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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Answer key: Algebra worksheet — GCSE Higher
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- (c) 10x − 15 — Expand the bracket first: 3(2x − 5) = 6x − 15. Then add the 4x: 6x − 15 + 4x = 10x − 15. The option 10x − 5 comes from forgetting to multiply the 5 inside the bracket by 3 (treating it as 6x − 5), then adding 4x. The option 10x + 15 comes from a sign error when expanding, treating 3 × (−5) as +15 instead of −15, then adding 4x. The option 6x − 15 comes from expanding the bracket correctly but forgetting to add the 4x term at all.
- (b) 12n — Distributing the minus sign across the second bracket gives n² + 6n + 9 − n² + 6n − 9, and the n² terms and the +9/−9 cancel, leaving 6n + 6n = 12n. Writing 18 comes from only negating the first term of the second bracket, n², and treating the −6n and +9 as unchanged, which gives n² + 6n + 9 − n² − 6n + 9 = 18. Writing 2n² + 18 comes from adding the two brackets instead of subtracting them, (n² + 6n + 9) + (n² − 6n + 9) = 2n² + 18. Writing 6n comes from correctly negating the bracket but then only counting one of the two 6n terms, missing that they add rather than cancel.
- (d) x = −2, x = 6; y-intercept (0, 12) — Reflecting y = f(x) in the x-axis, to get y = −f(x), negates every y-value but leaves every x-value fixed. The x-intercepts happen where y = 0, and −0 = 0, so they are unaffected: y = −f(x) still crosses the x-axis at x = −2 and x = 6. The y-intercept is the value at x = 0: f(0) = −12, so −f(0) = 12, giving the point (0, 12) — the sign flips because the y-intercept is a nonzero y-value, unlike the roots. Writing 'x = 2, x = −6; y-intercept (0, −12)' comes from confusing −f(x) with f(−x) — reflecting in the y-axis instead of the x-axis, which negates the x-values of the intercepts (turning −2 into 2 and 6 into −6) but leaves f(0) unchanged, since f(−0) = f(0) = −12. Writing 'x = −2, x = 6; y-intercept (0, −12)' comes from forgetting that −f(x) is a reflection at all, and assumes both intercepts stay exactly as they were. Writing 'x = 2, x = −6; y-intercept (0, 12)' correctly negates the y-intercept but wrongly negates the x-intercepts too, as if a reflection in the x-axis also flipped the sign of every x-value.
- (b) 5 and −6 — Method: turn the sentence into an equation in one letter, collect every term on one side so it reads as a quadratic equal to zero, then factorise. Working: if the number is x then x² + x = 30, which rearranges to x² + x − 30 = 0; two numbers that multiply to −30 and add to 1 are 6 and −5, so (x + 6)(x − 5) = 0 and x = −6 or x = 5. Both values work: 5 + 25 = 30 and −6 + 36 = 30. Answer: 5 and −6. The distractors: 6 and −5 comes from reading the numbers inside the brackets as the solutions without changing their signs; 5 only comes from discarding the negative solution, although nothing in the question rules out a negative number; 5 and 6 comes from hunting for a factor pair of 30 instead of forming and solving the quadratic.
- (d) (−1, 4) — In parallelogram ABCD the side DC is parallel and equal to the side AB, so D = C − AB. The vector from A to B is (2 − (−3), 1 − 1) = (5, 0), so D = (4 − 5, 4 − 0) = (−1, 4). A candidate who adds this vector to C instead of subtracting it gets (4 + 5, 4 + 0) = (9, 4). A candidate who subtracts A's coordinates from C's rather than the vector AB, and drops the minus sign on −3 while doing so, works out (4 − 3, 4 − 1) and gets (1, 3). A candidate who makes only the y-part of that slip, working out 4 − 1 instead of 4 − 0, gets (−1, 3).
- (c) 12 cm — For a square, area = side². So side² = 144, giving side = ±12. Since a length must be positive, the side length is 12 cm. A candidate who gives both square roots without rejecting the negative one, which cannot be a length, answers 12 cm or −12 cm. A candidate who halves 144 instead of taking its square root gets 72 cm. A candidate who divides 144 by 4, confusing the area formula with a perimeter calculation, gets 36 cm.
- (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.
- (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.
- (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) 45 — v = s ÷ t = 180 ÷ 4 = 45 km/h. 720 comes from multiplying s and t instead of dividing, 180 × 4. 184 comes from adding s and t instead of dividing, 180 + 4. 176 comes from subtracting t from s instead of dividing, 180 − 4.
- (b) (2, 4) and (−2, −4) — Substitute y = 2x into x² + y² = 20: x² + (2x)² = 20, which gives x² + 4x² = 20, so 5x² = 20, x² = 4, and x = 2 or x = −2. Using y = 2x for each x-value: x = 2 gives y = 4; x = −2 gives y = −4. The path meets the pond's edge at (2, 4) and (−2, −4). Distractor routes: (2, −4) and (−2, 4) swaps the sign pairing, matching each x-value with the wrong sign of y instead of keeping each x with its own correctly-signed y. (2, 1) and (−2, −1) comes from using y = x/2 instead of y = 2x when finding the y-coordinates. (2, 4) alone stops after the positive square root of x² = 4 and never finds the second point from x = −2.
- (b) x ≤ 4 — Method: undo the number subtracted from 3x first, then divide by 3; the direction of the sign changes only for division by a negative number. Working: adding 1 to both sides of 3x − 1 ≤ 11 gives 3x ≤ 12; dividing both sides by 3, which is positive, gives x ≤ 4. Answer: x ≤ 4. The distractors: x ≥ 4 comes from turning the sign round while dividing by 3; x ≤ 10/3 comes from subtracting 1 from both sides instead of adding it, giving 3x ≤ 10; x ≤ 12 comes from stopping at 3x ≤ 12 and reading off the 12 without dividing by 3.
- (a) x = 6 or x = −4 — We need two numbers that multiply to −24 and add to −2: these are −6 and 4, since −6 × 4 = −24 and −6 + 4 = −2. So x² − 2x − 24 = (x − 6)(x + 4) = 0, giving x = 6 or x = −4. A candidate who swaps the signs, using 6 and −4 the wrong way round in the brackets, gets x = −6 or x = 4. A candidate who picks the wrong factor pair, 8 and −3 (which multiply to −24 but add to +5, not −2), gets (x + 8)(x − 3) = 0 and answers x = −8 or x = 3. A candidate who makes both factors negative gets x = −6 or x = −4, which would require the constant term to be +24, not −24.
- (c) Line D — Method: the steeper of two lines is the one whose gradient is greater, and the gradient of a line through two points is the change in y divided by the change in x. Working: line C rises 2 for a run of 4, so its gradient is 2 ÷ 4 = 1/2; the other line rises 2 for a run of 2, so its gradient is 2 ÷ 2 = 1; since 1 is greater than 1/2, it is line D that is the steeper. Answer: Line D. The distractors: Line C is chosen by comparing the x-coordinates and calling the line that reaches further along the x-axis the steeper one; They are equally steep comes from comparing only the y-coordinates, which are both 2, without dividing by the two different runs; There is not enough information comes from believing that a gradient can only be measured off a drawn graph, when two points on a line are enough to work it out.
- (a) −3/4 — Method: the tangent at a point on a circle is perpendicular to the radius drawn to that point, so find the gradient of the radius and then take its negative reciprocal. Working: the radius joins (0, 0) to (6, 8), so its gradient is 8 ÷ 6, which cancels to 4/3. Turning 4/3 upside down gives 3/4, and changing the sign gives −3/4. Answer: the gradient of the tangent is −3/4. The distractors: 4/3 is the gradient of the radius itself, quoted without taking the perpendicular at all; −4/3 changes the sign but leaves the fraction the same way up, so the two gradients do not multiply to give −1; 3/4 turns the fraction upside down but keeps it positive, which is the other half of the same rule left undone.
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