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.A transversal crosses a pair of parallel lines. At one line, the angle is (5x + 4)°. The corresponding angle at the other line is (3x + 24)°. Work out the value of x.
- 2.2.4 kg of cheese costs £36. Work out the cost of 1.5 kg of the same cheese.
- 3.Two vertical masts stand on level ground, 40 m apart. One mast is 30 m tall and the other is 50 m tall. Work out the straight-line distance between the tops of the two masts. Give your answer to 1 decimal place.
- 4.Work out the median of these six numbers: 13, 21, 22, 36, 37, 47
- 5.Solve x/4 + 1 = 3
- 6.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.
- 7.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.
- 8.Four angles meet at a point. Three of them measure 82°, 105° and 96°. Work out the size of the fourth angle.
- 9.A jumper costs £45 at Shop A, where it is reduced by 20%. The same jumper costs £34 at Shop B, where a further 10% reduction is then applied. Work out the difference between the two reduced prices.
- 10.A right-angled triangle has angles of 90°, 40° and x°. Work out the value of x.
- 11.A straight line crosses two parallel lines. An angle of 65° is formed at one crossing. At the other crossing, the angle that lies between the two parallel lines on the opposite side of the crossing line is also 65°. Write down the angle fact that explains why these two angles are equal.
- 12.Factorise x² − 64.
- 13.Expand 2(x + 12)
- 14.A regular polygon has an interior angle of 156°. Work out the number of sides of the polygon.
- 15.A straight line has equation y = 2x − 7. Write down the coordinates of the point where the line crosses the y-axis.y = 2x − 7
- 16.Solve (3x + 2)/5 = (x + 4)/3
- 17.Solve 2x + 1 = 9
- 18.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.
- 19.A ribbon of length 90 cm is cut into two pieces in the ratio 4:5. Work out the length of the shorter piece.
- 20.A regular polygon has 12 sides. Work out the size of one exterior angle of the polygon.
Answer key
- (b) 10 — Corresponding angles are equal, so 5x + 4 = 3x + 24. Subtracting 3x from both sides gives 2x + 4 = 24, then subtracting 4 gives 2x = 20, so x = 10. 14 comes from adding the constants, 4 + 24, instead of subtracting them when rearranging. 19 comes from treating the angles as co-interior (summing to 180°): 5x + 4 + 3x + 24 = 180 gives 8x = 152, so x = 19. 20 correctly reaches 2x = 20 but stops without dividing by 2.
- (c) £22.50 — Method: find the cost of 1 kg, then multiply by the mass wanted. Working: £36 ÷ 2.4 = £15 per kilogram, and 1.5 × £15 = £22.50. Answer: £22.50. The distractors: £15 is the price of 1 kg, which is the first step and not what the question asks for; £54 comes from multiplying the £36 by 1.5 without first reducing it to a price per kilogram; £57.60 comes from using the scale factor upside down, multiplying £36 by 2.4 ÷ 1.5 = 1.6 instead of by 1.5 ÷ 2.4.
- (a) 44.7 m — Method: the line joining the two tops is the hypotenuse of a right-angled triangle whose horizontal side is the gap between the masts and whose vertical side is the difference in their heights, so Pythagoras' theorem applies. Working: the difference in heights is 50 − 30 = 20 m, so d² = 40² + 20² = 1600 + 400 = 2000 and d = √2000 = 44.721…, which is 44.7 m to 1 decimal place. Answer: 44.7 m. The distractors: 34.6 m comes from subtracting the squares, √(40² − 20²), instead of adding them; 60.0 m comes from adding the two sides of the triangle, 40 + 20, rather than using Pythagoras' theorem; 50.0 m is the height of the taller mast, copied from the question in place of the distance asked for.
- (d) 29 — Method: with an even number of values there is no single middle value, so the median is the mean of the two values either side of the middle. Working: the six numbers are already in order and 6 ÷ 2 = 3, so the middle pair are the third and fourth values, 22 and 36; their mean is (22 + 36) ÷ 2 = 58 ÷ 2 = 29. Answer: 29, which lies between the two middle values as a median of an even data set must. The distractors: 22 comes from taking the lower of the two middle values and stopping there instead of averaging the pair; 36 comes from taking the larger value of that pair because it sits just past the halfway point of the list; 34 comes from working out the range, 47 − 13, instead of a measure of centre.
- (b) x = 8 — Method: undo the addition first, then undo the division by 4. Working: subtracting 1 from both sides gives x/4 = 2, and multiplying both sides by 4 gives x = 8. Answer: x = 8. The distractors: x = 2 comes from stopping at x/4 = 2 and writing 2 as the value of x; x = 16 comes from adding 1 to both sides instead of subtracting it, giving x/4 = 4; x = 0.5 comes from dividing by 4 instead of multiplying by 4 at the last step.
- (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.
- (d) 77 — Method: angles that meet at a point add up to 360°. Working: 82 + 105 + 96 = 283; 360 − 283 = 77. Answer: 77°. A candidate who gives the sum of the three known angles and forgets to subtract it from 360° gets 283. A candidate who leaves out the 105° angle, working out 360 − 82 − 96, gets 182. A candidate who leaves out the 82° angle, working out 360 − 105 − 96, gets 159.
- (d) £5.40 — Method: work out the reduced price at each shop separately, then subtract the smaller from the larger. Working: Shop A's reduced price is £45 × 0.8 = £36, and Shop B's reduced price is £34 × 0.9 = £30.60, so the difference is £36 − £30.60 = £5.40. Answer: £5.40. £11.00 comes from comparing the two ORIGINAL prices, £45 − £34, without applying either shop's reduction at all. £1.60 comes from finding Shop A's reduced price correctly, £36, but then subtracting Shop B's original (unreduced) price of £34 instead of its reduced price. £66.60 comes from adding the two reduced prices together, £36 + £30.60, instead of subtracting them.
- (d) 50 — The angles in any triangle add up to 180°, so x = 180 − 90 − 40 = 50. "90" comes from subtracting only the 90° angle from 180° and forgetting to also subtract the 40°. "130" comes from adding 90° and 40° together instead of subtracting them from 180°. "45" comes from halving the 90° angle instead of using the angle sum of the triangle at all.
- (d) Alternate angles are equal — The two 65° angles are on opposite sides of the line that crosses the parallel lines, in the shape of a Z, so they are alternate angles, and alternate angles between parallel lines are always equal. Corresponding angles are equal too, but they sit in matching positions at each crossing point, in the shape of an F — a different pair from the one shown here. Co-interior angles add up to 180°, not to each other's value, and they lie between the parallel lines on the same side, in the shape of a C. Angles on a straight line add up to 180°, but that rule is about two angles at a single point on one line, not about a pair of angles formed where a line crosses two parallel lines.
- (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) 2x + 24 — Method: multiply each term inside the bracket by the 2 in front of it. Working: 2 × x = 2x and 2 × 12 = 24, and the products are added because the bracket contains an addition. Answer: 2x + 24. The distractors: 2x + 12 comes from multiplying only the x by 2 and copying the 12 across unchanged; 24x comes from multiplying 2 by 12 and then attaching the letter to that product, as though the two terms inside the bracket could be joined into one; 2x + 14 comes from adding 2 and 12 instead of multiplying them.
- (c) 15 — The exterior angle is 180° − 156° = 24°, and the number of sides of a regular polygon is 360° divided by the exterior angle, so 360 ÷ 24 = 15. 24° is the exterior angle itself, stopping one step before the final division. 17 comes from finding 15 correctly and then adding 2, muddling the exterior angle rule with the (n − 2) that appears in the interior angle sum formula. 7.5 comes from dividing 180 by the exterior angle instead of 360, using the angles on a straight line rather than the total of the exterior angles of a polygon.
- (d) (0, −7) — At the y-axis, x = 0, so y = 2(0) − 7 = −7, giving the point (0, −7). A candidate who drops the negative sign on the constant term would write (0, 7). A candidate who swaps the coordinates, confusing the y-intercept with an x-intercept, would write (−7, 0). A candidate who uses the gradient, 2, as the x-coordinate instead of 0 would write (2, −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.
- (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.
- (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.
- (c) 40 cm — Method: split the total length into the number of parts shown by the ratio, then find the value of the shorter share. Working: the ratio 4:5 has 4 + 5 = 9 parts, so one part is 90 ÷ 9 = 10 cm, and the shorter piece is 4 × 10 = 40 cm. So the shorter piece is 40 cm. Distractor 50 cm is the length of the LONGER piece, not the shorter one. Distractor 45 cm comes from splitting the ribbon into two equal halves, ignoring the ratio. Distractor 10 cm is the value of one part, found correctly but never multiplied by 4.
- (a) 30° — Method: the exterior angles of any convex polygon add up to 360°, and in a regular polygon they are all equal, so divide 360° by the number of sides. Working: 360 ÷ 12 = 30. Answer: 30°. The distractors: 150° is the interior angle, 180 − 30, which answers for the wrong angle at the vertex; 15° comes from dividing 180 by 12, using the angles on a straight line instead of the full turn; 36° comes from dividing 360 by 12 − 2 = 10, carrying the subtraction of 2 out of the interior angle sum formula into a calculation that does not need it.
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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