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 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.
- 2.A taxi firm charges a fixed fee of £3.50 plus £2.20 per mile. Work out the total cost of a journey of 6 miles.
- 3.A cyclist's speed rises from rest to a maximum — quickly at first, then more and more slowly — so her speed-time graph is a curve that is concave down (its gradient decreases as time goes on). A trapezium estimate is made for the distance travelled during this phase, using two points on the curve. Is the trapezium estimate an overestimate or an underestimate of the true distance, and why?
- 4.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.
- 5.A tram travels between two stops. Its velocity-time graph consists of straight line segments joining the points (0, 0), (5, 20), (12, 20), (16, 4) and (20, 4), where time is in seconds and velocity is in m/s. Work out the average speed of the tram over the whole 20 seconds. Give your answer to 1 decimal place.
- 6.A phone plan costs a fixed £15 per month plus 8p per minute of calls. In one month, Jamal's bill was £23.80. Form an equation using m for the number of minutes, and solve it to find how many minutes Jamal used that month.
- 7.The graph of y = x² − 2x − 8 takes these values: when x = −2, y = 0; when x = −1, y = −5; when x = 0, y = −8; when x = 3, y = −5; when x = 4, y = 0. Use these values to write down the two solutions of x² − 2x − 8 = 0.y = x² − 2x − 8
- 8.f(x) = x³ − 3x² − 4. Work out the pair of consecutive integers between which the solution of f(x) = 0 lies.y = x
- 9.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
- 10.Mia buys 3 identical notebooks and a pen costing £1.50. She pays a total of £9.90. Form an equation using n for the cost of one notebook, and solve it to find n.
- 11.A ball is thrown in the air. Its height, h metres, above the ground after t seconds is given in this table: when t = 0, h = 0; when t = 1, h = 15; when t = 2, h = 20; when t = 3, h = 15; when t = 4, h = 0. Use the table to find the two times, in seconds, at which the ball is at ground level.
- 12.A cyclist's velocity increases from 3 m/s to 11 m/s over 5 seconds, at a constant rate. Work out the cyclist's acceleration, in m/s².
- 13.The point (9, 12) lies on the circle x² + y² = 225, which has centre (0, 0). The tangent to the circle at (9, 12) crosses the x-axis at the point P. Work out the x-coordinate of P.
- 14.A bath is filled with water; the graph of volume (litres) against time (minutes) is a curve, since the flow rate changes. The tangent to the curve at t = 10 minutes passes through the points (8, 130) and (12, 190), and the volume in the bath at t = 10 minutes is 160 litres. Use the gradient of this tangent to estimate the volume in the bath 5 minutes after t = 10 minutes.
- 15.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?
Answer key
- (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) £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.
- (a) Underestimate: the chord lies below the curve — For a curve that is concave down (bending downward), the straight line joining any two points on the curve lies BELOW the curve. The trapezium's top edge is this straight line, so the trapezium's area is smaller than the true area under the curve — the trapezium rule UNDERESTIMATES the distance in this case. This is because a curve whose gradient keeps decreasing bends away from any chord joining two of its points, dropping the chord below every point of the curve in between. The reverse, a chord lying above the curve, is the rule for a concave-up curve, not this one — check which way the curve bends before deciding. The trapezium rule does not always overestimate or underestimate, and it is not exact except when the graph really is a straight line.
- (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.
- (b) 12.7 m/s — Find the total distance from the area under the graph, then divide by the total time. Break the graph into its four straight sections: a triangle from (0, 0) to (5, 20), where 5 × 20 = 100 gives an area of 50 m when halved; a rectangle from (5, 20) to (12, 20), area 7 × 20 = 140 m; a trapezium from (12, 20) to (16, 4), where (20 + 4) × 4 = 96 gives an area of 48 m when halved; and a rectangle from (16, 4) to (20, 4), area 4 × 4 = 16 m — a total distance of 50 + 140 + 48 + 16 = 254 m. Dividing by the 20 seconds gives an average speed of 254 ÷ 20 = 12.7 m/s. Writing 15.1 m/s comes from forgetting to halve the trapezium area for the third section, using 96 instead of 48: a total of 302 m, and 302 ÷ 20 = 15.1 m/s. Writing 11.9 m/s comes from leaving out the final section, from (16, 4) to (20, 4), entirely: a total of 238 m, and 238 ÷ 20 = 11.9 m/s. Writing 12.1 m/s comes from dividing the correct total distance by 21 instead of 20 — a fencepost slip, counting the whole seconds from t = 0 to t = 20 inclusive as 21 seconds of travel rather than reading the journey as a duration of 20 seconds: 254 ÷ 21 = 12.1 m/s (1 d.p.).
- (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 = −2 or x = 4 — The solutions of x² − 2x − 8 = 0 are the x-values where y = 0. From the table, y = 0 when x = −2 and when x = 4, so those are the two solutions. Distractor origins: x = −1 or x = 3 picks the pair of x-values that give equal (but non-zero) y-values instead of y = 0; x = 0 or x = −8 mixes up an x-value with its corresponding y-value; x = −2 only reads off one of the two roots and misses the other.
- (d) 3 and 4 — f(3) = 3³ − 3 × 3² − 4 = 27 − 27 − 4 = −4, and f(4) = 4³ − 3 × 4² − 4 = 64 − 48 − 4 = 12. Since f(3) is negative and f(4) is positive, there is a change of sign, so the solution lies between 3 and 4. A sign slip when expanding −3x², treating it as +3x², gives f(3) = 27 + 27 − 4 = 50, a positive value; taken with the correctly negative f(2) = 8 − 12 − 4 = −8, this reads as a change of sign between 2 and 3. The same sign slip applied at x = 1 gives f(1) = 1 + 3 − 4 = 0, read as the point the sign changes, against f(0) = 0 + 0 − 4 = −4, giving 0 and 1. Shifting the correctly found interval up by one integer, an indexing slip, reports the change as lying between 4 and 5 instead of 3 and 4.
- (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.
- (c) £2.80 — 3n + 1.50 = 9.90, so 3n = 8.40 and n = 2.80, so each notebook costs £2.80. A candidate who forgets to subtract the cost of the pen and divides the total by 3 gets n = 9.90 ÷ 3 = £3.30. A candidate who adds the cost of the pen instead of subtracting it gets 3n = 11.40 and n = £3.80. A candidate who treats the pen as a fourth notebook and divides the total by 4 gets n = 9.90 ÷ 4 = £2.48 (2 d.p.).
- (a) t = 0 or t = 4 — The ball is at ground level exactly when h = 0. From the table, h = 0 at t = 0 and at t = 4, so those are the two times. Distractor origins: t = 1 or t = 3 picks the times with equal (but non-zero) height instead of ground level; t = 2 picks the time of maximum height instead of ground level; t = 0 finds only the starting time and misses the second one.
- (b) 1.6 m/s² — Acceleration is the change in velocity divided by the time taken: (11 − 3) ÷ 5 = 8 ÷ 5 = 1.6 m/s². Forgetting to subtract the initial velocity and dividing the final velocity by the time instead gives 11 ÷ 5 = 2.2 m/s². Inverting the fraction, dividing the time by the change in velocity, gives 5 ÷ 8 = 0.625 m/s². Finding the change in velocity, 8 m/s, but stopping without dividing by the time gives 8 m/s².
- (b) 25 — The tangent at (9, 12) is 9x + 12y = 225. Setting y = 0 (the x-axis): 9x = 225, so x = 25. Choosing 18.75 comes from swapping the coefficients in the tangent equation (using 12x + 9y = 225) before setting y = 0. Choosing 15 is where the circle itself meets the x-axis (from x² = 225), not where the tangent does. Choosing 9 is just the x-coordinate of the original point (9, 12), not the point P.
- (b) 235 litres — The flow rate at t = 10 is the gradient of the tangent: change in volume ÷ change in time = (190 − 130) ÷ (12 − 8) = 60 ÷ 4 = 15 litres per minute. Treating this rate as roughly constant for a short interval, the volume 5 minutes after t = 10 is estimated as 160 + 5 × 15 = 235 litres. Using the tangent's own point spacing — 2 minutes, from t = 10 to t = 12 — instead of the 5 minutes actually asked for gives 160 + 2 × 15 = 190, which is just the volume already given at one of the tangent's own points, not an answer to the question asked. Multiplying the gradient by the time WITHOUT adding the starting volume, 5 × 15 = 75, forgets that a rate estimates a CHANGE, which must be added to the starting volume, not given as the answer on its own. Subtracting instead of adding, 160 − 5 × 15 = 85, extrapolates backward in time rather than forward.
- (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.
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