked. The total bill for a job was £130. Work out how many hours the plumber worked.
- 9.A cyclist's distance-time graph shows the following: she travels 12 km in the first 30 minutes at a steady speed, then rests for 30 minutes, then travels a further 8 km in the next 15 minutes. Work out her average speed, in km/h, for the whole journey.
- 10.The graph of y = mx + 6 passes through the point (−2, 0). Work out the value of m.
- 11.A charity's fundraising total, T pounds, over d days follows T = (d − 3)(30 − d) for 3 ≤ d ≤ 30, where T = 0 marks the start and end of the campaign. Work out how many days the campaign runs for, from start to end.
- 12.Which of these is an equation, rather than an expression, a formula or an identity?
- 13.p = 4. Work out the value of 2p³.
- 14.The graph of y = x² − 7x + 2 crosses the x-axis at two points. One root, read from the graph, is approximately x = 0.30. Using the fact that the sum of the two roots of x² − 7x + 2 = 0 is 7, estimate the other root, correct to 2 decimal places.y = x² − 7x + 2
- 15.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.
Answer key
- (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.
- (d) x = 1 — A quadratic graph is symmetrical about its turning point, so the turning point's x-coordinate is the midpoint of the two roots: (−2 + 4) ÷ 2 = 2 ÷ 2 = 1. The option x = 2 comes from adding the two roots but forgetting to divide by 2. The option x = −1 makes a sign error when adding the roots, treating −2 + 4 as −2. The option x = 3 comes from subtracting the roots instead of adding them: (4 − (−2)) ÷ 2 = 3.
- (a) 52 — Test successive terms: n=6 gives 6²+3=39, which is not greater than 50. n=7 gives 7²+3=52, which is greater than 50, so the first term greater than 50 is 52. A candidate who stops at n=6, before checking whether 39 actually exceeds 50, would give 39. A candidate who solves n²>50 instead of n²+3>50, ignoring the +3 in the search, would find n=8 is the first value with n²>50 (since 7²=49) and compute 8²+3=67. A candidate who computes n² by doubling n instead of squaring it would compute 2×7+3=17.
- (a) x + 3 — Method: "years older" is an addition, and the number of years is added to the age that is known in terms of the letter. Working: Jack's age is x, and Kwame is 3 years older, so Kwame's age is x + 3. Answer: x + 3. The distractors: x − 3 comes from reading "older" as "younger" and subtracting instead of adding; 3x comes from reading "3 years older" as "3 times as old" and multiplying; 3 − x comes from writing the calculation the wrong way round, taking Jack's age from the 3 rather than adding.
- (b) 3, 6, 9, 12 — Method: substitute the positions n = 1, 2, 3 and 4 into the rule in turn, because a position-to-term rule gives each term from its own position number. Working: 3 × 1 = 3, 3 × 2 = 6, 3 × 3 = 9 and 3 × 4 = 12. Answer: 3, 6, 9, 12. The distractors: 3, 9, 27, 81 comes from reading 3n as 3 multiplied by itself n times and so multiplying by 3 at every step; 0, 3, 6, 9 comes from starting the count at n = 0, which shifts every term one place; 4, 5, 6, 7 comes from reading 3n as n + 3 and adding 3 to each position number instead of multiplying by 3.
- (a) −2 and 2 — Method: the graph crosses the x-axis where y = 0, so read the two crossing points straight off the curve. Working: the curve meets the horizontal axis two squares to the left of the origin and two squares to the right, at x = −2 and x = 2. Answer: −2 and 2. Distractor refutation: −4 and 4 comes from reading the value where the curve crosses the y-axis, −4, and using that value and its positive partner as the x-axis crossings instead. −2 and 3 comes from misreading the right-hand crossing point one square out along the grid, taking it where the curve is already above the axis. 0 and 2 comes from confusing the lowest point of the curve, which sits on the y-axis at x = 0, with one of the crossing points.
- (b) 20 — F = 4s, so when s = 5, F = 4 × 5 = 20. 9 comes from adding 4 and 5 instead of multiplying, the mistake described in the question. 25 comes from squaring s (5²) instead of multiplying by 4. 45 comes from writing the digits 4 and 5 next to each other instead of carrying out the multiplication.
- (b) 4 — Method: form the equation 30 + 25h = 130, where h is the number of hours, then solve for h. Working: subtract the call-out fee from the total bill: 25h = 130 − 30 = 100. Divide by the hourly rate: h = 100 ÷ 25 = 4. Answer: 4 hours. 5.2 comes from dividing the whole bill by the hourly rate without subtracting the fixed fee first, 130 ÷ 25. 3.5 comes from swapping the fee and the rate, subtracting the rate from the bill and dividing by the fee, (130 − 25) ÷ 30. 6.4 comes from adding the call-out fee to the bill instead of subtracting it, before dividing by the rate, (130 + 30) ÷ 25.
- (c) 16 km/h — Method: total distance = 12 + 8 = 20 km. Total time = 30 + 30 + 15 = 75 minutes = 1.25 hours. Average speed = total distance ÷ total time = 20 ÷ 1.25 = 16 km/h. Distractor origins: 20 km/h gives the total distance without ever dividing by the total time; 24 km/h uses only the speed of the first stage (12 km in 30 minutes), ignoring the second stage and the rest; 28 km/h averages the two separate stage speeds, 24 km/h and 32 km/h, instead of using total distance over total time.
- (c) 3 — Method: a point that lies on a graph makes its equation true, so substituting the coordinates into y = mx + 6 leaves an equation in m alone. Working: substituting x = −2 and y = 0 gives 0 = m × (−2) + 6, which rearranges to −2m = −6, so m = (−6) ÷ (−2) = 3. Answer: 3. The distractors: −3 comes from dividing −6 by 2 and keeping the minus sign, overlooking that the divisor is negative too, so the two signs cancel; −2 comes from writing down the x-coordinate of the given point in place of the gradient; 6 comes from reading the constant in y = mx + 6 as the gradient, confusing m with c.
- (c) 27 days — The campaign starts at d = 3 and ends at d = 30, so it runs for 30 − 3 = 27 days. Getting 33 days comes from adding the two values, 3 + 30 = 33, instead of subtracting them. Getting 30 days uses only the end day and ignores that the campaign did not start at day 0. Getting 24 days comes from subtracting the start day twice, 30 − 3 − 3 = 24, instead of once.
- (c) 3x + 5 = 17 — An equation contains an equals sign and is true only for particular value(s) of the unknown — solving 3x + 5 = 17 gives the single value x = 4. 3x + 5 is an expression: it has no equals sign, so it cannot be solved, only simplified or evaluated. A = πr² is a formula: it shows the general relationship between different quantities (area and radius), rather than asking for one unknown value. 3(x + 5) ≡ 3x + 15 is an identity: the ≡ sign shows it is true for every value of x, not just one. The equation is 3x + 5 = 17.
- (d) 128 — In 2p³ the index belongs to p only, so cube p first and multiply by the coefficient afterwards. Cubing gives 4 × 4 × 4 = 64, and then 2 × 64 = 128. Cubing the coefficient as well would mean working out (2 × 4)³, which is 512. Reading the index as an instruction to multiply by 3 gives 2 × 4 × 3 = 24, and ignoring the coefficient altogether leaves 64.
- (a) x ≈ 6.70 — Since the two roots sum to 7, the other root is 7 − 0.30 = 6.70. The option 7.30 comes from adding the given root to 7 instead of subtracting it. The option 6.30 comes from subtracting 0.70 (one minus the given root) rather than the given root itself. The option 0.70 confuses the required root with the amount by which the given root falls short of 1.
- (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.
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