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 straight lines cross at a point. Which statement about a pair of vertically opposite angles is always true?
- 2.Work out the equation of the straight line that passes through (−2, 3) and (4, −9).
- 3.A straight line has gradient 3 and passes through the point (1, 4). Work out the equation of the line.
- 4.Expand 2(x − 13)
- 5.Four books have 120, 200, 160 and 80 pages. Work out the mean number of pages.
- 6.In a right-angled triangle, θ is one of the acute angles. Write down the trigonometric ratio, in terms of the opposite and adjacent sides, that is used to find θ.
- 7.A taxi company charges a £2.50 booking fee plus £1.80 per mile. A journey costs £15.10 in total. Work out the number of miles travelled.
- 8.The marks scored by four pupils in a quiz were 8, 8, 8, 8. Work out the mean, the median and the mode of these marks.
- 9.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.
- 10.In a right-angled triangle, the side opposite angle θ is 6 cm and the hypotenuse is 10 cm. Work out the size of angle θ. Give your answer correct to 1 decimal place.
- 11.A straight line has gradient −3 and passes through the point (0, 1). Work out the value of y when x = 2.
- 12.A box holds 80 chocolates. 75% of them are milk chocolates. Work out how many milk chocolates are in the box.
- 13.Solve 3(2x − 1) = 27
- 14.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.
- 15.A regular nonagon has 9 sides. Work out the sum of its interior angles.
- 16.Solve 4(x − 3) = 2x + 6.
- 17.A regular polygon has an exterior angle of 45°. Work out the number of sides of the polygon.
- 18.There are 400 students at a school. 25% of them have a brother, 40% have a sister and 15% have both a brother and a sister. Work out how many of the students have neither a brother nor a sister.
- 19.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.
- 20.The solution to an inequality is n ≤ 5. Write down the largest integer value of n that satisfies this inequality.
Answer key
- (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.
- (a) y = −2x − 1 — Method: the gradient is the change in y divided by the change in x with both differences taken in the same order, and the constant then comes from substituting either point into y = mx + c. Working: m = (−9 − 3) ÷ (4 − (−2)) = (−12) ÷ 6 = −2, so the line is y = −2x + c; substituting (−2, 3) gives 3 = −2 × (−2) + c = 4 + c, so c = 3 − 4 = −1. Answer: y = −2x − 1. The distractors: y = −2x + 1 comes from rearranging 3 = 4 + c the wrong way round and taking the constant as 4 − 3; y = 2x + 7 comes from losing the minus sign when −12 is divided by 6 and then substituting correctly, 3 = 2 × (−2) + c; y = −(1/2)x + 2 comes from writing the gradient upside down as the change in x over the change in y, 6 ÷ (−12).
- (c) y = 3x + 1 — Method: a line of known gradient m has equation y = mx + c, and c is found by substituting the coordinates of a point known to lie on it. Working: the gradient is 3, so the line is y = 3x + c; substituting x = 1 and y = 4 gives 4 = 3 × 1 + c, so c = 4 − 3 = 1. Answer: y = 3x + 1. The distractors: y = 3x − 1 comes from working out the constant as mx − y, 3 − 4 = −1, instead of y − mx; y = x + 3 comes from swapping the two numbers over, putting the gradient 3 in the constant position and the x-coordinate 1 in front of x; y = 3x + 4 comes from using the y-coordinate 4 as the constant without substituting.
- (d) 2x − 26 — Method: multiply each term inside the bracket by the 2 in front of it, and keep the sign that belongs to each term. Working: 2 × x = 2x and 2 × 13 = 26; the bracket contains a subtraction, so the second term is subtracted. Answer: 2x − 26. The distractors: 2x − 13 comes from multiplying only the x by 2 and copying the 13 across unchanged; 2x + 26 comes from multiplying both terms correctly but losing the minus sign that belongs to the second term; 2x − 11 comes from subtracting 2 from 13 instead of multiplying 13 by 2.
- (c) 140 — Method: the mean shares the total equally between the items, so add the values and then divide by how many there are. Working: the total is 120 + 200 + 160 + 80 = 560 pages and there are 4 books, so the mean is 560 ÷ 4 = 140 pages. Answer: 140. The distractors: 560 comes from stopping at the total number of pages and never dividing by 4; 280 comes from dividing the total by 2 instead of by the 4 books; 120 comes from working out the range, 200 − 80, which measures spread instead of centre.
- (c) tan θ = opposite/adjacent — The tangent ratio is defined as tan θ = opposite/adjacent, so this is the ratio that connects the opposite and adjacent sides. "sin θ = opposite/adjacent" wrongly labels this ratio as sine, when sine actually connects the opposite side and the hypotenuse. "cos θ = opposite/adjacent" wrongly labels it as cosine, when cosine connects the adjacent side and the hypotenuse. "tan θ = adjacent/opposite" uses the correct ratio name but has the opposite and adjacent sides the wrong way round.
- (d) 7 — Method: set up the equation 2.50 + 1.80m = 15.10, then subtract the booking fee and divide by the cost per mile. Working: 1.80m = 15.10 − 2.50 = 12.60; m = 12.60 ÷ 1.80 = 7. Answer: 7 miles. 8.39 comes from dividing the whole £15.10 by £1.80 without first subtracting the booking fee: 15.10 ÷ 1.80 ≈ 8.39. 5.32 comes from swapping the two amounts round, subtracting £1.80 and dividing by £2.50: (15.10 − 1.80) ÷ 2.50 ≈ 5.32. 9.78 comes from adding the booking fee instead of subtracting it: (15.10 + 2.50) ÷ 1.80 ≈ 9.78.
- (c) mean = 8, median = 8, mode = 8 — Method: work out each measure separately — the mean is the total divided by how many values there are, the median is the middle value once the data are in order, and the mode is the value that occurs most often. Working: the total is 8 + 8 + 8 + 8 = 32 and there are 4 marks, so the mean is 32 ÷ 4 = 8; in order the marks read 8, 8, 8, 8, and the mean of the middle pair is (8 + 8) ÷ 2 = 8; the value 8 occurs 4 times and no other value occurs at all, so the mode is 8. Answer: mean = 8, median = 8, mode = 8 — when every value in a data set is the same, all three measures of central tendency take that value. The distractors: a mean of 32 comes from stopping at the total and never dividing by 4; a mode of 4 comes from writing down how many times 8 occurs instead of the value that occurs; a mean of 2 comes from dividing a single value, 8, by the 4 marks instead of dividing the total by 4.
- (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) 36.9° — sin θ = opposite/hypotenuse = 6/10 = 0.6, so θ = sin⁻¹(0.6) = 36.86...° ≈ 36.9°. "53.1°" finds the OTHER acute angle in the triangle, 90° − 36.9°, instead of θ itself, as if the two acute angles had been swapped. "31.0°" comes from using the tangent ratio instead of sine, working out tan⁻¹(6/10) = 31.0° with the wrong ratio for the two sides given. "36.8°" rounds sin⁻¹(0.6) = 36.86...° down to 36.8° instead of correctly rounding it up to 36.9°.
- (b) −5 — Method: a point whose x-coordinate is 0 lies on the y-axis, so its y-coordinate is the constant c; once m and c are both known the equation can be written down and x substituted into it. Working: the line passes through (0, 1), so c = 1 and the equation is y = −3x + 1; substituting x = 2 gives y = −3 × 2 + 1 = −6 + 1 = −5. Answer: −5. The distractors: 7 comes from ignoring the minus sign on the gradient and working out 3 × 2 + 1; 5 comes from working out 3 × 2 = 6 and then using the minus sign to take the constant off the product, 6 − 1 = 5; −7 comes from subtracting the constant instead of adding it, −6 − 1 = −7.
- (d) 60 — Method: 75% is three quarters, so divide by 4 to find one quarter and then multiply by 3. Working: 80 ÷ 4 = 20 for one quarter, and 3 × 20 = 60. Answer: 60 milk chocolates. The distractors: 20 is one quarter of 80, which is 25% and not 75%; 55 comes from subtracting 25 from 80 rather than taking 25% of 80 away from it; 75 comes from quoting the percentage itself as a number of chocolates.
- (b) 5 — Method: expand the bracket first, then solve the resulting linear equation. Working: 3(2x − 1) = 6x − 3, so 6x − 3 = 27. Add 3 to both sides: 6x = 30. Divide by 6: x = 5. Answer: 5. 4 2/3 comes from multiplying only the 2x inside the bracket by 3 and forgetting to multiply the −1 as well, giving 6x − 1 = 27. 4 comes from a sign error when expanding, writing 6x + 3 instead of 6x − 3. 4.5 comes from dividing 27 by 3 to get 2x − 1 = 9, then forgetting to add 1 back before dividing by 2, using 2x = 9 instead of 2x = 10.
- (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.
- (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) 9 — Method: expand the brackets fully first, then collect the x-terms and constants before dividing. Working: 4(x − 3) = 4x − 12, so 4x − 12 = 2x + 6; 4x − 2x = 6 + 12; 2x = 18; x = 9. Answer: x = 9. 4.5 comes from expanding only the x-term in the bracket and forgetting to multiply the 3, giving 4x − 3 = 2x + 6, then 2x = 9, x = 4.5. −3 comes from a sign error when expanding, giving 4x + 12 = 2x + 6, then 2x = −6, x = −3. 3 comes from a sign error collecting the x-terms, adding instead of subtracting: 4x + 2x = 6 + 12, so 6x = 18, x = 3.
- (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.
- (c) 200 students — Method: find the percentage who have a brother or a sister, taking care that the students with both are not counted twice, then take that percentage from 100% and apply the result to the 400 students. Working: 25% + 40% = 65%, but the 15% with both has been counted in each of those figures, so 65% − 15% = 50% have a brother or a sister; that leaves 100% − 50% = 50%, and 50% of 400 = 200. Answer: 200 students. The distractors: 260 students is 65% of 400, the number with a brother or a sister when the 15% overlap is counted twice; 140 students comes from taking that same uncorrected 65% away from the 400; 300 students comes from subtracting only the 25% with a brother and ignoring the sisters altogether.
- (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) 5 — The symbol ≤ means n can equal 5 or any number less than 5, so 5 is included and is the largest integer value. A candidate who treats the inequality as strict, as if it were n < 5, answers 4. A candidate who confuses ≤ with ≥ and looks for a value just above the boundary answers 6. A candidate who makes a sign error and reads the inequality as n ≤ −5 answers −5.
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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