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.For any two whole numbers, the product of the numbers is equal to the product of their highest common factor and their lowest common multiple. The highest common factor of 6 and 8 is 2, and 6 × 8 = 48. Work out the lowest common multiple of 6 and 8.
- 2.A line has equation y = −4x + 1. Work out the equation of the line perpendicular to it that passes through the point (0, 3).y = -4x + 1
- 3.In a mix of mortar the ratio of the mass of sand to the mass of cement is 5 : 2. Write the mass of cement as a fraction of the total mass of the mix.
- 4.A rectangle has vertices A(1, 1), B(4, 1), C(4, 5) and D(1, 5). Work out the perimeter of the rectangle.
- 5.On the probability scale from 0 (impossible) to 1 (certain), evens is exactly halfway. An event is 3/10 more likely than evens. Work out the probability of that event, giving your answer as a decimal.
- 6.A factory makes 4,000 light bulbs a day. In a random sample of 80 of one day's bulbs, 3 were faulty. Work out an estimate for the number of faulty bulbs the factory makes in a day.
- 7.The diameter of an artificial silk fibre is 4 × 10⁻⁶ metres. One nanometre is 10⁻⁹ metres. Work out the diameter of the fibre in nanometres.
- 8.Solve (3x + 2)/5 = (x + 4)/3
- 9.Priya is paid £58.50 for working 7.5 hours on a Saturday. Work out her rate of pay, in £ per hour.
- 10.Shape S is enlarged by a scale factor of −3 about a fixed centre. Which statement correctly describes the image compared to the original shape?
- 11.A box holds 5 blue pens and 7 black pens. Two pens are taken at random, one at a time, and are not put back. The first pen taken is black. Work out the probability that the second pen taken is blue.
- 12.At a book fair the ratio of fiction books to non-fiction books is 4 : 5. Of the non-fiction books, 3/5 are about history. Work out the fraction of all the books at the fair that are history books.
- 13.The diagram shows the graph of a quadratic function. Which equation could it represent?
- 14.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.
- 15.A construction for the perpendicular bisector of AB begins by opening a pair of compasses to more than half the length of AB, then drawing an arc centred at A. What is the next step?
- 16.The first five terms of a sequence are 7, 9, 13, 19, 27. By finding the second difference, work out the coefficient of n² in the nth term.
- 17.In a fruit drink, 3/8 of the total volume is orange juice and the rest is apple juice. Write down the ratio of orange juice to apple juice, in its simplest form.
- 18.A scale drawing of a bridge uses a scale of 1 : 25. A support beam on the drawing measures 4.4 cm. What is the real length of the beam, in metres?
- 19.The nth term of a sequence is 4n − 3. Work out the 7th term of the sequence.
- 20.A geometric sequence has first term 3 and common ratio 2. Work out the 5th term of the sequence.
Answer key
- (a) 24 — Method: rearrange the relationship so that the lowest common multiple stands alone; it is the product of the two numbers divided by their highest common factor. Working: 48 = 2 × the lowest common multiple, so the lowest common multiple is 48 ÷ 2 = 24. Checking, 24 is in the 6 times table and in the 8 times table. Answer: 24. The distractors: 48 comes from giving the product of the two numbers and never dividing by the highest common factor; 96 comes from multiplying by the highest common factor instead of dividing by it; 12 comes from dividing by the highest common factor twice, once for each of the two numbers.
- (d) y = (1/4)x + 3 — The given line has gradient −4, so the perpendicular gradient is the negative reciprocal, 1/4. Since (0, 3) is the y-intercept, the perpendicular line is y = (1/4)x + 3. Distractor routes: y = −4x + 3 uses the original line's own gradient instead of the perpendicular gradient, only changing the intercept for the new point. y = (−1/4)x + 3 takes the reciprocal of −4 but forgets to change its sign, so it is not the true negative reciprocal. y = 4x + 3 negates the gradient −4 but does not take its reciprocal, giving 4 instead of 1/4.
- (c) 2/7 — Method: each number in a ratio counts parts, so the whole is all the parts added together, and the fraction wanted is the cement parts over that total. Working: the mix has 5 + 2 = 7 parts altogether, and 2 of those parts are cement, so the cement is 2/7 of the mix. Answer: 2/7. The distractors: 2/5 comes from writing the cement parts over the sand parts, comparing one part with the other part instead of with the whole; 5/7 comes from writing the sand parts over the total, which answers about the wrong material; 7/2 comes from writing the total over the cement parts, turning the fraction upside down.
- (c) 14 units — Method: the perimeter of a rectangle is the distance all the way round its outside, 2 × (length + width), so the two side lengths must be found first; on a coordinate grid a side's length is the difference between the coordinates that change along it. Working: along AB, from (1, 1) to (4, 1), only x changes, so AB = 4 − 1 = 3. Along BC, from (4, 1) to (4, 5), only y changes, so BC = 5 − 1 = 4. Perimeter = 2 × (3 + 4) = 2 × 7 = 14. Answer: 14 units. The distractors: 18 units comes from reading the vertex numbers 4 and 5 as the side lengths instead of subtracting, giving 2 × (4 + 5); 12 units comes from working out the area, 3 × 4, in place of the perimeter; 7 units comes from adding one length to one width and stopping there, without doubling for the opposite pair of sides.
- (d) 0.80 — Evens is exactly halfway along the scale, at 0.5, and 3/10 written as a decimal is 0.3. A probability 3/10 higher than evens is 0.5 + 0.3 = 0.80. Giving 0.30 as the answer converts the increase but never adds it to the value of evens. Adding the increase to 1, the 'certain' end of the scale, instead of to evens gives 1 + 0.3 = 1.30, which cannot be a probability. Writing 3/10 as 0.03 instead of 0.3, a place-value slip, gives 0.5 + 0.03 = 0.53.
- (a) 150 bulbs — Method: assume the proportion faulty in a random sample is the proportion faulty in the whole day's output, and scale the sample up to the population. Working: the sample of 80 has to be scaled up to 4,000 bulbs, and 4,000 ÷ 80 = 50, so the day's output is 50 sample-sized batches. Each batch is expected to contain the same 3 faulty bulbs, so the estimate is 3 × 50 = 150. Answer: 150 bulbs, and it is an estimate, because another sample of 80 would probably contain a different number of faulty bulbs. The distractors: 50 bulbs is the scale factor 4,000 ÷ 80 written down as though it were the answer, so it reports how many batches there are rather than how many faulty bulbs; 120 bulbs comes from reading 3 out of 80 as 3%, then taking 0.03 × 4,000 = 120, but 3 out of 80 is 3.75%; 240 bulbs comes from 3 × 80 = 240, multiplying the faulty bulbs by the size of the sample instead of by the scale factor, which uses the 80 twice and the 4,000 not at all.
- (a) 4,000 nanometres — Method: the number of nanometres is the diameter divided by the length of one nanometre, and dividing powers of ten means subtracting the indices. Working: −6 − (−9) = 3, so 10⁻⁶ ÷ 10⁻⁹ = 10³, and the diameter is 4 × 10³ nanometres. Answer: 4,000 nanometres. The distractors: 400 nanometres comes from taking the difference between the indices as 2 instead of 3; 4 nanometres comes from changing the name of the unit without converting, leaving the coefficient untouched; 0.004 nanometres comes from dividing by 10³ instead of multiplying by it, as though a nanometre were the larger of the two units.
- (a) x = 3.5 — Method: clear the fractions by multiplying both sides by 15, expand both brackets, then collect the x terms on one side and the numbers on the other. Working: multiplying both sides by 15 gives 3(3x + 2) = 5(x + 4), which expands to 9x + 6 = 5x + 20; subtracting 5x and 6 from both sides gives 4x = 14, and dividing both sides by 4 gives x = 3.5. Answer: x = 3.5. The distractors: x = 1 comes from collecting the x terms by adding the 5x instead of subtracting it, giving 9x + 5x = 20 − 6 and so 14x = 14; x = 4.5 comes from expanding 3(3x + 2) as 9x + 2, multiplying only the x term by the 3, which leads to 4x = 18; x = 10 comes from reaching 4x = 14 correctly and then subtracting 4 instead of dividing by 4.
- (a) £7.80 — Rate of pay = total pay ÷ number of hours. £58.50 ÷ 7.5 = £7.80 per hour. £438.75 comes from multiplying the pay by the hours instead of dividing (£58.50 × 7.5). £0.13 comes from dividing the hours by the pay instead of the pay by the hours (7.5 ÷ 58.50). £51.00 comes from subtracting the hours from the pay (£58.50 − 7.5) instead of dividing.
- (a) 3 times the size, opposite side, rotated 180° — Method: for any enlargement, the MAGNITUDE of the scale factor gives the size ratio between image and object, while the SIGN decides which side of the centre the image falls on; a negative scale factor puts the image on the opposite side, which is equivalent to a 180° rotation about the centre. Working: the scale factor is −3, so the size ratio is the magnitude, which is 3, and the negative sign puts the image on the opposite side of the centre, rotated 180° relative to the original. Answer: 3 times the size, opposite side, rotated 180°. The magnitude of the scale factor controls the SIZE only: do not let the sign leak into it and turn 3 into 1/3. The sign controls the SIDE and orientation, which a positive-only view of enlargement, just 'further away' with the same orientation, misses entirely.
- (d) 5/11 — Method: the pen already taken was black, so update the contents of the box before working out the second probability. Working: the box held 12 pens and one black pen has gone, so 11 pens remain. None of the blue pens has been taken, so all 5 are still there, and the probability is 5/11, which will not cancel. Answer: the probability is 5/11. The distractors: 5/12 uses the box as it was at the start, which is only correct if the first pen is put back; 4/11 takes one off the blue count as well as the total, as though the pen removed had been blue; 6/11 gives the probability that the second pen is black, carrying on with the colour of the first pen instead of the colour asked for.
- (c) 1/3 — Method: find non-fiction's fraction of the whole, then multiply by the fraction of non-fiction that is history. Working: total parts = 4 + 5 = 9, so non-fiction is 5/9 of all books. History books are 3/5 of the non-fiction books: 3/5 × 5/9 = 15/45 = 1/3. Answer: 1/3. 3/5 comes from giving the fraction of non-fiction books that are history, without relating it to all the books at the fair. 5/9 comes from stopping after finding the fraction of all books that are non-fiction, without finding the history books within that. 4/15 comes from multiplying 3/5 by the fraction that is fiction (4/9) instead of the fraction that is non-fiction (5/9).
- (d) $y = x^2 - 2x - 3$ — Method: read the two x-intercepts (roots) from the graph, write the quadratic as the product of the corresponding factors, then expand. Working: the curve crosses the x-axis at x = −1 and x = 3, so the equation factorises as (x + 1)(x − 3), which expands to x² − 2x − 3. Answer: y = x² − 2x − 3. Distractor refutation: y = x² − x − 6 comes from misreading the left-hand crossing point as x = −2 instead of x = −1, giving factors (x + 2)(x − 3). y = x² − x − 2 comes from misreading the right-hand crossing point as x = 2 instead of x = 3, giving factors (x + 1)(x − 2). y = x² + 2x − 3 comes from writing the factors as (x − 1)(x + 3), swapping which root gets the plus sign and which gets the minus sign, giving the wrong sign on the x term.
- (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.
- (b) an arc of the same radius, centred at B — The perpendicular bisector construction needs two arcs of equal radius, one centred at each end of the segment — after the arc from A, the next step is an arc of exactly the same radius from B, so the two arcs cross at two points; the line through those two crossing points is the perpendicular bisector. "an arc of the same radius, centred at the midpoint of AB" is not possible yet, since the midpoint is only found once both arcs are drawn — it is the RESULT of the construction, not a step in it. "a smaller arc, centred at A again" gives two different-sized arcs from the same point, which never cross to give the bisector. "a straight line joining the ends of the first arc" only connects points on one arc, and locates nothing.
- (b) 1 — First differences: 2, 4, 6, 8. Second differences: 2, 2, 2, a constant, confirming the sequence is quadratic. The coefficient of n² is half the constant second difference: 2 ÷ 2 = 1. Using the second difference itself as the coefficient, without halving it, gives 2. Halving twice, dividing the second difference by 4 instead of by 2, gives 0.5. Doubling the second difference instead of halving it, on the mistaken rule that the n² coefficient is twice the second difference, gives 2 × 2 = 4.
- (c) 3 : 5 — If orange juice is 3/8 of the total, apple juice is the remaining 1 − 3/8 = 5/8. The ratio of orange to apple is therefore 3 : 5. Inverting gives 5 : 3, apple to orange instead of orange to apple. Using the denominator 8 as the second part of the ratio, 3 : 8, compares orange juice to the whole drink rather than to the apple juice alone. Pairing the total 8 with the apple fraction's numerator 5 gives 8 : 5, which mixes a whole-total figure with a part figure.
- (d) 1.1 — Method: multiply the drawing length by the scale factor to get the real length, then convert to the units asked for. Working: 4.4 cm × 25 = 110 cm = 1.1 m. A student who answers 4.4 has forgotten to use the scale at all. A student who answers 110 has correctly worked out the real length in centimetres but forgotten to convert it to metres. A student who answers 11 has used a scale factor of 2.5 instead of 25 by misreading the scale. Answer: 1.1 m.
- (b) 25 — Substitute n = 7: 4 × 7 − 3 = 28 − 3 = 25. Forgetting to subtract 3 gives 4 × 7 = 28. Subtracting 3 from 7 before multiplying by 4, 4 × (7 − 3) = 16, applies the operations in the wrong order. Substituting n = 8 by miscounting the position gives 4 × 8 − 3 = 29.
- (d) 48 — Each term is found by multiplying the previous term by the common ratio, 2: 3, 6, 12, 24, 48 — the 5th term is 48. A candidate who mistakes the common ratio for a common difference, and adds 2 four times, would reach 3+4×2=11. A candidate who works out the multiplier 2⁴=16 but forgets to multiply it by the first term would give 16. A candidate who multiplies one time too many (finding the 6th term instead of the 5th) would reach 3×2⁵=96.
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.