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 candle is 30 cm tall and burns down at a steady rate of 1 cm per hour. Write down the function for the height of the candle y, in centimetres, after x hours.
- 2.A plumber charges a call-out fee plus an hourly rate. The total cost, y in pounds, of a job lasting x hours is given by y = 45x + 60. Work out the total cost of a job that lasts 3 hours.y = 45x + 60
- 3.After a 20% discount, a jacket costs £48. Work out the original price of the jacket.
- 4.Two of the angles in a triangle are 50° and 70°. Work out the size of the third angle.
- 5.A straight line has gradient −2 and passes through the point (3, 1). Work out the equation of the line.
- 6.Three angles lie on a straight line. They measure 38°, 95° and x°. Work out the value of x.
- 7.A fruit punch is made from orange juice, pineapple juice and lemonade in the ratio 5:3:2. A jug holds 3.5 litres of punch in total. Work out the volume of pineapple juice needed.
- 8.Expand 7(a + 8)
- 9.Solve 3x − 5 = 4.
- 10.Simplify 2a + 3b − a + 2b − b
- 11.Priya says that x = 6 is the solution of the equation 4x + 8 = 24. Priya is wrong. Work out the correct value of x.
- 12.A right-angled triangle has a hypotenuse of 29 cm and one shorter side of 20 cm. Work out the length of the other shorter side.
- 13.Noah wrote down the numbers 9, 3, 7, 1, 5 and said, “The median is 7, because 7 is in the middle of my list.” Is Noah right? Give a reason for your answer.
- 14.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
- 15.A straight line passes through the points (1, 3) and (2, 5). Work out the equation of the line.
- 16.A rectangular garden has length (x + 7) m and width (x − 7) m. Work out an expression for the area of the garden, giving your answer in its simplest form.
- 17.A plumber charges a £35 call-out fee plus £20 for each hour worked, h. On a certain job the total charge was £115. Work out how many hours the plumber worked.
- 18.A company's profit this year is 130% of last year's profit. Last year's profit was £40,000. Work out this year's profit.
- 19.A straight line crosses a pair of parallel lines. At one of the crossing points, one of the four angles measures 55°. Work out the size of the angle next to it on a straight line at that crossing point.
- 20.The price of a games console is reduced by 10%. In a later sale the reduced price is reduced by 10% again. Work out the overall percentage decrease.
Answer key
- (c) y = −x + 30 — Method: in a linear model the value at the start is the constant term and the steady rate of change is the gradient, which is negative when the quantity is falling. Working: at x = 0 the candle is 30 cm tall, so the constant term is 30; it loses 1 cm every hour, so the gradient is −1 and the height after x hours is y = −x + 30. Answer: y = −x + 30. The distractors: y = x + 30 comes from taking the rate as +1 and making the candle grow rather than shrink; y = 30x − 1 comes from building the model correctly as 30 − 1x and then copying it into the form y = mx + c with the two numbers left where they stood, so the starting height 30 ends up multiplying x and the hourly 1 is left behind as the constant being taken away; y = −30x + 1 comes from reading the two numbers the other way round, taking 30 cm per hour as the rate and 1 cm as the starting height, which burns 30 cm an hour from a candle only 1 cm tall.
- (d) £195 — Substituting x = 3 into y = 45x + 60 gives y = 45 × 3 + 60 = 135 + 60 = 195. A candidate who forgets to add the call-out fee would get only 45 × 3 = £135. A candidate who adds the hours to the fee and the rate instead of multiplying would get 45 + 60 + 3 = £108. A candidate who multiplies both the hourly rate and the call-out fee by the number of hours would get 45 × 3 + 60 × 3 = £315.
- (b) £60 — £48 represents 100% − 20% = 80% of the original price. 1% = £48 ÷ 80 = £0.60, so 100% = £0.60 × 100 = £60.
- (a) 60° — Method: the three angles of a triangle add up to 180°, so add the two known angles and subtract the total from 180°. Working: 50 + 70 = 120, then 180 − 120 = 60. Answer: 60°. The distractors: 120° is the sum of the two known angles, given as the answer instead of being subtracted from 180°; 110° comes from subtracting only the 70° angle from 180° and forgetting the 50° one; 240° comes from subtracting the sum from 360°, using the angles at a point rather than the angle sum of a triangle.
- (a) y = −2x + 7 — Using y = −2x + c, and substituting the point (3, 1): 1 = −2(3) + c, so 1 = −6 + c, and c = 7, giving y = −2x + 7. A candidate who uses the y-coordinate of the point directly as the y-intercept, instead of solving for c, would write y = −2x + 1. A candidate who drops the negative sign on the gradient would write y = 2x + 7. A candidate who makes a sign error when isolating c, writing c = 1 − 6 = −5 instead of c = 1 + 6 = 7, would write y = −2x − 5.
- (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.
- (c) 1.05 litres — Method: find the value of one part of the ratio from the total volume, then find the share for pineapple juice. Working: the ratio 5:3:2 has 5 + 3 + 2 = 10 parts, so one part is 3.5 ÷ 10 = 0.35 litres, and the pineapple juice is 3 × 0.35 = 1.05 litres. So 1.05 litres of pineapple juice is needed. Distractor 1.75 litres is the volume of orange juice, not pineapple juice. Distractor 0.7 litres is the volume of lemonade, not pineapple juice. Distractor 0.35 litres is the value of one part, found correctly but never multiplied by 3.
- (d) 7a + 56 — Method: multiplying a single term over a bracket means every term inside the bracket is multiplied by the term outside. Working: 7 × a = 7a and 7 × 8 = 56, and the two products are added because the bracket contains an addition. Answer: 7a + 56. The distractors: 7a + 15 comes from adding 7 and 8 instead of multiplying them; 7a + 8 comes from multiplying only the first term inside the bracket and copying the 8 across unchanged; a + 56 comes from multiplying only the 8 and copying the a across unchanged.
- (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) a + 4b — Method: collect the terms in a and the terms in b separately, keeping the sign that stands in front of each term. Working: the a terms give 2a − a = a; the b terms give 3b + 2b − b = 4b; a and b are unlike terms, so the two results cannot be joined. Answer: a + 4b. The distractors: 3a + 4b comes from reading −a as +a, so that 2a + a = 3a; a + 6b comes from reading −b as +b, so that 3b + 2b + b = 6b; 5ab comes from combining the unlike terms a and 4b into one term by adding 1 and 4 and joining the letters.
- (b) 4 — Method: subtract the constant term from both sides first, then divide by the coefficient of x. Working: 4x = 24 − 8 = 16; x = 16 ÷ 4 = 4. Answer: x = 4. Priya's 6 comes from dividing 24 by 4 without first subtracting the 8, ignoring the constant term; substituting it back gives 4 × 6 + 8 = 32, not 24, which is why she is wrong. 16 comes from correctly finding 4x = 16 but forgetting to divide by 4. 12 comes from subtracting the coefficient 4 instead of dividing by it: 16 − 4 = 12. 8 comes from correctly finding 4x = 16 but then halving instead of dividing by 4.
- (b) 21 cm — By Pythagoras' theorem, the other side = √(29² − 20²) = √(841 − 400) = √441 = 21 cm. "9 cm" comes from subtracting the two given lengths directly, 29 − 20 = 9, instead of subtracting their squares. "441 cm" is the value under the square root sign, correct as far as it goes but with the final square root step left out. "35 cm" comes from adding the squares of the two given lengths instead of subtracting them, √(29² + 20²) = √1241 ≈ 35, treating both given lengths as if they were the two shorter sides rather than a shorter side and the hypotenuse.
- (b) No — in order the numbers are 1, 3, 5, 7, 9, so the median is 5. — Method: the median is the middle value of the data in order of size, so the data must be sorted before any position is read off. Working: Noah's list 9, 3, 7, 1, 5 is not in order; sorted it becomes 1, 3, 5, 7, 9, and with 5 values the middle position is the third, which now holds 5 rather than 7. Noah has read the third value of the unsorted list. Answer: no — in order the numbers are 1, 3, 5, 7, 9, so the median is 5. The distractors: the reply giving 3 as the median sorts the data correctly but then reads the value in the second place instead of the third; the reply that 7 is the third number he wrote accepts a position in the unsorted list, which is exactly the mistake the question is about; the reply using the mean claims a value of 7 for it, but the mean is 25 ÷ 5 = 5, so that reasoning is false as well.
- (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.
- (c) y = 2x + 1 — Method: the gradient of the line through two points is the change in y divided by the change in x, and the constant is then found by substituting one of the points into y = mx + c. Working: m = (5 − 3) ÷ (2 − 1) = 2 ÷ 1 = 2, so the line is y = 2x + c; substituting x = 1 and y = 3 gives 3 = 2 × 1 + c, so c = 3 − 2 = 1. Answer: y = 2x + 1. The distractors: y = x + 2 comes from taking the gradient as the change in x, 2 − 1 = 1, and then substituting (1, 3) to reach a constant of 2; y = 2x − 1 comes from working out the constant as mx − y, 2 × 1 − 3 = −1, instead of y − mx; y = 2x + 3 comes from using the y-coordinate of (1, 3) as the constant without substituting at all.
- (d) x² − 49 — Area = length × width = (x + 7)(x − 7). Expanding: x × x = x², x × (−7) = −7x, 7 × x = 7x, 7 × (−7) = −49. The two middle terms −7x and 7x cancel, leaving x² − 49. A candidate who misremembers the difference-of-two-squares result as a sum gets x² + 49. A candidate who makes a sign error and treats both middle terms as −7x instead of cancelling gets x² − 14x − 49. A candidate who confuses area with perimeter and simply adds the length and width gets 2x.
- (a) 4 — Method: set up the equation 35 + 20h = 115, then subtract the fixed fee and divide by the hourly rate. Working: 20h = 115 − 35 = 80; h = 80 ÷ 20 = 4. Answer: 4 hours. 5.75 comes from dividing the whole £115 by £20 without first subtracting the fixed fee: 115 ÷ 20 = 5.75. 2.71 comes from swapping the fee and the rate round, subtracting £20 and dividing by £35: (115 − 20) ÷ 35 ≈ 2.71. 7.5 comes from adding the fixed fee instead of subtracting it: (115 + 35) ÷ 20 = 7.5.
- (b) £52,000 — Method: convert 130% to a decimal multiplier and multiply it by last year's profit. Working: 130% = 1.3, so this year's profit is £40,000 × 1.3 = £52,000. Answer: £52,000. £12,000 comes from using only the extra 30% (130% − 100%) and forgetting to include the original 100%, £40,000 × 0.3 = £12,000. £40,130 comes from simply adding 130 onto £40,000, treating the percentage as an amount of money rather than a multiplier. £5,200 comes from misreading 130% as 13%, giving £40,000 × 0.13 = £5,200.
- (d) 125° — Method: two angles that sit next to each other at a crossing point lie on a straight line, so they add up to 180°; the angle that faces the given one across the point is the equal one, and that is not the angle asked for here. Working: 180° − 55° = 125°. Answer: 125°. The distractors: 55° is the angle vertically opposite the given one, taken by a candidate who reads “next to” as the facing angle and applies the equal-angles rule to the wrong pair; 35° comes from subtracting from 90°, treating the pair as complementary instead of as angles on a straight line; 90° comes from assuming that a line crossing a pair of parallel lines must meet them at right angles, which the question never says.
- (d) 19% — Method: write each decrease as a multiplier, multiply the multipliers, then compare the result with 100%. Working: a 10% decrease is a multiplier of 0.9, so the two reductions together give 0.9 × 0.9 = 0.81; the final price is 81% of the original, so the price has fallen by 100% − 81% = 19%. Answer: an overall decrease of 19%. The distractors: 20% comes from adding the two reductions, 10% + 10%, which charges the second 10% against the original price instead of against the already reduced price; 21% comes from using the increase multiplier by mistake, since 1.1 × 1.1 = 1.21, and reading that 21% as a decrease; 81% is the percentage of the original price still being paid, not the percentage taken off.
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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