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.Work out 6² − 4².
- 2.A circle has centre (0, 0) and equation x² + y² = 50. Work out the length of the diameter of the circle, correct to 1 decimal place.
- 3.The height of a ball, in metres, above the ground is plotted against time, in seconds, after it is thrown. The tangent to the graph at t = 1.5 seconds has gradient 2. Which of these four statements about the ball at t = 1.5 seconds is correct?
- 4.Work out the exact value of sin 30° + cos 60°.
- 5.The probability that Priya scores with a netball shot is 3/4. She takes two shots, and the shots are independent. Work out the probability that she scores with her first shot but misses her second shot.
- 6.A two-way table records the favourite subject, Maths or Art, of 60 pupils in Year 10, and whether each pupil is left-handed or right-handed. 9 of the 60 pupils are left-handed, and 6 of those left-handed pupils prefer Art. In total, 24 of the 60 pupils prefer Art. A pupil is chosen at random from the 60. Work out the probability that the pupil is right-handed and prefers Art.
- 7.Write these fractions in order, starting with the smallest: 2/3, 3/5, 5/6, 1/2
- 8.f(x) = 2x − 1. Work out ff(x).y = 2x − 1
- 9.A fixed job of fitting a solar array is shared between installers, and the time taken is inversely proportional to the number of installers working on it at once. With 4 installers the job takes 18 hours. Construct the equation connecting time T and number of installers n, then work out how long it would take with 6 installers.
- 10.Triangle DEF has a right angle at E. Angle DFE = 60° and EF = 9 cm. Work out the exact length of DE.
- 11.A group of 80 pupils was asked whether they walk to school. 45 of the pupils are girls and 35 are boys. 30 of the girls walk to school and 14 of the boys walk to school. One of the girls is picked at random. Work out the probability that she walks to school. Give your answer in its simplest form.
- 12.Work out 3 1/5 + 1 2/3. Give your answer as a mixed number in its simplest form.
- 13.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.
- 14.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.
- 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.Solve the simultaneous equations 2x + y = 7 and x + 2y = 8.
- 17.Write 250 g : 2 kg as a ratio in its simplest form.
- 18.In triangle ABC, AB = 15.6 cm, AC = 8.9 cm and angle BAC = 112°, the angle between them. Work out the area of triangle ABC. Give your answer to 1 decimal place.
- 19.Using the table of values of f(x) (x = 0, 1, 2, 3 gives f(x) = 5, 8, 4, 1), work out the value of −f(x) when x = 1.
- 20.A car's fuel tank contains 50 litres when it starts a journey. After travelling for 2 hours, it contains 38 litres, and fuel is used at a constant rate. Work out the rate at which fuel is used, in litres per hour.
Answer key
- (c) 20 — Method: work out each power separately before subtracting. Working: 6² = 36 and 4² = 16, so 6² − 4² = 36 − 16 = 20. Answer: 20. (4 comes from subtracting first, 6 − 4 = 2, and then squaring that result, instead of squaring each number first. 52 comes from adding the two squares, 36 + 16, instead of subtracting them. 2 comes from subtracting the two numbers, 6 − 4, and forgetting to square at all.)
- (c) 14.1 — Method: in x² + y² = r² the right-hand side is the square of the radius, so take its square root to find the radius, then double the radius because the diameter is twice the radius. Working: r² = 50, so r = √50 = 7.07106…, and the diameter is 2 × 7.07106… = 14.14213…, which is 14.1 correct to 1 decimal place. Answer: 14.1. The distractors: 7.1 is the radius, worked out correctly but never doubled, so it answers a question about the radius rather than the diameter; 100.0 comes from doubling the 50 on the right-hand side, treating r² as though it were already the radius; 25.0 comes from halving the 50, treating r² as though it were already the diameter.
- (c) Height rising at 2 m/s at t = 1.5 s — A tangent's gradient on a height-time graph is the instantaneous rate of change of height, in metres per second, so gradient 2 means the ball's height is increasing at 2 m/s at t = 1.5 s. Saying the height 'is 2 m' confuses the gradient, a rate, with the y-value on the graph, which is the ball's height itself. Saying the ball 'travelled 2 m from t = 1 to t = 2' treats the instantaneous gradient at one instant as if it were the total distance risen over a whole one-second interval, which is a different quantity found from two height readings, not from one tangent. Saying the speed 'is 2 m/s²' uses the wrong units — m/s² measures acceleration, the rate of change of speed, not speed itself. Always check that the units quoted match what a height-time graph's gradient can actually give you: metres per second.
- (c) 1 — sin 30° = 1/2 and cos 60° = 1/2, so sin 30° + cos 60° = 1/2 + 1/2 = 1. 0 comes from subtracting the two values instead of adding them (1/2 − 1/2). 1/2 comes from writing down only one of the two exact values and forgetting to add the other. √3 comes from swapping the two angles and working out sin 60° + cos 30° = √3/2 + √3/2 = √3.
- (c) 3/16 — The probability that Priya misses a shot is 1 − 3/4 = 1/4. The two shots are independent, so multiply the probability of scoring the first shot by the probability of missing the second shot: 3/4 × 1/4 = 3/16. Choosing 9/16 comes from using the probability of scoring, 3/4, for both shots instead of switching to the miss probability for the second shot, 3/4 × 3/4 = 9/16, which is the probability that she scores with both shots. Choosing 1/2 comes from subtracting the two probabilities instead of multiplying them, 3/4 − 1/4 = 1/2. Choosing 3/4 comes from giving only the probability that she scores with the first shot, forgetting to combine it with what happens on the second shot.
- (a) 3/10 — There are 60 − 9 = 51 right-handed pupils. Of the 24 pupils who prefer Art, 6 are left-handed, so 24 − 6 = 18 are right-handed and prefer Art. The probability that a randomly chosen pupil is right-handed and prefers Art is 18/60, which simplifies to 3/10. Giving 2/5 is 24/60 simplified — the probability of preferring Art, ignoring the right-handed condition entirely. Giving 17/20 is 51/60 simplified — the probability of being right-handed, ignoring the Art condition entirely. Giving 1/10 is 6/60 simplified — the probability of being left-handed and preferring Art, the wrong hand condition.
- (a) 1/2, 3/5, 2/3, 5/6 — Convert all four fractions to a common denominator of 30: 2/3 is 20/30, 3/5 is 18/30, 5/6 is 25/30, and 1/2 is 15/30. Ordering by these numerators, smallest to largest, gives 15/30, 18/30, 20/30, 25/30, which is 1/2, 3/5, 2/3, 5/6. Ordering by the size of the numerator in the original fractions, 1, 2, 3, 5, rather than converting to a common denominator, gives the wrong order 1/2, 2/3, 3/5, 5/6, because it ignores that the denominators are different. Ordering largest to smallest instead of smallest to largest, as the question asks, gives 5/6, 2/3, 3/5, 1/2. Using the rule "the bigger the denominator, the smaller the fraction" to place the last two, so that 5/6 is put below 2/3 because 6 is bigger than 3, gives 1/2, 3/5, 5/6, 2/3 — that rule only holds when the numerators are the same, and here 20/30 really is smaller than 25/30. So the correct order, smallest to largest, is 1/2, 3/5, 2/3, 5/6.
- (c) 4x − 3 — ff(x) means f(f(x)): substitute f(x) into f in place of x. f(f(x)) = 2 × f(x) − 1 = 2 × (2x − 1) − 1. Expanding the bracket: 2 × (2x − 1) = 4x − 2. Combining the constant terms: −2 − 1 = −3, so f(f(x)) = 4x − 3. Writing 4x − 2 comes from expanding 2(2x − 1) correctly to get 4x − 2, then forgetting to subtract the outer 1 at all. Writing 4x² − 4x + 1 comes from reading ff(x) as f(x) multiplied by itself, (2x − 1)(2x − 1) = 4x² − 4x + 1, instead of substituting f(x) into f. Writing 4x − 1 comes from doubling the coefficient of x in the original rule directly, without actually substituting f(x) into f at all.
- (d) 12 hours — Since time is inversely proportional to the number of installers, T = k/n. Using n = 4, T = 18: 18 = k ÷ 4, so k = 18 × 4 = 72. The equation is T = 72/n. When n = 6: T = 72 ÷ 6 = 12. Using the original number of installers instead of the new one gives T = 72 ÷ 4 = 18, the wrong value substituted. Treating more installers as needing more time, as if T were directly proportional to n, gives k = 18 ÷ 4 = 4.5 and then T = 4.5 × 6 = 27, the opposite relationship to the one described. Stopping at k = 72 and reporting it gives the time the job would take a single installer working alone — the constant still has to be divided by the new number of installers before it answers the question asked. With 6 installers, the job takes 12 hours.
- (d) 9√3 cm — DE is opposite the 60° angle at F, and EF is adjacent to it, so DE = EF × tan 60° = 9 × √3 = 9√3 cm. 9√3/2 cm comes from using sin 60° = √3/2 instead of tan 60°. 3√3 cm comes from using tan 30° = 1/√3 instead of tan 60° (9 × 1/√3 = 9/√3 = 3√3). 18 cm is the hypotenuse DF, not DE: it comes from using cos 60° = 1/2 and working out 9 ÷ 1/2 = 18, which finds the wrong side of the triangle.
- (d) 2/3 — Method: the pupil picked is known to be a girl, so the sample space is the 45 girls and not all 80 pupils; divide the number of girls who walk by the number of girls. Working: 30 of the 45 girls walk to school, which gives 30/45. Dividing the numerator and the denominator by 15 gives 2/3. Answer: the probability is 2/3. The distractors: 3/8 is 30/80, dividing the girls who walk by every pupil in the group, which is the commonest slip on a conditional probability because it keeps the restricted numerator but the whole denominator; 1/3 is 15/45, counting the 15 girls who do not walk to school and so answering the opposite event inside the correct group; 11/20 is 44/80, adding the 30 girls and the 14 boys who walk and dividing by the whole group, which throws away the information that the pupil picked is a girl.
- (d) 4 13/15 — Convert to fifteenths: 1/5 is equivalent to 3/15 (multiply by 3/3), and 2/3 is equivalent to 10/15 (multiply by 5/5), so 3 1/5 is equivalent to 3 3/15 and 1 2/3 is equivalent to 1 10/15. Add the whole numbers (3 + 1 = 4) and the fractions (3/15 + 10/15 = 13/15), giving 4 13/15. A candidate who adds the numerators and denominators straight across, treating 1/5 + 2/3 as (1+2)/(5+3), gets a fraction part of 3/8, giving 4 3/8. A candidate who adds the fraction parts correctly but forgets to add the second whole number gets 3 13/15. A candidate who adds the whole numbers but copies the first fraction across without ever adding 2/3 to it gets 4 1/5.
- (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.
- (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.
- (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.
- (c) x = 2, y = 3 — Method: no letter cancels straight away, so make one letter the subject of one equation and substitute that expression into the other. Working: 2x + y = 7 gives y = 7 − 2x, so x + 2y = 8 becomes x + 2(7 − 2x) = 8, that is x + 14 − 4x = 8, so −3x = −6 and x = 2; substituting back gives y = 7 − 4 = 3. Answer: x = 2, y = 3, which checks in 2 + 6 = 8. The distractors: x = 3, y = 2 comes from adding the equations to x + y = 5 and then subtracting them the wrong way round, writing x − y = 1 instead of x − y = −1; x = 2, y = 5 comes from substituting x = 2 into 2x + y = 7 as 2 + y = 7, dropping the coefficient; x = −2, y = 11 comes from dividing −3x = −6 as though two minus signs gave a negative answer.
- (a) 1:8 — Convert 2 kg to grams: 2 kg = 2000 g. The ratio is 250 : 2000. The highest common factor of 250 and 2000 is 250. Divide both parts by 250: 250 ÷ 250 = 1 and 2000 ÷ 250 = 8, so the ratio is 1 : 8. Leaving the kilograms unconverted gives 250 : 2, which simplifies to 125 : 1 — the units on each side are different, so this does not compare like with like. Dividing by 50 instead of 250 gives 5 : 40, which still shares a common factor of 5, so it is not fully simplified. Swapping the order gives 8 : 1, grams to kilograms the wrong way round.
- (c) 64.4 cm² — Method: with two sides and the angle between them, use Area = (1/2)ab sin C. Working: Area = 1/2 × 15.6 × 8.9 × sin 112° = 64.4 cm² (1 d.p.). Answer: 64.4 cm². Leaving out the 1/2 gives 128.7 cm²; using cos 112° instead of sin 112° gives a negative value, which a candidate who drops the minus sign reads as 26.0 cm²; and squaring one side instead of multiplying the two different given sides together gives 112.8 cm². Sin C is never negative for an angle between 0° and 180°, so a negative area is always a sign that cos was used by mistake — check you used sin before you trust your answer.
- (a) −8 — −f(x) means take the output value from the table and change its sign, without changing which x-value is looked up. From the table, f(1) = 8, so −f(1) = −8. Reading f(1) = 8 from the table but forgetting to apply the negative sign gives 8. Misreading the row and using f(0) = 5 instead of f(1) = 8, then negating it, gives −5. Confusing −f(x) with f(x) − 1 — taking f(1) = 8 and subtracting 1 instead of negating — gives 7.
- (c) 6 — Method: rate = amount of fuel used ÷ time taken. Working: fuel used = 50 − 38 = 12 litres, time taken = 2 hours, so rate = 12 ÷ 2 = 6 litres per hour. Answer: the rate is 6 litres per hour. 44 comes from finding the average of the two fuel amounts, (50 + 38) ÷ 2, instead of the fuel used. 12 comes from working out the fuel used but forgetting to divide by the time taken. 19 comes from dividing the remaining fuel, 38, by the time taken instead of the fuel used.
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.