Printable · GCSE Higher · ages 14-16
Algebra worksheet — GCSE Higher
Fifteen questions across the algebra statements at Higher 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 Higher
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- 1.Work out the value of m² + 3m when m = −2
- 2.A quadratic curve has a root at x = −2 and its turning point has x-coordinate 3. Work out the curve's other root, using the symmetry of the graph.
- 3.Solve 5(x + 3) = 40
- 4.A straight line has equation y = 4 − 3x. Work out the gradient of the line.
- 5.Expand and simplify 2(a + 6) − 2(a − 7)
- 6.A gym charges a joining fee plus a monthly fee. Anna paid £100 in total after 3 months of membership. Ben paid £160 in total after 6 months of membership (same joining fee and monthly fee as Anna). Work out the monthly fee.
- 7.The graph of y = 20/x is drawn for x > 0. As the value of x increases, what happens to the value of y?
- 8.A rule turns each input x into an output y. The inputs are x = −1, 0, 1, 2 and the matching outputs are y = 5, 3, 1, −1. Work out the rule.
- 9.A car's fuel tank contains 50 litres when it starts a journey. After travelling for 2 hours, it contains 38 litres, and fuel is used at a constant rate. Work out the rate at which fuel is used, in litres per hour.
- 10.Which of these rules generates the sequence 6, 11, 16, 21, …?
- 11.The diagram shows the graph of a function. Which equation could it represent?
- 12.A graph shows the total monthly cost of a mobile phone tariff against the number of minutes of calls. The line starts at £12 and stays level until 200 minutes, then rises by 5p for each further minute of calls. Use the graph to work out the total cost of a month in which 250 minutes of calls are made.
- 13.A distance-time graph is a straight line from (0, 0) to (4, 100), where time is in hours and distance is in kilometres. Work out the gradient of the line.
- 14.An exponential model has equation y = A × bˣ, where b > 1. Its graph passes through the points (1, 135) and (2, 405). Work out the value of A.
- 15.A quadratic graph y = ax² + bx + c has its turning point on the y-axis. Which statement about its roots must be true?
Answer key
- (d) −2 — Method: substitute the value into both terms, working out the index and the multiplication before the addition. Working: m² = (−2) × (−2) = 4 and 3m = 3 × (−2) = −6, so the expression becomes 4 + (−6), which is −2. Answer: −2. The distractors: −10 comes from squaring −2 as −4, giving −4 + (−6); 10 comes from working out 3m as +6 and losing the minus sign, giving 4 + 6; −14 comes from working from left to right instead of multiplying first, giving (4 + 3) × (−2).
- (c) x = 8 — The turning point lies exactly halfway between the two roots. If the other root is r, the midpoint of −2 and r must be 3, so (−2 + r) ÷ 2 = 3, giving r = 8. Choosing x = 5 comes from adding 2 and 3 rather than using the midpoint relationship correctly. Choosing x = 1 comes from subtracting 2 from 3 instead of reflecting −2 across the turning point. Choosing x = −8 finds the right distance but then reflects in the y-axis instead of in the line of symmetry x = 3, so the sign of the answer is flipped.
- (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.
- (d) −3 — Method: rewrite the equation in the form y = mx + c, then read off the gradient. Working: y = 4 − 3x can be written as y = −3x + 4, so comparing with y = mx + c gives m = −3. Answer: the gradient is −3. The value 3 comes from ignoring the negative sign on the x term. The value 4 comes from reading off the y-intercept instead of the gradient. The value −4 comes from a sign error, applying the negative sign to the intercept instead of the gradient.
- (d) 26 — Method: expand both brackets, treating the second as multiplication by −2 so that both of its terms change sign, then collect like terms. Working: 2(a + 6) = 2a + 12 and −2(a − 7) = −2a + 14, so the expression becomes 2a + 12 − 2a + 14; the a terms give 2a − 2a = 0, so no term in a survives, and the numbers give 12 + 14 = 26. Answer: 26. The distractors: −2 comes from expanding the second bracket as −2a − 14, so that the numbers give 12 − 14; 4a − 2 comes from adding 2(a − 7) instead of subtracting it, giving 2a + 12 + 2a − 14; 13 comes from multiplying the 2 over only the first term of each bracket, giving 2a + 6 − 2a + 7.
- (d) £20 — Let f be the joining fee and m the monthly fee: f + 3m = 100 and f + 6m = 160. Subtracting the first equation from the second eliminates f: 3m = 60, so m = 20. A candidate who finds the joining fee instead of the monthly fee would get f = 100 − 3(20) = £40. A candidate who divides Ben's total by his number of months, ignoring that part of the cost is a fixed joining fee, would get 160 ÷ 6 ≈ £26.67. A candidate who divides the difference in cost by the total number of months instead of the difference in months would get (160 − 100) ÷ 9 ≈ £6.67.
- (b) y decreases towards zero but never reaches it — Method: as x gets larger, dividing 20 by a bigger number gives a smaller result, so y decreases; because 20/x can never be exactly zero for any positive x, the curve gets closer to zero without ever reaching it, so y decreases towards zero but never reaches it. Distractor origins: 'y increases towards a limit but never reaches it' has the relationship backwards, treating y as increasing when it is actually decreasing; 'y decreases at a steady rate and reaches zero' wrongly assumes the graph behaves like a straight line that eventually hits zero; 'y stays the same however large x becomes' wrongly assumes there is no change in y at all.
- (a) y = −2x + 3 — Method: find the step in the outputs for each step of 1 in the input — falling outputs mean a negative multiplier — then read off the output when the input is 0, because that is the number added on. Working: the outputs 5, 3, 1, −1 fall by 2 each time x rises by 1, so x is multiplied by −2; the output at x = 0 is 3, so 3 is added. Answer: y = −2x + 3, checked at x = 2 by −2 × 2 + 3 = −1. The distractors: y = 2x + 3 comes from taking the size of the step, 2, as the multiplier and ignoring the fact that the outputs are falling; y = −2x − 3 comes from using the correct multiplier but writing the number added on as −3 instead of the output 3 listed at x = 0; y = −x + 4 comes from taking the multiplier as −1, its size read from the step of 1 in the inputs instead of the step of 2 in the outputs and its sign from the fact that the outputs fall, and then fitting the number added on to the pair x = −1, y = 5.
- (c) 6 — Method: rate = amount of fuel used ÷ time taken. Working: fuel used = 50 − 38 = 12 litres, time taken = 2 hours, so rate = 12 ÷ 2 = 6 litres per hour. Answer: the rate is 6 litres per hour. 44 comes from finding the average of the two fuel amounts, (50 + 38) ÷ 2, instead of the fuel used. 12 comes from working out the fuel used but forgetting to divide by the time taken. 19 comes from dividing the remaining fuel, 38, by the time taken instead of the fuel used.
- (c) 5n + 1 — Method: find the common difference, then use it as the coefficient of n in the position-to-term rule, and find the constant by checking against the first term. Working: the common difference is 5, so the rule has the form 5n + c. Using the 1st term: 5(1) + c = 6, so c = 1. The rule is 5n + 1. Answer: 5n + 1. 5n − 1 uses the correct coefficient but the wrong sign for the constant. 6n comes from using the first term as the coefficient of n instead of the common difference — it matches the 1st term by coincidence but fails from the 2nd term onward. n + 5 swaps the coefficient and the constant around, using the common difference as the constant instead of the coefficient of n.
- (b) $y = x^3 - 4x$ — Method: count how many times the curve crosses the x-axis and check whether it is a cubic (an S-shape with up to three crossing points) rather than a lower power, and note which way it runs overall from bottom-left to top-right or the reverse. Working: the curve crosses the x-axis at three points, x = −2, 0 and 2, and runs from bottom-left to top-right, which matches y = x³ − 4x = x(x − 2)(x + 2). Answer: y = x³ − 4x. Distractor refutation: y = −x³ + 4x comes from a sign error on every term, which would flip the curve so it ran from top-left to bottom-right instead. y = x³ + 4x comes from a sign error on the x term only, which removes two of the three crossing points, since x(x² + 4) has only x = 0 as a real root. y = x² − 4x comes from dropping the cubic term altogether, giving a parabola with only two crossing points instead of an S-shaped curve with three.
- (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).
- (a) 25 — Gradient = change in y ÷ change in x = (100 − 0) ÷ (4 − 0) = 100 ÷ 4 = 25. Dividing time by distance instead of distance by time gives 0.04; multiplying the two values instead of dividing gives 400; stopping at the change in distance, 100, forgets to divide by the change in time.
- (a) 45 — The ratio between the two given points, one power of x apart, gives b: 405 ÷ 135 = 3, so b = 3. Substituting back, at x = 1, y = A × b, so 135 = A × 3, giving A = 45. Giving the common ratio b itself instead of A confuses which unknown was asked for and produces 3 — wrong, because the question asks for A, not b. Giving 135 instead treats the first given point as the y-intercept and reads A off it directly — wrong, because that point is at x = 1, not x = 0, so 135 is A × b, not A. Assuming A equals b⁰ = 1 by itself, rather than substituting a known point to solve for A, gives 1 — wrong, because b⁰ is always 1 regardless of A; A must be found using an actual (x, y) pair from the graph.
- (d) If the graph has two real roots, they are equal and opposite in value, so they sum to zero. — A turning point on the y-axis means the graph's axis of symmetry is the line x = 0, so any two roots must be symmetrical about x = 0 — equal in size but opposite in sign, summing to zero. The graph could still have no real roots, but that is not guaranteed just from the turning point's position, so the option claiming it must have none is too strong. The roots do not have to both be positive — if real, one is positive and one negative (or both are zero). The graph does not have to touch the x-axis at exactly one point either; it could cross at two symmetrical points, touch at one point, or miss the x-axis entirely.
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