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.In a year group of 60 pupils, 25 play football, 18 play tennis and 8 play both football and tennis. A pupil who plays football or tennis or both is picked at random. Work out the probability that this pupil plays both sports.
(a)8/43
(b)8/35
(c)8/25
(d)2/15
2.Marta is drawing a cumulative frequency diagram for the times, t seconds, of 100 telephone calls. The grouped frequencies are: 0 ≤ t < 10, 7 calls; 10 ≤ t < 20, 19 calls; 20 ≤ t < 30, 34 calls; 30 ≤ t < 40, 40 calls. Write down the coordinates of the point Marta should plot for the class 20 ≤ t < 30.
(a)(30, 60)
(b)(30, 34)
(c)(25, 60)
(d)(20, 60)
3.Two expressions are 4(x + 3) and 4x + 3. A student checks whether they are equivalent by substituting x = 2. Which statement correctly interprets the result?
(a)4(x + 3) = 20 and 4x + 3 = 11 when x = 2, so the two expressions are not equivalent, because the bracket means the 3 must be added before multiplying by 4.
(b)4(x + 3) = 11 and 4x + 3 = 11 when x = 2, since the bracket has no effect on the multiplication, so the two expressions are equivalent.
(c)4(x + 3) = 20 and 4x + 3 = 11 when x = 2, but the two expressions are equivalent because both involve the same terms, 4x and 3.
(d)4(x + 3) = 20 and 4x + 3 = 11 when x = 2, so the two expressions are not equivalent, but they would become equal for some larger value of x.
4.A student is asked whether 3(x − 4) = 3x − 4 is an identity. Which statement gives the correct verdict and reason?
(a)It is an identity, because both sides begin with the term 3x, so they must be equivalent.
(b)It is not even an ordinary equation with a solution: expanding the left-hand side gives 3x − 12, and 3x − 12 = 3x − 4 would require −12 = −4, which is never true.
(c)It is not an identity, because the student forgot to multiply the 4 by 3 on both sides of the equation.
(d)It is not an identity, because the student needs to substitute a specific value of x before comparing the two sides.
5.Which expression is equivalent to 3(2x − 5) + 4x?
(a)10x + 15
(b)6x − 15
(c)10x − 15
(d)10x − 5
6.The graph of y = f(x) crosses the x-axis at x = −3 and x = 7, and crosses the y-axis at (0, 21). A second graph crosses the x-axis at x = −7 and x = 3, and crosses the y-axis at the same point, (0, 21). The second graph is y = g(x). Which of these could be the equation of g(x)?
(a)y = f(x) − 4
(b)y = f(−x)
(c)y = −f(x)
(d)y = −f(−x)
7.f(x) = 2x + 1. Work out the value of x for which f⁻¹(x) = 5.
y = 2x + 1
(a)6
(b)9
(c)2
(d)11
8.At a garden centre, 3/5 of the plants for sale are perennials. 1/4 of the perennials are in flower. Work out the probability that a plant picked at random from the garden centre is a perennial and is in flower.
(a)4/9
(b)17/20
(c)3/20
(d)1/4
9.The graph of y = cos x is transformed onto the graph of y = cos(x − 90°). State the direction of the translation and which standard graph the image is.
y = cos(x)
(a)Positive x-direction, 90°; image is y = sin x.
(b)Negative x-direction, 90°; image is y = sin x.
(c)Positive x-direction, 90°; image is y = −sin x.
(d)Positive y-direction, 90°; image is y = sin x.
10.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
11.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
12.The point (5, −12) lies on the circle x² + y² = 169, which has centre (0, 0). Work out the equation of the tangent to the circle at (5, −12), giving your answer in the form y = mx + c.
(a)y = −(12/5)x
(b)y = (5/12)x − 25/12
(c)y = (5/12)x − 169/12
(d)y = −(5/12)x − 119/12
13.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
14.Point A has coordinates (2, 5). Point A is translated to point B using the column vector with top number 6 and bottom number −3. Write down the coordinates of point B.
(a)(8, 2)
(b)(−1, 11)
(c)(−4, 8)
(d)(8, 8)
15.A histogram shows the times, t minutes, taken by 120 visitors to complete an escape room. The bar for 0 ≤ t < 10 has a frequency density of 5 visitors per minute, the bar for 10 ≤ t < 20 has a frequency density of 2 visitors per minute, the bar for 20 ≤ t < 40 has a frequency density of 1.5 visitors per minute, and the bar for 40 ≤ t < 60 has a frequency density of 1 visitor per minute. Work out which class contains the median time.
(a)The class with times from 0 up to 10
(b)The class with times from 20 up to 40
(c)The class with times from 10 up to 20
(d)The class with times from 40 up to 60
16.The equation x² − x − 6 = 0 has roots x = 3 and x = −2. It can be rearranged as xₙ₊₁ = xₙ² − 6. This formula is used with starting value x₀ = 2.9, close to the root x = 3. Work out what happens to the sequence of values as n increases.
(a)The sequence settles towards x = 3
(b)The sequence stays constant at x = 2.9
(c)The sequence diverges, moving away from x = 3
(d)The sequence settles towards x = −2
17.The point A(−6, 8) lies on the circle x² + y² = 100, whose centre is the origin O. The tangent to the circle at A crosses the y-axis at the point B. Work out the length of OB.
(a)12.5
(b)8
(c)10
(d)3.5
18.In a histogram of the lengths, x cm, of some rods, the bar for 10 ≤ x < 30 has a frequency density of 3 per cm. The bar for 30 ≤ x < 45 is twice as tall as the bar for 10 ≤ x < 30. Work out the number of rods with a length in the class 30 ≤ x < 45.
(a)6
(b)45
(c)90
(d)120

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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