18 demanding questions: algebraic proof, functions, vectors, histograms, conditional probability and iteration.

🔺 Higher stretch — grade 8 and 9 topics

Maths · GCSE Higher · 18 questions · ~50 min

MathsUKwww.geekhero.co.uk

The top of a Higher paper is not the same content done faster — it is the statements that only appear at the top, and questions that hide two steps inside one instruction. This sheet gathers eighteen of them: algebraic proof and showing two expressions are equivalent, inverse and composite functions, transformations of graphs, the equation of a circle and its tangent, iteration, vectors used to construct a geometric argument, histograms with unequal class intervals, and conditional probability from a tree or a Venn diagram. It is deliberately hard and deliberately short. If you get half of it out, you are working where grades 8 and 9 are decided — and the worked answers are written to show the step that was actually being tested, not just the arithmetic around it.

1.Triangle ABC is translated by the column vector with top number 3 and bottom number −5 to form triangle A′B′C′. Triangle A′B′C′ is then translated by the column vector with top number −7 and bottom number 2 to form triangle A″B″C″. Work out the single column vector that translates triangle ABC directly to triangle A″B″C″.
(a)(10, −7)
(b)(4, −3)
(c)(−4, 3)
(d)(−4, −3)
2.The graph of y = f(x) passes through the point (0, 4). Work out the y-coordinate of the point where the graph of y = f(x) − 6 crosses the y-axis.
(a)10
(b)−6
(c)4
(d)−2
3.Which expression is equivalent to 6x − (2x − 5)?
(a)8x + 5
(b)8x − 5
(c)4x − 5
(d)4x + 5
4.A factory tests components from a large batch in which 6% are defective. Two components are selected at random, and the batch is large enough that the selections can be treated as independent. Given that at least one of the two components is defective, work out the probability that both are defective.
(a)0.0036
(b)0.0309
(c)0.0600
(d)0.5000
5.A rectangular garden has width w metres and length (w + 3) metres. A gardener writes its perimeter as 2w + 3. Which statement corrects the gardener's mistake?
(a)It is 2(w + (w + 3)) = 4w + 6, not 2w + 3.
(b)It is 3w + 6, since the length doubles but not the width.
(c)It is w + (w + 3) = 2w + 3, matching the gardener.
(d)It is 4(w + 3) = 4w + 12, treating it as a square.
6.f(x) = x³ − 3x² − 4. Work out the pair of consecutive integers between which the solution of f(x) = 0 lies.
y = x
(a)4 and 5
(b)2 and 3
(c)0 and 1
(d)3 and 4
7.A circle has centre (0, 0) and passes through the point (12, 35). Work out the equation of the circle.
(a)x² + y² = 1225
(b)x² + y² = 1369
(c)x² + y² = 144
(d)x² + y² = 2209
8.A proof sets out to show that the sum of the squares of two consecutive odd numbers, written as 2n + 1 and 2n + 3, is always 2 more than a multiple of 8. Four attempts to expand (2n + 1)² + (2n + 3)² and reach a conclusion are shown below. Which attempt correctly proves this claim?
(a)(2n + 1)² + (2n + 3)² = (4n² + 4n + 1) + (4n² + 9) = 8n² + 4n + 10 = 4(2n² + n + 2) + 2, so the sum is always 2 more than a multiple of 4.
(b)(2n + 1)² + (2n + 3)² = (4n² + 4n + 1) + (4n² + 12n + 9) = 8n² + 16n + 10 = 2(4n² + 8n + 5), and 4n² + 8n + 5 is an integer, so the sum is always even, which means it is a multiple of 8.
(c)(2n + 1)² + (2n + 3)² = (4n² + 4n + 1) + (4n² + 12n + 9) = 8n² + 16n + 10 = 8(n² + 2n + 1) + 2, and n² + 2n + 1 is an integer, so the sum is always 2 more than a multiple of 8.
(d)(2n + 1)² + (2n + 3)² = (4n² + 4n + 1) + (4n² + 12n + 9) = 8n² + 16n + 10 = 8(n² + 2n) + 10, so the sum is always 10 more than a multiple of 8.
9.A circle has centre (0, 0) and equation x² + y² = 8. Work out the radius of the circle, giving your answer as a surd in its simplest form.
(a)√2
(b)8
(c)4
(d)2√2
10.The times, t seconds, taken by 142 competitors to complete a lap are grouped like this: 0 ≤ t < 10, 20 competitors; 10 ≤ t < 25, 12 competitors; 25 ≤ t < 45, 50 competitors; 45 ≤ t < 75, 60 competitors. A histogram is drawn. Write down the class whose bar is the tallest.
(a)10 ≤ t < 25
(b)45 ≤ t < 75
(c)0 ≤ t < 10
(d)25 ≤ t < 45
11.The equation x³ = 6x + 20 can be solved using the iterative formula xₙ₊₁ = ∛(6xₙ + 20). Taking x₀ = 3, x₁ = 3.3620 correct to 4 decimal places. Using the full unrounded value of x₁, work out x₂ correct to 3 decimal places.
(a)3.362
(b)2.722
(c)0.556
(d)3.425
12.A scientist has grouped the lifetimes, in hours, of 300 batteries into classes of unequal width. She wants a diagram in which the number of batteries in a class is given by the area of its bar. Write down the type of diagram she should draw.
(a)A cumulative frequency diagram, with running totals plotted
(b)A bar chart, with frequency up the vertical axis
(c)A histogram, with frequency density up the vertical axis
(d)A pie chart, with one sector for each class interval
13.The times, t minutes, taken by 120 runners to finish a fun run are summarised by these cumulative frequencies: t < 20, 8 runners; t < 30, 26 runners; t < 40, 74 runners; t < 50, 110 runners; t < 60, 120 runners. Work out the number of runners who took 40 minutes or longer to finish.
(a)36
(b)74
(c)46
(d)10
14.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
15.The equation x³ − 2x − 7 = 0 has exactly one solution. It can be found using the iterative formula xₙ₊₁ = ∛(2xₙ + 7), with starting value x₀ = 2, so that x₁ is the value after the formula has been used once. Work out the solution correct to 2 decimal places, iterating until two consecutive values round to the same 2 decimal places.
(a)2.25
(b)1.91
(c)2.26
(d)2.22
16.m is the column vector with top number 4 and bottom number 6. n is the column vector with top number −6 and bottom number −9. Given that n = k × m for some number k, work out the value of k.
(a)1.5
(b)−10
(c)−1.5
(d)−24
17.A company has 400 employees. 150 of them work part time and 160 of them cycle to work. Working part time and cycling to work are independent. Work out how many of the employees you would expect both to work part time and to cycle to work.
(a)10
(b)60
(c)310
(d)75
18.Which expression is equivalent to 7x − 3(2x − 6)?
(a)13x − 18
(b)x − 18
(c)x + 18
(d)x + 6

What is on this worksheet?

The sheet holds 18 questions drawn from the MathsUK bank — the content areas covered: Algebra, Geometry and measures, Probability, Statistics (statements A6, A7, A13, A16, A20, G25, S3, P9). It is pitched at GCSE Higher and takes about 50 minutes to work through in full. It is built for independent practice, with full answers at the end for self-marking.

How to use the sheet well

Print it or open it on screen — both work. Printing is A4; the screen view fits phones and tablets.
Do all 18 questions before checking — about 50 minutes is the guide, but there is no time pressure.
Check the answers — press “Show answers” or print the answer page separately.
Redo the questions you got wrong — twice as effective as doing 18 fresh ones.
“New questions” — builds a fresh sheet on the same statements, so you can practise again without repeats.

Why this sheet helps

MathsUK worksheets use questions graded by difficulty and a fair spread of correct-answer positions (the answer is not always (a)) — so the student really has to think about each question rather than guess a pattern. Every question is tagged to a DfE content statement and checked before it enters the bank. The answers come with a step-by-step explanation, not just a value — so a wrong answer becomes a lesson.

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