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.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.
- 2.Five pupils spent these numbers of minutes on their homework: 50, 65, 55, 90, 60. Work out the median time.
- 3.The rent on a flat increases by 10% one year and by a further 10% the following year. Work out the overall percentage increase over the two years.
- 4.In a test, Amelia answered 18 of the 24 questions correctly. Work out the percentage of the questions she answered correctly.
- 5.A rectangular garden has its length and width in the ratio 5:3. Given that the perimeter of the garden is 64 m, work out the width of the garden.
- 6.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.
- 7.The price of a jacket increases by 50% and then decreases by 50%. Describe the overall change from the original price.
- 8.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.
- 9.The mean mass of four parcels is 17 kg. Three of the parcels have masses 12 kg, 16 kg and 18 kg. Work out the mass of the fourth parcel.
- 10.A netball team scored 50, 83 and 68 points in three matches. Work out the mean number of points scored.
- 11.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
- 12.Work out 25% of 200.
- 13.A regular polygon has an exterior angle of 45°. Work out the number of sides of the polygon.
- 14.Two angles are supplementary. One of them is twice the size of the other. Work out the size of the smaller angle.
- 15.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
- 16.Solve the inequality x − 10 < −3.
- 17.A shop recorded the number of books it sold on five days: 100, 40, 70, 20, 60. Work out the range of the numbers of books sold.
- 18.The sum of the interior angles of a polygon is 1980°. Work out the number of sides of the polygon.
- 19.A right-angled triangle has two sides of length 5 cm and 12 cm, and the angle between those two sides is 90°. Work out the length of the hypotenuse.
- 20.Noah says that y = 3 is the solution of the equation y + 10 = 12. Noah is wrong. Work out the correct value of y.
Answer key
- (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.
- (c) 60 minutes — Method: the median is the middle value once the data have been put in order of size, so the list must be sorted before any position is read. Working: in order the times are 50, 55, 60, 65, 90 minutes; there are 5 values, so the middle position is the third and the time sitting there is 60 minutes. Answer: 60 minutes. The distractors: 64 minutes comes from working out the mean, 320 ÷ 5, instead of the median; 70 minutes comes from taking the time halfway between the shortest and the longest, (50 + 90) ÷ 2; 40 minutes comes from working out the range, 90 − 50, which measures spread rather than centre.
- (c) 21% — Method: an increase of 10% is a multiplier of 1.1, and two successive increases are found by multiplying the multipliers. Working: 1.1 × 1.1 = 1.21, so the rent is 121% of the original, which is an increase of 21%. Answer: 21%. The distractors: 20% comes from adding the two percentages, which ignores that the second 10% is taken of a larger amount; 121% is the multiplier written as the change rather than the change itself; 11% comes from slipping in the multiplication and getting 1.11 instead of 1.21.
- (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.
- (c) 12 m — Method: the perimeter of a rectangle is twice the sum of the length and the width, so half the perimeter is one length plus one width; split that half using the ratio. Working: half of 64 is 64 ÷ 2 = 32 m, the ratio 5:3 has 5 + 3 = 8 parts, so one part is 32 ÷ 8 = 4 m, and the width is 3 × 4 = 12 m. So the width is 12 m. Distractor 20 m is the length, 5 parts, not the width. Distractor 24 m comes from splitting the whole perimeter, 64 m, into 8 parts and multiplying by 3, forgetting to halve the perimeter first. Distractor 8 m comes from the same slip stopped one step earlier: splitting the whole perimeter into 8 parts, 64 ÷ 8 = 8, and giving that instead of the width.
- (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) 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) 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.
- (c) 22 kg — Method: multiply the mean by the number of parcels to rebuild the total mass, then subtract the masses that are known. Working: four parcels with a mean mass of 17 kg have a total mass of 17 × 4 = 68 kg; the three known parcels total 12 + 16 + 18 = 46 kg; so the fourth parcel has mass 68 − 46 = 22 kg. Answer: 22 kg. The distractors: 68 kg comes from stopping at the total mass of all four parcels; 17 kg comes from assuming the missing parcel must have the mean mass; 5 kg comes from multiplying the mean by 3, the number of parcels whose mass is given, leaving 51 − 46 = 5.
- (c) 67 — Method: the mean is the total of the values divided by how many values there are, so add first and divide second. Working: the total is 50 + 83 + 68 = 201 points and three matches were played, so the mean is 201 ÷ 3 = 67 points. Answer: 67. The distractors: 68 comes from writing down the median, the middle value of 50, 68, 83, instead of the mean; 33 comes from working out the range, 83 − 50, which measures spread and not centre; 100.5 comes from dividing the total by 2 instead of by the 3 matches played.
- (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) 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) 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.
- (d) 60 — Method: write the two supplementary angles as x and 2x, since one is twice the other, and solve x + 2x = 180. Working: 3x = 180, so x = 60. Answer: the smaller angle is 60°. A candidate who gives the larger angle, 2x, instead of the smaller angle gets 120. A candidate who uses a complementary sum of 90° instead of a supplementary sum of 180° gets 30. A candidate who divides 180 by 2 instead of by 3 gets 90.
- (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) x < 7 — Method: add 10 to both sides to leave x on its own; adding the same number to both sides never changes the direction of an inequality. Working: adding 10 to both sides of x − 10 < −3 leaves x on the left and −3 + 10 on the right, and −3 + 10 = 7, so x < 7. Answer: x < 7. The distractors: x > 7 comes from turning the sign round while adding, as though every move flipped it; x < −13 comes from subtracting 10 from both sides instead of adding it, giving −3 take away 10; x < 13 comes from ignoring the minus sign on −3 and working out 3 + 10 instead.
- (c) 80 — Method: the range is a measure of spread and is found by subtracting the smallest value from the largest. Working: the largest number sold is 100 and the smallest is 20, so the range is 100 − 20 = 80. Answer: 80. The distractors: 100 comes from writing down the largest value and never subtracting the smallest; 60 comes from working out the median, the middle value of 20, 40, 60, 70, 100, instead of the range; 58 comes from working out the mean, 290 ÷ 5, which measures centre rather than spread.
- (b) 13 — Using the sum of interior angles formula, (n − 2) × 180° = 1980°, so n − 2 = 1980 ÷ 180 = 11, and n = 11 + 2 = 13. 11 stops after the division, forgetting to add 2 back to find n. 15 adds 2 twice by mistake, giving 11 + 2 + 2. 22 divides 1980 by 90 instead of 180.
- (c) 13 cm — Method: the two given sides meet at the right angle, so they are the shorter pair and the hypotenuse comes from Pythagoras' theorem, a² + b² = c². Working: c² = 5² + 12² = 25 + 144 = 169, so c = √169 = 13. Answer: 13 cm. The distractors: 17 cm comes from adding the two sides, 5 + 12, rather than adding their squares; 60 cm comes from multiplying them, 5 × 12, which is twice the area of the triangle and not a length; 7 cm comes from subtracting, 12 − 5, as though the hypotenuse were the difference of the two shorter sides.
- (c) y = 2 — Method: substituting Noah's value shows why it fails, and the equation is then solved by undoing the addition. Working: substituting y = 3 gives 3 + 10 = 13, which is not 12, so Noah's value is not a solution; subtracting 10 from both sides of y + 10 = 12 gives y = 2, and substituting back gives 2 + 10 = 12. Answer: y = 2. The distractors: y = 22 comes from adding 10 to both sides instead of subtracting it; y = −2 comes from carrying out the subtraction the wrong way round, 10 − 12 rather than 12 − 10; y = 12 comes from copying the right-hand side as the value of y and ignoring the 10 that is added to 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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