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.After a 20% discount, a jacket costs £48. Work out the original price of the jacket.
- 2.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.
- 3.Solve 5(x + 3) = 40
- 4.At an animal shelter, the ratio of cats to dogs is 2:5. There are 18 more dogs than cats. Work out the number of cats.
- 5.A jug of squash is made by mixing water and syrup in the ratio 6:1. Nia wants to make 8.4 litres of squash. Work out how much syrup she needs, in litres.
- 6.Solve the inequality 5 − x ≤ 2.
- 7.Three business partners share a profit of £48,000 in the ratio 3:5:4. Work out how much the partner with 5 parts receives.
- 8.Solve 2x + 1 = 9
- 9.Priya says that x = 6 is the solution of the equation 4x + 8 = 24. Priya is wrong. Work out the correct value of x.
- 10.A laptop priced at £520 is first increased by 15%, and then the new price is decreased by 20%. Work out the final price of the laptop.
- 11.The mean of three numbers is 50. A fourth number, 100, is added to the set. Work out the mean of the four numbers.
- 12.Expand 2(x − 13)
- 13.Work out the median of these six numbers: 13, 21, 22, 36, 37, 47
- 14.Expand 4(2x − 3).
- 15.Work out the equation of the straight line through the points (−3, 4) and (1, −8).
- 16.A school trip costs a £15 deposit plus £9 per student for the coach. The total cost for a class is £186. Work out how many students went on the trip.
- 17.Simplify 5x + 3y − 2x + y.
- 18.Harry counted the coins he found on each of six days: 5, 7, 8, 9, 11, 30. Write down the outlier.
- 19.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.
- 20.In triangle ABC the angle at C is 90°, AC = 8 cm and the angle at A is 30°. Work out the length of BC. Give your answer to 1 decimal place.
Answer key
- (b) £60 — £48 represents 100% − 20% = 80% of the original price. 1% = £48 ÷ 80 = £0.60, so 100% = £0.60 × 100 = £60.
- (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.
- (d) x = 5 — Method: expand the bracket by multiplying both terms inside it by 5, then undo the addition and the multiplication in turn. Working: expanding gives 5x + 15 = 40; subtracting 15 from both sides gives 5x = 25; dividing both sides by 5 gives x = 5. Answer: x = 5. The distractors: x = 8 comes from dividing both sides by 5 first, reaching x + 3 = 8 and writing 8 as the value of x without taking the 3 away; x = 11 comes from adding 15 to both sides instead of subtracting it, giving 5x = 55; x = 7.4 comes from expanding 5(x + 3) as 5x + 3, multiplying only the x by the 5, which leads to 5x = 37.
- (b) 12 — The difference between the parts of the ratio is 5 − 2 = 3 parts, and this is worth 18. Divide to find one part: 18 ÷ 3 = 6. Cats have 2 parts: 2 × 6 = 12. (30 is the number of dogs, using 5 parts instead of 2. 6 is the value of one part — the number of cats is 2 lots of this, not just one. 9 comes from dividing 18 by 2 and stopping there, instead of dividing by the difference in parts, 3, and then multiplying by 2.)
- (c) 1.2 — Add the parts of the ratio: 6 + 1 = 7. Divide the total amount by the number of parts: 8.4 ÷ 7 = 1.2 litres, which is the value of one part and also the amount of syrup, since syrup is 1 part. (7.2 litres is the amount of water, using 6 parts instead of 1. 1.4 comes from dividing 8.4 by 6 — the water's part of the ratio — instead of dividing by the total number of parts, 7. 0.84 comes from dividing 8.4 by 10 instead of by 7.)
- (c) x ≥ 3 — Method: collect the number terms first; the x term is negative, so the final step multiplies both sides by −1, and that is the one step that turns the inequality sign round. Working: subtracting 5 from both sides of 5 − x ≤ 2 gives −x ≤ −3; multiplying both sides by −1 turns −x into x and −3 into 3, and because the multiplier is negative the ≤ becomes ≥, so x ≥ 3. Answer: x ≥ 3. The distractors: x ≤ 3 comes from multiplying by −1 without turning the sign round, the commonest slip on this type; x ≤ −3 comes from reading −x ≤ −3 as though the minus sign could simply be rubbed off the left-hand side; x ≥ −3 comes from turning the sign round correctly but leaving the right-hand side at −3 instead of multiplying it by −1 as well.
- (a) £20,000 — Method: add the parts of the ratio, divide the profit by the number of parts to find the value of one part, then multiply by the parts belonging to the partner asked about. Working: 3 + 5 + 4 = 12 parts, £48,000 ÷ 12 = £4,000 for one part, and that partner has 5 parts, so 5 × £4,000 = £20,000. Answer: £20,000. The distractors: £4,000 is the value of one part; £12,000 is the 3-part share; £16,000 is the 4-part share, which is what a candidate reaches by assuming the biggest share must go with the number written last in the ratio.
- (c) x = 4 — Method: undo the addition of 1 first, then undo the multiplication by 2. Working: subtracting 1 from both sides gives 2x = 8, and dividing both sides by 2 gives x = 4. Answer: x = 4. The distractors: x = 8 comes from stopping at 2x = 8 and writing 8 as the value of x; x = 5 comes from adding 1 to both sides instead of subtracting it, giving 2x = 10; x = 16 comes from multiplying 8 by 2 instead of dividing by 2.
- (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.
- (c) £478.40 — Method: apply the percentage increase, then apply the percentage decrease to the new price. Working: after the increase, the laptop costs £520 × 1.15. Multiplying this result by 0.80 gives the final price, £478.40. Answer: £478.40. £494 comes from combining the two percentages into a single net change (15% − 20% = −5%) and applying it directly, £520 × 0.95 = £494, instead of applying the two changes one after the other. £416 comes from applying only the 20% decrease to the original price, £520 × 0.80 = £416, forgetting the increase entirely. £598 comes from applying only the 15% increase and stopping there, forgetting to apply the decrease at all.
- (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) 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.
- (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) 8x − 12 — Multiply each term inside the bracket by 4: 4 × 2x = 8x and 4 × (−3) = −12, so 4(2x − 3) = 8x − 12. A candidate who forgets to multiply the second term by 4 gets 8x − 3. A candidate who makes a sign error, treating 4 × (−3) as +12, gets 8x + 12. A candidate who adds 4 to the bracket instead of multiplying gets 2x + 1.
- (a) y = −3x − 5 — Gradient = (−8 − 4) ÷ (1 − (−3)) = −12 ÷ 4 = −3. Using the point (1, −8): −8 = −3(1) + c, so c = −5, giving y = −3x − 5. A candidate who drops the negative sign on the gradient, using m = 3 instead, would then solve −8 = 3(1) + c to get c = −11, writing y = 3x − 11. A candidate who makes a sign error isolating c, writing c = 5 instead of −5, would write y = −3x + 5. A candidate who mixes up both mistakes — keeping the correct gradient but the wrong, positive value of c from the flipped-gradient calculation — would write y = −3x + 11.
- (c) 19 — Method: set up the equation 15 + 9n = 186, then subtract the deposit and divide by the cost per student. Working: 9n = 186 − 15 = 171; n = 171 ÷ 9 = 19. Answer: 19 students. 20.67 comes from dividing the whole £186 by £9 without first subtracting the deposit: 186 ÷ 9 ≈ 20.67. 11.8 comes from swapping the two amounts round, subtracting £9 and dividing by £15: (186 − 9) ÷ 15 = 11.8. 22.33 comes from adding the deposit instead of subtracting it: (186 + 15) ÷ 9 ≈ 22.33.
- (c) 3x + 4y — Collect the x terms: 5x − 2x = 3x. Collect the y terms: 3y + y = 4y. So 5x + 3y − 2x + y = 3x + 4y. A candidate who subtracts the y terms instead of adding them (3y − y) gets 3x + 2y. A candidate who adds 2x instead of subtracting it (5x + 2x) gets 7x + 4y. A candidate who wrongly combines the x and y terms into a single term gets 6xy.
- (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.
- (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.
- (d) 4.6 cm — Method: BC is opposite the 30° angle and AC is next to it, so the ratio that links the two is tan θ = opposite ÷ adjacent. Working: tan 30° = BC ÷ 8, so BC = 8 × tan 30° = 4.6188…, which is 4.6 to 1 decimal place. Answer: 4.6 cm. The distractors: 13.9 cm comes from dividing by tan 30° instead of multiplying by it; 4.0 cm comes from using sin 30°, which treats the 8 cm side as the hypotenuse when it is the side next to the 30° angle; 6.9 cm comes from using cos 30° in place of tan 30°, which gives the wrong pair of sides.
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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