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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Algebra worksheet — GCSE Higher
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- 1.The equation x³ + 4x − 9 = 0 is to be solved by iteration. Work out which one of these iterative formulas comes from a correct rearrangement of that equation.
- 2.The equation x³ − 3x − 4 = 0 has a root near x = 2. Four students each try a different iterative formula, all starting from x₀ = 2: xₙ₊₁ = ∛(3xₙ + 4); xₙ₊₁ = (xₙ³ − 4) ÷ 3; xₙ₊₁ = 4 ÷ (xₙ² − 3); xₙ₊₁ = xₙ³ − 2xₙ − 4. Only one of these formulas keeps producing values that settle near the root when it is repeated. Work out x₁, correct to 3 decimal places, for the formula that does this.
- 3.A circular pond has equation x² + y² = 20, with lengths in metres from the centre of the garden. A straight path runs along the line y = 2x, entering the pond and leaving it again. Work out the coordinates of the two points where the path meets the edge of the pond.y = 2x
- 4.A tap's flow rate, in litres per minute, is plotted against time, in minutes, on a graph. What are the units of the area under this graph?
- 5.A car's speed-time graph shows the following: its speed increases steadily from 0 m/s to 20 m/s over the first 10 seconds, then stays constant at 20 m/s for the next 15 seconds. Work out the total distance travelled in the first 25 seconds.
- 6.Two brothers have ages that add up to 30. The elder brother is 6 years older than the younger brother. Work out the age of the younger brother.
- 7.An allotment is in the shape of a rectangle. Its length is 5 m more than its width, x metres, and its area is 20 m². This gives x² + 5x − 20 = 0, which can be solved using the iterative formula xₙ₊₁ = 20 ÷ (xₙ + 5). Taking x₀ = 2, so that x₁ is the value found after the formula has been used once, work out x₂ correct to 2 decimal places.
- 8.The graph of W = 840 ÷ L shows how the width, W metres, of a rectangular field of area 840 m² depends on its length, L metres. Use the relationship to work out the width when the length is 35 m, and hence the perimeter of the field.
- 9.A tree is 2 metres tall. Each year its height increases by 10% of its height at the start of that year. Work out the height of the tree after 3 years, giving your answer to 1 decimal place.
- 10.A ball's height, h metres, t seconds after being thrown follows h = (t − 1)(9 − t). Given that the ball is at ground level at t = 1 and t = 9, work out at what time t the ball reaches its maximum height, using symmetry.
- 11.The formula for converting a temperature in Fahrenheit, F, to Celsius, C, is C = 5(F − 32) ÷ 9. Work out C, to 1 decimal place, when F = 98.6.
- 12.A cuboid has a square base of side x metres and a height that is 3 m more than x. Its volume is 150 m³. This gives the equation x³ + 3x² − 150 = 0, which can be solved using the iterative formula xₙ₊₁ = ∛(150 − 3xₙ²). Taking x₀ = 4, work out x₂ correct to 2 decimal places.
- 13.A ferry's distance-time graph is a straight line from (0 minutes, 0 km) to (25 minutes, 15 km). Work out the ferry's speed in kilometres per hour.
- 14.The iterative formula xₙ₊₁ = √(2xₙ + 15) is used repeatedly, starting from x₀ = 1. Work out the value that xₙ approaches, correct to 2 decimal places.
- 15.A car accelerates uniformly from rest at 2.5 m/s² until it reaches a speed of 20 m/s, then travels at this constant speed for a further 30 seconds. Work out the total distance travelled.
Answer key
- (c) xₙ₊₁ = ∛(9 − 4xₙ) — Method: a formula xₙ₊₁ = f(xₙ) is a correct rearrangement when the equation x = f(x) turns back into the equation you started with, so rearrange x³ + 4x − 9 = 0 by making the cube the subject. Working: x³ + 4x − 9 = 0 gives x³ = 9 − 4x, because the 4x and the 9 each change sign as they cross the equals sign; taking the cube root of both sides gives x = ∛(9 − 4x), which is the formula xₙ₊₁ = ∛(9 − 4xₙ). Answer: xₙ₊₁ = ∛(9 − 4xₙ). The distractors: ∛(9 + 4xₙ) comes from writing x³ = 9 + 4x, moving the 4x across the equals sign without changing its sign; (9 + xₙ³)/4 comes from making the linear term the subject but keeping the sign of the cube, writing 4x = 9 + x³ when the equation gives 4x = 9 − x³; ∛(9 − 4xₙ³) cubes the x in the linear term as well, changing a term the original equation never cubed.
- (a) 2.154 — Continuing xₙ₊₁ = ∛(3xₙ + 4) from x₁ = 2.154 gives x₂ = 2.187, x₃ = 2.195, settling towards the root near 2.196 — this is the formula that converges. Continuing xₙ₊₁ = (xₙ³ − 4) ÷ 3 from x₁ = 1.333 gives x₂ = −0.543 and then x₃ = −1.387, moving further from the root each time. Continuing xₙ₊₁ = 4 ÷ (xₙ² − 3) from x₁ = 4 gives x₂ = 0.308 and then x₃ = −1.377, swinging wildly rather than settling. Continuing xₙ₊₁ = xₙ³ − 2xₙ − 4 from x₁ = 0 gives x₂ = −4 and then x₃ = −60, running away from the root entirely.
- (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.
- (a) Litres — The area under a graph is found by multiplying a y-value by an x-value, so its units are the y-axis units multiplied by the x-axis units: litres per minute × minutes = litres, since the 'per minute' cancels with the 'minutes'. Answering 'litres per minute' keeps the y-axis units unchanged, as if multiplying by time did nothing to the units at all. Answering 'litres per minute squared' treats the x-axis as also being measured 'per minute', squaring a unit that should instead cancel. Answering 'minutes' keeps only the x-axis units and drops the rate altogether.
- (d) 400 m — The distance travelled equals the area under the speed-time graph. The first 10 seconds form a triangle with base 10 and height 20, giving an area of 0.5 × 10 × 20 = 100 m. The next 15 seconds form a rectangle with base 15 and height 20, giving an area of 15 × 20 = 300 m. The total distance is 100 + 300 = 400 m. The option 500 m treats the whole 25 seconds as travelled at the constant 20 m/s, ignoring that the speed was building up during the first 10 seconds: 25 × 20 = 500. The option 300 m only counts the constant-speed section and forgets the triangle section entirely. The option 200 m comes from working out the triangle's area without halving it (10 × 20 = 200) and forgetting the rectangle altogether.
- (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.
- (a) 2.55 — x₁ = 20 ÷ (2 + 5) = 20 ÷ 7 = 2.857142857. x₂ = 20 ÷ (2.857142857 + 5) = 20 ÷ 7.857142857 = 2.545454545, which rounds to 2.55. Reporting x₁ instead of x₂ gives 2.857142857, which rounds to 2.86. Dropping the +5 in the denominator, using xₙ₊₁ = 20 ÷ xₙ, gives x₁ = 20 ÷ 2 = 10 and x₂ = 20 ÷ 10 = 2, which is 2.00. A sign error in the denominator, using xₙ₊₁ = 20 ÷ (xₙ − 5), gives x₁ = 20 ÷ (2 − 5) = −6.666666667 and x₂ = 20 ÷ (−6.666666667 − 5) = −1.714285714, which rounds to −1.71.
- (b) 118 m — Width = 840 ÷ 35 = 24 m. Perimeter = 2 × (length + width) = 2 × (35 + 24) = 2 × 59 = 118 m. The option 59 m gives the sum of the length and width but forgets to double it for the perimeter. The option 70 m doubles only the length (2 × 35 = 70) and leaves out the width entirely. The option 48 m doubles only the width (2 × 24 = 48) and leaves out the length entirely.
- (d) 2.7 — The height after 3 years of 10% compound growth is 2 × 1.1³ = 2.662 m, which rounds to 2.7 m. A candidate who adds 10% of the original height (0.2 m) in each of the 3 years, instead of compounding on the new height each time, would reach 2 + 3×0.2 = 2.6 m. A candidate who compounds for only 2 years would reach 2 × 1.1² = 2.42 m, rounding to 2.4 m. A candidate who compounds for 4 years instead of 3 would reach 2 × 1.1⁴ = 2.928 m, rounding to 2.9 m.
- (d) t = 5 — The maximum height occurs halfway between the two times when the ball is at ground level: the midpoint of t = 1 and t = 9 is (1 + 9) ÷ 2 = 5, so the ball reaches its maximum height at t = 5 seconds. Getting t = 4 comes from halving the DIFFERENCE between the times, 9 − 1 = 8, then 8 ÷ 2 = 4, instead of finding their midpoint. Getting t = 8 uses that difference, 9 − 1 = 8, as if the gap between the two times were itself the time of the maximum. Getting t = 10 adds the two times, 1 + 9 = 10, but forgets to divide by 2.
- (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.).
- (b) 4.39 — x₁ = ∛(150 − 3 × 4²) = ∛(150 − 48) = ∛102 = 4.672 (unrounded). x₂ = ∛(150 − 3 × 4.672²) = ∛(150 − 65.49) = ∛84.51 = 4.39 (2 d.p.). Choosing 4.67 stops after only one iteration, giving x₁ instead of x₂. Choosing 84.51 finds the value inside the cube root for x₂ but never takes the cube root. Choosing 6.32 comes from adding 3xₙ² instead of subtracting it inside the root, which does not match the given formula.
- (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.
- (b) 5.00 — Continuing the iteration: x₁ = √(2 × 1 + 15) = √17 = 4.1231, x₂ = √(2 × 4.1231 + 15) = √23.2462 = 4.8214, x₃ = √(2 × 4.8214 + 15) = √24.6428 = 4.9642, x₄ = √(2 × 4.9642 + 15) = √24.9284 = 4.9928, and the values keep climbing towards 5.00 as n increases (the limit L satisfies L² = 2L + 15, so L² − 2L − 15 = 0, giving L = 5). Choosing 4.99 stops after x₄, one iteration before the value has settled fully to 5.00. Choosing 17.00 uses the value under the very first square root (2 × 1 + 15 = 17) as if that number itself were the limit. Choosing 1.00 assumes the sequence never moves from the starting value x₀.
- (d) 680 m — First find how long the acceleration takes: acceleration = change in speed ÷ time, so 2.5 = 20 ÷ t, giving t = 20 ÷ 2.5 = 8 seconds. The distance during this phase is the area of a triangle with base 8 and height 20: 1/2 × 8 × 20 = 80 m. The distance during the constant-speed phase is 20 × 30 = 600 m, since distance = speed × time at a constant speed. The total distance is 80 + 600 = 680 m. Using the given 30 seconds for the acceleration phase as well, 1/2 × 30 × 20 = 300, plus the correct 600, gives 900 m — but 30 seconds is only stated for the constant-speed phase. Leaving out the 1/2 and using the full rectangle for the acceleration phase, 8 × 20 = 160, plus the correct 600, gives 760 m — the speed is not constant during acceleration, so this area is a triangle, not a rectangle. Swapping the two times round, and using 8 seconds for the constant-speed distance instead of 30, 20 × 8 = 160, plus the correct triangle area of 80, gives 240 m.
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