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)
MathsUKwww.geekhero.co.uk
- 1.Work out (2/3)² + (1/2)². Give your answer as a fraction.
- 2.Which of these statements is an identity?
- 3.A footpath is drawn 3 cm long on a map with a scale of 1 : 50 000. Work out the real length of the footpath, in metres.
- 4.In triangle ABC, AB = 8 cm, AC = 5 cm and angle BAC = 60°. Work out the value of BC².
- 5.The probability that Isla answers a quiz question correctly is 0.8. She answers three questions, and her answers are independent of each other. Work out the probability that she answers all three correctly.
- 6.There are 10,000 pupils in a city. A researcher picks a sample of 500 of them by drawing names at random from a list of every pupil in the city. Give the reason why this method gives a representative sample.
- 7.2/3 of an amount of money is £30. Work out the amount.
- 8.A ribbon of length L metres is cut into 5 equal pieces, and then 3 metres is removed from one piece. Write down an expression, in metres, for the length of that piece after 3 metres is removed.
- 9.Line P passes through the origin and the point (3, 21). Line Q passes through the origin and the point (5, 45). Both lines represent the distance, in metres, run by an athlete against the time, in seconds. Work out the gradient of the steeper line.
- 10.Triangle D has vertices (3, 1), (3, 5) and (6, 1). It is reflected in the line x = 3. Write down the number of vertices of the triangle that are invariant under this reflection.
- 11.A machine produces bolts. In a sample of 250 bolts, 15 are faulty. Work out the relative frequency of a bolt being faulty, giving your answer as a fraction in its simplest form.
- 12.A number, x, is truncated (not rounded) to 1 decimal place and the result is 6.2. Write down the error interval for x.
- 13.A region R satisfies y ≤ x + 1, x + y ≤ 5 and y ≥ 0. Which of these correctly describes R in words?
- 14.A map has a scale of 2 cm : 5 km. A footpath measures 8 cm on the map. Work out the real length of the footpath, in kilometres.
- 15.Triangles LMN and XYZ have LM = XY, MN = YZ and LN = XZ. Give a reason why triangle LMN is congruent to triangle XYZ.
- 16.The graph of y = sin x is reflected in the x-axis to give the graph of y = g(x). Work out the value of g(90), where x is measured in degrees.y = sin(x)
- 17.A textbook is reduced from £60 to £45. Work out the percentage reduction.
- 18.In cyclic quadrilateral ABCD, with vertices in that order around the circle, the diagonal BD is drawn. In triangle ABD, angle ABD = 35° and angle ADB = 65°. Side DC is extended beyond C to a point E. Work out the size of angle BCE.
- 19.The cost, in pounds, of hiring a van for d days is given by the formula C = 25 + 18d. A customer's hire cost £133. Work out how many days, d, they hired the van for, by first rearranging the formula to make d the subject.
- 20.Which of these equations represents the graph of y = 2ˣ translated by 3 units in the positive y-direction?
Answer key
- (c) 25/36 — Method: square a fraction by squaring its numerator and its denominator separately, then add the two results over a common denominator. Working: (2/3)² = 4/9 and (1/2)² = 1/4; the lowest common denominator of 9 and 4 is 36, so 4/9 = 16/36 and 1/4 = 9/36, and 16 + 9 = 25. Answer: 25/36. The distractors: 49/36 comes from adding the two fractions first and squaring the total, giving (7/6)²; 5/13 comes from squaring correctly but then adding the numerators and the denominators, as (4 + 1)/(9 + 4); 7/3 comes from doubling each fraction instead of squaring it, giving 4/3 + 1.
- (c) 2(x + 4) ≡ 2x + 8 — 2(x + 4) ≡ 2x + 8 is an identity because expanding the bracket on the left gives exactly the right-hand side for every value of x — try any number and both sides match. 2(x + 4) = 20 is an equation: expanding gives 2x + 8 = 20, which is only true for the single value x = 6. 2(x + 4) = 2x + 4 is not true for any value of x at all: expanding the left side gives 2x + 8, which can never equal 2x + 4 since 8 ≠ 4. x + 4 = 2x is also an equation, true only for the single value x = 4. The identity is 2(x + 4) ≡ 2x + 8.
- (b) 1500 m — Method: multiply the map length by the scale to get the real length in centimetres, then divide by 100 to change centimetres into metres. Working: 3 × 50 000 = 150 000 cm, and 150 000 ÷ 100 = 1500. Answer: 1500 m. The distractors: 15000 m comes from dividing the 150 000 cm by 10 instead of 100; 150 m comes from dividing by 1000, the conversion for kilometres; 50000 m comes from writing the scale itself as the answer and ignoring the 3 cm measured on the map.
- (a) 49 cm² — Method: two sides and the angle between them are given, so use the cosine rule a² = b² + c² − 2bc cos A with BC facing the 60° angle. By hand, cos 60° = 0.5. Working: BC² = 8² + 5² − 2 × 8 × 5 × cos 60° = 64 + 25 − 80 × 0.5 = 89 − 40 = 49. Answer: BC² = 49 cm². The distractors: 129 cm² comes from adding the final term instead of subtracting it, 89 + 40; 89 cm² comes from leaving the cosine term out altogether and treating the 60° as though it were a right angle, so that Pythagoras applies; 69 cm² comes from forgetting the factor 2 in 2bc cos A and subtracting only 8 × 5 × 0.5 = 20.
- (c) 0.512 — Method: independent events that must all happen are combined by multiplying their probabilities. Working: the first two questions give 0.8 × 0.8 = 0.64. Bringing in the third gives 0.64 × 0.8, and since 64 × 8 = 512 with three decimal places in the product, this is 0.512. Answer: the probability is 0.512. The distractors: 0.64 comes from multiplying only two of the three probabilities and stopping; 0.8 comes from reading 'independent' as meaning the probability never changes and writing down the single-question figure; 0.0512 comes from a place-value slip in the last multiplication, counting four decimal places instead of three.
- (c) Because every pupil has an equal chance of being picked — Method: whether a sample represents its population is decided by the selection method, not by the size of the sample, so ask whether the method gives every member of the population the same chance of being chosen. Working: the names are drawn at random from a list of all 10,000 pupils, so each pupil has the same chance, 500 out of 10,000, of being drawn, and no group of pupils is more likely to appear than any other; that is what keeps bias out of the sample. Answer: because every pupil has an equal chance of being picked. The distractors: the reply about 5% treats the sampling fraction as the test of fairness, but a badly chosen 5% is still biased and a well chosen 1% is not; the reply about 500 being large enough makes size the test instead, which is the same mistake in another form, since a large sample drawn from one school would still misrepresent the city; the reply about the most willing pupils describes self-selection, which hands the choice of who is in the sample to the pupils who feel most strongly about the question.
- (b) £45 — Since 2/3 of the amount is £30, one third is £30 ÷ 2 = £15, and the whole amount is three thirds: £15 × 3 = £45. Applying the fraction forwards to £30 instead of reversing it, £30 × 2/3 = £20, treats the given amount as the whole rather than as two thirds of it. Finding one third correctly as £15 but forgetting to multiply by 3 to get the whole amount leaves £15 as the final answer. Reading £30 as one third of the amount rather than as two thirds, and so multiplying straight by 3, gives £30 × 3 = £90.
- (c) L/5 − 3 — Each of the 5 equal pieces is L/5 metres long, and removing 3 metres from one piece gives L/5 − 3. Subtracting the 3 metres before dividing by 5, (L − 3)/5, divides the removed length between all 5 pieces instead of taking it from just one. Dividing only the 3 by 5 instead of dividing L by 5, L − 3/5, divides the wrong number. Writing 5/L − 3 inverts the fraction, swapping which number is the numerator.
- (b) 9 m/s — The gradient of line P is 21 ÷ 3 = 7, so P has a rate of 7 m/s. The gradient of line Q is 45 ÷ 5 = 9, so Q has a rate of 9 m/s. Because 9 is greater than 7, line Q is the steeper line, with gradient 9 m/s. Taking line P's gradient instead of Q's gives 7 m/s, the less steep line. Subtracting the two lines' coordinates directly, (45 − 21) ÷ (5 − 3) = 24 ÷ 2 = 12 m/s, mixes points from different lines rather than using one line's own two points. Adding the two gradients, 7 + 9 = 16 m/s, treats 'steeper' as a total rather than a comparison.
- (d) 2 — Method: a point is invariant under a reflection exactly when it lies on the mirror line, so check each vertex's coordinate against the line's equation. Working: the line of reflection is x = 3, so a vertex is invariant only if its x-coordinate equals 3. (3, 1) has x = 3, so it is invariant. (3, 5) has x = 3, so it is also invariant. (6, 1) has x = 6, so it is not invariant, since 6 is not equal to 3. Two of the three vertices are invariant. Answer: 2. Check EVERY vertex against the mirror line's equation rather than assuming a shape either keeps all its vertices fixed or none of them: a vertex is invariant only when it sits exactly on the line, and here two do and one does not.
- (d) 3/50 — Relative frequency is the number of faulty bolts divided by the total sample size: 15/250, which simplifies to 3/50 by dividing both the numerator and the denominator by 5. Using 235, the number of bolts that were NOT faulty, as the denominator instead of the total 250 gives 15/235, which simplifies to 3/47. Inverting the fraction, dividing the total by the number of faulty bolts instead of the other way round, gives 250/15, which simplifies to 50/3 — a value greater than 1, which cannot be a probability. Simplifying by dividing the numerator and the denominator by different numbers, 15 ÷ 15 = 1 and 250 ÷ 25 = 10, gives 1/10.
- (b) 6.2 ≤ x < 6.3 — Truncating simply cuts off the digits after the required decimal place instead of rounding them, so every value from 6.2 up to (but not reaching) 6.3 truncates to 6.2. This gives the error interval 6.2 ≤ x < 6.3, with no allowance made on the lower side because truncation never rounds a smaller value up into this interval. Using 6.15 ≤ x < 6.25 applies the rounding rule of going half a unit either side, which does not apply to truncation. Writing 6.1 < x ≤ 6.2 puts the interval below 6.2 instead of above it. Writing 6.2 ≤ x ≤ 6.3 wrongly includes 6.3, which truncates down to itself, not to 6.2.
- (d) Below y = x + 1, below x + y = 5, above y = 0 — For y ≤ x + 1, R lies on or below the line y = x + 1. For x + y ≤ 5 (that is, y ≤ 5 − x), R lies on or below that line too. For y ≥ 0, R lies on or above the x-axis. Combining all three: R is below y = x + 1, below x + y = 5, and above y = 0. Distractor routes: "Above y = x + 1" flips the first inequality, describing the wrong side of that line. "Above x + y = 5" flips the second inequality, describing the wrong side of that line. "Below y = 0" flips the third inequality, describing the wrong side of the x-axis.
- (b) 20 km — The scale 2 cm : 5 km means each 1 cm on the map represents 5 ÷ 2 = 2.5 km in real life. The footpath is 8 cm on the map, so its real length is 8 × 2.5 = 20 km. 40 km comes from multiplying 8 by 5 directly, ignoring that the scale's '2 cm' has to be divided out first: 8 × 5 = 40. 3.2 km comes from dividing 8 by 2.5 instead of multiplying: 8 ÷ 2.5 = 3.2. 5 km comes from multiplying 2.5 by the scale's '2' instead of by the footpath's 8 cm: 2.5 × 2 = 5.
- (a) SSS – all three corresponding sides are equal — All three pairs of corresponding sides are stated as equal — LM = XY, MN = YZ and LN = XZ — with no angle mentioned. This matches the SSS condition, so triangle LMN is congruent to triangle XYZ.
- (a) −1 — Reflecting y = sin x in the x-axis gives y = −sin x, so g(x) = −sin x. Since sin 90° = 1, g(90) = −1. Reading sin 90° = 1 and forgetting to apply the reflection gives 1. Misreading the angle as 0° instead of 90° gives sin 0° = 0, so 0. Confusing sin 90° with sin 30° = 0.5, then reflecting it, gives −0.5.
- (d) 25% — Method: percentage decrease = decrease ÷ original amount × 100. Working: the reduction is £60 − £45 = £15, and 15 ÷ 60 = 0.25, so 0.25 × 100 = 25%. Answer: 25%. The distractors: 15% comes from quoting the £15 reduction as though pounds and per cent were the same thing; 33% comes from dividing the £15 by the new price £45 instead of by the original £60, which gives 33% to the nearest per cent; 75% is the new price written as a percentage of the old one, which is what is still paid rather than what has been taken off.
- (a) 80° — In triangle ABD, the three angles sum to 180°: angle BAD = 180° − 35° − 65° = 80°. ABCD is a cyclic quadrilateral, so the exterior angle at C, angle BCE, is equal to the interior angle at the opposite vertex, angle BAD: angle BCE = 80°. Adding the two given angles in triangle ABD instead of subtracting them from 180°, 35° + 65° = 100°, and then applying the exterior-angle rule correctly still gives the wrong angle BAD and so the wrong exterior angle, 100°. Halving the correct angle BAD, 80° ÷ 2 = 40°, confuses the exterior-angle rule with a different circle theorem in which one angle is half of another. Doubling it instead, 80° × 2 = 160°, makes the same kind of confusion in the other direction.
- (d) 6 — Method: rearrange the formula to make d the subject, then substitute C = 133. Working: C = 25 + 18d, so subtracting 25 from both sides gives C − 25 = 18d, then dividing by 18 gives d = (C − 25) / 18. Substituting C = 133: d = (133 − 25) / 18 = 108 / 18 = 6. The value 7.4 comes from dividing 133 by 18 without subtracting the fixed £25 first (133 / 18 ≈ 7.4). The value 4.6 comes from pairing the numbers the wrong way round, working out (133 − 18) / 25 = 4.6. The value 90 comes from subtracting both 25 and 18 from 133 instead of dividing by 18.
- (a) y = 2ˣ + 3 — A translation of 3 units in the positive y-direction shifts the whole graph up, which means adding to the output: y = f(x) + k with k = 3, so the image is y = 2ˣ + 3. Adding the 3 inside the power instead of outside it, which translates the graph horizontally instead of vertically, gives y = 2ˣ⁺³. Using a negative 3, which moves the graph down instead of up, gives y = 2ˣ − 3. Mistaking 2ˣ for the linear expression 2x and adding 3 inside brackets gives y = 2(x + 3).
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.