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.
Algebra worksheet — GCSE Higher
MathsUKwww.geekhero.co.uk
- 1.The first four terms of a sequence are 4, 9, 14, 19. Work out an expression, in terms of n, for the nth term.
- 2.The graph of y = f(x) has a maximum turning point at (−1, 6). Write down the coordinates of the maximum turning point of the graph of y = f(x − 3).
- 3.A circle has centre (0, 0) and equation x² + y² = 25. Work out the x-coordinates of the two points where the circle crosses the line y = 3.
- 4.A circle has centre (0, 0) and passes through the point (5, 12). Work out the equation of the circle.
- 5.The formula for the area of a trapezium is A = (a + b)h / 2, where a and b are the parallel side lengths and h is the height. Make h the subject of the formula.
- 6.Solve 3x + 14 = 8x − 6
- 7.Write y = x² + 12x + 40 in the form (x + a)² + b, and hence write down the minimum value of y.y = x² + 12x + 40
- 8.A pattern is made from tiles. Pattern 1 uses 4 tiles, pattern 2 uses 7 tiles, pattern 3 uses 10 tiles and pattern 4 uses 13 tiles, with each pattern using 3 more tiles than the one before. Work out an expression, in terms of n, for the number of tiles used in pattern n.
- 9.Work out the coordinates of the point where the graph of y = 2x − 6 crosses the x-axis.y = 2x − 6
- 10.The equation x² + 2x − 5 = 0 can be solved using the iterative formula xₙ₊₁ = 5/(xₙ + 2). The starting value is x₀ = 1, so x₁ is the value after the formula has been used once. Work out x₂ correct to 2 decimal places.
- 11.A circle has centre (0, 0) and equation x² + y² = 36. Work out the coordinates of the two points where the circle crosses the y-axis.
- 12.A car's speed increases steadily while it accelerates. Its speed v m/s after t seconds is given by v = u + at, where u is the starting speed in m/s and a is the acceleration in m/s². A car starts at u = 4 m/s and accelerates at a = 2 m/s² until it reaches v = 20 m/s. Work out t, the time taken in seconds.
- 13.Work out the integer values of x that satisfy x² − 2x − 8 ≤ 0.
- 14.A student attempts to prove that the product of two consecutive integers is always even: (i) Let the two consecutive integers be n and n + 1. (ii) Since n(n + 1) is even, one of n and n + 1 must be an even number. (iii) Therefore, n(n + 1) is even. At which statement does the proof first assume the very fact it is trying to prove?
- 15.The graph of P = 480 ÷ V shows how the pressure, P pascals, of a gas depends on its volume, V cm³, at constant temperature. Use the relationship to work out the volume when the pressure is 32 pascals, and then work out by how much the volume increases when the pressure drops to 24 pascals.
Answer key
- (d) 5n − 1 — The common difference is 5 (9−4=5), so the expression starts 5n. To match the first term when n=1, 5×1+c=4, so c=−1: the nth term is 5n−1. A candidate who uses the first term itself as the constant, instead of first term minus the common difference, would write 5n+4 (giving 9, 14, 19, 24 — one term too high throughout). A candidate who omits the constant term altogether would write just 5n (giving 5, 10, 15, 20, not matching the sequence). A candidate who adds the common difference to n instead of multiplying would write n+5 (giving 6, 7, 8, 9, far too small).
- (b) (2, 6) — y = f(x − 3) translates y = f(x) horizontally by 3 units to the RIGHT — inside the brackets, subtracting moves the graph in the positive x-direction. Turning point (−1, 6) → (−1 + 3, 6) = (2, 6). The common slip is to move LEFT instead, since the sign inside the bracket is negative — that gives (−4, 6). Changing the y-coordinate instead of the x-coordinate, as in (−1, 3) or (−1, 9), treats this as a vertical shift, which y = f(x − 3) is not.
- (d) x = 4 and x = −4 — Substituting y = 3 gives x² + 9 = 25, which simplifies to x² = 16, so x = 4 or x = −4. Choosing 'x = 3 and x = −3' uses the given value y = 3 as if it were the x-coordinate. Choosing 'x = 4' alone finds the positive square root of 16 but forgets the negative root. Choosing 'x = 5 and x = −5' skips subtracting 3² = 9 from 25 and takes the square root of 25 directly.
- (b) x² + y² = 169 — Method: a circle centred on the origin has equation x² + y² = r², and every point on it satisfies that equation, so substituting the coordinates of a point that lies on the circle gives r² directly. Working: substituting x = 5 and y = 12 gives 5² + 12² = 25 + 144 = 169, so r² = 169 and the circle is x² + y² = 169. Answer: x² + y² = 169. The distractors: x² + y² = 13 uses the radius, √169 = 13, where r² belongs, which is the confusion between r and r² made in the other direction; x² + y² = 17 adds the two coordinates, 5 + 12, instead of adding their squares; x² + y² = 119 subtracts the squares, 144 − 25, treating 12 as the hypotenuse of the right-angled triangle rather than as one of the shorter sides.
- (d) h = 2A / (a + b) — Method: undo the multiplication by 1/2 by multiplying both sides by 2, then undo the multiplication by (a + b) by dividing both sides by it. Working: A = (a + b)h / 2, so multiplying both sides by 2 gives 2A = (a + b)h, then dividing both sides by (a + b) gives h = 2A / (a + b). The value h = A / (a + b) comes from forgetting to multiply by 2 to clear the 1/2 first. The value h = 2A / a + b comes from dividing only by a and leaving b outside the fraction, instead of dividing by the whole bracket (a + b). The value h = (a + b) / (2A) comes from inverting the fraction.
- (c) x = 4 — Method: with an unknown on both sides, subtract the smaller x term from both sides so that all the x is on one side, then collect the numbers on the other. Working: subtracting 3x from both sides gives 14 = 5x − 6; adding 6 to both sides gives 20 = 5x; dividing both sides by 5 gives x = 4. Checking: 3 × 4 + 14 = 26 and 8 × 4 − 6 = 26. Answer: x = 4. The distractors: x = 2.5 comes from taking 3x off the left-hand side only, leaving 14 = 8x − 6 and so 8x = 20; x = −1.6 comes from changing the sign of the 8x when it is moved across but leaving the sign of the 14 unchanged, giving 3x − 8x = −6 + 14 and so −5x = 8; x = 15 comes from reaching 5x = 20 correctly and then subtracting 5 instead of dividing by 5.
- (c) 4 — x² + 12x + 40 = (x + 6)² − 6² + 40 = (x + 6)² + 4, so a = 6 and b = 4. Since (x + 6)² can never be negative, y = (x + 6)² + 4 is smallest when the bracket is zero, so the minimum value of y is 4. Giving 40 instead reads off the ORIGINAL constant term and ignores the completing-the-square step entirely — wrong, because 40 is the value of y when x = 0, not the minimum value. Giving −6 instead answers with the x-coordinate of the turning point (where the bracket is zero) rather than the minimum y-value itself — wrong, because the question asks for the minimum value of y, not the value of x that produces it. Giving 36 instead stops after squaring half the coefficient, 6² = 36, without combining it with the 40 already in the expression — wrong, because the minimum value is 40 minus 36, not 36 on its own.
- (b) 3n + 1 — Method: find how many more tiles each pattern uses, then find the constant by adjusting pattern 1's total. Working: each pattern uses 3 more tiles than the last, so the coefficient of n is 3. The constant is pattern 1's total minus the common difference: 4 − 3 = 1. Answer: the nth term is 3n + 1. 3n + 4 comes from using pattern 1's total, 4, as the constant without subtracting the common difference. 3n − 2 comes from a slip in working out the constant, subtracting the common difference twice (4 − 3 − 3 = −2) instead of once. n + 3 comes from swapping the common difference and the constant.
- (d) (3, 0) — Method: every point on the x-axis has y-coordinate 0, so substituting y = 0 into the equation and solving gives the x-coordinate of the crossing point. Working: 0 = 2x − 6 gives 2x = 6, so x = 6 ÷ 2 = 3 and the graph crosses the x-axis at (3, 0). Answer: (3, 0). The distractors: (0, −6) is where the graph crosses the y-axis, found by substituting x = 0 rather than y = 0; (−3, 0) comes from moving the 6 across the equals sign without changing its sign, giving 2x = −6; (6, 0) comes from reading the constant straight off as the crossing point and never dividing by the gradient 2.
- (a) 1.36 — Method: put the starting value into the right-hand side to get x₁, feed that value back in to get x₂, and round only once the second value has been found. Working: x₁ = 5 ÷ (1 + 2) = 5 ÷ 3 = 1.66666…; x₂ = 5 ÷ (1.66666… + 2) = 5 ÷ 3.66666… = 1.36363…. The digit in the third decimal place is 3, so x₂ = 1.36 correct to 2 decimal places. Answer: 1.36. The distractors: 1.67 is x₁, the value after a single use of the formula, given by a candidate who counts the starting value itself as x₁; 1.49 is x₃ = 1.48648…, one use of the formula too many; 1.37 comes from writing x₁ down as 1.66, truncating the display instead of keeping it in full, and then working out 5 ÷ 3.66 = 1.36612…, which rounds up to 1.37.
- (b) (0, 6) and (0, −6) — Method: every point on the y-axis has x-coordinate 0, so substitute x = 0 into the equation of the circle and solve for y, remembering that a square root has a negative value as well as a positive one. Working: putting x = 0 into x² + y² = 36 leaves y² = 36, so y = 6 or y = −6, and the two crossings are (0, 6) and (0, −6). Answer: (0, 6) and (0, −6). The distractors: (0, 36) and (0, −36) use 36 itself as the distance from the centre, which reads r² as r; (6, 0) and (−6, 0) are the right distance from the centre but are the crossings of the x-axis, found by setting y = 0 instead of x = 0; (0, 18) and (0, −18) halve 36, treating the right-hand side of the equation as a diameter.
- (b) 8 s — Rearranging v = u + at for t: subtract u from both sides to get v − u = at, then divide by a: t = (v − u)/a. Substituting u = 4, a = 2, v = 20: t = (20 − 4)/2 = 16/2 = 8 s. Answering 12 s comes from adding u instead of subtracting it: (20 + 4)/2 = 12. Answering 32 s comes from multiplying (v − u) by a instead of dividing by it: 16 × 2 = 32. Answering 6 s divides v by a first and then subtracts u, in the wrong order: 20/2 − 4 = 10 − 4 = 6. The time taken is 8 s.
- (b) −2, −1, 0, 1, 2, 3, 4 — Factorise x² − 2x − 8 = (x − 4)(x + 2), giving roots x = 4 and x = −2. Since the coefficient of x² is positive and the inequality is ≤ 0, the solution is the closed interval between the roots, −2 ≤ x ≤ 4, with the roots included because the inequality is not strict. The integers in this interval are −2, −1, 0, 1, 2, 3, 4. Distractor routes: −1, 0, 1, 2, 3 drops both endpoints, treating ≤ as if it were the strict inequality <. −2, −1, 0, 1, 2, 3, 4, 5 comes from mis-factorising as (x − 5)(x + 2), giving an upper root of 5 instead of 4. −3, −2, −1, 0, 1, 2, 3 comes from mis-factorising as (x − 4)(x + 3), giving a lower root of −3 instead of −2.
- (c) Statement (ii) — Statement (ii) opens with 'Since n(n + 1) is even', treating the very fact the proof is meant to establish as if it were already known — that is circular reasoning, assuming the conclusion to help derive itself. Statement (i) only names the two consecutive integers as n and n + 1; it makes no claim about whether their product is even, so it introduces nothing circular. Statement (iii) states the conclusion, and would be a valid final step if statement (ii) had reached 'one of n and n + 1 is even' by a genuine argument, such as considering the cases where n is even or odd separately. Saying the proof assumes nothing circular is wrong, because statement (ii)'s opening clause is exactly that assumption.
- (b) 5 cm³ — When P = 32, V = 480 ÷ 32 = 15 cm³. When P = 24, V = 480 ÷ 24 = 20 cm³. The increase in volume is 20 − 15 = 5 cm³. Giving only the second volume, 20 cm³, forgets to subtract the first volume. Giving only the first volume, 15 cm³, answers the wrong part of the question. Subtracting the two pressures instead of the two volumes, 32 − 24 = 8, mixes up which quantity the question asks for.
Build your own mix at the worksheet builder.