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) y = 1/x — A graph with two separate curved branches must have a value of x that is excluded from it, creating a break in the curve. For y = 1/x, x = 0 is excluded, so the graph splits into one branch where x and y are both positive, and one where x and y are both negative. y = x² is a single U-shaped curve with no break, even though it has two 'arms'. y = x³ is one continuous curve that passes through the origin without any break. y = 2x + 1 is a single straight line; even though it passes through regions where x and y share the same sign, it is one continuous line, not two separate branches.
- (a) (0, 7) and (0, −7) — A tangent is horizontal where the radius to that point is vertical, i.e. where the point lies on the y-axis. On x² + y² = 49, setting x = 0 gives y² = 49, so y = 7 or y = −7. The points are (0, 7) and (0, −7). (7, 0) and (−7, 0) comes from swapping the condition — these are the points where the tangent is VERTICAL, not horizontal (the radius there is horizontal). (0, 7) only comes from finding one valid point but forgetting that y² = 49 also gives the negative root, y = −7. (7, 0) only combines both mistakes: the wrong axis, and only one of the two roots.
- (b) n² + 2n + 3 — First differences: 5, 7, 9, 11. Second differences: 2, 2, 2, so a = 2 ÷ 2 = 1. Subtracting n² (1, 4, 9, 16, 25) from the terms (6, 11, 18, 27, 38) leaves 5, 7, 9, 11, 13, which is the linear expression 2n + 3. So the nth term is n² + 2n + 3. Using the second difference itself as a, without halving it, gives 2n² + 2n + 3. Finding a = 1 correctly but then dropping the linear part 2n, keeping only the constant, gives n² + 3. Finding a = 1 correctly but dropping the constant +3 gives n² + 2n.
- (a) 12.5 — Method: the tangent at A is perpendicular to the radius OA, so find the gradient of OA, take its negative reciprocal, write the equation of the tangent and find where it meets the y-axis; the length of OB is then the distance of that crossing from the origin. Working: OA runs from (0, 0) to (−6, 8), so its gradient is 8 ÷ (−6), which cancels to −4/3; the negative reciprocal of −4/3 is 3/4. Substituting into y − 8 = 3/4(x + 6) gives y = 0.75x + 4.5 + 8, so y = 0.75x + 12.5 and B is (0, 12.5). The length OB is therefore 12.5. Answer: 12.5. The distractors: 10 is the radius of the circle, quoted on the assumption that the tangent always meets an axis one radius from the centre, which is only true when the radius itself lies along that axis; 8 is the y-coordinate of A, quoted by treating the tangent as horizontal so that it keeps the height of A; 3.5 comes from turning the gradient of OA upside down without changing its sign, which gives y = −0.75x + 3.5.
- (d) 8 — Method: write the nth term of the sequence, 2000 + 800(n − 1), and find the smallest whole n for which it is greater than 7000. Working: 2000 + 800(n − 1) > 7000, so 800(n − 1) > 5000, giving n − 1 > 6.25. Since n − 1 must be a whole number, the smallest value is 7, so n = 8. Check: year 8's total is 2000 + 800 × 7 = 7600, which is more than £7000, while year 7's total is 2000 + 800 × 6 = 6800, which is not. Answer: year 8. 7 comes from rounding 6.25 to the nearest whole number, 6, and then adding 1, instead of rounding up to the next whole number before adding 1. 6 comes from using 6.25 rounded down to 6 as the year number directly, without adding the 1 needed to convert from the number of increases to the year number. 9 comes from adding one extra year beyond the year that already satisfies the condition.
- (a) x = 4 or x = −3 — Method: find two numbers that multiply to give −12 and add to give −1 — these are −4 and 3. So x² − x − 12 = (x − 4)(x + 3) = 0, giving x = 4 or x = −3. Distractor origins: x = −4 or x = 3 has the signs the wrong way round; x = 4 or x = 3 makes both roots positive, ignoring the sign of −12; x = 12 or x = −1 comes from reading off the coefficient and the constant directly instead of factorising.
- (d) x² − 6x + 9 — Method: squaring a bracket means multiplying that bracket by itself, so expand (x − 3)(x − 3) term by term and then collect like terms. Working: x × x = x², x × (−3) = −3x, (−3) × x = −3x and (−3) × (−3) = 9, giving x² − 3x − 3x + 9, and the two middle terms collect to −6x. Answer: x² − 6x + 9. The distractors: x² − 3x + 9 comes from writing down only one of the two middle products instead of both; x² + 6x + 9 comes from treating (−3) × x as +3x, so the middle terms are added rather than subtracted; x² − 9 comes from treating the square as the difference of two squares (x − 3)(x + 3).
- (d) 1 — Method: find the step in the outputs for each step of 1 in the input, write the rule from that step and from one pair of values, then apply the rule to the last input. Working: the outputs −5, −3, −1 rise by 2 while x rises in ones, so x is multiplied by 2; at x = 1, 2 × 1 = 2 while y = −5, so 7 is subtracted, giving y = 2x − 7; at x = 4 the rule gives 2 × 4 = 8 and 8 − 7 = 1. Answer: y = 1. The distractors: 3 comes from carrying the outputs on one step too far, to x = 5; 0 comes from assuming the outputs −5, −3, −1 carry on by adding 1 rather than by adding 2; 8 comes from doubling the input and forgetting to subtract the 7.
- (d) 2√2 — The radius satisfies r² = 8, so r = √8 = √(4 × 2) = √4 × √2 = 2√2. Choosing 8 forgets to take the square root of r² at all. Choosing 4 comes from halving 8 instead of finding its square root. Choosing √2 splits off the factor of 4 correctly but forgets to multiply the 2 back in front of the root.
- (b) x = 3, y = 4 — −x² + 6x − 5 = −(x² − 6x) − 5 = −[(x − 3)² − 9] − 5 = −(x − 3)² + 9 − 5 = −(x − 3)² + 4. Because the coefficient of x² is negative, −(x − 3)² is at most zero, so this turning point is a maximum. Substituting x = 3 gives (x − 3)² = 0, so y = 4, confirming the maximum value 4 at x = 3: turning point (3, 4). Using 6 instead of half of 6 inside the bracket — forgetting to halve before completing the square — lands on turning point (6, 31), wrong, because only half the coefficient belongs inside the bracket. Reading the bracket's sign directly as the turning point's x-coordinate gives (−3, 4) — wrong, because (x − 3)² is zero at x = 3, not x = −3. Computing 5 − 9 = −4 instead of 9 − 5 = 4 flips the sign of the constant, giving (3, −4) — wrong, since the constant must be evaluated as 9 minus 5, not 5 minus 9. Whenever the leading coefficient is negative, the turning point is a maximum, not a minimum — check by substituting back into the original equation.
- (a) y = −2x + 7 — Using y = −2x + c, and substituting the point (3, 1): 1 = −2(3) + c, so 1 = −6 + c, and c = 7, giving y = −2x + 7. A candidate who uses the y-coordinate of the point directly as the y-intercept, instead of solving for c, would write y = −2x + 1. A candidate who drops the negative sign on the gradient would write y = 2x + 7. A candidate who makes a sign error when isolating c, writing c = 1 − 6 = −5 instead of c = 1 + 6 = 7, would write y = −2x − 5.
- (c) 5n + 1 — Method: find the common difference, then use it as the coefficient of n in the position-to-term rule, and find the constant by checking against the first term. Working: the common difference is 5, so the rule has the form 5n + c. Using the 1st term: 5(1) + c = 6, so c = 1. The rule is 5n + 1. Answer: 5n + 1. 5n − 1 uses the correct coefficient but the wrong sign for the constant. 6n comes from using the first term as the coefficient of n instead of the common difference — it matches the 1st term by coincidence but fails from the 2nd term onward. n + 5 swaps the coefficient and the constant around, using the common difference as the constant instead of the coefficient of n.
- (c) 5n + 7 — Method: find the common difference between the rows, then find the constant by adjusting the first row's total. Working: each row has 5 more seats than the last, so the coefficient of n is 5. The constant is the first row's total minus the common difference: 12 − 5 = 7. Answer: the nth term is 5n + 7. 5n + 12 comes from using row 1's total, 12, as the constant without subtracting the common difference. 5n + 2 comes from a slip in working out the constant, subtracting the common difference twice (12 − 5 − 5 = 2) instead of once. 7n + 5 comes from swapping the common difference and the constant.
- (a) x + 14 — Expand each bracket separately: 3(x + 4) = 3x + 12, and −2(x − 1) = −2x + 2 (multiply −2 by both x and −1). Combine: 3x + 12 − 2x + 2 = x + 14. Writing x + 10 comes from taking −2(x − 1) as −2x − 2, not flipping the sign of the −1 inside the bracket. Writing 5x + 10 comes from treating the second bracket as +2(x − 1) instead of subtracting it, so the x-terms are added rather than subtracted. Writing x + 13 comes from only multiplying the 2 by the x, and carrying the −1 across unmultiplied.
- (a) 45 — The ratio between the two given points, one power of x apart, gives b: 405 ÷ 135 = 3, so b = 3. Substituting back, at x = 1, y = A × b, so 135 = A × 3, giving A = 45. Giving the common ratio b itself instead of A confuses which unknown was asked for and produces 3 — wrong, because the question asks for A, not b. Giving 135 instead treats the first given point as the y-intercept and reads A off it directly — wrong, because that point is at x = 1, not x = 0, so 135 is A × b, not A. Assuming A equals b⁰ = 1 by itself, rather than substituting a known point to solve for A, gives 1 — wrong, because b⁰ is always 1 regardless of A; A must be found using an actual (x, y) pair from the graph.
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