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%.
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GCSE Higher sample Paper 2 (calculator)
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- 1.Four friends share a restaurant bill of £53.90 equally. A calculator gives each share as 13.475. Work out how much each friend should pay.
- 2.The perimeter of a rectangle is given by the formula P = 2(l + w), where l is the length and w is the width. A rectangular garden has a perimeter of 46 m and a length of 14 m. Rearrange the formula to make w the subject, then work out the width of the garden.
- 3.A room is drawn on a plan with a scale of 1 : 200. On the plan the room has an area of 12 cm². Work out the real area of the room, in square metres.
- 4.Work out the exact value of (sin 30°)² + (cos 30°)².
- 5.Two ordinary fair dice are rolled and the two scores are added together. Given that the total is an even number, work out the probability that the total is 8.
- 6.Grace asked 12 children how many brothers and sisters they have. Her results, in order, were 0, 0, 1, 1, 1, 1, 2, 2, 3, 3, 4, 6. Work out the median number of brothers and sisters.
- 7.Work out the value of .
- 8.A hiker's distance-time graph shows the following: he walks 5 km in the first 1 hour at a steady speed, rests for 1 hour, then walks a further 9 km in the next 2 hours, also at a steady speed. Work out his average speed for the whole journey, in km/h.
- 9.The time taken for a train journey is inversely proportional to the average speed of the train. At an average speed of 60 km/h the journey takes 2 hours. Work out the time taken at an average speed of 40 km/h.
- 10.A ship's radio can be heard up to 30 km from the ship. A lighthouse's light can be seen up to 20 km from the lighthouse. The ship and the lighthouse are 40 km apart along the coast. Describe the region where BOTH the radio can be heard AND the light can be seen.
- 11.In a game, Zara says the probability of scoring is 3/8, the probability of missing is 5/12 and the probability of a rebound is 1/6, and that these are the only three outcomes. Is Zara correct that her three probabilities are valid?
- 12.Light travels at 3 × 10⁸ metres per second. Work out how far light travels in 2 × 10⁻⁶ seconds. Give your answer in standard form.
- 13.A circle has centre (0, 0) and passes through the point (5, 12). Work out the equation of the circle.
- 14.A shop's profit, P pounds, is in direct proportion to the number of items sold, n. When 15 items are sold, the profit is £45. Write down the ratio n : P in its simplest form for this shop.
- 15.Work out the exact value of tan 30° + tan 30°.
- 16.The first five terms of a quadratic sequence are −1, 4, 13, 26, 43. Work out an expression, in terms of n, for the nth term.
- 17.A train travels at a constant speed of 90 km/h. Work out the speed in m/s.
- 18.A straight line touches the edge of a circle at exactly one point and does not cross into the circle at all. Write down the term for this line.
- 19.Solve the simultaneous equations 3x + y = 14 and x + y = 6. Work out the value of x.
- 20.By completing the square, find the turning point of the curve y = x² − 4x + 9.y = x² − 4x + 9
Answer key
- (b) £13.48 — Method: an amount of money is written to the nearest penny, which is 2 decimal places, so the calculator display has to be rounded to 2 decimal places. Working: 53.90 ÷ 4 = 13.475, and the digit in the third decimal place is 5, so the penny digit goes up from 7 to 8. Answer: £13.48. The distractors: £13.47 comes from chopping the third decimal place off instead of rounding with it; £13.50 comes from rounding to the nearest 10p rather than to the nearest penny; £13.40 comes from cutting the display short at 1 decimal place, which is both the wrong degree of accuracy and a truncation rather than a rounding.
- (a) 9 m — Undo the multiplication by the bracket first: dividing both sides by 2 gives P/2 = l + w. Subtracting the length from both sides gives w = P/2 − l. Substituting the measurements, 46 ÷ 2 = 23, and 23 − 14 = 9, so the width is 9 m. Taking the length off before halving gives (46 − 14) ÷ 2 = 16, which halves the length as well; expanding to P = 2l + 2w and then forgetting to divide by 2 gives 46 − 28 = 18; subtracting the length in the wrong direction gives 23 + 14 = 37.
- (b) 48 m² — Method: lengths are multiplied by the scale factor, but areas are multiplied by its square. Working: 1 cm on the plan stands for 200 cm = 2 m, so 1 cm² on the plan stands for 2 × 2 = 4 m², and 12 × 4 = 48. Answer: 48 m². The distractors: 24 m² comes from scaling the area by the length factor 2 instead of by its square; 2400 m² comes from multiplying the area by the scale 200 as though it were a length; 4800 m² comes from working in centimetres, 12 × 200² = 480 000 cm², and then dividing by 100 instead of by 10 000 to reach square metres.
- (d) 1 — sin 30° = 1/2, so (sin 30°)² = 1/4. cos 30° = √3/2, so (cos 30°)² = 3/4. Adding these gives 1/4 + 3/4 = 1. '1/4' only calculates (sin 30°)² and forgets to add the cos 30° term. '3/4' only calculates (cos 30°)² and forgets to add the sin 30° term. '−1/2' comes from subtracting the two squared values instead of adding them: 1/4 − 3/4 = −1/2.
- (c) 5/18 — Method: knowing the total is even cuts the 36 equally likely outcomes down to the even ones, so count those first and then count how many of them give 8. Working: the even totals occur as 2 once, 4 three times, 6 five times, 8 five times, 10 three times and 12 once, which is 18 outcomes. The total is 8 for 2 and 6, 3 and 5, 4 and 4, 5 and 3, and 6 and 2, which is 5 outcomes. The probability is 5/18, which will not cancel. Answer: the probability is 5/18. The distractors: 5/36 keeps the right count of ways to make 8 but divides by all 36 outcomes, ignoring the fact that the odd totals have already been ruled out; 1/6 treats the six even totals 2, 4, 6, 8, 10 and 12 as equally likely and picks one of them, which they are not; 1/11 treats the eleven possible totals from 2 to 12 as equally likely and uses neither the counting nor the condition.
- (a) 1.5 — Method: with an even number of values the median is the mean of the two middle values, which for 12 values are the 6th and the 7th once the data are in order. Working: the results are already in order, and 12 ÷ 2 = 6, so the middle pair are the 6th value, 1, and the 7th value, 2; the median is (1 + 2) ÷ 2 = 1.5. Answer: 1.5 brothers and sisters. The distractors: 1 comes from reading the 6th value and stopping there instead of averaging the middle pair; 2 comes from working out the mean, 24 ÷ 12, instead of the median; 6 comes from working out the range, 6 − 0, which measures spread rather than centre.
- (d) 8 — Method: write $16^{3/4}$ as $(\sqrt[4]{16})^3$ — the denominator of the index gives the root, the numerator gives the power. Working: $\sqrt[4]{16} = 2$, so $16^{3/4} = 2^3 = 8$. Answer: 8. A candidate who multiplies 16 by 3/4 is treating the index as an ordinary factor and gets 12 — a fractional index is not a multiplier. A candidate who takes the square root instead of the fourth root and then cubes it works out $(\sqrt{16})^3 = 4^3$ and gets 64; the denominator 4 names a fourth root, not a square root. A candidate who takes the fourth root of 16 correctly but stops there, without cubing it, gets 2.
- (a) 3.5 km/h — Average speed is total distance ÷ total time. Total distance = 5 + 9 = 14 km. Total time, including the rest, = 1 + 1 + 2 = 4 hours. So average speed = 14 ÷ 4 = 3.5 km/h. Leaving out the 1 hour rest and dividing by only the 3 hours of walking gives 14 ÷ 3 = 4.67 km/h. Averaging the two separate speeds, 5 km/h and 4.5 km/h, instead of using total distance over total time, gives (5 + 4.5) ÷ 2 = 4.75 km/h. Dividing only the second leg's distance by the total time, 9 ÷ 4 = 2.25 km/h, ignores the first leg's distance.
- (c) 3 hours — Method: inverse proportion means speed × time is constant for the journey, so find that constant and divide it by the new speed. Working: 60 × 2 = 120, which is the distance in kilometres; at 40 km/h the time is 120 ÷ 40 = 3 hours. Answer: 3 hours. The distractors: 1.5 hours is the ratio of the speeds, 60 ÷ 40, given as a time instead of being used to scale the original 2 hours; 1 hour 20 minutes comes from treating time as directly proportional to speed, 2 × 40 ÷ 60, which has the slower train arriving sooner; 2 hours comes from finding the constant 120 and then dividing it by the original 60 km/h again, so the time never changes.
- (a) the lens-shaped region where the two circles overlap — The region within 30 km of the ship is a circle of radius 30 km around the ship; the region within 20 km of the lighthouse is a circle of radius 20 km around the lighthouse. Since 30 + 20 = 50 km is more than the 40 km between them, and 30 − 20 = 10 km is less than 40 km, the two circles genuinely overlap, in a lens-shaped region where both conditions hold at once. "the whole smaller circle, since it fits inside the larger one" would only be true if the 40 km distance between the centres were small enough for one circle to swallow the other, which it is not here. "no overlap — 30 km and 20 km don't add to 40 km" wrongly assumes the two radii must add up to exactly the distance between the centres for an overlap to exist; any total greater than that distance is enough. "a straight strip, exactly halfway between the two" confuses this with an equidistant locus, which is a different problem from the one asked here.
- (d) No, because 3/8 + 5/12 + 1/6 = 23/24 — Using a common denominator of 24: 3/8 = 9/24, 5/12 = 10/24 and 1/6 = 4/24. Adding these numerators gives 9 + 10 + 4 = 23, so the three probabilities sum to 23/24, which is less than 1 — Zara is not correct. Adding the original numerators (3 + 5 + 1 = 9) over a denominator of 12 instead of converting each fraction properly gives 9/12 = 3/4, still less than 1 but the wrong fraction. Converting 1/6 to 5/24 instead of 4/24 (using the wrong scaling) makes the total 9/24 + 10/24 + 5/24 = 24/24 = 1, wrongly suggesting the probabilities are valid. Judging validity from the fact that each individual fraction lies between 0 and 1 ignores that an exhaustive set must sum to exactly 1, not merely contain valid individual values.
- (a) 6 × 10² metres — Method: distance = speed × time, so multiply the coefficients and add the indices. Working: 3 × 2 = 6 for the coefficients, and 8 + (−6) = 2 for the indices; 6 lies between 1 and 10, so the coefficient needs no adjustment. Answer: 6 × 10² metres, which is 600 metres. The distractors: 5 × 10² metres comes from adding the coefficients, 3 + 2, instead of multiplying them; 6 × 10¹⁴ metres comes from subtracting the indices, 8 − (−6), which is the rule for dividing rather than for multiplying; 6 × 10⁻⁴⁸ metres comes from multiplying the indices, 8 × (−6), instead of adding them.
- (b) x² + y² = 169 — Method: a circle centred on the origin has equation x² + y² = r², and every point on it satisfies that equation, so substituting the coordinates of a point that lies on the circle gives r² directly. Working: substituting x = 5 and y = 12 gives 5² + 12² = 25 + 144 = 169, so r² = 169 and the circle is x² + y² = 169. Answer: x² + y² = 169. The distractors: x² + y² = 13 uses the radius, √169 = 13, where r² belongs, which is the confusion between r and r² made in the other direction; x² + y² = 17 adds the two coordinates, 5 + 12, instead of adding their squares; x² + y² = 119 subtracts the squares, 144 − 25, treating 12 as the hypotenuse of the right-angled triangle rather than as one of the shorter sides.
- (c) 1 : 3 — n : P = 15 : 45. Dividing both parts by their highest common factor, 15, gives 1 : 3. Inverting the ratio, 3 : 1, swaps profit and number of items. Dividing only the n-part by 15, getting 1, but leaving the P-part as 45 gives 1 : 45 — only one side has been simplified. Dividing only the P-part by 15, getting 3, but leaving the n-part as 15 gives 15 : 3, the opposite partial mistake.
- (a) 2√3/3 — tan 30° = √3/3, so tan 30° + tan 30° = 2 × √3/3 = 2√3/3. √3 comes from wrongly treating tan 30° + tan 30° as tan(30° + 30°) = tan 60° = √3 — adding angles is not the same as adding ratios. √3/3 comes from forgetting to double the value and just writing down tan 30° on its own. 2√3 comes from doubling the numerator of √3/3 but forgetting to keep the denominator of 3.
- (d) 2n² − n − 2 — First differences: 5, 9, 13, 17. Second differences: 4, 4, 4, so a = 4 ÷ 2 = 2. Subtracting 2n² (2, 8, 18, 32, 50) from the terms (−1, 4, 13, 26, 43) leaves −3, −4, −5, −6, −7, which is the linear expression −n − 2. So the nth term is 2n² − n − 2. Using the second difference itself as a, without halving it, gives 4n² − n − 2. Finding a = 2 correctly but dropping the linear remainder −n − 2 entirely leaves 2n². Treating the first first difference (5) as a common difference and building a + (n − 1)d = −1 + 5(n − 1) gives 5n − 6, which only matches the first two terms.
- (b) 25 m/s — Convert km/h to m/s by multiplying by 1000 (km to m) and dividing by 3600 (hours to seconds): 90 × 1000 ÷ 3600 = 25 m/s. Working out 90 ÷ 60 = 1.5 converts using 60, as if going from hours to minutes rather than to seconds. Working out 90 × 3.6 = 324 multiplies by 3.6 instead of dividing by it, going the wrong way between the units. Working out 90 × 1000 = 90000 converts kilometres to metres but forgets to convert hours to seconds at all. The train's speed is 25 m/s.
- (d) tangent — A line that touches a circle at exactly one point, without crossing into the circle, is called a tangent. A chord is a straight line joining two points ON the circle, so it touches at two points, not one. A radius runs from the centre to the circle's edge, not along the outside of it. A diameter is a chord that passes through the centre, also touching the circle at two points.
- (d) 4 — Subtracting the second equation from the first: the y-terms cancel, and the x-terms combine as 3x − x = 2x; the right-hand sides give 14 − 6 = 8. This gives 2x = 8, so x = 4. A candidate who subtracts in the wrong order would get 2x = 6 − 14 = −8, so x = −4. A candidate who uses only the first equation, dividing 14 by 3 as if y were 0, would get x ≈ 4.67. A candidate who reports the value of y instead of x would get y = 6 − 4 = 2.
- (c) x = 2, y = 5 — x² − 4x + 9 = (x − 2)² − 2² + 9 = (x − 2)² + 5. Substituting x = 2: 2² = 4, 4 × 2 = 8, so 4 − 8 + 9 = 5, confirming the minimum value 5 at x = 2: turning point (2, 5). Using −4 instead of half of it inside the bracket gives (x − 4)² − 7, turning point (4, −7) — wrong, because only half the coefficient of x belongs inside the bracket. Reading the bracket's sign directly as the turning point's x-coordinate gives (−2, 5) — wrong, because (x − 2)² is zero at x = 2, not x = −2. Computing 4 − 9 = −5 instead of 9 − 4 = 5 flips the sign of the constant, giving (2, −5) — wrong, since the completed square's constant must be evaluated as 9 minus 4, not 4 minus 9. Substitute the x-value back into the original equation whenever you are unsure of a sign.
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.