Sample paper · GCSE Higher · grades 4–9
GCSE Higher sample Paper 1 (non-calculator)
The real Paper 1 is 1.5 hour 30 minutes and 80 marks, non-calculator, and all three papers carry equal weight. This sample is 20 original questions in the same content proportions as the Higher qualification — number 15%, algebra 30%, ratio, proportion and rates of change 20%, geometry and measures 20%, probability 7.5%, statistics 7.5% — with no calculator-only items. AO1 / AO2 / AO3 at this tier: 40% / 30% / 30%.
Non-calculatorGCSE Higher
GCSE Higher sample Paper 1 (non-calculator)
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- 1.A window display has red baubles, gold baubles and green baubles in the ratio 4 : 5 : 6. What fraction of the baubles are not green?
- 2.The equation 3(2x − 1) = 4x + 9 is rearranged by expanding the brackets. Which of these is the correctly expanded equation?
- 3.A conversion graph between kilometres and miles passes through the origin and the point (50, 31), with the number of kilometres on the horizontal axis and the number of miles on the vertical axis. Work out the number of miles equivalent to 1 kilometre.
- 4.Two similar hexagonal tiles have lengths in the ratio 4 : 7. A side of the smaller tile is 8.4 cm long. Work out the length of the corresponding side of the larger tile.
- 5.A train company runs 25 trains a day, every day. The probability that any one train is late is 0.08. Work out how many late trains the company should expect over a period of 4 weeks.
- 6.The times, in seconds, taken by 11 runners to finish a race are given in order: 40, 42, 45, 47, 50, 52, 55, 58, 60, 63, 65. Work out the upper quartile of these times.
- 7.Work out 3³ + 2⁴.
- 8.To rearrange y = 2x − 5 to make x the subject, Sam writes: 'Add 5 to both sides to get y + 5 = 2x, then divide both sides by 2 to get x = (y + 5)/2.' Which statement about Sam's method is correct?y = 2x − 5
- 9.A road sign 3 m tall casts a shadow 5 m long. At the same time, a nearby lamppost casts a shadow 10 m long. Work out the height of the lamppost.
- 10.A right-angled triangle has a hypotenuse of 10 cm. One of its other angles is 45°. Work out the exact length of one of the two shorter sides.
- 11.In a class of 32 pupils, 19 study geography and 15 study history. 8 study both subjects. Work out n(G ∪ H), the number of pupils who study geography or history or both.
- 12.A digital timer truncates every time to 1 decimal place. It shows a swimmer's time for one length as 12.3 seconds. Using t for the swimmer's actual time in seconds, write down the error interval for t.
- 13.The solution set of an inequality is described as: x is less than −3, or x is greater than or equal to 1. Write this solution set using set notation.
- 14.5 decorators paint 3 rooms in 6 hours. All the decorators work at the same rate. Work out how long it takes 10 decorators to paint 5 rooms.
- 15.Three of these are rational numbers and one is irrational. Write down the one that is irrational.
- 16.A point has coordinates (x, y). In which two quadrants is the product x × y positive?
- 17.A furniture designer makes a scale model of a wardrobe using a scale of 1 : 8. The real wardrobe is 2.4 m tall and 1.2 m wide. Work out the perimeter of the front face of the model wardrobe, in centimetres.
- 18.A quadrilateral has two pairs of parallel sides. All four of its interior angles are right angles, but its sides are not all the same length. Which quadrilateral is this?
- 19.A proof that (n + 3)² − (n − 3)² is always a multiple of a certain number begins: Line 1: (n + 3)² − (n − 3)² = (n² + 6n + 9) − (n² − 6n + 9). Which expression correctly completes Line 2?
- 20.A model rocket's height is given by h = −5t² + 20t, where h is in metres and t is in seconds. A student says the graph of h against t is n-shaped. Which statement gives the correct verdict on the SHAPE and the reason that settles it from the equation?
Answer key
- (b) 3/5 — Method: add all three parts for the total, add together the parts that are not green, then write this over the total. Working: total parts = 4 + 5 + 6 = 15. Not green = 4 + 5 = 9. Fraction = 9/15 = 3/5. Answer: 3/5. 2/5 comes from finding the fraction that IS green (6/15 = 2/5) instead of not green. 4/15 comes from only counting the red baubles as 'not green' and forgetting the gold ones. 9/10 comes from adding only two of the three ratio parts to find the total (4+6=10), missing out the gold part, while still using 9 for the numerator.
- (b) 6x − 3 = 4x + 9 — Method: multiply every term inside the bracket by the number outside it; the right-hand side stays as it is given. Working: 3 × 2x = 6x and 3 × (−1) = −3, so 3(2x − 1) = 6x − 3. Answer: 6x − 3 = 4x + 9. 6x − 1 = 4x + 9 comes from multiplying only the 2x by 3 and leaving the −1 unchanged. 5x − 3 = 4x + 9 comes from adding the 3 to the 2 instead of multiplying, treating 3 × 2x as (3 + 2)x = 5x. 6x − 4 = 4x + 9 comes from working out 3 × (−1) as −1 − 3 = −4 instead of 3 × (−1) = −3.
- (a) 0.62 miles — The gradient of the line is the change in miles divided by the change in kilometres: 31 ÷ 50 = 0.62, so 1 kilometre converts to 0.62 miles. Dividing the wrong way round, 50 ÷ 31 = 1.612..., rounds to 1.61 miles — that finds how many kilometres are in 1 mile, not the reverse. Doubling the gradient, 1.24 miles, comes from using 62 ÷ 50 instead of 31 ÷ 50. Reading off the y-coordinate of the given point without dividing by the x-coordinate gives 31.00 miles, which is the number of miles for 50 kilometres, not for 1 kilometre.
- (c) 14.7 cm — The scale factor from the smaller tile to the larger tile is 7 ÷ 4 = 1.75, so the larger side is 8.4 × 1.75 = 14.7 cm. '4.8 cm' comes from scaling by 4 ÷ 7 instead, using the ratio the wrong way round. '11.4 cm' comes from adding the difference between the ratio numbers, 7 − 4 = 3, onto 8.4, instead of scaling. '58.8 cm' comes from multiplying by 7 on its own, using a ratio number as the scale factor instead of working out 7 ÷ 4 = 1.75 first.
- (d) 56 — Method: count the trials over the whole period first, then multiply the number of trials by the probability. Working: 4 weeks is 4 × 7 = 28 days, and at 25 trains a day that is 25 × 28 = 700 trains. The expected number of late trains is 700 × 0.08 = 56. Answer: about 56 late trains over the 4 weeks. The distractors: 2 is the expected number for a single day, 25 × 0.08 = 2, with the 28 days never brought in; 14 uses one week instead of four, 25 × 7 × 0.08 = 14; 644 is 700 − 56 and counts the trains expected to be on time.
- (b) 60 — Method: for n ordered values, the upper quartile sits at position 3(n + 1) ÷ 4, counting from the smallest. Working: n + 1 = 11 + 1 = 12; 3 × 12 = 36 and 36 ÷ 4 = 9, so the upper quartile is the 9th value in the list 40, 42, 45, 47, 50, 52, 55, 58, 60, 63, 65, which is 60. Answer: the upper quartile is 60 seconds. Watch which quartile you find: counting to the 6th value gives the median, 52, not the upper quartile; using 3 × 11 = 33 and 33 ÷ 4 = 8.25 without adding 1 to n first, then rounding down, reaches the 8th value, 58, not the 9th; and counting to the 3rd value uses the lower quartile's position, 45, the wrong end of the list.
- (c) 43 — Method: work out each power separately before adding. Working: 3³ = 27 and 2⁴ = 16, so 3³ + 2⁴ = 27 + 16 = 43. Answer: 43. (25 comes from using 3² instead of 3³, giving 9 + 16. 35 comes from working out 2⁴ as 2 × 4 = 8 instead of 2 × 2 × 2 × 2, giving 27 + 8. 432 comes from multiplying the two powers together instead of adding them.)
- (c) Yes; x = (y + 5)/2 is right — Sam's method is correct throughout: adding 5 to both sides gives y + 5 = 2x, and dividing both sides by 2 gives x = (y + 5)/2, so Sam is right. The option giving x = (y − 5)/2 assumes 5 should be subtracted again, but 2x − 5 = y means 5 has already been subtracted, so it must be added back, not taken away a second time. The option giving x = y/2 + 5 divides only the y term by 2 and leaves the 5 unhalved, which is not a valid rearrangement. The option agreeing Sam is correct but changing step 2 to x = 2(y + 5) confuses '2x' with 'x divided by 2' — since x is multiplied by 2, the inverse is division, not multiplication. Sam's working, and his final formula x = (y + 5)/2, are both correct.
- (c) 6.00 m — Method: the ratio of height to shadow length is the same for both objects. Working: road sign height ÷ shadow = 3 ÷ 5 = 0.6. Lamppost height = 0.6 × 10 = 6.00 m. Wrong options: 16.67 m comes from inverting the ratio, using shadow ÷ height instead of height ÷ shadow (10 × 5 ÷ 3); 8.00 m comes from adding the difference between the two shadow lengths to the road sign's height instead of scaling (3 + (10 − 5)); 1.50 m comes from multiplying by the ratio of the two shadow lengths the wrong way round (3 × 5 ÷ 10).
- (a) 5√2 cm — Method: the angles of a triangle add to 180°, so the third angle is 45° as well and the two shorter sides are equal. Take one of them as the side opposite a 45° angle and use sin 45° = opposite ÷ hypotenuse. Working: the exact value of sin 45° is √2/2, so the shorter side = 10 × √2 ÷ 2, and half of 10 is 5. Answer: 5√2 cm, which is about 7.07 cm. Remembering sin 45° as √2 rather than as √2 halved gives 10√2 cm, which is longer than the hypotenuse. Halving the hypotenuse because 45° is half of 90° gives 5 cm. Taking the value from the other special triangle, sin 60° = √3/2, gives 5√3 cm.
- (b) 26 — Method: n(G ∪ H) = n(G) + n(H) − n(G ∩ H), taking off the overlap once so the pupils who study both subjects are not counted twice. Working: 19 + 15 − 8 = 26. Answer: 26. Watch out: adding 19 and 15 without taking off the overlap gives 34, which counts the 8 pupils who study both subjects twice. Taking the 8 off both totals before adding, 19 − 8 + 15 − 8 = 18, counts only the pupils who study exactly one of the two subjects and leaves out the 8 who study both. And writing down 11, which is 19 − 8, gives the number who study geography only, not the number who study geography or history or both.
- (c) 12.3 ≤ t < 12.4 — Method: truncating cuts the later digits off instead of rounding them, so nothing is ever pushed upwards. The displayed value is therefore the smallest the time can be, and the time can run up to, but not reach, the next value the display can show. Working: the display reads 12.3, so the actual time is at least 12.3 seconds; as soon as the time reaches 12.3 + 0.1 = 12.4 seconds the display would read 12.4, so 12.4 is not included. Answer: 12.3 ≤ t < 12.4. The distractors: 12.25 ≤ t < 12.35 is the interval for a time rounded to 1 decimal place, and this display does not round; 12.3 < t ≤ 12.4 excludes the one value the display certainly allows and includes the one it rules out; 12.3 ≤ t ≤ 12.4 treats 12.4 seconds as possible, but at 12.4 seconds the display would no longer read 12.3.
- (a) {x : x < −3} ∪ {x : x ≥ 1} — "Less than −3" stays strict, since the wording never says "or equal to": x < −3. "Greater than or equal to 1" is inclusive: x ≥ 1. These are two separate, non-overlapping ranges joined with "or", so in set notation they are combined with the union symbol: {x : x < −3} ∪ {x : x ≥ 1}. Distractor routes: {x : x ≤ −3} ∪ {x : x > 1} swaps the strict and inclusive signs, marking −3 as included and 1 as excluded, the opposite of the wording. {x : −3 < x ≤ 1} treats "or" as "and", joining the two conditions into one continuous interval between the values instead of a union of two separate ranges. {x : x > −3} ∪ {x : x ≤ 1} reverses both inequality directions; the two reversed ranges then overlap and between them cover every number on the number line, so that set is the whole of the real line rather than the two separate ranges the description asks for.
- (c) 5 hours — Method: measure the job in decorator-hours, which is in the same ratio as the number of rooms, then share the decorator-hours between the decorators available. Working: 5 decorators × 6 hours = 30 decorator-hours for 3 rooms, so one room takes 30 ÷ 3 = 10 decorator-hours; 5 rooms take 5 × 10 = 50 decorator-hours; shared between 10 decorators that is 50 ÷ 10 = 5 hours. Answer: 5 hours. The distractors: 3 hours comes from halving the 6 hours because the number of decorators doubles, while forgetting that there are also more rooms to paint; 10 hours comes from scaling the 6 hours up for the rooms only, 6 × 5 ÷ 3, and leaving the workforce at 5 decorators; 6 hours comes from assuming that doubling the decorators and increasing the rooms cancel each other out, which they do not, because the rooms rise by a factor of 5/3 and the workforce by a factor of 2.
- (c) cos 45° — cos 45° = √2/2, and √2 cannot be written as an exact fraction, so this value is irrational. tan 0° = 0, sin 90° = 1 and sin 30° = 1/2 are all rational, since each can be written as an exact whole number or fraction.
- (b) the first and the third — Method: the sign of a product depends only on the signs of the two numbers multiplied: like signs give a positive product and unlike signs a negative one, so look for the quadrants in which both coordinates carry the same sign. Working: in the first quadrant x and y are both positive, and positive times positive is positive; in the third quadrant both are negative, and negative times negative is positive as well; in the second quadrant x is negative while y is positive, and in the fourth x is positive while y is negative, so each of those gives a negative product. Answer: the first and the third. The distractors: 'the second and the fourth' comes from applying the rule for a NEGATIVE product, unlike signs, to a positive one; 'the first and the second' comes from testing only the y-coordinate and keeping the quadrants where the height is positive; 'the first and the fourth' comes from testing only the x-coordinate in the same way.
- (d) 90 cm — Method: scale each dimension by the scale factor, then find the perimeter. Working: model height = 240 ÷ 8 = 30 cm; model width = 120 ÷ 8 = 15 cm. Perimeter = 2 × (30 + 15) = 90 cm. Wrong options: 11.25 cm comes from squaring the scale factor as if finding an area (720 ÷ 64); 510 cm comes from scaling only one dimension and leaving the other at full size; 720 cm comes from finding the real perimeter (2 × (240 + 120)) but forgetting to scale it down at all.
- (c) Rectangle — A rectangle has two pairs of parallel sides and four right angles, but does not require all sides to be equal — this matches exactly, so Rectangle is correct. A square also has four right angles and parallel sides, but additionally requires all four sides to be equal, which contradicts 'not all the same length', so it is wrong. A rhombus has two pairs of parallel sides and all four sides equal, but its angles are not generally 90° unless it is also a square, so it does not match the right-angle condition here. A kite has no pairs of parallel sides at all, so it does not match the first condition given.
- (b) 12n — Distributing the minus sign across the second bracket gives n² + 6n + 9 − n² + 6n − 9, and the n² terms and the +9/−9 cancel, leaving 6n + 6n = 12n. Writing 18 comes from only negating the first term of the second bracket, n², and treating the −6n and +9 as unchanged, which gives n² + 6n + 9 − n² − 6n + 9 = 18. Writing 2n² + 18 comes from adding the two brackets instead of subtracting them, (n² + 6n + 9) + (n² − 6n + 9) = 2n² + 18. Writing 6n comes from correctly negating the bracket but then only counting one of the two 6n terms, missing that they add rather than cancel.
- (a) Yes — the t² coefficient is negative, giving an n-shape. — The coefficient of t² in h = −5t² + 20t is −5, which is negative, so the graph is n-shaped with a maximum point — this matches the physical story of the rocket rising then falling, but the shape itself is decided by the negative coefficient of t², not by the story alone. Saying the shape comes from the story rather than the coefficient gets the reasoning backwards — the algebra determines the shape, and the story happens to agree with it. Saying it is U-shaped because height starts by increasing confuses the early part of the curve with its overall shape; a U-shaped curve would mean the rocket's height eventually increases again forever, which does not happen here. Saying it is n-shaped only because the rocket lands treats a consequence of the shape as if it were the cause.
How the 20 questions are shared out
- Number — 3 questions (15% of the qualification)
- Algebra — 6 questions (30% of the qualification)
- Ratio, proportion and rates of change — 4 questions (20% of the qualification)
- Geometry and measures — 4 questions (20% of the qualification)
- Probability — 2 questions (7.5% of the qualification)
- Statistics — 1 question (7.5% of the qualification)
Where an area has fewer printable questions than its share, the shortfall is filled from the other areas. These are original questions, not past papers.