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.The price of a cycling helmet rises from £80 to £116. Work out the percentage increase.
- 2.Which of these equations is equivalent to 5x − 2 = 3x + 8?
- 3.Solve 3x − 1 = 8
- 4.Harry counted the coins he found on each of six days: 5, 7, 8, 9, 11, 30. Write down the outlier.
- 5.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.
- 6.The numbers 2, 4, 6, 8, 10 and 12 have a mean of 7 and a median of 7. The value 100 is now added to the list. Which measure is changed more by adding 100, and why?
- 7.Write 3/4 as a percentage.
- 8.A support strut for a shed roof is in the shape of a right-angled triangle. The two shorter sides of the strut are 5 m and 9 m. Work out the length of the sloping strut (the hypotenuse). Give your answer correct to 1 decimal place.
- 9.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
- 10.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.
- 11.The ratio of Josh's savings to Mia's savings is 4:7. Mia has £39 more than Josh. Work out Josh's savings.
- 12.Solve the inequality 3x + 6 ≤ 0.
- 13.Two of the angles in a triangle are 50° and 70°. Work out the size of the third angle.
- 14.Simplify 2b − 1 + 2b − 4 + 2b
- 15.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.
- 16.The mean of three numbers is 50. A fourth number, 100, is added to the set. Work out the mean of the four numbers.
- 17.Factorise x² − 64.
- 18.The angles of a triangle are in the ratio 2:3:4. Work out the size of the smallest angle.
- 19.A straight line passes through the points (−1, 2) and (3, 14). Work out the equation of the line.
- 20.Solve 5x − 9 = 16
Answer key
- (a) 45% — Method: percentage increase = increase ÷ original amount × 100. Working: the increase is £116 − £80 = £36, and 36 ÷ 80 = 0.45, so 0.45 × 100 = 45%. Answer: 45%. The distractors: 36% comes from quoting the £36 increase as though pounds and per cent were the same thing; 31% comes from dividing the £36 increase by the new price £116 instead of by the original £80, which gives 31% to the nearest per cent; 145% is the new price written as a percentage of the original price, which is the whole of the new price rather than the increase.
- (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) x = 3 — Method: undo the operations on the left in reverse order — deal with the subtraction of 1 first, then with the multiplication by 3. Working: adding 1 to both sides gives 3x = 9, and dividing both sides by 3 gives x = 3. Answer: x = 3. The distractors: x = 9 comes from stopping at 3x = 9 and writing 9 as the value of x; x = 6 comes from subtracting 3 from 9 instead of dividing by 3; x = 27 comes from multiplying 9 by 3 instead of dividing by 3.
- (a) 30 — Method: an outlier is a value that lies far away from the pattern set by the rest of the data, so compare each value with the group the others form. Working: five of the counts, 5, 7, 8, 9 and 11, lie within 6 of one another and the steps between them are 2, 1, 1 and 2; the remaining count of 30 is 19 above the nearest of them, so it is the value that does not belong to the pattern. Answer: 30. The distractors: 5 comes from picking the smallest value, on the idea that the odd one out must be at the bottom of the list; 11 comes from ordering the data and stopping one value short, taking the largest of the counts that sit close together; 8.5 comes from working out the median, (8 + 9) ÷ 2, and giving a measure of centre where a value standing apart was asked for.
- (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.
- (c) The mean, because every value counts towards it, so 100 pulls it from 7 up to about 20.3. — Method: work each measure out before the extra value is added and again afterwards, then compare the size of the two changes. Working: before, the six values total 42, so the mean is 42 ÷ 6 = 7, and the middle pair 6 and 8 give a median of (6 + 8) ÷ 2 = 7; after, the seven values total 142, so the mean is 142 ÷ 7 = 20.29 to 2 decimal places, while the median is now the 4th of the seven ordered values, which is 8; the mean has moved by about 13.3 and the median by 1. Answer: the mean, because every value counts towards it, so 100 pulls it from 7 up to about 20.3 — this is why the median is often preferred when a data set contains an outlier. The distractors: the reply that the mean rises by 100 adds the extra value to the mean instead of adding it to the total; the reply that the median moves to 12 takes the largest of the original values as the new middle instead of counting to the 4th of the seven values; the reply about even and odd counts quotes a rule that does not exist, since the median moved because a very large value was added, not because the count of values changed.
- (d) 75% — Method: change a fraction to a percentage by dividing the numerator by the denominator and multiplying by 100. Working: 3 ÷ 4 = 0.75, and 0.75 × 100 = 75. Answer: 75%. The distractors: 34% comes from reading the digits 3 and 4 straight off as a percentage; 0.75% comes from dividing but forgetting to multiply by 100; 133% comes from inverting the fraction and working out 4 ÷ 3 instead.
- (d) 10.3 m — By Pythagoras' theorem, the hypotenuse = √(5² + 9²) = √(25 + 81) = √106 = 10.29...≈ 10.3 m. "106 m" is the value under the square root sign, correct as far as it goes but with the final square root step left out. "14 m" comes from adding the two shorter sides, 5 + 9, instead of using Pythagoras' theorem at all. "10.2 m" comes from rounding 10.29...m down to 10.2 instead of correctly rounding it up to 10.3.
- (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).
- (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) £52 — Method: the difference between the two ratio numbers tells you how many parts the £39 difference represents. Working: the difference in parts is 7 − 4 = 3, and this represents £39, so one part is £39 ÷ 3 = £13. Josh's savings are 4 × £13 = £52. So Josh has £52. Distractor £91 is Mia's savings, not Josh's. Distractor £39 comes from using the given £39 difference as the final answer, without scaling it to Josh's number of parts. Distractor £13 is the value of one part, found correctly but never multiplied by 4.
- (b) x ≤ −2 — Method: take the number term off both sides, then divide by the coefficient of x; the sign turns round only when you divide BY a negative number, and here you divide by 3. Working: subtracting 6 from both sides of 3x + 6 ≤ 0 gives 3x ≤ −6; dividing both sides by 3, which is positive, gives x ≤ −2. Answer: x ≤ −2. The distractors: x ≥ −2 comes from turning the sign round because the right-hand side has become negative, which is not the rule; it is the sign of the divisor that matters; x ≤ 2 comes from moving the 6 across without changing its sign, giving 3x ≤ 6; x ≤ −18 comes from multiplying both sides by 3 instead of dividing by it.
- (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) 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.
- (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.
- (a) 62.5 — Method: a mean cannot be averaged with a new value — rebuild the total, add the new value to it, then divide by the new count. Working: three numbers with a mean of 50 have a total of 50 × 3 = 150; adding 100 makes the total 150 + 100 = 250; there are now 4 numbers, so the new mean is 250 ÷ 4 = 62.5. Answer: 62.5. The distractors: 75 comes from averaging the old mean with the new value, (50 + 100) ÷ 2, which ignores that three numbers pull against one; 50 comes from assuming an extra value leaves the mean unchanged; 37.5 comes from dividing the old total of 150 by the new count of 4, adding the new value to the count but not to the total.
- (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.
- (c) 40° — Method: the three angles of a triangle always add up to 180°, so split 180° into the number of parts shown by the ratio. Working: the ratio 2:3:4 has 2 + 3 + 4 = 9 parts, so one part is 180 ÷ 9 = 20°, and the smallest angle is 2 × 20 = 40°. So the smallest angle is 40°. Distractor 60° is the middle-sized angle, 3 parts. Distractor 80° is the largest angle, 4 parts. Distractor 20° is the value of one part, found correctly but never multiplied by 2.
- (c) y = 3x + 5 — Method: find the gradient from the two points, then substitute one point into y = mx + c to find c. Working: gradient = (14 − 2) ÷ (3 − (−1)) = 12 ÷ 4 = 3. Using the point (3, 14): 14 = 3(3) + c, so 14 = 9 + c, giving c = 5. Answer: y = 3x + 5. y = 6x − 4 comes from mishandling the negative x-coordinate, treating 3 − (−1) as 3 − 1 = 2, so the gradient becomes 12 ÷ 2 = 6, and then c = 14 − 18 = −4. y = 3x + 23 comes from a sign error isolating c, adding 9 to 14 instead of subtracting it: c = 14 + 9 = 23. y = x/3 + 13 comes from dividing the change in x by the change in y instead of the other way round, giving a gradient of 4 ÷ 12 = 1/3, and then c = 14 − 1 = 13.
- (b) 5 — Method: add 9 to both sides, then divide by 5. Working: 5x = 16 + 9 = 25, so x = 25 ÷ 5 = 5. Answer: 5. 3.2 comes from dividing 16 by 5 directly, without adding 9 first. 1.4 comes from a sign error, subtracting 9 from 16 instead of adding it, then dividing by 5. 25 comes from correctly working out 5x = 25 but stopping there, without dividing by 5 to find x.
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.
Similar worksheets worth a look
- ✏️ Paper 1 non-calculator warm-up — Foundation · 20 questions · ~25 min
- 🎯 Grade 4 pass booster · 25 questions · ~40 min
- 📈 Quadratics: factorise, complete the square, formula · 24 questions · ~45 min
- 💷 Percentages and compound interest · 22 questions · ~40 min