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.Two identical ladders lean against the same vertical wall from opposite sides, each making an angle of 58.2° with the ground. The two ladders and the ground form a triangle. Work out the size of the angle between the two ladders at the top, where they meet.
- 2.Simplify 2b − 1 + 2b − 4 + 2b
- 3.40% of a number is 12 more than 25% of the same number. Work out the number.
- 4.The masses of eight school bags, in kilograms, are 3, 4, 4, 5, 6, 7, 8 and 11. Work out the median mass.
- 5.A charity fun run raises money through entry fees and donations. Entry fees raise £1,260, which is 60% of the total amount raised. Work out how much money was raised through donations.
- 6.In a right-angled triangle one of the other two angles is 45°, and the side opposite that 45° angle is 7 cm. Work out the length of the hypotenuse. Give your answer to 1 decimal place.
- 7.A zip wire is fixed at an angle of 50° to the horizontal, running from a high platform down to the ground. The zip wire itself, from the platform to the ground, is 12 m long. Work out the horizontal distance covered by the zip wire. Give your answer correct to 1 decimal place.
- 8.Work out 50% of 60.
- 9.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
- 10.Which of these equations is equivalent to 5x − 2 = 3x + 8?
- 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.In a spelling test the 20 pupils in Group A had a mean mark of 80, and the 30 pupils in Group B had a mean mark of 70. Work out the mean mark of all 50 pupils.
- 13.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.
- 14.A metal alloy is made from copper and tin in the ratio 7:3. Work out the mass of tin in 250 g of the alloy.
- 15.Solve the inequality 9 − 2x ≥ 1.
- 16.Work out the median of these six numbers: 13, 21, 22, 36, 37, 47
- 17.Seven pupils were asked how many books they had read last month. Their answers were 3, 5, 5, 7, 8, 5, 3. Write down the mode.
- 18.Solve 3x − 5 = 4.
- 19.A conservatory has a roof panel shaped like a regular octagon. Work out the interior angle of the octagon, then use it to work out the size of the reflex angle at the same vertex, on the outside of the roof panel.
- 20.Two straight lines cross at a point. Which statement about a pair of vertically opposite angles is always true?
Answer key
- (d) 63.6 — Method: the three angles inside the triangle formed by the two ladders and the ground add up to 180°. Working: 180 − 58.2 − 58.2 = 63.6. Answer: 63.6°. A candidate who assumes the top angle equals the base angles gives 58.2. A candidate who subtracts only one base angle from 180°, working out 180 − 58.2, gets 121.8. A candidate who doubles the base angle instead of subtracting it twice from 180°, working out 2 × 58.2, gets 116.4.
- (a) 6b − 5 — Method: sort the expression into terms in b and number terms, then collect each kind, keeping the sign in front of every term. Working: the b terms give 2b + 2b + 2b = 6b; the number terms are −1 and −4, and −1 − 4 = −5. Answer: 6b − 5. The distractors: 6b − 3 comes from working out the difference 4 − 1 instead of adding two negative numbers; 6b + 5 comes from adding 1 and 4 while ignoring both minus signs; 8b − 5 comes from multiplying the coefficients 2, 2 and 2 instead of adding them.
- (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) 5.5 kg — Method: with an even number of values the median is the mean of the two middle values, taken once the data are in order of size. Working: the eight masses are already in order and 8 ÷ 2 = 4, so the middle pair are the 4th and 5th values, 5 kg and 6 kg; the median is (5 + 6) ÷ 2 = 5.5 kg. Answer: 5.5 kg. The distractors: 5 kg comes from reading the 4th value and stopping there instead of averaging the middle pair; 8 kg comes from working out the range, 11 − 3, which measures spread rather than centre; 4 kg comes from writing down the modal mass, the only value that occurs twice, instead of the median.
- (c) £840 — Method: find the total amount raised using the reverse percentage, then subtract the entry fees to find the donations. Working: £1,260 is 60% of the total, so the total is £1,260 ÷ 0.6, and subtracting the entry fees from this total leaves £840 raised through donations. Answer: £840. £2,100 comes from correctly finding the total amount raised but then forgetting to subtract the entry fees, giving the total instead of the donations alone. £504 comes from working out 40% of the entry fees themselves, £1,260 × 0.4 = £504, instead of first finding the total amount raised. £1,890 comes from treating £1,260 as 40% of the total instead of 60%, dividing by 0.4 to get a total of £3,150, and then subtracting the entry fees from that incorrect total.
- (c) 9.9 cm — Method: the 7 cm side is opposite the 45° angle and the hypotenuse is wanted, so use sin θ = opposite ÷ hypotenuse and rearrange it for the hypotenuse. Working: sin 45° = 7 ÷ h, so h = 7 ÷ sin 45° = 9.899…, which is 9.9 to 1 decimal place. Answer: 9.9 cm. The distractors: 5.0 cm comes from multiplying by sin 45° instead of dividing by it; 14.0 cm comes from doubling the 7 cm side, which is the rule for a side opposite 30° and not one opposite 45°; 7.0 cm comes from reading the two equal sides of a 45° right-angled triangle as including the hypotenuse, when the equal pair is the two shorter sides.
- (d) 7.7 m — The horizontal distance is adjacent to the 50° angle and the zip wire is the hypotenuse, so horizontal distance = 12 × cos 50° = 12 × 0.6428... = 7.71...≈ 7.7 m. "9.2 m" uses the sine ratio instead of cosine, 12 × sin 50° = 9.19...≈ 9.2 m, which actually finds the vertical drop of the zip wire, not the horizontal distance. "15.7 m" comes from dividing by the sine ratio instead of multiplying by the cosine ratio, 12 ÷ sin 50° = 15.66...≈ 15.7 m, both the wrong operation and the wrong ratio. "12.0 m" simply uses the length of the zip wire itself as the horizontal distance, ignoring the angle of 50° altogether.
- (d) 30 — Method: 50% is one half, so 50% of a quantity is the quantity divided by 2. Working: 60 ÷ 2 = 30. Answer: 30. The distractors: 120 comes from multiplying by 2 instead of dividing; 3000 comes from multiplying by 50 without dividing by 100; 6 comes from finding 10% instead of 50%.
- (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.
- (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.
- (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.
- (b) 74 marks — Method: the two groups are different sizes, so their means cannot simply be averaged — rebuild each group's total mark, add the totals and divide by all 50 pupils. Working: Group A scored 20 × 80 = 1600 marks and Group B scored 30 × 70 = 2100 marks, giving 1600 + 2100 = 3700 marks altogether, so the overall mean is 3700 ÷ 50 = 74 marks. Answer: 74 marks, which sits nearer to 70 than to 80 because the larger group scored 70. The distractors: 75 marks comes from averaging the two group means, (80 + 70) ÷ 2, as though the groups were the same size; 76 marks comes from attaching each mean to the other group's size, (20 × 70 + 30 × 80) ÷ 50; 150 marks comes from adding the two means together and never dividing at all.
- (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) 75 g — Method: split the total mass into the number of parts shown by the ratio, then find the mass of tin. Working: the ratio 7:3 has 7 + 3 = 10 parts, so one part is 250 ÷ 10 = 25 g, and the mass of tin is 3 × 25 = 75 g. So the alloy contains 75 g of tin. Distractor 175 g is the mass of copper, not tin. Distractor 125 g comes from splitting the alloy into two equal halves, ignoring the ratio. Distractor 25 g is the value of one part, found correctly but never multiplied by 3.
- (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) 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.
- (c) 5 — Method: the mode is the value that occurs most often, so count how many times each different value appears and compare the counts. Working: 3 appears twice, 5 appears three times, 7 appears once and 8 appears once, so the highest frequency is three and the value carrying it is 5. Answer: 5. The distractors: 3 comes from writing down the frequency of the most common answer instead of the answer itself; 7 comes from taking the middle number of the list as it was written, which applies the median without ordering the data and without answering the question asked; 8 comes from picking the largest value, which confuses the mode with the maximum.
- (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.
- (a) 225 — Method: find the interior angle of the regular octagon, then subtract it from 360° to find the reflex angle at the same vertex, since the interior angle and the reflex angle together make a full turn. Working: there are 8 − 2 = 6 triangles' worth of angle in the octagon, so the interior angle = 6 × 180 ÷ 8 = 135; reflex angle = 360 − 135 = 225. Answer: 225°. A candidate who stops after finding the interior angle gives 135. A candidate who works out the exterior angle instead, 360 ÷ 8 = 45, gives 45. A candidate who subtracts the exterior angle from 360° instead of the interior angle, working out 360 − 45, gets 315.
- (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.
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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