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

Maths · GCSE Higher · Algebra · Calculator · 15 questions

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1.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.
(a)36 km/h
(b)14.4 km/h
(c)6.25 km/h
(d)0.6 km/h
2.The iterative formula xₙ₊₁ = 5 − 3/xₙ is used with starting value x₀ = 2.5, so that x₁ is the value after the formula has been used once. Work out x₄ correct to 3 significant figures.
(a)3.80
(b)4.29
(c)4.30
(d)2.50
3.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.
(a)84.51
(b)4.39
(c)4.67
(d)6.32
4.A recipe gives the cooking time, in minutes, for a chicken as T = 40m + 20, where m is the mass in kg. A chicken has a mass of 1.8 kg. Work out the cooking time in hours and minutes.
(a)1 hour 12 minutes
(b)1 hour 48 minutes
(c)1 hour 40 minutes
(d)1 hour 32 minutes
5.The formula for the energy stored in a stretched spring is E = (1/2)kx², where E is in joules, k is the spring constant in N/m and x is the extension in m. A spring with a spring constant of 40 N/m is stretched so that its extension is 0.3 m. Work out the energy stored in the spring.
(a)1.8
(b)72
(c)3.6
(d)6
6.Which expression is equivalent to 0.5(4x + 6) − x?
(a)3x + 3
(b)2x + 3
(c)x + 3
(d)x + 6
7.A closed cylinder has radius r cm and height (r + 5) cm. Its volume is 300 cm³, giving the equation πr²(r + 5) = 300, which can be solved using the iterative formula rₙ₊₁ = √(300 ÷ (π(rₙ + 5))). Taking r₀ = 3, work out r₃ correct to 2 decimal places.
(a)3.45
(b)3.38
(c)3.36
(d)4.82
8.A cafe sells coffees at £c each and pastries at £p each. Three coffees and two pastries cost £9.60. Two coffees and two pastries cost £7.60. Work out the price of one coffee.
(a)£0.20
(b)£2.00
(c)£3.20
(d)£1.80
9.The mean of four numbers is 12.5. Three of the numbers are 8, 15 and 19. Form an equation using n for the fourth number, and solve it to find n.
(a)−4.5
(b)50
(c)29.5
(d)8
10.The diagram shows a distance–time graph for a cyclist travelling at a constant speed, for the first 6 minutes of a journey. Distance is in kilometres and time is in minutes. Work out how far the cyclist would travel in 20 minutes at the same speed.
(a)20 km
(b)10 km
(c)40 km
(d)3 km
11.A distance-time graph is a straight line from (0, 0) to (4, 100), where time is in hours and distance is in kilometres. Work out the gradient of the line.
(a)25
(b)400
(c)0.04
(d)100
12.A runner's journey is shown on a distance-time graph: she runs 8 km in the first hour, rests for 30 minutes with no distance gained, then runs a further 4 km in the next 30 minutes. Work out her average speed for the whole journey, in kilometres per hour.
(a)12 km/h
(b)8 km/h
(c)6 km/h
(d)4 km/h
13.A quadratic graph y = ax² + bx + c has its turning point on the y-axis. Which statement about its roots must be true?
(a)The two roots must both be positive.
(b)The graph has no roots at all.
(c)The graph must touch the x-axis at exactly one point.
(d)If the graph has two real roots, they are equal and opposite in value, so they sum to zero.
14.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
(a)x = −2
(b)x = −1 or x = 3
(c)x = −2 or x = 4
(d)x = 0 or x = −8
15.The tangent to a curve at the point where x = 4 passes through the points (2, 5) and (6, 21). Use these two points to estimate the gradient of the curve at x = 4.
(a)4
(b)16
(c)0.25
(d)8

Answer key

(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.
(c) 4.30 — Method: substitute the starting value into the right-hand side to get x₁, then feed each value back in, keeping the whole display and respecting the order of operations, which divides before it subtracts. Working: x₁ = 5 − 3 ÷ 2.5 = 5 − 1.2 = 3.8; x₂ = 5 − 3 ÷ 3.8 = 5 − 0.78947… = 4.21052…; x₃ = 5 − 3 ÷ 4.21052… = 5 − 0.7125 = 4.2875; x₄ = 5 − 3 ÷ 4.2875 = 5 − 0.69970… = 4.30029…, which is 4.30 correct to 3 significant figures. Answer: 4.30. The distractors: 4.29 is x₃ = 4.2875 rounded, reached by counting the starting value itself as the first iterate and so stopping one use of the formula early; 3.80 is x₁, the value after a single use of the formula; 2.50 comes from working out (5 − 3) ÷ xₙ instead of 5 − (3 ÷ xₙ), subtracting before dividing, which produces the sequence 0.8, 2.5, 0.8, 2.5 and lands on 2.5 at the fourth step.
(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.
(d) 1 hour 32 minutes — T = 40 × 1.8 + 20 = 72 + 20 = 92 minutes. Since 92 = 60 + 32, the cooking time is 1 hour 32 minutes. A candidate who rounds the mass to 2 kg before substituting gets 40 × 2 + 20 = 100 minutes = 1 hour 40 minutes. A candidate who forgets to add the 20 minutes gets 40 × 1.8 = 72 minutes = 1 hour 12 minutes. A candidate who multiplies the mass by (40 + 20) = 60 instead of substituting into the formula gets 1.8 × 60 = 108 minutes = 1 hour 48 minutes.
(a) 1.8 — E = (1/2)kx² = 0.5 × 40 × 0.3² = 0.5 × 40 × 0.09 = 1.8 joules. 3.6 comes from forgetting the (1/2) at the front, 40 × 0.09. 72 comes from squaring kx together instead of squaring only x, 0.5 × (40 × 0.3)² = 0.5 × 144 = 72. 6 comes from using x instead of x², 0.5 × 40 × 0.3.
(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.
(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.
(d) 8 — Since the mean of the four numbers is 12.5, their total is 4 × 12.5 = 50. The three known numbers add up to 8 + 15 + 19 = 42, so n = 50 − 42 = 8. A candidate who multiplies the mean by 3 instead of 4 gets a total of 37.5, giving n = 37.5 − 42 = −4.5. A candidate who forgets to subtract the three known numbers and gives the total itself as n states n = 50. A candidate who subtracts the mean from the total of the three known numbers instead of the other way round gets n = 42 − 12.5 = 29.5.
(b) 10 km — Method: find the constant speed from the graph (distance ÷ time for any point on the line), then multiply that speed by 20 minutes. Working: the line passes through (4 minutes, 2 km), so the speed is 2 ÷ 4 = 0.5 km per minute; in 20 minutes the cyclist travels 0.5 × 20 = 10 km. Answer: 10 km. Distractor refutation: 3 km comes from reading off the distance shown at the end of the plotted section (6 minutes) and stopping there, instead of extending the line to 20 minutes. 20 km comes from misreading the speed as 1 km per minute instead of 0.5 km per minute, doubling the true rate. 40 km comes from dividing 20 by the speed instead of multiplying by it, a reciprocal mix-up.
(a) 25 — Gradient = change in y ÷ change in x = (100 − 0) ÷ (4 − 0) = 100 ÷ 4 = 25. Dividing time by distance instead of distance by time gives 0.04; multiplying the two values instead of dividing gives 400; stopping at the change in distance, 100, forgets to divide by the change in time.
(c) 6 km/h — Total distance = 8 + 4 = 12 km. Total time = 1 hour + 0.5 hours resting + 0.5 hours = 2 hours. Average speed = total distance ÷ total time = 12 ÷ 2 = 6 km/h. A speed of 8 km/h comes from leaving the resting time out of the total time: 12 ÷ 1.5 = 8. A speed of 4 km/h comes from dividing by too much time, such as double-counting the rest period: 12 ÷ 3 = 4. A speed of 12 km/h simply gives the total distance and forgets to divide by the total time at all.
(d) If the graph has two real roots, they are equal and opposite in value, so they sum to zero. — A turning point on the y-axis means the graph's axis of symmetry is the line x = 0, so any two roots must be symmetrical about x = 0 — equal in size but opposite in sign, summing to zero. The graph could still have no real roots, but that is not guaranteed just from the turning point's position, so the option claiming it must have none is too strong. The roots do not have to both be positive — if real, one is positive and one negative (or both are zero). The graph does not have to touch the x-axis at exactly one point either; it could cross at two symmetrical points, touch at one point, or miss the x-axis entirely.
(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.
(a) 4 — The gradient of a straight line through two points is the change in y divided by the change in x. Change in y = 21 − 5 = 16. Change in x = 6 − 2 = 4. Gradient = 16 ÷ 4 = 4. Dividing x by y instead of y by x gives 0.25; forgetting to divide by the change in x at all leaves 16; dividing by only one of the two x-coordinates, 16 ÷ 2 = 8, uses the wrong denominator.

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