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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- (d) 6n + 3 — Method: find the price per passenger, then find the booking fee that fits a booking for 1 passenger. Working: the cost rises by £6 for each extra passenger (15 − 9 = 6, 21 − 15 = 6, 27 − 21 = 6), so the cost has the form 6n + c. Substituting n = 1: 6(1) + c = 9, so c = 3. Answer: the cost in pounds is 6n + 3. 6n ignores the booking fee altogether. 6n + 9 uses the cost of one passenger, £9, as the fee without first subtracting the £6 per-passenger price. 9n + 6 swaps the two numbers, using the first cost as the price per passenger and the difference as the fee.
- (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.
- (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.
- (c) −8 — The nth term is the first term plus (n − 1) lots of the common difference: 40 + 8 × (−6) = 40 − 48 = −8. A candidate who uses 9 lots of the common difference instead of 8 gets 40 + 9 × (−6) = −14. A candidate who treats the common difference as +6 instead of −6 gets 40 + 8 × 6 = 88. A candidate who uses only 7 lots of the common difference gets 40 + 7 × (−6) = −2.
- (d) 12 — Method: call the younger brother's age x, write the elder brother's age in terms of x, and form an equation from the total. Working: the elder brother is x + 6, so x + (x + 6) = 30; simplifying gives 2x + 6 = 30, subtracting 6 from both sides gives 2x = 24, and dividing by 2 gives x = 12. Checking: 12 and 18 add up to 30 and differ by 6. Answer: 12. The distractors: 18 comes from solving correctly and then giving the elder brother's age, which is not the age asked for; 15 comes from halving 30 and ignoring the 6-year difference altogether; 24 comes from taking 6 off the total, 30 − 6 = 24, and giving that as an age.
- (d) y = x² − 10x + 21 — A translation by the vector (3, 0) moves the graph 3 units in the positive x-direction, which means replacing every x in the equation with (x − 3). Substitute into x² − 4x: (x − 3)² − 4(x − 3). Expand (x − 3)² to x² − 6x + 9, and expand −4(x − 3) to −4x + 12. Collecting like terms, x² − 6x + 9 − 4x + 12 = x² − 10x + 21, so the image is y = x² − 10x + 21. Substituting (x + 3) instead of (x − 3) — translating in the wrong direction — gives y = x² + 2x − 3. Adding 3 straight onto the original equation, treating the translation as vertical instead of horizontal, gives y = x² − 4x + 3. Expanding (x − 3)² as x² − 3x + 9, using −3x instead of −6x for the middle term, and then combining with −4(x − 3) gives y = x² − 7x + 21.
- (d) 8(7x − 3) — Method: find the highest common factor of the two terms, write it in front of a bracket and divide each term by it. Working: 56 = 8 × 7 and 24 = 8 × 3, so the highest common factor is 8; dividing gives 56x ÷ 8 = 7x and 24 ÷ 8 = 3, and the subtraction sign stays between them. Answer: 8(7x − 3), which multiplies back out to 56x − 24. The distractors: 8(7x + 3) comes from dropping the minus sign of −24 while dividing; 8(56x − 3) comes from dividing only the number term by 8 and leaving 56x untouched inside the bracket; 8(7x − 16) comes from subtracting 8 from 24 instead of dividing 24 by 8.
- (d) 7n − 2 — Method: find how much the total cost rises each month, then find the constant that fits the cost for one month. Working: the cost rises by £7 for each extra month (12 − 5 = 7, 19 − 12 = 7, 26 − 19 = 7), so the cost has the form 7n + c. Substituting n = 1: 7(1) + c = 5, so c = −2. Answer: the total cost in pounds is 7n − 2. The value 7n comes from leaving out the constant. The value 7n + 5 comes from using the cost of one month as the constant directly, without subtracting the monthly rise first. The value 5n + 7 comes from swapping the roles of the cost of one month, £5, and the monthly rise, £7 — using the cost of one month as the coefficient of n and the rise as the constant.
- (d) 11 — f⁻¹(x) = 5 means x = f(5), since applying f to both sides undoes the inverse. f(5) = 2 × 5 + 1 = 11. Writing 2 comes from confusing f⁻¹(x) = 5 with f(x) = 5, and solving 2x + 1 = 5 instead: 2x = 4, x = 2. Writing 9 comes from finding f⁻¹(x) with a sign error, f⁻¹(x) = (x + 1)/2 instead of (x − 1)/2, then setting this equal to 5: x + 1 = 10, x = 9. Writing 6 comes from finding f⁻¹(x) without dividing by 2 at all, f⁻¹(x) = x − 1, then setting this equal to 5: x = 6.
- (a) Overestimates — trapezium edges lie above the curve. — For a concave-up curve, each chord connecting two points on the curve lies ABOVE the curve between those points, because the curve bends away from the chord underneath it. The trapezium rule uses these chords as the top edges of each trapezium, so the trapeziums cover a larger region than the true area under the curve — the estimate is an overestimate. Saying it underestimates because the edges lie below the curve gets the direction backwards: for a concave-up curve the edges lie above, not below. Saying it overestimates 'because more strips always increase it' gives the right verdict for the wrong reason — using MORE, narrower strips makes a trapezium estimate more accurate and brings it closer to the true area, it does not simply increase the total. Saying the estimate is exact ignores that straight edges cannot follow a curved line perfectly; there is always a gap between the chord and the curve on a bend, which is exactly what makes the rule an estimate rather than an exact calculation.
- (b) x = 2, y = −5 — 2x² − 8x + 3 rewrites as 2(x² − 4x) + 3, then as 2[(x − 2)² − 4] + 3, which simplifies to 2(x − 2)² − 5, since −2 × 4 + 3 = −5. Substituting x = 2: 2 × 2² = 8, 8 × 2 = 16, so 8 − 16 + 3 = −5, confirming the minimum value −5 at x = 2: turning point x = 2, y = −5. Halving b instead of halving b/a — using 4 as the shift instead of 2 — lands on turning point x = 4, y = −29, which is wrong because when a ≠ 1 the shift inside the bracket is b/(2a), not b/a alone. Reading the bracket's sign directly as the turning point's x-coordinate gives x = −2, y = −5 — wrong, because (x − 2)² is zero at x = 2, not x = −2. Computing 8 − 3 = 5 instead of 3 − 8 = −5 flips the sign of the constant, giving x = 2, y = 5 — wrong, since the completed square's constant must be evaluated as 3 minus 8, not 8 minus 3. Whenever a ≠ 1, factor a out of the x² and x terms first, and always check a turning point by substituting back into the original equation.
- (c) x ≥ 3 — Method: collect the number terms first; the x term is negative, so the final step multiplies both sides by −1, and that is the one step that turns the inequality sign round. Working: subtracting 5 from both sides of 5 − x ≤ 2 gives −x ≤ −3; multiplying both sides by −1 turns −x into x and −3 into 3, and because the multiplier is negative the ≤ becomes ≥, so x ≥ 3. Answer: x ≥ 3. The distractors: x ≤ 3 comes from multiplying by −1 without turning the sign round, the commonest slip on this type; x ≤ −3 comes from reading −x ≤ −3 as though the minus sign could simply be rubbed off the left-hand side; x ≥ −3 comes from turning the sign round correctly but leaving the right-hand side at −3 instead of multiplying it by −1 as well.
- (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.
- (b) 14 m/s — Average speed for a whole journey is total distance divided by total time: 350 ÷ 25 = 14 m/s. Multiplying the distance and time instead of dividing gives 350 × 25 = 8750 m/s. Adding the distance and time instead of dividing gives 350 + 25 = 375 m/s. Inverting the division, working out time divided by distance, gives 25 ÷ 350, which rounds to 0.07 m/s.
- (c) 5 — Method: let a be the number of adult tickets, so the number of child tickets is 9 − a. Form the equation 10a + 6(9 − a) = 74. Working: expand the bracket: 10a + 54 − 6a = 74, so 4a = 20, giving a = 5. Answer: 5 adult tickets. 4 comes from correctly finding the number of child tickets but reporting it instead of the number of adult tickets asked for. 1.25 comes from a sign error expanding the bracket, writing 10a + 54 + 6a = 74 instead of subtracting, which gives 16a = 20. 7.4 comes from dividing the total takings by the adult ticket price only, ignoring the children entirely, 74 ÷ 10.
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