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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- (b) x = −4, y = −3 and x = 3, y = 4 — Substitute y = x + 1 into x² + y² = 25: x² + (x + 1)² = 25, which expands to x² + x² + 2x + 1 = 25, giving 2x² + 2x − 24 = 0, or x² + x − 12 = 0. Factorise: (x + 4)(x − 3) = 0, so x = −4 or x = 3. Using y = x + 1: when x = −4, y = −3; when x = 3, y = 4. Distractor routes: x = 4, y = 5 and x = −3, y = −2 comes from mis-factorising x² + x − 12 as (x − 4)(x + 3), reversing the sign of each root. x = −4, y = −3 alone stops after finding the first root of the quadratic and never goes back for the second pair. x = 12, y = 13 comes from misreading x² + y² = 25 as the linear equation x + y = 25 and solving that together with y = x + 1.
- (a) (15, 0) — Method: the tangent is perpendicular to the radius at the point of contact, so find the gradient of the radius, take its negative reciprocal, write the equation of the tangent, then substitute y = 0 because every point on the x-axis has y-coordinate 0. Working: the radius from (0, 0) to (3, 6) has gradient 6 ÷ 3 = 2, so the tangent has gradient −1/2. Substituting into y − 6 = −1/2(x − 3) gives y = −0.5x + 7.5. Setting y = 0 gives 0.5x = 7.5, so x = 15 and P is (15, 0). Answer: (15, 0). The distractors: (0, 7.5) is where the same tangent crosses the y-axis, reached by setting x = 0 instead of y = 0; (0, 0) comes from using the gradient of the radius, 2, for the tangent, which gives the line y = 2x through the centre and so crosses the x-axis at the origin; (6, 0) comes from changing the sign of the radius gradient without turning it upside down, which gives y = −2x + 12.
- (a) 0 — the line does not intersect the circle — Substitute y = 2x + 7 into x² + y² = 4: x² + (2x + 7)² = 4, which expands to x² + 4x² + 28x + 49 = 4, giving 5x² + 28x + 45 = 0. The discriminant is b² − 4ac = 28² − 4 × 5 × 45. Since 28² = 784 and 4 × 5 × 45 = 900, the discriminant is 784 − 900 = −116. Since the discriminant is negative, the quadratic has no real solutions, so the line does not meet the circle at all: 0 intersection points. Distractor routes: "1 — the line is a tangent" confuses a negative discriminant with a zero one; a discriminant of exactly zero gives one point, a tangent, but −116 is not zero. "2 — the line crosses the circle at two points" assumes a positive discriminant without actually working it out. "It cannot be found without solving the quadratic" is the whole point the discriminant exists to avoid — its sign alone, without finding x, tells you the number of real solutions.
- (a) 5x − 3 = 12 — 5x − 3 = 12 is an equation with exactly one solution: adding 3 and dividing by 5 gives x = 3, and no other value works. 5x − 3 = 5x − 3 is true for every value of x, since both sides are identical — it has infinitely many solutions, not one. 5x − 3 > 12 is an inequality: any value of x greater than 3 satisfies it, so it has a whole range of solutions, not a single one. 5x − 3 = 5x + 2 has no solution at all, since subtracting 5x from both sides leaves −3 = 2, which is never true. The equation with exactly one solution is 5x − 3 = 12.
- (c) P — Sequence P has a common ratio of 2 (4 × 2 = 8, 8 × 2 = 16, 16 × 2 = 32), so it is geometric. Sequence Q has a common difference of 4, which is arithmetic, not geometric — a candidate who confuses a constant difference with a constant ratio picks Q. Sequence R has increasing differences of 3, 5, 7, a quadratic sequence, not geometric — a candidate who assumes any fast-growing sequence must be geometric picks R. Sequence S is the square numbers from 2² onwards (2², 3², 4², 5²), which is also quadratic, not geometric — a candidate who thinks squaring always means geometric growth picks S.
- (c) 2 — For a geometric sequence, a₄ = a₂ × r². So r² = a₄ ÷ a₂ = 24 ÷ 6 = 4, giving r = 2 or r = −2. Since all the terms are positive, r = 2. 4 comes from computing a₄ ÷ a₂ = 4 and stopping there, without taking the square root to find r. 18 comes from subtracting instead of dividing: a₄ − a₂ = 24 − 6 = 18, taken directly as the common ratio. −2 is the other square root of 4, but it is rejected because all the sequence's terms are given as positive.
- (c) 2 — At x = 1, y = 1² − 3(1) = 1 − 3 = −2. At x = 4, y = 4² − 3(4) = 16 − 12 = 4. Gradient of the chord = change in y ÷ change in x = (4 − (−2)) ÷ (4 − 1) = 6 ÷ 3 = 2. Writing down the change in y, 6, and stopping there without dividing by the change in x gives 6. Losing the negative sign on y = −2 at x = 1 and treating it as +2 gives (4 − 2) ÷ (4 − 1) = 2 ÷ 3 = 2/3. Dividing the wrong way round, change in x ÷ change in y, gives (4 − 1) ÷ (4 − (−2)) = 3 ÷ 6 = 1/2.
- (b) s = d / t — Method: undo the multiplication by t by dividing both sides by t. Working: d = st, so dividing both sides by t gives s = d / t. The value s = dt comes from multiplying by t instead of dividing. The value s = t / d comes from inverting the fraction, dividing t by d instead of d by t. The value s = d − t comes from subtracting t instead of dividing by it.
- (d) £40 — Method: add the two prices to make one expression in x, turn 'at least' into ≥, solve the inequality, then read the lowest possible price off the boundary of the solution set. Working: the two items cost x + (2x + 30) = 3x + 30 pounds, so 3x + 30 ≥ 150; subtracting 30 from both sides gives 3x ≥ 120; dividing both sides by 3 gives x ≥ 40, and the smallest value the shirt price is allowed to take is the boundary, £40. Answer: £40. The distractors: £50 comes from leaving the £30 out of the total and solving 3x ≥ 150; £60 comes from using the trousers expression on its own, 2x + 30 ≥ 150; £120 comes from stopping at 3x ≥ 120 and reading the 120 as the price of the shirt without dividing by 3.
- (b) x = 7, y = 3 — Method: one equation contains +y and the other −y, so adding them removes y; the value found is then substituted back to get the other letter. Working: adding x + y = 10 and x − y = 4 gives 2x = 14, so x = 7; substituting into x + y = 10 gives 7 + y = 10, so y = 3. Answer: x = 7, y = 3, and 7 − 3 = 4 as required. The distractors: x = 7, y = 4 comes from finding x correctly and then taking the 4 in x − y = 4 to be the value of y; x = 5, y = 5 comes from splitting the total of 10 equally and never using the difference; x = 14, y = −4 comes from adding the equations to 2x = 14 and forgetting to halve, so that x is taken as 14 and y as 10 − 14.
- (a) 3 and 7 — Method: one equation is linear and the other is not, so rearrange the linear equation and substitute it into the other to leave a single quadratic in one letter. Working: x + y = 10 gives y = 10 − x, so xy = 21 becomes x(10 − x) = 21, which rearranges to x² − 10x + 21 = 0; factorising gives (x − 3)(x − 7) = 0, so the two values are 3 and 7, and 3 + 7 = 10 with 3 × 7 = 21. Answer: 3 and 7. The distractors: 1 and 21 comes from using only the product and taking the first factor pair of 21; −3 and −7 comes from factorising as (x + 3)(x + 7), which reverses the sign of both values; 5 and 5 comes from using only the sum and splitting 10 into two equal parts.
- (b) 6 — 3n + 4.5 = 22.5, so 3n = 18 and n = 6. A candidate who divides 22.5 by 3 first and ignores the 4.5 gets n = 7.5. A candidate who makes a sign error and forms the equation 3n − 4.5 = 22.5 gets 3n = 27 and n = 9. A candidate who divides by 3 before subtracting the 4.5, working out 22.5 ÷ 3 + 4.5, gets n = 12.
- (a) (40, 0) — Method: a graph crosses the x-axis where y = 0, which in this context is the moment the tank holds no water, so substituting y = 0 and solving gives the time. Working: 0 = −5x + 200 gives 5x = 200, so x = 200 ÷ 5 = 40 and the crossing point is (40, 0); the tank is empty after 40 minutes. Answer: (40, 0). The distractors: (0, 200) is the y-axis crossing, the 200 litres in the tank at the start, found by substituting x = 0 instead of y = 0; (200, 0) comes from reading the constant 200 as the x-coordinate without dividing by the 5 litres lost each minute; (0, 40) has the right number in the wrong place, with the time written as a y-coordinate.
- (d) 4x + 5 — A minus sign directly before a bracket changes the sign of both terms inside it: 6x − (2x − 5) = 6x − 2x + 5 = 4x + 5. The option 4x − 5 comes from only changing the sign of the 2x term and not the −5, giving 6x − 2x − 5. The option 8x − 5 comes from adding 2x instead of subtracting it, as if the minus sign did not apply to the bracket, giving 6x + 2x − 5. The option 8x + 5 repeats that same addition mistake and also changes the sign of the −5 term.
- (d) x = −4, y = −19 — x² + 8x − 3 = (x + 4)² − 4² − 3 = (x + 4)² − 19. Substituting x = −4: (−4)² = 16, 8 × (−4) = −32, so 16 − 32 − 3 = −19, confirming the minimum value −19 at x = −4: turning point (−4, −19). Using 8 instead of half of it inside the bracket gives (x + 8)² − 67, turning point (−8, −67) — wrong, because only half the coefficient of x belongs inside the bracket. Reading the bracket's sign directly as the turning point's x-coordinate gives (4, −19) — wrong, because (x + 4)² is zero at x = −4, not x = 4. Computing 16 + 3 = 19 instead of −3 − 16 = −19 flips the sign of the constant, giving (−4, 19) — wrong, since the completed square's constant must be evaluated as −3 minus 16, not 16 plus 3. Check by substitution whenever the sign of a constant feels uncertain.
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