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 2x² when x = −3
- 2.Kai writes the expression 8 − 3x + 5x² and says it has three terms. Ryan says it only has two terms because a number on its own is not a term. Work out the correct number of terms in the expression.
- 3.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.
- 4.Work out the value of 3a − 2b when a = 5 and b = 4
- 5.The curve y = x² − 1 and the line y = 3x − 3 meet at two points. Work out the x-coordinates of those two points.y = 3x − 3y = x² − 1
- 6.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.
- 7.Which of these is a formula, rather than an expression, an equation, or an identity?
- 8.Expand and simplify 2(a + 6) − 2(a − 7)
- 9.A ramp rises 3 m over a horizontal distance of 12 m. Work out the gradient of the ramp.
- 10.f(x) = x + 3 and g(x) = 2x. Work out fg(x).y = x + 3
- 11.A region S consists of every point that lies on or above the line y = 2, on or below the line y = x, and on or to the left of the line x = 6. Which of these is the system of inequalities that defines S?y = x
- 12.Two numbers have a sum of 45. The larger number is 9 more than the smaller number. Form an equation using n for the smaller number, and solve it to find the larger number.
- 13.Describe the single transformation that maps the graph of y = x² onto the graph of y = x² + 3.y = x²y = x² + 3
- 14.Aisha saves money each week. In week 1 she has saved £15 in total, in week 2 she has saved £23 in total, in week 3 she has saved £31 in total and in week 4 she has saved £39 in total, with the total increasing by the same amount each week. Work out an expression, in terms of n, for the total amount she has saved, in pounds, after week n.
- 15.The first four terms of a sequence are 2, 2√3, 6, 6√3, ... Work out the next term.
Answer key
- (a) 18 — Method: substitute the value, apply the index before the multiplication, and remember that a negative number multiplied by itself gives a positive result. Working: x² = (−3) × (−3) = 9, and then 2 × 9 = 18. Answer: 18. The distractors: −18 comes from squaring only the 3 and leaving the minus sign outside the index, giving 2 × (−9); 36 comes from multiplying 2 by −3 first and squaring afterwards, giving (−6)²; −12 comes from reading x² as 2x, so that the calculation becomes 2 × 2 × (−3).
- (b) 3 — Method: a term is any part of an expression separated from the rest by a + or − sign, including a number on its own. Working: 8 − 3x + 5x² splits at the + and − signs into 8, −3x and 5x² — that is three separate terms. Answer: 3. Ryan's answer of 2 comes from wrongly excluding the number 8, thinking a term must contain a letter. 5 comes from miscounting by treating the coefficients and powers as separate terms as well as the letters. 1 comes from treating the whole expression as a single term because it is written without brackets.
- (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.
- (d) 7 — Method: substitute both values, work out the two multiplications first, and only then subtract. Working: 3a = 3 × 5 = 15 and 2b = 2 × 4 = 8, so the expression becomes 15 − 8 = 7. Answer: 7. The distractors: 23 comes from adding the two products instead of subtracting, giving 15 + 8; −7 comes from subtracting the wrong way round and working out 8 − 15; 52 comes from working from left to right instead of following the order of operations, giving 3 × 5 = 15, then 15 − 2 = 13, then 13 × 4.
- (d) x = 1 and x = 2 — Method: where a line meets a curve the two expressions for y are equal, so set them equal and solve the quadratic that results. Working: x² − 1 = 3x − 3 collects to x² − 3x + 2 = 0; factorising gives (x − 1)(x − 2) = 0, so x = 1 or x = 2, and each value gives the same y on both graphs. Answer: x = 1 and x = 2. The distractors: x = −1 and x = −2 come from factorising as (x + 1)(x + 2) and so reversing the sign of both roots; x = −1 and x = 4 come from moving the −3 across the equals sign without changing its sign, which gives x² − 3x − 4 = 0; x = 1 and x = −1 come from setting each expression equal to zero separately instead of equal to each other.
- (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) V = lwh — V = lwh is a formula: it relates one quantity, the volume V, to others, the length, width and height, in a way that is true generally. A candidate who picks lwh has chosen the expression, not a full statement relating two quantities. A candidate who picks lwh = 60 has chosen an equation, since it is only true for particular values of l, w and h that multiply to 60. A candidate who picks 2(l + w) ≡ 2l + 2w has chosen an identity, mistaking the identity symbol ≡ for a sign that makes a statement a formula.
- (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.
- (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) 2x + 3 — fg(x) means f(g(x)): apply g first, then apply f to the result. g(x) = 2x, so f(g(x)) = f(2x) = 2x + 3. Writing 2x + 6 comes from working out gf(x) instead — g(f(x)) = g(x + 3) = 2(x + 3) = 2x + 6 — which applies the functions in the wrong order. Writing 3x + 3 comes from adding f(x) and g(x) together, (x + 3) + 2x = 3x + 3, instead of composing them. Writing 2x² + 6x comes from multiplying f(x) and g(x) together, (x + 3)(2x) = 2x² + 6x, instead of substituting one into the other.
- (b) y ≥ 2, y ≤ x, x ≤ 6 — "On or above the line y = 2" means y ≥ 2. "On or below the line y = x" means y ≤ x. "On or to the left of the line x = 6" means x ≤ 6. Together these give y ≥ 2, y ≤ x, x ≤ 6. Distractor routes: y ≤ 2, y ≤ x, x ≤ 6 flips the first inequality, describing "on or below" y = 2 instead of "on or above". y ≥ 2, y ≥ x, x ≤ 6 flips the second, describing "on or above" y = x instead of "on or below". y ≥ 2, y ≤ x, x ≥ 6 flips the third, describing "on or to the right of" x = 6 instead of "on or to the left".
- (a) 27 — Method: let the smaller number be n, so the larger number is n + 9. Form the equation n + (n + 9) = 45. Working: simplify: 2n + 9 = 45, so 2n = 36, giving n = 18. The larger number is 18 + 9 = 27. Answer: 27. 18 is the smaller number, found correctly, but given as the answer to the question asking for the larger number. 22.5 comes from ignoring the "9 more" part and simply halving the total, 45 ÷ 2. 36 comes from a sign error when isolating the constant, treating the equation as 2n = 45 + 9 instead of 2n = 45 − 9.
- (a) A translation by vector (0, 3) — y = x² + 3 adds a constant outside the squaring, so it is a vertical translation of y = x² — every point moves the same distance parallel to the y-axis, with no change in x. The vector is (0, 3), matching the +3. A vector of (3, 0) confuses this with a horizontal shift; (0, −3) has the right axis but the wrong sign, since the graph moves up, not down; a stretch changes the shape of the curve, which adding a constant term never does.
- (c) 8n + 7 — Method: find the weekly increase, then find the constant by adjusting the week 1 total. Working: the total goes up by £8 each week (23 − 15 = 8), so the coefficient of n is 8. The constant is the week 1 total minus the common difference: 15 − 8 = 7. Answer: the nth term is 8n + 7. 8n + 15 comes from using the week 1 total, 15, as the constant without subtracting the common difference. 8n − 1 comes from a slip in working out the constant, subtracting the common difference twice (15 − 8 − 8 = −1) instead of once. 7n + 8 comes from swapping the weekly increase and the constant.
- (a) 18 — The common ratio is 2√3 ÷ 2 = √3. Checking: 6 ÷ 2√3 = √3 and 6√3 ÷ 6 = √3, so the ratio is consistent throughout. The next term is 6√3 × √3 = 6 × 3 = 18, since √3 × √3 = 3. Looking only at the coefficients 2, 2, 6, 6 and continuing them by doubling the last one gives 6 × 2 = 12, which is wrong because the step from each term to the next is a multiplication by √3, not a pattern in the coefficients alone. Doubling the previous term instead of multiplying by the surd ratio √3 gives 6√3 × 2 = 12√3, which is wrong because the common ratio is √3, not 2. Using 3 instead of √3 as the common ratio — squaring the true ratio by mistake — gives 6√3 × 3 = 18√3, which is wrong because 3 is the SQUARE of the common ratio, not the ratio itself.
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