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%.

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GCSE Higher sample Paper 1 (non-calculator)

Maths · GCSE Higher · Paper 1 · Non-calculator · 20 questions

MathsUKwww.geekhero.co.uk

1.Round 0.006852 to 2 significant figures.
(a)0.0069
(b)0.0068
(c)0.007
(d)0.00685
2.A parcel delivery company's cost graph shows the following: the cost is a flat £5 for parcels weighing up to 2 kg, and then rises by £2 for each additional kg above 2 kg. Use the graph to work out the weight of a parcel that costs £17.
(a)8.5 kg
(b)8 kg
(c)6 kg
(d)7.5 kg
3.A map has a scale of 1 : 25 000. A footpath measures 6 cm on the map. Work out the real length of the footpath, in kilometres.
(a)1.5 km
(b)0.15 km
(c)150 km
(d)150000 km
4.A trapezium has vertices (2, 1), (6, 1), (5, 4) and (3, 4). It is rotated 180° about the vertex (2, 1). Work out the number of points on the trapezium — including its vertices, edges and interior — that are invariant under this rotation.
(a)0 points
(b)1 point
(c)2 points
(d)4 points
5.An ordinary six-sided dice, numbered 1 to 6, is rolled 30 times and lands on a 6 seven times. Ravi says the theoretical probability of rolling a 6 and the relative frequency of rolling a 6 in this trial are the same number. Is Ravi right?
(a)Yes — relative frequency always equals theory.
(b)Yes — the theoretical probability has now become 7/30.
(c)No — 7/30 is the relative frequency; theory stays 1/6.
(d)Neither can be found — 30 rolls is too few to tell.
6.A scatter graph shows the height, x cm, and the mass, y kg, of 20 pupils in Year 10. The heights on the graph run from 150 cm to 180 cm, and the line of best fit is y = 0.9x − 85. Nadia puts x = 90 into this equation to estimate the mass of a two-year-old child who is 90 cm tall. Is her estimate reliable? Give a reason for your answer.
y = 0.9x − 85
(a)Yes, the line passes through all 20 of the points
(b)Yes, the equation works for any value put into it
(c)No, 90 cm is far outside the heights on the graph
(d)No, a line of best fit cannot be used to predict at all
7.A car travels 180 km using 6 litres of fuel. Work out the car's fuel consumption in kilometres per litre, then work out how many kilometres it can travel on a full tank of 12 litres at this rate.
(a)2160 km
(b)30 km
(c)90 km
(d)360 km
8.Work out the value of a² − 2b when a = −3 and b = 4.
(a)1
(b)−17
(c)−11
(d)17
9.A kettle uses 2.1 kWh of energy. A toaster uses 0.9 kWh of energy. Write the energy used by the kettle as a fraction of the energy used by the toaster, giving your answer in its simplest form.
(a)7/10
(b)4/3
(c)3/7
(d)7/3
10.A warehouse stores identical cube-shaped crates. Its plan view is a 2 by 4 rectangle of crate positions, and every position is filled to a height of 3 crates, except one corner position, which has only 2 crates stacked on it because a delivery was incomplete. How many crates are there in total?
(a)22
(b)23
(c)21
(d)24
11.Two ordinary fair dice are rolled. Work out the probability that both dice show the same score.
(a)1/3
(b)1/6
(c)1/36
(d)1/2
12.p = 24 and q = 6, each correct to the nearest integer. Work out the greatest degree of accuracy to which p ÷ q can be guaranteed correct.
(a)the nearest whole number
(b)it cannot be given to any degree of accuracy
(c)the nearest 0.1
(d)the nearest 10
13.A scout leader is 44 years old and one of the scouts is 16 years old. Work out how many years it will be until the leader is exactly twice as old as the scout.
(a)6 years
(b)12 years
(c)14 years
(d)28 years
14.A straight-line graph shows the cost, C pounds, of hiring a bike for h hours. The line passes through the points (1, 12) and (4, 27). Which of these statements about the line is true?
(a)The hourly rate is £5 and the fixed fee is £7
(b)The hourly rate is £5 and the fixed fee is £12
(c)The hourly rate is £7 and the fixed fee is £5
(d)The hourly rate is £15 and the fixed fee is £7
15.A cheese counter sells cheddar at £8.40 per kilogram. Work out the cost of a 350 g piece of cheddar.
(a)£24.00
(b)£2.94
(c)£29.40
(d)£2940.00
16.Solve the simultaneous equations 3x + 2y = 16 and x + y = 7. Work out the value of y.
(a)2
(b)5
(c)9
(d)−5
17.A novel has 312 pages. A comic has 48 pages. Write the number of pages in the novel as a fraction of the number of pages in the comic, giving your answer in its simplest form.
(a)11/2
(b)13/2
(c)2/13
(d)13/15
18.Work out the exact value of tan 30° + tan 30°.
(a)2√3/3
(b)√3/3
(c)√3
(d)2√3
19.Solve 6x − 5 = 3x + 13.
(a)18
(b)2
(c)2.67
(d)6
20.A straight line passes through the points (2, 7) and (5, 16). Work out the value of y when x = 0.
(a)1
(b)−1
(c)3
(d)4

Answer key

(a) 0.0069 — Leading zeros are not significant, so the significant figures in 0.006852 start at 6: 6, 8, 5, 2. Rounding to 2 significant figures means keeping 6 and 8, and looking at the next digit, 5, to decide whether to round up. Since 5 rounds up, the second significant figure increases from 8 to 9: 0.006852 rounds to 0.0069. A candidate who rounded to 1 significant figure instead of 2 wrote 0.007. A candidate who rounded to 3 significant figures instead of 2 wrote 0.00685. A candidate who did not round up despite the next digit being 5 wrote 0.0068.
(b) 8 kg — The cost above the flat £5 charge is 17 − 5 = £12. At £2 per kg, this covers 12 ÷ 2 = 6 kg above the first 2 kg, so the total weight is 2 + 6 = 8 kg. Dividing the full £17 by £2 per kg without first taking off the £5 flat charge gives 17 ÷ 2 = 8.5 kg. Taking off the £5 flat charge and dividing by £2 per kg, but forgetting to add back the 2 kg that the flat charge covers, gives 12 ÷ 2 = 6 kg. Taking off £2 instead of £5 as the flat charge, (17 − 2) ÷ 2 = 7.5 kg, swaps which number is the fixed fee.
(a) 1.5 km — Multiply the map length by the scale: 6 × 25 000 = 150 000 cm. Convert to kilometres: 150 000 cm = 1.5 km. Dividing by only 1000 instead of the full conversion when changing units gives 150 km, a hundred times too large. Misreading the scale as 1 : 2500 instead of 1 : 25 000 gives 6 × 2500 = 15 000 cm = 0.15 km, a hundred times too small. Leaving the answer as 150 000 without converting units at all, and calling it 150 000 km, mistakes centimetres for kilometres completely.
(b) 1 point — A rotation about a point P always leaves P itself unchanged, and — unless the shape has rotational symmetry about that exact point — no other point of the shape maps onto itself. The centre of rotation here is the vertex (2, 1), so exactly one point of the trapezium, that vertex, is invariant. Thinking that a rotation always has no invariant points ignores the centre of rotation itself, which is always fixed, and gives 0 points. Assuming every vertex of the shape is invariant confuses a rotation with the identity transformation and gives 4 points, the total number of vertices. Believing that the point diametrically opposite the centre is also fixed applies point symmetry of the whole coordinate grid rather than checking whether that point actually lies on this particular trapezium, and gives 2 points.
(c) No — 7/30 is the relative frequency; theory stays 1/6. — The theoretical probability of rolling a 6 on an ordinary dice is fixed at 1/6, worked out from the number of equally likely outcomes, and does not change however the dice is actually rolled. The relative frequency from this trial is 7/30, found from what happened in these particular 30 rolls. Since 7/30 and 1/6 are different numbers, the correct statement is 'No — 7/30 is the relative frequency; theory stays 1/6.' Assuming the two values must always match because they describe the same event gives 'Yes — relative frequency always equals theory.' Believing that an observed result redefines the theoretical probability gives 'Yes — the theoretical probability has now become 7/30.' Refusing to work out either value at all gives 'Neither can be found — 30 rolls is too few to tell', which ignores that both numbers CAN be calculated from the information given.
(c) No, 90 cm is far outside the heights on the graph — Method: a line of best fit describes the trend only across the stretch of data it was drawn through; predicting beyond that stretch is extrapolation, and nothing in the data supports it. Working: the heights used to draw this line run from 150 cm to 180 cm, all of them Year 10 pupils, while 90 cm is 60 cm below the shortest of them and belongs to a two-year-old child, whose build follows no trend the graph has measured. Substituting anyway gives 0.9 × 90 − 85 = −4, a mass of −4 kg, which cannot exist. Answer: no, because 90 cm is far outside the heights on the graph. The distractors: saying a line of best fit cannot be used to predict at all throws away its main purpose, since a prediction made between the plotted values is perfectly sound; saying the line passes through all 20 points misdescribes a line of best fit, which is drawn to follow the trend of the points and will normally pass through few of them; saying the equation works for any value put into it treats an equation fitted to Year 10 heights as a law of nature, and the mass of −4 kg shows what that assumption produces.
(d) 360 km — Method: first find the kilometres per litre by dividing distance by fuel used, then multiply this rate by the new tank size. Working: 180 ÷ 6 = 30 km per litre; 30 × 12 = 360 km. Answer: 360 km. 30 km comes from finding the correct fuel consumption but stopping there, without scaling it up to the full tank. 2160 km comes from multiplying the original distance (180) by the tank size (12) directly, skipping the unit rate. 90 km comes from pairing the numbers the wrong way round: dividing the distance by the new tank size, 180 ÷ 12 = 15, and then multiplying by the original 6 litres, 15 × 6 = 90.
(a) 1 — a² − 2b = (−3)² − 2(4) = 9 − 8 = 1. A candidate who squares −3 but keeps the negative sign gets −9 − 8 = −17. A candidate who forgets to square a and substitutes it as −3 gets −3 − 8 = −11. A candidate who adds instead of subtracting 2b gets 9 + 8 = 17.
(d) 7/3 — Put the kettle's energy over the toaster's energy: 2.1/0.9. Multiply both numbers by 10 to clear the decimals: 21/9. Divide both by their highest common factor, 3: 21÷3 = 7, 9÷3 = 3, giving 7/3. (3/7 comes from writing the energy values the wrong way round. 4/3 comes from finding the difference, 2.1 − 0.9 = 1.2 kWh, and writing it as a fraction of the toaster's energy, 1.2/0.9. 7/10 comes from comparing the kettle's energy to the total energy used by both appliances, 2.1/3.0.)
(b) 23 — If every position were filled to the full height of 3, the total would be 2 × 4 × 3 = 24 crates. One corner position has only 2 crates instead of 3, one crate short of full height there, so the actual total is 24 − 1 = 23. "24" comes from using the full height everywhere and forgetting the one incomplete corner. "22" comes from removing 2 crates for the incomplete corner instead of the 1 that is actually missing (3 − 2 = 1, not 2). "21" comes from removing all 3 crates at that corner, as though the position were completely empty rather than 2 crates short.
(b) 1/6 — Method: list the ordered pairs where the two scores match, and divide by the 36 equally likely pairs. Working: the matching pairs are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6), which is 6 pairs out of 36, cancelling down to 1/6. Answer: 1/6. Watch out: writing down 1/36 finds the probability of one particular double, such as (6, 6), rather than any double at all. Guessing 1/2 treats 'same' and 'different' as equally likely events, when there are only 6 matching pairs against 30 non-matching ones. And writing down 1/3 comes from listing each double twice, once for each order of the two dice, giving 12 pairs out of 36 — but (1, 1) is a single outcome, and swapping the two dice over does not produce a second one.
(a) the nearest whole number — The error intervals are 23.5 ≤ p < 24.5 and 5.5 ≤ q < 6.5. The minimum of p ÷ q is 23.5 ÷ 6.5 ≈ 3.615, and the maximum is 24.5 ÷ 5.5 ≈ 4.455. Both of these round to 4 at the nearest whole number, so the answer is guaranteed correct to the nearest whole number — but not to the nearest 0.1, since 3.615 rounds to 3.6 while 4.455 rounds to 4.5, which do not agree. Claiming the nearest 0.1 assumes every figure a calculator shows is trustworthy, without checking whether the bounds actually agree that far. Claiming only the nearest 10 badly understates how much can be guaranteed here, since both bounds already round to 4, not merely to 0. Saying no degree of accuracy can be guaranteed gives up before checking whether the bounds agree at any level at all.
(b) 12 years — Method: call the number of years t, add t to both ages, and form an equation from the comparison at that future time. Working: in t years the leader will be 44 + t and the scout will be 16 + t, so 44 + t = 2(16 + t); expanding gives 44 + t = 32 + 2t, and subtracting t and 32 from both sides gives t = 12. Checking: in 12 years the leader will be 56 and the scout 28, and 56 = 2 × 28. Answer: 12 years. The distractors: 6 years comes from halving the leader's present age instead, so that the scout has to reach 22, which takes 6 years; 14 years comes from halving the 28-year gap between the two ages; 28 years comes from giving the age gap itself as the number of years.
(a) The hourly rate is £5 and the fixed fee is £7 — The gradient is (27 − 12) ÷ (4 − 1) = 15 ÷ 3 = £5, the hourly rate. Using the point (1, 12): 12 = 5 × 1 + fee, so the fee is 12 − 5 = £7. That gives 'The hourly rate is £5 and the fixed fee is £7'. Swapping the two figures gives the statement with £7 as the rate and £5 as the fee, which has them the wrong way round. Taking the C-value of the first point, £12, as the fixed fee ignores that 1 hour of hire is already included in that £12. Using 15, the change in C, as the hourly rate without dividing by the change in h (3 hours) gives the statement claiming a £15 hourly rate.
(b) £2.94 — Method: the price is quoted for each kilogram, so the mass has to be written in kilograms before it is multiplied by the price. Working: 1 kg = 1000 g, so 350 ÷ 1000 = 0.35 and the piece weighs 0.35 kg. The cost is then 8.40 × 0.35 = 2.94. Answer: £2.94. Treating 350 g as 3.5 kg, a division by 100 rather than by 1000, gives 8.40 × 3.5 = 29.40. Multiplying the price by the number of grams gives 8.40 × 350 = 2940. Dividing the price by the mass instead of multiplying gives 8.40 ÷ 0.35 = 24.
(b) 5 — From x + y = 7, x = 7 − y. Substituting into 3x + 2y = 16: 3(7 − y) + 2y = 16, so 21 − 3y + 2y = 16, giving 21 − y = 16, so y = 5 (then x = 2). A candidate who forgets to multiply the y-term inside the bracket by 3 would write 21 − y + 2y = 16, giving 21 + y = 16, so y = −5. A candidate who subtracts the two equations directly, (3x + 2y) − (x + y) = 16 − 7, gets 2x + y = 9, and if they wrongly treat this as giving y alone, ignoring the x term, they would answer 9. A candidate who reports the value of x instead of y would answer 2.
(b) 13/2 — Put the number of pages in the novel over the number of pages in the comic: 312/48. Divide both numbers by their highest common factor, 24: 312÷24 = 13, 48÷24 = 2, giving 13/2. (2/13 comes from writing the page counts the wrong way round. 11/2 comes from finding the difference in the page counts, 312 − 48 = 264, and writing it as a fraction of the comic's page count, 264/48. 13/15 comes from comparing the novel's page count to the total number of pages in both books, 312/360.)
(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) 6 — Method: collect the x-terms on one side and the constants on the other, then divide by the remaining coefficient of x. Working: 6x − 3x = 13 + 5, so 3x = 18, x = 18 ÷ 3 = 6. Answer: x = 6. 2.67 comes from a sign error when moving the 5, subtracting instead of adding: 3x = 13 − 5 = 8, x = 8 ÷ 3 ≈ 2.67. 2 comes from a sign error when moving the x-term, adding instead of subtracting: 9x = 18, x = 2. 18 comes from correctly finding 3x = 18 but forgetting to divide by 3.
(a) 1 — Method: the value of y when x = 0 is where the line meets the y-axis, which is the constant c in y = mx + c, so the gradient is worked out from the two given points first and the constant follows by substituting one of them. Working: m = (16 − 7) ÷ (5 − 2) = 9 ÷ 3 = 3, so the line is y = 3x + c; substituting x = 2 and y = 7 gives 7 = 3 × 2 + c, so c = 7 − 6 = 1, and the value of y when x = 0 is that constant. Answer: 1. The distractors: 3 comes from stopping at the gradient and offering it as the intercept; 4 comes from stepping back from x = 2 to x = 0 by one unit of x instead of two, 7 − 3 = 4; −1 comes from working the constant out as mx − y, 3 × 2 − 7 = −1, instead of y − mx.

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.