Sample paper · GCSE Higher · grades 4–9
GCSE Higher sample Paper 2 (calculator)
The real Paper 2 is 1.5 hour 30 minutes and 80 marks, 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 a calculator allowed. AO1 / AO2 / AO3 at this tier: 40% / 30% / 30%.
CalculatorGCSE Higher
GCSE Higher sample Paper 2 (calculator)
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- 1.A circle has a radius of 3 cm. Which of these is the exact area of the circle?
- 2.A market trader builds a display of oranges in a rectangular block. The block is n oranges wide and 4 oranges longer than it is wide. The number of oranges needed is 5 when n = 1, 12 when n = 2, 21 when n = 3, 32 when n = 4, and 45 when n = 5. Work out an expression, in terms of n, for the number of oranges needed for a display of width n.
- 3.The ratio of y to x is always 3 : 4. Write down which equation could represent this relationship.
- 4.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?
- 5.A bag contains 3 red beads and 7 blue beads. A bead is taken out at random and not put back. A second bead is then taken out at random. Work out the probability that the first bead is red and the second bead is also red.
- 6.In a random sample of 40 pupils at a school, 6 are left-handed. The school has 900 pupils. Work out an estimate for the number of left-handed pupils in the school.
- 7.Two fair six-sided dice are rolled together. One die is red and the other is blue. Work out how many different outcomes are possible.
- 8.The formula for the area of a trapezium is A = (a + b)h / 2, where a and b are the parallel side lengths and h is the height. Make h the subject of the formula.
- 9.A cordial drink is made by mixing cordial concentrate and water in the ratio 1 : 6. Freya says that a 700 ml jug of this drink contains 100 ml of concentrate. Is she correct? Give a reason for your answer.
- 10.A roof truss is shaped like an isosceles triangle resting on a horizontal ceiling joist. The truss's sloping edges are equal in length, and the angle at its apex is 40°. The ceiling joist continues in a straight line beyond the foot of the truss. Work out the angle between the truss's sloping edge and the extended joist, on the outside of the truss.
- 11.In a survey, 120 adults were asked whether they have a driving licence. 70 of the adults are women and 50 are men. 45 of the women and 35 of the men have a driving licence. One of the adults who has a driving licence is picked at random. Work out the probability that this adult is a man.
- 12.A plumber charges a call-out fee of £45 and then £28 for each hour she works. She is at a house from 09:15 until 12:45. Work out the total charge.
- 13.Solve x² + 7x = 0.
- 14.A beaker holds 500 g of a salt solution of concentration 12%. Pure water is added until the concentration falls to 8%. Work out the mass of water added.
- 15.Triangle PQR and triangle XYZ have angle P = angle X, angle Q = angle Y and angle R = angle Z, with no side lengths given for either triangle. Which of the following correctly describes the relationship between the two triangles?
- 16.A tank already contains some water and is then filled at a constant rate. After 2 minutes it contains 50 litres in total; after 7 minutes it contains 150 litres in total. Work out the rate at which water is added, in litres per minute.
- 17.Two printers' running costs are shown on separate straight-line graphs of total cost against number of pages printed. Printer A's graph has gradient £0.04 per page. Printer B's graph has gradient £0.05 per page. Work out how much cheaper it is to print 200 pages on Printer A than on Printer B, based on these running costs alone.
- 18.A rectangular garden gate is 3 m wide and 4 m high. A straight brace runs from one corner of the gate to the opposite corner. Work out the length of the brace.
- 19.A student claims: 'For every positive integer n, n² + n + 1 is a prime number.' Which value of n shows that this claim is false?
- 20.The diagram shows a distance–time graph for a cyclist travelling at a constant speed, for the first 6 minutes of a journey. Distance is in kilometres and time is in minutes. Work out how far the cyclist would travel in 20 minutes at the same speed.
Answer key
- (c) 9π cm² — The area of a circle is π × r². With a radius of 3 cm this is π × 3² = 9π cm², and this is exact because π has not been replaced by any approximation. Writing 28.3 cm² replaces π with a rounded decimal value, 3.14, and then rounds the result again, so it is only an approximation. Writing 28.26 cm² uses π ≈ 3.14 without a final rounding step, but this is still only an approximation of 9π, not the exact value. Writing 27 cm² comes from replacing π with the rough approximation 3, which is even further from the true value.
- (b) n² + 4n — First differences of 5, 12, 21, 32, 45 are 7, 9, 11, 13. Second differences are 2, 2, 2, so a = 2 ÷ 2 = 1. Subtracting n² (1, 4, 9, 16, 25) from the totals (5, 12, 21, 32, 45) leaves 4, 8, 12, 16, 20, the linear expression 4n. So the number of oranges is n² + 4n. Using the second difference itself as the coefficient of n², without halving it, gives 2n² + 4n. Finding a = 1 correctly but dropping the linear remainder 4n entirely leaves n². Treating the first first difference (7) as a constant common difference and building a linear formula 5 + 7(n − 1) = 7n − 2 fits only the first two totals, and gives 19 for n = 3 instead of 21.
- (d) y = 3x/4 — y : x = 3 : 4 means y/x = 3/4. Rearranging to make y the subject gives y = (3/4)x = 3x/4. A student who mixes up which quantity goes on top gets y = 4x/3. A student who treats the ratio numbers as the coefficient and constant of a linear equation instead of a proportional relationship gets y = 3x + 4. A student who mistakes the relationship for inverse proportion gets y = 3/(4x).
- (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.
- (d) 1/15 — The probability that the first bead is red is 3/10. Since the first bead is not put back, there are now only 2 red beads left out of 9 beads in total, so the probability that the second bead is also red is 2/9. Multiplying these, 3/10 × 2/9 = 6/90 = 1/15. A candidate who answers 9/100 has treated the beads as replaced, using 3/10 twice. A candidate who answers 3/50 has correctly reduced the red count to 2 for the second pick but forgotten that the total also falls to 9, using 2/10 instead. A candidate who answers 5/19 has added the numerators and added the denominators, (3+2)/(10+9), instead of multiplying.
- (b) 135 — Method: use the sample to find the PROPORTION of left-handed pupils, then apply that same proportion to the whole school population. Working: in the sample, 6 out of 40 pupils are left-handed, a proportion of 6 ÷ 40 = 0.15. Applying that proportion to the school's 900 pupils gives an estimate of 0.15 × 900 = 135 pupils. Giving 6 simply repeats the number of left-handed pupils IN THE SAMPLE, without scaling up to the whole school at all. Multiplying the population by the number of left-handed pupils in the sample without first dividing by the sample size, 900 × 6 = 5400, badly overestimates — that is more pupils than the whole school has. Dividing the population by the sample size but forgetting to multiply by the number of left-handed pupils found, 900 ÷ 40 = 22.5, finds the scale factor but stops one step short of using it. Always find the proportion in the sample first, then scale that same proportion up to the population.
- (c) 36 — Method: set the outcomes out in a grid with one die along the top and the other down the side, so that every cell of the grid is one outcome, and count the cells. Working: the red die can land in 6 ways, so the grid has 6 columns, and the blue die can also land in 6 ways, so the grid has 6 rows; the number of cells is 6 × 6 = 36. Answer: 36. The distractors: 12 comes from adding 6 and 6 instead of multiplying them; 6 comes from counting the outcomes of a single die and forgetting that the second die also has to land; 21 comes from treating the two dice as indistinguishable, so that a red 2 with a blue 3 and a red 3 with a blue 2 are counted as one outcome.
- (d) h = 2A / (a + b) — Method: undo the multiplication by 1/2 by multiplying both sides by 2, then undo the multiplication by (a + b) by dividing both sides by it. Working: A = (a + b)h / 2, so multiplying both sides by 2 gives 2A = (a + b)h, then dividing both sides by (a + b) gives h = 2A / (a + b). The value h = A / (a + b) comes from forgetting to multiply by 2 to clear the 1/2 first. The value h = 2A / a + b comes from dividing only by a and leaving b outside the fraction, instead of dividing by the whole bracket (a + b). The value h = (a + b) / (2A) comes from inverting the fraction.
- (c) Yes — 700 ÷ 7 = 100 ml for the 1 part of concentrate — Method: add the ratio parts to find the total number of parts, divide the total volume by this, then use the ratio to find concentrate's share. Working: 1 + 6 = 7 parts. 700 ÷ 7 = 100 ml per part. Concentrate = 1 part = 100 ml, so Freya is correct. Wrong options: 'divide 700 by 6' uses only one of the ratio numbers instead of the total of 7 parts, giving about 117 ml; '600 ml is concentrate' swaps which ratio number belongs to the concentrate and which belongs to the water; 'half of 700 ml should be concentrate' ignores the ratio altogether and assumes an equal split.
- (a) 110° — The truss is isosceles, so its two base angles are equal; since the three angles of the triangle add up to 180° and the apex is 40°, each base angle is (180 − 40) ÷ 2 = 70°. The angle between the sloping edge and the joist, extended beyond the base of the truss, sits on a straight line with that 70° base angle, and angles on a straight line add up to 180°, so the required angle is 180 − 70 = 110°. "70°" comes from stopping after finding the base angle and giving it directly, without also using the straight-line fact to find the angle on the OUTSIDE of the truss. "140°" comes from doubling the base angle instead of using the straight-line fact. "20°" comes from taking half of the apex angle (40° ÷ 2), mistaking it for the required angle instead of properly using the triangle's angle sum.
- (a) 7/16 — Method: the adult picked is known to have a driving licence, so the sample space is everyone with a licence; divide the number of men with a licence by that total. Working: 45 women and 35 men have a licence, so 80 adults have one. The men with a licence give 35/80, and dividing the numerator and the denominator by 5 gives 7/16. Answer: the probability is 7/16. The distractors: 7/10 is 35/50, the probability that an adult has a licence given that he is a man, which is the condition and the event the wrong way round; 7/24 is 35/120, dividing by all 120 adults surveyed instead of by the 80 who have a licence; 5/12 is 50/120, the probability that an adult picked from the whole survey is a man, which uses none of the licence information the question supplies.
- (d) £143 — From 09:15 to 12:45 is 3 hours 30 minutes, which is 3.5 hours. The charge for the time is 3.5 × £28 = £98, and the call-out fee is added on once: £98 + £45 = £143. £98 leaves out the call-out fee altogether, £129 charges for 3 hours instead of 3.5 hours, and £255.50 adds the fee to the hourly rate before multiplying, which charges the £45 for every hour.
- (c) x = 0 or x = −7 — Factorising: x² + 7x = x(x + 7) = 0, so x = 0 or x + 7 = 0, giving x = 0 or x = −7. A candidate who divides both sides of the original equation by x, which loses the solution x = 0, gets only x = −7. A candidate who makes a sign error solving x + 7 = 0 gets x = 0 or x = 7. A candidate who misreads the coefficient and doubles it gets x = 0 or x = −14.
- (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.
- (a) Similar, but AAA alone does not prove congruence — Three equal corresponding angles (AAA) show that the two triangles are similar — the same shape — but says nothing about their size, so it does not prove congruence. They could be congruent, or one could simply be an enlargement of the other; without matching side lengths, congruence is not established, so 'congruent because AAA proves congruence' is wrong. Equal angles do not force equal sides — a triangle can be enlarged to any size while keeping the same angles, so that option is also wrong. Something CAN be said here — that the triangles are similar — so 'no relationship can be determined' is wrong too.
- (b) 20 — The rate is the gradient: the change in litres divided by the change in time. From 2 to 7 minutes, the tank gains 150 − 50 = 100 litres over 7 − 2 = 5 minutes, so the rate is 100 ÷ 5 = 20 litres per minute. 100 comes from working out the change in litres but forgetting to divide by the change in time. 25 comes from using only the first reading, 50 ÷ 2, and ignoring the second reading entirely. 0.05 comes from dividing the change in time by the change in litres instead of the other way round, 5 ÷ 100.
- (d) £2.00 — Cost for Printer A = 200 × £0.04 = £8. Cost for Printer B = 200 × £0.05 = £10. Difference = £10 − £8 = £2.00.
- (b) 5 m — Method: the brace, the width and the height form a right-angled triangle in which the brace faces the right angle, so it is the hypotenuse and Pythagoras' theorem applies, a² + b² = c². Working: c² = 3² + 4² = 9 + 16 = 25, so c = √25 = 5. Answer: 5 m. The distractors: 7 m comes from adding the two sides, 3 + 4, instead of adding their squares; 25 m comes from stopping at c² = 25 and forgetting to take the square root; 12 m comes from multiplying 3 × 4, which gives the area of the gate in square metres and not a length across it.
- (c) n = 4 — Testing n = 4: 4² = 16, and 16 + 4 + 1 = 21, and 21 = 3 × 7 is not a prime number, so this value disproves the claim. Testing n = 1: 1² = 1, and 1 + 1 + 1 = 3, which is prime, so it does not disprove the claim. Testing n = 2: 2² = 4, and 4 + 2 + 1 = 7, which is prime. Testing n = 3: 3² = 9, and 9 + 3 + 1 = 13, which is also prime — a counterexample has to give a result that isn't prime, and only n = 4 does that.
- (b) 10 km — Method: find the constant speed from the graph (distance ÷ time for any point on the line), then multiply that speed by 20 minutes. Working: the line passes through (4 minutes, 2 km), so the speed is 2 ÷ 4 = 0.5 km per minute; in 20 minutes the cyclist travels 0.5 × 20 = 10 km. Answer: 10 km. Distractor refutation: 3 km comes from reading off the distance shown at the end of the plotted section (6 minutes) and stopping there, instead of extending the line to 20 minutes. 20 km comes from misreading the speed as 1 km per minute instead of 0.5 km per minute, doubling the true rate. 40 km comes from dividing 20 by the speed instead of multiplying by it, a reciprocal mix-up.
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.