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.A graph has equation y = 3x + 2. Which word or phrase best describes its shape?y = 3x + 2
- 2.The diagram shows the graph of the cost, in pounds, of a taxi journey plotted against the distance travelled, in miles, for journeys of up to 4 miles. The same fixed charge and the same cost per mile apply to longer journeys. Work out the cost of a 6-mile journey.
- 3.Which of these equations is equivalent to 5x − 2 = 3x + 8?
- 4.Three vertices of a rectangle are (−4, −1), (2, −1) and (2, 3). The sides of the rectangle are parallel to the axes. Write down the coordinates of the fourth vertex.
- 5.Solve 5(x + 3) = 40
- 6.A candle is 30 cm tall and burns down at a steady rate of 1 cm per hour. Write down the function for the height of the candle y, in centimetres, after x hours.
- 7.Which expression is equivalent to 4(x − 2) + 10?
- 8.A colony of bacteria doubles in number every hour. At 9am there are 5 bacteria in the colony. Work out how many bacteria there will be at 12 noon.
- 9.Ben writes n + 4 ≤ 9. Which statement describes what Ben has written? Give a reason for your answer.
- 10.The formula for the circumference of a circle is C = 2πr, where r is the radius. Make r the subject of the formula.
- 11.Solve 2(x + 3) = 10
- 12.A graph has equation y = −2x² + 5. Which statement about its shape is correct?y = -2x² + 5
- 13.A ramp rises 3 m over a horizontal distance of 12 m. Work out the gradient of the ramp.
- 14.Which of these equations is true for every value of x, making it an identity rather than an equation with just one solution?
- 15.A ball is thrown in the air. Its height, h metres, above the ground after t seconds is given in this table: when t = 0, h = 0; when t = 1, h = 15; when t = 2, h = 20; when t = 3, h = 15; when t = 4, h = 0. Use the table to find the two times, in seconds, at which the ball is at ground level.
Answer key
- (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.
- (b) £14 — Method: read the fixed charge (the cost at 0 miles) and the rate (the cost per extra mile) from the graph, then use them to work out the cost for a distance beyond the part that is plotted. Working: the graph shows a fixed charge of £2 at 0 miles, and the cost rises by £2 for every extra mile, so for 6 miles the cost is £2 + (£2 × 6) = £2 + £12 = £14. Answer: £14. Distractor refutation: £12 comes from multiplying the rate by the distance and leaving out the £2 fixed charge. £8 comes from misreading the rate as £1 per mile instead of £2 per mile. £24 comes from adding the fixed charge to the rate first and then multiplying the total by the distance, instead of multiplying the rate by the distance and then adding the fixed charge.
- (d) 2x = 10 — Method: collect the x-terms on one side and the constants on the other, keeping the equals sign balanced. Working: 5x − 3x = 8 + 2, giving 2x = 10. Answer: 2x = 10. 2x = 6 comes from a sign error moving the constant, subtracting instead of adding: 8 − 2 = 6. 8x = 6 comes from adding the x-terms instead of subtracting them (5x + 3x = 8x), alongside the constant sign error. 2x = −10 comes from a sign error moving the constant the other way: −2 − 8 = −10.
- (b) (−4, 3) — Method: in a rectangle whose sides are parallel to the axes only two different x-coordinates and two different y-coordinates appear, and each of them is shared by a pair of vertices, so the missing vertex takes the x-coordinate and the y-coordinate that so far appear only once. Working: the x-coordinates given are −4, 2 and 2, so 2 is already used twice and −4 is used once; the y-coordinates given are −1, −1 and 3, so −1 is already used twice and 3 is used once; the fourth vertex therefore has x = −4 and y = 3. Answer: (−4, 3). The distractors: (3, −4) comes from picking the two unpaired coordinates correctly and then writing them in the wrong order; (−4, −5) comes from matching the 4-unit vertical side but measuring it downwards from (−4, −1) instead of upwards; (8, 3) comes from carrying on round the shape with the horizontal step used earlier, adding 6 to the x-coordinate of (2, 3) instead of closing the rectangle.
- (d) x = 5 — Method: expand the bracket by multiplying both terms inside it by 5, then undo the addition and the multiplication in turn. Working: expanding gives 5x + 15 = 40; subtracting 15 from both sides gives 5x = 25; dividing both sides by 5 gives x = 5. Answer: x = 5. The distractors: x = 8 comes from dividing both sides by 5 first, reaching x + 3 = 8 and writing 8 as the value of x without taking the 3 away; x = 11 comes from adding 15 to both sides instead of subtracting it, giving 5x = 55; x = 7.4 comes from expanding 5(x + 3) as 5x + 3, multiplying only the x by the 5, which leads to 5x = 37.
- (c) y = −x + 30 — Method: in a linear model the value at the start is the constant term and the steady rate of change is the gradient, which is negative when the quantity is falling. Working: at x = 0 the candle is 30 cm tall, so the constant term is 30; it loses 1 cm every hour, so the gradient is −1 and the height after x hours is y = −x + 30. Answer: y = −x + 30. The distractors: y = x + 30 comes from taking the rate as +1 and making the candle grow rather than shrink; y = 30x − 1 comes from building the model correctly as 30 − 1x and then copying it into the form y = mx + c with the two numbers left where they stood, so the starting height 30 ends up multiplying x and the hourly 1 is left behind as the constant being taken away; y = −30x + 1 comes from reading the two numbers the other way round, taking 30 cm per hour as the rate and 1 cm as the starting height, which burns 30 cm an hour from a candle only 1 cm tall.
- (a) 4x + 2 — Expand the bracket: 4(x − 2) = 4x − 8. Then add the 10 that follows the bracket: 4x − 8 + 10 = 4x + 2. The option 4x − 8 comes from expanding the bracket correctly but then forgetting to add the 10. The option 4x + 8 comes from forgetting to multiply the 2 inside the bracket by 4 (treating it as 4x − 2 instead of 4x − 8), then adding 10. The option 4x + 18 comes from a sign error when expanding — treating 4 × (−2) as +8 instead of −8, giving 4x + 8, then adding 10.
- (a) 40 — From 9am to 12 noon is 3 hours, so the population doubles three times: 5 × 2³ = 40. A candidate who counts the elapsed time as 4 hours (an off-by-one counting error) would reach 5 × 2⁴ = 80. A candidate who counts it as only 2 hours would reach 5 × 2² = 20. A candidate who misreads 'doubles' as 'increases by 2' each hour would compute 5 + 3 × 2 = 11.
- (d) An inequality, because ≤ compares the two sides — The symbol ≤ means 'is less than or equal to', so the statement compares the sizes of the two sides instead of saying they are equal: that makes it an inequality. Solving it gives n ≤ 5, a whole range of values rather than the single value an equation would give. An identity has to be true for every value of the letter, and this fails at n = 6, so it is not one. A formula works one quantity out from another, and there is only one letter here.
- (a) r = C / (2π) — Method: undo the multiplication by 2π by dividing both sides by 2π. Working: C = 2πr, so dividing both sides by 2π gives r = C / (2π). The value r = C / π comes from dividing by π only and forgetting the factor of 2 in 2π. The value r = 2πC comes from multiplying by 2π instead of dividing. The value r = C − 2π comes from subtracting 2π instead of dividing by it.
- (d) x = 2 — Method: expand the bracket by multiplying both terms inside it by 2, then undo the addition and the multiplication in turn. Working: expanding gives 2x + 6 = 10; subtracting 6 from both sides gives 2x = 4; dividing both sides by 2 gives x = 2. Answer: x = 2. The distractors: x = 5 comes from dividing both sides by 2 first, reaching x + 3 = 5 and writing 5 without taking the 3 away; x = 8 comes from adding 6 to both sides instead of subtracting it, giving 2x = 16; x = 3.5 comes from expanding 2(x + 3) as 2x + 3, multiplying only the x by the 2, which leads to 2x = 7.
- (d) It is n-shaped, since the x² coefficient is negative. — The coefficient of x² is −2, which is negative, so the quadratic curve opens downward — shaped like an n, with a maximum turning point. Saying it is U-shaped focuses only on x² being non-negative and ignores that the −2 in front of it flips the whole curve to open downward. Saying it is a straight line confuses having a constant term with being linear — any equation with an x² term is a curve, not a line. Saying it repeatedly rises and falls like a wave describes a trigonometric graph such as y = sin x, not a quadratic.
- (a) 0.25 — Method: gradient = vertical rise ÷ horizontal distance. Working: gradient = 3 ÷ 12 = 0.25. Answer: the gradient of the ramp is 0.25. 4 comes from inverting the gradient, dividing the horizontal distance by the vertical rise instead of the other way round. 9 comes from subtracting the rise from the horizontal distance instead of dividing. 36 comes from multiplying the rise and the horizontal distance instead of dividing.
- (b) 2(x + 3) = 2x + 6 — An identity is true for every value of x, not just one. Expanding 2(x + 3) gives 2x + 6, which matches the right-hand side exactly — so the equation holds for every value of x, and it is an identity. Each of the other three is only true for one particular value of x: 5x − 3 = 12 gives x = 3, x + 7 = 15 gives x = 8, and 3x = x + 10 gives x = 5 — these are ordinary equations, not identities.
- (a) t = 0 or t = 4 — The ball is at ground level exactly when h = 0. From the table, h = 0 at t = 0 and at t = 4, so those are the two times. Distractor origins: t = 1 or t = 3 picks the times with equal (but non-zero) height instead of ground level; t = 2 picks the time of maximum height instead of ground level; t = 0 finds only the starting time and misses the second one.
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