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
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- 1.Line L has equation y = 2x + 3. Work out the equation of the line perpendicular to L that passes through the point (4, 1), giving your answer in the form x + 2y = c.y = 2x + 3
- 2.A geometric sequence has first term 3 and common ratio √3. Work out the position of the first term of the sequence that exceeds 30.
- 3.Work out the value of a² − 2b when a = −3 and b = 4.
- 4.The graph of y = f(x) has a minimum point at (4, −1). Write down the coordinates of the corresponding turning point on the graph of y = −f(x).
- 5.One solution of the equation x² − (k + 1)x + k = 0 is x = 3. Work out the value of k.
- 6.The first four terms of a sequence are 2, 6, 10, 14. Priya says the nth term is 4n. Work out the correct expression for the nth term.
- 7.The formula y = x² + 1 gives y in terms of x. Work out the value of y when x = −2.y = x² + 1
- 8.A candidate solves the inequality x² > 9 and writes: "x² > 9, so x > 3." Which of these is a correct comment on the candidate's work?
- 9.A lift has a weight limit: it must carry no more than 630 kg. Let w be the total weight, in kg, carried by the lift. Which inequality describes this?
- 10.A number machine multiplies its input by 3 and then adds 7. The output is 1. Work out the input.
- 11.The graph of y = cos x is transformed onto the graph of y = cos(x − 90°). State the direction of the translation and which standard graph the image is.y = cos(x)
- 12.The graph of y = f(x) crosses the x-axis at x = −3 and x = 7, and crosses the y-axis at (0, 21). A second graph crosses the x-axis at x = −7 and x = 3, and crosses the y-axis at the same point, (0, 21). The second graph is y = g(x). Which of these could be the equation of g(x)?
- 13.The first four terms of a sequence are 6, 13, 20, 27. Work out an expression, in terms of n, for the nth term.
- 14.A sequence has the position-to-term rule n² + 1, where n is the position number. Work out how much bigger the 5th term is than the 4th term.
- 15.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.
Answer key
- (c) x + 2y = 6 — L has gradient 2, so the perpendicular gradient is −1/2. Substituting (4, 1) into y − 1 = −(1/2)(x − 4): y = −(1/2)x + 2 + 1 = −(1/2)x + 3, so 2y = −x + 6, giving x + 2y = 6. Distractor routes: x − 2y = 2 comes from using gradient +1/2 instead of −1/2, forgetting the negative sign the perpendicular rule needs. x + 2y = 9 comes from swapping the point's coordinates, substituting (1, 4) instead of (4, 1). x + 2y = −2 comes from a sign error distributing the negative gradient over the bracket, computing −(1/2)(x − 4) as −(1/2)x − 2 instead of −(1/2)x + 2.
- (d) 6th term — The terms are aₙ = 3 × (√3)ⁿ⁻¹: a₁ = 3, a₂ = 3√3 ≈ 5.196, a₃ = 9, a₄ = 9√3 ≈ 15.588, a₅ = 27, a₆ = 27√3 ≈ 46.765. Since a₅ = 27 is still below 30 and a₆ ≈ 46.8 is above 30, the 6th term is the first to exceed 30. 4th term comes from treating the common ratio as 3 instead of √3: 3, 9, 27, 81, … — under that wrong ratio, 81, the 4th term, is the first over 30. 5th term comes from misjudging a₅ = 27 as already greater than 30. 7th term comes from tracking only the whole-number terms (3, 9, 27, 81, …) and reporting the ORIGINAL position of 81 in the full sequence — 81 is indeed a₇, but a₆ ≈ 46.8 already exceeds 30 first.
- (a) 1 — a² − 2b = (−3)² − 2(4) = 9 − 8 = 1. A candidate who squares −3 but keeps the negative sign gets −9 − 8 = −17. A candidate who forgets to square a and substitutes it as −3 gets −3 − 8 = −11. A candidate who adds instead of subtracting 2b gets 9 + 8 = 17.
- (a) (4, 1) — y = −f(x) reflects the graph of y = f(x) in the x-axis: every point (x, y) maps to (x, −y). Applying this to (4, −1): the x-coordinate stays 4, and the y-coordinate −1 becomes its negative, 1. Leaving the y-coordinate unchanged skips the reflection entirely, giving (4, −1); reflecting the x-coordinate instead, or reflecting both, mixes this up with a reflection in the y-axis or a rotation, giving (−4, −1) or (−4, 1).
- (a) k = 3 — Method: a solution of an equation makes both sides balance, so substitute it in and solve the equation in k that is left. Working: putting x = 3 gives 3² − (k + 1) × 3 + k = 0, that is 9 − 3k − 3 + k = 0, so 6 − 2k = 0 and k = 3; the equation is then x² − 4x + 3 = 0, whose solutions are 3 and 1. Answer: k = 3. The distractors: k = −3 comes from solving 6 − 2k = 0 as though it gave 2k = −6; k = 4 comes from expanding −3(k + 1) as −3k − 1, multiplying only the k by 3; k = 1.5 comes from working 3² as 3 × 2 = 6, which leaves 3 − 2k = 0.
- (c) 4n − 2 — Method: find the common difference, then find the constant that fits the first term. Working: 6 − 2 = 4, 10 − 6 = 4, 14 − 10 = 4, so the terms increase by 4 each time and the nth term has the form 4n + c. Substituting n = 1: 4(1) + c = 2, so c = −2. Answer: the correct nth term is 4n − 2. The value 4n is Priya's value, which comes from using only the common difference and leaving out the constant. The value 4n + 2 comes from a sign error when finding the constant. The value 2n + 2 comes from using the first term, 2, as the coefficient of n instead of the common difference, and then attaching +2 rather than working the constant out.
- (c) 5 — Method: substitute the value into the formula, work out the index first and the addition afterwards. Working: x² = (−2) × (−2) = 4, so y = 4 + 1 = 5. Answer: 5. The distractors: −3 comes from squaring only the 2 and keeping the minus sign outside the index, giving −4 + 1; 1 comes from adding the 1 to x before squaring, giving (−2 + 1)²; 4 comes from squaring correctly but stopping there and never adding the 1.
- (d) Wrong — the correct solution is x < −3 or x > 3. — Squaring or unsquaring an inequality is not a safe one-step move: the solution must be split into two branches, since a number less than −3 also squares to more than 9. So x > 3 finds only half the solution set. The full answer is x < −3 or x > 3. Distractor routes: "Right — square rooting both sides gives x > 3" treats the square root of an inequality the same as the square root of an equation and misses the negative branch entirely. "Wrong — the correct solution is −3 < x < 3" applies the between-the-roots pattern that belongs to the opposite inequality, x² < 9. "Right, but x = −3 and x = 3 should also work" treats the inequality as if it also allowed equality, when x² > 9 is strict and excludes both x = 3 and x = −3.
- (d) w ≤ 630 — 'No more than 630 kg' means the weight can be exactly 630 kg or anything less, so the correct inequality is w ≤ 630, using 'less than or equal to' to include the limit itself. Writing w < 630 excludes 630 kg itself, as though the limit could not be reached exactly. Writing w ≥ 630 reverses the direction, describing a minimum weight rather than a maximum. Writing w > 630 both reverses the direction and excludes the boundary value. The inequality describing the lift's weight limit is w ≤ 630.
- (d) −2 — Method: run the machine backwards, undoing the operations in the opposite order and swapping each one for its inverse. Working: the machine added 7 last, so take 7 off the output: 1 − 7 = −6; before that the machine had multiplied by 3, so divide: −6 ÷ 3, and a negative divided by a positive stays negative. Answer: −2, which checks because 3 × (−2) + 7 = −6 + 7 = 1. The distractors: 2 comes from dividing 6 by 3 and losing the minus sign; −6 comes from taking the 7 off and stopping there, never undoing the multiplication; −18 comes from multiplying −6 by 3 instead of dividing by 3.
- (a) Positive x-direction, 90°; image is y = sin x. — Writing cos(x − 90°) as cos(x − a) with a = 90 shows this is a horizontal translation, y = f(x − a), which moves the graph 90° in the positive x-direction; the identity cos(x − 90°) = sin x confirms the image is y = sin x. Choosing the negative x-direction reverses the sign inside the bracket — subtracting inside the bracket always translates in the positive x-direction, not the negative one, so that statement is wrong on direction. Getting the direction right but conflating the subtraction inside the bracket with an extra reflection of the output flips the sign of the resulting graph, wrongly giving y = −sin x. Treating the subtraction as if it changed the output directly, rather than the input, wrongly calls this a vertical translation even while still correctly recalling that the image simplifies to y = sin x.
- (b) y = f(−x) — y = f(−x) reflects the graph of y = f(x) in the y-axis: every x-coordinate changes sign. The x-intercepts −3 and 7 become 3 and −7, matching the second graph's intercepts −7 and 3. A point already on the y-axis is unaffected, since −0 = 0, so the y-intercept (0, 21) stays exactly where it is — matching the second graph as well. y = −f(x) leaves the x-intercepts unchanged at −3 and 7, since f(x) = 0 exactly where −f(x) = 0, which does not match; it also sends the y-intercept to (0, −21), a second mismatch. y = −f(−x) does send the x-intercepts to the right places, −7 and 3, but it sends the y-intercept to (0, −21) instead of (0, 21), so it fails the second clue. y = f(x) − 4 moves every point down 4, sending the y-intercept to (0, 17) instead of (0, 21), so it fails the y-axis clue. Test each option against BOTH clues — the pair of x-intercepts and the point on the y-axis — because more than one option gets only one of the two right.
- (c) 7n − 1 — Method: find the common difference, then find the constant that fits the first term. Working: 13 − 6 = 7, 20 − 13 = 7, 27 − 20 = 7, so the terms increase by 7 each time and the nth term has the form 7n + c. Substituting n = 1: 7(1) + c = 6, so c = −1. Answer: the nth term is 7n − 1. The value 7n comes from leaving out the constant altogether. The value 7n + 6 comes from using the first term as the constant directly, without subtracting the common difference first. The value 6n + 1 comes from using the first term, 6, as the coefficient of n instead of the common difference, and then attaching +1 with the sign of the constant flipped.
- (a) 9 — Method: the gaps in this sequence are not constant, so work out each of the two terms named from the rule and then subtract the earlier from the later. Working: the 5th term is 5² + 1 = 25 + 1 = 26 and the 4th term is 4² + 1 = 16 + 1 = 17, so the difference is 26 − 17. Answer: 9. The distractors: 7 comes from using the 3rd and 4th terms, one position too early, 17 − 10; 11 comes from using the 5th and 6th terms, one position too late, 37 − 26; 1 comes from subtracting the position numbers, 5 − 4, instead of the terms themselves.
- (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.
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