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.The graph of P = 480 ÷ V shows how the pressure, P pascals, of a gas depends on its volume, V cm³, at constant temperature. Use the relationship to work out the volume when the pressure is 32 pascals, and then work out by how much the volume increases when the pressure drops to 24 pascals.
(a)8 cm³
(b)5 cm³
(c)15 cm³
(d)20 cm³
2.Point R has coordinates (3, −5). Point R is reflected in the line y = 2 to point S. Write down the coordinates of S.
(a)(3, 5)
(b)(−3, 9)
(c)(3, 9)
(d)(3, 7)
3.The line y = 2x + 7 and the circle x² + y² = 4 are given. By finding the discriminant of the resulting quadratic, without solving it fully, work out how many points the line and the circle intersect at.
y = 2x + 7
(a)0 — the line does not intersect the circle
(b)It cannot be found without solving the quadratic.
(c)1 — the line is a tangent to the circle
(d)2 — the line crosses the circle at two points
4.A cinema sells adult tickets for £10 and child tickets for £6. A family group buys 9 tickets in total, spending £74. Work out how many adult tickets were bought.
(a)7.4
(b)1.25
(c)5
(d)4
5.Solve the inequality 2(3x − 1) ≥ 4x + 8.
(a)x ≥ 4.5
(b)x ≥ 5
(c)x ≥ 1
(d)x ≥ 20
6.A taxi fare, in pounds, for a journey of m miles is given by the formula F = 3 + 2.5m. A journey costs £15.50. Work out the distance travelled, m, by first rearranging the formula to make m the subject, then substituting F = 15.50.
(a)7.4
(b)6.2
(c)3.2
(d)5
7.The iterative formula xₙ₊₁ = 12 ÷ (xₙ + 2) is used repeatedly, starting from x₀ = 1. Work out the value that xₙ approaches, correct to 2 decimal places.
(a)3.00
(b)6.00
(c)2.61
(d)4.00
8.A phone company works out a monthly bill, £B, using the formula B = 18 + 0.05t, where £18 is the fixed monthly charge and t is the number of extra text messages sent beyond the free allowance, each charged at 5p. Farida's bill for one month is £24.50. Work out the number of extra text messages, t, she sent.
(a)130
(b)13
(c)65
(d)490
9.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
10.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.
(a)xₙ₊₁ = (9 + xₙ³)/4
(b)xₙ₊₁ = ∛(9 − 4xₙ³)
(c)xₙ₊₁ = ∛(9 − 4xₙ)
(d)xₙ₊₁ = ∛(9 + 4xₙ)
11.By completing the square, find the turning point of the curve y = x² + 8x − 3.
y = x² + 8x − 3
(a)x = −8, y = −67
(b)x = −4, y = 19
(c)x = 4, y = −19
(d)x = −4, y = −19
12.On Monday, a runner covers 15 km in 2.5 hours. On Tuesday, she covers 12 km in 1.5 hours. Using the formula speed = distance ÷ time, work out on which day she ran faster, and by how much.
(a)Tuesday, by 2 km/h
(b)Tuesday, by 8 km/h
(c)Tuesday, by 3 km/h
(d)Monday, by 2 km/h
13.Show that the equation x³ − x − 3 = 0 has a solution between x = 1 and x = 2, by working out f(1) and f(2), where f(x) = x³ − x − 3.
y = x
(a)f(1) = −3 and f(2) = −1
(b)f(1) = −3 and f(2) = 3
(c)f(1) = 3 and f(2) = −3
(d)f(1) = −2 and f(2) = 5
14.A tram sets off from a stop. Its velocity-time graph rises in a straight line from 0 m/s to 12 m/s over the first 8 seconds, then stays constant at 12 m/s for a further 10 seconds. Work out the total distance the tram travels in these 18 seconds, using the area under the graph.
(a)120 m
(b)216 m
(c)108 m
(d)168 m
15.A plumber charges a call-out fee of £30 plus £25 for each hour worked. The total bill for a job was £130. Work out how many hours the plumber worked.
(a)6.4
(b)4
(c)5.2
(d)3.5

Answer key

(b) 5 cm³ — When P = 32, V = 480 ÷ 32 = 15 cm³. When P = 24, V = 480 ÷ 24 = 20 cm³. The increase in volume is 20 − 15 = 5 cm³. Giving only the second volume, 20 cm³, forgets to subtract the first volume. Giving only the first volume, 15 cm³, answers the wrong part of the question. Subtracting the two pressures instead of the two volumes, 32 − 24 = 8, mixes up which quantity the question asks for.
(c) (3, 9) — Reflecting in the horizontal line y = 2 keeps the x-coordinate the same and maps y to 2 × 2 − y = 4 − (−5) = 9, so S = (3, 9). A candidate who uses k − y instead of 2k − y gets 2 − (−5) = 7, giving (3, 7). A candidate who reflects in the x-axis instead of the line y = 2, simply changing the sign of y, gets (3, 5). A candidate who also changes the sign of the x-coordinate, as if reflecting in both axes, gets (−3, 9).
(a) 0 — the line does not intersect the circle — Substitute y = 2x + 7 into x² + y² = 4: x² + (2x + 7)² = 4, which expands to x² + 4x² + 28x + 49 = 4, giving 5x² + 28x + 45 = 0. The discriminant is b² − 4ac = 28² − 4 × 5 × 45. Since 28² = 784 and 4 × 5 × 45 = 900, the discriminant is 784 − 900 = −116. Since the discriminant is negative, the quadratic has no real solutions, so the line does not meet the circle at all: 0 intersection points. Distractor routes: "1 — the line is a tangent" confuses a negative discriminant with a zero one; a discriminant of exactly zero gives one point, a tangent, but −116 is not zero. "2 — the line crosses the circle at two points" assumes a positive discriminant without actually working it out. "It cannot be found without solving the quadratic" is the whole point the discriminant exists to avoid — its sign alone, without finding x, tells you the number of real solutions.
(c) 5 — Method: let a be the number of adult tickets, so the number of child tickets is 9 − a. Form the equation 10a + 6(9 − a) = 74. Working: expand the bracket: 10a + 54 − 6a = 74, so 4a = 20, giving a = 5. Answer: 5 adult tickets. 4 comes from correctly finding the number of child tickets but reporting it instead of the number of adult tickets asked for. 1.25 comes from a sign error expanding the bracket, writing 10a + 54 + 6a = 74 instead of subtracting, which gives 16a = 20. 7.4 comes from dividing the total takings by the adult ticket price only, ignoring the children entirely, 74 ÷ 10.
(b) x ≥ 5 — Expand the bracket: 2(3x − 1) = 6x − 2, so the inequality is 6x − 2 ≥ 4x + 8. Subtract 4x from both sides and add 2 to both sides: 2x ≥ 10. Divide both sides by 2: x ≥ 5. A candidate who only multiplies the 3x by 2 and forgets to multiply the −1 gets 6x − 1 ≥ 4x + 8, leading to x ≥ 4.5. A candidate who adds 4x instead of subtracting it gets 10x ≥ 10, leading to x ≥ 1. A candidate who multiplies by 2 instead of dividing gets x ≥ 20.
(d) 5 — Method: rearrange the formula to make m the subject, then substitute F = 15.50. Working: F = 3 + 2.5m, so subtracting 3 from both sides gives F − 3 = 2.5m, then dividing by 2.5 gives m = (F − 3) / 2.5. Substituting F = 15.50: m = (15.50 − 3) / 2.5 = 12.50 / 2.5 = 5. The value 6.2 comes from dividing 15.50 by 2.5 without subtracting the fixed £3 first. The value 3.2 comes from dividing first and subtracting afterwards, in the wrong order: (15.50 / 2.5) − 3 = 3.2. The value 7.4 comes from adding £3 instead of subtracting it before dividing: (15.50 + 3) / 2.5 = 7.4.
(c) 2.61 — The limit L satisfies L = 12 ÷ (L + 2), so L(L + 2) = 12, giving L² + 2L − 12 = 0 and L = (−2 + √52) ÷ 2 = 2.6056, which is 2.61 to 2 decimal places (the early iterates 4, 2, 3, 2.4, 2.73, ... oscillate around this value before settling). Choosing 3.00 reads off x₃, one of the early oscillating values, before the sequence has settled close to the limit. Choosing 4.00 reads off x₁, the very first iterate, not the value the sequence approaches. Choosing 6.00 comes from writing the limit equation as L = 12 ÷ 2, leaving L itself out of the denominator.
(a) 130 — Method: substitute the total bill into the formula, then subtract the fixed charge and divide by the cost per text message. Working: 24.50 = 18 + 0.05t, so 0.05t = 24.50 − 18 = 6.50, t = 6.50 ÷ 0.05 = 130. Answer: 130 extra text messages. 490 comes from dividing the whole bill by 0.05 without first subtracting the £18 fixed charge: 24.50 ÷ 0.05 = 490. 13 comes from dividing the £6.50 by 0.5 instead of 0.05, moving the decimal point one place too far: 6.50 ÷ 0.5 = 13. 65 comes from dividing the £6.50 by 0.1 instead of 0.05.
(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.
(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.
(d) x = −4, y = −19 — x² + 8x − 3 = (x + 4)² − 4² − 3 = (x + 4)² − 19. Substituting x = −4: (−4)² = 16, 8 × (−4) = −32, so 16 − 32 − 3 = −19, confirming the minimum value −19 at x = −4: turning point (−4, −19). Using 8 instead of half of it inside the bracket gives (x + 8)² − 67, turning point (−8, −67) — wrong, because only half the coefficient of x belongs inside the bracket. Reading the bracket's sign directly as the turning point's x-coordinate gives (4, −19) — wrong, because (x + 4)² is zero at x = −4, not x = 4. Computing 16 + 3 = 19 instead of −3 − 16 = −19 flips the sign of the constant, giving (−4, 19) — wrong, since the completed square's constant must be evaluated as −3 minus 16, not 16 plus 3. Check by substitution whenever the sign of a constant feels uncertain.
(a) Tuesday, by 2 km/h — Monday's speed is 15 ÷ 2.5 = 6 km/h and Tuesday's speed is 12 ÷ 1.5 = 8 km/h, so Tuesday was faster, by 8 − 6 = 2 km/h. A candidate who works out the correct speeds but mislabels which day is faster gets Monday, by 2 km/h. A candidate who divides 15 ÷ 2.5 incorrectly as 5 instead of 6 gets a difference of 8 − 5 = 3 km/h, still crediting Tuesday. A candidate who forgets to find Monday's speed and gives Tuesday's speed itself as the difference states Tuesday, by 8 km/h.
(b) f(1) = −3 and f(2) = 3 — f(1) = 1³ − 1 − 3 = 1 − 1 − 3 = −3. f(2) = 2³ − 2 − 3 = 8 − 2 − 3 = 3. Since f(1) is negative and f(2) is positive, there is a change of sign, so a solution lies between x = 1 and x = 2. Dropping the −x term entirely gives f(1) = 1 − 3 = −2 and f(2) = 8 − 3 = 5. Using x² instead of x³ throughout gives f(1) = 1 − 1 − 3 = −3, which happens to coincide with the correct value at x = 1, but f(2) = 4 − 2 − 3 = −1, which does not show a change of sign at all. Working out −f(x) instead of f(x), a sign-flip error, gives f(1) = 3 and f(2) = −3, the correct sizes but with both signs reversed.
(d) 168 m — Split the area under the graph into two parts. The rising section (0 to 8 s) is a triangle: area = 0.5 × 8 × 12 = 48. The constant section, from 8 s to 18 s (10 s), is a rectangle: area = 12 × 10 = 120. Total distance = 48 + 120 = 168 m. Leaving out the 0.5 doubles the triangle, giving 8 × 12 + 120 = 216 m; ignoring the triangle section altogether gives only 120 m; averaging the start and final speeds over the whole 18 seconds, (0 + 12) ÷ 2 = 6, then 6 × 18 = 108 m, wrongly treats the tram as accelerating the whole time, when the graph is flat for the last 10 seconds.
(b) 4 — Method: form the equation 30 + 25h = 130, where h is the number of hours, then solve for h. Working: subtract the call-out fee from the total bill: 25h = 130 − 30 = 100. Divide by the hourly rate: h = 100 ÷ 25 = 4. Answer: 4 hours. 5.2 comes from dividing the whole bill by the hourly rate without subtracting the fixed fee first, 130 ÷ 25. 3.5 comes from swapping the fee and the rate, subtracting the rate from the bill and dividing by the fee, (130 − 25) ÷ 30. 6.4 comes from adding the call-out fee to the bill instead of subtracting it, before dividing by the rate, (130 + 30) ÷ 25.

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