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 25% of 200.
- 2.The price of a jacket increases by 50% and then decreases by 50%. Describe the overall change from the original price.
- 3.Solve the inequality 9 − 2x ≥ 1.
- 4.Factorise x² − 64.
- 5.A regular nonagon has 9 sides. Work out the sum of its interior angles.
- 6.Expand and simplify (x − 3)²
- 7.The mean of three numbers is 50. A fourth number, 100, is added to the set. Work out the mean of the four numbers.
- 8.A jug of squash is made by mixing water and syrup in the ratio 6:1. Nia wants to make 8.4 litres of squash. Work out how much syrup she needs, in litres.
- 9.Solve (3x + 2)/5 = (x + 4)/3
- 10.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.
- 11.Solve 5(x + 3) = 40
- 12.In a shop a shirt costs £x and a pair of trousers costs £(2x + 30). Together the two items cost at least £150. Work out the lowest possible price of the shirt.
- 13.In a test, Amelia answered 18 of the 24 questions correctly. Work out the percentage of the questions she answered correctly.
- 14.To solve 5x − 4 = 2x + 8, Chloe's working is shown. Line 1: 5x − 4 = 2x + 8. Line 2: 3x − 4 = 8. Line 3: 3x = 4. Line 4: x = 4/3. Chloe has made a mistake in her working. Which line contains the mistake?
- 15.Work out the equation of the straight line through the points (−2, 1) and (0, 7).
- 16.After a 20% discount, a jacket costs £48. Work out the original price of the jacket.
- 17.A ladder of length 7 m leans against a vertical wall. The foot of the ladder is 3 m from the base of the wall. Work out how high up the wall the ladder reaches. Give your answer correct to 1 decimal place.
- 18.Three angles lie on a straight line. They measure 38°, 95° and x°. Work out the value of x.
- 19.Solve 3x − 5 = 4.
- 20.40% of a number is 12 more than 25% of the same number. Work out the number.
Answer key
- (a) 50 — Method: 25% is one quarter, so 25% of a quantity is the quantity divided by 4. Working: 200 ÷ 4 = 50. Answer: 50. The distractors: 25 comes from writing the percentage itself as the answer; 100 comes from halving, which is 50% not 25%; 800 comes from multiplying by 4 instead of dividing.
- (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) x ≤ 4 — Subtract 9 from both sides: −2x ≥ 1 − 9 = −8. Divide both sides by −2, flipping the inequality: x ≤ 4. A candidate who divides by −2 but forgets to flip the inequality gets x ≥ 4. A candidate who divides −8 by −2 but keeps a negative sign gets x ≤ −4. A candidate who miscalculates 1 − 9 as −10 gets x ≤ 5.
- (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.
- (a) 1260° — The sum of the interior angles of a polygon with n sides is (n − 2) × 180°. For a nonagon, n = 9, so the sum is (9 − 2) × 180° = 7 × 180° = 1260°. 1620° uses 9 × 180° without subtracting 2 from n first. 140° is the size of a single interior angle of a regular nonagon (1260° ÷ 9), not the sum of all nine. 1440° uses (n − 1) × 180° = 8 × 180° instead of (n − 2) × 180°.
- (d) x² − 6x + 9 — Method: squaring a bracket means multiplying that bracket by itself, so expand (x − 3)(x − 3) term by term and then collect like terms. Working: x × x = x², x × (−3) = −3x, (−3) × x = −3x and (−3) × (−3) = 9, giving x² − 3x − 3x + 9, and the two middle terms collect to −6x. Answer: x² − 6x + 9. The distractors: x² − 3x + 9 comes from writing down only one of the two middle products instead of both; x² + 6x + 9 comes from treating (−3) × x as +3x, so the middle terms are added rather than subtracted; x² − 9 comes from treating the square as the difference of two squares (x − 3)(x + 3).
- (a) 62.5 — Method: a mean cannot be averaged with a new value — rebuild the total, add the new value to it, then divide by the new count. Working: three numbers with a mean of 50 have a total of 50 × 3 = 150; adding 100 makes the total 150 + 100 = 250; there are now 4 numbers, so the new mean is 250 ÷ 4 = 62.5. Answer: 62.5. The distractors: 75 comes from averaging the old mean with the new value, (50 + 100) ÷ 2, which ignores that three numbers pull against one; 50 comes from assuming an extra value leaves the mean unchanged; 37.5 comes from dividing the old total of 150 by the new count of 4, adding the new value to the count but not to the total.
- (c) 1.2 — Add the parts of the ratio: 6 + 1 = 7. Divide the total amount by the number of parts: 8.4 ÷ 7 = 1.2 litres, which is the value of one part and also the amount of syrup, since syrup is 1 part. (7.2 litres is the amount of water, using 6 parts instead of 1. 1.4 comes from dividing 8.4 by 6 — the water's part of the ratio — instead of dividing by the total number of parts, 7. 0.84 comes from dividing 8.4 by 10 instead of by 7.)
- (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.
- (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) x = 5 — Method: expand the bracket by multiplying both terms inside it by 5, then undo the addition and the multiplication in turn. Working: expanding gives 5x + 15 = 40; subtracting 15 from both sides gives 5x = 25; dividing both sides by 5 gives x = 5. Answer: x = 5. The distractors: x = 8 comes from dividing both sides by 5 first, reaching x + 3 = 8 and writing 8 as the value of x without taking the 3 away; x = 11 comes from adding 15 to both sides instead of subtracting it, giving 5x = 55; x = 7.4 comes from expanding 5(x + 3) as 5x + 3, multiplying only the x by the 5, which leads to 5x = 37.
- (d) £40 — Method: add the two prices to make one expression in x, turn 'at least' into ≥, solve the inequality, then read the lowest possible price off the boundary of the solution set. Working: the two items cost x + (2x + 30) = 3x + 30 pounds, so 3x + 30 ≥ 150; subtracting 30 from both sides gives 3x ≥ 120; dividing both sides by 3 gives x ≥ 40, and the smallest value the shirt price is allowed to take is the boundary, £40. Answer: £40. The distractors: £50 comes from leaving the £30 out of the total and solving 3x ≥ 150; £60 comes from using the trousers expression on its own, 2x + 30 ≥ 150; £120 comes from stopping at 3x ≥ 120 and reading the 120 as the price of the shirt without dividing by 3.
- (a) 75% — Method: to express one quantity as a percentage of another, divide the part by the whole and multiply by 100. Working: 18 ÷ 24 = 0.75, and 0.75 × 100 = 75. Answer: 75%. The distractors: 25% is the percentage she got wrong, 6 out of 24; 133% comes from dividing the whole by the part, 24 ÷ 18; 18% comes from writing the number of correct answers with a percent sign.
- (d) Line 3 — Method: check each line of Chloe's working against the correct algebraic step. Working: Line 1 to Line 2 is correct, subtracting 2x from both sides gives 3x − 4 = 8. But Line 2 to Line 3 should add 4 to both sides, giving 3x = 12, not 3x = 4 — the constant −4 has been dropped rather than removed correctly. Line 4 follows correctly from Chloe's own, incorrect, Line 3. Answer: Line 3. Line 1 is simply the original equation, copied out correctly. Line 2 correctly subtracts 2x from both sides of Line 1. Line 4 divides Chloe's own Line 3 by 3 correctly — the arithmetic there is fine, the mistake happened one line earlier.
- (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) £60 — £48 represents 100% − 20% = 80% of the original price. 1% = £48 ÷ 80 = £0.60, so 100% = £0.60 × 100 = £60.
- (c) 6.3 m — By Pythagoras' theorem, the height = √(7² − 3²) = √(49 − 9) = √40 = 6.32...≈ 6.3 m. "6.4 m" rounds 6.32...m up to 6.4 instead of correctly rounding it down to 6.3. "4.0 m" comes from subtracting the two given lengths directly, 7 − 3 = 4, instead of subtracting their squares. "10.0 m" comes from adding the two given lengths, 7 + 3 = 10, instead of using Pythagoras' theorem at all.
- (b) 47 — Method: angles on a straight line add up to 180°. Working: 180 − 38 − 95 = 47. Answer: x = 47. A candidate who uses 360° instead of 180° gets 227. A candidate who subtracts only one of the two given angles from 180° gets 142. A candidate who adds the two given angles instead of subtracting them from 180° gets 133.
- (d) 3 — Method: add the constant term to both sides first, then divide by the coefficient of x. Working: 3x = 4 + 5 = 9; x = 9 ÷ 3 = 3. Answer: x = 3. −1/3 comes from a sign error when moving the 5, subtracting instead of adding: 3x = 4 − 5 = −1, then x = −1/3. 6 comes from subtracting the coefficient 3 instead of dividing by it: 9 − 3 = 6. 9 comes from correctly finding 3x = 9 but forgetting to divide by 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.
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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