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.Describe a sequence of two transformations that maps the graph of y = x² onto the graph of y = −(x − 5)².
y = x²
(a)Reflect in the y-axis, then translate +5 in x.
(b)Translate +5 in x, then reflect in the x-axis.
(c)Translate −5 in x, then reflect in the x-axis.
(d)Translate +5 in y, then reflect in the x-axis.
2.The graph of y = 2x² + 6x − 1 is translated by the vector (−2, 0). Work out the equation of the image, giving your answer in the form y = 2x² + bx + c.
y = 2x² + 6x − 1y = 2x²
(a)y = 2x² + 10x + 15
(b)y = 2x² + 14x + 19
(c)y = 2x² − 2x − 5
(d)y = 2x² + 14x + 20
3.A company's weekly profit, in £, is modelled by y = f(x), where x is the number of weeks since launch. The graph of y = f(x) has a maximum at (10, 45000). A rival company uses the same marketing strategy but starts trading 6 weeks later and has fixed costs £8000 higher every week, so its profit is modelled by y = f(x − 6) − 8000. In which week does the rival's maximum weekly profit occur, and what is it?
(a)Week 16, £53,000
(b)Week 4, £37,000
(c)Week 16, £37,000
(d)Week 4, £53,000
4.The graph of y = f(x) has roots at x = −1 and x = 4, and crosses the y-axis at (0, −8). Which statement about the graph of y = f(x − 3) is correct?
(a)x = 2, x = 7; y-intercept cannot be found here
(b)x = 2, x = 7; y-intercept stays at (0, −8)
(c)x = −4, x = 1; y-intercept stays at (0, −8)
(d)x = −4, x = 1; y-intercept cannot be found here
5.A photo printing service has two adverts for its price. Advert A: cost in pounds = 3(2n + 4) for n photos. Advert B: cost in pounds = 6n + 12. A customer says the two adverts always charge the same amount. Is the customer correct?
(a)They always charge the same, since 3(2n + 4) = 6n + 12.
(b)It depends on the value of n, so it cannot be decided.
(c)Advert A is cheaper, since 3(2n + 4) = 6n + 4.
(d)Advert A is cheaper, since 3(2n + 4) = 2n + 7.
6.A school buys pens from two suppliers and has 1000 pens in stock. Supplier X provided 70% of the pens and supplier Y provided the other 30%. 2% of supplier X's pens are faulty and 8% of supplier Y's pens are faulty. A pen picked at random from the stock is found to be faulty. Work out the probability that it came from supplier Y. Give your answer as a fraction.
(a)3/125
(b)12/19
(c)4/5
(d)3/10
7.Describe the single transformation that maps the graph of y = f(x) onto the graph of y = f(−x).
(a)A rotation of 180° about the origin
(b)A reflection in the x-axis
(c)A reflection in the y-axis
(d)A reflection in the line y = x
8.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
9.A student is proving that (n + 1)² − n² is always an odd number. Which of these correctly completes the first line of algebra?
(a)(n + 1)² − n² = 1
(b)(n + 1)² − n² = 2n + 1
(c)(n + 1)² − n² = 2n
(d)(n + 1)² − n² = n² + 2n + 1
10.The graph of y = f(x) has a maximum turning point at (5, 8). Which of these correctly gives the corresponding turning point on the graph of y = −f(x + 1), and its type?
(a)(4, −8), a minimum point
(b)(4, 8), a minimum point
(c)(6, −8), a minimum point
(d)(4, −8), a maximum point
11.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.
12.At a sports centre, 45% of the members are aged under 18. 60% of the members aged under 18 swim each week. 20% of the members aged 18 or over swim each week. Work out the percentage of all the members who swim each week.
(a)40%
(b)38%
(c)80%
(d)42%
13.A vet records the masses, m kg, of the dogs seen in one week as a histogram. The bar for 0 ≤ m < 5 has a frequency density of 4 per kg, the bar for 5 ≤ m < 15 has a frequency density of 2.6 per kg, and the bar for 15 ≤ m < 40 has a frequency density of 1.2 per kg. Work out the total number of dogs seen that week.
(a)7.8
(b)39
(c)76
(d)107
14.The equation x² + 2x − 5 = 0 can be solved using the iterative formula xₙ₊₁ = 5/(xₙ + 2). The starting value is x₀ = 1, so x₁ is the value after the formula has been used once. Work out x₂ correct to 2 decimal places.
(a)1.36
(b)1.49
(c)1.37
(d)1.67
15.Which of these is an identity?
(a)2(3x + 1) = 6x + 2
(b)5x + 1 = 5(x + 1)
(c)7 − x = x − 7
(d)4x − 3 = 3x + 5
16.f(x) = x³ − 3x − 20, and the equation f(x) = 0 has exactly one solution. Work out the pair of consecutive integers between which that solution lies.
y = x
(a)2 and 3
(b)3 and 4
(c)4 and 5
(d)6 and 7
17.A rectangular sheet of metal measures 20 cm by 12 cm. A square of side x cm is cut from each corner and the sides are folded up to make an open box of volume 200 cm³. This gives x³ − 16x² + 60x − 50 = 0, which can be solved using the iterative formula xₙ₊₁ = (16xₙ² − xₙ³ + 50)/60. The starting value is x₀ = 1, so x₁ is the value after the formula has been used once. Work out x₃, and use it to find the longer side of the base of the box correct to 1 decimal place.
(a)1.1 cm
(b)9.7 cm
(c)17.7 cm
(d)18.9 cm
18.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

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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