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.Expand and simplify √3(2 + √12).
- 2.On a distance-time graph, a horizontal line segment shows a period when the graph's gradient is 0. What does this tell you about the journey during that time?
- 3.The speed of a car is given by v = d ÷ t. A car travels 180 km in 3 hours. Work out the average speed of the car in km/h.
- 4.Two similar triangular tiles have lengths in the ratio 3 : 5. The longest side of the smaller tile is 9 cm. Work out the length of the longest side of the larger tile.
- 5.At a coffee shop, each customer buys tea, coffee, water or juice, and never more than one of these. The probability that a customer buys tea is 0.36 and the probability that they buy juice is 0.04. The probability that a customer buys coffee is three times the probability that they buy water. Work out the probability that a customer buys water.
- 6.A town council wants to find out about the eating habits of the people who live in the town. It asks only the members of a local sports club. Give a reason why this sample is biased.
- 7.p = 24 and q = 6, each correct to the nearest integer. Work out the greatest degree of accuracy to which p ÷ q can be guaranteed correct.
- 8.Which expression is equivalent to 3(x + 4) − 2(x − 1)?
- 9.The price of a cycling helmet rises from £80 to £116. Work out the percentage increase.
- 10.A shop stacks tins in a display block. The front elevation of the block is 4 tins wide and 3 tins high. The side elevation shows the block is 2 tins deep. Every position in the block is filled. How many tins are used in total?
- 11.On Amelia's walk to school there are two sets of traffic lights. The probability that the first set is green when she reaches it is 0.4. The probability that the second set is green when she reaches it is 0.5. The two sets work independently of each other. Work out the probability that both sets are green.
- 12.A photography studio offers 4 backdrops and 3 outfits for a portrait session. The plain grey backdrop, one of the 4 backdrops, cannot be used with the formal suit outfit, one of the 3 outfits. Work out how many different backdrop-and-outfit combinations are possible.
- 13.x = 5. Work out the value of 3x² − 4.
- 14.A recipe for 8 people needs 320 g of rice. Sanjay scales the recipe down to serve 5 people, then wants to scale his new amount back up to serve 8 people again. Work out the multiplier he should use to scale his amount for 5 people back up to the amount for 8 people.
- 15.Triangle ABC is similar to triangle PQR. AB corresponds to PQ, and BC corresponds to QR. AB = 5 cm, PQ = 15 cm and BC = 7 cm. Work out the length of QR.
- 16.A plumber charges a call-out fee of £30 plus £25 per hour worked. Work out the total charge for a job that takes 3 hours.
- 17.5 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.
- 18.A shape is reflected in the x-axis and the image is then reflected in the y-axis. Which single transformation is equivalent to this combination for every point?
- 19.The height, h metres, of a ball t seconds after a stopwatch is started follows h = (t − 1)(5 − t) for 1 ≤ t ≤ 5, where h = 0 means the ball is at ground level. Work out the two times at which the ball is at ground level.
- 20.The graph of y = f(x) has a maximum turning point at (5, 8). Which of these correctly gives the corresponding turning point on the graph of y = −f(x + 1), and its type?
Answer key
- (d) 6 + 2√3 — Multiply √3 by each term in the bracket separately. First term: √3 × 2 = 2√3. Second term: √3 × √12 = √(3 × 12) = √36 = 6. Adding the two results in the order they were found, and writing the whole-number term first, gives 6 + 2√3. Adding the numbers under the root for the second term instead of multiplying them (3 + 12 = 15) gives √15 in place of 6, leading to √15 + 2√3. Multiplying √3 by the 2 but never distributing to the √12 term at all leaves just 2√3. Treating √3 × 2 as if the 3 were multiplied by the 2 inside the root, √3 × 2 → √6, while still getting the second term correct, gives 6 + √6.
- (c) The object was stationary (not moving) — On a distance-time graph, the gradient at any point represents the speed at that point. A gradient of 0 means distance is not changing over time, so the object is stationary. A straight, sloped line (not flat) shows constant nonzero speed; a curve bending one way shows acceleration or deceleration; a flat section is not a maximum speed — it is no speed at all. Read the shape of the graph, not just how steep it looks.
- (d) 60 km/h — Method: use the formula v = d ÷ t with the distance and time given. Working: 180 ÷ 3 = 60 km/h. So the average speed is 60 km/h. Distractor 540 km/h comes from multiplying the distance and time instead of dividing. Distractor 90 km/h comes from dividing by 2 instead of 3. Distractor 18 km/h comes from dividing by 10 instead of 3, a decimal-point slip.
- (c) 15 cm — Method: going from the smaller tile to the larger one, every length is multiplied by the scale factor 5 ÷ 3. Working: 9 ÷ 3 = 3, and 3 × 5 = 15. Answer: 15 cm. The distractors: 11 cm comes from adding the difference between the parts of the ratio, 5 − 3 = 2, to the 9 cm side, treating a ratio as a gap rather than a multiplier; 45 cm comes from multiplying by 5 and forgetting to divide by 3; 27 cm comes from multiplying by 3, which is the part of the ratio belonging to the smaller tile.
- (a) 0.15 — Let P(water) = x, so P(coffee) = 3x. The four outcomes are exhaustive: 0.36 + 0.04 + x + 3x = 1, so 0.4 + 4x = 1, giving 4x = 0.6 and x = 0.15. So P(water) = 0.15. Reporting 3x, the coffee probability, instead of water gives 0.45. Splitting the remaining 0.6 evenly between coffee and water, ignoring the 3:1 ratio, gives 0.30. Stopping once the remaining probability 0.6 is found, without dividing by the four equal shares, gives 0.60.
- (c) Club members probably eat differently from most people — Method: a sample is biased when the group it is drawn from differs from the population in the very thing the survey is measuring, so compare the subgroup with the population on that quantity. Working: the survey measures eating habits, and people who join a sports club take more exercise than average and are known to eat differently from the town as a whole, so their replies pull the results away from the true picture for the town however many of them are asked. Answer: club members probably eat differently from most people. The distractors: the reply about the number of members treats bias as a question of size, but a large biased sample is still biased; the reply that the members were picked at random is false, since the council picked a club rather than picking residents, and it confuses bias with non-response; the reply that everyone asked lives in the town notes something true of the members but draws the false conclusion that the sample therefore covers the town, when a sample must reflect a population and not merely be taken from inside it.
- (a) the nearest whole number — The error intervals are 23.5 ≤ p < 24.5 and 5.5 ≤ q < 6.5. The minimum of p ÷ q is 23.5 ÷ 6.5 ≈ 3.615, and the maximum is 24.5 ÷ 5.5 ≈ 4.455. Both of these round to 4 at the nearest whole number, so the answer is guaranteed correct to the nearest whole number — but not to the nearest 0.1, since 3.615 rounds to 3.6 while 4.455 rounds to 4.5, which do not agree. Claiming the nearest 0.1 assumes every figure a calculator shows is trustworthy, without checking whether the bounds actually agree that far. Claiming only the nearest 10 badly understates how much can be guaranteed here, since both bounds already round to 4, not merely to 0. Saying no degree of accuracy can be guaranteed gives up before checking whether the bounds agree at any level at all.
- (a) x + 14 — Expand each bracket separately: 3(x + 4) = 3x + 12, and −2(x − 1) = −2x + 2 (multiply −2 by both x and −1). Combine: 3x + 12 − 2x + 2 = x + 14. Writing x + 10 comes from taking −2(x − 1) as −2x − 2, not flipping the sign of the −1 inside the bracket. Writing 5x + 10 comes from treating the second bracket as +2(x − 1) instead of subtracting it, so the x-terms are added rather than subtracted. Writing x + 13 comes from only multiplying the 2 by the x, and carrying the −1 across unmultiplied.
- (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.
- (a) 24 — The display is 4 tins wide, 3 tins high and 2 tins deep, and every position in that block is filled, so the total number of tins is the product of all three measurements: 4 × 3 × 2 = 24. "12" comes from multiplying only the width and height shown in the front elevation (4 × 3), forgetting the depth entirely. "9" comes from adding the three measurements (4 + 3 + 2) instead of multiplying them. "6" comes from multiplying only the height and depth (3 × 2), forgetting the width shown by the front elevation.
- (a) 0.2 — Method: for independent events, the probability that both happen is the product of the two probabilities. Working: the first set is green with probability 0.4 and the second with probability 0.5, so the calculation is 0.4 × 0.5. Since 4 × 5 = 20 and the two factors carry one decimal place each, the product carries two. Answer: the probability is 0.2. The distractors: 0.9 comes from adding 0.4 and 0.5 instead of multiplying them; 0.45 comes from averaging the two probabilities; 0.1 comes from subtracting 0.4 from 0.5, treating the question as a difference.
- (c) 11 — Method: work out the total number of combinations as if there were no restriction, then subtract the one combination that is not allowed. Working: without any restriction there are 4 backdrops × 3 outfits = 12 combinations. The grey backdrop with the formal suit is not allowed, removing 1 combination: 12 − 1 = 11. Answer: 11. 12 comes from forgetting to remove the combination that is not allowed. 8 comes from removing the entire formal suit outfit from the count instead of just the one combination with the grey backdrop. 10 comes from removing two combinations instead of just the one that is not allowed.
- (c) 71 — Method: square x first, then multiply by 3, then subtract 4, following the order of operations. Working: x² = 5² = 25; 3 × 25 = 75; 75 − 4 = 71. Answer: 71. 221 comes from squaring (3x) as a whole first: (3 × 5)² = 225, then − 4 = 221, squaring the coefficient along with x. 75 comes from correctly working out 3x² but forgetting to subtract the 4. 3 comes from subtracting the 4 from x before squaring: (5 − 4)² × 3 = 3, doing the operations in the wrong order.
- (a) 8/5 — Scaling down from 8 people to 5 people uses the multiplier 5/8. To reverse this and scale back up from 5 people to 8 people, use the reciprocal of that multiplier: flip 5/8 to get 8/5. 5/8 comes from using the forward (scaling down) multiplier again, instead of reversing it. 3/5 comes from writing the difference in people (8 − 5 = 3) over 5, instead of using the reciprocal of 5/8. 25/64 comes from multiplying the forward multiplier by itself (5/8 × 5/8), instead of finding its reciprocal.
- (d) 21 cm — The scale factor from triangle ABC to triangle PQR is PQ ÷ AB = 15 ÷ 5 = 3. QR = BC × scale factor = 7 × 3 = 21 cm. A student who divides BC by the scale factor instead of multiplying gets 7 ÷ 3 = 2.33 cm. A student who multiplies BC by AB instead of by the scale factor gets 7 × 5 = 35 cm.
- (b) £105 — The hourly charge is 25 × 3 = £75. Adding the call-out fee: £75 + £30 = £105. A candidate who forgets the call-out fee gives just the hourly charge, £75. A candidate who adds the call-out fee to the hourly rate before multiplying by the hours, (30 + 25) × 3, gets £165. A candidate who multiplies the call-out fee by the number of hours instead of the hourly rate, 30 × 3, gets £90.
- (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.
- (a) A rotation of 180° about the origin — Method: composing two reflections in lines that cross is always a single rotation about the point where the lines meet, through twice the angle between them. Working: the x-axis and y-axis meet at the origin at an angle of 90°, so the combined transformation is a rotation about the origin through 2 × 90 = 180 degrees. Answer: a rotation of 180° about the origin. The rotation angle is TWICE the angle between the mirror lines, not the angle itself, and the centre is always where the two lines cross, not some other point, and the result of two reflections in intersecting lines is a rotation, never another reflection.
- (c) t = 1 second and t = 5 seconds — The ball is at ground level when h = 0, which happens when either bracket is zero. t − 1 = 0 gives t = 1, and 5 − t = 0 gives t = 5. The option with t = −1 makes a sign error solving the first bracket. The option with t = −5 makes a sign error solving the second bracket, treating 5 − t = 0 as if it gave a negative solution. The option with t = 4 seconds is an arithmetic slip in solving 5 − t = 0.
- (a) (4, −8), a minimum point — The transformation x → x + 1 inside f translates the graph 1 unit to the LEFT, so the x-coordinate becomes 5 − 1 = 4. The minus sign in front of f reflects the graph in the x-axis, so the y-coordinate becomes −8, and a reflection in the x-axis turns every maximum into a minimum, so (4, −8) is a minimum point. Writing '(4, −8), a maximum point' gets the coordinates right but forgets that a reflection in the x-axis swaps maximum and minimum points. Writing '(6, −8), a minimum point' comes from translating 1 unit to the RIGHT instead of the left — f(x + 1) always moves the graph in the negative x-direction. Writing '(4, 8), a minimum point' gets the x-coordinate and the type right, but forgets to actually negate the y-coordinate, even though it does correctly reclassify the point as a minimum.
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.