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.Solve 2x + 1 = 9
- 2.A ribbon of length 90 cm is cut into two pieces in the ratio 4:5. Work out the length of the shorter piece.
- 3.A necklace is made using gold beads and silver beads in the ratio 7:3. There are 40 more gold beads than silver beads. Work out the total number of beads in the necklace.
- 4.A ramp's sloped surface crosses two horizontal parallel rails. At the top rail, the angle between the ramp and the rail on the right of the ramp is (3x + 10)°. At the bottom rail, the angle between the ramp and the rail on the left of the ramp is (5x − 30)°. Work out the value of x.
- 5.The angles of a triangle are in the ratio 2:3:4. Work out the size of the smallest angle.
- 6.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.
- 7.A conservatory has a roof panel shaped like a regular octagon. Work out the interior angle of the octagon, then use it to work out the size of the reflex angle at the same vertex, on the outside of the roof panel.
- 8.Solve 4x + 3 = 19
- 9.Solve 3(2x − 1) = 27
- 10.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.
- 11.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.
- 12.A bag contains red counters and blue counters in the ratio 5:3. There are 56 counters in the bag altogether. Work out how many counters are blue.
- 13.A right-angled triangle has a hypotenuse of 10 cm and one of its other angles is 45°. Work out the length of one of the two shorter sides. Give your answer to 1 decimal place.
- 14.Work out 50% of 60.
- 15.The mean of the four numbers 10, 15, 20 and x is 18. Work out the value of x.
- 16.Solve the inequality 2(3x − 1) ≥ 4x + 8.
- 17.Amelia's mean mark over four tests is 29. Her first three marks are 31, 26 and 26. Work out her fourth mark.
- 18.Solve the inequality x + 5 < 12.
- 19.A straight line passes through the points (2, 5) and (4, 11). Work out the gradient of the line.
- 20.A flagpole is 12 m tall. Freya stands 9 m from the base of the flagpole on level ground. Work out the angle of elevation of the top of the flagpole from where Freya stands. Give your answer correct to 1 decimal place.
Answer key
- (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.
- (c) 40 cm — Method: split the total length into the number of parts shown by the ratio, then find the value of the shorter share. Working: the ratio 4:5 has 4 + 5 = 9 parts, so one part is 90 ÷ 9 = 10 cm, and the shorter piece is 4 × 10 = 40 cm. So the shorter piece is 40 cm. Distractor 50 cm is the length of the LONGER piece, not the shorter one. Distractor 45 cm comes from splitting the ribbon into two equal halves, ignoring the ratio. Distractor 10 cm is the value of one part, found correctly but never multiplied by 4.
- (a) 100 — The difference between the parts of the ratio is 7 − 3 = 4 parts, and this is worth 40 beads. Divide to find one part: 40 ÷ 4 = 10. The total number of parts is 7 + 3 = 10, so the total number of beads is 10 × 10 = 100. (40 is just the given difference between gold and silver, not the total. 70 is the number of gold beads only, using 7 parts. 30 is the number of silver beads only, using 3 parts.)
- (d) 20 — Alternate angles between parallel lines are equal, so 3x + 10 = 5x − 30. Rearranging, 10 + 30 = 5x − 3x, so 40 = 2x, and x = 20. −10 comes from a sign error when rearranging, moving a term to the wrong side and getting −20 = 2x instead. 25 comes from wrongly treating the two angles as co-interior and adding them to 180°: (3x + 10) + (5x − 30) = 180 gives 8x − 20 = 180, so x = 25. 47.5 makes the same co-interior mistake but sets the sum equal to 360° instead of 180°, giving 8x − 20 = 360 and x = 47.5.
- (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) 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) 225 — Method: find the interior angle of the regular octagon, then subtract it from 360° to find the reflex angle at the same vertex, since the interior angle and the reflex angle together make a full turn. Working: there are 8 − 2 = 6 triangles' worth of angle in the octagon, so the interior angle = 6 × 180 ÷ 8 = 135; reflex angle = 360 − 135 = 225. Answer: 225°. A candidate who stops after finding the interior angle gives 135. A candidate who works out the exterior angle instead, 360 ÷ 8 = 45, gives 45. A candidate who subtracts the exterior angle from 360° instead of the interior angle, working out 360 − 45, gets 315.
- (c) 4 — Method: subtract 3 from both sides, then divide by 4. Working: 4x = 19 − 3 = 16, so x = 16 ÷ 4 = 4. Answer: 4. 4.75 comes from dividing 19 by 4 directly, without subtracting 3 first. 3.75 comes from subtracting 4 instead of 3 before dividing. 5.5 comes from a sign error, adding 3 to 19 instead of subtracting it.
- (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.
- (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) 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.
- (d) 21 — Add the parts: 5 + 3 = 8. Divide the total by the number of parts: 56 ÷ 8 = 7, so one part is worth 7 counters. Blue has 3 parts: 3 × 7 = 21. (35 is the number of red counters, using 5 parts instead of 3. 28 comes from splitting 56 counters in half instead of in the ratio 5:3. 7 is the value of one part — the number of blue counters is 3 lots of this, not just one.)
- (c) 7.1 cm — Method: a shorter side is opposite the 45° angle and the hypotenuse is known, so sin θ = opposite ÷ hypotenuse gives that side directly. Working: sin 45° = x ÷ 10, so x = 10 × sin 45° = 7.071…, which is 7.1 to 1 decimal place. Answer: 7.1 cm. The distractors: 5.0 cm comes from halving the hypotenuse, which is the rule for the side opposite a 30° angle and not a 45° one; 14.1 cm comes from dividing by sin 45° instead of multiplying by it, which makes a shorter side longer than the hypotenuse; 10.0 cm comes from taking tan 45° = 1 and concluding that the shorter side matches the hypotenuse.
- (d) 30 — Method: 50% is one half, so 50% of a quantity is the quantity divided by 2. Working: 60 ÷ 2 = 30. Answer: 30. The distractors: 120 comes from multiplying by 2 instead of dividing; 3000 comes from multiplying by 50 without dividing by 100; 6 comes from finding 10% instead of 50%.
- (a) 27 — Method: turn the mean into a total using total = mean × number of values, then subtract the numbers that are already known. Working: four numbers with a mean of 18 have a total of 18 × 4 = 72; the three known numbers give 10 + 15 + 20 = 45; so x = 72 − 45 = 27. Answer: 27, and checking, (10 + 15 + 20 + 27) ÷ 4 = 72 ÷ 4 = 18. The distractors: 72 comes from stopping at the total the four numbers must reach and never subtracting the known three; 18 comes from assuming the missing number must equal the mean; 45 comes from stopping at the total of the three known numbers.
- (b) x ≥ 5 — Expand the bracket: 2(3x − 1) = 6x − 2, so the inequality is 6x − 2 ≥ 4x + 8. Subtract 4x from both sides and add 2 to both sides: 2x ≥ 10. Divide both sides by 2: x ≥ 5. A candidate who only multiplies the 3x by 2 and forgets to multiply the −1 gets 6x − 1 ≥ 4x + 8, leading to x ≥ 4.5. A candidate who adds 4x instead of subtracting it gets 10x ≥ 10, leading to x ≥ 1. A candidate who multiplies by 2 instead of dividing gets x ≥ 20.
- (c) 33 — Method: multiply the mean by the number of tests to get the total marks, then subtract the marks that are already known. Working: four tests with a mean of 29 give a total of 29 × 4 = 116 marks; the first three marks total 31 + 26 + 26 = 83; so the fourth mark is 116 − 83 = 33. Answer: 33, and checking, (31 + 26 + 26 + 33) ÷ 4 = 116 ÷ 4 = 29. The distractors: 116 comes from stopping at the total for all four tests; 29 comes from assuming the missing mark must be the mean itself; 4 comes from multiplying the mean by 3, the number of marks given, leaving 87 − 83 = 4.
- (a) x < 7 — Subtract 5 from both sides: x < 12 − 5, so x < 7. A candidate who adds 5 instead of subtracting gets x < 17. A candidate who subtracts the wrong way round gets x < −7. A candidate who correctly finds 7 but wrongly flips the inequality (as if dividing by a negative had happened) writes x > 7.
- (c) 3 — Gradient = (change in y) ÷ (change in x) = (11 − 5) ÷ (4 − 2) = 6 ÷ 2 = 3. A candidate who puts the change in x over the change in y instead would get 2 ÷ 6 = 1/3. A candidate who subtracts the y-coordinates in the reverse order, but not the x-coordinates, would get (5 − 11) ÷ (4 − 2) = −3. A candidate who adds the coordinates instead of subtracting them would get (11 + 5) ÷ (4 + 2) = 16/6 = 8/3.
- (b) 53.1° — The angle of elevation is opposite the height of the flagpole, 12 m, and adjacent to the distance from its base, 9 m, so tan θ = 12/9 = 1.333..., giving θ = tan⁻¹(1.333...) = 53.13...° ≈ 53.1°. "36.9°" finds the OTHER acute angle of the triangle, 90° − 53.1°, the angle at the top of the flagpole rather than the angle of elevation at Freya's position. "48.6°" comes from wrongly treating 9/12 as a sine ratio and finding sin⁻¹(0.75) = 48.6°, when neither side here is the hypotenuse. "41.4°" comes from wrongly treating 9/12 as a cosine ratio and finding cos⁻¹(0.75) = 41.4°, again without a hypotenuse in the ratio at all.
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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