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.Solve (3x + 2)/5 = (x + 4)/3
- 2.Solve the inequality x + 5 < 12.
- 3.Which of these equations describes a vertical line?
- 4.The price of a jacket increases by 50% and then decreases by 50%. Describe the overall change from the original price.
- 5.Work out the equation of the straight line through the points (−2, 1) and (0, 7).
- 6.Expand 7(a + 8)
- 7.A company's profit this year is 130% of last year's profit. Last year's profit was £40,000. Work out this year's profit.
- 8.Factorise fully 5x + 5y − 5
- 9.Solve the inequality 5 − x ≤ 2.
- 10.A right-angled triangle has its two shorter sides equal to 8 cm and 15 cm. Work out the value of cos θ, where θ is the angle opposite the 8 cm side. Give your answer as a fraction.
- 11.40% of a number is 12 more than 25% of the same number. Work out the number.
- 12.The ratio of Josh's savings to Mia's savings is 4:7. Mia has £39 more than Josh. Work out Josh's savings.
- 13.A straight line passes through the points (2, 5) and (4, 11). Work out the gradient of the line.
- 14.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.
- 15.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.
- 16.Three business partners share a profit of £48,000 in the ratio 3:5:4. Work out how much the partner with 5 parts receives.
- 17.Priya has a budget of £50 for a school trip. The coach costs £14 and each student ticket costs £4. Using the inequality 14 + 4s ≤ 50, work out the greatest number of student tickets, s, she can buy.
- 18.The price of a share falls by 10% on Monday and then rises by 10% on Tuesday. Work out the overall percentage change from Monday's starting price.
- 19.Which of these equations is equivalent to 5x − 2 = 3x + 8?
- 20.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.
Answer key
- (a) x = 3.5 — Method: clear the fractions by multiplying both sides by 15, expand both brackets, then collect the x terms on one side and the numbers on the other. Working: multiplying both sides by 15 gives 3(3x + 2) = 5(x + 4), which expands to 9x + 6 = 5x + 20; subtracting 5x and 6 from both sides gives 4x = 14, and dividing both sides by 4 gives x = 3.5. Answer: x = 3.5. The distractors: x = 1 comes from collecting the x terms by adding the 5x instead of subtracting it, giving 9x + 5x = 20 − 6 and so 14x = 14; x = 4.5 comes from expanding 3(3x + 2) as 9x + 2, multiplying only the x term by the 3, which leads to 4x = 18; x = 10 comes from reaching 4x = 14 correctly and then subtracting 4 instead of dividing by 4.
- (a) x < 7 — Subtract 5 from both sides: x < 12 − 5, so x < 7. A candidate who adds 5 instead of subtracting gets x < 17. A candidate who subtracts the wrong way round gets x < −7. A candidate who correctly finds 7 but wrongly flips the inequality (as if dividing by a negative had happened) writes x > 7.
- (c) x = 5 — Method: every point on a vertical line has the same x-coordinate however far up or down the line it lies, so the equation of a vertical line fixes x at a number and does not involve y at all. Working: of the four equations only x = 5 fixes x; it is satisfied by (5, 0), (5, 1), (5, 7) and by every other point whose x-coordinate is 5, and those points form a vertical line. Answer: x = 5. The distractors: y = 5 fixes the y-coordinate instead of the x-coordinate, which gives a horizontal line; y = 5x is a line through the origin with gradient 5, steep but not vertical, and it has a different y-value for every x; x = y fixes neither coordinate and is the line through the origin with gradient 1.
- (c) a decrease of 25% — Method: use multipliers. An increase of 50% is × 1.5 and a decrease of 50% is × 0.5. Working: 1.5 × 0.5 = 0.75, so the final price is 75% of the original. Answer: a decrease of 25%. The distractors: no change comes from assuming +50% and −50% cancel; a decrease of 50% comes from applying only the second change; an increase of 25% has the direction wrong.
- (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.
- (d) 7a + 56 — Method: multiplying a single term over a bracket means every term inside the bracket is multiplied by the term outside. Working: 7 × a = 7a and 7 × 8 = 56, and the two products are added because the bracket contains an addition. Answer: 7a + 56. The distractors: 7a + 15 comes from adding 7 and 8 instead of multiplying them; 7a + 8 comes from multiplying only the first term inside the bracket and copying the 8 across unchanged; a + 56 comes from multiplying only the 8 and copying the a across unchanged.
- (b) £52,000 — Method: convert 130% to a decimal multiplier and multiply it by last year's profit. Working: 130% = 1.3, so this year's profit is £40,000 × 1.3 = £52,000. Answer: £52,000. £12,000 comes from using only the extra 30% (130% − 100%) and forgetting to include the original 100%, £40,000 × 0.3 = £12,000. £40,130 comes from simply adding 130 onto £40,000, treating the percentage as an amount of money rather than a multiplier. £5,200 comes from misreading 130% as 13%, giving £40,000 × 0.13 = £5,200.
- (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.
- (c) x ≥ 3 — Method: collect the number terms first; the x term is negative, so the final step multiplies both sides by −1, and that is the one step that turns the inequality sign round. Working: subtracting 5 from both sides of 5 − x ≤ 2 gives −x ≤ −3; multiplying both sides by −1 turns −x into x and −3 into 3, and because the multiplier is negative the ≤ becomes ≥, so x ≥ 3. Answer: x ≥ 3. The distractors: x ≤ 3 comes from multiplying by −1 without turning the sign round, the commonest slip on this type; x ≤ −3 comes from reading −x ≤ −3 as though the minus sign could simply be rubbed off the left-hand side; x ≥ −3 comes from turning the sign round correctly but leaving the right-hand side at −3 instead of multiplying it by −1 as well.
- (c) 15/17 — Method: cos θ = adjacent ÷ hypotenuse, so find the hypotenuse with Pythagoras' theorem first and then decide which short side is next to θ. Working: the hypotenuse is √(8² + 15²) = √(64 + 225) = √289 = 17 cm. The angle θ is opposite the 8 cm side, so the side next to it is the 15 cm side, and cos θ = 15 ÷ 17. Answer: 15/17. The distractors: 8/17 is sin θ, opposite over hypotenuse, used in place of the cosine; 8/15 is tan θ, opposite over adjacent; 17/15 comes from writing the cosine ratio upside down, as hypotenuse over adjacent.
- (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.
- (b) £52 — Method: the difference between the two ratio numbers tells you how many parts the £39 difference represents. Working: the difference in parts is 7 − 4 = 3, and this represents £39, so one part is £39 ÷ 3 = £13. Josh's savings are 4 × £13 = £52. So Josh has £52. Distractor £91 is Mia's savings, not Josh's. Distractor £39 comes from using the given £39 difference as the final answer, without scaling it to Josh's number of parts. Distractor £13 is the value of one part, found correctly but never multiplied by 4.
- (c) 3 — Gradient = (change in y) ÷ (change in x) = (11 − 5) ÷ (4 − 2) = 6 ÷ 2 = 3. A candidate who puts the change in x over the change in y instead would get 2 ÷ 6 = 1/3. A candidate who subtracts the y-coordinates in the reverse order, but not the x-coordinates, would get (5 − 11) ÷ (4 − 2) = −3. A candidate who adds the coordinates instead of subtracting them would get (11 + 5) ÷ (4 + 2) = 16/6 = 8/3.
- (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.
- (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.
- (a) £20,000 — Method: add the parts of the ratio, divide the profit by the number of parts to find the value of one part, then multiply by the parts belonging to the partner asked about. Working: 3 + 5 + 4 = 12 parts, £48,000 ÷ 12 = £4,000 for one part, and that partner has 5 parts, so 5 × £4,000 = £20,000. Answer: £20,000. The distractors: £4,000 is the value of one part; £12,000 is the 3-part share; £16,000 is the 4-part share, which is what a candidate reaches by assuming the biggest share must go with the number written last in the ratio.
- (b) 9 — Subtract 14 from both sides: 4s ≤ 36. Divide both sides by 4: s ≤ 9, so the greatest number of tickets is 9. A candidate who forgets the £14 coach cost solves 4s ≤ 50, getting s ≤ 12.5, rounded down to 12. A candidate who adds the £14 instead of subtracting it solves 4s ≤ 64, getting s = 16. A candidate who miscalculates 50 − 14 as 32 solves 4s ≤ 32, getting s = 8.
- (b) −1% — Method: write each change as a multiplier and multiply them. A 10% fall is × 0.9 and a 10% rise is × 1.1. Working: 0.9 × 1.1 = 0.99, so the final price is 99% of the original, which is 1% less. Answer: an overall change of −1%. The distractors: 0% comes from assuming a 10% fall and a 10% rise cancel — they do not, because the rise is 10% of a smaller amount; +1% has the size right but the sign wrong, from reading the multiplier 0.99 as 1% above 1 instead of 1% below it; −2% comes from finding the 1% fall and then counting it once for each of the two changes.
- (d) 2x = 10 — Method: collect the x-terms on one side and the constants on the other, keeping the equals sign balanced. Working: 5x − 3x = 8 + 2, giving 2x = 10. Answer: 2x = 10. 2x = 6 comes from a sign error moving the constant, subtracting instead of adding: 8 − 2 = 6. 8x = 6 comes from adding the x-terms instead of subtracting them (5x + 3x = 8x), alongside the constant sign error. 2x = −10 comes from a sign error moving the constant the other way: −2 − 8 = −10.
- (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.
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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