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.Work out the median of these five numbers: 2, 4, 7, 12, 26
- 2.Expand and simplify 2(a + 6) − 2(a − 7)
- 3.Solve 2x + 1 = 9
- 4.A regular polygon has an exterior angle of 45°. Work out the number of sides of the polygon.
- 5.A charity shop and a school share collection-box money in the ratio 5 : 8. The charity shop receives £47.50. Work out how much the school receives.
- 6.Solve (2x + 1)/3 = 5
- 7.A rectangle has length (2x + 5) cm and width (x − 2) cm. Work out an expression, in terms of x, for the perimeter of the rectangle. Give your answer in its simplest form.
- 8.Solve the inequality 2(x − 1) ≤ 8.
- 9.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.
- 10.Two angles are complementary. One of them is 35°. Work out the size of the other angle.
- 11.A rectangular garden has length (x + 7) m and width (x − 7) m. Work out an expression for the area of the garden, giving your answer in its simplest form.
- 12.The equation 3(2x − 1) = 4x + 9 is rearranged by expanding the brackets. Which of these is the correctly expanded equation?
- 13.40% of a number is 12 more than 25% of the same number. Work out the number.
- 14.Two straight lines cross at a point. Which statement about a pair of vertically opposite angles is always true?
- 15.A straight line crosses two parallel lines. One of the angles formed is (2x + 10)°, and the angle alternate to it is 74°. Work out the value of x.
- 16.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
- 17.Factorise fully 5x + 5y − 5
- 18.Work out the difference between 45% of 70 and 35% of 80.
- 19.Simplify x + x + 4
- 20.A right-angled triangle has a hypotenuse of 10 cm and one of its other angles is 45°. Work out the length of one of the two shorter sides. Give your answer to 1 decimal place.
Answer key
- (c) 7 — Method: the median is the middle value when the data are written in order of size, and with an odd number of values there is exactly one middle value. Working: the numbers are already in order, 2, 4, 7, 12, 26, and there are 5 of them, so the middle position is the third and the value sitting there is 7. Answer: 7, with two values below it and two above it. The distractors: 10.2 comes from working out the mean, 51 ÷ 5, instead of the median; 14 comes from taking the value halfway between the smallest and the largest, (2 + 26) ÷ 2; 24 comes from working out the range, 26 − 2, which measures spread rather than centre.
- (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) x = 4 — Method: undo the addition of 1 first, then undo the multiplication by 2. Working: subtracting 1 from both sides gives 2x = 8, and dividing both sides by 2 gives x = 4. Answer: x = 4. The distractors: x = 8 comes from stopping at 2x = 8 and writing 8 as the value of x; x = 5 comes from adding 1 to both sides instead of subtracting it, giving 2x = 10; x = 16 comes from multiplying 8 by 2 instead of dividing by 2.
- (c) 8 — Method: the exterior angles of a polygon add up to 360°, so divide 360° by the size of one exterior angle. Working: 360 ÷ 45 = 8. Answer: 8 sides. A candidate who divides into a half turn instead of a full turn, working out 180 ÷ 45, gets 4. A candidate who reads off the given exterior angle as if it were the number of sides gets 45. A candidate who subtracts instead of dividing, working out 360 − 45, gets 315.
- (b) £76.00 — One part of the ratio is £47.50 ÷ 5 = £9.50. The school receives 8 parts, so its share is 9.50 × 8 = £76.00. Dividing £47.50 by 8 instead of 5, treating the charity's amount as if it were 8 parts, gives 47.50 ÷ 8 = 5.9375, then × 5 = £29.69. Adding the charity's amount to the school's amount instead of stopping at the school's own share gives the total collected, 9.50 × 13 = £123.50. Adding one part to the charity's amount instead of multiplying one part by 8 gives 47.50 + 9.50 = £57.00.
- (a) 7 — Method: multiply both sides by 3 to clear the fraction, then solve the resulting equation. Working: 2x + 1 = 5 × 3 = 15. Subtract 1: 2x = 14. Divide by 2: x = 7. Answer: 7. 2 comes from ignoring the denominator altogether, treating the equation as 2x + 1 = 5 without multiplying by 3 first. 8 comes from a sign error, adding 1 to 15 instead of subtracting it, giving 2x = 16. 14 comes from correctly reaching 2x = 14 but stopping there, without dividing by 2 to find x.
- (b) 6x + 6 — Perimeter = 2[(2x + 5) + (x − 2)] = 2(3x + 3) = 6x + 6. A candidate who adds the length and width but forgets to double for the perimeter gets 3x + 3. A candidate who makes a sign error and adds 2 instead of subtracting it before doubling gets 2[(2x + 5) + (x + 2)] = 6x + 14. A candidate who multiplies the length and width instead of adding them, confusing the perimeter formula with the area formula, and then doubles that product, gets 2(2x + 5)(x − 2) = 4x² + 2x − 20.
- (b) x ≤ 5 — Method: divide out the bracket first, then undo the number term; the inequality sign turns round only if both sides are multiplied or divided by a negative number. Working: dividing both sides of 2(x − 1) ≤ 8 by 2 gives x − 1 ≤ 4, and 2 is positive so the ≤ is unchanged; adding 1 to both sides gives x ≤ 5. Answer: x ≤ 5. The distractors: x ≤ 3 comes from subtracting 1 from 4 instead of adding 1 to both sides; x ≤ 4 comes from stopping at 8 ÷ 2 = 4 and never undoing the −1 inside the bracket; x < 5 comes from reading ≤ as a strict inequality, which wrongly leaves the boundary value out of the solution set.
- (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.
- (b) 55° — Method: complementary angles are a pair that add up to 90°, so subtract the known angle from 90°. Working: 90 − 35 = 55. Answer: 55°. The distractors: 145° comes from subtracting from 180°, which is the rule for supplementary angles and not for complementary ones; 325° comes from subtracting from 360°, the total of the angles at a point; 90° is the total that the pair must make, written down in place of the missing angle.
- (d) x² − 49 — Area = length × width = (x + 7)(x − 7). Expanding: x × x = x², x × (−7) = −7x, 7 × x = 7x, 7 × (−7) = −49. The two middle terms −7x and 7x cancel, leaving x² − 49. A candidate who misremembers the difference-of-two-squares result as a sum gets x² + 49. A candidate who makes a sign error and treats both middle terms as −7x instead of cancelling gets x² − 14x − 49. A candidate who confuses area with perimeter and simply adds the length and width gets 2x.
- (b) 6x − 3 = 4x + 9 — Method: multiply every term inside the bracket by the number outside it; the right-hand side stays as it is given. Working: 3 × 2x = 6x and 3 × (−1) = −3, so 3(2x − 1) = 6x − 3. Answer: 6x − 3 = 4x + 9. 6x − 1 = 4x + 9 comes from multiplying only the 2x by 3 and leaving the −1 unchanged. 5x − 3 = 4x + 9 comes from adding the 3 to the 2 instead of multiplying, treating 3 × 2x as (3 + 2)x = 5x. 6x − 4 = 4x + 9 comes from working out 3 × (−1) as −1 − 3 = −4 instead of 3 × (−1) = −3.
- (b) 80 — Method: the difference between 40% and 25% of the number is 15% of the number, and that difference is 12. Working: 15% of the number is 12, so 1% of the number is 12 ÷ 15 = 0.8, and the number is 0.8 × 100 = 80. Check: 40% of 80 is 32, 25% of 80 is 20, and 32 − 20 = 12. Answer: 80. The distractors: 30 comes from solving 40% of the number = 12; 48 comes from solving 25% of the number = 12; 15 is the percentage difference written as the answer.
- (d) They are always equal to each other — Method: label the four angles a, b, c, d in order around the crossing point, then use the fact that neighbouring angles lie on a straight line. Working: a and b lie on a straight line, so a + b = 180°; b and c also lie on a straight line, so b + c = 180°. Both a and c are therefore 180° minus b, which forces a = c. Answer: a pair of vertically opposite angles is always equal to each other. The distractors: the claim that each is 90° holds only when the two lines happen to be perpendicular, so it is not always true; the claim that they add to 180° confuses the opposite pair with the neighbouring pair that lies along a straight line, and again holds only in the perpendicular case; the claim that they add to 360° uses the total of all four angles at the point rather than of the opposite pair.
- (c) 32 — Method: alternate angles between parallel lines are equal, so 2x + 10 = 74. Working: subtracting 10 from both sides gives 2x = 64; dividing by 2 gives x = 32. Answer: x = 32. A candidate who forgets to subtract 10 first and divides 74 by 2 directly gets 37. A candidate who treats the angles as co-interior instead of alternate, so that the two expressions add to 180° rather than being equal, gets 48 after solving. A candidate who makes a sign error and treats the equation as 2x equalling 10 minus 74 instead of 74 minus 10 gets −32.
- (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.
- (c) 5(x + y − 1) — Method: take out the highest common factor of all three terms and divide every term by it, the number term included. Working: the highest common factor of 5x, 5y and −5 is 5; dividing gives 5x ÷ 5 = x, 5y ÷ 5 = y and −5 ÷ 5 = −1, so the bracket holds x + y − 1. Answer: 5(x + y − 1), which multiplies back out to 5x + 5y − 5. The distractors: 5(x + y + 1) comes from dividing −5 by 5 and losing the minus sign; 5(x + y − 5) comes from dividing only the terms containing a letter by 5 and carrying the −5 into the bracket unchanged; 5(xy − 1) comes from collecting the unlike terms 5x and 5y as 5xy before factorising.
- (b) 3.5 — Method: work out each percentage of its number separately, then subtract the smaller result from the larger one. Working: 45% of 70 = 31.5, and 35% of 80 = 28, so the difference is 31.5 − 28 = 3.5. Answer: 3.5. 11.5 comes from pairing the percentages with the wrong numbers, working out 35% of 70 = 24.5 and 45% of 80 = 36, and finding their difference. 59.5 comes from adding the two correct results, 31.5 + 28, instead of subtracting them. 10 comes from simply subtracting the two percentages themselves, 45 − 35, without applying them to the numbers at all.
- (d) 2x + 4 — Method: collect the like terms, which here are the two terms in x, and leave the number term on its own because a number and a term in x are unlike. Working: x + x is one lot of x added to one more lot of x, which is 2x; the 4 has nothing like it to join with, so it is written after the 2x. Answer: 2x + 4. The distractors: x² + 4 comes from multiplying the two x terms instead of adding them; x + 4 comes from treating x + x as a single x, as though the repeated letter counted only once; 6x comes from collecting the unlike terms together, adding 1 + 1 + 4 and attaching the letter to that total.
- (c) 7.1 cm — Method: a shorter side is opposite the 45° angle and the hypotenuse is known, so sin θ = opposite ÷ hypotenuse gives that side directly. Working: sin 45° = x ÷ 10, so x = 10 × sin 45° = 7.071…, which is 7.1 to 1 decimal place. Answer: 7.1 cm. The distractors: 5.0 cm comes from halving the hypotenuse, which is the rule for the side opposite a 30° angle and not a 45° one; 14.1 cm comes from dividing by sin 45° instead of multiplying by it, which makes a shorter side longer than the hypotenuse; 10.0 cm comes from taking tan 45° = 1 and concluding that the shorter side matches the hypotenuse.
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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