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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- (b) y = 2x² + 14x + 19 — A translation by the vector (−2, 0) moves the graph 2 units in the negative x-direction, which means replacing x with (x + 2): 2(x + 2)² + 6(x + 2) − 1. Expanding 2(x + 2)² gives 2x² + 8x + 8, and 6(x + 2) gives 6x + 12; collecting terms 2x² + 8x + 8 + 6x + 12 − 1 simplifies to 2x² + 14x + 19. Substituting (x − 2) instead of (x + 2) — translating in the wrong direction — gives 2(x−2)² + 6(x−2) − 1, which simplifies to 2x² − 2x − 5. Distributing the outer 2 only onto the x² term of the expansion, instead of onto every term of (x + 2)², gives 2x² + 10x + 15. Dropping the −1 when collecting the constant terms, treating 8 + 12 as 20 rather than 8 + 12 − 1 as 19, gives 2x² + 14x + 20.
- (b) (−2, 0) — Method: a graph meets the x-axis where the y-value is 0, so setting y = 0 turns the equation into a linear equation in x. Working: 0 = 3x + 6 gives 3x = −6, so x = (−6) ÷ 3 = −2 and the meeting point is (−2, 0). Answer: (−2, 0). The distractors: (2, 0) comes from solving 3x = −6 and then dropping the minus sign from the result; (0, 6) is the y-axis crossing, found by substituting x = 0 instead of y = 0; (6, 0) comes from reading the constant 6 straight off as the x-coordinate, without dividing by 3 and without changing its sign.
- (d) −2 — y = f(x) − 6 is f(x) shifted down by 6, so every y-value on the graph decreases by 6. At x = 0, f(0) = 4, so the new y-value is 4 − 6 = −2. Adding 6 instead of subtracting gives 10; writing down the shift itself, −6, or leaving the original value 4 unchanged both skip the translation altogether.
- (b) The total volume of water, in litres, that has flowed in. — On a rate-time graph, the y-axis is in litres per minute and the x-axis is in minutes; multiplying a rate by a time gives litres per minute × minutes = litres, a total volume. So the area under the graph represents the total volume of water that has flowed in. Thinking the area itself represents the rate, rather than what the rate accumulates to, gives the wrong claim about the average rate of flow. Confusing the area with the gradient of the graph — which measures how the rate is changing — gives the wrong claim about litres per minute squared. Ignoring the flow-rate axis and focusing only on the time axis gives the wrong claim that the area is simply the total time.
- (c) 21.6 — m² = (−2.4)² = 5.76, since a negative number squared is positive, so 5m² = 28.8. Then 3m = 3 × (−2.4) = −7.2. Total: 28.8 + (−7.2) = 21.6. −36 comes from taking m² as negative, 5 × (−5.76) − 7.2 = −36, forgetting that squaring removes the negative sign. 36 comes from ignoring the negative sign on m altogether and using m = 2.4 throughout. 29.4 comes from treating 3m as 3 + m instead of 3 × m, giving 28.8 + (3 + (−2.4)).
- (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.
- (d) 30 — Method: give each group a letter, turn each sentence of the stem into an equation in those letters, then eliminate one letter by adding the two equations. Working: let m be the number of men and w the number of women. The 50 adults give m + w = 50, and 10 more men than women gives m − w = 10. Adding the two equations term by term cancels w: (m + w) + (m − w) = 50 + 10, so 2m = 60 and m = 30. Substituting back into m + w = 50 gives w = 20, and 30 − 20 = 10 as required. Answer: 30. The distractors: 20 comes from subtracting the equations instead of adding them, which cancels m and gives 2w = 40, so w = 20 — the other unknown, not the one asked for; 40 comes from forming the second equation as w = 10, reading '10 more men than women' as '10 women', and then getting m = 50 − 10 from the first equation; 35 comes from using m + w = 50 alone with m = w, giving 25 each, and then adding the whole difference of 10 to the men instead of splitting it between the two groups, which leaves a difference of 20 rather than 10.
- (d) 44 m — The trapezium rule between t = 2 and t = 5 uses only the speeds at t = 2, 3, 4 and 5 — 12, 15, 16 and 14 — with three strips of width 1: adding the first and last readings and twice the sum of the middle readings gives 12 + 14 + 2 × (15 + 16) = 88, and half of that is 44 m. Writing 68.5 m comes from applying the trapezium rule to the whole table, from t = 0 to t = 6, instead of restricting it to the interval t = 2 to t = 5 that the question asks for. Writing 28.5 m comes from forgetting to double the two middle readings, 15 and 16, in the trapezium rule formula. Writing 39 m comes from averaging only the first and last speeds in the interval, 12 and 14, and multiplying by the 3-second interval, ignoring the readings at t = 3 and t = 4 in between.
- (a) Underestimate: the chord lies below the curve — For a curve that is concave down (bending downward), the straight line joining any two points on the curve lies BELOW the curve. The trapezium's top edge is this straight line, so the trapezium's area is smaller than the true area under the curve — the trapezium rule UNDERESTIMATES the distance in this case. This is because a curve whose gradient keeps decreasing bends away from any chord joining two of its points, dropping the chord below every point of the curve in between. The reverse, a chord lying above the curve, is the rule for a concave-up curve, not this one — check which way the curve bends before deciding. The trapezium rule does not always overestimate or underestimate, and it is not exact except when the graph really is a straight line.
- (b) 6x − 3 = 4x + 9 — Method: multiply every term inside the bracket by the number outside it; the right-hand side stays as it is given. Working: 3 × 2x = 6x and 3 × (−1) = −3, so 3(2x − 1) = 6x − 3. Answer: 6x − 3 = 4x + 9. 6x − 1 = 4x + 9 comes from multiplying only the 2x by 3 and leaving the −1 unchanged. 5x − 3 = 4x + 9 comes from adding the 3 to the 2 instead of multiplying, treating 3 × 2x as (3 + 2)x = 5x. 6x − 4 = 4x + 9 comes from working out 3 × (−1) as −1 − 3 = −4 instead of 3 × (−1) = −3.
- (b) 8 — The perimeter is 2(x + (x + 3)) = 4x + 6, so 4x + 6 = 38, which gives 4x = 32 and x = 8. A candidate who forgets the '+3' and treats the rectangle as a square, solving 2(2x) = 38, gets x = 9.5. A candidate who forgets to double the sum of the sides, solving 2x + 3 = 38, gets x = 17.5. A candidate who uses 3x instead of x + 3 for the length, solving 2(x + 3x) = 38, gets x = 4.75.
- (c) x² − 2x − 15 ≥ 0 — The critical values are x = −3 and x = 5, so the quadratic factorises as (x + 3)(x − 5). Expanding: (x + 3)(x − 5) = x² − 2x − 15. Since the solution set is OUTSIDE the roots (x ≤ −3 or x ≥ 5), the quadratic must be ≥ 0 there, since a positive U-shape sits above the axis outside its roots. So the inequality is x² − 2x − 15 ≥ 0. Distractor routes: x² + 2x − 15 ≥ 0 comes from factorising as (x − 3)(x + 5), swapping the sign of each root when forming the brackets. x² − 2x − 15 ≤ 0 keeps the correct expansion but uses ≤ 0, which gives the region between the roots instead of outside them. x² − 8x + 15 ≥ 0 comes from using roots x = 3 and x = 5, dropping the negative sign on −3 before expanding.
- (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).
- (c) x = 2y + 10 — Method: undo the operations done to x in reverse order — add 5, then multiply by 2. Working: y = x/2 − 5, so y + 5 = x/2, so x = 2(y + 5) = 2y + 10. Answer: x = 2y + 10. x = 2y + 5 comes from multiplying only the x/2 term by 2 and forgetting to multiply the 5 as well. x = 2y − 10 comes from a sign error, subtracting 5 instead of adding it before multiplying by 2. x = (y + 5)/2 comes from dividing by 2 instead of multiplying, the wrong operation to undo a division.
- (c) 3.153 — x₁ = √(4 × 1 + 1) = √5 = 2.236067977. x₂ = √(4 × 2.236067977 + 1) = √9.944271908 = 3.153453965, which rounds to 3.153. Reporting x₁ instead of x₂ gives 2.236067977, which rounds to 2.236. A sign error inside the root, using xₙ₊₁ = √(4xₙ − 1) instead of √(4xₙ + 1), gives x₁ = √3 = 1.732050808 and x₂ = √(4 × 1.732050808 − 1) = √5.928203232 = 2.434790182, which rounds to 2.435. Applying the formula in the wrong order, working out √(4xₙ) + 1 at every step instead of √(4xₙ + 1), gives x₁ = √4 + 1 = 3 and x₂ = √(4 × 3) + 1 = 4.464101615, which rounds to 4.464.
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