20 questions on the grade 4-5 content both tiers share — a fair test of whether a Higher entry is the right call.
⚖️ Foundation to Higher crossover check
The tier decision is made by the school, but it is made on evidence, and this sheet is designed to produce some. Every question on it sits in the crossover band — the grade 4 to 5 content that appears on both the Foundation and the Higher papers: multiplying out and factorising, solving linear equations and inequalities, straight-line graphs and gradients, ratio and proportion, Pythagoras' theorem, angle reasoning, and averages from a frequency table. A student who works through this comfortably has the foundation a Higher entry needs. A student who is guessing on half of it will almost certainly come away with a better grade from a Foundation paper they can attempt in full. Useful for a parents' evening conversation, and for a department deciding entries.
- 1.Expand 4(2x − 3).
- 2.Factorise fully 12x² + 18x.
- 3.In triangle ABC the angle at A and the angle at C are equal. The side AB is extended beyond B, and the exterior angle formed at B measures 98°. Work out the size of the angle at A.
- 4.A gas company charges a standing charge plus a rate per unit used. The total cost, y in pounds, for using x units is given by y = 0.15x + 20. Work out the cost of using 200 units.y = 0.15x + 20
- 5.Simplify (5x² + 3x − 2) − (2x² − x + 5)
- 6.A recipe for pastry uses flour and butter in the ratio 3:2. A baker has 180 g of butter and wants to make pastry using all of it. Work out the total mass of pastry the baker can make.
- 7.Expand and simplify 2(a + 6) − 2(a − 7)
- 8.A straight line has equation y = 4x + 3. A second line is parallel to the first line and passes through the point (0, −5). Work out the equation of the second line.y = 4x + 3
- 9.In a school choir the ratio of boys to girls is 3:4. When 6 more boys join the choir, the ratio of boys to girls becomes 1:1. Work out how many girls are in the choir.
- 10.A television cable runs in a straight line from the edge of a flat roof 20 m above the ground to a bracket on a wall 5 m above the ground. The horizontal distance between the roof edge and the bracket is 15 m. Work out the length of the cable. Give your answer to 1 decimal place.
- 11.Grace asked 12 children how many brothers and sisters they have. Her results, in order, were 0, 0, 1, 1, 1, 1, 2, 2, 3, 3, 4, 6. Work out the median number of brothers and sisters.
- 12.Factorise x² − 64.
- 13.Harry will spend at most £150 on a party. The cake costs £60 and each helium balloon costs £3. Solve an inequality to find all the possible numbers of balloons, x, that he can buy.
- 14.At a youth club the ratio of juniors to seniors is 3:5. There are 40 members altogether. Work out how many seniors there are.
- 15.Work out the equation of the straight line through the points (−2, 1) and (0, 7).
- 16.Isla and her brother share £60 in the ratio 1:2. Work out Isla's share.
- 17.At an animal shelter, the ratio of cats to dogs is 2:5. There are 18 more dogs than cats. Work out the number of cats.
- 18.The solution to an inequality is n ≤ 5. Write down the largest integer value of n that satisfies this inequality.
- 19.In a right-angled triangle, the two shorter sides are a and b, and the hypotenuse is c. Write down the correct statement of Pythagoras' theorem.
- 20.A regular hexagon is divided into six identical triangles by joining its centre to each of the six vertices. Work out the size of the angle of one of these triangles at the centre of the hexagon.
Answer key
- (c) 8x − 12 — Multiply each term inside the bracket by 4: 4 × 2x = 8x and 4 × (−3) = −12, so 4(2x − 3) = 8x − 12. A candidate who forgets to multiply the second term by 4 gets 8x − 3. A candidate who makes a sign error, treating 4 × (−3) as +12, gets 8x + 12. A candidate who adds 4 to the bracket instead of multiplying gets 2x + 1.
- (c) 6x(2x + 3) — The highest common factor of 12x² and 18x is 6x. Dividing each term by 6x gives 12x² ÷ 6x = 2x and 18x ÷ 6x = 3, so 12x² + 18x = 6x(2x + 3). A candidate who only takes out the number 6 (missing the x) gets 6(2x² + 3x), which is not fully factorised. A candidate who only takes out 2x (missing the extra factor of 3 in 6) gets 2x(6x + 9), also not fully factorised — the bracket still shares a common factor. A candidate who takes out 3x instead of the full 6x gets 3x(4x + 6), which again is not fully factorised since 4x + 6 shares a common factor of 2.
- (a) 49° — Method: an exterior angle of a triangle equals the sum of the two interior angles that are not next to it, which here are the angles at A and at C; since those two are equal, the exterior angle is twice the angle at A. Working: 2 × angle A = 98, so angle A = 98 ÷ 2 = 49. Answer: 49°. The distractors: 82° is the interior angle at B, 180 − 98, given in place of the angle at A; 41° comes from finding that interior angle of 82° and halving it, 82 ÷ 2, instead of halving the exterior angle; 98° comes from taking the exterior angle to be equal to the angle at A on its own, with no halving at all.
- (a) £50 — Substituting x = 200 into y = 0.15x + 20 gives y = 0.15 × 200 + 20 = 30 + 20 = £50. A candidate who forgets to add the standing charge would get only 0.15 × 200 = £30. A candidate who misplaces the decimal point in the rate, using 1.5 instead of 0.15, would get 1.5 × 200 + 20 = £320. A candidate who swaps the roles of the rate and the number of units would work out 0.15 × 20 + 200 = £203.
- (d) 3x² + 4x − 7 — Method: the minus sign in front of the second bracket changes the sign of every term inside it; then collect like terms. Working: removing the brackets gives 5x² + 3x − 2 − 2x² + x − 5; the squared terms give 5x² − 2x² = 3x², the x terms give 3x + x = 4x, and the number terms give −2 − 5 = −7. Answer: 3x² + 4x − 7. The distractors: 7x² + 2x + 3 comes from adding the two brackets instead of subtracting, giving 5x² + 2x², 3x − x and −2 + 5; 3x² + 2x + 3 comes from applying the minus sign to 2x² only, leaving −x and +5 unchanged so that 3x − x = 2x and −2 + 5 = 3; 3x² + 4x + 3 comes from changing the signs of the terms with letters but leaving +5 as it stood, so the number terms give −2 + 5 = 3.
- (c) 450 g — Method: use the amount of butter given to find the value of one part of the ratio, then find the mass of flour, and finally add flour and butter to get the total. Working: 180 g of butter is 2 parts, so one part is 180 ÷ 2 = 90 g. The flour is 3 parts, so 3 × 90 = 270 g, and the total mass is 270 + 180 = 450 g. So the baker can make 450 g of pastry. Distractor 270 g is only the mass of flour, forgetting to add the butter back on. Distractor 300 g comes from treating the 180 g as 3 parts instead of 2, swapping which ratio number matches the butter. Distractor 540 g comes from multiplying 180 by 3 directly instead of first finding the value of one part.
- (d) 26 — Method: expand both brackets, treating the second as multiplication by −2 so that both of its terms change sign, then collect like terms. Working: 2(a + 6) = 2a + 12 and −2(a − 7) = −2a + 14, so the expression becomes 2a + 12 − 2a + 14; the a terms give 2a − 2a = 0, so no term in a survives, and the numbers give 12 + 14 = 26. Answer: 26. The distractors: −2 comes from expanding the second bracket as −2a − 14, so that the numbers give 12 − 14; 4a − 2 comes from adding 2(a − 7) instead of subtracting it, giving 2a + 12 + 2a − 14; 13 comes from multiplying the 2 over only the first term of each bracket, giving 2a + 6 − 2a + 7.
- (c) y = 4x − 5 — Parallel lines have the same gradient, so the new line has gradient 4; since it passes through (0, −5), its y-intercept is −5, giving y = 4x − 5. A candidate who drops the negative sign on the y-intercept would write y = 4x + 5. A candidate who changes the sign of the gradient, instead of keeping it the same for a parallel line, would write y = −4x − 5. A candidate who confuses m and c, using the y-intercept of the first line (3) as the gradient of the second, would write y = 3x − 5.
- (b) 24 — Method: let one part of the ratio be worth x, write both groups in terms of x, and use the fact that the two groups end up equal. Working: the boys are 3x and the girls are 4x; after the 6 boys join, 3x + 6 = 4x, so x = 6; the girls are 4 parts, so 4 × 6 = 24. Answer: 24 girls. The distractors: 18 is the number of boys before the 6 join, which is 3 × 6; 30 comes from adding the 6 new members to the girls as well as to the boys; 42 is the total number of members in the choir before the 6 boys join, the 18 boys and the girls together.
- (d) 21.2 m — Method: the cable is the hypotenuse of a right-angled triangle whose vertical side is the drop from the roof to the bracket and whose horizontal side is 15 m, so use Pythagoras' theorem. Working: the drop is 20 − 5 = 15 m, so c² = 15² + 15² = 225 + 225 = 450 and c = √450 = 21.213…, which is 21.2 m to 1 decimal place. Answer: 21.2 m. The distractors: 25.0 m comes from using the whole 20 m height of the roof as the vertical side and forgetting that the bracket is already 5 m up; 30.0 m comes from adding the two sides of the triangle, 15 + 15, instead of using Pythagoras' theorem; 15.0 m is the horizontal distance on its own, which would be the length of the cable only if it ran level.
- (a) 1.5 — Method: with an even number of values the median is the mean of the two middle values, which for 12 values are the 6th and the 7th once the data are in order. Working: the results are already in order, and 12 ÷ 2 = 6, so the middle pair are the 6th value, 1, and the 7th value, 2; the median is (1 + 2) ÷ 2 = 1.5. Answer: 1.5 brothers and sisters. The distractors: 1 comes from reading the 6th value and stopping there instead of averaging the middle pair; 2 comes from working out the mean, 24 ÷ 12, instead of the median; 6 comes from working out the range, 6 − 0, which measures spread rather than centre.
- (d) (x − 8)(x + 8) — x² − 64 = x² − 8², a difference of two squares, which factorises as (x − 8)(x + 8). A candidate who treats it as a perfect square with a repeated negative factor gets (x − 8)(x − 8), which expands to x² − 16x + 64 — wrong on both the middle and constant terms. A candidate who uses a repeated positive factor gets (x + 8)(x + 8), which expands to x² + 16x + 64. A candidate who picks a different factor pair of 64, such as 4 and 16, without checking that the middle term cancels, gets (x − 4)(x + 16), which expands to x² + 12x − 64 — the wrong middle term.
- (d) x ≤ 30 — Method: add the fixed cost to the cost of x balloons, set that total against the £150 limit with the sign that 'at most' calls for, then solve. Working: the total spend is 60 + 3x pounds, so 60 + 3x ≤ 150; subtracting 60 from both sides gives 3x ≤ 90; dividing both sides by 3, a positive number, gives x ≤ 30. Answer: x ≤ 30. The distractors: x ≤ 90 comes from taking the cake off the budget and stopping at 3x ≤ 90, reading the 90 as a number of balloons when it is the money left for them; x ≤ 50 comes from dividing the whole £150 by 3 and leaving the cake out of the calculation altogether; x ≥ 30 comes from reading 'at most' as 'at least', which reverses the condition.
- (d) 25 — Method: add the parts of the ratio, divide the total membership by the number of parts to find the value of one part, then multiply by the parts belonging to the group asked for. Working: 3 + 5 = 8 parts, 40 ÷ 8 = 5 members in one part, and the seniors are 5 parts, so 5 × 5 = 25. Answer: 25 seniors. The distractors: 15 is the number of juniors, which is the 3-part group; 5 is the size of one part only; 24 comes from dividing the 40 by 5, the seniors' number in the ratio, to get 8 and then multiplying that by 3.
- (c) y = 3x + 7 — Method: the gradient is the change in y divided by the change in x with both taken in the same order, and a point whose x-coordinate is 0 gives the constant straight away because it lies on the y-axis. Working: m = (7 − 1) ÷ (0 − (−2)) = 6 ÷ 2 = 3; the point (0, 7) lies on the y-axis, so c = 7 and the line is y = 3x + 7. Answer: y = 3x + 7. The distractors: y = −3x + 7 comes from taking the y-difference as 1 − 7 while taking the x-difference as 0 − (−2), so the two subtractions run in opposite orders; y = 3x + 1 comes from using the y-coordinate of (−2, 1) as the constant instead of the point that actually lies on the y-axis; y = (1/3)x + 7 comes from writing the gradient upside down as the change in x over the change in y, 2 ÷ 6.
- (b) £20 — Method: add the parts of the ratio, divide the amount by the number of parts to find the value of one part, then take the parts belonging to the person asked about. Working: 1 + 2 = 3 parts, £60 ÷ 3 = £20 for one part, and Isla has 1 part. Answer: £20. The distractors: £40 is the 2-part share, which belongs to her brother; £30 comes from reading 1:2 as one part out of two and halving the £60; £120 comes from multiplying the £60 by 2 instead of dividing it into 3 parts.
- (b) 12 — The difference between the parts of the ratio is 5 − 2 = 3 parts, and this is worth 18. Divide to find one part: 18 ÷ 3 = 6. Cats have 2 parts: 2 × 6 = 12. (30 is the number of dogs, using 5 parts instead of 2. 6 is the value of one part — the number of cats is 2 lots of this, not just one. 9 comes from dividing 18 by 2 and stopping there, instead of dividing by the difference in parts, 3, and then multiplying by 2.)
- (a) 5 — The symbol ≤ means n can equal 5 or any number less than 5, so 5 is included and is the largest integer value. A candidate who treats the inequality as strict, as if it were n < 5, answers 4. A candidate who confuses ≤ with ≥ and looks for a value just above the boundary answers 6. A candidate who makes a sign error and reads the inequality as n ≤ −5 answers −5.
- (d) a² + b² = c² — Pythagoras' theorem states that the square of the hypotenuse equals the sum of the squares of the other two sides, so a² + b² = c². "a + b = c" adds the sides directly without squaring them at all. "a² − b² = c²" subtracts the squares instead of adding them. "a² + b² = c" adds the squares correctly but forgets to square the hypotenuse on the other side of the equation.
- (d) 60° — Method: the six angles at the centre together make one complete turn of 360°, and because the hexagon is regular they are all equal, so divide 360° by 6. Working: 360 ÷ 6 = 60. Answer: 60°. The distractors: 120° is the interior angle of a regular hexagon, 720 ÷ 6, which is the angle at a vertex and not the angle at the centre; 45° comes from dividing 360 by 8, treating the hexagon as though it had eight sides; 30° comes from halving the angle at the centre, as though each of the six triangles were split again by a line of symmetry.
What is on this worksheet?
The sheet holds 20 questions drawn from the MathsUK bank — the content areas covered: Algebra, Ratio, proportion and rates of change, Geometry and measures, Statistics (statements A4, A17, A22, A9, R5, R9, G20, G3, S4). It is pitched at GCSE Foundation and takes about 35 minutes to work through in full. It works as a class handout, a homework, or a warm-up before a test.
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 20 questions before checking — about 35 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 20 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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