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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- (c) 6 — The units digit must be even, so it can be 2 or 8, giving 2 choices. The tens digit can then be any of the remaining 3 digits, since one digit has been used for the units. Multiply: 2 × 3 = 6. 12 comes from working out how many two-digit numbers can be made in total, 4 × 3 = 12, ignoring the requirement that the number is even. 8 comes from choosing the units digit from 2 options and then wrongly allowing any of the 4 digits again for the tens digit, 2 × 4 = 8, which lets a digit repeat. 2 comes from counting only the choices for the units digit and forgetting the tens digit.
- (a) (15, 0) — Method: the tangent is perpendicular to the radius at the point of contact, so find the gradient of the radius, take its negative reciprocal, write the equation of the tangent, then substitute y = 0 because every point on the x-axis has y-coordinate 0. Working: the radius from (0, 0) to (3, 6) has gradient 6 ÷ 3 = 2, so the tangent has gradient −1/2. Substituting into y − 6 = −1/2(x − 3) gives y = −0.5x + 7.5. Setting y = 0 gives 0.5x = 7.5, so x = 15 and P is (15, 0). Answer: (15, 0). The distractors: (0, 7.5) is where the same tangent crosses the y-axis, reached by setting x = 0 instead of y = 0; (0, 0) comes from using the gradient of the radius, 2, for the tangent, which gives the line y = 2x through the centre and so crosses the x-axis at the origin; (6, 0) comes from changing the sign of the radius gradient without turning it upside down, which gives y = −2x + 12.
- (d) £3.60 per component — The gradient of a cost-against-components graph has units of pounds per component, since cost is measured in pounds and the horizontal axis counts components. So 3.60 means it costs an extra £3.60 to produce one more component at that point. Calling it '£3.60 total cost' confuses the gradient, a rate, with the y-value on the graph, which is the total cost itself. Giving it as 3.60 components per pound swaps which axis is on top, giving the units of the reciprocal gradient, not the gradient itself. Turning 3.60 into a percentage invents a unit that has no basis in the graph's axes — a gradient here is a number of pounds, not a percentage. Always build the gradient's units from the two axes' own units, in the order y-axis over x-axis.
- (a) Kite — Method: check each named quadrilateral's properties against the three facts given, one at a time. Working: a kite has two pairs of adjacent sides equal (not opposite pairs), one pair of opposite angles equal (the two angles where an unequal pair of sides meet), and exactly one line of symmetry — matching all three facts. Options: a rhombus does have equal adjacent sides, but all four of its sides are equal, both pairs of its opposite angles are equal, and it has two lines of symmetry rather than exactly one; a parallelogram has its opposite sides equal rather than adjacent pairs, both pairs of opposite angles equal, and no line of symmetry at all; a trapezium does not generally have any pair of equal adjacent sides or a line of symmetry. Answer: kite.
- (c) 7/8 — Method: 'at least one head' is the opposite of 'no heads at all', so work out the probability of three tails and take it away from 1. Working: a flip that is not a head has probability 1/2, and the flips are independent, so three tails in a row has probability 1/2 × 1/2 × 1/2 = 1/8. Taking this from 8/8 leaves 7/8. Answer: the probability is 7/8. The distractors: 1/8 is the probability of three tails, written down without the final subtraction; 3/8 is the probability of exactly one head, which comes from reading 'at least one' as 'exactly one'; 1/2 comes from giving the probability of a head on a single flip and ignoring that three flips are made.
- (c) 76 — Method: the frequency of each class is the area of its bar, frequency density × class width, so work out all three frequencies and add them. Working: the widths are 5, 10 and 25 kg, so the frequencies are 4 × 5 = 20, 2.6 × 10 = 26 and 1.2 × 25 = 30; the total is 20 + 26 + 30 = 76. Answer: 76 dogs. The distractors: 7.8 comes from adding the three frequency densities, 4 + 2.6 + 1.2, treating each height as though it were a count; 107 comes from using each upper class boundary as the width, giving 4 × 5, 2.6 × 15 and 1.2 × 40; 39 comes from using the first class width, 5, for every bar, which ignores the unequal intervals and gives 20 + 13 + 6.
- (a) 2 × 3² × 5 — Method: divide repeatedly by the smallest prime number until only prime factors remain. Working: 90 ÷ 2 = 45, 45 ÷ 3 = 15, 15 ÷ 3 = 5, and 5 is prime, so 90 = 2 × 3 × 3 × 5, written as 2 × 3² × 5. 2 × 3 × 15 stops before the 15 is broken down into 3 × 5, so it is not fully factorised. 3 × 3 × 10 stops before the 10 is broken down into 2 × 5. 2 × 45 stops after only one division. Answer: 2 × 3² × 5.
- (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.
- (b) 2/5 — The ratio red : yellow is 6:15, so write red over yellow: 6/15. Divide both numbers by their highest common factor, 3: 6÷3 = 2, 15÷3 = 5, giving 2/5. (5/2 comes from writing the ratio the wrong way round, yellow over red, 15/6, which simplifies to 5/2. 2/7 comes from comparing the red paint to the total amount of paint, 6 parts out of 21. 5/7 comes from comparing the yellow paint to the total amount of paint, 15 parts out of 21.)
- (b) 10.5 m² — The area of a trapezium is half of the sum of the parallel sides, multiplied by the width. Add the parallel sides: 2.5 + 4.5 = 7. Multiply by the width: 7 × 3 = 21. Half of 21 is 10.5 m². 21 m² forgets to halve, using the full (2.5+4.5)×3. 3.5 m² averages the two parallel sides, (2.5+4.5)÷2 = 3.5, but forgets to multiply by the width. 6.75 m² treats the bed as a triangle, using only the longer parallel side: half of 4.5 × 3.
- (b) 10 — Method: A′ means everything in the universal set that is NOT in A, so n(A′) = n(universal set) − n(A). Working: the universal set has 15 elements. A = {3, 6, 9, 12, 15}, so n(A) = 5. n(A′) = 15 − 5 = 10. Answer: 10. Watch out: writing down 5 gives n(A) itself, the size of the multiples-of-3 set, which is the opposite of its complement. Writing down 11 comes from missing 15 off the list of multiples of 3, treating A as only {3, 6, 9, 12}, so A is undercounted as 4 and A′ is overstated as 15 − 4. And writing down 12 comes from only listing the multiples of 3 up to 9 — 3, 6 and 9 — and missing that 12 and 15 also belong to A, undercounting A as 3 rather than 5.
- (a) 2√3 — Multiply the top and bottom of the fraction by √3, since √3 × √3 = 3: 6/√3 = (6 × √3)/(√3 × √3) = 6√3/3. Dividing 6 by 3 gives 2, so the fraction simplifies to 2√3. Multiplying only the numerator by √3 and then cancelling the surd in the denominator against it as if they were the same term, without properly squaring the denominator, leads to 6. Dividing 6 by 3 as 3 instead of 2 gives 3√3 — a slip in the final division. Simplifying 6√3/3 by cancelling the whole numerator's 3 with the denominator's 3, including the surd, gives 2, which loses the surd altogether.
- (d) (6, 6) — y = f(x − 4) + 1 combines two translations: the −4 inside the brackets moves the graph 4 units to the RIGHT, and the +1 outside the brackets moves it 1 unit UP. Applying both to (2, 5): add 4 to the x-coordinate 2, and add 1 to the y-coordinate 5, giving (6, 6). Moving left instead of right, as the minus sign inside the bracket might suggest, gives (−2, 6) or (−2, 4); forgetting the horizontal shift altogether and only applying the vertical one gives (2, 6).
- (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.
- (b) 7 — Method: each coordinate of a midpoint is the mean of the matching pair of coordinates, so the y-coordinate of the midpoint depends on the two y-coordinates alone. Working: the y-coordinates are 4 and 10, so the mean is (4 + 10) ÷ 2 = 14 ÷ 2 = 7. Answer: 7. The distractors: 5 comes from working out the x-coordinate of the midpoint, (2 + 8) ÷ 2 = 5, and writing that down in place of the y-coordinate the question asked for; 3 comes from halving the difference of the y-coordinates, (10 − 4) ÷ 2 = 3, which is half the vertical gap rather than a position; 14 comes from adding the two y-coordinates and forgetting to halve the total.
- (c) −0.5 — fg(x) = f(g(x)) = f(x²) = x² + 2. gf(x) = g(f(x)) = g(x + 2) = (x + 2)² = x² + 4x + 4. Setting fg(x) = gf(x): x² + 2 = x² + 4x + 4. Subtract x² from both sides: 2 = 4x + 4. Subtract 4 from both sides: −2 = 4x, so x = −0.5. Writing 1.5 comes from adding the 4 instead of subtracting it: 4x = 2 + 4 = 6, giving x = 1.5. Writing 'no solution' comes from expanding (x + 2)² as x² + 4 using (a + b)² = a² + b², losing the middle term — the equation then reads x² + 2 = x² + 4, which has no solution, but the expansion itself is wrong. Writing 0 comes from treating gf(x) as g(x) + f(x) instead of g(f(x)): x² + (x + 2) = x² + 2 gives x = 0, but that adds the two functions rather than composing them.
- (d) 20 — Method: equivalent ratios are linked by a single multiplier, so find it from the part you know and apply it to the other part. Working: 15 ÷ 3 = 5, so the multiplier is 5, and 4 × 5 = 20. Answer: 20. The distractors: 16 comes from adding the difference between the ratio parts, 4 − 3 = 1, to 15, treating the ratio as a difference; 60 comes from multiplying 15 by 4 without first dividing by 3; 11.25 comes from using the ratio the wrong way round, working out 15 × 3 ÷ 4.
- (d) 13 — Method: OP and PQ meet at a right angle because of the tangent–radius fact, so triangle OPQ is right-angled at P; use Pythagoras' theorem. Working: OQ² = OP² + PQ² = 5² + 12² = 25 + 144 = 169; OQ = √169 = 13. A student who answers 17 has simply added the two given lengths (5 + 12) instead of using Pythagoras' theorem. A student who answers 7 has subtracted the two given lengths (12 − 5) instead of using Pythagoras' theorem. A student who answers 144 has correctly squared 12 but stopped there, forgetting to add 5² and take the square root. Answer: 13 cm.
- (c) y = 4x − 5 — Parallel lines have the same gradient, so the new line has gradient 4; since it passes through (0, −5), its y-intercept is −5, giving y = 4x − 5. A candidate who drops the negative sign on the y-intercept would write y = 4x + 5. A candidate who changes the sign of the gradient, instead of keeping it the same for a parallel line, would write y = −4x − 5. A candidate who confuses m and c, using the y-intercept of the first line (3) as the gradient of the second, would write y = 3x − 5.
- (d) (4, 3) — The sides are parallel to the axes: the missing vertex must share the y-coordinate 3 with (−2, 3) and the x-coordinate 4 with (4, −3), giving (4, 3). (−4, 3) comes from a sign error on the x-coordinate. (4, −9) comes from continuing the pattern of the given points by subtracting 6 from the y-coordinate again instead of matching it to (−2, 3). (3, 4) comes from swapping the x- and y-coordinates.
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.