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
Answer key: GCSE Higher sample Paper 1 (non-calculator)
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- (a) 2, 3, 4, 5 — Method: work out which whole numbers satisfy both parts of the inequality. Working: n ≥ 2 means n can be 2 or more; n < 6 means n must be less than 6, so 6 itself is not included. The whole numbers that fit both conditions are 2, 3, 4 and 5. Answer: 2, 3, 4, 5. 2, 3, 4, 5, 6 treats < 6 as ≤ 6 and wrongly includes 6. 3, 4, 5 treats ≥ 2 as > 2 and wrongly leaves out 2. 1, 2, 3, 4, 5 wrongly includes 1, which does not satisfy n ≥ 2.
- (b) 6 — Method: for any point that lies on y = k/x, the value of k is found by multiplying the x-coordinate and the y-coordinate together, since k = x × y. Working: k = 2 × 3 = 6. Answer: k = 6. Distractor refutation: 1.5 comes from dividing the y-coordinate by the x-coordinate instead of multiplying them. 5 comes from adding the two coordinates instead of multiplying them. 9 comes from misreading the point's x-coordinate as 3 instead of 2, then multiplying 3 × 3.
- (b) 2/5 — Method: a fraction taken 'of the total' has the whole batch as its denominator, so add the two masses first and then write the butter over that total. Working: the total mass is 450 g + 300 g = 750 g, so the fraction is 300/750; the highest common factor of 300 and 750 is 150, and 300 ÷ 150 = 2 while 750 ÷ 150 = 5. Answer: 2/5 of the batch. The distractors: 2/3 comes from comparing the butter with the flour, 300/450, a part-to-part fraction when the question asks for a part compared with the whole; 3/5 comes from writing the flour over the total, 450/750, which answers about the wrong ingredient; 3/2 comes from writing the flour over the butter, 450/300, which both uses the wrong denominator and reverses the order.
- (b) 2x + 2y = 180° — Method: use the fact that OA, OB and OC are all radii, so triangles OAC and OBC are each isosceles, and set up the angle sum of triangle ABC as a whole. Working: since OA = OC, triangle OAC is isosceles, so angle OAC = angle OCA = x. Since OB = OC, triangle OBC is isosceles, so angle OBC = angle OCB = y. The angle at A in triangle ABC is x, the angle at B is y, and the angle at C is angle OCA plus angle OCB, which is x + y. The three angles of triangle ABC sum to 180°, giving x + y + (x + y) = 180°, which is 2x + 2y = 180°. Answer: 2x + 2y = 180°. Dividing this by 2 gives x + y = 90°, and since angle ACB = x + y, this proves angle ACB = 90°: the point C was never restricted to any particular position on the circle, so the proof holds for every choice of C. Include ALL THREE angles of the triangle in the sum, do not use a quadrilateral's 360° total, and remember angle ACB is the SUM x + y, not x alone.
- (b) 0.150 — The six scores are exhaustive, so their probabilities sum to 1. The probability of not landing on 6 is 1 − 0.25 = 0.75, and this is shared equally between the other five scores, so each has probability 0.75 ÷ 5 = 0.150. Giving 0.750 as the answer stops after finding the probability of not landing on 6, without sharing it out between the five remaining scores. Dividing 0.75 by 6 instead of by 5 gives 0.125, wrongly including the score of 6 among the equally likely scores. Ignoring the bias completely and dividing 1 by all six scores gives 1 ÷ 6 = 0.167.
- (a) 1.5 — Method: with an even number of values the median is the mean of the two middle values, which for 12 values are the 6th and the 7th once the data are in order. Working: the results are already in order, and 12 ÷ 2 = 6, so the middle pair are the 6th value, 1, and the 7th value, 2; the median is (1 + 2) ÷ 2 = 1.5. Answer: 1.5 brothers and sisters. The distractors: 1 comes from reading the 6th value and stopping there instead of averaging the middle pair; 2 comes from working out the mean, 24 ÷ 12, instead of the median; 6 comes from working out the range, 6 − 0, which measures spread rather than centre.
- (b) 30 — Method: the smallest matching total is the lowest common multiple of the two pack sizes. Working: multiples of 6 are 6, 12, 18, 24, 30 …; multiples of 10 are 10, 20, 30 …. The lowest common multiple is 30. 60 comes from working out 6 × 10 = 60, the product of the pack sizes rather than their lowest common multiple. 16 comes from working out 6 + 10 = 16, which is not a common multiple at all. 2 is the highest common factor of 6 and 10, not a total of tickets. Answer: 30.
- (b) x = −4, y = −3 and x = 3, y = 4 — Substitute y = x + 1 into x² + y² = 25: x² + (x + 1)² = 25, which expands to x² + x² + 2x + 1 = 25, giving 2x² + 2x − 24 = 0, or x² + x − 12 = 0. Factorise: (x + 4)(x − 3) = 0, so x = −4 or x = 3. Using y = x + 1: when x = −4, y = −3; when x = 3, y = 4. Distractor routes: x = 4, y = 5 and x = −3, y = −2 comes from mis-factorising x² + x − 12 as (x − 4)(x + 3), reversing the sign of each root. x = −4, y = −3 alone stops after finding the first root of the quadratic and never goes back for the second pair. x = 12, y = 13 comes from misreading x² + y² = 25 as the linear equation x + y = 25 and solving that together with y = x + 1.
- (b) 250 g — Method: adding water changes the total mass but not the mass of salt, so find the salt, hold it fixed, use the new ratio to find the new total mass and subtract the mass already in the beaker. Working: 12:100 = x:500 gives 12 ÷ 100 × 500 = 60 g of salt; that 60 g must be 8% of the new mixture, so 8:100 = 60:y gives y = 60 ÷ 8 × 100 = 750 g; the water added is 750 − 500 = 250 g. Answer: 250 g. The distractors: 750 g is the mass of the diluted solution, given without taking away the 500 g that was in the beaker to start with; 60 g is the mass of salt, the quantity that stays the same, given instead of the mass of water; 20 g comes from treating the fall from 12% to 8% as 4% of the original 500 g, which measures a change in concentration as though it were a mass of water.
- (b) a rectangle — Lying on its side, the cylinder's curved surface touches the table along a straight line, and the two flat circular ends face sideways rather than up or down; viewed from directly above, the outline traced is a rectangle — as long as the cylinder and as wide as its diameter. "a circle" would be correct if the cylinder stood upright on one of its circular ends instead of lying on its side. "a triangle" belongs to a cone lying or standing so that it narrows to a point in that view, which a cylinder never does. "an oval" is a common guess from picturing the round ends, but from directly above those ends are edge-on and contribute to the rectangle's short sides, not a curved outline.
- (d) 1/8 — The multiples of 3 from 1 to 8 are 3 and 6, so the probability of that event is 2/8, which simplifies to 1/4. The probability of the coin landing on heads is 1/2. Since the spin and the toss are independent, multiply the two probabilities: 1/4 × 1/2 = 1/8. A candidate who answers 1/4 has considered only the spinner and forgotten to combine it with the coin toss. A candidate who answers 1/2 has considered only the coin and forgotten the spinner condition entirely. A candidate who answers 1/16 has counted only one number, 6, as a multiple of 3 instead of two, giving 1/8 × 1/2.
- (c) 5 × 10⁶ — Method: standard form is written as A × 10ⁿ, where A is at least 1 and less than 10 and n counts the places the decimal point moves. Working: the digits of 5,000,000 give a coefficient of A = 5, and the decimal point travels from the end of 5,000,000 until it sits just after the 5, a move of 6 places, so n = 6. Answer: 5 × 10⁶. The distractors: 50 × 10⁵ comes from stopping before the coefficient has been brought into range, and 50 is not less than 10, so it is not standard form; 5 × 10⁷ comes from counting the seven digits of 5,000,000 instead of the six places the decimal point moves; 5 × 10⁻⁶ comes from making the index negative because the decimal point was carried to the left, when a negative index belongs to a number smaller than 1.
- (a) x = 4 ± √13 — Method: halve the coefficient of x to form the bracket, subtract the square of that number to keep the expression equal to the original, then rearrange and take the square root of both sides. Working: half of −8 is −4, so x² − 8x + 3 = (x − 4)² − 16 + 3 = (x − 4)² − 13; setting this equal to zero gives (x − 4)² = 13, so x − 4 = ±√13 and x = 4 ± √13. Answer: x = 4 ± √13. The distractors: x = 8 ± √13 comes from putting the whole coefficient 8 inside the bracket instead of half of it; x = 4 ± √19 comes from adding 16 and 3 rather than subtracting 3 from 16; x = −4 ± √13 comes from writing the bracket as (x + 4)², which reverses the sign of the number that comes out of it.
- (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.
- (a) 080° — The back bearing differs from the given bearing by 180°. Because 260° is greater than 180°, subtract 180°: 260 − 180 = 80°, so the bearing of A from B is 080°. Choosing 440° adds 180° instead of subtracting it, even though the result would be more than a full turn (260 + 180 = 440). Choosing 180° assumes the back bearing is always exactly 180°, ignoring the original bearing altogether. Choosing 100° comes from measuring the reflex angle the other way round the circle (360 − 260 = 100) instead of applying the 180° back-bearing rule.
- (c) x = 4 — Method: with an unknown on both sides, subtract the smaller x term from both sides so that all the x is on one side, then collect the numbers on the other. Working: subtracting 3x from both sides gives 14 = 5x − 6; adding 6 to both sides gives 20 = 5x; dividing both sides by 5 gives x = 4. Checking: 3 × 4 + 14 = 26 and 8 × 4 − 6 = 26. Answer: x = 4. The distractors: x = 2.5 comes from taking 3x off the left-hand side only, leaving 14 = 8x − 6 and so 8x = 20; x = −1.6 comes from changing the sign of the 8x when it is moved across but leaving the sign of the 14 unchanged, giving 3x − 8x = −6 + 14 and so −5x = 8; x = 15 comes from reaching 5x = 20 correctly and then subtracting 5 instead of dividing by 5.
- (a) 1/9 — The ratio copper : tin is 9:1, so write tin over copper: 1/9. (9/1 comes from writing the ratio the wrong way round, copper over tin. 1/10 comes from comparing the tin to the total mass of the alloy, 1 part out of 10. 9/10 comes from comparing the copper to the total mass of the alloy, 9 parts out of 10.)
- (b) 2.8 m — Use Pythagoras' Theorem: the plank is the hypotenuse (3.5 m) of a right-angled triangle formed with the wall and the ground (2.1 m). height² = 3.5² − 2.1² = 12.25 − 4.41 = 7.84. height = √7.84 = 2.8 m. A student who subtracts the two given lengths directly instead of using Pythagoras gets 3.5 − 2.1 = 1.4 m. A student who doubles the distance from the wall by mistake gets 2.1 × 2 = 4.2 m.
- (a) (0, 7) and (0, −7) — A tangent is horizontal where the radius to that point is vertical, i.e. where the point lies on the y-axis. On x² + y² = 49, setting x = 0 gives y² = 49, so y = 7 or y = −7. The points are (0, 7) and (0, −7). (7, 0) and (−7, 0) comes from swapping the condition — these are the points where the tangent is VERTICAL, not horizontal (the radius there is horizontal). (0, 7) only comes from finding one valid point but forgetting that y² = 49 also gives the negative root, y = −7. (7, 0) only combines both mistakes: the wrong axis, and only one of the two roots.
- (b) The total volume of water, in litres, that has flowed in. — On a rate-time graph, the y-axis is in litres per minute and the x-axis is in minutes; multiplying a rate by a time gives litres per minute × minutes = litres, a total volume. So the area under the graph represents the total volume of water that has flowed in. Thinking the area itself represents the rate, rather than what the rate accumulates to, gives the wrong claim about the average rate of flow. Confusing the area with the gradient of the graph — which measures how the rate is changing — gives the wrong claim about litres per minute squared. Ignoring the flow-rate axis and focusing only on the time axis gives the wrong claim that the area is simply the total time.
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.