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.
Answer key: Algebra worksheet — GCSE Higher
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- (a) 120 m — The gradient of the second line is (0 − V)/(9 − 6) = −V/3, and this equals −8, so V = 24. The distance from t = 0 to t = 6 is the area of a trapezium with parallel sides 4 and 24 and width 6: 1/2 × (4 + 24) × 6 = 84. The distance from t = 6 to t = 9 is the area of a triangle with base 3 and height 24: 1/2 × 3 × 24 = 36. The total distance is 84 + 36 = 120 m. Using V = 8, treating the gradient's number as the missing velocity itself rather than solving −V/3 = −8 for V, gives a trapezium area of 1/2 × (4 + 8) × 6 = 36 and a triangle area of 1/2 × 3 × 8 = 12, a total of 48 m. Leaving out the 1/2 in the trapezium formula, (4 + 24) × 6 = 168, plus the correct triangle of 36, gives 204 m. Using the full 6 seconds as the triangle's base instead of the 3 seconds the second line actually lasts, 1/2 × 6 × 24 = 72, plus the correct trapezium of 84, gives 156 m.
- (a) 130 — Method: substitute the total bill into the formula, then subtract the fixed charge and divide by the cost per text message. Working: 24.50 = 18 + 0.05t, so 0.05t = 24.50 − 18 = 6.50, t = 6.50 ÷ 0.05 = 130. Answer: 130 extra text messages. 490 comes from dividing the whole bill by 0.05 without first subtracting the £18 fixed charge: 24.50 ÷ 0.05 = 490. 13 comes from dividing the £6.50 by 0.5 instead of 0.05, moving the decimal point one place too far: 6.50 ÷ 0.5 = 13. 65 comes from dividing the £6.50 by 0.1 instead of 0.05.
- (a) 3 — y = f(x + 2) is f(x) translated 2 units to the LEFT (inside the bracket, adding moves the graph in the negative x-direction). The root moves with the whole graph: 5 − 2 = 3. Moving right instead of left gives 7; assuming a bracket shift leaves the root unchanged gives 5; writing down the shift amount 2 itself skips the translation altogether.
- (a) 4 — a₅ = a₄ + a₃, so a₄ = a₅ − a₃ = 29 − 11 = 18. a₄ = a₃ + a₂, so a₂ = a₄ − a₃ = 18 − 11 = 7. a₃ = a₂ + a₁, so a₁ = a₃ − a₂ = 11 − 7 = 4. Checking forwards: 4, 7, 11, 18, 29. Answering 7 stops one step early, reporting a₂ = 7 instead of continuing back one more step to a₁ — wrong, because the question asks for a₁, not a₂. Answering 18 reports a₄ = 18, an intermediate value found along the way, instead of a₁ — wrong, because a₄ is a term used to reach the answer, not the term the question asks for. Answering 3 takes one backward step too many, working out a further term a₀ = a₂ − a₁ = 7 − 4 = 3 — wrong, because the sequence starts at a₁, so a₁ = 4 is as far back as the question goes.
- (d) −11/60 — The radius from the origin to (11, 60) has gradient 60/11. The tangent is perpendicular to this radius, so its gradient is the negative reciprocal: −1 ÷ (60/11) = −11/60. Choosing 60/11 uses the radius's gradient unchanged, without applying perpendicularity. Choosing −60/11 negates the radius's gradient but forgets to take its reciprocal. Choosing 11/60 takes the reciprocal correctly but keeps the gradient positive instead of negative.
- (d) x = 0, y = −1 and x = 3, y = 5 — Set the two expressions for y equal: 2x − 1 = x² − x − 1. Rearranging, subtracting 2x and adding 1 to both sides: 0 = x² − x − 1 − 2x + 1 = x² − 3x, so x² − 3x = 0. Factorise: x(x − 3) = 0, giving x = 0 or x = 3. Using y = 2x − 1: x = 0 gives y = −1; x = 3 gives y = 5. Distractor routes: x = 0, y = −1 alone stops after the factor x = 0 and never checks the second factor, x − 3 = 0. x = 0, y = −1 and x = −3, y = −7 comes from mis-factorising x² − 3x as x(x + 3), a sign error that gives a second root of −3 instead of 3. x = −2, y = −5 and x = 1, y = 1 comes from adding 2x to both sides instead of subtracting it when rearranging, giving x² + x − 2 = 0 instead of x² − 3x = 0.
- (c) x = 0 or x = −7 — Factorising: x² + 7x = x(x + 7) = 0, so x = 0 or x + 7 = 0, giving x = 0 or x = −7. A candidate who divides both sides of the original equation by x, which loses the solution x = 0, gets only x = −7. A candidate who makes a sign error solving x + 7 = 0 gets x = 0 or x = 7. A candidate who misreads the coefficient and doubles it gets x = 0 or x = −14.
- (d) L/6 − 4 — Method: find the length of one equal piece first (divide by 6), then apply the later change (subtract 4) to that piece. Working: one piece is L/6 metres; removing 4 metres from it gives L/6 − 4. Answer: L/6 − 4. L/6 + 4 comes from adding the 4 metres instead of removing it. (L − 4)/6 comes from removing the 4 metres from the whole rope before cutting it into pieces, the wrong order. 4 − L/6 comes from subtracting the piece length from 4 instead of the other way round.
- (d) x = 2 or x = −5 — Method: find two numbers that multiply to give −10 and add to give 3 — these are 5 and −2. So x² + 3x − 10 = (x + 5)(x − 2) = 0, giving x = −5 or x = 2. Distractor origins: x = −2 or x = 5 swaps the signs of the two roots; x = 2 or x = 5 makes both roots positive, ignoring the sign of −10; x = −5 or x = −2 makes both roots negative.
- (b) An identity, since both sides are equal for every x. — Expanding the left-hand side: 5(2x−3)=10x−15, which is exactly the same as the right-hand side for every value of x, so it is an identity, not an equation that is only true for one particular x. A candidate who treats every equals-sign statement as an equation, without checking whether it holds for all values of x, would choose the equation option. A candidate who mistakes it for a formula is assuming it relates two different letters or quantities, but only x appears — there is no second variable such as area or cost — so it is not a formula. A candidate who mistakes it for an inequality is assuming the two sides are only equal for particular values of x, but expanding shows they are identical for every value, not just some.
- (c) y = 3x + 1 — Method: a line of known gradient m has equation y = mx + c, and c is found by substituting the coordinates of a point known to lie on it. Working: the gradient is 3, so the line is y = 3x + c; substituting x = 1 and y = 4 gives 4 = 3 × 1 + c, so c = 4 − 3 = 1. Answer: y = 3x + 1. The distractors: y = 3x − 1 comes from working out the constant as mx − y, 3 − 4 = −1, instead of y − mx; y = x + 3 comes from swapping the two numbers over, putting the gradient 3 in the constant position and the x-coordinate 1 in front of x; y = 3x + 4 comes from using the y-coordinate 4 as the constant without substituting.
- (b) 1 at x = 90° — The graph of y = sin x rises from (0°, 0) to its maximum value of 1 at x = 90°, a quarter of the way through the period. Thinking the maximum occurs where the graph crosses the y-axis, at the start of the curve, gives 1 at x = 0°, but sin 0° = 0, not the maximum. Thinking the maximum occurs halfway to x = 360°, rather than a quarter of the way, gives 1 at x = 180°, but sin 180° = 0, not the maximum. Swapping the roles of the maximum value and the x co-ordinate at which it occurs gives 90 at x = 1, but the maximum value of sin x is 1, not 90.
- (c) 17.7 cm — Method: the iteration converges on x, the depth of the box, which is also the side of each square cut away; a square is removed from both ends of the 20 cm side, so the longer side of the base is 20 − 2x. Run the formula three times, then carry out that subtraction. Working: x₁ = (16 × 1² − 1³ + 50) ÷ 60 = 65 ÷ 60 = 1.08333…; x₂ = 67.50636… ÷ 60 = 1.12510…; x₃ = 68.82959… ÷ 60 = 1.14715…. The longer side of the base is 20 − 2 × 1.14715… = 17.70568…, which is 17.7 cm correct to 1 decimal place. Answer: 17.7 cm. The distractors: 1.1 cm is x₃ itself rounded, the depth of the box, given by a candidate who runs the iteration correctly and then stops before the step the question actually asks for; 18.9 cm is 20 − 1.14715…, taking away one square instead of two and forgetting that a corner is cut from each end of that side; 9.7 cm is 12 − 2 × 1.14715…, the shorter side of the base, which measures the wrong edge of the sheet.
- (d) 30° and 150° — The first solution is x = 30°, since sin 30° = 0.5. The graph of y = sin x is symmetrical about x = 90° between 0° and 180°, so the second solution is 180° − 30° = 150°. Adding 180° to the first solution instead of subtracting it from 180° gives 30° and 210°, but sin 210° = −0.5, not 0.5. Reflecting the first solution about x = 90° by adding 30° to 90° instead of subtracting from 180° gives 30° and 120°, but sin 120° = √3/2, not 0.5. Misremembering the standard value and using sin 45° = 0.5 instead of sin 30° = 0.5 gives 45° and 135°, but sin 45° = √2/2, not 0.5.
- (c) 4(n + 3) — 'Add 3 to n' must happen before 'multiply the result by 4', so the addition needs brackets to show it happens first: 4(n + 3). Writing 4n + 3 multiplies n by 4 immediately and only adds the 3 afterwards, which reverses the order the words describe. Writing n + 3 × 4 only multiplies the 3 by 4, and adds n as a separate, unmultiplied term — it treats 'the result' as just the 3, not the whole of n + 3. Writing 3(n + 4) keeps the correct structure but swaps which number is added and which is multiplied. The expression for 'add 3 to n, then multiply the result by 4' is 4(n + 3).
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