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 top of a clock tower is 25 m above level ground. Oliver stands on the ground 25 m from the foot of the tower. Work out the angle of elevation of the top of the tower from the point where Oliver stands.
- 2.A cycle route is 84 km long. Freya sets off along it at a steady 14 km/h. Write down the function for the distance y, in kilometres, that is still to be cycled after x hours.
- 3.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.
- 4.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.
- 5.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.
- 6.The price of a jacket increases by 50% and then decreases by 50%. Describe the overall change from the original price.
- 7.The angles of a triangle are in the ratio 2:3:4. Work out the size of the smallest angle.
- 8.Divide 84 in the ratio 3:4. Work out the smaller share.
- 9.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.
- 10.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.
- 11.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.
- 12.An amount of money is shared in the ratio 1:2:3. The largest share is £90 more than the smallest share. Work out the total amount that was shared.
- 13.The mean of three numbers is 50. A fourth number, 100, is added to the set. Work out the mean of the four numbers.
- 14.A regular polygon has an exterior angle of 45°. Work out the number of sides of the polygon.
- 15.A transversal crosses a pair of parallel lines. At one line, the angle is (5x + 4)°. The corresponding angle at the other line is (3x + 24)°. Work out the value of x.
- 16.In a recipe the mass of chocolate to the mass of milk is in the ratio 1:4. Amelia uses 200 g of milk. Work out the mass of chocolate she needs.
- 17.Which of these equations describes a vertical line?
- 18.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.
- 19.Amelia's mean mark over four tests is 29. Her first three marks are 31, 26 and 26. Work out her fourth mark.
- 20.Freya is designing a triangular flower bed. Two of its corners have equal angles, and the angle at the third corner is 40°. Work out the size of one of the two equal corner angles.
Answer key
- (c) 45° — Method: the tower, the ground and the line of sight form a right-angled triangle in which the 25 m height is opposite the angle of elevation and the 25 m along the ground is adjacent to it, so use tan θ = opposite ÷ adjacent. Working: tan θ = 25 ÷ 25 = 1, so θ = tan⁻¹(1). Answer: 45°. The distractors: 90° comes from using sin θ = 25 ÷ 25 = 1, which treats the 25 m along the ground as the hypotenuse when it is the side next to the angle; 1° comes from writing down the value of tan θ as though it were the angle itself; 50° comes from adding the two given lengths, 25 + 25, instead of comparing them.
- (b) y = 84 − 14x — Method: the distance still to go is the whole route minus the distance already covered, and at a steady speed the distance covered after x hours is the speed multiplied by x. Working: after x hours Freya has cycled 14x km, so the distance remaining is the 84 km route take away 14x, giving y = 84 − 14x, a straight line with intercept 84 and gradient −14. Answer: y = 84 − 14x. The distractors: y = 84 + 14x comes from adding the distance cycled to the length of the route rather than taking it away, so the ride would grow longer the further she goes; y = 14x − 84 comes from carrying out the subtraction the wrong way round, which is negative for the whole of the ride; y = 14 − 84x comes from swapping the two numbers over, treating 84 km as the hourly speed and 14 km as the length of the route.
- (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) 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.)
- (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) 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) 36 — Add the parts of the ratio: 3 + 4 = 7. Divide the total by the number of parts: 84 ÷ 7 = 12, so one part is worth 12. The smaller share has 3 parts: 3 × 12 = 36. (48 is the larger share, using 4 parts instead of 3. 42 comes from splitting 84 in half, treating the ratio as if it were 1:1. 28 comes from dividing 84 by 3 — one of the ratio numbers — instead of dividing by the total number of parts, 7.)
- (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.
- (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.
- (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) £270 — Method: the £90 is a difference between two shares, so turn it into a number of parts before finding the value of one part. Working: the largest share is 3 parts and the smallest is 1 part, so the difference is 3 − 1 = 2 parts and 2 parts are worth £90; one part = £90 ÷ 2 = £45; the whole amount is 1 + 2 + 3 = 6 parts, so 6 × £45 = £270. Answer: £270. The distractors: £540 comes from treating the £90 as the value of one part and multiplying it by the 6 parts; £180 comes from finding the £45 correctly but adding only the 1-part and 3-part shares and forgetting the middle share; £135 comes from multiplying £45 by 3 and giving the largest share instead of the total.
- (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.
- (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.
- (b) 10 — Corresponding angles are equal, so 5x + 4 = 3x + 24. Subtracting 3x from both sides gives 2x + 4 = 24, then subtracting 4 gives 2x = 20, so x = 10. 14 comes from adding the constants, 4 + 24, instead of subtracting them when rearranging. 19 comes from treating the angles as co-interior (summing to 180°): 5x + 4 + 3x + 24 = 180 gives 8x = 152, so x = 19. 20 correctly reaches 2x = 20 but stops without dividing by 2.
- (d) 50 g — Method: the milk is 4 parts of the ratio, so use the milk to find the value of one part, then read off the chocolate, which is 1 part. Working: one part = 200 ÷ 4 = 50, and the chocolate is one part. Answer: 50 g. The distractors: 40 g comes from treating the 200 g as the total mass of the mixture and splitting it into 1 + 4 = 5 parts; 250 g is the total mass of the finished mixture, the 200 g of milk plus the chocolate, rather than the chocolate on its own; 800 g comes from multiplying 200 by 4 instead of dividing, which scales the milk up rather than down to the chocolate.
- (c) x = 5 — Method: every point on a vertical line has the same x-coordinate however far up or down the line it lies, so the equation of a vertical line fixes x at a number and does not involve y at all. Working: of the four equations only x = 5 fixes x; it is satisfied by (5, 0), (5, 1), (5, 7) and by every other point whose x-coordinate is 5, and those points form a vertical line. Answer: x = 5. The distractors: y = 5 fixes the y-coordinate instead of the x-coordinate, which gives a horizontal line; y = 5x is a line through the origin with gradient 5, steep but not vertical, and it has a different y-value for every x; x = y fixes neither coordinate and is the line through the origin with gradient 1.
- (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.
- (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) 70° — Method: the three angles of a triangle add up to 180°, so take the known corner away from 180° to find what is left for the other two corners, then share that remainder equally because those two corners are equal. Working: 180 − 40 = 140, and the two equal corners share that 140° between them, so 140 ÷ 2 = 70. Answer: 70°. The distractors: 140° comes from stopping after 180 − 40 and offering the combined total of the two equal corners as though it were the size of one of them; 20° comes from halving the 40° corner that was given instead of halving the 140° that the other two corners have to share; 50° comes from 90 − 40, using the two acute angles of a right-angled triangle as the fixed total rather than the 180° angle sum of the whole triangle.
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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