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) 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) 3.5 km/h — Average speed is total distance ÷ total time. Total distance = 5 + 9 = 14 km. Total time, including the rest, = 1 + 1 + 2 = 4 hours. So average speed = 14 ÷ 4 = 3.5 km/h. Leaving out the 1 hour rest and dividing by only the 3 hours of walking gives 14 ÷ 3 = 4.67 km/h. Averaging the two separate speeds, 5 km/h and 4.5 km/h, instead of using total distance over total time, gives (5 + 4.5) ÷ 2 = 4.75 km/h. Dividing only the second leg's distance by the total time, 9 ÷ 4 = 2.25 km/h, ignores the first leg's distance.
- (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.
- (a) £50 — Substituting x = 200 into y = 0.15x + 20 gives y = 0.15 × 200 + 20 = 30 + 20 = £50. A candidate who forgets to add the standing charge would get only 0.15 × 200 = £30. A candidate who misplaces the decimal point in the rate, using 1.5 instead of 0.15, would get 1.5 × 200 + 20 = £320. A candidate who swaps the roles of the rate and the number of units would work out 0.15 × 20 + 200 = £203.
- (a) 36 km/h — Speed is distance ÷ time. Convert 25 minutes to hours: 25 ÷ 60 = 5/12 hours. Then 15 ÷ (5/12) = 15 × 12/5 = 36 km/h. Dividing 15 by 25 without converting minutes to hours gives 0.6 km/h. Multiplying 15 by 25/60 instead of dividing by it gives 6.25 km/h. Working out the speed in km per minute, 0.6, and then multiplying by 24 instead of 60 gives 14.4 km/h.
- (d) 4.521 — x₁ = √(3 × 4 + 7) = √19 = 4.358898944. x₂ = √(3 × 4.358898944 + 7) = √20.076696833 = 4.480702716. x₃ = √(3 × 4.480702716 + 7) = √20.442108148 = 4.521294964, which rounds to 4.521. Mislabelling the starting value x₀ as x₁, so that the working stops one iteration too early, reports the true x₂ = 4.480702716, which rounds to 4.481. Working out one iteration too many reports the true x₄ = √(3 × 4.521294964 + 7) = 4.534741987, which rounds to 4.535. Applying the formula in the wrong order, calculating √(3xₙ) + 7 at every step instead of √(3xₙ + 7), gives, from x₀ = 4: √12 + 7 = 10.464101615, then √(3 × 10.464101615) + 7 = 12.602883619, then √(3 × 12.602883619) + 7 = 13.148873950, which rounds to 13.149.
- (b) 3.38 — r₁ = √(300 ÷ (π × 8)) = √11.9366 = 3.4550. r₂ = √(300 ÷ (π × 8.4550)) = √11.2947 = 3.3608. r₃ = √(300 ÷ (π × 8.3608)) = √11.4232 = 3.3798, which rounds to 3.38. Choosing 3.36 stops at r₂, one iteration too early. Choosing 4.82 leaves out the '+ 5' inside the bracket, dividing by π × rₙ instead of π × (rₙ + 5). Choosing 3.45 comes from using π ≈ 3 instead of the calculator's π key throughout.
- (b) £2.00 — Subtracting the second equation from the first eliminates the pastries: (3c + 2p) − (2c + 2p) = 9.60 − 7.60, so c = 2.00. A candidate who divides the first total by the number of coffees alone, ignoring the pastries, would get 9.60 ÷ 3 = £3.20. A candidate who finds the price of a pastry instead of a coffee — using c = 2.00 in 2c + 2p = 7.60 to get p = 1.80 — would answer £1.80. A candidate who reaches the correct difference of £2.00 but then mistakenly divides again or misplaces the decimal point would get £0.20.
- (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.
- (d) 12 — Since a₃ = a₂ + a₁, rearranging gives a₂ = a₃ − a₁ = 17 − 5 = 12. 22 comes from adding instead of subtracting: a₃ + a₁ = 17 + 5 = 22. 11 comes from treating the sequence as if it were arithmetic and averaging the two given terms: (5 + 17) ÷ 2 = 11 — but a Fibonacci-type sequence isn't arithmetic, so this doesn't apply. −12 comes from subtracting in the wrong order: a₁ − a₃ = 5 − 17 = −12.
- (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.
- (c) 14.1 — Method: in x² + y² = r² the right-hand side is the square of the radius, so take its square root to find the radius, then double the radius because the diameter is twice the radius. Working: r² = 50, so r = √50 = 7.07106…, and the diameter is 2 × 7.07106… = 14.14213…, which is 14.1 correct to 1 decimal place. Answer: 14.1. The distractors: 7.1 is the radius, worked out correctly but never doubled, so it answers a question about the radius rather than the diameter; 100.0 comes from doubling the 50 on the right-hand side, treating r² as though it were already the radius; 25.0 comes from halving the 50, treating r² as though it were already the diameter.
- (c) 72 m — The first section is a triangle with base 4 s and height 8 m/s, area = 1/2 × 4 × 8 = 16 m. The middle section is a rectangle with base 6 s and height 8 m/s, area = 6 × 8 = 48 m. The final section is a triangle with base 2 s and height 8 m/s, area = 1/2 × 2 × 8 = 8 m. Total distance = 16 + 48 + 8 = 72 m. Treating both sloping sections as full rectangles instead of triangles — forgetting the factor of a half — gives 4 × 8 + 6 × 8 + 2 × 8 = 96 m, which is wrong because a sloping speed-time section is a triangle, not a rectangle. Leaving out the final, decreasing section entirely gives 16 + 48 = 64 m, which is wrong because the cyclist is still moving and covering distance while slowing down. Leaving out the middle, constant-speed section gives 16 + 8 = 24 m, which is wrong because the cyclist covers the most distance of all during the 6 seconds of constant speed.
- (c) 12 cm — For a square, area = side². So side² = 144, giving side = ±12. Since a length must be positive, the side length is 12 cm. A candidate who gives both square roots without rejecting the negative one, which cannot be a length, answers 12 cm or −12 cm. A candidate who halves 144 instead of taking its square root gets 72 cm. A candidate who divides 144 by 4, confusing the area formula with a perimeter calculation, gets 36 cm.
- (b) (2, 4) and (−2, −4) — Substitute y = 2x into x² + y² = 20: x² + (2x)² = 20, which gives x² + 4x² = 20, so 5x² = 20, x² = 4, and x = 2 or x = −2. Using y = 2x for each x-value: x = 2 gives y = 4; x = −2 gives y = −4. The path meets the pond's edge at (2, 4) and (−2, −4). Distractor routes: (2, −4) and (−2, 4) swaps the sign pairing, matching each x-value with the wrong sign of y instead of keeping each x with its own correctly-signed y. (2, 1) and (−2, −1) comes from using y = x/2 instead of y = 2x when finding the y-coordinates. (2, 4) alone stops after the positive square root of x² = 4 and never finds the second point from x = −2.
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