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) 3 and 4 — Method: the graph of f(x) is continuous, so where it crosses the x-axis the value of f(x) changes sign; substitute consecutive integers until one value is negative and the next is positive. Working: f(2) = 8 − 6 − 20 = −18, f(3) = 27 − 9 − 20 = −2 and f(4) = 64 − 12 − 20 = 32. The sign changes from negative to positive between x = 3 and x = 4, so the solution lies there. Answer: 3 and 4. The distractors: 2 and 3 comes from ignoring the −3x term and solving x³ = 20, whose root is 2.71, one interval to the left; 6 and 7 comes from reading x³ as x² and solving x² − 3x − 20 = 0, whose positive root is 6.22; 4 and 5 is the interval immediately after the change of sign, named by a candidate who finds f(4) positive and quotes the interval beginning there instead of the one across which the sign actually turned.
- (d) 37.0 — C = 5(98.6 − 32) ÷ 9 = 5 × 66.6 ÷ 9 = 333 ÷ 9 = 37.0. A candidate who forgets to subtract 32 first gets 5 × 98.6 ÷ 9 = 54.8 (1 d.p.). A candidate who forgets the 5 ÷ 9 factor entirely and just works out F − 32 gets 66.6. A candidate who multiplies by 9 ÷ 5 instead of 5 ÷ 9 gets 66.6 × 9 ÷ 5 = 119.9 (1 d.p.).
- (c) 81 m — Split the velocity-time graph into its three phases. The accelerating phase (0 to 4 s) is a triangle: 1/2 × 4 × 6 = 12. The constant phase (4 s at 6 m/s, for 10 s) is a rectangle: 10 × 6 = 60. The decelerating phase (3 s) is a triangle: 1/2 × 3 × 6 = 9. Total distance: 12 + 60 + 9 = 81 m. Forgetting the final decelerating phase entirely and adding only the first two areas gives 12 + 60 = 72 m. Treating all three phases as rectangles, forgetting to halve either triangular phase, gives 4 × 6 + 10 × 6 + 3 × 6 = 102 m. Halving only the deceleration triangle correctly but treating the acceleration phase as a rectangle too gives 4 × 6 + 60 + 9 = 93 m.
- (a) Subtracting 4x gives 3 = 10, which is never true. — Method: try to solve the equation as normal and see what happens. Working: subtract 4x from both sides: 4x + 3 − 4x = 4x + 10 − 4x, giving 3 = 10. This statement is false for every value of x, so the equation has no solution. Answer: subtracting 4x gives 3 = 10, which is never true. "x would have to be negative" invents a constraint on x that the equation never states. "It's true for every x" confuses this equation with an identity, where both sides would simplify to the same expression. "x = 7" misreads the false statement 3 = 10 as something to solve for x, rather than recognising it means no solution exists.
- (d) £16.70 — The mileage charge is 2.20 × 6 = £13.20. Adding the fixed fee: £13.20 + £3.50 = £16.70. A candidate who forgets the fixed fee gives just the mileage charge, £13.20. A candidate who adds the fixed fee to the per-mile rate before multiplying by the number of miles, (3.50 + 2.20) × 6, gets £34.20. A candidate who rounds £2.20 down to £2 gets 2 × 6 + 3.50 = £15.50.
- (d) 110 — Method: form the equation 15 + 0.08m = 23.80, where m is the number of minutes, then solve for m. Working: subtract the fixed fee: 0.08m = 23.80 − 15 = 8.80. Divide by the cost per minute: m = 8.80 ÷ 0.08 = 110. Answer: 110 minutes. 1.1 comes from using 8 instead of 0.08 as the cost per minute, forgetting to convert pence to pounds. 297.5 comes from dividing the whole bill by the cost per minute without subtracting the fixed fee first, 23.80 ÷ 0.08. 485 comes from adding the fixed fee to the bill instead of subtracting it, before dividing by the cost per minute, (23.80 + 15) ÷ 0.08.
- (c) x + 3 — Expand the bracket: 0.5(4x + 6) = 2x + 3. Then subtract the x: 2x + 3 − x = x + 3. The option 2x + 3 comes from expanding the bracket correctly but then forgetting to subtract the x at all. The option x + 6 comes from forgetting to multiply the 6 inside the bracket by 0.5 (treating it as 2x + 6), then subtracting x. The option 3x + 3 comes from adding the x instead of subtracting it: 2x + 3 + x = 3x + 3.
- (a) 48 litres — Split the area into three sections. The rectangle (t = 0 to 2) has area 2 × 12 = 24. The first trapezium (t = 2 to 5, parallel sides 12 and 3, width 3) has area 1/2 × (12 + 3) × 3 = 22.5. The second trapezium (t = 5 to 6, parallel sides 3 and 0, width 1) has area 1/2 × (3 + 0) × 1 = 1.5. Total volume: 24 + 22.5 + 1.5 = 48 litres. Forgetting the rectangle and adding only the two trapeziums gives 22.5 + 1.5 = 24 litres. Using a single trapezium across the whole 6 minutes, with parallel sides 12 and 0, ignoring that the first 2 minutes are constant, gives 1/2 × (12 + 0) × 6 = 36 litres. Correctly finding the rectangle and the first trapezium but forgetting to halve the second trapezium, using (3 + 0) × 1 = 3 instead of 1.5, gives 24 + 22.5 + 3 = 49.5 litres.
- (d) 6.0 — Subtract 1.8 from both sides: 6.3x ≤ 38.2. Divide both sides by 6.3: x ≤ 6.0634… . The question asks for the greatest value to 1 decimal place that still satisfies the inequality. Testing 6.1: 6.3 × 6.1 + 1.8 = 40.23, which is more than 40, so 6.1 fails. Testing 6.0: 6.3 × 6.0 + 1.8 = 39.6, which is no more than 40, so 6.0 works and is the greatest such value. A candidate who simply rounds 6.0634… to 1 decimal place answers 6.1, without checking that it satisfies the inequality. A candidate who adds 1.8 instead of subtracting works out 41.8 ÷ 6.3 and answers 6.6. A candidate who forgets the 1.8 altogether divides 40 by 6.3 and answers 6.3.
- (a) 1.861 — x₁ = ∛(10 − 2²) = ∛6 = 1.817120593. x₂ = ∛(10 − 1.817120593²) = ∛6.698072751 = 1.885022855. x₃ = ∛(10 − 1.885022855²) = ∛6.446688837 = 1.861139399, which rounds to 1.861. Reporting x₂ instead of x₃ gives 1.885022855, which rounds to 1.885. Stopping after the first iteration and reporting x₁ instead of x₃ gives 1.817120593, which rounds to 1.817. A sign error inside the cube root, using xₙ₊₁ = ∛(10 + xₙ²) instead of ∛(10 − xₙ²), gives x₁ = ∛14 = 2.410142264, x₂ = ∛(10 + 2.410142264²) = 2.509763724, and x₃ = ∛(10 + 2.509763724²) = 2.535437381, which rounds to 2.535.
- (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 = 5 — Area = length × width, so x(x + 3) = 40, which rearranges to x² + 3x − 40 = 0. This factorises as (x + 8)(x − 5) = 0: the two numbers in the brackets must multiply to −40 and add to +3, and the pair 8 and −5 does both. This gives x = −8 or x = 5. Since x is a length, it cannot be negative, so x = 5. A candidate who gives both solutions without rejecting the negative one, which cannot be a length, answers x = 5 or x = −8. A candidate who picks the wrong factor pair of 40, such as 10 and −4 instead of 8 and −5, gets (x + 10)(x − 4) = 0 and answers x = 4. A candidate who rejects the wrong root, keeping the negative solution instead of the positive one, answers x = −8.
- (d) x + y = 7 — The midpoint of AB is (−1 + 5)/2, (2 + 8)/2, which is (2, 5). The gradient of AB is (8 − 2)/(5 − (−1)) = 6/6 = 1. The perpendicular bisector has gradient −1 and passes through (2, 5): y − 5 = −(x − 2), which rearranges to x + y = 7. Distractor routes: x − y = −3 uses the gradient of AB itself, 1, rather than its negative reciprocal, giving a line PARALLEL to AB through the midpoint instead of perpendicular to it. x + y = 12 comes from adding the y-coordinates, 2 + 8 = 10, but forgetting to divide by 2, using the midpoint (2, 10) instead of (2, 5). x + y = 4 comes from using A's x-coordinate, −1, directly instead of averaging it with B's, giving the point (−1, 5) instead of the true midpoint (2, 5).
- (d) x = −2, y = −5 — 3x² + 12x + 7 rewrites as 3(x² + 4x) + 7, then as 3[(x + 2)² − 4] + 7, which simplifies to 3(x + 2)² − 5, since −3 × 4 + 7 = −5. Substituting x = −2: 3 × (−2)² = 12, 12 × (−2) = −24, so 12 − 24 + 7 = −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 = −41, 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 12 − 7 = 5 instead of 7 − 12 = −5 flips the sign of the constant, giving x = −2, y = 5 — wrong, since the completed square's constant must be evaluated as 7 minus 12, not 12 minus 7. 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.
- (b) x² + y² = 1369 — For a circle centred at the origin, the radius squared equals the sum of the squares of the coordinates of any point on it: r² = 12² + 35² = 144 + 1225 = 1369. The equation is x² + y² = 1369. x² + y² = 2209 comes from adding the coordinates first and then squaring the sum: (12 + 35)² = 47² = 2209, instead of squaring each coordinate separately. x² + y² = 1225 comes from using only 35² and leaving out the 12² term. x² + y² = 144 comes from using only 12² and leaving out the 35² term.
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